4, Topological measure spaces

MEASURE THEORY Volume 4

D.H.Fremlin

By the same author: Topological Riesz Spaces and Measure Theory, Cambridge University Press, 1974. Consequences of Martin’s Axiom, Cambridge University Press, 1982. Companions to the present volume: Measure Theory, vol. 1, Torres Fremlin, 2000. Measure Theory, vol. 2, Torres Fremlin, 2001. Measure Theory, vol. 3, Torres Fremlin, 2002.

First printing November 2003

MEASURE THEORY Volume 4 Topological Measure Spaces

D.H.Fremlin Research Professor in Mathematics, University of Essex

Dedicated by the Author to the Publisher

This book may be ordered from the publisher at the address below. For price and means of payment see the author’s Web page http://www.essex.ac.uk/maths/staff/fremlin/mtsales.htm, or enquire from [email protected].

First published in 2003 by Torres Fremlin, 25 Ireton Road, Colchester CO3 3AT, England c D.H.Fremlin 2003 ° The right of D.H.Fremlin to be identified as author of this work has been asserted in accordance with the Copyright, Designs and Patents Act 1988. This work is issued under the terms of the Design Science License as published in http://dsl.org/copyleft/dsl.txt. For the source files see http://www.essex. ac.uk/maths/staff/fremlin/mt4.2003/index.htm. Library of Congress classification QA312.F72 AMS 2000 classification 28A99 ISBN 0-9538129-4-4 Typeset by AMS-TEX Printed in England by Biddles Short Run Books, King’s Lynn

5

Contents General Introduction

10

Introduction to Volume 4

11

Chapter 41: Topologies and measures I Introduction 411 Definitions

13 13

Topological, inner regular, τ -additive, outer regular, locally finite, effectively locally finite, quasi-Radon, Radon, completion regular, Baire, Borel and strictly positive measures; measurable and almost continuous functions; self-supporting sets and supports of measures; Stone spaces; Dieudonn´ e’s measure.

412 Inner regularity

19

Exhaustion; Baire measures; Borel measures on metrizable spaces; completions and c.l.d. versions; complete locally determined spaces; inverse-measure-preserving functions; subspaces; indefinite-integral measures; products; outer regularity.

413 Inner measure constructions

31

Inner measures; constructing a measure from an inner measure; the inner measure defined by a measure; complete locally determined spaces; extension of functionals to measures; countably compact classes; constructing measures dominating given functionals.

414 τ -additivity

50

Semi-continuous functions; supports; strict localizability; subspace measures; regular topologies; density topologies; lifting topologies.

415 Quasi-Radon measure spaces

58

Strict localizability; subspaces; regular topologies; hereditarily Lindel¨ of spaces; products of separable metrizable spaces; comparison and specification of quasi-Radon measures; construction of quasi-Radon measures extending given functionals; indefinite-integral measures; Lp spaces; Stone spaces.

416 Radon measure spaces

73

Radon and quasi-Radon measures; specification of Radon measures; c.l.d. versions of Borel measures; locally compact topologies; constructions of Radon measures extending or dominating given functionals; additive functionals on Boolean algebras and Radon measures on Stone spaces; subspaces; products; Stone spaces of measure algebras; compact and perfect measures; representation of homomorphisms of measure algebras; the split interval.

417 τ -additive product measures

88

The product of two effectively locally finite τ -additive measures; the product of many τ -additive probability measures; Fubini’s theorem; generalized associative law; measures on subproducts as image measures; products of strictly positive measures; quasi-Radon and Radon product measures; when ‘ordinary’ product measures are τ -additive; continuous functions and Baire σ-algebras in product spaces.

418 Measurable functions and almost continuous functions

110

Measurable functions; into (separable) metrizable spaces; and image measures; almost continuous functions; continuity, measurability, image measures; expressing Radon measures as images of Radon measures; Prokhorov’s theorem on projective limits of Radon measures; representing measurable functions into L0 spaces.

419 Examples

126

A nearly quasi-Radon measure; a Radon measure space in which the Borel sets are inadequate; a nearly Radon measure; the Stone space of the Lebesgue measure algebra; measures with domain Pω1 ; notes on Lebesgue measure.

Chapter 42: Descriptive set theory Introduction 421 Souslin’s operation

138 138

Souslin’s operation; is idempotent; as a projection operator; Souslin-F sets; *constituents.

422 K-analytic spaces

148

Usco-compact relations; K-analytic sets; and Souslin-F sets; *First Separation Theorem.

423 Analytic spaces

155

Analytic spaces; are K-analytic spaces with countable networks; Souslin-F sets; Borel measurable functions; injective images of Polish spaces; non-Borel analytic sets; von Neumann-Jankow selection theorem; *constituents of coanalytic sets.

424 Standard Borel spaces Elementary properties; isomorphism types; subspaces; Borel measurable actions of Polish groups.

165

6

Chapter 43: Topologies and measures II Introduction 431 Souslin’s operation

172 172

The domain of a complete locally determined measure is closed under Souslin’s operation; the kernel of a Souslin scheme is approximable from within.

432 K-analytic spaces

176

Topological measures on K-analytic spaces; extensions to Radon measures; expressing Radon measures as images of Radon measures.

433 Analytic spaces

180

Measures on spaces with countable networks; inner regularity of Borel measures; expressing Radon measures as images of Radon measures; measurable and almost continuous functions; the von NeumannJankow selection theorem; products; extension of measures on σ-subalgebras; standard Borel spaces.

434 Borel measures

184

Classification of Borel measures; Radon spaces; universally measurable sets and functions; Borel-measurecompact, Borel-measure-complete and pre-Radon spaces; countable compactness and countable tightness; quasi-dyadic spaces and completion regular measures; first-countable spaces and Borel product measures.

435 Baire measures

203

Classification of Baire measures; extension of Baire measures to Borel measures (Maˇrik’s theorem); measure-compact spaces; sequential spaces and Baire product measures.

436 Representation of linear functionals

209

Smooth and sequentially smooth linear functionals; measures and sequentially smooth functionals; Baire measures; products of Baire measures; quasi-Radon measures and smooth functionals; locally compact spaces and Radon measures.

437 Spaces of measures

219

Smooth and sequentially smooth duals; signed measures; embedding spaces of measurable functions in the bidual of Cb (X); vague and narrow topologies; product measures; extreme points; uniform tightness; Prokhorov spaces.

438 Measure-free cardinals

240

Measure-free cardinals; point-finite families of sets with measurable unions; measurable functions into metrizable spaces; Radon and measure-compact metric spaces; metacompact spaces; hereditarily weakly θ-refinable spaces; when c is measure-free.

439 Examples

254

Measures with no extensions to Borel measures; universally negligible sets; Hausdorff measures are rarely semi-finite; a smooth linear functional not expressible as an integral; a first-countable non-Radon space; Baire measures not extending to Borel measures; N c is not Borel-measure-compact; the Sorgenfrey line; Q is not a Prokhorov space.

Chapter 44: Topological groups Introduction 441 Invariant measures on locally compact spaces

271 271

Measures invariant under homeomorphisms; Haar measures; measures invariant under isometries.

442 Uniqueness of Haar measure

280

Two (left) Haar measures are multiples of each other; left and right Haar R measures; R Haar measurable and Haar negligible sets; the modular function of a group; formulae for f (x−1 )dx, f (xy)dx.

443 Further properties of Haar measure

287

The Haar measure algebra of a group carrying Haar measures; actions of the group on the Haar measure algebra; locally compact groups; actions of the group on L0 and Lp ; the bilateral uniformity; Borel sets are adequate; completing the group; expressing an arbitrary Haar measure in terms of a Haar measure on a locally compact group; completion regularity of Haar measure; invariant measures on the set of left cosets of a closed subgroup of a locally compact group; modular functions of subgroups and quotient groups; transitive actions of compact groups on compact spaces.

444 Convolutions Convolutions of quasi-Radon measures; the Banach algebra of signed τ -additive measures; continuous actions and corresponding actions on L0 (ν) for an arbitrary quasi-Radon measure ν; convolutions of measures and functions; indefinite-integral measures over a Haar measure µ; convolutions of functions; Lp (µ); approximate identities.

311

7

445 The duality theorem

334

Dual groups; Fourier-Stieltjes transforms; Fourier transforms; identifying the dual group with the maximal ideal space of L1 ; the topology of the dual group; positive definite functions; Bochner’s theorem; the Inversion Theorem; the Plancherel Theorem; the Duality Theorem.

446 The structure of locally compact groups

357

Finite-dimensional representations separate the points of a compact group; groups with no small subgroups have B-sequences; chains of subgroups.

447 Translation-invariant liftings

372

Translation-invariant liftings and lower densities; Vitali’s theorem and a density theorem for groups with B-sequences; Haar measures have translation-invariant liftings.

448 Invariant measures on Polish spaces

383

Countably full local semigroups of Aut A; σ-equidecomposability; countably non-paradoxical groups; G-invariant additive functions from A to L∞ (C); measures invariant under Polish group actions (the Nadkarni-Becker-Kechris theorem).

449 Amenable groups

393

Amenable groups; permanence properties; locally compact amenable groups; Tarski’s theorem; discrete amenable groups.

Chapter 45: Perfect measures, disintegrations and processes Introduction 451 Perfect, compact and countably compact measures

413 414

Basic properties of the three classes; subspaces, completions, c.l.d. versions, products; measurable functions from compact measure spaces to metrizable spaces; *weakly α-favourable spaces.

452 Integration and disintegration of measures

427

Integrating families of probability measures; τ -additive and Radon measures; disintegrations and regular conditional probabilities; disintegrating countably compact measures; disintegrating Radon measures; *images of countably compact measures.

453 Strong liftings

441

Strong and almost strong liftings; existence; on product spaces; disintegrations of Radon measures over spaces with almost strong liftings; Stone spaces; Losert’s example.

454 Measures on product spaces

454

Perfect, compact and countably compact measures on product spaces; extension of finitely additive functions with perfect countably additive marginals; Kolmogorov’s extension theorem; measures defined from conditional distributions; distributions of random processes; measures on C(T ) for Polish T .

455 Markov process and Brownian motion

464

Definition of Markov process from conditional distributions; existence of a measure representing Brownian motion; continuous sample paths.

456 Gaussian distributions

472

Gaussian distributions; supports; universal Gaussian distributions; cluster sets of n-dimensional processes; τ -additivity; Gaussian processes.

457 Simultaneous extension of measures

488

Extending families of finitely additive functionals; Strassen’s theorem; extending families of measures; examples.

458 Relative independence and relative products

497

Relatively independent families of σ-algebras and random variables; relative distributions; relatively independent families of closed subalgebras of a probability algebra; relative free products of probability algebras; relative products of probability spaces; existence of relative products.

459 Symmetric measures and exchangeable random variables

510

Exchangeable families of inverse-measure-preserving functions; de Finetti’s theorem; countably compact symmetric measures on product spaces disintegrate into product measures; symmetric quasi-Radon measures.

Chapter 46: Pointwise compact sets of measurable functions Introduction 461 Barycenters and Choquet’s theorem

522 522

Barycenters; elementary properties; sufficient conditions for existence; closed convex hulls of compact sets; Kreˇın’s theorem; measures on sets of extreme points.

462 Pointwise compact sets of continuous functions Angelic spaces; the topology of pointwise convergence on C(X); weak convergence and weakly compact sets in C0 (X); Radon measures on C(X); separately continuous functions; convex hulls.

532

8

463 Tp and Tm

538

Pointwise convergence and convergence in measure on spaces of measurable functions; compact and sequentially compact sets; perfect measures and Fremlin’s Alternative; separately continuous functions.

464 Talagrand’s measure

550

The usual measure on PI; the intersection of a sequence of non-measurable filters; Talagrand’s measure; the L-space of additive functionals on PI; measurable and purely non-measurable functionals.

465 Stable sets

563

Stable sets of functions; elementary properties; pointwise compactness; pointwise convergence and convergence in measure; a law of large numbers; stable sets and uniform convergence in the strong law of large numbers; stable sets in L0 and L1 ; *R-stable sets.

466 Measures on linear topological spaces

590

Quasi-Radon measures for weak and strong topologies; Kadec norms; constructing weak-Borel measures; characteristic functions of measures on locally convex spaces; universally measurable linear operators.

*467 Locally uniformly rotund norms

598

Locally uniformly rotund norms; separable normed spaces; long sequences of projections; K-countably determined spaces; weakly compactly generated spaces; Banach lattices with order-continuous norms; Eberlein compacta.

Chapter 47: Geometric measure theory Introduction 471 Hausdorff measures

610 610

Metric outer measures; Increasing Sets Lemma; analytic spaces; inner regularity; Vitali’s theorem and a density theorem; Howroyd’s theorem.

472 Besicovitch’s Density Theorem

626

Besicovitch’s Covering Lemma; Besicovitch’s Density Theorem; *a maximal theorem.

473 Poincar´e’s inequality

633

Differentiable and Lipschitz functions; smoothing by convolution; the Gagliardo-Nirenberg-Sobolev inequality; Poincar´ e’s inequality for balls.

474 The distributional perimeter

647

The divergence of a vector field; sets with locally finite perimeter, perimeter measures and outwardnormal functions; the reduced boundary; invariance under isometries; isoperimetric inequalities; Federer exterior normals; the Compactness Theorem.

475 The essential boundary

670

Essential interior, closure and boundary; the reduced boundary; perimeter measures; characterizing sets with locally finite perimeter; the Divergence Theorem; calculating perimeters from cross-sectional counts; Cauchy’s Perimeter Theorem; the Isoperimetric Theorem for convex sets.

476 Concentration of measure

690

Hausdorff metrics; Vietoris topologies; concentration by partial reflection; concentration of measure in R r ; the Isoperimetric Theorem; concentration of measure on spheres.

Chapter 48: Gauge integrals Introduction 481 Tagged partitions

704 704

Tagged partitions and Riemann sums; gauge integrals; gauges; residual sets; subdivisions; examples (the Riemann integral, the Henstock integral, the symmetric Riemann-complete integral, the McShane integral, box products, the approximately continuous Henstock integral).

482 General theory

714

Saks-Henstock lemma; when gauge-integrable functions are measurable; when integrable functions are gauge-integrable; Iν (f × χH); integrating derivatives; B.Levi’s theorem; Fubini’s theorem.

483 The Henstock integral

729

The Henstock and Lebesgue integrals; indefinite Henstock integrals; Saks-Henstock lemma; fundamental theorem of calculus; the Perron integral; ACG∗ functions.

484 The Pfeffer integral The Tamanini-Giacomelli theorem; a family of tagged-partition structures; the Pfeffer integral; the SaksHenstock indefinite integral of a Pfeffer integrable function; Pfeffer’s Divergence Theorem; differentiating the indefinite integral; invariance under lipeomorphisms.

746

9

Chapter 49: Further topics Introduction 491 Equidistributed sequences

765 765

The asymptotic density ideal Z; equidistributed sequences; when equidistributed sequences exist; Z = PN/Z; effectively regular measures; equidistributed sequences and induced embeddings of measure algebras in Z.

492 Combinatorial concentration of measure

782

Concentration of measure in product spaces; concentration of measure in permutation groups.

493 Extremely amenable groups

789

Extremely amenable groups; concentrating additive functionals; measure algebras under 4 ; L0 ; automorphism groups of measure algebras; isometry groups of spheres in inner product spaces; locally compact groups.

494 Cubes in product spaces

800

Subsets of measure algebras with non-zero infima; product sets included in given sets of positive measure.

495 Poisson point processes

802

Poisson distributions; Poisson point processes; disintegrations; transforming disjointness into stochastic independence; representing Poisson point processes by Radon measures; exponential distributions and Poisson point processes on [0, ∞[.

Appendix to Volume 4 Introduction 4A1 Set theory

823 823

Cardinals; closed cofinal sets and stationary sets; ∆-system lemma; free sets; Ramsey’s theorem; the Marriage Lemma; filters; normal ultrafilters; Ostaszewski’s ♣; cardinals of σ-algebras.

4A2 General topology

827

Glossary; general constructions; Fσ , Gδ , zero and cozero sets; countable chain condition; separation ˇ axioms; compact and locally compact spaces; Lindel¨ of spaces; Stone-Cech compactifications; uniform spaces; first-countable, sequential, countably tight, metrizable spaces; countable networks; secondcountable spaces; separable metrizable spaces; Polish spaces; order topologies.

4A3 Topological σ-algebras

848

Borel σ-algebras; measurable functions; hereditarily Lindel¨ of spaces; second-countable spaces; Polish spaces; ω1 ; Baire σ-algebras; product spaces; compact spaces; Baire property algebras; cylindrical σalgebras.

4A4 Locally convex spaces

857

Linear topological spaces; locally convex spaces; Hahn-Banach theorem; normed spaces; inner product spaces; max-flow min-cut theorem.

4A5 Topological groups

863

Group actions; topological groups; uniformities; quotient groups; metrizable groups.

4A6 Banach algebras

869

Stone-Weierstrass theorem (fourth form); multiplicative linear functionals; spectral radius; invertible elements; exponentiation.

Concordance

873

References for Volume 4

874

Index to Volumes 1-4 Principal topics and results General index

881 893

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General introduction In this treatise I aim to give a comprehensive description of modern abstract measure theory, with some indication of its principal applications. The first two volumes are set at an introductory level; they are intended for students with a solid grounding in the concepts of real analysis, but possibly with rather limited detailed knowledge. As the book proceeds, the level of sophistication and expertise demanded will increase; thus for the volume on topological measure spaces, familiarity with general topology will be assumed. The emphasis throughout is on the mathematical ideas involved, which in this subject are mostly to be found in the details of the proofs. My intention is that the book should be usable both as a first introduction to the subject and as a reference work. For the sake of the first aim, I try to limit the ideas of the early volumes to those which are really essential to the development of the basic theorems. For the sake of the second aim, I try to express these ideas in their full natural generality, and in particular I take care to avoid suggesting any unnecessary restrictions in their applicability. Of course these principles are to to some extent contradictory. Nevertheless, I find that most of the time they are very nearly reconcilable, provided that I indulge in a certain degree of repetition. For instance, right at the beginning, the puzzle arises: should one develop Lebesgue measure first on the real line, and then in spaces of higher dimension, or should one go straight to the multidimensional case? I believe that there is no single correct answer to this question. Most students will find the one-dimensional case easier, and it therefore seems more appropriate for a first introduction, since even in that case the technical problems can be daunting. But certainly every student of measure theory must at a fairly early stage come to terms with Lebesgue area and volume as well as length; and with the correct formulations, the multidimensional case differs from the one-dimensional case only in a definition and a (substantial) lemma. So what I have done is to write them both out (§§114-115). In the same spirit, I have been uninhibited, when setting out exercises, by the fact that many of the results I invite students to look for will appear in later chapters; I believe that throughout mathematics one has a better chance of understanding a theorem if one has previously attempted something similar alone. As I write this Introduction (September 2003), the plan of the work is as follows: Volume Volume Volume Volume Volume

1: 2: 3: 4: 5:

The Irreducible Minimum Broad Foundations Measure Algebras Topological Measure Spaces Set-theoretic Measure Theory.

Volume 1 is intended for those with no prior knowledge of measure theory, but competent in the elementary techniques of real analysis. I hope that it will be found useful by undergraduates meeting Lebesgue measure for the first time. Volume 2 aims to lay out some of the fundamental results of pure measure theory (the Radon-Nikod´ ym theorem, Fubini’s theorem), but also gives short introductions to some of the most important applications of measure theory (probability theory, Fourier analysis). While I should like to believe that most of it is written at a level accessible to anyone who has mastered the contents of Volume 1, I should not myself have the courage to try to cover it in an undergraduate course, though I would certainly attempt to include some parts of it. Volumes 3 and 4 are set at a rather higher level, suitable to postgraduate courses; while Volume 5 will assume a wide-ranging competence over large parts of analysis and set theory. There is a disclaimer which I ought to make in a place where you might see it in time to avoid paying for this book. I make no attempt to describe the history of the subject. This is not because I think the history uninteresting or unimportant; rather, it is because I have no confidence of saying anything which would not be seriously misleading. Indeed I have very little confidence in anything I have ever read concerning the history of ideas. So while I am happy to honour the names of Lebesgue and Kolmogorov and Maharam in more or less appropriate places, and I try to include in the bibliographies the works which I have myself consulted, I leave any consideration of the details to those bolder and better qualified than myself. The work as a whole is not yet complete; and when it is finished, it will undoubtedly be too long to be printed as a single volume in any reasonable format. I am therefore publishing it one part at a time. However, drafts of most of the rest are available on the Internet; see http://www.essex.ac.uk/ maths/staff/fremlin/mt.htm for detailed instructions. For the time being, at least, printing will be in short runs. I hope that readers will be energetic in commenting on errors and omissions, since it should be possible to correct these relatively promptly. An inevitable consequence of this is that paragraph references may go out of date rather quickly. I shall be most flattered if anyone chooses to rely on this book as a source

Introduction to Volume 4

11

for basic material; and I am willing to attempt to maintain a concordance to such references, indicating where migratory results have come to rest for the moment, if authors will supply me with copies of papers which use them. I mention some minor points concerning the layout of the material. Most sections conclude with lists of ‘basic exercises’ and ‘further exercises’, which I hope will be generally instructive and occasionally entertaining. How many of these you should attempt must be for you and your teacher, if any, to decide, as no two students will have quite the same needs. I mark with a > those which seem to me to be particularly important. But while you may not need to write out solutions to all the ‘basic exercises’, if you are in any doubt as to your capacity to do so you should take this as a warning to slow down a bit. The ‘further exercises’ are unbounded in difficulty, and are unified only by a presumption that each has at least one solution based on ideas already introduced. Occasionally I add a final ‘problem’, a question to which I do not know the answer and which seems to arise naturally in the course of the work. The impulse to write this book is in large part a desire to present a unified account of the subject. Cross-references are correspondingly abundant and wide-ranging. In order to be able to refer freely across the whole text, I have chosen a reference system which gives the same code name to a paragraph wherever it is being called from. Thus 132E is the fifth paragraph in the second section of the third chapter of Volume 1, and is referred to by that name throughout. Let me emphasize that cross-references are supposed to help the reader, not distract her. Do not take the interpolation ‘(121A)’ as an instruction, or even a recommendation, to lift Volume 1 off the shelf and hunt for §121. If you are happy with an argument as it stands, independently of the reference, then carry on. If, however, I seem to have made rather a large jump, or the notation has suddenly become opaque, local cross-references may help you to fill in the gaps. Each volume will have an appendix of ‘useful facts’, in which I set out material which is called on somewhere in that volume, and which I do not feel I can take for granted. Typically the arrangement of material in these appendices is directed very narrowly at the particular applications I have in mind, and is unlikely to be a satisfactory substitute for conventional treatments of the topics touched on. Moreover, the ideas may well be needed only on rare and isolated occasions. So as a rule I recommend you to ignore the appendices until you have some direct reason to suppose that a fragment may be useful to you. During the extended gestation of this project I have been helped by many people, and I hope that my friends and colleagues will be pleased when they recognise their ideas scattered through the pages below. But I am especially grateful to those who have taken the trouble to read through earlier drafts and comment on obscurities and errors. Introduction to Volume 4 I return in this volume to the study of measure spaces rather than measure algebras. For fifty years now measure theory has been intimately connected with general topology. Not only do a very large proportion of the measure spaces arising in applications carry topologies related in interesting ways to their measures, but many questions in abstract measure theory can be effectively studied by introducing suitable topologies. Consequently any course in measure theory at this level must be frankly dependent on a substantial knowledge of topology. With this proviso, I hope that the present volume will be accessible to graduate students, and will lead them to the most important ideas of modern abstract measure theory. The first and third chapters of the volume seek to provide a thorough introduction into the ways in which topologies and measures can interact. They are divided by a short chapter on descriptive set theory, on the borderline between set theory, logic, real analysis and general topology, which I single out for detailed exposition because I believe that it forms an indispensable part of the background of any measure theorist. Chapter 41 is dominated by the concepts of inner regularity and τ -additivity, coming together in Radon measures (§416). Chapter 43 concentrates rather on questions concerning properties of a topological space which force particular relationships with measures on that space. But plenty of side-issues are treated in both, such as Lusin measurability (§418), the definition of measures from linear functionals (§436) and measure-free cardinals (§438). Chapters 45 and 46 continue some of the same themes, with particular investigations into ‘disintegrations’ or regular conditional probabilities (§§452-453), the abstract theory of stochastic processes (§§454-455), Talagrand’s theory of Glivenko-Cantelli classes (§465) and the theory of measures on normed spaces (§§466-467). In contrast with the relatively amorphous structure of Chapters 41, 43, 45 and 46, four chapters of this volume have definite topics. I have already said that Chapter 42 is an introduction to descriptive set theory;

12

Introduction to Volume 4

like Chapters 31 and 35 in the last volume, it is a kind of appendix brought into the main stream of the argument. Chapter 44 deals with topological groups. Most of it is of course devoted to Haar measure, giving the Pontryagin-van Kampen duality theorem (§445) and the Ionescu Tulcea theorem on the existence of translation-invariant liftings (§447). But there are also sections on Polish groups (§448) and amenable groups (§449), and some of the general theory of measures on measurable groups (§444). Chapter 47 is a second excursion, after Chapter 26, into geometric measure theory. It starts with Hausdorff measures (§471), gives a proof of the Di Giorgio-Federer Divergence Theorem (§475), and then examines a number of examples of ‘concentration of measure’ (§476). In Chapter 48, I set out the elementary theory of gauge integrals, with sections on the Henstock and Pfeffer integrals (§§483-484). Finally, in Chapter 49, I give notes on five special topics: equidistributed sequences (§491), combinatorial forms of concentration of measure (§492), extremely amenable groups (§493), subproducts in product spaces (§494) and Poisson point processes (§495). I had better mention prerequisites, as usual. To embark on this material you will certainly need a solid foundation in measure theory. Since I do of course use my own exposition as my principal source of references to the elementary ideas, I advise readers to ensure that they have easy access to all three previous volumes before starting serious work on this one. But you may not need to read very much of them. It might be prudent to glance through the detailed contents of Volume 1 and the first five chapters of Volume 2 to check that most of the material there is more or less familiar. But Volume 3, and the last three chapters of Volume 2, can probably be left on one side for the moment. Of course you will need the Lifting Theorem (Chapter 34) for §§447, 452 and 453, and Chapter 26 is essential background for Chapter 47, while Chapter 28 (on Fourier analysis) may help to make sense of Chapter 44, and parts of Chapter 27 (on probability theory) are necessary for §§455-456. And measure algebras are mentioned in every chapter except (I think) Chapter 48; but I hope that the cross-references are precise enough to lead you to just what you need to know at any particular point. Even Maharam’s theorem is hardly used in this volume. What you will need, apart from any knowledge of measure theory, is a sound background in general topology. This volume calls on a great many miscellaneous facts from general topology, and the list in §4A2 is not a good place to start if continuity and compactness and the separation axioms are unfamiliar. My primary reference for topology is Engelking 89. I do not insist that you should have read this book (though of course I hope you will do so sometime); but I do think you should make sure that you can use it. In the general introduction to this treatise, I wrote ‘I make no attempt to describe the history of the subject’, and I have generally been casual – some would say negligent – in my attributions of results to their discoverers. Through much of the first three volumes I did at least have the excuse that the history exists in print in far more detail than I am qualified to describe. In the present volume I find my position more uncomfortable, in that I have been watching the evolution of the subject relatively closely over the last thirty years, and ought to be able to say something about it. Nevertheless I remain reluctant to make definite statements crediting one person rather than another with originating an idea. My more intimate knowledge of the topic makes me even more conscious than elsewhere of the danger of error and of the breadth of reading that would be necessary to produce a balanced account. In some cases I do attach a result to a specific published paper, but these attributions should never be regarded as an assertion that any particular author has priority; at most, they declare that a historian should examine the source cited before coming to any decision. I assure my friends and colleagues that my omissions are not intended to slight either them or those we all honour. What I have tried to do is to include in the bibliography to this volume all the published work which (as far as I am consciously aware) has influenced me while writing it, so that those who wish to go into the matter will have somewhere to start their investigations.

411B

Definitions

13

Chapter 41 Topologies and Measures I I begin this volume with an introduction to some of the most important ways in which topologies and measures can interact, and with a description of the forms which such constructions as subspaces and product spaces take in such contexts. By far the most important concept is that of Radon measure (411H, §416). In Radon measure spaces we find both the richest combinations of ideas and the most important applications. But, as usual, we are led both by analysis of these ideas and by other interesting examples to consider wider classes of topological measure space, and the greater part of the chapter, by volume, is taken up by a description of the many properties of Radon measures individually and in partial combinations. I begin the chapter with a short section of definitions (§411), including a handful of more or less elementary examples. The two central properties of a Radon measure are ‘inner regularity’ (411B) and ‘τ -additivity’ (411C). The former is an idea of great versatility which I look at in an abstract setting in §412. I take a section (§413) to describe some methods of constructing measure spaces, extending the rather limited range of constructions offered in earlier volumes. There are two sections on τ -additive measures, §§414 and 417; the former covers the elementary ideas, and the latter looks at product measures, where it turns out that we need a new technique to supplement the purely measure-theoretic constructions of Chapter 25. On the way to Radon measures in §416, I pause over ‘quasi-Radon’ measures (411H, §415), where inner regularity and τ -additivity first come effectively together. The possible interactions of a topology and a measure on the same space are so varied that even a brief account makes a long chapter; and this is with hardly any mention of results associated with particular types of topological space, most of which must wait for later chapters. But I include one section on the two most important classes of functions acting between topological measure spaces (§418), and another describing some examples to demonstrate special phenomena (§419).

411 Definitions In something of the spirit of §211, but this time without apologising, I start this volume with a list of definitions. The rest of Chapter 41 will be devoted to discussing these definitions and relationships between them, and integrating the new ideas into the concepts and constructions of earlier volumes; I hope that by presenting the terminology now I can give you a sense of the directions the following sections will take. I ought to remark immediately that there are many cases in which the exact phrasing of the definitions is important in ways which may not be immediately apparent. 411A I begin with a phrase which will be a useful shorthand for the context in which most, but not all, of the theory here will be developed. Definition A topological measure space is a quadruple (X, T, Σ, µ) where (X, Σ, µ) is a measure space and T is a topology on X such that T ⊆ Σ, that is, every open set (and therefore every Borel set) is measurable. 411B Now I come to what are in my view the two most important concepts to master; jointly they will dominate the chapter. Definition Let (X, Σ, µ) be a measure space and K a family of sets. I say that µ is inner regular with respect to K if µE = sup{µK : K ∈ Σ ∩ K, K ⊆ E} for every E ∈ Σ. (Cf. 256Ac, 342Aa.) Remark Note that in this definition I do not assume that K ⊆ Σ, nor even that K ⊆ PX. But of course µ will be inner regular with respect to K iff it is inner regular with respect to K ∩ Σ. It is convenient in this context to interpret sup ∅ as 0, so that we have to check the definition only when µE > 0, and need not insist that ∅ ∈ K.

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Topologies and measures I

411C

411C Definition Let (X, Σ, µ) be a measure space and T a topology on X. I say that µ is τ -additive (the phrase τ -regular hasSalso been used)Sif whenever G is a non-empty upwards-directed family of open sets such that G ⊆ Σ and G ∈ Σ then µ( G) = supG∈G µG. Remark Note that in this definition I do not assume that every open set is measurable. Consequently we cannot take it for granted that an extension of a τ -additive measure will be τ -additive; on the other hand, the restriction of a τ -additive measure to any σ-subalgebra will be τ -additive. 411D Complementary to 411B we have the following. Definition Let (X, Σ, µ) be a measure space and H a family of subsets of X. Then µ is outer regular with respect to H if µE = inf{µH : H ∈ Σ ∩ H, H ⊇ E} for every E ∈ Σ. 411E I delay discussion of most of the relationships between the concepts here to later in the chapter. But it will be useful to have a basic fact set out immediately. Proposition Let (X, Σ, µ) be a measure space and T a topology on X. If µ is inner regular with respect to the compact sets, it is τ -additive. S proof Let G be a non-empty upwards-directed family of measurable open sets such that H = G ∈ Σ. If γ < µH, there is a compact set K ⊆ H such that µK ≥ γ; now there must be a G ∈ G which includes K, so that µG ≥ γ. As γ is arbitrary, supG∈G µG = µH. 411F In order to deal efficiently with measures which are not totally finite, I think we need the following ideas. Definitions Let (X, Σ, µ) be a measure space and T a topology on X. (a) I say that µ is locally finite if every point of X has a neighbourhood of finite measure, that is, if the open sets of finite outer measure cover X. (b) I say that µ is effectively locally finite if for every non-negligible measurable set E ⊆ X there is a measurable open set G ⊆ X such that µG < ∞ and E ∩ G is not negligible. Note that an effectively locally finite measure must measure many open sets, while a locally finite measure need not. (c) This seems a convenient moment at which to introduce the following term. A real-valued function f defined R on a subset of X is locally integrable if for every x ∈ X there is an open set G containing x such that G f is defined (in the sense of 214D) and finite. 411G Elementary facts (a) If µ is a locally finite measure on a topological space X, then µ∗ K < ∞ for every compact set K ⊆ X. P P The family G of open sets of finite outer measure is upwards-directed and covers X, so there must be some G ∈ G including K, in which case µ∗ K ≤ µ∗ G is finite. Q Q (b) A measure µ on R r is locally finite iff every bounded set has finite outer measure (cf. 256Ab). P P (i) If every bounded set has finite outer measure then, in particular, every open ball has finite outer measure, so that µ is locally finite. (ii) If µ is locally finite and A ⊆ Rr is bounded, then its closure A is compact Q (2A2F), so that µ∗ A ≤ µ∗ A is finite, by (a) above. Q (c) I should perhaps remark immediately that a locally finite topological measure need not be effectively locally finite (419A), and an effectively locally finite measure need not be locally finite (411P). (d) An effectively locally finite measure must be semi-finite.

411K

Definitions

15

(e) A locally finite measure on a Lindel¨of space X (definition: 4A2A) is σ-finite. P P Let G be the family of open sets of finite outer measure. Because µ is locally finite, G is a cover of X. Because X is Lindel¨of, there is a sequence hGn in∈N in G covering X. For each n ∈ N, there is a measurable set En ⊇ Gn of finite measure, and now hEn in∈N is a sequence of sets of finite measure covering X. Q Q (f ) Let (X, T, Σ, µ) be a topological measure space such that µ is locally finite and inner regular with respect to the compact sets. Then µ is effectively locally finite. P P Suppose that µE > 0. Then there is a measurable compact set K ⊆ E such that µK > 0. As in the argument for (a) above, there is an open set G of finite measure including K, so that µ(E ∩ G) > 0. Q Q (g) Corresponding to (a) above, we R have the following fact. If µ is a measure on a topological space and f ∈ L0 (µ) is locally integrable, then K f dµR is finite for every compact K ⊆ X, because K can be covered by a finite family of open sets G such that G |f |dµ < ∞. (h) If µ is a locally finite measure space X, and f ∈ Lp (µ) for some p ∈ [1, ∞], then f is R on a topological R locally integrable; this is because G |f | ≤ E f ≤ kf kp kχEkq is finite whenever G ⊆ E and µE < ∞, where 1 1 older’s inequality (244Eb). p + q = 1, by H¨ (i) If (X, T) is a completely regular space and µ is a locally finite topological measure on X, then the set of open sets with negligible boundaries is a base for T. P P If x ∈ G ∈ T, let H ⊆ G be an open set of finite measure containing x, and f : X → [0, 1] a continuous function such that f (x) = 1 and f (y) = 0 for y ∈ X \ H. Then {f −1 [{α}] : 0 < α < 1} is an uncountable disjoint family of measurable subsets of H, so there must be some α ∈ ]0, 1[ such that f −1 [{α}] is negligible. Set U = {y : f (y) > α}; then U is an open neighbourhood of x included in G and ∂U ⊆ f −1 [{α}] is negligible. Q Q 411H Two particularly important combinations of the properties above are the following. Definitions (a) A quasi-Radon measure space is a topological measure space (X, T, Σ, µ) such that (i) (X, Σ, µ) is complete and locally determined (ii) µ is τ -additive, inner regular with respect to the closed sets and effectively locally finite. (b) A Radon measure space is a topological measure space (X, T, Σ, µ) such that (i) (X, Σ, µ) is complete and locally determined (ii) T is Hausdorff (iii) µ is locally finite and inner regular with respect to the compact sets. 411I Remarks(a) You may like to seek your own proof that a Radon measure space is always quasiRadon, before looking it up in §416 below. (b) Note that a measure on Euclidean space R r is a Radon measure on the definition above iff it is a Radon measure as described in 256Ad. P P In 256Ad, I said that a measure µ on R r is ‘Radon’ if it is a locally finite complete topological measure, inner regular with respect to the compact sets. (The definition of ‘locally finite’ in 256A was not the same as the one above, but I have already covered this point in 411Gb.) So the only thing to add is that µ is necessarily locally determined, because it is σ-finite (256Ba). Q Q 411J The following special types of inner regularity are of sufficient importance to have earned separate names. Definitions (a) If (X, T) is a topological space, I will say that a measure µ on X is tight if it is inner regular with respect to the closed compact sets. (b) If (X, T, Σ, µ) is a topological measure space, I will say that µ is completion regular if it is inner regular with respect to the zero sets (definition: 3A3Pa). 411K Borel and Baire measures If (X, T) is a topological space, I will call a measure with domain (exactly) the Borel σ-algebra of X (4A3A) a Borel measure on X, and a measure with domain (exactly) the Baire σ-algebra of X (4A3K) a Baire measure on X. Of course a Borel measure is a topological measure in the sense of 411A. On a metric space, the Borel and Baire measures coincide (4A3Kb). The most important measures in this chapter will be c.l.d. versions of Borel measures.

16

Topologies and measures I

411L

411L When we come to look at functions defined on a topological measure space, we shall have to relate ideas of continuity and measurability. Two basic concepts are the following. Definition Let X be a set, Σ a σ-algebra of subsets of X and (Y, S) a topological space. I will say that a function f : X → Y is measurable if f −1 [G] ∈ Σ for every open set G ⊆ Y . Remarks (a) Note that a function f : X → R is measurable on this definition (when R is given its usual topology) iff it is measurable according to the familiar definition in 121C, which asks only that sets of the form {x : f (x) < α} should be measurable (121Ef). (b) For any topological space (Y, S), a function f : X → Y is measurable iff f is (Σ, B(Y ))-measurable, where B(Y ) is the Borel σ-algebra of Y (4A3Cb). 411M Definition Let (X, Σ, µ) be a measure space, T a topology on X, and (Y, S) another topological space. I will say that a function f : X → Y is almost continuous or Lusin measurable if µ is inner regular with respect to the family of subsets A of X such that f ¹A is continuous. 411N Finally, I introduce some terminology to describe ways in which (sometimes) measures can be located in one part of a topological space rather than another. Definitions Let (X, Σ, µ) be a measure space and T a topology on X. (a) I will call a set A ⊆ X self-supporting if µ∗ (A ∩ G) > 0 for every open set G such that A ∩ G is non-empty. (Such sets are sometimes called of positive measure everywhere.) (b) A support of µ is a closed self-supporting set F such that X \ F is negligible. (c) Note that µ can have at most one support. P P If F1 , F2 are supports then µ∗ (F1 \F2 ) ≤ µ∗ (X \F2 ) = 0 so F1 \ F2 must be empty. Similarly, F2 \ F1 = ∅, so F1 = F2 . Q Q (d) If µ is a τ -additive topological measure it has a support. P P Let G be the familySof negligible open S sets, and F the closed set X \ G. Then G is an upwards-directed family in T ∩ Σ and G ∈ T ∩ Σ, so S µ(X \ F ) = µ( G) = supG∈G µG = 0. If G is open and G ∩ F 6= ∅ then G ∈ / G so µ∗ (G ∩ F ) = µ(G ∩ F ) = µG > 0; thus F is self-supporting and is the support of µ. Q Q (e) Let X and Y be topological spaces with topological measures µ, ν respectively and a continuous inverse-measure-preserving function f : X → Y . Suppose that µ has a support E. Then f [E] is the support of ν. P P We have only to observe that for an open set H ⊆ Y νH > 0 ⇐⇒ µf −1 [H] > 0 ⇐⇒ f −1 [H] ∩ E 6= ∅ ⇐⇒ H ∩ f [E] 6= ∅ ⇐⇒ H ∩ f [E] 6= ∅. Q Q (f ) µ is strictly positive (with respect to T) if µ∗ G > 0 for every non-empty open set G ⊆ X, that is, X itself is the support of µ. *(g) If (X, T) is a topological space, and µ is a strictly positive σ-finite measure on X such that the domain Σ of µ includes a π-base U for T, then X is ccc. P P Let hEn in∈N be a sequence of sets of finite measure covering X. Let G be a disjoint family of non-empty open sets. For each G ∈ G, take P UG ∈ U \ {∅} such that UG ⊆ G; then µUG > 0, so there is an n(G) such that µ(En(G) ∩ UG ) > 0. Now G∈G,n(G)=k µ(Ek ∩ UG ) ≤ µEk is finite for every k, so {G : n(G) = k} must be countable and G is countable. Q Q 411O Example Lebesgue measure on R r is a Radon measure (256Ha); in particular, it is locally finite and tight. It is therefore τ -additive and effectively locally finite (411E, 411Gf). It is completion regular (because every compact set is a zero set, see 4A2Lc), outer regular with respect to the open sets (134F) and strictly positive.

411Q

Definitions

17

411P Example: Stone spaces (a) Let (Z, T, Σ, µ) be the Stone space of a semi-finite measure algebra (A, µ ¯), so that (Z, T) is a zero-dimensional compact Hausdorff space, (Z, Σ, µ) is complete and semi-finite, the open-and-closed sets are measurable, the negligible sets are the nowhere dense sets, and every measurable set differs by a nowhere dense set from an open-and-closed set (311I, 321K, 322Bd, 322Qa). (b) µ is inner regular with respect to the open-and-closed sets (322Qa); in particular, it is completion regular and tight. Consequently it is τ -additive (411E). (c) µ is strictly positive, because the open-and-closed sets form a base for T (311I) and a non-empty openand-closed set has non-zero measure. µ is effectively locally finite. P P Suppose that E ∈ Σ is not negligible. There is a measurable set F ⊆ E such that 0 < µF < ∞; now there is a non-empty open-and-closed set G included in F , in which case µG < ∞ and µ(E ∩ G) > 0. Q Q (d) The following are equiveridical, that is, if one is true so are the others: (i) (A, µ ¯) is localizable; (ii) µ is strictly localizable; (iii) µ is locally determined; (iv) µ is a quasi-Radon measure. P P The equivalence of (i)-(iii) is Theorem 322N. (iv)⇒(iii) is trivial. If one, therefore all, of (i)-(iii) are true, then µ is a topological measure, because if G ⊆ Z is open, then G is open-and-closed, by 314S, therefore measurable, and G \ G is nowhere dense, therefore also measurable. We know already that µ is complete, effectively locally finite and τ -additive, so that if it is also locally determined it is a quasi-Radon measure. Q Q (e) The following are equiveridical: (i) µ is a Radon measure; (ii) µ is totally finite; (iii) µ is locally finite; (iv) µ is outer regular with respect to the open sets. P P (ii)⇒(iv) If µ is totally finite and E ∈ Σ, then for any ² > 0 there is a closed set F ⊆ Z \ E such that µF ≥ µ(Z \ E) − ², and now G = Z \ F is an open set including E with µG ≤ µE + ². (iv)⇒(iii) Suppose that µ is outer regular with respect to the open sets, and z ∈ Z. Because Z is Hausdorff, {z} is closed. If it is open it is measurable, and because µ is semi-finite it must have finite measure. Otherwise it is nowhere dense, therefore negligible, and must be included in open sets of arbitrarily small measure. Thus in both cases z belongs to an open set of finite measure; as z is arbitrary, µ is locally finite. (iii)⇒(ii) Becasue Z is compact, this is a consequence of 411Ga. (i)⇒(iii) is part of the definition of ‘Radon measure’. Finally, (ii)+(iii)⇒(i), again directly from the definition and the facts set out in (a)-(b) above. Q Q 411Q Example: Dieudonn´ e’s measure Recall that a set E ⊆ ω1 is a Borel set iff either E or its complement includes a cofinal closed set (4A3J). So we may define a Borel measure µ on ω1 by saying that µE = 1 if E includes a cofinal closed set and µE = 0 if E is disjoint from a cofinal closed set. If E is disjoint from some cofinal closed set, so is any subset of E, so µ is complete. Since µ takes only the values 0 and 1, it is a purely atomic probability measure. µ is a topological measure; being totally finite, it is surely locally finite and effectively locally finite. It is inner regular with respect to the closed sets (because if µE > 0, there is a cofinal closed set F ⊆ E, and now F is a closed set with µF = µE), therefore outer regular with respect to the open sets. It is not τ -additive (because ξ = [0, ξ[ is an open set of zero measure for every ξ < ω1 , and the union of these sets is a measurable open set of measure 1). µ is not completion regular, because the set of countable limit ordinals is a closed set (4A1Bb) which does not include any uncountable zero set (see 411Ra below). The only self-supporting subset of ω1 is the empty set (because there is a cover of ω1 by negligible open sets). In particular, µ does not have a support. Remark There is a measure of this type on any ordinal of uncountable cofinality; see 411Xj.

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Topologies and measures I

411R

411R Example: The Baire σ-algebra of ω1 The Baire σ-algebra Ba(ω1 ) of ω1 is the countablecocountable algebra (4A3P). The countable-cocountable measure µ on ω1 is therefore a Baire measure on the definition of 411K. Since all sets of the form ]ξ, ω1 [ are zero sets, µ is inner regular with respect to the zero sets and outer regular with respect to the cozero sets. Since sets of the form [0, ξ[ (= ξ) form a cover of ω1 by measurable open sets of zero measure, µ is not τ -additive. 411X Basic exercises (a) Let (X, Σ, µ) be a totally finite measure space and T a topology on X. Show that µ is inner regular with respect to the closed sets iff it is outer regular with respect to the open sets, and is inner regular with respect to the zero sets iff it is outer regular with respect to the cozero sets. (b) Let µ be a Radon measure on R r , where r ≥ 1, and f ∈ L0 (µ). Show that f is locally integrable in R the sense of 411Fc iff it is locally integrable in the sense of 256E, that is, E f dν < ∞ for every bounded set E ⊆ Rr. (c) Let µ be a measure on a topological space, µ ˆ its completion and µ ˜ its c.l.d. version. Show that µ is locally finite iff µ ˆ is locally finite, and in this case µ ˜ is locally finite. > (d) Let µ be an effectively locally finite measure on a topological space X. (i) Show that the completion and c.l.d. version of µ are effectively locally finite. (ii) Show that if µ is complete and locally determined, then the union of the measurable open sets of finite measure is conegligible. (iii) Show that if X is hereditarily Lindel¨of then µ must be σ-finite. (e) Let X be a topological space and µ a measure on X. Let U ⊆ L0 (µ) be the set of equivalence classes of locally integrable functions in L0 (µ). Show that U is a solid linear subspace of L0 (µ). Show that if µ is locally finite then U includes Lp (µ) for every p ∈ [0, ∞]. (f ) Let X be a topological space. (i) Let µ, ν be two totally finite Borel measures which agree on the closed sets. Show that they are equal. (Hint: 136C.) (ii) Let µ, ν be two totally finite Baire measures which agree on the zero sets. Show that they are equal. (g) Let (X, T) be a topological space, µ a measure on X, and Y a subset of X; let TY , µY be the subspace topology and measure. Show that if µ is a topological measure, or locally finite, or a Borel measure, so is µY . (h) Let h(Xi , Σi , µi )ii∈I be a family of measure spaces, with direct sum (X, Σ, µ); suppose that we are given a topology Ti on each Xi , and let T be the disjoint union topology on X. Show that µ is a topological measure, or locally finite, or effectively locally finite, or a Borel measure, or a Baire measure, or strictly positive, iff every µi is. (i) Let (X, Σ, µ) and (Y, T, ν) be two measure spaces, with c.l.d. product measure λ on X × Y . Suppose we are given topologies T, S on X, Y respectively, and give X × Y the product topology. Show that λ is locally finite, or effectively locally finite, if µ and ν are. (j) Let κ be any cardinal of uncountable cofinality (definition: 3A1Fb). Show that there is a complete topological probability measure µ on κ defined by saying that µE = 1 if E includes a cofinal closed set in κ, 0 if E is disjoint from some cofinal closed set. Show that µ is inner regular with respect to the closed sets but is not completion regular. 411Y Further exercises (a) Show that a function f : R r → R s is measurable iff it is almost continuous (where R r is endowed with Lebesgue measure and its usual topology, of course). (Hint: 256F.) (b) Let (X, ρ) be a metric space, r ≥ 0, and write µHr for r-dimensional Hausdorff measure on X (264K, §471). (i) Show that µHr is a topological measure, outer regular with respect to the Borel sets. (ii) Show that if X is complete then the c.l.d. version of µHr is tight, therefore completion regular.

412A

Inner regularity

19

(c) Let (X, T, Σ, µ) be a topological measure space. Set E = {E : E ⊆ X, µ(∂E) = 0}, where ∂E is the boundary of A. (i) Show that E is a subalgebra of PX, and that every member of E is measured by the completion of µ. (E is sometimes called the Jordan algebra of (X, T, Σ, µ). Do not confuse with the ‘Jordan algebras’ of abstract algebra.) (ii) Suppose that µ is complete and totally finite and inner regular with respect to the closed sets, and that T is normal. Show that {E • : E ∈ E} is dense in the measure algebra of µ endowed with its usual topology. (Hint: if f : X → R is continuous, then {x : f (x) ≤ α} ∈ E for all but countably many α.) (iii) Suppose that µ is a quasi-Radon measure and T is completely regular. Show that {E • : E ∈ E} is dense in the measure algebra of µ. (Hint: 414Aa.) 411 Notes and comments Of course the list above can give only a rough idea of the ways in which topologies and measures can interact. In particular I have rather arbitrarily given a sort of priority to three particular relationships between the domain Σ of a measure and the topology: ‘topological measure space’ (in which Σ includes the Borel σ-algebra), ‘Borel measure’ (in which Σ is precisely the Borel σ-algebra) and ‘Baire measure’ (in which Σ is the Baire σ-algebra). Abstract topological measure theory is a relatively new subject, and there are many technical questions on which different authors take different views. For instance, the phrase ‘Radon measure’ is commonly used to mean what I would call a ‘tight locally finite Borel measure’ (cf. 416F); and some writers enlarge the definition of ‘topological measure’ to include Baire measures as defined above. I give very few examples at this stage, two drawn from the constructions of Volumes 1-3 (Lebesgue measure and Stone spaces, 411O-411P) and one new one (‘Dieudonn´e’s measure’, 411Q), with a glance at the countable-cocountable measure of ω1 (411R). The most glaring omission is that of the product measures on {0, 1}I and [0, 1]I . I pass these by at the moment because a proper study of them requires rather more preparation than can be slipped into a parenthesis. (I return to them in 416U.) I have also omitted any discussion of ‘measurable’ and ‘almost continuous’ functions, except for a reference to a theorem in Volume 2 (411Ya), which will have to be repeated and amplified later on (§418). There is an obvious complementarity between the notions of ‘inner’ and ‘outer’ regularity (411B, 411D), but it works well only for totally finite spaces (411Xa); in other cases it may not be obvious what will happen (411O, 411Pe, 412W).

412 Inner regularity As will become apparent as the chapter progresses, the concepts introduced in §411 are synergic; their most interesting manifestations are in combinations of various kinds. Any linear account of their properties will be more than usually like a space-filling curve. But I have to start somewhere, and enough results can be expressed in terms of inner regularity, more or less by itself, to be a useful beginning. After a handful of elementary basic facts (412A) and a list of standard applications (412B), I give some useful sufficient conditions for inner regularity of topological and Baire measures (412D, 412E, 412G), based on an important general construction (412C). The rest of the section amounts to a review of ideas from Volume 2 and Chapter 32 in the light of the new concept here. I touch on completions (412H), c.l.d. versions and complete locally determined spaces (412H, 412J, 412L), strictly localizable spaces (412I), inverse-measurepreserving functions (412K, 412M), measure algebras (412N), subspaces (412O, 412P), indefinite-integral measures (412Q) and product measures (412R-412V), with a brief mention of outer regularity (412W); most of the hard work has already been done in Chapters 21 and 25. 412A I begin by repeating a lemma from Chapter 34, with some further straightforward facts. Lemma (a) Let (X, Σ, µ) be a measure space and K a family of sets such that whenever E ∈ Σ and µE > 0 there is a K ∈ K ∩ Σ such that K ⊆ E and µK > 0. ThenPwhenever E ∈ Σ there is a countable disjoint family hKi ii∈I in K ∩ Σ such that Ki ⊆ E for every i and i∈I µKi = µE. If moreover (†) K ∪ K 0 ∈ K whenever K, K 0 are disjoint members of K, S then µ is inner regular with respect to K. If i∈I Ki ∈ K for every countable disjoint family hKi ii∈I in K, then for every E ∈ Σ there is a K ∈ K ∩ Σ such that K ⊆ E and µK = µE.

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Topologies and measures

412A

(b) Let (X, Σ, µ) be a measure space, T a σ-subalgebra of Σ, and K a family of sets. If µ is inner regular with respect to T and µ¹ T is inner regular with respect to K, then µ is inner regular with respect to K. (c) Let (X, Σ, µ) be a semi-finite measure space and hKn in∈N a sequence of families of sets such that µ is inner regular with respect to Kn and T (‡) if hKi ii∈N is a non-increasing sequence in Kn , then i∈N Ki ∈ Kn T for every n ∈ N. Then µ is inner regular with respect to n∈N Kn . proof (a) This is 342B-342C. (b) If E ∈ Σ and γ < µE, there are an F ∈ T such that F ⊆ E and µF > γ, and a K ∈ K ∩ T such that K ⊆ F and µK ≥ γ. (c) Suppose that E ∈ Σ and that 0 ≤ γ < µE. Because µ is semi-finite, there is an F ∈ Σ such that F ⊆ E and γ < µF < ∞ (213A). Choose hKi ii∈N inductively, as follows. Start with K0 = F . Given that Ki ∈ Σ and γ < µKi , then let ni ∈ N be such that 2−ni (i + 1) is an odd integer, and choose Ki+1 ∈ Kni such that Ki+1T⊆ Ki and µKi+1 > γ; this will be possible because µ is inner regular with respect to Kni . Consider K = i∈N Ki . Then K ⊆ E and µK = limi→∞ µKi ≥ γ. But also T K = j∈N K2n (2j+1) ∈ Kn T because hK2n (2j+1) ij∈N is a non-increasing sequence in Kn , for each n. So K ∈ n∈N Kn . As E and γ are T arbitrary, µ is inner regular with respect to n∈N Kn . 412B Corollary Let (X, Σ, µ) be a measure space and T a topology on X. Suppose that K is either the family of Borel subsets of X or the family of closed subsets of X or the family of compact subsets of X or the family of zero sets in X, and suppose that whenever E ∈ Σ and µE > 0 there is a K ∈ K ∩ Σ such that K ⊆ E and µK > 0. Then µ is inner regular with respect to K. proof In every case, K satisfies the condition (†) of 412Aa. 412C The next lemma provides a particularly useful method of proving that measures are inner regular with respect to ‘well-behaved’ families of sets. Lemma Let (X, Σ, µ) be a semi-finite measure space, and suppose that A ⊆ Σ and K are such that ∅ ∈ A ⊆ Σ, (†) K ∪ K 0 ∈ K whenever K, K 0 ∈ K, T (‡) n∈N Kn ∈ K for every sequence hKn in∈N in K, X \ A ∈ A for every A ∈ A, whenever A ∈ A, F ∈ Σ and µ(A ∩ F ) > 0, there is a K ∈ K ∩ A such that K ⊆ A and µ(K ∩ F ) > 0. Let T be the σ-subalgebra of Σ generated by A. Then µ¹ T is inner regular with respect to K. proof (a) Write A for the measure algebra of (X, Σ, µ), and L = K ∩ T, so that L is also closed under finite unions and countable intersections. Set H = {E : E ∈ Σ, supL∈L,L⊆E L• = E • } in A, T0 = {E : E ∈ H, X \ E ∈ H}, so that the last two conditions tell us that A ⊆ T0 . (b) The intersection of any sequence in H belongs to H. P P Let hHn in∈N be a sequence in H with intersection H. Write An for {L• : L ∈ L, L ⊆ Hn } ⊆ A for each n ∈ N. Since A is weakly (σ, ∞)distributive (322F), An is upwards-directed, and sup An = Hn• for each n ∈ N,

412F

Inner regularity

21

H • = inf Hn• n∈N

(because F 7→ F : Σ → A is sequentially order-continuous, by 321H) = inf sup An = sup{ inf an : an ∈ An for every n ∈ N} •

n∈N

(316J) = sup{(

n∈N

\

Ln )• : Ln ∈ L, Ln ⊆ Hn for every n ∈ N}

n∈N

⊆

{L• : L ∈ L, L ⊆ H}

⊆

H •,

(by (‡))

and H ∈ H. Q Q (c) The union of any sequence in H belongs to H. P P If hHn in∈N is a sequence in H with union H then supL∈L,L⊆H L• ⊇ supn∈N supL∈L,L⊆En L• = supn∈N Hn• = H • , so H ∈ H. Q Q (d) T0 is a σ-subalgebra of Σ. P P (i) ∅ and X belong to A ⊆ H, so ∅ ∈ T0 . (ii) Obviously X \ E ∈ T0 0 wheneverTE ∈ T . (iii) If hEn in∈N is a sequence in T0 with union E then E ∈ H, by (c); but also Q X \ E = n∈N (X \ En ) belongs to H, by (b). So E ∈ T0 . Q (e) Accordingly T ⊆ T0 , and E • = supL∈L,L⊆E L• for every E ∈ T. It follows at once that if E ∈ T and µE > 0, there must be an L ∈ L such that L ⊆ E and µL > 0; since (†) is true, and L ⊆ T, we can apply 412Aa to see that µ¹ T is inner regular with respect to L, therefore with respect to K. 412D As corollaries of the last lemma I give two-and-a-half basic theorems. Theorem Let (X, T) be a topological space and µ a semi-finite Baire measure on X. Then µ is inner regular with respect to the zero sets. proof Write Σ for the Baire σ-algebra of X, the domain of µ, K for the family of zero sets, and A for K ∪ {X \ K : K ∈ K}. Since the union of two zero sets is a zero set (4A2C(b-ii)), the intersection of a sequence of zero sets is a zero set (4A2C(b-iii)), and the complement of a zero set is the union of a sequence of zero sets (4A2C(b-vi)), the conditions of 412C are satisfied; and as the σ-algebra generated by A is just Σ, µ is inner regular with respect to K. 412E Theorem Let (X, T) be a perfectly normal topological space (e.g., any metrizable space). Then any semi-finite Borel measure on X is inner regular with respect to the closed sets. proof Because the Baire and Borel σ-algebras are the same (4A3Kb), this is a special case of 412D. 412F Lemma Let (X, Σ, µ) be a measure space and T a topology on X such that µ is effectively locally finite with respect to T. Then µE = sup{µ(E ∩ G) : G is a measurable open set of finite measure} for every E ∈ Σ. proof Apply 412Aa with K the family of subsets of measurable open sets of finite measure.

22

Topologies and measures

412G

412G Theorem Let (X, Σ, µ) be a measure space with a topology T such that µ is effectively locally finite with respect to T and Σ is the σ-algebra generated by T ∩ Σ. If µG = sup{µF : F ∈ Σ is closed, F ⊆ G} for every measurable open set G of finite measure, then µ is inner regular with respect to the closed sets. proof In 412C, take K to be the family of measurable closed subsets of X, and A to be the family of measurable sets which are either open or closed. If G ∈ Σ ∩ T, F ∈ Σ and µ(G ∩ F ) > 0, then there is an open set H of finite measure such that µ(H ∩ G ∩ F ) > 0, because µ is effectively locally finite; now there is a K ∈ K such that K ⊆ H ∩ G and µK > µ(H ∩ G) − µ(H ∩ G ∩ F ), so that µ(K ∩ F ) > 0. This is the only non-trivial item in the list of hypotheses in 412C, so we can conclude that µ¹ T is inner regular with respect to K, where T is the σ-algebra generated by A; but of course this is just Σ. Remark There is a similar result in 416F(iii) below. 412H Proposition Let (X, Σ, µ) be a measure space and K a family of sets. (a) If µ is inner regular with respect to K, so are its completion µ ˆ (212C) and c.l.d. version µ ˜ (213E). (b) Now suppose that T (‡) n∈N Kn ∈ K whenever hKn in∈N is a non-increasing sequence in K. If either µ ˆ is inner regular with respect to K or µ is semi-finite and µ ˜ is inner regular with respect to K, then µ is inner regular with respect to K. proof (a) If F belongs to the domain of µ ˆ, then there is an E ∈ Σ such that E ⊆ F and µ ˆ(F \ E) = 0. So if 0 ≤ γ < µ ˆF = µE, there is a K ∈ K ∩ Σ such that K ⊆ E ⊆ F and µ ˆK = µK ≥ γ. If H belongs to the domain of µ ˜ and 0 ≤ γ < µ ˜H, there is an E ∈ Σ such that µE < ∞ and µ ˆ(E ∩ H) > γ (213D). Now there is a K ∈ K ∩ Σ such that K ⊆ E ∩ H and µK ≥ γ. As µK < ∞, µ ˜K = µK ≥ γ. (b) Write µ ˇ for whichever of µ ˆ, µ ˜ is supposed to be inner regular with respect to K. Then µ ˇ is inner regular with respect to Σ (212Ca, 213Fc), so is inner regular with respect to K ∩ Σ (412Ac). Also µ ˇ extends µ (212D, 213Hc). Take E ∈ Σ and γ < µE = µ ˇE. Then there is a K ∈ K ∩ Σ such that K ⊆ E and γ<µ ˇK = µK. As E and γ are arbitrary, µ is inner regular with respect to K. 412I Lemma Let (X, Σ, µ) be a strictly localizable measure space and K a family of sets such that whenever E ∈ Σ and µE > 0 there is a K ∈ K ∩ Σ such that K ⊆ E and µK > 0. (a) There is a decomposition hXi ii∈I of X such that at most one Xi does not belong to K, and that exceptional one, if any, is negligible. P (b) There is a disjoint family L ⊆ K ∩ Σ such that µ∗ A = L∈L µ∗ (A ∩ L) for every A ⊆ X. (c) If µ is σ-finite then the family hXi ii∈I of (a) and the set L of (b) can be taken to be countable. proof (a) Let hEj ij∈J be any decomposition of X. For each j ∈ J, let Kj be a maximal disjoint subset of {K : K ∈ K ∩ Σ, K ⊆ Ej , µK > 0}. S Because µEj < ∞, Kj must be countable. Set Ej0 = Ej \ Kj . By the maximality of Kj , Ej0 cannot include S any non-negligible set in K ∩ Σ; but this means that µEj0 = 0. Set X 0 = j∈J Ej0 . Then P P µX 0 = j∈J µ(X 0 ∩ Ej ) = j∈J µEj0 = 0. S Note that if j, j 0 ∈ J are distinct, and K ∈ Kj , K 0 ∈ Kj 0 , then K ∩ K 0 = ∅; thus L = j∈J Kj is disjoint. Let hXi ii∈I be any indexing of {X 0 } ∪ L. This is a partition (that is, disjoint cover) of X into sets of finite measure. If E ⊆ X and E ∩ Xi ∈ Σ for every i ∈ I, then for every j ∈ J S E ∩ Ej = (E ∩ X 0 ∩ Ej ) ∪ K∈Kj E ∩ K belongs to Σ, so that E ∈ Σ and µE =

P

P j∈J

K∈Kj

µ(E ∩ K) =

P i∈I

µ(E ∩ Xi ).

Thus hXi ii∈I is a decomposition of X, and it is of the right type because every Xi but one belongs to L ⊆ K.

412K

Inner regularity

(b) If now A ⊆ X is any set, µ∗ A = µA A =

P i∈I

µA (A ∩ Xi ) =

23

P i∈I

µ∗ (A ∩ Xi )

by 214Ia, writing µA for the subspace measure on A. So we have P P µ∗ A = µ∗ (A ∩ X 0 ) + L∈L µ∗ (A ∩ L) = L∈L µ∗ (A ∩ L), while L ⊆ K is disjoint. (c) If µ is σ-finite we can take J to be countable, so that I and L will also be countable. 412J Proposition Let (X, Σ, µ) be a complete locally determined measure space, and K a family of sets such that µ is inner regular with respect to K. (a) If E ⊆ X is such that E ∩ K ∈ Σ for every K ∈ K ∩ Σ, then E ∈ Σ. (b) If E ⊆ X is such that E ∩ K is negligible for every K ∈ K ∩ Σ, then E is negligible. (c) For any A ⊆ X, µ∗ A = supK∈K∩Σ µ∗ (A ∩ K). R (d) Let f be a non-negative [0, ∞]-valued function defined R R on a subset of X. If K f is defined in [0, ∞] for every K ∈ K, then f is defined and equal to supK∈K K f . R (e) If f is a µ-integrable function and ² > 0, there is a K ∈ K such that X\K |f | ≤ ². Remark In (c), we must interpret sup ∅ as 0 if K ∩ Σ = ∅. proof (a) If F ∈ Σ and µF < ∞, then E ∩ F ∈ Σ. P P If µF = 0, this is trivial, because µ is complete and E ∩ F is negligible. Otherwise, there is a sequence hKn in∈N in K ∩ Σ such that Kn ⊆ F for each n and S supn∈N µKn = µF . Now E ∩ F \ n∈N Kn is negligible, therefore measurable, while E ∩ Kn is measurable for every n ∈ N, by hypothesis; so E ∩ F is measurable. Q Q As µ is locally determined, E ∈ Σ, as claimed. (b) By (a), E ∈ Σ; and because µ is inner regular with respect to K, µE must be 0. (c) Let µA be the subspace measure on A. Because µ is complete and locally determined, µA is semi-finite (214Ic). So if 0 ≤ γ < µ∗ A = µA A, there is an H ⊆ A such that µA H is defined, finite and greater than γ. Let E ∈ Σ be a measurable envelope of H (132Ee), so that µE = µ∗ H > γ. Then there is a K ∈ K ∩ Σ such that K ⊆ E and µK ≥ γ. In this case µ∗ (A ∩ K) ≥ µ∗ (H ∩ K) = µ(E ∩ K) = µK ≥ γ. As γ is arbitrary, µ∗ A ≤ supK∈K∩Σ µ∗ (A ∩ K); but the reverse inequality is trivial, so we have the result. (d) Applying (b) with E = X \ dom f , we see that f is defined almost everywhere R in X. Applying (a) with E = {x : Rx ∈ dom f, f (x) ≥ α} for each α ∈ R, we see that f is measurable. So f is defined in [0, ∞], R R and of course f ≥ sup f . If γ < f , there is a non-negative simple function g such that g ≤a.e. K∈K K R R f and g R> γ; taking E = {x : g(x) > 0}, there R R is a K ∈ K Rsuch that K ⊆ E and µ(E \ K)kgk∞ ≤ γ − g, so that K f ≥ K g ≥ γ. As γ is arbitrary, f = supK∈K K f . R R (e) By (d), there is a K ∈ K such that K |f | ≥ |f | − ². Remark See also 413F below. 412K Proposition Let (X, Σ, µ) be a complete locally determined measure space, (Y, T, ν) a measure space and f : X → Y a function. Suppose that K ⊆ T is such that (i) ν is inner regular with respect to K; (ii) f −1 [K] ∈ Σ and µf −1 [K] = νK for every K ∈ K; (iii) whenever E ∈ Σ and µE > 0 there is a K ∈ K such that νK < ∞ and µ(E ∩ f −1 [K]) > 0. Then f is inverse-measure-preserving for µ and ν. proof (a) If F ∈ T, E ∈ Σ and µE < ∞, then E ∩ f −1 [F ] ∈ Σ. P P Let H1 , H2 ∈ Σ be measurable envelopes for E ∩ f −1 [F ] and E \ f −1 [F ] respectively. ?? If µ(H1 ∩ H2 ) > 0, there is a K ∈ K such that νK is finite

24

Topologies and measures

412K

and µ(H1 ∩ H2 ∩ f −1 [K]) > 0. Because ν is inner regular with respect to K, there are K1 , K2 ∈ K such that K1 ⊆ K ∩ F , K2 ⊆ K \ F and νK1 + νK2 > ν(K ∩ F ) + ν(K \ F ) − µ(H1 ∩ H2 ∩ f −1 [K]) = νK − µ(H1 ∩ H2 ∩ f −1 [K]). Now µ(H1 ∩ f −1 [K2 ]) = µ∗ (E ∩ f −1 [F ] ∩ f −1 [K2 ]) = 0, µ(H2 ∩ f −1 [K1 ]) = µ∗ (E ∩ f −1 [K1 ] \ f −1 [F ]) = 0, so µ(H1 ∩ H2 ∩ f −1 [K1 ∪ K2 ]) = 0 and µ(H1 ∩ H2 ∩ f −1 [K]) ≤ µ(f −1 [K] \ f −1 [K1 ∪ K2 ]) = µf −1 [K] − µf −1 [K1 ] − µf −1 [K2 ] = νK − νK1 − νK2 < µ(H1 ∩ H2 ∩ f −1 [K]), which is absurd. X X Now (E ∩ H1 ) \ (E ∩ f −1 [F ]) ⊆ H1 ∩ H2 is negligible, therefore measurable (because µ is complete), and E ∩ f −1 [F ] ∈ Σ, as claimed. Q Q (b) It follows (because µ is locally determined) that f −1 [F ] ∈ Σ for every F ∈ T. (c) If F ∈ T and νF = 0 then µf −1 [F ] = 0. P P?? Otherwise, there is a K ∈ K such that νK < ∞ and 0 < µ(f −1 [F ] ∩ f −1 [K]) = µf −1 [F ∩ K]. Let K 0 ∈ K be such that K 0 ⊆ K \ F and νK 0 > νK − µf −1 [F ∩ K]. Then f −1 [K 0 ] ∩ f −1 [F ∩ K] = ∅, so νK = µf −1 [K] ≥ µf −1 [K 0 ] + µf −1 [F ∩ K] > νK 0 + νK − νK 0 = νK, which is absurd. X XQ Q (d) Finally, µf −1 [F ] P = νF for every F ∈ T. P P Let hKi ii∈I be S a countable disjoint family in K such that Ki ⊆ F for every i and i∈I νKi = νF (412Aa). Set F 0 = F \ i∈I Ki . Then P P µf −1 [F ] = µf −1 [F 0 ] + i∈I µf −1 [Ki ] = µf −1 [F 0 ] + i∈I νKi = µf −1 [F 0 ] + νF . If νF = ∞ then surely µf −1 [F ] = ∞ = νF . Otherwise, νF 0 = 0 so µf −1 [F 0 ] = 0 (by (c)) and again µf −1 [F ] = νF . Q Q Thus f is inverse-measure-preserving. 412L Corollary Let X be a set and K a family of subsets of X. Suppose that µ, ν are two complete locally determined measures on X, with domains including K, agreeing on K, and both inner regular with respect to K. Then they are identical (and, in particular, have the same domain). proof Apply 412K with X = Y and f the identity function to see that µ extends ν; similarly, ν extends µ and the two measures are the same. 412M Corollary Let (X, Σ, µ) be a complete probability space, (Y, T, ν) a probability space and f : X → Y a function. Suppose that whenever F ∈ T and νF > 0 there is a K ∈ T such that K ⊆ F , νK > 0, f −1 [K] ∈ Σ and µf −1 [K] ≥ νK. Then f is inverse-measure-preserving. proof Set K∗ = {K : K ∈ T, f −1 [K] ∈ Σ, µf −1 [K] ≥ νK}. Then K∗ is closed under countable disjoint unions and includes K, so for every F ∈ T there is a K ∈ K∗ such that K ⊆ F and νK = νF , by 412Aa. But this means that µf −1 [K] = νK for every K ∈ K∗ . P P There is a K 0 ∈ K∗ such that K 0 ⊆ Y \ K and 0 νK = 1 − νK; but in this case µf −1 [K 0 ] + µf −1 [K] ≤ 1 = νK 0 + νK, so µf −1 [K] must be equal to νK. Q Q Moreover, there is a K ∗ ∈ K∗ such that νK ∗ = νY = 1, so −1 ∗ −1 µf [K ] = µX = 1 and µ(E ∩ f [K ∗ ]) > 0 whenever µE > 0. Applying 412K to K∗ we have the result.

412Q

Inner regularity

25

412N Lemma Let (X, Σ, µ) be a measure space and K a family of subsets of X such that µ is inner regular with respect to K. Then E • = sup{K • : K ∈ K ∩ Σ, K ⊆ E} in the measure algebra A of µ, for every E ∈ Σ. In particular, {K • : K ∈ K ∩ Σ} is order-dense in A; and if K is closed under finite unions, then {K • : K ∈ K ∩ Σ} is topologically dense in A for the measure-algebra topology. proof ?? If E • 6= sup{K • : K ∈ K ∩ Σ, K ⊆ E}, there is a non-zero a ∈ A such that a ⊆ E • \ K • whenever K ∈ K ∩ Σ and K ⊆ E. Express a as F • where F ⊆ E. Then µF > 0, so there is a K ∈ K ∩ Σ such that K ⊆ F and µK > 0. But in this case 0 6= K • ⊆ a, while K ⊆ E. X X It follows at once that D = {K • : K ∈ K ∩ Σ} is order-dense. If K is closed under finite unions, and a ∈ A, then Da = {d : d ∈ D, d ⊆ a} is upwards-directed and has supremum a, so a ∈ Da ⊆ D (323D(a-ii)). 412O Lemma Let (X, Σ, µ) be a measure space and K a family of subsets of X such that µ is inner regular with respect to K. (a) If E ∈ Σ, then the subspace measure µE (131B) is inner regular with respect to K. (b) Let Y ⊆ X be any set such that the subspace measure µY (214A-214B) is semi-finite. Then µY is inner regular with respect to KY = {K ∩ Y : K ∈ K}. proof (a) This is elementary. (b) Suppose that F belongs to the domain ΣY of µY and 0 ≤ γ < µY F . Because µY is semi-finite there is an F 0 ∈ ΣY such that F 0 ⊆ F and γ < µY F 0 < ∞. Let E be a measurable envelope for F 0 with respect to µ, so that µE = µ∗ F 0 = µY F 0 > γ. There is a K ∈ K ∩ Σ such that K ⊆ E and µK ≥ γ, in which case K ∩ Y ∈ KY ∩ ΣY and µY (K ∩ Y ) = µ∗ (K ∩ Y ) = µ∗ (K ∩ F 0 ) = µ(K ∩ E) = µK ≥ γ. As F and γ are arbitrary, µY is inner regular with respect to KY . Remark Recall from 214I that if (X, Σ, µ) has locally determined negligible sets (in particular, is either strictly localizable or complete and locally determined), then all its subspaces are semi-finite. 412P Proposition Let (X, Σ, µ) be a measure space, T a topology on X and Y a subset of X; write TY for the subspace topology of Y and µY for the subspace measure on Y . Suppose that either Y ∈ Σ or µY is semi-finite. (a) If µ is a topological measure, so is µY . (b) If µ is inner regular with respect to the Borel sets, so is µY . (c) If µ is inner regular with respect to the closed sets, so is µY . (d) If µ is inner regular with respect to the zero sets, so is µY . (e) If µ is effectively locally finite, so is µY . proof (a) is an immediate consequence of the definitions of ‘subspace measure’, ‘subspace topology’ and ‘topological measure’. The other parts follow directly from 412O if we recall that (i) a subset of Y is Borel for TY whenever it is expressible as Y ∩ E for some Borel set E ⊆ X (4A3Ca); (ii) a subset of Y is closed in Y whenever it is expressible as Y ∩ F for some closed set F ⊆ X; (iii) a subset of Y is a zero set in Y whenever it is expressible as Y ∩ F for some zero set F ⊆ X (4A2C(b-v)); (iv) µ is effectively locally finite iff it is inner regular with respect to subsets of open sets of finite measure. 412Q Proposition Let (X, Σ, µ) be a measure space, and ν an indefinite-integral measure over µ (definition: 234B). If µ is inner regular with respect to a family K of sets, so is ν. proof Because µ and its completion µ ˆ give the same integrals, ν is an indefinite-integral measure over µ ˆ; and as µ ˆ is still inner regular with respect to K (412H), we may suppose that µ itself is complete. Let f

26

Topologies and measures

412Q

be a Radon-Nikod´ ym derivative of ν with respect to µ; by 234Ca, we may suppose that f : X → [0, ∞[ is Σ-measurable. Suppose that F ∈ dom(ν) and that γ < νF . Set G = {x : f (x) > 0}, so that F ∩ G ∈ Σ (234D). For n ∈ N, set Hn = {x : x ∈ F, 2−n ≤ f (x) ≤ 2n }, so that Hn ∈ Σ and R R νF = f × χF dµ = limn→∞ f × χHn dµ. R Let n ∈ N be such that f × χHn dµ > γ. n If µHn = ∞, there is a K ∈ K such that R K ⊆ Hn and µK ≥ 2 Rγ, so that νK ≥ γ. If µHnR is finite, there is a K R∈ K such that 2n (µHn − µK) ≤ f × χHn dµ − γ, so that f × χ(Hn \ K)dµ + γ ≤ f × χHn and νK = f × χK dµ ≥ γ. Thus in either case we have a K ∈ K such that K ⊆ F and νK ≥ γ; as F and γ are arbitrary, ν is inner regular with respect to K. 412R Lemma Let (X, Σ, µ) and (Y, T, ν) be measure spaces, with c.l.d. product space (X × Y, Λ, λ) (251F). Suppose that K ⊆ PX, L ⊆ PY , M ⊆ P(X × Y ) are such that (i) µ is inner regular with respect to K; (ii) ν is inner regular with respect to L; (iii) K × L ∈ M for all K ∈ K, L ∈ L; (iv) T M ∪ M 0 ∈ M whenever M , M 0 ∈ M; (v) n∈N Mn ∈ M for every sequence hMn in∈N in M. Then λ is inner regular with respect to M. proof Write A = {E × Y : E ∈ Σ} ∪ {X × F : F ∈ T}. Then if V ∈ A, W ∈ Λ and λ(W ∩ V ) > 0, there is an M ∈ M ∩ A such that M ⊆ W and λ(M ∩ V ) > 0. P P Suppose that V = E × Y where E ∈ Σ. There must be E0 ∈ Σ and F0 ∈ T, both of finite measure, such that λ(W ∩ V ∩ (E0 × F0 )) > 0 (251F). Now there is a K ∈ K such that K ⊆ E ∩ E0 and µ((E ∩ E0 ) \ K) · νF0 < λ(W ∩ V ∩ (E0 × F0 )); but this means that M = K × Y is included in V and µ(W ∩ M ) > 0. Reversing the roles of the coordinates, the same argument deals with the case in which V = X × F for some F ∈ T. Q Q b is inner regular with respect to K. But λ is inner regular with respect to Σ⊗T b By 412C, λ0 = λ¹Σ⊗T (251Ib) so is also inner regular with respect to M (412Ab). 412S Proposition Let (X, Σ, µ) and (Y, T, ν) be measure spaces, with c.l.d. product space (X × Y, Λ, λ). Let T, S be topologies on X and Y respectively, and give X × Y the product topology. (a) If µ and ν are inner regular with respect to the closed sets, so is λ. (b) If µ and ν are tight (that is, inner regular with respect to the closed compact sets), so is λ. (c) If µ and ν are inner regular with respect to the zero sets, so is λ. (d) If µ and ν are inner regular with respect to the Borel sets, so is λ. (e) If µ and ν are effectively locally finite, so is λ. proof We have only to read the conditions (i)-(v) of 412R carefully and check that they apply in each case. (In part (e), recall that ‘effectively locally finite’ is the same thing as ‘inner regular with respect to the subsets of open sets of finite measure’.) 412T Lemma Let h(Xi , Σi , µi )ii∈I be a family of probability spaces, with product probability space (X, Λ, λ) (§254). Suppose that Ki ⊆ PXi , M ⊆ PX are such that (i) µi is inner regular with respect to Ki for each i ∈ I; (ii) πi−1 [K] ∈ M for every i ∈ I, K ∈ Ki , writing πi (x) = x(i) for x ∈ X; (iii) T M ∪ M 0 ∈ M whenever M , M 0 ∈ M; (iv) n∈N Mn ∈ M for every sequence hMn in∈N in M. Then λ is inner regular with respect to M. proof The argument is nearly identical to that of 412R. Write A = {πi−1 [E] : i ∈ I, E ∈ Σi }. Then if V ∈ A, W ∈ Λ and λ(W ∩ V ) > 0, express V as πi−1 [E], where i ∈ I and E ∈ Σi , and take K ∈ Ki such that K ⊆ E and µi (E \ K) < λ(W ∩ V ); then M = πi−1 [K] belongs to M ∩ A, is included in W , and meets V in a non-negligible set. So, just as in 412R, the conditions of 412C are met. N It follows that λ0 = λ¹ c i∈I Σi is inner regular with respect to K. But λ is the completion of λ0 (254Fd, 254Ff), so is also inner regular with respect to M (412Ha).

*412W

Inner regularity

27

412U Proposition Let h(Xi , Σi , µi )ii∈I be a family of probability spaces, with product probability space (X, Λ, λ). Suppose that we are given a topology Ti on each Xi , and let T be the product topology on X. (a) If every µi is inner regular with respect to the closed sets, so is λ. (b) If every µi is inner regular with respect to the zero sets, so is λ. (c) If every µi is inner regular with respect to the Borel sets, so is λ. proof This follows from 412T just as 412S follows from 412R. 412V Corollary Let h(Xi , Σi , µi )ii∈I be a family of probability spaces, with product probability space (X, Λ, λ). Suppose that we are given a Hausdorff topology Ti on each Xi , and let T be the product topology on X. Suppose that every µi is tight, and that Xi is compact for all but countably many i ∈ I. Then λ is tight. proof By 412Ua, λ is inner regular with respect to the closed sets. If W ∈ Λ and γ < λW , let V ⊆ W be a measurable closed set such that λV > γ. Let J be the set of those i ∈ I such that Xi is not compact; we P are supposing that J is countable. Let h²i ii∈J be a family of strictly positive real numbers such that i∈J ²j ≤ λV − γ (4A1P). For each i ∈ J, let KiQ⊆ Xi be a compact measurable set such that µi (Xi \ Ki ) ≤ ²i ; and for i ∈ I \ J, set Ki = Xi . Then K = i∈I Ki is a compact measurable subset of X, and P λ(X \ K) ≤ i∈J µ(Xi \ Ki ) ≤ λV − γ, so λ(K ∩ V ) ≥ γ; while K ∩ V is a compact measurable subset of W . As W and γ are arbitrary, λ is tight. *412W Outer regularity I have already mentioned the complementary notion of ‘outer regularity’ (411D). In this book it will not be given much prominence. It is however a useful tool when dealing with Lebesgue measure (see, for instance, the proof of 225K), for reasons which the next proposition will make clear. Proposition Let (X, Σ, µ) be a measure space and T a topology on X. (a) Suppose that µ is outer regular with respect to the open sets. Then for any integrable function f : X → [0,R ∞] and ²R > 0, there is a lower semi-continuous measurable function g : X → [0, ∞] such that f ≤ g and g ≤ ² + f . (b) Now suppose that there is a sequence of measurable open sets of finite measure covering X. Then the following are equiveridical: (i) µ is inner regular with respect to the closed sets; (ii) µ is outer regular with respect to the open sets; (iii) for any measurable set E ⊆ X and ² > 0, there are a measurable closed set F ⊆ E and a measurable open set H ⊇ E such that µ(H \ F ) ≤ ²; (iv) for every measurable function f : X → R [0, ∞[ and ² > 0, there is a lower semi-continuous measurable function g : X → [0, ∞] such that f ≤ g and g − f ≤ ²; (v) for every measurable function f : X → R and ² > 0, there is a lower semi-continuous measurable function g : X → ]−∞, ∞] such that f ≤ g and µ{x : g(x) ≥ f (x) + ²} ≤ ². R proof (a) Let η ∈ ]0, 1] be such that η(7 + f dµ) ≤ ². For n ∈ Z, set En = {x : (1 + η)n ≤ f (x) < (1 + η)n+1 }, and let En0 ∈ Σ be a measurable cover of En ; let Gn ⊇ En0 be a measurable open set such P∞ −|n| 0 that µGn ≤ 3 η + µEn . Set g = n=−∞ (1 + η)n+1 χGn . Then g is lower semi-continuous (4A2B(d-iii), 4A2B(d-v)), f ≤ g and Z gdµ =

∞ X

(1 + η)n+1 µGn

n=−∞

≤ (1 + η)

∞ X

(1 + η)n µEn0 +

n=−∞

Z

≤ (1 + η)

f dµ + 7η ≤

Z

∞ X n=−∞

f dµ + ²,

(1 + η)n+1 3−|n| η

28

Topologies and measures

*412W

as required. (b) Let hGn in∈N be a sequence of open sets of finite measure covering X; replacing it by h if necessary, we may suppose that hGn in∈N is non-decreasing and that G0 = ∅.

S i
Gi in∈N

(i)⇒(iii) Suppose that µ is inner regular with respect to the closed sets, and that E ∈ Σ, ² > 0. −n−2 For each ². Then S n ∈ N let Fn ⊆ Gn \ E be a measurable closed set such that µF1n ≥ µ(Gn \ E) − 2 H = n∈N (Gn \ Fn ) is a measurable open set including E and µ(H \ E) ≤ 2 ². Applying the same argument to X \ E, we get a closed set F ⊆ E such that µ(E \ F ) ≤ 12 ², so that µ(H \ F ) ≤ ². (ii)⇒(iii) The same idea works. Suppose that µ is outer regular with respect to the open sets, and that E ∈ Σ, that µ(Hn \ En ) ≤ 2−n−2 ²; S ² > 0. For each n ∈ N, let Hn ⊇ Gn ∩ En be an open set such 1 then H = n∈N Hn is a measurable open set including E, and µ(H \ E) ≤ 2 ². Now repeat the argument on X \ E to find a measurable closed set F ⊆ E such that µ(E \ F ) ≤ 12 ². (iii)⇒(iv) Assume (iii), and let f : X S → [0, ∞[ be a measurable function, ² > 0. Set ηn = 2−n ²/(16 + 4µGn ) for each n ∈ N. For k ∈ N set Ek = n∈N {x : x ∈ Gn , kηn ≤ f (x) < (k + 1)ηn }, and choose an open set Hk ⊇ Ek such that µ(Hk \ Ek ) ≤ 2−k . Set g = supk,n∈N (k + 1)ηn χ(Gn ∩ Hk ). Then g : X → [0, ∞] is lower semi-continuous (4A2B(d-v) again). Since supk,n∈N kηn χ(Gn ∩ Ek ) ≤ f ≤ supk,n∈N (k + 1)ηn χ(Gn ∩ Ek ), f ≤ g and g − f ≤ supk,n∈N (k + 1)ηn χ(Gn ∩ Hk \ Ek ) + supk,n∈N ηn χ(Gn ∩ Ek ) has integral at most

P∞ P∞ k=0

n=0 (k

+ 1)ηn 2−k +

P∞ n=0

ηn µGn ≤ ².

(i)⇒(v) Assume (i), and suppose that f : X → R is measurable and ² > 0. For each n ∈ N, let αn ≥ 0 be such that µEn < 2−n−1 ², where En = {x : x ∈ Gn+1 \ Gn , f (x) ≤ −αn }. Let Fn ⊆ (Gn+1 \ G ) \ En be a Pn∞ measurable closed set such that µ((Gn+1 \Gn )\Fn ) ≤ 2−n−2 ². Because hFn in∈N is disjoint, h =S n=0 αn χFn is defined as a function from X to R. {Fn : n ∈ N} is locally finite, so {x : h(x) ≥ α} = n∈N,αn ≥α Fn is closed for every α > 0 (4A2B(h-ii)), and h is upper semi-continuous. Now f1 = f + h is a measurable function. Since (i)⇒(iii)⇒(iv), there is a measurable lower semi-continuous function g1 : X → [0, ∞] such R that f1+ ≤ g1 and g1 − f1+ ≤ 21 ²2 , where f1+ = max(0, f1 ). But if we now set g = g1 − h, g is lower semi-continuous, f ≤ g and {x : f (x) + ² ≤ g(x)} ⊆ {x : f1+ (x) + ² ≤ g1 (x)} ∪ {x : f1 (x) < 0} [ ⊆ {x : f1+ (x) + ² ≤ g1 (x)} ∪ (Gn+1 \ Gn ) \ Fn n∈N

has measure at most ², as required. (iv)⇒(ii) and (v)⇒(ii) Suppose that either (iv) or (v) is true, and that E ∈ Σ, ² > 0. Then there is a measurable lower semi-continuous ≤ g and µ{x : χE(x) + 21 ≤ g(x)} ≤ R function g1 : X → ]0, ∞] such that χE 1 ², since this is certainly true if g − χE ≤ 2 ². Set G = {x : g(x) > 2 }; then E ⊆ G and µ(G \ E) ≤ ². (iii)⇒(i) is trivial. Assembling these fragments, the proof is complete. 412X Basic exercises (a) Let (X, Σ, µ) be a measure space and T a topology on X such that µ is inner regular with respect to the closed sets and with respect to the compact sets. Show that it is tight. (b) Explain how 213A is a special case of 412Aa. >(c) Let (X, Σ, µ) be a measure space, and Σ0 a σ-subalgebra of Σ such that µ is inner regular with respect to Σ0 . Show that if 1 ≤ p < ∞ then every member of Lp (µ) is of the form f • for some Σ0 -measurable f : X → R.

412Xp

Inner regularity

29

> (d) Let (X, Σ, µ) be a semi-finite T measure space and A ⊆ Σ an algebra of sets such that the σ-algebra generated by A is Σ. Write K for { n∈N En : En ∈ A for every n ∈ N}. Show that µ is inner regular with respect to K. (e) Let (X, T, Σ, µ) be an effectively locally finite Hausdorff topological measure space such that µ is inner regular with respect to the Borel sets. Suppose that µG = sup{µK : K ⊆ G is compact} for every open set G ⊆ X. Show that µ is tight. (f ) Let (X, T) be a topological space such that every open set is an Fσ set. Show that any effectively locally finite Borel measure on X is inner regular with respect to the closed sets. (g) Let (X, T) be a normal topological space and µ a topological measure on X which is inner regular with respect to the closed sets. Show that µG = sup{µH : H ⊆ G is a cozero set} for every open set G ⊆ X. Show that if µ is totally finite, then µF = inf{µH : H ⊇ F is a zero set} for every closed set F ⊆ X. (h) Let (X, Σ, µ) be a complete locally determined measure space, and suppose that µ is inner regular with respect to a family K of sets. Let Σ0 be the σ-algebra of subsets of X generated by K ∩ Σ. (i) Show that µ is the c.l.d. version of µ¹Σ0 . (Hint: 412J-412L.) (ii) Show that if µ is σ-finite, it is the completion of µ¹Σ0 . > (i) (i) Let (X, Σ, µ) be a σ-finite measure space and T a σ-subalgebra of Σ. Show that if µ is inner regular with respect to T then the completion of µ¹T extends µ, so that µ and µ¹T have the same negligible sets. (ii) Show that if µ is a σ-finite topological measure which is inner regular with respect to the Borel sets, then every µ-negligible set is included in a µ-negligible Borel set. (j) Devise a direct proof of 412L, not using 412K, by (i) showing that µ∗ (A ∩ K) = ν ∗ (A ∩ K) whenever A ⊆ X, K ∈ K (ii) showing that µ∗ = ν ∗ (iii) quoting 213C. (k) Let (X, Σ, µ) be a complete locally determined measure space, Y a set and f : X → Y a function. Show that the following are equiveridical: (i) µ is inner regular with respect to {f −1 [B] : B ⊆ Y } (ii) f −1 [f [E]] \ E is negligible for every E ∈ Σ. (l) Let h(Xi , Σi , µi )ii∈I be a family of measure spaces, with direct sum (X, Σ, µ). Suppose that for each i ∈ I we are given a topology Ti on Xi , and let T be the corresponding disjoint union topology on X. Show that (i) µ is inner regular with respect to the closed sets iff every µi is (ii) µ is inner regular with respect to the compact sets iff every µi is (iii) µ is inner regular with respect to the zero sets iff every µi is (iv) µ is inner regular with respect to the Borel sets iff every µi is. (m) Use 412L and 412Q to shorten the proof of 253I. (n) Let hXi ii∈I be a family of sets, and suppose that we are given, for each i ∈ I, a σ-algebra Σi of subsets Q N of Xi and a topology Ti on Xi . Let T be the product topology on X = i∈I Xi , and Σ = c i∈I Σi . Let µ be a totally finite measure with domain Σ, and set µi = µπi−1 for each i ∈ I, where πi (x) =Sx(i) for Q i ∈ I, x ∈ X. (i) Show that µ is inner regular with respect to the family K of sets expressible as X \ n∈N i∈I Eni where Eni ∈ Σi for every n, i and {i : Eni 6= Xi } is finite for each n. (ii) Show that if every µi is inner regular with respect to the closed sets, so is µ. (iii) Show that if every µi is inner regular with respect to the zero sets, so is µ. (iv) Show that if every µi is inner regular with respect to the Borel sets, so is µ. (v) Show that if every µi is tight, and all but countably many of the Xi are compact, then µ is tight. (o) Let (X, Σ, µ) be a measure space and T a Lindel¨of topology on X such that µ is locally finite. (i) Show that µ is σ-finite. (ii) Show that µ is inner regular with respect to the closed sets iff it is outer regular with respect to the open sets. (p) Let X be a topological space and µ a measure on X which is outer regular with respect to the open sets. Show that for any Y ⊆ X the subspace measure on Y is outer regular with respect to the open sets.

30

Topologies and measures

412Xq

(q) Let X be a topological space and µ a measure on X which is outer regular with respect to the open sets. Show that if f : RX → R is integrable and ² > 0 then there is a lower semi-continuous g : X → ]−∞, ∞] such that f ≤ g and g − f ≤ ². (r) Let (X, Σ, µ) be a semi-finite measure space and hfn in∈N a sequence in L0 (µ) which converges almost everywhere to f ∈ L0 (µ). Show that µ is inner regular with respect to {E : hfn ¹Ein∈N is uniformly convergent}. (Cf. 215Yb.) (s) In 216E, give {0, 1}I its usual compact Hausdorff topology. Show that the measure µ described there is inner regular with respect to the zero sets. 412Y Further exercises (a) Let K be the family of subsets of R which are homeomorphic to the Cantor set. Show that Lebesgue measure is inner regular with respect to K. (Hint: show that if F ⊆ R \ Q is an uncountable compact set, then {x : [x − δ, x + δ] ∩ F is uncountable for every δ > 0} belongs to K.) (b) (i) Show that if X is a perfectly normal space then any semi-finite topological measure on X which is inner regular with respect to the Borel sets is inner regular with respect to the closed sets. (ii) Show that any subspace of a perfectly normal space is perfectly normal. (iii) Show that the split interval I k (343J, 343Yc, 419L) is perfectly normal. (iv) Show that ω1 , with its order topology, is completely regular, normal and Hausdorff, but not perfectly normal. (v) Show that I k × I k is not perfectly normal. (vi) Show that [0, 1]I is perfectly normal iff I is countable. (c) Let (X, Σ, µ) be a measure space, and suppose that µ is inner regular with respect to a family K ⊆ Σ such that K ∪ K 0 ∈ K for all K, K 0 ∈ K. Write Σf for {E : E ∈ Σ, µE < ∞}. Show that {E • : E ∈ K ∩ Σf } is dense in {E • : E ∈ Σf } for the strong measure-algebra topology. (d) Let X be a normal topological space and Y a closed subset of X. Show that every Baire subset of Y is the intersection of Y with a Baire subset of X. (Hint: use Tietze’s theorem.) (e) Let (X, Σ, µ) be [0, 1] with Lebesgue measure, and Y = [0, 1] with counting measure ν; give X its usual topology and Y the discrete topology, and let λ be the c.l.d. product measure on X × Y . (i) Show that µ, ν and λ are all tight (for the appropriate topologies) and therefore completion regular. (ii) Let λ0 be the primitive product measure on X × Y (definition: 251C). Show that λ0 is not tight. (Hint: 252Yf.) Remark : it is undecidable in ZFC whether λ0 is inner regular with respect to the closed sets. (f ) Give an example of a Hausdorff topological measure space (X, T, Σ, µ) such that µ is complete, strictly localizable and outer regular with respect to the open sets, but not inner regular with respect to the closed sets. 412 Notes and comments In this volume we are returning to considerations which have been left on one side for almost the whole of Volume 3 – the exceptions being in Chapter 34, where I looked at realization of homomorphisms of measure algebras by functions between measure spaces, and was necessarily dragged into an investigation of measure spaces which had enough points to be adequate codomains (343B). The idea of ‘inner regularity’ is to distinguish families K of sets which will be large enough to describe the measure entirely, but whose members will be of recognisable types. For an example of this principle see 412Ya. Of course we cannot always find a single type of set adequate to fill a suitable family K, though this happens oftener than one might expect, but it is surely easier to think about an arbitrary zero set (for instance) than an arbitrary measurable set, and whenever a measure is inner regular with respect to a recognisable class it is worth knowing about it. I have tried to use the symbols † and ‡ (412A, 412C) consistently enough for them to act as a guide to some of the ideas which will be used repeatedly in this chapter. Note the emphasis on disjoint unions and countable intersections; I mentioned similar conditions in 136Xi-136Xj. You will recognise 412Aa as an exhaustion principle; note that it is enough to use disjoint unions, as in 313K. In the examples of this section this disjointness is not important. Of course inner regularity has implications for the measure algebra (412N), but it is important to recognise that ‘µ is inner regular with respect to K’ is saying much more than

§413 intro.

Inner measure constructions

31

‘{K • : K ∈ K} is order-dense in the measure algebra’; the latter formulation tells us only that whenever µE > 0 there is a K ∈ K such that K \ E is negligible and µK > 0, while the former tells us that we can take K to be actually a subset of E. 412D, 412E and 412G are all of great importance. 412D looks striking, but of course the reason it works is just that the Baire σ-algebra is very small. In 412E the Baire and Borel σ-algebras coincide, so it is nothing but a special case of 412D; but as metric spaces are particularly important it is worth having it spelt out explicitly. In 412D and 412E the hypothesis ‘semi-finite’ is sufficient, while in 412G we need ‘effectively locally finite’; this is because in both 412D and 412E the open sets we are looking at are countable unions of measurable closed sets. There are interesting non-metrizable spaces in which the same thing happens (412Yb). As you know, I am strongly prejudiced in favour of complete and locally determined measures, and the Baire and Borel measures dealt with in these three results are rarely complete; but they can still be applied to completions and c.l.d. versions of these measures, using 412Ab or 412H. 412O-412V are essentially routine. For subspace measures, the only problem we need to come to terms with is the fact that subspaces of semi-finite measure spaces need not be semi-finite (216Xa). For product measures the point is that the c.l.d. product of two measure spaces, and the product of any family of probability spaces, as I defined them in Chapter 25, are inner regular with respect to the σ-algebra of sets generated by the cylinder sets. This is not in general true of the ‘primitive’ product measure (412Ye), which is one of my reasons for being prejudiced against it. I should perhaps warn you of a trap in the language I use here. I say that if the factor measures are inner regular with respect to the closed sets, so is the c.l.d. product measure. But I do not say that all closed sets in the product are measurable for the product measure, even if closed sets in the factors are measurable for the factor measures. So the path is open for a different product measure to exist, still inner regular with respect to the closed sets; and indeed I shall be going down that path in §417. The uniqueness result in 412L specifically refers to complete locally determined measures defined on all sets of the family K. ThereQis one special difficulty in 412V: in order to ensure that there are enough compact measurable sets in X = i∈I Xi , we need to know that all but countably many of the Xi are actually compact. When we come to look more closely at products of Radon probability spaces we shall need to consider this point again (417Q, 417Xq). In fact some of the ideas of 412U-412V are not restricted to the product measures considered there. Other measures on the product space will have inner regularity properties if their images on the factors, their ‘marginals’ in the language of probability theory, are inner regular; see 412Xn. I will return to this in §454. This section is almost exclusively concerned with inner regularity. The complementary notion of outer regularity is not much use except in σ-finite spaces (415Xh), and not always then (416Yd). In totally finite spaces, of course, and some others, any version of inner regularity corresponds to a version of outer regularity, as in 412Wb(i)-(ii); and when we have something as strong as 412Wb(iii) available it is worth knowing about it.

413 Inner measure constructions I now turn in a different direction, giving some basic results on the construction of inner regular measures. The first step is to describe ‘inner measures’ (413A) and a construction corresponding to the Carath´eodory construction of measures from outer measures (413C). Just as every measure gives rise to an outer measure, it gives rise to an inner measure (413D). Inner measures form an effective tool for studying complete locally determined measures (413F). The most substantial results of the section concern the construction of measures as extensions of functionals defined on various classes K of sets. Typically, K is closed under finite unions and countable intersections, though it is sometimes possible to relax the hypotheses to some extent. The methods here make it possible to distinguish arguments which produce finitely additive functionals (413H, 413N, 413P, 413Q) from the succeeding steps to countably additive measures (413I, 413O, 413S). 413H-413M investigate conditions on a functional φ : K → [0, ∞[ sufficient to produce a measure extending φ, necessarily unique, which is inner regular with respect to K or Kδ , the set of intersections of sequences in K. 413N-413O look instead at functionals defined on sublattices of the class K of interest, and at sufficient conditions to ensure the existence of a measure, not normally unique, defined on the whole of K, inner regular with respect to K and extending

32

Topologies and measures I

§413 intro.

the given functional. Finally, 413P-413S are concerned with majorizations rather than extensions; we seek a measure µ such that µK ≥ λK for K ∈ K, while µX is as small as possible. 413A I begin with some material from the exercises of earlier volumes. Definition Let X be a set. An inner measure on X is a function φ : PX → [0, ∞] such that φ∅ = 0; (α) φ(A ∪ B) ≥ φA + φB for all disjoint A, B ⊆ X; T (β) if hAn in∈N is a non-increasing sequence of subsets of X and φA0 < ∞ then φ( n∈N An ) = inf n∈N φAn ; (∗) φA = sup{φB : B ⊆ A, φB < ∞} for every A ⊆ X. 413B The following fact will be recognised as an element of Carath´eodory’s method. There will be an application later in which it will be useful to know that it is not confined to proving countable additivity. Lemma Let X be a set and φ : X → [0, ∞] any function such that φ∅ = 0. Then Σ = {E : E ⊆ X, φA = φ(A ∩ E) + φ(A \ E) for every A ⊆ X} is an algebra of subsets of X, and φ(E ∪ F ) = φE + φF for all disjoint E, F ∈ Σ. proof The symmetry of the definition of Σ ensures that X \ E ∈ Σ whenever E ∈ Σ. If E, F ∈ Σ and A ⊆ X, then

φ(A ∩ (E∪F )) + φ(A \ (E ∪ F )) = φ(A ∩ (E ∪ F ) ∩ E) + φ(A ∩ (E ∪ F ) \ E) + φ(A \ (E ∪ F )) = φ(A ∩ E) + φ((A \ E) ∩ F ) + φ((A \ E) \ F ) = φ(A ∩ E) + φ(A \ E) = φA. As A is arbitrary, E ∪ F ∈ Σ. Finally, if A ⊆ X, φ(A ∩ ∅) + φ(A \ ∅) = φ∅ + φA = φA because φ∅ = 0; so ∅ ∈ Σ. Thus Σ is an algebra of sets. If E, F ∈ Σ and E ∩ F = ∅, then φ(E ∪ F ) = φ((E ∪ F ) ∩ E) + φ((E ∪ F ) \ E) = φE + φF . 413C Measures from inner measures I come now to a construction corresponding to Carath´eodory’s method of defining measures from outer measures. Theorem Let X be a set and φ : X → [0, ∞] an inner measure. Set Σ = {E : E ⊆ X, φ(A ∩ E) + φ(A \ E) = φA for every A ⊆ X}. Then (X, Σ, φ¹Σ) is a complete measure space. proof (Compare 113C.) (a) The first step is to note that if A ⊆ B ⊆ X then φB ≥ φA + φ(B \ A) ≥ φA. Next, a subset E of X belongs to Σ iff φA ≤ φ(A ∩ E) + φ(A \ E) whenever A ⊆ X and µA < ∞. P P Of course any element of Σ satisfies the condition. If E satisfies the condition and A ⊆ X, then φA = sup{φB : B ⊆ A, φB < ∞} ≤ sup{φ(B ∩ E) + φ(B \ E) : B ⊆ A} = φ(A ∩ E) + φ(A \ E) ≤ φA,

413D

Inner measure constructions

33

so E ∈ Σ. Q Q (b) By 413B, Σ is an algebra of subsets of X. Now suppose that hEn in∈N is a non-decreasing sequence in Σ, with union E. If A ⊆ X and φA < ∞, then φ(A \ E) = inf n∈N φ(A \ En ) = limn→∞ φ(A \ En ) because hA \ En in∈N is non-increasing and φ(A \ E0 ) is finite; so φ(A ∩ E) + φ(A \ E) ≥ limn→∞ φ(A ∩ En ) + φ(A \ En ) = φA. By (a), E ∈ Σ. So Σ is a σ-algebra. (c) If E, F ∈ Σ and E ∩ F = ∅ then φ(E ∪ F ) = φE + φF , by 413B. If hEn in∈N is a disjoint sequence in Σ with union E, then S Pn µE ≥ µ( i≤n Ei ) = i=0 µEi P∞ P∞ P∞ for every n, so µE ≥ Si=0 µEi . ?? If µE > i=0 µEi , there is an A ⊆ E such that i=0 µEi < φA < ∞. But now, setting Fn = i≤n Ei for each n, we have limn→∞ φ(A \ Fn ) = 0, so that P∞ φA = limn→∞ φ(A ∩ Fn ) + φ(A \ Fn ) = i=0 φ(A ∩ Ei ) < φA, P∞ which is absurd. X X Thus µE = i=0 µEi . As hEn in∈N is arbitrary, µ is a measure. (d) Finally, suppose that B ⊆ E ∈ Σ and µE = 0. Then for any A ⊆ X we must have φ(A ∩ B) + φ(A \ B) ≥ φ(A \ E) = φ(A ∩ E) + φ(A \ E) = φA, so B ∈ Σ. Thus µ is complete. Remark For a simple example see 213Yc. 413D The inner measure defined by a measure Let (X, Σ, µ) be any measure space. has an associated outer measure µ∗ defined by the formula

Just as µ

µ∗ A = inf{µE : A ⊆ E ∈ Σ} (132A-132B), it gives rise to an inner measure µ∗ defined by the formula µ∗ A = sup{µE : E ∈ Σf , E ⊆ A}, where I write Σf for {E : E ∈ Σ, µE < ∞}. P P µ∗ ∅ = µ∅ = 0. (†) If A ∩ B = ∅, and E ⊆ A, F ⊆ B belong to Σf , then E ∪ F ⊆ A ∪ B also has finite measure, so µ∗ (A ∪ B) ≥ µ(E ∪ F ) = µE + µF ; taking the supremum over E and F , µ∗ (A ∪ B) ≥ µ∗ A + µ∗ B. (‡) If hAn in∈N is a non-increasing sequence of sets with intersection A and µ∗ A0 < ∞, then for each n ∈ N we can find an En ⊆ An such that µEn ≥ µ∗ An − 2−n . In this case, S S µ( m∈N Em ) = supn∈N µ( m≤n Em ) ≤ µ∗ A0 < ∞. Set E=

T

S n∈N

m≥n

Em ⊆ A.

Then E ∈ Σf , so µ∗ A ≥ µE ≥ lim supn→∞ µEn = limn→∞ µ∗ An ≥ µ∗ A. (*) If A ⊆ X and µ∗ A = ∞ then sup{µ∗ B : B ⊆ A, µ∗ B < ∞} ≥ sup{µE : E ∈ Σf , E ⊆ A} = ∞. Q Q Warning Many authors use the formula µ∗ A = sup{µE : A ⊇ E ∈ Σ}. In ‘ordinary’ cases, when (X, Σ, µ) is semi-finite, this agrees with my usage (413Ed); but for non-semi-finite spaces there is a difference.

34

Topologies and measures I

413D

For elementary properties of the construction here see 213Xe. 413E I note the following elementary facts concerning inner measures defined from measures. Proposition Let (X, Σ, µ) be a measure space. Write Σf for {E : E ∈ Σ, µE < ∞}. (a) For every A ⊆ X there is an E ∈ Σ such that E ⊆ A and µE = µ∗ A. (b) µ∗ A ≤ µ∗ A for every A ⊆ X. (c) If E ∈ Σ and A ⊆ X, then µ∗ (E ∩ A) + µ∗ (E \ A) ≤ µE, with equality if either (i) µE < ∞ or (ii) µ is semi-finite. (d) In particular, µ∗ E ≤ µE for every E ∈ Σ, with equality if either µE < ∞ or µ is semi-finite. (e) If µ is inner regular with respect to K, then µ∗ A = sup{µK : K ∈ K ∩ Σf , K ⊆ A} for every A ⊆ X. (f) If A ⊆ X is such that µ∗ A = µ∗ A < ∞, then A is measured by the completion of µ. (g) If µ ˆ, µ ˜ are the completion and c.l.d. version of µ, then µ ˆ∗ = µ ˜ ∗ = µ∗ . (h) If (Y, T, ν) is another measure space, and f : X → Y is an inverse-measure-preserving function, then µ∗ (f −1 [B]) ≤ ν ∗ B,

µ∗ (f −1 [B]) ≥ ν∗ B

for every B ⊆ Y , and ν ∗ (f [A]) ≥ µ∗ A for every A ⊆ X. (i) Suppose that µ is semi-finite. If A ⊆ E ∈ Σ, then E is a measurable envelope of A iff µ∗ (E \ A) = 0. f proof (a) S There is a sequence hEn in∈N in Σ such that En ⊆ A for each n and limn→∞ µEn = µ∗ A; now set E = n∈N En .

(b) If E ⊆ A ⊆ F we must have µE ≤ µF . (c) If F ⊆ E ∩ A and F ∈ Σf , then µF + µ∗ (E \ A) ≤ µF + µ(E \ F ) = µE; taking the supremum over F , µ∗ (E ∩ A) + µ∗ (E \ A) ≤ µE. If µE < ∞, then µ∗ (E ∩ A) = sup{µF : F ∈ Σ, F ⊆ E ∩ A} = µE − inf{µ(E \ F ) : F ∈ Σ, F ⊆ E ∩ A} = µE − inf{µF : F ∈ Σ, E \ A ⊆ F ⊆ E} = µE − µ∗ (E \ A). If µ is semi-finite, then µ∗ (E ∩ A) + µ∗ (E \ A) ≥ sup{µ∗ (F ∩ A) + µ∗ (F \ A) : F ∈ Σf , F ⊆ E} = sup{µF : F ∈ Σf , F ⊆ E} = µE. (d) Take A = E in (c). (e) µ∗ A = sup{µE : E ∈ Σf , E ⊆ A} = sup{µK : K ∈ K ∩ Σ, ∃ E ∈ Σf , K ⊆ E ⊆ A} = sup{µK : K ∈ K ∩ Σf , K ⊆ A}. (f ) By (a) above and 132Aa, there are E, F ∈ Σ such that E ⊆ A ⊆ F and µE = µ∗ A = µ∗ A = µF < ∞; now µ(F \ E) = 0, so F \ A and A are measured by the completion of µ. ˇ for its domain, and let A ⊆ X. (i) If γ < µ∗ A, there is an E ∈ Σ such (g) Write µ ˇ for either µ ˆ or µ ˜, and Σ that E ⊆ A and γ ≤ µE < ∞; now µ ˇE = µE (212D, 213Fa), so µ ˇ∗ A ≥ γ. As γ is arbitrary, µ∗ A ≤ µ ˇ∗ A.

413F

Inner measure constructions

35

ˇ such that E ⊆ A and γ ≤ µ (ii) If γ < µ ˇ∗ A, there is an E ∈ Σ ˇE < ∞. Now there is an F ∈ Σ such that F ⊆ E and µF = µ ˇE (212C, 213Fc), so that µ∗ A ≥ γ. As γ is arbitrary, µ∗ A ≥ µ ˇ∗ A. (h) This is elementary; all we have to note is that if F , F 0 ∈ T and F ⊆ B ⊆ F 0 , then f −1 [F ] ⊆ f −1 [B] ⊆ f [F 0 ], so that −1

νF = µf −1 [F ] ≤ µ∗ f −1 [B] ≤ µ∗ f −1 [B] ≤ µf −1 [F 0 ] = νF 0 . Now, for A ⊆ X, µ∗ A ≤ µ∗ (f −1 [f [A]]) ≤ ν ∗ (f [A]). (i)(i) If E is a measurable envelope of A and F ∈ Σ is included in E \ A, then µF = µ(F ∩ E) = µ∗ (F ∩ A) = 0; as F is arbitrary, µ∗ (E \ A) = 0. (ii) If E is not a measurable envelope of A, there is an F ∈ Σ such that µ∗ (F ∩ A) < µ(F ∩ E). Let G ∈ Σ be such that F ∩ A ⊆ G and µG = µ∗ (F ∩ A). Then µ(F ∩ E \ G) > 0; because µ is semi-finite, µ∗ (E \ A) ≥ µ∗ (F ∩ E \ G) > 0. 413F The language of 413D makes it easy to express some useful facts about complete locally determined measure spaces, complementing 412J. Lemma Let (X, Σ, µ) be a complete locally determined measure space and K a family of subsets of X such that µ is inner regular with respect to K. Then for E ⊆ X the following are equiveridical: (i) E ∈ Σ; (ii) E ∩ K ∈ Σ whenever K ∈ Σ ∩ K; (iii) µ∗ (K ∩ E) + µ∗ (K \ E) = µ∗ K for every K ∈ K; (iv) µ∗ (K ∩ E) + µ∗ (K \ E) = µ∗ K for every K ∈ K; (v) µ∗ (E ∩ K) = µ∗ (E ∩ K) for every K ∈ K ∩ Σ; (vi) min(µ∗ (K ∩ E), µ∗ (K \ E)) < µK whenever K ∈ K ∩ Σ and 0 < µK < ∞; (vii) max(µ∗ (K ∩ E), µ∗ (K \ E)) > 0 whenever K ∈ K ∩ Σ and µK > 0. proof (a) Assume (i). Then of course E ∩ K ∈ Σ for every K ∈ Σ ∩ K, and (ii) is true. For any K ∈ K there is an F ∈ Σ such that F ⊇ K and µF = µ∗ K (132Aa); now µ∗ K ≤ µ∗ (K ∩ E) + µ∗ (K \ E) ≤ µ(F ∩ E) + µ(F \ E) = µF = µ∗ K, so (iii) is true. Next, for any K ∈ K, µ∗ (K ∩ E) + µ∗ (K \ E) ≤ µ∗ K = sup{µF : F ∈ Σf , F ⊆ K} (writing Σf for {F : F ∈ Σ, µF < ∞}) = sup{µ(F ∩ E) + µ(F \ E) : F ∈ Σf , F ⊆ K} ≤ µ∗ (K ∩ E) + µ∗ (K \ E). So (iv) is true. If K ∈ K ∩ Σ, then µ∗ (E ∩ K) = sup{µF : F ∈ Σf , F ⊆ E ∩ K} = µ(E ∩ K) = µ∗ (E ∩ K) because µ is semi-finite. So (v) is true. Since (iii)⇒(vi) and (iv)⇒(vii), we see that all the conditions are satisfied. (b) Now suppose that E ∈ / Σ; I have to show that (ii)-(vii) are all false. Because µ is locally determined, there is an F ∈ Σf such that E ∩ F ∈ / Σ. Take measurable envelopes H, H 0 of F ∩ E, F \ E respectively 0 (132Ee). Then F \ H ⊆ F ∩ E ⊆ F ∩ H, so G = (F ∩ H) \ (F \ H 0 ) = F ∩ H ∩ H 0 cannot be negligible. Take K ∈ K ∩ Σ such that K ⊆ G and µK > 0. As G ⊆ F , µK < ∞. Now µ∗ (K ∩ E) = µ∗ (K ∩ F ∩ E) = µ(K ∩ H) = µK,

36

Topologies and measures I

413F

µ∗ (K \ E) = µ∗ (K ∩ F \ E) = µ(K ∩ H 0 ) = µK. But this means that µ∗ (K ∩ E) = µK − µ∗ (K \ E) = 0,

µ∗ (K \ E) = µK − µ∗ (K ∩ E) = 0,

by 413Eb. Now we see that this K witnesses that (ii)-(vii) are all false. 413G The ideas of 413F can be used to give criteria for measurability of real-valued functions. I spell out one which is particularly useful. Lemma Let (X, Σ, µ) be a complete locally determined measure space and suppose that µ is inner regular with respect to K ⊆ Σ. Suppose that f : X → R is a function, and for α ∈ R set Eα = {x : f (x) ≤ α}, Fα = {x : f (x) ≥ β}. Then f is Σ-measurable iff min(µ∗ (Eα ∩ K), µ∗ (Fβ ∩ K)) < µK whenever K ∈ K, 0 < µK < ∞ and α < β. proof (a) If f is measurable, then µ∗ (Eα ∩ K) + µ∗ (Fβ ∩ K) = µ(Eα ∩ K) + µ(Fβ ∩ K) ≤ µK whenever K ∈ Σ and α < β, so if 0 < µK < ∞ then we must have min(µ∗ (Eα ∩ K), µ∗ (Fβ ∩ K)) < µK. (b) If f is not measurable, then there is some α ∈ R such that Eα is not measurable. 413F(iii) tells us that there is a K ∈ K such that 0 < µK < ∞ and µ∗ (Eα ∩ K) = µ∗ (K \ Eα ) = µK. Note that K is a measurable envelope of K ∩ Eα (132Eb). Now hK ∩ Fα+2−n in∈N is a non-decreasing sequence with union K \ Eα , so there is some β > α such that K ∩ Fβ is not negligible. Let H ⊆ K be a measurable envelope of K ∩ Fβ , and K 0 ∈ K such that K 0 ⊆ H and µK 0 > 0; then µ∗ (K 0 ∩ Eα ) = µ∗ (K 0 ∩ K ∩ Eα ) = µ(K 0 ∩ K) = µK 0 , µ∗ (K 0 ∩ Fβ ) = µ∗ (K 0 ∩ H ∩ Fβ ) = µ(K 0 ∩ H) = µK 0 , so K 0 , α and β witness that the condition is not satisfied. 413H Inner measure constructions based on 413C are important because they offer an efficient way of setting up measures which are inner regular with respect to given families of sets. Two of the fundamental results are 413I and 413J. I proceed by means of a lemma on finitely additive functionals. Lemma Let X be a set and K a family of subsets of X such that ∅ ∈ K, (†) K ∪ K 0 ∈ K whenever K, K 0 ∈ K are disjoint, (‡) K ∩ K 0 ∈ K for all K, K 0 ∈ K. Let φ0 : K → [0, ∞[ be a functional such that (α) φ0 K = φ0 L + sup{φ0 K 0 : K ∈ K, K 0 ⊆ K \ L} whenever K, L ∈ K and L ⊆ K. Set φA = sup{φ0 K : K ∈ K, K ⊆ A} for A ⊆ X, Σ = {E : E ⊆ X, φA = φ(A ∩ E) + φ(A \ E) for every A ⊆ X}. Then Σ is an algebra of subsets of X, including K, and φ¹Σ : Σ → [0, ∞] is an additive functional extending φ0 . proof (a) To see that Σ is an algebra of subsets and φ¹Σ is additive, all we need to know is that φ∅ = 0 (413B); and this is because, applying hypothesis (α) with K = L = ∅, φ0 ∅ = φ0 ∅ + φ0 ∅, so φ0 ∅ = 0. (α) also assures us that φ0 L ≤ φ0 K whenever K, L ∈ K and L ⊆ K, so φK = φ0 K for every K ∈ K. (b) To check that K ⊆ Σ, we have a little more work to do. First, observe that (†) and (α) together tell us that φ0 (K ∪ K 0 ) = φ0 K + φ0 K 0 for all disjoint K, K 0 ∈ K. So if A, B ⊆ X and A ∩ B = ∅ then

413I

Inner measure constructions

φA + φB =

sup K∈K,K⊆A

=

φ0 K +

sup K,L∈K,K⊆A,L⊆B

sup L∈K,L⊆A

37

φ0 L

φ0 (K ∪ L) ≤ φ(A ∪ B).

(c) K ⊆ Σ. P P Take K ∈ K, A ⊆ X. If L ∈ K and L ⊆ A, then φ0 L = φ0 (K ∩ L) + sup{φ0 L0 : L0 ∈ K, L0 ⊆ L \ K} ≤ φ(A ∩ K) + φ(A \ K). (Note the use of the hypothesis (‡).) As L is arbitrary, φA ≤ φ(A ∩ K) + φ(A \ K). We already know that φ(A ∩ K) + φ(A \ K) ≤ φA; as A is arbitrary, K ∈ Σ. Q Q This completes the proof. 413I Theorem Let X be a set and K a family of subsets of X such that ∅ ∈ K, (†) K ∪ K 0 ∈ K whenever K, K 0 ∈ K are disjoint, T (‡) n∈N Kn ∈ K whenever hKn in∈N is a sequence in K. Let φ0 : K → [0, ∞[ be a functional such that (α) φ0 K = φ0 L + sup{φ0 K 0 : K ∈ K, K 0 ⊆ K \ L} whenever K, L ∈ K and L ⊆ K, (β) inf n∈N φ0 Kn = 0 whenever hKn in∈N is a non-increasing sequence in K with empty intersection. Then there is a unique complete locally determined measure µ on X extending φ0 and inner regular with respect to K. proof (a) Set φA = sup{φ0 K : K ∈ K, K ⊆ A} for A ⊆ X, Σ = {E : E ⊆ X, φA = φ(A ∩ E) + φ(A \ E) for every A ⊆ X}. Then 413H tells us that Σ is an algebra of subsets of X, including K, and µ = φ¹Σ is an additive functional extending φ0 . T P Set L = T (b) Now µ( n∈N Kn ) = inf n∈N µKn whenever hKn in∈N is a non-increasing sequence in K. P K . Of course µL ≤ inf µK . For the reverse inequality, take ² > 0. Then (α) tells us that there n n∈N n n∈N is a K 0 ∈ K such that K 0 ⊆ K0 \ L and µK0 ≤ µL + µK 0 + ². Since hKn ∩ K 0 in∈N is a non-increasing sequence in K with empty intersection, (β) tells us that there is an n ∈ N such that µ(Kn ∩ K 0 ) ≤ ². Now µK0 − µL = µ(K0 \ L) = µ(K0 \ (K 0 ∪ L)) + µK 0 ≤ ² + µ(Kn ∩ K 0 ) + µ(K 0 \ Kn ) ≤ 2² + µ(K0 \ Kn ) = 2² + µK0 − µKn . (These calculations depend, of course, on the additivity of µ and the finiteness of µK0 .) So µL ≥ µKn − 2². As ² is arbitrary, µL = inf n∈N µKn . Q Q (c) If hAn in∈N is a non-increasing sequence of subsets of X, with intersection A, and φA0 < ∞, then φA = inf n∈N φAn . P P Of course φA ≤ φAn for every n. Given ² > 0, then for each n ∈ N choose Kn ∈ K such that K ⊆ A and φ0 KnT≥ φAn − 2−n ² (this is where I use the hypothesis that φA0 is finite); set n n T Ln = i≤n Ki for each n, L = n∈N Ln . Then we have φAn+1 − µLn+1 = φAn+1 − µ(Kn+1 ∩ Ln ) = φAn+1 − µKn+1 − µLn + µ(Kn+1 ∪ Ln ) ≤ 2−n−1 ² − µLn + φAn because Kn+1 ⊆ An+1 ⊆ An and Ln ⊆ Kn ⊆ An . Inducing on n, we see that µLn ≥ φAn − 2² + 2−n ² for every n. So φA ≥ µL = inf n∈N µLn ≥ inf n∈N φAn − 2²,

38

Topologies and measures I

413I

using (b) above for the middle equality. As ² is arbitrary, φA = inf n∈N φAn . Q Q (d) It follows that φ is an inner measure. P P The arguments of parts (a) and (b) of the proof of 413H tellTus that φ∅ = 0 and φ(A ∪ B) ≤ φA + φB whenever A, B ⊆ X are disjoint. We have just seen that φ( n∈N An ) = inf n∈N φAn whenever hAn in∈N is a non-increasing sequence of sets and φA0 < ∞. Finally, φK = φ0 K is finite for every K ∈ K, so φA = sup{φB : B ⊆ A, φB < ∞} for every A ⊆ X. Putting these together, φ is an inner measure. Q Q (e) So 413C tells us that µ is a complete measure, and of course it is inner regular with respect to K, by the definition of φ. It is semi-finite because µK = φ0 K is finite for every K ∈ K. Now suppose that E ⊆ X and that E ∩ F ∈ Σ whenever µF < ∞. Take any A ⊆ X. If L ∈ K and L ⊆ A, we have L ∈ Σ and µL < ∞, so φ0 L = µL = µ(L ∩ E) + µ(L \ E) = φ(L ∩ E) + φ(L \ E) ≤ φ(A ∩ E) + φ(A \ E); taking the supremum over L, φA ≤ φ(A ∩ E) + φ(A \ E). As A is arbitrary, E ∈ Σ; as E is arbitrary, µ is locally determined. (f ) Finally, µ is unique by 412L. 413J Theorem Let X be a set and K a family of subsets of X such that ∅ ∈ K, (†) K ∪ K 0 ∈ K whenever K, K 0 ∈ K are disjoint, (‡) K ∩ K 0 ∈ K whenever K, K 0 ∈ K. Let φ0 : K → [0, ∞[ be a functional such that (α) φ0 K = φ0 L + sup{φ0 K 0 : K ∈ K, K 0 ⊆ K \ L} whenever K, L ∈ K and L ⊆ K, (β) inf n∈N φ0 Kn = 0 whenever hKn in∈N is a non-increasing sequence in K with empty intersection. Then there is a unique complete locally determined measure µ on X extending φ0 and inner regular with respect to Kδ , the family of sets expressible as intersections of sequences in K. proof (a) Set ψA = sup{φ0 K : K ∈ K, K ⊆ A} for A ⊆ X, T = {E : E ⊆ X, ψA = ψ(A ∩ E) + ψ(A \ E) for every A ⊆ X}. Then 413H tells us that T is an algebra of subsets of X, including K, and ν = ψ¹ T is an additive functional extending φ0 . (b) Write Tf for {E : E ∈ T, νE < ∞}. If hEn in∈N is a non-increasing sequence in Tf with empty intersection, limn→∞ νEn = 0. P P Given ² > 0, we can choose a sequence hKn in∈N in K such that Kn ⊆ En and for each n. Set Ln =

νKn = φ0 Kn ≥ νEn − 2−n ²

T i≤n

Ki for each n; then limn→∞ νLn = limn→∞ φ0 Ln = 0

by hypothesis (β). But also, for each n, νEn ≤ νLn +

Pn

ν(Ei \ Ki ) ≤ νLn + 2², S because ν is additive and non-negative and En ⊆ Ln ∪ i≤n (Ei \ Ki ). So lim supn→∞ νEn ≤ 2²; as ² is arbitrary, limn→∞ νEn = 0. Q Q i=0

(c) Write Tfδ for the family of sets expressible as intersections of sequences in Tf , and for H ∈ Tfδ set φ1 H = inf{νE : H ⊆ E ∈ T}. Note that because E ∩ F ∈ Tf for every E, F ∈ Tf , every member of Tfδ can be expressed as the intersection of a non-increasing sequence in Tf . (i) If hEn in∈N is a non-increasing sequence in Tf with intersection H ∈ Tfδ , φ1 H = limn→∞ νEn . P P Of course

413J

Inner measure constructions

39

φ1 H ≤ inf n∈N νEn = limn→∞ νEn . On the other hand, if H ⊆ E ∈ T, then hEn \ Ein∈N is a non-increasing sequence in Tf with empty intersection, and νE ≥ limn→∞ ν(En ∩ E) = limn→∞ νEn − limn→∞ ν(En \ E) = limn→∞ νEn T by (b) above. As E is arbitrary, φ1 ( n∈N En ) = limn→∞ νEn . Q Q (ii) Because K ⊆ Tf , Kδ ⊆ Tfδ . Now for any H ∈ Tfδ , φ1 H = sup{φ1 L : L ∈ Kδ , L ⊆ H}. P P Express T H as n∈N En where hEn in∈N is a non-increasing sequence in Tf . Given ² > 0, weTcan choose a sequence hKn in∈N in K such that Kn ⊆ En and νKn ≥ νEn − 2−n ² for each n. Setting Ln = i≤n Ki for each n and T L = n∈N Ln , we have L ∈ Kδ , L ⊆ H and Pn φ1 H = limn→∞ νEn ≤ limn→∞ (νLn + i=0 ν(Ei \ Ki )) ≤ φ1 L + 2². As ² is arbitrary, this gives the result. Q Q (d) We find that Tfδ , φ1 satisfy the conditions of 413I. P P Of course ∅ ∈ Tfδ . If G, H ∈ Tfδ and G ∩ H = ∅, T T express them as n∈N En , n∈N Fn where hEn in∈N , hFn in∈N are non-increasing sequences in Tf . Then T G ∪ H = n∈N En ∪ Fn belongs to Tfδ , and φ1 (G ∪ H) = lim ν(En ∪ Fn ) = lim νEn + νFn − ν(En ∩ Fn ) n→∞

n→∞

= lim νEn + νFn n→∞

(by (b)) = φ1 G + φ1 H. The definition of Tfδ as the set of intersections of sequences in Tf ensures that the intersection of any sequence in Tfδ will belong to Tfδ . T T Now suppose that G, H ∈ Tfδ and that G ⊆ H. Express them as intersections n∈N En , n∈N Fn of non-increasing sequences in Tf , so that φ1 G = limn→∞ νEn , φ1 H = limn→∞ νFn . For each n, set T Hn = m∈N Fm \ En , so that Hn ∈ Tfδ , Hn ⊆ H \ G, and φ1 Hn = lim ν(Fm \ En ) = lim νFm − ν(Fm ∩ En ) m→∞

m→∞

≥ lim νFm − νEn = φ1 H − νEn . m→∞

Accordingly sup{φ1 G0 : G0 ∈ Tfδ , G0 ⊆ H \ G} ≥ supn∈N φ1 H − νEn = φ1 H − φ1 G. On the other hand, if G0 ∈ Tfδ and G0 ⊆ H \ G, then φ1 G + φ1 G0 = φ1 (G ∪ G0 ) ≤ φ1 H because of course φ1 is non-decreasing, as well as being additive on disjoint sets. So sup{φ1 G0 : G0 ∈ Tfδ , G0 ⊆ H \ G} = φ1 H − φ1 G as required by condition (α) of 413I. Finally, suppose that hHn in∈N is a non-increasing sequence in Tfδ with empty intersection. For each n ∈ N, let hEni ii∈N be a non-increasing sequence in Tf with intersection T Hn , and set Fm = n≤m Enn for each m. Then hFm im∈N is a non-increasing sequence in Tf with empty intersection, while Hm ⊆ Fm for each m, so limm→∞ φ1 Hm ≤ limm→∞ νFm = 0. Thus condition 413I(β) is satisfied, and we have the full list. Q Q

40

Topologies and measures I

413J

(e) By 413I, we have a complete locally determined measure µ, extending φ1 , and inner regular with respect to Tfδ . Since φ1 K = νK = φ0 K for K ∈ K, µ extends φ0 . If G belongs to the domain of µ, and γ < µG, there is an H ∈ Tfδ such that H ⊆ G and γ < µH = φ1 H; by (c-ii), there is an L ∈ Kδ such that L ⊆ H and γ ≤ φ1 L = µL. Thus µ is inner regular with respect to Kδ . To see that µ is T unique, observe that if µ0 is any other measure with these properties, and L ∈ Kδ , then L is expressible as n∈N Kn where hKn in∈N is a sequence in K. Now T T µL = limn→∞ µ( i≤n Ki ) = limn→∞ φ0 ( i≤n Ki ) = µ0 L. So µ and µ0 must agree on Kδ , and by 412L they are identical. 413K Corollary (a) Let X be a set, Σ a subring of PX, and ν : Σ → [0, ∞[ a non-negative finitely additive functional such that limn→∞ νEn = 0 whenever hEn in∈N is a non-increasing sequence in Σ with empty intersection. Then ν has a unique extension to a complete locally determined measure on X which is inner regular with respect to the family Σδ of intersections of sequences in Σ. (b) Let X be a set, Σ a subalgebra of PX, and ν : Σ → [0, ∞[ a non-negative finitely additive functional such that limn→∞ νEn = 0 whenever hEn in∈N is a non-increasing sequence in Σ with empty intersection. Then ν has a unique extension to a measure defined on the σ-algebra of subsets of X generated by Σ. proof (a) Take Σ, ν in place of K, φ0 in 413J. (b) Let ν1 be the complete extension as in (a), and let ν10 be the restriction of ν1 to the σ-algebra Σ0 generated by Σ; this is the extension required here. To see that ν10 is unique, use the Monotone Class Theorem (136C). Remark You will sometimes see (b) above stated as ‘an additive functional on an algebra of sets extends to a measure iff it is countably additive’. But this formulation depends on a different interpretation of the phrase ‘countably additive’ from the one used in this book; see the note after the definition in 326E. 413L It will be useful to have a definition extending an idea in §342. Definition A countably compact class (or semicompact T paving) is a family K of sets such that T i≤n Ki 6= ∅ for every n ∈ N. n∈N Kn 6= ∅ whenever hKn in∈N is a sequence in K such that 413M Corollary Let X be a set and K a countably compact class of subsets of X such that ∅ ∈ K, (†) K ∪ K 0 ∈ K whenever K, K 0 ∈ K are disjoint, (‡) K ∩ K 0 ∈ K whenever K, K 0 ∈ K. Let φ0 : K → [0, ∞[ be a functional such that (α) φ0 K = φ0 L + sup{φ0 K 0 : K ∈ K, K 0 ⊆ K \ L} whenever K, L ∈ K and L ⊆ K. Then there is a unique complete locally determined measure µ on X extending φ0 and inner regular with respect to Kδ , the family of sets expressible as intersections of sequences in K. proof The point is that the hypothesis (β) of 413J is necessarily satisfied: if hKn in∈N is a non-increasing sequence in K with empty intersection, then, because K is countably compact, there must be some n such that Kn = ∅. Since hypothesis (α) here is already enough to ensure that φ0 ∅ = 0, φ0 K ≥ 0 for every K ∈ K, we must have inf n∈N φ0 Kn = 0. So we apply 413J to get the result. 413N I now turn to constructions of a different kind, being extension theorems in which the extension is not uniquely defined. Again I start with a theorem on finitely additive functionals. Theorem Let X be a set, T0 a subring of PX, and ν0 : T0 → [0, ∞[ a finitely additive functional. Suppose that K ⊆ PX is a family of sets such that (†) K ∪ K 0 ∈ K whenever K, K 0 ∈ K are disjoint, (‡) K ∩ K 0 ∈ K for all K, K 0 ∈ K, every member of K is included in some member of T0 ,

413N

Inner measure constructions

41

and ν0 is inner regular with respect to K in the sense that (α) ν0 E = sup{ν0 K : K ∈ K ∩ T0 , K ⊆ E} for every E ∈ T0 . Then ν0 has an extension to a non-negative finitely additive functional ν1 , defined on a subring T1 of PX including T0 ∪ K, inner regular with respect to K, and such that whenever E ∈ T1 , ² > 0 there is an E0 ∈ T0 such that ν1 (E4E0 ) ≤ ². proof (a) Let P be the set of all non-negative additive real-valued functionals ν, defined on subrings of PX, inner regular with respect to K, and such that (∗) whenever E ∈ dom ν, ² > 0 there is an E0 ∈ T0 such that ν(E4E0 ) ≤ ². Order P by extension of functions, so that P is a partially ordered set. (b) It will be convenient to borrow some notation from the theory of countably additive functionals. If T is a subring of PX and ν : T → [0, ∞[ is a non-negative additive functional, set ν ∗ A = inf{νE : A ⊆ E ∈ T},

ν∗ A = sup{νE : A ⊇ E ∈ T}

for every A ⊆ X (interpreting inf ∅ as ∞ if necessary). Now if A ⊆ X and E, F ∈ T are disjoint, ν ∗ (A ∩ (E ∪ F )) = ν ∗ (A ∩ E) + ν ∗ (A ∩ F ), ν∗ (A ∩ (E ∪ F )) = ν∗ (A ∩ E) + ν∗ (A ∩ F ).

