Topics in measure theory and real analysis

ATLANTIS S TUDIES IN M ATHEMATICS VOLUME 2 S ERIES E DITOR : J. VAN M ILL

Atlantis Studies in Mathematics Series Editor: J. van Mill, VU University Amsterdam, Amsterdam, the Netherlands (ISSN: 1875-7634)

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c ATLANTIS PRESS / WORLD SCIENTIFIC 

Topics in Measure Theory and Real Analysis Alexander B. Kharazishvili A. Razmadze Mathematical Institute Tbilisi Republic of Georgia

A MSTERDAM – PARIS

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Atlantis Studies in Mathematics Volume 1: Topological Groups and Related Structures - A. Arhangel’skii, M. Tkachenko

ISBN: 978-90-78677-20-8 ISSN: 1875-7634

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c 2009 ATLANTIS PRESS / WORLD SCIENTIFIC 

Preface

This book is concerned with questions of classical measure theory and related topics of real analysis. At the beginning, it should be said that the choice of material included in the present book was completely dictated by research interests and preferences of the author. Nevertheless, we hope that this material will be of interest to a wide audience of mathematicians and, primarily, to those who are working in various branches of modern mathematical analysis, probability theory, the theory of stochastic processes, general topology, and functional analysis. In addition, we touch upon deep set-theoretical aspects of the topics discussed in the book; consequently, set-theorists may detect nontrivial items of interest to them and find out new applications of set-theoretical methods to various problems of measure theory and real analysis. It should also be noted that questions treated in this book are related to material found in the following three monographs previously published by the author. 1) Transformation Groups and Invariant Measures, World Scientific Publ. Co., LondonSingapore, 1998. 2) Nonmeasurable Sets and Functions, North-Holland Mathematics Studies, Elsevier, Amsterdam, 2004. 3) Strange Functions in Real Analysis, 2nd edition, Chapman and Hall/CRC, Boca Raton, 2006. For the convenience of our readers, we will first, briefly and schematically, describe the scope of this book. In Chapter 1, we consider the general problem of extending partial real-valued functions which, undoubtedly, is one of the most important problems in all of contemporary mathematics and which deserves to be discussed thoroughly. Since the satisfactory solution to this task requires a separate monograph, we certainly do not intend on entering deeply into v

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various aspects of the problem of extending partial functions, but rather we restrict ourselves to several examples that are important for real analysis and classical measure theory and vividly show the fundamental character of this problem. The corresponding examples are given in Chapter 1 and illustrate different approaches and appropriate research methods. Notice that some of the examples presented in this chapter are considered in more details in subsequent sections of the book. Chapter 2 is devoted to a special, but very important, case of the extension problem for real-valued partial functions. Namely, we discuss therein several variants of the so-called measure extension problem and we pay our attention to purely set-theoretical, algebraic and topological aspects of this problem. In the same chapter, the classical method of extending measures, developed by Marczewski (see [234] and [235]), is presented. Also, a useful theorem is proved which enables us to extend any σ -finite measure μ on a base set E to a measure μ  on the same E, such that all members of a given family of pairwise disjoint subsets of E become μ  -measurable (see [1] and [13]). This theorem is then repeatedly applied in further sections of the book. In Chapters 3 and 4 we primarily deal with those measures on E which are invariant or quasi-invariant with respect to a certain group of transformations of E. It is widely known that invariant and quasi-invariant measures play a central role in the theory of topological groups, functional analysis, and the theory of dynamical systems. We discuss some general properties of invariant and quasi-invariant measures that are helpful in various fields of mathematics. First of all, we mean the existence and uniqueness properties of such measures. The problem of the existence and uniqueness of an invariant measure naturally arises for a locally compact topological group endowed with the group of all its left (right) translations. In this way, we come to the well-known Haar measure. The theory of Haar measure is thoroughly covered in many text-books and monographs (see, for instance, [80], [83], [182], [202]), so we leave aside the main aspects of this theory. But we present the classical Bogoliubov-Krylov theorem on the existence of a dynamical system for a oneparameter group of homeomorphisms of a compact metric space E. More precisely, we formulate and prove a significant generalization of the Bogoliubov-Krylov statement: the so-called fixed-point theorem of Markov and Kakutani ([93], [168]) for a solvable group of affine continuous transformations of a nonempty compact convex set in a Hausdorff topological vector space. In the same chapters, we distinguish the following two situations: the case when a given topological space E is locally compact and the case when E is not locally compact. The latter case involves the class of all infinite-dimensional Hausdorff topolog-

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ical vector spaces for which the problem of the existence of a nonzero σ -finite invariant (respectively, quasi-invariant) Borel measure needs a specific formulation. Some results in this direction are presented with necessary comments. Chapter 5 is concerned with measurability properties of real-valued functions defined on an abstract space E, when a certain class M of measures on E is determined. We introduce three notions for a given function f acting from E into the real line R. Namely, f may be (a) absolutely nonmeasurable with respect to M, (b) relatively measurable with respect to M, (c) absolutely (or universally) measurable with respect to M. We examine these notions and show their close connections with some classical constructions in measure theory. It should be pointed out that the standard concept of measurability of f with respect to a fixed measure μ on E is a particular case of the notions (b) and (c). In this case, the role of M is played by the one-element class {μ }. In Chapter 6 we discuss, again from the measure-theoretical point of view, some properties of the so-called step-functions. Since step-functions are rather simple representatives of the class of all functions (namely, the range of a step-function is at most countable), it is reasonable to consider them in connection with the measure extension problem. It turns out that the behavior of such functions is essentially different in the case of ordinary measures and in the case of invariant (quasi-invariant) measures. In Chapter 7, we introduce and investigate the class of almost measurable real-valued functions on R. This class properly contains the class of all Lebesgue measurable functions on R and has certain interesting features. A characterization of almost measurable functions is given and it is shown that any almost measurable function becomes measurable with respect to a suitable extension of the standard Lebesgue measure λ on R. Chapter 8 focuses on several important facts from general topology. In particular, Kuratowski’s theorem (see, for instance, [58], [101], [149]) on closed projections is presented with some of its applications among which we especially examine the existence of a comeager set of continuous nowhere differentiable functions in the classical Banach space C[0, 1]. Also, we prove a deep theorem on the existence of Borel selectors for certain partitions of a Polish topological space, which is essentially used in the sequel. In Chapter 9 the concept of the weak transitivity of an invariant measure is considered and its influence on the existence of nonmeasurable sets is underlined. Here it is vividly shown that some old ideas of Minkowski [173] which were successfully applied by him in

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convex geometry and geometric number theory, are also helpful in constructions of various paradoxical (e.g., nonmeasurable) sets. Actually, Minkowski had at hand all the needed tools to prove the existence of those subsets of the Euclidean space Rn (n  1), which are nonmeasurable with respect to the classical Lebesgue measure λn on Rn . Chapter 10 covers bad subgroups of an uncountable solvable group (G, ·). The term ”bad”, of course, means the nonmeasurability of a subgroup with respect to a given nonzero σ finite invariant (quasi-invariant) measure μ on G. We establish the existence of such subgroups of G and, moreover, show that some of them can be applied to obtain invariant (quasi-invariant) extensions of μ . So, despite their bad structural properties, certain nonmeasurable subgroups of G have a positive side from the view-point of the general measure extension problem. The next two chapters (i.e., Chapters 11-12) are devoted to the structure of algebraic sums of small (in a certain sense) subsets of a given uncountable commutative group (G, +). Recall that the first deep result in this direction was obtained by Sierpi´nski in his classical work [219] where he stated that there are two Lebesgue measure zero subsets of the real line R, whose algebraic (i.e., Minkowski’s) sum is not Lebesgue measurable. Let us stress that in [219] the technique of Hamel bases was heavily exploited and in the sequel such an approach became a powerful research tool for further investigations. We develop Sierpi´nski’s above-mentioned result and generalize it in two directions. Namely, we consider the purely algebraic aspect of the problem and its topological aspect as well. The difference between these two aspects is primarily caused by two distinct concepts of “smallness” of subsets of R. In Chapters 13 and 14 we turn our attention to Sierpi´nski-Zygmund functions [225] and study them from the point of view of the measure extension problem. It is well known that the restriction of a Sierpi´nski-Zygmund function to any subset of R of cardinality continuum is discontinuous. This circumstance directly implies that a Sierpi´nski-Zygmund function is nonmeasurable in the Lebesgue sense and, moreover, is nonmeasurable with respect to the completion of an arbitrary nonzero σ -finite diffused Borel measure on R. In addition, no Sierpi´nski-Zygmund function has the Baire property. We give two new constructions of Sierpi´nski-Zygmund functions. (1) The construction of a Sierpi´nski-Zygmund function which is absolutely nonmeasurable with respect to the class of all nonzero σ -finite diffused measures on R. (Notice that this result needs some extra set-theoretical axioms.)

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(2) The construction of a Sierpi´nski-Zygmund function which is relatively measurable with respect to the class of all translation-invariant extensions of the Lebesgue measure λ on R. (This result does not need any additional set-theoretical hypotheses.) Chapters 15-17 are similar to each other in the sense that the main topics discussed therein are connected with different constructions of nonseparable extensions of σ -finite measures. Among the results presented in these chapters, let us especially mention: (i) the construction (assuming the Continuum Hypothesis) of a nonseparable extension of the Lebesgue measure without producing new null-sets; (ii) the construction (also under some additional set-theoretical axioms) of nonseparable invariant extensions of σ -finite invariant measures by using their nontrivial ergodic components; (iii) the construction (assuming again the Continuum Hypothesis) of a nonseparable nonatomic left invariant σ -finite measure on any uncountable solvable group. In Chapter 18, we consider universally measurable additive functionals. The universal measurability is treated here in a generalized sense, namely, a real-valued functional f on a Hilbert space E is universally measurable if and only if for any σ -finite Borel measure

μ given on E, there exists an extension μ  of μ such that f becomes measurable with respect to μ  . It is established that there are universally measurable additive functionals which are everywhere discontinuous. This result may be regarded as a counter-version to the well-known statement (see, e.g., [97], [153], [154]), according to which any universally measurable (in the usual sense) additive functional on E is necessarily continuous. Chapter 19 is devoted to certain strange subsets of the Euclidean plane R2 . We discuss various properties of these paradoxical sets from the measure-theoretical view-point. In particular, a subset Z of R2 is constructed which is almost invariant under the group of all translations of R2 , is λ2 -thick (where λ2 stands for the two-dimensional Lebesgue measure on R2 ) and, in addition, has the property that for each straight line l in R2 , the intersection l ∩ Z is of cardinality strictly less than the cardinality of the continuum. By using this set Z, we define translation-invariant extensions of λ2 for which no analogue of the classical Fubini theorem can be valid. The final chapter is connected with certain restrictions of functions acting from R into R. The first examples of those restrictions of measurable functions, which are defined on sufficiently large subsets of R and have various nice properties, were given in widely known

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statements of real analysis. For instance, in accordance with the classical Luzin theorem (see, e.g., [16], [26], [65], [80], [161], [183], and [192]), every real-valued Lebesgue measurable function restricted to a certain set of strictly positive λ -measure becomes continuous (and an analogous purely topological result holds true in terms of the Baire property and category). We touch upon some other results in this direction. In particular, it is proved that every Lebesgue measurable function g : R → R is monotone on a nonempty perfect subset of R. At the same time, such a perfect set does not need to be of strictly positive λ -measure. The last circumstance is shown by considering Jarnik’s [88] continuous nowhere approximately differentiable function whose existence is a rather deep theorem of real analysis (cf. [33], [34], and [167]). In general, we tried to present the material in a self-contained form completely accessible to graduate and post-graduate students. Moreover, for the reader’s convenience, six Appendices are attached to the main text of this book, which can be read independently of the material covered in the chapters. In Appendix 1 some auxiliary set-theoretical facts and constructions are considered, which are essential in various sections of the book. Namely, elements of infinite combinatorics (e.g., infinite trees and K¨onig’s lemma), several delicate set-theoretical statements, and the existence of an uncountable universal measure zero subset of R are discussed. Appendix 2 is devoted to various theorems on the existence of measurable selectors. Results of this type are important and attractive and have found applications in numerous branches of modern mathematics. We begin with the Choquet theorem on capacities (see [24], [52], [187]) and show its close connection with statements about measurable selectors. Also, we present the following two fundamental results in this topic: the theorem of Kuratowski and Ryll-Nardzewski [151] and, as one of its consequences, the Luzin-Jankov-von Neumann theorem (see, e.g., [99]). In Appendix 3 deep properties of σ -finite Borel measures on metrizable topological spaces are examined. It is proved that if the topological weight of a metric space E is not measurable in the Ulam sense, then any σ -finite Borel measure on E admits a separable support (cf. [192]). This important fact is essentially used in Chapter 12. Appendix 4 contains a detailed proof of the existence of a continuous function f : R → R which is nowhere approximately differentiable. As mentioned above, the first example of such a function was constructed by Jarnik in his remarkable work [88]. A function of this type is needed for considerations in Chapter 20.

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In Appendix 5 general properties of commutative groups (primarily, infinite commutative groups) are examined and one useful theorem, due to Kulikov, on the algebraic structure of such groups is proved. This theorem is utilized in several parts of the main text of the book; see especially Chapters 10 and 11. Appendix 6 presents elements of classical descriptive set theory. Namely, we touch upon certain properties of Borel and analytic (Suslin) subsets of uncountable Polish spaces and apply those properties to the question of measurability of sets or functions. (In this context, Appendix 2 is also helpful.) The so-called separation principle, first introduced and extensively studied by Luzin and Sierpi´nski, receives special attention. Of course, our presentation of this material is concise and superficial. The standard monographs or text-books devoted to classical descriptive set theory are [99], [148], [150], [160], and [162]; see also Martin’s article in [10]. Finally, we would like to note that all sections of the book, including the Appendices, are provided with exercises which contain additional information concerning the questions under discussion. Some of the exercises are quite easy but some of them involve difficult mathematical facts and need intensive efforts for their solution. These more difficult exercises are marked by the symbol ∗ and we recommend that the reader solve them in order to better understand the subject. Finally, we also state in the text several unsolved problems which are motivated by (or closely connected with) topics presented in this book. The Bibliography consists of 251 titles and contains only the most relevant ones. Of course, it is far from being complete but rather provides a basic orientation to the subject in order to stimulate further interest of our readers in various questions of measure theory and real analysis. A.B. Kharazishvili

Contents

Preface

v

1.

The problem of extending partial functions . . . . . . . . . . . . . . . . . .

1

2.

Some aspects of the measure extension problem . . . . . . . . . . . . . . . 19

3.

Invariant measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4.

Quasi-invariant measures . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

5.

Measurability properties of real-valued functions . . . . . . . . . . . . . . . 79

6.

Some properties of step-functions connected with extensions of measures . . 97

7.

Almost measurable real-valued functions . . . . . . . . . . . . . . . . . . . 111

8.

Several facts from general topology . . . . . . . . . . . . . . . . . . . . . . 125

9.

Weakly metrically transitive measures and nonmeasurable sets . . . . . . . 145

10. Nonmeasurable subgroups of uncountable solvable groups . . . . . . . . . 159 11. Algebraic sums of measure zero sets . . . . . . . . . . . . . . . . . . . . . 177 12. The absolute nonmeasurability of Minkowski’s sum of certain universal measure zero sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 13. Absolutely nonmeasurable additive Sierpi´nski-Zygmund functions . . . . . 215 14. Relatively measurable Sierpi´nski-Zygmund functions . . . . . . . . . . . . 227 15. A nonseparable extension of the Lebesgue measure without new null-sets . 241 16. Metrical transitivity and nonseparable extensions of invariant measures . . 257 17. Nonseparable left invariant measures on uncountable solvable groups . . . 269 18. Universally measurable additive functionals . . . . . . . . . . . . . . . . . 281 19. Some subsets of the Euclidean plane . . . . . . . . . . . . . . . . . . . . . 297 20. Restrictions of real-valued functions . . . . . . . . . . . . . . . . . . . . . 313 Appendix 1. Some set-theoretical facts and constructions . . . . . . . . . . . . 339 Appendix 2. The Choquet theorem and measurable selectors . . . . . . . . . . 359 xiii

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Appendix 3. Borel measures on metric spaces . . . . . . . . . . . . . . . . . . 383 Appendix 4. Continuous nowhere approximately differentiable functions . . . 397 Appendix 5. Some facts from the theory of commutative groups . . . . . . . . 411 Appendix 6. Elements of descriptive set theory . . . . . . . . . . . . . . . . . 423 Bibliography

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Chapter 1

The problem of extending partial functions

There are several general concepts and ideas in contemporary mathematics which play a fundamental role in almost all of its branches. Among ideas of this kind, the concept of extending a given partial function is of undoubted interest and of paramount importance for various domains of mathematics. For instance, every working mathematician knows that the problem of extending partial functions is considered and intensively studied in universal algebra, general and algebraic topology, mathematical and functional analysis, as well as other fields. Of course, this problem has specific features in any of the above-mentioned disciplines and it frequently needs special approaches or appropriate research tools which are suitable only for a given situation and are applicable to concrete mathematical objects, for example, groups, topological spaces, ordered sets, differentiable manifolds, and other structures. However, this problem can also be examined from the abstract view-point and methodological conclusions of a general character can be made. Below, we touch upon different aspects of the problem and illustrate them by relevant examples. Some of those examples will be envisaged more thoroughly in subsequent sections of this book. The main goal of our preliminary consideration is to demonstrate how the problem of extending partial functions accumulates ideas from different areas of modern mathematics. The best known example of this type is the famous Tietze-Urysohn theorem which states that every real-valued continuous function defined on a closed subset of a normal topological space (E, T ) can be extended to a real-valued continuous function defined on the whole space E (see, for instance, [58], [101], and [148]). Another example of this kind is the classical Hahn-Banach theorem which states that a continuous linear functional defined on a vector subspace of a given normed vector space (E, || · ||) can be extended to a continuous linear functional having the same norm and defined on the whole E (see any text-book of functional analysis, for instance, [56], [57], A.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7_1, © 2009 Atlantis Press/World Scientific

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or [209]). The third example of this sort, although far from the main topics of topology and analysis, is the problem of extending a given partially recursive function to a recursive function. As known, the latter should be defined on the set N (= ω ) of all natural numbers. The existence of a partially recursive function that does not admit an extension to a recursive function is crucial for basic statements of mathematical logic and the theory of algorithms. It suffices to mention G¨odel’s incompleteness theorem of the formal arithmetic (see, for instance, [10] and [215]). Obviously, many other interesting and important examples can be pointed out in this context. The present book contains selected topics of measure theory which is a necessary part of modern mathematical and functional analysis. As is well known, ordinary measures are real-valued functions defined on certain classes of subsets of a given base set E and having the countable additivity property. Of course, in contemporary mathematics the so-called vector-valued measures and operator-valued measures are also extremely important and are used in many questions of analysis and the theory of stochastic processes, but we do not touch them in our further considerations. Here we would like to stress especially that topics presented in this book are primarily concentrated around the measure extension problem which plays a significant role in numerous questions of real analysis, probability theory, and set-theoretical topology. Actually, the measure extension problem will be central for us in most sections of the book. Consequently, it is reasonable to begin our preliminary discussion by considering several facts from mathematical analysis, which are closely connected to extensions of partial real-valued functions. Some of the facts listed below are fairly standard and accessible to average-level students. But among the presented facts the reader will also encounter those which are more important and deeper and which find applications in various domains of modern mathematics. Let R denote the real line and let X be an arbitrary subset of R. A function f : X → R is called a partial function acting from R into R. For this f , we may write f : R → R saying that f is a partial function whose domain is contained in R. As usual, we denote dom( f ) = X. If X = R, then we obviously have an ordinary function f : R → R.

The problem of extending partial functions

3

The symbol ran( f ) denotes the range of a partial function f , i.e., ran( f ) = { f (x) : x ∈ dom( f )}. If Y is any subset of R, then the symbol f |Y stands for the restriction of a partial function f to Y . As a rule, people working in classical mathematical analysis are often interested in the following general question. Does there exist an extension f ∗ : R → R of a partial function f : R → R such that f ∗ is defined on the whole R and has certain ”nice” properties? In particular, we may require that f ∗ should be differentiable or continuous or semicontinuous or monotone or convex or Borel measurable or Lebesgue measurable or should have the Baire property. (Notice that the Baire property can be regarded as a topological version of measurability; extensive material about this property is contained in remarkable books [148], [176], and [192].) An analogous question arises in a more general situation, e.g., for partial functions f acting from subsets of an abstract set E into R, where E is assumed to be endowed with some additional structure. In such a case an extension f∗ : E → R with dom( f ∗ ) = E must preserve a given structure on E or should be compatible, in an appropriate sense, with this structure. It is clear that questions of the above-mentioned kind quite frequently arise in mathematical analysis, general topology, and abstract algebra. Therefore, this topic is of interest for large groups of the working mathematicians. Below, we have made a small list of results in this direction, have commented on each of them or have given a necessary explanation, and have referred the reader to other related works in which extensions of partial functions are considered more thoroughly (see, [58], [70], [83], [101], and [148]). For the convenience of potential readers, the material below is presented in the form of examples of statements about extensions of real-valued partial functions. By the way, we think that in various lecture courses for students it is useful to provide them with additional information concerning extensions of partial functions. Such an approach essentially helps them to see more vividly deep connections and interactions between distinct fields of contemporary mathematics. In addition, the students should know that the general problem of extending partial functions is important for all mathematics because this problem almost permanently occurs in different mathematical branches and finds numerous applications.

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We start with a fundamental result of purely set-theoretical flavor. Example 1. Let {Xi : i ∈ I} be an arbitrary family of nonempty sets. A partial selector of (for) {Xi : i ∈ I} is any family {xi : i ∈ J} where J is a subset of I and xi ∈ Xi for all indices i ∈ J. If J = I, then {xi : i ∈ J} is said to be a selector of (for) the given family {Xi : i ∈ I}. The natural question arises whether every partial selector of {Xi : i ∈ I} can be extended to a selector of the same family. As is well known, this question is solved positively if and only if the Axiom of Choice (denoted by AC) is assumed. At the same time, even in a very special case card(I) = 2c , (∀i ∈ I)(card(Xi ) = 2), where c denotes the cardinality of the continuum, the question of the existence of a selector of {Xi : i ∈ I} is highly nontrivial. For instance, as shown by Sierpi´nski, in this case the positive answer to the question necessarily implies the existence of a subset of R which is not measurable in the Lebesgue sense (see [223] and references therein; cf. also [90]). The formal Zermelo-Fraenkel set theory (denoted usually by ZF) is the standard system of set-theoretical statements (axioms) without the Axiom of Choice (see, e.g., [10], [91], [145], [150], and [215]). The symbol ZFC stands for the theory ZF & AC. There are many set-theoretical assertions equivalent (within ZF) to AC, for instance, the Zorn lemma, Zermelo’s theorem on the existence of a well-ordering of any set, the equality a2 = a for all infinite cardinal numbers a, and so on. (In this connection, see especially [90], [150], and [223].) It is remarkable that some nontrivial and interesting equivalents of AC can be formulated in concrete mathematical disciplines. For example, in general topology we have the fundamental Tychonoff theorem stating that the product of any family of quasicompact spaces is a quasicompact space, too. As demonstrated by Kelley [100], this theorem is equivalent to AC within ZF theory. In linear algebra we have a very important theorem stating that every vector space (over an arbitrary field) possesses at least one basis. As shown by Blass [14], this theorem is also equivalent to AC within ZF theory. Additionally, the importance of AC in classical mathematical analysis is well known (see, e.g., the old extensive work by Sierpi´nski [217]). It suffices to remind that even a proof of the equivalence of the two standard definitions, due to Cauchy and Heine respectively, of the continuity at a point x ∈ dom( f ) of a partial function f : R → R needs some weak form of AC. The following simple result can be included in a beginner lecture course of real analysis.

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Example 2. Let a partial function f : R → R be given. Then it admits a continuous extension f ∗ : R → R defined on R if and only if for each open interval ]a, b[ ⊂ R, the restriction of f to the set dom( f ) ∩ ]a, b[ is uniformly continuous. More generally, let {]ai , bi [ : i ∈ I} be a family of open intervals in R such that R = ∪{]ai , bi [ : i ∈ I}. We can assert that a partial function f : R → R admits a continuous extension f ∗ with dom( f ∗ ) = R if and only if the restrictions of this f to all sets dom( f ) ∩ ]ai , bi [

(i ∈ I)

are uniformly continuous. The proof of this fact is quite easy. It suffices to take into account that any continuous real-valued function defined on a closed bounded subinterval of R is uniformly continuous. Also, it is not hard to give an example of a partial function f : R → R for which there exists a countable family {[ai , bi ] : i ∈ I} of segments such that R = ∪{[ai , bi ] : i ∈ I} and all restrictions f |(dom( f ) ∩ [ai , bi ]) are uniformly continuous, but f does not admit a continuous extension f ∗ with dom( f ∗ ) = R. The next example is far from trivial, however. Example 3. Let X and Y be two metric spaces and let Y be complete. Suppose also that f : X → Y is a continuous partial mapping. Then there exists a continuous partial mapping f ∗ : X → Y extending f and defined on some Gδ -subset of X; consequently, f ∗ can be extended to a Borel mapping acting from the whole X into Y . This result is due to Lavrentiev and has numerous applications in descriptive set theory and general topology (see [58], [148], [157], and Exercise 13 for Chapter 8 of this book). Example 4. Recall that a partial function f : R → R is upper (respectively, lower) semicontinuous if for each t ∈ R, the set {x ∈ dom( f ) : f (x) < t} (respectively, the set {x ∈ dom( f ) : f (x) > t}) is open in dom( f ). For f with dom( f ) = R, this definition is equivalent to the following: f is upper (respectively, lower) semicontinuous if and only if limsupy→x f (y) = f (x)

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(respectively, liminfy→x f (y) = f (x)) for all x ∈ R (see, e.g., [58], [101], [148], and [183]). It is interesting to notice that every bounded upper (lower) semicontinuous partial function admits a bounded upper (lower) semicontinuous extension defined on R. Let us formulate a more precise result in this direction. First, recall that a partial function g : R → R is locally bounded from above (from below) if for each point x ∈ R, there exists a neighborhood U(x) such that g|U(x) is bounded from above (from below). A partial function g : R → R is locally bounded if it is locally bounded from above and from below simultaneously. We also recall that an upper semicontinuous function can take its values from the set R ∪ {−∞} and a lower semicontinuous function can take its values from the set R ∪ {+∞}. (These assumptions are convenient in numerous topics of mathematical analysis.) Now, let f : R → R ∪ {−∞, +∞} be any partial function. The following two assertions are equivalent: (a) f admits an upper (lower) semicontinuous extension f ∗ whose domain coincides with the closure of dom( f ); (b) f is upper (lower) semicontinuous and locally bounded from above (from below). The equivalence of (a) and (b) implies the validity of the next two statements. (i) Let f : R → [a, b] be a partial upper semicontinuous function. Then there exists an upper semicontinuous function f ∗ : R → [a, b] extending f . (ii) Let f : R → [a, b] be a partial lower semicontinuous function. Then there exists a lower semicontinuous function f ∗ : R → [a, b] extending f . It should be mentioned that the same results hold true in a more general situation, namely, for partial semicontinuous bounded functions acting from a normal topological space E into the real line R. Of course, in this generalized case, the Tietze-Urysohn theorem should be applied to E in order to obtain the required result. The next example deals with monotone extensions of partial functions acting from R into R. Example 5. Let f : R → R be a partial function increasing on its domain. It is easy to show that f can always be extended to an increasing function f ∗ defined on some maximal (with respect to inclusion) subinterval of R. Let us denote the above-mentioned maximal subinterval by T and let a = inf(T ), b = sup(T ). If a = −∞ and b = +∞, then f ∗ is the required increasing extension of f defined on the

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whole R. If a = −∞, then in view of the maximality of T , we must have inf{ f (t) : t ∈ T } = −∞ and, therefore, f cannot be extended to an increasing function acting from R into R. Analogously, if b = +∞, then in view of the maximality of T , we must have sup{ f (t) : t ∈ T } = +∞ and therefore f cannot be extended to an increasing function acting from R into R. We see that in both of these cases our partial function f is not locally bounded. A similar result holds true for any decreasing partial function f : R → R. We thus obtain a necessary and sufficient condition for extending a given partial function to a monotone function acting from R into R. Namely, the following two assertions are equivalent: (a) f is extendable to a monotone function f ∗ with dom( f ∗ ) = R; (b) f is monotone and locally bounded. Of course, there is no problem connected with extending monotone partial functions if we admit infinite values of functions under consideration (cf. Example 4). Indeed, in such a case any monotone partial function f : R → R ∪ {−∞, +∞} can be extended to a monotone function f ∗ with dom( f ∗ ) = R. Example 6. Let f : R → R be a partial function. Suppose that f is Borel on its domain, i.e., for every Borel set B ⊂ R, the pre-image f −1 (B) is a Borel subset of dom( f ) where dom( f ) is assumed to be endowed with the induced topology. It can be proved that f always admits a Borel extension f ∗ : R → R with dom( f ∗ ) = R. However, the proof of this fact is far from being easy. It needs a certain classification of all Borel partial functions acting from R into R. This classification is due to Baire (see [3], [4], and [5]). According to it, any Borel partial function f has its own Baire order α = α ( f ), where α is some countable ordinal number. For instance, the equality α ( f ) = 0 simply means that f is continuous on its domain. Taking into account the above-mentioned classification and Lavrentiev’s result mentioned earlier (see Example 3), the existence of f ∗ can be established by using the method of transfinite induction on α (for more details, see [148] or Exercise 16 for Chapter 8). Example 7. Let f : R → R be a partial function. The following two assertions are equivalent: (a) f admits an extension f ∗ defined on R and measurable in the Lebesgue sense;

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(b) there exists a Lebesgue measure zero set A ⊂ R such that the restriction of f to the set dom( f ) \ A is a Borel function on dom( f ) \ A. This fact can be proved by using the well-known Luzin criterion for the Lebesgue measurability of real-valued functions (see, e.g., [161], [183], and [192]). A parallel fact holds true for partial functions having the Baire property. Similarly to the measurability in the Lebesgue sense, for a partial function f : R → R, the following two assertions are equivalent: (c) f admits an extension f ∗ defined on R and having the Baire property; (d) there exists a first category set B ⊂ R such that the restriction of f to the set dom( f ) \ B is a Borel function on dom( f ) \ B. In connection with the latter fact, let us remark that if X is a second category subset of R, then there always exists a function f : X → R which cannot be extended to a function f ∗ : R → R having the Baire property. This deep result is due to Novikov (see [188]). It is essentially based on the Axiom of Choice and some special facts from descriptive set theory (e.g., the separation principle for analytic sets). A detailed discussion of this result is also given in Chapter 14 of [122]. Example 8. If g : R → R is a Lebesgue measurable function (respectively, function having the Baire property), then there exists a nonempty perfect set P ⊂ R such that g|P is monotone on P (see Chapter 20 of this book). Sierpi´nski and Zygmund proved in their joint work [225] that there exists a function h:R→R satisfying the following condition: for any set X ⊂ R of cardinality continuum, the restriction h|X is not continuous. In particular, this condition readily implies that for any set X ⊂ R of cardinality continuum, the restriction h|X is not monotone on X. Indeed, it suffices to use the fact that the set of all discontinuity points of any monotone partial function acting from R into R is at most countable. Assuming the Continuum Hypothesis or, more generally, Martin’s Axiom (see [10], [40], [67], [91], and [145]), we have the following two statements. (a) If X ⊂ R is of second category, then h|X cannot be extended to a function on R having the Baire property (cf. Novikov’s result mentioned above). (b) If X ⊂ R is of strictly positive outer Lebesgue measure, then h|X cannot be extended to a function on R measurable in the Lebesgue sense.

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Notice that the validity of (a) and (b) does not need the full power of Martin’s Axiom. Actually, it suffices to assume that any subset of R of cardinality strictly less than c is of first category (respectively, of Lebesgue measure zero) in R. Statement (a) directly implies that no Sierpi´nski-Zygmund function has the Baire property. Statement (b) directly implies that no Sierpi´nski-Zygmund function is measurable in the Lebesgue sense. Moreover, we can assert that a Sierpi´nski-Zygmund function is nonmeasurable with respect to the completion of any nonzero σ -finite diffused (i.e., vanishing at all singletons) Borel measure on R (see Exercise 2 for Chapter 13). In addition to the above, no Sierpi´nski-Zygmund function is countably continuous. (A partial function f : R → R is called countably continuous if dom( f ) admits a representation in the form dom( f ) = ∪{Xn : n < ω }, where all restrictions f |Xn (n < ω ) are continuous.) Sierpi´nski-Zygmund functions have other interesting and important properties. Many works were devoted to functions of Sierpi´nski-Zygmund type (see [7], [42], [123], [124], [148], [185], and [199]). In the sequel, we will be dealing with Sierpi´nski-Zygmund functions possessing some additional properties which are of interest from the measuretheoretical point of view and are closely connected with the measure extension problem (see Chapters 13 and 14). Example 9. Consider the set R as a vector space over the field Q of all rational numbers. Let f : Q → Q denote the identity mapping. Since Q is a vector subspace of R (actually, Q can be treated as a ”line” in the infinite-dimensional vector space R) and f is a partial linear functional, it admits a linear extension f∗ : R → Q with dom( f ∗ ) = R. Of course, the construction of f ∗ is not effective because it is based on the Axiom of Choice or on the Zorn lemma (cf. Exercise 5 for this chapter). In fact, the obtained extension f ∗ is a solution of the Cauchy functional equation, that is we have f ∗ (x + y) = f ∗ (x) + f ∗ (y) for all x ∈ R and y ∈ R. At the same time, taking into account the relation ran( f ∗ ) = Q, we see that f ∗ is discontinuous at all points of R or, equivalently, f ∗ is a nontrivial solution of the Cauchy functional equation (cf. [18], [82], [108], [143], and [176]). It is well known that all nontrivial solutions of the Cauchy functional equation are nonmeasurable

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with respect to the Lebesgue measure on R and do not possess the Baire property (see, for instance, [143]). In our case, it can be observed that the set X = {x ∈ R : f ∗ (x) = 0} is not Lebesgue measurable and does not have the Baire property. Notice also that among nontrivial solutions of the Cauchy functional equation, we can encounter some Sierpi´nskiZygmund functions (see Chapters 13 and 14). In addition, it should be pointed out that certain nontrivial solutions of the Cauchy functional equation are successfully applied in some deep geometrical questions concerning equidecomposability of polyhedra lying in a finite-dimensional Euclidean space (see [18] where Hilbert’s third problem and related topics are discussed in detail). Example 10. In fact, the preceding example is purely algebraic. Another example of a similar type is the following. Let (G, +) be a commutative group and let (H, +) be a divisible commutative group, i.e., any equation of the form nx = h

(n ∈ N \ {0}, h ∈ H)

has a solution in H. Suppose that a partial homomorphism φ : G → H is given, which means that φ is a homomorphism from some subgroup of G into H. Then we can assert that there always exists a homomorphism φ ∗ : G → H extending φ (see [70], [83], [152], or Exercise 2 in Appendix 5). The above-mentioned result is extremely useful in all theory of commutative groups. For instance, by applying it, we may readily prove that every commutative group can be isomorphically embedded in some divisible commutative group. Since all divisible commutative groups admit a visual algebraic characterization (cf. [70] or [152]), the importance of this result becomes quite evident. An analogous statement concerning extensions of partial continuous homomorphisms can be formulated in the case of a compact commutative group (G, +) and the one-dimensional unit torus (S1 , ·) which is a canonical representative of the class of all divisible commutative compact groups (see [83], [177], and [202]). Example 11. We may expect that if a partial function f : R → R is defined on a small (in an appropriate sense) subset of R, then f admits extensions with ”nice” properties as well as extensions with ”bad” properties. Indeed, if dom( f ) is of Lebesgue measure zero (respectively, is of first category), then f trivially can be extended to a Lebesgue measurable function (respectively, to a function possessing the Baire property). Thus, in both of these cases, we come to extensions with “nice” properties.

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On the other hand, it is possible to present an example of an extension of a partial function which is defined on a small subset of R, with an extremely “bad” property from the measure-theoretical view-point. For this purpose, let us first introduce two auxiliary notions which play a significant role in the sequel. We recall that a set X ⊂ R is universal measure zero if there exists no nonzero σ -finite diffused Borel measure on X. Evidently, if X ⊂ R is countable, then X is universal measure zero. The question of the existence of uncountable universal measure zero subsets of R was posed many years ago and turned out to be closely connected with some rather delicate set-theoretical constructions due to Hausdorff, Mahlo, Luzin and other authors. For instance, every Luzin subset of R is uncountable and universal measure zero (various interesting properties of Luzin sets are considered in [143], [148], [159], [176], and [192]; see also Chapters 5, 12 and Appendix 1). We shall say that a function g : R → R is if there exists no nonzero σ -finite diffused measure

μ on R such that g is measurable with respect to μ . (It is assumed in this definition that the domain of μ may be an arbitrary σ -algebra of subsets of R, containing all singletons.) We thus see that absolutely nonmeasurable functions (if they exist) are of more pathological nature than well-known examples of Lebesgue nonmeasurable real-valued functions. It can be proved that the following two assertions are equivalent: (a) a function g : R → R is absolutely nonmeasurable; (b) the set ran(g) is universal measure zero and for each t ∈ R, the set g−1 (t) is at most countable. For the proof of the equivalence (a) ⇔ (b), see Chapter 5. Starting with this result, it is not hard to show that an absolutely nonmeasurable function acting from R into R exists if and only if there exists a universal measure zero subset of R of cardinality continuum. Also, we can infer the validity of the next statement. Under the Continuum Hypothesis (or, more generally, under Martin’s Axiom), for a partial function f : R → R, the following two assertions are equivalent: (c) f admits an absolutely nonmeasurable extension f ∗ with dom( f ∗ ) = R; (d) the set ran( f ) is universal measure zero and the sets f −1 (t) are countable for all t ∈ R. From the equivalence of (c) and (d) we also obtain, under the same additional set-theoretical assumptions, that any partial function f : R → R defined on a countable subset of R admits an extension to an absolutely nonmeasurable function defined on the whole R. Let us remark that according to Blumberg’s fundamental theorem [15], any function acting

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from R into R can be regarded as an extension of some continuous partial function whose domain is a countable everywhere dense subset of R (for the proof, see [15] or Exercises 21 and 22 of Chapter 8). The existence of a Sierpi´nski-Zygmund function h : R → R shows that under the Continuum Hypothesis, there is no uncountable set X ⊂ R such that h|X is continuous and, consequently, h cannot be considered as an extension of a continuous partial function defined on an uncountable subset of R. On the other hand, according to a recent result of Roslanowski and Shelah [208], it is consistent with ZFC that any function f : R → R may be regarded as an extension of a continuous function defined on a Lebesgue nonmeasurable subset of R which, obviously, is necessarily uncountable. Let us also notice that under Martin’s Axiom, there exist additive absolutely nonmeasurable functions which simultaneously are Sierpi´nski-Zygmund functions. One construction of such functions will be given later in this book (see Chapter 13). It is based on the fact that there exists a generalized Luzin subset of R which simultaneously is a vector space over the field Q of all rational numbers. The next example is concerned with extensions of real-valued partial functions of two variables. Example 12. Let λ (= λ1 ) denote the Lebesgue measure on the real line R (= R1 ). Consider a function of two real variables Φ : R × [0, 1] → R. Recall that Φ satisfies the Carath´eodory conditions if the following two relations hold: (i) for each x ∈ R, the function Φ(x, ·) : [0, 1] → R is continuous; (ii) for each y ∈ [0, 1], the function Φ(·, y) : R → R is λ -measurable. Functions of this type play a prominent role in mathematical analysis, the theory of ordinary differential equations, optimization theory, and probability theory. It is well known that if Φ satisfies the Carath´eodory conditions, then Φ is measurable with respect to the product σ -algebra dom(λ ) ⊗ B([0, 1]), where B([0, 1]) denotes the

σ -algebra of all Borel subsets of [0, 1]. Now, take a partial function of the form F : R × [0, 1] → R, i.e., suppose that dom(F) ⊂ R × [0, 1]. In addition, suppose that F is measurable with respect to the product σ -algebra dom(λ ) ⊗ B([0, 1]). Then the following two assertions are equivalent:

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(a) F admits an extension F ∗ : R × [0, 1] → R satisfying the Carath´eodory conditions and such that dom(F ∗ ) = R × [0, 1]; (b) for each x ∈ R, the partial function F(x, ·) is uniformly continuous on its domain. This result is rather deep. (The reader may compare it with the simplest Example 2.) Indeed, to establish the equivalence of (a) and (b), we have to apply the Choquet theorem on capacities and the theorem on measurable selectors due to Kuratowski and Ryll-Nardzewski (see Appendix 2). It should be mentioned that the same result remains true if we replace the Lebesgue measure space (R, dom(λ ), λ ) by an arbitrary σ -finite complete measure space (Ω, dom(μ ), μ ), simultaneously replace the segment [0, 1] by a compact metric space Y and consider partial functions of the form F : Ω × Y → R. For more details, see Exercise 13 of Appendix 2. Notice also that the standard Wiener process W : R[0,1] × [0, 1] → R, which is regarded as a reasonable mathematical model of the Brownian motion, yields a good example of a function of two variables, satisfying the Carath´eodory conditions. Here R[0,1] stands, as usual, for the space of all real-valued functions defined on the segment [0, 1], and this space is assumed to be equipped with the completion of the Wiener probability measure μw . More precisely, we are dealing with a certain set Ω ⊂ R[0,1] such that: (c) μw (Ω) = 1; (d) for any t ∈ [0, 1], the partial function W (·,t) is measurable with respect to μw ; (e) for any ω ∈ Ω, the trajectory W (ω , ·) is continuous on [0, 1]. Extensive information about stochastic processes and corresponding probability measures may be found in [16], [26], [52], [187], [205], and [226]. Now, we would like to consider a more difficult special case of the problem of extending partial functions. In fact, our last example will be concerned with extensions of real-valued partial set-functions. We recall that a set-function is any function whose domain is some family of sets. Equivalently, we may say that a set-function f is a function whose domain is a subset of the power set P(E) of some base set E. Thus, f can be treated as a partial function acting from the set E  = P(E). Of course, partial functions of the form f : P(E) → R

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are of prime interest in this book because they include measures and measure type functionals on E, for instance, capacities in the Choquet sense (see Appendix 2). Example 13. The Lebesgue measure λ is a real-valued function defined on some class of subsets of R. It is well known that the class dom(λ ) is properly contained in P(R), i.e., there are λ -nonmeasurable sets in R. However, the proof of this fact needs uncountable forms of the Axiom of Choice. Thus, λ may be regarded as a partial function acting from the power set of R into R ∪ {+∞}, i.e., λ is a partial function of the form

λ : P(R) → R ∪ {+∞}. Several constructions of λ -nonmeasurable subsets of R are presented in the literature which give us Vitali sets, Bernstein sets, nontrivial ultrafilters in the set N of all natural numbers, and other pathological sets of real numbers. Compare also Example 9, where the

λ -nonmeasurable set X associated with a certain nontrivial solution of the Cauchy functional equation was pointed out. It is natural to ask whether there exists a measure μ on R extending λ and defined on the family of all subsets of R. This problem was originally posed by Banach and then was studied by many famous mathematicians. Under some additional set-theoretical assumptions (e.g., the Continuum Hypothesis or Martin’s Axiom), the answer is negative (see, for instance, [67], [69], [78], [79], [91], [192], [222], [224], and [238]). But it seems that this question is undecidable within the standard system of axioms of contemporary set theory, e.g., within the Zermelo-Fraenkel formal system ZFC. It can be shown (see [1], [13], and Chapter 2) that for any countable disjoint family {Xi : i ∈ I} of subsets of R, there exists a measure λ  on R extending λ and such that {Xi : i ∈ I} ⊂ dom(λ  ). We thus see that the countable additivity property can be preserved by λ  under the assumption that the ”points” Xi (i ∈ I) onto which we extend λ are pairwise disjoint. Actually, it was established in [1] and [13] that the same result remains valid for any σ -finite measure

μ given on an abstract set E and for any disjoint family {Y j : j ∈ J} of subsets of E. In particular, we easily obtain from this result that for every finite family {Z1 , Z2 , ..., Zn } of subsets of E, there exists a measure μ  extending μ and satisfying the relation {Z1 , Z2 , ..., Zn } ⊂ dom(μ  ). However, if we have an arbitrary infinite sequence {Z1 , Z2 , ..., Zn , ...} of subsets of E, then we cannot assert, in general, that μ is extendable to a measure μ  for which all these subsets

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become μ  -measurable. In other words, sometimes there are countably many ”points” in the power set of E, which all together do not admit further extensions of a given nonzero

σ -finite measure μ . For instance, we always have such ”bad” points Z1 , Z2 , ..., Zn , ... in the power set of E where E is an arbitrary uncountable universal measure zero subset of R. It should be noticed that the existence of an uncountable universal measure zero set E ⊂ R is well known and can be proved within ZFC theory. In Appendix 1 of the present book we give a construction of E starting with the classical Sierpi´nski partition of the product set ω1 × ω1 , where ω1 denotes, as usual, the least uncountable cardinal number. In this context, the works [197] and [250] should also be mentioned, in which analogous and stronger results are obtained stating the existence of uncountable universally small subsets of R. Finishing this chapter, we hope that the examples just considered and concerning extensions of partial real-valued functions are sufficiently illustrative to show the reader the importance of the problem of extending partial functions. We will continue our discussion of this problem in the following sections. As already said, the measure extension problem touched upon in Example 13 will be of special interest in our further considerations. EXERCISES 1. Give a proof of the statement presented in Example 1. Namely, verify that the following two assertions are equivalent within ZF theory: (a) the Axiom of Choice; (b) if {Xi : i ∈ I} is an arbitrary family of nonempty sets, then any partial selector of {Xi : i ∈ I} can be extended to a selector of this family. Moreover, by using AC demonstrate that every infinite set contains a countably infinite subset. (Note that this fact is deducible with the aid of some weak forms of AC but is not provable within ZF theory.) 2. Let ]a, b[ be an open subinterval of R and let f : ]a, b[ → R be a convex function on ]a, b[. Show that the following two statements are equivalent: (1) there exists a convex extension f ∗ : R → R of f defined on the whole R; (2) f admits a continuous extension defined on [a, b] and this extension has the right-hand side derivative at a and the left-hand side derivative at b. Notice that the above-mentioned derivatives are assumed to be finite. 3. Give a proof of the equivalence (a) ⇔ (b), where (a) and (b) are assertions from Example 4.

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4. Give a proof of the equivalence (a) ⇔ (b), where (a) and (b) are assertions from Example 5. 5∗ . Show that the set X = {x ∈ R : f ∗ (x) = 0} described in Example 9 is not Lebesgue measurable and does not possess the Baire property. 6∗ . Let (E, d) be a metric space and let f : E → R be a partial function satisfying the Lipschitz condition with a Lipschitz constant L, that is | f (x) − f (y)|  Ld(x, y)

(x ∈ dom( f ), y ∈ dom( f )).

Prove that there exists an extension f ∗ : E → R of f also satisfying the Lipschitz condition with the same constant L and defined on the whole space E. 7. Let f : R → R be a function and let x ∈ R. Recall that t ∈ R is a Dini right derived number of f at x if there exists a sequence {hn : n ∈ N} of strictly positive real numbers such that limn→+∞ hn = 0, t = limn→+∞ ( f (x + hn ) − f (x))/hn . In this case, the notation t ∈ D+f (x) is frequently used. Let f : R → R be a continuous function. Suppose that for any x ∈ R, this f admits a nonnegative Dini right derived number at x. Show that f is increasing on R. Starting with this result, prove that the following two assertions are equivalent for a continuous function f acting from R into R: (a) f satisfies the Lipschitz condition; (b) there is a constant L  0 such that for any point x ∈ R, the function f has at least one Dini right derived number whose absolute value does not exceed this L. 8. Let Bn denote the unit ball in the Euclidean space Rn and let Sn−1 denote the boundary of Bn . Show that the following two assertions are equivalent: (a) the identity embedding of Sn−1 into Bn does not admit a continuous extension defined on the whole Bn ; (b) Bn has the fixed-point property which means that for every continuous mapping f : Bn → Bn , there exists a point x ∈ Bn such that f (x) = x. As is well known, assertion (b) is valid and was first proved by Brower (for the proof, see [56], [58], and [148]). Consequently, (a) is valid, too. Brower’s theorem (b) is very deep, has numerous applications, and was generalized in many directions (see, e.g., [56] and [57]). One of important generalizations of this theorem is due to Kakutani and states the existence of fixed-points for certain set-valued mappings.

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9∗ . Show that every partial isometry f : Rn → Rn admits an extension to an isometry f ∗ : Rn → Rn . Find necessary and sufficient conditions under which f ∗ is uniquely determined by f . On the other hand, for any infinite-dimensional real Hilbert space E, give an example of a partial isometry g : E → E which does not admit an extension to an isometry of the whole E. 10. A partially ordered set (E, ) is called complete if for any set X ⊂ E, there exists sup(X) (equivalently, there exists inf(X)). Let (E, ) be a complete partially ordered set and let φ : E → E be a monotone mapping (i.e., φ is either increasing or decreasing). Show that there exists a fixed-point of φ ; in other words, show that (∃y ∈ E)(φ (y) = y). For this purpose, assume without loss of generality that φ is increasing and put Y = {x ∈ E : x  φ (x)}, y = sup(Y ). Check that φ (y) = y, i.e., y is the desired fixed-point of φ . This result is due to Tarski and is known as Tarski’s fixed-point theorem. 11∗ . Let a function Φ : R × [0, 1] → R satisfy the Carath´eodory conditions (see Example 12 of this chapter). Prove that Φ is measurable with respect to the product σ -algebra dom(λ ) ⊗ B([0, 1]). 12∗ . Let (E, S , μ ) be a σ -finite measure space and let S0 = {X ∈ S : μ (X) < +∞}. Consider in S0 (which is a σ -ring of subsets of E) the equivalence relation R(X,Y ) defined by the formula R(X,Y ) ⇔ μ (XY ) = 0, where the symbol  stands for the operation of symmetric difference of two sets, that is XY = (X \ Y ) ∪ (Y \ X). Let S0 /R denote the corresponding quotient set and let, for any X ∈ S0 , the symbol [X] denote the R-equivalence class containing X. Verify that the function d([X], [Y ]) = μ (XY ) ([X] ∈ S0 /R, [Y ] ∈ S0 /R) is a metric on S0 /R such that the metric space (S0 /R, d) is complete.

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This (S0 /R, d) is usually called the metric space canonically associated with the given measure μ . The topological weight of this space (see Appendix 3) is a certain inner characteristic of μ . Denote it by w(μ ). Check that if w(μ )  ω , then w(μ ) is equal to the topological weight of the Hilbert space L2 (μ ) consisting of all real-valued square integrable (with respect to μ ) functions on E. A measure μ is called separable if w(μ )  ω . Otherwise, μ is called a nonseparable measure. Demonstrate that if S is a countably generated σ -algebra of subsets of E, then μ is a separable measure. Also, give an example which shows that the converse assertion is not valid. Observe that the completion of a separable measure is separable, too. Infer from this fact that the Lebesgue measure λn on the Euclidean space Rn is separable. As mentioned in Example 13 of this chapter, the one-dimensional Lebesgue measure

λ = λ1 cannot be extended (within ZFC theory) to a universal measure defined on P(R). However, there exist nonseparable extensions of λ which are invariant under the group of all translations of R (and even under the group of all isometries of R). Nonseparable σ -finite measures with some specific properties will be considered in Chapters 15, 16, and 17.

Chapter 2

Some aspects of the measure extension problem

In this chapter, we concisely present several versions of the general measure extension problem which is a special case of the problem of extending partial real-valued functions, envisaged in the preceding chapter. Undoubtedly, the measure extension problem is important in many questions of analysis and probability theory. Among various aspects of this problem the following three should be especially distinguished: purely set-theoretical, algebraic, and topological. Below, we will touch upon each of the above-mentioned aspects and will show their specific features. Since the measure extension problem has its origins in real analysis, namely, in classical theory of the Lebesgue measure λ (= λ1 ) on the real line R (= R1 ), we also schematically consider several constructions of proper extensions of λ and compare such extensions to each other. However, a more detailed consideration of those constructions will be given in subsequent sections. Let E be a set, S be an algebra of subsets of E, and let μ be a nonzero σ -finite measure on S . The general measure extension problem is to extend μ onto a maximally large class of subsets of E. This problem was originally formulated at the end of the 19th century, within the theory of real-valued functions. As is well known, it was partially solved by Lebesgue [158] at the beginning of the 20th century. Namely, starting with the classical Jordan measure, Lebesgue gave a construction of his measure λ for the real line R and, in a similar way, he introduced his measure λn for the n-dimensional Euclidean space Rn (see [80], [158], [183], [192], and [210]). Actually, Lebesgue’s above-mentioned construction extends the one-dimensional (respectively, multi-dimensional) Riemann integral to the one-dimensional (respectively, multi-dimensional) Lebesgue integral. The latter integral allows operation with a sufficiently large class of real-valued functions and turns out to be a necessary research tool in various problems of modern analysis and probability theory. Then Carath´eodory’s fundamental theorem followed, which deals with σ -finite measures given on abstract measurable spaces of type (E, S ). Let us recall the formulation of this A.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7_2, © 2009 Atlantis Press/World Scientific

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classical result. Theorem 1. Any σ -finite measure μ defined on an algebra S of subsets of E admits a unique measure μ  extending μ and defined on the generated σ -algebra σ (S ). Here we omit the standard proof of Carath´eodory’s theorem (see [16], [56], [65], [80], [83], [194], and [210]). This theorem shows that, without loss of generality, we can consider only those σ -finite measures which are defined on σ -algebras of sets. Afterwards, the measure extension problem found important applications in many other domains of mathematics, such as axiomatic set theory, general topology, functional analysis, probability theory, and the theory of stochastic processes. A sufficiently general method of extending measures was suggested by Marczewski (Szpilrajn) in his classical works [234] and [235]. We would like to describe this method because it will be used many times in our further considerations. Namely, let E be a set, μ be a nonzero σ -finite complete measure on some σ -algebra of subsets of E, and let I be a σ -ideal of subsets of E such that (∀Y ∈ I )(μ∗ (Y ) = 0), where μ∗ usually stands for the inner measure associated with μ . Denote S = dom(μ ) and consider the σ -algebra S  of subsets of E, generated by S ∪ I , i.e., S  = σ (S ∪ I ). Obviously, any set Z ∈ S  can be represented in the form Z = (X ∪Y1 ) \ Y2, where X ∈ S and both Y1 and Y2 are some members of I . Let us put

μ  (Z) = μ  ((X ∪Y1 ) \ Y2) = μ (X). It can easily be checked that the functional μ  is well defined on S  in the sense that the value μ  (Z) does not depend on the above-mentioned representation of Z, and μ  turns out to be a measure on S  extending μ . Also, it directly follows from the definition of μ  that (∀Y ∈ I )(μ  (Y ) = 0). Clearly, the measure μ  is a proper extension of μ if and only if there exists a μ nonmeasurable set belonging to the σ -ideal I .

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Observe that this construction generalizes the standard procedure of obtaining the completion of μ . If we have a single set Y ⊂ E with μ∗ (Y ) = 0, then we can extend a given measure μ to a measure μ  in such a way that μ  (Y ) = 0. Indeed, it suffices to consider the σ -ideal I of subsets of E, generated by the one-element family {Y }, that is I = {Z : Z ⊂ Y }, and to apply to this I the Marczewski method described above. The construction of μ  is, in fact, purely set-theoretical because no specific properties of (E, S , μ ) are utilized here. Notice that a purely set-theoretical, or abstract, aspect of the measure extension problem was deeply investigated by famous representatives of the Polish mathematical school (Banach, Kuratowski, Sierpi´nski, Ulam, Tarski, and Marczewski). They obtained a number of fundamental results in this direction which have stimulated further remarkable investigations in the subject by other mathematicians (see, e.g., [69], [91], [144], [150], and [231]). The structure of so-called large cardinals was deeply studied in those investigations. In particular, according to Ulam’s classical theorem [238], it is consistent with ZFC theory that the domain of any extension μ  of a nonzero σ -finite diffused (i.e., vanishing at all singletons) measure μ given on an uncountable set E cannot coincide with the power set P(E) of E. Consequently, there always exists a set X ⊂ E such that X ∈ dom(μ  ). Then μ  can be extended to a measure μ  so that X becomes μ  -measurable and, in general, there are various possibilities to construct such an extension μ  (see Exercise 1 for this chapter). Thus, by starting with the assumption that there are no large cardinals, we can infer that there are no maximal extensions of the original nonzero σ -finite diffused measure μ . Extensive information about large cardinals and their combinatorial properties can be found in [10], [69], [91], [145], and [150] (see also Appendix 1). For our purposes, we need to look at one important result on extensions of σ -finite measures, obtained in the works [1] and [13]. Theorem 2. Let E be a set, μ be a σ -finite measure on E and let {Xi : i ∈ I} be an arbitrary disjoint family of subsets of E. Then there exists a measure ν on E extending μ and satisfying the relation {Xi : i ∈ I} ⊂ dom(ν ).

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Proof. Obviously, we may suppose (without loss of generality) that the given family {Xi : i ∈ I} is a partition of E. Let us first consider the case where the partition {Xi : i ∈ I} is countable, i.e., we have card(I)  card(N) = ω . Let {ti : i ∈ I} be an injective family of real numbers and let f : E → R be a step-function such that ran( f |Xi ) = {ti } for any i ∈ I. Clearly, it suffices to show that there exists an extension of μ for which f becomes measurable. Without loss of generality, we may assume that either I = {1, 2, ..., m} or I = N = ω . Under this assumption, let us put: Xn = a μ -measurable hull of Xn , where n ∈ I; Yn = Xn \ (∪{Xk : k < n}), where n ∈ I. Obviously, the family of sets {Yn : n ∈ I} is a disjoint covering of E. Define a new function f0 : E → R by putting f0 (x) = tn if and only if x ∈ Yn . Since I is countable and all sets Yn (n ∈ I) are

μ -measurable, it immediately follows from the definition of f0 that f0 is a μ -measurable function. Let us show that

μ∗ ({x ∈ E : f (x) = f0 (x)}) = 0. Suppose otherwise, that is

μ∗ ({x ∈ E : f (x) = f0 (x)}) > 0. Then there exists an index n ∈ I such that

μ∗ (Yn ∩ {x ∈ E : f (x) = f0 (x)}) > 0. On the other hand, it is easy to verify the following inclusion: Yn ∩ {x ∈ E : f (x) = f0 (x)} ⊂ Xn \ Xn, which gives a contradiction with the definition of the measurable hull Xn of Xn . The contradiction obtained establishes the required equality

μ∗ ({x ∈ E : f (x) = f0 (x)}) = 0. But, by virtue of the Marczewski method of extending σ -finite measures, the set X = {x ∈ E : f (x) = f0 (x)}

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can be made measurable with respect to some extension ν of μ and, moreover, this procedure can be made so that ν (X) = 0. Consequently, f becomes measurable with respect to the same ν . Let now {Xi : i ∈ I} be an arbitrary partition of E. Using the σ -finiteness of μ which implies the validity of the so-called countable chain condition, it is not difficult to show that there exists a set J ⊂ I satisfying the following two relations: (a) card(I \ J)  ω ; (b) for any countable set J0 ⊂ J, we have μ∗ (∪{X j : j ∈ J0 }) = 0. Starting with (b) and applying again the Marczewski method, we first make all the sets X j ( j ∈ J) to be measurable with respect to some extension μ  of μ . Notice, by the way, that according to this method, μ  (X j ) = 0 for all j ∈ J. Finally, we apply our previous argument to the countable disjoint family {Xi : i ∈ I \ J} and extend μ  to a measure μ  such that {Xi : i ∈ I \ J} ⊂ dom(μ  ). It is clear now that ν = μ  turns out to be the required extension of μ which ends the proof of the theorem. Remark 1. Theorem 2 shows, in particular, that having any finite family of subsets of E, we can always extend μ to a measure μ  , which makes all these subsets to be μ  -measurable (cf. also Exercise 1). On the other hand, it is well known that an analogous assertion fails to be true for some countable families of subsets of E, where card(E) = ω1 (see Appendix 1). Moreover, the existence of a Luzin set L ⊂ R with card(L) = c readily implies that there is a countably generated σ -algebra of subsets of R containing all Borel sets in R and not admitting a nonzero σ -finite continuous (i.e., diffused) measure. Recall that the existence of a Luzin set necessarily needs additional set-theoretical axioms. In this context, it should be underlined that the existence of other small subsets of R having cardinality ω1 , can be established within the theory ZFC (see [69], [79], [113], [148], [172], [197], and [250]). Some questions of this type will be discussed in more details in our further considerations (see especially Appendix 1). Another aspect of the measure extension problem has an algebraic (more precisely, grouptheoretical) flavor. Namely, suppose that E is an arbitrary set and denote by Sym(E) the family of all bijections of E onto itself. Obviously, Sym(E) becomes a group with respect to the standard composition operation ◦. This group is usually called the symmetric group of E. Any subgroup G of Sym(E) is called a group of transformations of E.