P P ν ∗ (A ∩ (E ∪ F )) = inf{νG : G ∈ T, A ∩ (E ∪ F ) ⊆ G} = inf{νG : G ∈ T, A ∩ (E ∪ F ) ⊆ G ⊆ E ∪ F } = inf{ν(G ∩ E) + ν(G ∩ F ) : G ∈ T, A ∩ (E ∪ F ) ⊆ G ⊆ E ∪ F } = inf{νG1 + νG2 : G1 , G2 ∈ T, A ∩ E ⊆ G1 ⊆ E, A ∩ F ⊆ G2 ⊆ F } = inf{νG1 : G1 ∈ T, A ∩ E ⊆ G1 ⊆ E} + inf{νG2 : G2 ∈ T, A ∩ F ⊆ G2 ⊆ F } = ν ∗ (E ∩ A) + ν ∗ (F ∩ A), ν∗ (A ∩ (E ∪ F )) = sup{νG : G ∈ T, A ∩ (E ∪ F ) ⊇ G} = sup{ν(G ∩ E) + ν(G ∩ F ) : G ∈ T, A ∩ (E ∪ F ) ⊇ G} = sup{νG1 + νG2 : G1 , G2 ∈ T, A ∩ E ⊇ G1 , A ∩ F ⊇ G2 } = sup{νG1 : G1 ∈ T, A ∩ E ⊇ G1 } + sup{νG2 : G2 ∈ T, A ∩ F ⊇ G2 } = ν∗ (E ∩ A) + ν∗ (F ∩ A). Q Q (c) The key to the proof is the following fact: if ν ∈ P and M ∈ K, there is a ν 0 ∈ P such that ν 0 extends ν and M ∈ dom ν 0 . P P Set T = dom ν, T0 = {(E ∩ M ) ∪ (F \ M ) : E, F ∈ T}. For H ∈ T0 , set ν 0 H = ν ∗ (H ∩ M ) + ν∗ (H \ M ). Now we have to check the following. (i) T0 is a subring of PX, because if E, F , E 0 , F 0 ∈ T then ((E ∩ M ) ∪ (F \ M )) ∗ ((E 0 ∩ M ) ∪ (F 0 \ M )) = ((E ∗ E 0 ) ∩ M ) ∪ ((F ∗ F 0 ) \ M ) for both the Boolean operations ∗ = 4 and ∗ = ∩. T0 ⊇ T because E = (E ∩ M ) ∪ (E \ M ) for every E ∈ T. (Cf. 312M.) M ∈ T0 because there is some E ∈ T0 such that M ⊆ E, so that M = (E ∩ M ) ∪ (∅ \ M ) ∈ T0 . (ii) ν 0 is finite-valued because if H = (E ∩ M ) ∪ (F \ M ), where E, F ∈ T, then ν 0 H ≤ νE + νF . If H, H ∈ T are disjoint, they can be expressed as (E ∩ M ) ∪ (F \ M ), (E 0 ∩ M ) ∪ (F 0 \ M ) where E, F , E 0 , F 0 belong to T; replacing E 0 , F 0 by E 0 \ E and F 0 \ F if necessary, we may suppose that E ∩ E 0 = F ∩ F 0 = ∅. Now 0

42

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ν 0 (H ∪ H 0 ) = ν ∗ ((E ∪ E 0 ) ∩ M ) + ν∗ ((F ∪ F 0 ) ∩ (X \ M )) = ν ∗ (E ∩ M ) + ν ∗ (E 0 ∩ M ) + ν∗ (F ∩ (X \ M )) + ν∗ (F 0 ∩ (X \ M )) (by (b) above) = ν 0H + ν 0H 0. Thus ν 0 is additive. (iii) If E ∈ T, then ν∗ (E \ M ) = sup{νF : F ∈ T, F ⊆ E \ M } = sup{νE − ν(E \ F ) : F ∈ T, F ⊆ E \ M } = sup{νE − νF : F ∈ T, E ∩ M ⊆ F ⊆ E} = νE − inf{νF : F ∈ T, E ∩ M ⊆ F ⊆ E} = νE − ν ∗ (E ∩ M ). So ν 0 E = ν ∗ (E ∩ M ) + ν∗ (E \ M ) = νE. Thus ν 0 extends ν. (iv) If H ∈ T0 and ² > 0, express H as (E ∩ M ) ∪ (F \ M ), where E, F ∈ T. Then we can find (α) a K ∈ K ∩ T such that K ⊆ E and ν(E \ K) ≤ ² (β) an F 0 ∈ T such that F 0 ⊆ F \ M and νF 0 ≥ ν∗ (F \ M ) − ² (γ) a K 0 ∈ K ∩ T such that K 0 ⊆ F 0 and νK 0 ≥ νF 0 − ². Set L = (K ∩ M ) ∪ K 0 ∈ T0 ; by the hypotheses (†) and (‡), L ∈ K. Now L ⊆ H and ν 0 L = ν 0 (K ∩ M ) + ν 0 K 0 = ν 0 (E ∩ M ) − ν 0 ((E \ K) ∩ M ) + νK 0 = ν ∗ (H ∩ M ) − ν ∗ ((E \ K) ∩ M ) + νK 0 ≥ ν ∗ (H ∩ M ) − ν(E \ K) + νF 0 − ² ≥ ν ∗ (H ∩ M ) + ν∗ (F \ M ) − 3² = ν 0 H − 3². As H and ² are arbitrary, ν is inner regular with respect to K. (v) Finally, given H ∈ T0 and ² > 0, take E, F ∈ T such that H ∩ M ⊆ E, F ⊆ H \ M , νE ≤ ν (H ∩ M ) + ² and νF ≥ ν∗ (H \ M ) − ². In this case, ∗

ν 0 (E \ (H ∩ M )) = ν 0 E − ν 0 (H ∩ M ) = νE − ν ∗ (H ∩ M ) ≤ ², ν 0 ((H \ M ) \ F ) = ν 0 (H \ M ) − ν 0 F = ν∗ (H \ M ) − νF ≤ ². But as H4(E ∪ F ) ⊆ (E \ (H ∩ M )) ∪ ((H \ M ) \ F ), ν 0 (H4(E ∪F )) ≤ 2². Now ν satisfies the condition (∗), so there is an E0 ∈ T0 such that ν((E ∪F )4E0 ) ≤ ², and ν 0 (H4E0 ) ≤ 3². As H and ² are arbitrary, ν 0 satisfies (∗). This completes the proof that ν 0 is a member of P extending ν. Q Q (d) It is easy to check that if Q ⊆ P is a non-empty totally ordered subset, the smallest common extension ν 0 of the functions in Q belongs to P . (To see that ν 0 is inner regular with respect to K, observe that if E ∈ dom ν 0 and γ < ν 0 E, there is some ν ∈ Q such that E ∈ dom ν; now there is a K ∈ K ∩ dom ν such that K ⊆ E and νK ≥ γ, so that K ∈ K ∩ dom ν 0 and ν 0 K ≥ γ.) And of course P is not empty, because ν0 ∈ P . So by Zorn’s Lemma P has a maximal element ν1 say; write T1 for the domain of ν1 . If M ∈ K there is an element of P , with a domain containing M , extending ν1 ; as ν1 is maximal, this must be ν1 itself, so M ∈ T1 . Thus K ⊆ T1 , and ν1 has all the required properties.

413P

Inner measure constructions

43

413O Corollary Let (X, Σ0 , µ0 ) be a measure space and K a countably compact class of subsets of X such that (†) K ∪ K 0 ∈ K whenever K, K 0 ∈ K are disjoint, T (‡) n∈N Kn ∈ K for every sequence hKn in∈N in K, µ∗0 K < ∞ for every K ∈ K, µ0 is inner regular with respect to K. Then µ0 has an extension to a complete locally determined measure µ, defined on every member of K, inner regular with respect to K, and such that whenever E ∈ dom µ and µE < ∞ there is an E0 ∈ Σ0 such that µ(E4E0 ) = 0. proof (a) Set T0 = {E : E ∈ Σ0 , µ0 E < ∞}, ν0 = µ0 ¹ T0 . Then ν0 , T0 satisfy the conditions of 413N; take ν1 , T1 as in 413N. If K, L ∈ K and L ⊆ K, then ν1 L + sup{ν1 K 0 : K 0 ∈ K, K 0 ⊆ K \ L} = ν1 L + ν1 (K \ L) = ν1 K. So ν1 ¹K satisfies the conditions of 413M and there is a complete locally determined measure µ, extending ν1 ¹K, and inner regular with respect to K. (b) Write Σ for the domain of µ. Then T1 ⊆ Σ. P P If E ∈ T1 and K ∈ K, µ∗ (K ∩ E) + µ∗ (K \ E) ≥ sup{µK 0 : K 0 ∈ K, K 0 ⊆ K ∩ E} + sup{µK 0 : K 0 ∈ K, K 0 ⊆ K \ E} = sup{ν1 K 0 : K 0 ∈ K, K 0 ⊆ K ∩ E} + sup{ν1 K 0 : K 0 ∈ K, K 0 ⊆ K \ E} = ν1 (K ∩ E) + ν1 (K \ E) = ν1 K = µK. By 413F(iv), E ∈ Σ. Q Q It follows at once that µ extends ν1 , since if E ∈ T1 ν1 E = sup{ν1 K : K ∈ K, K ⊆ E} = sup{µK : K ∈ K, K ⊆ E} = µE. (c) In particular, µ agrees with µ0 on T0 . Now in fact µ extends µ0 . P P Take E ∈ Σ0 . If K ∈ K, there is an E0 ∈ Σ0 such that K ⊆ E0 and µ0 E0 < ∞. Since E ∩ E0 ∈ T0 ⊆ Σ, E ∩ K = E ∩ E0 ∩ K ∈ Σ. As K is arbitrary, E ∈ Σ, by 413F(ii). Next, because every member of K is included in a member of T0 , µ0 E = sup{µ0 K : K ∈ K ∩ Σ0 , K ⊆ E} = sup{µ0 (E ∩ E0 ) : E0 ∈ T0 } = sup{µ(E ∩ E0 ) : E0 ∈ T0 } = sup{µK : K ∈ K, K ⊆ E} = µE. Q Q (d) Finally, suppose that E ∈ Σ and µE < ∞. For each n ∈ N P we can find Kn ∈ K, En ∈ Σ0 such that ∞ 0 −n −n Kn ⊆ E, µ(E \ K ) ≤ 2 and ν (K 4E ) ≤ 2 . In this case n 1 n n n=0 µ(En 4E) < ∞, so µ(E4E ) = 0, S T 0 where E = n∈N m≥n Em ∈ Σ0 . Thus µ has all the required properties. 413P I now describe an alternative route to some of the applications of 413N. As before, I do as much as possible in the context of finitely additive functionals. Lemma Let X be a set and K a sublattice of PX containing ∅. Let λ : K → [0, ∞[ be a bounded functional such that λ∅ = 0,

λK ≤ λK 0 whenever K, K 0 ∈ K and K ⊆ K 0 ,

λ(K ∪ K 0 ) + λ(K ∩ K 0 ) ≥ λK + λK 0 for all K, K 0 ∈ K. Then there is a finitely additive functional ν : PX → [0, ∞[ such that νX = supK∈K λK,

νK ≥ λK for every K ∈ K.

0 0 proof (a) The key to the proof isPthe followingPfact: if K0 , . . P . , Kn ∈ K, then Pn there are K0 , . . . , Kn ∈ K n n n 0 0 0 0 0 such that K0 ⊆ K1 ⊆ . . . ⊆ Kn , i=0 χKi = i=0 χKi and i=0 λKi ≥ i=0 λKi . P P Induce on n. If

44

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n = 0 the result is trivial. For the inductive step to n + 1, given K0 , . . . , Kn+1 ∈ K, set Ln+1 = Kn+1 and use the inductive hypothesis to find L0 , . . . , Ln ∈ K such that Pn Pn Pn Pn L0 ⊆ . . . ⊆ Ln , i=0 χLi = i=0 χKi and i=0 λLi ≥ i=0 λKi . Now set L0n = Ln+1 ∩ Ln , L0i = Li for i < n, and use the inductive hypothesis again to find K00 , . . . , Kn0 ∈ K such that Pn Pn Pn Pn K00 ⊆ . . . ⊆ Kn0 , i=0 χKi0 = i=0 χL0i and i=0 λKi0 ≥ i=0 λL0i . Set 0 Kn+1 = L0n+1 = Ln ∪ Ln+1 . Pn Then all the Li , L0i and Ki0 belong to K. We know that χKn0 ≤ i=0 χL0i , so S 0 Kn0 ⊆ i≤n L0i ⊆ Ln ⊆ Kn+1 ; 0 accordingly we have K00 ⊆ K10 ⊆ . . . ⊆ Kn0 ⊆ Kn+1 . Next, n+1 X

χKi0 =

i=0

n X

χL0i + χL0n+1 =

i=0

=

n−1 X

n−1 X

χLi + χ(Ln ∩ Ln+1 ) + χ(Ln ∪ Ln+1 )

i=0

χLi + χLn + χLn+1 =

i=0

n+1 X

λKi0 ≥

i=0

n X

≥

χLi + χKn+1 =

i=0

λL0i + λL0n+1 =

i=0 n−1 X

n X

n−1 X

n+1 X

χKi ,

i=0

λLi + λ(Ln ∩ Ln+1 ) + λ(Ln ∪ Ln+1 )

i=0

λLi + λLn + λLn+1

i=0

(using the hypothesis on λ) =

n X

λLi + λKn+1 ≥

n+1 X

i=0

λKi ,

i=0

0 , and the induction continues. Q Q so we have an appropriate family K00 , . . . , Kn+1

(b) For the moment, suppose that supK∈K P λK = 1. In T this case, if hKi ii∈I is a finite indexed family in P If I = ∅ this is trivial, so we may K, there is a set J ⊆ I such that #(J) ≥ i∈I λKi and i∈J Ki 6= ∅. P 0 0 , . . . , K as in (a). If Kn0 = ∅ then every Ki is empty so suppose that I = {0, . . . , n} for some n. Take K n 0 Pn (because λ∅ = 0) i=0 λKi = 0 and we may take J = ∅. TOtherwise, let m be the first number such that 0 0 Km 6= ∅, and take x ∈ Km , J = {i : i ≤ n, x ∈ Ki }. Then i∈J Ki 6= ∅ and #(J) = ≥

n X i=0 n X i=m

χKi (x) =

n X

χKi0 (x) = n − m + 1

i=0

λKi0 =

n X i=0

λKi0 ≥

n X

λKi ,

i=0

as claimed. Q Q By 391F, there is an additive functional ν : PX → [0, 1] such that νX = 1 and νK ≥ λK for every K ∈ K, as required. (c) For the general case, set γ = supK∈K λK. If γ = 0, take ν to be the zero functional; if γ > 0, apply (a)-(b) to the functional γ −1 λ. Remark If P is a lattice, a function f : P → R such that f (p ∨ q) + f (p ∧ q) ≥ f (p) + f (q) for all p, q ∈ P is called supermodular.

413Q

Inner measure constructions

45

413Q Theorem Let X be a set and K a sublattice of PX containing ∅. Let Σ be the algebra of subsets of X generated by K, and ν0 : Σ → [0, ∞[ a finitely additive functional. Then there is a finitely additive functional ν : Σ → [0, ∞[ such that (i) νX = supK∈K ν0 K (ii) νK ≥ ν0 K for every K ∈ K (iii) ν is inner regular with respect to K in the sense that νE = sup{νK : K ∈ K, K ⊆ E} for every E ∈ Σ. proof (a) Set γ = supK∈K ν0 K. Let P be the set of all functionals λ : K → [0, γ] such that λK + λK 0 ≤ λ(K ∪ K 0 ) + λ(K ∩ K 0 ) for every K, K 0 ∈ K. Give P the natural partial ordering inherited from R K . Note that ν0 ¹K belongs to P . If Q ⊆ P is non-empty and upwards-directed, then sup Q, taken in R K , belongs to P ; so there is a maximal λ ∈ P such that ν0 ¹K ≤ λ. By 413P, there is a non-negative additive functional ν on PX such that νK ≥ λK for every K ∈ K and νX = γ. Since ν¹K also belongs to P , we must have νK = λK for every K ∈ K. (b) Now for any K0 ∈ K, νK0 + sup{νL : L ∈ K, L ⊆ X \ K0 } = γ. P P (i) Set L = {L : L ∈ K, L ⊆ X \ K0 }. For A ⊆ X, set θ0 A = supL∈L ν(A ∩ L). Because L is upwards-directed, θ0 : PX → R is additive, and of course 0 ≤ θ0 ≤ ν. Set θ1 = ν − θ0 , so that θ1 is another additive functional, and write λ0 K = θ0 K + sup{θ1 M : M ∈ K, M ∩ K0 ⊆ K} for K ∈ K. (ii) If K, K 0 ∈ K and ² > 0, there are M , M 0 ∈ K such that M ∩ K0 ⊆ K, M 0 ∩ K0 ⊆ K 0 and θ0 K + θ1 M ≥ λ0 K − ²,

θ0 K 0 + θ1 M 0 ≥ λ0 K 0 − ².

Now M ∪ M 0 , M ∩ M 0 ∈ K, (M ∪ M 0 ) ∩ K0 ⊆ K ∪ K 0 ,

(M ∩ M 0 ) ∩ K0 ⊆ K ∩ K 0 ,

so λ0 (K ∪ K 0 ) + λ0 (K ∩ K 0 ) ≥ θ0 (K ∪ K 0 ) + θ1 (M ∪ M 0 ) + θ0 (K ∩ K 0 ) + θ1 (M ∩ M 0 ) = θ0 K + θ1 M + θ0 K 0 + θ1 M 0 ≥ λ0 K + λ0 K 0 − 2². As ² is arbitrary, λ0 (K ∪ K 0 ) + λ0 (K ∩ K 0 ) ≥ λK + λK 0 . (iii) Suppose that K, M ∈ K are such that M ∩ K0 ⊆ K. If L ∈ L, then ν(K ∩ L) + θ1 M = ν(K ∩ L) + νM − θ0 M = ν(M ∩ K ∩ L) + ν(M ∪ (K ∩ L)) − θ0 M ≤ γ because K ∩ L ∈ L; taking the supremum over L and M , λ0 K ≤ γ. As K is arbitrary, λ0 ∈ P . (iv) If K ∈ K, then of course K ∩ K0 ⊆ K, so λ0 K ≥ θ0 K + θ1 K = νK = λK. Thus λ0 ≥ λ. Because λ is maximal, λ0 = λ. But this means that λK0 = λ0 K0 = θ0 K0 + sup{θ1 M : M ∈ K, M ∩ K0 ⊆ K0 } = supM ∈K θ1 M . Now given ² > 0 there is an M ∈ K such that γ − ² ≤ ν0 M ≤ λM = νM , so that νK0 = λK0 ≥ θ1 M = νM − θ0 M ≥ γ − ² − θ0 M ≥ γ − ² − supL∈L νL,

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413Q

and νK0 + supL∈L νL ≥ γ − ². As ² is arbitrary, νK0 + supL∈L νL ≥ γ. But of course νK0 + νL ≤ νX = γ for every L ∈ L, so νK0 + supL∈L νL = γ, as claimed. Q Q (c) It follows that if K, L ∈ K and L ⊆ K, νK = νL + sup{νK 0 : K 0 ∈ K, K 0 ⊆ K \ L}. P P Because ν is additive and non-negative, we surely have νK ≥ νL + sup{νK 0 : K 0 ∈ K, K 0 ⊆ K \ L}. On the other hand, given ² > 0, there is an M ∈ K such that M ⊆ X \ L and νL + νM ≥ γ − ², so that M ∩ K ∈ K, M ∩ K ⊆ K \ L and νL + ν(M ∩ K) = νL + νK + νM − ν(M ∪ K) ≥ νK + γ − ² − γ = νK − ². As ² is arbitrary, νK ≤ νL + sup{νK 0 : K 0 ∈ K, K 0 ⊆ K \ L} and we have equality. Q Q (d) By 413H, we have an additive functional ν 0 : Σ → [0, ∞[ such that ν 0 E = sup{νK : K ∈ K, K ⊆ E} for every E ∈ Σ. It is easy to show that ν 0 and ν must agree on Σ, but even without doing so we can see that ν 0 has the properties (i)-(iii) required in the theorem. 413R The following lemma on countably compact classes, corresponding to 342Db, will be useful. Lemma (Marczewski 53) Let K be a countably compact class of sets. Then there is a countably compact T class K∗ ⊇ K such that K ∪ L ∈ K∗ and n∈N Kn ∈ K∗ whenever K, L ∈ K∗ and hKn in∈N is a sequence in K∗ . proof (a) Write Ks for {K0 ∪ . .T . ∪ Kn : K0 , . . . , Kn ∈ K}. Then Ks is countably compact. P P Let hLn iS n∈N 6 ∅ for each n ∈ N. Then there is an ultrafilter F on X = K be a sequence in Ks such that i≤n Li = containing every Ln . For each n, T Ln is a finite union of members T of K, so thereTmust be a Kn ∈ K such that Kn ⊆ Ln and Kn ∈ F. Now i≤n Ki 6= ∅ for every n, so n∈N Kn 6= ∅ and n∈N Ln 6= ∅. As hLn in∈N is arbitrary, Ks is countably compact. Q Q Note that L ∪ L0 ∈ Ks for all L, L0 ∈ Ks . (b) Write K∗ for

T

{

L0 : L0 ⊆ Ks is non-empty and countable}.

T Then K∗ is countably compact. P P If hMn in∈N is any sequence in K∗ such that i≤n Mi 6= ∅ for every n ∈ N, T then for each n ∈ N letSLn ⊆ Ks be aTcountable non-empty set such = Ln . Let hLn in∈N be a T that Mn T sequence running over n∈N Ln ; then i≤n Li 6= ∅ for every n, so n∈N Ln = n∈N Mn is non-empty. As hMn in∈N is arbitrary, K∗ is countably compact. Q Q (c) Of course K ⊆ Ks ⊆ K∗ . It is immediate from the definition of K∗ that it is closed under T countable ∗ intersections. Finally, if M , M ∈ K , let L , L ⊆ K be countable sets such that M = 1 2 1 2 s 1 T T L1 and M2 = L2 ; then L = {L1 ∪ L2 : L1 ∈ L1 , L2 ∈ L2 } is a countable subset of Ks , so M1 ∪ M2 = L belongs to K∗ . 413S Corollary Let X be a set and K a countably compact class of subsets of X. Let T be a subalgebra of PX and ν : T → R a non-negative finitely additive functional. (a) There is a complete measure µ on X such that µX ≤ νX, K ⊆ dom µ and µK ≥ νK for every K ∈ K ∩ T. (b) If (†) K ∪ K 0 ∈ K whenever K, K 0 ∈ K, T (‡) n∈N Kn ∈ K for every sequence hKn in∈N in K, we may arrange that µ is inner regular with respect to K.

413Xj

Inner measure constructions

47

proof By 413R, there is always a countably compact class K∗ ⊇ K satisfying (†) and (‡); for case (b), take K∗ = K. By 391G, there is an extension of ν to a finitely additive functional ν 0 : PX → R. Let T1 be the subalgebra of PX generated by K∗ . By 413Q, there is a non-negative additive functional ν1 : T1 → R such that ν1 X ≤ ν 0 X = νX, ν1 K ≥ ν 0 K = νK for every K ∈ K∗ ∩ T and ν1 E = sup{ν1 K : K ∈ K∗ , K ⊆ E} for every E ∈ T1 . In particular, if K, L ∈ K∗ , ν1 L + sup{ν1 K 0 : K 0 ∈ K∗ , K 0 ⊆ K \ L} = ν1 L + ν1 (K \ L) = ν1 K. So K∗ and ν1 ¹K∗ satisfy the hypotheses of 413M. Accordingly we have a complete measure µ extending ν1 ¹K∗ and inner regular with respect to K∗ = Kδ∗ ; in which case µK = ν1 K ≥ νK for every K ∈ K ∩ T, µX = supK∈K∗ µK = supK∈K∗ ν1 K ≤ ν1 X ≤ νX, as required. 413X Basic exercises (a) Define φ : PN → [0, ∞[ by setting φA = 0 if A is finite, ∞ otherwise. Check that φ satisfies conditions (α) and (β) of 413A, but that if we attempt to reproduce the construction of 413C then we obtain Σ = PN, µ = φ, so that µ is not countably additive. (b) Let φ1 , φ2 be two inner measures on a set X, inducing measures µ1 , µ2 by the method of 413C. (i) Show that φ = φ1 + φ2 is an inner measure. (ii) Show that the measure µ induced by φ extends the measure µ1 + µ2 defined on dom µ1 ∩ dom µ2 . > (c) Let X be a set, φ an inner measure on X, and µ the measure constructed from it by the method of 413C. (i) Let Y be a subset of X. Show that φ¹ PY is an inner measure on Y , and that the measure on Y defined from it extends the subspace measure µY induced on Y by µ. (ii) Let Y be a set and f : X → Y a function. Show that B 7→ φf −1 [B] is an inner measure on Y , and that it defines a measure on Y which extends the image measure µf −1 . (d) Let (X, Σ, µ) be any measure space. Set θA = 21 (µ∗ A + µ∗ A) for every A ⊆ X. Show that θ is an outer measure on X, and that the measure defined from θ by Carath´eodory’s method extends µ. S > (e) Show that there is a partition hAn in∈N of [0, 1] such that µ∗ ( i≤n Ai ) = 0 for every n, where µ∗ is Lebesgue inner measure. (Hint: set An = (A + qn ) ∩ [0, 1] where hqn in∈N is an enumeration of Q and A is a suitable set; cf. 134B.) (f ) Let (X, Σ, µ) be a measure space. (i) Show that µ∗ ¹Σ is the semi-finite version µsf of µ as constructed in 213Xc. (ii) Show that if A is any subset of X, and ΣA the subspace σ-algebra, then µ∗ ¹ΣA is a semi-finite measure on A. > (g) Let (X, Σ, µ) and (Y, T, ν) be two measure spaces, and λ the c.l.d. product measure on X × Y . Show that λ∗ (A × B) = µ∗ A · ν∗ B for all A ⊆ X and B ⊆ Y . (Hint: use Fubini’s theorem to show that λ∗ (A × B) ≤ µ∗ A · ν∗ B.) (h) Let X be a set and µ, ν two complete locally determined measures on X with domains Σ, T respectively, both inner regular with respect to K ⊆ Σ ∩ T. Suppose that, for K ∈ K, µK = 0 iff νK = 0. Show that Σ = T and that µ and ν give rise to the same negligible sets. > (i) Let (X, T, ν) be a measure space. (i) Show that the measure constructed by the method of 413C from the inner measure ν∗ is the c.l.d. version of ν. (ii) Set K = {E : E ∈ T, νE < ∞}, φ0 = ν¹K. Show that K, φ0 satisfy the conditions of 413I, and that the measure constructed by the method there is again the c.l.d. version of ν. (j) Let (X, Σ, µ) be a complete locally determined measure space and L a family of subsets of X such that µ is inner regular with respect to L. Set K = {K : K ∈ L ∩ Σ, µK < ∞} and φ0 = µ¹K. Show that K, φ0 satisfy the conditions of 413I and that the measure constructed from K, φ0 by the method there is just µ.

48

Topologies and measures I

413Xk

> (k) Let K be the family of subsets of R expressible as finite unions of bounded closed intervals. (i) Show from first principles that there is a unique functional φ0 : K → [0, ∞[ such that φ0 [α, β] = β − α whenever α ≤ β and φ0 satisfies the conditions of 413J. (ii) Show that the measure on R constructed from φ0 by the method of 413J is Lebesgue measure. (l) Let X be a set, Σ a subring of PX, and ν : Σ → [0, ∞[ a non-negative additive functional such that limn→∞ νEn = 0 whenever hEn in∈N is a non-increasing sequence in Σ with empty intersection, as in 413K. Define θ : PX → [0, ∞] by setting P∞ θA = inf{ n=0 νEn : hEn in∈N is a sequence in Σ covering A} for A ⊆ X, interpreting inf ∅ as ∞ if necessary. Show that θ is an outer measure. Let µθ be the measure defined from θ by Carath´eodory’s method. Show that the measure defined from ν by the process of 413K is the c.l.d. version of µθ . (Hint: the c.l.d. version of µθ is inner regular with respect to Σδ .) > (m) Let X be a set, Σ a subring of PX, and ν : Σ → [0, ∞[ a non-negative additive functional. Show that the following are equiveridical: (i) ν has an extension to a measure on X; n = 0 whenever S (ii) limn→∞ PνE ∞ hEn in∈N is a non-increasing sequence in Σ withSempty intersection; (iii) ν( n∈N En ) = n=0 νEn whenever hEn in∈N is a disjoint sequence in Σ such that n∈N En ∈ Σ. (n) Let h(X Q n , Σn , µn )in∈N be a sequence of probability spaces, and F a non-principal ultrafilter on N. equivalence relation; For x, y ∈ n∈N Xn , write x ∼ y if {n : x(n) = y(n)} ∈ F. (i) Show that ∼ is an Q (Compare write X for the set of equivalence classes, and x• ∈ X for the equivalence class of x ∈ n∈N Xn .Q 351M.) (ii) Let Σ be the set of subsets of X expressible in the form Q(hEn in∈N ) = {x• : x ∈ n∈N En }, where En ∈ Σn for each n ∈ N. Show that Σ is an algebra of subsets of X, and that there is a well-defined additive functional ν : Σ → [0, 1] defined by setting ν(Q(hEn in∈N )) = limn→F T µn En . (iii) Show that for any non-decreasing sequence hHi ii∈N in Σ there is an H ∈ Σ such that H ⊆ i∈N Hi and νH = limn→∞ νHn . (Hint: express each Hi as Q(hEin in∈N ). Do this in such a way that Ei+1,n ⊆ Ein for all i, n. Take a decreasing sequence hJi ii∈N in F, with empty intersection, such that νHi ≤ µEin + 2−i for n ∈ Ji . Set En = Ein for n ∈ Ji \ Ji+1 .) (iv) Show that there is a unique extension of ν to a complete probability measure µ on X which is inner regular with respect to Σ. (This is a kind of Loeb measure.) (o) Let A be a Boolean algebra and K ⊆ A a sublattice containing 0. Suppose that λ : K → [0, ∞[ is a bounded functional such that λ0 = 0, λa ≤ λa0 whenever a, a0 ∈ K and a ⊆ a0 , and λ(a ∪ a0 ) + λ(a ∩ a0 ) ≥ λa + λa0 for all a, a0 ∈ K. Show that there is a non-negative additive functional ν : A → R such that νa ≥ λa for every a ∈ K and ν1 = supa∈K λa. (p) Let X be a set and K a sublattice of PX containing ∅. Let λ : K → R be a bounded order-preserving function such that λ∅ = 0 and λ(K ∪ K 0 ) + λ(K ∩ K 0 ) = λK + λK 0 for all K, K 0 ∈ K. Show that there is a non-negative additive functional ν : PX → R extending λ. (Hint: for finite K, induce on #(K). For the general case, recall that if γ = supK∈K λK, the additive functionals form a compact subset of [0, γ]PX .) (If P is a lattice, a functional f : P → R such that f (p ∨ q) + f (p ∧ q) = f (p) + f (q) for all p, q ∈ P is called modular.) (q) Let X be a set and K a sublattice of PX containing K. Let λ : K → [0, 1] be a functional such that λK ≤ λK 0 whenever K, K 0 ∈ K and K ⊆ K 0 ,

inf K∈K λK = 0,

λ(K ∪ K 0 ) + λ(K ∩ K 0 ) ≤ λK + λK 0 for all K, K 0 ∈ K (λ is submodular). Show that there is a finitely additive functional ν : PX → [0, 1] such that νX = supK∈K λK,

νK ≤ λK for every K ∈ K.

413Y Further exercises (a) Give an example of two inner measures φ1 , φ2 on a set X such that the measure defined by φ1 + φ2 strictly extends the sum of the measures defined by φ1 , φ2 .

413 Notes

Inner measure constructions

49

Q (b) Let h(XQ be any family of probability spaces, and λ the product measure on X = i∈I Xi . i , Σi , µi )ii∈IQ Show that λ∗ ( i∈I Ai ) ≤ i∈I (µi )∗ Ai whenever Ai ⊆ Xi for every i, with equality if I is countable. (c) Let (X, Σ, µ) be a totally finite measure space, and Z the Stone space of the Boolean algebra Σ. For b for the corresponding open-and-closed subset of Z. Show that there is a unique function E ∈ Σ write E b = E for every E ∈ Σ. Show that there is a measure ν on Z, inner regular with f : X → Z such that f −1 [E] respect to the open-and-closed sets, such that f is inverse-measure-preserving with respect to µ and ν, and that f represents an isomorphism between the measure algebras of µ and ν. Use this construction to prove (vi)⇒(i) in Theorem 343B without appealing to the Lifting Theorem. (d) Let X be a set, T a subalgebra of PX, and ν : T → [0, ∞[ a finitely additive functional. Suppose that there is a set K ⊆ T, containing ∅, such that (i) µF = sup{µK : K ∈ K, K ⊆ F } for every F ∈ T (ii) T K is monocompact, that is, n∈N Kn 6= ∅ for every non-decreasing sequence in K. Show that ν extends to a measure on X. (e) (i) Let X be a topological space. Show that the family of closed countably compact subsets of X is a countably compact class. (ii) Let X be a Hausdorff space. Show that the family of sequentially compact subsets of X is a countably compact class. 413 Notes and comments I gave rather few methods of constructing measures in the first three volumes of this treatise; in the present volume I shall have to make up for lost time. In particular I used Carath´eodory’s construction for Lebesgue measure (Chapter 11), product measures (Chapter 25) and Hausdorff measures (Chapter 26). The first two, at least, can be tackled in quite different ways if we choose. The first alternative approach I offer is the ‘inner measure’ method of 413C. Note the exact definition in 413A; I do not think it is an obvious one. In particular, while (α) seems to have something to do with subadditivity, and (β) is a kind of sequential order-continuity, there is no straightforward way in which to associate an outer measure with an inner measure, unless they both happen to be derived from measures (132B, 413D), even when they are finite-valued; and for an inner measure which is allowed to take the value ∞ we have to add the semi-finiteness condition (∗) of 413A (see 413Xa). Once we have got these points right, however, we have a method which rivals Carath´eodory’s in scope, and in particular is especially well adapted to the construction of inner regular measures. As an almost trivial example, we have a route to the c.l.d. version of a measure µ (413Xi(i)), which can be derived from the inner measure µ∗ defined from µ (413D). Henceforth µ∗ will be a companion to the familiar outer measure µ∗ , and many calculations will be a little easier with both available, as in 413D-413F. The intention behind 413I-413J is to find a minimal set of properties of a functional φ0 which will ensure that it has an extension to a measure. Indeed it is easy to see that, in the context of 413I, given a family K with the properties (†) and (‡) there, a functional φ0 on K can have an extension to an inner regular measure iff it satisfies the conditions (α) and (β), so in this sense 413I is the best possible result. Note that while Carath´eodory’s construction is liable to produce wildly infinite measures (like Hausdorff measures, or primitive product measures), the construction here always gives us locally determined measures, provided only that φ0 is finite-valued. We have to work rather hard for the step from 413I to 413J. Of course 413I is a special case of 413J, and I could have saved a little space by giving a direct proof of the latter result. But I do not think that this would have made it easier; 413J really does require an extra step, because somehow we have to extend the functional φ0 from K to Kδ . The method I have chosen uses 413B and 413H to cast as much of the argument as possible into the context of algebras of sets with additive functionals, where I hope the required manipulations will seem natural. (But perhaps I should insist that you must not take them too much for granted, as some of the time we have a finitely additive functional taking infinite values, and must take care not to subtract illegally, as well as not to take limits in the wrong places.) Note that the progression φ0 → φ1 → µ in the proof of 413J involves first an approximation from outside (if K ∈ Kδ , then φ1 K will be inf{φ0 K 0 : K ⊆ K 0 ∈ K}) and then an approximation from inside (if E ∈ Σ, then µE = sup{φ1 K : K ∈ Kδ , K ⊆ E}). The essential difficulty in the proof is just that we have to take successive non-exchangeable limits. I have slipped 413K in as a corollary of 413J; but it can be regarded as one of the fundamental results of measure theory. A non-negative finitely additive functional ν on an algebra Σ of sets can be extended to a countably additive

50

Topologies and measures

413 Notes

S P∞ measure iff it is ‘relatively countably additive’ in the sense that ν( n∈N En ) = n=0 µEn whenever hEn in∈N S is a disjoint sequence in Σ such that n∈N En ∈ Σ (413Xm). Of course the same result can easily be got from an outer measure construction (413Xl). Note that the outer measure construction also has repeated limits, albeit simpler ones: in the formula P∞ θA = inf{ n=0 νEn : hEn in∈N is a sequence in Σ covering A} P∞ Pn the sum n=0 νEn = supn∈N i=0 νEi can be regarded as a crude approximation from inside, while the infimum is an approximation from outside. To get the result as stated in 413K, of course, the outer measure construction needs a third limiting process, to obtain the c.l.d. version automatically provided by the inner measure method, and the inner regularity with respect to Σδ , while easily checked, also demands a few words of argument. Many applications of the method of 413I-413J pass through 413M; if the family K is a countably compact class then the sequential order-continuity hypothesis (β) of 413I or 413J becomes a consequence of the other hypotheses. The essence of the method is the inner regularity hypothesis (α). I have tried to use the labels †, ‡, α and β consistently enough to suggest the currents which I think are flowing in this material. In 413N we strike out in a new direction. The object here is to build an extension which is not going to be unique, and for which choices will have to be made. As with any such argument, the trick is to specify the allowable intermediate stages, that is, the partially ordered set P to which we shall apply Zorn’s Lemma. But here the form of the theorem makes it easy to guess what P should be: it is the set of functionals satisfying the hypotheses of the theorem which have not wandered outside the boundary set by the conclusion, that is, which satisfy the condition (∗) of part (a) of the proof of 413N. The finitistic nature of the hypotheses makes it easy to check that totally ordered subsets of P have upper bounds (that is to say, if we did this by transfinite induction there would be no problem at limit stages), and all we have to prove is that maximal elements of P are defined on adequately large domains; which amounts to showing that a member of P not defined on every element of K has a proper extension, that is, setting up a construction for the step to a successor ordinal in the parallel transfinite induction (part (c) of the proof). Of course the principal applications of 413N in this book will be in the context of countably additive functionals, as in 413O. It is clear that 413N and 413Q overlap to some extent. I include both because they have different virtues. 413N can be applied to infinite measures in a way that 413Q, as given, cannot; but its chief advantage, from the point of view of the work to come, is the approximation of members of T1 , in measure, by members of T0 . This will eventually enable us to retain control of the Maharam types of measures constructed by the method of 413O. In 413S we have a different kind of control; we can specify a lower bound for the measure of each member of our basic class K, provided only that our specifications are consistent with some finitely additive functional.

414 τ -additivity The second topic I wish to treat is that of ‘τ -additivity’. Here I collect results which do not depend on any strong kind of inner regularity. I begin with what I think of as the most characteristic feature of τ -additivity, its effect on the properties of semi-continuous functions (414A), with a variety of corollaries, up to the behaviour of subspace measures (414K). A very important property of τ -additive topological measures is that they are often strictly localizable (414J). The theory of inner regular τ -additive measures belongs to the next section, but here I give two introductory results: conditions under which a τ -additive measure will be inner regular with respect to closed sets (414M) and conditions under which a measure which is inner regular with respect to closed sets will be τ -additive (414N). I end the section with notes on ‘density’ and ‘lifting’ topologies (414P-414R). 414A Theorem Let (X, T) be a topological space and µ an effectively locally finite τ -additive measure on X with domain Σ. S (a) Suppose that G is a non-empty family in Σ ∩ T such that H = G also belongs to Σ. Then supG∈G G• = H • in the measure algebra A of µ.

414D

τ -additivity

51

(b) Write L for the family of Σ-measurable lower semi-continuous functions from X to R. Suppose that ∅ 6= A ⊆ L and set g(x) = supf ∈A f (x) for every x ∈ X. If g is Σ-measurable and finite almost everywhere, then g˜• = supf ∈A f • in L0 (µ), where g˜(x) = g(x) whenever g(x) is finite. T • T(c) •Suppose that F is a non-empty family of measurable closed sets such that F ∈ Σ. Then inf F ∈F F = ( F) in A. (d) Write U for the family of Σ-measurable upper semi-continuous functions from X to R. Suppose that A ⊆ U is non-empty and set g(x) = inf f ∈A f (x) for every x ∈ X. If g is Σ-measurable and finite almost everywhere, then g˜• = inf f ∈A f • in L0 (µ), where g˜(x) = g(x) whenever g(x) is finite. proof (a) ?? If H • 6= supG∈G G• , there is a non-zero a ∈ A such that a ⊆ H • but a ∩ G• = 0 for every G ∈ G. Express a as E • where E ∈ Σ and E ⊆ H. Because µ is effectively locally finite, there is a measurable open set H0 of finite measure such that µ(H0 ∩ E) > 0. Now {H0 ∩ G : G ∈ G} is an upwards-directed family of measurable open sets with union H0 ∩ H ⊇ H0 ∩ E; as µ is τ -additive, there is a G ∈ G such that µ(H0 ∩ G) > µH0 − µ(H0 ∩ E). But in this case µ(G ∩ E) > 0, which is impossible, because G• ∩ E • = 0. X X (b) For any α ∈ R, {x : g(x) > α} =

S

f ∈A {x

: f (x) > α},

and these are all measurable open sets. Identifying {x : g(x) > α}• ∈ A with [[˜ g • > α]] (364Jb), we see from • • • (a) that [[˜ g > α]] = supf ∈A [[f > α]] for every α. But this means that g˜ = supf ∈A f • , by 364Mb. (c) Apply (a) to G = {X \ F : F ∈ F}. (d) Apply (b) to {−f : f ∈ A}. 414B Corollary Let X be a topological space and µ an effectively locally finite τ -additive topological measure on X. (a) Suppose that A is a non-empty upwards-directed family of lower semi-continuous functions from X R R to [0, ∞]. Set g(x) = supf ∈A f (x) in [0, ∞] for every x ∈ X. Then g = supf ∈A f in [0, ∞]. (b) Suppose that A is a non-empty downwards-directed family of non-negative continuous real-valued functions on RX, and that g(x) = inf x∈A f (x) for every x ∈ X. If any member of A is integrable, then R g = inf f ∈A f . proof (a) Of course all the f ∈ A, and also g, are measurable functions. Set gn = g ∧ nχX for every n ∈ N. Then gn (x) = supf ∈A (f ∧ nχX)(x) for every x ∈ X, so gn = supf ∈A (f ∧ nχX)• , by 414Ab, and •

R

gn =

R

gn• = supf ∈A

by 365Dh. But now, of course,

R

g = supn∈N

R

R

(f ∧ nχX)• = supf ∈A

gn = supn∈N,f ∈A

R

R

f ∧ nχX

f ∧ nχX = supf ∈A

R

f,

as claimed. (b) Take an integrable f0 ∈ A, and apply (a) to {(f0 − f )+ : f ∈ A}. 414C Corollary Let (X, T, Σ, µ) be an effectively locally finite τ -additive topological measure spaceTand F a non-empty downwards-directed family of closed sets. If inf F ∈F µF is finite, this is the measure of F. T proof Setting F0 = F, then F0• = inf F ∈F F • , by 414Ac; now µF0 = µ ¯F0• = inf F ∈F µ ¯F • = inf F ∈F µF by 321F. 414D Corollary Let µ be an effectively locally finite τ -additive measure on a topological space X. If ν is a totally finite measure with the same domain as µ, truly continuous with respect to µ, then ν is τ -additive. In particular, if µ is σ-finite and ν is absolutely continuous with respect to µ, then ν is τ -additive.

52

Topologies and measures

414D

proof We have a functional ν¯ : A → [0, ∞[, where A is the measure algebra of µ, such that ν¯E • = νE for every E in the common domain Σ of µ and ν. Now ν¯ is continuous for the measure-algebra topology of A (327Cd), therefore completely additive (327Ba), therefore order-continuous (326Kc). So if G is an upwards-directed family of open sets belonging to Σ with union G0 ∈ Σ, supG∈G νG = supG∈G ν¯G• = ν¯G•0 = νG0 because G•0 = supG∈G G• . The last sentence follows at once, because on a σ-finite space an absolutely continuous countably additive functional is truly continuous (232Bc). 414E Corollary Let (X, T, Σ, µ) be an effectively locally finiteS τ -additive topological measure space. Suppose that G ⊆ T is non-empty and upwards-directed, and H = G. Then (a) µ(E ∩ H) = supG∈G µ(E ∩ G) for every E ∈ Σ; virtually measurable real-valued function defined almost everywhere on X, then R (b) if f is a non-negative R f = sup f in [0, ∞]. G∈G H proof (a) In the measure algebra (A, µ ¯) of µ, (E ∩ H)• = E • ∩ H • = E • ∩ sup G• G∈G

= sup E ∩ G = sup (E ∩ G)• , •

•

G∈G

G∈G

using 414Aa and the distributive law 313Ba. So µ(E ∩ H) = µ ¯(E ∩ H)• = supG∈G µ ¯(E ∩ G)• = supG∈G µ(E ∩ G) by 321D, because G and {(E ∩ G)• : G ∈ G} are upwards-directed. (b) For each G ∈ G,

R G

f=

R

f × χG =

R

(f × χG)• =

R

f • × χG• ,

where χG• can be interpreted either as (χG)• (in L0 (µ)) or as χ(G• ) (in L0 (A), where A is the measure algebra of µ); see 364K. Now H • = supG∈G G• (414Aa); since χ and × are order-continuous (364Kc, 364P), f • × χH • = supG∈G f • × χG• ; so

R

H

f=

R

f • × χH • = supG∈G

R

f • × χG• = supG∈G

R

G

f

by 365Dh. 414F Corollary Let (X, T, Σ, µ) be an effectively locally finite τ -additive topological measure space. Then for every E ∈ Σ there is a unique relatively closed self-supporting set F ⊆ E such that µ(E \ F ) = 0. proof Let G be the set {G : G S∈ T, µ(G ∩ E) = 0}. Then G is upwards-directed, so µ(E ∩ G∗ ) = supG∈G µ(E ∩ G) = 0, where G∗ = G. Set F = E \ G∗ . Then F ⊆ E is relatively closed, and µ(E \ F ) = 0. If H ∈ T and H ∩ F 6= ∅, then H ∈ / G so µ(F ∩ H) = µ(E ∩ H) > 0; thus F is self-supporting. If F 0 ⊆ E is another self-supporting relatively closed set such that µ(E \ F 0 ) = 0, then µ(F \ F 0 ) = µ(F 0 \ F ) = 0; but as F \ F 0 is relatively open in F , and F 0 \ F is relatively open in F 0 , these must both be empty, and F = F 0 . 414G Corollary If (X, T, Σ, µ) is a Hausdorff effectively locally finite τ -additive topological measure space and E ∈ Σ is an atom for µ (definition: 211I), then there is an x ∈ E such that E \ {x} is negligible. proof Let F ⊆ E be a self-supporting set such that µ(E \ F ) = 0. Since µF = µE > 0, F is not empty; take x ∈ F . ?? If F 6= {x}, let y ∈ F \ {x}. Because T is Hausdorff, there are disjoint open sets G, H containing x, y respectively; and in this case µ(E ∩ G) = µ(F ∩ G) and µ(E ∩ H) = µ(F ∩ H) are both non-zero, which is impossible, since E is an atom. X X So F = {x} and E \ {x} is negligible. 414H Corollary If (X, T, Σ, µ) is an effectively locally finite τ -additive topological measure space and ν is an indefinite-integral measure over µ (definition: 234B), then ν is a τ -additive topological measure.

414K

τ -additivity

53

proof Because ν measures every set in Σ (234Da), it is a topological measure. To see that it is τ -additive, apply 414Eb to a Radon-Nikod´ ym derivative of ν. 414I Proposition Let (X, T, Σ, µ) be a complete locally determined effectively locally finite τ -additive S topological measure space. If E ⊆ X, G ⊆ T are such that E ⊆ G and E ∩ G ∈ Σ for every G ∈ G, then E ∈ Σ. proof Set K = {K : K ∈ Σ, E ∩ K S ∈ Σ}. Then whenever F ∈ Σ and µF > 0 there is a K ∈ K included in F with µK > 0. P P Set K1 = F \ G. Then K1 is a member of K included in F . If µK1 > 0 then we can stop. Otherwise, G ∗ = {G0 ∪ . . . ∪ Gn : G0 , . . . , Gn ∈ G} is an upwards-directed family of open sets, and S supG∈G ∗ µ(F ∩ G) = µ(F ∩ G ∗ ) = µF > 0, by 414E. So there is a G ∈ G ∗ such that µ(F ∩ G) > 0; but now E ∩ G ∈ Σ so F ∩ G ∈ K. Q Q By 412Aa, µ is inner regular with respect to K; by 412Ja, E ∈ Σ. 414J Theorem Let (X, T, Σ, µ) be a complete locally determined effectively locally finite τ -additive topological measure space. Then µ is strictly localizable. proof Let F be a maximal disjoint family of self-supporting measurable sets of finite measure. Then whenever E ∈ Σ and µE > 0, there is an F ∈ F such that µ(E ∩ F ) > 0. P P?? Otherwise, let G be an open set of finite measure such that µ(G ∩ E) > 0, and set F0 = {F : F ∈ F, F ∩SG 6= ∅}. Then µ(F ∩ G) S> 0 0 for every F ∈ F0 , while µG < ∞ and F is disjoint, so F is countable and F ∈ Σ. Set E = E \ F0 ; 0 0 0 S then E \ E 0 = E ∩ F0 is negligible, so µ(G ∩ E 0 ) > 0. By 414F, there is a self-supporting set F 0 ⊆ G ∩ E 0 such that µF 0 > 0. But in this case F 0 ∩ F = ∅ for every F ∈ F, so we ought to have added F 0 to F. X XQ Q This means that F satisfies the criterion of 213O. Because (X, Σ, µ) is complete and locally determined, it is strictly localizable. 414K Proposition Let (X, Σ, µ) be a measure space and T a topology on X, and Y ⊆ X a subset such that the subspace measure µY is semi-finite (see the remark following 412O). If µ is an effectively locally finite τ -additive topological measure, so is the subspace measure µY . proof By 412P, µY is an effectively locally finite topological measure. Now suppose that H is a non-empty upwards-directed family in TY with union H ∗ . Set S G = {G : G ∈ T, G ∩ Y ∈ H}, G∗ = G, so that G is upwards-directed and H ∗ = Y ∩G∗ . Let K be the family of sets K ⊆ X such that K ∩G∗ \G = ∅ for some G ∈ G. If E ∈ Σ, µE = µ(E \ G∗ ) + µ(E ∩ G∗ ) = µ(E \ G∗ ) + sup µ(E ∩ G) G∈G

(414Ea) = sup µ(E \ (G∗ \ G)), G∈G

so µ is inner regular with respect to K. By 412Ob, µY is inner regular with respect to {K ∩ Y : K ∈ K}. So if γ < µY H ∗ , there is a K ∈ K such that K ∩ Y ⊆ H ∗ and µY (K ∩ Y ) ≥ γ. But now there is a G ∈ G such that K ∩ G∗ \ G = ∅, so that K ∩ Y ⊆ G ∩ Y ∈ H and supH∈H µH ≥ γ. As H and γ are arbitrary, µ is τ -additive. Remarks Recall from 214I that if (X, Σ, µ) has locally determined negligible sets (in particular, is either strictly localizable or complete and locally determined), then all its subspaces are semi-finite. In 419C below I describe a tight locally finite Borel measure with a subset on which the subspace measure is not semi-finite, therefore not effectively locally finite or τ -additive. In 419A I describe a σ-finite locally finite τ -additive topological measure, inner regular with respect to the closed sets, with a closed subset on which the subspace measure is totally finite but not τ -additive.

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414L

414L Lemma Let (X, T) be a topological space, and µ, ν two effectively locally finite Borel measures on X which agree on the open sets. Then they are equal. proof Write Tf for the family of open sets of finite measure. (I do not need to specify which measure I am using here.) For G ∈ Tf , set µG E = µ(G ∩ E), νG E = ν(G ∩ E) for every Borel set E. Then µG and νG are totally finite Borel measures which agree on T. By the Monotone Class Theorem (136C), µG and νG agree on the σ-algebra generated by T, that is, the Borel σ-algebra B. Now, for any E ∈ B, µE = supG∈Tf µG E = supG∈Tf νG E = νE, by 412F. So µ = ν. 414M Proposition Let (X, Σ, µ) be a measure space with a regular topology T such that µ is effectively locally finite and τ -additive and Σ includes a base for T. (a) µG = sup{µF : F ∈ Σ is closed, F ⊆ G} for every open set G ∈ Σ. (b) If µ is inner regular with respect to the σ-algebra generated by T ∩ Σ, it is inner regular with respect to the closed sets. proof (a) For U ∈ Σ ∩ T, the set HU = {H : H ∈ Σ ∩ T, H ⊆ U } S is an upwards-directed family of open sets, and HU = U because T is regular and Σ includes a base for T. Because µ is τ -additive, µU = sup{µH : H ∈ HU }. Now, given γ < µG, we can choose hUn in∈N in Σ ∩ T inductively, as follows. Start by taking U0 ⊆ G such that γ < µU0 < ∞ (using the hypothesis that µ is effectively locally finite). Given Un ∈ Σ ∩ T and µUn > γ, take Un+1 ∈ Σ ∩ T such that U n+1 ⊆ Un and µUn+1 > γ. On completing the induction, set T T F = n∈N Un = n∈N U n ; then F is a closed set belonging to Σ, F ⊆ G and µF ≥ γ. As γ is arbitrary, we have the result. (b) Let Σ0 be the σ-algebra generated by Σ ∩ T and set µ0 = µ¹Σ0 . Then Σ0 ∩ T = Σ ∩ T is still a base for T and µ0 is still τ -additive and effectively locally finite, so by (a) and 412G it is inner regular with respect to the closed sets. Now we are supposing that µ is inner regular with respect to Σ0 , so µ is inner regular with respect to the closed sets, by 412Ab. 414N Proposition Let (X, Σ, µ) be a measure space and T a topology on X. Suppose that (i) µ is semifinite and inner regular with respect to the closed sets (ii) whenever F is a non-empty downwards-directed family of measurable closed sets with empty intersection and inf F ∈F µF < ∞, then inf F ∈F µF = 0. Then µ is τ -additive. proof Let G be a non-empty upwards-directed family of measurable open sets with measurable union H. Take any γ < µH. Because µ is semi-finite, there is a measurable set E ⊆ H such that γ < µE < ∞. Now there is a measurable closed set F ⊆ E such that µF ≥ γ. Consider F = {F \ G : G ∈ G}. This is a downwards-directed family of closed sets of finite measure with empty intersection. So inf G∈G µ(F \ G) = 0, that is, γ ≤ µF = supG∈G µ(F ∩ G) ≤ supG∈G µG. As γ is arbitrary, µH = supG∈G µG; as G is arbitrary, µ is τ -additive. 414O The following elementary result is worth noting. Proposition If X is a hereditarily Lindel¨of space (e.g., if it is separable and metrizable) then every measure on X is τ -additive. proof If µ is a measure on X, with domain Σ, and G ⊆ Σ is a non-empty S Supwards-directed family of measurable open sets, then there is a sequence hGn in∈N in G such that G = n∈N Gn . Now S S µ( G) = limn→∞ µ( i≤n Gi ) ≤ supG∈G µG. As G is arbitrary, µ is τ -additive.

414Xb

τ -additivity

55

414P Density topologies Recall that a lower density of a measure space (X, Σ, µ) is a function φ : Σ → Σ such that φE = φF whenever E, F ∈ Σ and µ(E4F ) = 0, µ(E4φE) = 0 for every E ∈ Σ, φ∅ = ∅ and φ(E ∩ F ) = φE ∩ φF for all E, F ∈ Σ (341C). Proposition Let (X, Σ, µ) be a complete locally determined measure space and φ : Σ → Σ a lower density such that φX = X. Set T = {E : E ∈ Σ, E ⊆ φE}. Then T is a topology on X, the density topology associated with φ, and (X, T, Σ, µ) is an effectively locally finite τ -additive topological measure space; µ is strictly positive and inner regular with respect to the open sets. proof (a)(i) For any E ∈ Σ, φ(E ∩ φE) = φE because E \ φE is negligible; consequently E ∩ φE ∈ T. In particular, ∅ = ∅ ∩ φ∅ and X = X ∩ φX belong to T. If E, F ∈ T then φ(E ∩ F ) = φE ∩ φF ⊇ E ∩ F , so E ∩ F ∈ T.

S (ii) Suppose that G ⊆ T and H = G. By 341M, µ is (strictly) localizable, so G has an essential supremum F ∈ Σ such that F • = supG∈G G• in the measure algebra A of µ; that is, for E ∈ Σ, µ(G \ E) = 0 for every G ∈ G iff µ(F \ E) = 0. Now F \ H is negligible, by 213K. On the other hand, G ⊆ φG = φ(G ∩ F ) ⊆ φF for every G ∈ G, so H ⊆ φF , and H \ F ⊆ φF \ F is negligible. But as µ is complete, this means that H ∈ Σ. Also φH = φF ⊇ H, so H ∈ T. Thus T is closed under arbitrary unions and is a topology. (b) By its definition, T is included in Σ, so µ is a topological measure. If E ∈ Σ then E ∩ φE belongs to T, is included in E and has the same measure as E; so µ is inner regular with respect to the open sets. If E ∈ T is non-empty, then φE ⊇ E is non-empty, so µE > 0; thus µ is strictly positive. Finally, if G is a S non-emptySupwards-directed family in T, then the argument of (a-ii) shows that ( G)• = supG∈G G• in A, so that µ( G) = supG∈G µG. Thus µ is τ -additive. If E ∈ Σ and µE > 0 then there is an F ⊆ E such that 0 < µF < ∞, and now E ∩ φF is an open set of non-zero finite measure included in E; so µ is effectively locally finite. 414Q Lifting topologies Let (X, Σ, µ) be a measure space and φ : Σ → Σ a lifting, that is, a Boolean homomorphism such that φE = ∅ whenever µE = 0 and µ(E4φE) = 0 for every E ∈ Σ (341A). The lifting topology associated with φ is the topology generated by {φE : E ∈ Σ}. Note that {φE : E ∈ Σ} is a topology base, so is a base for the lifting topology. 414R Proposition Let (X, Σ, µ) be a complete locally determined measure space and φ : Σ → Σ a lifting with lifting topology S and density topology T. Then S ⊆ T ⊆ Σ, and µ is τ -additive, effectively locally finite and strictly positive with respect to S. Moreover, S is zero-dimensional. proof Of course φ is a lower density, so we can talk of its density topology, and since φ2 E = φE, φE ∈ T for every E ∈ Σ, so S ⊆ T. Because µ is τ -additive and strictly positive with respect to T, it must also be τ -additive and strictly positive with respect to S. If E ∈ Σ and µE > 0 there is an F ⊆ E such that 0 < µF < ∞, and now φF is an S-open set of finite measure meeting E in a non-negligible set; so µ is effectively locally finite with respect to S. Of course S is zero-dimensional because φ[Σ] is a base for S consisting of open-and-closed sets. 414X Basic exercises (a) Let (X, Σ, µ) and (Y, T, ν) be measure spaces with topologies T and S, and f : X → Y a continuous inverse-measure-preserving function. Show that if µ is τ -additive with respect to T then ν is τ -additive with respect to S. Show that if ν is locally finite, so is µ. (b) Let h(Xi , Σi , µi )ii∈I be a family of measure spaces, with direct sum (X, Σ, µ); suppose that we are given a topology Ti on Xi for each i, and let T be the disjoint union topology on X. Show that µ is τ -additive iff every µi is.

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414Xc

> (c) Let (X, T) be a topological space and µ a totally finite measure on X which is inner regular with respect to the closed sets. Suppose that µX = supG∈G µG whenever G is an upwards-directed family of measurable open sets covering X. Show that µ is τ -additive. (d) Let µ be an effectively locally finite τ -additive σ-finite measure on a topological space X, and ν : Σ → [0, ∞[ a countably additive functional which is absolutely continuous with respect to µ. Show from first principles that ν is τ -additive. (e) Give an example of an indefinite-integral measure over Lebesgue measure on R which is not effectively locally finite. (Hint: arrange for every non-trivial interval to have infinite measure.) (f ) Let (X, T) be a topological space and µ a complete locally determined effectively locally finite τ additive topological measure on X. Show that if f is a real-valued function, defined on a subset of X, which is locally integrable in the sense of 411Fc, then f is measurable. (g) Let (X, T) be a topological space and µ an effectively locally finite τ -additive measure on X. Let G be a cover of X consisting of measurable open sets, and K the ideal of subsets of X generated by G. Show that µ is inner regular with respect to K. (h) Let (X, T, Σ, µ) be a complete locally determined effectively locally finite τ -additive topological measure space, and A a subset of X. Suppose that for every x ∈ X there is an open set G containing x such that A ∩ G is negligible. Show that A is negligible. (i) Give an alternative proof of 414K based on the fact that the canonical map from the measure algebra of µ to the measure algebra of µY is order-continuous (322Yd). > (j) (i) If µ is an effectively locally finite τ -additive Borel measure on a regular topological space, show that the c.l.d. version of µ is a quasi-Radon measure. (ii) If µ is a locally finite, effectively locally finite τ -additive Borel measure on a locally compact Hausdorff space, show that µ is tight, so that the c.l.d. version of µ is a Radon measure. > (k) Let (X, Σ, µ) be a complete locally determined measure space and φ a lower density on X such that φX = X; let T be the corresponding density topology. (i) Show that a dense open subset of X must be conegligible. (ii) Show that a subset of X is nowhere dense for T iff it is negligible iff it is meager for T. (iii) Show that a function f : X → R is Σ-measurable iff it is T-continuous at almost every point of X. (Hint: if f is measurable, set Eq = {x : f (x) > q}, Fq = {x : f (x) < q}; show that f is continuous at every point S of X \ q∈Q ((Eq \ φEq ) ∪ (Fq \ φFq )).) (l) Let (X, Σ, µ) be a complete locally determined measure space and φ : Σ → Σ a lower density such that φX = X, with density topology T. Show that if A ⊆ X and E is a measurable envelope of A then the T-closure of A is just A ∪ (X \ φ(X \ E)). (m) Let µ be Lebesgue measure on R r , Σ its domain, φ : Σ → Σ lower Lebesgue density (341E) and T the corresponding density topology. Show that for any A ⊆ R, the closure of A for T is just A ∪ {x : lim supδ↓0

µ∗ (A∩B(x,δ)) µB(x,δ)

> 0}, and the interior is A ∩ {x : limδ↓0

µ∗ (A∩B(x,δ)) µB(x,δ)

= 1}.

(n) Let (X, Σ, µ) be a complete locally determined measure space and φ : Σ → Σ a lower density such that φX = X; let T be the associated density topology. Let A be a subset of X and E a measurable envelope of A; let ΣA be the subspace σ-algebra and µA the subspace measure on A. (i) Show that we have a lower density φA : ΣA → ΣA defined by setting φA (F ∩ A) = A ∩ φ(E ∩ F ) for every F ∈ Σ. (ii) Show that φA A = A iff A ⊆ φE, and that in this case the density topology on A derived from φA is just the subspace topology. (o) Let (X, Σ, µ) be a complete locally determined measure space and φ : Σ → Σ a lifting, with density topology T and lifting topology S. (i) Show that T = {H ∩ G : G ∈ S, H is conegligible} = {H ∩ φE : E ∈ Σ, H is conegligible}. (ii) Show that if A ⊆ X and E is a measurable envelope of A then the T-closure of A is A ∪ φE.

414Yf

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57

(p) Let (X, Σ, µ) be a complete locally determined measure space and φ : Σ → Σ a lifting; let S be its lifting topology. Let A be a subset of X such that A ⊆ φE for some (therefore any) measurable envelope E of A. Let ΣA be the subspace σ-algebra and µA the subspace measure on A. (i) Show that we have a lifting φA : ΣA → ΣA defined by setting φA (F ∩ A) = A ∩ φF for every F ∈ Σ. (ii) Show that the lifting topology on A derived from φA is just the subspace topology. (q) Let (X, Σ, µ) and (Y, T, ν) be complete locally determined measure spaces and f : X → Y an inversemeasure-preserving function. (i) Suppose that we have lower densities φ : Σ → Σ and ψ : T → T such that φX = X, ψY = Y and φf −1 [F ] = f −1 [ψF ] for every F ∈ T. Show that f is continuous for the density topologies of φ and ψ. (ii) Show that if φ and ψ are liftings then f is continuous for the lifting topologies. (r) Let (X, Σ, µ) be a complete locally determined measure space and φ : Σ → Σ a lifting, with associated lifting topology S. Show that a function f : X → R is Σ-measurable iff there is a conegligible set H such that f ¹H is S-continuous. (Compare 414Xk.) (s) Let (X, Σ, µ) be a complete locally determined measure space and φ : Σ → Σ a lifting. Let (Z, T, ν) be the Stone space of the measure algebra of µ, and f : X → Z the inverse-measure-preserving function associated with φ (341P). Show that the lifting topology on X is just {f −1 [G] : G ⊆ Z is open}. (t) Let (X, Σ, µ) be a strictly localizable measure space and φ : Σ → Σ a lifting. Write L∞ for the Banach lattice of bounded Σ-measurable real-valued functions on X, identified with L∞ (Σ) (363H); let T : L∞ → L∞ be the Riesz homomorphism associated with φ (363F). (i) Show that T 2 = T . (ii) Show that if X is given the lifting topology S defined by φ, then T [L∞ ] is precisely the space of bounded continuous real-valued functions on X. (iii) Show that if f ∈ L∞ , x ∈ X and ² > 0 there is an S-open set U containing 1 R x such that |(T f )(x) − f dµ| ≤ ² for every non-negligible measurable set V included in U . V µV

414Y Further exercises (a) Let (X, T, Σ, µ) be a totally finite topological measure space. For E ∈ Σ set S µτ E = inf{µ(E \ G) + supG∈G µ(E ∩ G) : G ⊆ T is an upwards-directed set}. Suppose either that µ is inner regular with respect to the closed sets or that T is regular. Show that µτ is a τ -additive measure, the largest τ -additive measure with domain Σ which is dominated by µ. (b) Let X be a set, Σ an algebra of subsets of X, and T a topology on X. Let M be the L-space of bounded finitely additive real-valued functionals on Σ (362B). Let N ⊆ M be the set of those functionals ν such that inf G∈G |ν|(H \ G) = 0 whenever G ⊆ T ∩ Σ is a non-empty upwards-directed family with union H ∈ Σ. Show that N is a band in M . (Cf. 362Xi.) (c) Find a probability space (X, Σ, µ) and a topology T on X such that Σ includes a base for T and µ is τ -additive, but there is a set E ∈ Σ such that the subspace measure µE is not τ -additive. (d) Let (X, Σ, µ) be a complete locally determined measure space and φ a lower density on X such that φX = X; let T be the density topology. Show that Σ is precisely the Baire property algebra for T, so that (X, T) is a Baire space. (e) Let φ be lower Lebesgue density on R r , and T the associated density topology. Show that every T-Borel set is an Fσ set for T. (f ) Let (X, ρ) be a metric space and µ a strictly positive locally finite quasi-Radon measure on X; write T for the topology of X and Σ for the domain of µ. For E ∈ Σ set φ(E) = {x : x ∈ X, limδ↓0

µ(E∩B(x,δ)) µB(x,δ)

= 1}.

Suppose that E \ φ(E) is negligible for every E ∈ Σ (cf. 261D, 472D). (i) Show that φ is a lower density for µ, with φ(X) = X. Let Td be the associated density topology. (ii) Suppose that H ∈ Td and that K ⊆ H is T-closed and ρ-totally bounded. Show that there is a T-closed, ρ-totally bounded K 0 ⊆ H such that K is ´ & Zaj´ıc ˇek included in the Td -interior of K 0 . (iii) Show that Td is completely regular. (Hint: Lukeˇ s Maly 86.)

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414Yg

(g) Show that the density topology on R associated with lower Lebesgue density is not normal. (h) Let µ be Lebesgue measure on R r , Σ its domain, φ : Σ → Σ lower Lebesgue density and T the corresponding density topology. (i) Show that if f : Rr → Rr is a bijection such that f and f −1 are both differentiable everywhere, with continuous derivatives, then f is a homeomorphism for T. (Hint: 263D.) (ii) Show that if φ : Σ → Σ is a lifting and S the corresponding lifting topology, then x 7→ −x is not a homeomorphism for S. (Hint: 345Xc.) 414 Notes and comments I have remarked before that it is one of the abiding frustrations of measure theory, at least for anyone ambitious to apply the power of modern general topology to measure-theoretic problems, that the basic convergence theorems are irredeemably confined to sequences. In Volume 3 I showed that if we move to measure algebras and function spaces, we can hope that the countable chain condition or the countable sup property will enable us to replace arbitrary directed sets with monotonic sequences, thereby giving theorems which apply to apparently more general types of convergence. In 414A and its corollaries we come to a quite different context in which a measure, or integral, behaves like an order-continuous functional. Of course the theorems here depend directly on the hypothesis of τ -additivity, which rather begs the question; but we shall see in the rest of the chapter that this property does indeed often appear. For the moment, I remark only that as Lebesgue measure is τ -additive we certainly have a non-trivial example to work with. The hypotheses of the results above move a touch awkwardly between those with the magic phrase ‘topological measure’ and those without. The point is that (as in 412G, for instance) it is sometimes useful to be able to apply these ideas to Baire measures on completely regular spaces, which are defined on a base for the topology but may not be defined on every open set. I hope that no confusion will arise between the two topologies associated with a lifting on a complete locally determined space. I have called them the ‘density topology’ and the ‘lifting topology’ because the former can be defined directly from a lower density; but it would be equally reasonable to call them the ‘fine’ and ‘coarse’ lifting topologies. The density topology has the apparent advantage of giving us a measure which is inner regular with respect to the Borel sets, but at the cost of being wildly irregular regarded as a topological space. It has the important advantage that there are densities (like the Lebesgue lower density) which have some claim to be called canonical, and others with useful special properties, as in §346, while liftings are always arbitrary and invariance properties for them sometimes unachievable. So, for instance, the Lebesgue density topology on R r is invariant under diffeomorphisms, which no lifting topology can be (414Yh). The lifting topology is well-behaved as a topology, but only in special circumstances (as in 453Xd) is the measure inner regular with respect to its Borel sets, and even the closure of a set can be difficult to determine. As with inner regularity, τ -additivity can be associated with the band structure of the space of bounded additive functionals on an algebra (414Yb); there will therefore be corresponding decompositions of measures into τ -additive and ‘purely non-τ -additive’ parts (cf. 414Ya).

415 Quasi-Radon measure spaces We are now I think ready to draw together the properties of inner regularity and τ -additivity. Indeed this section will unite several of the themes which have been running through the treatise so far: (strict) localizability, subspaces and products as well as the new concepts of this chapter. In these terms, the principal results are that a quasi-Radon space is strictly localizable (415A), any subspace of a quasi-Radon space is quasi-Radon (415B), and the product of a family of strictly positive quasi-Radon probability measures on separable metrizable spaces is quasi-Radon (415E). I describe a basic method of constructing quasi-Radon measures (415K), with details of one of the standard ways of applying it (415L, 415N) and some notes on how to specify a quasi-Radon measure uniquely (415H-415I). I spell out useful results on indefinite-integral measures (415O) and Lp spaces (415P), and end the section with a discussion of the Stone space Z of a localizable measure algebra A and an important relation in Z × X when A is the measure algebra of a quasi-Radon measure space X (415Q-415R). It would be fair to say that the study of quasi-Radon spaces for their own sake is a minority interest. If you are not already well acquainted with Radon measure spaces, it would make good sense to read this

415D

Quasi-Radon measure spaces

59

section in parallel with the next. In particular, the constructions of 415K and 415L derive much of their importance from the corresponding constructions in §416. 415A Theorem A quasi-Radon measure space is strictly localizable. proof This is a special case of 414J. 415B Theorem Any subspace of a quasi-Radon measure space is quasi-Radon. proof Let (X, T, Σ, µ) be a quasi-Radon measure space and (Y, TY , ΣY , µY ) a subspace with the induced topology and measure. Because µ is complete, locally determined and localizable (by 415A), so is µY (214Id). Because µY is semi-finite and µ is an effectively locally finite τ -additive topological measure, so is µY (414K). Because µ is inner regular with respect to the closed sets and µY is semi-finite, µY is inner regular with respect to the relatively closed subsets of Y (412Pc). So µY is a quasi-Radon measure. 415C In regular topological spaces, the condition ‘inner regular with respect to the closed sets’ in the definition of ‘quasi-Radon measure’ can be weakened or omitted. Proposition Let (X, T) be a regular topological space. (a) If µ is a complete locally determined effectively locally finite τ -additive topological measure on X, inner regular with respect to the Borel sets, then it is a quasi-Radon measure. (b) If µ is an effectively locally finite τ -additive Borel measure on X, its c.l.d. version is a quasi-Radon measure. proof (a) By 414Mb, µ is inner regular with respect to the closed sets, which is the only feature missing from the given hypotheses. (b) The c.l.d. version of µ satisfies the hypotheses of (a). 415D In separable metric spaces, among others, we can even omit τ -additivity. Proposition Let (X, T) be a hereditarily Lindel¨of topological space; e.g., a separable metrizable space (4A2P(a-iii)). (i) If µ is a complete effectively locally finite measure on X, inner regular with respect to the Borel sets, and its domain includes a base for T, then it is a quasi-Radon measure. (ii) If µ is an effectively locally finite Borel measure on X, then its completion is a quasi-Radon measure. (iii) Any quasi-Radon measure on X is σ-finite. (iv) If X is regular, any quasi-Radon measure on X is completion regular. proof (a) The basic S fact we S need is that if G is any family of open sets in X, then there is a countable G0 ⊆ G such that G0 = G (4A2H(c-i)). Consequently any effectively locally finite measure µ on X is σ-finite. P P Let G be the family of measurable S open sets of finite measure. Let G0 ⊆ G be a countable set with the same union as G. Then E = X \ G0 is measurable, and E ∩ G = ∅ for every G ∈ G, so µE = 0; accordingly G0 ∪ {E} is a countable cover of X by sets of finite measure, and µ is σ-finite. Q Q Moreover, any measure on X is τ -additive. P P If G is a non-empty upwards-directed family of open S measurable sets, there is a sequence hG i in G with union G. If n ∈ N there is a G ∈ G such that n n∈N S G ⊆ G, so i i≤n S S S µ( G) = µ( n∈N Gn ) = supn∈N µ( i≤n Gi ) ≤ supG∈G µG. As G is arbitrary, µ is τ -additive. Q Q (b)(i) Now let µ be a complete effectively locally finite measure on X, inner regular with respect to the Borel sets, and with domain Σ including a base for the topology of X. If H ∈ T, then G = {G : G ∈ Σ ∩ T, G ⊆ H} S has union H, because Σ ∩ T is a base for T; but in this case there is a countable G0 ⊆ G such that H = G0 , so that H ∈ Σ. Thus µ is a topological measure. We know also from (a) that it is τ -additive and σ-finite, therefore locally determined. By 415Ca, it is a quasi-Radon measure. (ii) If µ is an effectively locally finite Borel measure on X, then its completion µ ˆ satisfies the conditions of (i), so is a quasi-Radon measure.

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415D

(iii) If µ is a quasi-Radon measure on X, it is surely effectively locally finite, therefore σ-finite. (iv) If X is regular, then every closed set is a zero set (4A2H(c-iii)), so any measure which is inner regular with respect to the closed sets is completion regular. 415E I am delaying most of the theory of products of (quasi-)Radon measures to §417. However, there is one result which is so important that I should like to present it here, even though some of the ideas will have to be repeated later. Theorem Let h(Xi , Ti , Σi , µi )ii∈I be a family of separable metrizableQquasi-Radon probability spaces such that every µi is strictly positive, and λ the product measure on X = i∈I Xi . Then (i) λ is a completion regular quasi-Radon measure; (ii) if F ⊆ X is a closed self-supporting set, there is a countable set J ⊆ I such that F is determined by coordinates in J, so F is a zero set. Q proof (a) Write Λ for the domain of λ, and U for the family of subsets of X of the form i∈I Gi where Gi ∈ Ti for every i ∈ I and {i : Gi 6= Xi } is finite. Then U is a base for the topology of X, included in Λ. Q (J) For J ⊆ I let λJ be the product measure on X = X for the family J i and ΛJ its domain. Write U i∈J Q of subsets of XJ of the form i∈J Gi where Gi ∈ Ti for each i ∈ J and {i : Gi 6= Xi } is finite. (b) Consider first the case in which I is countable. In this case X also is separable and metrizable (4A2P(a-v)), while Λ includes a base for its topology. Also λ is a complete probability measure and inner regular with respect to the closed sets (412Ua), so must be a quasi-Radon measure, by 415D(i). (c) Now consider uncountable I. The key to the proof is the following fact: if V ⊆ U has union W , then W ∈ Λ and λW = supV ∈V ∗ λV , where V ∗ is the set of unions of finite subsets of V. P P (i) S By 215B(iv), there is a countable set V1 ⊆ V such that λ(U \ W1 ) = 0 for every U ∈ V, where W1 = V1 . Every member of U is determined by coordinates in some finite set (see 254M for this concept), so there is a countable set J ⊆ I such that every member of V1 is determined by coordinates in J, and W1 also is determined by coordinates in J. Let πJ : X → XJ be the canonical map. Because it is an open map (4A2B(f-i)), πJ [W ] and πJ [W1 ] are open in XJ , and belong to ΛJ , by (b). S (ii) ?? Suppose, if possible, that λJ πJ [W ] > λJ πJ [W1 ]. Since πJ [W ] = {πJ [U ] : U ∈ V}, while λJ is quasi-Radon and all the sets πJ [U ] are open, there must be some U ∈ V such that λJ (πJ [U ] \ πJ [W1 ]) > 0 (414Ea). Now πJ is inverse-measure-preserving (254Oa), so 0 < λπJ−1 [πJ [U ] \ πJ [W1 ]] = λ(πJ−1 [πJ [U ]] \ πJ−1 [πJ [W1 ]]) = λ(πJ−1 [πJ [U ]] \ W1 ), because W1 is determined by coordinates in J. Q At this point note that U is of the form i∈I Gi , where Gi ∈ Ti for each I, so we can express U as −1 U 0 ∩ U 00 , where U 0 = πJ−1 [πJ [U ]] and U 00 = πI\J [πI\J [U ]]. U 0 depends on coordinates in J and U 00 depends on coordinates in I \ J. In this case λ(U \ W1 ) = λ(U 00 ∩ U 0 \ W1 ) = λU 00 · λ(U 0 \ W1 ), because U 00 depends on coordinates in I \ J and U 0 \ W1 depends on coordinates in J, and we can identify λ with the product λI\J × λJ (254N). But now recall that every µQ i is strictly positive. Since U is surely not empty, no Gi can be empty and no µi Gi can be 0. Consequently i∈I µi Gi > 0 (because only finitely many terms in the product are less than 1) and λU > 0; more to the point, λU 00 > 0. Since we already know that λ(U 0 \ W1 ) > 0, we have λ(U \ W1 ) > 0. But this contradicts the first sentence of (i) just above. X X (iii) Thus λJ πJ [W ] = λJ πJ [W1 ]. But this means that λπJ−1 [πJ [W ]] = λW1 . Since λ is complete and W1 ⊆ W ⊆ πJ−1 [πJ [W ]], λW is defined and equal to λW1 . Taking hVn in∈N to be a sequence running over V1 ∪ {∅}, we have S S λW = λW1 = λ( n∈N Vn ) = supn∈N λ( i≤n Vi ) ≤ supV ∈V ∗ λV ≤ λW, so λW = supV ∈V ∗ λV , as required. Q Q (d) Thus we see that λ is a topological measure. But it is also τ -additive. P P If W is an upwards-directed family of open sets in X with union W ∗ , set

415G

Quasi-Radon measure spaces

∗

S

61

V = {U : U ∈ U , ∃ W ∈ W, U ⊆ W }. ∗

Then W = V, so λW = supV ∈V ∗ λV , where V ∗ is the set of finite unions of members of V. But because W is upwards-directed, every member of V ∗ is included in some member of W, so λW ∗ = supV ∈V ∗ λV ≤ supW ∈W λW ≤ λW ∗ . As W is arbitrary, λ is τ -additive. Q Q (e) As in (b) above, we know that λ is a complete probability measure and is inner regular with respect to the closed sets, so it is a quasi-Radon measure. Because λ is inner regular with respect to the zero sets (412Ub), it is completion regular. (f ) Now suppose that F ⊆ X is a closed self-supporting set. By 254Oc, there is a set W ⊆ X, determined by coordinates in some countable set J ⊆ I, such that W 4F is negligible. ?? Suppose, if possible, that x ∈ F and y ∈ X \ F are such that x¹J = y¹J. Then there is a U ∈ U such that y ∈ U ⊆ X \ F . As in (b-ii) above, we can express U as U 0 ∩ U 00 where U 0 , U 00 ∈ U are determined by coordinates in J and I \ J respectively. In this case, λ(F ∩ U ) = λ(W ∩ U ) = λ(W ∩ U 0 ) · λU 00 = λ(F ∩ U 0 ) · λU 00 > 0, because x ∈ F ∩ U 0 and F is self-supporting, while U 00 6= ∅ and λ is strictly positive. But F ∩ U = ∅, so this is impossible. X X Thus F is determined by coordinates in the countable set J. Consequently it is of the form πJ−1 [πJ [F ]]. But πJ [X \ F ] is open (4A2B(f-i)), so its complement πJ [F ] is closed. Now XJ is metrizable (4A2P(a-v)), so πJ [F ] is a zero set (4A2Lc) and F is a zero set (4A2C(b-iv)). 415F Corollary(a) If Y is either [0, 1[ or ]0, 1[, endowed with Lebesgue measure, and I is any set, then Y I , with the product topology and measure, is a quasi-Radon measure space. (b) If hνi ii∈I is a family of probability distributions on R, in the sense of §271 (that is, Radon probability measures), and every νi is strictly positive, then the product measure on R I is a quasi-Radon measure. Remark See also 416U below, and 453I, where there is an alternative proof of the main step in 415E, I applicable to some further cases. Yet another approach, most immediately applicable to [0, 1[ , is in 443Xq. For further facts about these product measures, see §417, particularly 417G and 417M. 415G Comparing quasi-Radon measures: Proposition Let X be a topological space, and µ, ν two quasi-Radon measures on X. Then the following are equiveridical: (i) µF ≤ νF for every closed set F ⊆ X; (ii) dom ν ⊆ dom µ and µE ≤ νE for every E ∈ dom ν. If ν is locally finite, we can add (iii) µG ≤ νG for every open set G ⊆ X; (iv) there is a base U for the topology of X such that G ∪ H ∈ U for all G, H ∈ U and µG ≤ νG for G ∈ U. proof (a) Of course (ii)⇒(i). Suppose that (i) is true. Observe that if E ∈ dom µ ∩ dom ν (for instance, if E ⊆ X is Borel), then µE = sup{µF : F ⊆ E is closed} ≤ sup{νF : F ⊆ E is closed} = νE. S Set H = {H : H ⊆ X is open, νH < ∞}, and W = H. Then ν(X \ W ) = 0, because ν is effectively locally finite, so µ(X \ W ) = 0. Set F = {F : F ⊆ X is closed, µF < ∞}. Take any E ∈ dom ν. If F ∈ F and ² > 0, then µ(F ∩ W ) = µF , so there is an H ∈ H such that µ(F ∩ H) ≥ µF − ². Now there are closed sets F1 ⊆ F ∩ H ∩ E, F2 ⊆ F ∩ H \ E such that νF1 + νF2 ≥ ν(F ∩ H) − ², that is, ν((F ∩ H) \ (F1 ∪ F2 )) ≤ ², so that µ((F ∩ H) \ (F1 ∪ F2 )) ≤ ² and µF1 + µF2 ≥ µ(F ∩ H) − ². This means that µ∗ (F ∩ E) + µ∗ (F \ E) ≥ µ(F ∩ H) − ² ≥ µF − ².

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415G

As ² is arbitrary, µ∗ (F ∩ E) + µ∗ (F \ E) ≥ µF ; as µ is inner regular with respect to F, µ measures E, by 413F(vii). Thus dom ν ⊆ dom µ; and we have already observed that µE ≤ νE whenever E is measured by both. (b) The first sentence in the proof of (a) shows that (i)⇒(iii), and (iii)⇒(iv) is trivial. If (iv) is true and G ⊆ X is open, then V = {V : V ∈ U , V ⊆ G} is upwards-directed and has union G, so µG = supV ∈V µV ≤ supV ∈V νV = νG. Thus (iv)⇒(iii). Now assume that ν is locally finite and that (iii) is true. ?? Suppose, if possible, that F ⊆ X is a closed set such that νF < µF . Then H, as defined in part (a) of the proof, is upwards-directed and has union X, so there is an H ∈ H such that νF < µ(F ∩ H). Now there is a closed set F 0 ⊆ H \ F such that νF 0 > ν(H \ F ) − µ(F ∩ H) + νF ≥ νH − µ(F ∩ H). Set G = H \ F 0 , so that F ∩ H ⊆ G and νG = νH − νF 0 < µ(F ∩ H) ≤ µG, which is impossible. X X This shows that (provided that ν is locally finite) (iii)⇒(i). 415H Uniqueness of quasi-Radon measures: Proposition Let (X, T) be a topological space and µ, ν two quasi-Radon measures on X. Then the following are equiveridical: (i) µ = ν; (ii) µF = νF for every closed set F ⊆ X; (iii) µG = νG for every open set G ⊆ X; (iv) there is a base U for the topology of X such that G ∪ H ∈ U for every G, H ∈ U and µ¹U = ν¹U ; (v) there is a base U for the topology of X such that G ∩ H ∈ U for every G, H ∈ U and µ¹U = ν¹U. proof Of course (i) implies all the others. (ii)⇒(i) is immediate from 415G (see also 412L). If (iii) is true, then, for any closed set F ⊆ X, µF = sup{µ(G ∩ F ) : G ∈ T, µG < ∞} = sup{µG − µ(G \ F ) : G ∈ T, µG < ∞} = sup{νG − ν(G \ F ) : G ∈ T, νG < ∞} = νF ; so (iii)⇒(ii). (iv)⇒(iii) by the argument of (iv)⇒(iii) in the proof of 415G. Finally, suppose that (v) is true. Then µ(G0 ∪ . . . ∪ Gn ) = ν(G0 ∪ . . . ∪ Gn ) for all G0 , . . . , Gn ∈ U. P P Induce on n. For the inductive step to n ≥ 1, if any Gi has infinite measure (for either measure) the result is trivial. Otherwise, [ [ µ(G0 ∪ . . . ∪ Gn ) = µ( Gi ) + µGn − µ( (Gn ∩ Gi )) i
= ν(

[

i
i
Gi ) + νGn − ν(

[

(Gn ∩ Gi )) = ν(G0 ∪ . . . ∪ Gn ). Q Q

i
So µ and ν agree on the base {G0 ∪ . . . ∪ Gn : G0 , . . . , Gn ∈ U}, and (iv) is true. 415I Proposition Let RX be a completely regular topological space and µ, ν two quasi-Radon measures R on X such that f dµ = f dν whenever f : X → R is a bounded continuous function integrable with respect to both measures. Then µ = ν. proof ?? Otherwise, there is an open set G ⊆ X such that µG 6= νG; suppose µG < νG. Because ν is effectively locally finite, there is an open set G0 ⊆ G such that µG < νG0 < ∞. Now the cozero sets form a base for the topology of X, so H = {H : H ⊆ G0 is a cozero set} has union G0 ; as ν is τ -additive, there is an H ∈ H such that νH > µG. Express H as {x : g(x) > 0} where g : X → [0, ∞[ is continuous. For each n ∈ N, setRfn = ng ∧ χX; then hfn in∈N is a non-decreasing sequence with limit χH, so there is an n ∈ N R such that fn dν > µG ≥ fn dµ. But fn is both µ-integrable and ν-integrable because µG and νH are both finite. X X

415K

Quasi-Radon measure spaces

63

415J Proposition Let X be a regular topological space, Y a subspace of X, and ν a quasi-Radon measure on Y . Then there is a quasi-Radon measure µ on X such that µE = ν(E ∩ Y ) whenever µ measures E, that is, Y has full outer measure in X and ν is the subspace measure on Y . proof Write B for the Borel σ-algebra of X, and set µ0 E = ν(E ∩ Y ) for every E ∈ B. Then it is easy to see that µ0 is a τ -additive Borel measure on X. Moreover, µ0 is effectively locally finite. P P If E ∈ B and µ0 E > 0, there is a relatively open set H ⊆ Y such that νH < ∞ and ν(H ∩ E ∩ Y ) > 0. Now H is of the form G ∩ Y where G ⊆ X is open, and we have µ0 G = νH < ∞, µ0 (E ∩ G) = ν(H ∩ E ∩ Y ) > 0. Q Q By 415Cb, the c.l.d. version µ of µ0 is a quasi-Radon measure on X. If E ∈ dom µ, then E ∩ Y ∈ dom ν. P P Let FY be the set of relatively closed subsets of Y of finite measure for ν. If F ∈ FY , it is expressible as F 0 ∩ Y where F 0 is a closed subset of X, and µF 0 = µ0 F 0 = νF is finite. So there are E1 , E2 ∈ B such that E1 ⊆ E ∩ F 0 ⊆ E2 and µE1 = µ(E ∩ F 0 ) = µE2 . Accordingly E1 ∩ Y ⊆ E ∩ Y ∩ F ⊆ E2 ∩ Y and ν(E1 ∩ Y ) = ν(E2 ∩ Y ) = µ(E ∩ F 0 ) is finite. This means that E ∩ Y ∩ F ∈ dom ν; because ν is complete and locally determined and inner regular with respect to FY , E ∩ Y ∈ dom ν, by 412Ja. Q Q If E ∈ dom µ, then µE = sup{µF : F ⊆ E is closed} = sup{ν(F ∩ Y ) : F ⊆ E is closed} ≤ ν(E ∩ Y ). On the other hand, if γ < ν(E ∩ Y ), there is a relatively open set H ⊆ Y such that νH < ∞ and ν(E ∩ Y ∩ H) ≥ γ (412F). Let G ⊆ X be an open set such that G ∩ Y = H. Then µE ≥ µG − µ(G \ E) = νH − ν(H \ E) = ν(E ∩ Y ∩ H) ≥ γ. As γ is arbitrary, µE = ν(E ∩ Y ). Thus µE = ν(E ∩ Y ) whenever µ measures E. So if E, F ∈ dom µ and E ∩ Y ⊆ F , µE = ν(E ∩ Y ) ≤ ν(F ∩ Y ) = µF ; ∗

as F is arbitrary, µ (E ∩ Y ) = µE; as E is arbitrary, Y has full outer measure in X. Moreover, if µY is the subspace measure on Y , µY H = µ∗ H = νH whenever H ∈ dom µY , that is, H = E ∩Y for some E ∈ dom µ. Now µY , like ν, is a quasi-Radon measure on Y (415B), and they agree on the (relatively) closed subsets of Y , so are equal, by 415H. 415K I come now to a couple of basic results on the construction of quasi-Radon measures. The first follows 413J. Theorem Let X be a topological space and K a family of closed subsets of X such that ∅ ∈ K, (†) K ∪ K 0 ∈ K whenever K, K 0 ∈ K are disjoint, (‡) F ∈ K whenever K ∈ K and F ⊆ K is closed. Let φ0 : K → [0, ∞[ be a functional such that (α) φ0 K = φ0 L + sup{φ0 K 0 : K 0 ∈ K, K 0 ⊆ K \ L} whenever K, L ∈ K and L ⊆ K, (β) inf K∈K0 φ0 K = 0 whenever K0 is a non-empty downwards-directed subset of K with empty intersection, (γ) whenever K ∈ K and φ0 K > 0, there is an open set G such that the supremum supK 0 ∈K,K 0 ⊆G φ0 K 0 is finite, while φ0 K 0 > 0 for some K 0 ∈ K such that K 0 ⊆ K ∩ G. Then there is a unique quasi-Radon measure on X extending φ0 and inner regular with respect to K. proof By 413J, there is a complete locally determined measure µ on X, inner regular with respect to K, and extending φ0 ; write Σ for the domain of µ. If F ⊆ X is closed, then K ∩ F ∈ K ⊆ Σ for every K ∈ K, so F ∈ Σ, by 413F(ii); accordingly every open set is measurable. Because µ is inner regular with respect to

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Topologies and measures

415K

K it is surely inner regular with respect to the closed sets. If E ∈ Σ and µE > 0, there is a K ∈ K such that K ⊆ E and µK > 0; now (γ) tells us that there is an open set G such that µG < ∞ and µ(G ∩ K) > 0, so that µ(G ∩ E) > 0. As E is arbitrary, µ is effectively locally finite. Now suppose that G is a non-empty upwards-directed family of open sets with union H, and that γ < µH. Then there is a K ∈ K such that K ⊆ H and µK > γ. Applying the hypothesis (β) to K0 = {K \G : G ∈ G}, we see that inf G∈G µ(K \G) = 0, so that supG∈G µG ≥ supG∈G µ(K ∩ G) = µK ≥ γ. As G and γ are arbitrary, µ is τ -additive. So µ is a quasi-Radon measure. 415L Proposition Let (X, Σ0 , µ0 ) be a measure space and T a topology on X such that µ0 is τ -additive, effectively locally finite and inner regular with respect to the closed sets, and Σ0 includes a base for T. Then µ0 has a unique extension to a quasi-Radon measure µ on X such that (i) µF = µ∗0 F whenever F ⊆ X is closed and µ∗0 F < ∞, (ii) µG = (µ0 )∗ G whenever G ⊆ X is open, (iii) the embedding Σ0 ⊆ Σ identifies the measure algebra (A0 , µ ¯0 ) of µ0 with an order-dense subalgebra of the measure algebra (A, µ ¯) of µ, so that the subrings Af0 , Af of elements of finite measure coincide, and Lp (µ0 ) may be identified with Lp (µ) for 1 ≤ p < ∞, (iv) whenever E ∈ Σ and µE < ∞, there is an E0 ∈ Σ0 such that µ(E4E0 ) = 0, (v) for every µ-integrable real-valued function f there is a µ0 -integrable function g such that f = g µ-a.e. If µ0 is complete and locally determined, then we have (i)0 µF = µ∗0 F for every closed F ⊆ X. If µ0 is localizable, then we have (iii)0 A0 = A, so that L0 (µ) ∼ = L0 (µ0 ), L∞ (µ) ∼ = L∞ (µ0 ), (iv)0 for every E ∈ Σ there is an E0 ∈ Σ0 such that µ(E4E0 ) = 0, (v)0 for every Σ-measurable real-valued function f there is a Σ0 -measurable real-valued function g such that f = g µ-a.e. proof (a) Let K be the set of closed subsets of X of finite outer measure for µ0 . Note that µ0 is inner regular with respect to K, because it is inner regular with respect to the closed sets and also with respect to the sets of finite measure. It is obvious from its definition that K satisfies (†) and (‡) of 415K. For K ∈ K, set φ0 K = µ∗0 K. Then φ0 satisfies (α)-(γ) of 415K. α) If K, L ∈ K and L ⊆ K, take measurable envelopes E0 , E1 ∈ Σ0 of K, L respectively. (i) Let P P (α ² > 0. Because µ0 is inner regular with respect to the closed sets, there is a closed set F ∈ Σ0 such that F ⊆ E0 \ E1 and µF ≥ µ0 (E0 \ E1 ) − ². Set K 0 = F ∩ K. Then K 0 ∈ K and φ0 K 0 = µ∗0 (F ∩ K) = µ0 (F ∩ E0 ) = µ0 F ≥ µ0 E0 − µ0 E1 − ² = φ0 K − φ0 L − ². As ² is arbitrary, we have φ0 K ≤ φ0 L + sup{φ0 K 0 : K 0 ∈ K, K 0 ⊆ K \ L}. (ii) On the other hand, ?? suppose, if possible, that there is a closed set K 0 ⊆ K \ L such that µ∗0 L + µ∗0 K 0 > µ∗ K. Let E2 be a measurable envelope of K 0 , so that µ0 E1 + µ0 E2 > µ0 E0 ; since µ0 (E1 \ E0 ) = µ∗0 (L \ E0 ) = µ∗0 ∅ = 0,

µ0 (E2 \ E0 ) = µ∗0 (K 0 \ E0 ) = 0,

µ0 (E1 ∩ E2 ) > 0. Because µ0 is effectively locally finite, there is a measurable open set G0 , of finite measure, such that µ0 (G0 ∩ E1 ∩ E2 ) > 0. Set G = {G ∪ G0 : G, G0 ∈ Σ0 ∩ T, G ⊆ G0 \ L, G0 ⊆ G0 \ K 0 }. Then G is an upwards-directed family of measurable open sets, and because Σ0 includes a base for the topology of X, its union is (G0 \L)∪(G0 \K 0 ) = G0 . So there is an H ∈ G such that µ0 H > µ0 G0 −µ0 (E1 ∩E2 ),

415L

Quasi-Radon measure spaces

65

that is, there are open sets G, G0 ∈ Σ0 such that G ⊆ G0 \ L, G0 ⊆ G0 \ K 0 and µ0 ((G ∪ G0 ) ∩ E1 ∩ E2 )) > 0. But we must have µ0 (G ∩ E1 ) = µ∗0 (G ∩ L) = 0,

µ0 (G0 ∩ E2 ) = µ∗0 (G0 ∩ K 0 ) = 0,

so this is impossible. X X Accordingly φ0 K ≥ φ0 L + sup{φ0 K 0 : K 0 ∈ K, K 0 ⊆ K \ L}, so that φ0 satisfies condition (α) of 415K. β ) Let K0 ⊆ K be a non-empty downwards-directed family with empty intersection. Fix K0 ∈ K0 and (β ² > 0. Let E0 be a measurable envelope of K0 and G0 a measurable open set of finite measure such that µ0 (G0 ∩ E0 ) ≥ µ0 E0 − ². Then G = {G : G ∈ Σ0 ∩ T, G ⊆ G0 \ K for some K ∈ K0 such that K ⊆ K0 } T is an upwards-directed family of measurable open sets, and its union is G0 \ K0 = G0 , again because Σ0 includes a base for the topology T. So there is a G ∈ G such that µ0 G ≥ µ0 G0 − ². Let K ∈ K0 be such that K ⊆ K0 and G ∩ K = ∅; then φ0 K = µ∗0 K ≤ µ0 (E0 \ G) ≤ µ0 (E0 \ G0 ) + µ0 (G0 \ G) ≤ 2². As ² is arbitrary, inf K∈K0 φ0 K = 0. (γγ ) If K ∈ K and φ0 K > 0, let E0 be a measurable envelope of K. Then there is a measurable open set G of finite measure such that µ0 (G ∩ E0 ) > 0. Of course supK 0 ∈K,K 0 ⊆G φ0 K 0 ≤ µ0 G < ∞; but also there is a measurable closed set K 0 ⊆ G ∩ E0 such that µ0 K 0 > 0, in which case φ0 (K ∩ K 0 ) = µ0 (E0 ∩ K 0 ) > 0. So φ0 satisfies condition (γ). Q Q (b) By 415K, φ0 has an extension to a quasi-Radon measure µ on X which is inner regular with respect to K. Write Σ for the domain of µ. Note that, for K ∈ K, µK = φ0 K = µ∗0 K, so we can already be sure that the conclusion (i) of the proposition is satisfied. Now µ extends µ0 . P P(i) Take any K ∈ K. Let E0 ∈ Σ0 be a measurable envelope of K for the measure µ0 . If E ∈ Σ0 , then surely µ∗ (K ∩ E) = sup{µK 0 : K 0 ∈ K, K 0 ⊆ K ∩ E} = sup{µ∗0 K 0 : K 0 ∈ K, K 0 ⊆ K ∩ E} ≤ µ∗0 (K ∩ E). On the other hand, given γ < µ∗0 (K ∩ E) = µ0 (E0 ∩ E), there is a closed set F ∈ Σ0 such that F ⊆ E0 ∩ E and µ0 F ≥ γ, so that µ∗ (K ∩ E) ≥ µ(K ∩ F ) = µ∗0 (K ∩ F ) = µ0 (E0 ∩ F ) ≥ γ. Thus µ∗ (K ∩ E) = µ∗0 (K ∩ E) for every K ∈ K, E ∈ Σ0 . (ii) If K ∈ K, E ∈ Σ0 then µ∗ (K ∩ E) + µ∗ (K \ E) = µ∗0 (K ∩ E) + µ∗0 (K \ E) = µ∗0 K = µK. Because µ is complete and locally determined and inner regular with respect to K, E ∈ Σ (413F(iv)). Thus Σ0 ⊆ Σ. (iii) For any E ∈ Σ0 , we now have µE = sup{µK : K ∈ K, K ⊆ E} = sup{µ∗0 K : K ∈ K, K ⊆ E} ≤ µ0 E = sup{µ0 K : K ∈ K ∩ Σ0 , K ⊆ E} ≤ µE. As E is arbitrary, µ extends µ0 . Q Q

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415L

(c) Because Σ0 ∩ T is a base for T, closed under finite unions, µ is unique, by 415Hc. (d) Now for the conditions (i)-(v). I have already noted that (i) is guaranteed by the construction. Concerning (ii), if G ⊆ X is open, we surely have (µ0 )∗ G ≤ µ∗ G = µG because µ extends µ0 . On the other hand, writing G = {G0 : G0 ∈ Σ0 ∩ T, G0 ⊆ G}, G is upwards-directed and has union G, so µG = supG0 ∈G µG0 = supG0 ∈G µ0 G0 ≤ (µ0 )∗ G. So (ii) is true. Because µ extends µ0 , the embedding Σ0 ⊆ Σ corresponds to a measure-preserving embedding of A0 as a σ-subalgebra of A. To see that A0 is order-dense in A, take any non-zero a ∈ A. This is expressible as E • for some E ∈ Σ with µE > 0. Now there is a K ∈ K such that K ⊆ E and µK > 0. There is an E0 ∈ Σ0 which is a measurable envelope for K with respect to µ0 , so that µE0 = µ0 E0 = µ∗0 K = µK. But this means that 0 6= E0• = K • ⊆ E • = a in A, while E0• ∈ A0 . As a is arbitrary, A0 is order-dense in A. If a ∈ Af , then B = {b : b ∈ A0 , b ⊆ a} is upwards-directed and supb∈B µ ¯0 b ≤ µ ¯a is finite; accordingly B has a supremum in A0 (321C), which must also be its supremum in A, which is a (313O, 313K). So a ∈ A0 . Thus Af can be identified with Af0 . But this means that, for any p ∈ [1, ∞[, Lp (µ) ∼ ¯) is identified = Lp (A, µ p p ∼ with L (A0 , µ ¯0 ) = L (µ) (366H). This proves (iii). Of course (iv) and (v) are just translations of this. If E ∈ Σ and µE < ∞, then E • ∈ Af ⊆ A0 , that is, there is an E0 ∈ Σ0 such that µ(E4E0 ) = 0. If f is µ-integrable, then f • ∈ L1 (µ) = L1 (µ0 ), that is, there is a µ0 -integrable function f0 such that f = f0 µ-a.e. (e) If µ0 is complete and locally determined and F ⊆ X is an arbitrary closed set, then µ∗0 F = supK∈K µ∗0 (F ∩ K) = supK∈K µ(F ∩ K) = supK∈K,K⊆F µK = µF by 412Jc, because µ and µ0 are both inner regular with respect to K. (f ) If µ0 is localizable, A0 is Dedekind complete; as it is order-dense in A, the two must coincide (314Ia). Consequently L0 (µ) ∼ = L0 (A) = L0 (A0 ) ∼ = L0 (µ0 ), L∞ (µ) ∼ = L∞ (A) = L∞ (A0 ) ∼ = L∞ (µ0 ). So (iii)0 is true; now (iv)0 and (v)0 follow at once. 415M Corollary Let (X, T) be a regular topological space and µ0 an effectively locally finite τ -additive measure on X, defined on the σ-algebra Σ0 generated by a base for T. Then µ0 has a unique extension to a quasi-Radon measure on X. proof By 414Mb, µ0 is inner regular with respect to the closed sets. So 415L gives the result. 415N Corollary Let (X, T) be a completely regular topological space, and µ0 a τ -additive effectively locally finite Baire measure on X. Then µ0 has a unique extension to a quasi-Radon measure on X. proof This is a special case of 415M, because the domain Σ0 of µ0 , the Baire σ-algebra, is generated by the family of cozero sets, which form a base for T (4A2Fc). 415O Proposition (a) Let (X, T) be a topological space, and µ, ν two quasi-Radon measures on X. Then ν is an indefinite-integral measure over µ iff νF = 0 whenever F ⊆ X is closed and µF = 0. (b) Let (X, T, Σ, µ) be a quasi-Radon measure space, and ν an indefinite-integral measure over µ (definition: 234B). If ν is effectively locally finite it is a quasi-Radon measure. proof (a) If ν is an indefinite-integral measure over µ, then of course it is zero on all µ-negligible closed sets. So let us suppose that the condition is satisfied. Write Σ = dom µ and T = dom ν. (i) If E ⊆ X is a µ-negligible Borel set it is ν-negligible, because every closed subset of E must be µ-negligible, therefore ν-negligible, and ν is inner regular with respect to the closed sets. In particular,

415P

Quasi-Radon measure spaces

67

taking U ∗ to be the union of the family U = {U : U ∈ T, µU < ∞}, ν(X \ U ∗ ) = µ(X \ U ∗ ) = 0 because µ is effectively locally finite. Also, of course, taking V ∗ to be the union of the family V = S {V : V ∈ T, νV < ∞}, ν(X \ V ∗ ) = 0 because ν is effectively locally finite. Setting G = U ∩ V and G∗ = G, we have G∗ = U ∗ ∩ V ∗ , so G∗ is ν-conegligible. (ii) In fact, every µ-negligible set E is ν-negligible. P P?? Otherwise, ν ∗ (E ∩ G∗ ) > 0. Because the subspace measure νE is quasi-Radon (415B), there is a G ∈ G such that ν ∗ (E ∩ G) > 0. But there is an Fσ set H ⊆ G \ E such that µH = µ(G \ E), and now E ∩ G is included in the µ-negligible Borel set G \ H, so that ν(E ∩ G) = ν(G \ H) = 0. X XQ Q (iii) Let K be the family of closed subsets F of X such that either F is included in some member of G or F ∩ G∗ = ∅. If E ∈ dom µ and µE > 0, then there is an F ∈ K such that F ⊆ E and µF > 0. P P If µ(E \ G∗ ) > 0 take any closed set F ⊆ E \ G∗ with µF > 0. Otherwise, µ(E ∩ G∗ ) > 0. Because the subspace measure µE is quasi-Radon, there is a G ∈ G such that µ(E ∩ G) > 0; and now we can find a closed set F ⊆ E ∩ G with µF > 0, and F ∈ K. Q Q (iv) By 412I, there is a decomposition hXi ii∈I for µ such that every Xi except perhaps one belongs to K and that exceptional one, if any, is µ-negligible. Now hXi ii∈I is a decomposition for ν. P P Every Xi is measured by ν because it is either closed or µ-negligible, and of finite measure for ν because it is included in either a member of G or the ν-negligible set X \ G∗ . If E ⊆ X and νE > 0, then ν(E ∩ G∗ ) > 0, so there must be some G ∈SG such that ν(E ∩ G) > 0. Now J = {i : i ∈ I, µ(Xi ∩ G) > 0} is countable, and S ν(G \ i∈J Xi ) = µ(G \ i∈J Xi ) = 0, so there is an i ∈ J such that ν(Xi ∩ E) > 0. By 213O, hXi ii∈I is a decomposition for ν. Q Q (v) It follows that Σ ⊆ T. P P If E ∈ Σ, then for every i ∈ I there is an Fσ set H ⊆ E ∩ Xi such that E ∩ Xi \ H is µ-negligible, therefore ν-negligible, and E ∩ Xi ∈ T. As i is arbitrary, E ∈ T. Q Q In fact, ν is the completion of ν¹Σ. P P If F S ∈ T, then for every i ∈ I there is an Fσ set Hi ⊆ F ∩ Xi such that F ∩ Xi \ Hi isPν-negligible. Set H = i∈I Hi ; because H ∩ Xi = Hi belongs to Σ for every i, H ∈ Σ; 0 0 and ν(F \ H) = i∈I ν(F ∩ Xi \ H) = 0. Similarly, there is an H ∈ Σ such that H ⊆ X \ F and 0 0 0 ν((X \ F ) \ H ) = 0, so that H ⊆ F ⊆ X \ H and ν((X \ H ) \ H) = 0. So F is measured by the completion of ν¹Σ. Since ν itself is complete, it must be the completion of ν¹Σ. Q Q (vi) By (iv), ν is inner regular with respect to {E : E ∈ Σ, µE < ∞}. By 234G, ν is an indefiniteintegral measure over µ. R (b) Let f ∈ L0 (µ) be a non-negative function such that νF = f × χF dµ whenever this is defined. Because µ is complete and locally determined, so is ν (234Fb). Because µ is an effectively locally finite τ -additive topological measure, ν is a τ -additive topological measure (414H). Because µ is inner regular with respect to the closed sets, so is ν (412Q). Since we are assuming in the hypotheses that ν is effectively locally finite, it is a quasi-Radon measure. 415P Proposition Let (X, T, Σ, µ) be a quasi-Radon measure space. (a) Suppose that (X, T) is completely regular. If 1 ≤ p < ∞ and f ∈ Lp (µ), then for any ² > 0 there is a bounded continuous function g : X → R such that µ{x : g(x) 6= 0} < ∞ and kf − gkp ≤ ². (b) Suppose that (X, T) is regular and Lindel¨of. Let f ∈ L0 (µ) be locally integrable. Then for any ² > 0 there is a continuous function g : X → R such that kf − gk1 ≤ ². proof (a) Write C for the set of bounded continuous functions g : X → R such that {x : g(x) 6= 0} has finite measure. Then C is a linear subspace of RX included in Lp = Lp (µ). Let U be the closure of C in Lp , that is, the set of h ∈ Lp such that for every ² > 0 there is a g ∈ C such that kh − gkp ≤ ². Then U is closed under addition and scalar multiplication. Also χE ∈ U whenever µE < ∞. P P Let ² > 0. Set δ = 41 ²1/p . Write G for the family of open sets of finite measure. Because µ is effectively locally finite, there is a G ∈ G such that µ(E \ G) ≤ δ. Let F ⊆ G \ E be a closed set such that µF ≥ µ(G \ E) − δ; then µ(E4(G \ F )) ≤ 2δ. Write H for the family of cozero sets. Because T is completely regular, H is a base for T; because H is closed under finite unions (4A2C(b-iii)) and µ is τ -additive, there is an H ∈ H such that H ⊆ G \ F and µH ≥ µ(G \ F ) − δ, so that µ(E4H) ≤ 3δ. Express H as {x : g(x) > 0} where g : X → R is a continuous function. For each n ∈ N, set gn = ng ∧ χX ∈ C; then

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415P

|χE − gn |p ≤ χ(E4H) + (χH − gn )p for every n, so

R

|χE − gn |p ≤ µ(E4H) +

R

(χH − gn )p → µ(E4H) R as n → ∞, because gn → χH. So there is an n ∈ N such that |χE − gn |p ≤ 4δ, that is, kχE − gn kp ≤ ². As ² is arbitrary, χE ∈ U. Q Q Accordingly every simple function belongs to U. But if f ∈ Lp and ² > 0, there is a simple function h such that kf − hkp ≤ 21 ² (244Ha); now there is a g ∈ C such that kh − gkp ≤ 21 ² and kf − gkp ≤ ², as claimed. R (b) This time, write G for the family of open subsets of X such that G f is finite, so that G is an open cover of X. As X is paracompact (4A2H(b-i)), there is a locally finite family G0 ⊆ G covering X, which must be countable (4A2H(b-ii)). P Let h²G iG∈G0 be a family of strictly positive real numbers such that G∈G0 ²G ≤ ² (4A1P). Since X is completely regular (4A2H(b-i)),R we can apply (a) to see that, for each G ∈ G0 , there is a continuous function gG : X → R such that |gG − f × χG| ≤ ²G . Next, because X is normal (4A2H(b-i)), there is P a family hhG iG∈G0 of continuous functions from X to [0, 1] such that hG ≤ χG for every G ∈ G0 and G∈G0 hG (x) = P1 for every x ∈ X (4A2F(d-ix)). Set g(x) = G∈G0 gG (x)hG (x) for every x ∈ X. Because G0 is locally finite, g : X → R is continuous (4A2Bh). Now Z Z X X Z |f − g| = | (f − gG ) × hG | ≤ |(f − gG ) × hG | G∈G0

≤

X Z G∈G0

|f − gG | ≤ G

X

G∈G0

²G ≤ ²,

G∈G0

as required. 415Q Recall (411P) that if (A, µ ¯) is a localizable measure algebra, with Stone space (Z, S, T, ν), then ν is a strictly positive completion regular quasi-Radon measure, inner regular with respect to the open-and-closed sets (which are all compact). The following construction is primarily important for Radon measure spaces (see 416V), but is also of interest for general quasi-Radon spaces. Proposition Let (X, T, Σ, µ) be a quasi-Radon measure space and (A, µ ¯) its measure algebra. Let (Z, S, T, ν) be the Stone space of (A, µ ¯). For E ∈ Σ let E ∗ ⊆ Z be the open-and-closed set corresponding to the image E • of E in A. Define R ⊆ Z × X by saying that (z, x) ∈ R iff x ∈ F whenever F ⊆ X is closed and z ∈ F ∗ . Set Q = R−1 [X]. (a) R is a closed subset of Z × X. (b) For any E ∈ Σ, R[E ∗ ] is the smallest closed set such that µ(E \ R[E ∗ ]) = 0. In particular, if F ⊆ X is closed then R[F ∗ ] is the self-supporting closed set included in F such that µ(F \ R[F ∗ ]) = 0; and R[Z] is the support of µ. (c) Q is of full outer measure in Z. (d) For any E ∈ Σ, R−1 [E]4(Q∩E ∗ ) is negligible; consequently ν ∗ R−1 [E] = µE and R−1 [E]∩R−1 [X \E] is negligible. (e) For any A ⊆ X, ν ∗ R−1 [A] = µ∗ A. (f) If (X, T) is regular, then R−1 [G] is relatively open in Q for every open set G ⊆ X, R−1 [F ] is relatively closed in Q for every closed set F ⊆ X and R−1 [X \ E] = Q \ R−1 [E] for every Borel set E ⊆ X. proof (a) R=

\

((Z \ F ∗ ) × X) ∪ (Z × F )

F ⊆X is closed

is an intersection of closed sets, therefore closed.

S (b) Let G be the family of open sets G ⊆ X such that µ(E ∩ G) = 0, and G0 = G; then G0 ∈ G (414Ea). Set F0 = X \ G0 , so that F0 is the smallest closed set such that E \ F0 is negligible, and F0∗ ⊇ E ∗ .

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If (z, x) ∈ R and z ∈ E ∗ we must have x ∈ F0 . Thus R[E ∗ ] ⊆ F0 . On the other hand, if x ∈ F0 , and G is an open set containing x, then G ∈ / G so µ(G ∩ E) > 0 and (E ∩ G)∗ 6= ∅. Accordingly {(G ∩ E)∗ : x ∈ G ∈ T} is a downwards-directed family of non-empty open-and-closed sets in the compact space Z and has non-empty intersection, containing a point z say. If H ⊆ X is closed and z ∈ H ∗ , then X \ H is open and z ∈ / (X \ H)∗ , ∗ so x cannot belong to X \ H, that is, x ∈ H; as H is arbitrary, (z, x) ∈ R and x ∈ R[E ]; as x is arbitrary, R[E ∗ ] = F0 , as claimed. Of course, when E is actually closed, R[E ∗ ] = F0 ⊆ E. Taking E = X we see that R[Z] = R[X ∗ ] is the support of µ. (c) If W ∈ T and νW > 0, there is a non-empty open-and-closed set U ⊆ W , by 322Qa, which must be of the form E ∗ for some E ∈ Σ. By (b), R[E ∗ ] cannot be empty; but E ∗ ⊆ W , so R[W ] 6= ∅, that is, W ∩ Q 6= ∅. As W is arbitrary, ν∗ (Z \ Q) = 0, that is, Z is a measurable envelope of Q (413Ei). • • (d)(i) S Let F be the set of closed subsets of X included in E. Then supF ∈F F = E in A (412N), so E ∗ \ F ∈F F ∗ is nowhere dense and negligible. Now for each F ∈ F, R[F ∗ ] ⊆ F , so Q ∩ F ∗ ⊆ R−1 [F ] ⊆ R−1 [E]. Accordingly S Q ∩ E ∗ \ R−1 [E] ⊆ E ∗ \ F ∈F F ∗

is nowhere dense and negligible. (ii) ?? Suppose, if possible, that ν ∗ (R−1 [E] \ E ∗ ) > 0. Then there is an open-and-closed set U of finite measure such that ν ∗ (R−1 [E] ∩ U \ E ∗ ) > 0 (use 412Jc). Express U as H ∗ , where µH < ∞, and let F ⊆ H \ E be a closed set such that µ((H \ E) \ F ) < ν ∗ (R−1 [E] ∩ H ∗ \ E ∗ ). Then we must have ν ∗ (R−1 [E] ∩ F ∗ ) > 0. But R[F ∗ ] ⊆ F ⊆ X \ E so F ∗ ∩ R−1 [E] = ∅, which is impossible. X X (iii) Putting these together, (Q ∩ E ∗ )4R−1 [E] is negligible. (iv) It follows at once that (because Z is a measurable envelope for Q) ν ∗ R−1 [E] = ν ∗ (Q ∩ E ∗ ) = νE ∗ = µE. Moreover, applying the result to X \ E, R−1 [X \ E] ∩ R−1 [E] ⊆ (R−1 [X \ E]4(Q ∩ (X \ E)∗ )) ∪ (R−1 [E]4(Q ∩ E ∗ )) is negligible. (e)(i) Take E ∈ Σ such that A ⊆ E and µE = µ∗ A; then R−1 [A] ⊆ R−1 [E], so ν ∗ R−1 [A] ≤ ν ∗ R−1 [E] ≤ µE = µ∗ A. (ii) ?? Suppose, if possible, that ν ∗ R−1 [A] < µ∗ A. Let W ∈ T be such that R−1 [A] ⊆ W and νW = ν ∗ R−1 [A]. Then there is an F ∈ Σ such that ν(W 4F ∗ ) = 0. Since µF = νF ∗ = νW < µ∗ A, µ∗ (A \ F ) > 0; let G be a measurable envelope of A \ F disjoint from F . Then G∗ ∩ F ∗ = ∅ so ν(G∗ \ W ) = νG∗ = µG > 0 and there is a non-empty open-and-closed V ⊆ G∗ \ W ; let H ∈ Σ be such that H ⊆ G and V = H ∗ . In this case, R[V ] is closed and µ(H \ R[V ]) = 0, by (a), so that H ∩ R[V ] is measurable, not negligible, and included in G. But H ∩ R[V ] ∩ A is empty, because V ∩ R−1 [A] is empty, so µ∗ (H ∩ R[V ] ∩ A) < µ(H ∩ R[V ]), and G cannot be a measurable envelope of A \ F . X X Thus ν ∗ R−1 [A] = µ∗ A, as claimed. (f ) Suppose now that (X, T) is regular. (i) If G ⊆ X is open, R−1 [G] ∩ R−1 [X \ G] = ∅. P P If z ∈ R−1 [G], then there is an x ∈ G such that (z, x) ∈ R. Let H be an open set containing x such that H ⊆ G. Then x ∈ / X \ H so z ∈ / (X \ H)∗ , that is, ∗ z ∈ H . But ∗

R[H ∗ ] ⊆ R[H ] ⊆ H ⊆ G, ∗

so H ∩ R−1 [X \ G] = ∅ and z ∈ / R−1 [X \ G]. Q Q (ii) It is easy to check that

70

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415Q

E = {E : E ⊆ X, R−1 [E] ∩ R−1 [X \ E] = ∅} = {E : E ⊆ X, R−1 [X \ E] = Q \ R−1 [E]} is a σ-algebra of subsets of X (indeed, an algebra closed under arbitrary unions), just because R ⊆ Z × X and R−1 [X] = Q. Because it contains all open sets, E must contain all Borel sets. (iii) Now suppose once again that G ⊆ X is open and that z ∈ R−1 [G]. As in (i) above, there is an open set H ⊆ G such that z ∈ H ∗ ⊆ Z \ R−1 [X \ G], so that z ∈ H ∗ ∩ Q ⊆ R−1 [G]. As z is arbitrary, R−1 [G] is relatively open in Q. (iv) Finally, if F ⊆ X is closed, R−1 [F ] = Q \ R−1 [X \ F ] is relatively closed in Q. 415R Proposition Let (X, T, Σ, µ) be a Hausdorff quasi-Radon measure space and (Z, S, T, ν) the Stone space of its measure algebra. Let R ⊆ Z × X be the relation described in 415Q. Then (a) R is (the graph of) a function f ; (b) f is inverse-measure-preserving for the subspace measure νQ on Q = dom f , and in fact µ is the image measure νQ f −1 ; (c) if (X, T) is regular, then f is continuous. proof (a) If z ∈ Z and x, y ∈ X are distinct, let G, H be disjoint open sets containing x, y respectively. Then (X \ G)∗ ∪ (X \ H)∗ = ((X \ G) ∪ (X \ H))∗ = Z, defining ∗ as in 415Q, so z must belong to at least one of (X \ G)∗ , (X \ H)∗ . In the former case (z, x) ∈ /R and in the latter case (z, y) ∈ / R. This shows that R is a function; to remind us of its new status I will henceforth call it f . The domain of f is just Q = R−1 [X]. (b) By 415Qd, f is inverse-measure-preserving for νQ and µ. Suppose that A ⊆ X and f −1 [A] is in the domain TQ of νQ , that is, is of the form Q ∩ U for some U ∈ T. Take any E ∈ Σ such that µE > 0; then either ν(E ∗ ∩ U ) > 0 or ν(E ∗ \ U ) > 0. (α) Suppose that ν(E ∗ ∩ U ) > 0. Because ν is inner regular with respect to the open-and-closed sets, there is an H ∈ Σ such that H ∗ ⊆ E ∗ ∩ U and µH = νH ∗ > 0. Now there is a closed set F ⊆ E ∩H with µF > 0. In this case, f [F ∗ ] ⊆ F ⊆ E, by 415Qb, while F ∗ ∩U ⊆ f −1 [A], so f [F ∗ ] ⊆ E ∩ A. But this means that µ∗ (E ∩ A) ≥ µf [F ∗ ] = µF > 0. (β) If ν(E ∗ \ U ) > 0, then the same arguments show that µ∗ (E \ A) > 0. (γ) Thus µ∗ (E ∩ A) + µ∗ (E \ A) > 0 whenever µE > 0. Because µ is complete and locally determined, A ∈ Σ (413F(vii)). Thus we see that {A : A ⊆ X, f −1 [A] ∈ TQ } is included in Σ, and µ is the image measure νQ f −1 . (c) If T is regular, then 415Qf tells us that f is continuous. 415X Basic exercises >(a) Let (X, T, Σ, µ) be a quasi-Radon measure space and E ∈ Σ an atom for the measure. Show that there is a closed set F ⊆ E such that µF > 0 and F is an atom of Σ, in the sense that the only measurable subsets of F are ∅ and F . (Hint: 414G.) Show that µ is atomless iff all countable subsets of X are negligible. (b) Let h(Xi , Ti , Σi , µi )ii∈I be any family of quasi-Radon measure spaces. Show that the direct sum measure µ on X = {(x, i) : i ∈ I, x ∈ Xi } is a quasi-Radon measure when X is given its disjoint union topology. (c) Let S be the right-facing Sorgenfrey topology or lower limit topology on R, that is, the topology generated by the half-open intervals of the form [α, β[. Show that Lebesgue measure is completion regular and quasi-Radon for S. (Hint: 114Yj or 221Yb, or 419L.) (d) Let X be a topological space and µ a complete measure on X, and suppose that there is a conegligible closed measurable set Y ⊆ X such that the subspace measure on Y is quasi-Radon. Show that µ is quasiRadon.

415Yb

Quasi-Radon measure spaces

71

(e) Let (X, T, Σ, µ) be a quasi-Radon measure space. Show that µ is inner regular with respect to the family of self-supporting closed sets included in open sets of finite measure. (f ) Let (X, T, Σ, µ) be a quasi-Radon measure space. Show that for any E ∈ Σ, ² > 0 there is an open set G such that µG ≤ µE + ² and E \ G is negligible. (g) Find a compact Hausdorff quasi-Radon measure space which is not σ-finite. (h) Let (X, T, Σ, µ) be an atomless quasi-Radon measure space which is outer regular with respect to the open sets. Show that it is σ-finite. (Hint: if not, take a decomposition hXi ii∈I in which every Xi except one is self-supporting, and a set A meeting every Xi in just one point.) (i) Let (X, Σ, µ) be a σ-finite measure space in which Σ is countably generated as a σ-algebra. Show that, for a suitable topology on X, the completion of µ is a quasi-Radon measure. (Hint: take the topology generated by a countable subalgebra of Σ, and use the arguments of 415D.) (j) Let h(Xi , Ti , Σi , µi )ii∈I be a family of Q quasi-Radon probability spaces such that every µi is strictly positive, and λ the product measure on X = i∈I Xi . Show that if every Ti has a countable network, λ is a quasi-Radon measure. Q (k) Let hXi ii∈I be a family of separable metrizable spaces, and µ a quasi-Radon measure on X = i∈I Xi . Show that µ is completion regular iff every self-supporting closed set in X is determined by coordinates in a countable set. (Hint: 4A2Eb.) (l) Find two quasi-Radon measures µ, ν on the unit interval such that µG ≤ νG for every open set G but there is a closed set F such that νF < µF . (m) Let X be a topological space and µ, ν two quasi-Radon measures on X. (i) Suppose that µF = νF whenever F ⊆ X is closed and both µF and νF are finite. Show that µ = ν. (ii) Suppose that µG = νG whenever G ⊆ X is open and both µG and νG are finite. Show that µ = ν. (n) In 415L, write µ ˜0 for the c.l.d. version of µ0 (213E). Show that µ extends µ ˜0 . Show that µ ˜0 is τ -additive and inner regular with respect to the closed sets. (o) Let (X, T, Σ, µ) be a σ-finite paracompact Hausdorff quasi-Radon measure space, and f ∈ L0 (µ) a Rlocally integrable function. Show that for any ² > 0 there is a continuous function g : X → R such that |f − g| ≤ ². > (p) Let (X, T, Σ, µ) be a σ-finite completely regular quasi-Radon measure space. (i) Show that for every E ∈ Σ there is an F in the Baire σ-algebra Ba(X) of X such that µ(E4F ) = 0. (Hint: start with an open set E of finite measure.) (ii) Show that for every Σ-measurable function f : X → R there is a Ba(X)-measurable function equal almost everywhere to f . (q) Let (X, Σ, µ) be a measure space and f a µ-integrable real-valued function. Show that there is a unique quasi-Radon measure λ on R such that λ{0} = 0 and λ [α, ∞[ = µ∗ {x R : x ∈ dom R f , f (x) ≥ α}, λ ]−∞, −α] = µ∗ {x : x ∈ dom f , f (x) ≤ −α} whenever α > 0; and that h dλ = hf dµ whenever h ∈ L0 (λ) and h(0) = 0 and either integral is defined in [−∞, ∞]. (Hint: set λE = µ∗ f −1 [E \ {0}] for Borel sets E ⊆ R, and use 414Mb, 414O and 235Ib.) 415Y Further exercises (a) Give an example of two quasi-Radon measures µ, ν on R such that their sum, as defined in 112Xe, is not effectively locally finite, therefore not a quasi-Radon measure. (b) Show that any quasi-Radon measure space is isomorphic, as topological measure space, to a subspace ˆ be its of a compact quasi-Radon measure space. (Hint: if X is a T1 quasi-Radon measure space, let X Wallman compactification (Engelking 89, 3.6.21).)

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Topologies and measures

415Yc

(c) Let (X, T, Σ, µ) be a quasi-Radon measure space. Show that the following are equiveridical: (i) µ is outer regular with respect to the open sets; (ii) every negligible subset of X is included in an open set of finite measure; (iii) {x : µ{x} = 0} can be covered by a sequence of open sets of finite measure. (d) Show that + : R × R → R is continuous for the right-facing Sorgenfrey topology. (e) Let r ≥ 1. On R r let S be the topology generated by the half-open intervals [a, b[ where a, b ∈ R r (as defined in §115). (i) Show that S is the product topology if each factor is given the right-facing Sorgenfrey topology (415Xc). (ii) Show that Lebesgue measure is quasi-Radon for S. (Hint: induce on r. See also 417Yi.) (f ) Let Y ⊆ [0, 1] be a set of full outer measure and zero inner measure for Lebesgue measure µ. Give [0, 1] the topology T generated by S∪{Y } where S is the usual topology. Show that the subspace measure ν = µY is quasi-Radon for the subspace topology TY , but that there is no measure λ on X which is quasi-Radon for T and such that the subspace measure λY is equal to ν. (g) Find a base U for the topology of X = {0, 1}N and two totally finite (quasi-)Radon measures µ, ν on X such that G ∩ H ∈ U for all G, H ∈ U, µG ≤ νG for every G ∈ U , but νX < µX. (h) Let X be a topological space and G an open cover of X. Suppose that for each G ∈ G we are given a quasi-Radon measure µG on G such that µG (U ) = µH (U ) whenever G, H ∈ G and U ⊆ G ∩ H is open. Show that there is a unique quasi-Radon measure on X such that each µG is the subspace measure on G. (Hint: if hµG iG∈G is a maximal family with the given properties, then G is upwards-directed.) (i) Let (X, Σ, µ) be a measure space and T a topology on X, and suppose that there is a family U ⊆ Σ ∩ T such that (i) µU < ∞ for every U ∈ U (ii) for every U ∈ U, T S∩ Σ ∩ PU is a base S for the subspace topology of U (iii) if G is an upwards-directed family in T ∩ Σ and G ∈ U , then µ( G) = supG∈G µG (iv) µ is inner regular with respect to the closed sets (v) if E ∈ Σ and µE > 0 then there is a U ∈ U such that µ(E ∩ U ) > 0. Show that µ has an extension to a quasi-Radon measure on X. (j) Let (X, T, Σ, µ) be a quasi-Radon measure space such that T is normal (but not necessarily Hausdorff or regular). Show that if 1 ≤ p < ∞, f ∈ Lp (µ) and ² > 0, there is a bounded continuous function g : X → R such that kf − gkp ≤ ² and {x : g(x) 6= 0} has finite measure. (k) Let (X, T, Σ, µ) be a completely regular quasi-Radon measure space and suppose that we are given a uniformity defining the topology T. Show that if 1 ≤ p < ∞, f ∈ Lp (µ) and ² > 0, there is a bounded uniformly continuous function g : X → R such that kf − gkp ≤ ² and {x : g(x) 6= 0} has finite measure. (l) Let (X, T, Σ, µ) be a completely regular quasi-Radon measure space and τ an extended Fatou norm on L0 (µ) such that (i) τ ¹Lτ is an order-continuous norm (ii) whenever E ∈ Σ and µE > 0 there is an open set G such that µ(E ∩ G) > 0 and τ (χG• ) < ∞. Show that Lτ ∩ {f • : f : X → R is continuous} is norm-dense in Lτ . (m) Let (X, T, Σ, µ) be a quasi-Radon measure space. Show that µ is a compact measure in the sense of §342 iff there is a locally compact topology S on X such that (X, S, Σ, µ) is quasi-Radon. 415 Notes and comments 415B is particularly important because a very high proportion of the quasiRadon measure spaces we study are actually subspaces of Radon measure spaces. I would in fact go so far as to say that when you have occasion to wonder whether all quasi-Radon measure spaces have a property, you should as a matter of habit look first at subspaces of Radon measure spaces; if the answer is affirmative for them, you will have most of what you want, even if the generalization to arbitrary quasi-Radon spaces gives difficulties. Of course the reverse phenomenon can also occur. Stone spaces (411P) can be thought of as quasi-Radon compactifications of Radon measure spaces (416V). But this is relatively rare. Indeed the reason why I give so few examples of quasi-Radon spaces at this point is just that the natural ones arise from Radon measure spaces. Note however that the quasi-Radon product of an uncountable family of Radon

§416 intro.

Radon measure spaces

73

probability spaces need not be Radon (see 417Xq), so that 415E here and 417O below are sources of nonRadon quasi-Radon measure spaces. Density and lifting topologies can also provide us with quasi-Radon measure spaces (453Xd, 453Xg). 415K is the second in a series of inner-regular-extension theorems; there will be a third in 416J. I have been saying since Volume 1 that the business of measure theory, since Lebesgue’s time, has been to measure as many sets and integrate as many functions as possible. I therefore take seriously any theorem offering a canonical extension of a measure. 415L and its corollaries can all be regarded as improvement theorems, showing that a good measure can be made even better. We have already had such improvement theorems in Chapter 21: the completion and c.l.d. version of a measure (212C, 213E). In all such theorems we need to know exactly what effect our improvement is having on the other constructions we are interested in; primarily, the measure algebra and the function spaces. The machinery of Chapter 36 shows that if we understand the measure algebra(s) involved then the function spaces will give us no further surprises. Completion of a measure does not affect the measure algebra at all (322Da). Taking the c.l.d. version does not change Af = {a : µ ¯a < ∞} or L1 (213Fc, 213G, 322Db, 366H), but can affect the rest of the measure algebra and therefore L0 and L∞ . In this respect, what we might call the ‘quasi-Radon version’ behaves like the c.l.d. version (as could be expected, since the quasi-Radon version must itself be complete and locally determined; cf. 415Xn). The archetypal application of 415L is 415N. We shall see later how Baire measures arise naturally when studying Banach spaces of continuous functions (436E). 415N will be one of the keys to applying the general theory of topological measure spaces in such contexts. A virtue of Baire measures is that inner regularity with respect to closed sets comes almost free (412D); but there can be unsurmountable difficulties if we wish to extend them to Borel measures (439M), and it is important to know that τ -additivity, even in the relatively weak form allowed by the definition I use here (411C), is often enough to give a canonical extension to a well-behaved measure defined on every Borel set. In 415C we have inner regularity for a different reason, and the measure is already known to be defined on every Borel set, so in fact the quasi-Radon version of the measure is just the c.l.d. version (415Xd). For a volume and a half I have neglected indefinite-integral measures, though they are mentioned in the exercises; but we shall need them later, and in 415O I spell out a result which it will be useful to be able to quote. The exact hypotheses are not perhaps instantly predictable; see 414Xe. One interpretation of the Lifting Theorem is that for a complete strictly localizable measure space (X, Σ, µ) there is a function g : X → Z, where Z is the Stone space of the measure algebra of µ, such that E4g −1 [E ∗ ] is negligible for every E ∈ Σ, where E ∗ ⊆ Z is the open-and-closed set corresponding to the image of E in the measure algebra (341Q). For a Hausdorff quasi-Radon measure space we have a function f : Q → X, where Q is a dense subset of Z, such that (Q ∩ E ∗ )4f −1 [E] is negligible for every E ∈ Σ (415Qd, 415R); moreover, there is a canonical construction for this function. For completeness’ sake, I have given the result for general, not necessarily Hausdorff, spaces X (415Q); but evidently it will be of greatest interest for regular Hausdorff spaces (415Rc), especially if they happen to be ‘compact’ in the sense of §342. Perhaps I should remark that in the most important applications, Q is the whole of Z (416Xw). Of course the question arises, whether f g can be the identity. (Z typically has larger cardinal than X, so asking for gf to be the identity is a bit optimistic.) This is in fact an important question; I will return to it in 453M.

416 Radon measure spaces We come now to the results for which the chapter so far has been preparing. The centre of topological measure theory is the theory of ‘Radon’ measures (411Hb), measures inner regular with respect to compact sets. Most of the section is devoted to pulling the earlier work together, and in particular to re-stating theorems on quasi-Radon measures in the new context. Of course this has to begin with a check that Radon measures are quasi-Radon (416A). It follows immediately that Radon measures are (strictly) localizable (416B). After presenting a miscellany of elementary facts, I turn to the constructions of §413, which take on simpler and more dramatic forms in this context (416J-416P). I proceed to investigate subspace measures (416R-416T) and some special product measures (416U). I end the section with further notes on the forms which earlier theorems on Stone spaces (416V) and compact measure spaces (416W) take when applied to Radon measure spaces.

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416A

416A Proposition A Radon measure space is quasi-Radon. proof Let (X, T, Σ, µ) be a Radon measure space. Because T is Hausdorff, every compact set is closed, so µ is inner regular with respect to the closed sets. By 411E, µ is τ -additive; by 411Gf, it is effectively locally finite. Thus all parts of condition (ii) of 411Ha are satisfied, and µ is a quasi-Radon measure. 416B Corollary A Radon measure space is strictly localizable. proof Put 416A and 415A together. 416C In order to use the results of §415 effectively, it will be helpful to spell out elementary conditions ensuring that a quasi-Radon measure is Radon. Proposition Let (X, T, Σ, µ) be a locally finite Hausdorff quasi-Radon measure space. Then the following are equiveridical: (i) µ is a Radon measure; (ii) whenever E ∈ Σ and µE > 0 there is a compact set K such that µ(E ∩ K) > 0; (iii) sup{K • : K ⊆ X is compact} = 1 in the measure algebra of µ. If µ is totally finite we can add (iv) sup{µK : K ⊆ X is compact} = µX. proof (i)⇒(ii) and (ii) ⇐⇒ (iii) are trivial. For (ii)⇒(i), observe that if E ∈ Σ and µE > 0 there is a compact set K ⊆ E such that µK > 0. P P There is a compact set K 0 such that µ(E ∩ K 0 ) > 0, by hypothesis; now there is a closed set K ⊆ E ∩ K 0 such that µK > 0, because µ is inner regular with respect to the closed sets, and K is compact. Q Q By 412B, µ is tight. Being a complete, locally determined, locally finite topological measure, it is a Radon measure. When µX < ∞, of course, we also have (ii) ⇐⇒ (iv). 416D Some further elementary facts are worth writing out plainly. Lemma (a) In a Radon measure space, every compact set has finite measure. (b) Let (X, T, Σ, µ) be a Radon measure space, and E ⊆ X a set such that E ∩ K ∈ Σ for every compact K ⊆ X. Then E ∈ Σ. (c) A Radon measure is inner regular with respect to the self-supporting compact sets. (d) Let X be a Hausdorff space and µ a locally finite complete locally determined measure on X which is tight (that is, inner regular with respect to the compact sets). If every compact set belongs to the domain of µ, µ is a Radon measure. proof (a) 411Ga. (b) We have only to remember that µ is complete, locally determined and tight, and apply 413F(ii). (c) If (X, T, Σ, µ) is a Radon measure space, E ∈ Σ and γ < µE, there is a compact set K ⊆ E such that µK ≥ γ. By 414F, there is a self-supporting relatively closed set L ⊆ K such that µL = µK; but now of course L is compact, while L ⊆ E and µL ≥ γ. (d) Let K be the family of compact subsets of X; write Σ for the domain of µ. If F ⊆ X is closed, then F ∩ K ∈ K ⊆ Σ for every K ∈ K; accordingly F ∈ Σ. But this means that every closed set, therefore every open set, belongs to Σ, and µ is a Radon measure. 416E Specification of Radon measures In 415H I described some conditions which enable us to be sure that two quasi-Radon measures on a given topological space are the same. In the case of Radon measures we have a similar list. This time I include a note on the natural ordering of Radon measures. Proposition Let X be a Hausdorff space and µ, ν two Radon measures on X. (a) The following are equiveridical: (i) dom ν ⊆ dom µ and µE ≤ νE for every E ∈ dom ν; (ii) µK ≤ νK for every compact set K ⊆ X.

416E

Radon measure spaces

75

(iii) µG ≤ νG for every open set G ⊆ X; (iv) µF ≤ νF for every closed set F ⊆ X. If X is locally R compact, R we can add (v) f dµ ≤ f dν for every non-negative continuous function f : X → R with compact support. (b) The following are equiveridical: (i) µ = ν; (ii) µK = νK for every compact set K ⊆ X. (iii) µG = νG for every open set G ⊆ X; (iv) µF = νF for every closed set F ⊆ X. If X is locally R compact, R we can add (v) f dµ = f dν for every continuous function f : X → R with compact support. proof (a)(i)⇒(iv)⇒(ii) and (i)⇒(iii) are trivial. (ii)⇒(i) If (ii) is true, then µE = supK⊆E

µK ≤ supK⊆E

is compact

is compact

νK = νE

for every set E measured by both µ and ν. Also dom ν ⊆ dom µ. P P Suppose that E ∈ dom ν and that K ⊆ X is a compact set such that µK > 0. Then there are compact sets K1 ⊆ K ∩ E, K2 ⊆ K \ E such that νK1 + νK2 ≥ ν(K ∩ E) + ν(K \ E) − µK = νK − µK. So µ(K \ (K1 ∪ K2 )) ≤ ν(K \ (K1 ∪ K2 )) < µK and µK1 + µK2 > 0. This shows that µ∗ (K ∩ E) + µ∗ (K \ E) > 0. As K is arbitrary, E ∈ dom µ (413F(vii)). Q Q So (i) is true. (iii)⇒(ii)SThe point is that if K ⊆ X is compact, then µK = inf{µG : G ⊆ X is open, K ⊆ G}. P P Because X = {µG : G ⊆ X is open, µG < ∞}, there is an open set G0 of finite measure including K. Now, for any γ > µK, there is a compact set L ⊆ G0 \ K such that µL ≥ µG0 − γ, so that µG ≤ γ, where G = G0 \ L is an open set including K. Q Q The same is true for ν. So, if (iii) is true, µK = inf G⊇K

is open

µG ≤ inf G⊇K

is open

νG = νK

for every compact K ⊆ X, and (ii) is true. (iii)⇒(v) If (iii) is true and f : X → [0, ∞[ is a non-negative continuous function, then Z

Z

∞

f dµ =

µ{x : f (x) > t}dt 0

(252O)

Z

Z

∞

≤

ν{x : f (x) > t}dt =

f dν.

0

(v)⇒(iii) If X is locally compact and (v) is true, take any open set G ⊆ X, and consider A = {f : f is a continuous function with compact support from X to [0, 1] and f ≤ χG}. Then A is upwards-directed and supf ∈A f (x) = χG(x) for every x ∈ X, by 4A2G(e-i). So µG = supf ∈A by 414Ba. As G is arbitrary, (iii) is true. (b) now follows at once, or from 415H.

R

f dµ ≤ supf ∈A

R

f dν = νG

76

Topologies and measures I

416F

416F Proposition Let X be a Hausdorff space and µ a Borel measure on X. Then the following are equiveridical: (i) µ has an extension to a Radon measure on X; (ii) µ is locally finite and tight; (iii) µ is locally finite and effectively locally finite, and µG = sup{µK : K ⊆ G is compact} for every open set G ⊆ X; (iv) µ is locally finite, effectively locally finite and τ -additive, and µG = sup{µ(G ∩ K) : K ⊆ X is compact} for every open set G ⊆ X. In this case the extension is unique; it is the c.l.d. version of µ. proof (a)(i)⇒(iv) If µ = µ ˜¹B where µ ˜ is a Radon measure and B is the Borel σ-algebra of X, then of course µ is locally finite and effectively locally finite and τ -additive because µ ˜ is (see 416A) and every open set belongs to B. Also µG = sup{µK : K ⊆ G is compact} ≤ sup{µ(G ∩ K) : K ⊆ X is compact} ≤ µG for every open set G ⊆ X, because µ ˜ is tight and compact sets belong to B. (b)(iv)⇒(iii) Suppose that (iv) is true. Of course µ is locally finite and effectively locally finite. Suppose that G ⊆ X is open and that γ < µG. Then there is a compact K ⊆ X such that µ(G ∩ K) > γ. By 414K, the subspace measure µK is τ -additive. Now K is a compact Hausdorff space, therefore regular. By 414Ma there is a closed set F ⊆ G ∩ K such that µK F ≥ γ. Now F is compact, F ⊆ G and µF ≥ γ. As G and γ are arbitrary, (iii) is true. (c)(iii)⇒(ii) I have to show that if µ satisfies the conditions of (iii) it is tight. Let K be the family of compact subsets of X and A the family of subsets of X which are either open or closed. Then whenever A ∈ A, F ∈ Σ and µ(A ∩ F ) > 0, there is a K ∈ K such that K ⊆ A and µ(K ∩ F ) > 0. P P Because µ is effectively locally finite, there is an open set G of finite measure such that µ(G ∩ A ∩ F ) > 0. (α) If A is open, then there will be a compact set K ⊆ G ∩ A such that µK > µ(G ∩ A) − µ(G ∩ A ∩ F ), so that µ(K ∩ F > 0. (β) If A is closed, then let L ⊆ G be a compact set such that µL > µG − µ(G ∩ A ∩ F ); then K = L ∩ A is compact and µ(K ∩ F ) > 0. Q Q By 412C, µ is inner regular with respect to K, as required. (d)(ii)⇒(i) If µ is locally finite and tight, let µ ˜ be the c.l.d. version of µ. Then µ ˜ is complete, locally determined, locally finite (because µ is), a topological measure (because µ is) and tight (because µ is, using 412Ha); so is a Radon measure. Every compact set has finite measure for µ, so µ is semi-finite and µ ˜ extends µ (213Hc). (e) By 416Eb there can be at most one Radon measure extending µ, and we have observed in (c) above that in the present case it is the c.l.d. version of µ. 416G One of the themes of §434 will be the question: on which Hausdorff spaces is every locally finite quasi-Radon measure a Radon measure? I do not think we are ready for a general investigation of this, but I can give one easy special result. Proposition Let (X, T) be a locally compact Hausdorff space and µ a locally finite quasi-Radon measure on X. Then µ is a Radon measure. proof µ satisfies condition (ii) of 416C. P P Take E ∈ dom µ such that µE > 0. Let G be the family of relatively compact open subsets of X; then G is upwards-directed and has union X. By 414Ea, there is a G ∈ G such that µ(E ∩ G) > 0. But now G is compact and µ(E ∩ G) > 0. Q Q By 416C, µ is a Radon measure. 416H Corollary Let (X, T) be a locally compact Hausdorff space, and µ a locally finite, effectively locally finite, τ -additive Borel measure on X. Then µ is tight and its c.l.d. version is a Radon measure, the unique Radon measure on X extending µ. proof By 415Cb, the c.l.d. version µ ˜ of µ is a quasi-Radon measure extending µ. Because µ is locally finite, so is µ ˜; by 416G, µ ˜ is a Radon measure. By 416Eb, the extension is unique. Now

416K

Radon measure spaces

µE = µ ˜E = supK⊆E

is compact

µ ˜K = supK⊆E

77 is compact

µK

for every Borel set E ⊆ X, so µ itself is tight. 416I While on the subject of locally compact spaces, I mention an important generalization of a result from Chapter 24. Proposition Let (X, T, Σ, µ) be a locally compact Radon measure space. Write Ck for the space of continuous real-valued functions on X with compact supports. If 1 ≤ p < ∞, f ∈ Lp (µ) and ² > 0, there is a g ∈ Ck such that kf − gkp ≤ ². proof By 415Pa, there is a bounded continuous function h1 : X → R such that G = {x : h1 (x) 6= 0} has finite measure and kf − h1 kp ≤ 21 ². Let K ⊆ G be a compact set such that kh1 k∞ (µ(G \ K))1/p ≤ 21 ², and let h2 ∈ Ck be such that χK ≤ h2 ≤ χG (4A2G(e-i)). Set g = h1 × h2 . Then g ∈ Ck and

R

|h1 − g|p ≤

R

G\K

|h1 |p ≤ µ(G \ K)kh1 kp∞ ,

so kh1 − gkp ≤ 21 ² and kf − gkp ≤ ², as required. 416J I turn now to constructions of Radon measures based on ideas in §413. Theorem Let X be a Hausdorff space. Let K be the family of compact subsets of X and φ0 : K → [0, ∞[ a functional such that (α) φ0 K = φ0 L + sup{φ0 K 0 : K ∈ K, K 0 ⊆ K \ L} whenever K, L ∈ K and L ⊆ K, (γ) for every x ∈ X there is an open set G containing x such that sup{φ0 K : K ∈ K, K ⊆ G} is finite. Then there is a unique Radon measure on X extending φ0 . proof By 413M, there is a unique complete locally determined measure µ on X, extending φ0 , which is inner regular with respect to K. By (γ), µ is locally finite; by 416Dd, it is a Radon measure. 416K Proposition Let X be a regular Hausdorff space. Let K be the family of compact subsets of X, and φ0 : K → [0, ∞[ a functional such that (α1 ) φ0 K ≤ φ0 (K ∪ L) ≤ φ0 K + φ0 L for all K, L ∈ K, (α2 ) φ0 (K ∪ L) = φ0 K + φ0 L whenever K, L ∈ K and K ∩ L = ∅, (γ) for every x ∈ X there is an open set G containing X such that sup{φ0 K : K ∈ K, K ⊆ G} < ∞. Then there is a unique Radon measure µ on X such that µK = inf G⊆X

is open,K⊆G

supL⊆G is compact φ0 L

for every K ∈ K. proof (a) For open sets G ⊆ X set ψG = supL∈K,L⊆G φ0 L, and for compact sets K ⊆ X set φ1 K = inf{ψG : G ⊆ X is open, K ⊆ G}. Evidently ψG ≤ ψH whenever G ⊆ H. We need to know that ψ(G ∪ H) ≤ ψG + ψH for all open sets G, H ⊆ X. P P If L ⊆ G ∪ H is compact, then the disjoint compact sets L \ G, L \ H can be separated by disjoint open sets H 0 , G0 (4A2F(h-i)); now L \ G0 ⊆ H, L \ H 0 ⊆ G are compact and cover L, so φ0 L ≤ φ0 (L \ G0 ) + φ0 (L \ H 0 ) ≤ ψH + ψG. As L is arbitrary, ψ(G ∪ H) ≤ ψG + ψH. Q Q Moreover, ψ(G ∪ H) = ψG + ψH if G ∩ H = ∅. P P If K ⊆ G, L ⊆ H are compact, then φ0 K + φ0 L = φ0 (K ∪ L) ≤ ψ(G ∪ H). As K and L are arbitrary, ψG + ψH ≤ ψ(G ∪ H). Q Q

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416K

(b) It follows that φ1 K is finite for every compact K ⊆ X. P P Set G = {G : G ⊆ X is open, ψG < ∞}. Then (a) tells us that G is upwards-directed. But also we are supposing that G covers X, by (γ). So if K ⊆ X is compact there is a member of G including K and φ1 K < ∞. Q Q (c) Now φ1 satisfies the conditions of 416J. α) Suppose that K, L ∈ K and L ⊆ K. Set γ = sup{φ1 M : M ∈ K, M ⊆ K \ L}. Take any ² > 0. P P(α Let G be an open set such that K ⊆ G and ψG ≤ φ1 K + ². If M ∈ K and M ⊆ K \ L, there are disjoint open sets U , V such that L ⊆ U and M ⊆ V (4A2F(h-i) again); we may suppose that U ∪ V ⊆ G. In this case,

φ1 L + φ1 M ≤ ψU + ψV = ψ(U ∪ V ) (by the second part of (a) above) ≤ ψG ≤ φ1 K + ². As M is arbitrary, γ ≤ φ1 K − φ1 L + ². On the other hand, there is an open set H such that L ⊆ H and ψH ≤ φ1 L + ². Set F = K \ H, so that F is a compact subset of K \ L. Then there is an open set V such that F ⊆ V and ψV ≤ φ1 F + ². In this case K ⊆ H ∪ V , so φ1 K ≤ ψ(H ∪ V ) ≤ ψH + ψV ≤ φ1 L + φ1 F + 2² ≤ φ1 L + γ + 2², so γ ≥ φ1 K − φ1 L − 2². As ² is arbitrary, γ = φ1 K − φ1 L; as K and L are arbitrary, φ1 satisfies condition (α) of 416J. (γγ ) Any x ∈ X is contained in an open set G such that ψG < ∞; but now sup{φ1 K : K ∈ K, K ⊆ G} ≤ ψG is finite. So φ1 satisfies condition (γ) of 416J. Q Q (d) By 416J, there is a unique Radon measure on X extending φ1 , as claimed. 416L Corollary Let X be a locally compact Hausdorff space. Let K be the family of compact subsets of X, and φ0 : K → [0, ∞[ a functional such that φ0 K ≤ φ0 (K ∪ L) ≤ φ0 K + φ0 L for all K, L ∈ K, φ0 (K ∪ L) = φ0 K + φ0 L whenever K, L ∈ K and K ∩ L = ∅. Then there is a unique Radon measure µ on X such that µK = inf{φ0 K 0 : K 0 ∈ K, K ⊆ int K 0 } for every K ∈ K. proof Observe that φ0 satisfies the conditions of 416K; 416K(γ) is true because X is locally compact. Define ψ, φ1 as in the proof of 416K, and set φ01 K = inf{φ0 K 0 : K 0 ∈ K, K ⊆ int K 0 } for every K ∈ K. Then φ01 = φ1 . P P Let K ∈ K, ² > 0. (i) There is an open set G ⊆ X such that K ⊆ G and ψG ≤ φ1 K + ². Now the relatively compact open subsets with closures included in G form an upwards-directed cover of K, so there is a K 0 ∈ K such that K ⊆ int K 0 and K 0 ⊆ G. Accordingly φ01 K ≤ φ0 K 0 ≤ ψG ≤ φK + ². (ii) There is an L ∈ K such that K ⊆ int L and φ0 L ≤ φ01 K + ², so that φ1 K ≤ ψ(int L) ≤ φ0 L ≤ φ01 K + ². (iii) As K, ² are arbitrary, φ01 = φ1 . Q Q Now 416K tells us that there is a unique Radon measure extending φ1 , and this is the measure we seek.

416O

Radon measure spaces

79

416M The extension theorems in the second half of §413 also have important applications to Radon measures. Henry’s Theorem (Henry 69) Let X be a Hausdorff space and µ0 a measure on X which is locally finite and tight. Then µ0 has an extension to a Radon measure µ on X; and the extension may be made in such a way that whenever µE < ∞ there is an E0 ∈ Σ0 such that µ(E4E0 ) = 0. proof All the work has been done in §413; we need to check here only that the family K of compact subsets of X and the measure µ0 satisfy the hypotheses of 413O. But (†) and (‡) there are elementary, and µ∗0 K < ∞ for every K ∈ K by 411Ga. Now take the measure µ from 413O. It is complete, locally determined and inner regular with respect to K; also K ⊆ dom µ. Because µ0 is locally finite and µ extends µ0 , µ is locally finite. By 416Dd, µ is a Radon measure. And the construction of 413O ensures that every set of finite measure for µ differs from a member of Σ0 by a µ-negligible set. 416N Proposition Let X be a Hausdorff space and T a subalgebra of PX. Let ν : T → R be a non-negative finitely additive functional. Then there is a Radon measure µ on X such that µX ≤ νX and µK ≥ νK for every compact set K ∈ T. proof Use 413S, with K the family of compact subsets of X, and 416Dd. 416O Theorem Let X be a Hausdorff space and T a subalgebra of PX. Let ν : T → R be a non-negative finitely additive functional such that νE = sup{νF : F ∈ T, F ⊆ E, F is closed} for every E ∈ T, νX = supK⊆X is compact inf F ∈T,F ⊇K νF . Then there is a Radon measure µ on X extending ν. proof (a) For A ⊆ X, write ν ∗ A = inf F ∈T,F ⊇A νF .

S Let hKn in∈N be a sequence of compact subsets of X such that limn→∞ ν ∗ Kn = νX; replacing Kn by i
νn0 E

+

νn0 F.

As E and F are arbitrary, νn0 is additive. Q Q 0 (b) For each n ∈ N, set νn E = νn+1 E − νn0 E for every E ∈ T; then νn is additive. Because Kn+1 ⊇ Kn , νn is non-negative. 0 If E ∈ T and E ∩ Kn+1 = ∅, then νn E = νn+1 E = 0. So if we set Tn = {E ∩ Kn+1 : E ∈ T}, we have an additive functional ν˜n : Tn → [0, ∞[ defined by setting ν˜n (E ∩ Kn+1 ) = νn E for every E ∈ T. Also ν˜n H = sup{˜ νn K : K ∈ Tn , K ⊆ H, K is compact} for every H ∈ Tn . P P Express H as E ∩ Kn+1 where E ∈ T. Given ² > 0, there is a closed set F ∈ T such that F ⊆ E and νF ≥ νE − ²; but now K = F ∩ Kn+1 ∈ Tn is a compact subset of H, and 0 ν˜n (H \ K) = νn (E \ F ) ≤ νn+1 (E \ F ) ≤ ν(E \ F ) ≤ ²,

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so ν˜n K ≥ ν˜n H − ². Q Q (c) For each n ∈ N, we have a Radon measure µn on Kn+1 , with domain Σn say, such that µn Kn+1 ≤ ν˜n Kn+1 and µn K ≥ ν˜n K for every compact set K ⊆ Kn+1 (416N). Since Kn+1 is itself compact, we must have µn Kn+1 = ν˜n Kn+1 . But this means that µn extends ν˜n . P P If H ∈ Tn , ² > 0 there is a compact set K ∈ Tn such that ν˜n K ≥ ν˜n H − ², so that (µn )∗ H ≥ µn K ≥ ν˜n H − ²; as ² is arbitrary, (µn )∗ H ≥ ν˜n H. So there is an F1 ∈ Σn such that F1 ⊆ H and µn F1 ≥ ν˜n H. Similarly, there is an F2 ∈ Σn such that F2 ⊆ Kn+1 \ H and µn F2 ≥ ν˜n (Kn+1 \ H). But in this case H \ F1 ⊆ Kn+1 \ (F1 ∪ F2 ) is µn -negligible, because µn F1 + µn F2 ≥ ν˜n H + ν˜n (Kn+1 \ H) = ν˜n Kn+1 = µn Kn+1 . So H \ F1 and H belong to Σn and µn H = µn F1 = ν˜n H. Q Q (d) Set Σ = {E : E ⊆ X, E ∩ Kn+1 ∈ Σn for every n ∈ N}, P∞ µE = n=0 µn (E ∩ Kn+1 ) for every E ∈ Σ. Then µ is a Radon measure on X extending ν. P P (i) It is easy to check that Σ is a σ-algebra of subsets of X including T, just because each Σn is a σ-algebra of subsets of Kn+1 including Tn ; and that µ is a complete measure because every µn is. (ii) If E ∈ T, then µE =

∞ X

µn (E ∩ Kn+1 ) =

n=0

= lim

∞ X

ν˜n (E ∩ Kn+1 ) =

n=0 n X

n→∞

∞ X

νn E = lim

n→∞

n=0

n X

νi E

i=0

ν ∗ (E ∩ Ki+1 ) − ν ∗ (E ∩ Ki ) = lim ν ∗ (E ∩ Kn ) ≤ νE. n→∞

i=0

On the other hand, µX = limn→∞ ν ∗ Kn+1 = νX, so in fact µE = νE for every E ∈ T, that is, µ extends ν. In particular, µ is totally finite, therefore locally determined and locally finite. (iii) If G ⊆ X is open, then G ∩ Kn+1 ∈ Σn for every n, so G ∈ Σ; thus µ is a topological measure. If µE > 0, there is some n ∈ N such that µn (E ∩ Kn+1 ) > 0; now there is a compact set K ⊆ E ∩ Kn+1 such that µn K > 0, so that µK > 0. This shows that µ is tight, so is a Radon measure, as required. Q Q Remark Observe that in this construction µKn+1 = =

∞ X i=0 ∞ X

µi (Kn+1 ∩ Ki+1 ) =

∞ X

ν˜i (Kn+1 ∩ Ki+1 ) =

i=0

∞ X

νi (Kn+1 ∩ Ki+1 )

i=0

0 νi+1 (Kn+1 ∩ Ki+1 ) − νi0 (Kn+1 ∩ Ki+1 )

i=0

=

∞ X

ν ∗ (Kn+1 ∩ Ki+1 ) − ν ∗ (Kn+1 ∩ Ki )

i=0

=

n X

ν ∗ (Kn+1 ∩ Ki+1 ) − ν ∗ (Kn+1 ∩ Ki ) = ν ∗ Kn+1

i=0

for every n ∈ N. What this means is that if instead of the hypothesis νX = supK⊆X

is compact

inf F ∈T,F ⊇K νF

we are presented with a specified non-decreasing sequence hLn in∈N of compact subsets of X such that νX = supn∈N ν ∗ Ln , then we can take Kn+1 = Ln in the argument above and we shall have µLn = ν ∗ Ln for every n.

416P

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81

416P Theorem Let X be a Hausdorff space and µ a locally finite measure on X which is inner regular with respect to the closed sets. Then the following are equiveridical: (i) µ has an extension to a Radon measure on X; (ii) for every non-negligible measurable set E ⊆ X there is a compact set K ⊆ E such that µ∗ K > 0. If µ is totally finite, we can add (iii) sup{µ∗ K : K ⊆ X is compact} = µX. proof Write Σ for the domain of µ. (a)(i)⇒(ii) If λ is a Radon measure extending µ, and µE > 0, then λE > 0, so there is a compact set K ⊆ E such that λK > 0; but now, because λ is an extension of µ, µ∗ K ≥ λ∗ K = λK > 0. α) Let E be the family of measurable envelopes of compact sets. Then µE < ∞ (b)(ii)⇒(i) & (iii)(α for every E ∈ E. P P If E ∈ E, there is a compact set K such that E is a measurable envelope of K. Now µE = µ∗ K is finite by 411G. Q Q Next, E is closed under finite unions, by 132Ed. The hypothesis (ii) tells us that if µE > 0 then there is some F ∈ E such that F ⊆ E and µF > 0; for there is a compact set K ⊆ E such that µ∗ K > 0, K has a measurable envelope F0 , and F = E ∩ F0 is still a measurable envelope of K. So in fact µ is inner regular with respect to E (412Aa). In particular, µ is semi-finite. If γ < µX there is an F ∈ E such that µF ≥ ², and now there is a compact set K such that F is a measurable envelope of K, so that µ∗ K = µF ≥ γ. As γ is arbitrary, (iii) is true. β ) Because µ is inner regular with respect to E, D = {E • : E ∈ E} is order-dense in the measure (β algebra (A, µ ¯) of µ (412N), so there is a family hdi ii∈I in D which is a partition of unity in A (313K). For each i ∈ I, take Ei ∈ E such that Ei• = di . Then X

µ(E ∩ Ei ) =

i∈I

X

µ ¯(E • ∩ di ) = µ ¯E •

i∈I

(321E) = µE for every E ∈ Σ. (γγ ) For each i ∈ I, let µi be the subspace measure on Ei . Then there is a Radon measure λi on Ei extending µi . P P Because µ is inner regular with respect to the closed sets, µi is inner regular with respect to the relatively closed subsets of Ei (412O). Also there is a compact subset K ⊆ Ei such that µi Ei = µEi = µ∗ K = µ∗i K, so µi satisfies the conditions of 416O and has an extension to a Radon measure. Q Q (δδ ) Define λE =

P i∈I

λi (E ∩ Ei )

whenever E ⊆ X is such that λi measures E ∩ Ei for every i ∈ I. Then λ is a Radon measure on X extending µ. P P It is easy to check that it is a measure, just because every λi is a measure, and it extends µ by (β) above. If G ⊆ X is open, then G ∩ Ei is relatively open for every i ∈ I, so λ measures G; thus λ is a topological measure. If λE = 0 and A ⊆ E, then λi (A ∩ Ei ) ≤ λ(E ∩ Ei ) = 0 for every i, so λA = 0; thus λ is complete. For all distinct i, j ∈ I, λi (Ei ∩ Ej ) = µi (Ei ∩ Ej ) = µ(Ei ∩ Ej ) = µ ¯(di ∩ dj ) = 0, so λEi = λi Ei = µi Ei is finite. This means that if E ⊆ X is such that λ measures E ∩ F whenever λF < ∞, then λ must measure E ∩ Ei for every i, and λ measures E; thus λ is locally determined. If λE > 0 there are an i ∈ I such that λi (E ∩ Ei ) > 0 and a compact K ⊆ E ∩ Ei such that 0 < λi K = λK; consequently λ

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is tight. Finally, if x ∈ X, there is an E ∈ Σ such that x ∈ int E and λE = µE < ∞, so λ is locally finite. Thus λ is a Radon measure. Q Q So (i) is true. (c) Finally, suppose that µ is totally finite and (iii) is true. Then we can appeal directly to 416O to see that (i) is true. 416Q Proposition (a) Let X be a compact Hausdorff space and E the algebra of open-and-closed subsets of X. Then any non-negative finitely additive functional from E to R has an extension to a Radon measure on X. If X is zero-dimensional then the extension is unique. (b) Let A be a Boolean algebra, and Z its Stone space. Then there is a one-to-one correspondence between non-negative additive functionals ν on A and Radon measures µ on Z given by the formula νa = µb a for every a ∈ A, where for a ∈ A I write b a for the corresponding open-and-closed subset of Z. proof (a) Let ν : E → [0, ∞[ be a non-negative additive functional. Then ν satisfies the conditions of 416O (because every member of E is closed, while X is compact), so has an extension to a Radon measure µ. If X is zero-dimensional, E is a base for the topology of X closed under finite unions and intersections, so µ is unique, by 415H(iv) or 415H(v). (b) The map a 7→ b a is a Boolean isomorphism between A and the algebra E of open-and-closed subsets of Z, so we have a one-to-one correspondence between non-negative additive functionals ν on A and nonnegative additive functionals ν 0 on E defined by the formula ν 0 b a = νa. Now Z is compact, Hausdorff and zero-dimensional, so ν 0 has a unique extension to a Radon measure on Z, by part (a). And of course every Radon measure µ on Z gives us a non-negative additive functional µ¹E on E, corresponding to a non-negative additive functional on A. 416R Theorem (a) Any subspace of a Radon measure space is a quasi-Radon measure space. (b) A measurable subspace of a Radon measure space is a Radon measure space. (c) If (X, T, Σ, µ) is a Hausdorff complete locally determined topological measure space, and Y ⊆ X is such that the subspace measure µY on Y is a Radon measure, then Y ∈ Σ. proof (a) Put 416A and 415B together. (b) Let (X, T, Σ, µ) be a Radon measure space, and (E, TE , ΣE , µE ) a member of Σ with the induced topology and measure. Because µ is complete and locally determined, so is µE (214Ja). Because T is Hausdorff, so is TE (4A2F(a-i)). Because µ is locally finite, so is µE . Because µ is tight (and a subset of E is compact for TE whenever it is compact for T), µE is tight (412Oa). (c) ?? If Y ∈ / Σ, then there is a set F ∈ Σ such that µ∗ (Y ∩ F ) < µ∗ (Y ∩ F ) (413F(v)). But now µ (Y ∩ F ) = µY (Y ∩ F ), so there is a compact set K ⊆ Y ∩ F such that µY K > µ∗ (Y ∩ F ). When regarded as a subset of X, K is still compact; because T is Hausdorff, K is closed, so belongs to Σ, and ∗

µ∗ (Y ∩ F ) ≥ µK = µY K > µ∗ (Y ∩ F ), which is absurd. X X 416S Corresponding to 415O, we have the following. Proposition Let (X, T, Σ, µ) be a Radon measure space, and ν an indefinite-integral measure over µ (definition: 234B). If ν is locally finite, it is a Radon measure. proof Because µ is complete and locally determined, so is ν (234Fb). Because µ is tight (412Q). So if ν is also locally finite, it is a Radon measure. 416T I said in the notes to §415 that the most important quasi-Radon measure spaces are subspaces of Radon measure spaces. I do not know of a useful necessary and sufficient condition, but the following deals with completely regular spaces.

416V

Radon measure spaces

83

Proposition Let (X, T, Σ, µ) be a locally finite completely regular Hausdorff quasi-Radon measure space. Then it is isomorphic, as topological measure space, to a subspace of a locally compact Radon measure space. ˇ proof (a) Write βX for the Stone-Cech compactification of X (4A2I); I will take it that X is actually a subspace of βX. Let U be the set of those open subsets U of βX such S that µ(U ∩ X) < ∞; then U is upwards-directed and covers X, because µ is locally finite. Set W = U ⊇ X. Then W is an open subset of βX, so is locally compact. (b) Let B be the Borel σ-algebra of W . Then V ∩ X is a Borel subset of X for every V ∈ B (4A3Ca), so we have a measure ν : B → [0, ∞] defined by setting νV = µ(X ∩ V ) for every V ∈ B. Now ν satisfies the conditions of 415Cb. P P (α) If νV > 0, then, because µ is effectively locally finite, there is an open set G ⊆ X such that µ(G ∩ V ) > 0 and µG < ∞. There is an open set U ⊆ βX such that U ∩ X = G, in which case U ⊆ W , νU < ∞ and ν(U ∩ V ) > 0. Thus ν is effectively locally finite. (β) If U is an upwards-directed family of open subsets of W , then {U ∩ X : U ∈ U} is an upwards-directed family of open subsets of X, so [ [ [ ν( U) = µ(X ∩ U ) = µ( {U ∩ X : U ∈ U}) = sup µ(X ∩ U ) = sup νU. Q Q U ∈U

U ∈U

So the c.l.d. version ν˜ of ν is a quasi-Radon measure on W (415Cb). (c) The construction of W ensures that ν and ν˜ are locally finite. By 416G, ν˜ is a Radon measure. So the subspace measure ν˜X is a quasi-Radon measure on X (416Ra). But ν˜X G = µG for every open set G ⊆ X. P P Note first that as ν effectively locally finite, therefore semi-finite, ν˜ extends ν (213Hc). If K ⊆ W is a compact set not meeting X, then ν˜K = νK = µ(K ∩ X) = 0; accordingly ν˜∗ (W \ X) = 0, by 413Ee. Now there is an open set U ⊆ W such that G = X ∩ U , and ν˜X G = ν˜∗ G ≤ ν˜U = νU = µ(U ∩ X) = µG = ν˜∗ (U ∩ X) + ν˜∗ (U \ X) (by 413E(c-ii), because ν˜ is semi-finite) ≤ ν˜∗ (U ∩ X) + ν˜∗ (W \ X) = ν˜∗ G. Q Q So 415H(iii) tells us that µ = ν˜X is the subspace measure induced by ν. 416U Theorem If h(Xi , Ti , Σi , µi )ii∈I is a family of compact metrizable Radon probability spaces such Q that every µi is strictly positive, the product measure on X = i∈I Xi is a completion regular Radon measure. In particular, the usual measures on {0, 1}I and [0, 1]I and PI are completion regular Radon measures, for any set I. proof By 415E, it is a completion regular quasi-Radon probability measure; but X is a compact Hausdorff space, so it is a Radon measure, by 416G or otherwise. Remark I suppose it is obvious that by the ‘usual measure on [0, 1]I ’ I mean the product measure when each copy of [0, 1] is given Lebesgue measure. Recall also that the ‘usual measure on PI’ is just the copy of the usual measure on {0, 1}I induced by the standard bijection A ↔ χA (254Jb), which is a homeomorphism (4A2Ud). 416V Stone spaces The results of 415Q-415R become simpler and more striking in the present context. Theorem Let (X, T, Σ, µ) be a Radon measure space, and (Z, S, T, ν) the Stone space of its measure algebra (A, µ ¯). For E ∈ Σ let E ∗ be the open-and-closed set in Z corresponding to the image E • of E in A. Define R ⊆ Z × X by saying that (z, x) ∈ R iff x ∈ F whenever F ⊆ X is closed and z ∈ F ∗ .

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S (a) R is the graph of a function f : Q → X, where Q = R−1 [X]. If we set W = {K ∗ : K ⊆ X is compact}, then W ⊆ Q is a ν-conegligible open set, and the subspace measure νW on W is a Radon measure. (b) Setting g = f ¹ W , g is continuous and µ is the image measure νW g −1 . (c) If X is compact, W = Q = Z and µ = νg −1 . proof (a) By 415Ra, R is the graph of a function. If K ⊆ X is compact and z ∈ K ∗ , then F = {F : F ⊆ X is closed, z ∈ F ∗ } is a family of non-empty closed subsets of X, closed under finite intersections, and containing the compact set K; so it has non-empty intersection, and there is an x ∈ K such that (z, x) ∈ R, that is, z ∈ Q and f (z) ∈ K. Thus W ⊆ Q. Of course W is an open set, being the union of a family of open-and-closed sets; but it is also conegligible, because sup{K • : K ⊆ X is compact} = 1 in A (412N), so Z \ W must be nowhere dense, therefore negligible. Now the subspace measure νW is quasi-Radon because ν is (411P(d-iv), 415B); but W is a union of compact open sets of finite measure, so νW is locally finite and W is locally compact; by 416G, νW is a Radon measure. (b) g is continuous. P P Let G ⊆ X be an open set and z ∈ g −1 [G]. Let K ⊆ X be a compact set such ∗ that z ∈ K . As remarked above, g(z) = f (z) belongs to K. K, being a compact Hausdorff space, is regular (3A3Bb), so there is an open set H containing g(z) such that L = H ∩ K ⊆ G. Note that L is compact, so L∗ ⊆ W . Now g(z) does not belong to the closed set X \ H, so z ∈ / (X \ H)∗ and z ∈ H ∗ ; accordingly ∗ ∗ ∗ ∗ −1 z ∈ (H ∩ K) ⊆ L . If w ∈ L , g(w) ∈ L ⊆ G; so L ⊆ g [G], and z ∈ int g −1 [G]. As z is arbitrary, g −1 [G] is open; as G is arbitrary, g is continuous. Q Q By 415Rb, we know that µ = νQ f −1 , where νQ is the subspace measure on Q. But as ν is complete and both Q and W are conegligible, we have νQ f −1 [A] = νf −1 [A] = νg −1 [A] = νW g −1 [A] whenever A ⊆ X and any of the four terms is defined, so that µ = νQ f −1 = νW g −1 . (c) If X is compact, then Z = X ∗ ⊆ W , so W = Q = Z and νg −1 = νW g −1 = µ. 416W Compact measure spaces Recall that a semi-finite measure space (X, Σ, µ) is ‘compact’ (as T a measure space) if there is a family K ⊆ Σ such that µ is inner regular with respect to K and K0 6= ∅ whenever K0 ⊆ K has the finite intersection property (342A); while (X, Σ, µ) is ‘perfect’ if whenever f : X → R is measurable and µE > 0, there is a compact set K ⊆ f [E] such that µf −1 [K] > 0 (342K). In §342 I introduced these concepts in order to study the realization of homomorphisms between measure algebras. The following result is now very easy. Proposition (a) Any Radon measure space is a compact measure space, therefore perfect. (b) Let (X, T, Σ, µ) be a Radon measure space, with measure algebra (A, µ ¯), and (Y, T, ν) a complete strictly localizable measure space, with measure algebra (B, ν¯). If π : A → B is an order-continuous Boolean homomorphism, there is a function f : Y → X such that f −1 [E] ∈ T and f −1 [E]• = πE • for every E ∈ Σ. If π is measure-preserving, f is inverse-measure-preserving. proof (a) If (X, T, Σ, µ) is a Radon measure space, µ is inner regular with respect to the compact class consisting of the compact subsets of X, so (X, Σ, µ) is a compact measure space. By 342L, it is perfect. (b) Use (i)⇒(v) of Theorem 343B. (Of course f is inverse-measure-preserving iff π is measure-preserving.) 416X Basic exercises > (a) Let (X, T, Σ, µ) be a Radon measure space, and E ∈ Σ an atom for the measure. Show that there is a point x ∈ E such that µ{x} = µE. (b) Let X be a topological space and µ a point-supported measure on X, as described in 112Bd. (i) Show that µ is tight, so is a Radon measure iff it is locally finite. In particular, show that if X has the discrete topology then counting measure on X is a Radon measure. (ii) Show that every purely atomic Radon measure is of this type. (c) Let h(Xi , Ti , Σi , µi )ii∈I be a family of Radon measure spaces, with direct sum (X, Σ, µ) (214K). Give X its disjoint union topology. Show that µ is a Radon measure.

416Xr

Radon measure spaces

85

(d) Let (X, T) be a Hausdorff space, and µ1 , µ2 two Radon measures on X, with domains Σ1 , Σ2 . (i) Show that µ1 + µ2 , defined on Σ1 ∩ Σ2 (112Xe), is a Radon measure. (ii) Show that αµ1 , defined on Σ1 , is a Radon measure for any α > 0. (e) Let (X, T, Σ, µ) be a Radon measure space. (i) Show that µ has a decomposition hXi ii∈I in which every Xi except at most one is a self-supporting compact set, and the exceptional one, if any, is negligible. (ii) Show that µ has a decomposition hXi ii∈I in which every Xi is expressible as the intersection of a S closed set with an open set. (Hint: enumerate the open sets of finite measure as hGξ iξ<κ , and set Xξ = Gξ \ η<ξ Gη .) (f ) Let X be a Hausdorff space and µ, ν two Radon measures on X such that νG = µG whenever G ⊆ X is open and min(µG, νG) < ∞. Show that µ = ν. (g) Explain how to prove 416H from 416F, without appealing to §415. (h) Give a direct proof of 416I not relying on 415O. (i) Let (X, T) be a completely regular topological space and µ a locally finite topological measure on X which is inner regular with respect to the closed sets. Show that µK = inf{µG : G ⊇ K is a cozero set}

Z

= inf{µF : F ⊇ K is a zero set} = inf{

f dµ : χK ≤ f ∈ C(X)}

for every compact set K ⊆ X. (j) Let (X, T, Σ, µ) be a locally compact Radon measure space, and Ck the space of continuous realvalued functions on X with compact supports. Show that {f • : f ∈ Ck } is dense in L0 (µ) for the topology of convergence in measure. (k) Let X be a completely regular Hausdorff space and ν a locally finite Baire measure on X. (i) Show that ν ∗ K = inf{νG : G ⊆ X is a cozero set, K ⊆ G} for every compact set K ⊆ X. (ii) Show that there is a Radon measure µ on X such that µK = ν ∗ K for every compact set K ⊆ X. (Hint: in the language of the proof of 416K, φ1 = ν ∗ ¹K.) (l) Let X be a Hausdorff space and ν a non-negative finitely additive functional defined on some algebra of subsets of X. Show that there is a Radon measure µ on X such that µX ≤ νX and µK ≥ νE whenever E ∈ T, K ⊆ X is compact and E ⊆ K. (Hint: start by extending ν to PX.) (m) Let (X, T) be a Hausdorff space, Σ ⊇ T a σ-algebra of subsets of X, and ν : Σ → [0, ∞[ a finitely additive functional such that νE = sup{νK : K ⊆ E is compact} for every E ∈ Σ. Show that ν is countably additive and that its completion is a Radon measure on X. (n) Explain how to prove 416Rb from 416C and 415B. (o) Let X be a Hausdorff space, µ a complete locally finite measure on X, and Y a conegligible subset of X. Show that µ is a Radon measure iff the subspace measure on Y is a Radon measure. (p) Let X be a Hausdorff space, Y a subset of X, and ν a Radon measure on Y . Define a measure µ on X by setting µE = ν(E ∩ Y ) whenever ν measures E ∩ Y . Show that if either Y is closed or ν is totally finite, µ is a Radon measure on X. (Cf. 418I.) (q) Let (X, T, Σ, µ) be a Radon measure space and E ⊆ Σ a non-empty upwards-directed family. Set νF = supE∈E µ(E ∩ F ) whenever F ⊆ X is such that µ measures E ∩ F for every E ∈ E. Show that ν is a Radon measure on X. (r) Let hXn in∈N be a sequence of Hausdorff spaces with product X; write B(Xn ) for the Borel σ-algebra N of Xn . Let T be the σ-algebra c n∈N B(Xn ) (definition: 254E). Let ν : T → [0, ∞[ be a finitely additive functional such that E 7→ νπn−1 [E] : B(Xn ) → [0, ∞[ is countably additive and tight for each n ∈ N, writing πn (x) = x(n) for x ∈ X, n ∈ N. Show that there is a unique Radon measure on X extending ν.

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Topologies and measures I

416Xs

S (s) Set S2∗ = n∈N {0, 1}n , and let φ : S2∗ → [0, ∞[ be a functional such that φ(σ) = φ(σ a 0) + φ(σ a 1) for every σ ∈ S2∗ , writing σ a 0 and σ a 1 for the two members of {0, 1}n+1 extending any σ ∈ {0, 1}n . Show that there is a unique Radon measure µ on {0, 1}N such that µ{x : x¹{0, . . . , n − 1} = σ} = φ(σ) whenever n ∈ N, σ ∈ {0, 1}N . (Hint: use 416Xr or 416Q.) (t) Let (X, T, Σ, µ) be a Radon measure space. Show that a measure ν on X is an indefinite-integral measure over µ iff (α) ν is a complete, locally determined topological measure (β) ν is tight (γ) νK = 0 whenever K ⊆ X is compact and µK = 0. (u) Let (X, T, Σ, µ) be a compact Hausdorff quasi-Radon measure space. Let W ⊆ X be the union of the open subsets of X of finite measure. Show that the subspace measure on W is a Radon measure. (v) Let (X, T, Σ, µ) be a completely regular Radon measure space. Show that it is isomorphic, as topological measure space, to a measurable subspace of a locally compact Radon measure space. (w) Let (X, T, Σ, µ) be a compact Radon measure space and (Z, S, T, ν) the Stone space of its measure algebra. For E ∈ Σ let E ∗ be the corresponding open-and-closed subset of Z, as in 416V. Show that the function described in 416V is the unique continuous function h : Z → X such that ν(E ∗ 4h−1 [E]) = 0 for every E ∈ Σ. (x) Show that the right-facing Sorgenfrey line (415Xc), with Lebesgue measure, is a quasi-Radon measure space which, regarded as a measure space, is compact, but, regarded as a topological measure space, is not a Radon measure space. 416Y Further exercises (a) Let X be a Hausdorff space and ν a countably additive real-valued functional defined on a σ-algebra Σ of subsets of X. Show that the following are equiveridical: (i) |ν| : Σ → [0, ∞[, defined as in 362B, is a Radon measure on X; (ii) ν is expressible as µ1 − µ2 , where µ1 , µ2 are Radon measures on X and Σ = dom µ1 ∩ dom µ2 . (b) Let X be a topological space and µ0 a semi-finite measure on X which is inner regular with respect to the family Kccc of closed countably compact sets. Show that µ0 has an extension to a complete locally determined topological measure µ on X, still inner regular with respect to Kccc ; and that the extension may be done in such a way that whenever µE < ∞ there is an E0 ∈ Σ0 such that µ(E4E0 ) = 0. (Hint: use the argument of 416M, but with K = {K : K ∈ Kccc , µ∗0 K < ∞}.) (c) Let X be a topological space and µ0 a semi-finite measure on X which is inner regular with respect to the family Ksc of sequentially compact sets. Show that µ0 has an extension to a complete locally determined topological measure µ on X, still inner regular with respect to Ksc ; and that the extension may be done in such a way that whenever µE < ∞ there is an E0 ∈ Σ0 such that µ(E4E0 ) = 0. P 1 (d) Let X ⊆ βN be the union of all those open sets G ⊆ βN such that n∈G∩N n+1 is finite. For E ⊆ X P 1 set µE = n∈E∩N n+1 . Show that µ is a σ-finite Radon measure on the locally compact Hausdorff space X. Show that µ is not outer regular with respect to the open sets. S P∞ (e) Set S ∗ = n∈N N n , and let φ : S ∗ → [0, ∞[ be a functional such that φ(σ) = i=0 φ(σ a i) for every σ ∈ S ∗ , writing σ a i for the members of N n+1 extending any σ ∈ N n . Show that there is a unique Radon measure µ on N N such that µIσ = φ(σ)P for every σ ∈ S ∗ , where Iσ S = {x : x¹{0, . . . , n − 1} = σ} for any N n ∈ N, σ ∈ N . (Hint: set θA = inf{ σ∈R φ(σ) : R ⊆ S ∗ , A ⊆ σ∈R Iσ } for every A ⊆ N N , and use Carath´eodory’s method.) (f ) Let (X, T, Σ, µ) be a Radon measure space. Show that a measure ν on X is an indefinite-integral measure over µ iff (i) there is a topology S on X, including T, such that ν is a Radon measure with respect to S (ii) νK = 0 whenever K is a T-compact set and µK = 0. (g) Let hxn in∈N enumerate subset of X = {0, 1}c (4A2B(e-ii)). Let λ be the usual measure on P∞ a dense 1 −n−2 χE(xn ) for E ∈ dom λ. (i) Show that µ is a strictly positive Radon X, and set µE = 2 λE + n=0 2

416 Notes

Radon measure spaces

87

probability measure on X of Maharam type c. (ii) Let I ∈ [c]≤ω be such that xm ¹I 6= xn ¹I whenever I m that R 6= n. Set Z = {0, 1} and let π : X → Z be the canonical map. Show that if f ∈ C(X) is such f × gπ dµ = 0 for every g ∈ C(Z), then f R= 0. (Hint: otherwise, take n ∈ N Rsuch that |f (xn )| ≥ 12 kf k∞ , and let g ≥ 0 be such that g(πxn ) = 1 and g d(µπ −1 ) < 3 · 2−n−3 ; show that f × gπ dµ > 0.) (iii) Show that there is no orthonormal basis for L2 (µ) in {f • : f ∈ C(X)}. (See Hart & Kunen 99.) 416 Notes and comments The original measures studied by Radon (Radon 13) were, in effect, what I call differences of Radon measures on R r , as introduced in §256. Successive generalizations moved first to Radon measures on general compact Hausdorff spaces, then to locally compact Hausdorff spaces, and finally to arbitrary Hausdorff spaces, as presented in this section. I ought perhaps to remark that, following Bourbaki 65, many authors use the term ‘Radon measure’ to describe a linear functional on a space of continuous functions; I will discuss the relationship between such functionals and the measures of this chapter in §436. For the moment, observe that by 415I a Radon measure on a completely regular space can be determined from the integrals it assigns to continuous functions. It is also common for the phrase ‘Radon measure’ to be used for what I would call a tight Borel measure; you have to check each author to see whether local finiteness is also assumed. In my usage, a Radon measure is necessarily the c.l.d. version of a Borel measure. The Borel measures which correspond to Radon measures are described in 416F. In §256, I discussed Radon measures on R r as a preparation for a discussion of convolutions of measures. It should now be becoming clear why I felt that it was impossible, in that context, to give you a proper idea of what a Radon measure, in the modern form, ‘really’ is. In Euclidean space, too many concepts coincide. As a trivial example, the simplest definition of ‘local finiteness’ (256Ab, 411Fa) is not the right formulation in other spaces. Next, because every closed set is a countable union of compact sets, there is no distinction between ‘inner regular with respect to closed sets’ and ‘inner regular with respect to compact sets’, so one cannot get any intuition for which is important in which arguments. (When we come to subspace measures on non-measurable subsets, of course, this changes; quasi-Radon measures on subsets of Euclidean space are important and interesting.) Third, the fact that the c.l.d. product of two Radon measures on Euclidean space is already a Radon measure (256K) leaves us with no idea of what to do with a general product of Radon measures. (There are real difficulties at this point, which I will attack in the next section. For the moment I offer just 416U.) And finally, we simply cannot represent a product of uncountably many Radon probability measures on Euclidean spaces as a measure on Euclidean space. As you would expect, a very large proportion of the results of this chapter, and many theorems from earlier volumes, were originally proved for compact Radon measure spaces. The theory of general totally finite Radon measures is, in effect, the theory of measurable subspaces of compact Radon measure spaces, while the theory of quasi-Radon measures is pretty much the theory of non-measurable subspaces of Radon measure spaces. Thus the theorem that every quasi-Radon measure space is strictly localizable is almost a consequence of the facts that every Radon measure space is strictly localizable and any subspace of a strictly localizable space is strictly localizable. The cluster of results between 416J and 416P form only a sample, I hope a reasonably representative sample, of the many theorems on construction of Radon measures from functionals on algebras or lattices of sets. (See also 416Ye.) The essential simplification, compared with the theorems in §413 and §415, is that we do not need to mention any σ- or τ -additivity condition of the type 413I(β) or 415K(β), because we are dealing with a ‘compact class’, the family of compact subsets of a Hausdorff space. We can use this even at some distance, as in 416O (where the hypotheses do not require any non-empty compact set to belong to the domain of the original functional). The particular feature of 416O which makes it difficult to prove from such results as 413J and 413O above is that we have to retain control of the outer measures of a sequence hKn in∈N of non-measurable sets. In general this is hard to do, and is possible here principally because the sequence is non-decreasing, so that we can make sense of the functionals νn E = ν ∗ (E ∩ Kn+1 ) − ν ∗ (E ∩ Kn ).

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§417 intro.

417 τ -additive product measures The ‘ordinary’ product measures introduced in Chapter 25 have served us well for a volume and a half. But we come now to a fundamental obstacle. If we start with two Radon measure spaces, their product measure, as defined in §251, need not be a Radon measure (419E). Furthermore, the counterexample is one of the basic compact measure spaces of the theory; and while it is dramatically non-metrizable, there is no other reason to set it aside. Consequently, if we wish (as we surely do) to create Radon measure spaces as products of Radon measure spaces, we need a new construction. This is the object of the present section. It turns out that the construction can be adapted to work well beyond the special context of Radon measure spaces; the methods here apply to general effectively locally finite τ -additive topological measures (for the product of finitely many factors) and to τ -additive topological probability measures (for the product of infinitely many factors). Elsewhere in this volume I will show how these constructions are related to some others which have been described for special spaces. The fundamental theorems are 417C and 417E, listing the essential properties of what I call ‘τ -additive product measures’, which are extensions of the c.l.d. product measures and product probability measures of Chapter 25. They depend on a straightforward lemma on the extension of a measure to make every element of a given class of sets negligible (417A). It is relatively easy to prove that the extensions are more or less canonical (417D, 417F). We still have Fubini’s theorem for the new product measures (417H), and the basic operations from §254 still apply (417J, 417K, 417M). It is easy to check that if we start with quasi-Radon measures, then the τ -additive product measure is again quasi-Radon (417N, 417O). The τ -additive product of two Radon measures is Radon (417P), and the τ -additive product of Radon probability measures with compact supports is Radon (417Q). In the last part of the section I look at continuous real-valued functions and Baire σ-algebras; it turns out that for these the ordinary product measures are adequate (417U, 417V). 417A Lemma Let (X, Σ, µ) be a semi-finite measure space, and A ⊆ PX a family of sets such that S µ∗ ( n∈N An ) = 0 for every sequence hAn in∈N in A. Then there is a measure µ0 on X, extending µ, such that (i) µ0 A is defined and zero for every A ∈ A, (ii) µ0 is complete if µ is, (iii) for every F in the domain Σ0 of µ0 there is an E ∈ Σ such that µ0 (F 4E) = 0. In particular, µ and µ0 have isomorphic measure algebras, so that µ0 is localizable if µ is. proof Let A∗ be the collection of subsets of X which can be covered by a countable subfamily of A. Then A∗ is a σ-ideal of subsets of X and µ∗ A = 0 for every A ∈ A∗ . Set Σ0 = {E4A : E ∈ Σ, A ∈ A∗ }. Then Σ0 is a σ-algebra of subsets of X. P P (i) ∅ = ∅4∅ ∈ Σ0 . (ii) If E ∈ Σ, A ∈ A∗ then X \ (E4A) = 0 (X \ E)4A ∈ Σ . (iii) If hEn in∈N , hAn in∈N are sequences in Σ, A∗ respectively, then S S S E = n∈N En ∈ Σ, A = E4 n∈N (En 4An ) ⊆ n∈N An ∈ A∗ , S Q so n∈N (En 4An ) = E4A ∈ Σ0 . Q If E, E 0 ∈ Σ, A, A0 ∈ A∗ and E4A = E 0 4A0 , then E4E 0 = A4A0 ∈ A∗ and µ∗ (E4E 0 ) = 0; because µ is semi-finite, µ(E4E 0 ) = 0 and µE = µE 0 . Accordingly we can define µ0 : Σ0 → [0, ∞] by setting µ0 (E4A) = µE whenever E ∈ Σ, A ∈ A∗ . Evidently µ0 extends µ and µ0 A = 0 for every A ∈ A. Also µ0 is a measure. P P (i) µ0 ∅ = µ∅ = 0. (ii) If 0 ∗ hFn in∈N S is a disjoint sequence in ∗Σ , with0 union F , express each Fn as En 4An where En ∈ Σ, An ∈ A ; set E = n∈N En , so that F 4E ∈ A and µ F = µE. If m 6= n, then Em ∩ En ⊆ Am ∪ An , so µ(Em ∩ En ) = 0; accordingly P∞ P∞ µ0 F = µE = n=0 µEn = n=0 µ0 Fn . Q Q A subset of X is µ0 -negligible iff it can be included in a set of the form E4A where µE = 0 and A ∈ A∗ , so µ0 is complete if µ is. The embedding Σ ⊆ Σ0 induces a measure-preserving homomorphism from the measure algebra of µ to the measure algebra of µ0 which is an isomorphism just because every member of Σ0 is the symmetric difference of a member of Σ and a µ0 -negligible set.

417C

τ -additive product measures

89

417B Lemma Let X and Y be topological spaces, and ν a τ -additive topological measure on Y . (a) If W ⊆ X × Y is open, then x 7→ νW [{x}] : X → [0, ∞] is lower semi-continuous. (b) If ν is effectively locally finite and σ-finite and W ⊆ X × Y is a Borel set, then x 7→ νW [{x}] is Borel measurable. R (c) If f : X × Y → [0, ∞] is a lower semi-continuous function, then x 7→ f (x, y)ν(dy) : X → [0, ∞] is lower semi-continuous. R (d) If ν is totally finite and f : X × Y → R is a bounded continuous function, then x 7→ f (x, y)ν(dy) is continuous. (e) If ν is totally finite and W ⊆ X × Y is a Baire set, then x 7→ νW [{x}] is Baire measurable. proof (a) If x ∈ X and νW [{x}] > α, then H = {H : H ⊆ Y is open, there is an open set G containing x such that G × H ⊆ W } is an upwards-directed family of open sets with union W [{x}], so there is an H ∈ H such that νH ≥ α. Now there is an open set G containing x such that G × H ⊆ W , so that νW [{x0 }] ≥ α for every x0 ∈ G. (b)(i) Suppose to begin with that ν is totally finite. In this case, the set {W : W ⊆ X × Y, x 7→ νW [{x}] is a Borel measurable function defined everywhere on X} is a Dynkin class containing every open set, so contains every Borel set, by the Monotone Class Theorem (136B). (ii) For the general case, let hYn in∈N be a disjoint sequence of sets of finite measure covering Y , and for n ∈ N let νn be the subspace measure on Yn . Then νn is effectively locally finite and τ -additive (414K). If W ⊆ X × Y is a Borel set, then Wn = W ∩ (X × Yn ) isPa relatively Borel set for each n, so that ∞ x 7→ νn Wn [{x}] is Borel measurable, by (i). Since νW [{x}] = n=0 νn Wn [{x}] for every x, x 7→ νW [{x}] is Borel measurable. (c) For i, n ∈ N set Wni = {(x, y) : f (x, y) > 2−n i}, so that Wni ⊆ X × Y is open. Set fn = P 4n sequence with supremum f . For n ∈ N and x ∈ X, 2 i=1 χWni ; then hfnnin∈N is a non-decreasing R R P4 −n fn (x, y)ν(dy) is lower semi-continuous, by (a) and 4A2B(dfn (x, y)ν(dy) =R 2 i=1 νWni [{x}], soRx 7→ R iii). By 414Ba, f (x, y)ν(dy) = supn∈N fn (x, y)ν(dy) for every x, so x 7→ f (x, y)ν(dy) is lower semicontinuous (4A2B(d-v)). R (d) Applying (c) to f +kf k∞ χ(X ×Y ), we see that x 7→ f (x, y)ν(dy) is lower semi-continuous. Similarly, R R x 7→ − f (x, y)ν(dy) is lower semi-continuous, so x 7→ f (x, y)ν(dy) is continuous (4A2B(d-vi)). −n

(e) Suppose first that W is a cozero set; let f : X × Y → [0, 1] be a continuous function such that W = {(x, y) : f (x, y) > 0}. For n ∈ N set fn = nf ∧ χ(X × Y ). Then hfn in∈N R is a non-decreasing sequence of continuous functions with supremum χW . By (d), all the functions x 7→ fn (x, y)ν(dy) are continuous, so their limit x 7→ νW [{x}] is Baire measurable. Now {W : W ⊆ X × Y, x 7→ νW [{x}] is a Baire measurable function defined everywhere on X} is a Dynkin class containing every cozero set, so contains every Baire set, by the Monotone Class Theorem again. 417C Theorem Let (X, T, Σ, µ) and (Y, S, T, ν) be effectively locally finite τ -additive topological measure spaces, with c.l.d. product (X × Y, Λ, λ). Then λ has an extension to a τ -additive topological measure ˜ on X × Y . Moreover, we can arrange that: λ ˜ is complete, locally determined and effectively locally finite, therefore strictly localizable; (i) λ ˜ there is a Q1 ∈ Λ such that λ(Q4Q ˜ ˜ of λ, (ii) if Q belongs to the domain Λ 1 ) = 0; that is to say, the ˜ ˜ embedding Λ ⊆ Λ induces an isomorphism between the measure algebras of λ and λ;

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Topologies and Measures

417C

˜ then (iii) if Q ∈ Λ, ˜ = sup{λ(Q ˜ ∩ (G × H)) : G ∈ T, µG < ∞, H ∈ S, νH < ∞}; λQ ˜ , and (iv) if W ⊆ X × Y is open, then there is an open set W0 ∈ Λ such that W0 ⊆ W and λW0 = λW ˜ = λ∗ W = λW

R

νW [{x}]µ(dx) =

R

µW −1 [{y}]ν(dy);

˜ × H) = µG · νH for every G ∈ T, H ∈ S; in particular, λ(G (v) the support of λ is the product of the supports of µ and ν; ˜ (vi) if µ and ν are both inner regular with respect to the Borel sets, so is λ; ˜ (vii) if µ and ν are both inner regular with respect to the closed sets, so is λ; ˜ (viii) if µ and ν are both tight (that is, inner regular with respect to the closed compact sets), so is λ. proof Write Σf = {E : E ∈ Σ, µE < ∞}, Tf = T ∩ Σf ,

Tf = {F : F ∈ T, νF < ∞}, Sf = S ∩ Tf .

(a) Let U be {G×H : G ∈ Tf , H ∈ Sf }. Because T ⊆ Σ and S ⊆ T, U ⊆ Λ. U need not be a base for the topology of X × Y , unless µ and ν are locally finite, but if an open subset of X × Y is included in a member of U it is the union of the members of U it includes. Moreover, if Q ∈ Λ, then λQ = supU ∈U λ(Q ∩ U ). P P S Q By 412R, λ is inner regular with respect to U ∈U PU . Q Write Us for the set of finite unions of members of U, and V for the set of non-empty upwards-directed families V ⊆ Us such that supV ∈V λV < ∞. For each V ∈ V, fix on a countable V 0 ⊆ V such that supV ∈V 0 λV = supV ∈V λV ; because V is upwards-directed, we may suppose that V 0 = {Vn : n ∈ N} for some S S non-decreasing sequence hVn in∈N in V. Set A(V) = V \ V 0 . (b)(i) For V ∈ Us , set fV (x) = νV [{x}] for every x ∈ X. This is always defined because V is open; moreover,RfV is lower semi-continuous, by 417Ba. Because V is a finite union of products of sets of finite measure, fV dµ = λV . (ii) The key to the proof is the following fact: for any V ∈ V, almost every vertical section of A(V) is negligible. P P hfV iV ∈V is a non-empty upwards-directed set of lower semi-continuous functions. Set S S g(x) = ν( V ∈V V [{x}]), h(x) = ν( V ∈V 0 V [{x}]) for every x ∈ X. Because V is upwards-directed and ν is τ -additive, g(x) = supV ∈V νV [{x}] = supV ∈V fV (x) in [0, ∞] for each x, so, by 414Ba,

R

g dµ = supV ∈V

R

fV dµ = supV ∈V λV = supV ∈V 0 λV =

R

h dµ.

Since h ≤ g and supV ∈V λV is finite, g(x) = h(x) < ∞ for µ-almost every x. But for any such x, we must have S S ν( V)[{x}] = ν( V 0 )[{x}] < ∞, so that A(V)[{x}] = (

S

V)[{x}] \ (

S

V 0 )[{x}]

is negligible. Q Q S (c) ?? Suppose, if possible, that there is a sequence hVn in∈N in V such that λ∗ ( n∈N A(Vn )) > 0. Take S W ∈ Λ such that W ⊆ n∈N A(Vn ) and λW > 0. Because almost every vertical section of every A(Vn ) is negligible, almost every vertical section of W is negligible. But this contradicts Fubini’s theorem (252F). X X (d) We may therefore apply the construction of 417A to form a measure λ0 on the σ-algebra Λ0 = {W 4A : ˜ be the c.l.d. version of λ0 W ∈ Λ, A ∈ A∗ }, where A∗ is the σ-ideal generated by {A(V) : V ∈ V}. Let λ ˜ its domain. (213E), and Λ

417C

τ -additive product measures

91

˜ Also, because λ is semi-finite, (i) If W ∈ Λ, then W ∈ Λ0 ⊆ Λ. λ0 W = λW = sup{λW 0 : W 0 ⊆ W, W ∈ Λ, λW 0 < ∞} ˜ ≤ λ0 W. ≤ sup{λ0 W 0 : W 0 ⊆ W, W ∈ Λ0 , λ0 W 0 < ∞} = λW ˜ ; as W is arbitrary, λ ˜ extends λ. Thus λW = λW ˜ there is a U ∈ U such ˜ and γ < λQ, (ii) It will be useful if we go directly to one of the targets: if Q ∈ Λ 0 0 ˜ that λ(Q ∩ U ) ≥ γ. P P There is a Q1 ∈ Λ such that Q1 ⊆ Q and γ < λ Q1 < ∞. There is a Q2 ∈ Λ such that λ0 (Q1 4Q2 ) = 0, so that λQ2 = λ0 Q2 = λ0 Q1 > γ. There is a U ∈ U such that λ(Q2 ∩ U ) ≥ γ, by (a). Now ˜ ∩ U ) ≥ λ0 (Q1 ∩ U ) = λ0 (Q2 ∩ U ) = λ(Q2 ∩ U ) ≥ γ. Q λ(Q Q ˜ is a topological measure. P ˜ > 0. ˜ and λQ (iii) λ P Let W ⊆ X × Y be an open set. Suppose that Q ∈ Λ ˜ By (ii), there is a U ∈ U such that λ(Q ∩ U ) > 0. Let V be {V : V ∈ Us , V ⊆ W ∩ U }. Then V ∈ V, so S ˜ λA(V) = λ0 A(V) = 0; since V 0 ∈ Λ, S ˜ W ∩ U = V ∈ Λ0 ⊆ Λ. But this means that ˜ ∗ (Q ∩ W ) + λ ˜ ∗ (Q \ W ) ≥ λ ˜ ∗ (Q ∩ U ∩ W ) + λ ˜ ∗ (Q ∩ U \ W ) = λ(Q ˜ ∩ U ) > 0. λ ˜ is complete and locally determined, and Q is arbitrary, this is enough to ensure that W ∈ Λ ˜ Because λ (413F(vii)). Q Q ˜ is τ -additive. P (iv) λ P?? Suppose, if possible, otherwise; that there is a non-empty upwards-directed S ∗ ˜ ∗ > γ = sup ˜ family W of open sets in X × Y such that λW W. In this case, we W ∈W λW , where W = can find a Q0 ∈ Λ0 such that Q0 ⊆ W ∗ and λ0 Q0 > γ, a Q1 ∈ Λ such that λ0 (Q0 4Q1 ) = 0, and a U ∈ U such that λ(Q1 ∩ U ) > S γ (using (a) again). Let V ∈ V be the set of those V ∈ Us such that V ⊆ W ∩ U for some W ∈ W. Then V = W ∗ ∩ U , so

γ < λ(Q1 ∩ U ) = λ0 (Q1 ∩ U ) = λ0 (Q0 ∩ U ) [ [ ∗ ˜ ˜ ˜ ≤ λ(W ∩ U ) = λ( V) = λ( V 0) ˜ (because λA(V) = 0) ˜ = sup λV V ∈V 0 0

(because V is countable and upwards-directed) ˜ ≤ γ, ≤ sup λW W ∈W

which is absurd. X XQ Q (e) Now for the supplementary properties (i)-(viii), in order. ˜ was constructed to be complete and locally determined. If λQ ˜ > 0, then by (d-ii) there is a U ∈ U (i) λ ˜ ˜ ˜ is effectively locally finite. such that λ(Q ∩ U ) > 0; since U is open and λU = λU is finite, this shows that λ By 414J, it is strictly localizable. (ii) The point is that λ is also (strictly) localizable. P P Let µ ˜, ν˜ be the c.l.d. versions of µ and ν. These are τ -additive topological measures (because µ and ν are), complete and locally determined (by construction), and are still effectively locally finite (cf. 412Ha), so are strictly localizable (414J again). Now λ is the c.l.d. product of µ ˜ and ν˜ (251S), therefore strictly localizable (251N). Q Q

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417C

˜ ˜ differs by a λ-negligible As remarked in 417A, it follows that λ0 is localizable, so that every member Q of Λ 0 0 set from a member Q0 of Λ (213Hb). Now there is a Q1 ∈ Λ such that λ (Q1 4Q0 ) = 0, in which case ˜ 1 4Q) = 0. λ(Q (iii) This is just (d-ii) above. (iv) Let W ⊆ X × Y be an open set. Set V = {V : V ∈ Us , V ⊆ W }. Then V is upwards-directed and has union W ∩ (X ∗ × Y ∗ ), where X ∗ and Y ∗ are the unions of the open sets of finite measure in X and Y respectively. Because µ and ν are effectively locally finite, X ∗ and Y ∗ are both conegligible. Now, as before, ˜ = λ(W ˜ ˜ = sup ˜ λW ∩ (X ∗ × Y ∗ )) = supV ∈V λV V ∈V λV ≤ λ∗ W ≤ λW . S If we take a countable set V0 ⊆ V such that supV ∈V0 λV = supV ∈V λV , and set W0 = V0 , then W0 is an ˜ . open set, belonging to λ and included in W , and λW0 = λW R Next, defining the functions fV as in part (a) of this proof, we have fV dµ = λV for every V ∈ V; and setting g(x) = νW [{x}], we have g(x) = supV ∈V fV (x) for every x ∈ X ∗ . So 414Ba tells us that

R

νW [{x}]µ(dx) =

R

R

g dµ = supV ∈V

R

˜ . fV dµ = supV ∈V λV = λW

˜ . µW −1 [{y}]ν(dy) = λW

(The point here is that while the arguments of part (b) of this proof give different roles to µ and ν, the actual construction performed in part (d) is symmetric between them.) Now, given G ∈ T and H ∈ S, set W = G × H; then R R ˜ × H) = µW −1 [{y}]ν(dy) = λ(G µGν(dy) = µG · νH. H

(v) Let E and F be the supports of µ, ν respectively. (By 411Nd these are defined.) Then E × F is a closed subset of X × Y . Because X \ E and Y \ F are negligible, E × F is conegligible. If W ⊆ X × Y is an open set and (x, y) ∈ W ∩ (E × F ), there are open sets G ⊆ X, H ⊆ Y such that (x, y) ∈ G × H ⊆ W . Now λ(W ∩ (E × F )) ≥ λ((G × H) ∩ (E × F )) = µ(G ∩ E) · ν(H ∩ F ) > 0. This shows that E × F is self-supporting, so is the support of λ. (vi), (vii), (viii) The same method works for all of these. Take K to be either the family of Borel subsets of X × Y (for (vi)) or the family of closed subsets of X × Y (for (vii)) or the family of closed compact subsets of X × Y (for (viii)); the essential features of K, valid for all three cases, are that (α) K \ W ∈ K whenever K ∈ K and W ⊆ X × Y is open, (β) K ∪ K 0 ∈ K for all K, K 0 ∈ K, T (γ) n∈N Kn ∈ K for every sequence hKn in∈N in K. Now the hypotheses of each part are just what we need in order to be sure that λ is inner regular with ˜ also is, without assuming that λ ˜ is respect to K (412Sd, 412Sa, 412Sb), and I am trying to show that λ inner regular with respect to Λ, because this need not be true. 0 ˜ ˜ and γ < λQ; Take Q ∈ Λ then there is a Q0 ∈ Λ0 such that S Q0 ⊆ Q and γ < λ Q0 < ∞.0 We have a Q1 ∈ Λ and a sequence hVn in∈N in V such that Q0 4Q1 ⊆ n∈N A(Vn ). Of course λQ1 = λ Q0 > γ; set ² = 41 (λQ1 − γ) > 0. For each n ∈ N, take Kn , Ln ∈ K ∩ Λ such that S S Kn ⊆ Q1 ∩ Vn0 , λ((Q1 ∩ Vn0 ) \ Kn ) ≤ 2−n ²,

Set Kn0 = Kn ∪ (Ln \

S

λ(Q1 \ Ln ) ≤ 2−n ².

Ln ⊆ Q1 , Vn ), so that Kn0 ∈ K and

[ ˜ 1 \ K 0 ) ≤ λ(Q ˜ 1 \ Ln ) + λA(V ˜ ˜ λ(Q Vn0 \ Kn ) n ) + λ(Q1 ∩ n [ = λ(Q1 \ Ln ) + λ(Q1 ∩ Vn0 \ Kn ) ≤ 2−n+1 ². Now K =

T n∈N

Kn0 belongs to K and K ⊆ Q1 \

S n∈N

A(Vn ) ⊆ Q0 ⊆ Q

417D

τ -additive product measures

93

because Kn0 ⊆ Q1 \ A(Vn ) for every n. And λ(Q1 \ K) ≤

∞ X

˜ 1 \ K 0 ) ≤ 4², λ(Q n

n=0

˜ = λK ≥ γ. As Q and γ are arbitrary, λ ˜ is inner regular with respect to K. so λK 417D Multiple products Just as with the c.l.d. product measure (see 251W), we can apply the construction of 417C repeatedly to obtain measures on the products of finite families of τ -additive measure spaces. Proposition (a) Let h(Xi , Ti , Σi , µi )ii∈I be a finite family of effectively locally finite τ -additive topological measure spaces. Then there is a unique complete locally determined effectively locally finiteQτ -additive ˜ on X = Q Xi , inner regular with respect to the Borel sets, such that λ( ˜ topological measure λ i∈I i∈I Gi ) = Q µ G whenever G ∈ T for every i ∈ I. i i i∈I i i ˜ k is the product measure defined by the construction of (a) (b) If now hIk ik∈K is a partition of I, and λ Q Q ˜ with the on Zk = i∈Ik Xi for each k ∈ K, then the natural bijection between X and k∈K Zk identifies λ ˜ k defined by the construction of (a). product of the λ proof (a)(i) Suppose first that every µi is inner regular with respect to the Borel sets. Then a direct ˜ with the required induction on #(I), using 417C for the inductive step, tells us that there is a measure λ properties. Note that, in 417C, (vi) ensures that (in the present context) all our product measures will be inner regular with respect to the Borel sets. (ii) For the general case, apply (a) to µi ¹B(Xi ), where B(Xi ) is the Borel σ-algebra of Xi for each i. ˜ is unique, suppose that λ0 is another measure with the same properties. Let U be (iii) To see that λ Q ˜ the set { i∈I Gi : Gi ∈ Ti for every i ∈ SI}, and Q Us the set of finite unions of members of U. Then λ and λ0 agree on Us . P P Suppose that U = j≤n i∈I Gji , where Gji ∈ Ti for j ≤ n, i ∈ I. If there is any j Q Q 0 ˜ such that i∈I µi Gji > 0} and i∈I µi Gji = ∞, then λU = λ U = ∞. Otherwise, set L = {j : j ≤ n, S G∗i = j∈L Gji for i ∈ I. Then ˜ \ (Q G∗ )) = 0 = λ0 (U \ (Q G∗ )). λ(U i i i∈I i∈I On the other hand, µi Gji must be finite whenever j ∈ L and i ∈ I, so µi G∗i is finite for every i. Consider Q ˜ λ0 agree on I = { i∈I Gi : Gi ∈ Ti , Gi ⊆ G∗i for every i ∈ I}. Then V ∩ V 0 ∈ I for all V , V 0 ∈ I, and λ, 0 ˜ I. It follows from the Monotone Class Theorem (136C), or otherwise, that λ and λ agree on the algebra of Q ˜ ∩ (G∗ × H ∗ )) = λ0 (U ∩ (G∗ × H ∗ )), so that λU ˜ = λ0 U . subsets of i∈I G∗i generated by I. In particular, λ(U Q Q ˜ and λ0 are τ -additive, But now, because λ ˜ = sup{λU ˜ : U ∈ Us , U ⊆ W } = sup{λ0 U : U ∈ Us , U ⊆ W } = λ0 W λW ˜ for every open set W ⊆ X. Writing B(X) for the Borel σ-algebra of X, λ¹B(X) and λ0 ¹B(X) are effectively ˜ and λ0 locally finite Borel measures which agree on the open sets, so must be equal, by 414L. Since both λ are complete locally determined measures defined on B and inner regular with respect to B, they also are equal, by 412L. 0 ˜ k on Q (b) Let λ0 be the measure on X corresponding to the τ -additive product of the λ k∈K Zk . Then λ is an effectively locally finite complete locally determined τ -additive topological measure inner regular with respect to the zero sets, and if Gi ∈ Ti for every i ∈ I then Q Q Q Q Q ˜ k (Q λ0 ( Gi ) = λ Gi ) = µi Gi = µi Gi , i∈I

˜ so λ0 = λ.

k∈K

i∈Ik

k∈K

i∈Ik

i∈I

94

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417E

417E Theorem Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces, with product probability space (X, Λ, λ). Then λ is τ -additive, and has an extension to a τ -additive topological ˜ on X. Moreover, we can arrange that: measure λ ˜ is complete; (i) λ ˜ there is a Q1 ∈ Λ such that λ(Q4Q ˜ ˜ (ii) if Q is measured by λ, 1 ) = 0; that is to say, the embedding Λ ⊆ Λ ˜ induces an isomorphism between the measure algebras of λ and λ; ˜ = λ∗ W for every open set W ⊆ X, and λF ˜ = λ∗ F for every closed set F ⊆ X; (iii) λW (iv) the support of λ is the product of the supports of the µi ; ˜ (v) if λ is inner regular with respect to the Borel sets, so is λ; ˜ (vi) if λ is inner regular with respect to the closed sets, so is λ; ˜ (vii) if λ is tight, so is λ. proof The strategy of the proof is the same S as in 417C, subject to some obviously necessary modifications. The key step, showing that every union n∈N A(Vn ) has zero inner measure, is harder, but we do save a little work because we no longer have to worry about sets of infinite measure. Q (a) I begin by setting up some machinery. Let C be the family of subsets of X expressible in the form i∈I Ei , where Ei ∈ Σi for every i and {i : Ei 6=QXi } is finite. Let U ⊆ C be the standard basis for the topology T of X, consisting of sets expressible as i∈I Gi where Gi ∈ Ti for every i ∈ I and {i : Gi 6= Xi } is finite. Write Us for the set of finite unions of members of U , and V for the set of non-empty upwardsdirected families in Us . Note that every member of Us is determined by coordinates in some finite subset of I (definition: 254M). Q If J ⊆ I, write λJ for the product measure on i∈J Xi ; we shall Q need λ∅ , which is the unique probability Q measure on the single-point set {∅} = i∈∅ Xi . For J ⊆ I, v ∈ i∈J Xi , W ⊆ X set if this is defined, identifying

fW (v) = λI\J {w : (v, w) ∈ W } Q i∈I\J Xi with X. i∈J Xi ×

Q

(b) We need two easy facts. R Q N (i) fW (v) = Q fW (v a t)µj (dt) whenever W ∈ c i∈I Σi , J ⊆ I, v ∈ i∈J Xi and j ∈ I \ J, writing v a t P Let A be the for the member of i∈J∪{j} Xi extending v and taking the value t at the coordinate j. P family of sets W satisfying the property. Then A is a Dynkin class including C, so includes the σ-algebra N Q generated by C, which is c i∈I Σi . Q Q a (ii) If J ⊆ I, v ∈ i∈J Xi , j ∈ I \ J and S V ∈ QUs , and we set g(t) = fV (v t) for t ∈ Xj , then g is lower semi-continuous. P P We can express V as n≤m i∈I Gni , where Gni ⊆ Xi is open for every n ≤ m, i ∈ I. Now if t ∈ Xj , we shall have {w : (v a t, w) ∈ V } ⊆ {w : (v a t0 , w) ∈ V } whenever t0 ∈ H = Xj ∩

T

{Gnj : n ≤ m, t ∈ Gnj }.

So g(t0 ) ≥ g(t) for every t0 ∈ H, which is an open neighbourhood of t. As t is arbitrary, g is lower semi-continuous. Q Q (c) For each V ∈ V, fix, for the remainder of this proof, a countable V 0 ⊆ V such that supV ∈V 0 λV = 0 supV ∈V λV ; because V is upwards-directed, S Swe0may suppose that V = {Vn : n ∈ N} for some non-decreasing sequence hVn in∈N in V. Set A(V) = V \ V . S ?? Suppose, if possible, that there is a sequence hVn in∈N in V such that λ∗ ( n∈N A(Vn )) > 0. S (i) We have λ∗ (X \ n∈N A(Vn )) < 1; let hCn in∈N be a sequence in C such that S S P∞ X \ n∈N A(Vn ) ⊆ n∈N Cn , n=0 λCn = γ0 < 1 0 (see 254A-254C). S 0 S For each n ∈ N, express Vn as {Vnr : r ∈ N} where hVnr ir∈N is non-decreasing, and set Wn = Vn = r∈N Vnr . Let J S ⊆ I be a countable set such that every Cn and every Vnr is dependent on coordinates in J. Express J as k∈N Jk where J0 = ∅ and, for each k, Jk+1 is equal either to Jk or to Jk

417E

τ -additive product measures

95

with one point added. (As in the proof of 254Fa, I am using a formulation which will apply equally to finite and infinite I, though of course the case of finite I is elementary once we have 417C.) (ii) For each n ∈ N, set Wn0 =

S

k∈N {x

: x ∈ X, fWn (x¹Jk ) = 1}.

λ(Wn0

Then \ Wn ) = 0. P P and of fWn as a k Q For any k ∈ N, if we think of λ as the product of λJk and λI\JQ measurable function on i∈Jk Xi , we see that {x : fWn (x¹Jk ) = 1} is of the form Fk × i∈I\Jk Xi , where Q Fk ⊆ i∈Jk Xi is measurable; and λ((Fk ×

Q

i∈I\Jk

Xi ) \ Wn ) =

R

Fk

(1 − fWn (v))λJk (dv) = 0.

Summing over k, we see that Wn0 \ Wn is negligible. Q Q S 0 , like W , is determined by coordinates in J. So n∈N Wn0 \ Wn is of the form Observe that every W n n Q E × i∈I\J Xi where λJ E = 0 (254Ob). There is therefore a sequence hDn in∈N of measurable cylinders in Q S P∞ such that E ⊆ n∈N Dn and n=0 λJ Dn < 1 − γ0 . Set Cn0 = {x : x ∈ X, x¹J ∈ Dn } ∈ C for each i∈J Xi S S n. Then n∈N Wn0 \ Wn ⊆ n∈N Cn0 , so S S S S (X \ n∈N A(Vn )) ∪ n∈N Wn0 \ Wn ⊆ n∈N Cn ∪ n∈N Cn0 , γ=

P∞ n=0

λCn +

P∞ n=0

λCn0 < 1,

while each Cn and each Cn0 is dependent on coordinates in a finite subset of J. Q (iii) For k ∈ N, let Pk be the set of those v ∈ i∈Jk Xi such that P∞ 0 (v) ≤ γ, fV (v) ≤ fWn (v) for every n ∈ N, V ∈ Vn . n=0 fCn (v) + fCn Our hypothesis is that

P∞ n=0

fCn (∅) + fCn0 (∅) =

P∞ n=0

λCn + λCn0 ≤ γ,

and the Vn0 were chosen such that fV (∅) = λV ≤ λWn = fWn (∅) for every n ∈ N, V ∈ Vn ; that is, ∅ ∈ P0 . (iv) Now if k ∈ N, v ∈ Pk there is a v 0 ∈ Pk+1 extending v. P P If Jk+1 = Jk we can take v 0 = v. Otherwise, Jk+1 = Jk ∪ {j} for some j ∈ I \ Jk . Now

γ≥

∞ X

fCn (v) + fCn0 (v) =

n=0

((b-i) above) =

Z X ∞

∞ Z X

fCn (v a t) + fCn0 (v a t) µj (dt)

n=0

fCn (v a t) + fCn0 (v a t) µj (dt),

n=0

so H = {t : t ∈ Xj ,

P∞ n=0

fCn (v a t) + fCn0 (v a t)µj (dt) ≤ γ}

has positive measure. Next, for V ∈ Us , set gV (t) = fV (v a t) for each t ∈ Xj . Then gV is lower semi-continuous, by (b-ii) above. For each n ∈ N, {gV : V ∈ Vn } is an upwards-directed family of lower semi-continuous functions, so its supremum gn∗ is also lower semi-continuous, and because µj is τ -additive,

R

gn∗ dµj = supV ∈Vn

R

gV dµj = supV ∈Vn fV (v) ≤ fWn (v) =

R

fWn (v a t)µj (dt)

(using 414B and (b-i) again). But also, because hVnr ir∈N is non-decreasing and has union Wn , fWn (v a t) = supr∈N fVnr (v a t) ≤ gn∗ (t) for every t ∈ Xj . So we must have

96

Topologies and Measures

417E

fWn (v a t) = gn∗ (t) a.e.(t). And this is true for every n ∈ N. There is therefore a t ∈ H such that fWn (v a t) = gn∗ (v a t) for every n ∈ N. Q Fix on such a t and set v 0 = v a t ∈ i∈Jk+1 Xi ; then v 0 ∈ Pk+1 , as required. Q Q (v) We can therefore choose a sequence hvk ik∈N such that vk ∈ Pk and vk+1 extends vk for each k. Choose x ∈ X such that x(i) = vk (i) whenever k ∈ N, i ∈ Jk and x(i) belongs to the support of µi whenever i ∈ I \ J. (Recall from 411Nd that µi does have a support.) We need to know that if k, n ∈ N and V ∈ Vn then fV \Wn (vk ) = 0. P P For any r ∈ N there is a V 0 ∈ Vn 0 such that V ∪ Vnr ⊆ V , so fV ∪Vnr (vk ) ≤ fV 0 (vk ) ≤ fWn (vk ), and fV \Wn (vk ) ≤ fV \Vnr (vk ) = fV ∪Vnr (vk ) − fVnr (vk ) ≤ fWn (vk ) − fVnr (vk ) → 0 as r → ∞. Q Q P Cn and Cn0 are determined by coordinates in a finite subset (vi) If n ∈ N, then x ∈ / Cn ∪ Cn0 . P of J, so must be determined by coordinates in Jk for some k ∈ N. Now fCn (vk ) + fCn0 (vk ) ≤ γ < 1, so Q / Cn ∪ Cn0 . Q {y : y¹Jk = vk } cannot be included in Cn ∪ Cn0 , and must be disjoint from it; accordingly x ∈ (vii) Because (X \

S n∈N

there is some n ∈ N such that

A(Vn )) ∪

S n∈N

Wn0 \ Wn ⊆

S n∈N

Cn ∪

S n∈N

Cn0 ,

S x ∈ A(Vn ) \ (Wn0 \ Wn ) ⊆ ( Vn ) \ Wn0 ,

that is, there is some V ∈ Vn such that x ∈ V \ Wn0 . Let U ∈ U be such that x ∈ U ⊆ V . Express U as U 0 ∩ U 00 where U 0 ∈ U is determined by coordinates in a finite subset of J, and U 00 ∈ U is determined by coordinates in a finite subset of I \ J. Let k ∈ N be such that U 0 is determined by coordinates in Jk . Then fU \Wn (vk ) ≤ fV \Wn (vk ) = 0 by (v) above. Now {w : w ∈

Q i∈I\Jk

Xi , (vk , w) ∈ U \ Wn } = {w : (vk , w) ∈ U 00 \ Wn }

(because (vk , w) = (x¹Jk , w) ∈ U 0 for every w), while {w : (vk , w) ∈ U 00 },

{w : (vk , w) ∈ Wn }

are stochastically independent because the former depends on coordinates in I \ J, while the latter depends on coordinates in J \ Jk . So we must have 0 = fU \Wn (vk ) = λI\Jk {w : (vk , w) ∈ U \ Wn } = λI\Jk {w : (vk , w) ∈ U 00 \ Wn } = λI\Jk {w : (vk , w) ∈ U 00 }(1 − λI\Jk {w : (vk , w) ∈ Wn }). At this point, recall that x(i) belongs to the support of µi for every i ∈ I \ J, while x ∈ U 00 . So if U 00 = {y : y(i) ∈ Hi for i ∈ K}, where K ⊆ I \ J is finite and Hi ⊆ Xi is open for every i, we must have µi Hi > 0 for every i, and Q λI\Jk {w : (vk , w) ∈ U 00 } = i∈K µi Hi > 0. On the other hand, we are also supposing that x ∈ / Wn0 , so λI\Jk {w : (vk , w) ∈ Wn } = fWn (vk ) < 1. But this means that we have expressed 0 as the product of two non-zero numbers, which is absurd. X X

417F

τ -additive product measures

97

S (d) Thus λ∗ ( n∈N A(Vn )) = 0 for every sequence hVn in∈N in V. Accordingly there is an extension of λ ˜ on X such that λA(V) ˜ ˜ being to a measure λ = 0 for every V ∈ V, the domain of λ ˜ = {W 4A : W ∈ Λ, A ∈ A∗ }, Λ where A∗ is the σ-ideal generated by {A(V) : V ∈ V} (417A). ˜ Now measure. P P If W ⊆ X is open, then V = {V : V ∈ Us , V ⊆ W } belongs to V, S λ is a topological S and V = W . Now V 0 ∈ Λ (because V 0 is countable), so S W = V 0 ∪ A(V) ˜ Q is measured by λ. Q ˜ Also, λ is τ -additive. P P Let W be a non-empty upwards-directed family of open subsets of X with union W ∗ . Set Then V ∈ V and

S

V = {V : V ∈ Us , ∃ W ∈ W, V ⊆ W }. ˜ V = W ∗ , so λA(V) = 0 and S 0 ∗ ˜ ˜ ˜ ≤ sup ˜ ˜ ∗ λW = λ( V ) = supV ∈V 0 λV W ∈W λW ≤ λW

˜ is τ -additive. Q (using the fact that V 0 is upwards-directed). As W is arbitrary, λ Q Of course it follows at once that λ is also τ -additive. (e) Now for the supplementary properties (i)-(vi) listed in the theorem. ˜ = 0 and Q0 ⊆ Q, then Q is expressible as W 4A where λW = 0 and A ∈ A∗ . But in ˜ λQ (i) If Q ∈ Λ, ˜ is complete. ˜ Thus λ this case (because λ is complete) λ(Q0 ∩ W ) = 0 while Q0 \ W ⊆ A ∈ A∗ , so Q0 ∈ Λ. ˜ ˜ differs by a λ-negligible (ii) As always, the construction ensures that every member of Λ set from some member of Λ. (iii) Let W ⊆ X be an open set. Set V = {V : V ∈ Us , V ⊆ W }. Then ˜ = sup ˜ ˜ λW V ∈V λV = supV ∈V λV ≤ λ∗ W ≤ λW ˜ is a τ -additive extension of λ. Now if F ⊆ X is closed, just because λ ˜ = 1 − λ(X ˜ λF \ F ) = 1 − λ∗ (X \ F ) = λ∗ F . Q (iv) For each i ∈ I write Fi for the support of µi , and set F = i∈I Fi . This is closed because every Fi is. Its complement is covered by the negligible open sets {x : x ∈ X, x(i) ∈ Xi \ Fi } as i runs over I; as λ is τ -additive, the union of the negligible open sets is negligible, and Q F is conegligible. If W ⊆ X is open and x ∈ F ∩ W , let U ∈ U be such that x ∈ U ⊆ W . Express U as i∈I Gi where Gi ∈ Ti for every i ∈ I and J = {i : Gi 6= Xi } is finite. Then x(i) ∈ Gi ∩ Fi , so µi Gi > 0, for every i. Accordingly Q λ(W ∩ F ) = λW = λU = i∈J µi Ui > 0. Thus F is self-supporting and is the support of λ. (v), (vi), (vi) Use the same arguments as in the corresponding parts of 417C, this time using 412U to confirm that λ is inner regular with respect to the given family of sets. 417F Corollary Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces such that µi is inner regular with respect to the Borel sets for each i. Then there is a unique complete τ -additive ˜ on X = Q Xi which extends the ordinary product measure and is inner regular topological measure λ i∈I with respect to the Borel sets. ˜ with the right properties. If λ0 is any other complete τ -additive proof By 417E(v) we have a measure λ ˜ topological measure, extending λ and inner regular with respect to the family B of Borel sets, then λ0 W = λW ˜ .Q for every open set W ⊆ X. P P By the argument of (e-iii) of the proof of 417E, λ0 W = λ∗ W = λW Q By ˜ ˜ for every Borel set W . Now λ0 and λ ˜ are 414L, applied to the Borel measures λ0 ¹B and λ¹B, λ0 W = λW supposed to be complete topological probability measures inner regular with respect to B, so they must be identical, by 412L or otherwise.

98

Topologies and Measures

417G

˜ Q 417G Notation In the context of 417D or 417F, I will call λ the τ -additive product measure on i∈I Xi . Note that the uniqueness assertions in 417D and 417F mean that for the products of finitely many probability spaces we do not need to distinguish between the two constructions. The latter also shows that we can relate 415E to the new method: if every Ti is separable and metrizable and every µi is strictly positive, then the ‘ordinary’ product measure λ is a complete topological measure. Since it is also inner regular with respect to the Borel sets (412Uc), it must be exactly the τ -additive product measure as described here. 417H Fubini’s theorem for τ -additive product measures Let (X, T, Σ, µ) and (Y, S, T, ν) be two complete locally determined effectively locally finite τ -additive topological measure spaces such that both µ ˜ be the τ -additive product measure on X × Y , and ν are inner regular with respect to the Borel sets. Let λ ˜ and Λ its domain. R ˜ is defined in [−∞, ∞] and (X × Y ) \ {(x, y) : (a) Let f be a [−∞, ∞]-valued function such that f dλ S (x, y) ∈ dom f, f (x, y) = 0} can be covered by a set of the form X × n∈N Yn where νYn < ∞ for every RR R ˜ n ∈ N. Then the repeated integral f (x, y)ν(dy)µ(dx) is defined and equal to f dλ. (b) Let f : X × Y → [0, ∞] be lower semi-continuous. Then

RR

f (x, y)ν(dy)µ(dx) =

RR

f (x, y)µ(dx)ν(dy) =

R

˜ f dλ

in [0, ∞]. RR ˜ ˜ (c) Let f be a Λ-measurable real-valued function defined λ-a.e. on X × Y . If either |f (x, y)|ν(dy)µ(dx) RR ˜ or |f (x, y)|µ(dx)ν(dy) is defined and finite, then f is λ-integrable. proof (a) I use 252B. R ˜ such that νW [{x}]µ(dx) is defined in [0, ∞] and equal to (i) Write W for the set of those W ∈ Λ ˜ . Then open sets belong to W, by 417C(iv). Next, any Borel subset of an open set of finite measure λW belongs to W. P P If W0 is an open set of finite measure, then {W : W ⊆ X × Y, W ∩ W0 ∈ W} is a Dynkin class containing every open set, so contains all Borel subsets of X × Y . Q Q S ˜ Now suppose that W ⊆ X × Y is λ-negligible and included in X × n∈N Yn , where νYn < ∞ for every n. Then W ∈ W. P P Set A = {x : x ∈ X, ν ∗ W [{x}] > 0}. For each n, let Hn ⊆ Y be an openSset of finite −n measure such that ν(Y S n \ Hn ) ≤ 2 ; we may arrange that Hn+1 ⊇ Hn for each n. Set H = n∈N Hn , so that W [{x}] \ H ⊆ n∈N Yn \ H is negligible for every x ∈ X. ˜ is inner regular with Fix an open set G ⊆ X of finite measure and n ∈ N for the moment. Because λ ˜ ˜ respect to the Borel sets, there is a Borel set V ⊆ (G × Hn ) \ W such that λV = λ((G × Hn ) \ W ), that is, ˜ 0 = 0, where V 0 = (G × Hn ) \ V ⊇ (G × Hn ) ∩ W . We know that V 0 ∈ W, so λV

R

˜ 0 = 0, νV 0 [{x}]dx = λV

and νV 0 [{x}] = 0 for almost every x ∈ X; but this means that Hn ∩ W [{x}] is negligible for almost every x ∈ G. At this point, recall that n was arbitrary, so H ∩ W [{x}] and W [{x}] are negligible for almost every x ∈ G, that is, A ∩ G is negligible. This is true for every open set G ⊆ X of finite measure. Because µ is inner regular with respect to subsets of open sets R of finite measure, and is complete and locally determined, A is negligible (412Jb). But this means that νW [{x}]µ(dx) is defined and equal to zero, so that W ∈ W. Q Q R ˜ is defined in [−∞, ∞] and that there is a sequence hYn in∈N of sets of (ii) Now suppose that f dλ S finite Smeasure in Y such that f (x, y) is defined and zero whenever x ∈ X and y ∈ Y \ n∈N Yn . Set Z = n∈N Yn . Write λ for the c.l.d. product measure on X × Y and Λ for its domain. Then there is a ˜ Λ-measurable function g : X × Y → [−∞, ∞] which is equal λ-almost everywhere to f . P P For q ∈ Q set ˜ q 4Vq ) = 0 (417C(ii)); ˜ and choose Vq ∈ Λ such that λ(W Wq = {(x, y) : (x, y) ∈ dom f, f (x, y) ≥ q} ∈ Λ, set g(x, y) = sup{q : q ∈ Q, (x, y) ∈ Vq } for x ∈ X, y ∈ Y , interpreting sup ∅ as −∞. Q Q Adjusting g if necessary, we may suppose that it is zero on X × (Y \ Z). Set A = (X × Y ) \ {(x, y) : f (x, y) = g(x, y)},

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˜ so that A is λ-negligible and included in X × Z. By (i), νA[{x}] = 0, that is, y 7→ f (x, y) and y 7→ g(x, y) are equal ν-a.e., for µ-almost every x. Write λX×Z for the subspace measure induced by λ on X × Z; note that this is the c.l.d. product of µ with the subspace measure νZ on Z, by 251P(ii-α). Now we have Z

Z

Z

˜= f dλ

˜= g dλ

g dλ

˜ to (X × Y, λ) is inverse-measure-preserving) (by 235Ib, because the identity map from (X × Y, λ) Z Z ZZ = g dλ = g dλX×Z = g(x, y)νZ (dy)µ(dx) X×Z

X×Z

Z

(by 252B, because νZ is σ-finite) ZZ = g(x, y)ν(dy)µ(dx) (because g(x, y) = 0 if y ∈ Y \ Z) ZZ = f (x, y)ν(dy)µ(dx). (b) If f is non-negative and lower semi-continuous, set Wni = {(x, y) : f (x, y) > 2−n i} for n, i ∈ N, and fn = 2−n

P 4n i=1

χWni

for n ∈ N. Applying 417C(iv) we see that

R

˜= fn dλ

RR

fn (x, y)dydx =

in [0, ∞] for every n; taking the limit as n → ∞,

R

˜= f dλ

RR

f (x, y)dydx =

RR

RR

fn (x, y)dxdy f (x, y)dxdy,

because hfn in∈N is a non-decreasing sequence with limit f . RR ˜ is (c) ?? Suppose, if possible, that γ = |f (x, y)|dydx is finite, but that f is not integrable. Because λ R ˜ ˜ > γ (213B). semi-finite, there must be a non-negative λ-simple function g such that g ≤a.e. |f | and g dλ ˜ For each n ∈ N, there are open sets Gn ⊆ X, Hn ⊆ Y of finite measure such that λ({(x, y) : g(x, y) ≥ −n −n 2R } \ (Gn × Hn )) ≤ 2 , by 417C(iii); now g × χ(Gn × Hn ) → g a.e., so there is some n such that ˜ > γ. In this case, setting g 0 (x, y) = min(g(x, y), |f (x, y)|) for (x, y) ∈ (Gn × Hn ) ∩ dom f , 0 g dλ Gn ×Hn R ˜ > γ. But we can apply (a) to g 0 to see that otherwise, we have g = g 0 a.e. on Gn × Hn , so that g 0 dλ γ<

R

˜= g 0 dλ

RR

g 0 (x, y)dydx ≤

RR

|f (x, y)|dydx ≤ γ,

which isRR absurd. X X ˜ So if |f (x, y)|dydx is finite, f must be λ-integrable. Of course the same arguments, reversing the roles RR ˜ of X and Y , show that f is λ-integrable if |f (x, y)|dxdy is defined and finite. 417I The constructions here have most of the properties one would hope for. I give several in the exercises (417Xd-417Xf, 417Xj). One fact which is particularly useful, and also has a trap in it, is the following. Proposition Let (X, T, Σ, µ) and (Y, S, T, ν) be effectively locally finite τ -additive topological measure ˜ the τ -additive product spaces in which the measures are inner regular with respect to the Borel sets, and λ measure on X × Y . Suppose that A ⊆ X and B ⊆ Y , and write µA , νB for the corresponding subspace measures; assume that both µA and νB are semi-finite. Then these are also effectively locally finite, τ ˜ A×B induced by λ ˜ on additive and inner regular with respect to the Borel sets, and the subspace measure λ A × B is just the τ -additive product measure of µA and νB .

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proof (a) To check that µA and νB are effectively locally finite, τ -additive and inner regular with respect to ˜ A×B inherits the same properties from λ, ˜ and is also complete the Borel sets, see 414K and 412P. Of course λ and locally determined, by 214Id. ˜ ∗ (C × D) = µA C · νB D. P (b) Now if C ∈ dom µA and D ∈ dom νB , then λ P (α) There are E ∈ Σ, F ∈ T ∗ ∗ such that C ⊆ E, D ⊆ F , µE = µ C and νF = ν D; in which case ˜ ∗ (C × D) ≤ λ(E ˜ × F ) = λ(E × F ) = µE · νF λ (251J) = µ∗ C · ν ∗ D = µA C · νB D. (β) If γ < µA C ·νB D then, because µA and νB are semi-finite, there are C 0 ⊆ C, D0 ⊆ D such that both have finite outer measure and µ∗ C 0 · ν ∗ D0 ≥ γ. In this case, take E 0 ∈ Σ, F 0 ∈ T such that C 0 ⊆ E 0 , D0 ⊆ F 0 and ˜ and C × D ⊆ W , we have C 0 × D0 ⊆ W ∩ (E × F ), both E 0 and F 0 have finite measure. Now if W ∈ dom λ ∗ 0 0 so that ν(W ∩ (E × F ))[{x}] ≥ ν D for every x ∈ C , and ˜ ≥ λW

R

E

ν(W ∩ (E × F ))[{x}]µ(dx) ≥ µ∗ C 0 · ν ∗ D0 ≥ γ,

˜ ∗ (C × D) ≥ γ; as γ is arbitrary, λ ˜ ∗ (C × D) ≥ µA C · νB D. Q by 417Ha. As W is arbitrary, λ Q (c) In particular, if U ⊆ A and V ⊆ B are relatively open, ˜ A×B (U × V ) = λ ˜ ∗ (U × V ) = µA U · νB V . λ ˜ A×B must be exactly the τ -additive product measure of µA and νB . But now 417D tells us that λ 417J In order to use 417H effectively in the theory of infinite products, we need an ‘associative law’ corresponding to 254N. Theorem Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces such that every ˜j µi is inner regular with respect to the Borel sets, and hKj ij∈J a partition of I. For each j ∈ J let λ Q ˜ be the τ -additive product measure on Zj = i∈Kj Xi , and write λ for the τ -additive product measure on Q X = i∈I Xi . Then the natural bijection Q x 7→ φ(x) = hx¹Kj ij∈J : X → j∈J Zj ˜ with the τ -additive product of the family hλ ˜ j ij∈J . identifies λ ˜ In particular, if K Q ⊆ I is any set,Qthen λ can be identified with the τ -additive product of the τ -additive product measures on i∈K Xi and i∈I\K Xi . proof We have a lot of measures to keep track of; I hope that the following notation will not be too ˜ j for the ordinary and opaque. Write λ for the ordinary product measure on X, and for j ∈ J write λj , λ Q τ -additive product measures on Zj . Write θ for the ordinary product measure on Z = j∈J Zj of the ˜ j , and θ˜ for the τ -additive product of the λ ˜ j . Write λ ˜ # for the measure on X τ -additive product measures λ # −1 −1 ˜ ˜ ˜ corresponding to θ on Z. (If you like, λ is the image measure θ(φ ) defined from θ˜ and the function ˜ # , like θ, ˜ is a complete τ -additive topological measure, inner regular with respect φ−1 : Z → X.) Then λ to theQBorel sets, because φ : X → Z is a homeomorphism. If C ⊆ X is a measurable cylinder, itQ is of the form i∈I Ei where Ei ∈ Σi for each i and {i : i ∈ I, Ei 6= Σi } is finite. So φ[C] is of the form j∈J Cj , Q where Cj = i∈Kj Ei , and Y Y Y ˜ # C = θ( ˜ ˜ j Cj λ Cj ) = θ( Cj ) = λ j∈J

=

Y

j∈J

λj C j =

j∈J

Y Y j∈J i∈Kj

j∈J

Ei =

Y

Ei = λC.

i∈I

˜ # ) to (X, λ), that λ ˜ # extends λ. So it But this means, applying 254G to the identity map from (X, λ is a complete τ -additive topological measure, inner regular with respect to the Borel sets, extending the ordinary product measure, and by the uniqueness declared in 417F, must be identical to the τ -additive product measure on X, as claimed.

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417K Proposition Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces such ˜ be their τ -additive product. ˜ λ) that every µi is inner regular with respect to the Borel sets, and let (X, Λ, ˜ J be the τ -additive product measure on XJ = Q ˜ For J ⊆ I let λ X , and Λ i J its domain; let πJ : X → XJ i∈J −1 ˜ ˜ ˜ is determined by be the canonical map. Then λJ is the image measure λπJ . In particular, if W ∈ Λ ˜ ˜ ˜ J and λJ πJ [W ] = λW . coordinates in J ⊆ I, then πJ [W ] ∈ Λ ˜ is an extension of the ordinary product measure λ on X, λπ ˜ −1 is an extension of λπ −1 , proof Because λ J J ˜ is a τ -additive topological measure and which is the ordinary product measure on XJ (254Oa). Because λ ˜ −1 is a τ -additive topological measure; because λ ˜ is a complete probability measure, πJ is continuous, λπ J −1 −1 ˜ ˜ ˜ so is λπJ . Finally, λπJ is inner regular with respect to the Borel sets. P P Recall that we may identify λ −1 −1 ˜ J and λ ˜ I\J (417J). If V ∈ dom λπ ˜ ˜ we can think of with the τ -additive product of λ , that is, π [V ] ∈ Λ, πJ−1 [V ] ⊆ X as V × XI\J ⊆ XI × XJ . In this case, we must have ˜ −1 [V ] = λπ J

R

J

J

˜ J V dλI\J , λ

˜ J V must be defined and equal to λπ ˜ −1 [V ]. by Fubini’s theorem for τ -additive products (417Ha); that is, λ J ˜ −1 [V ], there must be a Borel set V 0 ⊆ V such that λ ˜ J V 0 ≥ γ. In this case, because πJ is Now if γ < λπ J ˜ −1 [V 0 ] is defined. As with V , this measure must be λ ˜ J V 0 ≥ γ. continuous, πJ−1 [V 0 ] is also Borel, and λπ J −1 ˜ Since V and γ are arbitrary, λπ Q J is inner regular with respect to the Borel sets, as claimed. Q −1 ˜ ˜ By the uniqueness assertion in 417F, λπJ must be λJ exactly. ˜ is determined by coordinates in J, then If now W ∈ Λ ˜ J πJ [W ] = λπ ˜ −1 [πJ [W ]] = λW ˜ . λ J

417L Corollary Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces such that ˜ their τ -additive product. Let hKj ij∈J ˜ λ) every µi is inner regular with respect to the Borel sets, and (X, Λ, ˜ j for the σ-algebra of members of Λ determined by be a disjoint family of subsets of I, and for j ∈ J write Λ ˜ j ij∈J is a stochastically independent family of σ-algebras (definition: 272Ab). coordinates in Kj . Then hΛ proof It is enough to consider the case in which J is finite (272Bb), no Kj is empty (since if Kj = ∅ ˜ j for ˜ j = {∅, X}) and S an extra term if necessary). In this case, if Wj ∈ Λ then Λ j∈J Kj = I (adding T Q Q each j, then the identification between X and j∈J i∈Kj Xi , as described in 417J, matches j∈J Wj with Q Q ˜ i∈Kj Xi , we j∈J πKj [Wj ], writing πKj (x) for x¹Kj . Now if λj is the τ -additive product measure on Zj = ˜ ˜ ˜ ˜ have λj πKj [Wj ] = λWj , by 417K. Since λ can be identified with the τ -additive product of hλj ij∈J (417J), Q Q ˜ ˜ T ˜ λ( j∈J λWj . j∈J Wj ) = j∈J λj πKj [Wj ] = ˜ j ij∈J is independent. As hWj ij∈J is arbitrary, hΛ 417M Proposition Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces such that every µi is inner regular withQ respect to the Borel sets and strictly positive. For J ⊆ I let πJ be the ˜ J for the ordinary and τ -additive product measures canonical map from X onto XJ = i∈J Xi ; write λJ , λ ˜ ˜ ˜ ˜ ˜ I , λ = λI , Λ = Λ I . on XJ , and ΛJ , ΛJ for their domains. Set λ = λI , Λ = Λ (a) Let F ⊆ X be a closed self-supporting set, and J the smallest subset of I such that F is determined by coordinates in J (4A2B(g-ii)). Then ˜ ˜ is such that W 4F is λ-negligible (i) if W ∈ Λ and determined by coordinates in K ⊆ I, then K ⊇ J; (ii) J is countable; ˜ (iii) there is a W ∈ Λ, determined by coordinates in J, such that W 4F is λ-negligible. T ˜ ˜ is (b) λ is inner regular with respect to the family of sets of the form n∈N Vn where each Vn ∈ Λ determined by finitely many coordinates. ˜ there are a countable J ⊆ I and sets W 0 , W 00 ∈ Λ, ˜ determined by coordinates in J, such (c) If W ∈ Λ, −1 0 00 00 0 ˜ ˜ ˜ . that W ⊆ W ⊆ W and λ(W \ W ) = 0. Consequently λπJ [πJ [W ]] = λW

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proof (a)(i) ?? Suppose, if possible, otherwise. Then F is not determined by coordinates in K, so there are x Q ∈ F , y ∈ X \ F such that x¹K = y¹K. Let U be an open set containing y, disjoint from F , and of the form i∈I Gi , where Gi ∈ Ti for every i and L = {i : Gi 6= Xi } is finite. Set U 0 = {z : z ∈ X, z(i) ∈ Gi for every i ∈ L ∩ K}, U 00 = {z : z(i) ∈ Gi for every i ∈ L \ K}. Then U 0 ∩ W is determined by coordinates in K, while U 00 is determined by coordinates in I \ K, so ˜ ∩ U ) = λ(W ˜ ˜ ˜ ˜ 00 0 = λ(F ∩ U ) = λ(W ∩ U 0 ∩ U 00 ) = λ(W ∩ U 0 ) · λU (by 417L) ˜ ∩ U 0 ) · λU ˜ 00 = λ(F ˜ ∩ U 0) · = λ(F

Y

µi Gi .

i∈L\K

˜ ∩ U 0 ) > 0. Because every µi is But y ∈ U 0 , and x¹K = y¹K, so x ∈Q F ∩ U 0 ; as F is self-supporting, λ(F strictly positive, and no Gi is empty, i∈L\K µi Gi > 0; and this is impossible. X X ˜ 4W0 ) = 0. By 254Oc there is a W1 ∈ Λ, determined (ii) By 417E(ii), there is a W0 ∈ Λ such that λ(F ˜ 4W1 ) = 0, so (i) tells us by coordinates in a countable subset K of I, such that λ(W0 4W1 ) = 0. Now λ(F that J ⊆ K is countable. ˜ J -negligible. ˜ J . By 417E(ii), there is a V ∈ ΛJ such that V 4πJ [F ] is λ (iii) By 417K, πJ [F ] ∈ Λ −1 −1 Set W = πJ [V ]. Then W ∈ Λ, W is determined by coordinates in J, and W 4F = πJ [V 4πJ [F ]] is ˜ λ-negligible. ˜ which are determined by finitely many coordinates, (b)(i) Write V for the set of those members of Λ and Vδ for the set of intersections of sequences in V. Because V is closed under finite unions, so is Vδ ; Vδ is surely closed under countable intersections, and ∅, X belong to Vδ . (ii) We need to know that every self-supporting closed set F ⊆ X belongs to Vδ . P P By (a), F is determined by a countable set J of coordinates. Express J as the union of a non-decreasing sequence T F ∈ V . Q Q hJn in∈N of finite sets. Then Fn = πJ−1 [π [F ]] ∈ V for each n, and F = δ Jn n∈N n n ˜ and (iii) Let A be the family of subsets of X which are either open or closed. Then if A ∈ A, V ∈ Λ ˜ ∩ V ) > 0, there is a K ∈ Vδ ∩ A such that K ⊆ A and λ(K ˜ λ(A ∩ V ) > 0. P P (α) If A is open, set U = {U : U ∈ V is open, U ⊆ A}. S ˜ is τ -additive and V is closed under Because V includes a base for the topology of X, U = A; because λ ˜ ˜ ˜ ˜ ∩ V ) > 0. (β) If A finite unions, there is a U ∈ U such that U ⊆ A and λU > λA − λ(A ∩ V ), so that λ(U is closed, then it includes a self-supporting closed set V of the same measure (414F), which belongs to Vδ , by (ii) just above. Q Q ˜ ˜ to the Borel σ-algebra of X is inner regular with respect to Vδ . (iv) By 412C, the restriction λ¹B of λ ˜ ˜ But λ is just the completion of λ¹B, so is also inner regular with respect to Vδ (412Ha). ˜ n ≥ λW ˜ − 2−n (c) By (b), we have sequences hVn in∈N , hVn0 in∈N in Vδ such that Vn ⊆ W , Vn0 ⊆ X \ W , λV 0 −n 0 ˜ ˜ and λVn ≥ λ(X \ W ) − 2 for every n ∈ N. Each Vn , Vn is determined by a countable set of coordinates, so there isSa single countable set every Vn and every Vn0 is determined by coordinates in J. Set S J ⊆ 0I such that 0 00 0 W = n∈N Vn , W = X \ n∈N Vn ; then W , W 00 are both determined by coordinates in J, W 0 ⊆ W ⊆ W 00 00 ˜ and λ(W \ W 0 ) = 0, as required. 417N Theorem Let (X, T, Σ, µ) and (Y, S, T, ν) be two quasi-Radon measure spaces. Then the τ ˜ on X × Y is a quasi-Radon measure, the unique quasi-Radon measure on X × Y additive product measure λ ˜ × F ) = µE · νF for every E ∈ Σ, F ∈ T. such that λ(E ˜ is a complete, locally determined, effectively locally finite, τ -additive topological measure, inner proof λ regular with respect to the closed sets (417C(vii)). But this says just that it is a quasi-Radon measure. By 417D, it is the unique quasi-Radon measure with the right values on measurable rectangles.

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103

417O Theorem Let h(Xi , Ti , Σi , µi )ii∈I be a family of quasi-Radon probability spaces. Then the τ ˜ on X = Q Xi is a quasi-Radon measure, the unique quasi-Radon measure additive product measure λ i∈I on X extending the ordinary product measure. ˜ is inner regular with respect to the closed sets, so is a quasi-Radon measure, which proof By 417E(vi), λ is unique by 417F. 417P Theorem Let (X, T, Σ, µ) and (Y, S, T, ν) be Radon measure spaces. Then the τ -additive product ˜ on X ×Y is a Radon measure, the unique Radon measure on X ×Y such that λ(E ˜ ×F ) = µE · νF measure λ whenever E ∈ Σ, F ∈ T. ˜ is locally finite (because λ(G ˜ × H) = µG · νH is finite whenever proof Of course X × Y is Hausdorff, and λ ˜ is tight, so is a Radon measure. As in 417N, it is uniquely defined µG and νH are finite). By 417C(viii), λ by its values on measurable rectangles. ˜ 417Q Theorem Let h(XQ i , Ti , Σi , µi )ii∈I be a family of Radon probability spaces, and λ the τ -additive product measure on X = i∈I Xi . For each i ∈ I, let Zi ⊆ Xi be the support of µi . Suppose that ˜ is a Radon measure, the unique Radon measure on J = {i : i ∈ I, Zi is not compact} is countable. Then λ X extending the ordinary product measure. ˜ being totally finite, is locally proof Of course X, being a product of Hausdorff spaces, is Hausdorff, and λ, P finite. Now, given ² ∈ ]0, 1], let h²j ij∈J be a family of strictly positive numbers such that j∈J ²j ≤ ², and for j ∈ J choose a compact set Kj ⊆ XQ j such that µj Kj ≥ 1 − ²j ; for i ∈ I \ J, set Ki = Zi , so that Ki is compact and µi Ki = 1. Consider K = i∈I Ki . Then, using 417E(iii) and 254Lb for the two equalities, ˜ = λ∗ K = Q µi Ki ≥ Q λK j∈J 1 − ²j ≥ 1 − ², i∈I ˜ satisfies the condition (iv) of 416C, and where λ is the ordinary product measure on X. As ² is arbitrary, λ is a Radon measure. As in 417F, it is the unique Radon measure on X extending λ. 417R Notation I will use the phrase quasi-Radon product measure for a τ -additive product measure which is in fact a quasi-Radon measure; similarly, a Radon product measure is a τ -additive product measure which is a Radon measure. 417S Later I will give an example in which a τ -additive product measure is different from the corresponding c.l.d. product measure (419E). In 415E-415F, 415Ye and 416U I have described cases in which c.l.d. measures are τ -additive product measures. It remains very unclear when to expect this to happen. I can however give a couple of results which show that sometimes, at least, we can be sure that the two measures coincide. Proposition (a) Let (X, T, Σ, µ) and (Y, S, T, ν) be effectively locally finite τ -additive topological measure spaces such that both µ and ν are inner regular with respect to the Borel sets, and λ the c.l.d. product measure on X ×Y . If every open subset of X ×Y is measured by λ, then λ is the τ -additive product measure on X × Y . (b) Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces suchQthat every µi is inner regular with respect to the Borel sets, and λ the ordinary product measure on X = i∈I Xi . If every open subset of X is measured by λ, then λ is the τ -additive product measure on X. Q ˜ J the τ -additive (c) In (b), let λJ be the ordinary product measure on XJ = i∈J Xi for each J ⊆ I, and λ ˜ ˜ I is the product measure. If λJ = λJ for every finite J ⊆ I, and every µi is strictly positive, then λ = λ τ -additive product measure on X. proof (a), (b) In both cases, λ is a complete locally determined effectively locally finite τ -additive measure which is inner regular with respect to the Borel sets (assembling facts from 251I, 254F, 412S, 412U, 417C and 417E). The extra hypothesis added here is that λ is a topological measure, so itself satisfies the conditions of 417D or 417F, and is the τ -additive product measure.

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S ˜ J for every countable J ⊆ I. P (c)(i) The first step is to note that λJ = λ P Express J as n∈N Jn where T hJn in∈N is a non-decreasing sequence of finite sets. If F ⊆ XJ is closed, then it is n∈N πn−1 [πn [F ]], where πn : XJ → XJn is the canonical map for each n. But every πn [F ] is a closed subset of XJn , therefore measured by λJn ; because πn is inverse-measure-preserving (417K), πn−1 [πn [F ]] ∈ dom λJ for each n, and F ∈ dom λJ . Thus every closed set, therefore every open set is measured by λJ , and λJ is a topological ˜J . Q measure; by (b), λJ = λ Q ˜ such that W 0 ⊆ W ⊆ W 00 , (ii) Suppose that W ⊆ X is open. By 417M, there are W 0 , W 00 measured by λ 00 0 00 ˜ I (W \ W 0 ) = 0. Let J ⊆ I be a both W and W are determined by coordinates in a countable set, and λ ˜ J measures πJ [W 0 ], by 417K, countable set such that W 0 and W 00 depend on coordinates in J. Then λJ = λ −1 0 00 ˜ I (W 00 \W 0 ) = 0, so λ measures W = πJ [πJ [W ]], by 254Oa. Similarly, λ measures W . Now λ(W 00 \W 0 ) = λ so λ measures W . As W is arbitrary, λ is a topological measure and must be the τ -additive product measure, by (a). 417T Proposition Let (X, T, Σ, µ) and (Y, S, T, ν) be effectively locally finite τ -additive topological measure spaces such that both µ and ν are inner regular with respect to the Borel sets, and λ the c.l.d. product measure on X × Y . If X has a conegligible subset with a countable network (e.g., if X is separable and metrizable), then λ is the τ -additive product measure on X × Y . proof (a) Suppose to begin with that µ and ν are totally finite, and that X has a countable network; let ˜ be the τ -additive ˆ its domain. Let λ hAn in∈N run over a network for X. Let µ ˆ be the completion of µ and Σ ˜ I will use a product measure on X × Y . (We are going to need Fubini’s theorem both for λ and for λ. sprinkling of references to §§251-252 to indicate which parts of the argument below depend on the properties of λ.) Let W ⊆ X × Y be an open set. For each n ∈ N, set S Hn = {H : H ∈ S, An × H ⊆ W }, S P Of course An × Hn ⊆ W for every n ∈ N. If (x, y) ∈ W , so that Hn is open. Then W = n∈N An × Hn . P there are open sets G ⊆ X, H ⊆ Y such that (x, y) ∈ G × H ⊆ W ; now there is an n ∈ N such that x ∈ An ⊆ G, so that H ⊆ Hn and (x, y) ∈ An × Hn . Q Q ˜ By 417C(iv), there is an open set W0 in the domain Λ of λ such that W0 ⊆ W and λ(W \ W0 ) = 0. By 417Ha, applied to χ(W \ W0 ), A = {x : ν(W [{x}] \ W0R[{x}]) > 0} is µ-negligible. For each n ∈ N, x ∈ X set fn (x) = ν(Hn ∩ W0 [{x}]); then 252B tells us that fn dµ is defined and equal to λ(W0 ∩ (X × Hn )). ˆ ˆ If x ∈ An , then Hn ⊆ W [{x}], so In particular, fn is Σ-measurable. Set En = {x : fn (x) = νHn } ∈ Σ. An \ En ⊆ A. Now, by 252B again, Z λ((En × Hn ) \ W0 ) = ν(Hn \ W0 [{x}])µ(dx) En Z = νHn − ν(Hn ∩ W0 [{x}])µ(dx) = 0. So if we set W1 =

En

S n∈N

En × Hn , W1 \ W ⊆ W1 \ W0 is λ-negligible. On the other hand, S W \ W1 ⊆ n∈N (An \ En ) × Hn ⊆ A × Y

is also λ-negligible. Because λ is complete, W ∈ Λ. As W is arbitrary, λ is a topological measure and is ˜ by 417Sa. equal to λ, (b) Now consider the general case. Let Z be a conegligible subset of X with a countable network; since any subset of a space with a countable network again has a countable network (4A2Na), we may suppose that Z ∈ Σ. Again let W be an open set in X ×Y . This time, take arbitrary E ∈ Σ, F ∈ T of finite measure, and consider the subspace measures µE∩Z and νF . These are still effectively locally finite and τ -additive (414K), and are now totally finite. Also E ∩ Z has a countable network. So (a) tells us that the relatively open set W ∩ ((E ∩ Z) × F ) is measured by the c.l.d. product of µE∩Z and νF , which is the subspace measure on (E ∩ Z) × F induced by λ (251P). Since λ surely measures E × F , it measures W ∩ (Z × Y ) ∩ (E × F ). As

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E and F are arbitrary, λ measures W ∩ (Z × Y ) (251H). But λ((X \ Z) × Y ) = µ(X × Z) · νY = 0 (251Ia), so λ also measures W . As W is arbitrary, λ is the τ -additive product measure. 417U Proposition Let h(Xi , Ti , Σi , µi )ii∈I be a family Q of τ -additive topological probability spaces. Let λ be the ordinary product probability measure on X = i∈I Xi and Λ its domain. Then every continuous function f : X → R is Λ-measurable, so Λ includes the Baire σ-algebra of X. ˜ be a τ -additive topological measure extending λ (417E), and Λ ˜ its domain; then f is proof (a) Let λ ˜ is a topological measure. For α ∈ R, set ˜ Λ-measurable, just because λ Gα = {x : x ∈ X, f (x) < α},

Hα = {x : x ∈ X, f (x) > α},

Fα = {x : x ∈ X, f (x) = α}. ˜ α > 0} is countable, and A0 = R \ A is dense in R; let Then hFα iα∈R is disjoint, so A = {α : α ∈ R, λF 0 Q ⊆ A be a countable dense set. For each q ∈ Q, let Vq ⊆ Gq , Wq ⊆ Hq be such that ˜ q, λVq = λ∗ Gq = λG

˜ q λWq = λ∗ Hq = λH

(413Ea, 417E(iii)). Then ˜ λ∗ (Gq \ Vq ) ≤ λ(X \ (Vq ∪ Wq )) = 1 − λVq − λWq = λ(X \ (Gq ∪ Hq )) = 0. Because λ is complete, Gq \ Vq and Gq belong to Λ. But now, if α ∈ R, S {x : f (x) < α} = q∈Q,q<α Gq ∈ Λ, so f is Λ-measurable. (b) It follows that every zero set belongs to Λ, so that Λ must include the Baire σ-algebra of X. 417V Proposition Let (X, T, Σ, µ) and (Y, S, T, ν) be effectively locally finite τ -additive topological measure spaces, and (X × Y, Λ, λ) their c.l.d. product. Then every continuous function f : X × Y → R is Λ-measurable, and the Baire σ-algebra of X × Y is included in Λ. proof Let Z ⊆ X × Y be a zero set. If E ∈ Σ, F ∈ T are sets of finite measure, then Z ∩ (E × F ) is a zero set for the relative topology of E × F . Now the subspace measures µE and νF are τ -additive topological measures (414K), so Z ∩ (E × F ) is measured by the c.l.d. product µE × νF of µE and νF . P P If either µE or νF is zero, this is trivial. Otherwise, they have scalar multiples µ0E , νF0 which are probability measures, and of course are still τ -additive topological measures. By 417U, Z ∩ (E × F ) is measured by µ0E × νF0 . Since µE × νF is just a scalar multiple of µ0E × νF0 , Z ∩ (E × F ) is measured by µE × νF . Q Q But µE × νF is the subspace measure λE×F (251P), so Z ∩ (E × F ) ∈ Λ. As E and F are arbitrary, Z ∈ Λ (251H). Thus every zero set belongs to Λ; accordingly Λ must include the Baire σ-algebra, and every continuous function must be Λ-measurable. 417X Basic exercises (a) Let (X, Σ, µ) be a semi-finite measure space and A a family of subsets of X. Show that the following S are equiveridical: (i) there is a measure µ0 on X, extending µ, such that µ0 A = 0 for every A ∈ A; (ii) µ∗ ( n∈N An ) = 0 for every sequence hAn in∈N in A. (b) Let (X, Σ, µ) and (Y, T, ν) be measure spaces with topologies with respect to which µ and ν are locally finite. Show that the c.l.d. product measure on X × Y is locally finite. > (c) Let (X, T, Σ, µ) and (Y, S, T, ν) be topological measure spaces such that µ and ν are both effectively locally finite τ -additive Borel measures. Show that there is a unique effectively locally finite τ -additive Borel measure λ0 on X × Y such that λ0 (G × H) = µG · νH for all open sets G ⊆ X, H ⊆ Y . >(d) Let h(Xi , Ti , Σi , µi )ii∈I be a family of topological probability spaces in whichQevery µi is a τ additive Borel Q measure. Show that there is a unique τ -additive Borel measure λ0 on X = i∈I Xi such that Q 0 λ ( i∈I Fi ) = i∈I µi Fi whenever Fi ⊆ Xi is closed for every i ∈ I.