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We shall say that the pair (E, G) is a space equipped with a transformation group if G is a subgroup of Sym(E). Spaces equipped with transformation groups can be frequently met in algebra, geometry, topology, mathematical analysis, the theory of dynamical systems, and in other fields of mathematics. Let G be a group of transformations of E and let L be a family of subsets of E. We shall say that L is invariant under G or, in short, G-invariant if (∀g ∈ G)(∀X ∈ L )(g(X) ∈ L ). We shall say that a measure μ on E is invariant under G or, in short, G-invariant if dom(μ ) is a G-invariant class of subsets of E and (∀g ∈ G)(∀X ∈ dom(μ ))(μ (g(X)) = μ (X)). More generally, we shall say that a measure μ on E is quasi-invariant under G or, in short, G-quasi-invariant if dom(μ ) is a G-invariant class of subsets of E and (∀g ∈ G)(∀X ∈ dom(μ ))(μ (g(X)) = 0 ⇔ μ (X) = 0). In many situations a base set E is a group with respect to some algebraic operation ·, i.e., the pair (E, ·) is an abstract group. Then the group G of all left (respectively, right) translations of E is canonically isomorphic to (E, ·) and can be identified with E. So we may write (G, ◦) instead of (E, ·). In such a case, we deal with left invariant (left quasi-invariant) measures on E and, analogously, with right invariant (right quasi-invariant) measures on the same E. Actually, in many cases it suffices to consider only left invariant (left quasiinvariant) measures on groups. Obviously, if (E, ·) is a commutative group, then the concepts of left invariant (left quasiinvariant) and right invariant (right quasi-invariant) measures do not differ from each other. Suppose that an uncountable group (G, ·) is given and suppose that G is equipped with a nonzero σ -finite left G-invariant (or, more generally, nonzero σ -finite left G-quasiinvariant) measure μ . As stated by Kharazishvili [104] and Erd¨os and Mauldin [60], the domain of such a μ cannot be identical with the family P(G) of all subsets of G. Notice especially that this statement does not need any additional set-theoretical assumptions (see proof sketched in Exercises 9, 10 and 11 of Appendix 1). In view of the above-mentioned statement, it is natural to ask whether there exists a left G-invariant (left G-quasi-invariant) measure μ  on G properly extending μ . It is consistent with ZFC theory that the answer to this question is positive (see Exercise 20 for Chapter 3) but it is still unknown whether the same result can be obtained within ZFC. However, without using additional

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set-theoretical hypotheses, for some sufficiently wide classes of uncountable groups this question has a positive answer even in terms of certain subgroups of the original group. In particular, if (G, ·) is an arbitrary uncountable solvable group, then there always exists a

μ -nonmeasurable subgroup H of G such that μ can be extended to a left G-invariant (left G-quasi-invariant) measure μ  for which we have H ∈ dom(μ  ) and μ  (H) = 0. A more detailed information about this relatively recent result can be found in Chapter 10. Marczewski’s method of extending σ -finite measures successfully works in the cases of invariant and quasi-invariant measures but needs a slight modification (cf. [41], [42], [45], [108], [115], [195], [242], [247], and [249]). Namely, let E be a set, G be a group of transformations of E, and let μ be a σ -finite complete G-invariant (G-quasi-invariant) measure on some G-invariant σ -algebra of subsets of E. Suppose that a G-invariant σ -ideal I of subsets of E is given such that (∀Y ∈ I )(μ∗ (Y ) = 0). Again, let us denote S = dom(μ ) and consider the σ -algebra S  of subsets of E, generated by S ∪ I . It can easily be seen that S  is also a G-invariant σ -algebra of sets. We already know that any set Z ∈ S  can be represented in the form Z = (X ∪Y1 ) \ Y2, where X ∈ S and both Y1 and Y2 are some members of I . Let us put

μ  (Z) = μ  ((X ∪Y1 ) \ Y2) = μ (X). Again, it is not difficult to check that the functional μ  is well defined in the sense that the value μ  (Z) does not depend on the above-mentioned representation of Z, and μ  turns out to be a G-invariant (G-quasi-invariant) measure on S  extending μ . In addition to this circumstance, the equality μ  (Y ) = 0 is valid for all sets Y belonging to I . Consider now a particular case, when we have a single set Y ⊂ E with μ∗ (Y ) = 0 which satisfies the relation (∀g ∈ G)(μ ∗ (g(Y )Y ) = 0), where μ ∗ stands for the outer measure associated with μ (any set Y ⊂ E for which this relation holds is usually called almost G-invariant with respect to μ ). Then we can extend a given σ -finite G-invariant (G-quasi-invariant) measure μ to a G-invariant (G-quasiinvariant) measure μ  in such a way that μ  (Y ) = 0. Indeed, it suffices to consider the

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G-invariant σ -ideal I of subsets of E, generated by the one-element family {Y }, and to apply to I the above-mentioned construction (cf. [85], [198], [234], and [235]). The third aspect of the measure extension problem is purely topological. Let us recall that most measures considered in mathematical analysis, general topology, and probability theory are regular in an appropriate sense (see, e.g., [16], [26], [56], [65], [71], [80], and [83]). For example, if a σ -finite measure μ is given on a Hausdorff topological space E and for each set X ∈ dom(μ ), the equality

μ (X) = sup{μ (K) : K ∈ dom(μ ), K ⊂ X, K is compact} holds true, then μ is said to be a Radon measure. Radon measures on locally compact topological spaces play an extremely important role in various questions of analysis and probability theory. A wider class of measures constitute the so-called perfect measures which were introduced by Gnedenko and Kolmogorov (see Exercise 17 for this chapter). There are several deep statements in topological measure theory concerning those extensions of measures which preserve the regularity property. In order to present one of the results, we need A.D. Alexandrov’s theorem stating the countable additivity of any finite nonnegative finitely additive Radon type functional given on a Hausdorff space (see, for instance, [56]). Theorem 3. Let E be a Hausdorff topological space and let μ be a finite nonnegative finitely additive functional on some algebra of subsets of E. Suppose that μ is regular in the Radon sense, i.e., for any set Z ∈ dom(μ ), we have

μ (Z) = sup{μ (K) : K ∈ dom(μ ), K ⊂ Z, K is compact}. Then μ is countably additive or, in other words, μ is a measure. Proof. Notice first that for every set Z ∈ dom(μ ), we have

μ (Z) = inf{μ (U) : U ∈ dom(μ ), Z ⊂ U, U is open}. This fact directly follows from our assumptions on μ if we apply them to the set E \ Z and take into account the circumstance that every compact set in the given Hausdorff space E is a closed subset of E. Let now {Xn : n < ω } be a disjoint countable family of sets which all belong to dom(μ ) and suppose that their union X = ∪{Xn : n < ω }

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also belongs to dom(μ ). The finite additivity and nonnegativity of μ readily imply the inequality

∑{μ (Xn) : n < ω }  μ (X). Therefore, it suffices to prove the reverse inequality

μ (X)  ∑{μ (Xn ) : n < ω }. Fix a strictly positive real number ε . Obviously, there exists a compact set K ⊂ X such that

μ (X) − μ (K) < ε . Further, for each n < ω , we can find an open set Un ⊂ E containing Xn and such that

μ (Un ) − μ (Xn) < ε /2n . Since K is compact and the family {Un : n < ω } is an open covering of K, there exists a natural number m satisfying the relation K ⊂ U0 ∪U1 ∪ ... ∪Um . This relation implies

μ (X)  ε + μ (K)  ε + (μ (X0) + μ (X1 ) + ... + μ (Xm)) + 2ε  3ε + ∑{μ (Xn ) : n < ω }. Since ε may be arbitrarily small, we obtain the required reverse inequality and, therefore, the countable additivity of μ which completes the proof. From Theorem 3 one can readily deduce the following statement due to Henry and concerning extensions of finite regular measures. Theorem 4. Let E be a Hausdorff topological space and let μ be any σ -finite Radon measure defined on a σ -subalgebra of the Borel σ -algebra B(E). Then μ can be extended to a Radon measure defined on the whole Borel σ -algebra B(E). Proof. It suffices to consider the case when μ is a finite measure. Denote by M the family of all those nonnegative finitely additive Radon type functionals which extend μ and are defined on algebras of subsets of E. A direct verification shows that M is an inductive partially ordered set with respect to the relation R(μ  , μ  ) : μ  is a restriction o f μ  . By virtue of the Zorn lemma, there exists a maximal element in the above-mentioned partially ordered set. Denote such a maximal element by ν . Using Theorem 1 and Theorem 3,

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we first derive that ν is a measure and dom(ν ) is a σ -algebra of subsets of E. Moreover, we may assume without loss of generality that ν is complete. Now, we assert that dom(ν ) contains the whole Borel σ -algebra B(E). Suppose otherwise, that is / B(E) \ dom(ν ) = 0. This assumption implies that there exists a closed subset Z of E such that Z ∈ dom(ν ) (because B(E) is generated by the family of all closed subsets of E). In view of Exercise 1, we may reduce our consideration to the case where

ν∗ (Z) = ν∗ (E \ Z) = 0. In this case, on the σ -algebra generated by dom(ν ) ∪ {Z} we define a functional ν  by the formula

ν  ((Z ∩ X) ∪ ((E \ Z) ∩Y )) = ν (X), where X and Y are arbitrary elements of dom(ν ). It can readily be checked that the definition of ν  is correct and ν  is a nonnegative finitely additive Radon type functional strictly extending ν . But this circumstance contradicts the maximality of ν . The contradiction obtained finishes the proof. Remark 2. Alexandrov’s and Henry’s theorems are important and are successfully applied in various questions of measure theory. One typical application of Henry’s theorem is given in Exercise 10 of this chapter. The natural and interesting question arises whether any σ finite Borel measure on a complete metric space is Radon. It turns out that this question is closely connected with some set-theoretical assumptions (for more details, see Appendix 3). We have mentioned three aspects of the measure extension problem. But, of course, various combinations of these aspects can be considered. For instance, in our further constructions, we will be dealing with certain algebraic-topological aspects of the problem (cf. also [249]). In particular, we will encounter some extensions of the Lebesgue measure λ on the real line R, which are nonseparable (a topological property) and, simultaneously, invariant under the group of all translations of R (an algebraic property). In this connection, let us briefly describe one useful method of extending measures by applying those mappings whose graphs are thick in the measure-theoretical sense. This method was originally introduced by Kodaira and Kakutani in their famous construction of a nonseparable translation-invariant extension of λ (see [141] and Chapter 16).

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Let (E1 , S1 , μ1 ) and (E2 , S2 , μ2 ) be two measure spaces such that μ1 is σ -finite and μ2 is a probability measure. Let f : E1 → E2 be a mapping, Gr( f ) denote the graph of f , and let μ1 ⊗ μ2 be the product measure whose corresponding factors are μ1 and μ2 . Suppose that the set Gr( f ) is (μ1 ⊗ μ2 )-thick in the product space E1 × E2, which means that (μ1 ⊗ μ2 )∗ ((E1 × E2) \ Gr( f )) = 0. For any set Z ∈ dom(μ1 × μ2 ), let us define Z  = {x ∈ E1 : (x, f (x)) ∈ Z}. Further, we introduce the class of sets S1 = {Z  : Z ∈ dom(μ1 ⊗ μ2 )}. A straightforward verification shows that the following auxiliary proposition holds true. Lemma 1. S1 is a σ -algebra of subsets of E1 , which contains S1 . Let us define a functional μ1 on S1 by the formula

μ1 (Z  ) = (μ1 ⊗ μ2 )(Z) (Z ∈ dom(μ1 ⊗ μ2 )). Taking into account the thickness of the graph Gr( f ) in the product space E1 × E2 and the equality μ2 (E2 ) = 1, we easily get the next auxiliary proposition. Lemma 2. The functional μ1 is well defined, countably additive, and extends the given measure μ1 . We thus conclude that μ1 is a measure on S1 extending the original measure μ1 . It can also be readily verified that the following statement is valid. Lemma 3. The mapping f turns out to be measurable with respect to the σ -algebras S2 and S1 . In other words, for any set Y ∈ S2 , we have f −1 (Y ) ∈ S1 . As already said, the proofs of Lemmas 1, 2, and 3 are not difficult and the reader may try to prove these auxiliary propositions himself (herself). We will return to these lemmas in the sequel and will present some of their variations. But here we would like to give one immediate application of the above-mentioned lemmas. For this purpose, let us consider the concrete particular case, where E1 = E2 = R,

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and let us put:

μ1 = the Lebesgue measure λ on R; μ2 = a diffused probability Borel measure on R. By using a Bernstein type transfinite argument (see, for instance, [12], [30], [40], [99], [148], [150], [176], [192], or Chapter 5), a function f : R → R can be constructed, whose graph is thick with respect to the product measure μ1 ⊗ μ2 and, consequently, f is nonmeasurable with respect to λ (cf. Exercise 8). On the other hand, by virtue of Lemma 3, we have the following statement. Theorem 5. f is measurable with respect to some proper extension λ  of λ . The theorem just formulated is a typical representative of those results which are connected with functions having thick graphs and with corresponding extensions of measures. This approach will be developed in our further considerations (see especially Chapters 5 and 7). Actually, Theorem 5 may be regarded as a first step in much more complicated constructions, so it plays only an illustrative role here. For a while, let us return to the algebraic aspect of the measure extension problem. In several sections of this book, we will need the next auxiliary proposition. Lemma 4. Let E be a set, G be a group of transformations of E, and let μ be a σ -finite G-quasi-invariant measure on E. If X is an arbitrary μ -measurable set, then there exists a countable family {gi : i ∈ I} ⊂ G such that the set Y = ∪{gi (X) : i ∈ I} is almost G-invariant with respect to μ , i.e., we have the relation (∀g ∈ G)(μ (g(Y )Y ) = 0). The proof of this lemma is based on the method of transfinite induction. A sketch of the corresponding argument is given in Exercise 13 below. We leave to the reader to complete all omitted details. Let E be a set and let G be a group of transformations of E. Consider the class of all nonzero σ -finite G-invariant (or, more generally, G-quasi-invariant) measures on E and pose for them the measure extension problem. In other words, we are interested whether any nonzero σ -finite G-invariant (respectively, G-quasi-invariant) measure on E admits a proper extension belonging to the same class. Of course, in this case the problem has specific features and its solution essentially depends on structural properties of the pair

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(E, G). In our further considerations, we will be dealing with this problem many times. For its solving within ZFC theory (in certain particular cases), we will need the notion of absolutely negligible sets. These sets are very small from the view-point of the theory of invariant (quasi-invariant) measures (see [108], [115], [119], and [198]). Let us give the precise definition of such sets. Let M be a class of σ -finite G-invariant (G-quasi-invariant) measures on E. We say that a set Z ⊂ E is absolutely negligible with respect to M if for every measure μ ∈ M, there exists a measure ν ∈ M satisfying the following two conditions: (1) ν is an extension of μ ; (2) Z ∈ dom(ν ) and ν (Z) = 0. It immediately follows from this definition that if Z is absolutely negligible with respect to M and μ is an arbitrary measure from M, then Z ∈ dom(μ ) ⇒ μ (Z) = 0. However, the validity of the above implication for all μ ∈ M is not sufficient to assert that Z is absolutely negligible with respect to M. Suppose now that M coincides with the class of all σ -finite G-invariant (G-quasi-invariant) measures on E and let Z ⊂ E. If Z is absolutely negligible with respect to this M, then we shall say that Z is a G-absolutely negligible subset of E. Fortunately, it is possible to give a characterization of all G-absolutely negligible subsets of E only in terms of the pair (E, G). Namely, we have the following statement which turns out to be rather useful in further considerations (see [108], [109], and [115]). Theorem 6. Let (E, G) be a space equipped with a transformation group and let Z be a subset of E. Then these two assertions are equivalent: 1) Z is a G-absolutely negligible set; 2) for any countable family {gn : n < ω } of transformations from the group G, there exists a countable family {hm : m < ω } of transformations from the same group, such that ∩{hm (∪{gn (Z) : n < ω }) : m < ω } = 0. / Proof. Let the given set Z satisfy 1) and suppose that 2) is not true. Then there exists a countable family {gn : n < ω } of transformations from G such that for the set Z  = ∪{gn (Z) : n < ω } and for an arbitrary countable family {hm : m < ω } of transformations from G, we have / ∩{hm (Z  ) : m < ω } = 0.

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Let us denote Z  = E \ Z  and consider the G-invariant σ -ideal I of subsets of E, generated by the one-element family {Z  }. Obviously, we can define a complete probability G-invariant measure μ on E such that I = I (μ ), where I (μ ) stands for the σ -ideal of all μ -measure zero subsets of E. In particular, we have the relations Z  ∈ dom(μ ), Z  ∈ dom(μ ),

μ (Z  ) = 0, μ (Z  ) = 1. According to 1), there exists a G-quasi-invariant extension ν of μ such that Z ∈ dom(ν ), ν (Z) = 0. Consequently, we get

ν (Z  ) = ν (∪{gn (Z) : n < ω }) = 0 which contradicts the relation

ν (Z  ) = μ (Z  ) = 1. This contradiction establishes the validity of the implication 1) ⇒ 2). Suppose now that the set Z satisfies 2). Let μ be an arbitrary σ -finite G-quasi-invariant measure on E. We shall prove that there exists a G-quasi-invariant measure ν on E extending μ and such that the set Z is measurable with respect to ν and ν (Z) = 0. In order to show this fact, let us denote by J the G-invariant σ -ideal of subsets of E, generated by the one-element family {Z}. Taking 2) into account and applying Lemma 4, we easily infer that for each set X ∈ J , the equality μ∗ (X) = 0 holds true. Consequently, according to the Marczewski method, the measure μ can be extended to some G-quasi-invariant measure ν on E such that J ⊂ dom(ν ) and ν (X) = 0 for all sets X ∈ J . In particular, we obtain that Z ∈ dom(ν ) and ν (Z) = 0. Finally, let us observe that if the initial measure μ is G-invariant, then the extended measure

ν can be chosen to be G-invariant, too. We will return to absolutely negligible sets in subsequent parts of the book. EXERCISES 1∗ . Let E be a set and let μ be a finite complete measure on E. Suppose that there exists a set X ⊂ E nonmeasurable with respect to μ . Show that there are continuumly many measures

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μ  extending μ for which X becomes measurable and the values μ  (X) are distinct for different μ  . Argue in the following manner. First, observe that μ∗ (X) < μ ∗ (X). Let Y be a μ measurable kernel of X and let Z be a μ -measurable hull of X. Consider the sets U = X \ Y, V = Z \ Y. Check that the following relations hold: V ∈ dom(μ ), U ⊂ V, μ∗ (U) = μ∗ (V \ U) = 0. Keeping in mind this circumstance, taking V as E, and taking U as X, reduce the argument to the case where

μ∗ (X) = μ∗ (E \ X) = 0. In the latter case, fix a real number t ∈ [0, 1] and put

μt ((A ∩ X) ∪ (B ∩ (E \ X))) = t μ (A) + (1 − t)μ (B), where A and B are arbitrary μ -measurable subsets of E. Verify that μt is a measure on the σ -algebra σ (dom(μ ) ∪ {X}) and that μt is an extension of μ . In addition to this fact, check that if r ∈ [0, 1] and t = r, then μt (X) = μr (X). 2. Let (E, S , μ ) be a finite measure space and let X ⊂ E. We say that X has the uniqueness property with respect to μ if the following two conditions hold: (a) there exists a measure ν on E extending μ and such that X ∈ dom(ν ); (b) for any two measures ν1 and ν2 on E which extend μ and satisfy the relation X ∈ dom(ν1 ) ∩ dom(ν2 ), the equality ν1 (X) = ν2 (X) is valid. Denote by the symbol U (μ ) the class of all those subsets of E which have the uniqueness property with respect to μ . Let μ  be the completion of μ . Prove that the equality U (μ ) = dom(μ  ) is true. On the other hand, show that this equality does not always hold for σ -finite measures μ on E. 3. Let E be a Hausdorff topological space and let μ be a finite nonnegative finitely additive functional on B(E) such that

μ (U) = sup{μ (K) : K is compact, K ⊂ U}, μ (X) = inf{μ (V ) : V is open, X ⊂ V } for each open set U ⊂ E and for each Borel set X ⊂ E.

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Show that μ is a Radon measure. In other words, demonstrate that the equality

μ (X) = sup{μ (K) : K is compact, K ⊂ X} holds true for every Borel set X ⊂ E. 4. Give a proof of Lemma 1. 5. Give a proof of Lemma 2. 6. Give a proof of Lemma 3. 7. Let (E, S ) be a measurable space equipped with a σ -finite measure μ . Let f : E → R be a μ -measurable function. In the product space E × R consider the product measure μ ⊗ λ where λ as usual stands for the Lebesgue measure on R. Show that the graph of f regarded as a subset of E × R is of (μ ⊗ λ )-measure zero. Does this result remain true if (R, λ ) is replaced by (F, ν ), where F is an arbitrary set equipped with a σ -finite measure ν ? 8∗ . Let E be a topological space. We recall that a set X ⊂ E is totally imperfect in E if X contains no nonempty perfect subset of E. A set B ⊂ E is a Bernstein set in E if both B and E \ B are totally imperfect subsets of E. By applying the method of transfinite induction, show that there exists a function f : R → R whose graph Gr( f ) is totally imperfect in the Euclidean plane R2 = R × R. In particular, deduce from this property of Gr( f ) that f is nonmeasurable with respect to the completion of any nonzero σ -finite diffused Borel measure μ on R. For this purpose, utilize Luzin’s classical criterion (the so-called C-property) of the measurability of real-valued functions with respect to the completion of μ . 9∗ . By using the method of transfinite recursion and applying the Fubini theorem, define a bijective function f : R → R whose graph is thick in R2 with respect to the two-dimensional Lebesgue measure λ2 = λ1 ⊗ λ1 . It should be noticed in this place that the first transfinite construction of a function f : R → R whose graph is thick with respect to λ2 was done by Sierpi´nski (cf. also [72], [144], and [192]). 10∗ . Let E be a topological space. Consider the family C(E, R) of all real-valued continuous functions on E. According to the standard definition, the Baire σ -algebra of E (denoted by B0 (E)) is the smallest (with respect to inclusion) σ -algebra S of subsets of E such that all functions from C(E, R) are measurable in the sense of S , i.e., (∀a ∈ R)(∀b ∈ R)(∀ f ∈ C(E, R))( f −1 ([a, b]) ∈ S ).

Some aspects of the measure extension problem

35

Obviously, the relation B0 (E) ⊂ B(E) holds true, where B(E) denotes the Borel σ algebra of E. Suppose that E is compact and hence Hausdorff. Prove that any finite (more generally,

σ -finite) measure μ defined on B0 (E) is Radon. For this purpose, first check that B0 (E) is generated by the family of all closed subsets of E of type Gδ . Then, assuming the finiteness of μ , consider the family L of all those subsets X of E for which there exist two sets X  and X  satisfying the following three relations: (a) X  ⊂ X ⊂ X  ; (b) X  is of type Fσ and X  is of type Gδ ; (c) μ ∗ (X \ X ) = μ ∗ (X  \ X) = 0. Verify that L is a σ -algebra of sets and B0 (E) ⊂ L . Then reduce the case of a nonzero σ -finite measure ν with dom(ν ) = B0 (E) to the case of a probability measure μ , replacing ν by an equivalent probability measure on the same

σ -algebra B0 (E). Applying Henry’s theorem, deduce from the said above that any σ -finite measure on B0 (E) admits an extension to a Radon measure on B(E). Formulate and prove an appropriate generalization of these results for the case of a locally compact topological space E. 11. Let {Ei : i ∈ I} be an arbitrary family of compact spaces. Show the validity of the following equality: ⊗{B0 (Ei ) : i ∈ I} = B0 (∏{Ei : i ∈ I}). For this purpose, apply the Stone-Weierstrass theorem on approximation of continuous real-valued functions (see, e.g., [58] and [101]). On the other hand, give an example of a family {Ei : i ∈ I} of compact spaces such that ⊗{B(Ei ) : i ∈ I} = B(∏{Ei : i ∈ I}). 12. Recall that a set X ⊂ R is universal measure zero if for every σ -finite diffused Borel measure μ on R, we have μ ∗ (X) = 0. Equivalently, X ⊂ R is universal measure zero if there exists no nonzero σ -finite diffused Borel measure on X (cf. Example 11 of Chapter 1). Check that the family of all universal measure zero subsets of R forms a σ -ideal in the Boolean algebra P(R). Obviously, any countable subset of R is universal measure zero and, as has already been mentioned in Chapter 1, the existence of an uncountable universal measure zero set in

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R is a deep fact of classical descriptive set theory (for more details about this fact, see Appendix 1). Let ω1 denote, as usual, the least uncountable cardinal number. Prove that the following two assertions are equivalent: (a) there exists an uncountable universal measure zero subset of R; (b) there exists a countably generated σ -algebra S of subsets of ω1 such that all singletons in ω1 belong to S , but S does not admit any nonzero σ -finite diffused measure. For establishing the equivalence of (a) and (b), use Marczewski’s characteristic function (cf. the proof of Theorem 5 from Appendix 1). 13∗ . Let E be a set, G be a group of transformations of E and let μ be a σ -finite G-quasiinvariant measure on E. Take any μ -measurable set X with μ (X) > 0. We shall say that a set Y is a countable G-configuration of X if there exists a countable family {gi : i ∈ I} ⊂ G for which we have the equality Y = ∪{gi (X) : i ∈ I}. Using the method of transfinite recursion, define a family {Yα : α < ω1 } of countable configurations of X. Namely, put: (a) Y0 = X; (b) let α < ω1 and suppose that Yα has already been defined; if Yα is such that there exists a transformation g ∈ G for which μ (g(Yα )Yα ) > 0, then take Yα +1 = Yα ∪ g(Yα ) ∪ g−1(Yα ); otherwise, take Yα +1 = Yα ; (c) if α < ω1 is a limit ordinal, then put Yα = ∪{Yβ : β < α }. Prove that the family {Yα : α < ω1 } is stationary or, in other words, there exists α0 < ω1 such that Yα = Yα0 for all α ∈ [α0 , ω1 [. Deduce from this fact the assertion of Lemma 4. 14. Let (E, G) be a nonempty space equipped with a transformation group. Verify that the family of all G-absolutely negligible subsets of E is a proper G-invariant ideal in the Boolean algebra P(E) of all subsets of E. Prove also that if this ideal is not a σ -ideal, then there exists a nonzero σ -finite G-quasi-invariant measure μ on E such that every G-quasiinvariant extension μ  of μ is not maximal, i.e., we can find a proper G-quasi-invariant extension of μ  .

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15∗ . Let E be a base set and let G1 and G2 be two groups of transformations of E. Show that the inclusion relation G1 ⊂ G2 does not imply, in general, any inclusion relation between the classes of all G1 -absolutely negligible and all G2 -absolutely negligible subsets of E. 16∗ . A Hausdorff topological space E is called Radon (see [26]) if every σ -finite Borel measure on E is Radon. Prove that: (a) any countable Hausdorff space is Radon; (b) a Borel subset of a Radon space is Radon; (c) any complete separable metric space (i.e., any Polish topological space) is Radon. Consider the set of ordinal numbers [0, ω1 ] in its order topology. Check that this topological space is compact but is not Radon. 17∗ . Let (E, S , μ ) be a σ -finite measure space. We say that μ is a perfect measure (in the sense of Gnedenko and Kolmogorov) if for each μ -measurable function f : E → R, there exists a set Z ∈ B(R) such that Z ⊂ ran( f ), μ (E \ f −1 (Z)) = 0. Verify that: (a) every two-valued probability measure is perfect; (b) every Radon probability measure is perfect. In order to show the validity of (b), check that a direct analogue of Luzin’s C-property holds true for any real-valued function measurable with respect to a Radon probability measure.

Chapter 3

Invariant measures

We would like to begin this chapter with a few historical remarks. The general concept of an invariant measure comes from two sources the first of which is classical Euclidean geometry where the notion of the volume of a geometric body is paramount, and this notion presents the most natural example of an invariant functional with respect to the group of all isometric transformations of Euclidean space. The second source is the theory of dynamical systems which currently is one of the main branches in contemporary mathematics. Actually, the extensive study of various dynamical systems was primarily dictated by the expansion of methods of classical mathematical analysis and especially by the development of the theory of ordinary differential equations. The most important and widely known objects in this theory are the so-called autonomous systems. They are closely connected with certain transformation groups of a phase space. A particular case of systems of this type are Hamiltonian systems for which the fundamental Liouville theorem holds true (see, for instance, [196]; cf. also Exercises 1 and 2 of this chapter). Let E be a base set, that is an abstract phase space. Any bijection of the form g : E →E is usually called a transformation of E. Evidently, the family of all transformations of E is a group with respect to the composition operation ◦. As was mentioned in Chapter 2, this group is denoted by the symbol (Sym(E), ◦) and is called the symmetric group of E. Any subgroup G of Sym(E) is treated as a certain group of transformations of E. The pair (E, G) is usually called a space equipped with a transformation group. Let G be a group of transformations of E and let S be a σ -algebra of subsets of E. Let us remember that S is invariant with respect to G (or, briefly, G-invariant) if (∀X ∈ S )(∀g ∈ G)(g(X) ∈ S ). Suppose, in addition, that μ is a measure defined on a G-invariant σ -algebra S of subsets A.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7_3, © 2009 Atlantis Press/World Scientific

39

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of E. We usually say that μ is invariant with respect to G (or, briefly, G-invariant) if (∀X ∈ S )(μ (g(X)) = μ (X)) or, in other words, the values of μ are preserved under the action of G in E. It should be observed that invariant measures can be met in various fields of modern mathematics and are intensively investigated from different points of view (see [6], [41], [45], [46], [47], [65], [80], [81], [83], [85], [86], [93], [95], [108], [115], [141], [142], [165], [168], [182], [191], [193], [195], [198], [229], [242], [243], [245], [246], [247], [248], and [249]). In particular, they play a significant role in mathematical analysis (especially, in various topics of harmonic analysis and functional analysis), probability theory, ergodic theory, and topological dynamics. Therefore, it is reasonable to touch upon (very briefly, of course) some general properties of such measures. But here we do not intend to discuss the fundamental question concerning the existence of a nonzero invariant (finite or σ -finite) measure μ for a given space (E, G) and for a Ginvariant σ -algebra S ⊂ P(E) on which μ is required to be defined. Questions of this type and related to them are thoroughly envisaged, for example, in the works by Zakrzewski (see [245] and [246]). Our goal is rather moderate and we only wish to recall to the reader certain classical results from the theory of invariant measures (e.g., the famous KrylovBogoliubov theorem and its generalization obtained by Markov and Kakutani) and provide the reader with essential information about σ -finite invariant measures on abstract spaces endowed with transformation groups. First of all, we would like to pay our attention to several important examples of invariant measures. Also, we want to present certain general constructions of such measures and demonstrate how those constructions lead to many other nontrivial examples of invariant measures. Undoubtedly, the best known examples of invariant measures are the standard Borel measure and the standard Lebesgue measure on the Euclidean space Rn (n  1). Actually, the main achievement of Borel and Lebesgue was the idea of constructing a positive (i.e., nonnegative) countably additive translation-invariant functional on Rn which significantly extends the classical Jordan volume on the same space. Denote by the usual symbol B(Rn ) the σ -algebra of all Borel subsets of Rn . The standard Borel measure on Rn can be defined as a (unique) measure λn satisfying the following three conditions: 1) dom(λn ) = B(Rn ); 2) λn ([0, 1]n ) = 1;

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3) λn is invariant under the group of all translations of Rn . Since the space Rn is locally compact, the measure λn may also be regarded as a very particular case of a Haar measure on a locally compact topological group. The corresponding definition of a Haar measure will be formulated below. The completion of λn is usually called the standard Lebesgue measure on the space Rn and is often denoted by the same symbol λn . Obviously, the standard Lebesgue measure on Rn can be described as a smallest (with respect to inclusion) measure λn on Rn satisfying the following three relations: (1) λn is a complete measure; (2) [0, 1]n ∈ dom(λn ) and λn ([0, 1]n ) = 1; (3) λn is invariant under the group of all translations of Rn . It should be noticed that the above-mentioned relations directly imply the invariance of λn under the group of all isometric transformations of Rn . This circumstance can be easily established by using a purely geometric argument starting with the fact that the family of all reflections (i.e., mirror symmetries with respect to affine hyperplanes in Rn ) generates the group of all isometries of Rn (see Exercise 3). Let E be a locally compact topological group and let G be the group of all left (right) translations of E. We recall that a left (right) Haar measure on E is a measure μ satisfying these four conditions: a) μ is not identically equal to zero; b) dom(μ ) = B(E) and μ is a Radon measure on E, i.e., for any Borel set X ⊂ E, we have

μ (X) = sup{μ (K) : K ⊂ X, K is compact}; c) μ is invariant with respect to G; d) for every compact subset K of E, we have μ (K) < +∞. Notice that condition d) is equivalent to the following condition: d’) for each point x ∈ E, there exists a neighborhood V (x) of x such that

μ (V (x)) < +∞. In fact, the latter condition simply says that μ is a locally finite measure. The famous theorem due to Haar states that there exists at least one left (right) Haar measure

μ on E and, in addition, μ is unique in a certain sense. More precisely, any two left (right) Haar measures on E are proportional to each other. The theory of Haar measure is thoroughly considered in numerous books and monographs (see, for instance, [80], [83], or [182]). In particular, the theorem mentioned above and

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concerning the existence and uniqueness of Haar measure can be found in the same books [80], [83], and [182]. A detailed proof of this theorem is also given therein as well as in various other manuals of measure theory. In our book, we do not need many facts from this theory. Let us emphasize that if E is a compact topological group, then the left and right Haar measures are proportional and, therefore, can be identified. This result is not true, in general, even for σ -compact locally compact topological groups. The corresponding examples of essentially distinct left and right Haar measures on some groups of this type are due to von Neumann and are presented, for instance, in [80] and [83]. Let us also remark that if E is a σ -compact locally compact topological group, then a Haar measure μ on E is σ -finite. Conversely, it is not difficult to show that if a Haar measure μ on a locally compact topological group E is σ -finite, then E is σ -compact (see Exercise 5). The case of a Polish locally compact topological group E is of paramount importance. In this case, taking into account the circumstance that E possesses a countable base consisting of open sets with compact closures, we get that E is simultaneously a σ -compact space. In view of the preceding remarks, we conclude that in such a situation the Haar measure μ on E is σ -finite. In connection with this fact, let us also recall that a rather general result due to Ulam holds true, namely, every σ -finite Borel measure on a Polish topological space is necessarily Radon (see Exercise 16 from Chapter 2). The same result remains valid for

σ -finite Borel measures on Suslin spaces (see Appendix 2). Example 1. Let G be the group of all isometric transformations of the n-dimensional Euclidean space Rn . Let us equip G with its natural topology which is induced by the topology of the Euclidean space of dimension n2 . Then G becomes a locally compact Polish topological group and, moreover, it turns out to be a Lie group. Consequently, we may consider a σ -finite Haar measure μ on G. More generally, let H be an arbitrary closed subgroup of G. Then H is also a locally compact Polish topological group and we may consider a σ -finite Haar measure ν on H. Actually, ν is isomorphic, in the purely measuretheoretical sense, to an appropriate combination of the following canonical measures: (a) the Haar measure on the discrete group Z of all integers; (b) the standard Borel measure on the real line R; (c) the Haar measure on the one-dimensional unit torus S1 (= T); (d) the Haar probability measure on a finite subgroup of S1 . Let (Γ, ·) be a topological group. We recall that Γ is monothetic if there exists an element g ∈ Γ such that the subgroup of Γ generated by the one-element set {g} is everywhere dense

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in Γ. Clearly, such a Γ is necessarily commutative. For instance, it can easily be checked that the torus S1 considered as a commutative compact topological group is monothetic. Indeed, any element of S1 which has infinite order generates an everywhere dense subgroup of S1 . Example 2. Let (G, ·) be an arbitrary compact metrizable monothetic topological group equipped with the Haar probability measure μ . Let h be an element of G generating an everywhere dense subgroup of G. We denote this countable everywhere dense subgroup by H. Suppose now that ν is a Borel probability measure on G left invariant with respect to H. Then it is not hard to show that ν coincides with the Haar measure μ . Indeed, let us define a function h∗ : G → G by putting h∗ (g) = h · g

(g ∈ G)

and consider an arbitrary real-valued continuous function p on G. Since ν is left invariant with respect to H, we have the equalities 

G

p((h∗ )n (g))d ν (g) =

 G

p(hn g)d ν (g) =

 G

p(g)d ν (g)

for every integer n. Further, let us fix an element f ∈ G. Since G is metrizable, there exists a sequence {nk : k ∈ N} of integers, such that limk→∞ hnk = f . Therefore, we obtain the following relation: 

G

p( f g)d ν (g) =



G

limk→∞ p(hnk g)d ν (g) =

 G

p(g)d ν (g),

which shows that for any f ∈ G, the ν -integrals of the functions p : G → R, p ◦ f ∗ : G → R are equal to each other. Since p and f were taken arbitrarily, it is not difficult to deduce from the obtained result that our measure ν is left invariant with respect to the whole group G. Now, applying the uniqueness theorem to the Haar measure μ , we get the required equality ν = μ . In particular, if ν is a Borel probability measure on S1 invariant under an infinite cyclic subgroup of S1 which automatically is everywhere dense in S1 , then ν coincides with the Haar probability measure on S1 .