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(e) Let (X, T, Σ, µ) and (Y, S, T, ν) be effectively locally finite τ -additive topological measure spaces in ˜ the τ -additive product measure which the measures are inner regular with respect to the Borel sets, and λ on X × Y . Let hXi ii∈I , hYj ij∈J be decompositions for µ, ν respectively (definition: 211E). Show that ˜ (Cf. 251N.) hXi × Yj ii∈I,j∈J is a decomposition for λ. > (f ) Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces such that µi is inner ˜ the τ -additive product measure on X = Q Xi . regular with respect to the Borel sets for every i, and λ i∈I ˜ Take Ai ⊆ Xi for each i ∈ I. (i) Show that if µ∗i Ai = 1 for every i, then the subspace measure induced by λ Q # ˜ on A = i∈I Ai is just the τ -additive product λ of the subspace measures on the Ai . (Hint: show that if ˜ # (W ∩ A) for Borel sets W ⊆ X, then λ0 satisfies the conditions of 417Xd.) (ii) Show that we set λ0 W = λ ∗ ˜ A = Q µ∗ Ai . (Cf. 254L.) in any case λ i∈I i (g) Let h(Xi , Ti , Σi , µi )ii∈I and h(Yi , Si , Ti , νi )ii∈I be two families of τ -additive topological probability ˜ λ ˜ 0 be the spaces in which every µi and every Q νi is inner regularQwith respect to the Borel sets. Let λ, τ -additive product measures on X = i∈I Xi and Y = i∈I Yi respectively. Suppose that for each i ∈ I we are given a continuous inverse-measure-preserving function φi : Xi → Yi . Show that the function φ : X → Y defined by setting φ(x)(i) = φi (x(i)) for x ∈ X, i ∈ I is inverse-measure-preserving. (h) Let (X, T, Σ, µ) and (Y, S, T, ν) be two complete locally determined effectively locally finite τ -additive ˜ topological measure spaces such that both µ and ν are inner regular with respect to the Borel sets. Let λ ˜ its domain. Suppose that ν is σ-finite. Show that for be the τ -additive product measure on X × Y , and Λ ˜ W [{x}] ∈ T for almost every x ∈ X, and x 7→ νW [{x}] is measurable. any W ∈ Λ, > (i) Let (X, T, Σ, µ) be [0, 1] with its usual topology and Lebesgue measure, and let (Y, S, T, ν) be [0, 1] with the discrete topology and counting measure. (i) Show that both are Radon measure spaces. (ii) Show that the c.l.d. product measure on X × Y is a Radon measure. (Hint: 252Kc, or use 417T and 417P.) (iii) Show that 417Ha can fail if we omit the hypothesis on {(x, y) : f (x, y) 6= 0}. (j) Let (X, T, Σ, µ) and (Y, S, T, ν) be two effectively locally finite τ -additive topological measure spaces. ˜ the τ -additive product measure on X ×Y . Show that λ∗ (A×B) = Let λ be the c.l.d. product measure and λ ∗ ˜ λ (A × B) for all sets A ⊆ X, B ⊆ Y . (Hint: start with A, B of finite outer measure, so that 417I applies.) (k) Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces with strictly positive ˜ their τ -additive product. For ˜ λ) measures all inner regular with respect to the Borel sets, and (X, T, Λ, ˜ J be the τ -additive product measure on XJ = Q ˜ X , and Λ J ⊆ I let λ i J its domain. (i) Show that if f is i∈J ˜ ˜ a real-valued Λ-measurable function defined λ-almost everywhere on X, we can find a countable set J ⊆ I ˜ J -almost everywhere on XJ , such that f extends gπJ . (ii) In (i), ˜ J -measurable function g, defined λ and a Λ R R ˜ ˜ show that f dλ = g dλJ if either is defined in [−∞, ∞]. (l) Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces such that every µi is ˜ their τ -additive product. Show that for any ˜ λ) inner regular with respect to the Borel sets, and (X, T, Λ, ˜ ˜ determined by coordinates in J, with W ∈ Λ there is a smallest set J ⊆ I for which there is a W 0 ∈ Λ, 0 ˜ λ(W 4W ) = 0. (Hint: 254R.) (m) What needs to be added to 417M and 415Xk to complete a proof of 415E? (n) Let (X, T, Σ, µ) be an atomless τ -additive topological probability space such that µ is inner regular with respect to the Borel sets, and I a set of cardinal at most that of the support of µ. Show that the set of injective functions from I to X has full outer measure for the τ -additive product measure on X I . > (o) Let (X, T, Σ, µ) and (Y, S, T, ν) be Radon measure spaces. Show that the Radon product measure ˜ such that λ(K ˜ × L) = µK · νL for all compact sets K ⊆ X, L ⊆ Y . on X × Y is the unique Radon measure λ >(p) Let I be an uncountable set, and for each i ∈ I let Xi be {0, 1}, Ti = P{0, 1} the usual topology, and µi the measure on Σi = P{0, 1} defined by saying that µi A = 1 if 0 ∈ A, 0 otherwise. Check that

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˜ be the ordinary and τ -additive product measures on (Xi , Ti , Σi , µi ) is a Radon probability space. Let λ, λ Q I ˜ is not determined by any X = i∈I Xi = {0, 1} . Show that they are different. Show that the support of λ ˜ countable set of coordinates. Find a λ-negligible open set W ⊆ X such that its projection onto {0, 1}J is conegligible for every countable J ⊆ I. ˜ the quasi-Radon product (q) Let h(Xi , Ti , Σi , µi )ii∈I be a family of Radon probability spaces, and λ Q ˜ is a Radon measure measure on X = i∈I Xi . For each i ∈ I, let Zi ⊆ Xi be the support of µi . Show that λ iff {i : i ∈ I, Zi is not compact} is countable. In particular, show that the ordinary product measure on I [0, 1[ , where I is uncountable and each copy of [0, 1[ is given Lebesgue measure, is a quasi-Radon measure, but not a Radon measure. (r) Let h(Xn , Tn , Σn , µn )in∈N be a sequence of Radon probability spaces. Show that the Radon product Q Q∞ ˜ on X such that λ( ˜ Q measure on X = n∈N Xn is the unique Radon measure λ n∈N Kn ) = n=0 µn Kn whenever Kn ⊆ Xn is compact for every n. (s) Let (X, T, Σ, ν) and (Y, S, T, ν) be two topological measure spaces, of which (X, T) has a countable network. (i) Show that the c.l.d. product measure λ on X × Y is a topological measure. (ii) Show directly, without relying on ideas from 417D, that λ is τ -additive if ν is. (t) Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces such that every µi ˜ the ordinary and τ -additive product measures on is inner sets, and λ, λ Q regular with respect to the Borel ˜ boundary, then A is measured by λ. X = i∈I Xi . Show that if A ⊆ X has λ-negligible (u) Let us say that a topological space X is chargeable if there is an additive functional ν : PX → [0, ∞[ such that νG > 0 for every non-empty open set G ⊆ X. (i) Show that if there is a σ-finite measure µ on X such that µ∗ G > 0 for every non-empty open set G, then X is chargeable. (Hint: 215B(vii), 391G.) (ii) Show that any separable space is chargeable. (iii) Show that X is chargeable iff its regular open algebra is chargeable in the sense of 391X. (Hint: see the proof of 314P.) (iv) Show that any open subspace of a chargeable space is chargeable. (v) Show that if Y ⊆ X is dense, then X is chargeable iff Y is chargeable. (vi) Show that if X is expressible as the union of countably many chargeable subspaces, then it is chargeable. (vii) Show that any product of chargeable spaces is chargeable. (Cf. 391Xb(iii).) (viii) Show that if hXi ii∈I is a family of chargeable spaces with product X, then every regular open subset of X and every Baire subset of X is determined by coordinates in a countable set. (Hint: 4A2Eb, 4A3Mb.) (v) Let (X, T, Σ, µ) and (Y, S, T, ν) be quasi-Radon measure spaces such that µX · νY > 0. Show that the quasi-Radon product measure on X × Y is completion regular iff it is equal to the c.l.d. product measure and µ and ν are both completion regular. (Hint: 412Sc; if µE, νF are finite and Z ⊆ E × F is a zero set of positive measure, use Fubini’s theorem to show that Z has sections of positive measure.) (w) Let h(Xi , Ti , Σ Qi , µi )ii∈I be a family of quasi-Radon probability spaces. Show that the quasi-Radon product measure on i∈I Xi is completion regular iff it is equal to the ordinary product measure and every µi is completion regular. (x) Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces such that Q every µi is inner regular with respect to the Borel sets, and λ the τ -additive product measure on X = i∈I Xi ; write Λ for its domain. (i) Show that if W ∈ Λ, λW > 0 and ² > 0 then there are a finite J ⊆ I and a W 0 ∈ Λ such that λW 0 ≥ 1 − ² and for every x ∈ W 0 there is a y ∈ W such that x¹I \ J = y¹I \ J. (Cf. 254Sb.) (ii) Show that if A ⊆ X is determined by coordinates in I \ {i} for every i ∈ I then λ∗ A ∈ {0, 1}. (Cf. 254Sa.) 417Y Further exercises (a) Give an example to show that, in 417A, λ can be strictly localizable while λ0 is not. (b) Let (X, T, Σ, µ) and (Y, S, T, ν) be effectively locally finite τ -additive topological measure spaces such that µ and ν are both inner regular with respect to the Borel sets. (i) Fix open Rsets G ⊆ X, H ⊆ Y of finite measure. Let WGH be the set of those W ⊆ X × Y such that θGH (W ) = G νˆ(W [{x}] ∩ H)dx is defined, where νˆ is the completion of ν. (α) Show that every

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S open set belongs to WGH . (β) Show that θGH is countably additive in the sense that θGH ( n∈N Wn ) = P∞ S n=0 θGH (Wn ) for every disjoint sequence hWn in∈N in WGH , and τ -additive in the sense that θGH ( V) = supV ∈V θGH (V ) for every non-empty upwards-directed family V of open sets in X × Y . (γ) Show that every Borel set belongs to WGH . (Hint: Monotone Class Theorem.) (δ) Writing B for the Borel σ-algebra of X × Y , show that θGH ¹B is a τ -additive Borel measure; let λGH be its completion. (²) Show that λGH = θGH ¹ΛGH , where ΛGH = dom λGH . (ζ) Show that λGH (E × F ) is defined and equal to µE · νF whenever E ∈ Σ, F ∈ T, E ⊆ G and F ⊆ H. (Hint: start with open E and F , move to Borel E and F with the Monotone Class Theorem.) (η) Writing λ for the c.l.d. product measure on X × Y , show that λGH (W ) is defined and equal to λ(W ∩ (G × H)) whenever W ∈ dom λ. ˜ = sup ˜ to be T{ΛGH : G ∈ T, H ∈ S, µG < ∞, νH < ∞} and λW (ii) Now take Λ G,H λGH (W ) for ˜ ˜ W ∈ Λ. Show that λ is an extension of λ to a complete locally determined effectively locally finite τ -additive topological measure on X × Y which is inner regular with respect to the Borel sets, so is the τ -additive product measure as defined in 417G. (c) Let (X, Σ, µ) and (Y, T, ν) be complete measure spaces with topologies T, S. Suppose that µ and ν are effectively locally finite and τ -additive and moreover that their domains include bases for the two topologies. Show that the c.l.d. product measure on X × Y has the same properties. (Hint: start by assuming that µX and νY are both finite. If V is an upwards-directed family of measurable open sets with measurable open union W , look at gV (x) = νV [{x}] for V ∈ V.) (d) Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces such that every µi ˜ their τ -additive product. (i) Show that the ˜ λ) is inner regular with respect to the Borel sets, and (X, T, Λ, ˜ following are equiveridical: (α) µi is strictly positive for all but countably many i ∈ I; (β) whenever W ∈ Λ ˜ there are a countable J ⊆ I and W1 , W2 ∈ Λ, determined by coordinates in J, such that W1 ⊆ W ⊆ W2 and ˜ 2 \ W1 ) = 0. (ii) Show that when these are false, λ ˜ cannot be equal to the ordinary product measure λ(W on X. (e) Let (X, Σ, µ) and (Y, T, ν) be measure spaces with topologies T, S such that both µ and ν are inner regular with respect to the families of sequentially compact sets in each space. Show that the c.l.d. product measure λ on X × Y is also inner regular with respect to the sequentially compact sets, so has an extension to a topological measure which is inner regular with respect to the sequentially compact sets. (Hint: 412R, 416Yc.) (f ) Let h(Xi , Σi , µi )ii∈I be a family of probability spaces with topologies Ti such that every µi is inner regular with respect to the family of closedQcountably compact sets in Xi and every Xi is compact. Show that the ordinary product measure λ on X = i∈I Xi is also inner regular with respect to the closed countably ˜ which is inner regular with respect to the compact sets, so has an extension to a topological measure λ ˜ closed countably compact sets in X. Show that this can be done in such a way that for every W ∈ dom λ ˜ there is a V ∈ dom λ such that λ(W 4V ) = 0. (Hint: 412T, 416Yb.) (g) Let h(Xn , Σn , µn )in∈N be a sequence of probability spaces with topologies Tn such that every µn is inner regular with respect to the family of sequentially compact sets in Xn . Show that the ordinary product Q measure λ on X = n∈N Xn is also inner regular with respect to the sequentially compact sets, so has an ˜ which is inner regular with respect to the sequentially compact sets in extension to a topological measure λ ˜ there is a V ∈ dom λ such that X. Show that this can be done in such a way that for every W ∈ dom λ ˜ λ(W 4V ) = 0. ˜ (h) Let h(Xi , Ti , Σi , µi )ii∈I be a family Q of quasi-Radon probability spaces, and λ, λ the ordinary and quasi-Radon product measures on X = i∈I Xi . Suppose that all but one of the Ti have countable networks ˜ and all but countably many of the µi are strictly positive. Show that λ = λ. (i) Let us say that a quasi-Radon measure space (X, T, Σ, µ) has the simple product property if the c.l.d. product measure on X × Y is equal to the quasi-Radon product measure for every quasi-Radon measure space (Y, S, T, ν). (i) Show that if (X, T) has a countable network then (X, T, Σ, µ) has the simple

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product property. (ii) Show that if a quasi-Radon measure space has the simple product property so do all its subspaces. (iii) Show that the quasi-Radon product of two quasi-Radon measure spaces with the simple product property has the simple product property. (iv) Show that the quasi-Radon product of any family of quasi-Radon probability spaces with the simple product property has the simple product property. (v) Show that the real line with the right-facing Sorgenfrey topology (415Xc) and Lebesgue measure has the simple product property. 417 Notes and comments The general problem of determining just when a measure can be extended to a measure with given properties is one which will recur throughout this volume. I have more than once mentioned the Banach-Ulam problem; if you like, this is the question of whether there can ever be an extension of the countable-cocountable measure on a set X to a measure defined on the whole algebra PX. This particular question appears to be undecidable from the ordinary axioms of set theory; but for many sets (for instance, if X = ω1 ) it is known that the answer is ‘no’. (See 419G, 438C.) This being so, we have to take each manifestation of the general question on its own merits. In 417C and 417E the challenge is to take a product measure λ defined in terms of the factor measures alone, disregarding their topological properties, and extend it to a topological measure, preferably τ -additive. Of course there are important cases in which λ is itself already a topological measure; for instance, we know that the c.l.d. product of Lebesgue measure on R with itself is Lebesgue measure on R 2 (251M), and other examples are in 415E, 415Ye, 416U, 417S-417T and 417Yk. But in general not every open set in the product belongs to the domain of λ, even when we have the product of two Radon measures on compact Hausdorff spaces (419E). Once we have resolved to grasp the nettle, however, there is a natural strategy for the proof. It is easy ˜ then we to see that if λ, in 417C or 417E, is to have an extension to a τ -additive topological measure λ, ˜ must have λA(V) = 0 for every V belonging to the class V. Now 417A descibes a sufficient (and obviously necessary) condition for there to be an extension of λ with this property. So all we have to do is check. The check is not perfectly straightforward; in 417E it uses all the resources of the original proof that there is a product measure on an arbitrary product of probability spaces (which I suppose is to be expected), with 414B (of course) to apply the hypothesis that the factor measures are τ -additive, and a couple of extra wrinkles (the Wn0 and Cn0 of part (c-ii) of the proof of 417E, and the use of supports in part (c-v)). ˜ are It is worth noting that (both for finite and for infinite products) the measure algebras of λ and λ ˜ and the identical (417C(ii), 417E(ii)), so there is no new work to do in identifying the measure algebra of λ associated function spaces. An obstacle we face in 417C-417E is the fact that not every τ -additive measure µ has an extension to a τ additive topological measure, even when µ is totally finite and its domain includes a base for the topology. (I give an example in 419H.) Consequently it is not enough, in 417C or 417E, to show that the ordinary product measure λ is τ -additive. But perhaps I should remark that if λ is inner regular with respect to the closed sets, this obstacle evaporates (415L). Accordingly, for the principal applications (to quasi-Radon and Radon product measures, and in particular whenever the topological spaces involved are regular) we have rather easier proofs available, based on the constructions of §415. For completely regular spaces, there is yet another approach, because the product measures can be described in terms of the integrals of continuous functions (415I), which by 417U and 417V can be calculated from the ordinary product measures. Of course the proof that λ itself is τ -additive is by no means trivial, especially in the case of infinite products, corresponding to 417E; but for finite products there are relatively direct arguments, applying indeed to slightly more general situations (417Yc). If we have measures which are inner regular with respect to countably compact classes of sets, then there may be other ways of approaching the extension, using theorems from §413 (see 417Ye-417Yg), and for compact Radon measure spaces, λ becomes tight (412Sb, 412V), so its τ -additivity is elementary. As always, it is important to recognise which constructions are in some sense canonical. The arguments of 417C and 417E allow for the possibility that the factor measures are defined on σ-algebras going well beyond the Borel sets. For all the principal applications, however, the measures will be c.l.d. versions of Borel measures, and in particular will be inner regular with respect to the Borel sets. In such a context it is natural to ask for product measures with the same property, and in this case we can identify a canonical τ -additive topological product measure, as in 417D and 417F. (If you prefer to restrict your measures to Borel σ-algebras, you again get canonical product Borel measures (417Xc-417Xd).) Having done so, we

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can reasonably expect ‘commutative’ and ‘associative’ and ‘distributive’ laws, as in 417D, 417J and 417Xe. Subspaces mostly behave themselves (417I, 417Xf). Of course extending the product measure means that we get new integrable functions on the product, so that Fubini’s theorem has to be renegotiated. Happily, it remains valid, at least in the contexts in which it was effective before (417Ha); we still need, in effect, one of the measures to be σ-finite. The theorem still fails for arbitrary integrable functions on products of Radon measure spaces, and the same example works as before (417Xi). In fact this means that we have an alternative route to the construction of the τ -additive product of two measures (417Yb). But note that on this route ‘commutativity’, the identification of the product measure on X × Y with thatR on Y × X, becomes something which can no longer be taken ˜ to be νW [{x}]dx we have to worry about when, and why, this will for granted, Rbecause if we define λW −1 be equal to µW [{y}]dy. A version of Tonelli’s theorem follows from Fubini’s theorem, as before (417Hc). We also have results corresponding to most of the theorems of §254. But note that there are two traps. In the theorem that a measurable set can be described in terms of a projection onto a countable subproduct (254O, 417M) we need to suppose that the factor measures are strictly positive, and in the theorem that a product of Radon measures is a Radon measure (417Q) we need to suppose that the factor measures have compact supports. The basic examples to note in this context are 417Xp and 417Xq. It is not well understood when we can expect c.l.d. product measures to be topological measures, even in the case of compact Radon probability spaces. Example 419E remains a rather special case, but of course much more effort has gone into seeking positive results. Note that the ordinary product measures of this section are always effectively locally finite and τ -additive (417C, 417E), so that they will be equal to the τ additive products iff they measure every open set (417S). Regarding infinite products, the τ -additive product measure can fail to be the ordinary product measure in just two ways: if one of the finite product measures is not a topological measure, or if uncountably many of the factor measures are not strictly positive (417Sc, 417Xp, 417Yd). So it is finite products which need to be studied. Q Whenever we have a subset F of an infinite product X = i∈I Xi , it is important to know when F is determined by coordinates in a proper subset of I; in measure theory, we are particularly interested in sets determined by coordinates in countable subsets of I (254Mb). It may happen that there is a smallest set J such that F is determined by coordinates in J; for instance, when we have a topological product and F is closed (4A2Bg). When we have a product of probability spaces, we sometimes wish to identify sets J such that F is ‘essentially’ determined by coordinates in J, in the sense that there is an F 0 , determined by coordinates in J, such that F 4F 0 is negligible. In this context, again, there is a smallest such set (254Rd), which can be identified in terms of the probability algebra free product of the measure algebras (325Mb). In 417Ma the two ideas come together: under the conditions there, we get the same smallest J by either route. In 417Ma, we have a product of strictly positive τ -additive topological probability measures. If we keep the ‘strictly positive’ but abandon everything else, we still have very striking results just because the product topology is ccc, so that we can apply 4A2Eb. An abstract expression of this idea is in 417Xu.

418 Measurable functions and almost continuous functions In this section I work through the basic properties of measurable and almost continuous functions, as defined in 411L and 411M. I give the results in the full generality allowed by the terminology so far introduced, but most of the ideas are already required even if you are interested only in Radon measure spaces as the domains of the functions involved. Concerning the codomains, however, there is a great difference between metrizable spaces and others, and among metrizable spaces separability is of essential importance. I start with the elementary properties of measurable functions (418A-418C) and almost continuous functions (418D). Under mild conditions on the domain space, almost continuous functions are measurable (418E); for a separable metrizable codomain, we can expect that measurable functions should be almost continuous (418J). Before coming to this, I spend a couple of paragraphs on image measures: a locally finite image measure under a measurable function is Radon if the measure on the domain is Radon and the function is almost continuous (418I).

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418L-418Q are important results on expressing given Radon measures as image measures associated with continuous functions, first dealing with ordinary functions f : X → Y (418L) and then coming to Prokhorov’s theorem on projective limits of probability spaces (418M). The machinery of the first part of the section can also be used to investigate representations of vectorvalued functions in terms of product spaces (418R-418T). 418A Proposition Let X be a set, Σ a σ-algebra of subsets of X, Y a topological space and f : X → Y a measurable function. (a) f −1 [F ] ∈ Σ for every Borel set F ⊆ Y . (b) If A ⊆ X is any set, endowed with the subspace σ-algebra, then f ¹A : A → Y is measurable. (c) Let (Z, T) be another topological space. Then gf : X → Z is measurable for every Borel measurable function g : Y → Z; in particular, for every continuous function g : Y → Z. proof (a) The set {F : F ⊆ Y, f −1 [F ] ∈ Σ} is a σ-algebra of subsets of Y containing every open set, so contains every Borel subset of Y . (b) is obvious from the definition of ‘subspace σ-algebra’ (121A). (c) If H ⊆ Z is open, then g −1 [H] is a Borel subset of Y so (gf )−1 [H] = f −1 [g −1 [H]] belongs to Σ. 418B Proposition Let X be a set and Σ a σ-algebra of subsets of X. (a) If Y is a metrizable space and hfn in∈N is a sequence of measurable functions from X to Y such that f (x) = limn→∞ fn (x) is defined in Y for every x ∈ X, then f : X → Y is measurable. (b) If Y is a topological space, Z is a separable metrizable space and f : X → Y , g : X → Z are functions, then x 7→ (f (x), g(x)) : X → Y × Z is measurable iff f and g are measurable. (c) If Y is a hereditarily Lindel¨of space, U a family of open sets generating its topology, and f : X → Y a function such that f −1 [U ] ∈ Σ for every U ∈ U, then f is measurable. (d) If hYi ii∈I is a countable family of separable metrizable spaces, with product Y , then a function f : X → Y is measurable iff πi f : X → Yi is measurable for every i, writing πi (y) = y(i) for y ∈ Y , i ∈ N. proof (a) Let ρ be a metric defining the topology of Y . Let G ⊆ Y be any open set, and for each n ∈ N set Fn = {y : y ∈ Y, ρ(y, z) ≥ 2−n for every z ∈ Y \ G}. Then Fn is closed, so fi−1 [Fn ] ∈ Σ for every n, i ∈ N. But this means that S T f −1 [G] = n∈N i≥n fi−1 [Fi ] ∈ Σ. As G is arbitrary, f is measurable. (b)(i) The functions (y, z) 7→ y, (y, z) 7→ z are continuous, so if x 7→ (f (x), g(x)) is measurable, so are f and g, by 418Ac. (ii) Now suppose that f and g are measurable, and that W ⊆ Y × Z is open. By 4A2P(a-i), the topology of Z has a countable base H; let hHn in∈N be a sequence running over H ∪ {∅}. For each n, set S Gn = {G : G ⊆ Y is open, G × Hn ⊆ W }; S S then Gn is open and Gn × Hn ⊆ W . Accordingly W ⊇ n∈N Gn × Hn . But in fact W = n∈N Gn × Hn . P P If (y, z) ∈ W , there are open sets G ⊆ Y , H ⊆ Z such that (y, z) ∈ G × H ⊆ W . Now there is an n ∈ N such that z ∈ Hn ⊆ H, in which case G × Hn ⊆ W and G ⊆ Gn and (y, z) ∈ Gn × Hn . Q Q Accordingly S {x : (f (x), g(x)) ∈ W } = n∈N f −1 [Gn ] ∩ g −1 [Hn ] ∈ Σ. As W is arbitrary, x 7→ (f (x), g(x)) is measurable. (c) This is just 4A3Db. (d) If f is measurable, so is every πi f , by 418Ac. If every πi f is measurable, set U = {πi−1 [H] : i ∈ I, H ⊆ Yi is open}. Then U generates the topology of Y , and if U = πi−1 [H] then f −1 [U ] = (πi f )−1 [H], so f −1 [U ] ∈ Σ for every U . Also Y is hereditarily Lindel¨of (4A2P(a-iii)), so f is measurable, by (c).

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418C Proposition Let (X, Σ, µ) be a measure space and Y a Polish space. Let hfn in∈N be a sequence of measurable functions from X to Y . Then {x : x ∈ X, limn→∞ fn (x) is defined in Y } belongs to Σ. proof (Compare 121H.) Let ρ be a complete metric on Y defining the topology of Y . (a) For m, n ∈ N and δ > 0, the set {x : ρ(fm (x), fn (x)) ≤ δ} belongs to Σ. P P The function x 7→ (fm (x), fn (x)) : X → Y 2 is measurable, by 418Bb, and the function ρ : Y 2 → R is continuous, so x → 7 ρ(fm (x), fn (x)) is measurable and {x : ρ(fm (x), fn (x)) ≤ δ} ∈ Σ. Q Q (b) Now hfn (x)in∈N is convergent iff it is Cauchy, because Y is complete. But {x : x ∈ X, hfn (x)in∈N is Cauchy} =

\ [ \

{x : ρ(fi (x), fm (x)) ≤ 2−n }

n∈N m∈N i≥m

belongs to Σ. 418D Proposition Let (X, Σ, µ) be a measure space and T a topology on X. (a) Suppose that Y is a topological space. Then any continuous function from X to Y is almost continuous. (b) Suppose that Y and Z are topological spaces, f : X → Y is almost continuous and g : Y → Z is continuous. Then gf : X → Z is almost continuous. (c) Suppose that (Y, S, T, ν) is a σ-finite topological measure space, Z is a topological space, g : Y → Z is almost continuous and f : X → Y is inverse-measure-preserving and almost continuous. Then gf : X → Z is almost continuous. (d) Suppose that µ is semi-finite, and that hYi ii∈I is a countable family of topological spaces with product Y . Then a function f : X → Y is almost continuous iff fi = πi f is almost continuous for every i ∈ I, writing πi (y) = y(i) for i ∈ I, y ∈ Y . proof (a) is trivial. (b) The set {A : A ⊆ X, gf ¹A is continuous} includes {A : A ⊆ X, f ¹A is continuous}; so if µ is inner regular with respect to the latter, it is inner regular with respect to the former. (c) Take E ∈ Σ and γ < µE; take ² > 0. We have a cover of Y by a non-decreasing sequence hYn in∈N of measurable sets of finite measure; now hf −1 [Yn ]in∈N is a non-decreasing sequence covering E, so there is an n ∈ N such that µ(E ∩ f −1 [Yn ]) ≥ γ. Because f is inverse-measure-preserving, E ∩ f −1 [Yn ] has finite measure. Now we can find measurable sets F ⊆ Yn , E1 ⊆ E ∩ f −1 [Yn ] such that f ¹E1 , g¹F are continuous and νF ≥ νYn − ², µE1 ≥ µ(E ∩ f −1 [Yn ] \ E1 ) − ². In this case E0 = E1 ∩ f −1 [F ] has measure at least γ − 2² and gf ¹E0 is continuous. As E, γ and ² are arbitrary, gf is almost continuous. (d)(i) If f is almost continuous, every fi must be almost continuous, by (b). (ii) Now suppose that every fi is almost continuous. Take E ∈ Σ and γ < µE. There is an E0 ⊆ E such P that E0 ∈ Σ and γ < µE0 < ∞. Let h²i ii∈I be a family of strictly positive real numbers such that i∈I ²i ≤ µE0 − γ. For each Ti ∈ I choose a measurable set Fi ⊆ E0 such that µFi ≥ µE0 − ²i and fi ¹Fi is continuous. Then F = E0 ∩ i∈I Fi is a subset of E with measure at least γ, and f ¹F is continuous because fi ¹F is continuous for every i (3A3Ib). 418E Theorem Let (X, T, Σ, µ) be a complete locally determined topological measure space, Y a topological space, and f : X → Y an almost continuous function. Then f is measurable. proof Set K = {K : K ∈ Σ, f ¹K is continuous}; then µ is inner regular with respect to K. If H ⊆ Y is open and K ∈ K, then K ∩ f −1 [H] is relatively open in K, that is, there is an open set G ⊆ X such that K ∩ f −1 [H] = K ∩ G. Because µ is a topological measure, G ∈ Σ so K ∩ f −1 [H] ∈ Σ. As K is arbitrary, and µ is complete and locally determined, f −1 [H] ∈ Σ (412Ja). As H is arbitrary, f is measurable.

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418F Proposition Let (X, T, Σ, µ) be a semi-finite topological measure space, Y a metrizable space, and f : X → Y a function. Suppose there is a sequence hfn in∈N of almost continuous functions from X to Y such that f (x) = limn→∞ fn (x) for almost every x ∈ X. Then f is almost continuous. proof Suppose that E ∈ Σ and that γ < µE, ² > 0. Then there is a measurable set F ⊆ E such that γ ≤ µF < ∞; discarding a negligible set if necessary, we may arrange that f (x) = limn→∞ fn (x) for every x ∈ F . Let ρ be a metric on Y defining its topology. For each T n ∈ N, let Fn ⊆ F be a measurable set such that fn ¹Fn is continuous and µ(Fn \ F ) ≤ 2−n ²; set G = n∈N Fn , so that µG ≥ γ − 2² and fn ¹G is continuous for every n ∈ N. For m, n ∈ N, the functions x 7→ (fm (x), fn (x)) : G → Y 2 and x 7→ ρ(fm (x), fn (x)) : G → R are continuous, therefore measurable, because µ is a topological measure. Also hfn (x)in∈N is a Cauchy sequence for every x ∈ G. So if we set Gkn = {x : x ∈ G, ρ(fi (x), fj (x)) ≤ 2−k for all i, j ≥ n}, hGkn in∈N is a nondecreasing sequence of measurable sets with union G for each k ∈ N, and T we can find a strictly increasing sequence hnk ik∈N such that µ(G\Gknk ) ≤ 2−k ² for every k. Setting H = k∈N Gknk , µH ≥ µG−2² ≥ γ −4² and ρ(fi (x), fnk (x)) ≤ 2−k whenever x ∈ H and i ≥ nk ; consequently ρ(f (x), fnk (x)) ≤ 2−k whenever x ∈ H and k ∈ N. But this means that hfnk ik∈N converges to f uniformly on H, while every fnk is continuous on H, so f ¹H is continuous (3A3Nb). And of course H ⊆ E. As E, γ and ² are arbitrary, f is almost continuous. 418G Proposition Let (X, T, Σ, µ) be a σ-finite quasi-Radon measure space, Y a metrizable space and f : X → Y an almost continuous function. Then there is a conegligible set X0 ⊆ X such that f [X0 ] is separable. proof (a) Let K be the family of self-supporting measurable sets K of finite measure such that f ¹K is continuous. Then µ is inner regular with respect to K. P P If E ∈ Σ and γ < µE, there is an F ∈ Σ such that F ⊆ E and γ < µF < ∞; there is an H ∈ Σ such that H ⊆ F , γ ≤ µH and f ¹H is continuous; and there is a measurable self-supporting K ⊆ H with the same measure as H (414F), in which case K ∈ K and K ⊆ E and µK ≥ γ. Q Q (b) Now f [K] is ccc for every K ∈ K. P P If G is a disjoint family of non-empty relatively open subsets of f [K], then hK ∩ P f −1 [G]iG∈G is a disjoint family of non-empty relatively open subsets of K, because f ¹K is continuous, and G∈G µ(K ∩ f −1 [G]) ≤ µK. Because K is self-supporting, µ(K ∩ f −1 [G]) > 0 for every G ∈ G; because µK is finite, G is countable. As G is arbitrary, f [K] is ccc. Q Q Because Y is metrizable, f [K] must be separable (4A2Pd). S (c) BecauseSµ is σ-finite, there is a countable family L ⊆ K such that X0 = L is conegligible (412Ic). Now f [X0 ] = L∈L f [L] is a countable union of separable spaces, so is separable (4A2B(e-i)). 418H Proposition (a) Let X and Y be topological spaces, µ an effectively locally finite τ -additive measure on X, and f : X → Y an almost continuous function. Then the image measure µf −1 is τ -additive. (b) Let (X, T, Σ, µ) be a totally finite quasi-Radon measure space, (Y, S) a regular topological space, and f : X → Y an almost continuous function. Then there is a unique quasi-Radon measure ν on Y such that f is inverse-measure-preserving for µ and ν. proof (a) Let of Y , all measured by µf −1 , and suppose S H be an upwards-directed family of open−1subsets ∗ ∗ that H = H is also measurable. Take any γ < (µf )(H ) = µf −1 [H ∗ ]. Then there is a measurable set E ⊆ f −1 [H ∗ ] such that µE ≥ γ and f ¹E is continuous. Consider {E ∩ f −1 [H] : H ∈ H}. This is an upwards-directed family of relatively open measurable subsets of E with measurable union E. By 414K, the subspace measure on E is τ -additive, so γ ≤ µE ≤ supH∈H µ(E ∩ f −1 [H]) ≤ supH∈H µf −1 [H]. As γ is arbitrary, µf −1 [H ∗ ] ≤ supH∈H µf −1 [H]; as H is arbitrary, µf −1 is τ -additive. (b) By 418E, f is measurable. Let ν0 be the restriction of µf −1 to the Borel σ-algebra of Y ; by (a), ν0 is τ -additive, and f is inverse-measure-preserving with respect to µ and ν0 . Because Y is regular, the completion ν of ν0 is a quasi-Radon measure (415Cb). Because µ is complete, f is still inverse-measurepreserving with respect to µ and ν (235Hc).

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To see that ν is unique, observe that its values on Borel sets are determined by the requirement that f be inverse-measure-preserving, so that 415H gives the result. 418I The next theorem is one of the central properties of Radon measures. I have already presented what amounts to a special case in 256G. Theorem Let (X, T, Σ, µ) be a Radon measure space, Y a Hausdorff space, and f : X → Y an almost continuous function. If the image measure ν = µf −1 is locally finite, it is a Radon measure. proof (a) By 418E, f is measurable, that is, f −1 [H] ∈ Σ for every open set H ⊆ Y ; but this means that the domain T of ν contains every open set, and ν is a topological measure. (b) ν is tight (that is, inner regular with respect to the compact sets). P P If F ∈ T and νF > 0, then µf −1 [F ] > 0, so there is an E ⊆ f −1 [F ] such that µE > 0 and f ¹E is continuous. Next, there is a compact set K ⊆ E such that µK > 0. In this case, L = f [K] is a compact subset of F , and νL = µf −1 [L] ≥ µK > 0. By 412B, this is enough to prove that ν is tight. Q Q Note that because ν is locally finite, νL < ∞ for every compact L ⊆ Y (411Ga). (c) Because µ is complete, so is ν (212Bd). Next, ν is locally determined. P P Suppose that H ⊆ Y is such that H ∩ F ∈ T whenever νF < ∞. Then, in particular, H ∩ f [K] ∈ T whenever K ⊆ X is compact and f ¹K is continuous. But setting K = {K : K ⊆ X is compact, f ¹K is continuous}, µ is inner regular with respect to K (412Ac). And if K ∈ K, K ∩ f −1 [H] = K ∩ f −1 [H ∩ f [K]] ∈ Σ. Because µ is complete and locally determined, this is enough to show that f −1 [H] ∈ Σ (412Ja), that is, H ∈ T. As H is arbitrary, ν is locally determined. Q Q (d) Thus ν is a complete locally determined tight locally finite topological measure; that is, it is a Radon measure. 418J Theorem Let (X, Σ, µ) be a semi-finite measure space and T a topology on X such that µ is inner regular with respect to the closed sets. Suppose that Y is a separable metrizable space and f : X → Y is measurable. Then f is almost continuous. proof Let H be a countable base for the topology of Y , and hHn in∈N a sequence running over H ∪ {∅}. Take E ∈ Σ and γ < µE. Choose hEn in∈N inductively, as follows. There is an E0 ∈ Σ such that E0 ⊆ E and γ < µE0 < ∞. Given En ∈ Σ with γ < µEn < ∞, En \ f −1 [Hn ] ∈ Σ, so there is a closed set Fn ∈ Σ such that Fn ⊆ En \ f −1 [Hn ],

µ((En \ f −1 [Hn ]) \ Fn ) < µEn − γ;

set En+1 = (En ∩ f −1 [Hn ]) ∪ Fn , so that En+1 ∈ Σ,

En+1 ⊆ En ,

µEn+1 > γ,

En+1 \ f −1 [Hn ] = Fn .

Continue. T At the end of the induction, set F = n∈N En . Then F ⊆ E, µF ≥ γ, and for every n ∈ N F ∩ f −1 [Hn ] = F ∩ En+1 ∩ f −1 [Hn ] = F \ Fn is relatively open in F . It follows that f ¹F is continuous (4A2B(a-ii)). As E, γ are arbitrary, f is almost continuous. Remark For variations on this idea, see 418Yg, 433E and 434Yb; also 418Yh. 418K Corollary Let (X, T, Σ, µ) be a quasi-Radon measure space and Y a separable metrizable space. Then a function f : X → Y is measurable iff it is almost continuous. proof Put 418E and 418J together. Remark This generalizes 256F.

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418L In all the results above, the measure starts on the left of the diagram f : X → Y ; in 418I-418K, it is transferred to an image measure on Y . If X has enough compact sets, a measure can move in the reverse direction, as follows. Theorem Let (X, T) be a Hausdorff space, (Y, S, T, ν) a Radon measure space and f : X → Y a continuous function such that whenever F ∈ T and νF > 0 there is a compact set K ⊆ X such that ν(F ∩ f [K]) > 0. Then there is a Radon measure µ on X such that ν is the image measure µf −1 and the inverse-measurepreserving function f induces an isomorphism between the measure algebras of ν and µ. proof (a) Note first that ν is inner regular with respect to L = {f [K] : K ∈ K}, where K is the family of compact subsets of X. P P If νF > 0, there is a K ∈ K such that ν(F ∩ f [K]) > 0; now there is a closed set F 0 ⊆ F ∩ f [K] such that νF 0 > 0, and K 0 = K ∩ f −1 [F 0 ] is compact, while f [K 0 ] ⊆ F has non-zero measure. As L is closed under finite unions, this is enough to show that ν is inner regular with respect to L (412Aa). Q Q (b) Consequently there is a disjoint set L0 ⊆ L such that every non-negligible F ∈ T meets some member of L0 S in a non-negligible set (412Ib). We can express L0 as {f [K] : K ∈ K0 } where K0 ⊆ K is disjoint. Set X0 = K0 . (c) Set Σ0 = {X0 ∩ f −1 [F ] : F ∈ T}. Then Σ0 is a σ-algebra of subsets of X0 . If F , F 0 ∈ T and νF 6= νF 0 , then there must be some K ∈ K0 such that f [K]∩(F 4F 0 ) 6= ∅, so that X0 ∩f −1 [F ] 6= X0 ∩f −1 [F 0 ]; we therefore have a functional µ0 : Σ0 → [0, ∞] defined by setting µ0 (X0 ∩ f −1 [F ]) = νF whenever F ∈ T. It is easy to check that µ0 is a measure on X0 . Now µ0 is inner regular with respect to K. P P If E ∈ Σ0 and µE > 0, there is an F ∈ T such that E = X0 ∩f −1 [F ] and νF > 0. There are a K ∈ K0 such that ν(F ∩f [K]) > 0, and a closed set F 0 ⊆ F ∩f [K] such that νF 0 > 0; now K ∩ f −1 [F 0 ] = X0 ∩ f −1 [F 0 ] belongs to Σ0 ∩ K, is included in E and has measure greater than 0. Because K is closed under finite unions, this is enough to show that µ0 is inner regular with respect to K. Q Q (c) Set Σ1 = {E : E ⊆ X, E ∩ X0 ∈ Σ0 },

µ1 E = µ0 (E ∩ X0 ) for every E ∈ Σ1 .

Then µ1 is a measure on X (being the image measure µ0 ι−1 , where ι : X0 → X is the identity map), and is inner regular with respect to K. If F ∈ T, then µ1 f −1 [F ] = µ0 (X0 ∩ f −1 [F ]) = νF , so f is inverse-measure-preserving for µ1 and ν. Consequently µ1 is locally finite. P P If x ∈ X, there is an open set H ⊆ Y such that f (x) ∈ H and νH < ∞; now f −1 [H] is an open subset of X of finite measure containing x. Q Q In particular, µ∗1 K < ∞ for every compact K ⊆ X (411Ga). (d) By 413O, there is an extension of µ1 to a complete locally determined measure µ on X which is inner regular with respect to K, defined on every member of K, and such that whenever E belongs to the domain Σ of µ and µE < ∞, there is an E1 ∈ Σ1 such that µ(E4E1 ) = 0. Now µ is locally finite because µ1 is, so µ is a Radon measure; and f is inverse-measure-preserving for µ and ν because it is inverse-measurepreserving for µ1 and ν. (e) The image measure µf −1 extends ν, so is locally finite, and is therefore a Radon measure (418I); since it agrees with ν on the compact subsets of Y , it must be identical with ν. (f ) I have still to check that the corresponding measure-preserving homomorphism π from the measure algebra B of ν to the measure algebra A of µ is actually an isomorphism, that is, is surjective. If a ∈ A and µ ¯a < ∞, we can find E ∈ Σ such that E • = a and E1 ∈ Σ1 such that µ(E4E1 ) = 0. Now E1 ∩ X0 = −1 f [F ] ∩ X0 for some F ∈ T; but in this case µ(E1 4f −1 [F ]) = µ1 (E1 4f −1 [F ]) = 0,

a = E1• = (f −1 [F ])• = πF • .

Accordingly π[B] includes {a : µ ¯a < ∞}, and is order-dense in A. But as π is injective and B is Dedekind complete (being the measure algebra of a Radon measure, which is strictly localizable), it follows that π[B] = A (314Ib). Thus π is an isomorphism, as required.

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Remarks Of course this result is most commonly applied when X and Y are both compact and f is a surjection, in which case the condition (*) whenever F ∈ T and νF > 0 there is a compact set K ⊆ X such that ν(F ∩ f [K]) > 0 is trivially satisfied. Evidently (*) is necessary if there is to be any Radon measure on X for which f is inverse-measurepreserving, so in this sense the result is best possible. In 433D, however, there is a version of the theorem in which f is not required to be continuous. 418M Prokhorov’s theorem Suppose that (I, ≤), h(Xi , Ti , Σi , µi )ii∈I , hfij ii≤j∈I , (X, T) and hgi ii∈I are such that (I, ≤) is a non-empty upwards-directed partially ordered set, every (Xi , Ti , Σi , µi ) is a Radon probability space, fij : Xj → Xi is an inverse-measure-preserving function whenever i ≤ j in I, (X, T) is a Hausdorff space, gi : X → Xi is a continuous function for every i ∈ I, gi = fij gj whenever i ≤ j in I. Suppose moreover that for every ² > 0 there is a compact set K ⊆ X such that µi gi [K] ≥ 1 − ² for every i ∈ I. Then there is a Radon probability measure µ on X such that every gi is inverse-measure-preserving for µ. proof (a) Set T = {gi−1 [E] : i ∈ I, E ∈ Σi } ⊆ PX. Then T is a subalgebra of PX. P P (i) There is an i ∈ I, so ∅ = gi−1 [∅] belongs to T. (ii) If H ∈ T there are −1 i ∈ I, E ∈ Σi such that H = gi [E]; now X \ H = gi−1 [Xi \ E] belongs to T. (iii) If G, H ∈ T, there are i, j ∈ I and E ∈ Σi , F ∈ Σj such that G = gi−1 [E] and H = gj−1 [F ]. Now I is upwards-directed, so there is −1 −1 a k ∈ I such that i ≤ k and j ≤ k. Because fik and fjk are inverse-measure-preserving, fik [E] and fjk [F ] belong to Σk , so that G ∩ H = gi−1 [E] ∩ gj−1 [F ] = (fik gk )−1 [E] ∩ (fjk gk )−1 [F ] −1 −1 = gk−1 [fik [E] ∩ fjk [F ]] ∈ T. Q Q

(b) There is an additive functional ν : T → [0, 1] defined by writing νgi−1 [E] = µi E whenever i ∈ I and E ∈ Σi . P P (i) Suppose that i, j ∈ I and E ∈ Σi , F ∈ Σj are such that gi−1 [E] = gj−1 [F ]. Let k ∈ I be such that i ≤ k and j ≤ k. Then −1 −1 gk−1 [fik [E]4fjk [F ]] = gi−1 [E]4gj−1 [F ] = ∅, −1 −1 so gk [X] ∩ (fik [E]4fjk [F ]) = ∅. But now remember that for every ² > 0 there is a set K ⊆ X such that −1 −1 µk gk [K] ≥ 1 − ². This means that µk gk [X] must be 1, so that fik [E]4fjk [F ] must be negligible, and −1 −1 µi E = µk fik [E] = µk fjk [F ] = µj F .

Thus the proposed formula for ν defines a function on T. (ii) Now suppose that G, H ∈ T are disjoint. Again, take i, j ∈ I and E ∈ Σi , F ∈ Σj such that G = gi−1 [E] and H = gj−1 [F ], and k ∈ I such that i ≤ k and j ≤ k. Then −1 −1 νG + νH = µi E + µj F = µk fik [E] + µk fjk [F ] −1 −1 −1 −1 = µk (fik [E] ∪ fjk [F ]) + µk (fik [E] ∩ fjk [F ]) −1 −1 −1 −1 = νgk−1 [fik [E] ∪ fjk [F ]] + νgk−1 [fik [E] ∩ fjk [F ]]

= ν(G ∪ H) + ν(G ∩ H).

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But as ν∅ is certainly 0, we get ν(G ∪ H) = νG + νH. As G, H are arbitrary, ν is additive. Q Q Note that νX = 1. (c) νG = sup{νH : H ∈ T, H ⊆ G, H is closed} for every G ∈ T. P P If γ < νG, there are an i ∈ I and an E ∈ Σi such that G = gi−1 [E]. In this case µi E = νG > γ; let L ⊆ E be a compact set such that µi L ≥ γ; then H = gi−1 [L] is a closed subset of G and νH = µi L ≥ γ. Q Q νX = supK⊆X is compact inf G∈T,G⊇K νG. P P If ² > 0, there is a compact K ⊆ X such that µi gi [K] ≥ 1 − ² for every i ∈ I, by the final hypothesis of the theorem. If G ∈ T and G ⊇ K, there are an i ∈ I and an E ∈ Σi such that G = gi−1 [E], in which case gi [K] ⊆ E, so that νG = µi E ≥ µi gi [K] ≥ 1 − ². Thus inf G∈T,G⊇K νG ≥ 1 − ²; as ² is arbitrary, we have the result. Q Q This means that the conditions of 416O are satisfied, and there is a Radon measure µ extending ν. Of course this means that every gi is inverse-measure-preserving. 418N Remarks (a) Taking I to be a singleton, we get a version of 418L in which Y is a probability space, and omitting the check that the function g induces an isomorphism of the Q measure algebras. Taking I to be the family of finite subsets of a set T , and every Xi to be a product t∈i Zt of Radon probability spaces with its product Radon measure, we obtain a method of constructing products of arbitrary families of compact probability spaces from finite products. (b) In the hypotheses of 418M, I asked only that the fij should be measurable, and omitted any check on the compositions fij fjk when i ≤ j ≤ k. But it is easy to see that the fij must in fact be almost continuous, and that fij fjk must be equal almost everywhere to fik (418Xu), just as in 418P below. (c) In the theorem as written out above, the space X and the functions gi : X → Xi are part of the data. Of course in many applications we start with a structure (h(Xi , Ti , Σi , µi )ii∈I , hfij ii≤j∈I ), and the first step is to find a suitable X and gi , as in 418O and 418P. (d) There are important questions concerning possible relaxations of the hypotheses in 418M, especially in Q Q the special case already mentioned, in which Xi = t∈i Zt , fij (x) = x¹i when i ⊆ j ∈ [T ]<ω , X = t∈T Zt , and gi (x) = x¹i for x ∈ X and i ∈ I, but there is no suggestion that the µi are product measures. For a case in which we can dispense with auxiliary topologies on the Xi , see 451Yb. (e) A typical class of applications of Prokhorov’s theorem is in the theory of stochastic processes, in which we have large families hXt it∈T of random variables; for definiteness, imagine that T = [0, ∞[, so that we are looking at a system evolving over time. Not infrequently our intuition leads us to a clear description of the joint distributions νJ of finite subfamilies hXt it∈J without providing any suggestion of a measure space on which the whole family hXt it∈T might be defined. (As I tried to explain in the introduction to Chapter 27, probability spaces themselves are often very shadowy things in true probability theory.) Each νJ can be thought of as a Radon measure on R J , and for I ⊆ J ∈ [T ]<ω we have a natural map fIJ : R J → R I , setting fIJ (y) = y¹I for y ∈ R J . If our distributions νJ mean anything at all, every fIJ will surely be inversemeasure-preserving; this is simply saying that νI is the joint distribution of a subfamily of hXt it∈J . If we can find a Hausdorff space Ω and a continuous function g : Ω → R T such that, for every finite J ⊆ T and ² > 0, there is a compact set K ⊆ Ω such that νJ gJ [K] ≥ 1 − ² (where gJ (x) = g(x)¹J), then Prokhorov’s theorem will give us a measure µ on Ω which will then provide us with a suitable realization of hXt it∈T as a family of random variables on a genuine probability space, writing Xt (ω) = g(ω)(t). That they become continuous functions on a Radon measure space is a valuable shield against irrelevant complications. Clearly, if this can be done at all it can be done with Ω = R T ; but some of the central results of probability theory are specifically concerned with the possibility of using other sets Ω (e.g., Ω = C(T ), as in 455D). (f ) In (e) above, we do always have the option of regarding each νJ as a measure on the compact space [−∞, ∞]J . In this case, by 418O or otherwise, we can be sure of finding a measure on [−∞, ∞]T to support functions Xt , at the cost of either allowing the values ±∞ or (as I should myself ordinarily do) accepting

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that each Xt would be undefined on a negligible set. The advantage of this is just that it gives us confidence in applying the Kolmogorov-Lebesgue theory to the whole family hXt it∈T at once, rather than to finite or countable subfamilies. For an example of what can happen if we try to do similar things with non-compact measures, see 419K. For an example of the problems which can arise with uncountable families, see 418Xv. 418O I mention two cases in which we can be sure that the projective limit (X, hgi ii∈I ) required in Prokhorov’s theorem will exist. Proposition Suppose that (I, ≤), h(Xi , Ti )ii∈I and hfij ii≤j∈I are such that (I, ≤) is a non-empty upwards-directed partially ordered set, every (Xi , Ti ) is a compact Hausdorff space, fij : Xj → Xi is a continuous surjection whenever i ≤ j in I, fij fjk = fik whenever i ≤ j ≤ k in I. Then there are a compact Hausdorff space (X, T) and a family hgi ii∈I such that gi : X → Xi is a continuous surjection and fij gj = fi whenever i ≤ j ∈ I; so that if we endow the Xi with Radon probability measures for which the fij are inverse-measure-preserving, (I, hXi ii∈I , hfij ii≤j∈I , X, hgi ii∈I ) will satisfy all the hypotheses of 418M. proof Set X = {x : x ∈

Q i∈I

Xi , fij x(j) = x(i) whenever i ≤ j ∈ I},

gi (x) = x(i) for x ∈ X, i ∈ I. Of course gi = fij gj whenever i ≤ j. To see that every gi is surjective, observe that if y ∈ Xi and J ⊆ I is finite, then Q FJ = {x : x ∈ i∈I Xi , x(i) = y, fjk x(k) = x(j) whenever j ≤ k ∈ J} is a closed set. FJ is always non-empty, because if k is an upper bound of J ∪ {i} there is a z ∈ Xk such that fik (z) = y, in which case x ∈ FJ whenever x(j) = fjk (z) for every j ∈ J ∪Q {i}. Now {FJ : J ∈ [I]<ω } is a downwards-directed family of non-empty closed sets in the compact space j∈I Xj , so has non-empty intersection, and if x is any point of the intersection then x ∈ X and gi (x) = y. 418P Proposition Let (I, ≤), h(Xi , Ti , Σi , µi )ii∈I and hfij ii≤j∈I be such that (I, ≤) is a countable non-empty upwards-directed partially ordered set, every (Xi , Ti , Σi , µi ) is a Radon probability space, fij : Xj → Xi is an inverse-measure-preserving almost continuous function whenever i ≤ j in I, fij fjk = fik µk -a.e. whenever i ≤ j ≤ k in I. Then there are a Radon probability space (X, T, Σ, µ) and continuous inverse-measure-preserving functions gi : X → Xi such that gi = fij gj whenever i ≤ j in I. proof (a) We can use the same formula as in 418O: Q X = {x : x ∈ i∈I Xi , fij x(j) = x(i) whenever i ≤ j ∈ I}, gi (x) = x(i) for x ∈ X, i ∈ I. As before, the consistency relation gi = fij gj is a trivial consequence of the definition of X. For the rest, we have to check that the final condition P of 418M is satisfied. Fix ² ∈ ]0, 1[. Start by taking a family h²ij ii≤j∈I of strictly positive numbers such that i≤j∈I ²ij ≤ 12 ². (This is where we need to know that I is countable.) P P Set ²j = i≤j ²ij for each j, so that j∈I ²j ≤ 12 ². For i ≤ j ≤ k in I, set Eijk = {x : x ∈ Xk , fik (x) = fij fjk (x)},

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T so that Eijk is µk -conegligible; set Ek = i≤j≤k Eijk , so that Ek is µk -conegligible. For i ≤ j ∈ I, choose compact sets Kij ⊆ Ej such that µj Kij ≥ 1 − ²ij and fij ¹Kij is continuous. Now we seem to need a three-stage construction, as follows: T for j ∈ I, set Kj = i≤j Kij ; T −1 for j ∈ I, set Kj∗ = Kj ∩ i≤j fij [Ki ]; Q ∗ finally, set K = X ∩ i∈I Ki . Let us trace the properties of these sets stage by stage. (b) For each j ∈ I, Kj ⊆ Kjj ⊆ Ej is compact and P P µj (Xj \ Kj ) ≤ i≤j µj (Xj \ Kij ) ≤ i≤j ²ij = ²j , so that µj Kj ≥ 1 − ²j . Note that fik agrees with fij fjk on Kk whenever i ≤ j ≤ k, and that fij ¹Kj is continuous whenever i ≤ j. (c) Every Kj∗ is compact, and if i ≤ j ≤ k then fik agrees with fij fjk on Kk∗ , while fij ¹Kj∗ is always continuous. Also X −1 µj (Xj \ Kj∗ ) ≤ µj (Xj \ Kj ) + µj (Xj \ fij [Ki ]) ≤ ²j +

X

i≤j

²i ≤ ²,

i≤j

so µj Kj∗ ≥ 1 − ², for every j ∈ I. −1 The point of moving from Kj to Kj∗ is that fjk [Kk∗ ] ⊆ Kj∗ whenever j ≤ k in I. P P Kk∗ ⊆ fjk [Kj ], so ∗ fjk [Kk ] ⊆ Kj . If i ≤ j, then −1 −1 −1 Kk∗ = Kk∗ ∩ fik [Ki ] = Kk∗ ∩ fjk [fij [Ki ]] −1 because fij fjk agrees with fik on Kk∗ . So fjk [Kk∗ ] ⊆ fij [Ki ]. As i is arbitrary, fjk [Kk∗ ] ⊆ Kj∗ . Q Q Again because fik agrees with fij fjk on Kk∗ , we have fik [Kk∗ ] = fij [fjk [Kk∗ ]] ⊆ fij [Kj∗ ] whenever i ≤ j ≤ k. And because fij ¹Kj∗ is always continuous, all the sets fij [Kj∗ ] are compact.

(d)(i) K is compact. P P Q

K = {x : x ∈

Q i∈I

Ki∗ , fij x(j) = x(i) whenever i ≤ j ∈ I}

is closed in i∈I Ki∗ because fij ¹Kj∗ is always continuous (and every Xi is Hausdorff). Since compact, so is K. Q Q

Q i∈I

Ki∗ is

(ii) µi gi [K] ≥ 1−² for every i ∈ I. P P By (c), fik [Kk∗ ] ⊆ fij [Kj∗ ] whenever i ≤ j ≤ k. So {fij [Kj∗ ] : j ≥ i} is a downwards-directed family of compact sets; write L for their intersection. Since −1 µi fij [Kj∗ ] = µj fij [fij [Kj∗ ]] ≥ µj Kj∗ ≥ 1 − ²

for every j ≥ i, µi L ≥ 1 − ² (414C). If z ∈ L, then for every k ≥ i the set Q Fk = {x : x ∈ j∈I Kj∗ , x(k) = z, fjk x(k) = x(j) whenever j ≤ k} Q is a non-empty closed set in j∈I Kj∗ , while Fk ⊆ Fj when j ≤ k; so that {Fk : k ≥ i} is a downwards-directed T family of non-empty closed sets in a compact space, and has non-empty intersection. But if x ∈ k≥i Fk , then x ∈ K and x(k) = z, so z ∈ gi [K]. Thus gi [K] ⊇ L and µi gi [K] ≥ 1 − ². Q Q (e) As ² is arbitrary, the final condition of 418M is satisfied. But now 418M tells us that (β) is true. 418Q Corollary Let h(Xn , Tn , Σn , µn )in∈N be a sequence of Radon probability spaces, and suppose we are given an inverse-measure-preserving almost continuous function fn : Xn+1 → Xn for each n. Set Q X = {x : x ∈ n∈N Xn , fn (x(n + 1)) = x(n) for every n ∈ N}. Then there is a unique Radon probability measure µ on X such that all the coordinate maps x 7→ x(n) : X → Xn are inverse-measure-preserving.

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proof (a) For i ≤ j ∈ N, define fij : Xj → Xi by writing fii (x) = x for every x ∈ Xi , fi,j+1 = fij fj for every j ≥ i. It is easy to check that fij fjk = fik whenever i ≤ j ≤ k, and that every fij is inverse-measure-preserving and almost continuous (using 418Dc). So we are exactly in the situation of 418P, and we know that there is a Radon probability measure on X for which every gi is inverse-measure-preserving. (b) To see that µ is unique, observe that if K ⊆ X is compact and x ∈ X \ K then there is some finite set I ⊆ N such that {y : y¹I = x¹I} is disjoint from K. But as gn = fn gn+1 for every n, this means that {y : gn (y) = gn (x)} is disjoint from K for all n large enough. As x is arbitrary, we see that hgn−1 [gn [K]]in∈N is a non-increasing sequence with intersection K, so that µK = limn→∞ µgn−1 [gn [K]] = limn→∞ µn gn [K]. Thus the values of µ on compact sets are determined by the construction. By 416E(b-ii), µ is uniquely defined. 418R I turn now to a special kind of measurable function, corresponding to a new view of product spaces. Theorem Let X be a set, Σ a σ-algebra of subsets of X, and (Y, T, ν) a σ-finite measure space. Give L0 (ν) b b the topology of convergence in measure (§245). Write L0 (Σ⊗T) for the space of Σ⊗T-measurable functions b is the σ-algebra of subsets of X × Y generated by {E × F : E ∈ Σ, F ∈ T}. h : X × Y → R, where Σ⊗T Then for a function f : X → L0 (ν) the following are equiveridical: (i) f [X] is separable and f is measurable; b (ii) there is an h ∈ L0 (Σ⊗T) such that f (x) = h•x for every x ∈ X, where hx (y) = h(x, y) for x ∈ X, y ∈Y. proof Let hYn in∈N be a non-decreasing sequence of subsets of Y of finite measure covering Y . (a)(i)⇒(ii) pseudometric on L0 (ν) defined by saying that R For each n ∈ N, let ρn be the continuous 0 • • ρn (g1 , g2 ) = Yn min(1, |g1 − g2 |)dν for g1 , g2 ∈ L (T), writing L0 (T) for the space of T-measurable realvalued functions on Y (245A). Then {ρn : n ∈ N} defines the topology of L0 (ν) (see the proof of 245Eb). Because f [X] is separable, there is a sequence hvk ik∈N in L0 (ν) such that f [X] ⊆ {vk : k ∈ N}. For each k, choose gk ∈ L0 (T) such that gk• = vk . For n, k ∈ N set Enk = {x : x ∈ X, ρn (f (x), vk ) < 2−n }, S Hnk = Enk \ i
Yn

≤ ρn (gk•m+1 , f (x)) + ρn (f (x), gk•m ) ≤ ρm+1 (gk•m+1 , f (x)) + ρm (f (x), gk•m ) ≤ 3 · 2−m−1 . But this means that

P∞

R

m=0 Yn

(m+1)

min(1, |hx

(m)

(m)

−hx |) is finite, so that hhx im∈N must be convergent almost (m)

everywhere on Yn . As this is true for every n, hhx im∈N is convergent a.e. on Y . Moreover, (m)

limm→∞ (hx )• = limm→∞ gk•m = f (x) in L0 (ν).

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Since this is true for every x, W = {(x, y) : hh(m) (x, y)im∈N converges in R} b because every h(m) is Σ⊗T-measurable b has conegligible vertical sections, while of course W ∈ Σ⊗T (418C). (m) b If we set h(x, y) = limm→∞ h (x, y) for (x, y) ∈ W , 0 for other (x, y) ∈ X × Y , then h ∈ L0 (Σ⊗T), while (by 245Ca) (m)

h•x = limm→∞ (hx )• = f (x) in L0 (ν) for every x ∈ X. So we have a suitable h. b (ii)⇒(i) Let Φ be the set of those h ∈ L0 (Σ⊗T) such that (i) is satisfied; that is, x 7→ h•x is measurable, and {h•x : x ∈ X} is separable. Then ˜ belong to Φ, set A = {h• : x ∈ X}, A˜ = {h ˜ • : x ∈ X}. Then α) Φ is closed under addition. P (α P If h, h x x ˜• ) : both A and A˜ are separable metrizable spaces, so A × A˜ is separable and metrizable and x 7→ (h•x , h x 0 ˜ X → A × A is measurable (418Bb). But addition on L (ν) is continuous (245D), so ˜ • = (h + h) ˜ • x 7→ h•x + h x x is measurable (418Ac), and ˜ • : x ∈ X} ⊆ {u + u ˜ ˜ : u ∈ A, u ˜ ∈ A} {(h + h) x ˜ ∈ Φ. Q is separable (4A2Be). Thus h + h Q β ) Φ is closed under scalar multiplication, just because u 7→ αu : L0 (ν) → L0 (ν) is always continuous. (β (γγ ) If hh(n) in∈N is a sequence in Φ and h(x, y) = limn→∞ h(n) (x, y) for all x ∈ X, y ∈ Y , then h ∈ Φ. S (n) P P Setting An = {(hx )• : x ∈ X} for each n, then A = {h•x : x ∈ X} is included in n∈N An , which is (n) separable (4A2B(e-i)), so A is separable (4A2P(a-iv)); moreover, h•x = limn→∞ (hx )• for every x ∈ X, so • Q x 7→ hx is measurable, by 418Ba. Q b χW ∈ Φ}, then W \ W 0 ∈ W whenever (δδ ) What this means is that if weSset W = {W : W ∈ Σ⊗T, 0 0 W , W ∈ W and W ⊆ W , and that n∈N Wn ∈ W whenever hWn in∈N is a non-decreasing sequence in W. Also, it is easy to see that E × F ∈ W whenever E ∈ Σ and F ∈ T. By the Monotone Class Theorem b (136B), W includes the σ-algebraP generated by {E × F : E ∈ Σ, F ∈ T}, that is, is equal to Σ⊗T. It follows n b at once, from (α) and (β), that i=0 αi χWi ∈ Φ whenever W0 , . . . , Wn ∈ Σ⊗T and α0 , . . . , αn ∈ R, and b hence (using (γ)) that L0 (Σ⊗T) ⊆ Φ, which is what we had to prove. 418S Corollary Let (X, Σ, µ) and (Y, T, ν) be σ-finite measure spaces with c.l.d. product (X × Y, Λ, λ). Give L0 (ν) the topology of convergence in measure. Write L0 (λ) for the space of Λ-measurable real-valued functions defined λ-a.e. on X × Y , as in §241. (a) If h ∈ L0 (λ), set hx (y) = h(x, y) whenever this is defined. Then {x : f (x) = h•x is defined in L0 (ν)} is µ-conegligible, and includes a conegligible set X0 such that f : X0 → L0 (ν) is measurable and f [X0 ] is separable. (b) If f : X → L0 (ν) is measurable and there is a conegligible set X0 ⊆ X such that f [X0 ] is separable, then there is an h ∈ L0 (λ) such that f (x) = h•x for almost every x ∈ X. b (251K). So there is a conegligible proof (a) The point is that λ is just the completion of its restriction to Σ⊗T ˜ b b set W ∈ Σ⊗T such that h¹ W is Σ⊗T-measurable (212Fa). Setting h(x, y) = h(x, y) for (x, y) ∈ W , 0 ˜ • for every x ∈ X, we see from 418R that f˜ is measurable and that f˜[X] is otherwise, and setting f˜(x) = h x separable. But 252D tells us that X0 = {x : ((X × Y ) \ W )[{x}] is negligible} ˜ x ν-a.e., so that f (x) is defined and equal to f˜(x). This proves is conegligible; and if x ∈ X0 then hx = h the result.

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b (b) f ¹X0 satisfies 418R(i). So, setting f1 (x) = f (x) for x ∈ X0 , 0 otherwise, there is some h ∈ L0 (Σ⊗T) such that f1 (x) = h•x for every x, so that f (x) = h•x for almost every x, and (ii) is true. 418T Corollary (Mauldin & Stone 81) Let (Y, T, ν) be a σ-finite measure space, and (B, ν¯) its measure algebra, with its measure-algebra topology (§323). (a) Let X be a set, Σ a σ-algebra of subsets of X, and f : X → B a function. Then the following are equiveridical: (i) f [X] is separable and f is measurable; b such that f (x) = W [{x}]• for every x ∈ X. (ii) there is a W ∈ Σ⊗T (b) Let (X, Σ, µ) be a σ-finite measure space and Λ the domain of the c.l.d. product measure λ on X × Y . (i) Suppose that ν is complete. If W ∈ Λ, then {x : f (x) = W [{x}]• is defined in B} is µ-conegligible, and includes a conegligible set X0 such that f : X0 → B is measurable and f [X0 ] is separable. (ii) If f : X → B is measurable and there is a conegligible set X0 ⊆ X such that f [X0 ] is separable, b such that f (x) = W [{x}]• for almost every x ∈ X. then there is a W ∈ Σ⊗T proof Everything follows directly from 418R and 418S if we observe that B is homeomorphically embedded in L0 (ν) by the function F • 7→ (χF )• for F ∈ T (323Xf, 367R). We do need to check, for (i)⇒(ii) of part b b such that (a), that if h ∈ L0 (Σ⊗T) and h•x is always of the form (χF )• , then there is some W ∈ Σ⊗T • • hx = (χW [{x}]) for every x; but of course this is true if we just take W = {(x, y) : h(x, y) = 1}. Now (b-ii) follows from (a) just as 418Sb followed from 418R. 418X Basic exercises > (a) Let (X, Σ, µ) be a measure space, Y a set and h : X → Y a function; give Y the image measure µh−1 . Show that for any function g from Y to a topological space Z, g is measurable iff gh : X → Z is measurable. > (b) Let X be a set, Σ a σ-algebra of subsets of X, hYn in∈N a sequence ofQ topological spaces with product Y , and f : X → Y a function. Show that f is measurable iff ψn f : X → i≤n Yi is measurable for every n ∈ N, where ψn (y) = (y(0), . . . , y(n)) for y ∈ Y , n ∈ N. (c) Let (X, Σ, µ) be a semi-finite measure space, (Y, S) a metrizable space, and hfn in∈N a sequence of measurable functions from X to Y such that hfn (x)in∈N is convergent for almost every x ∈ X. Show that µ is inner regular with respect to {E : hfn ¹Ein∈N is uniformly convergent}. (Cf. 412Xr.) > (d) Set Y = [0, 1][0,1] , with the product topology. For x ∈ [0, 1], n ∈ N define fn (x) ∈ Y by saying that fn (x)(t) = max(0, 1 − 2n |x − t|) for t ∈ [0, 1]. Check that (i) each fn is continuous, therefore measurable; (ii) f (x) = limn→∞ fn (x) is defined in Y for every x ∈ [0, 1]; (iii) for each t ∈ [0, 1], the coordinate functional x 7→ f (x)(t) is continuous except at t, and in particular is almost continuous and measurable; (iv) f ¹F is not continuous for any infinite closed set F ⊆ [0, 1], and in particular f is not almost continuous; (v) every subset of [0, 1] is of the form f −1 [H] for some open set H ⊆ Y ; (vi) f is not measurable; (vii) the image measure µf −1 , where µ is Lebesgue measure on [0, 1], is neither a topological measure nor tight. (e) Let (X, T, Σ, µ) be a quasi-Radon measure space, Y a topological space, and f : X → Y a function. Suppose that for every x ∈ X there is an open set G containing x such that f ¹G is almost continuous with respect to the subspace measure on G. Show that f is almost continuous. (f ) For i = 1, 2 let (Xi , Ti , Σi , µi ) and (Yi , Si , Ti , νi ) be quasi-Radon measure spaces, and fi : Xi → Yi an almost continuous inverse-measure-preserving function. Show that (x1 , x2 ) 7→ (f1 (x1 ), f2 (x2 )) is inversemeasure-preserving for the quasi-Radon product measures. (g) Let h(Xi , Ti , Σi , µi )ii∈I and h(Yi , Ti , Σi , νi )ii∈I be two families of topological Q spaces with τ -additive Q Borel probability measures, and let µ, ν be the τ -additive product measures on X = i∈I Xi , Y = i∈I Yi . Suppose that every νi is strictly positive. Show that if fi : Xi → Yi is almost continuous and inversemeasure-preserving for each i, then x 7→ hfi (x(i))ii∈I : X → Y is inverse-measure-preserving, but need not be almost continuous.

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(h) Let (X, T, Σ, µ) and (Y, S, T, ν) be quasi-Radon measure spaces, (Z, U) a topological space and f : X × Y → Z a function which is almost continuous with respect to the quasi-Radon product measure on X × Y . Suppose that ν is σ-finite. Show that y 7→ f (x, y) is almost continuous for almost every x ∈ X. (i) Let (X, T, Σ, µ) be an effectively locally finite τ -additive topological measure space, Y a topological space and f : X → Y an almost continuous function. (i) Show that the image measure µf −1 is τ -additive. (ii) Show that if µ is a totally finite quasi-Radon measure and the topology on Y is regular, then µf −1 is quasi-Radon. (j) Let (X, T, Σ, µ) be a topological measure space and U a linear topological space. Show that if f : X → U and g : X → U are almost continuous, then f + g : X → U is almost continuous. (k) Let (X, T, Σ, µ) and (Y, S, T, ν) be topological measure spaces, and (Z, U) a topological space; let f : X → Y be almost continuous and inverse-measure-preserving, and g : Y → Z almost continuous. Show that if either µ is a Radon measure and ν is locally finite or µ is τ -additive and effectively locally finite and ν is effectively locally finite, then gf : X → Z is almost continuous. (Hint: show that if µE > 0 there is a set F such that νF < ∞ and µ(E ∩ f −1 [F ]) > 0.) (l) Let (X, Σ, µ) be a complete strictly localizable measure space, φ : Σ → Σ a lower density such that φX = X, and T the associated density topology on X (414P). Let f : X → R be a function. Show that the following are equiveridical: (i) f is measurable; (ii) f is almost continuous; (iii) f is continuous at almost every point; (iv) there is a conegligible set H ⊆ X such that f ¹H is continuous. (Cf. 414Xk.) (m) Let (X, Σ, µ) be a complete strictly localizable measure space, φ : Σ → Σ a lifting, and S the lifting topology on X (414Q). Let f : X → R be a function. Show that the following are equiveridical: (i) f is measurable; (ii) f is almost continuous; (iii) there is a conegligible set H ⊆ X such that f ¹H is continuous. (Cf. 414Xr.) (n) Let (X, T, Σ, µ) be a quasi-Radon measure space, (Y, S) a regular topological space and f : X → Y an almost continuous function. Show S that there is a quasi-Radon measure ν on Y such that f is inversemeasure-preserving for µ and ν iff {f −1 [H] : H ⊆ Y is open, µf −1 [H] < ∞} is conegligible in X. (o) Let X and Y be Hausdorff spaces and f : X → Y a continuous injective function. Show that if µ1 and µ2 are distinct totally finite Radon measures on X then µ1 f −1 6= µ2 f −1 . (p) Let (X, T, Σ, µ) be a Radon measure space, (Y, S) and (Z, U) Hausdorff spaces, f : X → Y an almost continuous function such that ν = µf −1 is locally finite, and g : Y → Z a function. Show that g is almost continuous with respect to ν iff gf is almost continuous with respect to µ. (q) Let (X, T, Σ, µ) and (Y, S, T, ν) be topological probability spaces, and Rf : X → Y a measurable R function such that µf −1 [H] ≥ νH for every H ∈ S. Show that (i) gf dµ = g dν for every g ∈ Cb (Y ) (ii) µf −1 [F ] = νF for every Baire set F ⊆ Y (iii) if µ is a Radon measure and f is almost continuous, then µf −1 [F ] = νF for every Borel set F ⊆ Y , so that if in addition ν is complete and inner regular with respect to the Borel sets then it is a Radon measure. (r) Let (X, T, Σ, µ) be a totally finite topological measure space in which the topology T is normal and µ is inner regular with respect to the closed sets. Show that if f : X → R is a measurable function and ² > 0 there is a continuous g : X → R such that µ{x : g(x) 6= f (x)} ≤ ². (s) Let X and Y be Hausdorff spaces, ν a totally finite Radon measure on Y , and f : X → Y an injective continuous function. Show that the following are equiveridical: (i) there is a Radon measure µ on X such that f is inverse-measure-preserving; (ii) f [X] is conegligible and f −1 : f [X] → X is almost continuous. (t) Let (X, T, Σ, µ) and (Y, S, T, ν) be Radon measure spaces and f : X → Y an almost continuous inverse-measure-preserving function. Show that (i) µ∗ A ≤ ν∗ f [A] for every A ⊆ X (ii) ν is precisely the image measure µf −1 .

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418Xu

(u) In 418M, show that all the fij must be almost continuous. Show that if i ≤ j ≤ k then fij fjk = fik almost everywhere on Xk . > (v) Let I be the family of finite subsets of [0, 1], and let (XI , TI , ΣI , µI ) be [0, 1] \ I with its subspace topology and measure for each I ∈ I. For I ⊆ J ∈ I, y ∈ XJ set fIJ (y) = y. Show that these XI , fIJ satisfy nearly all the hypotheses of 418O, but that there are no X, gI which satisfy the hypotheses of 418M. (w) Let T be any set, and X the set of total orders on T . (i) Regarding each member of X as a subset of T × T , show that X is a closed subset of P(T × T ). (ii) Show that there is a unique Radon measure µ on X such that Pr(t1 ≤ t2 ≤ . . . ≤ tn ) =

1 n!

for all distinct t1 , . . . , tn ∈ T . (Hint: for I ∈ [T ]<ω , let XI be the

set of total orderings on I with the uniform probability measure giving the same measure to each singleton; show that the natural map from XI to XJ is inverse-measure-preserving whenever J ⊆ I.) (x) In 418S, suppose that f1 : X → L0 (ν) and f2 : X → L0 (ν) correspond to h1 , h2 ∈ L0 (λ). Show that f1 (x) ≤ f2 (x) µ-a.e.(x) iff h1 ≤ h2 λ-a.e. Hence show that (if we assign appropriate algebraic operations to the space of functions from X to L0 (ν)) we have an f -algebra isomorphism between L0 (λ) and the space of equivalence classes of measurable functions from X to L0 (ν) with separable ranges. 418Y Further exercises (a) Let X be a set, Σ a σ-algebra of subsets of X, Y a topological space and f : X → Y a function. Set T = {F : F ⊆ Y, f −1 [F ] ∈ Σ}. Suppose that Y is hereditarily Lindel¨of and its topology is generated by some subset of T. Show that f is measurable. (b) Let (X, Σ, µ) be a measure space, Y and Z topological spaces and f : X → Y , g : X → Z measurable functions. Show that if Z has a countable network consisting of Borel sets (e.g., Z is second-countable, or Z is regular and has a countable network), then x 7→ (f (x), g(x)) : X → Y × Z is measurable. (c) Let X be a set, Σ a σ-algebra of subsets of X, and hYi ii∈I a countable family of topological spaces with product Y . Suppose that every Yi has a countable network, and that f : X → Y is a function such that πi f is measurable for every i ∈ I, writing πi (y) = y(i). Show that f is measurable. (d) Find strictly localizable Hausdorff topological measure spaces (X, T, Σ, µ), (Y, S, T, ν) and (Z, U, Λ, λ) and almost continuous inverse-measure-preserving functions f : X → Y , g : Y → Z such that gf is not almost continuous. (e) Let (X, Σ, µ) be a σ-finite measure space and T a topology on X such that µ is effectively locally finite and τ -additive. Let Y be a topological space and f : X → Y an almost continuous function. Show that there is a conegligible subset X0 of X such that f [X0 ] is ccc. (f ) Show that if µ is Lebesgue measure on R, T is the usual topology on R and S is the right-facing Sorgenfrey topology, then the identity map from (R, T, µ) to (R, S) is measurable, but not almost continuous, and the image measure is not a Radon measure. (g) Let (X, Σ, µ) be a semi-finite measure space and T a topology on X such that µ is inner regular with respect to the closed sets. Suppose that Y is a topological space with a countable network consisting of Borel sets, and that f : X → Y is measurable. Show that f is almost continuous. (h) Find a topological probability space (X, T, Σ, µ) in which µ is inner regular with respect to the closed sets, a topological space Y with a countable network and a measurable function f : X → Y which is not almost continuous. (i) Let (X, Σ, µ) be a semi-finite measure space and T a topology on X such that µ is inner regular with respect to the closed sets. Let A ⊆ L∞ (µ) be a norm-compact set. Show that there is a set B of bounded real-valued measurable functions on X such that (i) A = {f • : f ∈ B} (ii) B is norm-compact in `∞ (X) (iii) µ is inner regular with respect to {E : f ¹E is continuous for every f ∈ B}.

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(j) Let µ be Lebesgue measure on [0, 1]. For t ∈ [0, 1] set ut = χ[0, t]• ∈ L0 (µ). Show that A = {ut : t ∈ [0, 1]} is norm-compact in Lp (µ) for every p ∈ [1, ∞[ and also compact for the topology of convergence in measure on L0 (µ). Show that if B is a set of measurable functions such that A = {f • : f ∈ B} then µ is not inner regular with respect to {E : f ¹E is continuous for every f ∈ B}. (k) Suppose that (I, ≤), h(Xi , Ti , Σi , µi )ii∈I and hfij ii≤j∈I are such that (α) (I, ≤) is a non-empty upwards-directed partially ordered set (β) every (Xi , Ti , Σi , µi ) is a completely regular Hausdorff quasiRadon probability space (γ) fij : Xj → Xi is a continuous inverse-measure-preserving function whenever i ≤ j in I (δ) fij fjk = fik whenever i ≤ j ≤ k in I. Let Xi0 be the support of µi for each i; show that ˇ fij [Xj0 ] is a dense subset of Xi0 whenever i ≤ j. Let Zi be the Stone-Cech compactification of Xi0 and let ˜i be the Radon probability measure on f˜ij : Zj → Zi be the continuous extension of fij ¹Xj0 for i ≤ j; let µ Zi corresponding to µi ¹ PXi0 (416V). Show that Zi , f˜ij satisfy the conditions of 418O, so that we have a projective limit Z, hgi ii∈I , µ as in 418M. (l) Suppose that (I, ≤), h(Xi , Σi , µi )ii∈I , X, hfi ii∈I and hfij ii≤j∈I are such that (α) (I, ≤) is a nonempty upwards-directed partially ordered set (β) every (Xi , Σi , µi ) is a probability space (γ) fij : Xj → Xi is inverse-measure-preserving whenever i ≤ j in I (δ) fij fjk = fik whenever i ≤ j ≤ k (²) fi : X → Xi is a function for every i ∈ I (ζ) fi = fij fj whenever i ≤ j (η) whenever hin in∈N , hxn in∈N are such that hin in∈N is a non-decreasing sequence in I, xn ∈ Xin for every n ∈ N and fin in+1 (xn+1 ) = xn for every n ∈ N, then there is an x ∈ X such that fin (x) = xn for every n. Show that there is a probability measure on X such that every fi is inverse-measure-preserving. 418 Notes and comments The message of this section is that measurable functions are dangerous, but that almost continuous functions behave themselves. There are two fundamental problems with measurable functions: a function x 7→ (f (x), g(x)) may not be measurable when the components f and g are measurable (419Xg), and an image measure under a measurable function can lose tightness, even when both domain and codomain are Radon measure spaces and the function is inverse-measure-preserving (419Xh). (This is the ‘image measure catastrophe’ mentioned in 235J and the notes to §343.) Consequently, as long as we are dealing with measurable functions, we often have to impose strong conditions on the range spaces – commonly, we have to restrict ourselves to separable metrizable spaces (418B, 418C), or something similar, which indeed often means that a measurable function is actually almost continuous (418J, 433E). Indeed, for functions taking values in metrizable spaces, ‘almost continuity’ is very close to ‘measurable with essentially separable range’ (418G, 418J). The condition ‘separable and metrizable’ is a little stronger than is strictly necessary (418Yb, 418Yc, 418Yg), but covers the principal applications other than 433E. If we keep the ‘metrizable’ we can very substantially relax the ‘separable’ (438E, 438F), and it is in fact the case that a measurable function from a Radon measure space to any metrizable space is almost continuous (451S). These extensions apply equally to the results in 418R-418T (438Xg-438Xh, 451Xn). But both take us deeper into set theory than seems appropriate at the moment. For almost continuous functions, the two problems mentioned above do not arise (418Dd, 418I, 418Xt). Indeed we rather expect almost continuous functions to behave as if they were continuous. But we still have to be careful. The limit of a sequence of almost continuous functions need not be almost continuous (418Xd), unless the codomain is metrizable (418F); and if we have a function f from a topological measure space to an uncountable product of topological spaces, it can happen that every coordinate of f is an almost continuous function while f is not (418Xd again). But for many purposes, intuitions gained from the study of measurable functions between Euclidean spaces can be transferred to general almost continuous functions. Theorems 418L and 418M are of a quite different kind, but seem to belong here as well as anywhere. Even in the simplest application of 418L (when Y = [0, 1] with Lebesgue measure, and X ⊆ [0, 1]2 is a closed set meeting every vertical line) it is not immediately obvious that there will be a measure with the right projection onto the horizontal axis, though there are at least two proofs which are easier than the general case treated in 413N-413O-418L. As I explain in 418N, the really interesting question concerning 418M is when, given the projective system h(Xi , Ti , Σi , µi )ii∈I , hfij ii≤j∈I , we can expect to find X and hgi ii∈I satisfying the rest of the hypotheses, and once past the elementary results 418O-418Q this can be hard to determine. I describe a method in

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418Yk which can sometimes be used, but (like the trick in 418Nf) it is too easy and too abstract to be often illuminating. See 454G below for something rather deeper. The results of 418R-418T stand somewhat aside from anything else considered in this chapter, but they form part of an important technique. A special case has already been mentioned in 253Yg. I do not discuss vector-valued measurable functions in this book, except incidentally, but 418R is one of the fundamental results on their representation; it means, for instance, that if V is any of the Banach function spaces of Chapter 36 we can expect to represent Bochner integrable V -valued functions (253Yf) in terms of functions on product spaces, because V will be continuously embedded in an L0 space (367P). The measure-algebra version in 418T will be very useful in Volume 5 when establishing relationships between properties of measure spaces and corresponding properties of measure algebras.

419 Examples In §216, I went much of the way to describing examples of spaces with all the possible combinations of the properties considered in Chapter 21. When we come to topological measure spaces, the number of properties involved makes it unreasonable to seek any such comprehensive list. I therefore content myself with seven examples to indicate some of the boundaries of the theory developed here. The first example (419A) is supposed to show that the hypothesis ‘effectively locally finite’ which appears in so many of the theorems of this chapter cannot as a rule be replaced by ‘locally finite’. The next two (419C-419D) address technical questions concerning the definition of ‘Radon measure’, and show how small variations in the definition can lead to very different kinds of measure space. The fourth example (419E) shows that the τ -additive product measures of §417 are indeed new constructions. 419H is there to show that extension theorems of the types proved in §415 and §417 cannot be taken for granted. The classical example 419K exhibits one of the obstacles to generalizations of Prokhorov’s theorem (418M, 418Q). Finally, I return to the split interval (419L) to describe its standard topology and its relation to the measure introduced in 343J. 419A Example There is a locally compact Hausdorff space X with a complete, σ-finite, locally finite, τ -additive topological measure µ, inner regular with respect to the closed sets, which has a closed subset Y , of measure 1, such that the subspace measure µY on Y is not τ -additive. In particular, µ is not effectively locally finite. proof (a) Let Q be a countably P 1 infinite set, not containing any ordinal. Fix an enumeration hqn in∈N of Q, and for A ⊆ Q set νA = { n+1 : qn ∈ A}. Let I be the ideal {A : A ⊆ Q, νA < ∞}. For any sets I, J, say that I ⊆∗ J if I \ J is finite; then ⊆∗ is a reflexive transitive relation. Let κ be the smallest cardinal of any family K ⊆ I for which there is no I ∈ I such that K ⊆∗ I for every K ∈ K. Then κ is uncountable. P P (i) Of course κ is infinite. (ii) If hIn in∈N is any S sequence in I, then for each n ∈ N we can find a finite In0 ⊆ In such that ν(In \ In0 ) ≤ 2−n ; setting I = n∈N In \ In0 , we have νI ≤ 2 < ∞, while In ⊆∗ I for every n. Thus κ > ω. Q Q (b) There is a family hIξ iξ<κ in I such that (i) Iη ⊆∗ Iξ whenever η ≤ ξ < κ (ii) there is no I ∈ I such that Iξ ⊆∗ I for every ξ < κ. P P Take a family hKξ iξ<κ in I such that there is no I ∈ I such that Kξ ⊆∗ I for every ξ < κ. Choose hIξ iξ<κ in I inductively in such a way that Kξ ⊆∗ Iξ ,

Iη ⊆∗ Iξ for every η < ξ.

(This can be done because {Kξ } ∪ {Iη : η < ξ} will always be a subset of I of cardinal less than κ.) If now Iξ ⊆∗ I for every ξ < κ, then Kξ ⊆∗ I for every ξ < κ, so I ∈ / I. Q Q Hence, or otherwise, we see that κ is regular. P P If A ⊆ κ and #(A) < κ, then there is an I ∈ I such that Iζ ⊆∗ I for every ζ ∈ A; now there must be a ξ < κ such that Iξ 6⊆∗ I, in which case ζ < ξ for every ζ ∈ A, and A is not cofinal with κ. Q Q (c) Set X = Q ∪ κ. (This is where it is helpful to have arranged at the start that no ordinal belongs to Q, so that Q ∩ κ = ∅.) Let T be the family of sets G ⊆ X such that G ∩ κ is open for the order topology of κ, for every ξ ∈ G ∩ κ \ {0} there is an η < κ such that Iξ \ Iη ⊆∗ G,

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if 0 ∈ G then I0 ⊆∗ G. S (i) This is a Hausdorff topology on X. P P (α) It is easy to check that X ∈ T, ∅ ∈ T and G ∈ T for every G ⊆ T. (β) Suppose that G, H ∈ T. Then (G ∩ H) ∩ κ = (G ∩ κ) ∩ (H ∩ κ) is open for the order topology of κ. If ξ ∈ G ∩ H ∩ κ \ {0} there are η, ζ < ξ such that Iξ \ Iη ⊆∗ G and Iξ \ Iζ ⊆∗ H, and now α = max(η, ζ) < ξ,

Iη ∪ Iζ ⊆∗ Iα ,

so Iξ \ Iα ⊆∗ (Iξ \ Iη ) ∩ (Iξ \ Iζ ) ⊆∗ G ∩ H. Finally, if 0 ∈ G ∩ H then I0 ⊆∗ G ∩ H. So G ∩ H ∈ T. Thus T is a topology on X. (γ) For any ξ < κ, the set Eξ = (ξ + 1) ∪ Iξ is open-and-closed for T; for any q ∈ Q, {q} is open-and-closed. Since these sets separate the points of X, T is Hausdorff. Q Q (ii) The sets Eξ of the last paragraph are all compact for T. P P Let F be an ultrafilter on X containing Eξ . (α) If a finite set K belongs to F, then F must contain {x} for some x ∈ K, and converges to x. So suppose henceforth that F contains no finite set. (β) If E0 ∈ F, then for any open set G containing 0, E0 \ G is finite, so does not belong to F, and G ∈ F; as G is arbitrary, F → 0. (γ) If E0 ∈ / F, let η ≤ ξ be the least ordinal such that Eη ∈ F. If G is an open set containing η, there are ζ 0 , ζ 00 < η such that Iη \ Iζ 0 ⊆∗ G, ]ζ 00 , η] ⊆ G; so that Eη \ Eζ ⊆∗ G, where ζ = max(ζ 0 , ζ 00 ) < η. Now Eη ∈ F, Eζ ∈ / F and (Eη \ Eζ ) \ G ∈ / F, so that G ∈ F . As G is arbitrary, F → η. (δ) As F is arbitrary, Eξ is compact. Q Q (iii) It follows that T is locally compact. P P For q ∈ Q, {q} is a compact open set containing q; for ξ < κ, Eξ is a compact open set containing ξ. Q Q (iv) The definition of T makes it clear that Q ∈ T, that is, that κ is a closed subset of X. We need also to check that the subspace topology Tκ on κ induced by T is just the order topology of κ. P P (α) By the definition of T, G ∩ κ is open for the order topology of κ for every G ∈ T. (β) For any ξ < κ, Eξ is open-and-closed for T so ξ + 1 = Eξ ∩ κ is open-and-closed for Tκ . But this means that all sets of the forms S [0, ξ[ = η<ξ η + 1 and ]ξ, κ[ = κ \ (ξ + 1) belong to Tκ ; as these generate the order topology, every open set for the order topology belongs to Tκ , and the two topologies are equal. Q Q (d) Now let F be the filter on X generated by the cofinal closed sets in κ. Because the intersection of any sequence of closed cofinal sets in κ is another (4A1Bc), the intersection of any sequence in F belongs to F. So Σ = F ∪ {X \ F : F ∈ F} is a σ-algebra of subsets of X, and we have a measure µ1 : Σ → {0, 1} defined by saying that µ1 F = 1, µ1 (X \ F ) = 0 if F ∈ F. (e) Set µE = ν(E ∩ Q) + µ1 E for E ∈ Σ. Then µ is a measure. Let us work through the properties called for. (i) If µE = 0 and A ⊆ E, then X \ A ⊇ X \ E ∈ F , so A ∈ Σ. Thus µ is complete. (ii) µ(κ) = 1 and µ{q} is finite for every q ∈ Q, so µ is σ-finite. (iii) If G ⊆ X is open, then κ \ G is closed, in the order topology of κ; if it is cofinal with κ, it belongs to F; otherwise, κ ∩ G ∈ F . Thus in either case G ∈ Σ, and µ is a topological measure. (iv) The next thing to note is that µG = ν(G ∩ Q) for every open set G ⊆ X. P P If G ∈ / F this is trivial. If G ∈ F, then κ \ G cannot be cofinal with κ, so there is a ξ < κ such that κ \ ξ ⊆ G. ?? If G ∩ Q ∈ I, then (G ∩ Q) ∪ Iξ ∈ I. There must be a least η < κ such that Iη 6⊆∗ (G ∩ Q) ∪ Iξ ; of course η > ξ, so η ∈ G. There is some ζ < η such that Iη \ Iζ ⊆∗ G; but as Iζ ⊆∗ G ∪ Iξ , by the choice of η, we must also have Iη ⊆∗ G ∪ Iξ , which is impossible. X X Thus G ∩ Q ∈ / I and µG = ν(G ∩ Q) = ∞. Q Q (v) It follows that µ is τ -additive. P P Suppose that G ⊆ T is a non-empty upwards-directed set with union H. Then µH = ν(H ∩ Q) = supG∈G ν(G ∩ Q) = supG∈G µG

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419A

because ν is τ -additive (indeed, is a Radon measure) with respect to the discrete topology on Q. Q Q (vi) µ is inner regular with respect to the closed sets. P P Take E ∈ Σ, γ < µE. Then γ − µ1 (E ∩ κ) < ν(E ∩ Q), so there is a finite I ⊆ E ∩ Q such that νI > γ − µ(E ∩ κ). If µ1 (E ∩ κ) = 0, then I ⊆ E is already a closed set with µI > γ. Otherwise, E ∩ κ ∈ F , so there is a cofinal closed set F ⊆ κ such that F ⊆ E; now F is closed in X (because κ is closed in X and the subspace topology on κ is the order topology), so I ∪ F is closed, and µ(I ∪ F ) > γ. As E and γ are arbitrary, µ is inner regular with respect to the closed sets. Q Q (vii) µ is locally finite. P P For any ξ < κ, Eξ is an open set containing ξ, and µEξ = νIξ is finite. For any q ∈ Q, {q} is an open set containing q, and µ{q} = ν{q} is finite. Q Q (viii) Now consider Y = κ. This is surely a closed set, and µκ = 1. I noted in (b-iv) above that the subspace topology Tκ is just the order topology of κ. But this means that {ξ : ξ < κ} is an upwards-directed family of negligible relatively open sets with union κ, so that the subspace measure µκ = µ1 is not τ -additive. (ix) It follows from 414K that µ cannot be effectively locally finite; but it is also obvious from the work above that κ is a measurable set, of non-zero measure, such that µ(κ ∩ G) = 0 whenever G is an open set of finite measure. 419B Lemma For any non-empty set I, there is a dense Gδ set in [0, 1]I which is negligible for the usual measure on [0, 1]I . proof Fix on some i0 ∈ I, and set π(x) = x(i0 ) for each x ∈ [0, 1]I , so that π is continuous and inversemeasure-preserving for the usual topologies and measures on [0, 1]I and [0, 1]. For each n ∈ N let Gn ⊇ [0, 1] ∩ Q be an open subset [0, 1] with measure at most 2−n , so that π −1 [Gn ] is an open set of measure T of −1 I −n is any non-empty open set, at most 2 , and E = n∈N π [Gn ] is a Gδ set of measure 0. If H ⊆ [0, 1]T its image π[H] is open in [0, 1], so contains some rational number, and meets n∈N Gn ; but this means that H ∩ E 6= ∅, so E is dense. 419C Example (Fremlin 75b) There is a completion regular Radon measure space (X, T, Σ, µ) such that (i) there is an E ∈ Σ such that µ(F 4E) > 0 for every Borel set F ⊆ X, that is, not every element of the measure algebra of µ can be represented by a Borel set; (ii) µ is not outer regular with respect to the Borel sets; (iii) writing ν for the restriction of µ to the Borel σ-algebra of X, ν is a locally finite, effectively locally finite, tight (that is, inner regular with respect to the compact sets) τ -additive topological measure, and there is a set Y ⊆ X such that the subspace measure νY is not semi-finite. proof (a) For each ξ < ω1 set Xξ = [0, 1]ω1 \ξ , and take µξ to be the usual measure on Xξ ; write Σξ for its domain. S Note that µξ is a completion regular Radon measure for the usual topology Tξ of Xξ (416U). Set X = ξ<ω1 Xξ , and let µ be the direct sum measure on X (214K), that is, write Σ = {E : E ⊆ X, E ∩ Xξ ∈ Σξ for every ξ < ω1 }, P µE = ξ<ω1 µξ (E ∩ Xξ ) for every E ∈ Σ. Then µ is a complete locally determined (in fact, strictly localizable) measure on X. Write Σ for its domain. (b) For each η < ω1 let hβξη iξ≤η be a summable family of strictly positive real numbers with βηη = 1 (4A1P). Define gη : X → R by setting gη (x) =

1 x(η) βξη

if x ∈ Xξ where ξ ≤ η,

= 0 if x ∈ Xξ where ξ > η. Now define f : X → ω1 × R

ω1

by setting f (x) = (ξ, hgη (x)iη<ω1 )

if x ∈ Xξ . Note that f is injective. Let T be the topology on X defined by f , that is, the family {f −1 [W ] : W ⊆ ω1 × R ω1 is open}, where ω1 and R ω1 are given their usual topologies (4A2S, 3A3K), and their product

419C

Examples

129

is given its product topology. Because f is injective, T can be identified with the subspace topology on f [X]; it is Hausdorff and completely regular. (c) For ξ, η < ω1 , gη ¹Xξ is continuous for the compact topology Tξ . Consequently f ¹Xξ is continuous, and the subspace topology on Xξ induced by T must be Tξ exactly. It follows that µ is a Radon measure for T. P P (i) We know already that µ is complete and locally determined. (ii) If G ∈ T then G ∩ Xξ ∈ Tξ ⊆ Σξ for every ξ < ω1 , so G ∈ Σ; thus µ is a topological measure. (iii) If E ∈ Σ and µE > 0, there is a ξ < ω1 such that µξ (E ∩ Xξ ) > 0. Because µξ is a Radon measure, there is a Tξ -compact set F ⊆ E ∩ Xξ such that µξ F > 0. Now F is T-compact and µF > 0. As E is arbitrary, µ is tight (using 412B). (iv) If x ∈ X, take that ξ < ω1 such that x ∈ Xξ , and consider G = f −1 [(ξ + 1) × {w : w ∈ Rω1 , w(ξ) < 2}]. Because ξ + 1 is open in ω1 , G ∈ T. Because gξ (x) = x(ξ) ≤ 1, x ∈ G. Now for ζ ≤ ξ, µζ (G ∩ Xζ ) = µζ {x : x ∈ Xζ , gξ (x) < 2} −1 = µζ {x : x ∈ Xζ , βζξ x(ξ) < 2}

= µζ {x : x ∈ Xζ , x(ξ) < 2βζξ } ≤ 2βζξ , so µG =

P ζ≤ξ

µζ (G ∩ Xζ ) ≤ 2

P ζ≤ξ

βζξ < ∞.

As x is arbitrary, µ is locally finite, therefore a Radon measure. Q Q We also find that µ is completion regular. P P If E ⊆ X and µE > 0, then there is a ξ < ω1 such that µ(E ∩ Xξ ) > 0. Because µξ is completion regular, there is a set F ⊆ E ∩ XSξ , a zero set for Tξ , such that µF > 0. Now Xξ is a Gδ set in X (being the intersection of the open sets η<ζ<ξ+1 Xζ for η < ξ, unless ξ = 0, in which case Xξ itself is open), so F is a Gδ set in X (4A2C(a-iv)); being a compact Gδ set in a completely regular space, it is a zero set (4A2F(h-v)). Thus every set of positive measure includes a zero set of positive measure. So µ is inner regular with respect to the zero sets (412B). Q Q (d) The key to the example is the following fact: if G ⊆ X is open, then either there is a cofinal closed set V ⊆ ω1 such that G ∩ Xξ = ∅ for every ξ ∈ V or {ξ : µ(G ∩ Xξ ) 6= 1} is countable. P P Suppose that A = {ξ : G ∩ Xξ 6= ∅} meets every cofinal closed set, that is, is stationary (4A1C). Then B = A ∩ Ω is stationary, where Ω is the set of non-zero countable limit ordinals (4A1Bb, 4A1Cb). Let H ⊆ ω1 × R ω1 be an open set such that G = f −1 [H]. For each ξ ∈ B choose xξ ∈ G ∩ Xξ . Then f (xξ ) ∈ H, so there must be a ζξ < ξ, a finite set Iξ ⊆ ω1 , and a δξ > 0 such that z ∈ H whenever z = (γ, htη iη∈ω1 ) ∈ ω1 × R ω1 , ζξ < γ ≤ ξ and |tη − gη (x)| < δξ for every η ∈ Iξ . Because ξ is a non-zero limit ordinal, ζξ0 = sup({ζξ } ∪ (Iξ ∩ ξ)) < ξ. By the Pressing-Down Lemma (4A1Cc), there is a ζ < ω1 such that C = {ξ : ξ ∈ B, ζξ0 = ζ} is uncountable. ?? Suppose, if possible, that ζ < η < ω1 and µ(G ∩ Xη ) < 1. Then there is a measurable subset F of Xη \ G, determined by coordinates in a countable set J ⊆ ω1 \ η, such that µF = µη F > 0 (254Ob). Let ξ ∈ C be such that η < ξ and J ⊆ ξ, and take any y ∈ F . If we define y 0 ∈ Xη by setting y 0 (γ) = y(γ) for γ ∈ ξ \ η, = xξ (γ) for γ ∈ ω1 \ ξ, then y 0 ∈ F . But also ζξ < η < ξ and ξ \ η ⊆ ξ \ ζξ0 is disjoint from Iξ , so gγ (y 0 ) = gγ (xξ ) for every γ ∈ Iξ , since both are zero if γ ≤ η and otherwise y 0 (γ) = xξ (γ). By the choice of ζξ , Iξ we must have f (y 0 ) ∈ H and y 0 ∈ F ∩ G; which is impossible. X X Thus µ(G ∩ Xη ) = 1 for every η > ζ, as required by the second alternative. Q Q (e) For each ξ < ω1 , let Iξ be the family of negligible meager subsets of Xξ . Then Iξ is a σ-ideal; note that it contains every closed negligible set, because µξ is strictly positive. Set Tξ = Iξ ∪ {Xξ \ F : F ∈ Iξ }, so that Tξ is a σ-algebra of subsets of Xξ , containing every conegligible open set, and µξ F ∈ {0, 1} for every F ∈ Tξ . Set

130

Topologies and measures

419C

T = {E : E ∈ Σ, {ξ : E ∩ Xξ ∈ / Tξ } is non-stationary}. Then T is a σ-subalgebra of Σ (because the non-stationary sets form a σ-ideal of subsets of ω1 , 4A1Cb), and contains every open set, by (d); so includes the Borel σ-algebra B of X. If we set S Eξ = {x : x ∈ Xξ , x(ξ) ≤ 21 } for each ξ < ω1 , E = ξ<ω1 Eξ , then E ∈ Σ. But if F ⊆ X is a Borel set, F ∈ T so µ(E4F ) = ∞. This proves the property (i) claimed for the example. S (f ) Next, for each ξ < ω1 , take a negligible dense Gδ set Eξ0 ⊆ Xξ (419B). Set Y = ξ<ω1 Eξ0 , so that µY = 0. If F ⊇ Y is a Borel set, then F ∩ Xξ ⊇ Eξ ∈ / Iξ for every ξ < ω1 , while F ∈ T, so {ξ : µξ (F ∩ Xξ ) = 0} is non-stationary and µF = ∞. Thus µ is not outer regular with respect to the Borel sets. Taking ν = µ¹B, the subspace measure νY is not semi-finite. P P We have just seen thatSνY Y = ν ∗ Y is infinite. If F ∈ B and νF < ∞, then A = {ξ : µξ (F ∩ Xξ ) > 0} is countable, so F0 = ξ∈A Eξ0 and S F1 = F \ ξ∈A Xξ are negligible Borel sets; since F ∩ Y ⊆ F0 ∪ F1 , νY (F ∩ Y ) = 0. But this means that νY takes no values in ]0, ∞[ and is not semi-finite. Q Q Remark X here is not locally compact. But as it is Hausdorff and completely regular, it can be embedded as a subspace of a locally compact Radon measure space (X 0 , T0 , Σ0 , µ0 ) (416T). Now µ0 still has the properties (i)-(iii). 419D Example (Fremlin 75b) There is a complete locally determined τ -additive completion regular topological measure space (X, T, Σ, µ) in which µ is tight and compact sets have finite measure, but µ is not localizable. proof (a) Let I be a set of cardinal greater than c. Set X = [0, 1]I . For i ∈ I, t ∈ [0, 1] set Xit = {x : x ∈ X, x(i) = t}. Give Xit its natural topology Tit and measure µit , with domain Σit , defined from the expression of Xit as [0, 1]I\{i} × {t}, each factor [0, 1] being given its usual topology and Lebesgue measure, and the singleton factor {t} being given its unique (discrete) topology and (atomic) probability measure. By 416U, µit is a completion regular Radon measure. Set T = {G : G ⊆ X, G ∩ Xit ∈ Tit for all i ∈ I, t ∈ [0, 1]}, Σ = {E : E ⊆ X, E ∩ Xit ∈ Σit for all i ∈ I, t ∈ [0, 1]}, P µE = i∈I,t∈[0,1] µit (E ∩ Xit ) for every E ∈ Σ. (Compare 216D.) Then it is easy to check that T is a topology. T is Hausdorff because it is finer(= larger) than the usual topology S on X; because each Tit is the subspace topology induced by S, it is also the subspace topology induced by T. Next, the definition of µ makes it a locally determined measure; it is a tight complete topological measure because every µit is. (b) If K ⊆ X is compact, µK < ∞. P P?? Otherwise, M = {(i, t) : i ∈ I, t ∈ [0, 1], µit (K ∩ Xit ) > 0} must be infinite. Take any sequence h(in , tn )in∈N of distinct elements of M . Choose a sequence hxn in∈N in K inductively, as follows. Given hxm im
419F

Examples

131

S point recall that #(I) > c, so there is some j ∈ I \ ({k} ∪ t∈[0,1] Jt ). Since Xj0 ∩ Xkt is negligible for every R1 t ∈ [0, 1], Xj0 ∩ E must be negligible, and 0 νHt dt = 0, where Ht = {y : y ∈ [0, 1]I\{j,k} , (y, 0, t) ∈ E} and ν is the usual measure on [0, 1]I\{j,k} , identifying X with [0, 1]I\{j,k} × [0, 1] × [0, 1]. But because Ft is determined by coordinates in I \ {j}, we can identify it with Ft0 × [0, 1] × {t} where Ft0 is a ν-conegligible subset of [0, 1]I\{j,k} , and Ft0 ⊆ Ht , so νHt = 1 for every t, which is absurd. X X Thus E has no essential supremum in Σ, and µ cannot be localizable. Q Q (d) I have still to check that µ is completion regular. P P If E ∈ Σ and µE > 0, there are i ∈ I, t ∈ [0, 1] such that µit (E ∩ Xit ) > 0, and an F ⊆ E ∩ Xit , a zero set for the subspace topology of Xit , such that µit F > 0. But now observe that Xit is a zero set in X for the usual topology S, so that F is a zero set for S (4A2Gc) and therefore for the finer topology T. By 412B, this is enough to show that µ is inner regular with respect to the zero sets. Q Q Remark It may be worth noting that the topology T here is not regular. See Fremlin 75b, p. 106. 419E Example (Fremlin 76) Let (Z, S, T, ν) be the Stone space of the measure algebra of Lebesgue measure on [0, 1], so that ν is a strictly positive completion regular Radon probability measure (411P). Then the c.l.d. product measure λ on Z × Z is not a topological measure, so is not equal to the τ -additive product ˜ and λ ˜ is not completion regular. measure λ, ˜ described in 346K. We have W ∈ Λ = dom λ and W ˜ = S V, where proof Consider the sets W , W V = {G × H : G, H ⊆ Z are open-and-closed, (G × H) \ W is negligible}. ˜ is a union of open sets, therefore must be open in Z 2 . And λ∗ W ˜ ≤ λW . P W P?? Otherwise, there is a ˜ V ∈ Λ such that V ⊆ W and λV > λW . Now λ is tight, by 412Sb, so there is a compact set K ⊆ V such S that K ∈ Λ and λK > λW . There must be U0 , . . . , Un ∈ V such that K ⊆ i≤n Ui . But λ(Ui \ W ) = 0 for every i, so λ(K \ W ) = 0 and λK ≤ λW . X XQ Q ˜ should be 1 and λW strictly less than 1. So However, the construction of 346K arranged that λ∗ W ∗ ˜ ˜ ˜ λ∗ W < λ W and W ∈ / Λ. Accordingly λ is not a topological measure and cannot be equal to the Radon ˜ of 417P. measure λ We know that λ is inner regular with respect to the zero sets (412Sc) and is defined on every zero set ˜ properly extends λ. But this means that λ ˜ cannot be inner regular with respect to the zero (417V), while λ sets, by 412L, that is, cannot be completion regular. b 419F Theorem (Rao 69) P(ω1 × ω1 ) = Pω1 ⊗Pω 1 , the σ-algebra of subsets of ω1 generated by {E × F : E, F ⊆ ω1 }. proof (a) Because ω1 ≤ c, there is an injection h : ω1 → {0, 1}N ; set Ei = {ξ : h(ξ)(i) = 1} for each i ∈ N. b (b) Suppose that A ⊆ ω1 has countable vertical sections. Then A ∈ Pω1 ⊗Pω P Set B = A−1 [ω1 ] 1. P and for ξ ∈ B let fξ : N → A[{ξ}] be a surjection. Set gn (ξ) = fξ (n) for ξ ∈ B and n ∈ N, and An = {(ξ, fξ (n)) : ξ ∈ B} for n ∈ N. Then An = {(ξ, η) : ξ ∈ B, η = gn (ξ)} = {(ξ, η) : ξ ∈ B, η < ω1 , h(gn (ξ)) = h(η)} \¡ ¢ = {(ξ, η) : ξ ∈ gn−1 [Ei ], η ∈ Ei } ∪ {(ξ, η) : ξ ∈ B \ gn−1 [Ei ], η ∈ ω1 \ Ei } i∈N

b ∈ Pω1 ⊗Pω 1. So A=

S n∈N

b An ∈ Pω1 ⊗Pω Q 1. Q

b (c) Similarly, if a subset of ω1 × ω1 has countable horizontal sections, it belongs to Pω1 ⊗Pω 1 . But for 0 00 any A ⊆ ω1 × ω1 , A = A ∪ A where

132

Topologies and measures

419F

A0 = {(ξ, η) : (ξ, η) ∈ A, η ≤ ξ} has countable vertical sections, A00 = {(ξ, η) : (ξ, η) ∈ A, ξ ≤ η} has countable horizontal sections, b so both A0 and A00 belong to Pω1 ⊗Pω 1 and A also does. 419G Corollary (Ulam 30) Let Y be a set of cardinal at most ω1 and µ a σ-finite measure with domain PY . Then µ is point-supported: in particular, there is a countable conegligible set A ⊆ Y . P proof (a) Let µ0 be the point-supported part of µ, that is, µ0 A = y∈A µ{y} for every A ⊆ Y ; then µ0 is a measure (112Bd) and so is ν = µ − µ0 . If A ⊆ Y is countable, then of course µ0 A = µA, so νA = 0. Because µ is σ-finite, so is ν. (b) If Y is countable, we can stop. For the case in which #(Y ) = ω1 , it is enough to consider the case in which Y is actually equal to ω1 . Let λ = ν × ν be the product measure on ω1 × ω1 . By 419F, the domain of λ is the whole of P(ω1 × ω1 ); in particular, it contains the set V = {(ξ, η) : ξ ≤ η < ω1 }. Now by Fubini’s theorem λV =

R

and also λV =

νV [{ξ}]ν(dξ) =

R

R

ν(ω1 \ ξ)ν(dξ) = (νω1 )2 ,

νV −1 [{η}]ν(dη) =

R

ν(η + 1)ν(dη) = 0.

So νω1 = 0 and µ = µ0 . Now {y : µ{y} > 0} is a countable set (because µ is σ-finite) and is conegligible. Remark I ought to remark that this result, though not 419F, is valid for many other cardinals besides ω1 ; see, in particular, 438C below. There will be more on this topic in Volume 5. 419H Example There is a complete probability space (X, Σ, µ) with a topology T such that µ is τ additive and inner regular with respect to the Borel sets, T is generated by T ∩ Σ, but µ has no extension to a topological measure. proof (a) Set Y = ω1 + 1 = ω1 ∪ {ω1 }. Let T be the σ-algebra of subsets of Y generated by the σ-ideal of countable subsets of ω1 . Let ν be the probability measure with domain T defined by the formula 1 2

νF = #(F ∩ {0, ω1 }) for every F ∈ T. Set S = {∅, Y } ∪ {H : 0 ∈ H ⊆ ω1 }. This is a topology on Y , and every subset of Y is a Borel set for S; so ν is surely inner regular with respect to the Borel sets. Note that {{0, α} : α < ω1 } ∪ {Y } is a base for S included in T. (b) Let λ be the product probability measure on Y N , and Λ its domain; let Λ0 ⊆ Λ be the σ-algebra of subsets of Y N generated by sets of the form {y : y(i) ∈ F }, where i ∈ N and F ∈ T. Let S∗ be the product topology on Y N , so that S∗ ∩ Λ0 is a base for S∗ . Then λ is inner regular with respect to the Borel sets (412Uc, or otherwise). Define φ : {0, 1}N → Y by setting φ(u)(n) = 0 if u(n) = 0, = ω1 if u(n) = 1. Let νω be the usual measure on {0, 1}N ; then φ is inverse-measure-preserving for νω and λ, by 254H. If V ∈ Λ0 there is an α < ω1 such that whenever x ∈ V , y ∈ Y N and y(i) = x(i) whenever min(x(i), y(i)) < α, then y ∈ V .

419I

Examples

133

P P Let W be the family of sets V ∈ Λ with this property. Then W is a σ-algebra of subsets of Y N including the family C of measurable cylinders, so includes Λ0 . Q Q (c) Let X be Y N \ {0, ω1 }N . Then λ∗ X = 1. P P Take V ∈ Λ such that V ⊇ X and λV = λ∗ X. 0 0 Then there is a V ∈ Λ0 such that V ⊆ V and λV 0 = λV (254Ff). As remarked in (b), there is an α ∈ ]0, ω1 [ such that y ∈ V 0 whenever x ∈ V 0 , y ∈ Y N and y(i) = x(i) whenever min(x(i), y(i)) < α. But as {0, α}N \ {0} ⊆ X ⊆ V 0 , {0, ω1 }N \ {0} ⊆ V 0 . Accordingly λ∗ X = λV = λV 0 = νω φ−1 [V 0 ] = νω ({0, 1}N \ {0}) = 1. Q Q Give X the subspace measure µ induced by λ, with domain Σ = {X ∩ W : W ∈ Λ}, and the subspace topology T induced by S∗ . Then µX = λ∗ X = 1, and µ(X ∩ V ) = λV for every V ∈ Λ. Σ ∩ T is a base for T, just because Λ ∩ S∗ is a base for S∗ ; and µ is inner regular with respect to the Borel sets, by 412Pb. (d) ?? Suppose, if possible, that µ is not τ -additive. Let G ⊆ Σ ∩ T be a non-empty upwards-directed set with S union W ∈ Σ ∩ T such that µW > supG∈GSµG. Let G0 Sbe a countable subset of G such that µ(G \ G0 ) = 0 for every G ∈ G0 , and set W1 = W \ G0 ; then µ( G0 ) ≤ supG∈G µG, so µW1 > 0, while µ(G ∩ W1 ) = 0 for every G ∈ G. Express W1 as V1 ∩ X where V1 ∈ Λ. Then there is a V2 ∈ Λ0 such that V2 ⊆ V1 and λ(V1 \ V2 ) = 0, so that λV2 = λV1 = µW1 > 0. Let α < ω1 be such that y ∈ V2 whenever x ∈ V2 , y ∈ Y N and y(i) = x(i) whenever min(x(i), y(i)) < α. Let F ⊆ φ−1 [V2 ] be a non-negligible measurable self-supporting set for the Radon measure νω (416U, 414F). By Lemma 345E, there are u, u0 ∈ F which differ at exactly one coordinate; let i ∈ N be that coordinate, and suppose that u(i) = 1, u0 (i) = 0. We know that φ(u) ∈ V2 , and φ(u)(i) = ω1 . Define x ∈ X S by saying that x(j) = φ(u)(j) for j 6= i, x(i) = α. Then x ∈ V2 , by the choice of α. Now V2 ∩ X ⊆ W1 ⊆ G, soQthere is a G ∈ G containing x. Let V ⊆ Y N be a basic open set such that x ∈ X ∩ V ⊆ G; express V as j∈N Hj where Hj ∈ S for every j and J = {j : Hj 6= Y } is finite. Observe that 0 ∈ Hj and x(j) < ω1 for every j ∈ J, just because 0 belongs to every non-empty open subset of Y and the only open set containing ω1 is Y itself. But this means that u(j) = 0 for every j ∈ J \ {i}, so u0 (j) = 0 for every j ∈ J, and φ(u0 ) ∈ V ; thus the open set U = φ−1 [V ] meets F . Because F is self-supporting, 0 < νω (F ∩ U ) ≤ λ(V2 ∩ V ) = µ(V2 ∩ V ∩ X) ≤ µ(W1 ∩ G), which is impossible. X X Thus µ is τ -additive. (e) ?? But suppose, if possible, that there were a topological measure µ ˜ on X agreeing with µ on every open set in the domain of µ. For each i ∈ N, set πi (x) = x(i) for x ∈ X. Every subset of Y is a Borel set for S; because πi is continuous, the image measure µ ˜πi−1 is defined on PY . Now #(Y ) = ω1 , so there must be a countable conegligible set (419G), and there must be some αi < ω1 such that µ ˜πi−1 (ω1 \ αi ) = 0. On the other hand, µ ˜πi−1 (αi \ {0}) = µπi−1 (αi \ {0}) = λ{y : 0 < y(i) < αi } = ν(αi \ {0}) = 0, so µ ˜πi−1 (ω1 \ {0}) = 0. But the definition of X was exactly devised so that S X = i∈N πi−1 (ω1 \ {0}), so this is impossible. X X So we have the required example. Remark I note that the topology of X is not regular. Of course the phenomenon here cannot arise with regular spaces, by 415M. 419I For the next example it will be helpful to know some basic facts about Lebesgue measure which seemed a little advanced for Volume 1 and for which I have not found a suitable place since.

134

Topologies and measures

419I

Lemma (a) If (X, T, Σ, µ) is an atomless Radon measure space and E ∈ Σ has non-zero measure, then #(E) ≥ c. (b) The number of closed subsets of R is c. proof (a) There must be a compact set K ⊆ E such that µK > 0, and a self-supporting closed K 0 ⊆ K such that µK 0 = µK. Because µ{x} = 0 for every x ∈ X, K 0 can have no isolated points. So #(K 0 ) ≥ c (4A2G(i-ii)) and #(E) ≥ c. (b) Write E for the family of closed subsets of R. Let hUn in∈N enumerate a base for the topology of R (4A2Ua). For each I ⊆ N, set S FI = R \ n∈I Un . Because every open set is expressible as a union of some Ui , the map I 7→ FI : PN → E is surjective. So #(E) ≤ #(PN) = c. On the other hand, the map x → 7 [0, x] : [0, 1] → E is injective, so #(E) ≥ #([0, 1]) = c. Remark In fact, of course, the number of Borel subsets of R is c; see 4A1O. 419J The next result is a strengthening of 134D. Lemma Let µ be Lebesgue measure on R, and H any measurable subset of R. Then there is a disjoint family hAα iα 0. Let E be the family of closed subsets of H of non-zero measure. By 419Ib, #(E) ≤ c; enumerate E × c as h(Fξ , αξ )iξ 0 (413Ei), so there is a non-negligible measurable set E ⊆ H \ Aα . Now there is an F ∈ E such that F ⊆ E. Let ξ < c be such that F = Fξ and α = αξ ; then xξ ∈ Aα ∩ F , which is impossible. X X Thus H is always a measurable envelope of Aα . It follows from the definition of ‘measurable envelope’ that µ∗ Aα = µH. But also, if α < c, µ∗ Aα ≤ µ∗ (H \ Aα+1 ), which is 0, as we have just seen. So we have a suitable family. 419K Example (Blackwell 56) There are sequences hXn in∈N , hTn in∈N and hνn in∈N such that (i) for Q each n, (Xn , Tn ) is a separable metrizable space and νn is a quasi-Radon probability measure on Zn = map πmn : Zn → Zm is inverse-measure-preserving (iii) there is no i≤n Xi (ii) for m ≤ n the canonical Q probability measure on Z = i∈N Xi such that all the canonical maps from Z to Zn are inverse-measurepreserving. proof Let hAn in∈N be a disjoint sequence of subsets of [0, 1] such thatS µ∗ ([0, 1] \ An ) = 0, that is, µ∗ An = 1 for every n, where µ is Lebesgue measure (using 419J). Set Xn = i≥n Ai , so that hXn in∈N is a nonincreasing sequence of sets of outer measure 1 with empty intersection. For each n ≥ 1, we have a map fn : Xn → Zn defined by setting fn (x)(i) = x for every i ≤ n, x ∈ Xn . Let νn be the image measure µXn fn−1 , where µXn is the subspace measure on Xn induced by µ. Note that fn is a homeomorphism between Xn and the diagonal ∆n = {z : z ∈ Zn , z(i) = z(j) for all i, j ≤ n}, which is a closed subset of Zn ; so that νn , like µXn , is a quasi-Radon probability measure. If m ≤ n, then πmn is inverse-measure-preserving, where πmn (z)(i) = z(i) for z ∈ Zn , i ≤ m. P P If −1 W ⊆ Zm is measured by νm , then fm [W ] is measured by µXm , so is of the form Xm ∩ E where E is −1 −1 Lebesgue measurable. But in this case fm [πmn [W ]] = Xn ∩ E, so that −1 −1 νn (πmn [W ]) = µXn (fn−1 [πmn [W ]]) = µ∗ (Xn ∩ E) = µE = µ∗ (Xm ∩ E) = νm W . Q Q Q ?? But suppose, if possible, that there is a probability measure ν on Z = i∈N Xi such that πn : Z → Zn is inverse-measure-preserving for every n, where πn (z)(i) = z(i) for z ∈ Z, i ≤ n. Then

419Xc

Examples

135

νπn−1 [∆n ] = νn ∆n = µXn fn−1 [∆n ] = 1 for each n, so because

T n∈N

T 1 = ν( n∈N πn−1 [∆n ]) = ν{z : z ∈ Z, z(i) = z(j) for all i, j ∈ N} = ν∅, Xn = ∅; which is impossible. X X

419L The split interval again (a) For the sake of an example in §343, I have already introduced the ‘split interval’ or ‘double arrow space’. As this construction gives us a topological measure space of great interest, I repeat it here. Let I k be the set {a+ : a ∈ [0, 1]} ∪ {a− : a ∈ [0, 1]}. Order it by saying that a+ ≤ b+ ⇐⇒ a− ≤ b+ ⇐⇒ a− ≤ b− ⇐⇒ a ≤ b,

a+ ≤ b− ⇐⇒ a < b.

Then it is easy to check that I k is a totally ordered space, and that it is Dedekind complete. (If A ⊆ [0, 1] is a non-empty set, then supa∈A a− = (sup A)− , while supa∈A a+ is either (sup A)+ or (sup A)− , depending on whether sup A belongs to A or not.) Its greatest element is 1+ and its least element is 0− . Consequently the order topology on I k is a compact Hausdorff topology (4A2Rc, 4A2Ri). Note that Q = {q + : q ∈ [0, 1] ∩ Q} ∪ {q − : q ∈ [0, 1] ∩ Q} is dense, because it meets every non-trivial interval in I k . By 4A2E(a-ii) and 4A2Rn, I k is ccc and hereditarily Lindel¨of. (b) If we define h : I k → [0, 1] by writing h(a+ ) = h(a− ) = a for every a ∈ [0, 1], then h is continuous, because {x : h(x) < a} = {x : x < a− }, {x : h(x) > a} = {x : x > a+ } for every a ∈ [0, 1]. Now we can describe the Borel sets of I k , as follows: a set E ⊆ I k is Borel iff there is a Borel set F ⊆ [0, 1] such that E4h−1 [F ] is countable. P P Write Σ0 for the family of subsets E of I k such that E4h−1 [F ] is countable for some Borel set F ⊆ [0, 1]. It is easy to check that Σ0 is a σ-algebra of subsets of I k . −1 (If is S countable, so is (I k \ E)4h−1 [[0, 1] \ F ]; if En 4h−1 [Fn ] is countable for every n, so is S E4h [F ]−1 ( n∈N En )4h [ n∈N Fn ].) Because the topology of I k is Hausdorff, every singleton set is closed, so every countable set is Borel. Also h−1 [F ] is Borel for every Borel set F ⊆ [0, 1], because h is continuous (4A3Cd). So if E4h−1 [F ] is countable for some Borel set F ⊆ [0, 1], E = h−1 [F ]4(E4h−1 [F ]) is a Borel set in I k . Thus Σ0 is included in the Borel σ-algebra B of I k . On the other hand, if J ⊆ I k is an interval, h[J] is also −1 k an interval, therefore a Borel S set, and h [h[J]] \ J can contain at most two points, so J ∈ Σ0 . If G ⊆ I is open, it is expressible as i∈I Ji , where hJi ii∈I is a disjoint family of non-empty open intervals (4A2Rj). As X is ccc, I must be countable. Thus G is expressed as a countable union of members of Σ0 and belongs to Σ0 . But this means that the Borel σ-algebra B must be included in Σ0 , by the definition of ‘Borel algebra’. So B = Σ0 , as claimed. Q Q (c) In 343J I described the standard measure µ on I k ; its domain is the set Σ = {h−1 [F ]4M : F ∈ ΣL , M ⊆ I k , µL h[M ] = 0}, where ΣL is the set of Lebesgue measurable subsets of [0, 1] and µL is Lebesgue measure, and µE = µL h[E] for E ∈ Σ. h is inverse-measure-preserving for µ and µL . The new fact I wish to mention is: µ is a completion regular Radon measure. P P I noted in 343Ja that it is a complete probability measure; a fortiori, it is locally determined and locally finite. If G ⊆ I k is open, then we can express it as h−1 [F ]4C for some Borel set F ⊆ [0, 1], countable set C ⊆ I k ((b) above), so it belongs to Σ; thus µ is a topological measure. If E ∈ Σ and µE > γ, then F = [0, 1] \ h[I k \ E] is Lebesgue measurable, and µE = µL F . So there is a compact set L ⊆ F such that µL L ≥ γ. But now K = h−1 [L] ⊆ E is closed, therefore compact, and µK ≥ γ. Moreover, L is a zero set, being a closed set in a metrizable space (4A2Lc), so K is a zero set (4A2C(b-iv)). As E and γ are arbitrary, µ is inner regular with respect to the compact zero sets, and is a completion regular Radon measure. Q Q 419X Basic exercises (a) Show that the topological space X of 419A is zero-dimensional. (b) Give an example of a compact Radon probability space in which every dense Gδ set is conegligible. (Hint: 411P.) (c) In 419E, show that we can start from any atomless probability measure in place of Lebesgue measure on [0, 1].

136

Topologies and measures

419Xd

> (d) (i) Show that if E ⊆ R 2 is Lebesgue measurable, with non-zero measure, then it cannot be covered by fewer than c lines. (Hint: if H = {t : µ1 E[{t}] > 0}, where µ1 is Lebesgue measure on R, then µ1 H > 0, so #(H) = c. So if we have a family L of lines, with #(L) < c, there must be a t ∈ H such that Lt = {t} × R does not belong to L. Now #(Lt ∩ E) = c and each member of L meets Lt ∩ E in at most one point.) (ii) Show that there is a subset A of R 2 , of full outer measure, which meets every vertical line and every horizontal line in exactly one point. (Hint: enumerate R as htξ iξ (h) (i) Again writing I k for the split interval, show that the function which exchanges x+ and x− for every x ∈ [0, 1] is a Borel automorphism and an automorphism for the usual Radon measure ν on I k , but is not almost continuous. (ii) Show that if we set f (x) = x+ for x ∈ [0, 1], then f is inverse-measure-preserving for Lebesgue measure µL on [0, 1], but the image measure µL f −1 is not ν (nor, indeed, a Radon measure). 419Y Further exercises (a) In the example of 419E, show that there is a Borel set V ⊆ Z 2 such that ˜ = 0 and λ∗ V = 1. λV (b) Show that if A ⊆ Pω1 is any family with #(A) ≤ ω1 , there is a countably generated σ-algebra Σ of subsets of ω1 such that A ⊆ Σ. (c) Show that the split interval with its usual topology and measure has the simple product property (417Yi). 419 Notes and comments The construction of the locally compact space X in 419A from the family hIξ iξ<κ is a standard device which has been used many times. The relation ⊆∗ also appears in many contexts. In effect, part of the argument is taking place in the quotient algebra A = PQ/[Q]<ω , since I ⊆∗ J iff I • ⊆ J • in A; setting I # = {I • : I ∈ I}, the cardinal κ is min{#(A) : A ⊆ I # has no upper bound in I # }, the ‘additivity’ of the partially ordered set I # . Additivities of partially ordered sets will be one of the important concerns of Volume 5. I remark that we do not need to know whether (for instance) κ = ω1 or κ = c. This is an early taste of the kind of manoeuvre which has become a staple of set-theoretic analysis. It happens that the cardinal κ here is one of the most important cardinals of set-theoretic measure

419 Notes

Examples

137

theory; it is ‘the additivity of Lebesgue measure’ (529Xa), and under that name will appear repeatedly in Chapter 52. Observe that the measure µ of 419A only just fails to be a quasi-Radon measure; it is locally finite instead of being effectively locally finite. And it would be a Radon measure if it were inner regular with respect to the compact sets, rather than just with respect to the closed sets. 419C and 419D are relevant to the question: have I given the ‘right’ definition of Radon measure space? 419C is perhaps more important. Here we have a Radon measure space (on my definition) for which the associated Borel measure is not localizable. (If A is the measure algebra of the measure µ, and B the measure algebra of µ¹B where B is the Borel σ-algebra of X, then the embedding B ⊆ Σ induces an embedding of B in A which represents B as an order-dense subalgebra of A, just because µ is inner regular with respect to B. Property (i) of 419C shows that B 6= A, so B cannot be Dedekind complete in itself, by 314Ia.) Since (I believe) localizable versions of measure spaces should almost always be preferred, I take this as strong support for my prejudice in favour of insisting that ‘Radon’ measure spaces should be locally determined as well as complete. Property (ii) of 419C is not I think of real significance, but is further evidence, to be added to 415Xh, that outer regularity is like an exoskeleton: it may inhibit growth above a certain size. In 419D I explore the consequences of omitting the condition ‘locally finite’ from the definition of Radon measure. Even if we insist instead that compact sets should have finite measure, we are in danger of getting a non-localizable measure. Of course this particular space is pathological in terms of most of the criteria of this chapter – for instance, every non-empty open set has infinite measure, and the topology is not regular. Perhaps the most important example in the section is 419E. The analysis of τ -additive product measures in §417 was long and difficult, and if these were actually equal to the familiar product measures in all important cases the structure of the theory would be very different. But we find that for one of the standard compact Radon probability spaces of the theory, the c.l.d. product measure on its square is not a Radon measure, and something has to be done about it. I present 419H here to indicate one of the obstacles to any simplification of the arguments in 417C and 417E. It is not significant in itself, but it offers a welcome excuse to describe some fundamental facts about ω1 (419F-419G). Similarly, 419K asks for some elementary facts about Lebesgue measure (419I-419J) which seem to have got left out. This example really is important in itself, as it touches on the general problem of representing stochastic processes, to which I will return in Chapter 45.

138

Descriptive set theory

Chapter 42 Descriptive set theory At this point, I interpolate an auxiliary chapter, in the same spirit as Chapters 31 and 35 in the last volume. As with Boolean algebras and Riesz spaces, it is not just that descriptive set theory provides essential tools for modern measure theory; it also offers deep intuitions, and for this reason demands study well beyond an occasional glance at an appendix. Several excellent accounts have been published; the closest to what we need here is probably Rogers 80; at a deeper level we have Moschovakis 80, and an admirable recent treatment is Kechris 95. Once again, however, I indulge myself by extracting those parts of the theory which I shall use directly, giving proofs and exercises adapted to the ideas I am trying to emphasize in this volume and the next. The first section describes Souslin’s operation and its basic set-theoretic properties up to the theory of ‘constituents’ (421N-421Q), mostly steering away from topological ideas, but with some remarks on σalgebras and Souslin-F sets. §422 deals with usco-compact relations and K-analytic spaces, working through the topological properties which will be useful later, and giving a version of the First Separation Theorem (422I-422J). §423 looks at ‘analytic’ or ‘Souslin’ spaces, treating them primarily as a special kind of Kanalytic space, with the von Neumann-Jankow selection theorem (423N). §424 is devoted to ‘standard Borel spaces’; it is largely a series of easy applications of results in §423, but there is one substantial theorem on Borel measurable actions of Polish groups (424H).

421 Souslin’s operation I introduce Souslin’s operation S (421B) and show that it is idempotent (421D). I describe alternative characterizations of members of S(E), where E ⊆ PX, as projections of sets in N N ×X (421G-421J). I briefly mention Souslin-F sets (421J-421L) and a special property of ‘inner Souslin kernels’ (421M). At the end of the section I set up an abstract theory of ‘constituents’ for kernels of Souslin schemes and their complements (421N-421Q). 421A Notation Throughout this chapter, and frequently in the next, I shall regard a member of N as the set of its predecessors, so that N k can be identified with the set of functions from k to N, and if φ ∈ N N and k ∈ N, we can speak of the restriction φ¹k ∈ N k . In the same spirit, identifying functions with their graphs, I can write ‘σ ⊆ φ’ when σ ∈ N k , φ ∈ N N and φ extends σ. On occasion I may write #(σ) for the ‘length’ of a finite function σ – again identifying σ with its graph – so that #(σ) = k if σ ∈ X k . And if k = 0, identified with ∅, then the only function from k to X is the empty function, so X 0 becomes {∅}. I shall sometimes refer to the ‘usual topology of N N ’; this is the product topology if each copy of N is S given its discrete topology. Writing Iσ = {φ : φ ∈ N N , φ ⊇ σ} for σ ∈ S ∗ = k∈N N k , {Iσ : σ ∈ S ∗ } is a base for the topology of N N consisting of open-and-closed sets. If σ ∈ N k , i ∈ N I write σ a i for the member τ of N k+1 such that τ (k) = i and τ (j) = σ(j) for j < k. 421B Definition Let S be the set expressible in the form

S k≥1

N k . If E is a family of sets, I write S(E) for the family of sets [ \

Eφ¹k

φ∈N N k≥1

for some family hEσ iσ∈S in E. S T A family hEσ iσ∈S is called a Souslin scheme; the corresponding set φ∈N N k≥1 Eφ¹k is its kernel; the operation [ \ hEσ iσ∈S 7→ Eφ¹k φ∈N N k≥1

is Souslin’s operation or operation A. Thus S(E) is the family of sets obtainable from sets in E by Souslin’s operation. If E = S(E), we say that E is closed under Souslin’s operation.

421Ce

Souslin’s operation

139

S ∗ k Remark I should perhaps warn S you that T some authors use S = k∈N N here in place of S; so that their Souslin kernels are of the form φ∈N N k≥0 Eφ¹k ⊆ E∅ . Consequently, for such authors, any member of S(E) is included in some member of E. If E has a greatest member (or, fractionally more generally, if any sequence in E is bounded above in E) this makes no difference; but if, for instance, E is the family of compact subsets of a topological space, the two definitions of S may not quite coincide. I believe that on this point, for once, I am following the majority. facts (a) It is worth noting straight away that if E is any family of sets, then S 421C Elementary T P Set n∈N En and n∈N En belong to S(E) for any sequence hEn in∈N in E. P Fσ = Eσ(0) for every σ ∈ S, Gσ = Ek whenever k ∈ N, σ ∈ Nk+1 ; then

S n∈N

T n∈N

En =

En =

S

T φ∈N N

S

k≥1

T φ∈N N

k≥1

Fφ¹k ∈ S(E),

Gφ¹k ∈ S(E). Q Q

In particular, E ⊆ S(E). But note that there is no reason why E \ F should belong to S(E) for E, F ∈ E. (b) Let X and Y be sets, and f : X → Y a function. Let hFσ iσ∈S be a Souslin scheme in PY , with kernel B. Then f −1 [B] is the kernel of the Souslin scheme hf −1 [Fσ ]iσ∈S . P P T S T S Q f −1 [B] = f −1 [ φ∈N N n≥1 Fφ¹n ] = φ∈N N n≥1 f −1 [Fφ¹n ]. Q (c) Let X and Y be sets, and f : X → Y a function. Let F be a family of subsets of Y . Then {f −1 [B] : B ∈ S(F)} = S({f −1 [F ] : F ∈ F}). P P For a set A ⊆ X, A ∈ S({f −1 [F ] : F ∈ F }) iff there is some Souslin scheme hEσ iσ∈S in {f −1 [F ] : F ∈ F} such that A is the kernel of hEσ iσ∈S , that is, iff there is some Souslin scheme hFσ iσ∈S in F such that A is the kernel of hf −1 [Fσ ]iσ∈S , that is, iff A = f −1 [B] where B is the kernel of some Souslin scheme in F. Q Q (d) Let X and Y be sets, and f : X → Y a surjective function. Let F be a family of subsets of Y . Then S(F) = {B : B ⊆ Y, f −1 [B] ∈ S({f −1 [F ] : F ∈ F})}. P P If B ∈ S(F), then f −1 [B] ∈ S({f −1 [F ] : F ∈ F }), by (c) above. If B ⊆ Y and f −1 [B] ∈ S({f −1 [F ] : F ∈ F }), then there is a Souslin scheme hFσ iσ∈S in F such that f −1 [B] is the kernel of hf −1 [Fσ ]iσ∈S , that is, f −1 [B] = f −1 [C] where C is the kernel of hFσ iσ∈S . Because f is surjective, B = C ∈ S(F). Q Q (e) Souslin’s operation can be thought of as a projection operator, as follows. Let hEσ iσ∈S be a Souslin scheme with kernel A. Set T S R = n≥1 σ∈N n Iσ × Eσ , writing Iσ = {φ : σ ⊆ φ ∈ N N } as usual. Then R[N N ] = A. P P For any x, and any φ ∈ N N , (φ, x) ∈ R ⇐⇒ for every n ≥ 1 there is a σ ∈ N n such that x ∈ Eσ , φ ∈ Iσ ⇐⇒ x ∈ Eφ¹n for every n ≥ 1. But this means that x ∈ R[N N ] ⇐⇒ there is a φ ∈ N N such that (φ, x) ∈ R \ ⇐⇒ there is a φ ∈ N N such that x ∈ Eφ¹n ⇐⇒ x ∈ A. Q Q n≥1

140

Descriptive set theory

421D

421D The first fundamental theorem is that the operation S is idempotent. Theorem (Souslin 17) For any family E of sets, S(E) is closed under Souslin’s operation. S T proof (a) Let hAσ iσ∈S be a family in S(E), and set A = φ∈N N k≥1 Aφ¹k ; I have to show that A ∈ S(E). S T For each σ ∈ S, let hEστ iτ ∈S be a family in E such that Aσ = ψ∈N N m≥1 Eσ,ψ¹m . Then [ \ [ \ [ \ Eφ¹k,ψ¹m = Eφ¹k,ψk ¹m , A= φ∈N N k≥1 ψ∈N N m≥1

k,m≥1 φ∈N N Ψ∈(N N )N\{0}

writing Ψ = hψk ik≥1 for Ψ ∈ (N N )N\{0} . The idea of the proof is simply that N N × (N N )N\{0} is essentially identical to N N , so that all we have to do is to organize new names for the Eστ . But as it is by no means a trivial matter to devise a coding scheme which really works, I give the details at length. (b) The first step is to note that S and S 2 are countable, so there is a sequence hHn in∈N running over {Eστ : σ, τ ∈ S}. Next, choose any injective function q : N × N → N \ {0} such that q(0, 0) = 1 and q(0, 1) = 2. For k, m ≥ 1 set Jkm = {(i, 0) : i < k} ∪ {(i, k) : i < m}, so that J11 = {(0, 0), (0, 1)}, and choose a family h(kn , mn )in≥3 running over (N \ {0})2 such that q[Jkn ,mn ] ⊆ n for every n ≥ 3. (The pairs (kn , mn ) need not all be distinct, so this is easy to achieve.) Now, for υ ∈ N n , where n ≥ 3, set Fυ = Eστ where σ ∈ N kn , σ(i) = υ(q(i, 0)) for i < kn , τ ∈ N mn , τ (i) = υ(q(i, kn )) for i < mn ; these are well-defined because q[Jkn ,mn ] ⊆ n. For υ ∈ N 1 ∪ N 2 , set Fυ = Hυ(0) . (c) This defines a Souslin scheme hFυ iυ∈S in E. Let A0 be its kernel, so that A0 ∈ S(E). The point is that A0 = A. T P P (i) If x ∈ A, there must be φ ∈ N N , Ψ ∈ (N N )N\{0} such that x ∈ k,m≥1 Eφ¹k,ψk ¹m . Choose θ ∈ N N such that Hθ(0) = Eφ¹1,ψ1 ¹1 , θ(q(i, 0)) = φ(i) for every i ∈ N, θ(q(i, k)) = ψk (i) for every k ≥ 1, i ∈ N. (This is possible because q : N 2 → N \ {0} is injective.) Now Fθ¹1 = Fθ¹2 = Hθ(0) = Eφ¹1,ψ1 ¹1 certainly contains x. And for n ≥ 3, Fθ¹n = Eστ where σ(i) = θ(q(i, 0)) for i < kn , τ (i) = θ(q(i, kn )) for i < mn , that is, σ = φ¹kn and τ = ψkn ¹mn , so again x ∈ Fθ¹n . Thus T x ∈ n≥1 Fθ¹n ⊆ A0 . As x is arbitrary, A ⊆ A0 . (ii) Now take any x ∈ A0 . Let θ ∈ N N be such that x ∈ setting

T n≥1

Fθ¹n . Define φ ∈ NN , Ψ ∈ (N N )N\{0} by

φ(i) = θ(q(i, 0)) for i ∈ N, ψk (i) = θ(q(i, k)) for k ≥ 1, i ∈ N. If k, m ≥ 1, let n ≥ 3 be such that k = kn , m = mn . Then x ∈ Fθ¹n = Eστ , where σ(i) = θ(q(i, 0)) for i < kn ,

τ (i) = θ(q(i, kn )) for i < mn ,

that is, σ = φ¹kn = φ¹k and τ = ψkn ¹mn = ψk ¹m. As m and n are arbitrary, T x ∈ m,n≥1 Eφ¹k,ψk ¹m ⊆ A. As x is arbitrary, A0 ⊆ A. Q Q Accordingly we must have A ∈ S(E), and the proof is complete.

421G

Souslin’s operation

141

421E Corollary For any family E of sets, S(E) is closed under countable unions and intersections. proof For 421Ca tells us that the union and intersection of any sequence in S(E) will belong to SS(E) = S(E). 421F Corollary Let X be a set and E a family of subsets of X. Suppose that X and ∅ belong to S(E) and that X \ E ∈ S(E) for every E ∈ E. Then S(E) includes the σ-algebra of subsets of X generated by E. proof The set Σ = {F : F ∈ S(E), X \ F ∈ S(E)} is closed under complements (necessarily), contains ∅ (because ∅ and X belong to S(E)), and is also closed under countable unions, by 421E. So it is a σ-algebra; but the hypotheses also ensure that E ⊆ Σ, so that the σ-algebra generated by E is included in Σ and in S(E). 421G Proposition Let E be a family of sets such that ∅ ∈ E. For σ ∈ S = N N , φ ⊇ σ}. Then

S k≥1

N k set Iσ = {φ : φ ∈

S(E) = {R[N N ] : R ∈ S({Iσ × E : σ ∈ S, E ∈ E})} = {R[N N ] : R ∈ S({Iσ × E : σ ∈ S, E ∈ E}), R−1 [{x}] is closed for every x}. proof Set F = {Iσ × E : σ ∈ S, E ∈ E}. (a) Suppose first that A ∈ S(E). Let hEσ iσ∈S be a Souslin scheme in E with kernel A. Set T S R = k≥1 σ∈N k Iσ × Eσ . Then R ∈ S(F), by 421E, and R[N N ] = A, by 421Ce. Also T S R−1 [{x}] = k≥1 {Iσ : σ ∈ N k , x ∈ Eσ } is closed, for every x. (b) Now suppose that A = R[N N ] for some R ∈ S(F). Let hIτ (σ) × Eσ iσ∈S be a Souslin scheme in F with kernel R. For k ≥ 1, σ ∈ Nk set Fσ = Eσ if

\

Iτ (σ¹n) 6= ∅,

1≤n≤k

= ∅ otherwise. Then hFσ iσ∈S is a Souslin scheme in E, so its kernel A0 belongs to S(E). The point T is that A0 = A. P P (i) If x ∈ A, there are a φ ∈ NN such that (φ, x) ∈ R and a ψ ∈ N N such that (φ, x) ∈ n≥1 Iτ (ψ¹n) × Eψ¹n . Now, for any k ≥ 1, we have T T φ ∈ 1≤n≤k Iτ (ψ¹n) = 1≤n≤k Iτ ((ψ¹k)¹n) , T so that Fψ¹k = Eψ¹k contains x; thus x ∈ k≥1 Fψ¹k ⊆ A0 . As x is arbitrary, A ⊆ A0 . (ii) If x ∈ A0 , take T T ψ ∈ N N such that x ∈ n≥1 Fψ¹n . In this case we must have Fψ¹k 6= ∅, so 1≤n≤k Iτ (ψ¹n) 6= ∅, for every k ≥ 1. But what this means is that, setting τn = τ (ψ¹n) for each n ≥ 1, τn (i) = τm (i) whenever i ∈ N is T such that both are defined. So {τn : n ≥ 1} must have a common extension φ ∈ N N , and φ ∈ n≥1 Iτ (ψ¹n) . Now T (φ, x) ∈ n≥1 Iτ (ψ¹n) × Eψ¹n ⊆ R, so x ∈ A. Thus A0 ⊆ A and the two are equal. Q Q This shows that {R[N N ] : R ∈ S(F)} ⊆ S(E), and the proof is complete.

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421H

421H When the class E is a σ-algebra, the last proposition can be extended. Proposition Let S X be a set, and Σ a σ-algebra of subsets of X. Let B be the algebra of Borel subsets of N N . For σ ∈ S = k≥1 N k , set Iσ = {φ : φ ∈ N N , φ ⊇ σ}. Then b S(Σ) = {R[N N ] : R ∈ B ⊗Σ} = {R[N N ] : R ∈ S({Iσ × E : σ ∈ S, E ∈ Σ})} b = {R[N N ] : R ∈ S(B ⊗Σ)}. b is the σ-algebra of subsets of N N × X generated by {H × E : H ∈ B, E ∈ Σ}. Notation Recall that B ⊗Σ proof (a) Suppose first that A ∈ S(Σ). As in 421G, let hEσ iσ∈S be a Souslin scheme in Σ with kernel A, and set T S R = k≥1 σ∈N k Iσ × Eσ , b so that A = R[N N ]. Because every Iσ is an open-and-closed set in N N , R ∈ B ⊗Σ. Thus b S(Σ) ⊆ {R[N N ] : R ∈ B ⊗Σ}. b = S(F). P (b) Set F = {Iσ × E : σ ∈ S, E ∈ Σ}. Then S(B⊗Σ) P If E ∈ Σ and σ ∈ N k then S (N N × X) \ (Iσ × E) = (Iσ × (X \ E)) ∪ τ ∈Nk ,τ 6=σ Iτ × X ∈ S(F). Also NN × X =

S σ∈N1

Iσ × X,

∅ = Iτ × ∅

(where τ is any member of S) belong to S(F). S By 421F, S(F) includes the σ-algebra Λ of sets generated by F. Now if E ∈ Σ and H ⊆ N N is open, H = σ∈T Iσ for some T ⊆ S; as T is necessarily countable, S H × E = σ∈T Iσ × E ∈ Λ. Since {F : F ⊆ N N , F × E ∈ Λ} is a σ-algebra of subsets of N N , and we have just seen that it contains all b ⊆ Λ ⊆ S(F), and the open sets, it must include B; thus F × E ∈ Λ for every F ∈ B, E ∈ Σ. So B ⊗Σ b ⊆ SS(F) = S(F) S(F) ⊆ S(B⊗Σ) (421D). Q Q (c) Now we have

b S(Σ) ⊆ {R[N N ] : R ∈ B ⊗Σ} (by (a)) b ⊆ {R[N N ] : R ∈ S(B ⊗Σ)} = {R[N N ] : R ∈ S(F)} (by (b)) = S(Σ) by 421G. 421I There is a particularly simple description of sets obtainable by Souslin’s operation from closed sets in a topological space. Lemma Let X be a topological space and R ⊆ N N × X a closed set. Then S T R[A] = φ∈A n≥1 R[Iφ¹n ]. for any A ⊆ N N , writing Iσ = {φ : σ ⊆ φ ∈ N N } as usual. In particular, R[N N ] is the kernel of the Souslin scheme hR[Iσ ]iσ∈S .

421M

Souslin’s operation

proof Set B=

S

T φ∈A

n≥1

143

R[Iφ¹n ].

(i) If x ∈ R[A], there is a φ ∈ A such that (φ, x) ∈ R. In this case, φ ∈ Iφ¹n so x ∈ R[Iφ¹n ] ⊆ R[Iφ¹n ] for every n, and x ∈ B. Thus R[A] ⊆ B. (ii) If x ∈ B, let φ ∈ A be such that x ∈ R[Iφ¹n ] for every n ∈ N. ?? If (φ, x) ∈ / R, then (because R is closed) there are a σ ∈ S and an open G ⊆ X such that φ ∈ Iσ , x ∈ G and (Iσ × G) ∩ R = ∅. But this means that G ∩ R[Iσ ] = ∅ so G ∩ R[Iσ ] = ∅ and x ∈ / R[Iσ ]; which is absurd, because σ = φ¹n for some n ≥ 1. X X Thus (φ, x) ∈ R and x ∈ R[A]. As x is arbitrary, B ⊆ R[A] and B = R[A], as required. 421J Proposition Let X be a topological space, and F the family of closed subsets of X. Then a set A ⊆ X belongs to S(F) iff there is a closed set R ⊆ N N × X such that A is the projection of R on X. proof (a) Suppose that A ∈ S(F). Let hFσ iσ∈S be a Souslin scheme in F with kernel A. Set T S R = n≥1 σ∈N n Iσ × Fσ . For each n ≥ 1,

S σ∈N n

N

Iσ × Fσ = (N N × X) \

S σ∈Nn

Iσ × (X \ Fσ )

N

is closed in N × X, so R is closed; and the projection R[N ] is A, by 421Ce. (b) Suppose that R ⊆ N N is a closed set with projection A. Then A is the kernel of the Souslin scheme hR[Iσ ]iσ∈S , by 421I, so belongs to S(F). 421K Definition Let X be a topological space. A subset of X is a Souslin-F set in X if it is obtainable from closed subsets of X by Souslin’s operation; that is, is the projection of a closed subset of N N × X. For a subset of R r , or, more generally, of any Polish space, it is common to say ‘Souslin set’ for ‘Souslin-F set’; see 421Xm. 421L Proposition Let X be any topological space. Then every Baire subset of X is Souslin-F. proof Let Z be the family of zero sets in X. If F ∈ Z then X \ F is a countable union of zero sets (4A2C(b-vi)), so belongs to S(Z). By 421F, the σ-algebra generated by Z is included in S(Z) ⊆ S(F), where F is the family of closed subsets of X; that is, every Baire set is Souslin-F. T 421M Proposition Let E be any family of sets such that ∅ ∈ E and E ∪ E 0 , n∈N En belong to E for every E, E 0 ∈ E and all sequences hEn in∈N in E. (For instance, E could be the family of closed subsets of a topological space, or a σ-algebra of sets.) Let hESσ iσ∈STbe a Souslin scheme in E, and K ⊆ N N a set which is compact for the usual topology on N N . Then φ∈K n≥1 Eφ¹n ∈ E. S T proof Set A = φ∈K n≥1 Eφ¹n . For k ∈ N, set Kk = {φ¹k : φ ∈ K}; note that Kk ⊆ N k is compact, because φ 7→ φ¹k is continuous, therefore finite, because the topology of N k is discrete. Set T S T H = k≥1 σ∈Kk 1≤n≤k Eσ¹n . Because E is closed under finite unions and countable intersections, H ∈ T E. Now A = H. P P (i) If x ∈ A, take φ ∈ K such that x ∈ Eφ¹n for every n ≥ 1; then φ¹k ∈ Kk and x ∈ 1≤n≤k E(φ¹k)¹n for every k ≥ 1, T so x ∈ H. Thus A ⊆ H. (ii) If x ∈ H, then for each k ∈ N we have a σk ∈ Kk such that x ∈ 1≤n≤k Eσk ¹n . Choose φk ∈ K such that φk ¹k = σk for each k. Now K is supposed to be compact, so the sequence hφk ik∈N has a cluster point φ in K. If n ≥ 1, then Iφ¹n is a neighbourhood of φ in N N , so must contain φk for infinitely many k; let k ≥ n be such that φk ¹n = φ¹n. In this case x ∈ Eσk ¹n = Eφk ¹n = Eφ¹n . As n is arbitrary,

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x∈

T n≥1

421M

Eφ¹n ⊆ A.

Q As x is arbitrary, H ⊆ A and H = A, as claimed. Q So A ∈ E. *421N I now embark on preparations for the theory of ‘constituents’ of analytic and coanalytic sets. It turns out that much of the work can be done in the abstract context of this section. S Trees and derived trees (a) Let T be the family of subsets T of S = n≥1 N n such that σ¹k ∈ T whenever σ ∈ T and 1 ≤ k ≤ #(σ). Note that the intersection and union of any non-empty family of members of T again belong to T . Members of T are often called trees. (b) For T ∈ T , set ∂T = {σ : σ ∈ S, ∃ i ∈ N, σ a i ∈ T }, so that ∂T ∈ T and ∂T ⊆ T . Of course ∂T0 ⊆ ∂T1 whenever T0 , T1 ∈ T and T0 ⊆ T1 . T (c) For T ∈ T , define h∂ ξ T iξ<ω1 inductively by setting ∂ 0 T = T and, for ξ > 0, ∂ ξ T = ∂( η<ξ ∂ η T ). An easy induction shows that ∂ ξ T ∈ T , ∂ ξ T ⊆ ∂ η T and ∂ ξ+1 T = ∂(∂ ξ T ) whenever η ≤ ξ < ω1 . T (d) For any T ∈ T , there is a ξ < ω1 such that ∂ ξ T = ∂ η T whenever ξ ≤ η < ω1 . P P Set T1 = ξ<ω1 ∂ ξ T . For each σ ∈ S \ T1 , there is a ξσ < ω1 such that σ ∈ / ∂ ξσ T . Set ξ = sup{ξσ : σ ∈ S \ T1 }; because S is ξ countable, ξ < ω1 , and we must now have ∂ T = T1 , so that ∂ ξ T = ∂ η T whenever ξ ≤ η < ω1 . Q Q (e) For T ∈ T , its rank is the first ordinal r(T ) < ω1 such that ∂ r(T ) T = ∂ r(T )+1 T ; of course ∂ r(T ) T = ∂ T whenever r(T ) ≤ η < ω1 , and ∂(∂ r(T ) T ) = ∂ r(T ) T . η

(f ) For T ∈ T , the following are equiveridical: (α) ∂ r(T ) T 6= ∅; (β) there is a φ ∈ N N such that φ¹n ∈ T for every n ≥ 1. P P (i) If σ ∈ ∂ r(T ) T then σ ∈ ∂(∂ r(T ) T ) so there is an i ∈ N such that σ a i ∈ ∂ r(T ) T . We can therefore choose hσn in∈N inductively so that σn ∈ ∂ r(T ) T and σn+1 properly extends σn for every n. At S the end of the induction, φ = n∈N σn belongs to N N and φ¹n = σn ¹n ∈ ∂ r(T ) T ⊆ T for every n ≥ 1. (ii) If φ ∈ N N is such that φ¹n ∈ T for every n ≥ 1, then an easy induction shows that φ¹n ∈ ∂ ξ T for every ξ < ω1 and every n ≥ 1, so that ∂ r(T ) T is non-empty. Q Q (g) Now suppose that hAσ iσ∈S is a Souslin scheme. For any x we have a tree Tx ∈ T defined by saying that T Tx = {σ : σ ∈ S, , x ∈ 1≤i≤#(σ) Aσ¹i }. Now the kernel of hAσ iσ∈S is just A = {x : ∃ φ ∈ N N , x ∈

\

Aφ¹n }

n≥1

= {x : ∃ φ ∈ N N , φ¹n ∈ Tx ∀ n ≥ 1} = {x : ∂ r(T ) T 6= ∅} by (f). The sets {x : x ∈ X \ A, r(Tx ) = ξ} = {x : x ∈ X, r(Tx ) = ξ, ∂ ξ Tx = ∅}, for ξ < ω1 , are called constituents of X \ A. (Of course they should properly be called ‘the constituents of the Souslin scheme hAσ iσ∈S ’.)

*421Q

Souslin’s operation

145

*421O Theorem Let X be a set and Σ a σ-algebra of subsets of X. Let hAσ iσ∈S be a Souslin scheme in Σ with kernel A, and for x ∈ X set T Tx = {σ : σ ∈ S, x ∈ 1≤i≤#(σ) Aσ¹i } ∈ T as in 421Ng. (a) For every ξ < ω1 and σ ∈ S, {x : x ∈ X, σ ∈ ∂ ξ Tx } ∈ Σ. (b) For every ξ < ω1 , {x : x ∈ A, r(Tx ) ≤ ξ} and {x : x ∈ X \ A, r(Tx ) ≤ ξ} belong to Σ. In particular, all the constituents of X \ A belong to Σ. proof (a) Induce on ξ. For ξ = 0, we have {x : x ∈ X, σ ∈ ∂ 0 Tx } = {x : x ∈ X, σ ∈ Tx } =

T 1≤i≤#(σ)

Aσ¹i ∈ Σ.

For the inductive step to ξ > 0, we have {x : σ ∈ ∂ ξ Tx } = {x : σ ∈ ∂( =

[ \

\

∂ η Tx )} =

[

{x : σ a i ∈

i∈N

η<ξ

\

∂ η Tx }

η<ξ

{x : σ a i ∈ ∂ η Tx } ∈ Σ

i∈N η<ξ

because ξ is countable and all the sets {x : σ a i ∈ ∂ η Tx } belong to Σ by the inductive hypothesis. (b) Now, given ξ < ω1 , we see that r(Tx ) ≤ ξ iff ∂ ξ+1 Tx ⊆ ∂ ξ Tx , so that if we set Eξ = {x : x ∈ X, r(Tx ) ≤ ξ} then T / ∂ ξ Tx } Eξ = σ∈S {x : x ∈ X, σ ∈ ∂ ξ+1 Tx or σ ∈ belongs to Σ. Now, given that x ∈ Eξ , so that ∂ r(Tx ) Tx = ∂ ξ Tx , 421Ng tells us that x ∈ A iff ∅ ∈ ∂ ξ Tx ; so that Eξ ∩ A and Eξ \ A both belong to Σ. S Now the constituents of X \ A are the sets (Eξ \ A) \ η<ξ Eη for ξ < ω1 , so all belong to Σ. *421P Corollary Let X be a set and Σ a σ-algebra of subsets of X. If A ∈ S(Σ) then both A and X \ A can be expressed as the union of at most ω1 members of Σ. proof In the language of 421O, we have S A = ξ<ω1 Eξ ∩ A,

X \A=

S ξ<ω1

Eξ \ A.

*421Q Lemma Let X be a set and hAσ iσ∈S and hB Tσ iσ∈S two Souslin schemes of subsets of X. Suppose that whenever φ, ψ ∈ N N there is an n ≥ 1 such that 1≤i≤n Aφ¹i ∩ Bψ¹i = ∅. For x ∈ X set S T Tx = n≥1 {σ : σ ∈ N n , x ∈ 1≤i≤n Aσ¹i } as in 421Ng, and let B be the kernel of hBσ iσ∈S . Then supx∈B r(Tx ) < ω1 . T T proof For σ ∈ S set A0σ = 1≤i≤#(σ) Aσ¹i , Bσ0 = 1≤i≤#(σ) Bσ¹i . Then Tx = {σ : σ ∈ S, x ∈ A0σ } for each 0 = ∅. x ∈ X, B is the kernel of hBσ0 iσ∈S , and for every φ, ψ ∈ N N there is an n ∈ N such that A0φ¹n ∩ Bψ¹n Define hQξ iξ<ω1 inductively by setting Q0 = {(σ, τ ) : σ, τ ∈ S, A0σ ∩ Bτ0 6= ∅}, and, for 0 < ξ < ω1 , Qξ = {(σ, τ ) : σ, τ ∈ S, ∃ i, j ∈ N, (σ a i, τ a j) ∈

T η<ξ

Qη }.

Then the same arguments as in 421Na-421Nd show that there is a ζ < ω1 such that Qζ+1 = Qζ . ?? If Qζ 6= ∅, then, just as in 421Nf, there must be φ, ψ ∈ N N such that (φ¹m, ψ¹n) ∈ Qζ ⊆ Q0 for every m, 0 6= ∅ for every n ≥ 1, which is supposed to be impossible. X X n ≥ 1; but this means that A0φ¹n ∩ Bψ¹n 0 for every n ≥ 1. But this Now suppose that x ∈ B. Then there is a ψ ∈ N N such that x ∈ Bψ¹n means that (σ, ψ¹n) ∈ Q0 for every σ ∈ Tx and every n ≥ 1. An easy induction shows that (σ, ψ¹n) ∈ Qξ whenever ξ < ω1 , σ ∈ ∂ ξ Tx and n ≥ 1. But as Qζ = ∅ we must have ∂ ζ Tx = ∅ and r(Tx ) ≤ ζ. Thus supx∈B r(Tx ) ≤ ζ < ω1 , and the proof is complete.

146

Descriptive set theory

421X

421X Basic exercises (a) Let X be a set and E a family of subsets of X. (i) Show that ∅ ∈ S(E) iff there is a sequence in E with empty intersection. (ii) Show that X ∈ S(E) iff there is a sequence in E with union X. (b) Let E be a family of sets and F any set. Show that S({E ∩ F : E ∈ E}) = {A ∩ F : A ∈ S(E)}, S({E ∪ F : E ∈ E}) = {A ∪ F : A ∈ S(E)}. (c) Let E be a family of sets and F any set. Show that S(E ∪ {F }) = {F } ∪ {A ∩ F : A ∈ S(E)} ∪ {B ∪ F : B ∈ S(E)} ∪ {(A ∩ F ) ∪ B : A, B ∈ S(E)}. (d) Suppose that E is a family of sets with #(E) ≤ c. Show that #(S(E)) ≤ c. (Hint: #(E S ) ≤ #((PN)S ) = #(P(N × S)).) (e) Let E be the family of half-open intervals [2−n k, 2−n (k + 1)[, where n ∈ N, k ∈ Z; let G be the set of open subsets of R; let F be the set of closed subsets of R; let K be the set of compact subsets of R; let B be the Borel σ-algebra of R. Show that S(E) = S(F) = S(G) = S(K) = S(B). (Hint: 421F.) S (f ) Let I be the family {Iσ : σ ∈ k∈N N k } (421A); let G be the set of open subsets of N N ; let F be the set of closed subsets of N N ; let K be the set of compact subsets of N N ; let B be the Borel σ-algebra of N N . Show that S(I) = S(F) = S(G) = S(B), but that S(K) is strictly smaller thanSthese. (Hint: if hKn in∈N is any sequence in K, set φ(i) = 1 + supψ∈Ki ψ(i) for each i ∈ N, so that φ ∈ / n∈N Kn ; hence show that NN ∈ / S(K).) (g) Let X be a separable metrizable space with at least two points; let U be any base for its topology, and B its Borel σ-algebra. Show that S(U) = S(B). What can happen if #(X) ≤ 1? (h) Let X be a topological space; let Z be the set of zero sets in X, G the set of cozero sets, and Ba the Baire σ-algebra. Show that S(Z) = S(G) = S(Ba). (i) Let X be a set, E a family of subsets of X, and Σ the σ-algebra of subsets of X generated by E. Show that if #(E) ≤ c then #(Σ) ≤ c. (Hint: #(cS ) = #(P(N × N)) = c and Σ ⊆ S(E ∪ {X \ E : E ∈ E}).) (j) Let X be a topological space such that every open set is Souslin-F. Show that every Borel set is Souslin-F. (k) Let X be a topological space and B(X) its Borel σ-algebra. Show that S(B(X)) is just the set of projections on X of Borel subsets of N N × X. (Hint: 4A3G.) (l) Let X and Y be topological spaces, f : X → Y a continuous function and F ⊆ Y a Souslin-F set. Show that f −1 [F ] is a Souslin-F set in X. (m) Let X be any metrizable space; let G be the set of open subsets of X, F the set of closed subsets, and B the Borel σ-algebra. Show that S(G) = S(F) = S(B). (n) Let us say that a Souslin scheme hEσ iσ∈S is regular if Eσ ⊆ Eτ whenever σ, τ ∈ S, #(τ ) ≤ #(σ) and σ(i) ≤ τ (i) for every i < #(σ). Let E be a family of sets such that E ∪ F , E ∩ F ∈ E for all E, F ∈ E. Show that every member of S(E) can be expressed as the kernelTof a regular Souslin S scheme in E. (Hint: if hEσ iσ∈S is any Souslin scheme in E with kernel A, set Fσ = τ ⊆σ Eτ , Gσ = τ ≤σ Fτ , where τ ≤ σ if τ (i) ≤ σ(i) for i < #(τ ) = #(σ); show that A is the kernel of hFσ iσ∈S and of hGσ iσ∈S , using an idea from 421M for the latter.)

421 Notes

Souslin’s operation

147

(o) Let X be a Hausdorff space and hKσ iσ∈S a Souslin scheme in which every Kσ is a compact S topological T subset in X. Show that φ∈K n≥1 Kφ¹n is compact for any compact K ⊆ N N . 421Y Further exercises (a) Let X be a topological space, Y a Hausdorff space and f : X → Y a continuous function. Let K be the family of closed countably compact subsets of X. Show that for any E ⊆ K, {f [A] : A ∈ S(E)} = S({f [E] : E ∈ E}). (b) Let X be a topological space, and Ba its Baire σ-algebra. Show that S(Ba) is just the family of sets expressible as f −1 [B] where f is a continuous function from X to some metrizable space Y and B ⊆ Y is Souslin-F. (c) Let X be a set, E a family of subsets of X, and Σ the smallest σ-algebra of subsets of X including E and closed under Souslin’s operation.S Show that if #(E) ≤ c then #(Σ) ≤ c. (Hint: define hEξ iξ<ω1 by setting Eξ = S({X \ E : E ∈ E ∪ η<ξ Eξ }) for each ξ. Show that #(Eξ ) ≤ c for every ξ and that S Σ = ξ<ω1 Eξ .) (d) Let X be a compact S space and A a Souslin-F set in X. Show that there is a family hFξ iξ<ω1 of Borel sets such that X \ A = ξ<ω Fξ and whenever B ⊆ X \ A is a Souslin-F set there is a ξ < ω1 such that B ⊆ Fξ . (Hint: take Fξ = {x : r(Tx ) ≤ ξ} \ A as in 421Ob, and apply 421Q.) 421 Notes and comments In 111G, I defined the Borel sets of R to be the members of the smallest σ-algebra containing every open set. In 114E, I defined a set to be Lebesgue measurable if it behaves in the right way with respect to Lebesgue outer measure. The latter formulation, at least, provides some sort of testing principle to determine whether a set is Lebesgue measurable. But the definition of ‘Borel set’ does not. The only tool so far available for proving that a set E ⊆ R is not Borel is to find a σ-algebra containing all open sets and not containing E; conversely, the only method we have for proving properties of Borel sets is to show that a property is possessed by every member of some σ-algebra containing every open set. The revolutionary insight of Souslin 17 was a construction which could build every Borel set from rational intervals. (See 421Xe.) For fundamental reasons, no construction of this kind can provide all Borel sets without also producing other sets, and to actually characterize the Borel σ-algebra a further idea is needed (423Fa); but the class of analytic sets, being those constructible by Souslin’s operation from rational intervals (or open sets, or closed sets, or Borel sets – the operation is robust under such variations), turns out to have remarkable properties which make it as important in modern real analysis as the Borel algebra itself. The guiding principle of ‘descriptive set theory’ is that the properties of a set may be analysed in the light of a construction for that set. Thus we can think of a closed set F ⊆ R as S R \ (q,q0 )∈I ]q, q 0 [ where I ⊆ Q × Q. The principle can be effective because we often have such descriptions in terms of objects fundamentally simpler than the set being described. In the formula above, for instance, Q × Q is simpler than the set F , being a countable set with a straightforward description from N. The set P(Q × Q) is relatively complex; but a single subset I of Q × Q can easily be coded as a single subset of N (taking some more or less natural enumeration of Q2 as a sequence h(qn , qn0 )in∈N , and matching I with {n : (qn , qn0 ) ∈ I}). So, subject to an appropriate coding, we have a description of closed subsets of R in terms of subsets of N. At the most elementary level, this shows that there are at most c closed subsets of R. But we can also set out to analyse such operations as intersection, union, closure in terms of these descriptions. The details are complex, and I shall go no farther along this path here; but investigations of this kind are at the heart of some of the most exciting developments of twentieth-century real analysis. The particular descriptive method which concerns us in the present section is Souslin’s operation. Starting from a relatively simple class E, we proceed to the larger class S(E). The most fundamental property of S is 421D: SS(E) = S(E). This means, for instance, that if E ⊆ S(F) and F ⊆ S(E), then S(E) will be equal to S(F); consequently, different classes of sets will often have the same Souslin closures, as in 421Xe-421Xh.

148

Descriptive set theory

421 Notes

After a little practice you will find that it is often easy to see when two classes E and F are at the same level in this sense; but watch out for traps like the class of compact subsets of N N (421Xf) and odd technical questions (421Xg). Souslin’s operation, and variations on it, will be the basis of much of the next chapter; it has dramatic applications in general topology and functional analysis as well as in real analysis and measure theory. An important way of looking at the kernel of a Souslin T scheme S hEσ iσ∈S is to regard it as the projection on the second coordinate of the corresponding set R = k≥1 σ∈N k Iσ × Eσ (421Ce). We find that many other sets R ⊆ N N × X will also have projections in S(E) (421G, 421H). Let me remark that it is essential here that the first coordinate should be of the right type. In one sense, indeed, N N is the only thing that will do; but its virtue transfers to analytic spaces, as we shall see in 423M-423O below. We shall often want to deal with members of S(E) which are most naturally defined in terms of some such auxiliary space. I have moved into slightly higher gear for 421N-421Q because these are not essential for most of the work of the next chapter. From the point of view of this section 421P is very striking but the significance of 421Q is unlikely to be apparent. It becomes important in contexts in which the condition T ∀ φ, ψ ∈ N N ∃ n ≥ 1, 1≤i≤n Aφ¹i ∩ Bψ¹i = ∅ is satisfied for natural reasons. I will expand on these in the next two sections. In the meantime, I offer 421Yd as an example of what 421O and 421Q together can tell us.

422 K-analytic spaces I introduce K-analytic spaces, defined in terms of usco-compact relations. The first step is to define the latter (422A) and give their fundamental properties (422B-422E). I reach K-analytic spaces themselves in 422F, with an outline of the most important facts about them in 422G-422K. 422A Definition Let X and Y be Hausdorff spaces. A relation R ⊆ X × Y is usco-compact if (α) R[{x}] is a compact subset of Y for every x ∈ X, (β) R−1 [F ] is a closed subset of X for every closed set F ⊆ Y . (Relations satisfying condition (β) are sometimes called ‘upper semi-continuous’.) 422B The following elementary remark will be useful. Lemma Let X and Y be Hausdorff spaces and R ⊆ X × Y an usco-compact relation. If H is an open subset of Y including R[{x}] where x ∈ X, there is an open set G ⊆ X, containing x, such that R[G] ⊆ H. proof Set G = X \ R−1 [Y \ H]. Because Y \ H is closed, so is R−1 [Y \ H], and G is open. Of course R[G] ⊆ H, and x ∈ G because R[{x}] ⊆ H. 422C Proposition Let X and Y be Hausdorff spaces. Then a subset R of X × Y is an usco-compact relation iff whenever F is an ultrafilter on X × Y , containing R, such that the first-coordinate image π1 [[F]] of F has a limit in X, then F has a limit in R. proof Recall that, writing π1 (x, y) = x and π2 (x, y) = y for (x, y) ∈ X × Y , π1 [[F]] = {A : A ⊆ X, π1−1 [A] ∈ F} = {A : A ⊆ X, A × Y ∈ F} (2A1Ib), and that F → (x, y) iff π1 [[F]] → x and π2 [[F]] → y (3A3Ic). (a) Suppose that R is usco-compact and that F is an ultrafilter on X × Y , containing R, such that π1 [[F]] has a limit x ∈ X. ?? If F has no limit in R, then, in particular, it does not converge to (x, y) for any y ∈ R[{x}]; that is, π2 [[F]] does not converge to any point of R[{x}], that is, every point of R[{x}] belongs to an open set not belonging to π2 [[F]]. Because R[{x}] is compact, it is covered by a finite union of open sets not belonging to π2 [[F]]; but as π2 [[F]] is an ultrafilter (2A1N), there is an open set H ⊇ R[{x}] such that Y \ H ∈ π2 [[F]]. Now 422B tells us that there is an open set G containing x such that R[G] ⊆ H. In this case, G ∈ π1 [[F]] so G × Y ∈ F; at the same time, X × (Y \ H) ∈ F. So

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R ∩ (G × Y ) ∩ (X × (Y \ H)) ∈ F. But this is empty, by the choice of G; which is intolerable. X X Thus F has a limit in R, as required. (b) Now suppose that R has the property described. (i) Let x ∈ X, and suppose that G is an ultrafilter on Y containing R[{x}]. Set h(y) = (x, y) for y ∈ Y ; then F = h[[G]] is an ultrafilter on X × Y containing R. The image π1 [[F]] is just the principal filter generated by {x}, so certainly converges to x; accordingly F must converge to some point (x, y) ∈ R, and G = π2 [[F]] converges to y ∈ R[{x}]. As G is arbitrary, R[{x}] is compact (2A3R). (ii) Let F ⊆ Y be closed, and x ∈ R−1 [F ] ⊆ X. Consider E = {R, X × F } ∪ {G × Y : G ⊆ X is open, x ∈ G}. Then E has the finite intersection property. P P If G0 , . . . , GTn are open sets containing x, then R−1 [F ] meets G0 ∩ . . . ∩ Gn in z say, and now (z, y) ∈ R ∩ (X × F ) ∩ i≤n (Gi × Y ) for some y ∈ F . Q Q Let F be an ultrafilter on X × Y including E (4A1Ia). Because G × Y ∈ E ⊆ F for every open set G containing x, π1 [[F]] → x, so F converges to some point (x, y) of R. Because X × F is a closed set belonging to E ⊆ F, y ∈ F and x ∈ R−1 [F ]. As x is arbitrary, R−1 [F ] is closed; as F is arbitrary, R satisfies condition (β) of 422A, and is usco-compact. 422D Lemma (a) Let X and Y be Hausdorff spaces. If R ⊆ X × Y is an usco-compact relation, then R is closed in X × Y . (b) Let X and Y be Hausdorff spaces. If R ⊆ X × Y is an usco-compact relation and R0 ⊆ R is a closed set, then R0 is usco-compact. (c) Let X and Y be Hausdorff spaces. If f : X → Y is a continuous function, then its graph is an usco-compact relation. (d) Let hXi ii∈I and of Hausdorff spaces, and Ri ⊆ Xi × Yi an usco-compact relation Q Q hYi ii∈I be families for each i. Set X = i∈I Xi , Y = i∈I Yi and R = {(x, y) : x ∈ X, y ∈ Y, (x(i), y(i)) ∈ Ri for every i ∈ I}. Then R is usco-compact in X × Y . (e) Let X and Y be Hausdorff spaces, and R ⊆ X × Y an usco-compact relation. Then (i) R[K] is a compact subset of Y for any compact subset K of X (ii) R[L] is a Lindel¨of subset of Y for any Lindel¨of subset L of X. (f) Let X, Y and Z be Hausdorff spaces, and R ⊆ X × Y , S ⊆ Y × Z usco-compact relations. Then the composition S ◦ R = {(x, z) : there is some y ∈ Y such that (x, y) ∈ R and (y, z) ∈ S} is usco-compact in X × Z. (g) Let X and Y be Hausdorff spaces and Y0 any subset of Y . Then a relation R ⊆ X ×Y0 is usco-compact when regarded as a relation between X and Y0 iff it is usco-compact when regarded as a relation between X and Y . proof (a) If (x, y) ∈ R, there is an ultrafilter F containing R and converging to (x, y) (4A2Bc). By 422C, F must have a limit in R; but as X × Y is Hausdorff, this limit must be (x, y), and (x, y) ∈ R. As (x, y) is arbitrary, R is closed. (b) It is obvious that R0 will satisfy the condition of 422C if R does. (c) f [{x}] = {f (x)} is surely compact for every x ∈ X, and f −1 [F ] is closed for every closed set F ⊆ Y because f is continuous. (d) For i ∈ I, x ∈ X, y ∈ Y set φi (x, y) = (x(i), y(i)). If F is an ultrafilter on X × Y containing R such that π1 [[F]] has a limit in X, then π1 φi [[F]] = ψi π1 [[F]]

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has a limit in Xi for every i ∈ I, writing ψi (x) = x(i) for i ∈ I, x ∈ X. But φi [[F]] contains Ri , so has a limit (x0 (i), y0 (i)) in Xi × Yi , for each i. Accordingly (x0 , y0 ) is a limit of F in X × Y (3A3Ic). As F is arbitrary, R is usco-compact. (e) For the moment, let L be any subset of X. Let H be a family of open sets in Y covering R[L]. Let G be the family of those open sets G ⊆ X such that R[G] can be covered S by finitely many members of H. Then G covers L. P P If x ∈ L, then R[{x}] is a compact subset of R[L] ⊆ H,Sso there is a finite set H0 ⊆ H covering R[{x}]. Now there is an open set G containing x such that R[G] ⊆ H0 , by 422B. Q Q S (i) If L is compact, then there must be a finite subfamily G 0 of G covering L; now R[L] ⊆ R[ G 0 ] is covered by finitely many members of H. As H is arbitrary, R[L] is compact. S (ii) If L is Lindel¨of, then there must be a countable subfamily G 0 of G covering L; now R[L] ⊆ R[ G 0 ] is covered by countably many members of H. As H is arbitrary, R[L] is Lindel¨of. (f ) If x ∈ X then R[{x}] ⊆ Y is compact, so (SR)[{x}] = S[R[{x}] is compact, by (e-i). If F ⊆ Z is closed then S −1 [F ] ⊆ Y is closed so (SR)−1 [F ] = R−1 [S −1 [F ]] is closed. (g)(i) Suppose that R is usco-compact when regarded as a subset of X × Y0 . Set S = {(y, y) : y ∈ Y0 }; by (c), S is usco-compact when regarded as a subset of Y0 × Y , so by (f) R = SR is usco-compact when regarded as a subset of X × Y . (ii) If R is usco-compact when regarded as a subset of X × Y , and x ∈ X, then R[{x}] is a subset of Y0 which is compact for the topology of Y , therefore for the subspace topology of Y0 . If F ⊆ Y0 is closed for the subspace topology, it is of the form F 0 ∩ Y0 for some closed F 0 ⊆ Y , so R−1 [F ] = R−1 [F 0 ] is closed in X. As x and F are arbitrary, R is usco-compact in X × Y0 . 422E The following lemma is actually very important in the structure theory of K-analytic spaces (see 422Yb). It will be useful to us in 423C below. Lemma Let X and Y be Hausdorff spaces, and R ⊆ X × Y an usco-compact relation. If X is regular, so is R (in its subspace topology). proof ?? Suppose, if possible, otherwise; that there are a closed set F ⊆ R and an (x, y) ∈ R \ F which cannot be separated from F by open sets (in R). If G, H are open sets containing x, y respectively, then R ∩ (G × H), R \ (G × H) are disjoint relatively open sets in R, so the latter cannot include F ; that is, F ∩ (G × H) 6= ∅ whenever G, H are open and x ∈ G, y ∈ H. Accordingly there is an ultrafilter F on X × Y such that F ∩ (G × H) ∈ F whenever G ⊆ X, H ⊆ Y are open sets containing x, y respectively. In this case R ∈ F, and G ∈ π1 [[F]] for every open set G containing x. Because the topology of X is regular, every open set containing x includes G for some smaller open set G containing x, and belongs to π1 [[F]]; thus π1 [[F]] → x in X. Because R is usco-compact, F has a limit in R, which must be of the form (x, y 0 ). Because F ∈ F is closed (in R), (x, y 0 ) ∈ F . But also y 0 ∈ H for every open set H containing y, since X × H is a closed set belonging to F; because the topology of Y is Hausdorff, y 0 must be equal to y, and (x, y) ∈ F , which is absurd. X X 422F Definition (Frol´ık 61) Let X be a Hausdorff space. Then X is K-analytic if there is an usco-compact relation R ⊆ N N × X such that R[N N ] = X. If X is a Hausdorff space, we call a subset of X K-analytic if it is a K-analytic space in its subspace topology. 422G Theorem (a) Let X be a Hausdorff space. Then a subset A of X is K-analytic iff there is an usco-compact relation R ⊆ N N × X such that R[N N ] = A. (b) N N is K-analytic. (c) Compact Hausdorff spaces are K-analytic. (d) If X and Y are Hausdorff spaces and R ⊆ X × Y is an usco-compact relation, then R[A] is K-analytic whenever A ⊆ X is K-analytic. In particular, a Hausdorff continuous image of a K-analytic Hausdorff space is K-analytic. (e) A product of countably many K-analytic Hausdorff spaces is K-analytic.

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(f) A closed subset of a K-analytic Hausdorff space is K-analytic. (g) A K-analytic Hausdorff space is Lindel¨of, so a regular K-analytic Hausdorff space is completely regular. proof (a) A is K-analytic iff there is an usco-compact relation R ⊆ N N × A with projection A. But a subset of N N × X with projection A is usco-compact in N N × A iff it is usco-compact in N N × X, by 422Dg. (b) The identity function from N N to itself is an usco-compact relation, by 422Dc. (c) If X is compact, then R = N N × X is an usco-compact relation (because R[{φ}] = X is compact for every φ ∈ N N , while R−1 [F ] is either N N or ∅ for every closed F ⊆ X), so X = R[N N ] is K-analytic. (d) By (a), there is an usco-compact relation S ⊆ N N × X such that S[N N ] = A. Now R ◦ S ⊆ N N × Y is usco-compact, by 422Df, and RS[N N ] = R[A]. In particular, if X itself is K-analytic and f : X → Y is a continuous surjection, f is an usco-compact relation (422Dc), so Y = f [X] is K-analytic. (e) Let hXi ii∈I be a countable family of K-analytic Hausdorff spaces with product X. If I = ∅ then X = {∅} is compact, therefore K-analytic. Otherwise, choose for each i ∈ I an usco-compact relation Ri ⊆ N N × Xi such that Ri [N N ] = Xi . Set R = {(φ, x) : φ ∈ (N N )I , x ∈ X, (φ(i), x(i)) ∈ Ri for every i ∈ I}. By 422Dd, R is an usco-compact relation in (N N )I × X, and it is easy to see that R[(N N )I ] = X. But (N N )I ∼ = N N×I is homeomorphic to N N , because I is countable, so we can identify R with a relation in N N × X which is still usco-compact, and X is K-analytic. (f ) Let X be a K-analytic Hausdorff space and F a closed subset. Let R ⊆ N N × X be an usco-compact relation such that R[N N ] = X. Set R0 = R ∩ (N N × F ). Then R0 is a closed subset of R, so is usco-compact (422Da). By (a), F = R0 [N N ] is K-analytic. (g) Let X be a K-analytic Hausdorff space. N N is Lindel¨of (4A2Ub), and there is an usco-compact relation R such that R[N N ] = X, so that X is Lindel¨of, by 422D(e-ii). 4A2H(b-i) now tells us that if X is regular it is completely regular. 422H Theorem (a) If X is a Hausdorff space, then any K-analytic subset of X is Souslin-F in X. (b) If X is a K-analytic Hausdorff space, then a subset of X is K-analytic iff it is Souslin-F in X. (c) For any Hausdorff space X, the family of K-analytic subsets of X is closed under Souslin’s operation. proof (a) If A ⊆ X is K-analytic, there is an usco-compact relation R ⊆ N N × X such that R[N N ] = A, by 422Ga. By 422Da, R is a closed set; so A is Souslin-F by 421J. (b) Now suppose that X itself is K-analytic, and that A ⊆ X is Souslin-F in X. Then there is a closed set R ⊆ N N × X such that R[N N ] = A (421J, in the other direction). N N × X is K-analytic (422Gb, 422Ge), and R is closed, therefore itself K-analytic (422Gf); so its continuous image A is K-analytic, by 422Gd. (c)(i) The first step is to show that the union of a sequence of K-analytic subsets of X is K-analytic. P P Let hAn in∈N be a sequence of K-analytic sets, with union A. For each n ∈ N, let Rn ⊆ N N × X be an usco-compact relation such that Rn [N N ] = An . In (N × N N ) × X let R be the set {((n, φ), x) : n ∈ N, (φ, x) ∈ Rn }. N

If (n, φ) ∈ N × N , then R[{(n, φ)}] = Rn [{φ}] is compact; if F ⊆ X is closed, then R−1 [F ] = {(n, φ) : n ∈ N, φ ∈ Rn−1 [F ]} is closed in N × N N . So R is usco-compact, and of course S R[N × N N ] = n∈N Rn [N N ] = A. As N × N N is K-analytic (in fact, homeomorphic to N N ), A is K-analytic. Q Q (ii) S Now suppose that hAσ iσ∈S is a Souslin scheme consisting of K-analytic sets with kernel A. Then X 0 = σ∈S Aσ is K-analytic, by (i). By (a), every Aσ is Souslin-F when regarded as a subset of X 0 . But since the family of Souslin-F subsets of X 0 is closed under Souslin’s operation, by 421D, A is also Souslin-F in X 0 . By (b) of this theorem, A is K-analytic. As hAσ iσ∈S is arbitrary, we have the result.

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422I It seems that for the measure-theoretic results of §432, at least, the following result (the ‘First Separation Theorem’) is not essential. However I do not think it possible to get a firm grasp on K-analytic and analytic spaces without knowing some version of it, so I present it here. It is most often used through the forms in 422J and 422Xd below. S T Lemma Let X be a Hausdorff space. Let E be a family of subsets of X such that (i) n∈N En , n∈N En ∈ E whenever hEn in∈N is a sequence in ∈ E (ii) whenever x, y are distinct points of X, there are disjoint E, F ∈ E such that x ∈ int E and y ∈ int F . Then whenever A, B are disjoint non-empty K-analytic subsets of X, there are disjoint E, F ∈ E such that A ⊆ E and B ⊆ F . proof (a) We need to know that if K, L are disjoint non-empty compact subsets of X, there are disjoint E, F ∈ E such that K ⊆ int E and L ⊆ int F . P P For any point (x, y) ∈ K × L, we can find disjoint Exy , Fxy ∈ E such that x ∈ int E and y ∈ int F . Because L is compact and S T non-empty, there S is for each x ∈ K a non-empty finite set Ix ⊆ L such that L ⊆ y∈Ix int Fxy . Set Ex = y∈Ix Exy , Fx = y∈Ix Fxy ; then Ex , Fx are disjoint members of E, x ∈ int Ex and K is compact and S L ⊆ int Fx . Because S T not empty, there is a non-empty finite set J ⊆ K such that K ⊆ x∈J int Ex . Set E = x∈J Ex , F = x∈J Fx ; then E, F ∈ E, E ∩ F = ∅, K ⊆ int E and L ⊆ int F , as required. Q Q (b) LetSQ, R ⊆ N N × X be usco-compact relations such that Q[N N ] = A and R[N N ] = B. For each σ ∈ S ∗ = n∈N N n , set Iσ = {φ : σ ⊆ φ ∈ N N }, S so that A = A∅ and Aσ = i∈N Aσa i for every σ.

Aσ = Q[Iσ ],

Bσ = R[Iσ ],

(c) Write T for the set of pairs {(σ, τ ) : σ, τ ∈ S ∗ and there are disjoint E, F ∈ E such that Aσ ⊆ E, Bτ ⊆ F }. If σ, τ ∈ S ∗ are such that (σ a i, τ a j) ∈ T for every i, j ∈ N, then (σ, τ ) ∈ T . P P For each i, j ∈ N take disjoint Eij , Fij ∈ E such that S

Then E = i∈N (σ, τ ) ∈ T . Q Q

T j∈N

Eij , F =

T

Aσa i ⊆ Eij , Bτ a j ⊆ Fij . S i∈N j∈N Fij are disjoint and belong to E, and Aσ ⊆ E, Bτ ⊆ F . So

(d) ?? Now suppose, if possible, that there are no disjoint E, F ∈ E such that A ⊆ E and B ⊆ F ; that is, that (∅, ∅) ∈ / T . By (c), used repeatedly, we can find sequences hφ(i)ii∈N , hψ(i)ii∈N such that (φ¹n, ψ¹n) ∈ /T for every n ∈ N. Set K = Q[{φ}], L = R[{ψ}]. These are compact (because R is usco-compact) and disjoint (because K ⊆ A and L ⊆ B). By (a), there are disjoint E, F ∈ E such that K ⊆ int E and L ⊆ int F . By 422B, there are open sets U , V ⊆ N N such that φ ∈ U,

Q[U ] ⊆ int E,

ψ ∈V,

R[V ] ⊆ int F .

But now there is some n ∈ N such that Iφ¹n ⊆ U and Iψ¹n ⊆ V , in which case Aφ¹n ⊆ E,

Bψ¹n ⊆ F ,

and (φ¹n, ψ¹n) ∈ T , contrary to the choice of φ and ψ. X X This contradiction shows that the lemma is true. 422J Corollary Let X be a Hausdorff space and A, B disjoint K-analytic subsets of X. Then there is a Borel set which includes A and is disjoint from B. proof Apply 422I with E the Borel σ-algebra of X. *422K I give the next step in the theory of ‘constituents’ begun in 412O-412R. Theorem Let X be a Hausdorff space. (i) Suppose that X is regular. Let A ⊆ X be a K-analytic set. Then there is a non-decreasing family hEξ iξ<ω1 of Borel sets in X, with union X \ A, such that every Souslin-F subset of X disjoint from A is included in some Eξ .

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(ii) Suppose that X is regular. Let A ⊆ X be a Souslin-F set. Then there is a non-decreasing family hEξ iξ<ω1 of Borel sets in X, with union X \ A, such that every K-analytic subset of X \ A is included in some Eξ . (iii) Let A ⊆ X be a K-analytic set. Then there is a non-decreasing family hEξ iξ<ω1 of Borel sets in X, with union X \ A, such that every K-analytic subset of X \ A is included in some Eξ . proof (a) The first two parts depend on the following fact: if X is regular, R ⊆ N N × X is usco-compact, hFσ iσ∈S is a Souslin scheme consisting of closed sets with kernel B, and R[N N ] ∩ B = ∅, then for any φ, T ψ ∈ N N there is an n ≥ 1 such that R[Iφ¹n ] ∩ 1≤i≤n Fψ¹i is empty, where I write Iσ = {θ : σ ⊆ θ ∈ N N } for S T σ ∈ S = n≥1 Nn . P P We know that K = R[{φ}] is a compact set disjoint from the closed set n≥1 Fψ¹n . T So there is some m ≥ 1 such that K ∩ F = ∅ where F = 1≤i≤m Fψ¹i . Because X is regular, there are disjoint open sets G, H ⊆ X such that K ⊆ G and F ⊆ H (4A2F(h-ii)). Now R−1 [X \ G] is a closed set not containing φ, so there is some n such that R[Iφ¹n ] ⊆ G. Of course we can take n ≥ m, and in this case T R[Iφ¹n ] ∩ 1≤i≤n Fψ¹i ⊆ G ∩ F = ∅, as required. Q Q (b)(i) Suppose that A ⊆ X is K-analytic. Then there is an usco-compact set R ⊆ N N × X such that R[N N ] = A, and R is closed (422Da), so that A is the kernel of the Souslin scheme hR[Iσ ]iσ∈S (421I). For x ∈ X set Tx = {σ : σ ∈ S, x ∈ R[Iσ ]}, as in 421Ng, and let r(Tx ) < ω1 be the rank of the tree Tx . Then Eξ = {x : x ∈ X \ A, r(Tx ) ≤ ξ} is a Borel set for every ξ < ω1 , by 421Ob. Now suppose that B ⊆ X \ A is a Souslin-F set. Then it is the kernel of a Souslin scheme hFσ iσ∈S consisting of closed sets. If φ, ψ ∈ N N T then by (a) above there is an n ≥ 1 such that R[Iφ¹n ] ∩ 1≤i≤n Fψ¹i is empty. By 421Q, there must be some ξ < ω1 such that B ⊆ Eξ . So hEξ iξ<ω1 is a suitable family. (ii) The other part is almost the same. Suppose that A ⊆ X is Souslin-F. Then it is the kernel of a Souslin scheme hFσ iσ∈S consisting of closed sets. For x ∈ X set T S Tx = n≥1 {σ : σ ∈ Nn , x ∈ 1≤i≤n Fσ¹i }, and let r(Tx ) < ω1 be the rank of the tree Tx . Then Eξ = {x : x ∈ X \ A, r(Tx ) ≤ ξ} is a Borel set for every ξ < ω1 . Now let B ⊆ X \ A be a K-analytic set. There is an usco-compact set R ⊆ N N × X such that R[N N ] = B, and B is the kernel of the Souslin scheme hR[Iσ ]iσ∈S . If φ, ψ ∈ N N then by (a) above there is T an n ≥ 1 such that 1≤i≤n Fψ¹i ∩ R[Iφ¹n ] is empty. So there must be some ξ < ω1 such that B ⊆ Eξ . Thus here again hEξ iξ<ω1 is a suitable family. (c) If X is not regular, we still have a version of the result in (a), as follows: if R, S ⊆ N N × X and R[N N ] ∩ S[N N ] = ∅, then for any φ, ψ ∈ N N there is an n ≥ 1 such that R[Iφ¹n] ∩ S[Iψ¹n ] is empty. P P This time, R[{φ}] and S[{ψ}] are disjoint compact sets, so there are disjoint open sets G, H with R[{φ}] ⊆ G and S[{ψ}] ⊆ H (4A2F(h-i)). Now R[Iφ¹n ] ⊆ G and S[Iψ¹n ] ⊆ H for all n large enough. Q Q Now the argument of (b-i), with Fσ = S[Iσ ], gives part (iii). 422X Basic exercises (a) Let X and Y be Hausdorff spaces and X0 a closed subset of X. Show that a relation R ⊆ X0 × Y is usco-compact when regarded as a relation between X0 and Y iff it is usco-compact when regarded as a relation between X and Y . (b) Show that a locally compact Hausdorff space is K-analytic iff it is Lindel¨of iff it is σ-compact. > (c) Prove 422Hc from first principles, without using 421D. (Hint: if hRσ iσ∈S is a Souslin scheme of usco-compact relations in N N × X, {((φ, hψσ iσ∈S ), x) : (ψφ¹n , x) ∈ Rφ¹n for every n} is usco-compact in (N N × (N N )S ) × X.) (d) Let X be a completely regular Hausdorff space and A, B disjoint K-analytic subsets of X. Show that there is a Baire set including A and disjoint from B.

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(e) Let X be a Hausdorff K-analytic space. (i) Show that every Baire subset of X is K-analytic. (Hint: 136G.) (ii) Show that if X is regular, it is perfectly normal iff it is hereditarily Lindel¨of iff every open subset of X is K-analytic. (f ) Let X be a Hausdorff space in which every open set is K-analytic. Show that every Borel set is K-analytic. (Hint: apply 136Xi to the family of K-analytic subsets of X.) ˇ 422Y Further exercises (a) Let X be a completely regular Hausdorff space, and βX its Stone-Cech compactification. Show that X is K-analytic iff it is a Souslin-F set in βX. (b) Show that a Hausdorff space is K-analytic iff T S it is a continuous image of a Kσδ set in a compact Hausdorff space, that is, a set expressible as m∈N n∈N Kmn where every Kmn is compact. (Hint: Write K∗ for the class of Hausdorff continuous images of Kσδ subsets of compact Hausdorff spaces. (i) Show that N N is a Kσδ set in Y N , where Y is the one-point compactification of N. (ii) Show that if X is a compact Hausdorff space and R ⊆ N N × X is closed, then R ∈ K∗ . (iii) Show that if X is a compact Hausdorff space, then every Souslin-F subset of X belongs to K∗ . (iv) Show that if X is a regular K-analytic Hausdorff space, then X ∈ K∗ . (v) Show that if X is any Hausdorff space and R ⊆ N N × X is an usco-compact relation, then R ∈ K∗ . See Jayne 76.) (c) Let X be a normal space and C the family of countably compact closed subsets of X. Let A, B be disjoint sets obtainable from C by Souslin’s operation. (For instance, if X itself is countably compact, A and B could be disjoint Souslin-F sets.) Show that there is a Borel set including A and disjoint from B. (Hint: adapt the proof of 422I.) T (d) Let X be a Hausdorff space and hAn in∈N a sequence of K-analytic subsets of X such that Tn∈N An = ∅. Show that there is a sequence hEn in∈N of Borel sets such that An ⊆ En for every n ∈ N and n∈N En = ∅. N (Hint: for each n ∈ N choose an usco-compact T Rn ⊆ N × X with projection An . Consider the set T = {hσn in∈N : ∃ Borel En , Rn [Iσn ] ⊆ En ∀ n, n∈N En = ∅}.) (e) Explain how to prove 422J from 421Q, without using 422I. (f ) Let X be a set and S, T two Hausdorff topologies on X such that S ⊆ T and (X, T) is K-analytic. Show that S, T yield the same K-analytic subspaces of X. 422 Notes and comments In a sense, this section starts at the deep end of its topic. ‘Descriptive set theory’ originally developed in the context of the real line and associated spaces, and this remains the centre of the subject. But it turns out that some of the same arguments can be used in much more general contexts, and in particular greatly illuminate the theory of Radon measures on Hausdorff spaces. I find that a helpful way to look at K-analytic spaces is to regard them as a common generalization of compact Hausdorff spaces and Souslin-F subsets of R; if you like, any theorem which is true of both these classes has a fair chance of being true of all K-analytic spaces. In the next section we shall come to the special properties of the more restricted class of ‘analytic’ spaces, which are much closer to the separable metric spaces of the original theory. The phrase ‘usco-compact’ is neither elegant nor transparent, but is adequately established and (in view of the frequency with which it is needed) seems preferable to less concise alternatives. If we think of a relation R ⊆ X × Y as a function x 7→ R[{x}] from x to PY , then an usco-compact relation is one which takes compact values and is ‘upper semi-continuous’ in the sense that {x : R[{x}] ⊆ H} is open for every open set H ⊆ Y ; just as a real-valued function is upper semi-continuous if {x : f (x) < α} is open for every α. This is not supposed to be a book on general topology, and in my account of the topological properties of K-analytic spaces I have concentrated on facts which are useful when proving that spaces are K-analytic, on the assumption that these will be valuable when we seek to apply the results of §432 below. Other properties are mentioned only when they are relevant to the measure-theoretic results which are my real concern, and readers already acquainted with this area may be startled by some of my omissions. For a proper treatment of the subject, I refer you to Rogers 80. As usual, however, I take technical details

423B

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155

seriously in the material I do cover. I hope you will not find that such results as 422Dg and 422Ga try your patience too far. I think a moment’s thought will persuade you that it is of the highest importance that K-analyticity (like compactness) is an intrinsic property. In contrast, the property of being ‘Souslin-F’, like the property of being closed, depends on the surrounding space. A completely regular Hausdorff space ˇ is compact iff it must be a closed set in any surrounding Hausdorff space iff it is closed in its Stone-Cech compactification; and it is K-analytic iff it must be a Souslin-F set in any surrounding Hausdorff space iff it ˇ is a Souslin-F set in its Stone-Cech compactification (422Ya). For regular spaces, 422K gives us another version of the First Separation Theorem. But this one is simultaneously more restricted in its scope (it does not seem to have applications to Baire σ-algebras, for instance) and very much more powerful in its application. When all Borel sets are Souslin-F, as in the next section, it tells us something very important about the cofinal structure of the Souslin-F subsets of the complement of a K-analytic set.

423 Analytic spaces We come now to the original class of K-analytic spaces, the ‘analytic’ spaces. I define these as continuous images of N N (423A), but move as quickly as possible to their characterization as K-analytic spaces with countable networks (423C), so that many other fundamental facts (423E-423G) can be regarded as simple corollaries of results in §422. I give two versions of Lusin’s theorem on injective images of Borel sets (423I), and a form of the von Neumann-Jankow measurable selection theorem (423N). I end with notes on constituents of coanalytic sets (423P-423Q). 423A Definition A Hausdorff space is analytic or Souslin if it is either empty or a continuous image of N N . 423B Proposition (a) A Polish space (definition: 4A2A) is analytic. (b) A Hausdorff continuous image of an analytic Hausdorff space is analytic. (c) A product of countably many analytic Hausdorff spaces is analytic. (d) A closed subset of an analytic Hausdorff space is analytic. (e) An analytic Hausdorff space has a countable network consisting of analytic sets. proof (a) Let X be a Polish space. If X = ∅, we canS stop. Otherwise, let ρ be a metric on X, inducing its topology, under which X is complete. For σ ∈ S ∗ = n∈N N n choose Xσ ⊆ X as follows. X∅ = X. Given that Xσ is a closed non-empty subset of X, where σ ∈ S ∗ , then Xσ is separable, because X is separable and metrizable (4A2P(a-iv)), and we can choose a sequence hxσi ii∈N in Xσ such that {xσi : i ∈ N} is dense in Xσ . Set Xσa i = Xσ ∩ B(xσi , 2−n ) for each i ∈SN, where B(x, δ) = {y : ρ(y, x) ≤ δ}, and continue. Note that because {xσi : i ∈ N} is dense in Xσ , Xσ = i∈N Xσa i , for every σ ∈ S ∗ . For each φ ∈ N N , hXφ¹n in∈N is a non-increasing sequence of non-empty closed sets, and diam(Xφ¹n+1 ) T ≤ 2−n+1 for every n. Because X is complete under ρ, n∈N Xφ¹n is a singleton {f (φ)} say. (f (φ) is the limit of the Cauchy sequence hxφ¹n,φ(n) in∈N .) Thus we have a function f : N N → X. f is continuous because ρ(f (ψ), f (φ)) ≤ 2−n+1 whenever φ¹n + 1 = ψ¹n + 1 (since in this case both f (ψ) and f (φ) belong to Xφ¹n+1 ). f is surjective because, given x ∈ X, we can S choose hφ(i)ii∈N inductively so that x ∈ Xφ¹n for every n; at the inductive step, we have x ∈ Xφ¹n = i∈N X(φ¹n)a i , so we can take φ(n) such that x ∈ X(φ¹n)a φ(n) = Xφ¹n+1 . Thus X is a continuous image of N N , as claimed. (b) If X is an analytic Hausdorff space and Y is a Hausdorff continuous image of X, then either X is a continuous image of N N and Y is a continuous image of N N , or X = ∅ and Y = ∅. (c) Let hXi ii∈I be a countable family of analytic Hausdorff spaces, with product X. Then X is Hausdorff (3A3Id). If I = ∅ then X = {∅} is a continuous image of N N , therefore analytic. If there is some i ∈ I such that Xi = ∅, then X = ∅ is analytic. Otherwise, we have for each i ∈ I a continuous surjection fi : N N → Xi . Setting f (φ) = hfi (φ(i))ii∈I for φ ∈ (N N )I , f : (N N )I → X is a continuous surjection. But (N N )I ∼ = N N×I is N homeomorphic to N , so X is analytic.

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(d) Let X be an analytic Hausdorff space and F a closed subset of X. Then F is Hausdorff in its subspace topology (4A2F(a-i)). If X = ∅ then F = ∅ is analytic. Otherwise, there is a continuous surjection f : N N → X. Now H = f −1 [F ] is a closed subset of the Polish space N N , therefore Polish in its induced topology (4A2Qd). By (a), H is analytic, so its continuous image F = f [H] is also analytic, by (b). (e) Let X be an analytic Hausdorff space. If it is empty then of course it has a countable network consisting of analytic sets. Otherwise, there is a continuous surjection f : N N → X. For σ ∈ S ∗ set Iσ = {φ : σ ⊆ φ ∈ N N }; then {Iσ : σ ∈ S ∗ } is a base for the topology of N N , so {f [Iσ ] : σ ∈ S ∗ } is a network for the topology of X (see the proof of 4A2Nd). But Iσ is homeomorphic to N N , so f [Iσ ] is analytic, for every σ ∈ S ∗ , and {f [Iσ ] : σ ∈ S ∗ } is a countable network consisting of analytic sets. 423C Theorem A Hausdorff space is analytic iff it is K-analytic and has a countable network. proof (a) Let X be an analytic Hausdorff space. By 423Be, it has a countable network. If X = ∅ then surely it is K-analytic. Otherwise, X is a continuous image of N N . But N N is K-analytic (422Gb), so X also is K-analytic, by 422Gd. (b) Now suppose that X is a K-analytic Hausdorff space and has a countable network. (i) If X ⊆ N N then X is analytic. P P Let R ⊆ N N ×X be an usco-compact relation such that R[N N ] = X. Then R is still usco-compact when regarded as a subset of N N ×N N (422Dg), so is closed in N N ×N N (422Da). But N N × N N ∼ = N N is analytic, so R is in itself an analytic space (423Bd), and its continuous image X is analytic, by 423Bb. Q Q (ii) Now suppose that X is regular. By 4A2Ng, X has a countable network E consisting of closed sets. Adding ∅ to E if need be, we may suppose that E = 6 ∅. Let hEn in∈N be a sequence running over E. For each n ∈ N, let hFni ii∈N be a sequence running over {En } ∪ {E : E ∈ E, E ∩ En = ∅}. Now consider the relation T R = {(φ, x) : φ ∈ N N , x ∈ n∈N Fn,φ(n) } ⊆ N N × X. α) R is closed in N N × X. P (α P Because every Fni is closed, S / Fni } (N N × X) \ R = i,n∈N {(φ, x): φ(n) = i and x ∈ is open. Q Q β ) R[N N ] = X. P (β P For every n ∈ N, X \ En = because E is a network and En is closed, so such that x ∈ Fn,φ(n) , and (φ, x) ∈ R. Q Q

S S

{E : E ∈ E, E ⊆ X \ En }

i∈N

Fni = X. So, given x ∈ X, we can find for each n a φ(n)

(γγ ) R is the graph of a function. P P?? Suppose that we have (φ, x) and (φ, y) in R where x 6= y. Because the topology of X is Hausdorff, there is an n ∈ N such that x ∈ En and y ∈ / En . But in this case x ∈ En ∩ Fn,φ(n) , so Fn,φ(n) = En , while y ∈ Fn,φ(n) \ En , so Fn,φ(n) 6= En ; which is absurd. X XQ Q (δδ ) Set A = R−1 [X], so that R is the graph of a function from A to X; in recognition of its new status, give it a new name f . Then f is continuous. P P Suppose that φ ∈ A and that x = f (φ) ∈ G, where G ⊆ X is open. Then there is an n ∈ N such that x ∈ En ⊆ G. In this case x ∈ Fn,φ(n) , because (φ, x) ∈ R, so Fn,φ(n) = En . Now if ψ ∈ A and ψ(n) = φ(n), we must have f (ψ) ∈ Fn,ψ(n) = En ⊆ G. Thus f −1 [G] includes a neighbourhood of φ in A. As φ and G are arbitrary, f is continuous. Q Q (²²) At this point recall that X is K-analytic. It follows that N N × X is K-analytic (422Ge), so that its closed subset R is K-analytic (422Gf) and A, which is a continuous image of R, is K-analytic (422Gd). But now A is a K-analytic subset of N N , so is analytic, by (i) just above. And, finally, X is a continuous image of A, so is analytic. (iii) Thus any regular K-analytic space with a countable network is analytic. Now suppose that X is an arbitrary K-analytic Hausdorff space with a countable network. Let R ⊆ N N × X be an usco-compact

423G

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157

relation such that R[N N ] = X. Then R is a closed subset of N N × X, so is itself a K-analytic space with a countable network. But it is also regular, by 422E. So R is analytic and its continuous image X is analytic. This completes the proof. 423D Corollary (a) Any analytic Hausdorff space is hereditarily Lindel¨of. (b) In a regular analytic Hausdorff space, closed subsets are zero sets and the Baire and Borel σ-algebras coincide. (c) A compact subset of an analytic Hausdorff space is metrizable. (d) A metrizable space is analytic iff it is K-analytic. proof (a)-(c) These are true just because there is a countable network (4A2Nb, 4A2Na, 4A2Nh, 4A3Kb). (d) Let X be a metrizable space. If X is analytic, of course it is K-analytic. If X is K-analytic, it is Lindel¨of (422Gg) therefore separable (4A2Pd) and has a countable network (4A2P(a-iii)), so is analytic. 423E Theorem (a) For any Hausdorff space X, the family of subsets of X which are analytic in their subspace topologies is closed under Souslin’s operation. (b) Let (X, T) be an analytic Hausdorff space. For a subset A of X, the following are equiveridical: (i) A is analytic; (ii) A is K-analytic; (iii) A is Souslin-F; (iv) A can be obtained by Souslin’s operation from the family of Borel subsets of X. In particular, all Borel sets in X are analytic. proof (a) Let X be a Hausdorff space and A the family of analytic subsets of X. Let hAσ iσ∈S be a Souslin scheme in A with kernel A. Then every Aσ is K-analytic, so A is K-analytic, by 422Hc. Also every Aσ has a S countable network, so A0 = σ∈S Aσ has a countable network (4A2Nc); as A ⊆ A0 , A also has a countable network (4A2Na) and is analytic. (b) Because X has a countable network, so does A. So 423C tells us at once that (i) ⇐⇒ (ii). In particular, X is K-analytic, so 422Hb tells us that (ii) ⇐⇒ (iii). Of course (iii)⇒(iv). Now suppose that G ⊆ X is open. Then G ∈ A. P P If X = ∅ then G = ∅ is open. Otherwise, there is a continuous surjection f : N N → X. Set H = f −1 [G], so that H ⊆ N N is open and G = f [H]. Being an open set in a metric space, H is Fσ (4A2Lc), so, in particular, is Souslin-F; but N N is analytic, so H is analytic and its continuous image G is analytic. Q Q We have already seen that closed subsets of X belong to A (423Bd). Because A is closed under Souslin’s operation, it contains every Borel set, by 421F. It therefore contains every set obtainable by Souslin’s operation from Borel sets, and (i)⇒(i). Remark See also 423Yb below. 423F Proposition Let (X, T) be an analytic Hausdorff space. (a) A set E ⊆ X is Borel iff both E and X \ E are analytic. (b) If S is a coarser(= smaller) Hausdorff topology on X, then S and T have the same Borel sets. proof (a) If E is Borel, then E and X \ E are analytic, by 423Eb. If E and X \ E are analytic, they are K-analytic (423Eb) and disjoint, so there is a Borel set F ⊇ E which is disjoint from X \ E (422J); but now of course F = E, so E must be Borel. (b) Because the identity map from (X, T) to (X, S) is continuous, S is an analytic topology (423Bb) and every S-Borel set is T-Borel. If E ⊆ X is T-Borel, then it and its complement are T-analytic, therefore S-analytic (423Bb), and E is S-Borel by (a). 423G Lemma Let X and Y be analytic Hausdorff spaces and f : X → Y a Borel measurable function. (a) (The graph of) f is an analytic set. (b) f [A] is an analytic set in Y for any analytic set (in particular, any Borel set) A ⊆ X. (c) f −1 [B] is an analytic set in X for any analytic set (in particular, any Borel set) B ⊆ Y .

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proof (a) Let E be a countable network for the topology of Y . Set T R = E∈E (X × E) ∪ ((X \ f −1 [E]) × Y ). Then R is a Borel set in X × Y . But also R is the graph of f . P P If f (x) = y, then surely y ∈ E whenever x ∈ f −1 [E], so (x, y) ∈ R. On the other hand, if x ∈ X, y ∈ Y and f (x) 6= y, there are disjoint open sets G, H ⊆ Y such that f (x) ∈ G and y ∈ H; now there is an E ∈ E such that f (x) ∈ E ⊆ G, so that f (x) ∈ E but y ∈ / E, and (x, y) ∈ / R. Q Q Because X × Y is analytic (423Bc), R is analytic (423Eb). (b) If A ⊆ X is analytic, then A × Y and R ∩ (A × Y ) are analytic (423Ea), so f [A] = R[A], which is a continuous image of R ∩ (A × Y ), is analytic. (c) Similarly, if B ⊆ Y is analytic, then f −1 [B] is a continuous image of R ∩ (X × B), so is analytic. 423H Lemma Let (X, T) be an analytic Hausdorff space, and hEn in∈N any sequence of Borel sets in X. Then the topology T0 generated by T ∪ {En : n ∈ N} is analytic. proof If X = ∅ this is trivial. Otherwise, there is a continuous surjection f : N N → X. Set Fn = f −1 [En ] for each n; then Fn is a Borel subset of N N , so there is a Polish topology S0 on N N , finer than the usual topology, for which every Fn is open, by 4A3I. But now f is continuous for S0 and T0 , so T0 is analytic, by 423Ba and 423Bb. (Of course T0 is Hausdorff, because it is finer than T.) 423I Theorem Let X be a Polish space, E ⊆ X a Borel set, Y a Hausdorff space and f : E → Y an injective function. (a) If f is continuous, then f [E] is Borel. (b) If Y has a countable network (e.g., is an analytic space or a separable metrizable space), and f is Borel measurable, then f [E] is Borel. proof (a)(i) Since there is a finer Polish topology on X for which E is closed (4A3I), therefore Polish in the subspace topology (4A2Qd), and f will still be continuous for this topology, we may suppose that E = X. (ii) Let hUn in∈N run over a base for the topology of X (4A2P(a-i)). For each pair m, n ∈ N such that Um ∩ Un is empty, f [Um ] and f [Un ] are analytic sets in Y (423Eb, 423Bb) and are disjoint (because f is injective), so there is a Borel set Hmn including f [Um ] and disjoint from f [Un ] (422J). Set T En = f [Un ] ∩ {Hnm \ Hmn : m ∈ N, Um ∩ Un = ∅} for each n ∈ N; then En is a Borel set in Y including f [Un ]. Note that if Um ∩ Un is empty, then Em ∩ En ⊆ (Hmn \ Hnm ) ∩ (Hnm \ Hmn ) is also empty. Fix a metric ρ on X, inducing its topology, for which X is complete, and for k ∈ N set S Fk = {En : n ∈ N, diam Un ≤ 2−k }, T so that Fk is Borel. Let F = k∈N Fk ; then F is also a Borel subset of Y . The point is that F = f [X]. P P (i) If x ∈ X, then for every k ∈ N there is an n ∈ N such that x ∈ Un ⊆ {y : ρ(y, x) ≤ 2−k−1 }; now diam Un ≤ 2−k , so f (x) ∈ f [Un ] ⊆ En ⊆ Fk . As k is arbitrary, f (x) ∈ Fk ; as x is arbitrary, f [X] ⊆ F . (ii) If y ∈ F , then for each k ∈ N we can find an n(k) such that y ∈ En(k) and diam Un(k) ≤ 2−k . Since f [Un(k) ] ⊇ En(k) is not empty, nor is Un(k) , and we can choose xk ∈ Un(k) . Indeed, for any j, k ∈ N, En(j) ∩ En(k) contains y, so is not empty, and Un(j) ∩ Un(k) cannot be empty; but this means that there is some x in the intersection, and ρ(xj , xk ) ≤ ρ(xj , x) + ρ(x, xk ) ≤ diam Un(j) + diam Un(k) ≤ 2−j + 2−k . This means that hxk ik∈N is a Cauchy sequence. But X is supposed to be complete, so hxk ik∈N has a limit x say.

423K

Analytic spaces

159

?? If f (x) 6= y, then (because Y is Hausdorff) there is an open set H containing f (x) such that y ∈ / H. Now f is continuous, so there is a δ > 0 such that f (x0 ) ∈ H whenever ρ(x0 , x) ≤ δ. There is a k ∈ N such that 2−k + ρ(xk , x) ≤ δ. If x0 ∈ Un(k) , then ρ(x0 , x) ≤ ρ(x0 , xk ) + ρ(xk , x) ≤ δ; thus f [Un(k) ] ⊆ H, and En(k) ⊆ f [Un(k) ] ⊆ H. But y ∈ En(k) \ H. X X Thus y = f (x) belongs to f [X]; as y is arbitrary, F ⊆ f [X]. Q Q Accordingly f [X] = F is a Borel subset of Y , as claimed. (b) By 4A2Nf, there is a countable family V of open sets in Y such that whenever y, y 0 are distinct points of Y there are disjoint V , V 0 ∈ V such that y ∈ V and y 0 ∈ V 0 . Let S0 be the topology generated by V; then S0 is Hausdorff. For each V ∈ V, f −1 [V ] is a Borel set in X, so there is a Polish topology T0 on X, finer than the original topology, for which every f −1 [V ] is open (4A3I). Now f is continuous for T0 and S0 (4A2B(a-ii)), and E is T0 -Borel, so f [E] is a S0 -Borel set in Y , by (a). Since S0 is coarser than the original topology S on Y , f [X] is also S-Borel. 423J Lemma If X is an uncountable analytic Hausdorff space, it has subsets homeomorphic to {0, 1}N and N N . S proof (a) Let f : N N → X be a continuous surjection. Write S ∗ = n∈N N n , Iσ = {φ : σ ⊆ φ ∈ N N } for σ ∈ S∗, T = {σ : σ ∈ S ∗ , f [Iσ ] is uncountable}. Then if σ ∈ T there are τ , τ 0 ∈ T , both extending σ, such that f [Iτ ] ∩ f [Iτ 0 ] = ∅. P P Set S ∗ A = {f [Iτ ] : τ ∈ S \ T }. Then A is a countable union of countable sets, so is countable. There must therefore be distinct points x, y of f [Iσ ] \ A; express x as f (φ) and y as f (ψ) where φ and ψ belong to Iσ . Because X is Hausdorff, there are disjoint open sets G, H such that x ∈ G and y ∈ H. Because f is continuous, there are m, n ∈ N such that Iφ¹m ⊆ f −1 [G] and Iψ¹n ⊆ f −1 [H]. Of course both τ = φ¹m and τ 0 = ψ¹n must extend σ, and they belong to T because x ∈ f [Iτ ] \ A, y ∈ f [Iτ 0 ] \ A. Q Q S (b) We can therefore choose inductively a family hτ (υ)iυ∈S2∗ in T , where S2∗ = n∈N {0, 1}n , such that τ (∅) = ∅, τ (υ a i) ⊇ τ (υ) whenever υ ∈ S2∗ , i ∈ {0, 1}, f [Iτ (υa 0 )] ∩ f [Iτ (υa 1 ] = ∅ for every υ ∈ S2∗ . Note that #(τ (υ)) ≥ #(υ) for every υ ∈ S2 . For each z ∈ {0, 1}N , hτ (z¹n)in∈N is a sequence in S ∗ in which each term strictly extends its predecessor, so there is a unique g(z) ∈ N N such that τ (z¹n) ⊆ g(z) for every n. Now g(z 0 )¹n = g(z)¹n whenever z¹n = z 0 ¹n, so g and f g : {0, 1}N → X are continuous. If w, z are distinct points of {0, 1}N , there is a first n such that w(n) 6= z(n), in which case f g(w) ∈ f [Iτ (w¹n)a w(n) ] and f g(z) ∈ f [Iτ (w¹n)a z(n) ] are distinct. So f g : {0, 1}N → X is a continuous injection, therefore a homeomorphism between {0, 1}N and its image, because {0, 1}N is compact (3A3Dd). (c) Thus X has a subspace homeomorphic to {0, 1}N . Now {0, 1}N has a subspace homeomorphic to N. P P For instance, setting dn (n) = 1, dn (i) = 0 for i 6= n, D = {dn : n ∈ N} is homeomorphic to N. Q Q Now DN is homeomorphic to N N and is a subspace of ({0, 1}N )N ∼ = {0, 1}N×N ∼ = {0, 1}N , so {0, 1}N has a subspace homeomorphic to N N . Accordingly X also has a subspace homeomorphic to N N . 423K Corollary Any uncountable Borel set in any analytic space has cardinal c.

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Descriptive set theory

423K

proof If X is an analytic space and E ⊆ X is an uncountable Borel set, then E is analytic (423E), so includes a copy of {0, 1}N and must have cardinal at least #({0, 1}N ) = c. On the other hand, E is also a continuous image of N N , so has cardinal at most #(N N ) = c. 423L Proposition Let X be an uncountable analytic Hausdorff space. Then it has a non-Borel analytic subset. proof (a) I show first that there is an analytic set A ⊆ N N × N N such that every analytic subset of N N is a vertical section of A. P P Let U be a countable base for the topology of N N × N N , containing ∅, and hUn in∈N an enumeration of U. Write S M = (N N × N N × N N ) \ m,n∈N ({x : x(m) = n} × Un ). Then M is a closed subset of (N N × NN ) × N N , therefore analytic (423Ba, 423Bd), so its continuous image A = {(x, z) : there is some y such that (x, y, z) ∈ M } is analytic (423Bb). Now let E be any analytic subset of N N . By 423E, E is Souslin-F; by 421J, there is a closed set F ⊆ N N × N N such that E = {z : ∃ y, (y, z) ∈ F }. Let hx(m)im∈N be a sequence running over {n : n ∈ N, Un ∩ F = ∅}, so that S F = (N N × N N ) \ m∈N Ux(m) = {(y, z) : (x, y, z) ∈ M }. Now {z : (x, z) ∈ A} = {z : there is some y such that (x, y, z) ∈ M } = {z : there is some y such that (y, z) ∈ F } = E, and E is a vertical section of A, as required. Q Q (b) It follows that there is a non-Borel analytic set B ⊆ N N . P P Take A from (a) above, and try B = {x : (x, x) ∈ A}. Because B is the inverse image of A under the continuous map x 7→ (x, x), it is analytic (423Gc). ?? If B were a Borel set, then B 0 = N N \ B would also be Borel, therefore analytic (423E), and there would be an x ∈ NN such that B 0 = {y : (x, y) ∈ A}. But in this case x ∈ B ⇐⇒ (x, x) ∈ A ⇐⇒ x ∈ B 0 , which is a difficulty you may have met before. X XQ Q (c) Now return to our arbitrary uncountable analytic Hausdorff space X. By 423J, X has a subset Z homeomorphic to N N . By (b), Z has an analytic subset A which is not Borel in Z, therefore cannot be a Borel subset of X. 423M I devote a few paragraphs to an important method of constructing selectors. Theorem Let X be an analytic Hausdorff space, Y a set, and C ⊆ PY . Write T for the σ-algebra of subsets of Y generated by S(C), where S is Souslin’s operation, and V for S({F × C : F ⊆ X is closed, C ∈ C}). If W ∈ V, then W [X] ∈ T and there is a T-measurable function f : W [X] → X such that (f (y), y) ∈ W for every y ∈ W [X]. proof Write F for {F × C : F ⊆ X is closed, C ∈ C}. (a) Consider first the case in which X = N N and all the horizontal sections W −1 [{y}] of W are closed. S ∗ Let E be the family of closed subsets of Y . For σ ∈ S = n∈N N n set Iσ = {φ : σ ⊆ φ ∈ N N }. Then W ∩ (Iσ × Y ) ∈ V. P P Because Souslin’s operation is idempotent (421D), S(V) = V. The set {V : V ∩ (Iσ × Y ) ∈ V} is therefore closed under Souslin’s operation (apply 421Cc to the identity map from Iσ × Y to X × Y , or otherwise); since it includes F, it is the whole of V, and contains W . Q Q By 421G, W [Iσ ] = (W ∩(Iσ ×Y ))[N N ] belongs to S(C) ⊆ T for every σ. In particular, W [N N ] = W [I∅ ] ∈ T. Choose hYσ iσ∈S ∗ in T inductively, as follows. Y∅ = W [N N ]. Given that Yσ ∈ T and that Yσ ⊆ W [Iσ ], set

423N

Analytic spaces

Yσa j = Yσ ∩ W [Iσa j ] \

161

S i<j

W [Iσa i ]

for every j ∈ N. Continue. At the end of the induction, we have S S j∈N Yσ a j = Yσ ∩ j∈N W [Iσ a j ] = Yσ ∩ W [Iσ ] = Yσ for every σ ∈ S ∗ , while hYσa j ij∈N is always disjoint. So for each y ∈ Y∅ = W [N N ] we have a unique f (y) ∈ N N such that y ∈ Yf (y)¹n for every n. Since f −1 [Iσ ] = Yσ ∈ T for every σ ∈ S ∗ , f is T-measurable (4A2H(c-ii)). Also (f (y), y) ∈ W for every y ∈ W [N N ]. P P For each n ∈ N, y ∈ Yf (y)¹n = W [If (y)¹n ], so there is an xn ∈ N N such that xn ¹n = f (y)¹n and (xn , y) ∈ W . But this means that f (y) = limn→∞ xn ; since we are supposing that the horizontal sections of W are closed, (f (y), y) ∈ W . Q Q Thus the theorem is true if X = N N and W has closed horizontal sections. (b) Now suppose that X = N N and that W ⊆ N N × Y is any set in V. Then there is a Souslin scheme hFσ × Cσ iσ∈S in F with kernel W ; of course I mean you to suppose that Fσ ⊆ N N is closed and Cσ ∈ C for every σ. Set S N ˜ =T W × NN × Y . k Iσ × Fσ × Cσ ⊆ N k≥1

σ∈N

˜ onto the last two coordinates, by 421Ce. If y ∈ Y , then Then W is the projection of W S ˜}=T {(φ, ψ) : (φ, ψ, y) ∈ W {Iσ × Fσ : σ ∈ N k , y ∈ Cσ } k≥1

N

N

k

is closed in N × N . (If J is any subset of N , then S S S (N N × N N ) \ σ∈J Iσ × Fσ = σ∈J Iσ × (N N \ Fσ ) ∪ σ∈N k \J Iσ × N N is open.) Also Iσ × Fσ is a closed subset of N N × N N for every σ, and N N × N N is homeomorphic to N N . We ˜ , regarded as a subset of (N N × N N ) × Y , to see that W [N N ] = W ˜ [N N × N N ] ∈ T can therefore apply (a) to W N N N ˜ for and that there is a T-measurable function h = (g, f ) : W [N ] → N × N such that (g(y), f (y), y) ∈ W every y ∈ W [N N ]. Now, of course, f : W [N N ] → N N is T-measurable and (f (y), y) ∈ W for every y ∈ W [N N ]. (c) Finally, suppose only that X is an analytic Hausdorff space and that W ∈ V. If X is empty, so is Y , ˜ y) = (h(φ), y) for and the result is trivial. Otherwise, there is a continuous surjection h : N N → X. Set h(φ, ˜ : N N × Y → X × Y is a continuous surjection, and W ˜ −1 [W ] is the kernel of ˜ =h φ ∈ N N and y ∈ Y ; then h a Souslin scheme in ˜ −1 [F × C] : F ⊆ N N is closed, C ∈ C} = {h−1 [F ] × C : F ⊆ N N is closed, C ∈ C} {h ˜ [N N ] ∈ T and there is a T-measurable g : W [X] → N N by 421Cb. So we can apply (b) to see that W [X] = W ˜ for every y ∈ Y . Finally f = hg : W [X] → X is T-measurable and (f (y), y) ∈ W such that (g(y), y) ∈ W for every y ∈ Y . This completes the proof. 423N The expression V = S({F × C : F ⊆ X is closed, C ∈ C}) in 423M is a new formulation, and I had better describe one of the basic cases in which we can use the result. Corollary Let X be an analytic Hausdorff space and Y any topological space. Let T be the σ-algebra of subsets of Y generated by S(B(Y )), where B(Y ) is the Borel σ-algebra of Y . If W ∈ S(B(X × Y )), then W [X] ∈ T and there is a T-measurable function f : W [X] → X such that (f (y), y) ∈ W for every y ∈ W [X]. proof (a) Suppose to begin with that X = N N . In 423M, set C = B(Y ). S Then every open subset and every closed subset of X ×Y belongs to V as defined in 423M. P P For σ ∈ S ∗ = k∈N N k , set Iσ = {φ : σ ⊆ φ ∈ N N }. If V ⊆ X × Y is open, set S Hσ = {H : H ⊆ Y is open, Iσ × Hσ ⊆ V } S for each σ ∈ S ∗ . Because {Iσ : σ ∈ S ∗ } is a base for the topology of N N , V = σ∈S ∗ Iσ × Hσ ∈ V. As for the complement of V , we have

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Descriptive set theory

(N N × Y ) \ V =

\

423N

(N N × Y ) \ (Iσ × Hσ )

k∈N σ∈N k

=

\

k∈N σ∈N k

=

\

((N N \ Iσ ) × Y ) ∪ (N N × (Y \ Hσ )) [

(Iτ × Y ) ∪ (N N × (Y \ Hσ ))

k∈N τ ∈N k σ∈N k τ = 6 σ

again belongs to V, because V is closed under countable unions and intersections and contains Iτ × Y and N N × (Y \ Hσ ) for all σ, τ ∈ S ∗ . Q Q By 421F, V contains every Borel subset of N N × Y , so includes S(B(N N × Y )). So in this case we can apply 423M directly to get the result. (b) Now suppose that X is any analytic space. If X is empty, the result is trivial. Otherwise, let ˜ y) = (h(φ), y) for φ ∈ N N , y ∈ Y , so that h ˜ : N N ×Y → X×Y h : N N → X be a continuous surjection. Set h(φ, −1 −1 N ˜ ˜ ˜ ˜ ∈ S(B(N N ×Y )) is continuous. Set W = h [W ]. If V ∈ B(X ×Y ) then h [V ] ∈ B(N ×Y ) (4A3Cd), so W N N N ˜ ˜ ˜ (421Cc). By (a), W [N ] ∈ T and there is a T-measurable function g : W [N ] → N such that (g(y), y) ∈ W N N ˜ ˜ for every y ∈ W [N ]. It is now easy to check that W [X] = W [N ] ∈ T (this is where we need to know that h is surjective), that f = hg : W [X] → X is T-measurable, and that (f (y), y) ∈ W for every y ∈ W [X], as required. Remark This is a version of the von Neumann-Jankow selection theorem. 423O Corollary Let X and Y be analytic Hausdorff spaces, and f : X → Y a Borel measurable surjection. Let T be the σ-algebra of subsets of Y generated by the Souslin-F sets in Y . Then there is a T-measurable function g : Y → X such that f g is the identity on Y . proof The point is that the graph Γ = {(x, f (x)) : x ∈ X} of f is analytic (423Ga), therefore a Souslin-F set in X × Y (423Eb). Also Γ[X] = f [X] = Y . By 423N, there is a T-measurable function g : Y → X such that (g(y), y) ∈ Γ, that is, f (g(y)) = y, for every y ∈ Y . *423P Constituents of coanalytic sets: Theorem Let X be a Hausdorff space, and A ⊆ X an analytic subset of X. Then there is a non-decreasing family hEξ iξ<ω1 of Borel subsets of X, with union X \ A, such that every analytic subset of X \ A is included in some Eξ . proof Put 422K(iii) and 423C together. *423Q Remarks (a) Let A be an analytic set in an analytic space X and hEξ iξ<ω1 a family of Borel sets as in 423P. There is nothing unique about the Eξ . But if hEξ0 iξ<ω1 is another such family, then every Eξ0 is an analytic subset of X \ A, by 423E, so is included in some Eη ; and, similarly, every Eξ is included in some Eη0 . We therefore have a function f : ω1 → ω1 such that Eξ0 ⊆ Ef (ξ) and Eξ ⊆ Ef0 (ξ) for every ξ < ω. If = {ξ : ξ < ω1 , f (η) < ξ for every η < ξ}, then C is a closed cofinal set in ω1 (4A1Bd), and S we set C S 0 η<ξ Eη = η<ξ Eη for every ξ ∈ C. If X \ A is itself analytic, that is, if A is a Borel set, then we shall have to have X \ A = Eξ = Eξ0 for some ξ < ω1 . Another way of expressing the result in 423P is to say that if we write I = {B : B ⊆ X \ A is analytic}, then {E : E ∈ I, E is Borel} is cofinal with I (this is the First Separation Theorem) and cf I ≤ ω1 . (b) It is a remarkable fact that, in some models of set theory, we can have non-Borel coanalytic sets in Polish spaces such that all their constituents are countable (Jech 78, p. 529, Cor. 2). (Note that, by (a), this is the same thing as saying that X \ A is uncountable but all its Borel subsets are countable.) But in ‘ordinary’ cases we shall have, for every Borel subset E of X \ A, an uncountable Borel subset of (X \ A) \ E; so that for any family hGξ iξ<ω1 of Borel constituents of X \ A, there must be uncountably many uncountable Gξ . To see that this happens at least sometimes, take any non-Borel analytic subset A0 of N N (423L), and consider A = A0 × N N ⊆ (N N )2 . Then A is analytic (423B). If E ⊆ (N N )2 \ A is Borel, then

423Yc

Analytic spaces

163

π1 [E] = {x : (x, y) ∈ E} is an analytic subset of N N \ A, so is not the whole of N N \ A0 (by 423Fa). Taking any x ∈ (N N \ A0 ) \ π1 [E], {x} × N N is an uncountable Borel subset of ((N N )2 \ A) \ E. For an alternative construction, see 423Ye. 423X Basic exercises > (a) For a Hausdorff space X, show that the following are equiveridical: (i) X is analytic; (ii) X is a continuous image of a Polish space; (iii) X is a continuous image of a closed subset of N N . (b) Write out a direct proof of 423Ea, not quoting 423C or 421D. > (c) Let X be a set and S a Hausdorff topology on X, T an analytic topology on X such that S ⊆ T. Show that S and T have the same analytic sets. (Hint: 423F.) > (d) Let X be an analytic Hausdorff space. (i) Show that its Borel σ-algebra B(X) is countably generated as σ-algebra. (Hint: use 4A2Nf and 423Fb.) (ii) Show that there is an analytic subset Y of R such that (X, AX ) is isomorphic to (Y, AY ), where AX , AY are the families of Souslin-F subsets of X, Y respectively. (Hint: show that there is an injective Borel measurable function from X to R (cf. 343E), and use 423G.) (iii) Show that (X, B(X)) is isomorphic to (Y, B(Y )), where B(Y ) is the Borel σ-algebra of Y . (Hint: 423Fa.) (iv) Let TX , TY be the σ-algebras generated by AX , AY respectively. Show that (X, TX ) and (Y, TY ) are isomorphic. (e) Let S be the right-facing Sorgenfrey topology on R (415Xc). Show that S has the same Borel sets as the usual topology T on R. (Hint: S is hereditarily Lindel¨of (419Xf) and has a base consisting of T-Borel sets.) Show that S is not analytic. (f ) Let X be an analytic Hausdorff space and Y any topological space. Let T be the σ-algebra of subsets of Y generated by the Souslin-F sets. Show that if W ⊆ X × Y is Souslin-F, then W [X] ∈ T and there is a T-measurable function f : W [X] → Y such that (f (y), y) ∈ W for every y ∈ W [X]. (Hint: start with X = N N , as in 423N.) (g) Let X and Y be Hausdorff spaces, and R ⊆ X × Y an analytic set such that R−1 [Y ] = X. Show that there is a function g : X → Y , measurable with respect to the σ-algebra generated by the Souslin-F subsets of X, such that (x, g(x)) ∈ R for every x ∈ X. > (h) (i) Show that the family of analytic subsets of [0, 1] has cardinal c. (Hint: 421Xd.) (ii) Show that the σ-algebra T of subsets of [0, 1] generated by the analytic sets has cardinal c. (Hint: 421Xi.) (iii) Show that there is a set A ⊆ [0, 1] which does not belong to T. (i) Let X = Y = [0, 1]. Give Y the usual topology, and give X the topology corresponding to the one-point compactification of the discrete topology on [0, 1[, that is, a set G ⊆ X is open if either 1 ∈ / G or G is cofinite. Show that the identity map f : X → Y is a Borel measurable bijection, but that f −1 is not measurable for the σ-algebra of subsets of Y generated by the Souslin-F sets. 423Y Further exercises (a) Show that a space with a countable network is hereditarily separable (that is, every subset is separable), therefore countably tight. (b) Show that if X is a Hausdorff space with a countable network, then every analytic subset of X is obtainable by Souslin’s operation from the open subsets of X. (c) Let (X, T) and (Y, S) be analytic Hausdorff spaces and f : X → Y a Borel measurable function. (i) Show that there is a zero-dimensional separable metrizable topology S0 on Y with the same Borel sets and the same analytic sets as S. (Hint: 423Xd.) (ii) Show that there is a zero-dimensional separable metrizable topology T0 on X, with the same Borel sets and the same analytic sets as T, such that f is continuous for the topologies T0 , S0 . (iii) Explain how to elaborate these ideas to deal with any countable family of analytic spaces and Borel measurable functions between them.

164

Descriptive set theory

423Yd

(d) Let X be an analytic Hausdorff space, Y a Hausdorff space with a countable network, and f : X → Y a Borel measurable surjection. Let T be the σ-algebra of subsets of Y generated by the Souslin-F sets in Y . Show that there is a T-measurable function g : Y → X such that f g is the identity on Y . S (e) Set S = n≥1 N n and consider PS with its usual topology. Let T ⊆ PS be the set of trees (412O); show that T is closed, therefore a compact metrizable space. Set Fσ = {T : σ ∈ T ∈ T } for σ ∈ S, and let A be the kernel of the Souslin scheme hFσ iσ∈s . Show that the constituents of T \ A for this scheme are just the sets Gξ = {T : r(T ) = ξ}, where r is the rank function of 421Ne. Show by induction on ξ that all the Gξ is non-empty, so that A is not a Borel set. Show that #(Gξ ) = c for 1 ≤ ξ < ω1 . Show that if X is any topological space and B ⊆ X is a Souslin-F set, there is a Borel measurable function f : X → T such that B = f −1 [A]. 423 Notes and comments We have been dealing, in this section and the last, with three classes of topological space: the class of analytic spaces, the class of K-analytic spaces and the class of spaces with countable networks. The first is more important than the other two put together, and I am sure many people would find it more comfortable, if more time-consuming, to learn the theory of analytic spaces thoroughly first, before proceeding to the others. This was indeed my own route into the subject. But I think that the theory of K-analytic spaces has now matured to the point that it can stand on its own, without constant reference to its origin as an extension of descriptive set theory on the real line; and that our understanding of analytic spaces is usefully advanced by seeing how easily their properties can be deduced from the fact that they are K-analytic spaces with countable networks. ‘Selection theorems’ appear everywhere in mathematics. The axiom of choice is a selection theorem: if hAi ii∈I is a family of non-empty sets, there is a selector for the relation {(i, x) : i ∈ I, x ∈ Ai }. The Lifting Theorem (§341) is a special kind of selection theorem, where we seek a selector which is a Boolean homomorphism. In general topology we look for continuous selectors; in measure theory, for measurable selectors. Any selection theorem will have expressions either as a theorem on right inverses of functions, as in 423O, or as a theorem on selectors for relations, as in 423M-423N. In the language of this section, however, the strongest results are most easily set out in terms of relations, because the essence of the method is that we can find selectors taking values in analytic spaces, and the relations of 423M-423N can be very far from being analytic spaces in themselves, even when we have a natural topology on the space Y . The value of these results in measure theory will become clearer in §431, where we shall see that there are important σ-algebras which are closed under Souslin’s operation, so that they can include the algebras T of 423M-423O. Typical applications are in As in §422, I have made no attempt to cover the general theory of analytic spaces, nor even to give a balanced introduction. I have tried instead to give a condensed account of the principal methods for showing that spaces are analytic, with some of the ideas which can be applied to make them more accessible to the imagination (423J, 423Xc-423Xd, 423Yb-423Yc). Lusin’s theorem 423I does not mention ‘analytic’ sets in its statement, but it depends essentially on the separation theorem 422J, so cannot really be put with the other results on Polish spaces in 4A2Q. You must of course know that not all analytic sets are Borel (423L) and that not all sets are analytic (423Xh). For further information about this fascinating subject, see Rogers 80, Kechris 95 and Moschovakis 80. ‘Selection theorems’ appear everywhere in mathematics. The axiom of choice is a selection theorem; it says that whenever R ⊆ X × Y is a relation and R[X] = Y , there is a function f : Y → X such that (f (y), y) ∈ R for every y ∈ Y . The Lifting Theorem (§341) asks for a selector which is a Borel homomorphism. In general topology we look for continuous selectors. In measure theory, naturally, we are interested in measurable selectors, as in 423M-423O. Any selection theorem will have expressions either as a theorem on right inverses of functions, as in 423O, or as a theorem on selectors for relations, as in 423M-423N. In the language here, however, we get better theorems by examining relations, because the essence of the method is that we can find measurable functions into analytic spaces, and the relations of 423M can be very far from being analytic, even when there is a natural topology on the space Y . The value of these results will become clearer in §431, when we shall see that the σ-algebras T of 423M-423O are often included in familiar σ-algebras. Typical applications are in 433F-433G below.

424C

Standard Borel spaces

165

424 Standard Borel spaces This volume is concerned with topological measure spaces, and it will come as no surprise that the topological properties of Polish spaces are central to the theory. But even from the point of view of unadorned measure theory, not looking for topological structures on the underlying spaces, it turns out that the Borel algebras of Polish spaces have a very special position. It will be useful later on to be able to refer to some fundamental facts concerning them. 424A Definition Let X be a set and Σ a σ-algebra of subsets of X. We say that (X, Σ) is a standard Borel space if there is a Polish topology on X for which Σ is the algebra of Borel sets. Warning! Many authors reserve the phrase ‘standard Borel space’ for the case in which X is uncountable. I have seen the phrase ‘Borel space’ used for what I call a ‘standard Borel space’. 424B Proposition (a) If (X, Σ) is a standard Borel space, then Σ is countably generated as σ-algebra of sets. Q N (b) If h(Xi , Σi )ii∈I is a countable family of standard Borel spaces, then ( i∈I Xi , c i∈I Σi ) (definition: 254E) is a standard Borel space. (c) Let (X, Σ) and (Y, T) be standard Borel spaces and f : X → Y a (Σ, T)-measurable surjection. Then (i) if E ∈ Σ is such that f [E] ∩ f [X \ E] = ∅, then f [E] ∈ T; (ii) T = {F : F ⊆ Y, f −1 [F ] ∈ Σ}; (iii) if f is a bijection it is an isomorphism. (d) Let (X, Σ) and (Y, T) be standard Borel spaces and f : X → Y a (Σ, T)-measurable injection. Then Z = f [X] ∈ T and f is an isomorphism between (X, Σ) and (Z, TZ ), where TZ is the subspace σ-algebra. proof (a) Let T be a Polish topology on X such that Σ is the algebra of Borel sets. Then T has a countable base U , which generates Σ (4A3Da/4A3E). (b) For each i ∈ I let Ti be a Polish topology on Xi such that Σi is the algebra of Ti -Borel sets. Then N Q X = i∈I Ti , with the product topology T, is Polish (4A2Qc). By 4A3Dc/4A3E, Σ = c i∈I Σi is just the Borel σ-algebra of X, so (X, Σ) is a standard Borel space. (c) Let T, S be Polish topologies on X, Y respectively for which Σ and T are the Borel σ-algebras. Then f is Borel measurable. (i) By 423Gb and 423Eb, f [E] and f [X \ E] are analytic subsets of Y . But they are complementary, so they are Borel sets, by 423Fa. (ii) f −1 [F ] ∈ Σ for every F ∈ T, just because f is measurable. On the other hand, if F ⊆ Y and E = f −1 [F ] ∈ Σ, then F = f [E] ∈ T by (i). (iii) follows at once. (d) Give X and Y Polish topologies for which Σ, T are the Borel σ-algebras. By 423Ib, f [E] ∈ T for every E ∈ Σ; in particular, Z = f [X] belongs to T. Also f −1 [F ] ∈ Σ for every F ∈ TZ , so f is an isomorphism between (X, Σ) and (Z, TZ ). 424C Theorem Let (X, Σ) be a standard Borel space. (a) If X is countable then Σ = PX. (b) If X is uncountable then (X, Σ) is isomorphic to (N N , B(N N )), where B(N N ) is the algebra of Borel subsets of N N . proof Let T be a Polish topology on X such that Σ is its Borel σ-algebra. (a) Every singleton subset of X is closed, so must belong to Σ. If X is countable, every subset of X is a countable union of singletons, so belongs to Σ. (b) (Rao & Srivastava 94) The strategy of the proof is to find Borel sets Z ⊆ X, W ⊆ N N such that (Z, ΣZ ) ∼ = (N N , B(N N )) and (W, B(W )) ∼ = (X, Σ) (writing ΣZ , B(W ) for the subspace σ-algebras), and use a form of the Schr¨oder-Bernstein theorem.

166

Descriptive set theory

424C

(i) By 423J, X has a subset Z homeomorphic to N N ; let h : N N → Z be a homeomorphism. By 424Bd, h is an isomorphism between (N N , B(N N )) and (Z, ΣZ ). (ii) Let hUn in∈N run over a base for the topology of X. Define g : X → {0, 1}N ⊆ N N by setting g(x) = hχUn (x)in∈N for every x ∈ N. Then g is injective, because X is Hausdorff. Also g is Borel measurable, by 4A3D(c-ii). By 424Bd, g is an isomorphism between (X, Σ) and (W, B(W )), where W = g[X] belongs to B(N N ). (iii) We have Z ∈ Σ, W ∈ B(N N ) such that (Z, ΣZ ) ∼ = (N N , B(N N )) and (W, B(W )) ∼ = (X, Σ). By N N ∼ 344D, (X, Σ) = (N , B(N )), as claimed. 424D Corollary (a) If (X, Σ) and (Y, T) are standard Borel spaces and #(X) = #(Y ), then (X, Σ) and (Y, T) are isomorphic. (b) If (X, Σ) is an uncountable standard Borel space then #(X) = #(Σ) = c. proof These follow immediately from 424C, if we recall that if B(N N ) is the Borel σ-algebra of N N then #(B(N N )) = c (4A3Fb). 424E Proposition Let X be a set and Σ a σ-algebra of subsets of X; suppose that (X, Σ) is countably separated in the sense that there is a countable set E ⊆ Σ separating the points of X. If A ⊆ X is such that (A, ΣA ) is a standard Borel space, where ΣA is the subspace σ-algebra, then A ∈ Σ. proof Give A a Polish topology T such that ΣA is the Borel σ-algebra of A, and let S be the topology on X generated by E ∪ {X \ E : E ∈ E}. Then S is second-countable (4A2Oa), so has a countable network (4A2Oc), and is Hausdorff because E separates the points of X. The identity map from A to X is Borel measurable for T and S, so 423Ib tells us that A is S-Borel; but of course the S-Borel σ-algebra is just the σ-algebra generated by E (4A3Da), so is included in Σ. 424F Corollary Let X be a Polish space and A ⊆ X any set which is not Borel. Let B(A) be the Borel σ-algebra of A. Then (A, B(A)) is not a standard Borel space. 424G Proposition Let (X, Σ) be a standard Borel space. Then (E, ΣE ) is a standard Borel space for every E ∈ Σ, writing ΣE for the subspace σ-algebra. proof Let T be a Polish topology on X for which Σ is the Borel σ-algebra. Then there is a Polish topology T0 ⊇ T for which E is closed (4A3I), therefore itself Polish in the subspace topology T0E (4A2Qd). But T0 and T have the same Borel sets (423Fb), so ΣE is just the Borel σ-algebra of E for T0E , and (E, ΣE ) is a standard Borel space. *424H For the full strength of a theorem in §448 we need a remarkable result concerning group actions on Polish spaces. Theorem (Becker & Kechris 96) Let G be a Polish group, (X, T) a Polish space and • a Borel measurable action of G on X. Then there is a Polish topology T0 on X, yielding the same Borel sets as T, such that the action is continuous for T0 and the given topology of G. proof (a) Fix on a right-translation-invariant metric ρ on G defining the topology of G (4A5Q), and let D be a countable dense subset of G; write e for the identity of G. Let Z be the set of 1-Lipschitz functions from G to [0, 1], that is, functions f : G → [0, 1] such that |f (g) − f (h)| ≤ ρ(g, h) for all g, h ∈ G. Then Z, with the topology of pointwise convergence inherited from the product topology of [0, 1]G , is a compact metrizable space. P P It is a closed subset of [0, 1]G , so is a compact Hausdorff space. Writing q(f ) = f ¹D for f ∈ Z, q : Z → [0, 1]D is injective, because D is dense and every member of Z is continuous; but this means that Z is homeomorphic to q[Z], which is metrizable, by 4A2Pc. Q Q We see also that the Borel σ-algebra of Z is the σ-algebra generated by sets of the form Wgα = {f : f (g) < α} where g ∈ G and α ∈ [0, 1]. P P This σ-algebra contains every set of the form {f : f ∈ Z, α < f (g) < β}, where g ∈ G and α, β ∈ R; since these sets generate the topology of Z, the σ-algebra they generate is the Borel σ-algebra of Z, by 4A3Da. Q Q

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(b) There is a continuous action of G on Z defined by setting (g •f )(h) = f (hg) for f ∈ Z and g, h ∈ G. P P (i) If f ∈ Z and g, h1 , h2 ∈ G, then |(g •f )(h1 ) − (g •f )(h2 )| = |f (h1 g) − f (h2 g)| ≤ ρ(h1 g, h2 g) = ρ(h1 , h2 ) because ρ is right-translation-invariant. So g •f ∈ Z for every f ∈ Z, g ∈ G. (ii) (e•f )(h) = f (he) = f (h) for every h ∈ G, so e•f = f for every f ∈ Z. (iii) If f ∈ Z and g1 , g2 , h ∈ G, then (g1 •(g2 •f ))(h) = (g2 •f )(hg1 ) = f (hg1 g2 ) = ((g1 g2 )•f )(h); as h is arbitrary, (g1 g2 )•f = g1 •(g2 •f ). Thus h ∈ G and ² > 0. Set

•

is an action of G on Z. (iv) Suppose that g0 ∈ G, f0 ∈ Z,

V = {g : g ∈ G, ρ(hg, hg0 ) < 21 ²}, W = {f : f ∈ Z, |f (hg0 ) − f0 (hg0 )| < 21 ²}. Then V is an open set in G containing g0 (because g 7→ ρ(hg, hg0 ) is continuous) and W is an open set in Z containing f0 . If g ∈ V and f ∈ W , |(g •f )(h) − (g0 •f0 )(h)| = |f (hg) − f0 (hg0 )| ≤ |f (hg) − f (hg0 )| + |f (hg0 ) − f0 (hg0 )| 1 2

≤ ρ(hg, hg0 ) + ² ≤ ². As f0 , g0 , ² are arbitrary, the map (g, f ) 7→ (g •f )(h) is continuous; as h is arbitrary, the map (g, f ) 7→ g •f is continuous. Q Q (c) Let B(X) be the Borel σ-algebra of X. For x ∈ X and B ∈ B(X), set PB (x) = {g : g ∈ G, g •x ∈ B}, QB (x) =

S

{V : V ⊆ G is open, V \ PB (x) is meager},

fB (x)(g) = inf({1} ∪ {ρ(g, h) : h ∈ G \ QB (x)}) for g ∈ G. It is easy to check that fB (x)(g 0 ) ≤ ρ(g, g 0 ) + fB (x)(g 0 ) for all g, g 0 ∈ G, so that every fB (x) belongs to Z. Every PB (x) is a Borel set, because • is Borel measurable. It follows that QB (x)4PB (x) is meager. P P By 4A3Ra, there is an open set V0 ⊆ G such that V0 4(X \ PB (x)) is meager and V0 ∩ V is empty whenever V ⊆ G is open and V \ PB (x) is meager. Now QB (x) ∩ V0 is empty, so QB (x) \ PB (x) ⊆ (X \ PB (x)) \ V0 is meager. Also (X \ V 0 ) \ PB (x) ⊆ (X \ PB (x)) \ V0 is meager, so X \ V 0 ⊆ QB (x) and PB (x) \ QB (x) ⊆ (V 0 \ V0 ) ∪ (V0 \ (X \ PB (x)) is meager. Q Q Consequently QB (x) \ PB (x) is empty, because the only meager open set in G is empty, by Baire’s theorem for complete metric spaces (4A2Ma). Let V be a countable base for the topology of G containing G. (d) For each B ∈ B(X), the map fB : X → Z is Borel measurable. P P Because the Borel σ-algebra of Z is generated by the sets Wgα of (a) above, it is enough to show that {x : fB (x) ∈ Wgα } = {x : fB (x)(g) < α}

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*424H

always belongs to B(X), because {W : fB−1 [W ] ∈ B} is surely a σ-algebra of subsets of Z. But if α > 1 this set is X, while if α ≤ 1 it is

{x : there is some h ∈ X \ QB (x) such that ρ(g, h) < α} = {x : there is some V ∈ V such that V \ QB (x) 6= ∅ and ρ(g, h) < α for every h ∈ V } [ = {x : V 6⊆ QB (x)} V ∈V 0 0

(where V = {V : V ∈ V, ρ(g, h) < α for every h ∈ V }) [ {x : V \ PB (x) is not meager}. = V ∈V 0

But for any fixed V ∈ V, {x : V \ PB (x) is not meager} = X \ {x : W [{x}] is meager} where W = {(y, g) : y ∈ X, g ∈ V, g •y ∈ X \ B} b is a Borel subset of X × G, because • is supposed to be Borel measurable; and therefore W ∈ B(X)⊗B(G), b writing B(G) for the Borel σ-algebra of G (4A3G). Now B(G) ⊆ B(G), the Baire property algebra of G b b b B(G). (4A3Rb), so W ∈ B(X)⊗ By 4A3Rc, the quotient algebra B/M has a countable order-dense set, where M is the σ-ideal of meager sets, so 4A3Sa tells us that {x : W [{x}] is meager} ∈ B(X). Accordingly {x : V \ PB (x) is not meager} is Borel for every V , and {x : fB (x)(g) < α} is Borel. Q Q (e) If g ∈ G, B ∈ B(X) and x ∈ X, then g •fB (x) = fB (g •x). P P PB (g •x) = {h : h•(g •x) ∈ B} = {h : hg ∈ PB (x)} = PB (x)g −1 . Because the map h 7→ hg −1 : G → G is a homeomorphism, QB (g •x) = QB (x)g −1 ; because it is an isometry, fB (g •x)(h) = min(1, ρ(h, X \ QB (g •x))) = min(1, ρ(h, X \ QB (x)g −1 )) = min(1, ρ(hg, X \ QB (x))) = fB (x)(hg) = (g •fB (x))(h) for every h ∈ G. Q Q (f ) Let hB0m im∈N run over a base for T containing X. We can now find a countable set E ⊆ B(X) such that (i) the topology T∗ generated by E is a Polish topology finer than T (ii) fB is T∗ -continuous for every B ∈ E (iii) X ∈ E. P P Let W be a countable base for the topology of Z. Enumerate N × N × W as h(kn , mn , Wn )in∈N in such a way that kn ≤ n for every n. Having chosen Borel sets Bij ⊆ X for i ≤ n, j ∈ N in such a way that the topology Sn generated by {Bij : i ≤ n, j ∈ N} is a Polish topology finer than T, consider the set Cn = {x : fBkn ,mn (x) ∈ Wn }. This is T-Borel, therefore Sn -Borel, so by 4A3H there is a Polish topology Sn+1 ⊇ Sn such that Cn ∈ Sn+1 . Let hBn+1,m im∈N run over a base for Sn+1 ; by 423Fb, every Bn+1,m belongs to B(X). Continue. ∗ Let T∗ be the topology S generated by E = {Bij : i, j ∈ N}. By 4A2Qf, T is Polish, because it is the topology generated by n∈N Sn . If W ∈ W, B ∈ E there are i, j ∈ N such that B = Bij and an n ∈ N such that (i, j, W ) = (kn , mn , Wn ); now fB−1 [W ] = Cn ∈ Sn+1 ⊆ T∗ . As W is arbitrary, fB is T∗ -continuous. Also X ∈ {B0m : m ∈ N} ⊆ E, as required. Q Q

*424H

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169

(g) Define θ : X → Z E by setting θ(x)(E) = fE (x) for x ∈ X, E ∈ E. Then θ is injective. P P Suppose that x, y ∈ X and that x 6= y. For every g ∈ G, g −1 •(g •x) = x 6= y = g −1 •(g •y), so g •x 6= g •S y and there is some m ∈ N such that g •x ∈ B0m while g •y ∈ / B0m , that is, g ∈ PB0m (x) \ PB0m (y). Thus G = m∈N PB0m (x) \ PB0m (y); by Baire’s theorem, there is some m ∈ N such that PB0m (x) \ PB0m (y) is non-meager. Because QB0m (x)4PB0m (x) and QB0m (y)4PB0m (y) are both meager, {g : fB0m (x)(g) > 0} = QB0m (x) 6= QB0m (y) = {g : fB0m (y)(g) > 0}, and θ(x)(B0m ) = fB0m (x) 6= fB0m (y) = θ(y)(B0m ). So θ(x) 6= θ(y). Q Q Because fE is T∗ -continuous for every E ∈ E, θ is T∗ -continuous. (h) Let T0 be the topology on X induced by θ; that is, the topology which renders θ a homeomorphism between X and θ[X]. Because θ[X] ⊆ Z E is separable and metrizable, T0 is separable and metrizable. Because θ is T∗ -continuous, T0 ⊆ T∗ . (i) The action of G on X is continuous for the given topology S on G and T0 on X. P P For any E ∈ E, (g, x) 7→ θ(g •x)(E) = fE (g •x) = g •fE (x) ((e) above) is S × T0 -continuous because the action of G on Z is continuous ((b) above) and fE : X → Z is T0 -continuous (by the definition of T0 ). But this means that (g, x) 7→ θ(g •x) is S × T0 -continuous, so that (g, x) 7→ g •x is (S × T0 , T0 )-continuous. Q Q (j) Let σ be a complete metric on G defining the topology S, and τ a complete metric on X defining the topology T∗ . (We do not need to relate σ to ρ in any way beyond the fact that they both give rise to the same topology S.) For E ∈ E, V ∈ V and n ∈ N let SEV n be the set of those φ ∈ Z E such that either φ(E)(g) = 0 for every g ∈ V or there is an F ∈ E such that F ⊆ E, diamτ (F ) ≤ 2−n and φ(F )(g) > 0 for some g ∈ V . Then SEV n is the union of a closed set and an open set, so is a Gδ set in Z E (4A2C(a-i)). Consequently T Y = {φ : φ ∈ Z E , φ(X) = χG} ∩ θ[X] ∩ E∈E,V ∈V,n∈N SEV n is a Gδ subset of Z E , being the intersection of countably many Gδ sets. (k) θ[X] ⊆ Y . P P Let x ∈ X. (i) PX (x) = G so QX (x) = G and θ(x)(X) = fX (x) = χG. (ii) Of course θ(x) ∈ θ[X]. (iii) Suppose that E ∈ E, V ∈ V and n ∈ N. If QE (x) ∩ V = ∅ then θ(x)(E)(g) = fE (x)(g) = 0 for every g ∈ V , and θ(x)(E) ∈ SEV n . Otherwise, V ∩ PE (x) is non-meager. But S E = {F : F ∈ E, F ⊆ E, diamτ (F ) ≤ 2−n }, so PE (x) =

S

{PF (x) : F ∈ E, F ⊆ E, diamτ (F ) ≤ 2−n };

because E is countable, this is a countable union and there is an F ∈ E such that F ⊆ E, diamτ (F ) ≤ 2−n and PF (x) ∩ V is non-meager. In this case QF (x) ∩ V is non-empty and θ(x)(F ) = fF (x) is non-zero at some point of V ; thus again θ(x) ∈ SEV n . As E, V and n are arbitrary, we have the result. Q Q (l) (The magic bit.) Y ⊆ θ[X]. P P Take any φ ∈ Y . Choose hEn in∈N in E, hVn in∈N in V and h˜ gn in∈N in G as follows. E0 = X and V0 = G. Given that φ(En ) is non-zero at some point of Vn , then, because φ ∈ SEn Vn n , there is an En+1 ∈ E such that En+1 ⊆ En , diamτ (En+1 ) ≤ 2−n and φ(En+1 ) is non-zero at some point of Vn ; say g˜n ∈ Vn is such that φ(En+1 )(˜ gn ) > 0. Now we can find a Vn+1 ∈ V such that g˜n ∈ Vn+1 ⊆ Vn and diamσ (Vn+1 ) ≤ 2−n . Continue. We are supposing also that φ ∈ θ[X], so we have a sequence hxi ii∈N in X such that hθ(xi )ii∈N → φ, that is, hfE (xi )ii∈N → φ(E) for every E ∈ E. In particular, limi→∞ fEn+1 (xi )(˜ gn ) = φ(En+1 )(˜ gn ) > 0 for every n. Let hin in∈N be a strictly increasing sequence in N such that fEn+1 (xin )(˜ gn ) > 0 for every n. Then

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*424H

g˜n ∈ Vn+1 ∩ QEn+1 (xin ) ⊆ Vn+1 ∩ PEn+1 (xin ); there is therefore some gn ∈ Vn+1 ∩ PEn+1 (xin ), so that gn •xin ∈ En+1 . hVn in∈N is a non-increasing sequence of sets with σ-diameters converging to 0, so hgn in∈N is a Cauchy sequence for the complete metric σ. Similarly, hgn •xin in∈N is a Cauchy sequence for the complete metric τ , because diamτ (En+1 ) ≤ 2−n . We therefore have g ∈ G, y ∈ X such that hgn in∈N → g for S and hgn •xin in∈N → y for T∗ . In this case, hgn •xin in∈N → y for the coarser topology T0 , while hgn−1 in∈N → g −1 for S. Because the action is (S × T0 , T0 )-continuous, hxin in∈N = hgn−1 •(gn •xin )in∈N → g −1 •y for T0 . But θ is continuous for T0 , so θ(g −1 •y) = limn→∞ θ(xin ) = φ, and φ ∈ θ[X]. As φ is arbitrary, Y ⊆ θ[X]. Q Q (m) Thus θ[X] = Y is a Gδ set in the compact metric space Z E , and is a Polish space in its induced topology (4A2Qd). But this means that (X, T0 ), which is homeomorphic to θ[X], is also Polish. (n) I have still to check that T0 has the same Borel sets as T. But T, T∗ and T0 are all Polish topologies and T∗ is finer than both the other two. By 423Fb, T∗ has the same Borel sets as either of the others. This completes the proof. 424X Basic exercises > (a) Let h(Xi , Σi )ii∈I be a countable family of standard Borel spaces, and (X, Σ) their direct sum, that is, X = {(x, i) : i ∈ I, x ∈ Xi }, Σ = {E : E ⊆ X, {x : (x, i) ∈ E} ∈ Σi for every i}. Show that (X, Σ) is a standard Borel space. > (b) Let (X, Σ) be a standard Borel space and T a countably generated σ-subalgebra of Σ. Show that there is an analytic Hausdorff space Z such that T is isomorphic to the Borel σ-algebra of Z. (Hint: by 4A3I, we can suppose that X is a Polish space and T is generated by a sequence of open-and-closed sets, corresponding to a continuous function from X to {0, 1}N .) > (c) Let (X, Σ) be a standard Borel space and T1 , T2 two countably generated σ-subalgebras of Σ which separate the same points, in the sense that if x, y ∈ X then there is an E ∈ T1 such that x ∈ E, y ∈ X \ E iff there is an E 0 ∈ T2 such that x ∈ E 0 , y ∈ X \ E 0 . Show that T1 = T2 . (Hint: 424Xb, 423Fb.) In particular, if T1 separates the points of X then T1 = Σ. (d) Let A be a Dedekind σ-complete Boolean algebra. Show that A is isomorphic to the Borel σ-algebra of an analytic Hausdorff space iff it is isomorphic to a countably generated σ-subalgebra of the Borel σ-algebra of [0, 1]. > (e) Let U be a separable Banach space. Show that its Borel σ-algebra is generated, as σ-algebra, by the sets of the form {u : h(u) ≤ α} as h runs over the dual U ∗ and α runs over R. > (f ) Let X be a compact metrizable space. Show that the Borel σ-algebra of C(X) (the Banach space of continuous real-valued functions on X) is generated, as σ-algebra, by the sets {u : u ∈ C(X), u(x) ≥ α} as x runs over X and α runs over R. (g) Let (X, Σ) be a standard Borel space, Y any set, and T a σ-algebra of subsets of Y . Write T∗ for the σ-algebra of subsets of Y generated by S(T), the family of sets obtainable from sets in T by Souslin’s b operation. Let W ∈ S(Σ⊗T). Show that W [X] ∈ T∗ and that there is a (T∗ , Σ)-measurable function f : W [X] → X such that (f (y), y) ∈ W for every y ∈ W [X]. (Hint: 423M.) (h) Let (X, Σ) be a standard Borel space, Y any set, and T a countably generated σ-algebra of subsets of Y . Let f : X → Y be a (Σ, T)-measurable function, and write T∗ for the σ-algebra of subsets of Y generated by S(T), the family of sets obtainable from sets in T by Souslin’s operation. Show that (i) f [X] ∈ T∗ (ii) there is a (T∗ , Σ)-measurable function g : f [X] → X such that gf is the identity on X. (Hint: start with b the case in which T separates the points of Y , so that the graph of f belongs to Σ⊗T.)

424 Notes

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171

> (i) Show that 424Xc and 424Xh are both false if we omit the hypothesis ‘countably generated’. (Hint: consider (i) the countable-cocountable algebra of R (ii) the split interval (iii) 423Xi.) (j) Let (X, Σ) be a standard Borel space. Show that if X is uncountable, Σ has a countably generated σ-subalgebra not isomorphic either to Σ or to PI for any set I. (Hint: 423L.) (k) Let (X, Σ) be a standard Borel space. (i) Show that the Boolean algebra A = Σ/[X]≤ω is homogeneous. (ii) Show that for every sequentially order-continuous Boolean homomorphism π : A → A there is a (Σ, Σ)-measurable f : X → X representing π in the sense that π(E • ) = (f −1 [E])• for every E ∈ Σ. (Hint: reduce to the case X = {0, 1}N .) (iii) Show that for every π ∈ Aut A there is an automorphism f of (X, Σ) representing π. (Hint: 344Ya.) 424Y Further exercises (a) Let (X, T) be a Polish space and F the family of closed subsets of X. Let Σ be the σ-algebra of subsets of F generated by the sets EH = {F : F ∈ F, F ∩ H 6= ∅} as H runs over the open subsets of X. (i) Show that S (F, Σ) is a standard Borel space. (Hint: take a complete metric ρ defining the topology of X. Set S ∗ = n∈N N n and choose a family hUσ iσ∈S ∗ of open sets in X such that U∅ = X, S diam Uσ ≤ 2−n whenever #(σ) = n + 1, Uσ = i∈N Uσa i for every σ, U σa i ⊆ Uσ for every σ, i. Define ∗ f : F → {0, 1}S by setting f (F )(σ) = 1 if F ∩ Uσ 6= ∅, 0 otherwise. Show that Z = f [F] is a Borel set and that f is an isomorphism between Σ and the Borel σ-algebra of Z.) (This is the Effros Borel structure on F.) (ii) Show that [X]n ∈ Σ for every n ∈ N. (b) Let (X, Σ) be a standard Borel space. Let T be the family of Polish topologies on X for which Σ is the Borel σ-algebra. Show that any sequence in T has an upper bound in T, and that any sequence with a lower bound has a least upper bound. (c) Let (X, Σ) be a standard Borel space. Say that C ⊆ X is coanalytic if its complement belongs to S(Σ). Show that for any such C the partially ordered set Σ ∩ PC has cofinality 1 if C ∈ Σ and cofinality ω1 otherwise. (Hint: 423P.) (d) Let I k be the split interval. Show that there is a σ-algebra Σ of subsets of I k such that (I k , Σ) is a b standard Borel space and {(x, y) : x, y ∈ I k , x ≤ y} ∈ Σ⊗Σ. 424 Notes and comments In this treatise I have generally indulged my prejudice in favour of ‘complete’ measures. Consequently Borel σ-algebras, as such, have taken subordinate roles. But important parts of the theory of Lebesgue measure, and Radon measures on Polish spaces in general, are associated with the fact that these are completions of measures defined on standard Borel spaces. Moreover, such spaces provide a suitable framework for a large part of probability theory. Of course they become deficient in contexts where we need to look at uncountable independent families of random variables, and there are also difficulties with σ-subalgebras, even countably generated ones, since these can correspond to the Borel algebras of general analytic spaces, which will not always be standard Borel structures (424F, 423L). 424Xf and 424Ya suggest the ubiquity of standard Borel structures; the former shows that they are not always presented as countably generated algebras, while the latter is an example in which we have to make a special construction in order to associate a topology with the algebra. The theory is of course dominated by the results of §423, especially 423Fb and 423I. I include 424H in this section because there is no other convenient place for it, but I have an excuse: the idea of ‘Borel measurable action’ can, in this context, be described entirely in terms of σ-algebras, since the Borel algebra of G × X is just the σ-algebra product of the Borel algebras of the factors (as in 424Bb). Of course for the theorem as expressed here we do need to know that G has a Polish group structure; but X could be presented just as a standard Borel space. The result is a dramatic expression of the fact that, given a standard Borel space (X, Σ), we have a great deal of freedom in defining a corresponding Polish topology on X.