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Let G be a Polish topological group. As already mentioned, if G is locally compact, then there exists a nonzero σ -finite Haar measure on G invariant with respect to the group of all left (right) translations of G. Suppose now that G is not locally compact. Then the following question naturally arises. Does there exist a nonzero σ -finite Borel measure on G invariant with respect to the group of all left (right) translations of G? As far as we know, the question formulated above was first raised by Gelfand (cf. the comments given in [226]). Several authors investigated this question and it was finally established that the answer to it is negative. Moreover, a much stronger result can be obtained by using some simple geometrical properties of σ -compact subsets of a nonlocally compact Polish group G (see the next chapter devoted to σ -finite quasi-invariant measures on groups). In connection with the just said, it is also reasonable to remark that some Polish non-locally compact topological groups admit nonzero left (right) invariant Borel measures (see, e.g., [6] and [191]). Of course, those measures are not σ -finite and, therefore, they are of less interest for mathematical analysis, probability theory, and their applications. As mentioned earlier, the standard Borel measure on the real line R is an essentially unique

σ -finite measure defined on the σ -algebra of all Borel subsets of R and invariant with respect to the group of all translations of R. By using this uniqueness property, it is not difficult to describe the class of all nonzero σ -finite R-quasi-invariant Borel measures on R (see Exercise 4 for Chapter 4). Now, let us consider some natural operations which enable us to produce a new space with a σ -finite invariant measure by starting from an initial family of spaces equipped with corresponding σ -finite invariant measures. First, let us recall a purely algebraic notion of the product (respectively, weak product) of a family of groups. Suppose that a family of groups {(Gi , ·i ) : i ∈ I} is given and let ei denote the neutral element of Gi for each index i ∈ I. As usual, the product group (G, ·) = (∏ Gi , ·) i∈I

of this family is defined by introducing the group operation · in G, according to the formula g · h = (gi ·i hi )i∈I for any two elements g = (gi )i∈I and h = (hi )i∈I from G = ∏i∈I Gi .

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Further, we put G∗ = {g ∈ G : {i ∈ I : gi = ei } is f inite}. It can easily be checked that G∗ is a subgroup of G. The group G∗ is usually called the weak product group of a given family {(Gi , ·i ) : i ∈ I}. If all groups (Gi , ·i ) (i ∈ I) are commutative, then the weak product group coincides with the direct sum of {(Gi , ·i ) : i ∈ I}. We need the following easy auxiliary proposition. Lemma 1. Let E be a set and let G be a group of transformations of E. Suppose that a G-invariant algebra S of subsets of E is given with a σ -finite G-invariant measure μ on S . Let S  = σ (S ) denote the σ -algebra generated by S and let μ  denote the unique extension of μ defined on S  . Then S  is a G-invariant σ -algebra and μ  is a G-invariant measure. Proof. Consider the family of sets T = {X ∈ S  : (∀g ∈ G)(g(X) ∈ S  )}. A straightforward verification shows that T is a G-invariant σ -algebra of subsets of E and S ⊂ T . Consequently, S  = σ (S ) ⊂ T . At the same time, according to the definition of T , we have T ⊂ S  . Therefore, S  = T and S  is a G-invariant σ -algebra of subsets of E. Now, let us introduce the class of sets L = {X ∈ S  : (∀g ∈ G)(μ (g(X)) = μ (X))}. In a similar way, we can verify that the class L contains S and is closed under the unions of increasing sequences of members from L and is also closed under the intersections of decreasing sequences of members from L . This circumstance implies that L = S  (see Exercise 7) and, therefore, μ  is a G-invariant measure on S  . Unfortunately, we do not have an analogous proposition for σ -finite quasi-invariant measures (see Exercise 8). The next simple statement concerns the behavior of invariant probability measures under the operation of their product. Theorem 1. Let {(Ei , Gi ) : i ∈ I} be a family of spaces equipped with transformation groups. Suppose that for each index i ∈ I, a probability measure μi is given on the set Ei

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and denote by E the product of the family of sets {Ei : i ∈ I}. Let μ be the product measure of the family {μi : i ∈ I} and let G be the product group of the family {Gi : i ∈ I}. If all measures μi (i ∈ I) are Gi -invariant, then the product measure μ is G-invariant. Proof. The assertion follows directly from the definition of the product measure μ and from Lemma 1. An analogous statement is true for any finite family of σ -finite invariant measures (see Exercise 9). A deeper result is contained in the next statement (see, for example, [115]). Theorem 2. Let E be a set, {Gn : n ∈ N} be a sequence of groups of transformations of E, and let {μn : n ∈ N} be a sequence of σ -finite measures given on E. Suppose also that these five conditions are satisfied: 1) the family {Gn : n ∈ N} is increasing with respect to inclusion; 2) the family {dom(μn ) : n ∈ N} is increasing with respect to inclusion; 3) for each n ∈ N, the measure μn is invariant under the group Gn ; 4) there exists a sequence {Yn : n ∈ N} of subsets of E such that for any n ∈ N, we have Yn ∈ dom(μn ),

Yn ⊂ Yn+1 ,

and Yn is a Gn -invariant support of μn , i.e.,

μn (E \ Yn ) = 0,

(∀g ∈ Gn )(g(Yn ) = Yn );

5) for each n ∈ N, if a set X belongs to dom(μn+1 ), then the set X ∩Yn belongs to dom(μn ) and we have the equality

μn (X ∩Yn ) = μn+1 (X ∩Yn ). Let us denote G = ∪{Gn : n ∈ N} and put S = {Z ⊂ E : (∀n ∈ N)(Z ∩Yn ∈ dom(μn ))}. Then S is a G-invariant σ -algebra of subsets of E. Further, for each set Z ∈ S , define

μ (Z) = limn→∞ μn (Z ∩Yn ). Then the definition of μ is correct, μ is a σ -finite G-invariant measure on the σ -algebra S and the set Y = ∪{Yn : n ∈ N}

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is a G-invariant support of μ . Furthermore, for any n ∈ N, we have

μn = μ |dom(μn ). Proof. First, let us observe that if Z ∈ S , then for all natural numbers n, the relation

μn (Z ∩Yn ) = μn+1 (Z ∩Yn )  μn+1 (Z ∩Yn+1 ) is fulfilled. Consequently, there exists a limit limn→∞ μn (Z ∩Yn ) ∈ [0, +∞] which will be denoted by μ (Z). In this way, we obtain a functional

μ : S → [0, +∞]. It is not difficult to check the countable additivity of μ , so μ is a measure on S . Also, it directly follows from the definition of μ that for every natural number n, the restriction of

μ to dom(μn ) coincides with the measure μn . All other above-mentioned properties of μ are easily verified and we leave the corresponding details to the reader. The measure μ described in Theorem 2 may be regarded as an inductive limit of the given countable family {μn : n ∈ N} of invariant measures. Example 3. We can directly apply Theorem 2 in certain concrete situations which are motivated by topics of classical mathematical analysis or probability theory. For instance, let us consider the product space E = RN = ∏{Rk : k ∈ N}, where Rk = R for all natural indices k. Furthermore, for each k ∈ N, let us denote by λk the standard Borel measure on Rk and let νk be the Borel probability measure on E whose restriction to the unit segment [0, 1] ⊂ Rk coincides with the standard Borel measure on [0, 1]. Finally, we put

μn = (⊗{λk : k  n}) ⊗ (⊗{νk : k > n}) and denote Gn = R0 × ... × Rn × {0} × {0} × {0} × ... for every natural number n. Now, it can easily be checked that the family {μn : n ∈ N} of σ -finite measures and the family {Gn : n ∈ N} of transformation groups satisfy the assumptions of Theorem 2. Consequently, for the set E equipped with the group G = ∪{Gn : n ∈ N} = R(N)

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of its transformations, there exists a nonzero σ -finite G-invariant measure μ having the properties described in Theorem 2. Also, in view of the definition, all σ -algebras dom(μn ) coincide with the Borel σ -algebra of E. Hence we have dom(μ ) = B(E) and

μ ([0, 1]N ) = 1. We thus conclude that there exists a nonzero σ -finite Borel measure on the infinitedimensional Polish topological vector space RN , invariant with respect to the everywhere dense vector subspace R(N) consisting of all eventually finite real-valued sequences. Now, we wish to return to a finite-dimensional Euclidean space and discuss one important geometric property of the classical Lebesgue measure given on this space. We mean here the so-called Steinhaus property which was first established by Steinhaus in his remarkable paper [233]. Let λn denote, as usual, the standard Lebesgue measure on the n-dimensional Euclidean space Rn and let G be the group of all isometric transformations of Rn . Obviously, we may treat G as a locally compact Polish topological group with respect to the natural group operation and topology. Consider an arbitrary Lebesgue measurable subset Z of Rn with

λn (Z) > 0. Then it is not hard to show that there exists a neighborhood U of the neutral element of G, such that g(Z) ∩ Z = 0/ for all transformations g ∈ U. This property of the set Z is usually called the Steinhaus property of Z. Accordingly, it is frequently said that λn possesses the Steinhaus property. In particular, if we restrict our consideration to the group of all translations of Rn (which is canonically isomorphic to Rn ), then we easily obtain that there exists a neighborhood W of zero in Rn such that (h + Z) ∩ Z = 0/ for any translation h ∈ W . In other words, we have the relation W ⊂ Z − Z = {x − y : x ∈ Z, y ∈ Z}, which can successfully be applied in some constructions of Lebesgue nonmeasurable subsets of the space Rn (see, for instance, [119]). Notice that the Steinhaus property of λn admits a direct analogue for sets measurable with respect to a σ -finite Haar measure given on a σ -compact locally compact topological group (see [80], [83], and Exercise 13).

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It is also reasonable to point out here that the Steinhaus property has an analogue in terms of the Baire category for arbitrary topological groups. Namely, let (G, ·) be a topological group and let Z be a subset of G having the Baire property and not belonging to the family of all first category subsets of G. Then the set Z · Z −1 contains a neighborhood of the neutral element of G. This fact is sometimes called the Banach-Kuratowski-Pettis theorem (cf. [101], [148], and Exercise 15 in Appendix 3). Now, we turn our attention to an important case of (E, G), where E is a compact metric space and G is a certain group of homeomorphisms of E onto itself. In this situation, several fundamental results have been obtained by famous authors, among which there were Krylov, Bogoliubov, Markov, Kakutani, and others. Let T be a nonempty compact metric space and let u : T → T be some homeomorphism. Since u is a certain transformation of T , the natural question arises whether there exists a probability Borel measure on T invariant with respect to the group of transformations generated by u. Evidently, the last group is cyclic and coincides with the family [u] = {um : m ∈ Z}, where Z denotes the set of all integers. In general, [u] is infinite but, sometimes, may be finite. Here we wish to present a direct proof of the remarkable result due to Krylov and Bogoliubov, according to which such a measure does always exist. Notice that the argument uses some basic concepts of functional analysis (see, e.g., [47]). First, let us make several preliminary remarks. As usual, we denote by the symbol C(T ) = C(T, R) the Banach space of all real-valued continuous functions on T . By virtue of the classical Riesz theorem (see, for instance, [56] and [57]), there is a canonical one-to-one correspondence between Radon probability measures on T and linear normalized positive functionals on C(T ). In other words, any Radon probability measure ν on T can be canonically identified with some linear normalized positive (hence continuous) functional on C(T ). For the sake of convenience and brevity, this functional will be denoted by the same symbol

ν . Obviously, the set M of all such functionals is convex in the conjugate space (C(T ))∗ . Moreover, M turns out to be compact with respect to the weak topology σ ((C(T ))∗ ,C(T )). The latter fact immediately follows from the classical Tychonoff theorem on products of quasicompact spaces by taking into account the circumstance that M is a closed convex subset of the topological product

∏

f ∈C(T )

[−|| f ||, || f ||].

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In addition, since T is a compact metric space, C(T ) is separable and hence M is metrizable as well, because M can also be regarded as a closed subset of the product

∏ [−|| f ||, || f ||],

f ∈D

where D is a fixed countable everywhere dense subset of C(T ). So we derive that M is a compact metrizable space which implies that any sequence of points from M contains at least one convergent subsequence (cf. Example 2 of this chapter). Now, take an arbitrary functional ν ∈ M and for each natural number n, define a functional

νn by the formula ν ( f ) + ν ( f ◦ u) + ... + ν ( f ◦ un ) ( f ∈ C(T )). n+1 It can easily be seen that νn ∈ M. Thus, we obtain the sequence {νn : n ∈ N} of points νn ( f ) =

from M. Let {n(k) : k ∈ N} be a strictly increasing family of natural numbers, such that the corresponding partial sequence {νn(k) : k ∈ N} converges to some point μ ∈ M. Let us verify that

μ ( f ) = μ ( f ◦ u)

( f ∈ C(T )).

Indeed, we may write

μ ( f ) = limk→∞ νn(k) ( f ), μ ( f ◦ u) = limk→∞ νn(k) ( f ◦ u). But a simple calculation shows that

νn(k) ( f ) − νn(k) ( f ◦ u) =

ν ( f ) − ν ( f ◦ un(k)+1) , n(k) + 1

which implies |νn(k) ( f ) − νn(k) ( f ◦ u)| 

2||ν || · || f || . n(k) + 1

Tending k to infinity and taking into account the fact that n(k)  k, we come to the equality

μ ( f ) = μ ( f ◦ u) for all functions f ∈ C(T ). In addition, since u is a homeomorphism of T onto itself, we have

μ ( f ◦ u−1) = μ ( f ◦ u−1 ◦ u) = μ ( f ) for all f ∈ C(T ). So we conclude that our measure μ is invariant under the cyclic group of transformations generated by u.

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However, this group is rather small because it is at most countable. Kakutani [93] and Markov [168] extended the above result to any commutative group of homeomorphisms of T . For this purpose, they proved the following fundamental statement (cf. [22] and [57]). Theorem 3. Let E be a Hausdorff topological vector space, K be a nonempty compact convex subset of E and let G be a family of affine continuous mappings acting from K into itself. As usual, denote by IK the identity transformation of K and suppose that the following condition is fulfilled: there exists a finite sequence {G0 , G1 , ..., Gn } such that {IK } = Gn ⊂ Gn−1 ⊂ ... ⊂ G1 ⊂ G0 = G and for each natural index i ∈ [1, n] and for any two elements u ∈ Gi−1 and v ∈ Gi−1 , there is an element w ∈ Gi satisfying the equality u ◦ v = v ◦ u ◦ w. Then there exists at least one point of K invariant under G, that is there exists a point x ∈ K for which the relation (∀u ∈ G)(u(x) = x) holds true. Proof. We use induction on n. The case n = 0 is trivial. Consider the case n = 1. In this situation, we have u ◦ v = v ◦ u for any two elements u and v from G0 = G, i.e., all elements from G pairwise commute. Take any u ∈ G and for a natural number r > 0, put ur = (IK + u + u2 + ... + ur−1)/r. In view of the convexity of K, we get ur (K) ⊂ K. Let G∗ denote the family of all finite compositions of mappings of the form ur

(u ∈ G, r ∈ N \ {0}).

It is clear that all elements of G∗ pairwise commute, are affine and continuous, and map K into itself. Consider now the family of nonempty sets {v(K) : v ∈ G∗ }. All members of this family are compact. In addition, if v ∈ G∗ and v ∈ G∗ , then (v ◦ v )(K) = (v ◦ v )(K), (v ◦ v )(K) ⊂ v (K), (v ◦ v )(K) ⊂ v (K), which readily implies that the above-mentioned family is centered. Therefore, / ∩{v(K) : v ∈ G∗ } = 0.

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Choose a point x0 ∈ ∩{v(K) : v ∈ G∗ }. We are going to show that u(x0 ) = x0 for any u ∈ G. Indeed, if r is a nonzero natural number, then x0 ∈ ur (K) which means that there exists y ∈ K such that x0 = (y + u(y) + ... + ur−1(y))/r. Consequently, we may write u(x0 ) − x0 = (ur (y) − y)/r ∈ (K − K)/r. Let C be an arbitrary star-like symmetric open neighborhood of the neutral element of E. Clearly, K − K ⊂ ∪{mC : m ∈ N \ {0}}. Since K − K is compact, we come to the inclusion K − K ⊂ mC (or, equivalently, to the inclusion (K − K)/m ⊂ C) for all sufficiently large natural numbers m and, in particular, for sufficiently large r. In other words, the relation u(x0 ) − x0 ∈ C is valid. Since E is Hausdorff, we infer at once that u(x0 ) − x0 = 0 and thus u(x0 ) = x0 . Suppose now that the assertion of the theorem has already been established for n and let us prove it for n + 1. Consider the case when a sequence {IK } = Gn+1 ⊂ Gn ⊂ ... ⊂ G1 ⊂ G0 = G is given with the corresponding properties. Applying the inductive assumption to G1 , we deduce that the set K1 = {x ∈ K : (∀u ∈ G1 )(u(x) = x)} is nonempty. Also, it can easily be verified that the same K1 is convex and compact. Take any two elements u1 ∈ G and u2 ∈ G. According to the condition of the theorem, there is w ∈ G1 such that u1 ◦ u2 = u2 ◦ u1 ◦ w. If x ∈ K1 , then we have (u1 ◦ u2 )(x) = (u2 ◦ u1)(w(x)) = (u2 ◦ u1)(x). This circumstance means that all elements from G pairwise commute on K1 . Finally, let us take any u ∈ G and any w1 ∈ G1 ⊂ G. Then, according to the same condition, we may write the equality w1 ◦ u = u ◦ w1 ◦ w2

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for some mapping w2 ∈ G1 . Consequently, for each x ∈ K1 , we get w1 (u(x)) = u((w1 ◦ w2 )(x)) = u(x), thus it follows that u(x) ∈ K1 . In other words, all mappings from G pairwise commute and transform K1 into itself. So we come again to the case, where n = 1. As already shown, there exists a point x0 ∈ K1 invariant under all mappings from G which completes the proof.

Remark 1. The theorem of Markov and Kakutani is usually applied in those situations, where G is a solvable group of continuous affine transformations of a Hausdorff topological vector space E, which map a nonempty compact convex set K ⊂ E into itself. Clearly, in such a case, the condition of Theorem 3 is automatically satisfied. In order to give an application of Theorem 3 to the existence of dynamical systems, we also need the following simple auxiliary proposition. Lemma 2. Let T be a nonempty compact topological space and let G be a solvable group of homeomorphisms acting from T onto itself. For each element g ∈ G, define the mapping g : (C(T ))∗ → (C(T ))∗ by the formula g (φ ) = φ 

(φ ∈ (C(T ))∗ ),

where

φ  ( f ) = φ ( f ◦ g)

( f ∈ C(T )).

Then the family G = {g : g ∈ G} is a solvable group of continuous affine mappings of (C(T ))∗ into itself for which we have g (M) ⊂ M

(g ∈ G ),

where M = M(T ) denotes the compact convex set of all Radon probability measures on the given space T . The proof reduces to a straightforward verification. The corresponding details are not difficult and are left to the reader. Applying Theorem 3 and Lemma 2 to G and to M, we obtain the following statement of Markov and Kakutani.

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Theorem 4. Let T be a nonempty compact space and let G be a solvable group of homeomorphisms of T onto itself. Then there exists a Radon probability measure μ on T invariant under G, that is

μ (g(X)) = μ (X)

(g ∈ G, X ∈ B0 (T )).

Remark 2. Let E be a set and let Φ be a homomorphism of the additive group R into the symmetric group (Sym(E), ◦). Such a homomorphism is usually called a one-parameter group of transformations of E. Note that the homomorphic image Φ(R) is also frequently called a one-parameter group of transformations of E. Obviously, any one-parameter group is automatically commutative. Consequently, if in Theorem 4 the group G is a oneparameter group of homeomorphisms of T onto itself, then there always exists a Radon probability G-invariant measure on T . This result is due to Krylov and Bogoliubov. Remark 3. Let G be a solvable compact topological group. Theorem 4 immediately implies the existence of a Haar probability measure on G. In addition, by using the classical Fubini theorem to the product space G × G, it is not hard to prove the uniqueness of a Haar probability measure on G (cf. [209] and Exercise 17). One important application of Theorem 3 is presented in Exercise 9 of Chapter 17. Of course, our survey of invariant measures is far from being a complete one. Further information about various properties of invariant and quasi-invariant measures can be found in the works which were already mentioned in this chapter (see also the next chapter). EXERCISES 1. Any mapping f : Rn → Rn is often called a vector field in Rn because f associates the vector f (x) ∈ Rn to every point x ∈ Rn . This terminology is especially convenient when the following autonomous system of first-order ordinary differential equations is studied: x1 = f1 (x1 , x2 , ..., xn ), x2 = f2 (x1 , x2 , ..., xn ), .................................. xn = fn (x1 , x2 , ..., xn ),

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where f = ( f1 , f2 , ..., fn ). It is usually assumed that the mapping f satisfies the Lipschitz condition with respect to each of its arguments (so f is necessarily continuous). Under this assumption for any initial value (x01 , x02 , ..., x0n ) ∈ Rn , there exists a unique local solution (local trajectory) (x1 (t), x2 (t), ..., xn (t)) (−a  t  a) of this system which has the initial value x1 (0) = x01 , x2 (0) = x02 , ... , xn (0) = x0n at t = 0. If the norm of f is bounded, then the local solution (local trajectory) turns out to be global, i.e., it is defined on the whole R. Actually, we may suppose without loss of generality that the norm of f is bounded because an appropriate replacement of f by a vector field with bounded norm is always possible in such a way that the trajectories (considered as point-sets) remain the same. We will assume that this replacement is done, so all solutions of our system are global. Verify that the vector field f induces a certain one-parameter group G = {gt : t ∈ R} of transformations of the space Rn . Namely, for each t ∈ R, the transformation gt : Rn → Rn is defined as follows: for any point (x01 , x02 , ..., x0n ) ∈ Rn , we put gt (x01 , x02 , ..., x0n ) = (x1 (t), x2 (t), ..., xn (t)), where (x1 (t), x2 (t), ..., xn (t)) is the point of the above-mentioned trajectory, corresponding to t. We shall say that the group G is associated with the vector field f . Conversely, suppose that a one-parameter group G = {gt : t ∈ R} of transformations of Rn is given. Under some smoothness assumptions on G, show that this group induces a certain vector field f : Rn → Rn whose corresponding autonomous system yields (by the procedure described above) the same G. 2∗ . We preserve the notation of the previous exercise. Assume that a vector field f = ( f1 , f2 , ..., fn ) : Rn → Rn is sufficiently smooth and denote div( f ) = ∂ f1 /∂ x1 + ∂ f2 /∂ x2 + ... + ∂ fn/∂ xn .

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Prove that the following two assertions are equivalent: (a) the Lebesgue measure λn is invariant under the one-parameter group G = {gt : t ∈ R} associated with f ; (b) div( f )(x) = 0 for all points x ∈ Rn . This classical result is due to Liouville (see, for example, [196]). In particular, take the Euclidean space R2n and consider the autonomous Hamiltonian system x1 = ∂ H/∂ y1 , x2 = ∂ H/∂ y2 , ..., xn = ∂ H/∂ yn , y1 = −∂ H/∂ x1 , y2 = −∂ H/∂ x2 , ..., yn = −∂ H/∂ xn , where H : R2n → R is a smooth real-valued function of 2n real variables x1 , x2 , ..., xn , y1 , y2 , ..., yn . Check that the Lebesgue measure λ2n on R2n is invariant under the one-parameter group associated with this system. 3. Show that the group of all isometric transformations of the space Rn is generated by the family of all mirror symmetries (i.e., orthogonal reflections with respect to affine hyperplanes). Show also that for any affine hyperplane Γ in Rn and for any open subset U of Rn , it is possible to represent U in the form U = ∪{Ki : i ∈ I} where {Ki : i ∈ I} is a countable family of closed n-dimensional cubes whose interiors are pairwise disjoint and, in addition, each cube Ki (i ∈ I) has a facet parallel to Γ. Infer from these two geometrical facts that the Lebesgue measure λn is invariant under the group of all isometric transformations of Rn (for n  1, this group is not commutative and hence is not a one-parameter group). 4∗ . Let κ be a cardinal number and let S1 be the unit torus in R2 regarded as a compact commutative group. Prove that the following three assertions are equivalent: (a) κ does not exceed c; (b) the topological product group Sκ1 is monothetic; (c) the topological product group Sκ1 is separable which means that it contains a countable everywhere dense subset. For establishing the implication (a) ⇒ (b), use the classical theorem of Kronecker (see, e.g., [23]), which states that for given real numbers θ1 , θ2 , ..., θn , the following two relations are equivalent:

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(d) the system of numbers {1, θ1 , θ2 , ..., θn } is linearly independent over the field Q; (e) for any real numbers x1 , x2 , ..., xn and for every real ε > 0, there exist integers p1 , p2 , ..., pn , q such that |qθ j − p j − x j | < ε

(1  j  n).

Starting with this theorem of Kronecker, consider R as a vector space over Q and fix a linearly independent system {1} ∪ {θ j : j ∈ J} in R, where card(J) = κ  c. Further, define the element s = {s j : j ∈ J} = {(cos(2πθ j ), sin(2πθ j )) : j ∈ J} ∈ SJ1 and verify that s generates the cyclic group which is everywhere dense in SJ1 . Conclude from the said above that Sκ1 is a nonmetrizable compact monothetic group whenever ω1  κ  c. 5. Let G be a locally compact topological group and let μ be the left Haar measure on G. Prove that the following two assertions are equivalent: (a) μ is σ -finite; (b) G is σ -compact. For this purpose, apply Exercise 13 from Chapter 2. Also, prove that the next two assertions are equivalent: (c) μ is finite; (d) G is compact. 6. Complete the details of Example 2. 7. Let E be a set and let L be a class of subsets of E. According to the standard definition, L is called a monotone class if L is closed under the unions of all increasing sequences of its members and is also closed under the intersections of all decreasing sequences of its members (see, e.g., [80] and [187]). Let S be an algebra of subsets of E. Show that any monotone class containing S also contains the σ -algebra σ (S ). Conclude from this fact that σ (S ) coincides with the monotone class generated by S . 8. Give an example of a probability quasi-invariant measure μ defined on an algebra of sets, such that the unique extension μ  of μ defined on the generated σ -algebra does not have the quasi-invariance property with respect to the same transformation group.

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9. Let {μi : i ∈ I} be a finite family of σ -finite measures such that every measure μi (i ∈ I) is invariant under some group Gi . Show that the product measure

μ = ⊗{μi : i ∈ I} is invariant under the product group G = ∏{Gi : i ∈ I}. 10. Complete the details of the proof of Theorem 2. 11∗ . Show that the measure μ constructed in Example 3 is metrically transitive (i.e., ergodic) with respect to R(N) . In other words, demonstrate that for any μ -measurable set Z with μ (Z) > 0, there exists a countable family {gn : n ∈ N} ⊂ R(N) such that

μ (RN \ ∪{gn + Z : n ∈ N}) = 0. 12∗ . We recall the notation l1 = {(xn )n∈N ∈ RN :

∑

|xn | < +∞},

∑

|xn |2 < +∞}.

n∈N

l2 = {(xn )n∈N ∈ RN :

n∈N

Observe that both l1 and l2 are Borel vector subspaces of the Polish topological vector space RN . In addition, l1 can be considered as a separable Banach space with respect to the natural norm ||x|| =

∑ |xn |

(x = (xn )n∈N ∈ l1 )

n∈N

and l2 can be considered as a separable Hilbert space with respect to the natural inner product < x, y > =

∑ xn yn

(x = (xn )n∈N ∈ l2 , y = (yn )n∈N ∈ l2 ).

n∈N

Prove that there exists a nonzero σ -finite Borel measure ν on the Banach space l1 , invariant with respect to R(N) . Using this result and taking into account the relation l1 ⊂ l2 , infer that for every infinite-dimensional separable Hilbert space E, there exists a nonzero σ -finite Borel measure on E invariant with respect to some everywhere dense vector subspace of E. 13∗ . Let G denote the group of all isometric transformations of Rn , equipped with its natural topology. By applying the classical Lebesgue theorem on density points of Lebesgue measurable sets (see, e.g., [183], [192], [210], or Appendix 4), prove that for any λn -measurable

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set Z ⊂ Rn with λn (Z) > 0, there exists a neighborhood V of the neutral element of G, such that

λn (g(Z) ∩ Z) > 0 for all elements g ∈ V ; in particular, we have g(Z) ∩ Z = 0/ for all g ∈ V . Moreover, using the fact that the standard Borel measure on Rn is Radon, show that limg→e λn (g(Z) ∩ Z) = λn (Z), where e denotes the neutral element of G. Formulate and prove an analogous result for a Haar measure on a σ -compact locally compact topological group. 14∗ . Let S be a sphere in Rn (n  2) equipped with the (n − 1)-dimensional Lebesgue probability measure μ which is invariant under the group G of all rotations of S about its center. Fix a natural number k > 0. Let A be an arbitrary μ -measurable subset of S with

μ (A) < 1/k and let B be an arbitrary subset of S with card(B)  k. Show that there exists at least one rotation g ∈ G for which we have g(A) ∩ B = 0. / 15∗ . Describe all those (sufficiently smooth) homeomorphisms g : R2 → R2 which preserve the two-dimensional Lebesgue measure λ2 on the plane R2 . 16. Consider the unit circle S1 equipped with a group G of its homeomorphisms onto itself, such that: (a) G contains at least one rotation of infinite order; (b) G contains at least one transformation which is not an isometry of S1 . Show that there exists no G-invariant Borel probability measure μ on S1 . Conclude from this fact that no such group G can be solvable. 17∗ . By using the Fubini theorem, prove the uniqueness of a left (right) Haar probability measure μ on a compact topological group (G, ·). In other words, demonstrate that if ν is any Borel probability left (right) invariant measure on G, then ν = μ . Show also that the left Haar measure on G coincides with the right Haar measure on G. Further, apply the uniqueness property of the left Haar measure μ on a σ -compact locally compact group (H, ·) and prove that μ is metrically transitive (ergodic) with respect to the group of all left translations of H, i.e., prove that for every μ -measurable set X ⊂ H with

μ (X) > 0, there exists a countable family {hi : i ∈ I} ⊂ H such that μ (H \ ∪{hi X : i ∈ I}) = 0.

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Obtain an analogous result for the right Haar measure on H. 18∗ . Let (E, G) be a space equipped with a transformation group and let μ be a σ -finite G-invariant measure on E. Suppose that a family {Zi : i ∈ I} of subsets of E is given, which satisfies the following conditions: (a) every set Zi (i ∈ I) is almost G-invariant with respect to μ ; (b) for any countable subset J of I and for any family {Z j : j ∈ J} such that (∀ j ∈ J)(Z j = Z j ∨ Z j = E \ Z j ), the intersection ∩{Z j : j ∈ J} is μ -thick in E, i.e.,

μ∗ (E \ (∩{Z j : j ∈ J})) = 0. Consider the family T of all those subsets X of E which admit a representation in the form f (1)

X = ∪{Zi1

f (2)

∩ Zi2

f (n)

∩ ... ∩ Zin

∩Y f : f ∈ {0, 1}n },

where n is an arbitrary natural number, {i1 , i2 , ..., in } is an injective sequence of elements f (k)

from I, all sets Y f ( f ∈ {0, 1}n) belong to dom(μ ) and Zik

f (k)

= Zik if f (k) = 0, and Zik

=

E \ Zik if f (k) = 1. Verify that T is an algebra of sets and the inclusion dom(μ ) ⊂ T holds true. Further, define a functional μ  on T by the formula

μ  (X) = (1/2n ) ∑{μ (Y f ) : f ∈ {0, 1}n}. Check that this definition of μ  is correct in the sense that the value μ  (X) does not depend on a representation of X in the above-mentioned form. Check also that μ  is finitely additive and extends μ . Moreover, applying the K¨onig lemma on ω -trees (see Appendix 1), show that μ  is countably additive and, therefore, μ  is a measure on T extending μ . Finally, using Lemma 1 of this chapter, conclude that μ  can be extended to a G-invariant measure defined on the σ -algebra σ (T ). For the sake of brevity, we denote this measure by the same symbol μ  . Observe that if the original measure μ is nonzero and card(I) > ω , then μ  is a nonseparable extension of μ . 19. Let (E, G) be a space equipped with a transformation group, for which the following conditions are satisfied: (i) E is uncountable and card(G)  card(E); (ii) G acts transitively on E, i.e., for any two points x ∈ E and y ∈ E, there exists g ∈ G such that g(x) = y.

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Let α denote the least ordinal number whose cardinality is equal to card(E). Prove that E admits a representation in the form E = ∪{Xξ : ξ < α } for which the following relations are valid: (a) card(Xξ )  card(ξ ) + ω for every ordinal ξ < α ; (b) Xξ ∩ Xζ = 0/ for any two distinct ordinals ξ < α and ζ < α ; (c) for each set Ξ ⊂ [0, α [, the associated union X(Ξ) = ∪{Xξ : ξ ∈ Ξ} is an almost G-invariant subset of E, i.e., we have (∀g ∈ G)(card(g(X(Ξ))X(Ξ)) < card(E)). In order to show the existence of {Xξ : ξ < α }, first observe that the conditions (i) and (ii) imply the equality card(G) = card(E). Then construct an α -sequence {Gξ : ξ < α } of subgroups of G satisfying the relations: (1) card(Gξ )  card(ξ ) + ω for any ξ < α ; (2) ∪{Gξ : ξ < α } = G; (3) Gξ ⊂ Gζ whenever ξ < ζ < α . Further, fix a point e ∈ E and for each ordinal ξ < α , put Xξ = Gξ (e) \ ∪{Gζ (e) : ζ < ξ }. Verify that the family of sets {Xξ : ξ < α } yields the desired result. 20∗ . Let (E, G) be a space equipped with a transformation group. Suppose that the following relations hold: (i) E is uncountable; (ii) G acts transitively in E; (iii) G acts freely in E, i.e., for any two points x ∈ E and y ∈ E, there exists at most one g ∈ G such that g(x) = y. Let μ be a nonzero σ -finite G-invariant (G-quasi-invariant) measure on E. Assuming that card(E) is not measurable in the Ulam sense (see Appendix 1), prove that there exists a G-invariant (G-quasi-invariant) measure μ  on E strictly extending μ . For this purpose, consider two possible cases.