172

Topological measure spaces II

Chapter 43 Topologies and measures II The first chapter of this volume was ‘general’ theory of topological measure spaces; I attempted to distinguish the most important properties a topological measure can have – inner regularity, τ -additivity – and describe their interactions at an abstract level. I now turn to rather more specialized investigations, looking for features which offer explanations of the behaviour of the most important spaces, radiating outwards from Lebesgue measure. In effect, this chapter consists of three distinguishable parts and two appendices. The first three sections are based on ideas from descriptive set theory, in particular Souslin’s operation (§431); the properties of this operation are the foundation for the theory of two classes of topological space of particular importance in measure theory, the K-analytic spaces (§432) and the analytic spaces (§433). The second part of the chapter, §§434-435, collects miscellaneous results on Borel and Baire measures, looking at the ways in which topological properties of a space determine properties of the measures it carries. In §436 I present the most important theorems on the representation of linear functionals by integrals; if you like, this is the inverse operation to the construction of integrals from measures in §122. The ideas continue into §437, where I discuss spaces of signed measures representing the duals of spaces of continuous functions. The first appendix, §438, looks at a special topic: the way in which the patterns in §§434-435 are affected if we assume that our spaces are not unreasonably complex in a rather special sense defined in terms of measures on discrete spaces. Finally, I end the chapter with a further collection of examples, mostly to exhibit boundaries to the theorems of the chapter, but also to show some of the variety of the structures we are dealing with. 431 Souslin’s operation I begin the chapter with a short section on Souslin’s operation (§421). The basic facts we need to know are that (in a complete locally determined measure space) the family of measurable sets is closed under Souslin’s operation (431A), and that the kernel of a Souslin scheme can be approximated from within in measure (431D). 431A Theorem Let (X, Σ, µ) be a complete locally determined measure space. Then Σ is closed under Souslin’s operation. proof Let hEσ iσ∈S be a Souslin scheme in Σ with kernel A. Write S S ∗ = k∈N N k = S ∪ {∅}. If F ∈ Σ and µF < ∞, then A ∩ F ∈ Σ. P P For each σ ∈ S ∗ , set S T Aσ = φ∈N N ,φ⊇σ n≥1 Eφ¹n , and let Gσ be a measurable envelope of Aσ ∩ F . Because Aσ ⊆ Eσ (writing E∅ = X), we may suppose that Gσ ⊆ Eσ ∩ F . Now, for any σ ∈ S ∗ , S S Aσ ∩ F = i∈N Aσa i ∩ F ⊆ i∈N Gσa i , so Hσ = Gσ \

S i∈N

Gσa i

is negligible. S Set H = σ∈S ∗ Hσ , so that H is negligible. Take any x ∈ G∅ \ H. Choose hφ(i)ii∈N inductively, as follows. Given that σ = hφ(i)ii
431D

Souslin’s operation

173

431B Corollary If (X, T, Σ, µ) is a complete locally determined topological measure space, every SouslinF set in X (definition: 421K) is measurable. 431C Corollary Let X be a set and θ an outer measure on X. Let µ be the measure defined by Carath´eodory’s method, and Σ its domain. Then Σ is closed under Souslin’s operation. proof Let hEσ iσ∈S be a Souslin scheme in Σ with kernel A. Take any C ⊆ X such that θC < ∞. Then θC = θ¹ PC is an outer measure on C; let µC be the measure on C defined from θC by Carath´eodory’s method, and ΣC its domain. If σ ∈ S, D ⊆ C then θC (D ∩ C ∩ Eσ ) + θC (D \ (C ∩ Eσ )) = θ(D ∩ Eσ ) + θ(D \ Eσ ) = θD = θC D; as D is arbitrary, C ∩ Eσ ∈ ΣC . µC is a complete totally finite measure, so 431A tells us that the kernel of the Souslin scheme hC ∩ Eσ iσ∈S belongs to ΣC . But this is just C ∩ A (applying 421Cb to the identity map from C to X.) So θ(C ∩ A) + θ(C \ A) = θC (C ∩ A) + θC (C \ A) = θC C = θC. As C is arbitrary, A ∈ Σ (113D). As hEσ iσ∈S is arbitrary, we have the result. 431D Theorem Let (X, Σ, µ) be a complete locally determined measure space, and hEσ iσ∈S a Souslin scheme in Σ with kernel A. Then [ \ Eφ¹n ) : K ⊆ N N is compact} µA = sup{µ( φ∈K n≥1

= sup{µ(

[ \

Eφ¹n ) : ψ ∈ N N },

φ≤ψ n≥1

writing φ ≤ ψ if φ(i) ≤ ψ(i) for every i ∈ N. S T proof (a) By 431A, A is measurable. For K ⊆ N N , set HK = φ∈K n≥1 Eφ¹n . Of course HK ⊆ A, and we know from 421M (or otherwise) that H QK ∈ Σ if K is compact. So surely µA ≥ µHK for every compact K ⊆ N N . If ψ ∈ N N , then {φ : φ ≤ ψ} = i∈N (ψ(i) + 1) is compact. We therefore have µA ≥ sup{µ(

[ \

Eφ¹n ) : K ⊆ N N is compact}

φ∈K n≥1

≥ sup{µ(

[ \

Eφ¹n ) : ψ ∈ N N }.

φ≤ψ n≥1

So what I need to prove is that

S T µA ≤ sup{µ( φ≤ψ n≥1 Eφ¹n ) : ψ ∈ N N }.

S (b) Fix on a set F ∈ Σ of finite measure. For σ ∈ S ∗ = k∈N N k set S T Aσ = φ∈N N ,φ⊇σ n≥1 Eφ¹n . We need to know that Aσ belongs to Σ; this follows from 431A, because writing Eτ0 = Eτ if τ ⊆ σ or σ ⊆ τ , ∅ otherwise, S T 0 ∈ S(Σ) = Σ, Aσ = φ∈N N n≥1 Eφ¹n writing S for Souslin’s operation, as in §421. P Let ² > 0, and take S a family h²σ iσ∈S ∗ of strictly positive real numbers such that σ∈S ∗ ²σ ≤ ². For each σ ∈ S ∗ we have Aσ = i∈N Aσa i , so there is an mσ ∈ N such that S µ(F ∩ Aσ \ i≤mσ Aσa i ) ≤ ²σ .

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431D

Define ψ ∈ N N by saying that ψ(k) = max{mσ : σ ∈ Nk , σ(i) ≤ ψ(i) for every i < k} for each k ∈ N. Set H= (c) Set G=

S

T φ∈N N ,φ≤ψ

S σ∈S ∗

n≥1

F ∩ Aσ \

Eφ¹n .

S i≤mσ

Aσ a i ,

so that µG ≤ ², by the choice of the ²σ and the mσ . Then F ∩ A \ G ⊆ H. P P If x ∈ F ∩ A \ G, choose hφ(i)ii∈N inductively, as follows. Given that φ(i) ≤ ψ(i) for i < k and x ∈ Aσ , where σ = hφ(i)ii
T

µ∗ E = µ ˜E = supK⊆N N

is compact

µFK ,

where FK = φ∈K n≥1 Eφ¹n for K ⊆ N N . But every FK is closed, by 421M. So µ∗ E ≤ supF ⊆E as the reverse inequality is trivial, we have the result.

is closed

µF ;

*431F There is a topological version of 431A, as follows. Theorem Let X be any topological space, and Bb its Baire property algebra. (a) For any A ⊆ X, there is a Baire property envelope of A, that is, a set E ∈ Bb such that A ⊆ E and b E \ F is meager whenever A ⊆ F ∈ B. (b) Bb is closed under Souslin’s operation. proof (a) By 4A3Ra, there is an open set H ⊆ X such that A \ H is meager and H ∩ G is empty whenever b G ⊆ X is open and A ∩ G is meager. Set E = A ∪ H; then E ⊇ A and E4H = A \ H is meager, so E ∈ B. b let G be an open set such that G4(X \ F ) is meager. Then G ∩ A ⊆ G ∩ F is meager, so If A ⊆ F ∈ B, G ∩ H is empty and E \ F ⊆ (E4H) ∪ (G4(X \ F )) is meager. Thus E is a Baire property envelope of A. (b) Let hEσ iσ∈S be a Souslin scheme in Bb with kernel A. Write S S ∗ = k∈N N k = S ∪ {∅}. For each σ ∈ S ∗ , set Aσ =

S

T φ∈N N ,φ⊇σ

n≥1

Eφ¹n ,

and let Gσ be a Baire property envelope of Aσ as described in (a). Because Aσ ⊆ Eσ (writing E∅ = X), we may suppose that Gσ ⊆ Eσ . Now, for any σ ∈ S ∗ , S S Aσ = i∈N Aσa i ⊆ i∈N Gσa i , so

431 Notes

Souslin’s operation

Hσ = Gσ \

S i∈N

175

Gσa i

is meager. S Set H = σ∈S ∗ Hσ , so that H is negligible. Take any x ∈ G∅ \ H. Choose hφ(i)ii∈N inductively, as follows. Given that σ = hφ(i)ii (d) Let (X, Σ, µ) be a measure space with locally determined negligible sets (definition: 213I), and hEσ iσ∈S a Souslin scheme in Σ with kernel A. Show that S T µ∗ A = supψ∈N N µ( φ≤ψ n≥1 Eφ¹n ). > (e) Let (X, Σ, µ) be a semi-finite measure space, and hEσ iσ∈S a Souslin scheme in Σ with kernel A. Show that S T µ∗ A = supψ∈N N µ( φ≤ψ n≥1 Eφ¹n ). 431Y Further exercises (a) Let (X, Σ, µ) be a complete measure space with the measurable envelope property (213Xl). Show that Σ is closed under Souslin’s operation. (b) Let X be a set, Σ a σ-algebra of subsets of X, and I a σ-ideal of subsets of X such that I ⊆ Σ. Suppose that for every A ⊆ X there is an F ∈ Σ such that A ⊆ F and F \ E ∈ I whenever A ⊆ E ∈ Σ. Show that Σ is closed under Souslin’s operation. (c) Let X be a set, Σ a σ-algebra of subsets of X, and I an ω1 -saturated σ-ideal of Σ; suppose that A ∈ Σ whenever A ⊆ B ∈ I. Show that Σ is closed under Souslin’s operation. 431 Notes and comments From the point of view of measure theory, the most important property of Souslin’s operation, after its idempotence, is the fact that (for many measure spaces) the family of measurable sets is closed under the operation (431A). The proof I give here is based on the concept of measurable envelope, which can be used in other cases of great interest (431Yb, 431Yc). But for some applications it is also very important to know that if A is the S kernel T of a Souslin scheme hEσ iσ∈S , then A can be approximated from inside by sets of the form H = φ≤ψ n≥1 Eφ¹n (431D, 431Xd), which belong to the σ-algebra generated by the Eσ (421M). A typical application of this idea is when every Eσ is a Borel subset of R; then we find not only that A is Lebesgue measurable (indeed, measurable for every Radon measure on R) but that (for any given Radon measure µ) the Souslin scheme itself provides a Borel subset H of A of measure approximating the measure of A. Let me repeat that the essence of descriptive set theory is that we are not satisfied merely to know that a set of a certain type exists. We want also to know how to build it, because we expect that an explicit

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construction will be valuable later on. For instance, the construction given in 431D shows that if the Souslin scheme consists of closed compact sets, the set H will be compact (421Xo). I mention 431B as a typical application of 431A, even though it is both obvious and obviously less than what can be said. The algebras Σ of this section are algebras closed under Souslin’s operation. In a complete locally determined topological measure space, the algebra Σ of measurable sets includes the open sets (by definition), therefore the Borel algebra B, therefore S(B); but now we can take the algebra A1 generated by S(B), and A1 and S(A1 ) will also be included in Σ, so that Σ will included the algebra A2 generated by S(A1 ), and so on. (Note that S(A1 ) includes the σ-algebra generated by A1 , by 421F, so I do not need to mention that separately.) We haveSto run through all the countable ordinals before we can be sure of getting to the smallest algebra Aω1 = ξ<ω1 Aξ which contains every open set and is closed under Souslin’s operation, and we shall then have Aω1 ⊆ Σ. The result in 431D is one of the special features of measures. (A similar result, based on rather different hypotheses, is in 432J.) But the argument of 431A can be applied in many other cases; see 431Yb-431Yc. A striking one is 431F, which will be useful in Volume 5.

432 K-analytic spaces I describe the basic measure-theoretic properties of K-analytic spaces (§422). I start with ‘elementary’ results (432A-432C), assembling ideas from §§421, 422 and 431. The main theorem of the section is 432D, one of the leading cases of the general extension theorem 416P. An important corollary (432G) gives a sufficient condition for the existence of pull-back measures. I briefly mention ‘capacities’ (432I-432J). 432A Proposition Let (X, T, Σ, µ) be a complete locally determined Hausdorff topological measure space. Then every K-analytic subset of X is measurable. proof If A ⊆ X is K-analytic, it is Souslin-F (422Ha), therefore measurable (431B). 432B Theorem Let X be a K-analytic Hausdorff space, and µ a semi-finite topological measure on X. Then µX = sup{µK : K ⊆ X is compact}. proof If γ < µX, there is an E ∈ dom µ such that γ < µE < ∞; set νF = µ(E ∩ F ) for every Borel set F ⊆ X, so that ν is a totally finite Borel measure on X, and νX > γ. Let νˆ be the completion S of ν. Let R ⊆ N N × X be an usco-compact relation such that R[N N ] = X. Set Fσ = R[Iσ ] for σ ∈ S = k≥1 N k , where Iσ = {φ : σ ⊆ φ ∈ N N }. Because R is closed in N N × X (422Da), S XTis the kernel of the Souslin scheme hFσ iσ∈S (421I). By 431D, there is a compact L ⊆ N N such that νˆ( φ∈L n∈N Fφ¹n ) ≥ γ. But, by 421I, this is just νˆ(R[L]); and R[L] is compact, by 422D(e-i). So µR[L] is defined, with µR[L] ≥ νR[L] = νˆR[L], and we have a compact subset of X of measure at least γ. As γ is arbitrary, the proposition is proved. 432C Proposition Let X be a Hausdorff space such that all its open sets are K-analytic, and µ a Borel measure on X. (a) If µ is semi-finite, it is tight (that is, inner regular with respect to the compact sets). (b) If µ is locally finite, its completion is a Radon measure on X. proof (a) By 422Hb, every open subset of X is Souslin-F. Applying 421F to the family E of closed subsets of X, we see that every Borel subset of X is Souslin-F, therefore K-analytic (by 422Hb in the opposite direction). Now suppose that E ⊆ X is a Borel set. Then the subspace measure µE is a semi-finite Borel measure on the K-analytic space E, so by 432B µE = supK⊆E is compact µK. As E is arbitrary, µ is tight. (b) Because X is Lindel¨of (422Gg), µ is σ-finite (411Ge), therefore semi-finite. So (a) tells us that µ is tight. By 416F, its c.l.d. version is a Radon measure. But (because µ is σ-finite) this is just its completion (213Ha).

432G

K-analytic spaces

177

432D Theorem (Aldaz & Render 00) Let X be a K-analytic Hausdorff space and µ a locally finite measure on X which is inner regular with respect to the closed sets. Then µ has an extension to a Radon measure on X. In particular, µ is τ -additive. proof The point is that if µE > 0 then there is a compact K ⊆ E such that µ∗ K > 0. P P Write Σ for the domain of µ. Take γ < µE. Because X is Lindel¨of (422Gg), µ is σ-finite (411Ge), therefore semi-finite. Let E 0 ⊆ E be such that γ < µE 0 < ∞. Because µ is inner regular with respect to the closed sets, there is a closed set F ⊆ E such that µF > γ. S F is K-analytic (422Gf); let R ⊆ N N × F be an usco-compact relation such that R[N N ] = F . For σ ∈ S ∗ = n∈N N n set Fσ = {x : (φ, x) ∈ R for some φ ∈ N N such that φ(i) ≤ σ(i) for every i < #(σ)}. Then hFσa i ii∈N is a non-decreasing sequence with union Fσ , so µ∗ Fσ = supi∈N µ∗ Fσa i for every σ ∈ S ∗ . We can therefore find a sequence ψ ∈ N N such that µ∗ Fψ¹n > γ for every n ∈ N. Set Q

K = {φ : φ ∈ N N , φ(i) ≤ ψ(i) for every i ∈ N};

then K = n∈N (ψ(n) + 1) is compact, so R[K] is compact (422D(e-i)). ?? Suppose, if possible, that µ∗ R[K] < γ. Then there is an H ∈ Σ such that R[K] ⊆ H ⊆ F and µ(F \ H) > µF − γ. Because µ is inner regular with respect to the closed sets, there is a closed set F 0 ∈ Σ such that F 0 ⊆ F \ H and µF 0 > µF − γ. Since R[K] ∩ F 0 = ∅, K ∩ R−1 [F 0 ] = ∅. R−1 [F 0 ] is closed, because R is usco-compact, so there is some n such that L = {φ : φ ∈ N N , φ¹n = φ0 ¹n for some φ0 ∈ K} does not meet R−1 [F 0 ] (4A2F(h-vi)), and R[L] ∩ F 0 = ∅. But L is just {φ : φ(i) ≤ ψ(i) for every i < n}, so R[L] = Fψ¹n , and γ < µ∗ Fψ¹n ≤ µ(F \ F 0 ) < γ, which is absurd. X X Thus µ∗ R[K] ≥ γ. As γ > 0, we have the result. Q Q Now the theorem follows at once from 416P(ii)⇒(i). 432E Corollary Let X be a K-analytic Hausdorff space, and µ a locally finite quasi-Radon measure on X. Then µ is a Radon measure. proof By 432D, µ has an extension to a Radon measure µ0 . But of course µ and µ0 must coincide, by 415H or otherwise. 432F Corollary Let X be a K-analytic Hausdorff space, and ν a locally finite Baire measure on X. Then ν has an extension to a Radon measure on X; in particular, it is τ -additive. If the topology of X is regular, the extension is unique. proof Because X is Lindel¨of (422Gg), ν is σ-finite, therefore semi-finite; by 412D, it is inner regular with respect to the closed sets. So 432D tells us that it has an extension to a Radon measure on X. Since the extension is τ -additive, so is ν. If X is regular, then it must be completely regular (4A2H(b-i)), and the family G of cozero sets is a base for the topology closed under finite unions. If µ, µ0 are Radon measures extending ν, they agree on G, and must be equal, by 415H(iv). 432G Corollary Let X be a K-analytic Hausdorff space, Y a Hausdorff space and ν a locally finite measure on Y which is inner regular with respect to the closed sets. Let f : X → Y be a continuous function such that f [X] has full outer measure in Y . Then there is a Radon measure µ on X such that f is inverse-measure-preserving for µ and ν. If ν is Radon, it is precisely the image measure µf −1 .

178

Topologies and measures II

432G

proof (a) Write T for the domain of ν, and set Σ0 = {f −1 [F ] : f ∈ T}, so that Σ0 is a σ-algebra of subsets of X, and we have a measure µ0 on X defined by setting µ0 f −1 [F ] = νF whenever F ∈ T (132Yd). (b) If E ∈ Σ0 and γ < µ0 E, there is an F ∈ T such that E = f −1 [F ]. Now there is a closed set F 0 ⊆ F such that νF 0 ≥ γ. Because f is continuous, f −1 [F 0 ] is closed, and we have f −1 [F 0 ] ⊆ E and µ0 f −1 [F 0 ] ≥ γ. As E and γ are arbitrary, µ0 is inner regular with respect to the closed sets. If x ∈ X, then (because ν is locally finite) there are an open set H ⊆ Y and an F ∈ T such that f (x) ∈ H ⊆ F ,

νF < ∞.

In this case f −1 [H] is an open set containing x and µ∗0 f −1 [H] ≤ µ0 f −1 [F ] = νF < ∞. As x is arbitrary, µ0 is locally finite. (c) By 432D, there is a Radon measure µ on X extending µ0 . Because f is inverse-measure-preserving for µ0 and ν, it is surely inverse-measure-preserving for µ and ν. The image measure µf −1 extends ν, so must be locally finite; it is therefore a Radon measure (418I). So if ν itself is a Radon measure, it must be identical with µf −1 , by 416Eb. 432H Corollary Suppose that X is a set and that S, T are Hausdorff topologies on X such that (X, T) is K-analytic and S ⊆ T. Then the totally finite Radon measures on X are the same for S and T. proof Write f for the identity function on X regarded as a continuous function from (X, T) to (X, S). If µ is a totally finite T-Radon measure on X, then µ = µf −1 is S-Radon, by 418I. If ν is a totally finite S-Radon measure on X, then 432G tells us that it is of the form µ = µf −1 for some T-Radon measure µ, that is, is itself T-Radon. 432I Capacitability The next result is not exactly measure theory as studied in most of this treatise; but it is clearly very close to the other ideas of this section, and it has important applications to measure theory in the narrow sense. Definition Let (X, T) be a topological space. A (Choquet) capacity on X is a function c : PX → [0, ∞] such that (i) c(A) ≤ c(B) whenever A ⊆ B ⊆ X; (ii) limn→∞ c(An ) = c(A) whenever hAn in∈N is a non-decreasing sequence of subsets of X with union A; (iii) c(K) = inf{c(G) : G ⊇ K is open} for every compact set K ⊆ X. 432J Theorem (Choquet 55) Let X be a Hausdorff space and c a capacity on X. If A ⊆ X is K-analytic, then c(A) = sup{c(K) : K ⊆ A is compact}. proof Take γ < c(A). Let R ⊆ N N × X be an usco-compact relation such that R[N N ] = A; for σ ∈ S ∗ = S n n∈N N set Aσ = {x : (φ, x) ∈ R for some φ ∈ N N such that φ(i) ≤ σ(i) for every i < #(σ)}. Then hAσa i ii∈N is a non-decreasing sequence with union Aσ , so c(Aσ ) = supi∈N c(Aσa i ) for every σ ∈ S ∗ . We can therefore find a sequence ψ ∈ N N such that c(Aψ¹n ) > γ for every n ∈ N. Set Q

K = {φ : φ ∈ N N , φ(i) ≤ ψ(i) for every i ∈ N};

then K = n∈N (ψ(n) + 1) is compact, so R[K] is compact (422D(e-i)). ?? Suppose, if possible, that c(R[K]) < γ. Then, by (iii) of 432I, there is an open set G ⊇ R[K] such that c(G) < γ. Set F = X \ G, so that F is closed and K ∩ R−1 [F ] = ∅. R−1 [F ] is closed, because R is usco-compact, so there is some n such that L = {φ : φ ∈ N N , φ¹n = φ0 ¹n for some φ0 ∈ K} does not meet R−1 [F ] (4A2F(h-vi)), and R[L] ∩ F = ∅, that is, R[L] ⊆ G. But L is just {φ : φ(i) ≤ ψ(i) for every i < n}, so R[L] = Aψ¹n , and

432 Notes

K-analytic spaces

179

γ < c(Aψ¹n ) ≤ c(G) < γ, which is absurd. X X Thus c(R[K]) ≥ γ. As γ is arbitrary and R[K] is compact, we have the result. 432X Basic exercises (a) Put 422Xf, 431Xb and 432D together to prove 432C. (b) Let X be a K-analytic Hausdorff space, and µ a measure on X which is outer regular with respect to the open sets. Show that µX = supK⊆X is compact µ∗ K. (Hint: use the argument of part (a) of the proof of 432D.) > (c) Let X be a K-analytic Hausdorff space, and µ a semi-finite topological measure on X. Show that if either µ is inner regular with respect to the closed sets or X is regular and µ is a τ -additive Borel measure, then µ is tight. (d) Use 422Gf, 432B and 416C to prove 432E. > (e) Suppose that X is a set and that S, T are Hausdorff topologies on X such that (X, T) is K-analytic and S ⊆ T. Let (Z, U, T, ν) be a Radon measure space and f : Z → X a function which is almost continuous for U and S. Show that f is almost continuous for U and T. (Hint: it is enough to consider totally finite ν; show that νf −1 is T-Radon, so is inner regular for {K : TK = SK }, writing TK for the subspace topology induced by T on K.) (f ) Let X be a topological space and µ a locally finite measure on X which is inner regular with respect to the closed sets. Show that µ∗ is a capacity. (g) Let X be a topological space and F a closed subset of X. Define c : PX → {0, 1} by setting c(A) = 1 if A meets F , 0 otherwise. Show that c is a capacity on X. (h) Let X and Y be Hausdorff spaces, and R ⊆ X × Y an usco-compact relation. Show that if c is a capacity on Y , then A 7→ c(R[A]) is a capacity on X. (i) Use 432J and 432Xf to shorten the proof of 432D. 432Y Further exercises (a) Show that there are a K-analytic Hausdorff space X and a probability measure µ on X such that (i) µ is inner regular with respect to the Borel sets (ii) the domain of µ includes a base for the topology of X (iii) every compact subset of X is negligible. Show that there is no extension of µ to a topological measure on X. 432 Notes and comments The measure-theoretic properties of K-analytic spaces can largely be summarised in the slogan ‘K-analytic spaces have lots of compact sets’. I said above that it is sometimes helpful to think of K-analytic spaces as an amalgam of compact Hausdorff spaces and Souslin-F subsets of R. For the former, it is obvious that they have many compact subsets; for the latter, it is not obvious, but is of course one of their fundamental properties, deducible from 422De. 432B and part (a) of the proof of 432D are typical manifestations of the phenomenon. The real point of these theorems is that we can extend a Borel or Baire measure to a Radon measure with no assumption of τ -additivity (432F). A Radon measure must be τ -additive just because it is tight. A (locally finite) Borel or Baire measure must be τ -additive whenever the measurable open sets are K-analytic. The condition ‘every open set is K-analytic’ in 432C is of course a very strong one in the context of compact Hausdorff spaces (422Xe). But for analytic spaces it is automatically satisfied (423E), and that is the side on which the principal applications of 432C appear. The results which I call corollaries of 432D can mostly be proved by more direct methods (see 432Xd), but the line I choose here seems to be the most powerful technique. Indeed it can be used to deal with 432C as well (432Xa). In §434 I will discuss ‘universally measurable’ sets in topological spaces. In fact K-analytic sets are universally measurable in a particularly strong sense (432A). The point here is that K-analyticity is intrinsic;

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432 Notes

a K-analytic space is measurable whenever embedded as a subspace of a (complete locally determined) topological measure space. The theorems here touch on two phenomena of particular importance. First, in 432G we have an example of ‘pulling back’ a measure, that is, we have a measure ν on a set Y and a function f : X → Y and seek a Radon measure µ on X such that f is inverse-measure-preserving, or, even better, such that ν = µf −1 . There was a similar result in 418L. In both cases we have to suppose that f is continuous and (in effect) that ν is a Radon measure. (This is not part of the hypotheses of 432G, but of course it is an easy consequence of them, using 432B.) In 418L, we need a special hypothesis to ensure that there are enough compact subsets of X to carry an appropriate Radon measure; in 432G, this is an automatic result of assuming that X is K-analytic. Both 418L and 432G can be regarded as consequences of Henry’s theorem (416M). The difficulty arises from the requirement that µ should be a Radon measure; if we do not insist on this there is a much simpler solution, since we need suppose only that f [X] is of full outer measure (132Yd). The next theme I wish to mention is a related one, the investigation of comparable topologies. If S and T are (Hausdorff) topologies on a set X, and S is coarser than T (so that (X, S) is a continuous image of (X, T)), then 418I tells us that any totally finite T-Radon measure is S-Radon. We very much want to know when the reverse is true, so that the (totally finite) Radon measures for the two topologies are the same. 432H provides one of the important cases in which this occurs. The hypothesis ‘(X, T) is K-analytic’ generalizes the alternative ‘(X, T) is compact’; in the latter case, S = T, so that the result is, from our point of view here, trivial. (But from the point of view of elementary general topology, of course, it is one of the pivots of the theory of compact Hausdorff spaces.) In a similar vein we have a variety of important topological consequences of the same hypotheses (422Ye, 423Fb). The paragraphs 432I-432J may appear to be no more that a minor extension of ideas already set out. I ought therefore to say plainly that the topological and measure theory of K-analytic spaces have co-evolved with the notion of capacity, and that 432J (‘K-analytic spaces are capacitable’) is one of the cornerstones of a theory of which I am giving only a minuscule part. For a idea of the vitality and scope of this theory, see Dellacherie 80.

433 Analytic spaces We come now to the special properties of measures on ‘analytic’ spaces, that is, continuous images of N N , as described in §423. I start with a couple of facts about spaces with countable networks. 433A Proposition Let (X, T) be a topological space with a countable network, and µ a localizable topological measure on X which is inner regular with respect to the Borel sets. Then µ has countable Maharam type. proof Let µ ˜ be the c.l.d. version of µ (213E). Then the measure algebra A of µ ˜ can be identified with the measure algebra of µ (322D(b-iii)). Also µ ˜ is complete, locally determined and localizable, so every subset ˜ be the domain of µ of X has a measurable envelope with respect to µ ˜ (213L). Let Σ ˜, and E a countable ˜ network for T. For each E ∈ E, let FE ∈ Σ be a measurable envelope of E. ˜ F • ∈ B}. Let B be the order-closed subalgebra of A generated by {FE• : E ∈ E}, and set T = {F : F ∈ Σ, ˜ Because B is an order-closed subalgebra P If G ∈ T, set S of A, T is a σ-subalgebra of Σ. Now T ⊆ T. P E0 = {E : E ∈ E, E ⊆ G}. Set F = E∈E0 FE , so that F ∈ T and G ⊆ F . For each E ∈ E0 , FE \ G is negligible, so F \ G is negligible, and G• = F • ∈ B, so G ∈ T. Q Q It follows that T includes the Borel σ-algebra of X. Because µ is inner regular with respect to the Borel sets, B is order-dense in A, and B = A. Thus the countable set {FE• : E ∈ E} τ -generates A, and the Maharam type of A, which is the Maharam type of µ, is countable. 433B Lemma If (X, T) is a Hausdorff space with a countable network, then any topological measure on X is countably separated in the sense of 343D. proof By 4A2Nf, there is a countable family of open sets separating the points of X.

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433C Theorem Let X be an analytic Hausdorff space, and µ a Borel measure on X. (a) If µ is semi-finite, it is tight. (b) If µ is locally finite, its completion is a Radon measure on X. proof X is K-analytic (423C); moreover, every open subset of X is again analytic (423Eb). So 432C gives the result at once. Remark Compare 256C. 433D Theorem Let X and Y be analytic Hausdorff spaces, ν a totally finite Radon measure on Y and f : X → Y a Borel measurable function such that f [X] is of full outer measure in Y . Then there is a Radon measure µ on X such that ν = µf −1 . proof By 423Ga, the graph R of f is an analytic set in X × Y , therefore K-analytic. Set π1 (x, y) = x, π2 (x, y) = y for (x, y) ∈ R, so that π1 and π2 are continuous. Now π2 [R] = f [X] is of full outer measure, so by 432G there is a Radon measure λ on R such that ν = λπ2−1 . Next, because π1 is continuous, the image µ = λπ1−1 is a Radon measure on X, by 418I. But π2 = f π1 , so µf −1 = (λπ1−1 )f −1 = λ(f π1 )−1 = λπ2−1 = ν, as required. 433E Proposition Let (X, Σ, µ) be a semi-finite measure space and T a topology on X such that µ is inner regular with respect to the closed sets. Let (Y, S) be an analytic space and f : X → Y a measurable function. Then f is almost continuous. proof Take E ∈ Σ and γ < µE. Then there is an F ∈ Σ such that F ⊆ E and γ < µF < ∞. For Borel sets H ⊆ Y , set νH = µ(F ∩ f −1 [H]). Then ν is a totally finite Borel measure on Y , so is tight (that is, inner regular with respect to the compact sets) (433C); let K ⊆ Y be a compact set such that νK > γ, so that µ(F ∩ f −1 [K]) > γ. The subspace measure on L = F ∩ f −1 [K] is still inner regular with respect to the (relatively) closed sets (412Pc), and f ¹L is still measurable; but f ¹L is a function from L to K, and K is metrizable, by 423Dc. So f ¹L is almost continuous, by 418J, and there is a set F 0 ⊆ L, of measure at least γ, such that f ¹F 0 is continuous. As E and γ are arbitrary, f is almost continuous. Remark Compare 418Yg. 433F I give two simple corollaries of the von Neumann-Jankow selection theorem (423N-423O). Proposition Let (X, T) and (Y, S) be analytic Hausdorff spaces, and f : X → Y a Borel measurable surjection. Let ν be a complete locally determined topological measure on Y , and T its domain. Then there is a T-measurable function g : Y → X such that gf is the identity on X. proof By 423O we know that there is a function g : Y → X such that gf is the identity and g is T1 measurable, where T1 is the σ-algebra generated by the Souslin-F subsets of Y . But T contains every Souslin-F subset of Y , by 431B, therefore includes T1 , and g is actually T-measurable. 433G Proposition Let (X, T) be an analytic Hausdorff space, (Y, T, ν) a complete locally determined measure space, and f : X → Y a surjection. Suppose that there is some countable family F ⊆ T such that F separates the points of Y (that is, whenever y, y 0 are distinct points of Y there is a member of F containing one and not the other) and f −1 [F ] is a Borel subset of X for every F ∈ F . Then there is a T-measurable function g : Y → X such that f g is the identity on Y . proof Set A = F ∪ {Y \ F : F ∈ F}. The topology T1 on X generated by T ∪ {f −1 [A] : A ∈ A} is still analytic (423H). If we take S to be the topology on Y generated by A, then S is Hausdorff and f is (T1 , S)-continuous, so S is analytic (423Bb). Because S is generated by a countable subset A of T, it is second-countable, and S ⊆ T (4A3Da/4A3E). So ν is a topological measure with respect to S. By 433F, there is a function g : Y → X, measurable for T and the topology T1 , such that gf is the identity on X; and of course g is still measurable for T and the coarser original topology T on X.

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433H Because analytic spaces have countable networks (423C), and their compact subsets are therefore metrizable (4A2Nh), their measure theory is very close to that of R or [0, 1] or {0, 1}N . I give some simple manifestations of this principle. Proposition Let hXi ii∈I be a family of analytic spaces, and forQeach i ∈ I let µi be a Radon probability measure on Xi . Let λ be the ordinary product measure on X = i∈I Xi , as defined in §254. (a) If I is countable then λ is a Radon measure. (b) If every µi is strictly positive, then λ is a quasi-Radon measure. proof (a) In this case, X is analytic (423Bc), therefore hereditarily Lindel¨of (423Da). Let Λ be the domain of λ and T the topology of X. Then Λ ∩ T is a base for T; by 4A3Da, T ⊆ Λ. By 417Sb, λ is the τ -additive product measure on X; by 417Q, this is a Radon measure. Q (b) By (a), the ordinary product measure on i∈J Xi is a Radon measure for every finite set J ⊆ I. So 417Sc tells us that λ is the τ -additive product measure on X; by 417O, this is a quasi-Radon measure. 433I Proposition Let X be an analytic Hausdorff space, and T a countably generated σ-subalgebra of the Borel σ-algebra B(X) of X. Then any locally finite measure with domain T has an extension to a Radon measure on X. proof Let µ0 be a locally finite measure with domain T. (a) Consider first the case in which µ0 is totally finite. Let hFn in∈N be a sequence in T generating T as σ-algebra. Define f : X → {0, 1}N by setting f (x)(n) = χFn (x) for n ∈ N, x ∈ X. Then f is T-measurable (use 418Bd), so we have a Borel measure ν0 on {0, 1}N defined by setting ν0 E = µ0 f −1 [E] for every Borel set E ⊆ {0, 1}N . Now the completion ν of ν0 is a Radon measure (433C). Also f [X] must be analytic, by 423Gb, because f is B(X)-measurable. So ν measures f [X] (432A), and νf [X] = ν0∗ f [X] = ν0 {0, 1}N , that is, f [X] is ν-conegligible. By 433D, there is a Radon measure µ on X such that ν = µf −1 . Because every Fn is expressible as f −1 [E] for some E ∈ B({0, 1}N ), so is every member of T. If F ∈ T, take H ∈ B({0, 1}N ) such that F = f −1 [H]; then µF = νH = ν0 H = µ0 F . Thus µ extends µ0 and µ¹Σ will serve. (b) In general, because X is Lindel¨of and µ0 is locally finite, µ0 is σ-finite. Let hXn in∈N be a partition (n) of X into members of T such that µ0 Xn is finite for every n, and set µ0 F = µ0 (F ∩ Xn ) for every n and P∞ (n) every F ∈ T; then every µ0 has an extension to a Radon measure µ(n) , and we can set µF = n=0 µ(n) F for every F ∈ Σ. Because µ0 is locally finite, so is µ, and it is now easy to check that µ is complete, locally determined and tight, just because every µ(n) is. 433J I turn now to a brief mention of ‘standard Borel spaces’. From the point of view of this chapter, it is natural to regard the following results as simple corollaries of theorems about Polish spaces. But, as remarked in §424, there are cases in which a standard Borel space is presented without any specific topology being attached; and in any case it is interesting to look at the ways in which we can express these ideas as theorems about σ-algebras rather than about topological spaces. Proposition Let (X, Σ) be a standard Borel space and T a countably generated σ-subalgebra of Σ. Then any σ-finite measure with domain T has an extension to Σ. proof Let µ0 be a σ-finite measure with domain T. (a) If µ0 is totally finite, give X a Polish topology for which Σ is the Borel σ-algebra of X, and use 433I. (b) In general, let hXn in∈N be a partition of X into members of T such that µ0 Xn < ∞ for every n, and (n) (n) set µ0 F = µ0 (F ∩ Xn ) for every n and every F ∈ T; then every µ0 has an extension to a measure µ(n) P∞ with domain Σ, and we can set µF = n=0 µ(n) F for every F ∈ Σ.

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433K Proposition Let h(Xn , Σn , µn )in∈N be a sequence of probability spaces such that (Xn , Σn ) is a standard Borel space for every n. Suppose that for each n ∈ N we are given an inverse-measure-preserving function fn : Xn+1 → Xn . Then we can find a standard Borel space (X, Σ), a probability measure µ with domain Σ, and inverse-measure-preserving functions gn : X → Xn such that fn gn+1 = gn for every n. proof For each n, choose a Polish topology Tn on Xn such that Σn is the algebra of Tn -Borel sets. Let µ ˆn be the completion of µn ; then µ ˆn is a Radon measure (433C). Every fn is inverse-measure-preserving for µ ˆn+1 and µ ˆn , by 235Hc, and almost continuous, by 418J. By 418Q, we have a Radon measure µ ˆ on Q X = {x : x ∈ n∈N Xn , fn (x(n + 1)) = x(n) for every n ∈ N} such that the continuous maps xQ7→ x(n) = gn (x) : X → Xn are inverse-measure-preserving for every n. Now X is a Borel subset of Z = n∈N Xn . P P For each n ∈ N, let Un be a countable base for Tn . Then S S Z \ X = n∈N U,V ∈Un ,U ∩V =∅ {z : z(n) ∈ U, fn (z(n + 1)) ∈ V } is a countable union of Borel sets because {z : z(n) ∈ U } is open and {z : z(n + 1) ∈ fn−1 [V ]} is Borel whenever n ∈ N and U , V ∈ Un . So Z \ X and X are Borel sets. Q Q Accordingly (X, Σ) is a standard Borel space, where Σ is the Borel σ-algebra of X (424G). So if we take µ=µ ˆ¹Σ, we shall have a suitable measure on X. 433X Basic exercises (a) Let (X, T) be a topological space with a countable network, and µ a topological measure on X which is inner regular with respect to the Borel sets and has the measurable envelope property (213Xl). Show that µ has countable Maharam type. (b) Show that an effectively locally finite measure on a hereditarily Lindel¨of space (in particular, on any analytic space) is σ-finite. (c) Let X ⊆ [0, 1] be a set of Lebesgue outer measure 1 and inner measure 0. Show that the subspace measure on X is a totally finite Borel measure which is not tight. (d) Let X be a Hausdorff space and µ a locally finite measure on X, inner regular with respect to the Borel sets, such that dom µ includes a base for the topology of X. Suppose that Y ⊆ X is an analytic set of full outer measure. Show that µ has a unique extension to a Radon measure µ ˜ on X, and that Y is µ ˜-conegligible. (e) Let (X, Σ) be a standard Borel space. (i) Show that any semi-finite measure with domain Σ is a compact measure (definition: 342Ac), therefore perfect. (Hint: if X is given a suitable topology, the measure is tight.) (ii) Show that any measure with domain Σ is countably separated. (f ) Let (X, Σ, µ) and (Y, T, ν) be atomless probability spaces such that (X, Σ) and (Y, T) are standard Borel spaces. Show that (X, Σ, µ) and (Y, T, ν) are isomorphic. (Hint: by 344I, their completions are isomorphic; by 344H, they have negligible sets of cardinal c; show that any isomorphism between the completions is (Σ, T)-measurable on a conegligible set; use 424Da to match residual negligible sets.) (g) Let X be [0, 1] × {0, 1}, with its usual topology, and I k the split interval (419L); define f : X → I k by setting f (t, 0) = t− , f (t, 1) = t+ for t ∈ [0, 1]. (i) Give I k its usual Radon measure ν (343J, 419Lc). Show that there is no Radon measure λ on X such that ν = λf −1 . (ii) Let µ be the product Radon probability measure on X, starting from Lebesgue measure on [0, 1] and the usual fair-coin measure on {0, 1}. Show that f is inverse-measure-preserving for µ and ν. Show that f is not almost continuous. 433Y Further exercises (a) Find a Hausdorff topological space X with a countable network and a semi-finite Borel measure on X which does not have countable Maharam type. (b) Let X be an analytic Hausdorff space and µ an atomless Radon measure on X. Show that (X, µ) is isomorphic to Lebesgue measure on some measurable subset of R. (Hint: 344I.)

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(c) Let (X, T) be a Polish space without isolated points, and µ a σ-finite topological measure on X. Show that there is a conegligible meager set. (Hint: Show that every non-empty open set is uncountable. Find a countable dense negligible set and a negligible Gδ set including it.) (d) Let hXn in∈N be a sequence of analytic spaces and for each n ∈ N let µn be a Borel probability measure on Xn . Suppose that for each n ∈ N we are given an inverse-measure-preserving function fn : Xn+1 → Xn . Show that we can find a standard Borel space (X, Σ), a probability measure µ with domain Σ, and inversemeasure-preserving functions gn : X → Xn such that fn gn+1 = gn for every n. 433 Notes and comments The measure-theoretic results of 433C-433E are of much the same type as those in §432. A characteristic difference is that Borel measurable functions between analytic spaces behave in many ways like continuous functions. (Compare 433D and 432G.) You may feel that 423Yc offers some explanation of this. For any question which refers to the Borel algebra of an analytic space X, or to the class of its analytic subsets, we can expect to be able to suppose that X is separable and metrizable (see 423Xd), and that any single Borel measurable function appearing is continuous. (424H is a particularly remarkable instance of this principle.) 433H here amounts to spelling out a special case of ideas already treated in 417S. As this territory is relatively unfamiliar, I give detailed examples (423Xi, 433Xc, 433Xg, 439A, 439K) to show that the theorems of this section are not generally valid for compact Hausdorff spaces (the archetype of K-analytic spaces which need not be analytic), nor for separable metric spaces (the archetypical spaces with countable network). They really do depend on the particular combination of properties possessed by analytic spaces. For large parts of probability theory, standard Borel spaces provide an adequate framework, and have a number of advantages; some of the technical problems concerning measurability which loom rather large in this treatise disappear in such contexts. Many authors accordingly give them great prominence. I myself believe that the simplifications are an entrapment rather than a liberation, that sooner or later everyone has to leave the comfortable environment of Borel algebras on Polish spaces, and that it is better to be properly equipped with a suitable general theory when one does. But it is surely important to know what the simplifications are, and the results in 433J-433K will I hope show at least that there are wonderful ideas here, even if my own presentation tends to leave them on one side. 434 Borel measures What one might call the fundamental question of topological measure theory is the following. What kinds of measures can arise on what kinds of topological space? Of course this question has inexhaustible ramifications, corresponding to all imaginable properties of measures and topologies and connexions between them. The challenge I face here is that of identifying particular ideas as being more important than others, and the chief difficulty lies in the bewildering variety of topological properties which have been studied, any of which may have implications for the measure theory of the spaces involved. In this section and the next I give a sample of what is known, necessarily biased and incomplete. I try however to include the results which are most often applied and enough others for the proofs to contain, between them, most of the non-trivial arguments which have been found effective in this area. In 434A I set out a crude classification of Borel measures on topological spaces. For compact Hausdorff spaces, at least, the first question is whether they carry Borel measures which are not, in effect, Radon measures; this leads us to the definition of ‘Radon’ space (434C) which is also of interest in the context of general Hausdorff spaces. I give a brief account of the properties of Radon spaces (434F, 434N). I look also at two special topics: ‘quasi-dyadic’ spaces (434O-434Q) and a construction of Borel product measures by integration of sections (434R). In the study of Radon spaces we find ourselves looking at ‘universally measurable’ subsets of topological spaces (434D). These are interesting in themselves, and also interact with constructions from earlier parts of this treatise (434S-434T). Three further classes of topological space, defined in terms of the types of topological measure which they carry, are the ‘Borel-measure-compact’, ‘Borel-measure-complete’ and ‘preRadon’ spaces; I discuss them briefly in 434G-434J. They provide useful methods for deciding whether Hausdorff spaces are Radon (434K).

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434A Types of Borel measures In §411 I introduced the following properties which a Borel measure may or may not have: (i) inner regularity with respect to closed sets; (ii) inner regularity with respect to zero sets; (iii) tightness (that is, inner regularity with respect to closed compact sets); (iv) τ -additivity. These are of course interrelated. (ii)⇒(i) just because zero sets are closed, and (iii)⇒(iv) by 411E; in a Hausdorff space, (iii)⇒(i); and for an effectively locally finite measure on a regular topological space, (iv)⇒(i) (414Mb). On a regular Hausdorff space, therefore, we can divide totally finite Borel measures into four classes: (A) measures which are not inner regular with respect to the closed sets; (B) measures which are inner regular with respect to the closed sets, but not τ -additive nor tight; (C) measures which are τ -additive and inner regular with respect to the closed sets, but not inner regular with respect to the compact sets; (D) tight measures; and each of the classes (B)-(D) can be further subdivided into those which are completion regular (B1 , C1 , D1 ) and those which are not (B0 , C0 , D0 ). Examples may be found in 434Xf (type A), 411Q and 439K (type B0 ), 439J (type B1 ), 415Xc and 434Xa (type C1 ) and 434Xb (type D0 ), while Lebesgue measure itself is of type D1 , and any direct sum of spaces of types D0 and C1 will have type C0 . (The space in 439J depends for its construction on supposing that there is a cardinal which is not measure-free. It seems that no convincing example of a space of class B1 , that is, a completion regular, non-τ -additive Borel probability measure on a completely regular Hausdorff space, is known which does not depend on some special axiom beyond ordinary ZFC. For one of the obstacles to finding such a space, see 434Q below.) Note that a totally finite Borel measure µ on a regular Hausdorff space can be extended to a quasi-Radon measure iff µ is of class C or D (415M), and that in this case the quasi-Radon measure must be just the completion µ ˆ of µ. µ ˆ will be of the same type, on the classification here, as µ; in particular, µ ˆ will be a Radon measure iff µ is of class D (416F). 434B Compact, analytic and K-analytic spaces For any class of topological spaces, we can enquire which of the seven types of measure described above can be realized by measures on spaces of that class. The enquiry is limited only by our enterprise and diligence in seeking out new classes of topological space. For the spaces studied in §§432-433, however, we have something worth repeating here. On a K-analytic Hausdorff space, a semi-finite Borel measure which is inner regular with respect to the closed sets is tight (432B, 432D); consequently classes B and C of 434A cannot appear, and we are left with only the types A, D0 and D1 , all of which appear on compact Hausdorff spaces (434Xb, 434Xf). On an analytic Hausdorff space we have further simplifications: every semi-finite Borel measure is tight (433C), and (if X is regular) every closed set is a zero set (423Db). Thus on an analytic regular Hausdorff space only type D1 , of the seven types in 434A, can appear. (If the topology is not regular, we may also get measures of type D0 ; see 434Ya.) 434C Radon spaces: Definition For K-analytic Hausdorff spaces, therefore, we have a large gap between the ‘bad’ measures of class A and the ‘good’ measures of class D; furthermore, we have an important class of spaces in which type A cannot appear. It is natural to enquire further into the spaces in which every (totally finite) Borel measure is of class D, and (given that no exact description can be found) we are led, as usual, to a definition. A Hausdorff space X is Radon if every totally finite Borel measure on X is tight. 434D Universally measurable sets Before going farther with the study of Radon spaces it will be useful to spend a couple of paragraphs on the following concept. Let X be a topological space. (a) I will say that a subset E of X is universally measurable (in X) if it is measured by the completion of every totally finite Borel measure on X; that is, for every totally finite Borel measure µ on X there is a Borel set F ⊆ X such that E4F is µ-negligible.

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(b) A subset of X is universally measurable iff it is measured by every complete locally determined topological measure on X. P P (i) Suppose that A ⊆ X is universally measurable and that µ is a complete locally determined topological measure on X. Let F ⊆ X be such that µF is defined and finite. Then we have a totally finite Borel measure ν on X defined by setting νE = µ(F ∩ E) for every Borel set E ⊆ X. Now there are Borel sets E, B ⊆ X such that A4E ⊆ B and νB = 0. In this case, (A∩F )4(E ∩F ) ⊆ B ∩F and µ(B ∩ F ) = 0, so that (because µ is complete) A ∩ F is measured by µ. Because F is arbitrary and µ is locally determined, A is measured by µ. (ii) Suppose that A ⊆ X is measured by every complete locally determined topological measure on X. Then, in particular, it is measured by the completion of any totally finite Borel measure, so is universally measurable. Q Q (c) If X is a topological space then the family Σum of universally measurable subsets of X is a σ-algebra closed under Souslin’s operation and including the Borel σ-algebra. (For it is the intersection of the domains of a family of complete totally finite measures, and all these are σ-algebras including the Borel σ-algebra and closed under Souslin’s operation, by 431A.) In particular, Souslin-F sets are universally measurable, so (if X is Hausdorff) K-analytic and analytic sets are (422Ha, 423C). (d) Note that a function f : X → R is Σum -measurable iff it is µ-virtually measurable for every totally finite Borel measure µ on X (122Q, 212F). Generally, if Y is another topological space, I will say that f : X → Y is universally measurable if f −1 [H] ∈ Σum for every open set H ⊆ Y ; that is, if f is (Σum , B(Y ))-measurable, where B(Y ) is the Borel σ-algebra of Y . (Y )

(Y )

(e) In fact, if f : X → Y is universally measurable, then it is (Σum , Σum )-measurable, where Σum is the (Y ) algebra of universally measurable subsets of Y . P P Take F ∈ Σum and a totally finite Borel measure µ on X. If µ ˆ is the completion of µ, then the image measure ν = µ ˆf −1 is a complete totally finite topological −1 measure on Y , so measures F , and f [F ] ∈ dom µ ˆ. As µ is arbitrary, f −1 [F ] ∈ Σum ; as F is arbitrary, f (Y ) Q is (Σum , Σum )-measurable. Q (f ) It follows that if Z is a third topological space and f : X → Y , g : Y → Z are universally measurable, then gf : X → Z is universally measurable. 434E Universally Radon-measurable sets A companion idea is the following. (a) Let X be a Hausdorff space. I will say that a subset E of X is universally Radon-measurable if it is measured by every Radon measure on X. (b) If X is a Hausdorff space, the family ΣuRm of universally Radon-measurable subsets of X is a σalgebra closed under Souslin’s operation and including the algebra of universally measurable subsets of X (and, a fortiori, including the Borel σ-algebra). (Use 434Db and the idea of 434Dc.) (c) If X is a Hausdorff space, I will say that a function f from X to a topological space Y is universally Radon-measurable if f −1 [H] ∈ ΣuRm for every open set H ⊆ Y . A function f : X → R is ΣuRm measurable iff it is µ-virtually measurable for every totally finite tight Borel measure µ on X. (Compare 434Dd.) 434F Elementary properties of Radon spaces: Proposition Let X be a Hausdorff space. (a) The following are equiveridical: (i) X is a Radon space; (ii) every semi-finite Borel measure on X is tight; (iii) if µ is a locally finite Borel measure on X, its c.l.d. version µ ˜ is a Radon measure; (iv) whenever µ is a totally finite Borel measure on X, and G ⊆ X is an open set with µG > 0, then there is a compact set K ⊆ G such that µK > 0; (v) whenever µ is a non-zero totally finite Borel measure on X, there is a Radon subspace Y of X such that µ∗ Y > 0. (b) If Y ⊆ X is a subspace which is a Radon space in its induced topology, then Y is universally measurable in X.

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(c) If X is a Radon space and Y ⊆ X, then Y is Radon iff it is universally measurable in X iff it is universally Radon-measurable in X. In particular, all Borel subsets and all Souslin-F subsets of X are Radon spaces. (d) The family of Radon subspaces of X is closed under Souslin’s operation and set difference. proof (a)(i)⇒(ii) Let µ be a semi-finite Borel measure on X, E ⊆ X a Borel set and γ < µE. Because µ is semi-finite, there is a Borel set H of finite measure such that µ(E ∩ H) > γ. Set νF = µ(F ∩ H) for every Borel set F ⊆ X; then ν is a totally finite Borel measure on X, and νE > γ. Because X is a Radon space, there is a compact set K ⊆ E such that νK ≥ γ, and now µK ≥ γ. As µ, E and γ are arbitrary, (ii) is true. (ii)⇒(i) and (i)⇒(v) are trivial. (v)⇒(iv) Assume (v), and let µ be a totally finite Radon measure on X and G a non-negligible open set. Set νE = µ(E ∩ G) for every Borel set E ⊆ X. Then ν is a non-zero totally finite Borel measure on X, so there is a Radon subspace Y of X such that ν ∗ Y > 0. The subspace measure νY on Y is a Borel measure on Y , so is tight. Since νY (Y \ G) = ν(X \ G) = 0, νY (Y ∩ G) > 0 and there is a compact set K ⊆ Y ∩ G such that νY K > 0. Now µK > 0. As µ and G are arbitrary, (iv) is true. not-(i)⇒not-(iv) If X is not Radon, there is a totally finite Borel measure µ on X which is not tight. By 416F(iii), there is an open set G ⊆ X such that µG > supK⊆G is compact µK = γ say. Let K be the family of compact subsets of G. By 215B(v), S there is a non-decreasing sequence hKn in∈N in K such that µ(K \ F ) = 0 for every K ∈ K, where F = n∈N Kn . Observe that µF = limn→∞ µKn ≤ γ < µG. Now set νE = µ(E ∩ G \ F ) for every Borel set E ⊆ X. Then ν is a Borel measure on X, and νG > 0. If K ⊆ G is compact, then νK = µ(K \ F ) = 0. So ν and G witness that (iii) is false. (i)⇒(iii) The point is that µ ˜ is tight. P P If µ ˜E > γ, then, because µ ˜ is semi-finite, there is a set E 0 ⊆ E such that γ < µ ˜E 0 < ∞; now there is a Borel set H ⊆ E 0 such that µH = µ ˜E 0 (213Fc). Setting νF = µ(H ∩ F ) for every Borel set F , ν is a totally finite Borel measure on X and νH > γ, so there is a compact set K ⊆ H such that νK ≥ γ. Now γ ≤ µK < ∞, so µ ˜K = µK ≥ γ (213Fa), while K ⊆ E. As E and γ are arbitrary, µ ˜ is tight. Q Q On the other hand, every point of X belongs to an open set of finite measure for µ, which is still of finite measure for µ ˜ (213Fa again). So µ ˜ is locally finite; since it is surely complete and locally determined, it is a Radon measure. (iii)⇒(i) Assume (iii), and let µ be a totally finite Borel measure on X. Then its c.l.d. version µ ˜ is tight. But µ ˜ extends µ (213Hc), so µ also is tight. As µ is arbitrary, X is a Radon space. (b) Let µ be a totally finite Borel measure on X, and µ ˆ its completion; let ² > 0. Let µY be the subspace measure on Y , so that µY is a totally finite Borel measure on Y , and is tight. There is a compact set K ⊆ Y such that νK ≥ µY Y − ². But this means that µ∗ Y = µY Y ≤ µY K + ² = µ∗ (K ∩ Y ) + ² = µK + ² ≤ µ∗ Y + ². As ² is arbitrary, µ∗ Y = µ∗ Y , and Y is measured by µ ˆ (413Ef); as µ is arbitrary, Y is universally measurable. (c) (i) If Y is Radon, it is universally measurable, by (b). (ii) If Y is universally measurable, it is universally Radon-measurable, by 434Eb. (iii) Suppose that Y is universally Radon-measurable, and that ν is a totally finite Borel measure on Y . For Borel sets E ⊆ X, set µE = ν(E ∩ Y ). Then µ is a totally finite Borel measure on X, so its c.l.d. version µ ˜ is a Radon measure on X, by (a-iii). We are supposing that Y is universally Radon-measurable, so, in particular, it must be measured by µ ˜. We have

µ ˜(X \ Y ) =

sup K⊆X\Y is compact

(213Ha, because µ is totally finite)

µ ˜K =

sup K⊆X\Y is compact

µK

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Topologies and measures II

=

sup

434F

ν(K ∩ Y ) = 0,

K⊆X\Y is compact

and Y is µ ˜-conegligible. Now suppose that E ⊆ Y is a (relatively) Borel subset of Y . Then E is of the form F ∩ Y where F is a Borel subset of X, so that νE = µF = µ ˜F = µ ˜(Y ∩ F ) = µ ˜E =

sup K⊆E is compact

µK =

sup

νK.

K⊆E is compact

As E is arbitrary, ν is tight; as ν is arbitrary, Y is a Radon space. By 434Dc, it follows that all Borel subsets and all Souslin-F subsets of X are Radon spaces. (d) The first step is to note that if hEn in∈N is a sequence of Radon subspaces of X with union E, then E is Radon; this is immediate from (a-v) above. Now S let hEσ iσ∈S be a Souslin scheme, consisting of Radon subsets of X, with kernel A. We know that E = σ∈S Eσ is a Radon space. Every Eσ is universally measurable in E, by (b), so A also is (434Dc), and must be Radon, by (c). Thus the family of Radon subspaces of X is closed under Souslin’s operation. If E and F are Radon subsets of X, then E ∪ F is Radon, and, just as above, F is universally measurable in E ∪ F . But this means that E \ F = (E ∪ F ) \ F is universally measurable in E ∪ F , so that E \ F is Radon. 434G Just as we can address the question ‘when can we be sure that every Borel measure is of class D’ in terms of the definition of ‘Radon’ space (434C), we can form other classes of topological space by declaring that the Borel measures they support must be of certain kinds. Three definitions which lead to interesting patterns of ideas are the following. Definitions (a) A topological space X is Borel-measure-compact (Gardner & Pfeffer 84) if every totally finite Borel measure on X which is inner regular with respect to the closed sets is τ -additive, that is, X carries no measure of class B in the classification of 434A. (b) A topological space X is Borel-measure-complete (Gardner & Pfeffer 84) if every totally finite Borel measure on X is τ -additive. (If X is regular and Hausdorff, this amounts to saying that X carries no measures of classes A or B in the classification of 434A.) (c) A Hausdorff space X is pre-Radon (also called ‘hypo-radonian’, ‘semi-radonian’) if every τ additive totally finite Borel measure on X is tight. (If X is regular, this amounts to saying that X carries no measure of class C in the classification of 434A.) 434H Proposition Let X be a topological space and B its Borel σ-algebra. (a) The following are equiveridical: (i) X is Borel-measure-compact; (ii) every semi-finite Borel measure on X which is inner regular with respect to the closed sets is τ -additive; (iii) every effectively locally finite Borel measure on X which is inner regular with respect to the closed sets has an extension to a quasi-Radon measure; (iv) every totally finite Borel measure on X which is inner regular with respect to the closed sets has a support; (v) if µ is a non-zero totally finite Borel measure on X, inner regular with respect to the closed sets, and G is an open cover of X, then there is some G ∈ G such that µG > 0. (b) If X is Lindel¨of (in particular, if X is a K-analytic Hausdorff space), it is Borel-measure-compact. (c) If X is Borel-measure-compact and A ⊆ X is a Souslin-F set, then A is Borel-measure-compact in its subspace topology. In particular, any Baire subset of X is Borel-measure-compact. proof (a)(i)⇒(ii) Assume (i), and let µ be a semi-finite Borel measure on X which is inner regular with respect to the closed sets. Let G be an upwards-directed family of open sets with union G0 , and γ < µG0 .

434I

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189

Because µ is semi-finite, there is an H ∈ B such that µH < ∞ and µ(H ∩ G0 ) ≥ γ. Set νE = µ(E ∩ H) for every E ∈ B; then ν is a totally finite Borel measure on X. For any E ∈ B, νE = µ(E ∩ H) = sup{µF : F ⊆ E ∩ H is closed} ≤ sup{νF : F ⊆ E is closed}, so ν is inner regular with respect to the closed sets, and must be τ -additive. Now γ ≤ νG0 = supG∈G νG ≤ supG∈G µG. As γ and G is arbitrary, µ is τ -additive. (ii)⇒(iii) Assume (ii), and let µ be an effectively locally finite Borel measure on X which is inner regular with respect to the closed sets. Then it is semi-finite (411Gd), therefore τ -additive. By 415L, it has an extension to a quasi-Radon measure on X. (iii)⇒(i) If (i) is true and µ is a totally finite Borel measure on X which is inner regular with respect to the closed sets, then µ has an extension to a quasi-Radon measure, which is τ -additive, so µ is also τ -additive (411C). (i)⇒(iv) Use 411Nd. (iv)⇒(v) Suppose that (iv) is true, that µ is a non-zero totally finite Borel measure on X which is inner regular with respect to the closed sets, and that G is an open cover of X. If F is the support of µ, then µF > 0 so F 6= ∅; there must be some G ∈ G meeting F , and now µG > 0. not-(i)⇒not-(v) Suppose that there is a totally finite Borel measure µ on X, inner regular with respect to the closed sets, which is not τ -additive. Let G be an upwards-directed family of open sets such S that µG∗ > γ, where G∗ = G and γ = supG∈G S µG. Let hGn in∈N be a non-decreasing sequence in G such that µ(G \ G∗0 ) for every G ∈ G, where G∗0 = n∈N Gn (215B(v)). Then µG∗0 ≤ γ, so there is a closed set F ⊆ G∗ \ G∗0 such that µF > 0. Let ν be the Borel measure on X defined by setting µE = µ(E ∩ F ) for every E ∈ B. As in the argument for (i)⇒(ii), ν is inner regular with respect to the closed sets. Consider H = G ∪ {X \ F }; this is an open cover of X. If G ∈ G then νG ≤ µ(G \ G∗0 ) = 0, so νH = 0 for every H ∈ H; thus ν and H witness that (iv) is false. (b) Use (a-v) and 422Gg. (c) Let µ be a Borel measure on A which is inner regular with respect to the closed sets, that is to say, the relatively closed sets in A. Let ν be the corresponding Borel measure on X, defined by setting νE = µ(A ∩ E) for every E ∈ B. Let νˆ be the completion of ν. Putting 431D and 421M together, we see that νˆA = sup{ˆ ν F : F ⊆ A is closed in X}, that is, νX = sup{µF : F ⊆ A is closed in X}. But this means that if E ∈ B and γ < νE, there is a closed set F in X such that F ⊆ A such that µ(E ∩ F ) > γ; now there is a relatively closed set F 0 ⊆ A such that F 0 ⊆ E ∩ F and µF 0 ≥ γ, and as F 0 must be relatively closed in F it is closed in X, while νF 0 ≥ γ. Since E and γ are arbitrary, ν is inner regular with respect to the closed sets, and will be τ -additive. Now suppose that G is an upwards-directed family of relatively open subsets of A. Set H = {H : H ⊆ X is open, H ∩ A ∈ G}. Then H is upwards-directed, so S S µ( G) = ν( H) = supH∈H νH = supG∈G µG. As µ and G are arbitrary, A is Borel-measure-compact. By 421L, it follows that any Baire subset of X is Borel-measure-compact. 434I Proposition Let X be a topological space. (a) The following are equiveridical: (i) X is Borel-measure-complete; (ii) every semi-finite Borel measure on X is τ -additive; (iii) every totally finite Borel measure on X has a support; (iv) S whenever µ is a totally finite Borel measure on X there is a base U for the topology of X such that µ( {U : U ∈ U , µU = 0}) = 0.

190

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434I

(b) If X is regular, it is Borel-measure-complete iff every effectively locally finite Borel measure on X has an extension to a quasi-Radon measure. (c) If X is Borel-measure-complete, it is Borel-measure-compact. (d) If X is Borel-measure-complete, so is every subspace of X. (e) If X is hereditarily Lindel¨of (for instance, if X is separable and metrizable, see 4A2P(a-iii)), it is Borel-measure-complete, therefore Borel-measure-compact. proof (a)(i)⇒(ii) Use the argument of (i)⇒(ii) of 434Ha; this case is simpler, because we do not need to check that the auxiliary measure ν is inner regular. (ii)⇒(i) is trivial. (i)⇒(iv) If X is Borel-measure-complete and µ is a totally finite Borel measure on X, take U to be the family of all open subsets of X. S This is surely a base for the topology, and setting U0 = {U : U ∈ U , µU = 0}, U0 is upwards-directed so µ( U0 ) = supU ∈U0 µU = 0, as required. (iv)⇒(iii) Assume (iv), and let µ be a totally finite Borel measure on X. Take S a base U as in (iv), S so that µ( U0 ) = 0, where U0 is the family of negligible members of U . Set F = X \ U0 , so that F is a conegligible closed set. If G ⊆ X is an open set meeting F , there is a member U of U such that U ⊆ G and U ∩ F 6= ∅; now U ∈ / U so µ(G ∩ F ) = µG ≥ µU > 0. As G is arbitrary, F is self-supporting and is the support of µ. (iii)⇒(i) Assume (iii), and let µ be a totally finite Borel measure on X. Let G be an upwards-directed family of open sets with union G∗ . Set γ = supG∈G S µG. Let hGn in∈N be a non-decreasing sequence in G such that µ(G \ G∗0 ) for every G ∈ G, where G∗0 = n∈N Gn (215B(v)). Then µG∗0 ≤ γ. Let ν be the Borel measure on X defined by setting µE = µ(E ∩ G∗ \ G∗0 ) for every E ∈ B. Then ν has a support F say. Now νG = 0 for every G ∈ G, so F ∩ G = ∅ for every G ∈ G, and F ∩ G∗ = ∅; but this means that µ(G∗ \ G∗0 ) = νX = νF = 0. Accordingly µG∗ = γ. As µ and G are arbitrary, X is Borel-measure-complete. (b) If X is Borel-measure-complete and µ is an effectively locally finite Borel measure on X, then µ is τ -additive, by (a-ii), so extends to a quasi-Radon measure on X, by 415Cb. If effectively locally finite Borel measures on X extend to quasi-Radon measures, then any totally finite Borel measure is τ -additive, by 411C, and X is Borel-measure-complete. (c) Immediate from the definitions. (d) If Y ⊆ X and µ is a totally finite Borel measure on Y , let ν be the Borel measure on X defined by setting νE = µ(E ∩ Y ) for every Borel set E ⊆ X. It is easy to check that ν is τ -additive. So if G is an upwards-directed family of relatively open subsets of Y , and we set H = {H : H ⊆ X is open, H ∩ Y ∈ G}, we shall get S S µ( G) = ν( H) = supH∈H νH = supG∈G µG. As µ and G are arbitrary, Y is Borel-measure-complete. (e) If µ is a totally finite Borel measure on X and G is a non-empty upwards-directed family of open subsets of X with union G∗ , then there is a sequence hGn in∈N in G with union G∗ , by 4A2H(c-i). Because G is upwards-directed, there is a non-decreasing sequence hG0n in∈N in G such that G0n ⊇ Gn for every n ∈ N, so that µG∗ = limn→∞ µG0n ≤ supG∈G µG. As µ and G are arbitrary, X is Borel-measure-complete.

434J

Borel measures

191

434J Proposition Let X be a Hausdorff space. (a) The following are equiveridical: (i) X is pre-Radon; (ii) every effectively locally finite τ -additive Borel measure on X is tight; (iii) whenever µ is a non-zero totally finite τ -additive Borel measure on X, there is a compact set K ⊆ X such that µK > 0; (iv) whenever µ is a totally finite τ -additive Borel measure on X, µX = supK⊆X is compact µK; (v) whenever µ is a locally finite effectively locally finite τ -additive Borel measure on X, the c.l.d. version of µ is a Radon measure on X. (b) If X is pre-Radon, then every locally finite quasi-Radon measure on X is a Radon measure. (c) If X is regular and every totally finite quasi-Radon measure on X is a Radon measure, then X is pre-Radon. (d) If X is pre-Radon, then any universally Radon-measurable subspace (in particular, any Borel subset or Souslin-F subset) A of X is pre-Radon. (e) If A ⊆ X is pre-Radon in its subspace topology, it is universally Radon-measurable in X. (f) If X is K-analytic (for instance, if it is compact), it is pre-Radon. ˇ (g) If X is completely regular and Cech-complete (for instance, if it is locally compact (4A2Gj), or metrizable and complete under a metric inducing its topology (4A2Md)), it is pre-Radon. Q (h) If X = i∈I Xi where hXi ii∈I is a countable family of pre-Radon Hausdorff spaces, then X is pre-Radon. (i) If every point of X belongs to a pre-Radon open subset of X, then X is pre-Radon. proof (a)(i)⇒(ii) Suppose that X is pre-Radon, that µ is an effectively locally finite τ -additive Borel measure on X, that E ⊆ X is Borel, and that γ < µE. Because µ is semi-finite, there is a Borel set H ⊆ X of finite measure such that µ(H ∩ E) > γ. Set νF = µ(F ∩ H) for every Borel set F ⊆ X; then ν is a totally finite Borel measure on X, and is τ -additive, by 414Ea. Now νE > γ, so there is a compact set K ⊆ E such that γ ≤ νK ≤ µK. As E is arbitrary, µ is tight. (ii)⇒(iii) is trivial. (iii)⇒(iv) Assume (iii), and let µ be a totally finite τ -additive Borel measure on X. Let K be the family of compact subsets of X and set α = supK∈K µK. ?? Suppose, if possible, that µX > α. Let hKn in∈N S be a sequence in K such that supn∈N µKn = α, and set L = n∈N Kn ; then S µL = limn→∞ µ( i≤n Ki ) = α. Set νE = µ(E \ L) for every Borel set E ⊆ X. Then ν is a non-zero totally finite Borel measure on X, and is τ -additive, by 414E again. So there is a K ∈ K such that νK > 0. But now there is an n ∈ N such that νK + µKn > α, and in this case K ∪ Kn ∈ K and µ(K ∪ Kn ) = µK + µKn − µ(K ∩ Kn ) ≥ νK + µKn > α, which is impossible. X X So µX = α, as required. (iv)⇒(i) Assume (iv), and let µ be a totally finite τ -additive Borel measure on X. Suppose that E ⊆ X is Borel and that γ < µE. By (iii), there is a compact set K ⊆ X such that µK > µX − µE + γ, so that µ(E ∩ K) > γ. Consider the subspace measure µK on K. By 414K, this is τ -additive, so inner regular with respect to the closed subsets of K (414Mb). There is therefore a relatively closed subset F of K such that F ⊆ K ∩ E and µK F ≥ γ; but now F is a compact subset of E and µF ≥ γ. As E and γ are arbitrary, µ is tight. As µ is arbitrary, X is pre-Radon. (ii)⇒(v) Assume (ii), and let µ be a locally finite effectively locally finite τ -additive Borel measure on X. Then µ is tight, so by 416F(ii) its c.l.d. version is a Radon measure. (v)⇒(iv) Assume (v), and let µ be a totally finite τ -additive Borel measure on X. Then the c.l.d. version µ ˜ of µ is a Radon measure; but µ ˜ extends µ (213Hc), so supK⊆X

is compact

µK supK⊆X

is compact

µ ˜K = µ ˜X = µX.

(b) Let µ be a locally finite quasi-Radon measure on X. By (a-ii), µ is tight; by 416C, µ is a Radon measure.

192

Topologies and measures II

434J

(c) Let µ be a totally finite τ -additive Borel measure on X. Because X is regular, its c.l.d. version µ ˜ is a quasi-Radon measure (415C), therefore a Radon measure; but µ ˜ extends µ (213Hc again), so µ, like µ ˜, must be tight. As µ is arbitrary, X is pre-Radon. (d) A be a universally Radon-measurable subset of X, and µ a totally finite τ -additive Borel measure on A. Set νE = µ(E ∩ A) for every Borel set E ⊆ X; then ν is a totally finite τ -additive Borel measure on X. So its c.l.d. version (that is, its completion νˆ, by 213Ha) is a Radon measure on X, by (a-v). Now νˆ measures A, so µA = ν ∗ A = νˆA = sup{ˆ ν K : K ⊆ A is compact} = sup{µK : K ⊆ A is compact}. By (b-iv), A is pre-Radon. (e) Let µ be a totally finite Radon measure on X. Then the subspace measure µA is τ -additive (414K), so its restriction ν to the Borel σ-algebra of A is still τ -additive. Because A is pre-Radon, µ∗ A = µA A = νA = sup{νK : K ⊆ A is compact} = sup{µK : K ⊆ A is compact} = µ∗ A, and µ measures A (413Ef). As µ is arbitrary, A is universally Radon-measurable. (f ) Put 432B and (a-iv) together. (g) If we identify X with a Gδ set in a compact Hausdorff space Z, then Z is pre-Radon, by (f), so X is pre-Radon, by (d). (h) Let µ be a totally finite τ -additive Borel measure on X, and ² > 0. Let h²i ii∈I be a family of strictly P positive real numbers such that i∈I ²i ≤ ² (4A1P). For each i ∈ I, Borel set F ⊆ Xi set µi F = µπi−1 [F ], where πi (x) = x(i) for x ∈ X; because πi : X → Xi is continuous, µi is a totally finite τ -additive Borel measure on Xi . Because S set Ki ⊆ Xi such that µi (Xi \ Ki ) ≤ ²i , Q Xi is pre-Radon, we can find a compact by (a-iv). Now K = i∈I Ki is compact (3A3J), and X \ K ⊆ i∈I πi−1 [Xi \ Ki ], so P P µ(X \ K) ≤ i∈I µi (Xi \ Ki ) ≤ i∈I ²i ≤ ². As ² and µ are arbitrary, X satisfies the condition of (b-iv), and is pre-Radon. (i) Let G be a cover of X by pre-Radon open sets. Let µ be a non-zero totally finite τ -additive Borel S measure on X. Then µX = sup{µ( G0 ) : G0 ⊆ G is finite}, so there is some G ∈ G such that µG > 0. Now the subspace measure µG is a non-zero totally finite τ -additive Borel measure on G, so there is a compact set K ⊆ G such that µG K > 0, in which case µK > 0. As µ is arbitrary, X is pre-Radon, by (a-iii). 434K I return to criteria for deciding whether Hausdorff spaces are Radon. Proposition (a) A Hausdorff space is Radon iff it is Borel-measure-complete and pre-Radon. (b) An analytic Hausdorff space is Radon. In particular, any compact metrizable space is Radon and any Polish space is Radon. (c) ω1 and ω1 + 1, with their order topologies, are not Radon. (d) For a set I, [0, 1]I is Radon iff {0, 1}I is Radon iff I is countable. (e) A hereditarily Lindel¨of K-analytic Hausdorff space is Radon; in particular, the split interval (343J, 419L) is Radon. proof (a) Put the definitions 434C, 434Gb and 434Gc together, recalling that a tight measure is necessarily τ -additive (411E). (b) 433Cb. (c) Dieudonn´e’s measure (411Q) is a Borel measure on ω1 which is not tight, so ω1 is certainly not a Radon space; as it is an open set in ω1 + 1, and the subspace topology inherited from ω1 + 1 is the order topology of ω1 (4A2S(a-iii)), ω1 + 1 cannot be Radon (434Fc). (d) If I is countable, then {0, 1}I and [0, 1]I are compact metrizable spaces, so are Radon. If I is uncountable, then ω1 + 1, with its order topology, is homeomorphic to a closed subset of {0, 1}I . P P Set

434M

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193

κ = #(I). For ξ ≤ ω1 , η < κ set xξ (η) = 1 if η < ξ, 0 if ξ ≤ η. The map ξ 7→ xξ : ω1 + 1 → {0, 1}κ is injective because κ ≥ ω1 , and is continuous because all the sets {ξ : xξ (η) = 0} = (ω1 + 1) ∩ (η + 1) are open-and-closed in ω1 + 1. Since ω1 + 1 is compact in its order topology (4A2S(a-i)), it is homeomorphic to its image in {0, 1}κ ∼ Q = {0, 1}I . Q I By 434Fc, {0, 1} cannot be a Radon space. Since {0, 1}I is a closed subset of [0, 1]I , [0, 1]I is also not a Radon space. (e) Suppose that X is a hereditarily Lindel¨of K-analytic Hausdorff space. Then it is Borel-measurecomplete by 434Ie and pre-Radon by 434Jf, so by (a) here it is Radon. Since the split interval is compact and Hausdorff and hereditarily Lindel¨of (419La), it is a Radon space. 434L It is worth noting an elementary special property of metric spaces. Proposition If (X, ρ) is a metric space, then any quasi-Radon measure on X is inner regular with respect to the totally bounded subsets of X. proof Let µ be a quasi-Radon measure on X and Σ its domain. Suppose that E ∈ Σ and γ < µE. Then there is an open S set G of finite measure such that µ(E ∩ G) > γ; set δ = µ(E ∩ G) − γ. For n ∈ N, I ⊆ X set H(n, I) = x∈I {y : ρ(y, x) < 2−n }. Then {H(n, I) : I ∈ [X]<ω } is an upwards-directed family of open −n−1 sets covering X. δ. T Because µ is τ -additive, there is a finite set In ⊆ X such that µ(G \ H(n, In )) ≤ 2 Consider F = n∈N H(n, In ). This is totally bounded and µ(G \ F ) ≤ δ, so E ∩ F is totally bounded and µ(E ∩ F ) ≥ γ. As E and γ are arbitrary, µ is inner regular with respect to the totally bounded sets. 434M I turn next to a couple of ideas depending on countable compactness. Lemma Let X be a countably compact topological space and E a non-empty family of closed subsets of X with the finite intersection property. Then there is a Borel probability measure µ on X, inner regular with respect to the closed sets, such that µF = 1 for every F ∈ E. proof (a) By Zorn’s lemma, E is included in a maximal family E ∗ of closed subsets of X with the finite intersection property. (i) If F ⊆ X is closed and F ∩ F0 ∩ . . . ∩ Fn 6= ∅ for every F0 , . . . , Fn ∈ E ∗ , then E ∗ ∪ {F } has the finite intersection property, so F ∈ E ∗ . (ii) If F , F 0 ∈ E ∗ , then F ∩ F 0 ∩ F0 ∩ . . . ∩ Fn 6= ∅ for all F0 , . . . , Fn ∈ E ∗ , so F ∩ F 0 ∈ E ∗ . (iii) If F ⊆ X is closed and F ∩ F 0 ∈ E ∗ for every F 0 ∈ E ∗ , then F ∩ F0 ∩ . . . ∩ Fn ∈ E ∗ for every F0 , . . . , Fn ∈ E ∗ (because F0 ∩ . . . ∩ Fn ∈ E ∗ , by (ii)), so F ∈ E ∗ . T (iv) If hFn in∈N is a sequence in E ∗ , with intersection F , and F 0 ∈ E ∗ , then F 0 ∩ i≤n Fi 6= ∅ for every n ∈ N. Because X is countably compact, F 0 ∩ F 6= ∅ (4A2G(f-ii)). As F 0 is arbitrary, F ∈ E ∗ , by (iii). Thus E ∗ is closed under countable intersections. (b) Set Σ = {E : E ⊆ X, there is an F ∈ E ∗ such that either F ⊆ E or F ∩ E = ∅}, and define µ ˆ : Σ → {0, 1} by saying that µ ˆE = 1 if there is some F ∈ E ∗ such that F ⊆ E, 0 otherwise. Then µ ˆ is a probability measure on X. P P (i) ∅ ∈ Σ because E ∗ ⊇ E is not empty. (ii) X \ E ∈ Σ whenever E ∈ Σ because the definition of Σ is symmetric between E and X \ E. (iii) If hEn in∈N is any sequence in Σ with union E, then either there are n ∈ N, F ∈ E ∗ such that F ⊆T En ⊆ E and E ∈ Σ, or for every n ∈ N there is an Fn ∈ E ∗ such that Fn ∩ En = ∅. In this case F = n∈N Fn ∈ E ∗ , by (a-iv), and E ∩ F = ∅, so again E ∈ Σ. Thus Σ is a σ-algebra of subsets of X. (iv) µ ˆ∅ = 0 because ∅ cannot belong to E ∗ . (v) If hEn in∈N is any disjoint sequence in Σ with union E, then either there is some n such thatP µ ˆEn = 1, ∞ in which case µ ˆEi = 0 for every i 6= n (because any two membersP of E ∗ must meet) and µ ˆE = 1 = i=0 µ ˆ Ei , ∞ or µ ˆEi = 0 for every i, in which case, just as in (iii), µ ˆE = 0 = i=0 µ ˆEi . Thus µ ˆ is a measure. (vi) Because E ∗ 6= ∅, µ ˆX = 1. Thus µ ˆ is a probability measure. Q Q (c) If F ⊆ X is a closed set, then either F itself belongs to E ∗ , so F ∈ Σ, or there is some F 0 ∈ E ∗ such that F ∩ F 0 = ∅, in which case again F ∈ Σ. So Σ contains every closed set, therefore every Borel set, and µ ˆ

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434M

is a topological measure. By construction, µ ˆ is inner regular with respect to E ∗ and therefore with respect to the closed sets. Finally, if F ∈ E then F ∈ E ∗ , so µ ˆF = 1. We may therefore take µ to be the restriction of µ ˆ to the Borel σ-algebra of X, and µ will be a Borel measure on X, inner regular with respect to the closed sets, such that µE = 1 for every E ∈ E. 434N Proposition (a) Let X be a Borel-measure-compact topological space. Then closed countably compact subsets of X are compact. (b) Let X be a Borel-measure-complete topological space. Then closed countably compact subsets of X are compact. (c) Let X be a Hausdorff Borel-measure-complete topological space. Then compact subsets of X are countably tight. (d) In particular, any Radon compact Hausdorff space is countably tight. proof (a) Let C be a closed countably compact subset of X. Let F be an ultrafilter on C. Let E be the family of closed subsets of C belonging to F. Then E has the finite intersection property, so by 434M there is a Borel probability measure µ on C, inner regular with respect to the closed sets, such that µE = 1 for every E ∈ E. Let ν be the Borel measure on X defined by setting νH = µ(C ∩ H) for every Borel set H ⊆ X. Then ν is also inner regular with respect to the closed sets (of either C or X); because X is Borel-measure-compact, ν has a support F (434H(a-iv)). Since νF = νX = 1, F ∩ C 6= ∅; take x ∈ F ∩ C. If G is any open set (in X) containing x, then ν(X \ G) < 1, so C \ G ∈ / F and C ∩ G ∈ F. As G is arbitrary, F → x; as F is arbitrary, C is compact. (b) Repeat the argument of (a). Let C be a countably compact subset of X and F be an ultrafilter on C. Let E be the family of relatively closed subsets of C belonging to F. Then there is a Borel probability measure µ on C such that µE = 1 for every E ∈ E. Let ν be the Borel measure on X defined by setting νH = µ(C ∩ H) for every Borel set H ⊆ X. Because X is Borel-measure-complete, ν has a support F (434I(a-iii)). Since νF = νX = 1, F ∩ C 6= ∅; take x ∈ F ∩ C. If G is any open set containing x, then ν(X \ G) < 1, so C \ G ∈ / F and C ∩ G ∈ F. As G is arbitrary, F → x; as F is arbitrary, C is compact. S (c) Again let C be a (countably) compact subset of X. Take A ⊆ C, and set C0 = {B : B ∈ [A]≤ω }. Then C0 is countably compact. P P If hyn in∈N is any sequence in C0 , itShas a cluster point y ∈ C. For each n ∈ N there is a countable set Bn ⊆ A such that yn ∈ Bn . Now B = n∈N Bn is a countable subset of A, and y ∈ B ⊆ C0 , so y is a cluster point of hyn in∈N in C0 . As hyn in∈N is arbitrary, C0 is countably compact. Q Q By (b), C0 is compact, therefore closed, and must include A. Thus every point of A is in the closure of some countable subset of A. As A is arbitrary, C is countably tight. (d) Finally, a compact Radon Hausdorff space is Borel-measure-complete (434Ka) and countably compact, therefore countably tight. 434O Quasi-dyadic spaces I wish now to present a result in an entirely different direction. Measures of type B1 in the classification of 434A (completion regular, but not τ -additive) seem to be hard to come by. The next theorem shows that on a substantial class of spaces they cannot appear. First, we need a definition. Definition A topological space X is quasi-dyadic if it is expressible as a continuous image of a product of separable metrizable spaces. I give some elementary results to indicate what kind of spaces we have here. 434P Proposition (a) A continuous image of a quasi-dyadic space is quasi-dyadic. (b) Any product of quasi-dyadic spaces is quasi-dyadic. (c) A space with a countable network is quasi-dyadic. (d) A Baire subset of a quasi-dyadic space is quasi-dyadic. (e) If X is any topological space, a countable union of quasi-dyadic subspaces of X is quasi-dyadic. proof (a) Immediate from the definition.

434Q

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195

Q (b) Again immediate; if Xi is a continuous image of j∈Ji Yij , where Yij is a separable metric space for Q Q every i ∈ I, j ∈ Ji , then i∈I Xi is a continuous image of i∈I,j∈Ji Yij . (c) Let E be a countable network for the topology of X. On X let ∼ be the equivalence relation in which x ∼ y if they belong to just the same members of E; let Y be the space X/ ∼ of equivalence classes, and φ : X → Y the canonical map. Y has a separable metrizable topology with base {φ[E] : E ∈ E} ∪ {φ[X \ E] : E ∈ E}. Let I be any set such that #({0, 1}I ) ≥ #(X), and for each y ∈ Y let fy : {0, 1}I → y be a surjection. Then we have a continuous surjection f : Y × {0, 1}I → X given by saying that f (y, z) = fy (z) for y ∈ Y , z ∈ {0, 1}I . (d) Let hYi ii∈I be a family of separable metric spaces with product Y and f : Y → X a continuous surjection. If W ⊆ Y is a Baire subset Q set, it is determined by coordinates in a countable Q Qof I (4A3Nb), so can be regarded as W 0 × i∈I\J Yi , where J ⊆ I is countable and W 0 ⊆ i∈J Yi ; as i∈J Yi and W 0 are separable metrizable spaces (4A2Pa), W can be thought of as a product of separable metrizable spaces. Now the set {E : E ⊆ X, f −1 [E] is a Baire set in Y } is a σ-algebra containing every zero set in X, so contains every Baire set. Thus every Baire subset of X is a continuous image of a Baire subset of Y , and is therefore quasi-dyadic. Q S (e) If En ⊆ X is quasi-dyadic for each n ∈ N, then Z = N ×S n∈N En is quasi-dyadic, and f : Z → n∈N En is a continuous surjection, where f (n, hxi ii∈N ) = xn . So n∈N En is quasi-dyadic. 434Q Theorem(Fremlin & Grekas 95) A semi-finite completion regular topological measure on a quasi-dyadic space is τ -additive. proof ?? Suppose, if possible, otherwise. (a) The first step is the standard reduction to the case in which µX = 1 and X is covered by open sets of zero measure. In detail: suppose that X is a quasi-dyadic space and µ0 is a semi-finite completion regular topological measure on X which is not τ -additive. Let G be an upwards-directed family of open sets in X S such that µ0 ( G) is strictly greater than supG∈G µ S S0 G = γ say. Let hGn in∈N be a non-decreasing sequence in G such that limn→∞ µ0 Gn = γ, and set H0 = n∈N Gn , so that µ0 H0 = γ; take a closed set Z ⊆ G such that γ < µ0 Z < ∞. Set µ1 E = µ0 (E ∩ Z \ H0 ) for every Borel set E ⊆ X. Then µ1 is a non-zero totally finite Borel measure on X, and is completion regular. P P If E ⊆ X is a Borel set and ² > 0, there is a zero set F ⊆ E ∩ Z \ H0 such that µ0 F ≥ µ0 (E ∩ X \ H0 ) − ², and now µ1 F ≥ µ1 E − ². Q Q Note that µ1 (X \ Z) = µ1 G = 0 for every G ∈ G. For Borel sets E ⊆ X, set µE = µ1 E/µ1 X; then µ is a completion regular Borel probability measure on X, and G ∪ {X \ Z} is a cover of X by open negligible sets. (b) Now let hYi ii∈I Q be a family of separable metric spaces such that there is a continuous surjection f : Y → X, where Y = i∈I Yi . For each i ∈ I let Bi be a countable base for the topology of Yi . For J ⊆ I let C(J) be the family of all open cylinders in Y expressible in the form {s : s(i) ∈ Bi ∀ i ∈ K}, where K is a finite subset of J and Bi ∈ Bi for each i ∈ K; thus C(I) is a base for the topology of Y . Set C0 (J) = {U : U ∈ C(J), µ∗ f [U ] = 0} for each J ⊆ I. Note that (because every Bi is countable) C(J) and C0 (J) are countable for every countable subset J of I. It is easy to see that C(J) ∩ C(K) = C(J ∩ K) for all J, K ⊆ I, because if U ∈ C(I) is not empty it belongs to C(J) iff its projection onto Xi is the whole of Xi for every i ∈ / J. For each negligible set E ⊆ X, let hFn (E)in∈N be a family of zero sets, subsets of X \ E, such that supn∈N µFn = 1. Then each f −1 [Fn (E)] is a zero set in Y , so there is a countable set M (E) ⊆ I such that all the sets f −1 [Fn (E)] are determined by coordinates in M (E) (4A3Nc). Let J be the family of countable subsets J of I such that M (f [U ]) ⊆ J for every U ∈ C0 (J); then J is cofinal with [I]≤ω , that is, every countable subset of I is included in some member of J . P P If we start from any countable subset J0 of I and set S Jn+1 = Jn ∪ {M (f [U ]) : U ∈ C0 (Jn )} S for S each n ∈ N, S then every Jn is countable, and n∈N Jn ∈ J , because hJn in∈N is non-decreasing, so C0 ( n∈N Jn ) = n∈N C0 (Jn ). Q Q

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(c) For each J ∈ J , set QJ =

434Q

T S { n∈N Fn (f [U ]) : U ∈ C0 (J)}.

Then µQJ = 1 and f −1 [QJ ] is determined by coordinates in J, while f −1 [QJ ] ∩ U = ∅ whenever U ∈ C0 (J). If G ⊆ X is an open set, then G ∩ QJ = ∅ whenever J ∈ J and there is a negligible Baire set Q Q⊇G such that f −1 [Q] is determined by coordinates in J. P P Set H = πJ−1 [πJ [f −1 [G]]], where πJ : Y → i∈J Yi is the canonical map; then H is a union of members of C(J), because f −1 [G] is open in Y and πJ [f −1 [G]] Q is open in i∈J Yi . Also, because f −1 [Q] is determined by coordinates in J, H ⊆ f −1 [Q], so f [H] ⊆ Q and µ∗ f [H] = 0; thus all the members of C(J) included in H actually belong to C0 (J), and H ∩ f −1 [QJ ] = ∅. But this means that f −1 [G] ∩ f −1 [QJ ] = ∅ and (because f is a surjection) G ∩ QJ = ∅, as claimed. Q Q In particular, if G is a negligible open set in X, then G ∩ QJ = ∅ whenever J ∈ J and J ⊇ M (G). (d) If J ∈ J , there are s, s0 ∈ f −1 [QJ ] such that s¹J = s0 ¹J and f (s), f (s0 ) can be separated by open sets in X. P P Start from any x ∈ QJ and take a negligible open set G including x (recall that our hypothesis is that X is covered by negligible open sets). For each n ∈ N let hn : X → S R be a continuous function such that Fn (G) = h−1 [{0}]. We know that G ∩ Q = 6 ∅, while G ⊆ X \ ( J n n∈N Fn (G) ∩ QJ ), which is a S negligible Baire set; by (c), f −1 [X \ ( n∈N Fn (G) ∩ QJ )] is not determined by coordinates in J, and there must be some n such that f −1 [Fn (G) ∩ QJ ] is not determined by coordinates in J. Accordingly there must be s, s0 ∈ Y such that s¹J = s0 ¹J, s ∈ f −1 [Fn (G) ∩ QJ ] and s0 ∈ / f −1 [Fn (G) ∩ QJ ]. Now s ∈ f −1 [QJ ], 0 0 −1 which is determined by coordinates in J; since s¹J = s ¹J, s ∈ f [QJ ] and s0 ∈ / f −1 [Fn (G)]. Accordingly 0 0 hn (f (s)) = 0 6= hn (f (s )) and f (s), f (s ) can be separated by open sets. Q Q (e) We are now ready to embark on the central construction of the argument. We may choose inductively, for ordinals ξ < ω1 , sets Jξ ∈ J , negligible open sets Gξ , G0ξ ⊆ X, points sξ , s0ξ ∈ Y and sets Uξ , Vξ , Vξ0 ∈ C(I) such that Jη ⊆ Jξ , Uη , Vη , Vη0 all belong to C(Jξ ) and Gη ∩ QJξ = ∅ whenever η < ξ < ω1 (using the results of (b) and (c) to choose Jξ ); sξ ¹Jξ = s0ξ ¹Jξ , sξ ∈ f −1 [QJξ ] and f (sξ ) and f (s0ξ ) can be separated by open sets in X (using (d) to choose sξ , s0ξ ); Gξ , G0ξ are disjoint negligible open sets containing f (sξ ), f (s0ξ ) respectively (choosing Gξ , G0ξ ); Uξ ∈ C(Jξ ), Vξ , Vξ0 ∈ C(I \ Jξ ), sξ ∈ Uξ ∩ Vξ ⊆ f −1 [Gξ ], s0ξ ∈ Uξ ∩ Vξ0 ⊆ f −1 [G0ξ ] (choosing Uξ , Vξ , Vξ0 , using the fact that sξ ¹Jξ = s0ξ ¹Jξ ). On completing this construction, take for each ξ < ω1 a finite set Kξ ⊆ Jξ+1 such that Uξ , Vξ and Vξ0 all belong to C(Kξ ). By the ∆-system Lemma (4A1Db), there is an uncountable A ⊆ ω1 such that hKξ iξ∈A is a ˜ξ ∩ U 0 where U ˜ξ ∈ C(K), U 0 ∈ C(Kξ \ K). Then there ∆-system with root K say. For ξ ∈ A, express Uξ as U ξ ξ ˜ξ , so there is an uncountable B ⊆ A such that U ˜ξ is constant for are only countably many possibilities for U ˜ for the constant value. Let C ⊆ B be an uncountable set, not containing min A, such that ξ ∈ B; write U Kξ \ K does not meet Jη whenever ξ, η ∈ C and η < ξ (4A1Eb). Let D ⊆ C be such that D and C \ D are both uncountable. Note that K ⊆ Kη ⊆ Jξ whenever η, ξ ∈ A and η < ξ, so that K ⊆ Jξ for every ξ ∈ C. Consequently Uξ0 , Vξ and Vξ0 all belong to C(Kξ \ K) for every ξ ∈ C. (f ) Consider the open set G=

S ξ∈D

Gξ ⊆ X.

At this point the argument divides. ˜ ]) > 0. Then there is a Baire set Q ⊆ G such that µ∗ (Q ∩ f [U ˜ ]) > 0. case 1 Suppose µ∗ (G ∩ f [U Let J ⊆ I be a countable set such that f −1 [Q] is determined by coordinates in J. Let γ ∈ C \ D be so ˜ ] is not empty; take large that Kξ \ K does not meet J for any ξ ∈ A with ξ ≥ γ. Then Q ∩ QJγ ∩ f [U −1 ˜ s ∈ U ∩ f [Q ∩ QJγ ]. Because the Kξ \ K are disjoint from each other and from J ∪ Jγ for ξ ∈ D, ξ > γ, we may modify s to form s0 such that s0 ¹J ∪ Jγ = s¹J ∪ Jγ and s0 ∈ Uξ0 ∩ Vξ0 for every ξ ∈ D, ξ > γ; now ˜ (because K ⊆ Jγ ), so s0 ∈ U ˜ ∩ U 0 ∩ V 0 ⊆ f −1 [G0 ] and f (s0 ) ∈ s0 ∈ U / Gξ whenever ξ ∈ D, ξ > γ. On the ξ ξ ξ

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Borel measures

197

other hand, if ξ ∈ D and ξ < γ, Gξ ∩ QJγ = ∅, while s0 ∈ f −1 [QJγ ] (because f −1 [QJγ ] is determined by coordinates in Jγ ), so again f (s0 ) ∈ / Gξ . Thus f (s0 ) ∈ / G. But s0 ¹J = s¹J so f (s) ∈ Q ⊆ G; which is impossible. ˜ ]) > 0. This contradiction disposes of the possibility that µ∗ (G ∩ f [U ˜ ]) = 0. In this case there is a negligible Baire set Q ⊇ G ∩ f [U ˜ ]. Let case 2 Suppose that µ∗ (G ∩ f [U −1 J ⊆ I be a countable set such that f [Q] is determined by coordinates in J. Let ξ ∈ D be such that Kξ \ K does not meet J. Then ˜ ∩ U 0 ∩ Vξ ⊆ f −1 [Gξ ] ∩ U ˜ ⊆ f −1 [G ∩ f [U ˜ ]] ⊆ f −1 [Q], U ξ

˜ ⊆ f −1 [Q], because U 0 ∩ Vξ is a non-empty member of C(I \ J). But this means that µ∗ f [U ˜ ] = 0 and so U ξ ∗ −1 ∗ µ f [Uξ ] = 0. On the other hand, we have sξ ∈ Uξ ∩ f [QJξ ], so Uξ ∈ / C0 (Jξ ) and µ f [Uξ ] > 0. X X Thus this route also is blocked and we must abandon the original hypothesis that there is a quasi-dyadic space with a semi-finite completion regular topological measure which is not τ -additive. 434R There is a useful construction of Borel product measures which can be fitted in here. Proposition Let X and Y be topological spaces with Borel measures µ and ν; write B(X), B(Y ) for the Borel σ-algebras of X and Y respectively. If either X is first-countable or ν is τ -additive and effectively locally finite, there is a Borel measure λB on X × Y defined by the formula λB W = supF ∈B(Y ),νF <∞

R

ν(W [{x}] ∩ F )µ(dx)

for every Borel set W ⊆ X × Y . Moreover b (i) if µ is semi-finite, then λB agrees with the c.l.d. product measure λ on B(X)⊗B(Y ), and the c.l.d. ˜ version λB of λB extends λ; R (ii) if ν is σ-finite, then λB W = νW [{x}]µ(dx) for every Borel set W ⊆ X × Y ; (iii) if both µ and ν are τ -additive and effectively locally finite, then λB is just the restriction of the ˜ ( 417D, 417G) to the Borel σ-algebra of X × Y ; in particular, λB is τ -additive. τ -additive product measure λ proof (a) The point is that x 7→ ν(W [{x}] ∩ F ) is lower semi-continuous whenever W ⊆ X × Y is open and νF < ∞. P P Of course W [{x}] is always open, so ν always measures W [{x}] ∩ F . Take any α ∈ R and set G = {x : x ∈ X, ν(W [{x}] ∩ F ) > α}; let x0 ∈ G. α) Suppose that X is first-countable. Let hUn in∈N be a non-increasing sequence running over a (α base of open neighbourhoods of x0 . For each n ∈ N, set S Vn = {V : V ⊆ Y is open, Un × V ⊆ W }. Then hVn in∈N is a non-decreasing sequence with union W [{x}], so there is an n ∈ N such that ν(Vn ∩F ) > α. Now Vn ⊆ W [{x}] for every x ∈ Un , so Un ⊆ G. β ) Suppose that ν is τ -additive and effectively locally finite. Set (β V = {V : V ⊆ Y is open, U × V ⊆ W for some open set U containing x0 }. Then V is an upwards-directed family of open sets with union W [{x0 }], so there is a V ∈ V such that ν(V ∩ F ) > α (414Ea). Let U be an open set containing x0 such that U × V ⊆ W ; then V ⊆ W [{x}] for every x ∈ U , so U ⊆ G. (γγ ) Thus in either case we have an open set containing x0 and included in G. As x0 is arbitrary, G is open; as α is arbitrary, x 7→ ν(W [{x}] ∩ F ) is lower semi-continuous. Q Q (b) It follows that x 7→ ν(W [{x}] ∩ F ) is Borel measurable whenever W ⊆ X × Y is a Borel set and νF < ∞. P P Let W be the family of sets W ⊆ X × Y such that W [{x}] is a Borel set for every x ∈ X and x 7→ ν(W [{x}] ∩ F ) is Borel measurable. Then every S open subset of X × Y belongs to W (by (a) above), W \W 0 ∈ W whenever W , W 0 ∈ W and W 0 ⊆ W , and n∈N Wn ∈ W whenever hWn in∈N is a non-decreasing sequence in W. By the Monotone Class Theorem (136B), W includes the σ-algebra generated by the open sets, that is, the Borel σ-algebra of X × Y . Q Q R (c) It is now easy to check that W 7→ ν(W [{x}] ∩ F )µ(dx) is a Borel measure on X × Y whenever νF < ∞, and therefore that λB , as defined here, is a Borel measure.

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434R

b (d) Now suppose that µ is semi-finite, and that W ∈ B(X)⊗B(Y ). Then

λW =

sup

λ(W ∩ (E × F ))

µE<∞,νF <∞

(by the definition of ‘c.l.d. product measure’, 251F) Z = sup ν(W [{x}] ∩ F )µ(dx) µE<∞,νF <∞

E

(by Fubini’s theorem, 252C, applied to the product of the subspace measures µE and νF ) Z = sup ν(W [{x}] ∩ F )µ(dx) νF <∞

(by 213B, because µ is semi-finite) = λB W. (e) If, on the other hand, ν is σ-finite, let hFn in∈N be a non-decreasing sequence of sets of finite measure covering Y ; then Z Z λB W = sup ν(W [{x}] ∩ F )µ(dx) ≥ sup ν(W [{x}] ∩ Fn )µ(dx) νF <∞ n∈N Z Z = sup ν(W [{x}] ∩ Fn )µ(dx) = νW [{x}]µ(dx) ≥ λB W n∈N

for any Borel set W ⊆ X. (f ) If both µ and ν are τ -additive and effectively locally finite, so that we have a τ -additive product ˜ then Fubini’s theorem for such measures (417H) tells us that λB W = λW ˜ at least when W ⊆ measure λ, ˜ is finite. If W is any Borel subset of X × Y , then, as in (d), X × Y is a Borel set and λW Z λB W = sup ν(W [{x}] ∩ F )µ(dx) µE<∞,νF <∞

=

sup

E

˜ ˜ λ(W ∩ (E × F )) = λW

µE<∞,νF <∞

by 417C(iii). Remark The case in which X is first-countable is due to Johnson 82. *434S The concept of ‘universally measurable’ set enables us to extend a number of ideas from earlier sections. First, recall a problem from the very beginning of measure theory on the real line: the composition of Lebesgue measurable functions need not be Lebesgue measurable (134Ib), while the composition of a Borel measurable function with a Lebesgue measurable function is measurable (121Eg). In fact we can replace ‘Borel measurable’ by ‘universally measurable’, as follows. Proposition Let (X, Σ, µ) be a complete locally determined measure space, Y and Z topological spaces, f : X → Y a measurable function and g : Y → Z a universally measurable function. Then gf : X → Z is measurable. In particular, f −1 [F ] ∈ Σ for every universally measurable set F ⊆ Y . proof Let H ⊆ Z be an open set and E ∈ Σ a set of finite measure. Let µE be the subspace measure on E. Then the image measure ν = µE (f ¹E)−1 is a complete totally finite topological measure on Y , so its domain contains g −1 [H], and E ∩ (gf )−1 [H] = (f ¹E)−1 [g −1 [H]] ∈ dom µE ⊆ Σ. As E is arbitrary and µ is locally determined, (gf )−1 [H] ∈ Σ; as H is arbitrary, gf is measurable. Applying this to g = χF , we see that f −1 [F ] ∈ Σ for every universally measurable F ⊆ Y .

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*434T The next remark concerns the concept [[u ∈ E]] of §364. Proposition Let (A, µ ¯) be a localizable measure algebra. Write Σum for the algebra of universally measurable subsets of R. (a) For any u ∈ L0 = L0 (A), we have a sequentially order-continuous Boolean homomorphism E 7→ [[u ∈ E]] : Σum → A defined by saying that [[u ∈ E]] = sup{[[u ∈ F ]] : F ⊆ E is Borel} = inf{[[u ∈ F ]] : F ⊇ E is Borel} for every E ∈ Σum . ¯ (b) For any u ∈ L0 and universally measurable function h : R → R we have a corresponding element h(u) 0 of L defined by the formula ¯ [[h(u) ∈ E]] = [[u ∈ h−1 [E]]] for every E ∈ Σum , u ∈ L0 . proof We can regard (A, µ ¯) as the measure algebra of a complete strictly localizable measure space (X, Σ, µ) (322N), in which case L0 can be identified with L0 (µ) (364J). Write B for the Borel σ-algebra of R. (a) If u ∈ L0 , let f : X → R be a Σ-measurable function representing u. Then f −1 [E] ∈ Σ for every E ∈ Σum , by 434S. Setting φE = (f −1 [E])• , φ : Σum → A is a sequentially order-continuous Boolean homomorphism. Now φE = sup{[[u ∈ F ]] : F ∈ B, F ⊆ E} = inf{[[u ∈ F ]] : F ∈ B, F ⊇ E} for every E ∈ Σum . P P If H ∈ Σ and µH < ∞, then (writing µH for the subspace measure on H) the image measure µH (f ¹H)−1 is a complete topological measure, and its restriction ν to the Borel σ-algebra B of R is a totally finite Borel measure. Now E is measured by the completion of ν, so there are F1 , F2 ∈ B such that F1 ⊆ E ⊆ F2 and ν(F2 \ F1 ) = 0. In this case, [[u ∈ F1 ]] = (f −1 [F1 ])• ⊆ φE ⊆ (f −1 [F2 ])• = [[u ∈ F2 ]], using the formula of 364Jb, while H • ∩ [[u ∈ F1 ]] = H • ∩ [[u ∈ F2 ]]. As (A, µ ¯) is semi-finite, sup{[[u ∈ F ]] : F ∈ B, F ⊆ E} and inf{[[u ∈ F ]] : F ∈ B, F ⊇ E} must both be equal to φE. Q Q We can therefore identify the sequentially order-continuous Boolean homomorphism φ with E 7→ [[u ∈ E]], as described. (b) If u ∈ L0 and h : R → R is universally measurable, then, once again identifying u with f • where f is Σ-measurable, we see that hf is Σ-measurable (by 434S), so we have a corresponding element (hf )• of L0 . If E ∈ Σum , then [[(hf )• ∈ E]] = ((hf )−1 [E])• = (f −1 [h−1 [E]])• = [[u ∈ h−1 [E]]], ¯ using 434De to check that h−1 [E] ∈ Σum , so that we can identify (hf )• with h(u), as described. 434X Basic exercises > (a) Let A ⊆ [0, 1] be any non-measurable set. Show that the subspace measure on A is completion regular and τ -additive but not tight. >(b) Let X be any Hausdorff space with a point x such that {x} is not a Gδ set; for instance, X = ω1 + 1 and x = ω1 , or X = {0, 1}I for any uncountable set I and x any point of X. Show that setting µE = χE(x) we get a tight Borel measure on X which is not completion regular. > (c) Let X be a topological space. (i) Show that if A ⊆ X is universally measurable in X, then A ∩ Y is universally measurable in Y for any set Y ⊆ X. (ii) Show that if Y ⊆ X is universally measurable in X, and A ⊆ Y is universally measurable in Y , then A is universally measurable in X. (iii) Suppose that X is the product of a countableQ family hXi ii∈I of topological spaces, and Ei ⊆ Xi is a universally measurable set for each i ∈ I. Show that i∈I Ei is universally measurable in X. (d) Let X be a topological space, Y an analytic Hausdorff space, and W ⊆ X × Y a Borel set. Show that W −1 [Y ] is a universally measurable subset of X. (Hint: 423N.) (e) Let X be a Hausdorff space. (i) Show that, for A ⊆ X, the following are equiveridical: (α) A is universally Radon-measurable in X; (β) A is measured by every atomless Radon probability measure on X;

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(γ) A ∩ K is universally Radon-measurable in K for every compact K ⊆ X. (ii) Show that if A ⊆ X is universally Radon-measurable in X, then A ∩ Y is universally Radon-measurable in Y for any set Y ⊆ X. (iii) Show that if Y ⊆ X is universally Radon-measurable in X, and A ⊆ Y is universally Radon-measurable in Y , then A is universally Radon-measurable in X. (iv) Show that if G is an open cover of X, and A ⊆ X is such that A ∩ G is universally Radon-measurable (in G or in X) for every G ∈ G, then A is universally (X) (Y ) Radon-measurable in X. (v) Show that if Y is another Hausdorff space, and ΣuRm , ΣuRm are the algebras of universally Radon-measurable subsets of X, Y respectively, then every continuous function from X to Y is (X) (Y ) hXi ii∈I of topological (ΣuRm , ΣuRm )-measurable. (vi) Suppose that X is the product of a countable family Q spaces, and Ei ⊆ Xi is a universally Radon-measurable set for each i ∈ I. Show that i∈I Ei is universally Radon-measurable in X. > (f ) (i) Let µ0 be Dieudonn´e’s measure on ω1 . Give ω1 + 1 = ω1 ∪ {ω1 } its compact Hausdorff order topology, and define a Borel measure µ on ω1 + 1 by setting µE = µ0 (E ∩ ω1 ) for every Borel set E ⊆ ω1 + 1. Show that µ is a complete probability measure and is neither τ -additive nor inner regular with respect to the closed sets. (ii) Show that the universally measurable subsets of ω1 + 1 are just its Borel sets. (Hint: 4A3J, 411Q.) (iii) Show that every totally finite τ -additive topological measure on ω1 + 1 has a countable support. (iv) Show that every subset of ω1 + 1 is universally Radon-measurable. (g) (i) Show that there is a set X ⊆ [0, 1] such that K ∩ X and K \ X are both of cardinal c for every uncountable compact set K ⊆ [0, 1]. (Hint: 419Ib, 423K.) (ii) Show that if we give X its subspace topology, then every subset of X is universally Radon-measurable, but not every subset is universally measurable. (Hint: every compact subset of X is countable, so every Radon measure on X is purely atomic, but X has full outer Lebesgue measure in [0, 1].) (h) Show that a Hausdorff space X is Radon iff (α) every compact subset of X is Radon (β) for every non-zero totally finite Borel measure µ on X there is a compact subset K of X such that µK > 0. (Hint: 434F(a-v).) > (i) (i) Let X and Y be K-analytic Hausdorff spaces and f : X → Y a continuous surjection. Suppose that F ⊆ Y and that f −1 [F ] is universally Radon-measurable in X. Show that F is universally Radonmeasurable in Y . (Hint: 432G.) (ii) Let X and Y be analytic Hausdorff spaces and f : X → Y a Borel measurable surjection. Suppose that F ⊆ Y and that f −1 [F ] is universally Radon-measurable in X. Show that F is universally Radon-measurable in Y . (Hint: 433D.) (j) Show that if X is a perfectly normal space then it is Borel-measure-compact iff it is Borel-measurecomplete. (k) Show that if we give ω1 + 1 its order topology, it is Borel-measure-compact but not Borel-measurecomplete or pre-Radon, and that its open subset ω1 is not Borel-measure-compact. (l) Set X = ω1 + 1, with its order topology, and let Σ be the σ-algebra of subsets of X generated by the countable sets and the set Ω of limit ordinals in X. Show that there is a unique probability measure µ on X with domain Σ such that µξ = µΩ = 0 for every ξ < ω1 . Show that µ is inner regular with respect to the Borel sets, is defined on a base for the topology of the compact Hausdorff space X, but has no extension to a topological measure on X. (m) Show that R, with the right-facing Sorgenfrey topology, is Borel-measure-complete and Borelmeasure-compact, but not Radon or pre-Radon. (n) Let X be a topological space. (i) Show that the family of Borel-measure-compact subsets of X is closed under Souslin’s operation. (ii) Show that the union of a sequence of Borel-measure-complete subsets of X is Borel-measure-complete. (iii) Show that if X is Hausdorff then the family of pre-Radon subsets of X is closed under Souslin’s operation. (Hint: in (i) and (iii), start by showing that the family under consideration is closed under countable unions.)

434Yd

(o) Show that ]0, 1[

Borel measures ω1

201

is not pre-Radon.

> (p) Show that a K-analytic Hausdorff space is Radon iff all its compact subsets are Radon. (Hint: 432B, 434Xh.) (q) Suppose that X is a K-analytic Hausdorff space such that every Radon measure on X is completion regular. Show that X is a Radon space. (r) Let X and Y be topological spaces, and suppose that Y has a countable network. (i) Show that if X is Borel-measure-complete, then X × Y is Borel-measure-complete. (ii) Show that if X and Y are Radon Hausdorff spaces, then X × Y is Radon. Q Q (s) Let hXn in∈N be a sequence of topological spaces; write X = n∈N Xn and Zn = i≤n Xi for each n. (i) Show that if every Zn is Borel-measure-complete, so is X. (ii) Show that if every Zn is Hausdorff and pre-Radon, so is X. (iii) Show that if every Zn is Hausdorff and Radon, so is X. Q (t) (i) Let hXn in∈N be a sequence of Radon Hausdorff spaces such that i≤n Ki is Radon whenever Q n ∈ N and Ki ⊆ Xi is compact for every i ≤ n. Show that X = n∈N XnQis Radon. (ii) Let hXn in∈N be a sequence of Radon Hausdorff spaces with countable networks. Show that n∈N Xn is Radon. (u) (i) Let X be a Radon Hausdorff space and Y a Borel-measure-compact space. Show that X × Y is Borel-measure-compact. (ii) Let X be a Radon Hausdorff space and Y a Borel-measure-complete space. Show that X × Y is Borel-measure-complete. R RR (v) Show that if, in 434R, ν is σ-finite, then g dλB = g(x, y)ν(dy)µ(dx) for every λB -integrable function g : X × Y → R. (w) Show that the product measure construction of 434R is ‘associative’ and ‘distributive’ in the sense that (under the product measures on (X × Y ) × Z and X × (Y × Z) agree, and S S S appropriate hypotheses) those on i,j∈N (Xi × Yj ) and ( i∈N Xi ) × ( j∈N Yj ) agree. > (x) Show that the product measure construction of 434R is not ‘commutative’; indeed, taking µ = ν to be Dieudonn´e’s measure on ω1 , show that the Borel measures λ1 , λ2 on ω12 defined by setting λ1 W =

R

νW [{ξ}]µ(dξ),

λ2 W =

R

µW −1 [{η}]ν(dη)

are different. X ∈ E)’ for random variables X and (y) Read through §271, looking for ways to apply the concept ‘Pr(X universally measurable sets E. 434Y Further exercises (a) Set X = N \ {0, 1}. For m, p ∈ X set Ump = m + pN; show that {Ump : m, p ∈ X are coprime} is a base for a connected Hausdorff topology on X. (Hint: pq ∈ U mp for every q ≥ 1. See Steen & Seebach 78, ex. 60.) Show that X is a second-countable analytic Hausdorff space and carries a Radon measure which is not completion regular. (b) Let (X, Σ, µ) be a semi-finite measure space and T a topology on X such that µ is inner regular with respect to the closed sets. Suppose that Y is a topological space with a countable network consisting of universally measurable sets, and that f : X → Y is measurable. Show that f is almost continuous. (c) Let X be a completely regular Hausdorff space. Show that the following are equiveridical: (α) X is ˇ Radon; (β) X is a universally measurable subset of its Stone-Cech compactification; (γ) whenever Y is a Hausdorff space and X 0 is a subspace of Y which is homeomorphic to X, then X 0 is universally measurable in Y . (d) Let X be a completely regular Hausdorff space. Show that the following are equiveridical: (α) ˇ X is pre-Radon; (β) X is a universally Radon-measurable subset of its Stone-Cech compactification; (γ) 0 whenever Y is a Hausdorff space and X is a subspace of Y which is homeomorphic to X, then X 0 is universally Radon-measurable in Y .

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(e) Let X be a metrizable space without isolated points, and µ a σ-finite Borel measure on X. Show that there is a conegligible meager set. (Hint: there is a dense set D ⊆ X such that {{d} : d ∈ D} is σ-metrically-discrete.) (f ) (Aldaz 97) A topological space X is countably metacompact if whenever G is a countable open cover of X then there is a point-finite open cover H of X refining G. (i) Show that X is countably metacompact iff whenever hFn in∈N is a non-increasing sequence of closed sets with empty intersection in X then there is a sequence hGn in∈N of open sets, with empty intersection, such that Fn ⊆ Gn for every n. (ii) Let X be any topological space and ν : PX → [0, 1] a finitely additive functional such that νX = 1. Show that there is a finitely additive ν 0 : PX → [0, 1] such that νF ≤ ν 0 F = inf{ν 0 G : G ⊇ F is open} for every closed F ⊆ X. (Hint: 413Q.) (iii) Show that if X is countably metacompact and µ is any Borel probability measure on X, there is a Borel probability measure µ0 on X, inner regular with respect to the closed sets, such that µF ≤ µ0 F for every closed set F ⊆ X; so that µ and µ0 agree on the Baire σ-algebra of X. (g) Show that βN is not countably tight, therefore not Borel-measure-compact. (h) Let X be a quasi-dyadic space. Suppose that h(Gξ , Hξ )iξ<ω1 T is a family of T pairs of disjoint non-empty open sets. Show that there is an uncountable A ⊆ ω1 such that ξ∈B Gξ ∩ ξ∈A\B Hξ is non-empty for every B ⊆ A. (Hint: start by supposing that X is itself a product of separable metric spaces, and that every Gξ , Hξ is an open cylinder set; use the ∆-system lemma.) (i) (i) Show that the split interval is not quasi-dyadic. (ii) Show that R, with the Sorgenfrey right-facing topology, is not quasi-dyadic. (iii) Show that ω1 and ω1 +1, with their order topologies, are not quasi-dyadic. (Hint: 434Yh.) (j) Show that a perfectly normal quasi-dyadic space is Borel-measure-compact. (k) Let hXn in∈N Q be a sequence of first-countable spaces, and µn a Borel measure on Xn for each n. For n ∈ N set Zn = i≤n Xi , and let λn be the product Borel measure on Zn constructed by repeatedly using Q the method of 434R (cf. 434Xw). Show that there is a unique Borel measure λ on Z = n∈N Xn such that all the canonical maps from Z to Zn are inverse-measure-preserving. Show that, for any n, λ can be Q identified with the product of λn and a suitable product measure on i>n Xi . (l) Let X be a totally ordered set with its order topology. Show that any τ -additive Borel probability • measure on X has countable Maharam type. (Hint: {]−∞, x] : x ∈ X} generates the measure algebra.) 434Z Problems (a) Must every Radon compact Hausdorff space be sequentially compact? (b) Must a Hausdorff continuous image of a Radon compact Hausdorff space be Radon? 434 Notes and comments I said that the fundamental question of topological measure theory is ‘which measures can appear on which topological spaces’ ? In this section I have concentrated on Borel measures, classified according to the scheme laid out in §411. (Of course there are other kinds of classification. One of the most interesting is the Maharam classification of Chapter 33: we can ask what measure algebras can appear from topological measures on a given topological space. But I pass this by for the moment, with only 434Yl to give a taste.) We can ask this question from either of two directions. The obvious approach is to ask, for a given class of topological spaces, which types of measure can appear. But having discovered that (for instance) there are several types of topological space on which all (totally finite) Borel measures are tight, we can use this as a definition of a class of topological spaces, and ask the ordinary questions about this class. Thus we have ‘Radon’, ‘Borel-measure-complete’, ‘Borel-measure-compact’ and ‘pre-Radon’ spaces (434C, 434H, 434I, 434J). I have given precedence to the first partly to honour the influence of Schwartz 73 and partly because a compact Hausdorff space is always Borel-measure-compact and pre-Radon (434Hb, 434Jf) and is Borel-measure-complete iff it is Radon (434Ka). In effect, ‘Borel-measure-complete’ means ‘Borel measures are quasi-Radon’ (434Ib), ‘pre-Radon’ means ‘quasi-Radon measures are Radon’ (434Jb), and ‘Radon’ means ‘Borel measures are Radon’ (434F(a-iii)). These slogans have to be interpreted with

435A

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care; but it is true that a Hausdorff space is Radon iff it is both Borel-measure-complete and pre-Radon (434Ka). The concept of ‘Radon’ space is in fact one of the important contributions of measure theory to general topology, offering a variety of challenging questions. One which has attracted some attention is the problem of determining when products of Radon spaces are Radon. Uncountable products hardly ever are (434Ke); for countable products it is enough to understand products of finitely many compact spaces (434Xt); but the product of two compact spaces already seems to lead us into undecidable questions (438Xn, Wage 80). Two more very natural questions are in 434Z. One of the obstacles to the investigation is the rather small number of Radon compact Hausdorff spaces which are known. I should remark that if the continuum hypothesis (for instance) is true, then every compact Hausdorff space in which countably compact sets are closed is sequentially compact (Ismail & Nyikos 80, or Fremlin 84, 24Nc), so that in this case we have a quick answer to 434Za. You will recognise the construction of 434M as a universal version of Dieudonn´e’s measure (411Q). ‘Tightness’ is of great interest for other reasons (Engelking 89), and here is very helpful in giving quick proofs that spaces are not Radon (434Yg). A large proportion of the definitions in general topology can be regarded as different abstractions from the concept of metrizability. Countable tightness is an obvious example; so is ‘first-countability’ (434R). In quite a different direction we have ‘metacompactness’ (438J, 434Yf). The construction of the product measure in 434R is an obvious idea, as soon as you have seen Fubini’s theorem, but it is not obvious just when it will work. ‘Quasi-dyadic’ spaces are a relatively recent invention; I introduce them here only as a vehicle for the argument of 434Q. Of course a dyadic space (4A2A) is quasi-dyadic; for basic facts on dyadic spaces, see 4A2Dd, 4A5T and Engelking 89, §3.12 and 4.5.9-4.5.11.

435 Baire measures Imitating the programme of §434, I apply a similar analysis to Baire measures, starting with a simpleminded classification (435A). This time the central section (435D-435H) is devoted to ‘measure-compact’ spaces, those on which all (totally finite) Baire measures are τ -additive. 435A Types of Baire measures In 434A I looked at a list of four properties which a Borel measure may or may not possess: inner regularity with respect to closed sets, inner regularity with respect to zero sets, tightness (that is, inner regularity with respect to closed compact sets), and τ -additivity. Since every (semi-finite) Baire measure is inner regular with respect to the zero sets (412D), only two of the four are important considerations for Baire measures: tightness and τ -additivity. On the other hand, there is a new question we can ask. Given a Baire measure on a topological space, when can it be extended to a Borel measure? And in the case of a positive answer, we can ask whether the extension is unique, and whether we can find extensions to Borel measures satisfying the properties considered in 434A. We already have some information on this. If X is a completely regular space, and µ is a τ -additive effectively locally finite Baire measure on X, then µ has a (unique) extension to a τ -additive Borel measure (415N). While if µ is tight, the extension will also be tight (cf. 416C). Perhaps I should remark immediately that while there can be only one τ -additive Borel measure extending µ, there might be another Borel measure, not τ -additive, also extending µ; see 435Xa. Of course if there is any completion regular Borel measure extending µ, there is only one; moreover, if µ is σ-finite, and there is a completion regular Borel measure extending µ, this is the only Borel measure extending µ. (For every Borel set will belong to the domain of the completion of µ.) A possible division of Baire measures is therefore into classes (E) measures which are not τ -additive, (F) measures which are τ -additive, but not tight, (G) tight measures, and within these classes we can distinguish measures with no extension to a Borel measure (type E0 ), measures with more than one extension to a Borel measure (types E1 , F1 and G1 ), measures with exactly one extension to a Borel measure which is not completion regular (types E2 , F2 and G2 ) and measures with

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435A

an extension to a completion regular Borel measure (types E3 , F3 and G3 ). For examples, see 439M and 439O (E0 ), 439N (E2 ), 439J (E3 ), 435Xc (F1 ), 435Xd (F2 ), 415Xc and 434Xa (F3 ), 435Xa (G1 ), 435Xb (G2 ) and the restriction of Lebesgue measure to the Baire subsets of R (G3 ); other examples may be constructed as direct sums of these. A separate question we can ask of a Baire measure is whether it can be extended to a Radon measure. For this there is a straightforward criterion (435B), which shows that (at least for totally finite measures on completely regular spaces) only the types F1 and F2 are divided by this question. (If a Baire measure µ can be extended to a Radon measure, it is surely τ -additive. If µ is tight, it satisfies the criteria of 435B, so has an extension to a Radon measure. If µ has an extension to a completion regular Borel measure µ1 and has an extension to a Radon measure µ2 , then the completion µ ˆ of µ extends µ1 , while µ2 extends µ ˆ; so µ1 is the restriction of µ2 to the Borel sets and µ2 = µ ˆ1 = µ ˆ and µ, like µ2 , is tight, by 412Hb or otherwise. Thus no measure of type F3 can be extended to a Radon measure.) As with the classification of Borel measures that I offered in §434, any restriction on the topology of the underlying space may eliminate some of these possibilities. For instance, because a semi-finite Baire measure is inner regular with respect to the closed sets, we can have no (semi-finite) measure of classes E or F on a compact Hausdorff space. On a locally compact Hausdorff space we can have no effectively locally finite Baire measure of class F (435Xe), while on a K-analytic Hausdorff space we can have no locally finite Baire measure of class E (432F). In a metric space, or a regular space with a countable network (e.g., a regular analytic Hausdorff space), the Baire and Borel σ-algebras coincide (4A3Kb), so we can have no measures of type E0 , E1 , F1 or G1 . 435B Theorem Let X be a Hausdorff space and µ a locally finite Baire measure on X. Then the following are equiveridical: (i) µ has an extension to a Radon measure on X; (ii) for every non-negligible Baire set E ⊆ X there is a compact set K ⊆ E such that µ∗ K > 0. If µ is totally finite, we can add (iii) sup{µ∗ K : K ⊆ X is compact} = µX. proof Because µ is inner regular with respect to the closed sets (412D), this is just a special case of 416P. ˇ´ık 57) Let X be a normal countably paracompact space. Then any semi-finite 435C Theorem (Mar Baire measure on X has an extension to a semi-finite Borel measure which is inner regular with respect to the closed sets. proof (a) Let ν be a semi-finite Baire measure on X. Let K be the family of those closed subsets of X which are included in zero sets of finite measure, and set φ0 K = ν ∗ K for K ∈ K. Then K and φ0 satisfy the conditions of 413I, that is, ∅ ∈ K, (†) K ∪ K 0 ∈ K whenever K, K 0 ∈ K are disjoint, T (‡) n∈N Kn ∈ K whenever hKn in∈N is a sequence in K, (α) φ0 K = φ0 L + sup{φ0 K 0 : K ∈ K, K 0 ⊆ K \ L} whenever K, L ∈ K and L ⊆ K, (β) inf n∈N φ0 Kn = 0 whenever hKn in∈N is a non-increasing sequence in K with empty intersection. P P The first three are trivial. α) Take K, L ∈ K with L ⊆ K, and set γ = sup{φ0 K 0 : K ∈ K, K 0 ⊆ K \ L}. (i) If K 0 ⊆ K \ L is (α closed, then (because X is normal) there is a zero set F including K 0 and disjoint from L (4A2F(d-iv)), so φ0 K 0 + φ0 L = ν ∗ ((K 0 ∪ L) ∩ F ) + ν ∗ ((K 0 ∪ L) \ F ) = ν ∗ (K 0 ∪ L) ≤ ν ∗ K. As K 0 is arbitrary, γ + φ0 L ≤ φ0 K. (ii) Let ² > 0. Let F0 be a zero set of finite measure including K. Because ν is inner regular with respect to the zero sets (412D), there is a zero set F ⊆ F0 \ L such that νF ≥ ν∗ (F0 \ L) − ² (413Ee), so that ν(F0 \ F ) ≤ ν ∗ L + ² (413Ec). Set K 0 = K ∩ F . Then ν ∗ K = ν ∗ (K \ F ) + ν ∗ (K ∩ F ) ≤ ν(F0 \ F ) + ν ∗ K 0 ≤ ν ∗ L + ² + γ. As ² is arbitrary, ν ∗ K ≤ ν ∗ L + γ.

435Fb

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β ) If hKn in∈N is a non-increasing sequence in K with empty intersection, then (because XTis countably (β paracompact) there is a sequence hGn in∈N of open sets such that Kn ⊆ Gn for every n and n∈N Gn = ∅ (4A2Ff). T Because X is normal, there are zero sets Fn such that Kn ⊆ Fn ⊆ Gn for each n (4A2F(d-iv)), so that n∈N Fn = ∅. We may suppose that F0 has finite measure. In this case, T limn→∞ ν ∗ Kn ≤ limn→∞ ν( i≤n Fi ) = 0. Thus K and φ0 satisfy the conditions (α) and (β) as well. Q Q (b) By 413I, there is a complete locally determined measure µ on X, extending φ0 and inner regular with respect to K. If F ⊆ X is closed, then F ∩ K ∈ K for every K ∈ K, so F ∈ dom µ (413F(ii)); accordingly µ is a topological measure, and because ν also is inner regular with respect to K, µ must extend ν. So the restriction of µ to the Borel sets is a Borel extension of ν which is inner regular with respect to the closed sets. Remark If X is normal, but not countably paracompact, the result may fail; see 439O. I have stated the result in terms of ‘countable paracompactness’, but the formally distinct ‘countable metacompactness’ is also sufficient (435Ya). If we are told that the Baire measure is τ -additive and effectively locally finite, we have a much stronger result (415M). 435D Just as with the ‘Radon’ spaces of §434, we can look at classes of topological spaces defined by the behaviour of the Baire measures they carry. The class which has aroused most interest is the following. Definition A completely regular topological space X is measure-compact (sometimes called almost Lindel¨ of) if every totally finite Baire measure on X is τ -additive, that is, has an extension to a quasiRadon measure on X (415N). 435E The following lemma will make our path easier. Lemma Let X be a completely regular topological space and ν a totally finite Baire measure on X. Suppose that supG∈G νG = νX whenever G is an upwards-directed family of cozero sets with union X. Then ν is τ -additive. S proof Let G be an upwards-directed family of open Baire sets such that G∗ = G is also a Baire set, and ² > 0. Because ν is inner regular with respect to the zero sets, there is a zero set F ⊆ G∗ such that νF ≥ νG∗ − ². Let G 0 be the family of cozero sets included S in members of G; because X is completely regular, so that the cozero sets are a base for its topology, G 0 = G∗ , and of course G 0 is upwards-directed. Now H = {G ∪ (X \ F ) : G ∈ G 0 } is an upwards-directed family of cozero sets with union X, so there is a G0 ∈ G 0 such that ν(G0 ∪ (X \ F )) ≥ νX − ². In this case supG∈G νG ≥ νG0 ≥ νX − ² − ν(X \ F ) ≥ νG∗ − 2². As G and ² are arbitrary, ν is τ -additive. 435F Elementary facts (a) If X is a completely regular space which is not measure-compact, there are a Baire probability measure µ on X and a cover of X by µ-negligible cozero sets. P P There is a totally finite Baire measure ν on X which is not τ -additive. By 435E, there is an upwards-directed family G of cozero sets, covering X, such that sup SG∈G νG < νX. Let hGn in∈N be a sequence in G such that supn∈N νGn = supG∈G νG. Then γ = ν(X \ n∈N Gn ) > 0. Set 1 γ

µH = ν(H \

S n∈N

Gn )

for Baire sets H ⊆ X; then µ is a Baire probability measure and G is a cover of X by µ-negligible cozero sets. Q Q

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(b) Regular Lindel¨of spaces are measure-compact. (For if a Lindel¨of space can be covered by negligible open sets, it can be covered by countably many negligible open sets, so is itself negligible.) In particular, compact Hausdorff spaces, indeed all regular K-analytic Hausdorff spaces (422Gg), are measure-compact. Note that regular Lindel¨of spaces are normal and paracompact (4A2H(b-i)), so their measure-compactness is also a consequence of 435C and 434Hb. (c) An open subset of a measure-compact space need not be measure-compact (435Xi(i)). A continuous image of a measure-compact space need not be measure-compact (435Xi(ii)). N c is not measure-compact (439P). The product of two measure-compact spaces need not be measure-compact (439Q). (d) If X is a measure-compact completely regular space it is Borel-measure-compact. P P Let µ be a non-zero totally finite Borel measure on X and G an open cover of X. Let ν be the restriction of µ to the Baire σ-algebra of X, so that ν is τ -additive. Let U be the set of cozero sets S U ⊆ X included in members of G; because the family of cozero sets is a base for the topology of X, U = X, and there is some U ∈ U such that νU > 0. This means that there is some G ∈ G such that µG > 0. By 434H(a-v), X is Borel-measure-compact. Q Q 435G Proposition A Souslin-F subset of a measure-compact completely regular space is measurecompact. proof (a) Let X be a measure-compact completely regular space, hFσ iσ∈S a Souslin scheme consisting of closed subsets of X with kernel A, ν a totally finite Baire measure on A, and G an upwards-directed family of (relatively) cozero subsets of A covering A. Let ν1 be the Baire measure on X defined by setting ν1 H = ν(A ∩ H) for every Baire subset H of X. Because X is measure-compact, ν1 has an extension to a quasi-Radon measure µ on X. Let µA be the subspace measure on A. (b) By 431A, A is measured by µ. In fact µA = νA. P P The construction of µ given in 415K-415N ensures that µF = ν1∗ F for every closed set F , and this is in any case a consequence of the facts that µ is τ -additive and dom ν1 includes a base for the topology. For each σ ∈ S, in particular, µFσ = ν1∗ Fσ ; let Fσ0 ⊇ Fσ be a Baire set such that ν1 Fσ0 = ν1∗ Fσ . Then µFσ0 = ν1 Fσ0 = ν1∗ Fσ = µFσ and µ(Fσ0 \ Fσ ) = 0 for every σ ∈ S. Let A0 be the kernel of the Souslin scheme hFσ0 iσ∈S . Then A ⊆ A0 and P µ(A0 \ A) ≤ σ∈S µ(Fσ0 \ Fσ ) = 0, so µA = µA0 . On the other hand, writing νˆ1 for the completion of ν1 , A0 is measured by νˆ1 , by 431A again, so that (because µ extends ν1 ) µA = µA0 = µ∗ A0 ≤ (ν1 )∗ A0 = ν1∗ A0 ≤ µ∗ A0 = µA. Thus µA = ν1∗ A0 . But of course νA = ν1 X = ν1∗ A = ν1∗ A0 , so that µA = νA. Q Q Since we surely have µX = ν1 X = νA, we see that µ(X \ A) = 0. (c) It follows that µF = νF for every (relatively) zero set F ⊆ A. P P There is a closed set F 0 ⊆ X such 0 0 that F = A ∩ F . Now if H ⊆ X is a Baire set including F , H ∩ A is a (relatively) Baire set including F , so νF ≤ ν(H ∩ A) = ν1 H; as H is arbitrary, νF ≤ ν1∗ F 0 . But ν1∗ F 0 = µF 0 , as remarked above, and µ(X \ A) = 0, so µF = µF 0 = ν1∗ F 0 ≥ νF . On the other hand, A \ F is (relatively) cozero, so there is a non-decreasing sequence hFn in∈N of (relatively) zero subsets of A with union A \ F , and µ(A \ F ) = limn→∞ µFn ≥ limn→∞ νFn = ν(A \ F ).

435Xi

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207

Since we already know that µA = νA, it follows that µF = µA − µ(A \ F ) ≤ νA − ν(A \ F ) = νF , and µF = νF . Q Q (d) The set {E : E ⊆ A is a (relative) Baire set, µE = νE} therefore contains every (relatively) zero set, and by the Monotone Class Theorem (136C) contains every (relatively) Baire set. What this means is that µ actually extends ν; so the subspace measure µA = µ¹ PA also extends ν. But µA is a quasi-Radon measure (415B), therefore τ -additive, and ν must also be τ -additive. 435H Corollary A Baire subset of a measure-compact completely regular space is measure-compact. proof Put 435G and 421L together. 435X Basic exercises >(a) Give ω1 + 1 its order topology. (i) Show that its Baire σ-algebra Σ is just the family of sets E ⊆ ω1 + 1 such that either E or its complement is a countable subset of ω1 . (ii) Show that there is a unique Baire probability measure ν on ω1 + 1 such that ν{ξ} = 0 for every ξ < ω1 . (iii) Show that ν is τ -additive. (iv) Show that there is exactly one Radon measure on ω1 + 1 extending ν, but that the measure µ of 434Xf is another Borel measure also extending ν. > (b) Let I be a set of cardinal ω1 , endowed with the discrete topology, and X = I ∪ {∞} its one-point compactification (3A3O). Let µ be the Radon probability measure on X such that µ{∞} = 1. (i) Show that every subset of X is a Borel set. (ii) Show that {∞} is not a zero set. (iii) Let ν be the restriction of µ to the Baire σ-algebra of X. Show that ν is tight. Show that µ is the unique Borel measure extending ν (hint: you will need 419G), but is not completion regular. (iv) Show that the subspace measure νI on I is the countable-cocountable measure on I, and is not a Baire measure, nor has any extension to a Baire measure on I. (v) Show that X is measure-compact. (c) On R ω1 let µ be the Baire measure defined by saying that µE = 1 if χω1 ∈ E, 0 otherwise. (i) Show that µ is τ -additive, but not tight. (Hint: 4A3P.) (ii) Show that the map ξ 7→ χξ : ω1 + 1 → R ω1 is continuous, so that µ has more than one extension to a Borel measure. (iii) Show that µ has an extension to a Radon measure. (d) Set X = ω1 + 1 with the topology Pω1 ∪ {X \ A : A ⊆ ω1 is countable}. Let µ be the Baire measure on X defined by saying that, for Baire sets E ⊆ X, µE = 1 if ω1 ∈ E, 0 otherwise. (i) Show that a function f : X → R is continuous iff {ξ : ξ ∈ X, f (ξ) 6= f (ω1 )} is countable; show that X is completely regular and Hausdorff. (ii) Show that µ is τ -additive. (iii) Show that every subset of X is Borel. (iv) Show that the only Borel measure extending µ is that which gives measure 1 to {ω1 }, and that this is a Radon measure. (v) Show that all compact subsets of X are finite, so that µ is not tight. *(vi) Show that X is Lindel¨of. (e) Let X be a locally compact Hausdorff space and µ an effectively locally finite τ -additive Baire measure on X. Show that µ is tight. (Hint: the relatively compact cozero sets cover X; use 414Ea and 412D.) > (f ) Let X be a completely regular space and µ a totally finite τ -additive Borel measure on X. Let µ0 be the restriction of µ to the Baire σ-algebra of X. Show that µF = µ∗0 F for every closed set F ⊆ X. (g) Show that if a semi-finite Baire measure ν on a normal countably paracompact space is extended to a Borel measure µ by the construction in 435C, then the measure algebra of ν becomes embedded as an order-dense subalgebra of the measure algebra of µ, so that L1 (µ) can be identified with L1 (ν). (h) Show that a Borel-measure-compact normal countably paracompact space is measure-compact. (i) (i) Show that ω1 + 1 is measure-compact, in its order topology, but that its open subset ω1 is not (cf. 434Xk). (ii) Show that a discrete space of cardinal ω1 is measure-compact, but that it has a continuous image which is not measure-compact.

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(j) Let X be a metacompact completely regular space and ν a totally finite strictly positive Baire measure on X. Show that X is Lindel¨of, so that ν has an extension to a quasi-Radon measure on X. (Hint: if H is a point-finite open cover of X, not containing ∅, then for each H ∈ H choose a non-empty cozero set GH ⊆ H; show that {H : νGH ≥ δ} is finite for every δ > 0.) (k) A completely regular space X is strongly measure-compact (Moran 69) if µX = sup{µ∗ K : K ⊆ X is compact} for every totally finite Baire measure µ on X. (i) Show that a completely regular Hausdorff space X is strongly measure-compact iff every totally finite Baire measure on X has an extension to a Radon measure. (ii) Show that a strongly measure-compact completely regular space is measurecompact. (iii) Show that a Souslin-F subset of a strongly measure-compact completely regular space is strongly measure-compact. (iv) Show that a discrete space of cardinal ω1 is strongly measure-compact. (v) Show that a countable product of strongly measure-compact completely regular spaces is strongly measurecompact. (vi) Show that N ω1 is not strongly measure-compact. (Hint: take a non-trivial probability measure on N and consider its power on N ω1 .) (vii) Show that if X and Y are completely regular spaces, X is measure-compact and Y is strongly measure-compact then X × Y is measure-compact. (viii) Show that if X is a strongly measure-compact completely regular Hausdorff space, then it is Borel-measure-compact and pre-Radon. 435Y Further exercises (a) Show that a normal countably metacompact space (434Yf) is countably paracompact. ˇ (b) Let X be a completely regular Hausdorff space and βX its Stone-Cech compactification. Show that X is measure-compact iff whenever ν is a Radon measure on βX such that νX = 0, there is a ν-negligible Baire subset of βX including X. 435 Notes and comments The principal reason for studying Baire measures is actually outside the main line of this chapter. For a completely regular Hausdorff space X, write Cb (X) for the M -space of bounded continuous real-valued functions on X. Then Cb (X)∗ = Cb (X)∼ is an L-space (356N), and inside Cb (X)∗ we have the bands generated by the tight, smooth and sequentially smooth functionals (see 437Xq and 437A below), all identifiable, if we choose, with spaces of ‘signed Baire measures’. Wheeler 83 argues convincingly that for the questions a functional analyst naturally asks, these Baire measures are often an effective aid. From the point of view of the arguments in this section, the most fundamental difference between ‘Baire’ and ‘Borel’ measures lies in their action on subspaces. If X is a topological space and A is a subset of X, then any Borel or Baire measure µ on A provides us with a measure µ1 of the same type on X, setting µ1 E = µ(A ∩ E) for the appropriate sets E. In the other direction, if µ is a Borel measure on X, then the subspace measure µA is a Borel measure on A, because the Borel σ-algebra of A is just the subspace σ-algebra derived from the Borel algebra of X (4A3Ca). But if µ is a Baire measure on X, it does not follow that µA is a Baire measure on A; this is because (in general) not every continuous function f : A → [0, 1] has a continuous extension to X, so that not every zero set in A is the intersection of A with a zero set in X (see 435Xb). The analysis of those pairs (X, A) for which the Baire σ-algebra of A is just the subspace algebra derived from the Baire sets in X is a challenging problem in general topology which I pass by here. For the moment I note only that avoiding it is the principal technical problem in the proof of 435G. I do not know if I ought to apologise for ‘countably tight’ spaces (434N), ‘first-countable’ spaces (434R), ‘metacompact’ spaces (438J), ‘normal countably paracompact’ spaces (435C), ‘quasi-dyadic’ spaces (434O) and ‘sequential’ spaces (436F). General topology is notorious for invoking arcane terminology to stretch arguments to their utmost limit of generality, and even specialists may find their patience tried by definitions which seem to have only one theorem each. In 438J, for instance, it is obvious that the original result concerned metrizable spaces (438H), and you may well feel at first that the extension is a baroque overelaboration. On the other hand, there are (if you look for them) some very interesting metacompact spaces (Engelking 89, §5.3), and metacompactness has taken its place in the standard lists. In this book I try to follow a rule of introducing a class of topological spaces only when it is both genuinely interesting, from the point of view of general topology, and also supports an idea which is interesting from the point of view of measure theory.

436C

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209

436 Representation of linear functionals I began this treatise with the three steps which make measure theory, as we know it, possible: a construction of Lebesgue measure, a definition of an integral from a measure, and a proof of the convergence theorems. I used what I am sure is the best route: Lebesgue measure from Lebesgue outer measure, and integrable functions from simple functions. But of course there are many other paths to the same ends, and some of them show us slightly different aspects of the subject. In this section I come – rather later than many authors would – to an account of a procedure for constructing measures from integrals. I start with three fundamental theorems, the first and third being the most important. A positive linear functional on a truncated Riesz space of functions is an integral iff it is sequentially smooth (436D); a smooth linear functional corresponds to a quasi-Radon measure (436H); and if X is a compact Hausdorff space, any positive linear functional on C(X) corresponds to a Radon measure (436J-436K). 436A Definition Let X be a set, U a Riesz subspace of RX , and f : U → R a positive linear functional. I say that f is sequentially smooth if whenever hun in∈N is a non-increasing sequence in U such that limn→∞ un (x) = 0 for every x ∈ X, then limn→∞ f (un ) = 0. If (X, Σ, µ) is a measure space Rand U is a Riesz subspace of the space of real-valued µ-integrable functions defined everywhere on X, then dµ : U → R is sequentially smooth, by Fatou’s Lemma or Lebesgue’s Dominated Convergence Theorem. Remark It is essential to distinguish between ‘sequentially smooth’, as defined here, and ‘sequentially ordercontinuous’, as in 313Hb or 355G. In the context here, a positive linear operator f : U → R is sequentially order-continuous if limn→∞ f (un ) = 0 whenever hun in∈N is a non-increasing sequence in U such that 0 is the greatest lower bound for {un : n ∈ N} in U ; while f is sequentially smooth if limn→∞ f (un ) = 0 whenever hun in∈N is a non-increasing sequence in U such that 0 is the greatest lower bound for {un : n ∈ N} in RX . So there can be sequentially smooth functionals which are not sequentially order-continuous, as in 436Xi. A sequentially order-continuous positive linear functional is of course sequentially smooth. 436B Definition Let X be a set. I will say that a Riesz subspace U of RX is truncated (or satisfies Stone’s condition) if u ∧ χX ∈ U for every u ∈ U . In this case, u ∧ γχX ∈ U for every γ ≥ 0 and u ∈ U (being −u− if γ = 0, γ(γ −1 u ∧ χX) otherwise). 436C Lemma Let X be a set and U a truncated Riesz subspace of RX . Write K for the family of sets of the form {x : x ∈ X, u(x) ≥ 1} as u runs over U . Let f : U → R be a sequentially smooth positive linear functional, andR µ a measure on X such that µK is defined and equal to inf{f (u) : χK ≤ u ∈ U } for every K ∈ K. Then u dµ exists and is equal to f (u) for every u ∈ U . R proof It is enough to deal with the case u ≥ 0, since U = U + − U + and both f and are linear. Note that if v ∈ U , K ∈ K and v ≤ χK, then v ≤ w whenever χK ≤ w ∈ U , so f (v) ≤ µK. For k, n ∈ N set Knk = {x : u(x) ≥ 2−n k},

unk = u ∧ 2−n kχX.

Then, for k ≥ 1, Knk = {x : 2n k −1 u ≥ 1} ∈ K, 2n (un,k+1 − unk ) ≤ χKnk ≤ 2n (unk − un,k−1 ). So 2n f (un,k+1 − unk ) ≤ µKnk ≤ 2n f (unk − un,k−1 ), and f (un,4n +1 − un1 ) ≤

P4n k=1

2−n µKnk ≤ f (u).

But setting wn = un,4n +1 − un1 , hwn in∈N is a non-decreasing sequence of functions in U and supn∈N wn (x) = P 4n u(x) for every x, so limn→∞ f (u − wRn ) = 0 and limn→∞ f (wn ) = f (u). Also, setting vn = k=1 2−n χKnk , we have wn ≤ vn ≤ u and f (wn ) ≤ vn ≤ f (u) for each n, so

R

u = limn→∞

R

vn = f (u)

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by B.Levi’s theorem. 436D Theorem Let X be a set and U a truncated Riesz subspace of RX . Let f : U → R be a positive linear functional. Then the following are equiveridical: (i) f is sequentially smooth; R (ii) there is a measure µ on X such that u dµ is defined and equal to f (u) for every u ∈ U . proof I remarked in 436A that (ii)⇒(i) is a consequence of Fatou’s Lemma. So the argument here is devoted to proving that (i)⇒(ii). (a) Let K be the family of sets K ⊆ X such that χK = inf n∈N un for some sequence hun in∈N in U , taking the infimum in RX , so that (inf n∈N un )(x) = inf n∈N un (x) for every x ∈ X. Then K is closed under finite unions and countable intersections. P P (i) If K, K 0 ∈ K take sequences hun in∈N , hu0n in∈N in U such that 0 0 χK = inf n∈N un and χK = inf n∈N un ; then χ(K ∪ K 0 ) = inf m,n∈N um ∨ u0n , so K ∪ K 0 ∈ K. (ii) If hKn in∈N is a sequence in K, then for each n ∈TN we can choose a sequence huni ii∈N in U such that χKn = inf i∈N uni ; T now χ( n∈N Kn ) = inf n,i∈N uni , so n∈N Kn ∈ K. Q Q Note that ∅ ∈ K because 0 ∈ U . (b) We need to know that if u ∈ U then K = {x : u(x) ≥ 1} belongs to K. P P Set un = 2n ((u ∧ χX) − (u ∧ (1 − 2−n )χX)). Because U is truncated, every un belongs to U , and it is easy to check that inf n∈N un = χK. Q Q It follows that 1 α

{x : u(x) ≥ α} = {x : u(x) ≥ 1} ∈ K whenever u ∈ U and α > 0. (c) For K ∈ K, set φ0 K = inf{f (u) : u ∈ U, u ≥ χK}. Then φ0 satisfies the conditions of 413I. P P I have already checked (†) and (‡) of 413I. α) Fix K, L ∈ K with L ⊆ K. Set γ = sup{φ0 K 0 : K 0 ∈ K, K 0 ⊆ K \ L}. (α (i) Suppose that K 0 ∈ K is included in K \ L, and ² > 0. Let hun in∈N , hu0n in∈N be sequences in U such that χL = inf n∈N un and χK 0 = inf n∈N u0n , and let u ∈ U be such that u ≥ χK and f (u) ≤ φ0 K + ². Set vn = u ∧ inf i≤n ui , vn0 = u ∧ inf i≤n u0i for each n. Then hvn ∧ vn0 in∈N is a non-increasing sequence in U with infimum χL ∧ χK 0 = 0, so there is an n such that f (vn ∧ vn0 ) ≤ ². In this case φ0 L + φ0 K 0 ≤ f (vn ) + f (vn0 ) = f (vn + vn0 ) = f (vn ∨ vn0 ) + f (vn ∧ vn0 ) ≤ f (u) + ² ≤ φ0 K + 2². As ² is arbitrary, φ0 L + φ0 K 0 ≤ φ0 K. As K 0 is arbitrary, φ0 L + γ ≤ φ0 K. (ii) Next, given ² ∈ ]0, 1[, there are u, v ∈ U such that u ≥ χK, v ≥ χL and f (v) ≤ φ0 L + ². Consider K 0 = {x : x ∈ K, min(1, u(x)) − v(x) ≥ ²} ⊆ K \ L. By (a) and (b), K 0 ∈ K. If w ∈ U and w ≥ χK 0 , then v(x) + w(x) ≥ 1 − ² for every x ∈ K, so φ0 K ≤

1 f (v 1−²

+ w) ≤

1 (φ0 L + ² + f (w)). 1−²

As w is arbitrary, (1 − ²)φ0 K ≤ φ0 L + ² + φ0 K 0 ≤ φ0 L + ² + γ. As ² is arbitrary, φ0 K ≤ φ0 L + γ and we have equality, as required by (α) in 413I. β ) Now suppose that hKn in∈N is a non-increasing sequence in K with empty intersection. For each (β n ∈ N let huni ii∈N be a sequence in U with infimum χKn in RX . Set vn = inf i,j≤n uji for each n; then hvn in∈N is a non-increasing sequence in U with infimum inf n∈N χKn = 0, so inf n∈N f (vn ) = 0. But vn ≥ inf j≤n χKj = χKn ,

φ0 Kn ≤ f (vn )

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211

for every n, so inf n∈N φ0 Kn = 0, as required by (β) of 413I. Q Q (d) By 413I, there is a complete locally determined measure µ on X, inner regular with respect to K, R extending φ0 . By 436C, f (u) = u dµ for every u ∈ U , as required. 436E Proposition Let X be any topological space, and Cb (X) the space of bounded continuous realvalued functions on X. Then there is a one-to-one correspondence between totally finite Baire measures µ on X and sequentially smooth positive linear functionals f : Cb (X) → R, given by the formulae f (u) =

R

u dµ for every u ∈ Cb (X),

µZ = inf{f (u) : χZ ≤ u ∈ Cb (X)} for every zero set Z ⊆ X. proof (a) If µ is a totally finite Baire measure on X, then every continuous bounded real-valued function R is integrable, and f = dµ is a sequentially smooth positive linear operator on Cb (X), by Fatou’s Lemma, as usual. (b) If f : Cb (X) → R is a Rsequentially smooth positive linear operator, then 436D tells us that there is a measure µ0 on X such that u dµ0 is defined and equal to f (u) for every u ∈ Cb (X). By the construction in 436D, or otherwise, we may suppose that µ0 is complete, so that every u ∈ Cb (X) is Σ-measurable, where Σ is the domain of µ0 . It follows by the definition of the Baire σ-algebra Ba of X (4A3K) that Ba ⊆ Σ, so R that µ = µ0 ¹Σ is a Baire measure; of course we still have f (u) = u dµ for every u ∈ Cb (X). Also, if Z ⊆ X is a zero set, µZ = inf{f (u) : χZ ≤ u ∈ Cb (X)}. P P Express Z as {x : v(x) = 0} where v : X → [0, 1] is continuous. Set un = (χX − 2n v)+ for n ∈ N; then hun in∈N is a non-decreasing sequence in Cb (X) and hun (x)in∈N → (χZ)(x) for every x ∈ X, so Z µZ ≤ inf{ u dµ : χZ ≤ u ∈ Cb (X)} = inf{f (u) : χZ ≤ u ∈ Cb (X)} Z ≤ inf f (un ) = lim un dµ = µZ. Q Q n→∞

n∈N

(c) The argument of (b) shows that if two totally finite Baire measures give the same integrals to every member of Cb (X), then they must agree on all zero sets. By the Monotone Class Theorem (136C) they agreeRon the σ-algebra generated by the zero sets, that is, Ba, and are therefore equal. Thus the operator µ 7→ dµ from the set of totally finite Baire measures on X to the set of sequentially smooth positive linear R operators on Cb (X) is a bijection, and if f = dµ then µZ = inf{f (u) : χZ ≤ u ∈ Cb (X)} for every zero set Z, as required. 436F Corresponding to 434R, we have the following construction for product Baire measures, applicable to a slightly larger class of spaces. Proposition Let X be a sequential space, Y a topological space, and µ, ν totally finite Baire measures on X, Y respectively. Then there is a Baire measure λ on X × Y such that λW =

R

νW [{x}]µ(dx),

R

f dλ =

RR

f (x, y)ν(dy)µ(dx)

for every Baire set W ⊆ X × Y and every bounded continuous function f : X × Y → R. RR proof (a) φ(f ) = f (x, P R y)dydx is defined in R for every bounded continuous function f : X × Y → R. P For each x ∈ X, g(x) = f (x, y)dy is defined because y 7→ f (x, y) is continuous. If hxn in∈N is any sequence in X converging to x ∈ X, then g(x) =

R

f (x, y)dy =

R

limn→∞ f (xn , y)dy = limn→∞

R

f (xn , y)dy = limn→∞ g(xn )

by Lebesgue’s Dominated Convergence Theorem. RR R So g is sequentially continuous; because X is sequential, g is continuous (4A2Kd). So f (x, y)dydx = g(x)dx is defined in R. Q Q

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(b) Of course φ is a positive linear functional on Cb (X × Y ), and B.Levi’s theorem shows that it is R sequentially smooth. By 436E, there is a Baire measure λ on X × Y such that f dλ = φ(f ) for every f ∈ Cb (X × Y ). (c) If W ⊆ X × Y is a zero set, there is a non-increasing sequence hfn in∈N in Cb (X × Y ) such that χW = inf n∈N fn . So B.Levi’s theorem tells us that

R

νW [{x}]dx = limn→∞

R

fn (x, y)dydx = limn→∞

R

fn dλ = λW .

Now the Monotone Class Theorem (136B) tells us that R {W : W ⊆ X × Y is Baire, νW [{x}]dx exists = λW } includes the σ-algebra generated by the zero sets, that is, contains every Baire set in X × Y . So λ has the required properties. 436G Definition Let X be a set, U a Riesz subspace of RX , and f : U → R a positive linear functional. I say that f is smooth if whenever A is a non-empty downwards-directed family in U such that inf u∈A u(x) = 0 for every x ∈ X, then inf u∈A f (u) = 0. Of course a smooth functional is sequentially smooth. If (X, T, Σ, µ) is an effectively locally finite τ additive topological measure space and U is a Riesz subspace of RX consisting of integrable continuous R functions, then dµ : U → R is smooth, by 414Bb. Corresponding to the remark in 436A, note that an order-continuous positive linear functional must be smooth, but that a smooth positive linear functional need not be order-continuous. 436H Theorem Let X be a set and U a truncated Riesz subspace of RX . Let f : U → R be a positive linear functional. Then the following are equiveridical: (i) f is smooth; (ii) there are a topology T and a measure µ on X such that µ is a quasi-Radon measure with respect to R T, U ⊆ C(X) and u dµ is defined and equal to f (u) for every u ∈ U ; (iii) writing S for the coarsest topology on X for which every member of U R is continuous, there is a measure µ on X such that µ is a quasi-Radon measure with respect to S, and u dµ is defined and equal to f (u) for every u ∈ U . proof As remarked in 436G, in a fractionally more general context, (ii)⇒(i) is a consequence of 414B. Of course (iii)⇒(ii). So the argument here is devoted to proving that (i)⇒(iii). Except for part (b) it is a simple adaptation of the method of 436D. (a) Let K be the family of sets K ⊆ X such that χK = inf A in RX for some non-empty set A ⊆ U . Then K is closed under finite unions and arbitrary intersections. P P (i) If K, K 0 ∈ K take A, A0 ⊆ U such 0 0 0 0 that χK = inf A, χK = A ; then χ(K ∪ K ) = inf{uS∨ u : u ∈ A, u0 ∈ A0 }, so K ∪ K 0 ∈ K. (ii) If L is a non-empty subset of K with intersection K, set A = L∈L {u : χL ≤ u ∈ U }; then χK = inf A, so K ∈ K. Q Q Note that ∅ ∈ K because 0 ∈ U . As in part (b) of the proof of 436D, {x : u(x) ≥ α} ∈ K whenever α > 0 and u ∈ U . (b) Every member of K is closed for S, being of the form {x : u(x) ≥ 1 for every u ∈ A} for some A ⊆ U . We need to know that if K ∈ K and G ∈ S, then K \ G ∈ K. P P Take a non-empty set B ⊆ U such that χK = inf B. Because K is closed under finite unions and arbitrary intersections, SK = {G : G ⊆ X, K \G ∈ K} is a topology on X. (i) If u ∈ U and G = {x : u(x) > 0}, then χ(K \ G) = inf{(v − ku)+ : v ∈ B, k ∈ N} so K \ G ∈ K and G ∈ SK . (ii) If u ∈ U and α > 0, then {x : u(x) > α} = {x : (u − u ∧ αχX)(x) > 0} belongs to SK , by (i). (iii) If u ∈ U and α > 0, set G = {x : u(x) < α}. Then K \ G = K ∩ {x : u(x) ≥ α} ∈ K, so G ∈ SK . (iv) Thus every member of U + is SK -continuous (2A3Bc), so every member of U is SK continuous (2A3Be), and S ⊆ SK , that is, K \ G ∈ K for every G ∈ S. Q Q (c) For K ∈ K, set φ0 K = inf{f (u) : u ∈ U, u ≥ χK}. Then φ0 satisfies the conditions of 415K. P PI have already checked (†) and (‡) of 415K.

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α) Fix K, L ∈ K with L ⊆ K. Set γ = sup{φ0 K 0 : K 0 ∈ K, K 0 ⊆ K \ L}. (α (i) Suppose that K 0 ∈ K is included in K \ L and ² > 0. Set A = {u : χL ≤ u ∈ U }, A0 = {u : χK 0 ≤ u ∈ U }, so that χL = inf A and χK 0 = inf A0 , and let v ∈ U be such that v ≥ χK and f (v) ≤ φ0 K + ². Then {u ∧ u0 : u ∈ A, u0 ∈ A0 } is a downwards-directed family with infimum 0 in RX , so (because f is smooth) there are u ∈ A, u0 ∈ A0 such that f (u ∧ u0 ) ≤ ². In this case φ0 L + φ0 K 0 ≤ f (v ∧ u) + f (v ∧ u0 ) = f (v ∧ (u ∨ u0 )) + f (v ∧ u ∧ u0 ) ≤ φ0 K + 2². As ² is arbitrary, φ0 L + φ0 K 0 ≤ φ0 K. As K 0 is arbitrary, φ0 L + γ ≤ φ0 K. (ii) Next, given ² ∈ ]0, 1[, there are u, v ∈ U such that u ≥ χK, v ≥ χL and f (v) ≤ φ0 L + ². Consider K 0 = {x : x ∈ K, min(1, u(x)) − v(x) ≥ ²}. By the last remark in (a), K 0 ∈ K. If w ∈ U and w ≥ χK 0 , then v(x) + w(x) ≥ 1 − ² for every x ∈ K, so φ0 K ≤

1 f (v 1−²

+ w) ≤

1 (φ0 L + ² + f (w)). 1−²

As w is arbitrary, (1 − ²)φ0 K ≤ φ0 L + ² + φ0 K 0 ≤ φ0 L + ² + γ. As ² is arbitrary, φ0 K ≤ φ0 L + γ and we have equality, as required by (α) in 415K. β ) Now suppose that K0 is a non-empty downwards-directed subset of K with empty intersection. Set (β S A = K∈K0 {u : χK ≤ u ∈ U }. Then A is a downwards-directed subset of U and inf A = 0 in RX . Because f is smooth, 0 = inf u∈A f (u) = inf K∈K0 φ0 K. Thus (β) of 415K is satisfied. (γγ ) If K ∈ K and φ0 K > 0, take u ∈ U such that u ≥ χK, and consider G = {x : u(x) > 21 }. Then K ⊆ G, while G ⊆ {x : 2u(x) ≥ 1}, so sup{φ0 K 0 : K 0 ∈ K, K 0 ⊆ G} ≤ 2f (u) < ∞. Thus φ0 satisfies (γ) of 415K. Q Q (d) By 415K, there is a quasi-Radon measure µ on X extending φ0 . By 436C, f (u) = u ∈ U.

R

u dµ for every

Remark It is worth noting explicitly that µ, as constructed here, is inner regular with respect to the family K of sets K ⊆ X such that χK = inf A for some set A ⊆ U . 436I Lemma Let X be a topological space. Let C0 (X) be the space of continuous functions u : X → R which ‘vanish at infinity’ in the sense that {x : |u(x)| ≥ ²} is compact for every ² > 0. (a) C0 (X) is a norm-closed solid linear subspace of Cb (X), so is a Banach lattice in its own right. (b) C0 (X)∗ = C0 (X)∼ is an L-space (definition: 354M). (c) If A ⊆ C0 (X) is a non-empty downwards-directed set such that inf u∈A u(x) = 0 for every x ∈ X, then inf u∈A kuk∞ = 0. proof (a)(i) If u ∈ C0 (X), then K = {x : |u(x)| ≥ 1} is compact, so kuk∞ ≤ sup({1} ∪ {|u(x)| : x ∈ K}) is finite, and u ∈ Cb (X). (ii) If u, v ∈ C0 (X) and α ∈ R and w ∈ Cb (X) and |w| ≤ |u|, then for any ² > 0 1 2

1 2

{x : |u(x) + v(x)| ≥ ²} ⊆ {x : |u(x)| ≥ ²} ∪ {x : |v(x)| ≥ ²}, {x : |αu(x)| ≥ ²} ⊆ {x : |u(x)| ≥

² }, 1+|α|

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{x : |w(x)| ≥ ²} ⊆ {x : |u(x)| ≥ ²} are closed relatively compact sets, so are compact, and u + v, αu, w belong to C0 (X). Thus C0 (X) is a solid linear subspace of Cb (X). (iii) If hun in∈N is a sequence in C0 (X) which k k∞ -converges to u ∈ Cb (X), then for any ² > 0 there is an n ∈ N such that ku − un k∞ ≤ 21 ², so that 1 2

{x : |u(x)| ≥ ²} ⊆ {x : |un (x)| ≥ ²} is compact, and u ∈ C0 (X). Thus C0 (X) is norm-closed in Cb (X). (iv) Being a norm-closed Riesz subspace of a Banach lattice, C0 (X) is itself a Banach lattice. (b) By 356Dc, C0 (X)∗ = C0 (X)∼ is a Banach lattice. Now kf + gk = kf k + kgk for all non-negative f , g ∈ C0 (X)∗ . P P Of course kf + gk ≤ kf k + kgk. On the other hand, for any ² > 0 there are u, v ∈ C0 (X) such that kuk∞ ≤ 1, kvk∞ ≤ 1 and |f (u)| ≥ kf k − ², |g(v)| ≥ kgk − ². Set w = |u| ∨ |v|; then w ∈ C0 (X) and kwk∞ = max(kuk∞ , kvk∞ ) ≤ 1. So kf + gk ≥ (f + g)(w) ≥ f (|u|) + g(|v|) ≥ |f (u)| + |g(v)| ≥ kf k + kgk − 2². As ² is arbitrary, kf + gk ≥ kf k + kgk. Q Q So C0 (X)∗ is an L-space. (c) Let ² > 0. For u ∈ A set Ku = {x : u(x) ≥ ²}. Then {Ku : u ∈ A} is a downwards-directed family of closed compact sets with empty intersection, so there must be some u ∈ A such that Ku = ∅, and kuk∞ ≤ ². As ² is arbitrary, we have the result. Remark (c) is a version of Dini’s theorem. 436J Riesz Representation Theorem (first form) Let (X, T) be a locally compact Hausdorff space, and Ck (X) the space of continuous real-valued functions on X with compact support. If f R: Ck (X) → R is any positive linear functional, there is a unique Radon measure µ on X such that f (u) = u dµ for every u ∈ Ck (X). proof (a) The point is that f is smooth. P P Suppose that A ⊆ Ck (X) is non-empty and downwards-directed and that inf A = 0 in RX . Fix u0 ∈ A and set K = {x : u0 (x) > 0}, so that K is compact. Because X is locally compact, there is an open relatively compact set G ⊇ K. Now there is a continuous function u1 : X → [0, 1] such that u1 (x) = 1 for x ∈ K and u1 (x) = 0 for x ∈ X \ G (4A2F(h-iii)). Because G is relatively compact, u1 ∈ Ck (X). Take any ² > 0. By 436Ic, there is a v ∈ A such that kvk∞ ≤ ². Now there is a v 0 ∈ A such that 0 v ≤ v ∧ u0 , so that v 0 (x) ≤ ² for every x ∈ K and v 0 (x) = 0 for x ∈ / K. In this case v 0 ≤ ²u1 , and inf u∈A f (u) ≤ f (v 0 ) ≤ ²f (u1 ). As ² is arbitrary, inf u∈A f (u) = 0; as A is arbitrary, f is smooth. Q Q (b) Note that because T is locally compact, it is the coarsest topology on X for which every function in Ck (X) is continuous (4A2G(e-ii)). Also Ck (X) is a truncated Riesz subspace of R X . So 436H tells us that R there is a quasi-Radon measure µ on X such that f (u) = u dµ for every u ∈ Ck (X). And µ is locally finite. P P If x0 ∈ X, then (as in (a) above) there is a u1 ∈ Ck (X)+ such that u1 (x0 ) = 1; now G = {x : u1 (x) > 12 } is an open set containing x0 , and µG ≤ 2f (u1 ) is finite. Q Q By 416G, or otherwise, µ is a Radon measure. (c) By 416E(b-v), µ is unique. 436K Riesz Representation Theorem (second form) Let (X, T) be a locally compact Hausdorff space. If f : C0 (X) → RR is any positive linear functional, there is a unique totally finite Radon measure µ on X such that f (u) = u dµ for every u ∈ C0 (X).

*436L

Representation of linear functionals

215

proof (a) As noted in 436Ib, C0 (X)∗ = C0 (X)∼ , so f is k k∞ -continuous. Ck (X) is a linear subspace of C0 (X), and f ¹CkR(X) is a positive linear functional; so by 436J there is a unique Radon measure µ on X such that f (u) = u dµ for every u ∈ Ck (X). Now µ is totally finite. P P By 414Ab, µX = sup{f (u) : u ∈ Ck (X), 0 ≤ u ≤ χX} ≤ kf k < ∞. Q Q R (b) Accordingly u dµ is defined for every u ∈ Cb (X), and in particular for every u ∈ C0 (X). Next, Ck (X) is norm-dense in C0 (X). P P If u ∈ C0 (X)+ , then un = (u − 2−n χX)+ belongs to Ck (X) and −n ku − un k∞ for every n ∈ N, so u ∈ C k ; accordingly C0 (X) = C0 (X)+ − C0 (X)+ is included in C k . R ≤2 Q Q Since dµ, regarded as a linear functional on C0R(X), is positive, therefore continuous, and agrees with f on Ck (X), it must be identical to f . Thus f (u) = u dµ for every u ∈ C0 (X). (c) Because there is only one Radon measure giving the right integrals to members of Ck (X) (436J), µ is unique. *436L The results here, by opening a path between measure theory and the study of linear functionals on spaces of continuous functions, provide an enormously powerful tool for the analysis of dual spaces C(X)∗ and their relatives. I will explore some of these ideas in the next section. Here I will give only a sample theorem to show how measure theory can tell us something about Banach lattices which seems difficult to reach by other methods. Proposition Let X be a topological space and U a norm-closed linear subspace of Cb (X)∗ such that the functional u 7→ f (u × v) : Cb (X) → R belongs to U whenever f ∈ U and v ∈ Cb (X). Then U is a band in the L-space Cb (X)∗ . proof (a) Let e = χX be the standard order unit of Cb (X), and if f ∈ Cb (X)∗ and u, v ∈ Cb (X) write fv (u) for f (u × v). By 356Na, Cb (X)∗ = Cb (X)∼ is an L-space. (b) I show first that U is a Riesz subspace of Cb (X)∼ . P P If f ∈ U and ² > 0, there is a v ∈ Cb (X) such that |v| ≤ e and f (v) ≥ |f |(e) − ² (356B). Now fv ≤ |f | and k|f | − fv k = (|f | − fv )(e) ≤ ² (356Nb), while fv ∈ U . As ² is arbitrary, |f | ∈ U = U ; as f is arbitrary, U is a Riesz subspace of Cb (X)∗ (352Ic). Q Q (c) Now suppose that X is a compact Hausdorff space. Then U is a solid linear subspace of Cb (X)∼ = C(X)∼ . P P Suppose that f ∈ U and that 0 ≤ R g ≤ f . Let ² >R0. By either 436J or 436K, there are Radon measures µ, ν on X such that f (u) = u dµ and g(u) = u dν for every u ∈ C(X). By 416Ea, dom µ ⊆ dom ν and νE ≤ µE for every E ∈ dom R µ, so ν is an indefinite-integral measure over µ (415Oa, or otherwise); let w : X → [0,R1] be such that E w dµ = νE for every E ∈ dom ν. There is a continuous function v : X → R such that |w − v|dµ ≤ ² (416I), and now Z |g(u) − fv (u)| = |

Z u dν −

(235M)

Z u × v dµ| = |

u × (w − v)dµ|

Z ≤ kuk∞

|w − v|dµ ≤ ²kuk∞

for every u ∈ C(X), so kg − fv k ≤ ², while fv ∈ U . As ² is arbitrary, g ∈ U ; as f and g are arbitrary (and U is a Riesz subspace of C(X)∗ ), U is a solid linear subspace of C(X)∗ . Q Q Since every norm-closed solid linear subspace of an L-space is a band (use 354Ea), it follows that (provided X is a compact Hausdorff space) U is actually a band. (c) For the general case, let Z be the set of all Riesz homomorphisms z : Cb (X) → R such that z(e) = 1. Then we have a Banach lattice isomorphism T : Cb (X) → C(Z) given by the formula (T u)(z) = z(u) for u ∈ Cb (X), z ∈ Z (see the proofs of 353M and 354K). But note also that T is multiplicative (353Pd), and

216

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*436L

T 0 : C(Z)∗ → Cb (X)∗ is a Banach lattice isomorphism. Let V = (T 0 )−1 [U ] ⊆ C(Z)∗ ; then V is a closed linear subspace of C(Z)∗ . If g ∈ V and v, w ∈ C(Z), then g(v × w) = (T 0 g)(T −1 v × T −1 w), so gw , defined in C(Z)∗ by the convention of (a) above, is just (T 0 )−1 ((T 0 g)T −1 w ), and belongs to V . By (b), V is a band in C(Z)∗ so U is a band in Cb (X)∗ , as required. 436X Basic exercises > (a) Let (X, Σ, µ0 ) be a measure space, and U the R set of µ0 -integrable Σmeasurable real-valued functions defined everywhere on X. For u ∈ U set f (u) = u dµ0 . Show that U and f satisfy the conditions of 436D, and that the measure µ constructed from f by the procedure there is just the c.l.d. version of µ0 . R R (b) Let µ and ν be two complete locally determined measures on a set X, and suppose that f dµ = f dν for every function f : X → R for which either integral is defined in R. Show that µ = ν. > (c) Let X be a set, U a truncated Riesz subspace of RX , and f : U → R a sequentially smooth positive linear functional. For A ⊆ X set θA = inf{sup f (un ) : hun in∈N is a non-decreasing sequence in U + , n∈N

lim un (x) = 1 for every x ∈ A},

n→∞

taking inf ∅ = ∞ if need be. Show that θ is an R outer measure on X. Let µ0 be the measure defined from θ by Carath´eodory’s method. Show that f (u) = u dµ0 for every u ∈ U . Show that the measure µ constructed in 436D is the c.l.d. version of µ0 . (d) Let X be a set and U a truncated Riesz subspace of RX . Let τ : U → [0, ∞[ be a seminorm such that (i) τ (u) ≤ τ (v) whenever |u| ≤ |v| (ii) limn→∞ τ (un ) = 0 whenever hun in∈N is a non-increasing sequence in U Rand limn→∞ un (x) = 0 for every x ∈ X. Show that for any u0 ∈ U + Rthere is a measure µ on x such that u dµ is defined, and less than or equal to τ (u), for every u ∈ U , and u0 dµ = τ (u0 ). (Hint: put the Hahn-Banach theorem together with 436D.) (e) Let X be any topological space. Show that every positive linear functional on C(X) is sequentially smooth (compare 375A), so corresponds to a totally finite Baire measure on X. (f ) Let X be a completely regular space. Show that it is measure-compact iff every sequentially smooth positive linear functional on Cb (X) is smooth. (Hint: 436Xj.) (g) A completely regular space is called realcompact if every Riesz homomorphism from C(X) to R is of the form u 7→ αu(x) for some x ∈ X and α ≥ 0. (i) Show that, for any topological space X, any Riesz homomorphism from C(X) to R is representable by a Baire measure on X which takes at most two values. (ii) Show that a completely regular space X is realcompact iff every {0, 1}-valued Baire measure on X is of the form E 7→ χE(x). (iii) Show that a completely regular space X is realcompact iff every purely atomic totally finite Baire measure on X is τ -additive. (iii) Show that a measure-compact completely regular space is realcompact. (iv) Show that the discrete topology on [0, 1] is realcompact. (Hint: if ν is a Baire measure taking only the values 0 and 1, set x0 = sup{x : ν[x, 1] = 1}.) (v) Show that any product of realcompact completely regular spaces is realcompact. (vi) Show that a Souslin-F subset of a realcompact completely regular space is realcompact. (For realcompact spaces which are not measure-compact, see 439Xm.) (h) In 436F, suppose that µ and ν are τ -additive. Let µ ˜ and ν˜ be the corresponding quasi-Radon measures ˜ the quasi-Radon product of µ ˜ to the Baire (415N), and λ ˜ and ν˜ (417N). Show that λ is the restriction of λ σ-algebra of X × Y . >(i) For u ∈ C([0, 1]) let f (u) be the Lebesgue integral of u. Show that f is smooth (therefore sequentially smooth) but not sequentially order-continuous (therefore not order-continuous). (Hint: enumerate Q ∩ [0, 1] as hqn in∈N , and set un (x) = mini≤n 2i+2 |x − qi | for n ∈ N, x ∈ [0, 1]; show that inf n∈N un = 0 in C([0, 1]) but limn→∞ f (un ) > 0.)

436Y

Representation of linear functionals

217

(j) In 436E, show that µ is τ -additive iff f is smooth. (k) Suppose that X is a set, U is a truncated Riesz subspace of RX and f : U → R is a smooth positive linear functional. For A ⊆ X set θA = inf{sup f (u) : B is an upwards-directed family in U + u∈B

such that sup u(x) = 1 for every x ∈ A}, u∈B

taking inf ∅ = ∞ if need be. Show that θ is an R outer measure on X. Let µ0 be the measure defined from θ by Carath´eodory’s method. Show that f (u) = u dµ0 for every u ∈ U . Show that the measure µ constructed in 436H is the c.l.d. version of µ0 . (l) Let X be a completely regular topological space and f a smooth positive linear functional on Cb (X). R Show that there is a unique totally finite quasi-Radon measure µ on X such that f (u) = u dµ for every u ∈ Cb (X). > (m) For u ∈ C([0, 1]) let f (u) be the Riemann integral of u (134K). Show that the Radon measure on [0, 1] constructed by the method of 436J is just Lebesgue measure on [0, 1]. Explain how to construct Lebesgue measure on R from an appropriate version of the Riemann integral on R. (n) Let X be a topological space. Let f : Cb (X) → R be a positive linear functional. Show that the following are equiveridical: (i) f is tight, that is, for every ² > 0 there is a compact K ⊆ X Rsuch that f (u) ≤ ² whenever 0 ≤ u ≤ χ(X \ K) (ii) there is a tight Borel measure µ on X such that f (u) = u dµ for every u ∈ Cb (X). (Hint: show that a tight functional is smooth.) > (o) Let (X, R T, Σ, µ) and (Y, S, T, ν) be locally compact Radon measure spaces. (i) Show that the function x 7→ w(x, y)ν(dy) belongs to Ck (X) for every w ∈ Ck (X × Y ), so we have a positive linear functional h : Ck (X × Y ) → R defined by setting h(w) =

RR

w(x, y)ν(dy)µ(dx)

for w ∈ Ck (X × Y ). (ii) Show that the corresponding Radon measure on X × Y is just the product Radon measure as defined in 417P/417R. > (p) Let A be a Boolean algebra and Z its Stone space; identify L∞ (A) with C(Z), as in 363A. Let ν be a non-negative finitely additive functional on A, f the corresponding positive linear R functional on L∞ (A) (363K), and µ the corresponding Radon measure on Z (416Qb). Show that f (u) = u dµ for every u ∈ L∞ (A). (q) Let X be a locally compact Hausdorff space. Show that a sequence hfn in∈N in C0 (X) converges to 0 for the weak topology on C0 (X) iff supn∈N kfn k∞ is finite and limn→∞ fn (x) = 0 for every x ∈ X. (r) Let A be a Boolean algebra, and M (A) the L-space of bounded finitely additive functionals on A. Let U ⊆ M (A) be a norm-closed linear subspace such that a 7→ ν(a ∩ b) belongs to U whenever ν ∈ U and b ∈ A. Show that U is a band in M (A). (Hint: identify L∞ (A) with Cb (X), where X is the Stone space of A, and M (A) with Cb (X)∗ .) (s) Let X be a non-empty compact Hausdorff space, and φ : X → X a continuous function. Show that there is a Radon probability measure µ on X such that φ is inverse-measure-preserving for µ.R (Hint: let F be a non-principal ultrafilter on X, x0 any point of X, and define µ by the formula f dµ = 1 Pn k −1 = µ.) limn→F k=0 f (φ (x0 )) for every f ∈ C(X). Use 416E(b-v) to show that µφ n+1

436Y Further exercises (a) Let X be a set, U a Riesz subspace of RX , and f : U → R a sequentially smooth positive linear functional. (i) Write Uσ for the set of functions from X to [0, ∞] expressible as the supremum of a non-decreasing sequence hun in∈N in U + such that supn∈N f (un ) < ∞. Show that there is a

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436Y

functional fσ : Uσ → [0, ∞[ such that fσ (u) = supn∈N f (un ) whenever hun in∈N is a non-decreasing sequence in U + with supremum u ∈ Uσ . (Compare 122I.) (ii) Show that u + v ∈ Uσ and fσ (u + v) = fσ (u) + fσ (v) for all u, v ∈ Uσ . (iii) Suppose that u, v ∈ Uσ , u ≤ v and u(x) = v(x) whenever v(x) is finite. Show that fσ (u) = fσ (v). (Hint: take non-decreasing sequences hun in∈N , hvn in∈N with suprema u, v. Consider hf (vk −un −δvn )+ in∈N where k ∈ N, δ > 0.) (iv) Let V be the set of functions v : X → R such that there are u1 , u2 ∈ Uσ such that v(x) = u1 (x) − u2 (x) whenever u1 (x), u2 (x) are both finite. Show that V is a linear subspace of RX and that there is a linear functional g : V → R defined by setting g(v) = fσ (u1 ) − fσ (u2 ) whenever v = u1 − u2 in the sense of the last sentence. (v) Show that V is a Riesz subspace of RX . (vi) Show that if hvn in∈N is a non-decreasing sequence in V and γ = supn∈N g(vn ) is finite, then there is a v ∈ V such that g(v) = supn∈N g(v ∧ vn ) = γ. (This is a version of the Daniell integral.) (b) Develop further the theory of 436Ya, finding a version of Lebesgue’s Dominated Convergence Theorem, a concept of ‘negligible’ subset of X, and an L-space of equivalence classes of ‘integrable’ functions. (c) Let X be a countably compact topological space. Show that every positive linear functional on Cb (X) is sequentially smooth, so corresponds to a totally finite Baire measure on X. (d) Let X be a sequential space, Y a topological space, µ a semi-finite Baire measure on X and ν a σ-finite Baire measure on Y . Let µ ˜ be the c.l.d. version of µ. Show that there is a semi-finite Baire measure λ on X × Y such that λW =

R

νW [{x}]˜ µ(dx),

R

f dλ =

RR

f (x, y)ν(dy)˜ µ(dx)

for every Baire set W ⊆ X × Y and every non-negative continuous function f : X × Y → R. Show that the c.l.d. version of λ extends the c.l.d. product measure of µ and ν. (e) Let X0 , . . . , Xn be sequential spaces and µi a totally finite Baire measure on Xi for each i. (i) Show that if f : X0 × . . . × Xn → R is a bounded separately continuous function, then φ(f ) =

R

...

R

f (x0 , . . . , xn )µn (dxn ) . . . µ0 (dx0 )

is defined, so that we have a corresponding Baire product measure on X0 × . . . × Xn . (ii) Show that this product is associative. (f ) Let X and Y be compact Hausdorff spaces. (i) Show that there is a unique bilinear map φ : C(X)∗ × C(Y )∗ → C(X × Y )∗ which is separately continuous for the weak* topologies and such that φ(δx , δy ) = δ(x,y) for all x ∈ X and y ∈ Y , setting δx (f ) = f (x) for f ∈ C(X) and x ∈ X.R(ii) Show thatR if µ R and ν are Radon measures on X, Y respectively with Radon measure product λ, then φ( dµ, dν) = dλ. (iii) Show that kφk ≤ 1 (definition: 253Ab). 436 Notes and comments From the beginning, integration has been at the centre of measure theory. My own view – implicit in the arrangement of this treatise, from Chapter 11 onward – is that ‘measure’ and ‘integration’ are not quite the same thing. I freely acknowledge that my treatment of ‘integration’ is distorted by my presentation of it as part of measure theory; on the other side of the argument, I hold that regarding ‘measure’ as a concept subsidiary to ‘integral’, as many authors do, seriously interferes with the development of truly penetrating intuitions for the former. But it is undoubtedly true that every complete locally determined measure can be derived from its associated integral (436Xb). Moreover, it is clearly of the highest importance that we should be able to recognise integrals when we see them; I mean, given a linear functional on a linear space of functions, then if it can be expressed as an integral with respect to a measure this is something we need to know at once. And thirdly, investigation of linear functionals frequently leads us to measures of great importance and interest. Concerning the conditions in 436D, an integral must surely be ‘positive’ (because measures in this treatise are always non-negative) and ‘sequentially smooth’ (because measures are supposed to be countably additive). But it is not clear that we are forced to restrict our attention to Riesz subspaces of RX , and even less clear that they have to be ‘truncated’. In 439I below I give an example to show that this last condition is essential for 436D and 436H as stated. However it is not necessary for large parts of the theory. In many cases, if U ⊆ RX is a Riesz subspace which is not truncated, we can take an element e ∈ U + and look at

437A

Spaces of measures

219

Y = {x : e(x) > 0}, Ve = {u : u(x) = 0 for every x ∈ X \ Y } ∼ = We , where We = {u/e : u ∈ Ve } is a truncated Riesz subspace of R Y . But there are applications in which this approach is unsatisfactory and a more radical revision of the basic theory of integration, as in 436Ya, is useful. I have based the arguments of this section on the inner measure constructions of §413. Of course it is also possible to approach them by means of outer measures (436Xc, 436Xk). I emphasize the exercise 436Xo because it is prominent in ‘Bourbakist’ versions of the theory of Radon measures, in which (following Bourbaki 65 rather than Bourbaki 69) Radon measures are regarded as linear functionals on spaces of continuous functions. By 436J, this is a reasonably effective approach as long as we are interested only in locally compact spaces, and there are parts of the theory of topological groups (notably the duality theory of §445 below) in which it even has advantages. The construction of 436Xo shows that we can find a direct approach to the tensor product of linear functionals which does not require any attempt to measure sets rather than integrate functions. I trust that the prejudices I am expressing will not be taken as too sweeping a disparagement of such methods. Practically all correct arguments in mathematics (and not a few incorrect ones) are valuable in some way, suggesting new possibilities for investigation. In particular, one of the challenges of measure theory (not faced in this treatise) is that of devising effective theories of vector-valued measures. Typically this is much easier with Riemann-type integrals, and any techniques for working directly with these should be noted. 436L revisits ideas from Chapter 35, and the result would be easier to find if it were in §356. I include it here as an example of the way in which familiar material from measure theory (in particular, the RadonNikod´ ym theorem) can be drafted to serve functional analysis. I should perhaps remark that there are alternative routes which do not use measure theory explicitly, and while longer are (in my view) more illuminating.

437 Spaces of measures Once we have started to take the correspondence between measures and integrals as something which operates in both directions, we can go a very long way. While ‘measures’, as dealt with in this treatise, are essentially positive, an ‘integral’ can be thought of as a member of a linear space, dual in some sense to a space of functions. Since the principal spaces of functions are Riesz spaces, we find ourselves looking at dual Riesz spaces as discussed in §356; while the corresponding spaces of measures are close to those of §362. Here I try to draw these ideas together with an examination of the spaces Uσ∼ and Uτ∼ of sequentially smooth and smooth functionals, and the matching spaces Mσ and Mτ of countably additive and τ -additive measures (437A-437I). Because a (sequentially) smooth functional corresponds to a countably additive measure, which can be expected to integrate many more functions than those in the original Riesz space (typically, a space of continuous functions), we find that relatively large spaces of bounded measurable functions can be canonically embedded into the biduals (Uσ∼ )∗ and (Uτ∼ )∗ (437C, 437H, 437I). The guiding principles of functional analysis encourage us not only to form linear spaces, but also to examine linear space topologies, starting with norm and weak topologies. The theory of Banach lattices described in §354, particularly the theory of M - and L-spaces, is an important part of the structure here. In addition, our spaces Uσ∼ have natural weak* topologies which can be regarded as topologies on spaces of measures; these are the ‘vague’ topologies of 437J, which have already been considered, in a special case, in §285. It turns out that on the positive cone of Mτ , at least, the vague topology may have an alternative description directly in terms of the behaviour of the measures on open sets (437L). This leads us to a parallel idea, the ‘narrow’ topology on non-negative additive functionals (437Jd). The second half of the section is devoted to the elementary properties of narrow topologies (437K-437R), with especial reference to compact sets in these topologies (437O-437U). Seeking to identify narrowly compact sets, we come to the concept of ‘uniform tightness’ (437T). Bounded uniformly tight sets are narrowly relatively compact (437U); in ‘Prokhorov spaces’ (437V) the converse is true; I end the section with a list of the best-known Prokhorov spaces (437W). 437A Smooth and sequentially smooth duals Let X be a set, and U a Riesz subspace of RX . Recall that U ∼ is the Dedekind complete Riesz space of order-bounded linear functionals on U , that Uc∼ is the band of differences of sequentially order-continuous positive linear functionals, and that U × is the band of

220

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437A

differences of order-continuous positive linear functionals (356A). A functional f ∈ (U ∼ )+ is ‘sequentially smooth’ if inf n∈N f (un ) = 0 whenever hun in∈N is a non-increasing sequence in U and limn→∞ un (x) = 0 for every x ∈ X, and ‘smooth’ if inf u∈A f (u) = 0 whenever A ⊆ U is a non-empty downwards-directed set and inf u∈A u(x) = 0 for every x ∈ X (436A, 436G). (a) Set Uσ∼ = {f : f ∈ U ∼ , |f | is sequentially smooth}, the sequentially smooth dual of U . Then Uσ∼ is a band in U ∼ . P P (i) If f ∈ Uσ∼ , g ∈ U ∼ and |g| ≤ |f |, then |g|(un ) ≤ |f |(un ) → 0 whenever hun in∈N is a non-increasing sequence in U and limn→∞ un (x) = 0 for every x, so |g| is sequentially smooth and g ∈ Uσ∼ . Thus Uσ∼ is a solid subset of U ∼ . (ii) If f , g ∈ Uσ∼ and α ∈ R, then |f + g|(un ) ≤ |f |(un ) + |g|(un ) → 0,

|αf |(un ) = |α||f |(un ) → 0

whenever hun in∈N is a non-increasing sequence in U and limn→∞ un (x) = 0 for every x, so |f + g| and |αf | are sequentially smooth. Thus Uσ∼ is a Riesz subspace of U ∼ . (iii) Now suppose that B ⊆ (Uσ∼ )+ is an upwards-directed set with supremum g ∈ U ∼ , and that hun in∈N is a non-increasing sequence in U such that limn→∞ un (x) = 0 for every x ∈ X. Then, given ² > 0, there is an f ∈ B such that (g − f )(u0 ) ≤ ² (355Ed), so that g(un ) ≤ f (un ) + ² for every n, and lim supn→∞ g(un ) ≤ ² + limn→∞ f (un ) ≤ ². As ² and hun in∈N are arbitrary, g ∈ Uσ∼ ; as B is arbitrary, Uσ∼ is a band (352Ob). Q Q As remarked in 436A, sequentially order-continuous positive linear functionals are sequentially smooth, so Uc∼ ⊆ Uσ∼ . P (i) (b) Set Uτ∼ = {f : f ∈ U ∼ , |f | is smooth}, the smooth dual of U . Then Uτ∼ is a band in U ∼ . P Suppose that f , g ∈ Uτ∼ , α ∈ R, h ∈ U ∼ and |h| ≤ |f |. If A ⊆ U is a non-empty downwards-directed set and inf u∈A u(x) = 0 for every x ∈ X, and ² > 0, then there are u0 , u1 ∈ A such that |f |(u0 ) ≤ ² and |g|(u1 ) ≤ ², and a u ∈ A such that u ≤ u0 ∧ u1 . In this case |h|(u) ≤ |f |(u) ≤ ², |f + g|(u) ≤ |f |(u) + |g|(u) ≤ 2², |αf |(u) = |α||f |(u) ≤ |α|². As A and ² are arbitrary, h, f + g and αf all belong to Uτ∼ ; so that Uτ∼ is a solid Riesz subspace of U ∼ . (ii) Now suppose that B ⊆ (Uτ∼ )+ is an upwards-directed set with supremum g ∈ U ∼ , and that A ⊆ U is a non-empty downwards-directed set such that inf u∈A u(x) = 0 for every x ∈ X. Fix any u0 ∈ A. Then, given ² > 0, there is an f ∈ B such that (g − f )(u0 ) ≤ ², so that g(u) ≤ f (u) + ² whenever u ∈ A and u ≤ u0 . But A0 = {u : u ∈ A, u ≤ u0 } is also a downwards-directed set, and inf u∈A0 u(x) = 0 for every x ∈ X, so inf u∈A g(u) ≤ ² + inf u∈A0 f (u) ≤ ². As ² and A are arbitrary, g ∈ Uτ∼ ; as B is arbitrary, Uτ∼ is a band. Q Q Just as Uc∼ ⊆ Uσ∼ , U × ⊆ Uτ∼ . 437B Signed measures Collecting these ideas together with those of §§362-363, we are ready to approach ‘signed measures’. Recall that if X is a set and Σ is a σ-algebra of subsets of X, we can identify L∞ = L∞ (Σ), as defined in §363, with the space L∞ of bounded Σ-measurable real-valued functions (363Ha). Now, because L∞ is sequentially order-closed in RX , sequentially smooth functionals on L∞ are actually sequentially order∞ ∼ ∞ ∼ continuous, so (L∞ )∼ σ = (L )c . Next, we can identify (L )c with the space Mσ of countably additive functionals, or ‘signed measures’, on Σ (363K); if ν ∈ Mσ , the corresponding member of (L∞ )∼ c is the unique order-bounded (or norm-continuous) linear functional f on L∞ such that f (χE) = νE for every E ∈ Σ. If ν ≥ 0, so that ν is a totally finite measure with domain Σ, then of course f , when interpreted as a functional on L∞ , must be just integration with respect to ν. The identification between (L∞ )∼ in 363K is an L-space isomorphism. So it tells us, c and Mσ described R for instance, that if we are willing to use the symbol for the duality between L∞ and the space of bounded finitely additive functionals on Σ, as in 363L, then we can write

437C

Spaces of measures

R ∞

for every u ∈ L

u d(µ + ν) =

R

u dµ +

221

R

u dν

and all µ, ν ∈ Mσ .

437C Theorem Let X be a set and U a Riesz subspace of `∞ (X), the M -space of bounded real-valued functions on X, containing the constant functions. (a) Let Σ be the smallest σ-algebra of subsets of X with respect to which every member of U is measurable. Let Mσ = Mσ (Σ) be the L-space of countably additive functionals on Σ (326E, 362B). Then we have a R Banach lattice isomorphism T : Mσ → Uσ∼ defined by saying that (T µ)(u) = u dµ whenever µ ∈ Mσ+ and u ∈ U. (b) We now have a sequentially order-continuous norm-preserving Riesz homomorphism S, embedding the M -space L∞ = L∞ (Σ) of bounded real-valued Σ-measurableRfunctions on X (363Ha) into the M -space (Uσ∼ )∼ = (Uσ∼ )∗ = (Uσ∼ )× , defined by saying that (Sv)(T µ) = v dµ whenever v ∈ L∞ and µ ∈ Mσ+ . If u ∈ U , then (Su)(f ) = f (u) for every f ∈ Uσ∼ . proof (a)(i) The norm k k∞ is an order-unit norm on U (354Ga), so U ∗ = U ∼ is an L-space (356N), and the band Uσ∼ (437Aa) is an L-space in its own right (354O). (ii) As noted Banach lattice isomorphism T0 : Mσ → (L∞ )∼ c defined by saying R in 437B, we have a ∞ that (T0 µ)(u) = u dµ whenever u ∈ L and µ ∈ Mσ+ . If we set T µ = T0 µ ¹ U , then T is a positive linear operator from Mσ to U ∼ , just because U is a linear subspace of L∞ ; and since T µ ∈ Uσ∼ for every µ ∈ Mσ+ , + ∼ + P By 436D, T is an operator from Mσ to Uσ∼ . Now R every f ∈ (Uσ ) is of the form T µ for some µ ∈ Mσ . P there is some measure λ such that u dλ = f (u) for every u ∈ U . Completing λ if necessary, we see that we may suppose that every member of U is (dom λ)-measurable, that is, that Σ ⊆ dom λ; take µ = λ¹Σ. Q Q So T is surjective. (iii) Write K for the family of sets K ⊆ X such that χK = inf n∈N un for some sequence hun in∈N in U . (See the proof of 436D.) We need to know the following. (α) K ⊆ Σ. (β) If K ∈ K, then there is a non-increasing sequence hun in∈N in U such that χK = inf n∈N un . (For if hu0n in∈N is any sequence in U such that χK = inf n∈N u0n , we can set un = inf i≤n u0i for each i.) (γ) The σ-algebra of subsets of X generated by K is Σ. P P Let T be the σ-algebra of subsets of X generated by K. T ⊆ Σ because K ⊆ Σ. If u ∈ U and α > 0 then {x : u(x) ≥ α} ∈ K (see part (b) of the proof of 436D). So every member of U + , therefore every member of U , is T-measurable, and Σ ⊆ T. Q Q − (iv) T is injective. P P If µ1 , µ2 ∈ Mσ and T µ1 = T µ2 , set νi = µi + µ− 1 + µ2 for each i, so that νi is non-negative and T ν1 = T ν2 . If K ∈ K then there is a non-increasing sequence hun in∈N in U such that χK = inf n∈N un in RX , so

ν1 K = inf n∈N

R

un dν1 = inf n∈N

R

un dν2 = ν2 K.

Now K contains X and is closed under finite intersections and ν1 and ν2 agree on K. By the Monotone Class Theorem (136C), ν1 and ν2 agree on the σ-algebra generated by K, which is Σ; so ν1 = ν2 and µ1 = µ2 . Q Q Thus T is a linear space isomorphism between Mσ and Uσ∼ . (v) As noted in (ii), T [Mσ+ ] = (Uσ∼ )+ ; so T is a Riesz space isomorphism. (vi) Now if µ ∈ Mσ ,

kT µk = |T µ|(χX) (356Nb) = (T |µ|)(χX) (because T is a Riesz homomorphism) = |µ|(X) = kµk (362Ba). So T is norm-preserving and is an L-space isomorphism, as claimed. (b)(i) By 356Pb, (Uσ∼ )∗ = (Uσ∼ )∼ = (Uσ∼ )× is an M -space.

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437C

× (ii) We have a canonical map S0 : L∞ → ((L∞ )∼ c ) defined by saying that (S0 v)(h) = h(v) for every ∞ ∞ ∼ v ∈ L , h ∈ (L )c ; and by 356F, S0 is a Riesz homomorphism. If hvn in∈N is a non-increasing sequence in + L∞ with infimum 0, then inf n∈N (S0 vn )(h) = inf n∈N h(vn ) = 0 for every h ∈ ((L∞ )∼ c ) , so inf n∈N S0 vn = 0 (355Ee); as hvn in∈N is arbitrary, S0 is sequentially order-continuous (351Gb). ∞ Also S0 is norm-preserving. P P (α) If h ∈ (L∞ )∼ c and v ∈ L , then

|(S0 v)(h)| = |h(v)| ≤ khkkvk∞ , so kS0 vk ≤ kvk∞ . (β) If v ∈ L∞ and 0 ≤ γ < kvk∞ , take x ∈ X such that |v(x)| > γ, and define ∞ hx ∈ (L∞ )∼ c by setting hx (w) = w(x) for every w ∈ L ; then khx k = 1 and |(S0 v)(hx )| = |hx (v)| = |v(x)| ≥ γ, so kS0 vk ≥ γ. As γ is arbitrary, kS0 vk ≥ kvk∞ and kS0 vk = kvk∞ . Q Q ∼ (iii) Now T0 : Mσ → (L∞ )∼ c and T : Mσ → Uσ are both norm-preserving Riesz space isomorphisms, ∗ ∼ ∗ ∞ ∼ −1 ∼ so T0 T : Uσ → (L )c is another, and its adjoint S1 : ((L∞ )∼ c ) → (Uσ ) must also be a norm-preserving Riesz space isomorphism. So if we set S = S1 S0 , S will be a norm-preserving sequentially order-continuous Riesz homomorphism from L∞ to (Uσ∼ )× = (Uσ∼ )∗ .

(iv) Setting the construction out in this way tells us a lot about the properties of the operator S, but undeniably leaves it somewhat obscure. So let us start again from v ∈ L∞ and µ ∈ Mσ+ , and seek to calculate (Sv)(T µ). We have (Sv)(T µ) = (S1 S0 v)(T µ) = (S0 v)(T0 T −1 T µ) (because S1 is the adjoint of T0 T −1 )

Z = (S0 v)(T0 µ) = (T0 µ)(v) =

v dµ,

as claimed. If u ∈ U , then (T µ)(u) = (T0 µ)(u) for every µ ∈ Mσ , so if f ∈ Uσ∼ then (Su)(f ) = (S1 S0 u)(f ) = (S0 u)(T0 T −1 f ) = (T0 T −1 f )(u) = (T T −1 f )(u) = f (u). This completes the proof. 437D Remarks What is happening here is that the canonical Riesz homomorphism u 7→ u ˆ from U to (Uσ∼ )∗ (356F) has a natural extension to L∞ (Σ). The original homomorphism u 7→ u ˆ is not, as a rule, sequentially order-continuous, just because Uσ∼ is generally larger than Uc∼ ; but the extension to L∞ is sequentially order-continuous. If you like, it is sequential smoothness which is carried over to the extension, and because the embedding of L∞ in RX is sequentially order-continuous, a sequentially smooth operator on L∞ is sequentially order-continuous. R R In the statement of 437C, I have used the formulae (T µ)(u) = u dµ and (Sv)(T µ) = v dµ onR the assumption that µ ∈ Mσ+ , so that µ is actually a measure on the definition used in this treatise, and dµ is the ordinary integral as constructed in §122. Since the functions u and v are bounded, measurable and defined everywhere, we can choose toR extend the notion of integration to signed measures, as in 363L, in R which case the formulae (T µ)(u) = u dµ and (Sv)(T µ) = v dµ become meaningful, and true, for all µ ∈ Mσ , u ∈ U and v ∈ L∞ . In fact the ideas here can be pushed farther, as in 437Ib, 437Xe and 437Yd. 437E Corollary Let X be a completely regular Hausdorff space, and Ba = Ba(X) its Baire σ-algebra. Then we can identify Cb (X)∼ σ with the L-space Mσ (Ba) of countably additive functionals on Ba, and we ∗ have a norm-preserving sequentially order-continuous Riesz homomorphism from L∞ (Ba) to (Cb (X)∼ σ) R ∞ ∼ + defined by setting (Sv)(f ) = v dµf for every v ∈ L and f ∈ (Cb (X)σ ) , where µf is the Baire measure associated with f . proof Put 437C and 436E together.

437G

Spaces of measures

223

437F Proposition Let X be a topological space and B = B(X) its Borel σ-algebra. Let Mσ be the L-space of countably additive functionals on B. (a) Write Mτ ⊆ Mσ for the set of differences of τ -additive totally finite Borel measures. Then Mτ is a band in Mσ , so is an L-space in its own right. (b) Write Mt ⊆ Mτ for the set of differences of totally finite Borel measures which are tight (that is, inner regular with respect to the closed compact sets). Then Mt is a band in Mσ , so is an L-space in its own right. proof (a)(i) Let µ1 , µ2 be totally finite τ -additive Borel measures on X, α ≥ 0, and µ ∈ Mσ such that 0 ≤ µ ≤ µ1 . Then µ1 + µ2 , αµ1 and µ are totally finite τ -additive Borel measures. P P They all belong to Mσ , that is, are Borel measures. Now let G be a non-empty upwards-directed family of open sets in X with union H, and ² > 0. Then there are G1 , G2 ∈ G such that µ1 G1 ≥ µ1 H − ² and µ2 G2 ≥ µ2 H − ², and a G ∈ G such that G ⊇ G1 ∪ G2 . In this case, (µ1 + µ2 )(G) ≥ (µ1 + µ2 )(H) − 2², (αµ1 )(G) ≥ (αµ1 )(H) − α² and µG = µH − µ(H \ G) ≥ µH − µ1 (H \ G) ≥ µH − ². As G and ² are arbitrary, µ1 + µ2 , αµ1 and µ are all τ -additive. Q Q It follows that Mτ is a solid linear subspace of Mσ . (ii) Now suppose that B ⊆ Mτ+ is non-empty and upwards-directed and has a supremum ν in Mσ . Then ν ∈ Mτ . P P If G is a non-empty upwards-directed family of open sets with union H, then νH = sup µH µ∈B

(362Be) =

sup µ∈B,G∈G

µG = sup νG; G∈G

as G is arbitrary, ν is τ -additive and belongs to Mτ . Q Q As B is arbitrary, Mτ is a band in Mσ . By 354O, it is itself an L-space. (b) We can use the same arguments. Suppose that µ1 , µ2 ∈ Mσ are tight, α ≥ 0, and µ ∈ Mσ is such that 0 ≤ µ ≤ µ1 . If E ∈ B and ² > 0, there are closed compact sets K1 , K2 ⊆ E such that µ1 (E \ K1 ) ≤ ² and µ2 (E \ K2 ) ≤ ²; now K = K1 ∪ K2 is a closed compact subset of E, and µK = µE − µ(E \ K) ≥ µE − µ1 (E \ K) − µ2 (E \ K) ≥ µE − 2², (αµ1 )(K) ≥ (αµ1 )(E) − α². As E and ² are arbitrary, µ and αµ1 are tight; as µ1 , µ2 and α are arbitrary, Mt is a solid linear subspace of Mσ . Now suppose that B ⊆ Mt+ is non-empty and upwards-directed and has a supremum ν in Mσ . Take any E ∈ B and ² > 0. Then there is a µ ∈ B such that µE ≥ νE − ²; there is a closed compact set K ⊆ E such that µK ≥ µE − ²; and now νK ≥ νE − 2². As E and ² are arbitrary, ν is tight; as B is arbitrary, Mt is a band in Mσ , and is in itself an L-space. 437G Definitions Let X be a topological space. A signed Baire measure on X will be a countably additive functional on the Baire σ-algebra Ba(X), which by the Jordan decomposition theorem (231F) is expressible as the difference of two totally finite Baire measures; a signed Borel measure will be a countably additive functional on the Borel σ-algebra B(X), that is, the difference of two totally finite Borel measures; a signed τ -additive Borel measure will be a member of the L-space Mτ as described in 437F, that is, the difference of two τ -additive totally finite Borel measures; and if X is Hausdorff, a signed tight Borel measure will be a member of the L-space Mt as described in 437F, that is, the difference of two tight totally finite Borel measures.

224

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437H

437H Theorem Let X be a set and U a Riesz subspace of `∞ (X) containing the constant functions. Let T be the coarsest topology on X rendering every member of U continuous, and B = B(X) the corresponding Borel σ-algebra. (a) Let Mτ be the L-space of signed τ -additive Borel measures on X. Then we have a Banach lattice R isomorphism T : Mτ → Uτ∼ defined by saying that (T µ)(u) = u dµ whenever µ ∈ Mτ+ and u ∈ U . (b) We now have a sequentially order-continuous norm-preserving Riesz homomorphism S, embedding the M -space L∞ = L∞ (B) of bounded on X into (Uτ∼ )∼ = (Uτ∼ )∗ = (Uτ∼ )× , R Borel measurable functions ∞ defined by saying that (Sv)(T µ) = v dµ whenever v ∈ L and µ ∈ Mτ+ . If u ∈ U , then (Su)(f ) = f (u) for every f ∈ Uτ∼ . proof The proof follows the same lines as that of 437C. (a)(i) As before, the norm k k∞ is an order-unit norm on U , so U ∗ = U ∼ is an L-space, and the band (437Ab) is an L-space in its own right, like Mτ (437F).

Uτ∼

(ii) Let Mσ be the L-space of all countably additive functionals on B, so that Mτ is a band in MσR . Let u dµ T0 : Mσ → (L∞ )∼ c be the canonical Banach lattice isomorphism defined by saying that (T0 µ)(u) = whenever u ∈ L∞ and µ ∈ Mσ+ . If we set T µ = T0 µ¹U for µ ∈ Mτ , then T is a positive linear operator from Mτ to U ∼ , just because U is a Riesz subspace of L∞ ; and since T µ ∈ Uτ∼ for every µ ∈ Mτ+ (436H), T is ∼ + + an operator from Mτ to Uτ∼ . Now every P By 436H, there R f ∈ (Uτ ) is of the form T µ for some µ ∈ Mτ . P is a quasi-Radon measure λ such that u dλ = f (u) for every u ∈ U ; set µ = λ¹B. Q Q So T is surjective. (iii) Let K be the family of subsets K of X such that χK = inf A in RX for some non-empty subset A of U . Then K is just the family of closed sets for T. P P As noted in part (b) of the proof of 436H, every member of K is closed, and K \ G ∈ K whenever K ∈ K and G ∈ T; but as, in the present case, X ∈ K, every closed set belongs to K. Q Q − (iv) T is injective. P P If µ1 , µ2 ∈ Mτ and T µ1 = T µ2 , set νi = µi + µ− 1 + µ2 for each i, so that νi is non-negative and T ν1 = T ν2 . If K ∈ K, set A = {u : u ∈ U , u ≥ χK}, so that χK = inf A in RX , and A is downwards-directed. By 414Bb,

ν1 K = inf u∈A

R

u dν1 = inf u∈A

R

u dν2 = ν2 K.

Now K contains X and is closed under finite intersections and ν1 and ν2 agree on K. By the Monotone Class Theorem, ν1 and ν2 agree on the σ-algebra generated by K, which is B; so ν1 = ν2 and µ1 = µ2 . Q Q Thus T is a linear space isomorphism between Mτ and Uτ∼ . (v) As noted in (ii), T [Mτ+ ] = (Uτ∼ )+ ; so T is a Riesz space isomorphism. (vi) Now if µ ∈ Mτ , kT µk = |T µ|(χX) = (T |µ|)(χX) = |µ|(X) = kµk. So T is norm-preserving and is an L-space isomorphism, as claimed. (b)(i) By 356Pb, (Uτ∼ )∗ = (Uτ∼ )∼ = (Uτ∼ )× is an M -space. (ii) Because T0 : Mσ → (L∞ )∼ c is a Banach lattice isomorphism, and Mτ is a band in Mσ , W = T0 [Mτ ] ∞ is a band in (L∞ )∼ → W × defined by writing c . We therefore have a Riesz homomorphism S0 : L ∞ (S0 v)(h) = h(v) for v ∈ L , h ∈ W (356F). Just as in (b-ii) of the proof of 437C, S0 is sequentially order-continuous and norm-preserving. (We need to observe that hx in the second half of the argument there always belongs to W ; this is because hx = T0 (δx ), where δx ∈ Mτ is defined by setting δx (E) = χE(x) for every Borel set E.) ∼ (iii) Now T0 : Mσ → (L∞ )∼ c and T : Mτ → Uτ are both norm-preserving Riesz space isomorphisms, −1 ∼ so T0 T : Uτ → W is another, and its adjoint S1 : W ∗ → (Uτ∼ )∗ must also be a norm-preserving Riesz space isomorphism. So if we set S = S1 S0 , S will be a norm-preserving sequentially order-continuous Riesz homomorphism from L∞ to (Uτ∼ )× = (Uτ∼ )∗ .

(iv) If v ∈ L∞ and µ ∈ Mτ+ ,

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(Sv)(T µ) = (S1 S0 v)(T µ) = (S0 v)(T0 T −1 T µ) Z = (S0 v)(T0 µ) = (T0 µ)(v) = v dµ; if u ∈ U and f ∈ Uτ∼ , then (T µ)(u) = (T0 µ)(u) for every µ ∈ Mτ , so (Su)(f ) = (S1 S0 u)(f ) = (S0 u)(T0 T −1 f ) = (T0 T −1 f )(u) = (T T −1 f )(u) = f (u). 437I Proposition Let X be a locally compact Hausdorff space, B(X) its Borel σ-algebra, and L∞ the M -space of bounded Borel measurable real-valued functions on X. (a) Let Mt be the L-space of signed tight Borel measuresR on X. Then we have a Banach lattice isomorphism T : Mt → C0 (X)∗ defined by saying that (T µ)(u) = u dµ whenever µ ∈ Mt+ and u ∈ C0 (X). (b) Let ΣuRm be the algebra of universally Radon-measurable subsets of X (definition: 434E), and L∞ (ΣuRm ) the M -space of bounded ΣuRm -measurable real-valued functions on X. Then we have a normpreserving sequentially order-continuous Riesz homomorphism S : L∞ (ΣuRm ) → C0 (X)∗∗ defined by saying R that (Sv)(T µ) = v dµ whenever v ∈ L∞ (ΣuRm ) and µ ∈ Mt+ ; and (Su)(f ) = f (u) for every u ∈ C0 (X), f ∈ C0 (X)∗ . proof (a) The point is just that in this context Mt is equal to Mτ , as defined in 437F-437H (416H), while C0 (X)∗ = C0 (X)∼ τ (see part (a) of the proof of 436J), and the topology of X is completely regular, so we just have a special case of 437Ha. (b)(i) Write B for the Borel σ-algebra of X, and L∞ (B) ⊆ L∞ (ΣuRm ) for the space of bounded Borel measurable functions on X. As in 437Hb, we have a sequentially order-continuous Riesz homomorphism R S0 : L∞ (B) → C0 (X)∗∗ defined by saying that (S0 v)(T ν) = v dν whenever v ∈ L∞ (B) and ν ∈ Mt+ . (ii) If v ∈ L∞ (ΣuRm ), then sup{S0 w : w ∈ L∞ (B), w ≤ v} = inf{S0 w : w ∈ L∞ (B), w ≥ v} in C0 (X)∗∗ . P P Set A = {w : w ∈ L∞ (B), w ≤ v},

B = {w : w ∈ L∞ (B), w ≥ v}.

Because the constant functions belong to L∞ (B), A and B are both non-empty; of course w ≤ w0 and S0 w ≤ S0 w0 for every w ∈ A and w0 ∈ B; because C0 (X)∗∗ is Dedekind complete, φ = sup S0 [A] and ψ = inf S0 [B] are both defined in C0 (X)∗∗ , and φ ≤ ψ. If f ≥ 0 in C0 (X)∗ , then there is a ν ∈ Mt+ such that T ν = f . Since v is ν-virtually measurable (see 434Ec), there are (bounded) Borel measurable functions w, w0 such that w ≤ v ≤ w0 and w = w0 ν-a.e., that is, w ∈ A, w0 ∈ B and (S0 w)(f ) =

R

w dν =

R

w0 dν = (S0 w0 )(f ).

But as (S0 w)(f ) ≤ φ(f ) ≤ ψ(f ) ≤ (S0 w0 )(f ), φ(f ) = ψ(f ); as f is arbitrary, φ = ψ. Q Q (iii) We can therefore define S : L∞ (ΣuRm ) → C0 (X)∗∗ by setting Sv = sup{S0 w : w ∈ L∞ (B), w ≤ v} = inf{S0 w : w ∈ L∞ (B), w ≥ v} R for every v ∈RL∞ . The argument in (ii) tells us also that (Sv)(T ν) = v dν for every ν ∈ Mt+ ; that is, that (Sv)(T µ) = v dµ for every Radon measure µ on X. (iv) Now S is a norm-preserving sequentially order-continuous Riesz homomorphism. P P (Compare 355F.) (α) The non-trivial part of this is actually the check that S is additive. But the formula Sv = sup{S0 w : w ∈ L∞ (B), w ≤ v} ensures that Sv1 + Sv2 ≤ S(v1 + v2 ) for all v1 , v2 ∈ L∞ , while the formula

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Sv = inf{S0 w : w ∈ L∞ (B), w ≥ v} ensures that Sv1 + Sv2 ≥ S(v1 + v2 ) for all v1 , v2 . (β) It is easy to check that S(αv) = αSv whenever v ∈ L∞ (ΣuRm ) and α > 0, so that S is linear. (γ) Because S0 is a Riesz homomorphism, S(v1 ∧v2 ) = 0 whenever v1 ∧ v2 = 0, so S is a Riesz homomorphism. (δ) Now suppose that hvn in∈N is a non-increasing R sequence in L∞ (ΣuRm ) with infimum 0 in L∞ (ΣuRm ). Then inf n∈N vn (x) = 0 for every x ∈ X, so inf n∈N vn dν = 0 for every ν ∈ Mt+ and inf n∈N (Svn )(f ) = 0 for every f ∈ (C0 (X)∗ )+ . So inf n∈N Svn = 0; as hvn in∈N is arbitrary, S is sequentially order-continuous. (iv) If v ∈ L∞ (ΣuRm ), then |v| ≤ kvk∞ χX, so kSvk ≤ kvk∞ kS(χX)k = kvk∞ . On the other hand, for any x ∈ X we have the corresponding point-supported probability measure δx , with the matching functional hx ∈ C0 (X)∗ , and kSvk = k|Sv|k = kS|v|k ≥ (S|v|)(hx ) =

R

|v|dδx = |v(x)|,

so kSvk ≥ kvk∞ ; thus S is norm-preserving. Q Q

R Remark Once again, if we interpret integrals in the way set out in 363L, we shall have (Sv)(T µ) = v dµ whenever v ∈ L∞ (B) and u ∈ U and µ ∈ Mτ (in 437H) or v ∈ L∞ (ΣuRm ) and u ∈ C0 (X) and µ ∈ Mt (in 437I). 437J Vague and narrow topologies We are ready for another look at ‘vague’ topologies on spaces of measures. Let X be a topological space. (a) Let Σ be an algebra of subsets of X. I will say that Σ separates zero sets if whenever F , F 0 ⊆ X are disjoint zero sets then there is an E ∈ Σ such that F ⊆ E and E ∩ F 0 = ∅. (b) If Σ is any algebra of subsets of X, we can identify the Banach algebra and Banach lattice L∞ (Σ), as defined in §363, with the k k∞ -closed linear subspace of `∞ (X) generated by {χE : E ∈ Σ} (363F, 363Ha). If we do this, then Cb (X) ⊆ L∞ (Σ) iff Σ separates zero sets. P P (i) Suppose that Cb (X) ⊆ L∞ (Σ) and that F1 , F2 ⊆ X are disjoint zero sets. Let u1 , u2 : X → R be continuous functions such that Fi = u−1 i [{0}] for both i; then |u1 (x)| + |u(2)(x)| > 0 for every x; set v =

|u1 | , |u1 |+|u2 |

so that v : X → [0, 1] is continuous,

∞

v(x) = 0 for x ∈ F1 and v(x) = 1 for x ∈ F2 . Now v ∈ Cb (X) ⊆ L (Σ), so there is a w ∈ S(Σ), the linear subspace of L∞ (Σ) generated by {χE : E ∈ Σ}, such that kv − wk∞ < 21 (363C). Set E = {x : w(x) ≤ 21 }; then E ∈ Σ and F1 ⊆ E ⊆ X \ F2 . As F1 and F2 are arbitrary, Σ separates zero sets. (ii) Now suppose that Σ separates zero sets, that u : X → [0, 1] is continuous, and that n ≥ 1 is an integer. For i ≤ n, set Fi = {x : x ∈ X, u(x) ≤ ni }, Fi0 = {x : x ∈ X, u(x) ≥ i+1 }. Then Fi and Fi0 are Pnn 1 0 disjoint zero sets so there is an Ei ∈ Σ such that Fi ⊆ Ei ⊆ X \ Fi . Set w = n i=1 χEi ∈ S(Σ). If x ∈ X, let j ≤ n be such that nj ≤ u(x) < j+1 n ; then for i ≤ n i < j ⇒ x ∈ Fi0 ⇒ x ∈ Ei ⇒ x ∈ / Fi ⇒ i ≤ j, 1 and w(x) = n1 #({i : i ≤ n, x ∈ Ei }) is either nj or j+1 n . Thus |w(x) − u(x)| ≤ n . As x is arbitrary, 1 ∞ ∞ ∞ ku − wk∞ ≤ n ; as n is arbitrary, u ∈ L (Σ). As L (Σ) is a linear subspace of ` (X), this is enough to show that Cb (X) ⊆ L∞ (Σ). Q Q

(c) It follows that if Σ is an algebra of subsets of X separating the zero sets, andR ν : Σ → R is a R bounded additive functional, we have an interpretation Rof u dν for any u ∈ Cb (X); dν is the unique norm-continuous linear functional on L∞ (Σ) such that χE dν = νE for every E ∈ Σ (363L). The map R ν 7→ dν is a Banach lattice isomorphism from the L-space M (Σ) of bounded additive functionals on Σ to ∗ L∞ (Σ)∗ = L∞ (Σ)∼ R (363K). We therefore have a positive linear operator T : M (Σ) → Cb (X) defined by setting (T ν)(u) = u dν for every ν ∈ M (Σ) and u ∈ Cb (X). Except in the trivial case X = ∅, kT k = 1 (if x ∈ X, we have δx ∈ M (Σ) defined by setting δx (E) = χE(x) for E ∈ Σ, and kT (δx )k = 1).R The vague topology on M (Σ) is now the topology generated by the functionals ν 7→ u dν as u runs ∗ over Cb (X); that is, the coarsest topology on M (Σ) such that the canonical map R T : M (Σ) → Cb (X) ∗ is continuous for the weak* topology of Cb (X) . Because the functionals ν 7→ | u dν| are seminorms on M (Σ), the vague topology is a locally convex linear space topology.

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(d) There is a variant of the vague topology which can be applied directly to spaces of (non-negative) ˜ + be the set of all non-negative real-valued additive functionals defined on totally finite measures. Let M ˜ + is that generated by algebras of subsets of X which contain every open set. The narrow topology on M sets of the form ˜ + , νG > α}, {ν : ν ∈ M ˜ + , νX < α} {ν : ν ∈ M for open sets G ⊆ X and real numbers α. ˜ + → [0, ∞[ is continuous for the narrow topology, and if G ⊆ X is open Observe that ν 7→ νX : M ˜ + for the set of totally finite then ν 7→ νG is lower semi-continuous for the narrow topology. Writing M σ + ˜ topological measures on X, then ν 7→ νE : Mσ → [0, ∞[ is Borel measurable, for the narrow topology on ˜ σ+ , for every Borel set E ⊆ X (because the set of E for which ν 7→ νE is Borel measurable is a Dynkin M class containing the open sets). Writing P˜ for the set of topological probability measures, then the narrow topology on P˜ is generated by sets of the form {µ : µ ∈ P˜ , µG > α} for open sets G ⊆ X. (e) Vague topologies, being linear space topologies, are necessarily completely regular (3A4Ad, 4A2Ja). In the very general context of (c) here, in which we have a space M (Σ) of all finitely additive functionals on an algebra Σ, we do not expect the vague topology to be Hausdorff. But if we look at particular subspaces, such as the space Mσ (Ba(X)) of signed Baire measures, or the space Mτ of signed τ -additive Borel measures on a completely regular space X, we may well have a Hausdorff vague topology (437Xf). ˜ + is rarely Hausdorff. But on important subspaces we can get Similarly, the narrow topology on M + Hausdorff topologies. In particular, the narrow topology on the space MqR of totally finite quasi-Radon + measures is Hausdorff. P P Take distinct µ0 , µ1 ∈ MqR . If µ0 X 6= µ1 X then they can be separated by open sets of the form {µ : µX < α}, {µ : µX > α}. Otherwise, set γ = µ0 X = µ1 X. There is certainly an open set G such that µ0 G 6= µ1 G (415H); suppose that µ0 G < µ1 G. Because µ1 is inner regular with respect to the closed sets, there is a closed set F ⊆ G such that µ0 G < µ1 F . Now µ1 G + µ0 (X \ F ) = µ1 G + γ − µ0 F ≥ µ1 F + γ − µ0 G > γ, so there are α < µ0 (X \ F ) and β < µ1 G such that α + β > γ. Now {µ : µ(X \ F ) > α, µX < α + β} and {µ : µG > β, µX < α + β} are disjoint open sets containing µ0 , µ1 respectively. Q Q (f ) For a variant of the narrow topology, adapted to the space of all Radon measures on a Hausdorff space, see 495O below. ˜ + the set of all non-negative real-valued additive 437K Proposition Let X be a topological space, and M functionals defined on algebras of subsets of X containing every open set. R ˜ + → Cb (X)∗ defined by the formula (T ν)(f ) = u dν for every ν ∈ M ˜ +, (a) We have a function T : M u ∈ Cb (X). ˜ + and the weak* topology on Cb (X)∗ . (b) T is continuous for the narrow topology on M ˜ + is a family of τ -additive totally finite (c) Suppose now that X is completely regular, and that W ⊆ M topological measures such that two members of W which agree on the Borel σ-algebra are equal. Then T ¹W is a homeomorphism between W , with the narrow topology, and T [W ], with the weak* topology. proof (a) We have only to assemble the operators of 437Jc, noting that if an algebra of subsets of X contains every open set then it certainly separates the zero sets (indeed, it actually contains every zero set). ˜ + . Suppose that u ∈ Cb (X)+ and that γ ∈ R. Then (b) Write S forR the narrow topology on M ˜ + , u dµ > γ} belongs to S. P V = {ν : ν ∈ M P If γ < 0 this is trivial. Otherwise, for n, i ∈ N, let Gni be P 4n −n the cozero set {x : x ∈ X, u(x) > 2 i}, so that Gni ∈ dom ν. For n ∈ N set un = 2−n i=1 χGni . Then R S T P 4n Vn = {ν : un dν > γ} = { 1≤i≤4n {ν : νGni > γi } : 2−n i=1 γi > γ} S belongs to S. Since hun in∈N is a non-decreasing sequence converging to u for k k∞ , V = n∈N Vn and V ∈ S. Q Q R Next, W = {ν : u dν < γ} belongs to S. P P If u = 0 this is trivial. Otherwise, set v = kuk∞ χX − u, so that v ∈ Cb (X)+ . Then

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Topologies and measures II

W =

S

β>0 {ν : νX <

β , kuk∞

R

437K

v dν > β − γ}

belongs to S. Q Q R (ii) This shows that ν 7→ (T ν)(u) = u dν is S-continuous for every u ∈ Cb (X)+ and therefore for every u ∈ Cb (X). Since the weak* topology on Cb (X)∗ is the coarsest topology on Cb (X)∗ for which all the functionals f 7→ f (u) are continuous, T is continuous. (c)(i) Write T for the topology on W induced by T , that is, the family of sets of the form W ∩ T −1 [V ] where V ⊆ Cb (X)∗ is weak*-open. If G ⊆ X is open, then A = {u : u ∈ Cb (X), 0 ≤ u ≤ χG} is R upwards-directed and has supremum χG, so µG = sup u dµ for every µ ∈ W (414Ba). Accordingly u∈A S {µ : µ ∈ W , µG > α} = u∈A {µ : (T µ)(u) > α} belongs to T for every α ∈ R. Also, of course, {µ : µX < α} = {µ : (T µ)(χX) < α} ∈ T for every α. So if S0 is the narrow topology on W , S0 ⊆ T. Putting this together with (b), we see that S0 = T. (ii) Now the same formulae show that T ¹W is injective. P P Suppose that µ1 , µ2 ∈ W and that T µ1 = T µ2 . Then µ1 G = µ2 G for every open set G ⊆ X. By the Monotone Class Theorem, µ1 and µ2 agree on all Borel sets; but our hypothesis is that this is enough to ensure that µ1 = µ2 . Q Q Since T : W → T [W ] is continuous and open, it is a homeomorphism. 437L Corollary Let X be a completely regular topological space, and Mτ the space of signed τ -additive Borel measures on X. Then the narrow and vague topologies on Mτ+ coincide. ˜ τ+ (X) for the space of totally finite τ -additive 437M Theorem For a topological space X, write M + topological measures on X, Mτ (X) for the space of totally finite τ -additive Borel measures on X, Mτ (X) for the L-space of signed τ -additive Borel measures on X, and P˜τ (X) for the space of τ -additive topological probability measures on X. ˜ τ+ (X) and ν ∈ M ˜ τ+ (Y ), write µ × ν ∈ M ˜ τ+ (X × Y ) for the (a) Let X and Y be topological spaces. If µ ∈ M τ -additive product measure on X × Y (417G). Then (µ, ν) 7→ µ × ν is continuous for the narrow topologies ˜ τ+ (X), M ˜ τ+ (Y ) and M ˜ τ+ (X × Y ). on M (b) Let hXi ii∈I be a family of topological spaces, with product X. If hµi ii∈I is a family of probability Q measures such that µi ∈ P˜τ (Xi ) for each i, write i∈I µi ∈ P˜τ (X) for their τ -additive product. Then Q hµi ii∈I 7→ i∈I µi is continuous for the narrow topology on P˜τ (X) and the product of the narrow topologies Q on i∈I P˜τ (Xi ). (c) Let X and Y be topological spaces. (i) We have a unique bilinear map ψ : Mτ (X)×Mτ (Y ) → Mτ (X ×Y ) such that ψ(µ, ν) is the restriction of µ × ν to the Borel σ-algebra of X × Y whenever µ ∈ Mτ+ (X) and ν ∈ Mτ+ (Y ). (ii) kψk ≤ 1 (definition: 253Ab). (iii) ψ is separately continuous for the vague topologies on Mτ (X), Mτ (Y ) and Mτ (X × Y ). (d) In (c), suppose that X and Y are compact. If B ⊆ Mτ (X) and B 0 ⊆ Mτ (Y ) are norm-bounded, then ψ¹B × B 0 is continuous for the vague topologies. ˜ τ+ (X) × proof (a)(i) If W ⊆ X × Y is open and α ∈ R, then Q = {(µ, ν) : (µ × ν)(W ) > α} is open in M + 0 ˜ Mτ (Y ). P PS Suppose that (µ0 , ν0 ) ∈ Q. Because µ0 × ν0 is τ -additive, there is a subset W ⊆ W , expressible in the form i≤n Gi × Hi where Gi ⊆ X and Hi ⊆ Y are open for every i, such that α < (µ0 × ν0 )(W 0 ) =

R

ν0 W 0 [{x}]µ0 (dx)

(417C(iv)). Set u(x) = ν0 W 0 [{x}] for x ∈ X, so that u is lower semi-continuousP(417Ba). LetR η > 0 be such R ∞ that u dµ0 > α+(1+2µ0 X)η, and set Ei = {x : u(x) > ηi} for i ∈ N, so that η i=1 µ0 Ei > u dµ0 −ηµ0 X. ˜ τ+ (X) such that Because every Ei is open, there is a neighbourhood U of µ0 in M R R P∞ u dµ0 − ηµ0 X ≤ η i=1 µEi ≤ u dµ = (µ × ν0 )(W 0 ) for every µ ∈ U ; shrinking U if necessary, we can arrangeSat the same time that µX < µ0 X + 1 for every µ ∈ U . Next, observe that H = {W 0 [{x}] : x ∈ X} ⊆ { i∈I Hi : I ⊆ {0, . . . , n}} is finite, so there is a

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˜ τ+ (Y ) such that νH ≥ ν0 H − η for every H ∈ H and ν ∈ V . If µ ∈ U and neighbourhood V of ν0 in M ν ∈ V , we have Z Z 0 0 (µ × ν)(W ) ≥ (µ × ν)(W ) = νW [{x}]µ(dx) ≥ u(x) − η µ(dx) Z Z = u dµ − ηµX ≥ u dµ0 − ηµ0 X − η(1 + µ0 X) > α. As µ0 and ν0 are arbitrary, Q is open. Q Q (ii) Since (µ × ν)(X × Y ) = µX · νY , the sets {(µ, ν) : (µ × ν)(X × Y ) < α} are also open for every α ∈ R. So (µ, ν) 7→ µ × ν is continuous (4A2B(a-ii)). (b) For finite sets I, this is a simple induction on #(I), using 417Db. For infinite I, let W ⊆ X be an open set and α ∈ R, and consider Q Q = {hµi ii∈I : µi ∈ P˜τ (Xi ) for each i, ( µi )(W ) > α}. i∈I

IfQhµi ii∈I ∈ Q, then there is an open set W 0 ⊆ W , determined Q by coordinates in a finite set J ⊆ I, such that ( i∈I µi )(W 0 ) > α. Setting V = {x¹J : x ∈ W 0 }, we have ( i∈J µi )(V ) > α. Now we can find open sets Ui Q Q in P˜τ (Xi ), for i ∈ J, such that ( i∈J νi )(V ) > α whenever νi ∈ Ui for i ∈ J. If now hνi ii∈I ∈ i∈I P˜τ (Xi ) is such that νi ∈ Ui for every i ∈ J, Q Q Q ( i∈I νi )(W ) ≥ ( i∈I νi )(W 0 ) = ( i∈J νi )(V ) > α, Q so i∈I νi ∈ Q. As hµi ii∈I is arbitrary, Q Qis open. As W and α are arbitrary, hµi ii∈I 7→ i∈I µi is continuous. (c)(i) Start by writing ψ(µ, ν) = (µ × ν)¹B(X × Y ) for µ ∈ Mτ+ (X) and ν ∈ Mτ+ (Y ), where B(X × Y ) is the Borel σ-algebra of X × Y . If µ, µ1 , µ2 ∈ Mτ+ (X) and ν, ν1 , ν2 ∈ Mτ+ (Y ) and α ≥ 0, then ψ(µ1 + µ2 , ν) = ψ(µ1 , ν) + ψ(µ2 , ν). P P On each side of the equation we have a τ -additive Borel measure, and the two measures agree on the standard base W for the topology of X × Y consisting of products of open sets; since W is closed under finite intersections, they agree on the algebra generated by W and therefore on all open sets and therefore (using the Monotone Class Theorem yet again) on all Borel sets. Q Q Similarly, ψ(µ, ν1 + ν2 ) = ψ(µ, ν1 ) + ψ(µ, ν2 ),

ψ(αµ, ν) = ψ(µ, αν) = αψ(µ, ν).

Now if µ01 , µ02 ∈ Mτ+ (X) and ν10 , ν20 ∈ Mτ+ (Y ) are such that µ1 − µ2 = µ01 − µ02 and ν1 − ν2 = ν10 − ν20 , we shall have ψ(µ1 , ν1 ) − ψ(µ1 ,ν2 ) − ψ(µ2 , ν1 ) + ψ(µ2 , ν2 ) = ψ(µ1 , ν1 + ν20 ) − ψ(µ1 + µ02 , ν20 ) + ψ(µ02 , ν20 ) − ψ(µ1 , ν10 + ν2 ) + ψ(µ1 + µ02 , ν10 ) − ψ(µ02 , ν10 ) − ψ(µ2 , ν1 + ν20 ) + ψ(µ2 + µ01 , ν20 ) − ψ(µ01 , ν20 ) + ψ(µ2 , ν10 + ν2 ) − ψ(µ2 + µ01 , ν10 ) + ψ(µ01 , ν10 ) = ψ(µ02 , ν20 ) − ψ(µ02 , ν10 ) − ψ(µ01 , ν20 ) + ψ(µ01 , ν10 ). We can therefore extend ψ to an operator on Mτ (X) × Mτ (Y ) by setting ψ(µ1 − µ2 , ν1 − ν2 ) = ψ(µ1 , ν1 ) − ψ(µ1 , ν2 ) − ψ(µ2 , ν1 ) + ψ(µ2 , ν2 ) whenever µ1 , µ2 ∈ Mτ+ (X) and ν1 , ν2 ∈ Mτ+ (Y ), and it is straightforward to check that ψ is bilinear. (ii) If µ ∈ Mτ (X), then kµk = µ+ (X) + µ− (X), where µ+ and µ− are evaluated in the Riesz space Mτ (X). Now if ν ∈ Mτ (Y ), |ψ(µ, ν)| = |ψ(µ+ , ν + ) − ψ(µ+ , ν − ) − ψ(µ− , ν + ) + ψ(µ− , ν − )| ≤ ψ(µ+ , ν + ) + ψ(µ+ , ν − ) + ψ(µ− , ν + ) + ψ(µ− , ν − ),

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so kψ(µ, ν)k = |ψ(µ, ν)|(X × Y ) ≤ ψ(µ+ , ν + )(X × Y ) + ψ(µ+ , ν − )(X × Y ) + ψ(µ− , ν + )(X × Y ) + ψ(µ− , ν − )(X × Y ) = µ+ (X) · ν + (Y ) + µ+ (X) · ν − (Y ) + µ− (X) · ν + (Y ) + µ− (X) · ν − (Y ) = kµkkνk. As µ and ν are arbitrary, kψk ≤ 1. (iii) Fix ν ∈ Mτ+ (Y ) and w ∈ Cb (X × Y )+ , and consider the map µ 7→ Note first that if µ ∈ Mτ+ (X),

R

w dψ(µ, ν) =

R

w d(µ × ν) =

RR

R

w dψ(µ, ν) : Mτ (X) → R.

w(x, y)ν(dy)µ(dx)

(417H). Since both sides of this equation are linear in µ, we have

R

R

w dψ(µ, ν) =

R

w(x, y)ν(dy)µ(dx)

for every µ ∈ Mτ (X). Now x 7→ w(x, y)ν(dy) is continuous. P P By 417Bc, it is lower semi-continuous; but R if α ≥ kwk∞ and w0 = αχ(X × Y ) − w, then x 7→ w0 (x, y)ν(dy) is lower semi-continuous, so x 7→ ανY −

R

w0 (x, y)ν(dy) =

R

w(x, y)ν(dy)

R is also upper semi-continuous, therefore continuous. Q Q It follows at once that µ 7→ w(x, y)ν(dy)µ(dx) is continuous for the vague topology on Mτ (X). The argument has R supposed that w and ν are positive; but taking positive and negative parts as usual, we see that µ 7→ w dψ(µ, ν) is vaguely continuous for every w ∈ Cb (X × Y ) and ν ∈ Mτ (Y ). As w is arbitrary, µ 7→ ψ(µ, ν) is vaguely continuous, for every ν. Similarly, ν 7→ ψ(µ, ν) is vaguely continuous for every µ, and ψ is separately continuous. (d) Now suppose that X and Y are compact. Let W be the linear subspace of C(X × Y ) generated by {u ⊗ v : u ∈ C(X), v ∈ C(Y )}, writing (u ⊗ v)(x, y) = u(x)v(y) as in 253B. Then W is a subalgebra of C(X × Y ) separating the points of X × Y and containing the constant functions, so is k k∞ -dense in C(X × Y ) (281E). Now (µ, ν) 7→

R

u ⊗ v dψ(µ, ν) =

is continuous whenever u ∈ C(X) and v ∈ C(Y ), so (µ, ν) 7→

R

R

u dµ ·

R

v dν

w dψ(µ, ν)

is continuous whenever w ∈ W . Now suppose that B ⊆ Mτ (X) and B 0 ⊆ Mτ (Y ) are bounded. Let γ ≥ 0 be such that kµk ≤ γ for every µ ∈ B and kνk ≤ γ for every ν ∈ B. If w ∈ C(X × Y ) and ² > 0, there is a w0 ∈ W such that kw − w0 k∞ ≤ ². In this case Z Z | w dψ(µ, ν) − w0 dψ(µ, ν)| ≤ kw − w0 k∞ kψ(µ, ν)k ≤ ²kµkkνk ≤ γ 2 ² R whenever µ ∈ B and ν ∈ B 0 . As ² is arbitrary, the function (µ, ν) 7→ w dψ(µ, ν) is uniformly approximated on B × B 0 by vaguely continuous functions, and is therefore itself vaguely continuous on B × B 0 . 437N (Quasi)-Radon measures Throughout the above discussion, I have tried to maintain the formal distinctions between ‘quasi-Radon measure’ and ‘τ -additive effectively locally finite Borel measure inner regular with respect to the closed sets’, and between ‘Radon measure’ and ‘tight locally finite Borel measure’. There are obvious problems in interpreting the sum and difference of measures with different domains, which are readily soluble (see, for instance, 416Ea and 416Xd) but in the context of this section are unilluminating. + If, however, we take MqR to be the set of totally finite quasi-Radon measures on X, and X is completely + regular, we have a canonical embedding of MqR into a cone in the L-space Cb (X)∗ ; more generally, even

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+ if our space X is not completely regular, the map µ 7→ µ¹B(X) : MqR → Mσ (B(X)) is still injective, and + we can identify MqR with a cone in the L-space Mτ of signed τ -additive Borel measures (often the whole positive cone of Mτ , as in 415M). Similarly, when X is Hausdorff, we can identify totally finite Radon measures with tight totally finite Borel measures (416F). It is even possible to extend these ideas to measures which are not totally finite (437Yj), though there may be new difficulties (415Ya).

437O As usual, the case in which X is compact and Hausdorff is particularly important. 0 for the set of Radon measures Proposition Let X be a compact Hausdorff space, and γ ∈ [0, ∞[. Write PRγ 0 µ on X such that µX ≤ γ, and PRγ for the set of Radon measures µ on X such that µX = γ. Then PRγ and PRγ are compact in their narrow topologies.

proof By 437Kc the narrow topologies on these sets can be identified with the weak* topologies on their ˜ + → C(X)∗ be the map described in 437K. Then images in Cb (X)∗ = C(X)∗ . Let T : M 0 T [PRγ ] = {f : f ∈ C(X)∗ , f (u) ≥ 0 for every u ∈ C(X)+ , f (χX) ≤ γ}. R 0 P P If µ ∈ PRγ , then of course (T µ)(u) = u dµ ≥ 0 whenever u ≥ 0, and (T µ)(χX) = µX ≤ γ. If f ∈ (C(X)∗ )+ and f (χX) ≤ γ, then the Riesz representation theorem (436J/436K) tells us that f = T µ for 0 some Radon measure µ, and µX = f (χX) ≤ γ, so µ ∈ PRγ .Q Q 0 This description of T [PRγ ] makes it plain that it is weak*-closed in C(X)∗ . But also, of course, kT µk = 0 0 0 µX ≤ γ for every µ ∈ PRγ ; since the unit ball of C(X)∗ is weak*-compact (3A5F), so is T [PRγ ]. So PRγ is compact for the narrow topology. R 0 Since PRγ = {µ : µ ∈ PRγ , χX dµ = γ}, it too is compact. 0 437P The sets PRγ , PRγ can be identified with convex sets in C(X)∗ , and can therefore have extreme points. These extreme points are easy to identify; I look at PR1 = PR .

Proposition Let X be a Hausdorff space, and PR the set of Radon probability measures on X. The natural embedding of PR into the linear space Mσ (B(X)) of signed Borel measures on X gives PR a convex structure, for which the extreme points are just the point-supported measures δx for x ∈ X, where δx (E) = χE(x) for every E ⊆ X. proof (a) Suppose that x ∈ X. If µ1 , µ2 ∈ PR are such that δx = 12 (µ1 + µ2 ), then (performing the linear operations in Mσ ) we must have µ1 E ≤ 2µE for every Borel set E; in particular, µ1 (X \ {x}) = 0 and µ1 {x} = 1, that is, µ1 = δx . Similarly, µ2 = δx ; as µ1 and µ2 are arbitrary, δx is an extreme point of PR . (b) Suppose that µ is an extreme point of PR . Let K be the support of µ. ?? If K has more than one point, take distinct x, y ∈ K. As X is Hausdorff, there are disjoint open sets G, H such that x ∈ G and y ∈ H. Set E = G ∩ K, α = µE. Because K is the support of µ, α > 0. But similarly µ(H ∩ K) > 0 and α < 1. Let µ1 , µ2 be the indefinite-integral measures defined over µ by α1 χE and β1 χ(X \ E) respectively. Then both are Radon probability measures on X (416S), so belong to PR . Now µF = αµ1 F + βµ2 F for every Borel set F , so (computing αµ1 + βµ2 in Mσ ) µ = αµ1 + βµ2 ; as neither µ1 nor µ2 is equal to µ, µ is not extreme in PR . X X Thus K = {x} for some x ∈ X. But this means that µ{x} = 1 and µ(X \ {x}) = 0, so µ = δx is of the declared form. 437Q For metrizable spaces we can go farther. It will help to have a straightforward lemma. Lemma (a) Let X and Y be Hausdorff spaces, and φ : X → Y a continuous function. Let MR+ (X), MR+ (Y ) ˜ be the spaces of totally finite Radon measures on X and Y respectively. Writing φ(µ) for the image measure + ˜ + + −1 µφ for µ ∈ MR , φ : MR (X) → MR (Y ) is continuous for the narrow topologies on MR+ (X) and MR+ (Y ). (b) If Y is a Hausdorff space, X a subset of Y , and φ : X → Y the identity map, then φ˜ is a homeomorphism between MR+ (X) and {ν : ν ∈ MR+ (Y ), ν(Y \ X) = 0}. proof (a) All we have to do is to recall from 418I that µφ−1 ∈ MR+ (Y ) for every µ ∈ MR+ (X), and observe that

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{µ : (µφ−1 )(H) > α} = {µ : µφ−1 [H] > α},

437Q

{µ : (µφ−1 )(Y ) < α} = {µ : µX < α}

are narrowly open in MR+ (X) for every open set H ⊆ Y and α ∈ R. ˜ (b) First note that if µ ∈ MR+ (X), then certainly φ(µ)(Y \ X) = 0; while if ν ∈ MR+ (Y ) and ν(Y \ X) = 0, ˜ then µ = ν¹PX is a Radon measure on X (416Rb) and ν = φ(µ). Thus φ˜ is a continuous bijection from MR+ (X) to {ν : ν ∈ MR+ (Y ), ν(Y \ X) = 0}. Now if G ⊆ X is relatively open and α ∈ R, there is an open set H ⊆ Y such that G = H ∩ X, so that ˜ {µ : µ ∈ M + (X), µG > α} = {µ : φ(µ)(H) > α} R

is the inverse image of a narrowly open set in MR+ (Y ); and of course ˜ {µ : µ ∈ MR+ (X), µX < α} = {µ : φ(µ)(Y ) < α} is also the inverse image of an open set. So φ˜ is a homeomorphism between MR+ (X) and {ν : ν ∈ MR+ (Y ), ν(Y \ X) = 0}. 437R Theorem For a Hausdorff space X, write PR (X) for the set of Radon probability measures on X, with its narrow topology. (a) If X is a compact metrizable space, PR (X) is compact and metrizable. (b) If X is a Polish space, PR (X) is Polish. proof (a) Let U be a countable base for the topology of X which contains ∅ and is closed under finite unions. For U ∈ U, q ∈ Q set HU q = {µ : µ ∈ PR (X), µU > q}. Then every HU q is open for the narrow topology T on PR (X); let S be the topology generated by {HU q : U ∈ U, q ∈ Q}, so that S is a second-countable topology coarser than T. If G ⊆ X is open and α ∈ R, then {U : U ∈ U, U ⊆ G} is an upwards-directed set with union G, so S S {µ : µ ∈ PR (X), µG > α} = U ∈U ,U ⊆G q∈Q,q≥α HU q ∈ S. Accordingly S must be actually equal to T, and T is second-countable; since we already know that T is compact (437O), it must be metrizable (4A2Nh). (b) By 4A2Qg, we can suppose that X is a Gδ subset of a compact metrizable space Z. Then we have a homeomorphism betweenTPR (X) and Q = {ν : ν ∈ PR (Z), ν(Z \ X) = 0} (437Qb). Now Q is a Gδ set in PR (Z). P P Express X as n∈N Hn where Hn ⊆ Z is open for each n. Then T T Q = n∈N {ν : νHn = 1} = m,n∈N {ν : νHn > 1 − 2−m } is a Gδ set in PR (Z). Q Q By (a), PR (Z) is a compact metrizable space; by 4A2Qd, its Gδ subset Q is Polish; so PR (X) also is Polish, as claimed. 437S We now have a language in which to express a fundamental result in the theory of dynamical systems. Theorem Let X be a non-empty compact Hausdorff space, and φ : X → X a continuous function. Write Qφ for the set of Radon probability measures on X for which φ is inverse-measure-preserving. Then Qφ is not empty, and is compact for the narrow topology; regarded as a subset of the linear space Mt of signed tight Borel measures on X, Qφ is convex. proof (a) Write PR for the set of Radon probability measures on X. For µ ∈ PR , write T µ for the correR sponding linear functional on C(X), so that (T µ)(u) = u dµ for u ∈ C(X). By the Riesz Representation Theorem (436J/436K), T is a bijection between PR and the set K of positive linear functionals f on C(X) such that f (χX) = 1. By 437H/437I, the linear structure of Mt matches that of C(X)∗ , so the convex structure of K, when transferred to PR by T , corresponds to the convex structure of {µ¹B(X) : µ ∈ PR } ⊆ Mt . By 437Kc, the narrow topology on PR corresponds to the weak* topology Ts (C(X)∗ , C(X)) on K. As noted in 437O, PR and K are compact. (b) If µ ∈ PR , then µ ∈ Qφ iff (T µ)(uφ) = (T µ)(u) for every u ∈ C(X). P P For µ ∈ PR , consider the image measure µφ−1 . This is a Radon measure (418I). Now

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233

µ ∈ Qφ ⇐⇒ µφ−1 [E] = µE for every E ∈ dom µ (because µφ−1

⇐⇒ µφ−1 = µ and µ are Radon measures, so if they agree on all Borel sets they must be identical) Z Z −1 ⇐⇒ u d(µφ ) = u dµ for every u ∈ C(X)

(416E(b-v))

Z ⇐⇒

Z uφ dµ =

u dµ for every u ∈ C(X)

(235A) ⇐⇒ (T µ)(uφ) = (T µ)(u) for every u ∈ C(X). Q Q (c) Since {f : f ∈ C(X)∗ , f (u) = f (uφ) for every u ∈ C(X)} is convex and closed in C(X)∗ for the weak* topology, Qφ is convex and closed in PR , and is therefore compact. (d) To see that Qφ is not empty, take anyPx0 ∈ X and a non-principal ultrafilter F on N. Define n 1 i f : C(X) → R by setting f (u) = limn→F n+1 i=0 u(φ (x0 )) for every u ∈ C(X) = Cb (X). Then f is a positive linear functional and f (χX) = 1. So there is a µ ∈ PR such that f = T µ. If u ∈ C(X), then f (u) = f (uφ). P P 1 n→F n+1

|f (uφ) − f (u)| = | lim =

n X

u(φi+1 (x0 )) − u(φi (x0 ))|

i=0

1 | lim u(φn+1 (x0 )) − u(x0 )| n→F n+1

≤ lim

1

n→F n+1

|u(φn+1 (x0 )) − u(x0 )| ≤ lim

n→F

2kuk∞ n+1

= 0. Q Q

By the characterization in (b), µ ∈ Qφ . 437T Prokhorov spaces For topological spaces X which are not compact, weak* compactness in Cb (X)∗ is still interesting, and we can look for measure-theoretic criteria which will help us to recognise sets of measures which are relatively compact in the vague topology. The most important of these seems to be the following. Definition Let X be a Hausdorff space. (a) If A is a norm-bounded subset of Cb (X)∗ , we say that A is uniformly tight if for every ² > 0 there is a compact set K ⊆ X such that |f (u)| ≤ ² whenever f ∈ A, u ∈ Cb (X) and |u| ≤ χ(X \ K). (Cf. 436Xn.) ˜ be the space of bounded additive functionals defined on subalgebras of PX containing every (b) Let M ˜ define |ν| ∈ M ˜ by saying that if Σ is the domain of ν, then |ν|E = sup{νF − ν(E \ F ) : open set. For ν ∈ M F ∈ Σ, F ⊆ E} for every E ∈ Σ; that is, we calculate |ν| in the L-space M (Σ) of bounded additive functionals on Σ (362Ba). (Warning! if T is a subalgebra of Σ, |ν¹T| may be different from |ν|¹T.) I ˜ is tight if |ν|(E) = sup{|ν|(K) : K ⊆ E is closed and compact} for every say that a functional ν ∈ M ˜ is uniformly tight if for every ² > 0 E ∈ dom ν, and that a norm-bounded set A of tight functionals in M there is a compact set K ⊆ X such that |ν|(X \ K) ≤ ² for every ν ∈ A. 437U Proposition Let X be a Hausdorff space, andR MR+ the set of totally finite Radon measures on X. For µ ∈ MR+ define T µ ∈ Cb (X)∗ by setting (T µ)(u) = u dµ for every u ∈ Cb (X). (a) Let A ⊆ Cb (X)∗ be a norm-bounded set. Then the following are equiveridical: (i) A is uniformly tight; (ii) there is a solid convex weak*-closed uniformly tight set A˜ ⊆ Cb (X)∗ including A;

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437U

(iii) there is a uniformly tight set B ⊆ MR+ such that {|f | : f ∈ A} = T [B]. (b) If A ⊆ MR+ is bounded and uniformly tight, then A is relatively compact in MR+ for the narrow topology on MR+ . proof (a)(i)⇒(ii) If A is uniformly tight, then for each n ∈ N we can choose a compact set Kn ⊆ X such that |f (u)| ≤ 2−n whenever u ∈ Cb (X) and |u| ≤ χ(X \ Kn ). Set A˜ = {f : f ∈ Cb (X)∗ , |f (u)| ≤ 2−n whenever u ∈ Cb (X), n ∈ N and |u| ≤ χ(X \ Kn )}. ˜ g ∈ Cb (X)∗ and |g| ≤ |f |, then If f ∈ A, |g(v)| ≤ |g|(|v|) ≤ |f |(|v|) = sup{|f (u)| : |u| ≤ |v|} ˜ Thus A˜ is solid (352Ja). It is easy to check that A˜ is for every v ∈ Cb (X); it follows at once that g ∈ A. convex, weak*-closed and uniformly tight. So (ii) is true. (ii)⇒(i) is trivial. (i)⇒(iii) Suppose that A is uniformly tight. For every n ∈ N choose a compact set Kn ⊆ X such S that |f (u)| ≤ 2−n whenever u ∈ Cb (X) and |u| ≤ χ(X \ Kn ); replacing Kn by i≤n Ki if necessary, we can arrange that hKn in∈N is non-decreasing. Let B1 be the set of Radon measures µ on X such that µX ≤ supf ∈A kf k and µ(X \ Kn ) ≤ 2−n for every n ∈ N; of course B1 is uniformly tight. α) Set A0 = {|f | : f ∈ A}. Then every g ∈ A0 is smooth. P (α P Let f ∈ A be such that g = |f |. Suppose that D ⊆ Cb (X)∗ is non-empty and downwards-directed, and that inf u∈D u(x) = 0 for every x ∈ X. Fix v0 ∈ D. Let ² > 0. Let n ∈ N be such that 2−n (kgk + kv0 k∞ ) ≤ ². Then there is a v ∈ D such that v ≤ v0 and v(x) ≤ 2−n whenever x ∈ Kn (apply 436Ic to {v¹Kn : v ∈ D, v ≤ v0 } ⊆ C(Kn )). Set v 0 = v ∧ 2−n χX, v 00 = v − v 0 . Then g(v 0 ) ≤ 2−n kgk. Also, if u ∈ Cb (X) and |u| ≤ v 00 , we have |u| ≤ kv0 k∞ χ(X \ Kn ), so |f (u)| ≤ 2−n kv0 k∞ ; as u is arbitrary, g(v 00 ) ≤ 2−n kv0 k∞ . Putting these together, g(v) ≤ 2−n (kgk + kv0 k∞ ≤ ². As D and ² are arbitrary, g is smooth. Q Q β ) For every g ∈ A0 there is a µ ∈ B1 such that g = T µ. P (β P ByR 436H there is a measure ν on X, quasi-Radon for the topology S generated by Cb (X), such that g(u) = u dν for every u ∈ Cb (X); and of course νX ≤ supf ∈A kf k. As noted in the remark following 436H, we can arrange that ν should be inner regular for the family L of sets L such that χL = inf D for some non-empty D ⊆ Cb (X). If L ∈ L and n ∈ N and Kn ∩ L = ∅ and ² > 0, set D = {u : u ∈ Cb (X), χL ≤ u ≤ χX}. Then D is downwards-directed and inf u∈D u(x) = 0 for every x ∈ Kn , so there is a u ∈ D such that u(x) ≤ ² for every x ∈ Kn ; in this case, (1 − ²)νL ≤ g((u − ²χX)+ ) ≤ 2−n . As ² is arbitrary, νL ≤ 2−n ; as L is arbitrary, ν∗ (X \ Kn ) ≤ 2−n and ν ∗ Kn ≥ νX − 2−n . What this means is that νX = supn∈N ν ∗ Kn , so by 416O there is a Radon measure µ on X extending ν such that µKn = ν ∗ Kn Rfor every Rn. In this case µ(X \ Kn ) = ν∗ (X \ Kn ) ≤ 2−n for every n, and µ ∈ B1 . Of course we now have u dµ = u dν = g(u) for every u ∈ Cb (X) (because the identity map from X to itself is inverse-measure-preserving for µ and ν), so T µ = g. Q Q (γγ ) So if we set B = {µ : µ ∈ B1 , T µ ∈ A0 }, then B witnesses that (iii) is true. (iii)⇒(i) If B ⊆ MR+ is uniformly tight, then {T µ : µ ∈ B} is uniformly tight. P P For every ² > 0 there is a compact set K ⊆ X such that µ(X \ K) ≤ ² for every µ ∈ B; now |(T µ)(u)| ≤ ² whenever µ ∈ B, u ∈ Cb (X) and |u| ≤ χ(X \ K). Q Q So (iii) implies that A0 = {|f | : f ∈ A} is uniformly tight; because (i)⇒(ii), it follows at once that A is uniformly tight. (b) Set γ = supµ∈A µX (taking γ = 0 in the trivial case A = ∅), and for each n ∈ N let Kn ⊆ X be a compact set such that µ(X \ Kn ) ≤ 2−n for every µ ∈ A. Let A˜ ⊇ A be the set of Radon measures µ on X ˜ and such that µX ≤ γ and µ(X \ Kn ) ≤ 2−n for every n ∈ N. Let F be an ultrafilter on MR+ containing A, set νE = limµ→F µE for every Borel set E ⊆ X. Then ν is a non-negative additive functional on the Borel σ-algebra, and νX ≤ γ, while ν(X \ Kn ) ≤ 2−n for every n. By 416N, there is a Radon measure λ on X such that λX ≤ νX and λK ≥ νK for every compact set K. Now since inf K⊆X is compact ν(X \ K) = 0, supK⊆X is compact ν(E ∩ K) = νE for every E ∈ dom ν. In particular, λX ≤ νX = supK⊆X

is compact

νK ≤ supK⊆X

is compact

λK = λX

437W

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235

and λX = νX. Now λKn ≥ νKn ≥ νX − 2−n = λX − 2−n ˜ Next, if G ⊆ X is open, λG ≤ νG. P for every n, so λ ∈ A. P If ² > 0, let K ⊆ X be a compact set such that νK ≥ νX − ². Then λG ≤ λX − λ(K \ G) ≤ νX − ν(K \ G) ≤ ² + νK − ν(K \ G) = ² + ν(K ∩ G) ≤ ² + νG; as ² is arbitrary, λG ≤ νG. Q Q Now suppose that U is a subset of MR+ which contains λ and is open for the narrow topology. Then there must be open sets G0 , . . . , Gn ⊆ X and α0 , . . . , αn , β ∈ R such that V = {µ : µ ∈ MR+ , µGi > αi for every i ≤ n, µX < β} contains λ and is included in U . In this case, however, νX < β and νGi > αi for every i, so V ∈ F and U ∈ F. Thus λ is a limit of F; as F is arbitrary, A˜ is compact and A is relatively compact. 437V In important cases, the narrowly compact subsets of MR+ (X) are exactly the bounded uniformly tight sets. Once again, it is worth introducing a word to describe when this happens. Definition Let X be a Hausdorff space and PR (X) the set of Radon probability measures on X. X is a Prokhorov space if every subset of PR (X) which is compact for the narrow topology is uniformly tight. 437W Theorem (a) Compact Hausdorff spaces are Prokhorov spaces. (b) A closed subspace of a Prokhorov Hausdorff spaces is a Prokhorov space. (c) An open subspace of a Prokhorov Hausdorff space is a Prokhorov space. (d) The product of a countable family of Prokhorov Hausdorff spaces is a Prokhorov space. (e) Any Gδ subset of a Prokhorov Hausdorff space is a Prokhorov space. ˇ (f) Cech-complete spaces are Prokhorov spaces. (g) Polish spaces are Prokhorov spaces. proof (a) This is trivial; on a compact Hausdorff space the set of all Radon probability measures is uniformly tight. (b) Let X be a Prokhorov Hausdorff space, Y a closed subset of X, and A ⊆ PR (Y ) a narrowly compact ˜ set. Taking φ to be the identity map from Y to X, and defining φ˜ : MR+ (Y ) → MR+ (X) as in 437Q, φ[A] is narrowly compact in PR (X), so is uniformly tight. For any ² > 0, there is a compact set K ⊆ X such that ˜ φ(µ)(X \ K) ≤ ² for every µ ∈ A. Now K ∩ Y is a compact subset of Y and µ(Y \ (K ∩ Y )) ≤ ² for every µ ∈ A. As ² is arbitrary, A is uniformly tight in PR (Y ). (c) Let X be a Prokhorov Hausdorff space, Y an open subset of X, and A ⊆ PR (Y ) a narrowly compact ˜ set. Once again, take φ to be the identity map from Y to X, so that φ[A] ⊆ PR (X) is narrowly compact and uniformly tight in PR (X). ?? Suppose, if possible, that A is not uniformly tight in PR (Y ). Then there is an ² > 0 such that AK = {µ : µ ∈ A, µ(Y \ K) ≥ 5²} is non-empty for every compact set K ⊆ Y . Note that AK ⊆ AK 0 whenever K ⊇ K 0 , so {AK : K ⊆ Y is compact} has the finite intersection property, and there is an ultrafilter F on PR (Y ) containing every AK . Because A is narrowly compact, there is a λ ∈ PR (Y ) such that F → λ. Let K ∗ ⊆ Y be a compact set such that λ(Y \ K ∗ ) ≤ ². ˜ ˜ As φ[A] is uniformly tight, there is a compact set L ⊆ X such that µ(Y \ L) = φ(µ)(X \ L) ≤ ² for every ∗ µ ∈ A. Now K and L \ Y are disjoint compact sets in the Hausdorff space X, so there are disjoint open sets G, H ⊆ X such that K ∗ ⊆ G and L \ Y ⊆ H (4A2Fh). Set K = L \ H ⊇ L ∩ G; then K is a compact subset of Y . As AK ∈ F, there must be a µ ∈ AK such that µY ≤ λY + ² and µ(G ∩ Y ) ≥ λ(G ∩ Y ) − ². Accordingly

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437W

µ(Y ∩ L) ≤ ², µ(Y \ G) = µY − µ(G ∩ Y ) ≤ λY + ² − λ(G ∩ Y ) + ² = λ(Y \ G) + 2² ≤ λ(Y \ K ∗ ) + 2² ≤ 3², µ((Y ∩ L) ∪ (Y \ G)) = µ(Y \ (L ∩ G)) ≥ µ(Y \ K) ≥ 5², which is impossible. X X Thus A is uniformly tight. As A is arbitrary, Y is a Prokhorov space. (d) Let hXn in∈N be a sequence of Prokhorov Hausdorff spaces with product X. Let A ⊆ PR (X) be a narrowly compact set. Let ² > 0. For each n ∈ N let πn : X → Xn be the canonical map and π ˜n : MR+ (X) → MR+ (Xn ) the associated function. Then π ˜n [A] is narrowly compact in PR (Xn ),Qtherefore uniformly tight, and there is a compact set Kn ⊆ Xn such that (˜ πn µ)(Xn \Kn ) ≤ 2−n−1 ². Set K = n∈N Kn , S −1 so that K is a compact subset of X and X \ K = n∈N πn [Xn \ Kn ]. If µ ∈ A, then P∞ P∞ µ(X \ K) ≤ n=0 µπn−1 [Xn \ Kn ] ≤ n=0 2−n−1 ² = ². As ² is arbitrary, A is uniformly tight; as A is arbitrary, X is a Prokhorov space.

T (e) Let X be a Prokhorov Hausdorff space and Y a Gδ subset of X. Express Y as n∈N Yn where every Q Yn ⊆ X is open. Set Q Z = {z : z ∈ n∈N Yn , z(m) = z(n) for all m, n ∈ N}. Because X is Hausdorff, Z is a closed subspace of n∈N Yn homeomorphic to Y . Putting (c), (d) and (b) together, Z and Y are Prokhorov spaces. ˇ (f ) Put (a), (e) and the definition of ‘Cech-complete’ together. (g) This is a special case of (f) (4A2Md). 437X Basic exercises (a) Let X be a set, U a Riesz subspace of RX and f ∈ U ∼ . (i) Show that f ∈ Uσ∼ iff limn→∞ f (un ) = 0 whenever hun in∈N is a non-increasing sequence in U such that limn→∞ un (x) = 0 for every x ∈ X. (Hint: show that in this case, if 0 ≤ vn ≤ un , we can find k(n) such that f (vn ∧ uk(n) ) ≤ f (vn )+2−n .) (ii) Show that f ∈ Uτ∼ iff inf u∈A |f (u)| = 0 whenever A ⊆ U is a non-empty downwards-directed set and inf u∈A u(x) = 0 for every x ∈ X. (Hint: given ² > 0, set B = {v : f (v) ≥ inf u∈A f + (u) − ², ∃ w ∈ A, v ≥ w} and show that B is a downwards-directed set with infimum 0 in RX .) (b) Let X be a set, U a Riesz subspace of `∞ (X) containing the constant functions, and Σ the smallest σ-algebra of subsets of X with respect to which every member of U is measurable. Let µ and ν be two totally finite measures on X with domain Σ, and f , g the corresponding linear functionals on U . Show that f ∧ g = 0 in U ∼ iff there is an E ∈ Σ such that µE = ν(X \ E) = 0. (Hint: 326I.) (c) (i) Show that S, in 437C, is the unique sequentially order-continuous positive linear operator from L∞ to (Uσ∼ )∗ which extends the canonical embedding of U in (Uσ∼ )∗ . (ii) Show that S, in 437H, is the unique sequentially order-continuous positive linear operator from L∞ to (Uτ∼ )∗ which extends the canonical embedding of U in (Uτ∼ )∗ and is ‘τ -additive’ in the sense that whenever G is a non-empty upwards-directed family of open sets with union H then S(χH) = supG∈G S(χG) in (Uτ∼ )∗ . (d) Let X and Y be completely regular topological spaces and φ : X → Y a continuous function. Define T : Cb (Y ) → Cb (X) by setting T (v) = vφ for every v ∈ Cb (Y ), and let T 0 : Cb (Y )∗ → Cb (X)∗ be its adjoint. ∼ (i) Show that T 0 is a norm-preserving Riesz homomorphism. (ii) Show that T 0 [Cb (Y )∼ σ ] ⊆ Cb (X)σ , and ∼ 0 that if f ∈ Cb (X)σ corresponds to a Baire measure µ on X, then T f corresponds to the Baire measure ∼ ∼ µφ−1 ¹Ba(Y ). (iii) Show that T 0 [Cb (Y )∼ τ ] ⊆ Cb (X)τ , and that if f ∈ Cb (X)τ corresponds to a Borel measure 0 −1 ∞ µ on X, then T f corresponds to the Borel measure µφ ¹B(Y ). (iv) Write L∞ X and LY for the M -spaces ∼ ∗ of bounded real-valued Borel measurable functions on X, Y respectively, and SX : L∞ X → (Cb (X)τ ) , ∞ ∼ ∗ SY : LY → (Cb (Y )τ ) for the canonical Riesz homomorphisms as constructed in 437Hb. Show that if ∗ ∼ ∗ 0 ∼ 00 ∞ T 00 : (Cb (Y )∼ τ ) → (Cb (X)τ ) is the adjoint of T ¹Cb (X)τ , then T SY (v) = SX (vφ) for every v ∈ L (Y ). (e) Let X be a topological space, L∞ (Σum ) the space of bounded universally measurable real-valued functions on X, and Mσ the space of countably additive functionals on the Borel σ-algebra of X. Show that we have a sequentially order-continuous Riesz homomorphism S : L∞ (Σum ) → Mσ∗ defined by the formula R (Sv)(µ) = v dµ whenever v ∈ L∞ (Σum ) and µ ∈ Mσ+ .