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1. The cofinality of card(E) is equal to ω , i.e., E admits a representation in the form E = ∪{Xn : n < ω }, where card(Xn ) < card(E) for all n < ω . Check that Xn ∈ dom(μ ) for some n < ω and apply Marczewski’s method to the G-invariant

σ -ideal generated by {Xn }. 2. The cofinality of card(E) is greater than or equal to ω1 . In this case, consider a representation of E in the form E = ∪{Xξ : ξ < α }, where the family of sets {Xξ : ξ < α } is as in Exercise 19. Preserving the notation of Exercise 19 and taking into account the assumption on card(E), verify that there exists a set Ξ ⊂ [0, α [ for which the associated almost G-invariant set X(Ξ) = ∪{Xξ : ξ ∈ Ξ} is nonmeasurable with respect to μ . Then apply to X(Ξ) Marczewski’s method of extending

σ -finite G-invariant (G-quasi-invariant) measures. In particular, let (G, ·) be an arbitrary uncountable group equipped with a nonzero σ -finite left invariant (left quasi-invariant) measure μ . If card(G) is not measurable in the Ulam sense, then μ can be strictly extended to a left invariant (left quasi-invariant) measure μ  on G. The result formulated in Exercise 20 is essentially due to Hulanicki [85] and Pkhakadze [198]. It shows that the assertion of the strict extendability of any nonzero σ -finite left invariant (left quasi-invariant) measure on an uncountable group does not contradict ZFC theory. At present, it is unknown whether the same assertion can be established without additional set-theoretical hypotheses. In the sequel, we will consider an analogous statement and prove it within ZFC for more concrete classes of uncountable groups (see Chapter 10).

Chapter 4

Quasi-invariant measures

Let E be an arbitrary set and let G be a group of transformations of E; in this case, we usually say that the pair (E, G) is a space equipped with a transformation group (see Chapters 2 and 3). Let S be a σ -algebra of subsets of E. We recall that S is invariant with respect to G (or invariant under G or, briefly, G-invariant) if (∀X ∈ S )(∀g ∈ G)(g(X) ∈ S ). Suppose that μ is a measure defined on a G-invariant σ -algebra S of subsets of E. We say that μ is quasi-invariant with respect to G (or quasi-invariant under G or, briefly, G-quasiinvariant) if (∀X ∈ S )(∀g ∈ G)(μ (X) = 0 ⇔ μ (g(X)) = 0). It immediately follows from this definition that every G-invariant measure on E is necessarily G-quasi-invariant. Obviously, the converse assertion is not true in general. Similarly to invariant measures, quasi-invariant measures can be met in many fields of mathematics and are objects of extensive investigations (see, for instance, [16], [80], [83], [115], [117], [165], and [226]). In particular, quasi-invariant measures play an essential role in various topics of mathematical analysis and probability theory. In this chapter we will concisely examine certain general properties of σ -finite (primarily, probability) quasi-invariant measures. We will recall several important examples of σ finite quasi-invariant measures. In addition, we will present some general constructions of

σ -finite quasi-invariant measures and demonstrate how those constructions lead to other interesting examples of measures of this type. If we are given a nonzero σ -finite G-quasi-invariant measure μ on E, then it is clear that any σ -finite measure ν which is equivalent to μ turns out to be G-quasi-invariant. Indeed, A.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7_4, © 2009 Atlantis Press/World Scientific

63

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from the classical Radon-Nikodym theorem (see, e.g., [16], [26], [56], [80], and [210]), we have a representation

ν (X) =

 X

f dμ

(X ∈ dom(μ )),

where f : E → R is a strictly positive Radon-Nikodym derivative of ν with respect to μ . In other words, f satisfies the equality f = d ν /d μ . This representation immediately yields the G-quasi-invariance of ν . Notice that among all those measures which are equivalent to μ there always exists a probability measure. In this context, an important problem concerning σ -finite invariant and probability quasiinvariant measures arises. Its formulation is as follows: Let E be a set, G be a group of transformations of E and let ν be a probability G-quasiinvariant measure on E. Does there exist a nonzero σ -finite G-invariant measure μ on E which is equivalent to ν ? Another important problem arising in connection with nonzero σ -finite quasi-invariant (invariant) measures is the problem of their existence for a given space E equipped with a transformation group G and with a G-invariant σ -algebra S of subsets of E. A particular case which is of special interest is when we are given a group (G, ·) endowed with the group of all its left translations and with a left-invariant σ -algebra of its subsets. However, we do not intend to discuss these two problems here (for more details, see [245], [246], and [249]). Let (G, ·) be a Polish topological group. If, in addition, G is locally compact, then there exists a nonzero σ -finite Haar measure on G invariant (hence, quasi-invariant) with respect to the group of all left (right) translations of G. Suppose now that G is not locally compact. Then the following question naturally arises. Does there exist a nonzero σ -finite Borel measure on G quasi-invariant with respect to the group of all left (right) translations of G? As already mentioned in Chapter 3, the answer to this question is negative (in this connection, see [226] and the references given therein). Moreover, as indicated in the same chapter, a much stronger result can be obtained by starting with some simple geometrical properties of σ -compact subsets of a non-locally compact Polish group G. For this purpose, we need one auxiliary proposition. Lemma 1. Let (G, ·) be a topological group and let H and K be subsets of G. Suppose that the following two conditions are fulfilled:

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1) H cannot be covered by a countable family of compact subsets of G; 2) K is σ -compact in G. Then there exists an uncountable disjoint family consisting of left (right) H-translates of K.

Proof. Obviously, it is sufficient to consider only the case when the elements of H act in G as left translations of G. As usual, let ω1 denote the least uncountable ordinal number. We are going to construct, by using the method of transfinite recursion, a disjoint ω1 -family {Kξ : ξ < ω1 } consisting of some left H-translates of the given set K. First of all, we put K0 = K. Suppose now that for a nonzero ordinal number ξ < ω1 , the partial disjoint family {Kζ : ζ < ξ } of left H-translates of K has already been constructed. Let us define Z = ∪{Kζ : ζ < ξ }. Then, taking into account condition 2) and the inequality card(ξ )  ω , we obtain that the set Z is σ -compact in G. Consequently, the set Z −1 which is a homeomorphic image of Z is σ -compact as well. Further, the set Z · Z −1 = {x · y−1 : x ∈ Z, y ∈ Z} as the image of Z × Z under the continuous mapping

φ (x, y) = x · y−1

(x ∈ G, y ∈ G)

is also σ -compact in G. According to condition 1), we may write / H \ (Z · Z −1 ) = 0. Let h be an arbitrary element from H \ (Z · Z −1 ). Evidently, we have hZ ∩ Z = 0. / Let us put Kξ = hK. Then it is clear that the partial family {Kζ : ζ  ξ } is disjoint and consists of some left H-translates of K. Proceeding in such a manner, we are able to define the required disjoint ω1 -family {Kξ : ξ < ω1 } of left H-translates of K. Thus, the proof of Lemma 1 is complete. This lemma immediately implies the following statement. Theorem 1. Let (G, ·) be a Polish topological group and let H be a subgroup of G which cannot be covered by countably many compact subsets of G. Then there exists no nonzero

σ -finite left (right) H-quasi-invariant Borel measure on G.

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In particular, if (G, ·) is a non-locally compact Polish topological group, then there exists no nonzero σ -finite left (right) G-quasi-invariant Borel measure on G. Proof. Suppose to the contrary that there exists a nonzero σ -finite left (or, respectively, right) H-quasi-invariant Borel measure μ on G. Since G is a Polish topological space, μ must be a nonzero Radon measure. Consequently, we can find a compact subset K of G such that μ (K) > 0. Now, according to Lemma 1, there exists an uncountable disjoint family of left (right) H-translates of K. Since μ is left (right) H-quasi-invariant, all sets from the above-mentioned family have strictly positive μ -measure, which yields a contradiction with the fact that μ satisfies the countable chain condition since it is σ -finite. The contradiction obtained finishes the proof of the first part of Theorem 1. The second part of Theorem 1 trivially follows from the remark that if a Polish topological group G is not locally compact, then it is not σ -compact either; otherwise, in view of the Baire classical theorem, at least one member of a countable covering of G by compact sets must have nonempty interior, which easily implies the local compactness of G. In connection with Theorem 1, it is reasonable to remark once more that some Polish nonlocally compact topological groups admit nonzero left (right) invariant Borel measures (see [6] and [191]). But, since such measures are not σ -finite, they are of less interest for purposes of mathematical analysis and probability theory. We also wish to point out the fundamental result which is due to Mackey [165] concerning the structure of those Borel subgroups of a Polish topological group that admit nonzero

σ -finite quasi-invariant Borel measures. Namely, let G be a Borel subgroup of some Polish topological group (quite frequently, such a G is called a standard group). Then, according to Mackey’s theorem, the following two assertions are equivalent: (i) there exists a nonzero σ -finite Borel measure on G which is quasi-invariant with respect to the group of all left translations of G; (ii) there exist a locally compact Polish topological group Γ and a group isomorphism Φ : G → Γ, such that Φ is simultaneously a Borel isomorphism between G and Γ. Moreover, if μ stands for some nonzero σ -finite Borel measure on G which is quasiinvariant under the group of all left translations of G, then the group isomorphism Φ can be chosen so that the image-measure Φ(μ ) = μ ◦ Φ−1

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turns out to be equivalent to the left Haar measure ν on Γ, that is the class of all Φ(μ )measure zero sets coincides with the class of all ν -measure zero sets. In particular, by starting with this deep and important result of Mackey, it is easy to demonstrate that for any Polish topological group H, the following two relations are equivalent: (1) H is locally compact; (2) there exists a nonzero σ -finite Borel measure on H quasi-invariant with respect to the group of all left translations of H. For the proof of this equivalence, see [165] and [115]. Notice that the implication (1) ⇒ (2) is trivial, so the main difficulty is concentrated in establishing the converse implication (2) ⇒ (1). As mentioned earlier, the standard Borel measure on the real line R is an essentially unique nonzero σ -finite measure defined on the σ -algebra of all Borel subsets of R and invariant under the group of all translations of R. Actually, this fact is a very particular case of the uniqueness theorem for a Haar measure (see Chapter 3). In the same Chapter 3 we have pointed out that by using this uniqueness property, it is not difficult to describe the class of all nonzero σ -finite R-quasi-invariant Borel measures on R (see Exercise 4). Now, let us consider some natural operations which enable us to produce a new space equipped with a probability quasi-invariant measure starting from an initial family of spaces equipped with corresponding probability quasi-invariant measures. The next simple statement concerns the behavior of the quasi-invariance property under the operation of taking arbitrary products of probability quasi-invariant measures. Theorem 2. Let {(Ei , Gi ) : i ∈ I} be a family of spaces equipped with transformation groups. Suppose that for each index i ∈ I, a probability Gi -quasi-invariant measure μi is given on the set Ei . Denote by E the product of the family of sets {Ei : i ∈ I} and denote by μ the product measure of the family { μi : i ∈ I}, i.e., put

μ = ⊗{μi : i ∈ I}. Also, let G∗ stand for the weak product group of the family of groups {Gi : i ∈ I}. Then the product measure μ is G∗ -quasi-invariant. Proof. The statement can easily be established by using the classical Fubini theorem and the corresponding details are left to the reader. As an immediate application of Theorem 2, we have the following example.

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Example 1. Let {μi : i ∈ I} be a family of probability measures such that for any i ∈ I, the measure μi is defined on the Borel σ -algebra B(R) of R and is R-quasi-invariant. As usual, we denote by R(I) the vector subspace of RI consisting of all those elements of RI whose supports are finite, i.e., R(I) = {(xi )i∈I ∈ RI : card({i ∈ I : xi = 0}) < ω }. Actually, R(I) coincides with the direct sum of a family of copies of R, whose cardinality is equal to card(I). Obviously, R(I) is everywhere dense in the topological vector space RI . According to Theorem 2, the product measure μ of the family {μi : i ∈ I} is defined on the canonical cylindrical σ -subalgebra of B(RI ) and is quasi-invariant with respect to the vector space R(I) . In particular, if card(I)  card(N) = ω , then dom(μ ) coincides with B(RI ) and, consequently, μ turns out to be a Borel probability measure on the space RI , quasi-invariant under R(I) . Thus, we have a large class of Borel probability measures on the infinite-dimensional Polish topological vector space RN , which are quasi-invariant with respect to the everywhere dense vector subspace R(N) of RN . Notice that R(N) is in some sense a maximal vector subspace of RN possessing the property described in Example 3 (in this connection, see Exercise 5). Having a Borel probability measure on RN quasi-invariant under an everywhere dense vector subspace of RN , we can try to construct analogous measures for other infinitedimensional topological vector spaces. In particular, the following question naturally arises. Does there exist for every separable Banach space (E, || · ||), a Borel probability measure μ on E such that μ is quasi-invariant with respect to some everywhere dense vector subspace of E? It turns out that the answer to this question is positive. In order to present the corresponding result, we need one simple lemma from the theory of Banach spaces. First, let us recall that l1 usually denotes the classical Banach space consisting of all those sequences (xn )n∈N ∈ RN for which the inequality

∑ |xn | < +∞

n∈N

is fulfilled. Of course, l1 is equipped with its standard norm ||x|| =

∑ |xn |

n∈N

(x = (xn )n∈N ∈ l1 ).

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The space l1 contains R(N) and is separable. Moreover, it is the initial space in the class of all infinite-dimensional separable Banach spaces. More precisely, we have the following auxiliary statement (cf. [56], [209]). Lemma 2. Let (E, || · ||) be an arbitrary separable Banach space. Then there exists a continuous linear surjection Φ : l1 → E. Proof. Denote by B the closed unit ball in E, i.e., put B = {e ∈ E : ||e||  1}. Since E is separable, we can choose a sequence of points {en : n ∈ N} ⊂ B which is everywhere dense in B. Now, take an arbitrary element x = (xn )n∈N ∈ l1 and define Φ(x) =

∑

xn en .

n∈N

Notice that this definition of Φ is correct because of the inequalities

∑ ||xnen ||  ∑ |xn| < +∞.

n∈N

n∈N

Also, it can easily be seen that Φ is a linear continuous mapping acting from l1 into E. It remains to check that Φ is a surjection. For this purpose, let us first show that B ⊂ Φ(l1 ). Let e be any point of B. We shall construct, by ordinary recursion, an injective infinite subfamily {en(k) : k ∈ N} of the sequence {en : n ∈ N} in such a way that ||e − en(0)/20 − ... − en(k) /2k ||  1/2k+1 for each k ∈ N. Evidently, we can find an element en(0) satisfying the inequality ||e − en(0)||  1/2. Suppose now that for a natural number k, the partial family {en(0) , en(1) , ... , en(k) } has already been defined. Then we may write ||2k+1 (e − en(0) − ... − en(k)/2k )||  1.

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Since the sequence {en : n ∈ N} is everywhere dense in B, we can choose an index n(k + 1) such that n(k + 1) > max{n(0), n(1), ..., n(k)} and ||2k+1 (e − en(0) − ... − en(k)/2k ) − en(k+1)||  1/2. From the latter inequality we immediately obtain ||e − en(0) − ... − en(k)/2k − en(k+1)/2k+1 ||  1/2k+2 . Proceeding in this manner, we come to the desired family {en(k) : k ∈ N}. Now, let us put y = (yn )n∈N ∈ RN , where yn = 0 if n differs from all n(k) (k ∈ N), and yn = 1/2k if n = n(k) for some k ∈ N. Then it is easy to see that y = (yn )n∈N ∈ l1 ,

Φ(y) = e.

Consequently, we get the relation B ⊂ Φ(l1 ). Taking into account the linearity of Φ, we conclude that Φ(l1 ) = E which finishes the proof of the lemma. Theorem 3. Let (E, || · ||) be an arbitrary separable Banach space. Then there exists a Borel probability measure on E quasi-invariant with respect to some everywhere dense vector subspace of E. Proof. We start with a Borel probability measure ν on l1 which is quasi-invariant with respect to an everywhere dense vector subspace of l1 . The existence of ν follows, e.g., from Exercise 6 where a much stronger result is formulated (see also Exercise 11 from Chapter 3). Further, according to Lemma 2, there exists a linear continuous surjection Φ : l1 → E. Let us put

μ (X) = ν (Φ−1 (X))

(X ∈ B(E)).

Then, as we know, μ is a Borel probability measure on E since it is the image of ν under Φ. Moreover, it can easily be verified that if the original measure ν is quasi-invariant with respect to an everywhere dense vector subspace U of l1 , then the measure μ is quasiinvariant with respect to the vector subspace V = Φ(U) of E. Taking into account the fact that Φ is a continuous surjection, we see that V is everywhere dense in E which ends the proof of Theorem 3.

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In connection with the preceding result, we can pose the general question concerning the existence of a Borel probability measure on a topological vector space, quasi-invariant with respect to an everywhere dense vector subspace. More exactly, we can formulate the following problem. Problem. Give a characterization of all those topological vector spaces E for which there exists at least one Borel probability measure quasi-invariant with respect to some everywhere dense vector subspace of E. This problem still remains open for the class of all topological vector spaces. However, for the class of all Banach spaces, we may assert that the above-mentioned problem is completely solved. Indeed, Theorem 3 states that every separable Banach space E admits a Borel probability measure quasi-invariant with respect to an everywhere dense vector subspace of E. On the other hand, if G is a nonseparable metrizable topological group whose topological weight is not a real-valued measurable cardinal number (see Appendix 3), then G does not admit a Borel probability measure quasi-invariant with respect to an everywhere dense subgroup of G (see Exercise 2). Thus, under the set-theoretical assumption that no cardinal number is real-valued measurable, we can infer that for a given Banach space E, these two assertions are equivalent: 1) E is a separable space; 2) there exists a Borel probability measure on E quasi-invariant with respect to some everywhere dense vector subspace of E. Fortunately, it turns out that an additional set-theoretical assumption can be omitted here. In this connection, see Exercise 7 in which the corresponding result is presented within the theory ZFC. Evidently, the problem posed above has a natural analogue for nonzero σ -finite invariant Borel measures. Namely, we can formulate the following problem. Problem. Give a characterization of all those topological vector spaces E for which there exists at least one nonzero σ -finite Borel measure invariant with respect to some everywhere dense vector subspace of E. This problem remains open, too, and we only want to recall the fact that there are infinitedimensional topological vector spaces E which admit a nonzero σ -finite Borel measure invariant under an everywhere dense vector subspace of E (see Example 3 from Chapter 3).

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For a while, let us return to Lemma 2 and recall that in accordance with this lemma for every separable Banach space E, there exists a linear continuous surjection Φ : l1 → E. Also, we know that there exists a nonzero σ -finite Borel measure ν on l1 invariant with respect to an everywhere dense vector subspace of l1 (see Exercise 11 of Chapter 3). Unfortunately, we cannot apply to Φ and to ν an argument similar to the proof of Theorem 3 in order to establish the existence of a nonzero σ -finite Borel measure on E invariant with respect to an everywhere dense vector subspace of E. The real reason of this fact can easily be explained. Indeed, simple examples show that a homomorphic image of a σ -finite measure is not, in general, a σ -finite measure. So in this situation, an argument analogous to the proof of Theorem 3 does not work. Now, let us discuss once more an important geometric property of the classical Lebesgue measure λn on the Euclidean space Rn , the so-called Steinhaus property (see Chapter 3). Our goal is to show that this property completely characterizes all those measures which are absolutely continuous with respect to λn . We restrict our further considerations to the case n = 1. A more general case of Rn may be considered analogously (cf. also Exercise 10). Let μ be a σ -finite Borel measure on the real line R. We shall say that μ has the Steinhaus property if for any set X ∈ dom(μ ) with μ (X) > 0, the difference set X − X = {x − x : x ∈ X, x ∈ X} is a neighborhood of zero in R (see Chapter 3). Below, we will give a characterization of all those σ -finite Borel measures on R which have the Steinhaus property (see [179]). First, let us describe one typical situation when the Steinhaus property fails to be true. Theorem 4. Let μ be a σ -finite Borel measure on R satisfying the following condition: there exist a Borel set X ⊂ R with μ (X) > 0 and a sequence of real numbers {tn : n < ω } such that limn→+∞tn = 0, limn→+∞ μ (X + tn ) = 0. Then μ does not possess the Steinhaus property. Proof. We may assume, without loss of generality, that 0 < μ (X) < +∞ and that the following relations are satisfied:

μ (X + tn ) < μ (X)/2n

(n = 2, 3, 4, ...).

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Let us define Y = X \ (∪{X + tn : 2  n < ω }). Then we have the inequality

μ (Y )  μ (X)/2 > 0. Also, by virtue of the definition of Y , we get (2  n < ω ),

Y ∩ (Y + tn ) = 0/ from which it immediately follows that tn ∈ Y − Y

(2  n < ω ).

Taking into account the equality limn→+∞tn = 0, we conclude that the difference set Y − Y is not a neighborhood of zero in R, so μ does not have the Steinhaus property. In order to characterize all those measures on R which possess the Steinhaus property, we need two auxiliary propositions. Lemma 3. Let μ be a σ -finite Borel measure on R absolutely continuous with respect to the Lebesgue measure λ . Then μ has the Steinhaus property. This lemma is trivial by virtue of the Radon-Nikodym theorem and in view of the wellknown fact that λ has the Steinhaus property (see Chapter 3). Lemma 4. Let μ be a σ -finite Borel measure on R singular with respect to λ . Denote by X a Borel subset of R such that

λ (X) = 0, μ (R \ X) = 0, and consider the function f (t) = μ (X + t)

(t ∈ R).

Then f (t) is equal to zero for λ -almost all points t ∈ R. Proof. The argument is fairly standard. Consider the set {(t, x) : x ∈ X + t} ⊂ R2 and denote by Φ its characteristic function which is Borel and hence measurable with respect to the product measure μ ⊗ λ . Using the Fubini theorem, we may write 

f (t)d λ (t) =



μ (X + t)d λ (t) =

 

Φ(t, x)d λ (t)d μ (x) =

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(

χx−X (t)d λ (t))d μ (x) =



λ (X)d μ (x) = 0,

which immediately yields the required result. Theorem 5. Let ν be a σ -finite Borel measure on R. The following two assertions are equivalent: (1) ν is absolutely continuous with respect to λ ; (2) ν possesses the Steinhaus property. Proof. The implication (1) ⇒ (2) is a direct consequence of Lemma 3. Let us establish the converse implication (2) ⇒ (1). Suppose that ν has the Steinhaus property but is not absolutely continuous with respect to

λ . Then ν can be represented in the form ν = ν1 + μ , where ν1 is a σ -finite Borel measure on R absolutely continuous with respect to λ and μ is a nonzero σ -finite Borel measure on R which is singular with respect to λ , i.e., there exists a Borel set X ⊂ R such that

μ (X) > 0, λ (X) = 0, μ (R \ X) = 0. By virtue of Lemma 4, the function f (t) = μ (X + t) (t ∈ R) is equal to zero for λ -almost all points t ∈ R. In particular, there exists a sequence {tn : n <

ω } ⊂ R of points tending to zero such that f (tn ) = 0 for any n < ω . Let us put Z = X \ (∪{X + tn : n < ω }). Then we get μ (Z) = μ (X) > 0. Further, since

ν (Z) = ν1 (Z) + μ (Z)  μ (Z), we also have ν (Z) > 0. At the same time, in view of the definition of Z, it is easy to see that Z ∩ (Z + tn ) = 0/ (n < ω ), thus it follows (by virtue of limn→+∞tn = 0) that ν does not possess the Steinhaus property, contradicting our assumption. The contradiction obtained finishes the proof. Example 2. A similar argument works for the left Haar measure on a Polish locally compact topological group (G, ·). We thus conclude that a σ -finite Borel measure ν on G has

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the Steinhaus property if and only if ν is absolutely continuous with respect to the left Haar measure μ on G (see Exercise 10 for this chapter). Moreover, the Steinhaus property implies at once that if A and B are Borel subsets of (G, ·) such that μ (A) > 0 and μ (B) > 0, then the set A · B = {a · b : a ∈ A, b ∈ B} contains in itself a nonempty open subset of G (see Exercise 11 below). In this context, it should be underlined that even in the case of the real line R and of the Lebesgue measure

λ on R, there are λ -measure zero sets C ⊂ R whose difference set C −C is a neighborhood of 0. For more details and further related results, see Chapters 11 and 12. As already mentioned in Chapter 3, the Steinhaus property admits an analogue in terms of the Baire category for arbitrary topological groups. Namely, let (G, ·) be a topological group and let Z be a subset of G having the Baire property and not belonging to the family of all first category subsets of G. Then the set Z ·Z −1 contains a neighborhood of the neutral element of G. This important fact also implies that if A and B are any two second category sets in G, both having the Baire property, then A · B contains a nonempty open subset of G.

EXERCISES 1. Let Γ be a σ -compact locally compact topological group equipped with a Haar measure

ν and let ν  denote the completion of ν . Suppose also that G is some separable topological group and let Ψ : Γ→G be a group homomorphism. By applying the Steinhaus property of ν , show that the following two assertions are equivalent: (a) Ψ is measurable with respect to ν  ; (b) Ψ is continuous. Start with this equivalence and prove by using the Mackey result formulated in this chapter and the classical Baire theorem on category that for any Polish topological group H, the next two relations are equivalent: (c) H is locally compact; (d) there exists a Borel probability measure on H quasi-invariant with respect to the group of all left translations of H.

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2. A well-known theorem of topological measure theory states that any σ -finite Borel measure given on a metric space E with non-real-valued measurable topological weight is concentrated on a closed separable subset of E (see Appendix 3). By starting with this theorem, establish the following result. Let G be a metrizable topological group whose topological weight is not a real-valued measurable cardinal. Suppose also that there exists a nonzero σ -finite Borel measure on G quasi-invariant with respect to an everywhere dense (in G) group of left (right) translations of G. Then G is a separable topological group. Consequently, if G is complete, then G is a Polish topological group. 3. Let G be an arbitrary σ -compact locally compact topological group satisfying the inequality card(G) > 2c and let μ be the left Haar measure on G. Then, by definition, μ is invariant with respect to the group of all left translations of G. On the other hand, show that G is not a separable topological space. In particular, suppose that c = ω1 , 2 c = ω2 , and let Γ = {0, 1}ω2 , where {0, 1} is equipped with the discrete topology. Then Γ is a commutative (with respect to addition (modulo 2)) nonseparable compact topological group, the weight of Γ is equal to ω2 , and according to the classical theorem of Ulam (see [238]),

ω2 is not a real-valued measurable cardinal. Conclude from the said above that the assumption of the metrizability of a topological group G is essential for the validity of the result formulated in Exercise 2. 4. Let λn denote the standard Borel measure on the n-dimensional Euclidean space Rn and let μ be an arbitrary nonzero σ -finite measure defined on the Borel σ -algebra of Rn . Demonstrate that these two assertions are equivalent: (a) μ is quasi-invariant with respect to the group of all translations of Rn ; (b) there exists a strictly positive Borel function φ : Rn → R such that

μ (Z) =



Z

φ (x)d λn (x)

for each Borel subset Z of Rn . Prove an analogous result in a more general situation where the space Rn is replaced by a locally compact Polish topological group G and the standard Borel measure λn is replaced by the left Haar measure on G.

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5∗ . Let G be a subgroup of the additive group RN , such that G \ R(N) = 0. / Prove that there exists a family { μn : n ∈ N} of probability measures satisfying these two conditions: (a) for each natural number n, the measure μn is defined on the Borel σ -algebra of the real line R and is quasi-invariant with respect to the group of all translations of R; (b) the product measure of the family { μn : n ∈ N} is not quasi-invariant with respect to G. 6. Consider the Banach space l1 as a Borel subset of the Polish topological vector space RN and show that there exists a family { μn : n ∈ N} of probability measures satisfying these two relations: (a) for each natural number n, the measure μn is defined on the Borel σ -algebra of R and is quasi-invariant with respect to the group of all translations of R; (b) for the product measure μ of the family { μn : n ∈ N}, the inequality μ (l1 ) > 0 holds true. Deduce from this fact that there exists a Borel probability measure ν on the Banach space l1 , quasi-invariant with respect to the vector space R(N) which is everywhere dense in l1 (cf. Exercise 11 from Chapter 3, where a much stronger result is formulated). 7. Let E be a nonseparable normed vector space, G denote the group of all translations of E, and let H be an everywhere dense subgroup of G. Further, let B be an arbitrary ball in E. Show that for any σ -finite H-quasi-invariant measure μ on E (not necessarily Borel), the implication B ∈ dom(μ ) ⇒ μ (B) = 0 holds true. Infer from this fact that there exists no nonzero σ -finite Borel measure on E quasi-invariant with respect to H. Finally, conclude that for a Banach space F, the following two assertions are equivalent: (a) F is separable; (b) there exists a Borel probability measure on F quasi-invariant with respect to some everywhere dense vector subspace of F. 8. Let κ be an arbitrary cardinal number. Prove that there exists a Borel probability measure on the topological vector space Rκ , quasi-invariant with respect to the everywhere dense vector subspace R(κ ) of Rκ . Also, check that if κ is strictly greater than the cardinality

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continuum, then the space Rκ is not separable. Deduce from these results that for a nonmetrizable topological vector space F, the assertions (a) and (b) of the preceding exercise can be non-equivalent. 9∗ . Let E be a normed vector space and let G be a vector subspace of E. We shall say that G is admissible if it is everywhere dense in E and there exists a nonzero σ -finite Borel measure on E quasi-invariant with respect to G. Show that the following two assertions are equivalent: (a) E is infinite-dimensional; (b) any admissible vector subspace of E is not minimal with respect to the inclusion relation. In order to establish the equivalence (a) ⇔ (b), utilize the fact that for any infinitedimensional normed vector space F, there exists a linear functional h : F → R which is discontinuous at all points of F (cf. Exercise 7 from Chapter 18). 10. Generalize Theorem 5 of this chapter to the case of a Polish locally compact group G equipped with its left Haar measure μ . In other words, show that for a σ -finite Borel measure ν on G, these two assertions are equivalent: (a) ν has the Steinhaus property; (b) ν is absolutely continuous with respect to μ . For this purpose, use an argument similar to the proof of Theorem 5 and apply the wellknown fact that if X is a Borel subset of G, then the equality μ (X) = 0 is equivalent to the equality μ (X −1 ) = 0 (see, e.g., [80] or [83]). 11. Let (G, ·) be a σ -compact locally compact topological group and let μ denote the left (right) Haar measure on G. By using the Steinhaus property, prove that for any two Borel sets A ⊂ G and B ⊂ G with μ (A) > 0 and μ (B) > 0, the set A · B = {a · b : a ∈ A, b ∈ B} has nonempty interior.

Chapter 5

Measurability properties of real-valued functions

In this chapter, two concepts associated with measurability properties of various real-valued functions are introduced and examined. These concepts are the notion of a relatively measurable real-valued function with respect to a given class M of measures and the notion of an absolutely nonmeasurable function with respect to M. Naturally, the usefulness of these concepts will be illustrated below by a number of relevant examples (cf. [130]). Further, a characterization of absolutely nonmeasurable functions will be established in the most important case when the role of M is played by the class ME of all nonzero σ -finite diffused measures on a given base set E. Also, it will be shown that the functions produced by Vitali’s classical partition of the real line R are relatively measurable with respect to the class of all extensions of the Lebesgue measure λ on this line. In order to obtain the latter result, Theorem 2 of Chapter 2 will be utilized. We begin with the precise definitions of the above-mentioned concepts. Let E be a set and let M be a class of measures on E. In general, we mean that the domains of measures from M are various σ -algebras of subsets of E. As usual, the real line R is assumed to be equipped with its Borel σ -algebra B(R), so we permanently will be dealing with the canonical measurable space (R, B(R)). We shall say that a function f : E → R is relatively measurable with respect to M if there exists at least one measure μ ∈ M such that f is measurable with respect to μ , i.e., (∀B ∈ B(R))( f −1 (B) ∈ dom(μ )). On the other hand, if there exists no measure μ ∈ M such that f is μ -measurable, then we shall say that f is absolutely nonmeasurable with respect to the given class M. Obviously, the introduced notion of a relatively measurable function generalizes the notion of a real-valued measurable function with respect to a concrete measure μ on E. Indeed, it we take the one-element class M = {μ }, then the relative measurability of a given function A.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7_5, © 2009 Atlantis Press/World Scientific

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f : E → R with respect to M is equivalent to the ordinary definition of the μ -measurability of f , and the absolute nonmeasurability of f with respect to the same M is equivalent to the standard definition of the μ -nonmeasurability of f . As soon as the notion of a relatively measurable (absolutely nonmeasurable) function is introduced, it trivially becomes possible to define an analogous notion of a relatively measurable (absolutely nonmeasurable) subset of E. A set X ⊂ E is called relatively measurable with respect to a class M of measures on E if the characteristic function χX of X is relatively measurable with respect to M. On the other hand, if there exists no measure μ ∈ M such that χX is μ -measurable, then X is called absolutely nonmeasurable with respect to M. For instance, let E be a base set equipped with a group G of its transformations and let M denote the class of all nonzero σ -finite G-invariant (or, more generally, G-quasi-invariant) measures on E. If a set X ⊂ E is absolutely nonmeasurable with respect to M, then we briefly say that X is G-absolutely nonmeasurable in E. Let μ be a measure on E such that (∀x ∈ E)({x} ∈ dom(μ )). As usual, we say that μ is diffused (or continuous) if it vanishes at all singletons in E, that is μ ({x}) = 0 for each x ∈ E. Example 1. For any set E, let ME be the class of all nonzero σ -finite diffused measures on E. Let f : E → R be a function and let for some t0 ∈ R, the relation card( f −1 (t0 )) > ω be satisfied, where ω denotes the first infinite cardinal number. In this case, we can assert that f is relatively measurable with respect to ME . Indeed, it is not difficult to define a complete diffused probability measure μ on E such that f −1 (t0 ) ∈ dom(μ ), μ ( f −1 (t0 )) = 1. Consequently, for any set T ⊂ R, we have μ ( f −1 (T )) = 1 if t0 ∈ T , and μ ( f −1 (T )) = 0 if t0 ∈ T . This circumstance, obviously, implies that f is measurable with respect to the measure μ and hence is relatively measurable with respect to the given class ME . We see, in particular, that if for an original set E the relation card(E) > 2ω = c holds true, then every function f : E → R is relatively measurable with respect to ME . Indeed, in this case there always exists a point t0 ∈ R for which we have card( f −1 (t0 )) > ω .

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Let F be a topological space all singletons of which belong to the Borel σ -algebra B(F). We recall that F is a universal measure zero space if there exists no nonzero σ -finite diffused Borel measure on F. Nontrivial examples of such spaces are given in exercises for this chapter, namely, see Exercises 15 and 16 where it is pointed out that some non-discrete Hausdorff universal measure zero spaces can be of arbitrarily large cardinality. We have already mentioned that the question is highly nontrivial whether or not there exist uncountable universal measure zero subspaces of the real line R. This question was investigated from different points of view. By using delicate arguments, several famous authors (Hausdorff, Luzin, Sierpi´nski, Marczewski and others) have established that there are uncountable universal measure zero subspaces of R. One classical construction of such a subspace of R, due to Luzin and Sierpi´nski, is presented in [113], [148], [150], and [172]. It is based on the canonical decomposition of a proper analytic subset of R into its Borel components (constituents). There are also other interesting constructions of uncountable universal measure zero subspaces of R (see, for instance, [79], [197], [250], and Appendix 1). By taking into account Example 1 and using the notion of a universal measure zero space, a characterization of absolutely nonmeasurable functions with respect to the class ME can be obtained. Theorem 1. A function f : E → R is absolutely nonmeasurable with respect to ME if and only if the following two conditions hold: 1) for each r ∈ R, the set f −1 (r) is at most countable; 2) the set ran( f ) is a universal measure zero subspace of R. Proof. Suppose first that f is absolutely nonmeasurable with respect to ME . Then the argument given in Example 1 shows that condition 1) must be satisfied. Let us verify that condition 2) is valid, too. Indeed, assuming that ran( f ) is not a universal measure zero subset of R, consider some Borel diffused probability measure ν on ran( f ) and denote S = { f −1 (Z) : Z ∈ dom(ν )}. Obviously, S is a σ -algebra of subsets of E and the family of countable sets { f −1 (r) : r ∈ ran( f )} forms a partition of E. We put

μ ( f −1 (Z)) = ν (Z)

(Z ∈ dom(ν )).

In this manner, the probability measure μ on the σ -algebra S is well defined and, according to its definition, the function f becomes μ -measurable. Clearly, the completion μ  of

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μ is a diffused measure and f remains measurable with respect to μ  . However, this circumstance contradicts our assumption that f is absolutely nonmeasurable with respect to the class ME . The contradiction obtained shows the necessity of conditions 1) and 2) for the absolute nonmeasurability of f with respect to ME . Now, assume that these two conditions are fulfilled for a given function f and let us establish that f is absolutely nonmeasurable with respect to ME . Suppose for a moment that there exists a measure μ belonging to ME such that f is measurable with respect to μ . We may assume, without loss of generality, that μ is a probability measure. Then, denoting by B(ran( f )) the Borel σ -algebra of ran( f ) ⊂ R, we may define

ν (Z) = μ ( f −1 (Z))

(Z ∈ B(ran( f ))).

So, we get a Borel probability measure ν on the space ran( f ) ⊂ R and we see that ν is diffused in view of condition 1). But this circumstance contradicts condition 2) and the obtained contradiction ends the proof of the theorem. Remark 1. Let ω1 denote, as usual, the least uncountable cardinal number and let E be a set such that card(E) > ω1 . It is impossible to establish within ZFC theory that there exist functions f : E → R absolutely nonmeasurable with respect to the class ME . Indeed, assuming the Continuum Hypothesis (c = ω1 ), we have card(E) > c and, as we already know, for such a set E there are no functions f : E → R absolutely nonmeasurable with respect to ME . Also, it is impossible to prove within ZFC theory the existence of functions f : R → R which are absolutely nonmeasurable with respect to the class MR . This fact directly follows from Theorem 1 and the circumstance that there are models of set theory in which the cardinality of the continuum c is strictly greater than ω1 and the cardinality of any universal measure zero subspace of R does not exceed ω1 . Models of this sort were first constructed by Baumgartner and Laver (for more details, see [172] and the references therein). On the other hand, under some additional set-theoretical hypotheses, it is not difficult to demonstrate that there exists a function g:R→R absolutely nonmeasurable with respect to the class MR . For instance, the existence of such a function follows from the existence of a Luzin subset of R whose cardinality is equal to c (more information about Luzin sets can be found, e.g., in [40], [143], [148], [172], [176], [192], and [222]; see also Chapter 12). The next example shows that, assuming

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Martin’s Axiom, absolutely nonmeasurable functions can be found even among injective homomorphisms of the additive group R into itself. Example 2. We recall that X ⊂ R is a generalized Luzin set if card(X) = c and for every first category set Y ⊂ R, the inequality card(X ∩Y ) < c holds true. Under Martin’s Axiom, there exists a generalized Luzin set L ⊂ R which is simultaneously a vector space over the field Q of all rational numbers. The construction of such a set L is fairly standard and is based on the method of transfinite recursion (cf. Chapter 12). It is well known that every Luzin set equipped with the induced topology is a universal measure zero space and, under Martin’s Axiom, every generalized Luzin set is a universal measure zero space, too. In particular, the above-mentioned set L is a universal measure zero subset of R. Let g:R→L be some isomorphism between the two sets R and L which both are regarded as vector spaces over Q. Then g can be considered as an injective group homomorphism from R into R with ran(g) = L and therefore g is a nontrivial solution of the classical Cauchy functional equation

φ (x + y) = φ (x) + φ (y)

(x ∈ R, y ∈ R).

Furthermore, according to Theorem 1, g turns out to be absolutely nonmeasurable with respect to the class MR . In this context, let us remind that any nontrivial solution of the Cauchy functional equation is necessarily nonmeasurable in the Lebesgue sense and does not possess the Baire property (see Exercise 2). It should also be underlined that the latter fact does not need any additional set-theoretical axioms which once again emphasizes the circumstance that the absolute nonmeasurability of a function (acting from R into itself) is a much stronger property than its nonmeasurability in the Lebesgue sense. It is reasonable to recall in this place that the first construction of a subset of R which is nonmeasurable with respect to the Lebesgue measure λ on R, was given by Vitali in his classical work [239]. Clearly, the characteristic function of a Vitali set yields the first example of a function acting from R into R and nonmeasurable in the Lebesgue sense. Now, we would like to introduce other real-valued functions of Vitali type and examine their measurability properties with respect to various classes of measures on R. Let W be an equivalence relation on R all equivalence classes of which are at most countable.

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We shall say that a mapping f : R → R is a Vitali type function for W if (r, f (r)) ∈ W for each r ∈ R and the set ran( f ) is a selector of the partition of R determined by W . It can be shown that if V is the classical Vitali equivalence relation on R, i.e., the equivalence relation defined by the formula (x, y) ∈ V ⇔ (x ∈ R & y ∈ R & x − y ∈ Q), then any Vitali type function for V is absolutely nonmeasurable with respect to the class of all translation-invariant extensions of the Lebesgue measure λ on R (see Exercise 2 of this chapter). However, the following somewhat surprising statement is valid. Theorem 2. Let M(λ ) denote the class of all those measures on R which extend λ . Then every Vitali type function for V is relatively measurable with respect to M(λ ). Proof. Our argument is essentially based on a result discussed in Chapter 2, which states that if E is any set, μ is a σ -finite measure on E and {Xn : n < ω } is a disjoint countable family of subsets of E, then there always exists a measure μ  on E extending μ and satisfying the relation {Xn : n < ω } ⊂ dom(μ  ). We are going to apply this result to the Lebesgue measure λ and to a certain partition of R. Let f : R → R be any Vitali type function for V . We put X = ran( f ). The family of sets {X + q : q ∈ Q} forms a countable partition of R. Therefore, according to the above-mentioned result, there exists a measure λ  on R extending λ and such that X + q ∈ dom(λ  ) for all q ∈ Q. We assert that f is measurable with respect to λ  . Indeed, for each Borel subset B of R, the equality f −1 (B) = ∪{X ∩ B + q : q ∈ Q} is easily verified. This equality can also be rewritten as f −1 (B) = ∪{(X + q) ∩ (B + q) : q ∈ Q}. The right-hand side of the latter relation directly indicates that the set f −1 (B) is λ  measurable, which yields at once the measurability of f with respect to λ  and hence the relative measurability of f with respect to the class M(λ ). This conclusion ends the proof of Theorem 2. Example 3. Assuming Martin’s Axiom, it is not difficult to construct a generalized Luzin set L ⊂ R and an equivalence relation W ⊂ R × R such that:

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1) (∀r ∈ R)(card(W (r)) = ω ); 2) L is a selector of the partition {W (r) : r ∈ R} of R. Let h : R → R be a Vitali type function for W such that ran(h) = L. Then, in view of Theorem 1, h is absolutely nonmeasurable with respect to the class MR and, consequently, h is absolutely nonmeasurable with respect to the class M(λ ). The previous example shows that the validity of Theorem 2 is essentially connected with some special (in fact, group-theoretical) properties of the Vitali partition V . Denote by M  (λ ) the class of all those measures on R which extend λ and are quasiinvariant under the group of all translations of R. Recall that quasi-invariant measures were defined in Chapter 4 and some of their general properties were also discussed therein. As observed before the formulation of Theorem 2, any Vitali type function for V is absolutely nonmeasurable with respect to the class of all translation-invariant extensions of λ . On the other hand, it can be demonstrated that there exists a Vitali type function for V which is relatively measurable with respect to M  (λ ) (see Theorem 3 below). At the same time, we do not know an answer to the following question. Problem. Does there exist a Vitali type function for V which is absolutely nonmeasurable with respect to M  (λ )? In this context, it makes sense to consider in more details Vitali’s classical construction [239] of a non-Lebesgue measurable subset of the real line R. Recall once more that the Vitali partition of R is determined by the following equivalence relation V (x, y): (x, y) ∈ V ⇔ (x ∈ R & y ∈ R & x − y ∈ Q). Let Z be an arbitrary selector of this partition. It can be shown that Z is nonmeasurable with respect to every measure on R extending λ and invariant under the group Q of all rational translations of R (cf. [46], [80], [83], [108], [143], [176], [183], [192], [240], and Exercise 2 for this chapter). According to our terminology, this circumstance immediately implies that Z is absolutely nonmeasurable with respect to the class of all translation-invariant extensions of λ . For translation-quasi-invariant extensions of λ , the situation is essentially different, which will be stated in the sequel. Among the selectors of the Vitali partition (which usually are called Vitali sets), we can encounter some subgroups of the additive group R. Indeed, if we treat R as a vector space over the field Q, then we may apply a well-known theorem from linear algebra which states

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that a vector subspace Q of R admits a complemented subspace in R, i.e., we come to a representation R = Q+H

(Q ∩ H = {0}),

where H is some vector space over Q. Actually, H is a hyperplane in R complemented to the ”line” Q in R. Consequently, H is also a subgroup of R such that no translationinvariant extension of λ can make H to be measurable. In addition, the relation card(R/H) = card(Q) = ω holds true, where the symbol R/H stands for the family of all translates of H. There are several other constructions, which establish that there exist subgroups G of R nonmeasurable with respect to λ but satisfying the inequality card(R/G) > ω . For instance, by applying the method of transfinite recursion, it is not difficult to construct a subgroup G of R satisfying the above-mentioned inequality and the relation

λ∗ (R \ G) = 0, where λ∗ stands for the inner measure associated with λ . Constructions of such groups G ⊂ R can be done similarly to the classical Bernstein construction (cf. [148], [176], [192], and Exercise 3). But the advantage of G over H is that G can be made measurable with respect to an appropriate translation-invariant extension of λ . This fact is not accidental. Its certain analogue remains valid in a more general situation, namely, in the case of an uncountable commutative group instead of R. In this connection, see especially Chapter 10 where we give an application of one purely algebraic statement to the question of the existence of nonmeasurable subgroups of uncountable commutative groups. In the same chapter, we solve the problem of extending nonzero σ -finite invariant (quasi-invariant) measures on such groups, by using their appropriate nonmeasurable subgroups. Let us return to the Vitali partition of R and to Vitali sets of special type. Let H ⊂ R be as earlier, that is let H be a hyperplane in R complemented to Q, where R is again regarded as a vector space over Q. We have already underlined that H, as a particular case of a Vitali set, is absolutely nonmeasurable with respect to the class of all translation-invariant extensions of λ . On the other hand, we can demonstrate that H becomes measurable with respect to a suitable translation-quasi-invariant extension of λ . The following statement contains a much stronger result which shows that there are many possibilities for obtaining such translation-quasi-invariant extensions of λ .

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Theorem 3. There exist continuumly many measures ν on R which extend λ , are quasiinvariant under the group of all translations of R, and satisfy the relation H ∈ dom(ν ). Proof. First, we would like to observe that H is λ -thick in R, i.e.,

λ∗ (R \ H) = 0 which is true because H is an everywhere dense subgroup of R and H is not of λ -measure zero. The details of checking the λ -thickness of H are left to the reader. Further, we introduce the disjoint family of sets {Hk : k < ω } = {q + H : q ∈ Q}. Obviously, for any h ∈ R, the family {h + Hk : k < ω } coincides with the family {Hψ (k) : k < ω }, where ψ is some permutation of ω . We thus derive that {Hk : k < ω } is a countable translation-invariant partition of R into λ -thick sets. Now, we introduce the class of sets S = {∪k<ω (Hk ∩ Xk ) : Xk ∈ dom(λ ) f or all k < ω }. It can easily be verified that S is a translation-invariant σ -algebra of subsets of R. Fix a sequence {ak : k < ω } of strictly positive real numbers such that

∑ ak = 1.

k<ω

Then take an arbitrary set ∪k<ω (Hk ∩ Xk ) from S and put

ν (∪k<ω (Hk ∩ Xk )) =

∑ ak λ (Xk ).

k<ω

In this manner we define a certain functional ν on S . Indeed, the definition of ν is correct in view of the λ -thickness of all sets Hk . By reason of the same circumstance, the functional

ν is countably additive, so ν is a σ -finite measure on S . If X ∈ dom(λ ), then ν (X) = ν (∪k<ω (Hk ∩ X)) =

∑ ak λ (X) = λ (X)

k<ω

which shows that ν extends λ . Finally, if we have

ν (∪k<ω (Hk ∩ Xk )) =

∑ ak λ (Xk ) = 0,

k<ω

then taking into account the inequalities ak > 0 for all k < ω , we get

λ (Xk ) = 0 (k < ω ), which implies for any h ∈ R that

ν (h + ∪k<ω (Hk ∩ Xk )) = ν (∪k<ω (Hψ (k) ∩ (h + Xk ))) =

∑ aψ (k) λ (h + Xk ) = 0,

k<ω

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where ψ is again some permutation of ω (of course, depending on h). Therefore, the measure ν is quasi-invariant under the group of all translations of R. Moreover, a similar argument yields that ν is also quasi-invariant under all central symmetries of R and hence

ν is quasi-invariant under the group of all isometric transformations of R. Finally, since ν depends on a choice of a sequence of strictly positive real numbers {ak : k <

ω } and there are continuumly many such sequences, we conclude that there exist at least continuumly many pairwise distinct translation-quasi-invariant extensions of λ for which the Vitali set H becomes measurable. Of course, it is important here that different choices of {ak : k < ω } produce different extensions of λ . Theorem 3 has thus been proved. Let E be a set equipped with a σ -finite measure μ and let f : E → R be a function satisfying the following condition: there exists a Borel probability measure ν on ran( f ) such that the graph of f is a (μ ⊗ ν )thick subset of the product space E × ran( f ). Then, applying the standard argument (cf. [141] and Lemmas 1,2,3 of Chapter 2), it is not hard to demonstrate that there exists a measure μ  on E such that: (1) μ  extends μ ; (2) f is measurable with respect to μ  . In other words, f turns out to be relatively measurable with respect to the class M(μ ) of all extensions of μ . In particular, if E = R and a function f : R → R has (λ ⊗ λ )-thick graph, then f turns out to be measurable with respect to an appropriate extension of λ and, consequently, f is relatively measurable with respect to the class M(λ ). Notice that there are various examples of functions f : R → R whose graphs are (λ ⊗ λ )thick subsets of the plane R2 (see, for instance, [72], [148], [192], and Exercise 8 of Chapter 2). Moreover, the following much stronger statement is valid. Theorem 4. There exists a function g : R → R having the property that for any nonzero

σ -finite diffused Borel measure μ on R and for any σ -finite measure ν on R, the graph of g is a (μ ⊗ ν )-thick subset of the plane R2 . Proof. In order to show the validity of this theorem, we start with a partition {Xt : t ∈ R} of R into Bernstein sets. Recall that a Bernstein set is any totally imperfect subset of R whose complement is also totally imperfect. The existence of such a partition is well known (see, e.g., [176], [192], and Exercise 5 for this chapter). Now, we define the function g as follows: for each x ∈ R, we put g(x) = t if and only if x ∈ Xt .

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It remains to check that g is the required function. Let μ be an arbitrary nonzero σ -finite diffused Borel measure on R and let ν be an arbitrary nonzero σ -finite measure on R. We must verify that the graph Gr(g) of g is (μ ⊗ ν )-thick in R2 . Indeed, if Z is any (μ ⊗ ν )-measurable set with (μ ⊗ ν )(Z) > 0, then by virtue of the classical Fubini theorem, there exists a point t0 ∈ R such that

μ ({x ∈ R : (x,t0 ) ∈ Z}) > 0. This circumstance implies that the set {x ∈ R : (x,t0 ) ∈ Z} contains a nonempty perfect subset of R (because μ is a diffused Radon measure). Keeping in mind the fact that Xt0 is a Bernstein subset of R, we get Xt0 ∩ {x ∈ R : (x,t0 ) ∈ Z} = 0, / (∃x)(g(x) = t0 & (x, g(x)) ∈ Z). Therefore, the set Gr(g) has nonempty intersection with Z and hence Gr(g) is (μ ⊗ ν )-thick in R2 which completes the proof of the theorem. Notice that in our further considerations we will be dealing with some other versions of Theorem 4 (see Chapter 18). We would like to finish this chapter with introducing one more notion closely connected with measurability properties of real-valued functions with respect to certain classes of measures. It should be mentioned that for a special class of measures, this notion was first introduced and thoroughly considered by Marczewski. Let E be a set and let M be a class of measures on E. We shall say that a function f : E → R is universally (or absolutely) measurable with respect to M if f turns out to be measurable with respect to all measures from M. In accordance with the above definition, we shall say that a set X ⊂ E is universally (or absolutely) measurable with respect to M if the characteristic function χX of X turns out to be universally measurable with respect to M. The concept of universal measurability of sets and functions will be discussed more thoroughly in our further considerations. Here we restrict ourselves only to giving one typical example from classical descriptive set theory. Example 4. Let E be a Polish topological space and let M denote the class of the completions of all σ -finite Borel measures on E. Let A (E) stand for the class of all Suslin subsets of E (see Chapter 8) and let σ (A (E)) denote the σ -algebra generated by this class. It is well known that any set X ∈ σ (A (E)) is universally measurable with respect to M (see

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[24], [26], [52], [99], [210], and Appendix 2). As shown by G¨odel and Novikov, an analogous fact fails to be true within ZFC theory for projective subsets of E of higher levels. For more details, see [10], [91], [99], and [188]. EXERCISES 1. Verify that every Luzin subset of R is universal measure zero. For this purpose, use the fact that any σ -finite diffused Borel measure on a separable metric space E is concentrated on some first category subset of E (in this connection, see Appendix 3). 2∗ . Let E be a set and let G be a group of transformations of E. We say that a set X ⊂ E is a G-selector if X has exactly one common point with each G-orbit in E. Now, let G be a countable everywhere dense subgroup of the additive group R. By using an argument similar to Vitali’s [239], prove that every G-selector is absolutely nonmeasurable with respect to the class of all G-invariant extensions of λ . In particular, all G-selectors are nonmeasurable with respect to λ . Apply this result to the case G = Q. Verify that any Vitali type function for the Vitali partition V of R is absolutely nonmeasurable with respect to the class of all translation-invariant extensions of λ . Show also that any nontrivial solution f : R → R of the Cauchy functional equation

φ (x) + φ (y) = φ (x + y) (x ∈ R, y ∈ R) is nonmeasurable with respect to λ and does not have the Baire property (cf. Example 9 from Chapter 1). 3∗ . By using the method of transfinite recursion, construct a subset G of the real line R that satisfies the following three conditions: (a) G is a vector space over Q (in particular, G is a subgroup of the additive group R); (b) card(R/G) > ω ; (c) λ∗ (R \ G) = 0. Show that any such set G is nonmeasurable with respect to λ but, simultaneously, is relatively measurable with respect to the class of all translation-invariant extensions of λ . 4. Let G be an uncountable subgroup of the additive group R. Prove that every G-selector is a G-absolutely negligible subset of R (see Chapter 2). In particular, conclude from this fact that every G-selector is relatively measurable with respect to the class of all G-invariant extensions of the Lebesgue measure λ .

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5∗ . Applying the method of transfinite recursion, construct a partition of R into continuumly many Bernstein sets. Moreover, prove the following general (purely set-theoretical) statement which trivially implies the existence of such a partition of R. Namely, let E be an arbitrary infinite set and let {Ai : i ∈ I} be a family of subsets of E satisfying the relations card(I)  card(E), (∀i ∈ I)(card(Ai ) = card(E)). Show that there exists a disjoint family {B j : j ∈ J} of subsets of E such that card(J) = card(E) and card(Ai ∩ B j ) = card(E) for all indices i ∈ I and for all indices j ∈ J. In particular, take: E = R; card(I) = card(R) = c; {Ai : i ∈ I} = the family of all nonempty perfect subsets of R. Then the corresponding family {B j : j ∈ J} produces the required partition of R into continuumly many Bernstein sets. 6∗ . Consider R as a vector space over the field Q. Prove that there exists a vector subspace of R which simultaneously is a Bernstein set in R. Also, assuming Martin’s Axiom, show that there exists a vector subspace of R which simultaneously is a generalized Luzin set in R. 7. By using Martin’s Axiom, give a precise construction of the set L described in Example 3. 8. Assuming Martin’s Axiom, prove that every generalized Luzin subset of R is universal measure zero. 9∗ . Let E be a set and let G be a group of transformations of E. We say that a set Z ⊂ E is G-negligible in E if the following two conditions hold: (a) Z is relatively measurable with respect to the class M of all nonzero σ -finite G-invariant measures on E; (b) for any measure μ ∈ M, we have Z ∈ dom(μ ) ⇒ μ (Z) = 0. Observe that every G-absolutely negligible set in E is also G-negligible. Consider the case E = G = R2 and applying the method of transfinite induction, give an example of a function f : R → R whose graph is G-negligible but is not G-absolutely negligible.

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10. Denote by M the class of all those translation-invariant measures on R2 which extend the two-dimensional Lebesgue measure λ2 . Let f be an arbitrary function acting from R into R. Show that the graph Gr( f ) of f is relatively measurable with respect to M. Moreover, prove that there exists a translation-invariant measure μ on R2 which extends λ2 and for which the graphs of all functions acting from R into R are μ -measurable. For this purpose, use the Marczewski method of extending invariant measures described in Chapter 2. 11. Let E be a set and let μ be a measure on E. Denote by M(μ ) the class of all those measures on E which extend μ . Let f : E → R be a function whose range is countable. Show that f is relatively measurable with respect to M(μ ). For this purpose, apply Theorem 2 from Chapter 2. 12∗ . Denote by MR the class of the completions of all nonzero σ -finite diffused Borel measures on R. Let f : R → R be a function. Show that the following two assertions are equivalent: (a) f is relatively measurable with respect to MR ; (b) f admits a representation in the form f = g ◦ h, where a function g : R → R is Lebesgue measurable and a function h : R → R is a Borel isomorphism of R onto itself. Also, verify that there exist functions belonging to the class MR which are not measurable in the Lebesgue sense. 13. Let Rn and Rm be two Euclidean spaces such that n = 0, m = 0, n = m and let f : R n → Rm be an isomorphism between the additive groups Rn and Rm . The existence of such an isomorphism can be established by using Hamel bases of Rn and Rm . Let Bn (respectively, Bm ) denote the unit ball in the space Rn (respectively, in the space Rm ). Show that: (a) the set f (Bn ) is nonmeasurable with respect to λm ; (b) the set f −1 (Bm ) is nonmeasurable with respect to λn . Infer from these facts that neither f nor f −1 are Lebesgue measurable. 14. Let E be a set whose cardinality is not real-valued measurable. Equip E with the discrete topology. Then E becomes a locally compact topological space. Denote by E ∗ the Alexandrov compactification of E (see, e.g., [58] and [101]). Check that E ∗ is a compact non-discrete universal measure zero space. 15∗ . Let a be an arbitrary infinite cardinal number, I be a set with card(I) = a and let Xi = {0, 1} for any i ∈ I. We equip each set Xi (i ∈ I) with its discrete topology. Further, in

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the product set X(a) = ∏{Xi : i ∈ I} consider the family B of all those sets Y which admit a representation Y = ∏{Yi : i ∈ I}, where Yi ⊂ Xi for any i ∈ I and card({i ∈ I : Yi = Xi }) < a. Verify that B is a base of some topology T on the set X(a) and show that T satisfies the following relations: (a) (X(a), T ) is an isodyne topological space, i.e., for every nonempty open set U ⊂ X(a), we have card(U) = card(X(a)) = 2a ; (b) if a is a regular cardinal number, then X(a) is an a-Baire space, i.e., no nonempty open set U ⊂ X(a) can be represented in the form U = ∪{Z j : j ∈ J}, where card(J)  a and all sets Z j ( j ∈ J) are of first category in X(a); (c) if ab = a for any nonzero cardinal b < a, then the topological weight w(X(a)) of X(a) is equal to a. Suppose, in addition, that the Generalized Continuum Hypothesis holds and a is a regular cardinal number nonmeasurable in the Ulam sense (see Appendix 1). Show that X(a) is a universal measure zero space, which means that no nonzero σ -finite diffused measure can be defined on the Borel σ -algebra of X(a). 16∗ . Assume that the Generalized Continuum Hypothesis holds and let a be an infinite regular cardinal number. Consider once again the space X(a) of the previous exercise. Prove that there exists a subset L(a) of X(a) satisfying the following three relations: (a) for every nonempty open set U ⊂ X(a), the equality card(U ∩ L(a)) = 2a is valid and, in particular, L(a) is an isodyne everywhere dense subspace of X(a); (b) if a set Z ⊂ X(a) is representable in the form Z = ∪{Z j : j ∈ J}, where card(J)  a and all sets Z j ( j ∈ J) are of first category in X(a), then card(Z ∩L(a))  a;

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(c) if a is not measurable in the Ulam sense, then L(a) is a universal measure zero space. Also, check that the set L(ω ) can be regarded as a certain Luzin subset of the standard Cantor space X(ω ) = {0, 1}ω . 17. Prove that the following two assertions are equivalent: (a) there exists a universal measure zero set X ⊂ R with card(X) = c; (b) any partial function f : R → R whose range is universal measure zero and for which all pre-images f −1 (t) (t ∈ R) are at most countable, can be extended to a function f ∗ : R → R absolutely nonmeasurable with respect to the class MR . 18. Let E be a topological space in which all singletons are Borel subsets of E. Show that the following two assertions are equivalent: (a) E is universal measure zero; (b) every subset of E is absolutely nonmeasurable with respect to the class ME of the completions of all nonzero σ -finite diffused Borel measures on E. Let us introduce the notation Z0 (E) = {Z ∈ B(E) : Z is universal measure zero}. Let X be a subset of E. Verify that the following two assertions are equivalent: (c) X is absolutely nonmeasurable with respect to the class ME ; (d) for any set Y ∈ B(E) \ Z0 (E), the relations Y ∩ X = 0, / Y ∩ (E \ X) = 0/ hold true. In particular, suppose that the conditions card(B(E) \ Z0 (E))  card(E), (∀Y ∈ B(E) \ Z0 (E))(card(Y ) = card(E)) are satisfied. Prove that there exists a subset of E absolutely nonmeasurable with respect to the class ME . Compare this exercise with the standard construction of Bernstein sets in uncountable Polish topological spaces. 19∗ . Equip R with the Sorgenfrey topology T whose base consists of all half-open intervals of the form [a, b[, where a ∈ R and b ∈ R (see [58] or [101]). Check that (R, T ) is a hereditarily Lindel¨of topological space whose Borel σ -algebra coincides with the standard Borel σ -algebra B(R). Infer from this fact that any Bernstein subset of R is absolutely

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nonmeasurable in the space (R, T ), with respect to the class of the completions of all nonzero σ -finite diffused Borel measures on (R, T ). Further, in the product space (R2 , T × T ) consider the set D = {(x, y) ∈ R2 : x + y = 1}. Check that D is a closed discrete subset of (R2 , T × T ). Conclude that if c is measurable in the Ulam sense, then this product space does not contain an absolutely nonmeasurable set with respect to the class of the completions of all nonzero σ -finite diffused Borel measures on (R2 , T × T ). 20∗ . Assume Martin’s Axiom and let E be a topological space satisfying the following three conditions: (a) E has a countable base; (b) all singletons in E are Borel subsets of E; (c) every closed set in E is of type Gδ . Prove that there exists a subset of E absolutely nonmeasurable with respect to the class ME . More precisely, establish the same result only supposing that all those subsets of E whose cardinalities are strictly less than c are universal measure zero. 21∗ . Assume Martin’s Axiom. Let E be a topological space such that: (a) card(E)  c; (b) E is metrizable. Show that there exists a subset of E absolutely nonmeasurable with respect to the class ME . Conclude the same result for any topological space which is Borel isomorphic to E. For this purpose, apply Theorem 1 from Appendix 3. 22. As shown by Roslanowski and Shelah [208], there exists a model of set theory in which for any function f : R → R, there is a set X ⊂ R such that λ ∗ (X) > 0 and f |X is continuous. Verify that in this model there are no absolutely nonmeasurable functions acting from R into R. 23. Suppose that every uncountable co-analytic subset of a Polish topological space contains a nonempty perfect set. As is well known, this assumption does not contradict ZFC theory (see [10], [91], and [99]). Let E be an uncountable Polish space and let M denote the class of the completions of all

σ -finite Borel measures on E. Consider any analytic set A ⊂ E which is not Borel in E. The existence of such an A is also well known (see, e.g., [99], [148], [150], [160], [162], and Appendix 6).

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Check that A cannot be represented in the form A = BX, where B ∈ B(E) and X is a universal measure zero set in E. Conclude from this fact that the σ -algebra of all those subsets of E which are universally measurable with respect to M is not generated by the Borel σ -algebra B(E) and the σ -ideal of all universal measure zero subsets of E. In connection with this exercise, let us also recall that, under the Constructibility Axiom of G¨odel, there exists an uncountable co-analytic subset of R which does not contain a nonempty perfect set (see [10], [91], and [188]).