437Xr

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(f ) Let X be a completely regular topological space. Show that the vague topology on the space of differences of τ -additive totally finite Borel measures on X is Hausdorff. (g) Let X be a topological space, and for x ∈ X set δx (E) = χE(x) for every E ∈ B = B(X). (i) Show that the linear span of {δx : x ∈ X} is dense in M (B) for the vague topology on M (B). (ii) Show that the convex hull of {δx : x ∈ X} is dense in {µ : µ ∈ M (B)+ , µX = 1} for the narrow topology. (Hint: for (ii), do not use the Hahn-Banach theorem.) > (h) Let X and Y be topological spaces, and φ : X → Y a continuous function. Write M# (X) for any of M (Ba(X)), Mσ (Ba(X)), M (B(X)), Mσ (B(X)), Mτ (X) or Mt (X), where Mτ (X) ⊆ Mσ (B(X)) is the space of signed τ -additive Borel measures and Mt (X) ⊆ Mτ (X) is the space of signed tight Borel measures; and M# (Y ) for the corresponding space based on Y . Show that there is a positive linear operator ˜ φ˜ : M# (X) → M# (Y ) defined by saying that φ(µ)(E) = µφ−1 [E] whenever µ ∈ M# (X) and E belongs to Ba(Y ) or B(Y ), as appropriate, and that φ˜ is continuous for the vague topologies on M# (X) and M# (Y ). (i) Let X be any topological space. Show that the narrow topology on the set of totally finite Borel measures on X is T0 . ˜ + the set of non-negative additive functionals defined on (j) Let X be any topological space and M ˜ + define µ + ν ∈ M ˜ + by setting (µ + ν)(E) = subalgebras of PX containing every open set. For µ, ν ∈ M ˜ + is continuous for the narrow topology. (ii) µE + νE for E ∈ dom µ ∩ dom ν. (i) Show that addition on M + + ˜ ˜ ˜ +. Show that (α, µ) 7→ αµ : [0, ∞[ × M → M is continuous for the narrow topology on M (k) Give X = ω1 + 1 its order topology. On X let µ be the Borel probability measure corresponding to Dieudonn´e’s measure on ω1 (see 434Xf) and ν the Borel probability measure such that ν{ω1 } = 1. (i) Show that µG ≥ νG for every open set G ⊆ X. (ii) Show that the narrow topology on the set of Borel probability measures on X is not T1 . > (l) Let X be a zero-dimensional compact Hausdorff space and E the algebra of open-and-closed subsets of X. (i) Show that E separates zero sets. (ii) Show that the vague topology on M (E) is just the pointwise topology induced by the usual topology of R E . (iii) Writing Mt for the space of signed tight Borel measures on X, show that µ 7→ µ¹E : Mt → M (E) is a Banach lattice isomorphism between the L-spaces Mt and M (E), and is also a homeomorphism when Mt and M (E) are given their vague topologies. ˜ + the space of non-negative real-valued additive functionals (m) Let X be a topological space, and M defined on algebras of subsets of X which contain every open set, R with its+narrow topology. Show that if ˜ → R is lower semi-continuous. u : X → R is a bounded lower semi-continuous function then ν 7→ u dν : M (n) In 437Mc, show that |ψ(µ, ν)| = ψ(|µ|, |ν|) for every µ ∈ Mτ (X) and ν ∈ Mτ (Y ). (o) (i) In 437P, give PR its narrow topology. Show that the map x 7→ δx is a homeomorphism between X and its image in PR . (ii) Show that if X is completely regular then x 7→ δx is a homeomorphism between X and its image in PR when PR is given the vague topology corresponding to its embedding in Mσ (B(X)). (p) Let X be a compact Hausdorff space and PR0 the set of Radon measures µ on X such that µX ≤ 1. Show that the extreme points of PR0 are the point-masses δx , as in 437P, together with the zero measure. (q) Let X be a topological space. Show that the set of tight functionals f ∈ Cb (X)∼ (436Xn) is a band in Cb (X)∼ included in Cb (X)∼ τ . ˜ + the set of non-negative additive functionals defined on subalgebras (r) Let X be a Hausdorff space and M of PX containing every open set, with the algebraic operations defined in 437Xj. Let A and B be norm˜ + and γ ≥ 0. Show that A ∪ B, A + B = {µ + ν : µ ∈ A, ν ∈ B} and bounded uniformly tight subsets of M {αµ : µ ∈ A, 0 ≤ α ≤ γ} are uniformly tight.

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> (s) Let X be a Prokhorov Hausdorff space, and A a set of Radon measures on X which is compact for the narrow topology. Show that A is uniformly tight. (Hint: (i) γ = supµ∈A µX is finite; (ii) for any 1 µ : µ ∈ A, µX ≥ ²} is narrowly compact, therefore uniformly tight; for any ² > 0 the set ² > 0 the set { µX {µ : µ ∈ A, µX ≥ ²} is uniformly tight.) (t) Give ω1 its order topology, and let Mt be the L-space of signed tight Borel measures on ω1 . (i) Show that ω1 is a Prokhorov space. (ii) For ξ < ω1 , define µξ ∈ Mt by setting µξ (E) = χE(ξ) − χE(ξ + 1) for every Borel set E ⊆ ω1 . Show that {µξ : ξ < ω1 } is relatively compact in Mt for the vague topology, but is not uniformly tight. 437Y Further exercises (a) Let X be a set and U a Riesz subspace of RX . Give formulae for the components of a given element of U ∼ in the bands Uσ∼ , (Uσ∼ )⊥ , Uτ∼ and (Uτ∼ )⊥ . (Hint: 356Yb.) (b) Let X be a compact Hausdorff space. Show that the dual C(X; C)∗ of the complex linear space of continuous functions from X to C can be identified with the space of ‘complex tight Borel measures’ on X, that is, the space of functionals µ : B(X) → C expressible as a complex linear combination of tight totally finite Borel measures; explain how this may be identified, as Banach space, with the complexification of the L-space Mt of signed tight Borel measures as described in 354Yk. Show that the complex Banach space ∗∗ L∞ C (B(X)) is canonically embedded in C(X; C) . (c) Write µc for counting measure on [0, 1], and µL for Lebesgue measure; write µc × µL for the product measure on [0, 1]2 , and µ for the direct sum of µc and µc ×µL . Show that the L-space C([0, 1])∼ is isomorphic, as L-space, to L1 (µ). (Hint: every Radon measure on [0, 1] has countable Maharam type.) (d) Let X be a set, U a Riesz subspace of `∞ (X) containing the constant functions, and Σ the smallest σ˜ for the intersection algebra of subsets of X with respect to which every member of U is measurable. Write Σ of the domains of the completions of the totally finite measures with domain Σ. Show that there is a unique ˜ to (Uσ∼ )∗ ∼ sequentially order-continuous norm-preserving Riesz homomorphism from L∞ (Σ) = Mσ∗ such that ∼ (Su)(f ) = f (u) whenever u ∈ U and f ∈ Uσ . (e) Let X be a completely regular Hausdorff space and Pτ the space of τ -additive Borel probability measures on X. Let B ⊆ Pτ be a non-empty set. Show that the following are equiveridical: (i) B is relatively compact in Pτ for the vague topology; (ii) whenever A ⊆ CRb (X) is non-empty and downwardsdirected and inf u∈A u(x) = 0 for every x ∈ A, then inf u∈A supµ∈B u dµ = 0; (iii) whenever G is an upwards-directed family of open sets with union X, then supG∈G inf µ∈B µG = 1. (f ) Explain how to express the proof of 285L(iii)⇒(ii) as (α) a proof that if the characteristic functions of a sequence hνn in∈N of Radon probability measures on R r converge pointwise to a characteristic function, then {νn : n ∈ N} is uniformly tight (β) the observation that any subalgebra of Cb (R r ) which separates the points of R r and contains the constant functions will define the vague topology on any vaguely compact set of measures. (g) Let X be a topological space. (i) Let Mσ (Ba) be the space of signed Baire measures on RX, and u : X → R a bounded Baire measurable function. Show that we have a linear functional µ 7→ u dµ : Mσ (Ba) → R agreeing with ordinary integration with respect to non-negative measures. Show that this functional is Baire measurable with respect to the vague topology on Mσ (Ba). (ii) Let RMτ be the space of signed τ -additive Borel measures on X. Show that we have a linear functional µ 7→ u dµ : Mτ → R agreeing with ordinary integration with respect to non-negative measures. Show that this functional is Borel measurable with respect to the vague topology on Mτ . (h) For a topological space X let Mτ (X) be the L-space of signed τ -additive Borel measures on X, and ψ : Mτ (X) × Mτ (X) → Mτ (X × X) the canonical bilinear map (437M); give Mτ (X) and Mτ (X × X) their vague topologies. (i) Show that if X = [0, 1] then ψ is not continuous. (ii) Show that X = Z and B is the unit ball of Mτ (X) then ψ¹B × B is not continuous.

437 Notes

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239

(i) Let (X, ρ) be a metric space. For f ∈ Cb (X)∼ σ , set kf kH = sup{|f (u)| : u ∈ Cb (X), kuk∞ ≤ 1, u is 1-Lipschitz} (definition: 471J). (i) Show that k kH is a norm on Cb (X)∼ σ . The metric it induces is sometimes called Huntingdon’s metric. (ii) Show that the topology induced by Huntingdon’s metric on the set Mτ of signed τ -additive Borel measures on X is the vague topology. (Hint: 434L.) (iii) Show that if (X, ρ) is complete then the set Mt of signed tight Borel measures is complete under Huntingdon’s metric. (Hint: show that any Cauchy sequence in Mt must be uniformly tight.) (iv) Show that if (X, ρ) is R with its usual metric, then Huntingdon’s metric, interpreted as a metric on the set of Radon probability measures on X, is not uniformly equivalent to L´evy’s metric as described in 274Ya. (j) Let X be any Hausdorff space, and MR∞+ the set of Radon measures on X. Define addition and scalar multiplication (by positive scalars) on MR∞+ as in 416Xd, and ≤ by the formulae of 416Ea. (i) Show that there is a Dedekind complete Riesz space V such that the positive cone of V is isomorphic to MR∞+ . (ii) Show that every principal band in V is an L-space. (iii) Show that if X is metrizable then V is perfect. (k) Let X be a compact Hausdorff space and PR the set of Radon probability measures on X, with its narrow topology. Show that the weight w(PR ) of PR is at most max(ω, w(X)). (Hint: the weight of C(X) in its norm topology is at most max(ω, w(X)).) ˇ (l) Let X be a Cech-complete completely regular Hausdorff space and PR the set of Radon probability ˇ measures on X, with its narrow topology. Show that PR is Cech-complete. (m) Let X and Y be topological spaces, and ψ : Mτ (X)×Mτ (Y ) → Mτ (X×Y ) the bilinear map of 437Mc. Write Mt (X), etc., for the spaces of signed tight Borel measures. (i) Show that ψ(µ, ν) ∈ Mt (X × Y ) for every µ ∈ Mt (X), ν ∈ Mt (Y ). (ii) Show that if B ⊆ Mt (X), B 0 ⊆ Mt (Y ) are norm-bounded and uniformly tight, then ψ¹B × B 0 is continuous for the vague topologies. (n) (i) X be a metrizable space, and A a narrowly compact subset of the set of Radon probability measures on X. Show that there is a separable subset Y of X which is conegligible for every measure in A. (ii) Show that a metrizable space is Prokhorov iff all its closed separable subspaces are Prokhorov. (o) I say that a completely regular Hausdorff space X is strongly Prokhorov if every vaguely compact subset of the space Mt (X) of signed tight Borel measures on X is uniformly tight. (i) Check that a strongly Prokhorov completely regular Hausdorff space is Prokhorov. (ii) Show that a closed subspace of a strongly Prokhorov completely regular Hausdorff space is strongly Prokhorov. (iii) Show that the product of a countable family of strongly Prokhorov completely regular Hausdorff spaces is strongly Prokhorov. (iv) Show that a Gδ subset of a strongly Prokhorov metrizable space is strongly Prokhorov. (v) Show that a Polish space is strongly Prokhorov. 437 Notes and comments The ramifications of the results here are enormous. For completely regular topological spaces X, the theorems of §436 give effective descriptions of the totally finite Baire, quasi-Radon and Radon measures on X as linear functionals on Cb (X) (436E, 436Xl, 436Xn). This makes it possible, and natural, to integrate the topological measure theory of X into functional analysis, through the theory of Cb (X)∗ . (See Wheeler 83 for an extensive discussion of this approach.) For the rest of this volume we shall never be far away from such considerations. In 437C-437I I give only a sample of the results, heavily slanted towards the abstract theory of Riesz spaces in Chapter 35 and the first part of Chapter 36. Note that while the constructions of the dual spaces U ∼ , Uc∼ and U × are ‘intrinsic’ to a Riesz space U , in that we can identify these functions as soon as we know the linear and order structure of U , the spaces Uσ∼ and Uτ∼ are definable only when U is presented as a Riesz subspace of RX . In the same way, while the space Mσ (Σ) of countably additive functionals on a σ-algebra Σ depends only on the Boolean algebra structure, the spaces Mτ here (not to be confused with the space of completely additive functionals considered in 362B) depend on the topology as well as the Borel algebra. (For an example in which radically different ´ sz Kunen & Rudin 76.) topologies give rise to the same Borel algebra, see Juha You may have been puzzled by the shift from ‘quasi-Radon’ measures in 436H to ‘τ -additive’ measures in 437H; somewhere the requirement of inner regularity has got lost. The point is that the topologies being

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considered here, being defined by declaring certain families of functions continuous, are (completely) regular; so that τ -additive measures are necessarily inner regular with respect to the closed sets (414Mb). The theory of ‘vague’ and ‘narrow’ topologies’ in 437J-437W here hardly impinges on the questions considered in §§274 and 285, where vague topologies first appeared. This is because the earlier investigation was dominated by the very special position of the functions x 7→ eiy . x (what we shall in §445 come to call the ‘characters’ of the additive group of R or R r ). One idea which does appear essentially in the proof of 285L, and has a natural interpretation in the general theory, is that of a ‘uniformly tight’ family of Radon measures; see 437T-437W. In 445Yh I set out a generalization of 285L to abelian locally compact groups. In §461 I will return to the general theory of extreme points in compact convex sets. Here I remark only that it is never surprising that extreme points should be special in some way, as in 437P; but the precise ways in which they are special are often unexpected. A good deal of work has been done on relationships between the topological properties of a topological space X and the space P of Radon probability measures on X with the vague topology. Here I show only that if X is compact, so is P (437O), that if X is compact and metrizable so is P (437Ra), and that if X is Polish so is P (437Rb), with a couple of hints at generalizations (437Yk-437Yl). The terms ‘vague’ and ‘narrow’ both appear in the literature on this topic, and I take the opportunity to use them both, meaning slightly different things. Vague topologies, in my usage, are linear space topologies on linear spaces of functionals; narrow topologies are topologies on spaces of (finitely additive) measures, which are not linear spaces, though we can make an attempt to define addition and multiplication by non-negative scalars (437Xj). I must warn you that this distinction is not standard. I see that the word ‘narrow’ appears a good deal oftener than the word ‘vague’, which is in part a reflection of a simple prejudice against signed measures; but from the point of view of this treatise as a whole, it is more natural to work with a concept well adapted to measures with variable domains, even if we are considering questions (like compactness of sets of measures) which originate in linear analysis. I should mention also that the definition in 437Jc includes a choice. The duality considered there uses the space Cb (X); for locally compact X, we have the rival spaces C0 (X) and Ck (X) (see 436J and 436K), and there are occasions when one of these gives a more suitable topology on a space of measures (as in 495Xl below). The elementary theory of uniform tightness and Prokhorov spaces (437T-437W) is both pretty and useful. The emphasis I give it here, however, is partly because it provides the background to a remarkable construction by D.Preiss (439S below), showing that Q is not a Prokhorov space.

438 Measure-free cardinals At several points in §418, and again in §434, we had theorems about separable metrizable spaces in which the proofs undoubtedly needed some special property of these spaces (e.g., the fact that they are Lindel¨of), but left it unclear whether something more general could be said. When we come to investigate further, asking (for instance) whether complete metric spaces in general are Radon (438H), we find ourselves once again approaching the Banach-Ulam problem, already mentioned at several points in previous volumes, and in particular in 363S. It seems to be undecidable, in ordinary set theory with the axiom of choice, whether or not every discrete space is Radon in the sense of 434C. On the other hand it is known that discrete spaces of cardinal at most ω1 (for instance) are indeed always Radon. While as a rule I am deferring questions of this type to Volume 5, this particular phenomenon is so pervasive that I think it is worth taking a section now to clarify it. The central definition is that of ‘measure-free cardinal’ (438A), and the basic results are 438B-438D. In particular, ‘small’ infinite cardinals are measure-free (438C). From the point of view of measure theory, a metrizable space whose weight is measure-free is almost separable, and most of the results in §418 concerning separable metrizable spaces can be extended (438E-438G). In fact ‘measure-free weight’ exactly determines whether a metrizable space is measure-compact (438J, 438Xk) and whether a complete metric space is Radon (438H). If c is measure-free, some interesting spaces of functions are Radon (438R). I approach these last spaces through the concept of ‘hereditary weak θ-refinability’ (438K), which enables us to do most of the work without invoking any special axiom.

438B

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241

438A Measure-free cardinals: Definition A cardinal κ is measure-free or of measure zero if whenever µ is a probability measure with domain Pκ then there is a ξ < κ such that µ{ξ} > 0. In 363S I discussed some statements equiveridical with the assertion ‘every cardinal is measure-free’. 438B It is worth getting some basic facts out into the open immediately. Lemma Let (X, Σ, µ) be a semi-finite measure space and hEi ii∈I a point-finite family of subsets of X such S S that #(I) is measure-free andS i∈J Ei ∈ Σ for every J ⊆ I. Set E = i∈I Ei . (a) µE = supJ⊆I is finite µ( i∈J Ei ). P (b) If hEi ii∈I is disjoint, then µE = i∈I µEi . In particular, if Σ = PX and A ⊆ X has measure-free P cardinal, then µA = x∈A µ{x}. S (c) If µ is σ-finite, then L = {i : i ∈ I, µEi > 0} is countable and i∈I\L Ei is negligible. proof (a)(i) The first step is to show, by induction on n, that the result is true if µX < ∞ and every Ei is negligible and #({i : i ∈ I, x ∈ Ei }) ≤ n for every x ∈ X. If n = 0 this is trivial, since S every Ei must be empty. For the inductive step to n ≥ 1, define ν : PI → [0, ∞[ by setting νJ = µ( i∈J Ei ) for every S J ⊆ I. Then ν is a measure on I. P P Write FJ = i∈J Ej for J ⊆ I. ( α) If J, K ⊆ I are disjoint, then for i ∈ I set Ei0 = Ei ∩ FK for i ∈ J, ∅ for i ∈ I \ J. In this case, hEi0 ii∈I is a family of negligible subsets of X, S 0 0 0 0 i∈J 0 Ei = FJ ∩J ∩ FK is measurable for every J ⊆ I, and #({i : x ∈ Ei }) ≤ n − 1 for every x ∈ X; so the inductive hypothesis tells us that S S µ( i∈I Ei0 ) = supJ 0 ⊆I is finite µ( i∈J 0 µEi0 ) = 0, that is, FJ ∩ FK is negligible. But this means that ν(J ∪ K) = µFJ∪K = µ(FJ ∪ FK ) = µFJ + µFK = νJ + νK. As J and K are arbitrary, ν is additive. (β) If hJn in∈N is a disjoint sequence in PI, then [ [ [ ν( Jn ) = µ( FJn ) = lim µ( FJm ) n∈N

n→∞

n∈N

= lim

n→∞

n X

νJm =

m=0

∞ X

m≤n

νJn ,

n=0

so ν is countably additive and is a measure. Q Q At the same time, ν{i} = µEi = 0 for every i. Because #(I) is measure-free, νI = 0. P P?? Otherwise, let f : I → κ = #(I) be any bijection and set λA =

1 νf −1 [A] νI

for every A ⊆ κ; then λ is a probability

measureSwith domain Pκ which is zero on singletons, and κ is not measure-free. X XQ Q But this means just that µ( i∈I Ei ) = 0. Thus the induction proceeds. S (ii) ?? Now suppose, if possible, that the general result is false. For finite sets J ⊆ I set FJ = i∈J Ei , as before, and consider E = {FJ : J ∈ [κ]<ω } (see 3A1J for this notation). Then E is upwards-directed and γ = supH∈E µH is finite, because it is less than µE; S let hHn in∈N be a non-decreasing sequence in E such that µ(H \ H ∗ ) = 0 for every H ∈ E, where H ∗ = n∈N Hn and µH ∗ = γ (215Ab). Because µ is semi-finite, there is an F ∈ Σ such that F ⊆ E and γ < µF < ∞. For each n ∈ N, set Yn = {x : x ∈ F \ H ∗ , #({i : x ∈ Ei }) ≤ n}. Then there is some n ∈ N such that µ∗ Yn > 0. Let ν be the subspace measure S on Yn , so that ν isSnon-zero and totally finite. Now hEi ∩ Yn ii∈I is a family of negligible subsets of Yn , i∈J Ei ∩ Yn = Yn ∩ i∈J Ei is measured by ν for every J ⊆ I, and #({i : x ∈ Ei ∩ Yn }) ≤ n for every x ∈ Yn . But this contradicts (i) above. X X This proves (a). (b) If hEi ii∈I is disjoint, then

S P P supJ∈[I]<ω µ( i∈J Ei ) = supJ∈[I]<ω i∈J µEi = i∈I µEi .

Setting I = A, Ex = {x} for x ∈ A, we get the special case.

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438B

(c) Let hXn in∈N be a non-decreasing sequence of measurable sets of finite measure covering X. For each −n n, set Ln = {i : i ∈ I, µ(Ei ∩ distinct elements SXn ) ≥ 2 }. ?? If Ln is infinite, take a sequence hik ik∈N of −n in Ln , and consider Gm = k≥m Eik for m ∈ N; then every Gm has measure at least 2 , G0 has finite T measure, hGm im∈N is non-increasing, and m∈N Gm is empty, because hEik ik∈N is point-finite. But this is impossible. X X S Thus every Ln is finite and L = n∈N Ln is countable. S Now (a), applied to hEi0 ii∈I where Ei0 = Ei if i ∈ I \ L, ∅ if i ∈ L, tells us that i∈I\L Ei is negligible. 438C I do not think we are ready for the most interesting set-theoretic results concerning measure-free cardinals. But the following facts may help to make sense of the general pattern. Theorem (Ulam 30) (a) ω is measure-free. (b) If κ is a measure-free cardinal and κ0 ≤ κ is a smaller cardinal, then κ0 is measure-free. (c) If hκξ iξ<λ is a family of measure-free cardinals, and λ is also measure-free, then κ = supξ<λ κξ is measure-free. (d) If κ is a measure-free cardinal so is κ+ . (e) The following are equiveridical: (i) c is not measure-free; (ii) there is a semi-finite measure space (X, PX, µ) which is not purely atomic; (iii) there is a measure µ on [0, 1] extending Lebesgue measure and measuring every subset of [0, 1]. (f) If κ ≥ c is a measure-free cardinal then 2κ is measure-free. proof (a) This is trivial. (b) If µ is a probability measure with domain Pκ0 , set νA = µ(κ0 ∩ A) for every A ⊆ κ. Then ν is a probability measure with domain Pκ, so there is a ξ < κ such that ν{ξ} > 0; evidently ξ < κ0 and µ{ξ} > 0. (c) Let µ be a probability measure on κ with domain Pκ. Define f : κ → λ by setting f (α) = min{ξ : α < κξ } for α < κ. Then the image measure µf −1 is a probability measure on λ with domain Pλ, so there is a ξ < λ such that µf −1 [{ξ}] > 0. Now µκξ > 0. Applying 438Bb to A = κξ , we see that there is an α < κξ such that µ{α} > 0. As µ is arbitrary, κ is measure-free. (d) By (a) and (b), we need consider only the case κ ≥ ω. ?? Suppose, if possible, that µ is a probability measure with domain Pκ+ such that µ{α} = 0 for every α < κ+ . For each α < κ+ , S let fα : α → κ be an injection. For β < κ+ , ξ < κ set A(β, ξ) = {α : β < α < κ+ , fα (β) = ξ}. Then κ \ ξ<κ A(β, ξ) = β + 1 S has cardinal at most κ, which is measure-free, so µ(β + 1) = 0 and µ( ξ<κ A(β, ξ)) > 0. Also hA(β, ξ)iξ<κ is disjoint. There is therefore a ξβ < κ such that µA(β, ξβ ) > 0, by 438Bb. Now κ+ > max(ω, κ), so there must be an η < κ such that B = {β : ξβ = η} is uncountable. In this case, however, hA(β, η)iβ∈B is an uncountable family of sets of measure greater than zero, and cannot be disjoint, because µ is totally finite (215B(iii)); but if α ∈ A(β, η) ∩ A(β 0 , η), where β 6= β 0 , then fα (β) = fα (β 0 ) = η, which is impossible, because fα is supposed to be injective. X X So there is no such measure µ, and κ+ is measure-free. (e)(i)⇒(ii) Suppose that c is not measure-free; let µ be a probability measure with domain Pc such that µ{ξ} = 0 for every ξ < c. Then µ is atomless. P P?? Suppose, if possible, that A ⊆ c is an atom for µ. Let f : c → PN be a bijection. For each n ∈ N, set En = {ξ : n ∈ f (ξ)}. Set D = {n T : µ(A ∩ ESn ) = µA}. Because A is an atom, µ(A ∩ En ) = 0 for every n ∈ N \ D. This means that B = n∈D En \ n∈N\D En has measure µA > 0; but f (ξ) = D for every ξ ∈ B, so #(B) ≤ 1, and µ{ξ} > 0 for some ξ, contrary to hypothesis. X XQ Q So (ii) is true. (ii)⇒(iii) Suppose that there is a semi-finite measure space (X, PX, µ) which is not purely atomic. Then there is a non-negligible set E ⊆ X which does not include any atom; let F ⊆ E be a set of non-zero finite measure. If we take ν to be

1 µF , µF

where µF is the subspace measure on F , then ν is an atomless

probability measure with domain PF . Consequently there is a function g : F → [0, 1] which is inversemeasure-preserving for ν and Lebesgue measure (343Cb). But this means that the image measure νg −1 is a measure defined on every subset of [0, 1] which extends Lebesgue measure.

438E

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not-(i)⇒not-(iii) Conversely, if c is measure-free, then any probability measure on [0, 1] measuring every subset must give positive measure to some singleton, and cannot extend Lebesgue measure. (f ) We are supposing that κ ≥ c is measure-free, so, in particular, c is measure-free. Let µ be a probability measure with domain P(2κ ). By (e), it cannot be atomless; let E ⊆ 2κ be an atom. Let f : 2κ → Pκ be a bijection, and for ξ < κ set Eξ = {α : α < 2κ , ξ ∈ f (α)}; set D = {ξ : ξ < κ, µ(E ∩ Eξ ) = µE}. Note that µ(E ∩ Eξ ) must be zero for every ξ ∈ κ \ D, so that E ∩ {α : ξ ∈ D4f (α)} is always negligible. Consider Aξ = {α : α ∈ E, ξ = min(D4f (α))} for ξ < κ. Then hAξ iξ<κ is a disjoint family of negligible sets, so its union A is negligible, by 438Bb, because κ is measure-free. But E \ A ⊆ f −1 [{D}] has at most one element, and is not negligible; so µ{α} > 0 for some α. As µ is arbitrary, 2κ is measure-free. Remark This extends the result of 419G, which used a different approach to show that ω1 is measure-free. We see from (d) above that ω2 , ω3 , . . . are all measure-free; so, by (c), ωω also is; generally, if κ is any measure-free cardinal, so is ωκ (438Xa). I ought to point out that there are even more powerful arguments showing that any cardinal which is not measure-free must be enormous (Solovay 71, Fremlin 93). In this context, however, c = 2ω can be ‘large’, at least in the absence of an axiom like the continuum hypothesis to locate it in the hierarchy hωξ iξ∈On ; it is generally believed that it is consistent to suppose that c is not measure-free. 438D I turn now to the contexts in which measure-free cardinals behave as if they were ‘small’. Proposition Let (X, Σ, µ) be a σ-finite measure space, Y a metrizable space with measure-free weight, and f : X → Y a measurable function. Then there is a closed separable set Y0 ⊆ Y such that f −1 [Y0 ] is conegligible; that is, there is a conegligible measurable set X0 ⊆ X such that f [X0 ] is separable. S proof Let U be a σ-disjoint base for the topology of Y (4A2L(h-ii)); express it as n∈N Un where each Un is a disjoint family of open sets. If n ∈ N, #(Un ) S ≤ w(Y ) (4A2Db) is S a measure-free cardinal (438Cb), and hf −1 [U ]iU ∈Un is a disjoint family in Σ such that u∈V f −1 [U ] = f −1 [ V] is measurable for every V ⊆ Un ; so 438Bc tells us that there is a countable set Vn ⊆ Un such that S S f −1 [ (Un \ Vn )] = U ∈Un \Vn f −1 [U ] is negligible. Set Then f

−1

[Y \ Y0 ] =

S n∈N

f

−1

S

Y0 = Y \

S

S n∈N

(Un \ Vn ).

[ (Un \ Vn )] is negligible. On the other hand, S {U ∩ Y0 : U ∈ U} ⊆ {∅} ∪ {V ∩ Y0 : V ∈ n∈N Vn }

is countable, and is a base for the subspace topology of Y0 (4A2B(a-iv)); so Y0 is second-countable and must be separable (4A2Oc). Thus we have an appropriate Y0 . Now X0 = f −1 [Y0 ] is conegligible and measurable and f [X0 ] ⊆ Y0 is separable (4A2P(a-iv)). 438E Proposition (cf. 418B) Let (X, Σ, µ) be a complete locally determined measure space. (a) If Y is a topological space, Z is a metrizable space, w(Z) is measure-free, and f : X → Y , g : X → Z are measurable functions, then x 7→ (f (x), g(x)) : X → Y × Z is measurable. (b) If hYn in∈N is a sequence of metrizable spaces, with product Y , and w(Yn ) is measure-free Q for every n ∈ N, and fn : X → Yn is measurable for every n ∈ N, then x 7→ f (x) = hfn (x)in∈N : X → n∈N Yn is measurable. proof (a)(i) Consider first the case in which µ is totally finite. Then there is a conegligible set X0 ⊆ X such that g[X0 ] is separable (438D). Applying 418Bb to f ¹X0 and g¹X0 , we see that x 7→ (f (x), g(x)) : X0 → Y × g[Z0 ] is measurable. As µ is complete, it follows that x 7→ (f (x), g(x)) : X → Y × Z is measurable. (ii) In the general case, take any open set W ⊆ Y × Z and any measurable set F ⊆ X of finite measure. Set Q = {x : (f (x), g(x)) ∈ W }. By (i), applied to f ¹F and g¹F , F ∩ Q ∈ Σ; as F is arbitrary and µ is locally determined, Q ∈ Σ; as W is arbitrary, x 7→ (f (x), g(x)) is measurable.

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(b) As in (a), it is enough to consider the case in which µ is totally finite. In this case, we have for each T n ∈ N a conegligible set Xn such that fn [Xn ] is separable. Set X 0 = n∈N Xn ; then 418Bd tells us that f ¹X 0 is measurable, so that f is measurable. 438F Proposition (cf. 418J) Let (X, Σ, µ) be a semi-finite measure space and T a topology on X such that µ is inner regular with respect to the closed sets. Suppose that Y is a metrizable space, w(Y ) is measure-free and f : X → Y is measurable. Then f is almost continuous. proof Take E ∈ Σ and γ < µE. Then there is a measurable set F ⊆ E such that γ < µF < ∞. Let F0 ⊆ F be a measurable set such that F \ F0 is negligible and f [F0 ] is separable (438D). By 412Pc, the subspace measure on F0 is still inner regular with respect to the closed sets, so f ¹F0 is almost continuous (418J), and there is a measurable set H ⊆ F0 , of measure at least γ, such that f ¹H is continuous. As E and γ are arbitrary, f is almost continuous. 438G Corollary (cf. 418K) Let (X, T, Σ, µ) be a quasi-Radon measure space and Y a metrizable space such that w(Y ) is measure-free. Then a function f : X → Y is measurable iff it is almost continuous. 438H Now let us turn to questions which arose in §434. Proposition A complete metric space is Radon iff its weight is measure-free. proof Let (X, ρ) be a complete metric space, and κ = w(X) its weight. (a) If κ is measure-free, let µ be any totally finite Borel measure on X. Applying 438D to the identity map from X to itself, we see that there is a closed separable conegligible subspace X0 . Now X0 is complete, so is a Polish space, and by 434Kb is a Radon space. The subspace measure µX0 is therefore tight (that is, inner regular with respect to the compact sets); as X0 is conegligible, it follows at once that µ also is. As µ is arbitrary, X is a Radon space. S (b) If κ is not measure-free, take any σ-disjoint base U for the topology of X. Express U as n∈N Un where every Un is disjoint. Then κ ≤ #(U) and there is a probability measure ν on U, with domain PU, such that ν{U } = 0 for every U ∈ U . Let n ∈ N be such that νUn > 0. For each U ∈ Un choose x SU ∈ U . For Borel sets E ⊆SX set µE = ν{U : U ∈ Un , xU ∈ E}; then µ is a Borel measure on X and µ( Un ) = νUn > 0, while µ( V) = νV = 0 for every finite V ⊆ Un . Thus µ is not τ -additive and cannot be tight, and X is not a Radon space. 438I Proposition Let X be a metrizable space and hFξ iξ<κ a non-decreasing family of closed subsets of X, where κ is a measure-free cardinal. Then P S S µ( ξ<κ Fξ ) = ξ<κ µ(Fξ \ η<ξ Fη ) for every semi-finite Borel measure µ on X. S proof (a) I had better begin by remarking that Hξ = η<ξ Fη is an Fσ set for every ordinal ξ ≤ κ, by P 4A2Ld and 4A2Ka. So, setting Eξ = Fξ \ Hξ , ξ<κ µEξ is defined. P (b) I show by induction on ζ that µHζ = ξ<ζ µEξ for every ζ ≤ κ. The induction starts trivially with µH0 = 0. The inductive step to a successor ordinal ζ + 1 is also immediate, as Hζ+1 = Hζ ∪ Eζ . For the inductive step to a limit ordinal ζ of countable cofinality, let hζn in∈N be a non-decreasing sequence in ζ with supremum ζ; then P P µHζ = supn∈N µHζn = supn∈N ξ<ζn µEξ = ξ<ζ µEξ , as required. (c) So we P are left with the case in which ζ is a limit ordinal of uncountable cofinality. In this case, µ(E ∩ Hζ ) ≤ ξ<ζ µEξ whenever µE is finite. P P Let U be a σ-disjoint base for the topology of X (4A2L(hS ii)), and express U as n∈N Un where each Un is disjoint. For n ∈ N, ξ ≤ ζ set S Vnξ = {U : U ∈ Un , U ∩ Hξ 6= ∅}, Vnξ = Vnξ .

438K

Measure-free cardinals

245

Define φn : Vnζ → ζ by saying that φn (x) = min({ξ : ξ < ζ, x ∈ Vnξ }). Then, for any D ⊆ ζ, S S S φ−1 (Vnξ \ η<ξ Vnη ) n [D] = ξ∈D is a union of members of Un , so is open. We therefore have a measure νn on Pζ defined by saying that νn D = µ(E ∩ φ−1 point, recall that we are supposing that κ is measure-free, n [D]) for every D ⊆ ζ. At this P so #(ζ) is also measure-free (438Cb) and νn ζ = ξ<ζ νn {ξ} = supξ<ζ νn ξ. Interpreting this in X, we have µ(E ∩ Vnζ ) = supξ<ζ µ(E ∩ Vnξ ). This is true for every n ∈ N. So there is a countable set C ⊆ ζ such that µ(E ∩ Vnζ ) = supξ∈C µ(E ∩ Vnξ ) for every n ∈ N. Because cf(ζ) > ω, there is an α < ζ such that C ⊆ α, and µ(E ∩ Vnζ ) = µ(E ∩ Vnα ), that is, E ∩ Vnζ \ Vnα is negligible, for every n ∈ N. Now note that Fα is closed. So Hζ \ F α ⊆

[

{U : U ∈ U , Hζ ∩ U 6= ∅, U ∩ Fα = ∅} [ = Vnζ \ Vn,α+1 , n∈N

and E ∩ Hζ \ Fα is negligible. Accordingly, using the inductive hypothesis, P P µ(E ∩ Hζ ) ≤ µFα = µHα+1 ≤ ξ≤α µEξ ≤ ξ<ζ µEξ , as claimed. Q Q P Because µ is semi-finite, and E is arbitrary, µHζ ≤ ξ<ζ Eξ ; but the reverse inequality is trivial, so we have equality, and the induction proceeds in this case also. P (d) At the end of the induction we have µHκ = ξ<κ µEξ , as stated. 438J So far we have been looking at metrizable spaces, the obvious first step. But it turns out that the concept of ‘metacompactness’ leads to generalizations of some of the results above. Proposition (Moran 70, Haydon 74) Let X be a metacompact space with measure-free weight. (a) X is Borel-measure-compact. (b) If X is normal, it is measure-compact. (c) If X is perfectly normal (for instance, if it is metrizable), it is Borel-measure-complete. proof (a) ?? If X is not Borel-measure-compact, there are a non-zero totally finite Borel measure µ on X and a cover G of X by negligible open sets (434H(a-v)). Let H be a point-finite open cover of X refining S G. By 4A2Dc, #(H) ≤ max(ω, w(X)), so is measure-free, by 438C.SBecause µ is a Borel measure, H0 is measurable for every H0 ⊆ H; µH = 0 for every H ∈ H; while µ( H) = µX > 0. But this contradicts 438Ba. X X (b) Now suppose that X is normal, and that µ is a totally finite Baire measure on X. Because a normal metacompact space is countably paracompact (4A2F(g-iii)), µ has an extension to a Borel measure µ1 which is inner regular with respect to the closed sets, by Maˇr´ık’s theorem (435C). Now µ1 is τ -additive, by (a) above, so µ also is (411C). As µ is arbitrary, X is measure-compact. (c) Since on a perfectly normal space the Baire and Borel measures are the same, X is Borel-measurecomplete iff it is measure-compact, and we can use (b). Remark The arguments here can be adapted in various ways, and in particular the hypotheses can be weakened; see 438Yd-438Yf. 438K Hereditarily weakly θ-refinable spaces A topological space X is hereditarily weakly θ-refinable (also called hereditarily σ-relatively metacompact, hereditarily weakly submetacompact) if forSevery family G of open subsets of X there is a σ-isolated family A of subsets of X, refining G, such S that A = G.

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438L

438L Lemma (a) Any subspace of a hereditarily weakly θ-refinable topological space is hereditarily weakly θ-refinable. (b) A hereditarily metacompact space (e.g., any metrizable space, see 4A2Lb) is hereditarily weakly θrefinable. (c) A hereditarily Lindel¨of space is hereditarily weakly θ-refinable. (d) A topological space with a σ-isolated network is hereditarily weakly θ-refinable. proof (a) If X is hereditarily weakly θ-refinable, Y is a subspace of X, and H is a family of open subsets of S Y , set G = {G : G ⊆ X is open, G ∩ Y ∈ H}. Then there is a σ-isolated familySA, refining G, with union G; and {A ∩ Y : A ∈ A} is σ-isolated (4A2B(a-viii)), refines H, and has union H. As H is arbitrary, Y is hereditarily weakly θ-refinable. (b) If X is hereditarily metacompact, and G is any family of open sets in X with union W , then G is an open cover of the metacompact space W , so has aT point-finite open refinement H with the same union. For each x ∈ W , set Hx = {H : x ∈ H ∈ H}, Vx = Hx , so that Hx is a non-empty finite set and Vx is an open set containing x. For n ≥ 1, set An = {x : x ∈ W, #(Hx ) = n}; then for any distinct x, y ∈ An , either Hx = Hy and Vx = Vy , or #(Hx ∪ Hy ) > n and Vx ∩ Vy ∩ An = ∅. This means that An = {Vx ∩ An : x ∈ An } is a partition of ASn into relatively open sets, and is an isolated family. Also, S An is a refinement of H and therefore of G; so n≥1 An is a σ-isolated refinement of G, and its union is n≥1 An = W . As G is arbitrary, X is hereditarily weakly θ-refinable. (c) If X is hereditarily Lindel¨of and G is a family of open subsets of X, there is a countable G0 ⊆ G with the same union; now G0 , being countable, is σ-isolated. As G is arbitrary, X is hereditarily weakly θrefinable. (d) If X has a σ-isolated network A, and G is any family of open subsets of X, then E = {A : A ∈ A, A is included in some member of G} S is a σ-isolated family (4A2B(a-viii)), refining G, with union G. 438M Proposition (Gardner 75) If X is a hereditarily weakly θ-refinable topological space with measure-free weight, it is Borel-measure-complete. proof Let µ be a Borel probability measure on X, and G the S S family of µ-negligible open sets. Let A be a σ-isolated family refining S G with union G. Express A as n∈N An where each An is an isolated family; for each n ∈ N, set Xn = An and let µn be the subspace measure on Xn . Then An is a disjoint family of relatively open µn -negligible sets; as #(An ) ≤ w(Xn ) ≤ w(X) (4A2D) is measure-free, and µn is a totally finite Borel measure on Xn , S µ∗ Xn = µn Xn = µn ( An ) = 0, S S by 438Bb. Now µ( G) = µ∗ ( n∈N Xn ) = 0. As µ is arbitrary, X is Borel-measure-complete (434I(a-iv)). 438N For the next few paragraphs, I will use the following notation. Let X be a topological space S and G a family of subsets of X. Then J (G) will be the family of subsets of X expressible as A for some S σ-isolated family A refining G. Observe that X is hereditarily weakly θ-refinable iff G belongs to J (G) for every family G of open subsets of X. (a) J (G) is always a σ-ideal of subsets of X. P P If A is a σ-isolated family of subsets of X, refining G, and B is any set, then {B ∩ A : A ∈ A} is still σ-isolated and still refines G; so any subset of a member of S J (G) belongs to J (G). If hAn in∈N is a sequence of σ-isolated families refining G, then n∈N An is σ-isolated and refines G; so the union of any sequence in J (G) belongs to J (G). Q Q (b) If H refines G, then J (H) ⊆ J (G). P P All we need to remember is that any family refining H also refines G. Q Q (c) If X and Y are topological spaces, A ⊆ X, f : A → Y is continuous, and H is a family of subsets of Y , set G = {f −1 [H] : H ∈ H}. Then J (G) ⊇ {f −1 [B] : B ∈ J (H)}. P P If B ∈ J (H), there is a

438P

Measure-free cardinals

247

σ-isolated family D of subsets of Y , refining H, S and with union B. Now A = {f −1 [D] : D ∈ D} refines −1 G and has union f [B]. We can express D as n∈N Dn , where each Dn is a relatively open family; set S An = {f −1 [D] : D ∈ Dn }, so that A = n∈N An . For each n, An is disjoint, because Dn is. Moreover, S S if D ∈ DS Dn for some open set H ⊆ Y , so that f −1 [D] = f −1 [H] ∩ An is relatively n , then D = H ∩ open in An ; this shows that An is an isolated family. Accordingly A is σ-isolated and witnesses that f −1 [B] ∈ J (G). As B is arbitrary, we have the result. Q Q (d)SIf X is a topological space, G is a family of subsets of X, and hDi ii∈I is an isolated family in J (G), then i∈I D P For each iS∈ I, let hAni in∈N be a sequence of isolated families, Si ∈ JS(G). P S Sall refining S G, such that Di = n∈N Ain . Set An = i∈I Ain for each n. Then A refines G, and A = n n n∈N i∈I Di . It S S is easy to check that every An is isolated, so that n∈N An witnesses that i∈I Di belongs to J (G). Q Q 438O Lemma Give R the topology S generated by the closed intervals ]−∞, t] for t ∈ R, and let r ≥ 1. Then R r , with the product topology corresponding to S, is hereditarily weakly θ-refinable. proof Induce on r. Write ≤ for the usual partial order of R r , and ]−∞, x] for {y : y ≤ x}; set VA = S r r r x∈A ]−∞, x] for A ⊆ R . The sets ]−∞, x], as x runs over R , form a base for the topology of R . The induction starts easily because S itself is hereditarily Lindel¨of. P P If G ⊆ S, set A = {x : x ∈ R, there is some G ∈ G such that ]−∞, x] ⊆ G}. Then A has a countable cofinal set D, so that there is a corresponding countable subset of G with the same union as G. Q Q By 438Lc, S is hereditarily weakly θ-refinable. For the inductive step to r + 1, where r ≥ 1, let G be a family of open subsets of R r+1 , and set A = {x : x ∈ R r , there is some G ∈ G such that ]−∞, x] ⊆ G}. For each k ≤ r, q ∈ Q set Kk = (r + 1) \ {k} and let Bkq ⊆ R Kk be the set {z : z a q ∈ VA }, writing z a q for that member x of R r+1 such that x¹Kk = z and x(k) = q. (I am thinking of members of R r+1 as functions from r + 1 = {0, . . . , r} to R.) Set Akq = {x : x ∈ VA , x(k) = q}, Gkq = {]−∞, x] : x ∈ Akq }. Then, in the notation of 438N, VAkq ∈ J (G). P P Set f (x) = x¹Kk for each x ∈ VAkq , so that f : VAkq → R Kk is continuous. For x ∈ Akq , ]−∞, x] = f −1 [ ]−∞, f (x)] ]. Now R KkSis hereditarily weakly θ-refinable, by the inductive hypothesis, so if we set Hkq = {]−∞, f (x)] : x ∈ Akq }, Hkq ∈ J (Hkq ) and (by 438Nc) S VAkq = f −1 [ Hkq ] ∈ J (Gkq ) ⊆ J (G). Q Q S Accordingly W ∈ J (G), where W = k≤r,q∈Q VAkq , by 438Na. Now consider VA \ W . If x, x0 ∈ VA \ W and x ≤ x0 then x = x0 . P P?? Otherwise, there are a k ≤ n and a q ∈ Q such that x(k) ≤ q ≤ x0 (k). In this case, setting y¹Kk = x¹Kk and y(k) = q, we have y ∈ Akq and x ∈ VAkq . X XQ Q But this means that the subspace topology of VA \ W is discrete, so Sthat {{x} : x ∈ VA \ W } is an isolated family covering VA \ W and refining G; thus VA \ W ∈ J (G) and G = VA belongs to J (G). As G is arbitrary, R r+1 is hereditarily weakly θ-refinable and the induction proceeds. 438P Proposition Let X ⊆ R R be the family of functions x : R → R such that lims↑t x(s) and lims↓t x(s) are defined in R for every t ∈ R. Then X, with its topology of pointwise convergence inherited from the product topology of R R , is a K-analytic hereditarily weakly θ-refinable space. proof (a) I start by showing that it is K-analytic. (i) The first step is to find an alternative characterization of X. If x ∈ X and U , V ⊆ R are disjoint, then I say that a finite set D ⊆ R is a back-and-forth set for x, U and V if whenever s, t are successive points of D (in the usual ordering of R) then either x(s) ∈ U and x(t) ∈ V , or x(s) ∈ V and x(t) ∈ U . Now, if x ∈ X and k ∈ N and U , V ⊆ R have disjoint closures, ωkU V (x) = sup{#(D) : D ⊆ [−k, k] is a back-and-forth set for x, U and V } is finite. P P For each t ∈ [−k, k] there is a δ > 0 such that {x(s) : t − δ < s < t} meets at most one of U and V , and {x(s) : t < s < t + δ} meets at most one of U and V . So if D is a back-and-forth set for x, U and V , D can meet each of the intervals ]t − δ, t[ and ]t, t + δ[ in at most one point, and #(D ∩ ]t − δ, t + δ[) ≤ 3.

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438P

By the Heine-Borel theorem, there is a cover of [−k, k] by m intervals of this type for some m ∈ N, so that ωkU V (x) ≤ 3m. Q Q (ii) If x ∈ X and htn in∈N is any bounded sequence in R, then it has a monotonic subsequence ht0n in∈N and hx(t0n )in∈N must be convergent in R. It follows that {x(t) : −k ≤ t ≤ k} is bounded in R for every k ∈ N. (iii) Conversely, let I be the family of non-empty open intervals in R with rational endpoints, and suppose that x ∈ R R is such that whenever U , V ∈ I have disjoint closures, and k ∈ N, then {x(t) : |t| ≤ k} is bounded and ωkU V (x) is finite. Then x ∈ X. P P Let t ∈ R. ?? If lims↓t x(s) is not defined in R, then (because {x(s) : t ≤ s ≤ t + 1} is bounded) there must be decreasing sequences htn in∈N , ht0n in∈N , both convergent to t, such that z = limn→∞ x(tn ), z 0 = limn→∞ x(t0n ) are both defined and different. Let U , V ∈ I be such that z ∈ U , z 0 ∈ V and U ∩ V = ∅; then for any m ∈ N there is a back-and-forth set D ⊆ {tn : n ∈ N} ∪ {t0n : n ∈ N} for x, U and V with #(D) ≥ m, so ωkU V (x) is infinite for some k. X X Thus lims↓t x(s) is defined in R. Similarly, lims↑t x(s) is defined in R; as t is arbitrary, x ∈ X. Q Q (iv) Now this means that X is a Souslin-F set in the compact space [−∞, ∞]R . P P For k, m ∈ N and disjoint U , V ∈ I, set Fkm = {x : x ∈ [−∞, ∞]R , |x(t)| ≤ m for every t ∈ [−k, k]}, 0 R FkU V m = {x : x ∈ [−∞, ∞] , ωkU V (x) ≤ m}. 0 0 Of course every Fkm is closed. But so is every FkU / FkU V m , because if x ∈ V m and D ⊆ [−k, k] is a back0 and-forth set for x, U and V of size greater than m, then D will witness that y ∈ / FkU V m for any y close enough to x. Now the characterization of X above tells us just that T T S S 0 X = k∈N m∈N Fkm ∩ k∈N,U,V ∈I,U ∩V =∅ m∈N FkU V m,

which is a countable intersection of Fσ sets, and is Souslin-F. Q Q So X is K-analytic (422Hb). (b) Now I turn to the proof that X is hereditarily weakly θ-refinable. Let G be a family of open sets in X. S (i) In the notation of S 438N, I seek to show that G belongs to J (G). Of course it will be enough to consider the case S in which G is non-empty. The following elementary remarks will be useful. (α) If D ⊆ G, and E is a partition of D into relatively open sets such that E refines G, then D ∈ J (G). (β) If hDi ii∈I S is any family in J (G), and hHi ii∈I is a family of open sets, and D ⊆ {x : #({i : x ∈ Hi }) = P hD ∩ Hi ∩ Di ii∈I is an isolated family in J (G); use 438Nd. 1}, then D ∩ i∈I Hi ∩ Di belongs to J (G). P Q Q (ii) Write Q for the family of all finite sequences q = (I0 , U0 , V0 , W0 , I1 , U1 , V1 , W1 , . . . , In , Un , Vn , Wn ) where I0 , I1 , . . . , In are disjoint members of I, all the Ui , Vi , Wi belong to I, and, for each i ≤ n, any pair of Ui , Vi , Wi are either equal or disjoint. Fix q = (I0 , . . . , Wn ) ∈ Q for the moment. Q (iii) Set Tq = i≤n Ii . For τ ∈ Tq , write Fq τ for the set of those x ∈ X such that, for every i ≤ n, S x(s) ∈ Ui for s ∈ Ii ∩ ]−∞, τ (i)[, x(τ (i)) ∈ Vi and x(s) ∈ Wi for s ∈ Ii ∩ ]τ (i), ∞[. Set Xq = {Fq τ : τ ∈ Tq }, and for τ ∈ Tq set Hq τ = {x : x ∈ Xq , x(τ (i)) ∈ Vi for every i ≤ n}. Finally, set Sq = {τ : τ ∈ Tq and Hq τ is included in some member of G}. (iv) If T ⊆ Sq then H =

S τ ∈T

Hq τ belongs to J (G). P P Induce on #(L(T )), where

L(T ) = {i : i ≤ n, there are τ , τ 0 ∈ T such that τ (i) 6= τ 0 (i)}. If L(T ) = ∅, then #(T ) ≤ 1, so H is either empty or included in some member of G, and the induction starts. For the inductive step to #(L(T )) = k ≥ 1, consider three cases.

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case 1 Suppose there is a j ∈ L(T ) such that Uj = Vj = Wj . Then x(t) ∈ Vj for every x ∈ Xq , t ∈ Ij . Fix any t∗ ∈ Ij and for τ ∈ Tq define τ ∗ ∈ Tq by setting τ ∗ (j) = t∗ , τ ∗ (i) = τ (i) for i 6= j; then Hq τ ∗ = Hq τ . Set T ∗ = {τ ∗ : τ ∈ST }; then L(T ∗ ) = L(T ) \ {j}, so #(L(T ∗ )) < #(L(T )), while T ∗ ⊆ Sq . By the inductive hypothesis, H = τ ∈T ∗ Hq τ is negligible. case 2 Suppose there is a j ∈ L(T ) such that Uj 6= Vj and Vj 6= Wj . Then Vj ∩ (Uj ∪ Wj ) = ∅. For s ∈ Ij set Ts∗ = {τ : S τ ∈ T, τ (j) = s}. Then #(L(Ts∗ )) < #(L(T )) so, by the inductive hypothesis, ∗ ∗ Hs ∈ J (G), where Hs = τ ∈T ∗ Hq τ . But, for τ ∈ T and x ∈ Hq τ , x(s) ∈ Vj iff s = τ (j); so Hs∗ = {x : x ∈ s H, x(s) ∈ Vj } and hHs∗ is∈Ij is a partition of H into relatively open sets. By (i-β), H ∈ J (G). case 3 Otherwise, L(T ) = J ∪ J 0 where J = {i : i ∈ L(T ), Ui = Vi } and J 0 = {i : i ∈ L(T ), Vi = Wi } are disjoint. For x ∈ H, i ∈ J we see that there is a largest t ∈ Ii such that x(t) ∈ Vi ; set φi (x) = −t. Similarly, for x ∈ H, i ∈ J 0 there is a smallest t ∈ Ii such that x(t) ∈ Vi ; in this case, set φi (x) = t. Observe that, for i ∈ J, s ∈ R, {x : x ∈ H, φi (x) ≤ s} = ∅ if s < −t for every t ∈ Ii , = H if − t < s for every t ∈ Ii , = {x : x ∈ H, x(−s) ∈ Vi } if − s ∈ Ii , so is always relatively open in H, and φi is continuous if R is given the topology S of Lemma 438O. Similarly, for i ∈ J 0 , s ∈ R, {x : x ∈ H, φi (x) ≤ s} = ∅ if s < t for every t ∈ Ii , = H if t < s for every t ∈ Ii , = {x : x ∈ H, x(s) ∈ Vi } if s ∈ Ii . So in this case also φi is continuous. Accordingly, giving R L(T ) the product topology corresponding to S, we have a continuous map φ : H → L(T ) R defined by setting φ(x) = hφi (x)ii∈L(T ) for x ∈ H. For τ ∈ T , set τ˜(i) = −τ (i) if i ∈ J, τ (i) if i ∈ J 0 , ˜ and Hτ = ]−∞, τ˜] ⊆ R L(T ) . Then Hq τ = {x : x ∈ H, x(τ (i)) ∈ Vi for every i ≤ n} = {x : x ∈ H, x(τ (i)) ∈ Vi for every i ∈ L(T )} (because if x ∈ H, i ≤ n, i ∈ / L(T ) then there is some τ 0 ∈ T such that x ∈ Hq τ 0 , so that x(τ 0 (i)) ∈ Vi and therefore x(τ (i)) ∈ Vi ) ˜ τ ]. = {x : x ∈ H, φi (x) ≤ τ˜(i) for every i ∈ L(T )} = φ−1 [H ˜ τ : τ ∈ T }. Because R L(T ) is hereditarily weakly θ-refinable (438O), and G˜ is a family of open Set G˜ = {H S˜ S −1 L(T ) ˜ By 438Nc, H = S [ G] belongs to J ({Hq τ : τ ∈ T } and subsets of R , G˜ =∈ J (G). τ ∈T Hq τ = φ therefore to J (G), by 438Nb. Thus in all three cases the induction proceeds. Q Q S (v) This means that, S for any q ∈SQ, Yq = {Hq τ : τ ∈ Sq } belongs to J (G). Since Q is countable, Y ∈ J (G), where Y = q ∈Q Yq . But G ⊆ Y . P P If x ∈ G ∈ G, there are t0 < . . . < tn and Vi0 ∈ I, for i ≤ n, such that x ∈ {y : y ∈ X, y(ti ) ∈ Vi0 for every i ≤ n} ⊆ G. Set zi = x(ti ), zi− = lims↑ti x(s), zi+ = lims↓ti x(s) for i ≤ n; let Ui , Vi , Wi ∈ I be such that zi− ∈ Ui , zi ∈ Vi ⊆ Vi0 , zi+ ∈ Wi and any pair of Ui , Vi , Wi are either equal or disjoint; and let I0 , . . . , In ∈ I be disjoint and such that ti ∈ Ii , x(s) ∈ Ui for s ∈ Ii ∩ ]−∞, ti [ and x(s) ∈ Wi for s ∈ Ii ∩ ]ti , ∞[ for each i ≤ n. Then, setting q = (I0 , . . . , Wn ) and τ (i) = ti for i ≤ n, x ∈ Xq τ ⊆ Hq τ ⊆ G, so that τ ∈ Sq and x ∈ Yq ⊆ Y . Q Q As G is arbitrary, X is hereditarily weakly θ-refinable.

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438Q Corollary (a) Let I k be the split interval (419L). Then any countable power of I k is a K-analytic hereditarily weakly θ-refinable space. (b) Let Y be the ‘Helly space’, the space of non-decreasing functions from [0, 1] to itself with the topology of pointwise convergence inherited from the product topology on [0, 1][0,1] (Kelley 55, Ex. 5M). Then Y is a K-analytic hereditarily weakly θ-refinable space. proof These are both (homeomorphic to) subspaces of the space X of Proposition 438P. To see this, argue as follows. For (a), observe that we have a function f : I k → X defined by setting f (t− )(s) = f (t+ )(s) = 1 if s < t, f (t− )(s) = f (t+ )(s) = 0 if s > t, and f (t− )(t) = 0, f (t+ )(t) = 1, and that f is a homeomorphism between I k and its image. Next, for any L ⊆ N, we can define g : (I k )L → X by setting g(t)(s) = f (tn )(s − 2n) if n ∈ L and 2n ≤ s ≤ 2n + 1 [ [2n, 2n + 1] = 0 if s ∈ R \ n∈L k L

for t = htn in∈N ∈ (I ) ; it is easy to check that g is a homeomorphism between (I k )L and its image in X. As for (b), if we take g(y) to be the extension of the function y : [0, 1] → [0, 1] to the function which is constant on each of the intervals ]−∞, 0] and [1, ∞[, then g : Y → X is a homeomorphism between Y and its image g[Y ]. Since both (I k )N and Y are compact, they are homeomorphic to closed subsets of X, and are K-analytic (422Gc) and hereditarily weakly θ-refinable (438La). 438R Proposition Assume that c is measure-free. Then the space X of 438P, (I k )N and the Helly space (438Qb) are all Radon spaces. proof By 438P and 438Q, they are K-analytic and hereditarily weakly θ-refinable, therefore pre-Radon and Borel-measure-complete (434Jf, 438M), therefore Radon (434Ka). 438S In 434R I described a construction of product measures. In accordance with my general practice of examining the measure algebra of any new measure, I give the following result. Proposition Let X and Y be topological spaces with σ-finite Borel measures µ, ν respectively. Suppose that either X is first-countable or ν is τ -additive and effectively locally finite. Write λ for the Borel measure on X × Y defined by the formula λW =

R

νW [{x}]µ(dx) for every Borel set W ⊆ X × Y

as in 434R(ii). If either the weight of X or the Maharam type of ν is a measure-free cardinal, then for every b Borel set W ⊆ X × Y there is a set W 0 ∈ B(X)⊗B(Y ) such that λ(W 4W 0 ) = 0; consequently, the measure algebra of λ can be identified with the localizable measure algebra free product of the measure algebras of µ and ν. proof (a) Write (B, ν¯) for the measure algebra of ν. With its measure-algebra topology, this is metrizable (323Gb). Let hYn in∈N be a non-decreasing sequence of Borel sets of finite measure in Y with union Y . (b) For the moment (down to the end of (e) below) fix on an open set W ⊆ X × Y . For x ∈ X, set f (x) = W [{x}]• in B. Then f : X → B is Borel measurable. P P (i) Let H ⊆ B be an open set. For k, n ∈ N set Enk = {x : x ∈ X, 2−n k ≤ ν(Yn ∩ W [{x}]) < 2−n (k + 1)}. Just as in part (a) of the proof of 434R, the function x 7→ ν(Yn ∩ W [{x}]) is lower semi-continuous, so Enk is a Borel set. Set S Gnk = {G : G ⊆ X is open, G ∩ Enk ⊆ f −1 [H]}; S then E = n,k∈N (Gnk ∩ Enk ) is a Borel set included in f −1 [H]. (ii) The point is that E = f −1 [H]. To see this, take any x such that f (x) ∈ H. Then there are b ∈ B, ² > 0 such that ν¯b < ∞ and c ∈ H whenever c ∈ B and ν¯(b ∩ (c 4 f (x))) ≤ 5². Since b = supn∈N b ∩ Yn• ,

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there is an n ∈ N such that ν¯(b \ Yn• ) ≤ ² and 2−n ≤ ². In this case c ∈ H whenever c ∈ B and ν¯(Yn• ∩ (c 4 f (x))) ≤ 4²; thus {x0 : ν(Yn ∩ (W [{x0 }]4W [{x}])) ≤ 4²} ⊆ f −1 [H]. Let k ∈ N be such that 2−n k ≤ ν(Yn ∩ W [{x}]) < 2−n (k + 1), that is, x ∈ Enk . Again using the ideas of part (a) of the proof of 434R, there are an open set G containing x and an open set V ⊆ Y such that G × V ⊆ W and ν(Yn ∩ V ) ≥ 2−n (k − 1). Now if x0 ∈ G ∩ Enk , V ⊆ W [{x0 }] ∩ W [{x}], so ν(Yn ∩ (W [{x0 }]4W [{x}])) ≤ ν(Yn ∩ W [{x0 }]) + ν(Yn ∩ W [{x}]) − 2ν(Yn ∩ V ) ≤ 2−n ((k + 1) + (k + 1) − 2(k − 1)) = 4 · 2−n ≤ 4². But this means that G ∩ Enk ⊆ f −1 [H], so G ⊆ Gnk and x ∈ G ∩ Enk ⊆ E. As x is arbitrary, f −1 [H] ⊆ E and E = f −1 [H]. (iii) Thus f −1 [H] is a Borel set. As H is arbitrary, f is Borel measurable. Q Q (c) We need to know also that if H is a disjoint family of open subsets of B all meeting f [X], then #(H) ≤ max(ω, min(w(X), τ (B))) P P Repeat the ideas of (b) above, setting S (H) Gnk = {G : G ⊆ X is open, G ∩ Enk ⊆ f −1 [H]} S (H) (H) for H ∈ H and k, n ∈ N, so that f −1 [H] = n,k∈N Gnk ∩Enk . For fixed n and k the family hGnk ∩Enk iH∈H is disjoint, so can have at most w(Enk ) ≤ w(X) non-empty members (4A2D). But this means that S (H) H = n,k∈N {H : Gnk ∩ Enk 6= ∅} has cardinal at most max(ω, w(X)). On the other hand, there is a set B ⊆ B, of cardinal τ (B), which τ -generates B. The algebra B0 generated by B has cardinal at most max(ω, #(B)) (331Gc), and B0 is topologically dense in B (323J), so every member of H meets B0 , and #(H) ≤ #(B0 ) ≤ max(ω, τ (B)). Putting these together, we have the result. Q Q In particular, under the hypotheses above, #(H) is measure-free whenever H is a disjoint family of open subsets of B all meeting f [X]. (d) The next step is to observe that there is a conegligible BorelSset Z ⊆ X such that f [Z] is separable. P P Let H be a σ-disjoint base for the topology of B; express it as n∈N Hn where each Hn is disjoint. Let hXm im∈N be a cover of X by Borel sets of finite measure. For n ∈ N consider Hn0 = {H : H ∈ Hn , H ∩f [X] 6= ∅}. For m ∈ N, we have a totally finite measure νnm with domain PHn0 defined by saying S νnm E = µ(Xm ∩ f −1 ( E)) for every E ⊆ Hn0 . Since Hn0 has measure-free cardinal, by (c), there must be a countable set Enm ⊆ Hn0 such that νnm (Hn0 \ Enm ) = 0. Set S S Z = X \ m,n∈N (Xm ∩ f −1 [ (Hn0 \ Enm )]); then Z is conegligible. If x ∈ Z and f (x) ∈ H ∈ Hn , then there is some m ∈ N such that x ∈ Xm , while H ∈ Hn0 , so H must belong to Enm . ButS this means that {f [Z] ∩ H : H ∈ H}, which is a base for the topology of f [Z], is just {f [Z] ∩ H : H ∈ m,n∈N Emn }, and is countable. So f [Z] is separable (4A2Oc), as required. Q Q b (e) 418T(a-ii) now tells us that there is a set W 0 ∈ B(X)⊗B(Y ) such that f (x) = W 0 [{x}]• for every 0 x ∈ Z, so that ν(W [{x}]4W [{x}]) = 0 for almost every x, that is, λ(W 4W 0 ) = 0. And this is true for every open set W ⊆ X × Y . b (f ) Now let W be the family of those Borel sets W ⊆ X × Y for which there is a W 0 ∈ B(X)⊗B(Y ) 0 such that λ(W 4W ) = 0. This is a σ-algebra containing every open set, so is the whole Borel σ-algebra, as required.

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b Since the c.l.d. product measure λ0 on X × Y is just the completion of its restriction to B(X)⊗B(Y ) b b (251K), and λ0 and λ agree on B(X)⊗B(Y ) (by Fubini’s theorem), the embedding B(X)⊗B(Y ) ⊆ B(X × Y ) induces an isomorphism between the measure algebras of λ and λ0 . As remarked in 325Eb, because µ and ν are strictly localizable, the latter may be identified with ‘the’ localizable measure algebra free product of the measure algebras of µ and ν. Remark The hypothesis on the weight of X can be slightly weakened; see 438Yg. To see that some restriction on (X, µ) and (Y, ν) is necessary, see 439L. 438X Basic exercises (a) Show that a cardinal κ is measure-free iff Mσ = Mτ , where Mσ , Mτ are the spaces of countably additive and completely additive functionals on the algebra Pκ, as in §362. (b) Let (X, Σ, µ) be a localizable measure space with magnitude (definition: 332Ga) which is either finite or a measure-free cardinal. Show that any absolutely continuous countably additive functional ν : Σ → R is truly continuous. (Hint: 363S.) (c) Let A be a Dedekind complete Boolean algebra with measure-free cellularity. Show that any countably additive functional ν : A → R is completely additive. (d) Let U be a Dedekind complete Riesz space such that any disjoint order-bounded family in U + has measure-free cardinal. Show that Uc∼ = U × . > (e) Show that if κ is a measure-free cardinal, so is ωκ . (Hint: show by induction on ordinals ξ that if #(ξ) is measure-free, then so is ωξ .) > (f ) Let (X, Σ, µ) be a complete locally determined measure space, (Y, ρ) a complete metric space, and hfn in∈N a sequence of measurable functions from X to Y . Suppose that w(Y ) is measure-free. Show that {x : limn→∞ fn (x) is defined in Y } is measurable. (Cf. 418C.) (g) Let (X, Σ, µ) and (Y, T, ν) be σ-finite measure spaces with c.l.d. product (X ×Y, Λ, λ). Give L0 (ν) the topology of convergence in measure. Suppose that f : X → L0 (ν) is measurable and there is a conegligible set X0 ⊆ X such that w(f [X0 ]) is measure-free. Show that there is an h ∈ L0 (λ) such that f (x) = h•x for every x ∈ X. (Cf. 418S.) (h) Let (Y, T, ν) be a σ-finite measure space, and (B, ν¯) its measure algebra, with its usual topology; assume that the Maharam type of B is measure-free. Let (X, Σ, µ) be a σ-finite measure space and Λ the domain of the c.l.d. product measure λ on X × Y . Show that if f : X → B is measurable, then there is a W ∈ Λ such that f (x) = Wx• for every x ∈ X. (Cf. 418T(b-ii).) (i) Let (X, Σ, µ) is a complete locally determined measure space and V a normed space such that w(V ) is measure-free. (i) Show that the space L of measurable functions from X to V is a linear space, setting (f + g)(x) = f (x) + g(x), etc. (ii) Show that if V is a Riesz space with a Riesz norm then L is a Riesz space under the natural operations. > (j) Let X be a topological space and G a point-finite open cover of X such that #(G) is measure-free. Suppose that E ⊆ X is such that E ∩G is universally measurable for every G ∈ G. Show that E is universally measurable. (Compare 434Xe(iv).) > (k) Show that for a metrizable space X, the following are equiveridical: (i) X is Borel-measure-compact; (ii) X is Borel-measure-complete; (iii) X is measure-compact; (iv) w(X) is measure-free. (l) Let Y be the Helly space. (i) Show that Y is a compact convex subset of R [0,1] with its usual topology. (ii) Show that there is a natural one-to-one correspondence between the split interval I k and the set of extreme points of Y , matching t− ∈ I k with the function χ [0, t[ and t+ with χ[0, t]. (iii) Let P be the set of Radon probability measures on I k with its narrow topology (437J). Show that there is a natural homeomorphism φ : P → Y defined by setting φ(µ)(t) = µ[0− , t− ] for µ ∈ P , t ∈ [0, 1].

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(m) Give R the right-facing Sorgenfrey topology (415Xc). Show that any countable power of R is hereditarily weakly θ-refinable. > (n) Let I k be the split interval. Show that I k × I k is a Radon space iff c is measure-free. (Hint: {(α+ , (1 − α)+ ) : α ∈ [0, 1]} is a discrete Borel subset of cardinal c.) (o) Give R the right-facing Sorgenfrey topology. Show that the following are equiveridical: (i) c is measure-free; (ii) R N , with the corresponding product topology, is Borel-measure-complete; (iii) R 2 , with the product topology, is Borel-measure-compact. (Compare 439Q.) (p) Suppose that c is measure-free. Let X ⊆ R R be the set of functions of bounded variation on R, with the topology of pointwise convergence inherited from the product topology of R R . Show that X is a Radon space. (Hint: X is an Fσ subset of the space of 438P.) 438Y Further exercises (a) Let S (X, Σ, µ) be a probability space and hEi ii∈I a point-finite family of measurable sets such that νJ = µ( i∈J Ei ) is defined for every J ⊆ I. Show directly that ν is a uniformly exhaustive Maharam submeasure on PI (definition: 392B), and use the Kalton-Roberts theorem (392J) to prove 438Ba. (b) Suppose that c is measure-free, but that κ > c is not measure-free. Show that there is a non-principal ω1 -complete ultrafilter on κ. (Hint: part (b) of the proof of 451P.) (c) Show that if X is a metrizable space and either c or w(X) is measure-free, then every σ-finite Borel measure on X has countable Maharam type. (d) Let X be a metacompact T1 space. Show that X is Borel-measure-compact iff every closed discrete subspace has measure-free cardinal. (e) Let X be a topological space such that every subspace of X is metacompact and has measure-free cellularity. Show that X is Borel-measure-complete. (f ) Let X be a normal metacompact Hausdorff space. Show that it is measure-compact iff every closed discrete subspace has measure-free cardinal. (g) In 438S, show that it would be enough to suppose that every discrete subset of X has cardinal of measure 0. ˜ ⊆ Z R be the set of functions x : R → Z such that lims↑t x(s), (h) Let Z be any Polish space, and let X ˜ with the topology of pointwise convergence lims↓t x(s) are defined in Z for every t ∈ R. Show that X, R inherited from Z , is a K-analytic hereditarily weakly θ-refinable space. (i) Suppose that c is measure-free. Let D be any subset of R and X ⊆ R D the set of functions of bounded variation on D, with the topology of pointwise convergence inherited from the product topology of R D . Show that X is a Radon space. (j) Let X be a totally ordered set with its order topology. Show that if c(X) is measure-free then every σ-finite Borel measure on X has countable Maharam type. (Cf. 434Yl.) 438 Notes and comments Since the axiom ‘every cardinal is measure-free’ is admissible – that is, will not lead to a paradox unless one is already latent in the Zermelo-Fraenkel axioms for set theory – it is tempting, in the context of this section, to assume it; so that ‘every complete metric space is Radon’ becomes a theorem, along with ‘every measurable function from a quasi-Radon measure space to a metrizable space is almost continuous’ (438G), ‘Uc∼ = U × for every Dedekind complete Riesz space U ’ (438Xd), ‘metacompact spaces are Borel-measure-compact’ (438J), ‘the sum of two measurable functions from a complete probability space to a normed space is measurable’ (438Xi) and ‘the Helly space is Radon’ (438R). Undoubtedly the consequent mathematical universe is tidier. In my view, the tidiness is the tidiness of poverty. Apart from

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438 Notes

anything else, it leads us to neglect such questions as ‘is every measurable function from a Radon measure space to a metrizable space almost continuous?’, which have answers in ZFC (451S). From the point of view of measure theory, the really interesting question is whether c is measure-free. It is not quite clear from the results above why this should be so; 438R is a very small part of the story. There is a larger hint in 438Ce-438Cf: if c is measure-free, but κ > c is not measure-free, then the witnessing measures will be purely atomic. I mean to return to this point in Volume 5. For a general exploration of universes in which c is not measure-free, see Fremlin 93. For fragments of what happens if we suppose that we have an atom for a measure which witnesses that κ is not measure-free, see 438Yb and the notes on normal filters in 4A1I-4A1L. There are many further applications of 438P besides those in 438Q and 438Xm-438Xp. But the most obvious candidate, the space C(R) of continuous real-valued functions on R, although indeed it is a Borel subset of the potentially Radon space of 438P, is in fact Radon whether or not c is measure-free (454Pa). As soon as we start using any such special axiom as ‘c = ω1 ’ or ‘c is measure-free’, we must make a determined effort to check, through such examples as 438Xn, that our new theorems do indeed depend on something more than ZFC.

439 Examples As in Chapter 41, I end this chapter with a number of examples, exhibiting some of the boundaries around the results in the rest of the chapter, and filling in a gap with basic facts about Lebesgue measure (439E). The first three examples (439A) are measures defined on σ-subalgebras of the Borel σ-algebra of [0, 1] which have no extensions to the whole Borel algebra. The next part of the section (439B-439G) deals with ‘universally negligible’ sets; I use properties of these to show that Hausdorff measures are generally not semi-finite (439H), closing some unfinished business from §§264, and that smooth linear functionals may fail to be representable by integrals in the absence of Stone’s condition (439I). In 439J-439R I set out some examples relevant to §§434-435, filling out the classification schemes of 434A and 435A, with spaces which just miss being Radon (439K) or measure-compact (439N, 439P, 439Q). In 439S I present the canonical example of a non-Prokhorov topological space, answering an obvious question from §437. 439A Example Let B be the Borel σ-algebra of [0, 1]. There is a probability measure ν defined on a σ-subalgebra T of B which has no extension to a measure on B. first construction Let A ⊆ [0, 1] be an analytic set which is not Borel (423L). Let I be the family of sets of the form E ∪ F where E, F are Borel sets, E ⊆ A and F ⊆ [0, 1] \ A. Then I is a σ-ideal of B not containing [0, 1]. Set T = I ∪ {[0, 1] \ H : H ∈ I}, and define ν : T → {0, 1} by setting νH = 0, ν([0, 1] \ H) = 1 for every H ∈ I; then ν is a probability measure (cf. countable-cocountable measures (211R) or Dieudonn´e’s measure (411Q)). ?? If µ : B → [0, 1] is a measure extending ν, then its completion µ ˆ measures A (432A). Also µ ˆ is a Radon measure (433Cb). Now every compact subset of A belongs to I, so µ ˆA = sup{ˆ µK : K ⊆ A is compact} = sup{νK : K ⊆ A is compact} = 0. Similarly µ ˆ([0, 1] \ A) = 0, which is absurd. X X second construction This time, let I be the family of meager Borel sets in [0, 1]. As before, let T be I ∪ {[0, 1] \ E : E ∈ I}, and set νE = 0, ν([0, 1] \ E) = 1 for E ∈ I. ?? If µ is a Borel measure extending ν, then µ([0, 1] \ Q) = 1, and µ is tight (that is, inner regular with respect to the compact sets), so there is a closed subset F of [0, 1] \ Q such that µF > 0. But F is nowhere dense, so νF = 0. X X third construction1 There is a function f : [0, 1] → {0, 1}c which is (B, Ba)-measurable, where Ba is the Baire σ-algebra of {0, 1}c , and such that f [ [0, 1] ] meets every non-empty member of Ba. P P Set X = C([0, 1])N with the product of the norm topologies, so that X is an uncountable Polish space (4A2Pe, 4A2Qc), and ([0, 1], B) is isomorphic to (X, B(X)), where B(X) is the Borel σ-algebra of X (424Da). Define g : X → {0, 1}[0,1] by saying that g(hui ii∈N )(t) = 1 iff limi→∞ ui (t) = 1. For each t ∈ [0, 1], 1I

am grateful to M.Laczkovich and D.Preiss for showing this to me.

439C

Examples

{hui ii∈N : lim ui (t) = 1} = i→∞

\ [

255

{hui ii∈N : |ui (t) − 1| ≤ 2−m for every i ≥ n}

m∈N n∈N

is a Borel subset of X, so g is (B(X), Ba({0, 1}[0,1] ))-measurable, where Ba({0, 1}[0,1] ) is the Baire σ-algebra of {0, 1}[0,1] (4A3Ne). If E ∈ Ba({0, 1}[0,1] ) is non-empty, there is a countable set I ⊆ [0, 1] such that E is determined by coordinates in I (4A3Nb), so that E ⊇ {w : w¹I = z} for some z ∈ {0, 1}I . Now we can find a sequence hui ii∈N in C([0, 1]) such that limi→∞ ui (t) = z(t) for every t ∈ I (if I ⊆ {tj : j ∈ N}, take ui such that |ui (tj ) − z(tj )| ≤ 2−i whenever j ≤ i), and in this case g(hui ii∈N ) ∈ E. Because (X, B(X)) ∼ = ([0, 1], B) and ({0, 1}[0,1] , Ba({0, 1}[0,1] )) ∼ = ({0, 1}c , Ba), we can copy g to a function f with the required properties. Q Q Now set T = {f −1 [H] : H ∈ Ba}, and let λ be the usual measure on {0, 1}c . Then T is a σ-subalgebra of B and there is a probability measure ν with domain T defined by saying that νf −1 [H] = λH for every H ∈ Ba. P P We need to know that if H, H 0 ∈ Ba and f −1 [H] = f −1 [H 0 ] then λH = λH 0 . But as 0 (H4H ) ∩ f [ [0, 1] ] = ∅, H = H 0 , so this is trivial. Thus the given formula defines a function ν : T → [0, 1]. It is now elementary to check that ν is a probability measure. Q Q In fact, because λ is completion regular (415E), f [ [0, 1] ] has full outer measure for λ, so the map H • 7→ f −1 [H]• from the measure algebra of λ to the measure algebra of ν is measure-preserving; since it is surely surjective, the measure algebras are isomorphic, and ν has Maharam type c. However, any probability measure on the whole algebra B has countable Maharam type (433A), so cannot extend ν. Remark Compare 433I-433J. 439B Definition Let X be a Hausdorff space. I will call X universally negligible if there is no Borel probability measure µ defined on X such that µ{x} = 0 for every x ∈ X. A subset of X will be ‘universally negligible’ if it is universally negligible in its subspace topology. 439C Proposition Let X be a Hausdorff space. (a) If A is a subset of X, the following are equiveridical: (i) A is universally negligible; (ii) µ∗ A = 0 whenever µ is a Borel probability measure on X such that µ{x} = 0 for every x ∈ X; (iii) µ∗ A = 0 whenever µ is a σ-finite topological measure on X such that µ{x} = 0 for every x ∈ A; (iv) for every σ-finite topological measure µ on X there is a countable set B ⊆ A such that µ∗ A = µB; (v) A is a Radon space and every compact subset of A is scattered. In particular, countable subsets of X are universally negligible. (b) The family of universally negligible subsets of X is a σ-ideal. (c) Suppose that Y is a universally negligible Hausdorff space and f : X → Y a Borel measurable function such that f −1 [{y}] is universally negligible for every y ∈ Y . Then X is universally negligible. (d) If the topology on X is discrete, X is universally negligible iff #(X) is measure-free. proof (a)(i)⇒(iii) If A is universally negligible and µ is a σ-finite topological measure on X such that µ{x} = 0 for every x ∈ A, let µA be the subspace measure on A. ?? If µ∗ A = α > 0, then (because µ is σ-finite) there is a measurable set E ⊆ X such that γ = µ∗ (E ∩ A) is finite and non-zero. The subspace measure µE∩A is a topological measure on E ∩ A; set νF = γ −1 µE∩A (E ∩ F ) for relatively Borel sets F ⊆ A; then ν is a Borel probability measure on A which is zero on singletons. X X So µ∗ A = 0. (iii)⇒(iv) If (iii) is true and µ is a σ-finite topological measure on X, set B = {x : x ∈ X, µ{x} > 0}. Because µ is σ-finite, B must be countable, therefore measurable, and if we set νE = µ(E \ B) for every Borel set E ⊆ X, ν is a σ-finite Borel measure on X and ν{x} = 0 for every x ∈ X. By (iii), ν ∗ A = 0, that is, there is a Borel set E ⊇ A such that µ(E \ B) = 0; in which case µ∗ A ≤ µ(E \ (B \ A)) = µ(E \ B) + µ(A ∩ B) = µ(A ∩ B) ≤ µ∗ A,

256

Topologies and measures II

439C

so µ∗ A = µ(A ∩ B). As µ is arbitrary, (iv) is true. (iv)⇒(ii) is trivial. not-(i)⇒not-(ii) If A is not universally negligible, let µ be a Borel probability measure on A which is zero on singletons. Set νE = µ(E ∩ A) for any Borel set E ⊆ X; then ν is a Borel probability measure on X which is zero on singletons, and ν ∗ A = 1. (i)⇒(v) Suppose that A is universally negligible. Let µ be a totally finite Borel measure on A. Applying (i)⇒(iv) with X = A, we see that there is a countable set B ⊆ A such that µB = µA; but this means that µ is inner regular with respect to the finite subsets of B, which of course are compact. As µ is arbitrary, A is a Radon space. ?? Suppose, if possible, that A has a compact set K which is not scattered. In this case there is a continuous surjection f : K → [0, 1] (4A2G(i-iv)). Now there is a Radon probability measure ν on K such that f is inverse-measure-preserving for ν and Lebesgue measure on [0, 1] and induces an isomorphism of the measure algebras, so that ν is atomless (418L). Accordingly we have a Borel probability measure µ on A defined by setting µE = ν(K ∩ E) for every relatively Borel set E ⊆ A, and µ{x} = 0 for every x ∈ A, so A is not universally negligible. X X Thus all compact subsets of A are scattered, and (v) is true. (v)⇒(i) Now suppose that (v) is true and that that µ is a Borel probability measure on A. Then µ has an extension to a Radon measure µ ˜ (434F(a-iii)). Let K ⊆ A be a non-empty compact set which is self-supporting for µ ˜ (416Dc). K is scattered, so has an isolated point {x}; because K is self-supporting, µ{x} = µ ˜{x} > 0. As µ is arbitrary, A is universally negligible. (b) This is immediate from (a-ii). (c) Let µ be a Borel probability measure on X. Then F 7→ νf −1 [F ] is a Borel probability measure on Y . Because Y is universally negligible, there must be a y ∈ Y such that µf −1 [{y}] > 0. Set E = f −1 [{y}] and let µE be the subspace measure on E. Then µE is a non-zero totally finite Borel measure on E. Since E is supposed to be universally negligible, there must be some x ∈ E such that 0 < µE {x} = µ{x}. (d) This is just a re-phrasing of the definition in 438A. 439D Remarks (a) The following will be useful when interpreting the definition in 439B. Let X be a hereditarily Lindel¨of Hausdorff space and µ a topological probability measure on X such that µ{x} = 0 for every x ∈ X. Then µ is atomless. P P Suppose that µH > 0. Write G = {G : G ⊆ X is open, µ(G ∩ H) = 0}. S S Then there is a countable G0 ⊆ G such that G0 = G (4A2H(c-i)), so S S µ(H ∩ G) = µ(H ∩ G0 ) = 0, S S and µ(H \ G) > 0. Because µ is zero on singletons, H \ G has at least two points x, y say. Now there are disjoint open sets G0 , G1 containing x, y respectively, and neither belongs to G, so H ∩ G0 , H ∩ G1 are disjoint subsets of H of positive measure. Thus H is not an atom. As H is arbitrary, µ is atomless. Q Q (b) The obvious applications of (a) are when X is separable and metrizable; but, more generally, we can use it on any Hausdorff space with a countable network, e.g., on any analytic space. 439E Lemma (a) Let E, B ⊆ R be such that E is measurable and µL E, µ∗L B are both greater than 0, where µL is Lebesgue measure. Then E − B = {x − y : x ∈ E, y ∈ B} includes a non-trivial interval. (b) If A ⊆ R and µ∗L A > 0, then A + Q is of full outer measure in R. proof (a) By 261Da, there are a ∈ E, b ∈ B such that lim

1

δ↓0 2δ

Let γ > 0 be such that

µ(E ∩ [a − δ, a + δ]) = lim

1

δ↓0 2δ

µ∗ (B ∩ [b − δ, b + δ]) = 1.

439F

Examples 3 2

257 3 2

µ∗L (B ∩ [b − δ, b + δ]) > δ

µL (E ∩ [a − δ, a + δ]) > δ,

whenever 0 < δ ≤ γ. Now suppose that 0 < δ ≤ γ. Then µL ((E + b) ∩ [a + b, a + b + δ]) = µL (E ∩ [a, a + δ]) 1 2

≥ µL (E ∩ [a − δ, a + δ]) − δ > δ, and similarly µ∗L ((B + a + δ) ∩ [a + b, a + b + δ]) = µ∗L (B ∩ [b − δ, b]) 1 2

≥ µ∗L (B ∩ [b − δ, b + δ]) − δ > δ. But this means that (E + b) ∩ (B + a + δ) cannot be empty. If u ∈ (E + b) ∩ (B + a + δ), then u − b ∈ E, u − a − δ ∈ B so a − b + δ = (u − b) − (u − a − δ) ∈ E − B. As δ is arbitrary, E − B includes the interval ]a − b, a − b + γ]. (b) ?? Suppose, if possible, otherwise; that there is a measurable set E ⊆ R such that µL E > 0 and E ∩ (A + Q) = ∅. Then E − A does not meet Q and cannot include any non-trivial interval. X X Remark There will be a dramatic generalization of (a) in 443Db. 439F Lemma Let κ be the least cardinal of any set of non-zero Lebesgue outer measure in R. (a) There is a set X ⊆ [0, 1] of cardinal κ and full outer Lebesgue measure. (b) If (Z, T, ν) is any atomless complete locally determined measure space and A ⊆ Z has cardinal less than κ, then ν ∗ A = 0. (c) (Grzegorek 81) There is a universally negligible set Y ⊆ [0, 1] of cardinal κ. proof (a) Take any set A ⊆ R such that #(A) = κ and µ∗L A > 0, where µL is Lebesgue measure. Set B = A + Q. Then (µL )∗ (R \ B) = 0, by 439Eb. Set X = [0, 1] ∩ B; then µ∗L X = 1 while #(X) ≤ #(B) = κ. By the definition of κ, #(X) must be exactly κ. (b) ?? Otherwise, by 412Jc, there is a set F ⊆ Z such that νF < ∞ and ν ∗ (F ∩ A) > 0. By 343Cc, there is a function f : F → [0, νZn ] which is inverse-measure-preserving for the subspace measure νF and Lebesgue measure on [0, νF ]. But f [A ∩ F ] has cardinal less than κ, so µL f [A ∩ F ] = 0 and 0 < ν ∗ (A ∩ F ) ≤ νf −1 [f [A ∩ F ]] = 0, which is absurd. X X (c)(i) Enumerate X as hxξ iξ<κ . For each ξ < κ, Aξ = {xη : η ≤ P ξ} has cardinal less than κ, so is Lebesgue ∞ negligible; let hIξn in∈N be a sequence of intervals covering Aξ with n=0 µL Iξn < 12 . Enlarging the intervals slightly if necessary, we may suppose that every Iξn has rational endpoints; let hJm im∈N enumerate the family of intervals in R with rational endpoints. Set Cmn = {ξ : ξ < κ, Iξn = Jm } for each m, n ∈ N. (ii) If ν is an atomless totally finite measure on κ which measures every Cmn , then νκ = 0. P P Note first that (by (b), applied to the completion of ν) ν ∗ ξ = 0 for every ξ < κ. Let λ be the (c.l.d.) product of µX , the subspace measure on X, with ν. Set S B = m,n∈N ((X ∩ Jm ) × Cmn ) ⊆ X × κ. Then B is measured by λ, so, by Fubini’s theorem,

R

νB[{x}]µX (dx) =

R

µX B −1 [{ξ}]ν(dξ)

258

Topologies and measures II

439F

(252D). Now look at the sectional measures νB[{x}], µX B −1 [{ξ}]. (Because B is actually a countable union of measurable rectangles, these are always defined.) For any x ∈ X, there is an η < κ such that x = xη , and now B[{x}] = {ξ : there are m, n ∈ N such that x ∈ Jm and ξ ∈ Cmn } = {ξ : there are m, n ∈ N such that x ∈ Jm and Iξn = Jm } = {ξ : there is an n ∈ N such that x ∈ Iξn } ⊇ κ \ η by the choice of the Iξn . But as ν ∗ η = 0, this means that νB[{x}] = νκ. On the other hand, if ξ < κ, then B −1 [{ξ}] = {x : there are m, n ∈ N such that x ∈ Jm and ξ ∈ Cmn } = {x : there are m, n ∈ N such that x ∈ Jm and Iξn = Jm } [ = {x : there is an n ∈ N such that x ∈ Iξn } = X ∩ Iξn , n∈N

so that µX B −1 [{ξ}] ≤ Returning to the integrals, we have νκ =

R

νB[{x}]µX (dx) =

P∞ n=0

R

1 2

µL Iξn ≤ .

1 2

µX B −1 [{ξ}]ν(dξ) ≤ νκ,

so that νκ must be 0, as claimed. Q Q (iii) Now there is an injective function g : κ → [0, 1] such that g[Cmn ] is relatively Borel in g[κ] for every m, n ∈ N. P P Define h : κ → {0, 1}N×N×N by setting h(ξ)(m, n, k) = 1 if ξ ∈ Cmn and xξ ∈ Jk = 0 otherwise. Then h is injective (because if ξ 6= η then xξ 6= xη , so there is some k such that xξ ∈ Jk and xη ∈ / Jk ), and h[Cmn ] = h[κ] ∩ {w : there is some k such that w(m, n, k) = 1} is relatively Borel in h[κ] for every m, n ∈ N. But now recall that {0, 1}N×N×N ∼ = {0, 1}N is homeomorphic to the Cantor set C ⊆ [0, 1] (4A2Uc). If φ : {0, 1}N×N×N → C is any homeomorphism, then φh has the required properties. Q Q (iv) Set Y = g[κ]. Because g is injective, #(Y ) = κ. Also Y is universally negligible. P P Suppose that ν˜ is a Borel measure on Y which is zero on singletons. Then it is atomless, because Y is separable and metrizable (439D). So its copy ν = ν˜(g −1 )−1 on κ is atomless. Because g[Cmn ] is a Borel subset of Y , ν measures Cmn for all m, n ∈ N, so ν˜Y = νκ = 0, by (ii) above. Q Q 439G Corollary A metrizable continuous image of a universally negligible metrizable space need not be universally negligible. proof Take X and Y from 439Fa and 439Fb above, and let f : X → Y be any bijection. Let Γ be the graph of f . The projection map (x, y) 7→ y : Γ → Y is continuous and injective, so Γ is universally negligible, by 439Cc. On the other hand, the projection map (x, y) 7→ x : Γ → X is continuous and surjective, and X is surely not universally negligible, since it is not Lebesgue negligible. 439H Corollary One-dimensional Hausdorff measure on R 2 is not semi-finite. proof Let µH1 be one-dimensional Hausdorff measure on R 2 . Let X, Γ be the sets described in 439F and the proof of 439G.

439K

Examples

259

(a) µ∗H1 Γ > 0. P P The first-coordinate map π1 : R 2 → R is 1-Lipschitz, so, writing µL for Lebesgue measure on R, 1 = µ∗L X = µ∗L π1 [Γ] ≤ µ∗H1 Γ by 264G and 264I. Q Q (b) If E ⊆ R 2 and µH1 E < ∞, then µH1 (E ∩ Γ) = 0, because E ∩ Γ is universally negligible (439Ca) and µH1 is a topological measure (264E) which is zero on singletons. (c) ?? Suppose, if possible, that Γ is not measured by µH1 . Then there is a set A ⊆ R 2 such that < µ∗H1 (A ∩ Γ) + µ∗H1 (A ∩ Γ) (264C, 264Fb). Let E be a Borel set including A such that µH1 E = µ∗H1 A (264Fa); then µH1 (E ∩ Γ) = 0, so µ∗H1 A

µ∗H1 (A ∩ Γ) + µ∗H1 (A ∩ Γ) ≤ µH1 (E ∩ Γ) + µ∗H1 A = µ∗H1 A. X X (d) Since Γ is measurable, not negligible, and meets every measurable set of finite measure in a negligible set, it is purely infinite, and µH1 is not semi-finite. 439I Example There are a set X, a Riesz subspace U of RX and a smooth positive linear functional h : U → R which is not expressible as an integral. proof Take X and Y from 439F. Replacing Y by Y \ {0} if need be, we may suppose that 0 ∈ / Y . Let f : X → Y be any bijection. Let U be the Riesz subspace {u × f : u ∈ Cb } ⊆ RX , where Cb is the space of bounded continuous functions from X to R. Because f is strictly positive, u 7→ u × f : Cb → U is a bijection, therefore a Riesz space isomorphism; moreover, for a non-empty set A ⊆ Cb , inf u∈A u(x) = 0 for every x ∈ X iff inf u∈A u(x)f (x) = R0 for every x ∈ X. We therefore have a smooth linear functional h : U → R defined by setting h(u × f ) = u dµX for every u ∈ Cb , where µX is the subspace measure on X induced by Lebesgue measure. (By 415B, µX is quasi-Radon, so the integral it defines on Cb is smooth, as noted in 436H.) ?? But suppose, if possible, that h is the integral with respect to some measure ν on X. Since f ∈ U , it must be T-measurable, where T is the domain of the completion νˆ of ν. Note that νˆ{x} = 0 for every x ∈ X. P P Set un (y) = max(0, 1 − 2n |y − x|) for y ∈ X. Then Z f (x)ˆ ν {x} = lim un × f dν = lim h(un × f ) n→∞ n→∞ Z = lim un dµX = µX {x} = 0, n→∞

so νˆ{x} = 0. Q Q ˆ of λ is a Radon measure on [0, 1] For Borel sets E ⊆ [0, 1] set λE = νˆf −1 [E]. Then the completion λ −1 ˆ (433Cb or 256C). If t ∈ [0, 1] then f [{t}] contains at most one point, so λ{t} = λ{t} = 0. But Y is ∗ ∗ ˆ Y = 0 (439Ca), that is, there is a Borel set E ⊇ Y with supposed to be universally negligible, so λ Y = λ λE = 0; in which case νX = νˆf −1 [E] = 0, which is impossible. X X Thus h is not an integral, despite being a smooth linear functional on a Riesz subspace of RX . Remark This example is adapted from Fremlin & Talagrand 78. 439J Example Assume that there is some cardinal κ which is not measure-free (definition: 438A). Give κ its discrete topology, and let µ be a probability measure with domain Pκ such that µ{ξ} = 0 for every ξ < κ. Now every subset of κ is open-and-closed, so µ is simultaneously a Baire probability measure and a completion regular Borel probability measure. Of course it is not τ -additive. In the classification schemes of 434A and 435A, we have a measure which is of type B1 as a Borel measure and type E3 as a Baire measure. 439K Example There is a first-countable compact Hausdorff space which is not Radon. proof The construction starts from a compact metrizable space (Z, S) with an atomless Radon probability measure µ. The obvious candidate is [0, 1] with Lebesgue measure; but for technical convenience in a later application I will instead use Z = {0, 1}N with its usual product topology and measure (254J).

260

Topologies and measures II

439K

(a) There is a topology Tc on Z such that (α) S ⊆ Tc ; (β) every point of Z belongs to a countable set which is compact and open for Tc ; T S (γ) if hFn in∈N is a sequence of Tc -closed sets with empty intersection, then n∈N F n is countable, where I write F

S

for the S-closure of F .

P P(i) Tc will be the last in a family hTξ iξ≤c of topologies. We must begin by enumerating Z as hzξ iξ
Tξ ∪ {{yξ }} ∪ {Ln : n ∈ N},

S m≥n

Km for each n.

(δδ ) Because Tξ ⊆ Tξ+1 , Xξ will be open in Xξ+1 . Because the Ki are always Tξ -open, and xξ , yξ are distinct points of Z \ Xξ , the topology on Xξ induced by Tξ+1 is just Tξ . Consequently (by the inductive hypothesis) the topology on Xη induced by Tξ+1 is Tη for every η ≤ ξ. We have t¹n = xξ ¹n for every t ∈ Ln , so Tξ+1 is finer than the usual topology on Xξ+1 . If x ∈ Xξ , then there is a countable Tξ -open Tξ -compact set containing x, which is still Tξ+1 -open and Tξ+1 -compact. Of course {yξ } is a countable Tξ+1 -open Tξ+1 -compact set containing yξ . As for xξ , L0 is surely countable and Tξ+1 -open. To see that it is Tξ+1 -compact, observe that any ultrafilter containing L0 either contains every Ln , and converges to xξ , or contains some Ki and converges to a point of Ki . Thus the induction proceeds at successor stages. S (iii) Inductive step to a limit ordinal If ξ ≤ κ is a non-zero limit ordinal, then we have Xξ = η<ξ Xη , S and can take Tξ to be the topology generated by η<ξ Tη . It is easy to check that this works (because the topologies Tη are consistent). (iv) At the end of the induction, we have Xc = Z because zξ ∈ Xξ+1 ⊆ Xc for every ξ. The final topology Tc on Z will have the properties (α) and (β) required. ?? Now suppose, if possible, that hFn in∈N is T S a sequence of Tc -closed sets with empty intersection, and that n∈N F n is S uncountable. For each n ∈ N, let Jn ⊆ Fn be a countable S-dense set. Then there is some ζ < c such that n∈N Jn ⊆ Xζ (because cf c > ω, see 4A1A(c-ii)). Let ξ ≥ ζ be such that Jn = Iξn for every n ∈ N. Then in the construction of Tξ+1 we must be in case (β) of (ii) above. Taking htm im∈N as described there, we have htm im∈N → xξ for TTξ+1 , and therefore for Tc . But for any n ∈ N, tm ∈ In ⊆ Fn for infinitely many m, so xξ ∈ Fn . Thus xξ ∈ n∈N Fn ; but this is impossible. X X

439K

Examples

261

So we have a topology of the type required. Q Q (b) There is a probability measure ν on Z, extending the usual measure µ, such that with respect to Tc ν is a topological measure inner regular with respect to the closed sets, but is not τ -additive. P P Let K be the family of Tc -closed subsets of Z. For F ∈ K, set φF = µ∗ F . S

(i) If E, F are disjoint Tc -closed sets, then E ∩ F (a-γ)). So

S

must be countable (take F2n = E, F2n+1 = F in S

S

φ(E ∪ F ) = µ∗ (E ∪ F ) = µ∗ ((E ∪ F ) ∩ E ) + µ∗ ((E ∪ F ) \ E ) = µ∗ E + µ∗ F = φE + φF. (ii) If E, F ∈ K, E ⊆ F and ² > 0, there is an S-open set G ⊇ E such that µG ≤ µ∗ E + ². Now F \ G ∈ K and φF = µ∗ (F ∩ G) + µ∗ (F \ G) ≤ µG + φ(F \ G) ≤ φE + φ(F \ G) + ². Putting this together with (i), we see that φF = φE + sup{φE 0 : E 0 ∈ K, E 0 ⊆ F \ E}. (iii) If hFn in∈N is a non-increasing sequence in K with empty intersection, then T S S limn→∞ φFn ≤ limn→∞ µF n = µ( n∈N F n ) = 0. (iv) Thus K and φ satisfy all the conditions of 413I, and there is a measure ν, extending φ, which is defined on every member of K and inner regular with respect to K, and therefore is (for Tc ) a topological measure inner regular with respect to the closed sets. If we write V for the family of Tc -compact Tc -open countable subsets of Z, then for any K ∈ V νK = φK = µK = 0, while V is upwards-directed and has union Z; so that ν is not τ -additive. Q Q (c) So far we seem to have very little more than is provided by ω1 with the order topology and Dieudonn´e’s measure. The point of doing all this work is the next step. Set X = Z × {0, 1} and give X the topology T generated by {G × {0, 1} : G ∈ S} ∪ {H × {1} : H ∈ Tc } ∪ {X \ (K × {1}) : K is Tc -compact}. (i) T is Hausdorff. P P If w, z are distinct points of X, then either their first coordinates differ and they are separated by sets of the form G0 × {0, 1}, G1 × {0, 1} where G0 , G1 belong to S, or they are of the form (x, 1), (x, 0) and are separated by open sets of the form K × {1}, X \ (K × {1}) for some set K which is compact and open for Tc . Q Q (ii) T is compact. P P Let F be an ultrafilter on X. Writing π1 (x, 0) = π1 (x, 1) = x for x ∈ Z, π1 [[F]] is S-convergent, to x0 say. If K × {1} ∈ F for some Tc -compact set K, then F is T-convergent to (x, 1); otherwise, it is T-convergent to (x, 0) (using 4A2B(a-v)). Q Q (iii) T is first-countable. P P If x ∈ Z, then {(x, 0), (x, 1)} = π1−1 [{x}] is a Gδ set in X because {x} is a Gδ set in Z and π1 is continuous (4A2C(a-iii)). Now {(x, 0)} and {(x, 1)} are relatively open in {(x, 0), (x, 1)}, so are Gδ sets in X (4A2C(a-iv)). Thus singletons are Gδ sets. Because T is compact and Hausdorff, it is first-countable (4A2Kf). Q Q (iv) (X, T) is not a Radon space. P P Z × {1} is an open subset of X, homeomorphic to Z with the topology Tc . But the measure ν of (b) above (or, if you prefer, its restriction to the Tc -Borel algebra) witnesses that Tc is not a Radon topology, so T also cannot be a Radon topology, by 434Fc. Q Q ´ sz Remark Aficionados will recognise Tc as a kind of ‘JKR-space’, derived from the construction in Juha Kunen & Rudin 76.

262

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439L

439L Example Suppose that κ is a cardinal which is not measure-free; let µ be a probability measure with domain Pκ which is zero on singletons. Give κ the discrete topology, so that µ is a Borel measure and κ is first-countable. Let ν be the restriction of the usual measure on Y = {0, 1}κ to the algebra B of Borel subsets of Y , so that ν is a τ -additive probability measure, and λ the product measure on κ × Y constructed by the method of 434R. Then S W = {(ξ, y) : ξ < κ, y(ξ) = 1} = ξ<κ {ξ} × {y : y(ξ) = 1} is open in κ × Y . b then λ(W 4W 0 ) = 21 . P If W 0 ∈ Pκ⊗B P There is a countable set E ⊆ B such that W 0 belongs to the σ-algebra generated by {A × E : A ⊆ κ, E ∈ E} (331Gd). For J ⊆ κ, write πJ (y) = y¹J for y ∈ Y , let νJ be the usual measure on {0, 1}J and ΛJ its domain, and let Λ0J be the family of sets E ⊆ Y such that there are H, H 0 ∈ ΛJ such that πJ−1 [H] ⊆ E ⊆ πJ−1 [H 0 ] and νJ (H 0 \ H) = 0. Then Λ0J ⊆ Λ0K whenever J ⊆ K ⊆ κ, and every set measured by ν belongs to Λ0J for some countable J (254Ob). There is therefore a countable set J ⊆ κ such that E ⊆ Λ0J . Also, of course, Λ0J is a σ-algebra of subsets of Y . The set {V : V ⊆ κ × Y, V [{ξ}] ∈ Λ0J for every ξ < κ} is a σ-algebra of subsets of κ × Y containing A × E whenever A ⊆ κ and E ∈ E, so contains W 0 . But this means that if ξ ∈ κ \ J, W [{ξ}] and W 0 [{ξ}] are stochastically independent, and ν(W [{ξ}]4W 0 [{ξ}]) = 21 . Since µ(κ \ J) = 1, λ(W 4W 0 ) =

R

1 2

ν(W [{ξ}]4W 0 [{ξ}])µ(dξ) = ,

as claimed. Q Q b In particular, W • in the the measure algebra of λ cannot be represented by a member of Pκ⊗B. 439M Example There is a first-countable locally compact Hausdorff space X with a Baire probability measure µ which is not τ -additive and has no extension to a Borel measure. In the classification of 435A, µ is of type E0 . proof Let Ω be the set of non-zero countable limit ordinals, and for each ξ ∈ Ω let hθξ (i)ii∈N be a strictly increasing sequence of ordinals with supremum ξ. Set X = ω1 × (ω + 1), and define a topology T on X by saying that G ⊆ X is open iff {ξ : (ξ, n) ∈ G} is open in the order topology of ω1 for every n ∈ N, whenever ξ ∈ Ω and (ξ, ω) ∈ G then there is some n < ω such that (η, i) ∈ G whenever n ≤ i < ω and θξ (i) < η ≤ ξ. This is finer than the product of the order topologies, so is Hausdorff. For every ξ < ω1 , n ∈ N, (ξ + 1) × {n} is a countable compact open set containing (ξ, n); for every ξ ∈ ω1 \ Ω, {(ξ, ω)} is a countable compact open set containg (ξ, ω); and for every ξ ∈ Ω, {(ξ, ω)} ∪ {(η, i) : i < ω, θξ (i) < η ≤ ξ} is a countable compact open subset of X containing (ξ, ω). Thus T is locally compact, and every singleton subset of X is Gδ , so T is first-countable (4A2Kf). If f : X → R is continuous, then for every n ∈ N there is a ζn < ω1 such that f is constant on {(ξ, n) : ζn ≤ ξ < ω1 } (4A2S(b-iii)). Setting ζ = supn∈N ζn , f must be constant on {(ξ, ω) : ξ ∈ Ω, ξ > ζ}. P P If ξ, η ∈ Ω\(ζ +1), then f (ξ, ω) = limi→∞ f (θξ (i)+1, i) and f (η, ω) = limi→∞ f (θη (i)+1, i). But there is some n such that both θξ (i) and θη (i) are greater than ζ for every i ≥ n, so that f (θξ (i)+1, i) = f (θη (i)+1, i) for every i ≥ n and f (ξ, ω) = f (η, ω). Q Q Writing Σ for the family of subsets E of X such that {ξ : ξ ∈ Ω, (ξ, ω) ∈ E} is either countable or cocountable in Ω, Σ is a σ-algebra of subsets of X such that every continuous function is Σ-measurable, so every Baire set belongs to Σ. We therefore have a Baire measure µ0 on X defined by saying that µ0 E = 0 if E ∩ (Ω × {ω}) is countable, 1 otherwise. {(ξ + 1) × (ω + 1) : ξ < ω1 } is a cover of X by negligible open-and-closed sets, so µ0 is not τ -additive. ?? Suppose, if possible, that µ were a Borel measure on X extending µ0 . Then we must have µ(ω1 ×{n}) = µ0 (ω1 × {n}) = 0 for every n ∈ N, so µ(ω1 × {ω}) = 1. Let λ be the subspace measure on ω1 × {ω} induced

439O

Examples

263

by µ. If A ⊆ ω1 , (A × {ω}) ∪ (ω1 × ω) is an open set, so λ is defined on every subset of ω1 × {ω}; and if ξ < ω1 , then µ0 ((ξ + 1) × (ω + 1)) = 0, so λ is zero on singletons. And this contradicts Ulam’s theorem (419G, 438Cd). X X 439N Example Give ω1 its order topology. (i) ω1 is a normal Hausdorff space which is not measure-compact. (ii) There is a Baire probability measure µ0 on ω1 which is not τ -additive and has a unique extension to a Borel measure, which is not completion regular; that is, µ0 is of type E2 in the classification of 435A. proof (a) As noted in 4A2Rc, order topologies are always normal and Hausdorff. (b) Let µ be Dieudonn´e’s measure on ω1 (411Q), and µ0 its restriction to the Baire σ-algebra. Then µ is the only Borel measure extending µ0 . P P Let ν be any Borel measure extending µ0 . Every set [0, ξ] = [0, ξ + 1[, where ξ < ω1 , is open-and-closed, so ν[0, ξ] = µ0 [0, ξ] = µ[0, ξ] = 0; also, of course, νX = 1. Let F ⊆ X be any closed set. If F is countable, then it is included in some initial segment [0, ξ], so νF = µF = 0. Now suppose that F is uncountable. Set G = ω1 \ F . For each ξ ∈ F , set ζξ = min{η : ξ < η ∈ F } and Gξ = ]ξ, ζξ [. Then hGξ iξ∈F is a disjoint family of open sets. By 438Bb and 419G/438Cd, P S ν( ξ∈F Gξ ) = ξ∈F νGξ = 0. But now

S 1 = νω1 = νF + ν [0, min F [ + ν( ξ∈F Gξ ) = νF = µF .

Thus µ and ν agree on the family E of closed sets. By the Monotone Class Theorem (136C), they agree on the σ-algebra generated by E, which is their common domain; so they are equal. Q Q (c) I have already remarked in 411Q-411R that µ and µ0 are not τ -additive and µ is not completion regular. So of course ω1 is not measure-compact. 439O In 439M I described a Baire measure with no extension to a Borel measure. In view of Maˇr´ık’s theorem (435C), it is natural to ask whether this can be done with a normal space. This leads us into relatively deep water, and the only examples known need special assumptions. Example Assume ♣. Then there is a normal Hausdorff space with a Baire probability measure µ which is not τ -additive and not extendable to a Borel measure. (In the classification of 435A, µ is of type E0 .) proof (a) ♣ implies that there is a family hCξ iξ<ω1 of sets such that (i) Cξ ⊆ ξ for every ξ < ω1 (ii) Cξ ∩ η is finite whenever η < ξ < ω1 (iii) for any uncountable sets A, B ⊆ ω1 there is a ξ < ω1 such that A ∩ Cξ and B ∩ Cξ are both infinite (4A1N). For A ⊆ ω1 , set A0 = {ξ : ξ < ω1 , A ∩ Cξ is infinite}; then A0 ∩ B 0 is non-empty whenever A, B ⊆ ω1 are uncountable. But this means that A0 ∩ B 0 is actually uncountable for uncountable A, B, since A0 ∩ B 0 \ γ ⊇ (A \ γ)0 ∩ (B \ γ)0 is non-empty for every γ < ω1 . Set X = ω1 × N. For x = (ξ, n) ∈ X, say that Ix = Cξ × {n − 1} if n ≥ 1, = ∅ otherwise . (b) Define a topology T on X by saying that a set G ⊆ X is open iff Ix \ G is finite for every x ∈ G. The form of the construction ensures that T is T1 . In fact, Ix ∩ Iy is finite whenever x 6= y in X. P P Express x as (ξ, m) and y as (η, n) where η ≤ ξ. If either m = 0 or n = 0 or m 6= n, Ix ∩ Iy = ∅. If n ≥ 1 and η < ξ, then Ix ∩ Iy ⊆ (Cξ ∩ η) × {m − 1} is finite. Similarly, Ix ∩ Iy is finite if m ≥ 1 and ξ < η. Q Q Consequently {x} ∪ J is closed for every x ∈ X, J ⊆ Ix .

264

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439O

Observe that (ξ + 1) × N is open and closed for every ξ < ω1 , again because Cη ∩ (ξ + 1) is finite whenever η ∈ Ω and η > ξ, while Cξ ⊆ ξ for every ξ. (c) The next step is to understand the uncountable closed subsets of X. First, if F ⊆ X is closed and n ∈ N, then F −1 [{n}]0 , as defined in (a), is a subset of F −1 [{n+1}], since if ξ ∈ F −1 [{n}]0 then I(ξ,n+1) ∩F is infinite. If F is uncountable, there is some n ∈ N such that F −1 [{n}] is uncountable, so that (inducing on m) F −1 [{m}] is uncountable for every m ≥ n. Finally, this means that if E, F ⊆ X are uncountable closed sets, there is an m ∈ N such that E −1 [{m}] and F −1 [{m}] are both uncountable, so that E −1 [{m}]0 ∩ F −1 [{m}]0 is non-empty and E ∩ F is non-empty. (d) It follows that X is normal. P P Let E and F be disjoint closed sets in X. By (c), at least one of them is countable; let us take it that E ⊆ ζ × N where ζ < ω1 . Enumerate the open-and-closed set W = (ζ + 1) × N as hxn in∈N . Choose hUn in∈N , hVn in∈N inductively, as follows. U0 = E, V0 = F ∩ W . If xn ∈ Un , then Un+1 = Un ∪ (Ixn \ Vn ) and Vn+1 = Vn ; if xn ∈ / Un , then Un+1 = Un and Vn+1 = Vn ∪ {xn } ∪ (Ixn \ Un ). An easy induction shows that, for every n, (α) Un ∩ Vn = ∅ (β) Un ∪ Vn ⊆ W (γ) Ix ∩ (Un ∪ Vn ) is finite for every x ∈ X \ (Un ∪ Vn ) (δ) Ix ∩ Vn is finite for every x ∈ Un (²) Ix ∩ Un is finite for every x ∈ Vn . S S At the end of the induction, set G = n∈N Un , H = n∈N Vn ∪ (X \ W ). Then E ⊆ G, F ⊆ H and G ∩ H = ∅. If x ∈ G, it is of the form xn for some n, in which case xn ∈ Un (because xn ∈ / Vn+1 ) and Ix \ Un+1 = Ix ∩ Vn is finite; thus G is open. If x ∈ H ∩ W , again it is of the form xn where this time xn ∈ / Un , so that Ix \ Vn+1 = Ix ∩ Un is finite; so H is open. Thus E and F are separated by open sets; since E and F are arbitrary, X is normal. Q Q Being T1 (see (b)), X is also Hausdorff. (e) Because disjoint closed sets in X cannot both be uncountable ((c) above), any bounded continuous function on X must be constant on a cocountable set. (Compare 4A2S(b-iii).) The countable-cocountable measure µ0 is therefore a Baire measure on X (cf. 411R). But it has no extension to a Borel measure. P P The point is that if A is any subset of ω1 , and n ∈ N, then (A × {n}) ∪ (ω1 × n),

ω1 × n

are both open, so A × {n} is Borel; accordingly every subset of X is a Borel set. But ω1 is measure-free (419G, 438Cd), so there can be no Borel probability measure on X which is zero on singletons. Q Q Of course µ0 is not τ -additive, because {(ξ + 1) × N : ξ < ω1 } is a cover of X by open-and-closed negligible sets. Remark Thus in Maˇr´ık’s theorem we really do need ‘countably paracompact’ as well as ‘normal’, at least if we want a theorem valid in ZFC. Observe that any example of this phenomenon must involve a Dowker space, that is, a normal Hausdorff space which is not countably paracompact. The one here is based on de Caux 76. Such spaces are hard to come by in ZFC if we do not allow ourselves to use special principles like ♣. ‘Real’ Dowker spaces have been described by Rudin 71 and Balogh 96; for a survey, see Rudin 84. I do not know if either of these can be adapted to provide a ZFC example to replace the one above. 439P Example (cf. Moran 68) N c is not Borel-measure-compact, therefore not Borel-measure-complete, measure-compact or Radon. proof Consider the topology Tc on Z = {0, 1}N , as constructed in 439K. Then (Z, Tc ) is homeomorphic to a closed subset of N Z × {0, 1}N , where in this product the second factor {0, 1}N is given its usual topology S. P P For each x ∈ Z, let Lx be a Tc -open Tc -compact subset of Z. The first thing to observe is that if x ∈ Z, and we write Vxm = {y : y ∈ Z, y¹m = x¹m} for each m ∈ N, then Ux = {Lx ∩ Vxm : m ∈ N} is a downwards-directed family of compact open neighbourhoods of x with intersection {x}, so is a base of neighbourhoods of x (4A2Gd); thus U = {Lx : x ∈ Z} ∪ S generates Tc . Now, for x ∈ Z, define φx : Z → N by setting φx (y) = 0 if y ∈ Lx , = m + 1 if y ∈ Vxm \ (Lx ∪ Vx,m+1 ).

439Q

Examples

265

Then every φx is Tc -continuous, so we have a Tc -continuous function φ : Z → N Z × {0, 1}N defined by setting φ(y) = (hφx (y)ix∈C , y) for y ∈ Z. Because every element of U is of the form φ−1 [H] for some open set H ⊆ N Z × {0, 1}N , Z is homeomorphic to its image φ[Z]. Now suppose that (w, z) ∈ φ[Z]. In this case, there is an ultrafilter containing φ[Z] which converges to (w, z) (4A2Bc), and this must be of the form φ[[F]] for some ultrafilter on Z (2A1Ib). Of course z = lim F for S. ?? If z is not the Tc -limit of F, then F can have no Tc -limit, and can contain no Tc -compact set (2A3R). In particular, Lz ∈ / F; but in this case Vzm \Lz ∈ F for every m, so that {(v, y) : v(x) > m} ∈ φ[[F]] for every m, and w(x) > m for every m, which is impossible. X X Thus F → z, and (as φ is continuous) (w, z) = φ(z). This shows that φ[Z] is closed, so we have the required homeomorphism between Z and a closed subset of N Z × {0, 1}N . Q Q Of course N Z × {0, 1}N is a closed subset of N Z × N N ∼ = N c . So Z is homeomorphic to a closed subset c of N . But Z, with Tc , carries a Borel probability measure ν which is inner regular with respect to the closed sets and is not τ -additive (439Kb). So (Z, Tc ) is not Borel-measure-compact. By 434Hc, N c is not Borel-measure-compact. By 434Ic, N c is not Borel-measure-complete; by 434K, it is not Radon; by 435Fd, it is not measure-compact. 439Q Example Let X be the real line with the right-facing Sorgenfrey topology, generated by sets of the form [a, b[ where a < b in R. Then X is measure-compact but X 2 is not. proof (a) Note that every set [a, b[ is open-and-closed in X, so that the topology is zero-dimensional, therefore completely regular; and it is finer than the usual topology of R, so is Hausdorff. X is Lindel¨of. P P Let G be an open cover of X. For each q ∈ Q, set S

Aq = {a : a ∈ ]−∞, q[ and there is some G ∈ G such that [a, q[ ⊆ G}.

Then q∈Q Aq = R. For each q ∈ Q, there is a countable set A0q ⊆ Aq such that inf A0q = inf Aq in [−∞, ∞] 0 and A0q contains min Aq if Aq has a least element. Now, Sfor each pair 0(a, q) where q ∈ Q and a ∈ Aq , choose Gaq ∈ G such that [a, q[ ⊆ Gaq . It is easy to see that {Gaq : a ∈ Aq } ⊇ Aq , so that the countable family {Gaq : q ∈ Q, a ∈ A0q } covers X. As G is arbitrary, X is Lindel¨of. Q Q It follows that X is measure-compact (435Fb). (b) Let S be the usual topology on R 2 , and T the product topology on X 2 . (i) Whenever G, H are disjoint T-open sets, there is an S-Borel set E such that G ⊆ E ⊆ X 2 \ H. P P For n ∈ N, set An = {(a, b) : [a, a + 2−n [ × [b, b + 2−n [ ⊆ G}. S

S

?? Suppose, if possible, that there is a point (x, y) ∈ An ∩ H, where I write to denote closure for the topology S. Let δ > 0 be such that [x, x + 2δ[ × [y, y + 2δ[ ⊆ H and 2δ < 2−n . Then there must be (a, b) ∈ An such that |a − x| ≤ δ and |b − y| ≤ δ. In this case, a ≤ x + δ < a + 2−n and b ≤ y + δ < b + 2−n , so (x + δ, y + δ) ∈ G; while δ was chosen so that (x + δ, y + δ) would belong to H. X X S S S Accordingly E = n∈N An is an S-Borel set disjoint from H. But G = n∈N An , so G ⊆ E. Q Q (ii) Consequently every T-continuous real-valued function is S-Borel measurable. P P If f : X 2 → R is T-continuous and α ∈ R, then there is an S-Borel set Eα such that {(x, y) : f (x, y) < α} ⊆ Eα ⊆ {(x, y) : f (x, y) ≤ α}. S But this means that {(x, y) : f (x, y) < α} = n∈N Eα−2−n is S-Borel. Q Q (iii) It follows that every T-Baire set is S-Borel. We therefore have a T-Baire probability measure ν on X defined by setting νE = µL {t : t ∈ [0, 1], (t, 1 − t) ∈ E} for every T-Baire subset of X 2 , where µL is Lebesgue measure on R. In this case every point (x, y) of X 2 belongs to a T-open set of zero measure for ν. P P Set K = {(t, 1−t) : t ∈ [0, 1]}. Then K is S-closed, therefore T-closed, and ν(X 2 \ K) = 0, so if (x, y) ∈ / K then we can stop. If (x, y) ∈ K, then [x, x + 1[ × [y, y + 1[ is a T-open T-closed set meeting K in the single point (x, y), so is a negligible T-neighbourhood of (x, y). Q Q

266

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439Q

Thus ν is not τ -additive and X 2 is not measure-compact. Remark Contrast this with 438Xo. 439R Example There are first-countable completely regular Hausdorff spaces X, Y with Baire probability measures µ, ν such that the Baire measures λ, λ0 on X × Y defined by the formulae

R

f dλW =

RR

f (x, y)ν(dy)µ(dx),

R

f dλ0 =

RR

f (x, y)µ(dx)ν(dy)

(436F) are different. proof Let X, Y be disjoint stationary subsets of ω1 (4A1Cd). Give each the topology induced by the order topology of ω1 . Let µ ˜ be Dieudonn´e’s measure on ω1 (411Q), and let µ ˜X , µ ˜Y be the subspace measures induced on X and Y by µ ˜; let µ and ν be the restrictions of µ ˜X , ν˜Y to the Baire σ-algebras of X, Y respectively. Then µX = µ ˜X X = µ ˜∗ X = 1 because X meets every cofinal closed set in ω1 ; similarly, νY = 1. Set W = {(x, y) : x ∈ X, y ∈ Y, x < y} = {(x, y) : x ∈ X, y ∈ Y, x ≤ y}. Then W is open-and-closed in X × Y (use 4A2Rl), so that f = χW is continuous. But

RR

RR

f (x, y)ν(dy)µ(dx) = f (x, y)µ(dx)ν(dy) =

R

R

ν{y : y ∈ Y, x < y}µ(dx) = 1, µ{x : x ∈ X, x < y}ν(dy) = 0.

Remark Contrast this with 434Xx and 439Yh. ˇ 439S The results of 437W leave open the question of which familiar spaces, beyond Cech-complete spaces, can be Prokhorov. In fact rather few are. The basis of any further investigation must be the following result. Theorem (Preiss 73) Q is not a Prokhorov space. proof (a) There is a non-decreasing sequence hXk ik∈N of non-empty compact subsets of X = Q ∩ [0, 1], with union X, such that whenever k ∈ N, x ∈ Xk and δ > 0, then Xk+1 ∩ [x − δ, x + δ] is infinite. P P Start by enumerating X as hqk ik∈N . Set X0 = {q0 }. Given that Xk ⊆ X is compact, then for each m ∈ N let Em be a finite cover of Xk by open intervals of length at most 2−m all meeting S Xk , and let Ikm be a finite subset of X \ Xk meeting every member of Em ; set Xk+1 = Xk ∪ {qk+1 } ∪ m∈N Ikm . If H is any cover of Xk+1 by open sets inSR, then there is a finite H0 ⊆ H covering XkS. There must be an m ∈ N suchSthat [x − 2−m , x + 2−m ] ⊆ H0 for every x ∈ Xk (2A2Ed), so that Ikl ⊆ H0 for every l ≥ m, and Xk+1 \ H0 is finite; accordingly there is a finite H1 ⊆ H covering Xk+1 . As H is arbitrary, Xk+1 is compact, and the induction can proceed. If x ∈ Xk and δ > 0, then for every m ∈ N there is an x0 ∈ Xk+1 \ Xk such that |x0 − x| ≤ 2−m , so that [x − δ, x + δ] ∩ Xk+1 must be infinite. Q Q (b) If h²k ik∈N is any sequence in ]0, ∞[, and F ⊆ [0, 1] is a countable closed set, then there is an x∗ ∈ X \ F such that ρ(x∗ , Xk ) < ²k for every k ∈ N. P P We can suppose that limk→∞ ²k = 0. Define hHk ik∈N inductively, as follows. H0 = R. Given Hk , set Hk+1 = Hk ∩ {x : ρ(x, Xk ∩ Hk ) < ²k }, where ρ(x, A) = inf y∈A |x − y| for x ∈ R, A ⊆ R. Observe that every THk is an open subset of R and that Xk ∩ Hk ⊆ Hk+1 ⊆ Hk for every k; consequently, setting E = k∈N Hk , E is a Gδ subset of R and Xk ∩Hk ⊆ E for every k. In particular, E∩X contains q0 and is not empty. Next, for each k, ρ(x, E∩Xk ) < ²k for every x ∈ Hk+1 and therefore for every x ∈ E; accordingly E ∩ X is dense in E. Moreover, if x ∈ E ∩ X, there is a k ∈ N such that x ∈ Xk ; we must have x ∈ Hk , and in this case Hk+1 is a neighbourhood of x. So every neighbourhood of x contains infinitely many points of Hk+1 ∩ Xk+1 ⊆ E ∩ X. Thus E ∩ X has no isolated points; it follows that E has no isolated points. By 4A2Mc and 4A2Me, E is uncountable. There is therefore a point z ∈ E \ F . Let m ∈ N be such that ρ(z, F ) ≥ ²m . As z ∈ Hm+1 , there is an x∗ ∈ Hm ∩ Xm such that |z − x∗ | < ²m and x∗ ∈ / F . Let k ∈ N. If k ≥ m then certainly ρ(x∗ , Xk ) = 0 < ²k . ∗ ∗ ∗ If k < m then x ∈ Hk+1 so ρ(x , Xk ) ≤ ρ(x , Hk ∩ Xk ) < ²k . So we have a suitable x∗ . Q Q

439Xe

Examples

267

(c) For n, k ∈ N set Gkn = {x : x ∈ R \ Xk , ρ(x, Xn ) > 2−k }. Then Gkn is an open subset of R. Let A be the set of Radon probability measures µ on X such that µ(Gkn ∩ X) ≤ 2−n for all n, k ∈ N. (d) Write A˜ for the set of Radon probability measures µ on [0, 1] such that µ(Gkn ∩ [0, 1]) ≤ 2−n for all k, n ∈ N. Then A˜ is a narrowly closed subset of the set of Radon probability measures on [0, 1], which is ˜ P itself narrowly compact (437O). Also µ([0, 1] \ X) = 0 for every µ ∈ A. P Let set K ⊆ [0, 1] \ X be compact, and n ∈ N. Then K and Xn are disjoint compact sets, so there is some k ∈ N such that |x − y| > 2−k for every x ∈ Xn and y ∈ K. In this case K ⊆ Gkn so µK ≤ 2−n . As n is arbitrary, µK = 0; as K is arbitrary, µ([0, 1] \ X) = 0. Q Q A is compact in the narrow topology. P P The identity map φ : X → [0, 1] induces a map φ˜ : MP+ (X) → + MP ([0, 1]) which is a homeomorphism between MP+ (X) and {µ : µ ∈ MP+ ([0, 1]), µ([0, 1] \ X) = 0} (437Qb). ˜ ˜ since A˜ ⊆ {µ : µ ∈ M + ([0, 1]), µ([0, 1] \ X) = 0}, φ¹A The definition of A makes it plain that it is φ˜−1 [A]; P ˜ is a homeomorphism between A and A, and A is compact. Q Q (e) A, regarded as a subset of MP+ (X), is not uniformly tight. P Let K ⊆ X be P compact. Consider the PP set C of those w ∈ [0, 1]X such that w(x) = 0 for every x ∈ K, x∈X w(x) ≤ 1 and x∈Gkn ∩X w(x) ≤ 2−n for all k, n ∈ N. Then C is a compact subset of [0, 1]X . If D ⊆ C is any non-empty upwards-directed X set, then sup D, taken in [0, 1] P , belongs to C. By Zorn’s Lemma, C has a maximal member w say. ?? Suppose, if possible, that x∈X w(x) = γ < 1. For each n ∈ N, let Ln ⊆ X be a finite set such P −n−1 w(x) ≥ γ − 2 , and mn ∈ N such that Ln ⊆ Xmn . By (b), there is an x∗ ∈ X \ K that x∈Ln ∗ −mn such that ρ(x , Xn ) < 2 for every n ∈ N. Let r ∈ N be such that x∗ ∈ Xr and γ + P 2−r ≤ 1, and set 0 ∗ ∗ −r 0 ∗ 0 X w (x ) = w(x )+2 , w (x) = w(x) w ∈ [0, 1] and x∈X w0 (x) ≤ 1. P }. Then certainly −n P for every x0 ∈ X \{x ∗ If k, n ∈ N and x ∈ / Gkn , then x∈Gkn ∩X w (x) = x∈Gkn ∩X w(x) ≤ 2 . If x∗ ∈ Gkn , then n < r and 2−k < ρ(x∗ , Xn ) < 2−mn , so mn < k and Ln ⊆ Xk and P P P −n−1 , x∈X\Ln w(x) ≤ 2 x∈X\Xk w(x) ≤ x∈Gkn ∩X w(x) ≤ P x∈Gkn ∩X

w0 (x) ≤ 2−n−1 + 2−r ≤ 2−n .

Thus w0 ∈ C and X Pw was not maximal. X Accordingly x∈X w(x) = 1 and the point-supported measure µ defined by w is a probability measure on X. By the definition of C, µ ∈ A and µ(X \ K) = 1. As K is arbitrary, A cannot be uniformly tight. Q Q (f ) Thus A witnesses that X = Q ∩ [0, 1] is not a Prokhorov space. Since X is a closed subset of Q, 437Wb tells us that Q is not a Prokhorov space. 439X Basic exercises (a) (i) Show that there is a set A ⊆ [0, 1] such that µ∗L A = 1, where µL is Lebesgue measure, and every member of [0, 1] is uniquely expressible as a + q where a ∈ A, q ∈ Q. (Hint: use an idea from 419J.) (ii) Define f : [0, 1] → A by setting f (x) = a when x ∈ a + Q. Show that the image measure µL f −1 takes only the values 0 and 1. (Aldaz 95. Compare 342Xg.) (b) Let X be a Radon Hausdorff space and A a subset of X. Show that A is universally negligible iff µA = 0 for every atomless Radon measure on X. > (c) Let X be a Hausdorff space. Show that a set A ⊆ X is universally negligible iff µA = 0 whenever µ is a complete locally determined topological measure on X such that µ{x} = 0 for every x ∈ X. (d) Let X be a Hausdorff space. Show that any universally negligible subset of X is universally measurable in the sense of 434D. (e) (i) Show that there is an analytic set A ⊆ R such that for any Borel subset E of R \ A there is an uncountable Borel subset of R \ (A ∪ E). (Hint: 423Qb, part (c) of the proof of 423L.) (ii) Show that A is universally measurable, but there is no Borel set E such that A4E is universally negligible.

268

Topologies and measures II

439Xf

(f ) Show that a first-countable compact Hausdorff space is universally negligible iff it is scattered iff it is countable. (g) Show that the product of two universally negligible Hausdorff spaces is universally negligible. (h) Let us say that a Hausdorff space X is universally τ -negligible if there is no τ -additive Borel probability measure on X which is zero on singletons. (i) Show that if X is a Hausdorff space and A ⊆ X, then A is universally τ -negligible iff µ∗ A = 0 for every τ -additive Borel probability measure on X such that µ{x} = 0 for every x ∈ X. (ii) Show that if X is a regular Hausdorff space, then a subset A of X is universally τ -negligible iff µA = 0 for every atomless quasi-Radon measure on X. (iii) Show that if X is a completely regular Hausdorff space, it is universally τ -negligible iff whenever µ is an atomless Radon measure on a space Z, and X 0 ⊆ Z is homeomorphic to X, then µX 0 = 0. (iv) Show that a Hausdorff space X is universally negligible iff it is Borel-measure-complete and universally τ -negligible. (v) Show that if X is a Hausdorff space, Y is a universally τ -negligible Hausdorff space, and f : X → Y is a continuous function such that f −1 [{y}] is universally τ -negligible for every y ∈ Y , then X is universally τ -negligible. (vi) Show that the product of two universally τ -negligible Hausdorff spaces is universally τ -negligible. (vi) Show that a scattered Hausdorff space (in particular, any discrete space) is universally τ -negligible. (vii) Show that a compact Hausdorff space is universally τ -negligible iff it is scattered. > (i) Let κ be the smallest cardinal of any subset of R which is not Lebesgue negligible. Show that if (Z, T, ν) is any complete locally determined atomless measure space and A ⊆ Z has cardinal less than κ, then νA = 0. (j) Let (X, ≤) be any well-ordered set and µ a non-zero σ-finite measure on X such that every singleton has measure 0. Show that {(x, y) : x ≤ y} is not measurable for the (c.l.d.) product measure on X × X. (Hint: Reduce to the case in which µ is complete and totally finite, X = ζ is an ordinal and µξ = 0 for every ξ < ζ. You will probably need 251P.) (k) Show that 1-dimensional Hausdorff measure on R 2 is not inner regular with respect to the closed sets. (Hint: 439H, 471R.) (l) Show that N I is not pre-Radon for any uncountable set I. (Hint: 417Xq.) (m) (i) Suppose that X is a completely regular Hausdorff space and there is a continuous function f from X to a separable metrizable space Z such that f −1 [{z}] is Lindel¨of for every z ∈ Z. Show that X is realcompact (definition: 436Xg). (ii) Show that the spaces X of 439K and X 2 of 439Q are realcompact. (iii) Show that c with the discrete topology, and N c with the product topology, are realcompact. 439Y Further exercises (a) (i) Show that a subset A of R is universally negligible iff f [A] is Lebesgue negligible for every continuous injective function f : R → R. (Hint: if ν is an atomless Borel probability measure on R, set f (x) = x + ν[0, x] for x ≥ 0, and show that µL f [E] ≥ νE for every Borel set E ⊆ [0, ∞[.) (ii) Let X be a separable metrizable space. Show that X is universally negligible iff f [X] is Lebesgue negligible for every injective Borel measurable function f : X → R. (b) For this exercise only, let us say that a ‘universally negligible measurable space’ is a pair (X, Σ) where X is a set and Σ a σ-algebra of subsets of X containing every countable subset of X such that there is no probability measure µ with domain Σ such that µ{x} = 0 for every x ∈ X. (i) Let X be a set, Σ a σ-algebra of subsets of X containing all countable subsets of X, A ⊆ X and ΣA the subspace σ-algebra. Show that (A, ΣA ) is universally negligible iff µ∗ A = 0 whenever µ is a probability measure with domain Σ which is zero on singletons. Show that if (X, Σ) is universally negligible so is (A, ΣA ). (ii) Let X and Y be sets, Σ and T σ-algebras of subsets of X and Y containing all appropriate countable sets, and f : X → Y a (Σ, T)-measurable function. Suppose that (Y, T) and (f −1 [{y}], Σf −1 [{y}] ) are universally negligible for every y ∈ Y . Show that (X, Σ) is universally negligible. (iii) Let X be a set and Σ a σ-algebra of subsets of X containing all countable subsets of X. Show that the set of those A ⊆ X such that (A, ΣA ) is universally negligible is a σ-ideal of subsets of X.

439 Notes

Examples

269

(c) Let X be an analytic space and A an analytic subset of X. Show that X \ A is universally negligible iff all the constituents of X \ A (for any Souslin scheme defining A) are countable. (d) (i) Let X be a metrizable space such that f [X] is Lebesgue negligible for every continuous function f : X → R. Show that X is universally negligible. (ii) Let X be a completely regular Hausdorff space such that f [X] is Lebesgue negligible for every continuous function f : X → R. Show that X is universally τ -negli