Chapter 6

Some properties of step-functions connected with extensions of measures

In this chapter we again will be concerned with the measure extension problem and will present one more application of Theorem 2 from Chapter 2. Primarily, we will be dealing here with certain measurability properties of the simplest real-valued functions - the socalled real-valued step-functions. Let E be a nonempty base set and let f be a function acting from E into the real line R. We recall that f is a step-function if the range of f is (at most) countable, i.e., we have card(ran( f ))  ω . Clearly, every step-function f : E → R produces a countable partition {Xi : i ∈ I} = { f −1 (t) : t ∈ ran( f )} of E. We shall say that this partition is canonically associated with f . Conversely, let {Xi : i ∈ I} be an arbitrary countable partition of E. We shall say that a step-function f : E → R is associated with this partition if the following two relations are satisfied: (a) the restriction of f to any set Xi is constant; (b) the restriction of f to any selector of {Xi : i ∈ I} is an injection. It is well known that real-valued step-functions, which have some additional properties, play an important role in different topics of mathematical analysis, especially in those questions which are connected with various kinds of approximation. For instance, if E is equipped with a σ -finite measure μ , then μ -measurable step-functions are needed for introducing the class of μ -integrable real-valued functions. Also, step-functions are essentially used in some questions concerning the sup-measurability of functions of two variables (see, for example, [120]) and in many other topics of real analysis and measure theory. Let E be equipped with a σ -finite measure μ and let f :E →R A.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7_6, © 2009 Atlantis Press/World Scientific

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be a μ -measurable step-function. It can easily be verified that, for any subset T of R, the pre-image f −1 (T ) is a μ -measurable subset of E. It is natural to conjecture that this property is closely connected with the notion of a step-function. Indeed, in the sequel it will be shown that a similar measurability property enables us to characterize step-functions in terms of extensions of measures (cf. [128]). We shall say that a function f : E → R is strongly measurable with respect to μ if for any T ⊂ R, the pre-image f −1 (T ) is a μ -measurable subset of E. It can readily be verified that for any step-function f : E → R, the following three relations are equivalent: (1) f is measurable with respect to μ ; (2) f is strongly measurable with respect to μ ; (3) for each t ∈ R, the pre-image f −1 (t) is measurable with respect to μ . For our further purposes, we need two auxiliary statements both of which have already been considered in preceding chapters. We recall their formulations here. Lemma 1. Let μ be a σ -finite measure on E and let {Xi : i ∈ I} be an arbitrary disjoint family of subsets of E. Then there exists a measure μ  on E extending μ and satisfying the relation {Xi : i ∈ I} ⊂ dom(μ  ). In particular, if the given family {Xi : i ∈ I} is a countable partition of E, then every stepfunction associated with {Xi : i ∈ I} becomes measurable with respect to the extension μ  .

The lemma just formulated is a direct consequence of Theorem 2 from Chapter 2. We also recall that a subset Y of R is universal measure zero if for any nonzero σ -finite diffused Borel measure ν on R, we have ν ∗ (Y ) = 0 where ν ∗ denotes the outer measure associated with ν . The next classical result is well known in descriptive set theory. Lemma 2. There are uncountable universal measure zero subsets of R. For the proof of Lemma 2, see Appendix 1. Notice once more that constructions of uncountable universal measure zero subsets of R and of other small subsets of R were presented by various authors and different ideas were used in those constructions: the existence of a canonical decomposition of a proper analytic set into its Borel components (constituents), Marczewski’s characteristic function, Ulam’s

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transfinite matrix, Fubini type argument, and other approaches. For more details about this topic, see especially [79], [162], [172], [197], and [250]. Theorem 1. Let E be an uncountable set and let f : E → R be a function. The following two relations are equivalent: (1) f is a step-function; (2) for any σ -finite measure μ on E, there exists a measure μ  on E extending μ and such that f is strongly measurable with respect to μ  . Proof. (1) ⇒ (2). Assume that (1) is valid and consider an arbitrary σ -finite measure μ on E. Let {Yi : i ∈ I} denote the countable partition of E produced by f . According to Lemma 1, there exists an extension μ  of μ such that {Yi : i ∈ I} ⊂ dom(μ  ). Obviously, f is strongly measurable with respect to μ  , so (2) is true. (2) ⇒ (1). Assume that (2) is valid and let us show that f is a step-function. Suppose otherwise, that is card(ran( f ))  ω1 . Clearly, we can find a set X ⊂ E such that f |X is an injection and card(ran( f |X)) = ω1 . Consider a complete diffused probability measure μ on E concentrated on the set X, which means that

μ (E \ X) = 0. The existence of such a μ is obvious. By virtue of (2), there exists an extension μ  of μ such that f becomes strongly measurable with respect to μ  . Now, for any set Z ⊂ ran( f |X), let us put

ν (Z) = μ  ( f −1 (Z)). So we get a diffused probability measure ν which is defined on the family of all subsets of ran( f |X). From the existence of ν we readily deduce that there is no universal measure zero subset of R whose cardinality is greater than or equal to ω1 . But this circumstance obviously contradicts Lemma 2. Under some additional set-theoretical assumptions, Theorem 1 can be essentially strengthened. For instance, let us consider the following set-theoretical assertion.

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(*) Any uncountable subset of R contains an uncountable universal measure zero set. It can easily be seen that (*) is implied by the conjunction of Martin’s Axiom with the negation of the Continuum Hypothesis, so (*) is consistent with ZFC theory. On the other hand, the existence of a Sierpi´nski subset of R readily implies that (*) is false (see Exercise 2 below; extensive information on Sierpi´nski sets may be found in [148], [172], [176], and [192]). So, the negation of (*) is also consistent with ZFC. We thus conclude that assertion (*) is independent of ZFC theory. The next statement is valid (cf. Theorem 1). Theorem 2. Suppose that assertion (*) holds true. Let E be an uncountable set and let f : E → R be a function. The following two relations are equivalent: (1) f is a step-function; (2) for any σ -finite measure μ on E, there exists a measure μ  on E extending μ and such that f is measurable with respect to μ  . Proof. The argument is very similar to the proof of Theorem 1. The implication (1) ⇒ (2) does not need an additional set-theoretical assumption and can be established in the same manner as before. Assume now that (2) is satisfied and let us show that f is a step-function. Again, suppose otherwise, i.e., suppose that card(ran( f ))  ω1 . According to (*), there exists an uncountable universal measure zero subset Y of ran( f ). Clearly, we can find a set X ⊂ E such that ran( f |X) = Y and the restriction f |X is an injective mapping. Consider an arbitrary complete diffused probability measure μ on E which is concentrated on the set X. By virtue of (2), there exists an extension μ  of μ such that f becomes μ  -measurable. Now, for every Borel subset Z of the space Y , let us put

ν (Z) = μ  ( f −1 (Z)). A straightforward verification shows that ν turns out to be a Borel diffused probability measure on Y , so we obtain a contradiction with the fact that Y is a universal measure zero subset of R which ends the proof of the theorem. In particular, Theorem 2 implies that it is impossible to define, within ZFC theory, a nonstep-function f : E → R having the measurability property (2). Now, let us consider step-functions for those nonzero σ -finite measures on E which are invariant (or, more generally, quasi-invariant) with respect to an uncountable group G of transformations of E.

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If we suppose that G acts freely in E, then the situation differs from the case of ordinary (i.e., non-invariant) measures. To see this circumstance, take an arbitrary uncountable set E with c f (card(E)) = ω . It is reasonable to recall here that the symbol c f (card(E)) denotes the cofinality of card(E) or, in other words, the least cardinal b such that there exists a partition {X j : j ∈ J} of E into sets X j ( j ∈ J) where card(J) = b and all card(X j ) are strictly less than card(E). Let G be a group of transformations of E acting freely in E and such that card(E) = card(G). Fix a countable partition {Xi : i ∈ I} of E satisfying the relation (∀i ∈ I)(card(Xi ) < card(E)). Let f : E → R be a step-function associated with this partition. It is easy to verify that f cannot be measurable with respect to a nonzero σ -finite G-quasi-invariant measure on E. To give deeper examples of this kind, let us recall two definitions (see Chapters 2 and 5). Let E be a set and let G be a group of transformations of E. We say that a set X ⊂ E is G-absolutely negligible if for every σ -finite G-invariant (Gquasi-invariant) measure μ on E, there exists a G-invariant (G-quasi-invariant) extension

μ  of μ such that μ  (X) = 0. We say that a set Y ⊂ E is G-absolutely nonmeasurable if for any nonzero σ -finite G-quasiinvariant measure μ on E, we have Y ∈ dom(μ ). Some properties of G-absolutely negligible and G-absolutely nonmeasurable subsets of E are discussed in [108], [115], and in Chapters 2, 10 and 11. In a particular case, where E is an uncountable commutative group (identified with the group G of all its translations), the following statement is valid. Lemma 3. Let (G, +) be an arbitrary uncountable commutative group. Then these two relations are satisfied: (1) there exists a G-absolutely nonmeasurable subset of G; (2) there exists a countable partition of G into G-absolutely negligible sets. The proof of (1) is given in [107] and [119]. For the proof of (2), see [115], [119], or Chapter 10 where an analogous statement is established in a more general case, namely, for all uncountable solvable groups (G, ·). From Lemma 3 we directly get two nontrivial examples.

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Example 1. Let X be a G-absolutely nonmeasurable subset of an uncountable commutative group (G, +) and let f = χX denote the characteristic function of X which trivially is a stepfunction. Then f is nonmeasurable with respect to any nonzero σ -finite G-quasi-invariant measure on G. In other words, Example 1 states that there are two-valued functions absolutely nonmeasurable with respect to the class of all nonzero σ -finite G-quasi-invariant measures on G. Example 2. For an uncountable commutative group (G, +), consider its countable partition {Yi : i ∈ I} into G-absolutely negligible sets. Let f :G→R denote an arbitrary step-function associated with this partition. Then we have: (1) f is nonmeasurable with respect to any nonzero σ -finite G-quasi-invariant measure on G; (2) for each t ∈ R, the set f −1 (t) is G-absolutely negligible. Indeed, (2) is obvious. To see (1), let us suppose that f is measurable with respect to some nonzero σ -finite G-quasi-invariant measure μ . Then all sets Yi which are the pre-images of certain singletons in R must be μ -measurable. Since all of them are also G-absolutely negligible, we must have μ (Yi ) = 0. From this circumstance it follows that

μ (G) = μ (∪{Yi : i ∈ I}) = ∑{μ (Yi ) : i ∈ I} = 0, which yields a contradiction. In other words, Example 2 states that there exist step-functions f on an uncountable commutative group (G, +), which are absolutely nonmeasurable with respect to the class of all nonzero σ -finite G-quasi-invariant measures on G, but each of the pre-images f −1 (t) (t ∈ R) is good for extending any σ -finite G-invariant (G-quasi-invariant) measure

μ on G. However, there are certain types of step-functions which also are good for obtaining invariant (respectively, quasi-invariant) extensions of σ -finite invariant (respectively, quasiinvariant) measures. To describe such functions, let us return to the general situation when an infinite set E is given with some group G of its transformations. For our further purposes, the notion of an almost G-invariant subset of E turns out to be helpful. It should be mentioned that this notion was first considered by Marczewski in connection with the algebraic aspect of the measure extension problem (cf. [234] and [235]). Later on, it was

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recognized that almost G-invariant sets play an important role in various topics of the theory of invariant and quasi-invariant measures (see [81], [83], [85], [95], [108], [115], and [198]). There are two definitions of almost invariant sets, which are rather similar to each other (cf. Chapter 2). A set Z ⊂ E is called almost G-invariant in E (in the set-theoretical sense) if card(g(Z)Z) < card(E) for each transformation g ∈ G. If μ is a measure on E, then a set Z ⊂ E is called almost G-invariant with respect to μ if

μ ∗ (g(Z)Z) = 0 for each transformation g ∈ G (where μ ∗ as usual denotes the outer measure associated with μ ). Notice that if a set Z ⊂ E is almost G-invariant with respect to μ , then any μ -measurable hull of Z is also almost G-invariant with respect to μ . The next lemma is probably well known (cf. [83], [85], [95], [108], [192], and [198]) but for the sake of completeness, we present its short proof here. Lemma 4. Let E be an uncountable set, G be a group of transformations of E with card(G)  card(E), and let I be a nonempty countable set. Then there exists a partition {Xi : i ∈ I} of E consisting of almost G-invariant subsets of E such that card(Xi ) = card(E) for all i ∈ I. Proof. We may assume, without loss of generality, that card(G) = card(E) and G acts transitively in E. Let α denote the least ordinal for which card(α ) = card(E) and let x be a fixed point of E. An increasing (by inclusion) transfinite sequence {Gξ : ξ <

α } of subgroups of G can readily be constructed satisfying the following conditions: (1) ∪{Gξ : ξ < α } = G; (2) card(Gξ )  card(ξ ) + ω for any ξ < α ; (3) Gξ (x) \ ∪{Gζ (x) : ζ < ξ } = 0/ for any ξ < α . Now, let {Ξi : i ∈ I} be a partition of [0, α [ such that (∀i ∈ I)(card(Ξi ) = card(α )). Putting Xi = ∪{(Gξ (x) \ ∪{Gζ (x) : ζ < ξ }) : ξ ∈ Ξi }

(i ∈ I),

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we come to the required partition {Xi : i ∈ I} of E. The details of checking this fact are left to the reader (cf. Exercise 19 for Chapter 3). Lemma 5. Let μ be a σ -finite G-quasi-invariant measure on E and let f : E → R be a step-function such that for any t ∈ R, the set f −1 (t) is almost G-invariant with respect to

μ . Then for any g ∈ G, the functions f and f ◦ g are equivalent with respect to μ , that is the equality

μ ∗ ({x ∈ E : f (x) = ( f ◦ g)(x)}) = 0 holds true. Proof. Denote by {Xi : i ∈ I} the countable partition of E associated with the given stepfunction f . It is clear that {x ∈ E : f (x) = ( f ◦ g)(x)} ⊂ ∪{Xi ∩ g−1 (X j ) : i ∈ I, j ∈ I, i = j}. Since the relations

μ ∗ (g−1 (X j )X j ) = 0, Xi ∩ X j = 0/ (i ∈ I, j ∈ I, i = j) are satisfied, we must have

μ ∗ (Xi ∩ g−1 (X j )) = 0. From this fact, taking into account the countability of I, we immediately obtain the required result. Theorem 3. Let E be a set with c f (card(E)) > ω , let G be a group of transformations of E which acts freely in E and whose cardinality is equal to card(E), and let {Xi : i ∈ I} be a countable partition of E into almost G-invariant sets. Denote by f : E → R any step-function associated with this partition. Then for every σ -finite G-invariant (G-quasiinvariant) measure μ on E, there exists a G-invariant (G-quasi-invariant) measure μ  on E such that: (1) μ  extends μ ; (2) f is measurable with respect to μ  . Proof. The argument presented below is rather similar to the proof of Theorem 2 from Chapter 2. However, there is also an essential difference implied by the circumstance that here we are dealing with invariant (quasi-invariant) measures. Since I is countable, we may suppose that either I = {1, 2, ..., n} or I = ω . For any i ∈ I, denote by ti the value of f at some point of Xi .

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Let μ be an arbitrary σ -finite G-invariant (G-quasi-invariant) measure on E.

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Since

c f (card(E)) > ω , we may assume, without loss of generality, that every set Z ⊂ E with card(Z) < card(E) belongs to the domain of μ and, consequently, μ (Z) = 0. Actually, in this situation the Marczewski method of extending invariant (quasi-invariant) measures works for the G-invariant σ -ideal consisting of the above-mentioned sets Z. Therefore, all sets Xi (i ∈ I) become almost G-invariant with respect to μ . For each index i ∈ I, denote by Yi a μ -measurable hull of Xi and define Zi = Yi \ ∪{Y j : j < i}. Notice that all sets Zi are pairwise disjoint, μ -measurable and almost G-invariant with respect to μ . Moreover, we have the equality E = ∪{Zi : i ∈ I}. Let f  : E → R be a step-function whose value on every nonempty set Zi is equal to ti . As established in Chapter 2,

μ∗ ({x ∈ E : f (x) = f  (x)}) = 0, where μ∗ stands for the inner measure associated with μ . But here we need a much stronger property of the set D = {x ∈ E : f (x) = f  (x)}. Namely, we must show that for any countable family {gk : k < ω } ⊂ G, the equality

μ∗ (∪{gk (D) : k < ω }) = 0 holds true. Indeed, the inclusion ∪{gk (D) : k < ω } ⊂ D ∪ (∪{{x ∈ E : ( f ◦ g−1 k )(x) = f (x)} : k < ω })  ∪(∪{{x ∈ E : ( f  ◦ g−1 k )(x) = f (x)} : k < ω })

is readily verified. Taking into account the relation μ∗ (D) = 0 and applying Lemma 5, we claim that

μ∗ (∪{gk (D) : k < ω }) = 0. Thus, the set D generates a G-invariant σ -ideal of subsets of E, all members of which have inner μ -measure zero. This circumstance enables us to extend the given measure μ to a G-invariant (G-quasi-invariant) measure μ  on E such that μ  (D) = 0 (cf. Chapter 2). Since the function f  is μ -measurable, it is also μ  -measurable. In view of the equality D = {x ∈ E : f (x) = f  (x)},

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we conclude that f is μ  -measurable, too. Remark 1. Actually, the preceding argument shows that if we have an arbitrary σ -finite G-invariant (G-quasi-invariant) measure μ on E and a step-function f : E → R such that all pre-images f −1 (t) (t ∈ R) are almost G-invariant with respect to μ , then there exists a G-invariant (G-quasi-invariant) extension μ  of μ for which f becomes μ  -measurable. In connection with Theorems 1 and 2, the following question arises. Does there exist a real-valued function f1 : E → R with card(ran( f1 )) > ω such that every

σ -finite G-invariant (G-quasi-invariant) measure μ on E admits a G-invariant (G-quasiinvariant) extension for which f1 becomes measurable? In order to give a positive answer to this question (in some natural situations), we need the next auxiliary proposition. Lemma 6. Let E be a set and let G be an uncountable group of transformations of E acting freely in E. Then there exists an uncountable G-absolutely negligible subset X of E. Proof. If card(G) > ω1 , then any set X ⊂ E with card(X) = ω1 is G-absolutely negligible in E (the checking this simple fact is left to the reader). Suppose now that card(G) = ω1 and fix a point x ∈ E. Let {Gξ : ξ < ω1 } be an increasing (by inclusion) ω1 -sequence of subgroups of G satisfying the following three relations: (a) ∪{Gξ : ξ < ω1 } = G; (b) card(Gξ )  ω for each ξ < ω1 ; (c) Gξ (x) \ ∪{Gζ (x) : ζ < ξ } = 0/ for each ξ < ω1 . Let X be a selector of the family of nonempty sets {(Gξ (x) \ ∪{Gζ (x) : ζ < ξ }) : ξ < ω1 }. Then, by using a characterization of absolutely negligible sets (see Theorem 6 of Chapter 2), it is not difficult to verify that X is a G-absolutely negligible subset of E. Since X is also uncountable, we get the required result. The next statement easily follows from Lemma 6. Theorem 4. Under the assumptions of Lemma 6, there exists a function f1 : E → R such that: (1) card(ran( f1 )) = ω1 ; (2) for any σ -finite G-invariant (G-quasi-invariant) measure μ on E, there exists a Ginvariant (G-quasi-invariant) extension of μ for which f1 becomes equivalent to zero and, consequently, becomes measurable.

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Proof. Using Lemma 6, we can find a G-absolutely negligible subset X of E with card(X) =

ω1 . Let f1 : E → R be a function defined as follows: f1 |X is injective and f1 |(E \ X) is equal to zero. A straightforward verification shows that f1 satisfies both relations (1) and (2). Comparing Theorem 2 with Theorem 4, we see that the case of ordinary measures essentially differs from the case of invariant (quasi-invariant) measures. Moreover, in view of Theorem 3, the following natural question arises. Problem. Let (E, G) be a space equipped with a transformation group. Give a characterization of all those step-functions f on E which have the generalized measurability property in the sense that for any σ -finite G-invariant (G-quasi-invariant) measure μ on E, there exists a G-invariant (G-quasi-invariant) extension μ  of μ such that f becomes μ  -measurable? This problem remains open. Example 3. Let f be a step-function of Theorem 3 such that ran( f ) ∩ {0} = 0/ and let f1 be a function of Theorem 4. Let us put f2 = f + f1 . Then for any nonzero σ -finite G-invariant (G-quasi-invariant) measure μ on E, there exists a G-invariant (G-quasi-invariant) extension μ  of μ such that f2 is μ  -measurable and, at the same time, f2 never becomes equivalent to zero (with respect to μ  ). In addition to this fact, if card(E) = ω1 and the Continuum Hypothesis holds, then the set ran( f2 ) ⊂ R can be as bad as possible (in the sense of its descriptive structure). Remark 2. If we deal with finitely additive G-invariant normalized measure type functionals on E, then it is reasonable to call a step-function any function f : E → R whose range is finite, i.e., card(ran( f )) < ω . In this case, corresponding measurability properties of f essentially depend on the algebraic structure of G. For instance, if G is amenable and μ is an arbitrary finitely additive G-invariant normalized measure type functional on E, then, according to von Neumann’s extension theorem (see, for instance, [83], [108], and [240]), every step-function on E becomes measurable with respect to an appropriate universal finitely additive G-invariant extension μ  of μ (as usual, the term ”universal” means here that dom(μ  ) coincides with

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the power set of E). On the other hand, if E = G and G admits finite paradoxical decompositions, then it is obvious that there exist step-functions on E which are absolutely nonmeasurable with respect to the class of all finitely additive left G-invariant normalized measure type functionals on E (cf. Example 1). For more details about finite paradoxical decompositions, see [155] and [240]. EXERCISES 1. Give a detailed explanation of the fact that Martin’s Axiom with the negation of the Continuum Hypothesis imply assertion (*) which was introduced in this chapter. 2∗ . Recall that S ⊂ R is a Sierpi´nski set in R if S is uncountable and the inequality card(S ∩ Y )  ω holds true for every Lebesgue measure zero subset Y of R. It is well known that the Continuum Hypothesis implies the existence of a Sierpi´nski set (see, for instance, [148], [172], [176], [192], [222], and Exercise 14 for Appendix 1). Check that any uncountable subset of a Sierpi´nski set S has strictly positive outer Lebesgue measure. Using this fact, establish that S does not contain an uncountable universal measure zero subset. This circumstance immediately shows that the existence of a Sierpi´nski subset of R implies the negation of the above-mentioned assertion (*). 3. Let E be a set, G be a group of transformations of E and let μ be a σ -finite G-quasiinvariant measure on E. Suppose that X ⊂ E is an almost G-invariant set with respect to μ . Verify that a μ -measurable hull of X is also almost G-invariant with respect to μ . 4∗ . Using the method of transfinite induction, prove that there exists a subset Z of R satisfying the following conditions: (a) card(Z) = card(R \ Z) = c; (b) λ∗ (Z) = λ∗ (R \ Z) = 0, i.e., both sets Z and R \ Z are thick with respect to the Lebesgue measure λ on R; (c) for any h ∈ R, the inequality card((h + Z)Z) < c holds true. Infer from this result that, under Martin’s Axiom, there exists a partition of R into two almost R-invariant subsets (with respect to λ ), none of which is of λ -measure zero. 5∗ . According to the classical result of Kunen, the real-valued measurability of c implies that there exists a set X ⊂ R such that card(X) < c and X is not measurable in the Lebesgue sense (for the proof, see [69], [79], and [119]).

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Deduce from this fact that under the same assumption, there exists a set Y ⊂ R such that card(Y ) < c and Y is thick with respect to λ . Conclude that: (a) Y is not measurable in the Lebesgue sense; (b) Y is not universal measure zero. 6. Complete the proof of Lemma 6, by showing that the selector X described therein is a G-absolutely negligible set in E. 7∗ . Let E be a set of cardinality ω1 and let G be a group of transformations of E satisfying the following conditions: (a) card(G) = ω1 ; (b) G acts freely in E. Applying the argument presented in the proof of Lemma 6, show that there exists a countable covering of E by G-absolutely negligible sets. Conclude from this fact that for any nonzero σ -finite G-invariant (G-quasi-invariant) measure μ on E, there always exists a G-invariant (G-quasi-invariant) measure on E strictly extending μ . 8. Complete the argument of Example 3 and show that under the Continuum Hypothesis, the set ran( f2 ) ⊂ R can be as bad as possible from the measure-theoretical and descriptive point of view.

Chapter 7

Almost measurable real-valued functions

In this chapter the notion of an almost measurable real-valued function is introduced and examined. Some properties of such functions are considered and their characterization is given by using the classical Fubini theorem and a theorem on the existence of measurable selectors (see Appendix 2). In addition to this result, close connections of almost measurable functions with extensions of the standard Lebesgue measure on the real line are pointed out and the structure of algebraic sums of almost measurable functions is studied. As usual, let R denote the real line and let f : R → R be a function. We say that f is almost continuous if for every open set U ⊂ R2 containing the graph Gr( f ) of f , there exists a continuous function g : R → R whose graph Gr(g) is also contained in U. The above notion was introduced by J. Stallings [232] in connection with some fixedpoint theorems. Notice that there are many works devoted to various properties of almost continuous functions (see, e.g., [184] and the list of references therein). The concept of almost measurable real-valued functions is relative to almost continuous functions and seems to be of some interest from the view-point of real analysis and classical measure theory. Even the definition of almost measurable functions is very similar to the definition of almost continuous functions. See below the definition of the almost measurability of functions. As usual, we denote by λ (= λ1 ) the one-dimensional Lebesgue measure on the real line R (= R1 ). The symbol λ2 stands for the two-dimensional Lebesgue measure on the Euclidean plane R2 (in fact, λ2 coincides with the completion of the product measure λ1 ⊗ λ1 ).

We shall say that a function f : R → R is almost measurable if for every λ2 -measurable set V ⊂ R2 containing the graph Gr( f ) of f , there exists a λ1 -measurable function g : R → R whose graph Gr(g) is also contained in V . A.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7_7, © 2009 Atlantis Press/World Scientific

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The following assertions are easy consequences of this definition. (a) Any λ1 -measurable function f : R → R is almost measurable. (b) If f : R → R is almost measurable and its graph is λ2 -measurable, then f is λ1 measurable. Indeed, (a) is trivial because for every λ2 -measurable set V ⊂ R2 containing the graph of f , we may put g = f . To show (b), it suffices to take as V the graph of a given function f ; then any Lebesgue measurable function g : R → R whose graph is contained in V necessarily coincides with f , which gives the required result. In connection with (b), let us also remark that if the graph of a function f :R→R is λ2 -measurable, then it is of λ2 -measure zero because in the plane R2 there are uncountably many pairwise disjoint translates of this graph. On the other hand, simple examples enable us to claim that the equality λ2 (Gr( f )) = 0 does not imply, in general, the λ1 measurability of f . It is well known (see, e.g., Exercise 9 from Chapter 2) that there exists a function f : R → R whose graph is a thick subset of the plane R2 , which means that (λ2 )∗ (R2 \ Gr( f )) = 0. Clearly, such an f is not Lebesgue measurable. As already mentioned in Chapter 2, one of the earliest examples of a function f of this type was constructed by W. Sierpi´nski, with the aid of the method of transfinite recursion (see, for instance, [72], [144], [148], and [192]). We shall establish below that all such functions f turn out to be almost measurable in the sense of our definition and, in this respect, they cannot be treated as extremely pathological functions. In the sequel, we will need an auxiliary proposition concerning the existence of measurable selectors. This proposition is a straightforward corollary of the classical theorem from descriptive set theory, which is due to Luzin, Jankov, and von Neumann (see Appendix 2). Lemma 1. Let A ⊂ R2 be an analytic (i.e., Suslin) set. There exists a Lebesgue measurable function g : pr1 (A) → R whose graph is contained in A. The proof of Lemma 1 (and of a more general statement) is presented in Appendix 2. Using this lemma, we can obtain a certain characterization of all almost measurable functions. Theorem 1. Let f : R → R be a function and let D denote some λ2 -measurable hull of the graph of f . The following two assertions are equivalent:

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(1) f is almost measurable; (2) there exists a disjoint covering {X1 , X2 } of R by two λ1 -measurable sets such that the restriction f |X1 is Lebesgue measurable and for each point x ∈ X2 , the inequality λ1∗ ({y ∈ R : (x, y) ∈ D}) > 0 is true. Proof. (1) ⇒ (2). Suppose that (1) is satisfied for the given function f . Let us put: X1 = {x ∈ R : λ1 ({y ∈ R : (x, y) ∈ D}) = 0}, X2 = R \ X1. Obviously, the sets X1 and X2 are disjoint, λ1 -measurable, and cover R. Moreover, in view of the classical Fubini theorem, we have

λ2 (D ∩ (X1 × R)) = 0. Now, let us define D = {(x, f (x)) : x ∈ X1 } ∪ (D ∩ (X2 × R)). Clearly, the set D ∩ (X2 × R) is λ2 -measurable. By virtue of the inclusion {(x, f (x)) : x ∈ X1 } ⊂ D ∩ (X1 × R), the set {(x, f (x)) : x ∈ X1 } is of λ2 -measure zero. Consequently, D is a λ2 -measurable subset of R2 containing the graph Gr( f ) of f . Since f is almost measurable, there exists a Lebesgue measurable function g : R → R whose graph is contained in D . Then we get the equality g|X1 = f |X1 which shows that the restriction f |X1 is Lebesgue measurable. At the same time, it directly follows from the definition of X2 that

λ1∗ ({y ∈ R : (x, y) ∈ D}) > 0 for any point x ∈ X2 . We thus conclude that the implication (1) ⇒ (2) holds true. Suppose now that (2) is satisfied for our function f and let {X1 , X2 } denote the corresponding disjoint covering of R by two λ1 -measurable sets. Let V ⊂ R2 be any λ2 -measurable set containing the graph of f . We may assume, without loss of generality, that V ⊂ D. This assumption implies the equality

λ2 (D \ V ) = 0

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because D is a measurable hull of Gr( f ). Consider a Borel subset V  of V such that λ2 (V \ V  ) = 0. Clearly, we have λ2 (D \ V  ) = 0. Further, take a Borel subset X2 of X2 such that

λ1 (X2 \ X2 ) = 0. Using the Fubini theorem once more, we easily deduce that

λ1 (X2 \ pr1 (V  ∩ (X2 × R))) = 0. Applying Lemma 1 to the Borel set V  ∩ (X2 × R), we can find a Lebesgue measurable function h : pr1 (V  ∩ (X2 × R)) → R whose graph Gr(h) is contained in the set V  ∩ (X2 × R). Let us put: g(x) = h(x) for all points x ∈ pr1 (V  ∩ (X2 × R)); g(x) = f (x) for all points x ∈ R \ pr1(V  ∩ (X2 × R)). So we come to the function g:R→R whose graph Gr(g) is contained in V . Finally, taking into account the Lebesgue measurability of f |X1 and the relation g|X1 = f |X1 , it is easy to see that g is also Lebesgue measurable. This establishes the implication (2) ⇒ (1) and ends the proof of the theorem. From Theorem 1 we immediately obtain the next statement. Theorem 2. Suppose that f : R → R is a function whose graph Gr( f ) is thick with respect to the measure λ2 . Then f is almost measurable. Proof. Indeed, using the notation of Theorem 1 and putting / X2 = R, D = R2 , X1 = 0, we see that assertion (2) of Theorem 1 is automatically satisfied for f . Consequently, f is almost measurable. Now, we are going to show that any almost measurable function turns out to be measurable with respect to an appropriate extension of the Lebesgue measure λ1 . For this purpose, we need the measure extension construction which is successfully applicable to a wide class of σ -finite measures (cf. [141] and Chapter 2). It is reasonable to recall here some details of the above-mentioned construction.

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Let E be a set, S be a σ -algebra of subsets of E and let μ be a σ -finite measure defined on S . Further, let F be another set and let ν be a probability measure defined on some

σ -algebra of subsets of F. Suppose that a function g:E →F is given, whose graph Gr(g) is thick with respect to the product measure μ ⊗ ν , i.e., we have the equality (μ ⊗ ν )∗ ((E × F) \ Gr(g)) = 0. For every set Z ∈ S ⊗ dom(ν ), let us put Z  = {x ∈ E : (x, g(x)) ∈ Z} and define S  = {Z  : Z ∈ S ⊗ dom(ν )},

μ  (Z  ) = (μ ⊗ ν )(Z) (Z  ∈ S  ). Taking into account the (μ ⊗ ν )-thickness of Gr(g), we can verify that: (i) S  is a σ -algebra of subsets of E containing S ; (ii) the functional μ  is well defined and is a measure on S  extending μ ; (iii) the function g is measurable with respect to the σ -algebras dom(ν ) and S  , which means that for every set Y ∈ dom(ν ), the pre-image g−1 (Y ) belongs to S  . The details of the verification of (i) - (iii) are left to the reader (cf. Lemmas 1, 2, and 3 from Chapter 2). The main idea of this construction will be used below in a very particular case, where we have E = F = R, μ = λ1 .

Lemma 2. Let X be a λ1 -measurable subset of R, let f : X → R be a function, and let D denote some λ2 -measurable hull of Gr( f ). Suppose that for each point x ∈ X, the relation 0 < λ1∗ ({y ∈ R : (x, y) ∈ D}) < +∞ holds true. Then there exists a measure λ1 on X such that: (1) λ1 extends the restriction of λ1 to the σ -algebra of all Lebesgue measurable subsets of X;

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(2) f is measurable with respect to λ1 . Proof. We may assume, without loss of generality, that X is a Borel subset of R and D is a Borel subset of R2 . Moreover, we may also assume that D ⊂ X × R. Then the function

φ (x) = λ1 ({y ∈ R : (x, y) ∈ D}) (x ∈ X) is Borel measurable and according to our assumption, we have 0 < φ (x) < +∞

(x ∈ R).

Let us put f1 (x) = f (x)/φ (x)

(x ∈ X).

Since f = f1 · φ and φ is Borel measurable, it suffices to find a measure λ1 which extends the restriction of λ1 to the σ -algebra of all Lebesgue measurable subsets of X and for which f1 becomes measurable. Let us define Φ(x, y) = (x, y/φ (x))

((x, y) ∈ X × R).

Notice that Φ is a Borel bijection of the set X × R onto itself. Putting D1 = Φ(D), we readily infer (by virtue of the Fubini theorem) that D1 is a measurable hull of Gr( f1 ). In addition to this fact, we have

λ1 ({y ∈ R : (x, y) ∈ D1 }) = 1 for all points x ∈ X. Now, we can apply the idea of the measure extension construction which was described earlier. Let S denote the σ -algebra of all Lebesgue measurable subsets of X. For any λ2 -measurable set Z ⊂ D1 , let us put Z  = {x ∈ X : (x, f1 (x)) ∈ Z} and define S  = {Z  : Z ⊂ D1 , Z ∈ dom(λ2 )},

λ1 (Z  ) = λ2 (Z) (Z  ∈ S  ). A straightforward verification shows that S  is a σ -algebra of subsets of X containing S , the functional λ1 is well defined and turns out to be a measure on X. As before, the details of this verification are left to the reader.

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Now, take any λ1 -measurable set T ⊂ X and consider the set D1 ∩ (T × R). Obviously, the relation T = {x ∈ X : (x, f1 (x)) ∈ (D1 ∩ (T × R))} holds true. Keeping in mind the equality λ1 ({y ∈ R : (x, y) ∈ D1 }) = 1 for each point x ∈ X, we may write

λ1 (T ) = λ2 (D1 ∩ (T × R)) =

 T

λ1 ({y ∈ R : (x, y) ∈ D1 })dx = λ1 (T )

which shows that λ1 extends the restriction of λ1 to the σ -algebra S . Finally, for any Borel set Y ⊂ R, we have f1−1 (Y ) = {x ∈ X : (x, f1 (x)) ∈ D1 ∩ (X × Y )} ∈ S  from which it follows that f1 is measurable with respect to λ1 . Lemma 2 has thus been proved. Theorem 3. Let f : R → R be an arbitrary almost measurable function. Then there exists a measure λ1 on R such that: (1) λ1 extends λ1 ; (2) f is measurable with respect to λ1 . Proof. Suppose first that ran( f ) ⊂ [0, 1]. Denote by D a λ2 -measurable hull of Gr( f ). Obviously, we may assume that D ⊂ R × [0, 1]. Since f is almost measurable, we can find a disjoint covering {X0 , X} of R satisfying the following relations: (a) both sets X0 and X are λ1 -measurable; (b) for each point x ∈ X0 , we have λ1 ({y ∈ R : (x, y) ∈ D}) = 0; (c) for each point x ∈ X, we have λ1∗ ({y ∈ R : (x, y) ∈ D}) > 0. Notice that the restriction of f to X0 is Lebesgue measurable (cf. the proof of Theorem 1). At the same time, for any point x ∈ X, the relation 0 < λ1∗ ({y ∈ R : (x, y) ∈ D})  1 is valid. So we may apply Lemma 2 to the function f |X. In this manner, we obtain an extension λ1 of λ1 such that f becomes λ1 -measurable. The case of an arbitrary almost measurable function f : R → R can readily be reduced to the previous case. Indeed, it is easy to construct a homeomorphism

φ : R → ]0, 1[

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which transforms the σ -ideal of all Lebesgue measure zero subsets of R onto the σ -ideal of all Lebesgue measure zero subsets of ]0, 1[. Let us put Φ(x, y) = (x, φ (y))

(x ∈ R, y ∈ R).

The mapping Φ transforms the set Gr( f ) onto the set Gr(φ ◦ f ). The above-mentioned property of φ directly implies that φ ◦ f is almost measurable and ran(φ ◦ f ) ⊂ [0, 1]. Consequently, there exists a measure λ1 on R extending λ1 for which φ ◦ f turns out to be measurable. It immediately follows from this fact that the original function f = φ −1 ◦ (φ ◦ f ) is measurable with respect to λ1 , too, which completes the proof. Example 1. Recall that B ⊂ R is a Bernstein set in R if for every nonempty perfect set P ⊂ R, both intersections P ∩ B and P ∩ (R \ B) are nonempty. Extensive information about Bernstein sets and their properties can be found in [12], [30], [40], [72], [138], [143], [148], [176], and [192] (cf. also Exercise 5 for Chapter 5). It is well known that the Bernstein sets are nonmeasurable with respect to the completion of any nonzero σ -finite diffused (i.e. vanishing at all singletons) Borel measure on R and, consequently, they are nonmeasurable with respect to λ1 . At the same time, if B is a Bernstein set, then there exists an extension

λ1 of λ1 for which B becomes measurable and, as explained in Chapter 2, this fact is true for any set X ⊂ R. Denoting by f = χB the characteristic function of B, we conclude that f is not almost measurable but is measurable with respect to λ1 . In particular, this result shows that the statement converse to Theorem 3 does not hold. Theorem 4. Let f : R → R and g : R → R be two functions. Suppose that f is almost measurable and g is Lebesgue measurable. Then their sum f + g is almost measurable. Proof. Taking into account Theorem 1 and the fact that the sum of any two Lebesgue measurable functions is also Lebesgue measurable, it suffices to establish the almost measurability of f |X + g|X where X is a λ1 -measurable subset of R and

λ1∗ ({y ∈ R : (x, y) ∈ D}) > 0 for all points x ∈ X (here D denotes a λ2 -measurable hull of the graph of f |X).

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Without loss of generality, we may assume that X is a Borel subset of R and D is a Borel subset of R2 . Let Φ be the characteristic function of D ⊂ X × R and let us define Ψ(x, y) = Φ(x, y − g(x)) (x ∈ X, y ∈ R). By using the Fubini theorem, it is not difficult to check that Ψ turns out to be the characteristic function of some λ2 -measurable hull C of the graph of ( f + g)|X and the inequality

λ1∗ ({y ∈ R : (x, y) ∈ C}) > 0 holds true for all points x ∈ X. By virtue of Theorem 1, this circumstance immediately yields that the function ( f + g)|X = f |X + g|X is almost measurable, which completes the proof. The next statement shows, in particular, that the sum of two almost measurable additive functions can be a function without the property of almost measurability. Theorem 5. There exist two functions f : R → R and g : R → R satisfying the following relations: (1) f and g are additive, that is they are homomorphisms of the additive group R into itself; (2) the graphs of f and g are thick with respect to λ2 ; (3) ran( f + g) = Q. Therefore, both f and g are almost measurable but their sum f + g is not almost measurable. Proof. Consider R as a vector space over the field Q. Let {ei : i ∈ I} be a Hamel basis of this space. Fix an index j ∈ I and denote by L the hyperplane in R generated by the family {ei : i ∈ I \ { j}}. As is well known, L is a thick subset of R with respect to the measure λ1 and, consequently, L is nonmeasurable with respect to λ1 . By using the standard transfinite construction, an additive function

φ :L→L can be defined such that its graph Gr(φ ) is thick in R2 with respect to the measure λ2 (cf. Exercise 9 from Chapter 2). Now, take an arbitrary x ∈ R. Then x admits a unique representation in the form x = qe j + e, where q ∈ Q and e ∈ L. Let us put: f (x) = f (qe j + e) = q/2 + φ (e),

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g(x) = g(qe j + e) = q/2 − φ (e). In this way, the functions f : R → R, g : R → R are completely determined. So, it remains to check the validity of the relations (1), (2), and (3). The additivity of both functions f and g follows directly from their definition. The λ2 -thickness of their graphs is easily implied by the λ2 -thickness of the set Gr(φ ), which is contained in Gr( f ) ∩ Gr(−g). Further, for any x ∈ R, consider its unique representation in the form x = qe j + e, where q ∈ Q and e ∈ L. Clearly, we have ( f + g)(x) = (q/2 + φ (e)) + (q/2 − φ (e)) = q. Consequently, the equality ran( f + g) = Q holds true. This equality shows that the additive function f +g : R → R is a nontrivial solution of the Cauchy functional equation, so f + g is not Lebesgue measurable. At the same time, the set Gr( f + g) is of λ2 -measure zero because Gr( f + g) is contained in the union of countably many horizontal lines in the plane R2 . Therefore, the function f + g is not almost measurable which completes the proof. The functions f and g described in Theorem 5 are almost measurable, so they become measurable with respect to appropriate extensions of the Lebesgue measure λ on R. It should be noticed that those extensions can be chosen to be quasi-invariant under the group of all translations of R (cf. Exercise 7 for this chapter). In view of Theorem 5, the following example seems to be relevant. Example 2. If E is an uncountable topological space, then we may introduce the class ME of the completions of all nonzero σ -finite diffused Borel measures on E. Consider more thoroughly the particular case when E = R. We know that a function f : R → R is relatively measurable with respect to MR if f admits a representation in the form f = g ◦ φ,

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where g : R → R is a Lebesgue measurable function and φ is a Borel isomorphism of R onto itself (see Exercise 12 for Chapter 5). The algebraic sum of two relatively measurable functions with respect to MR can be an absolutely nonmeasurable function with respect to the same class. To show this fact, consider two nonzero σ -finite diffused Borel measures μ1 and μ2 on R such that

μ1 ([0, +∞[) = 0, μ2 (] − ∞, 0]) = 0. Let B1 denote a Bernstein subset of the half-line [0, +∞[ and let B2 denote a Bernstein subset of the half-line ] − ∞, 0]. We define a function f1 : R → R by putting: f1 (x) = 1 if x ∈ B1 and f1 (x) = 0 if x ∈ R \ B1. Also, we define a function f2 : R → R by putting: f2 (x) = 1 if x ∈ B2 and f1 (x) = 0 if x ∈ R \ B2. It immediately follows from these definitions that the function f1 is measurable with respect to the completion of μ1 and the function f2 is measurable with respect to the completion of μ2 . At the same time, the sum f1 + f2 is the characteristic function of the subset B1 ∪ B2 of R. But it is easy to see that B1 ∪ B2 is a Bernstein subset of R. Therefore, f1 + f2 is absolutely nonmeasurable with respect to MR . Example 3. Let μ be a nonzero σ -finite diffused measure on R and let f : R → R be a function such that for some nonzero σ -finite Borel measure ν on ran( f ), the graph of f is (μ ⊗ ν )-thick in the product set R × ran( f ). Then we can assert that f is measurable with respect to an appropriate extension μ  of μ and hence f is not absolutely nonmeasurable. Indeed, we may suppose without loss of generality that ν is a Borel probability measure on ran( f ) and we may apply once again the method of extending σ -finite measures described in this chapter (see also Chapter 2). Namely, for each set Z ∈ dom(μ ⊗ ν ), we denote Z  = {x ∈ R : (x, f (x)) ∈ Z} and introduce the family of sets S  = {Z  : Z ∈ dom(μ ⊗ ν )}, which turns out to be a σ -algebra of subsets of R. In addition, if X ∈ dom(μ ), then we have X × ran( f ) ∈ dom(μ ⊗ ν ), X = {x ∈ R : (x, f (x)) ∈ X × ran( f )},

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thus it follows that X ∈ S  . Consequently, the inclusion dom(μ ) ⊂ S  holds true. Finally, for any Z ∈ dom(μ ⊗ ν ), we put

μ  (Z  ) = (μ ⊗ ν )(Z). As a direct verification shows, the functional μ  is well defined by virtue of the thickness of the graph of our function f and, moreover, this μ  is a measure on S  extending the initial measure μ . The definition of μ  also implies that the given function f is measurable with respect to μ  . In particular, if the graph of a function g : R → R is (λ ⊗ λ )-thick in the plane R2 , then, as we already know, g can be made measurable with respect to an appropriate extension of

λ . However, we cannot guarantee the quasi-invariance (under the group of all translations of R) of the obtained in this way extension. For the above-mentioned quasi-invariance, we need some additional properties of g. In particular, it suffices to suppose that g is a homomorphism of the additive group R into itself and that the graph of g is a λ2 -thick subset of R2 (see Exercise 7 below). In this context, the following problem naturally arises. Problem. Let μ be a measure on R extending λ and such that there exists a function acting from R into R whose graph is (μ ⊗ λ )-thick in the plane R2 . Does there exist a group homomorphism acting from R into R whose graph is also (μ ⊗ λ )-thick in R2 ? We do not know an answer to this question. EXERCISES 1. Define a function f : R → R as follows: f (x) = sin(1/x) if x = 0 and f (x) = 0 if x = 0. Check that f is not continuous but is almost continuous. Give other concrete examples of almost continuous functions acting from R into R which are not continuous. Show directly that the characteristic function of Q (the so-called Dirichlet function) is not almost continuous (cf. Exercise 4 below). 2∗ . Starting with the definition of almost continuous functions, demonstrate that if f : R → R is continuous and g : R → R is almost continuous, then their sum f + g is almost continuous. Is it true that if f : R → R and g : R → R are any two almost continuous functions, then their sum f + g is also almost continuous?

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3. The notion of an almost continuous function can be generalized in the following manner (see [232]). Let X and Y be two topological spaces and let f : X → Y be a mapping. We shall say that f is almost continuous if for any open set U ⊂ X × Y containing the graph of f , there exists a continuous mapping g : X → Y whose graph is also contained in U. Let f : X → Y be almost continuous and let h : Y → Z be continuous. Check that the mapping h ◦ f : X → Z is almost continuous. We shall say that a topological space E has the fixed-point property if for any continuous mapping f : E → E, there exists e ∈ E such that f (e) = e. Suppose that a Hausdorff space E has the fixed-point property (according to Brower’s theorem, E satisfies this condition if it is homeomorphic to a nonempty compact convex subset of a finite-dimensional Euclidean space). Let g : E → E be an almost continuous mapping. Show that there exists a fixed point for g, i.e., (∃e ∈ E)(g(e) = e). 4∗ . By using the result of the previous exercise, prove that every almost continuous function f : R → R has the Darboux property or the intermediate value property, which means that if a ∈ ran( f ) and b ∈ ran( f ), then the line segment with end-points a and b is contained in ran( f ). Infer from this fact that for a monotone function g : R → R, the following two conditions are equivalent: (a) g is continuous; (b) g is almost continuous. Conclude that there are many Borel measurable (hence Lebesgue measurable and almost measurable) functions which are not almost continuous. 5. Give a direct proof of Theorem 4 by starting with the definition of an almost measurable function. 6. Give a precise construction of an additive function φ : L → L mentioned in the proof of Theorem 5. 7. Let h : R → R be a function satisfying the following conditions: (a) h is additive; (b) the graph of h is λ2 -thick in the plane R2 . Prove that h is measurable with respect to some translation-quasi-invariant extension of the Lebesgue measure λ on R.

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In particular, let f and g be as in Theorem 5. Show that there exist two translation-quasiinvariant measures μ1 and μ2 on R both extending λ and such that f is measurable with respect to μ1 and g is measurable with respect to μ2 . 8∗ . Prove that for every function h : R → R, there exist two functions f : R → R and g : R → R satisfying the following conditions: (a) the graph of f is λ2 -thick in R2 ; (b) the graph of g is λ2 -thick in R2 ; (c) h = f + g. For this purpose, construct the required f and g by using the method of transfinite recursion. Derive from the above result that under Martin’s Axiom the sum of two almost measurable functions can be an absolutely nonmeasurable function (with respect to the class MR ). 9∗ . Prove that for every additive function h : R → R, there exist two functions f : R → R and g : R → R satisfying the following relations: (a) both f and g are additive functions; (b) the graph of f is λ2 -thick in R2 ; (c) the graph of g is λ2 -thick in R2 ; (d) h = f + g. For this purpose, consider R as a vector space over Q and use again the method of transfinite recursion. Deduce from the above result that under Martin’s Axiom the sum of two additive almost measurable functions can be an additive absolutely nonmeasurable function (with respect to the class MR ).

Chapter 8

Several facts from general topology

It is well known that close relationships between real analysis and general topology are very useful and fruitful for both of these mathematical disciplines. For instance, deep analogies between the Lebesgue measure and Baire category on R were extensively studied and underlined many times in the literature (see, e.g., [40], [148], [176], and [192]). In this chapter, we would like to discuss some topological facts and statements which play an auxiliary role in our further considerations. Recall that a topological space E is quasicompact if any open covering of E contains a finite subcovering. We begin with a simple but remarkable theorem of Kuratowski on closed projections (see [58] and [148]). Theorem 1. Let X be a topological space, Y be a quasicompact topological space, and let pr1 : X × Y → X denote the canonical projection defined by the formula pr1 ((x, y)) = x

((x, y) ∈ X × Y ).

Then pr1 is a closed mapping, i.e., for each closed subset A of X × Y , the image pr1 (A) is closed in X. Proof. Consider any point x ∈ X such that U(x) ∩ pr1 (A) = 0/ for every neighborhood U(x) of x. We must show that x ∈ pr1 (A). For this purpose, it suffices to establish that ({x} × Y ) ∩ A = 0. / Suppose otherwise, that is ({x} × Y ) ∩ A = 0. / Then for each point y from Y , there exists an open neighborhood W (x, y) of (x, y) satisfying the relation W (x, y) ∩ A = 0. / We may assume, without loss of generality, that W (x, y) has the form W (x, y) = U(x) × V(y), A.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7_8, © 2009 Atlantis Press/World Scientific

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where U(x) is an open neighborhood of x and V (y) is an open neighborhood of y. Since the space {x} × Y is quasicompact, there exist a finite sequence (x, y1 ), (x, y2 ), . . . , (x, yn ) of points from {x} × Y and the corresponding finite sequence U1 (x) × V1(y1 ), U2 (x) × V2(y2 ), . . . , Un (x) × Vn (yn ) of neighborhoods of these points such that {x} × Y ⊂ (U1 (x) × V1(y1 )) ∪ (U2 (x) × V2(y2 )) ∪ ... ∪ (Un (x) × Vn(yn )). Let us put U(x) = U1 (x) ∩U2 (x) ∩ ... ∩Un(x). Then U(x) is a neighborhood of x and it can readily be verified that U(x) ∩ pr1 (A) = 0, / contradicting the definition of x. The contradiction obtained finishes the proof. Theorem 1 has many applications in general topology and real analysis. For example, it can efficiently be applied in the proof of the existence of a continuous nowhere differentiable function f : [0, 1] → R (see Exercise 4 of this chapter). Moreover, this theorem turns out to be useful in some questions of the theory of ordinary differential equations (see [190] and [122]). It is interesting to notice that the statement converse to the Kuratowski theorem holds true, too (see Exercise 3). Let us briefly discuss some other useful facts about closed mappings (i.e., those mappings which preserve the closedness of sets) and their specific properties. For instance, it makes sense to study the structure of the graphs of such mappings. The following simple and well-known statement is concerned with the descriptive structure of graphs of continuous mappings acting from any topological space into a Hausdorff space.

Theorem 2. Let X be a topological space, Y be a Hausdorff topological space, and let f : X → Y be a continuous mapping. Then the graph Gr( f ) of f is a closed subset of the product space X × Y . Proof. We need one easy auxiliary proposition. Namely, if Z is a topological space, Y is a Hausdorff topological space and g : Z → Y, h : Z → Y

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are continuous mappings, then the set {z ∈ Z : g(z) = h(z)} is closed in Z because it coincides with the pre-image of the closed set {(y, y) : y ∈ Y } ⊂ Y × Y under the continuous mapping (g, h) : Z → Y × Y. Let now f : X → Y be an arbitrary continuous mapping. Put Z = X × Y and consider two continuous mappings g : X × Y → Y, h : X × Y → Y defined by the formulas g(x, y) = y, h(x, y) = f (x)

((x, y) ∈ X × Y ).

According to the said above, the set {(x, y) ∈ X × Y : g(x, y) = h(x, y)} = {(x, y) ∈ X × Y : y = f (x)} = Gr( f ) is closed in X × Y which completes the proof. Obviously, Theorem 2 enables us to construct many examples of those continuous mappings which are not closed but have closed graphs. However, the following statement is valid (cf. [58] and [101]). Theorem 3. Let X be a regular topological space, Y be a T1 -space and let f : X → Y be a closed continuous mapping. Then the graph Gr( f ) is a closed subset of the product space X ×Y. Proof. It suffices to show that the set (X × Y ) \ Gr( f ) is open in X × Y . Take any point (x, y) ∈ (X ×Y ) \ Gr( f ). We have x ∈ f −1 (y). Since f −1 (y) is closed in X, there exist open neighborhoods U(x) and V ( f −1 (y)) such that / U(x) ∩V ( f −1 (y)) = 0. The set Y \ f (X \ V ( f −1 (y))) is open and contains y. It is easy to verify that / Gr( f ) ∩ (U(x) × (Y \ f (X \ V ( f −1 (y))))) = 0. This fact immediately implies that the set (X × Y ) \ Gr( f ) is open which ends the proof. In general, the product of two closed continuous mappings does not need to be closed (see Exercise 5). On the other hand, we have the following statement.

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Theorem 4. The product of finitely many mappings whose graphs are closed is a mapping with closed graph. Proof. It suffices to consider two mappings f1 : X1 → Y1 ,

f2 : X2 → Y2 ,

whose graphs are closed. The product mapping f1 × f2 : X1 × X2 → Y1 × Y2 defined by the standard formula ( f1 × f2 )(x1 , x2 ) = ( f1 (x1 ), f2 (x2 )) ((x1 , x2 ) ∈ X1 × X2) has the graph Gr( f1 × f2 ) = {(x1 , x2 , y1 , y2 ) : f1 (x1 ) = y1 , f2 (x2 ) = y2 }. Clearly, the set Gr( f1 × f2 ) is homeomorphic to the set Gr( f1 ) × Gr( f2 ) under the canonical homeomorphism of X1 × X2 × Y1 × Y2 onto X1 × Y1 × X2 × Y2 . It remains to use the simple fact that the product of any two closed sets is closed, too, thus the required result immediately follows. Recall that a topological space E is compact if E is quasicompact and Hausdorff simultaneously. Theorem 5. Let X be a compact topological space, R be an equivalence relation on X and let G = Gr(R) denote the graph of R. The following three relations are equivalent: 1) G is a closed subset of X × X; 2) R is closed, i.e., the canonical surjection φ : X → X/R is a closed mapping; 3) the quotient space X/R is Hausdorff. Proof. 3) ⇒ 1). Let us show that this implication holds true without any assumption on X. Indeed, suppose that X/R is Hausdorff. This circumstance means that the diagonal  = {z ∈ X/R × X/R : pr1 (z) = pr2 (z)} is closed in the product space X/R × X/R. But we obviously have G = (φ × φ )−1 (). Thus G is closed, being the pre-image of  under the continuous mapping φ × φ . 1) ⇒ 2). Let G = Gr(R) be closed in the product space X × X. We must show that φ : X → X/R is a closed mapping. For this purpose, consider any closed set F ⊂ X. Clearly, we may write R(F) = pr2 ((F × X) ∩ G).

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The set (F × X) ∩ G is closed in X × X. Since X is compact, R(F) is also closed in view of the Kuratowski theorem on closed projections. But this fact precisely means that φ : X → X/R is a closed mapping. 2) ⇒ 3). Suppose that R is closed, i.e., φ : X → X/R is a closed mapping. As is well known, the compact X is also regular and, in view of the closedness of R, all of the singletons in the quotient space X/R are closed which means that X/R is a T1 -space. Applying Theorem 3, we derive that the graph Gr(φ ) of φ is a closed subset of the product space X × X/R which implies, by virtue of Theorem 4, that the product mapping φ × φ has closed graph. At the same time, for any closed set F ⊂ X × X, we may write the formula (φ × φ )(F) = pr2 ((F × ((X/R) × (X/R))) ∩ Gr(φ × φ )). Since F is quasicompact (as a closed subset of the quasicompact product space X × X), we can apply the Kuratowski theorem and infer from it that φ × φ is a closed mapping. Finally, it is clear that  = (φ × φ )({(x, x) : x ∈ X}) and since the set {(x, x) : x ∈ X} is closed,  turns out to be closed in the product space X/R × X/R. But the latter means that X/R is a Hausdorff topological space. Theorem 5 has thus been proved. Now, we would like to present one important statement concerning the existence of Borel selectors of certain partitions of a Polish topological space. This statement will be essentially utilized in Chapter 9 where we discuss the question of measurability of various selectors associated with countable transformation groups. Our presentation follows Bourbaki’s widely known book [24] with minor changes and simplifications. First of all, we need the next purely set-theoretical proposition. Theorem 6. Let E be an arbitrary set in which a countable system of its subsets (Fn1 n2 ...nk )n1 <ω ,n2 <ω ,...,nk <ω

(1  k < ω )

is given satisfying the following conditions: 1) E = ∪{Fn : n < ω }; 2) Fn1 n2 ...nk = ∪{Fn1 n2 ...nk n : n < ω } for any n1 < ω , n2 < ω , ..., nk < ω and for every nonzero natural number k. Let R be an equivalence relation in E. Define by ordinary recursion another countable system (Hn1 n2 ...nk )n1 <ω ,n2 <ω ,...,nk <ω

(1  k < ω )

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of subsets of E. Namely, put Hn = Fn ∩ (∩{E \ R(Fm ) : m < n}) (n < ω ), Hn1 n2 ...nk nk+1 = Fn1 n2 ...nk nk+1 ∩ Hn1 n2 ...nk ∩ (∩{E \ R(Fn1n2 ...nk m ) : m < nk+1 }). Then for each R-equivalence class X and for any natural number k  1, there exists a unique set of the form Hn1 n2 ...nk such that / Fn1 n2 ...nk ∩ X = Hn1 n2 ...nk ∩ X = 0. Proof. We argue by induction on k. First, consider the case when k = 1. Pick any R/ equivalence class X. Since X = 0, / there is a smallest natural number n for which X ∩Fn = 0. In accordance with the definition, we have Hn = Fn ∩ (∩{E \ R(Fm ) : m < n}). Taking into account the relations X = R(X), Fm ∩ X = 0/

(m < n),

we deduce that X ⊂ ∩{E \ R(Fm ) : m < n} from which it follows / Hn ∩ X = Fn ∩ X = 0. Further, since Hm ⊂ Fm , we readily infer that Hm ∩ X = 0/ whenever m < n. On the other hand, we have X ⊂ R(Fn ) and if a natural number m is strictly greater than n, the definition / Thus, the assertion of the theorem holds true for of Hm yields at once that Hm ∩ X = 0. k = 1. Suppose now that this assertion is valid for k and let us show that it remains true for k + 1. Let X be again an arbitrary R-equivalence class. According to the inductive assumption, there exists a unique set Hn1 n2 ...nk for which / Hn1 n2 ...nk ∩ X = Fn1 n2 ...nk ∩ X = 0. Using condition 2) of the theorem, we derive that there is a smallest natural number n such that / Fn1 n2 ...nk n ∩ X = 0. Let us put nk+1 = n and consider the set Hn1 n2 ...nk nk+1 = Fn1 n2 ...nk nk+1 ∩ Hn1 n2 ...nk ∩ (∩{E \ R(Fn1n2 ...nk m ) : m < nk+1 }).

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Similarly to the case k = 1, it can readily be seen that X ⊂ ∩{E \ R(Fn1n2 ...nk m ) : m < nk+1 } from which it follows that Hn1 n2 ...nk nk+1 ∩ X = Fn1 n2 ...nk nk+1 ∩ X = 0. / Again, according to the inductive assumption, if (m1 , m2 , ..., mk ) = (n1 , n2 , ..., nk ), then Hm1 m2 ...mk ∩ X = 0. / Consequently, only the intersection of the form Hn1 n2 ...nk m ∩ X, where m < ω , may turn out to be nonempty. But, as in the case k = 1, we easily conclude that this intersection is nonempty if and only if m = nk+1 (in view of the definitions of the natural index nk+1 and of the set Hn1 n2 ...nk nk+1 ). Thus, by induction, the theorem is proved. Now, we are ready to establish the next important statement. Theorem 7. Let E be a Polish topological space and let R be an equivalence relation in E satisfying the following conditions: 1) all R-equivalence classes are closed subsets of E; 2) for any closed subset F of E, the set R(F) is Borel in E. Then there exists a Borel selector of the partition E/R of E. Proof. Since E is a Polish space, we can associate to it a countable system (Fn1 n2 ...nk )n1 <ω ,n2 <ω ,...,nk <ω

(1  k < ω )

of closed subsets such that the conditions 1) and 2) of Theorem 6 are satisfied and, in addition, we may suppose that the diameter of each set Fn1 n2 ...nk is less than 2−k . For the above-mentioned system of closed subsets of E, define the corresponding countable system of sets (Hn1 n2 ...nk )n1 <ω ,n2 <ω ,...,nk <ω

(1  k < ω )

as in the formulation of Theorem 6. Using condition 2), it is easy to see that in our case all Hn1 n2 ...nk are Borel subsets of E. Now, for any natural number k  1, let us put Bk = ∪{Hn1 n2 ...nk : n1 < ω , n2 < ω , ..., nk < ω } and consider the set B = ∩{Bk : 1  k < ω }.

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We assert that B is the required Borel selector of E/R. Obviously, B is a Borel subset of E. Therefore, it only remains to show that the intersection of B with any R-equivalence class is a singleton. For this purpose, take an arbitrary R-equivalence class X. By virtue of Theorem 6, this X uniquely determines a decreasing sequence of sets Hn1 ⊃ Hn1 n2 ⊃ ... ⊃ Hn1 n2 ...nk ⊃ ... such that Hn1 n2 ...nk ∩ X = Fn1 n2 ...nk ∩ X = 0/ for each natural number k  1. In view of condition 1), the last relation shows, in particular, that all sets Hn1 n2 ...nk ∩ X are closed in E. Since their diameters tend to zero, their intersection has the form {x}, where x ∈ X. We thus get B ∩ X = {x}, which finishes the proof of Theorem 7. Some other statements concerning the existence of measurable selectors (for set-valued mappings) are discussed in Appendix 2. The preceding facts and constructions and especially the systems of sets described in Theorem 6 are closely related to the so-called A-operation from classical descriptive set theory. It makes sense to touch upon this fundamental set-theoretical operation and indicate its important properties. Let E be an arbitrary set in which a countable system of subsets (Fn1 n2 ...nk )n1 <ω ,n2 <ω ,...,nk <ω

(1  k < ω )

is given. Let us denote Ft = Ft1 ∩ Ft1t2 ∩ ... ∩ Ft1t2 ...tk ∩ ... (t ∈ ω ω \{0} ). Then the set A((Fn1 n2 ...nk )n1 <ω ,n2 <ω ,...,nk <ω ,k1 ) = ∪{Ft : t ∈ ω ω \{0} } is called the result of A-operation over the above-mentioned countable system of subsets of E. If L is some class of subsets of E, then the symbol A (L ) denotes the class of all those sets which can be obtained as a result of A-operation over all analogous countable systems of sets belonging to L . The members of A (L ) are usually called analytic sets over L .

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From the point of view of applications, the most important case is when E is a Polish topological space and an initial class L of subsets of E coincides with the family of all closed sets in E. In such a situation, the members of the class A (L ) are frequently called Suslin sets in E. It can be shown that every Borel subset of E is a Suslin set. The converse assertion is not true in general. Actually, the classical theorem of Suslin states that any uncountable Polish topological space E contains a Suslin set which is not Borel (see, e.g., [10], [99], [148], [150], [160], [162], or Appendix 6). The complements of analytic (Suslin) sets in E are called co-analytic (co-Suslin) sets. Analytic and co-analytic sets have rather good descriptive properties. They are universally (absolutely) measurable with respect to the class of the completions of all σ -finite Borel measures on E and possess the Baire property in the restricted sense. This circumstance is a direct consequence of the fact that the A-operation preserves the measurability and Baire property (cf. [10], [99], [148], [150], [160], [162], and [210]). Such a nice circumstance fails to be longer true for projective sets of higher levels. For instance, a continuous image (in R) of a co-analytic subset of R can be nonmeasurable in the Lebesgue sense (in some natural models of set theory). In connection with this unpleasant fact, see [10] and [188]. EXERCISES 1. Let Y be a σ -quasicompact topological space, that is Y can be covered by a countable family of its quasicompact subspaces. Show that for every topological space X, the canonical projection pr1 : X × Y → X has the following property: if A is an Fσ -subset of the product space X × Y , then the image pr1 (A) is an Fσ -subset of X. 2. Let Y be a metrizable topological space. Verify that the following two conditions are equivalent: (a) Y is compact; (b) the canonical projection pr1 : R × Y → R is a closed mapping. For this purpose, use the well-known criterion of the compactness of a metric space, which states that a metric space Y is compact if and only if it is complete and totally bounded. 3∗ . Let Y be a topological space having the property that for any topological space X, the canonical projection pr1 : X × Y → X is a closed mapping. Prove that Y is quasicompact.

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For this purpose, suppose otherwise and take a family {Fi : i ∈ I} of closed subsets of Y such that ∩{Fi : i ∈ I} = 0/ but Fi1 ∩ Fi2 ∩ ... ∩ Fin = 0/ for every finite subfamily {i1 , i2 , ..., in } of I. Fix an element x0 ∈ Y and put X = Y ∪ {x0 }. Let T denote the topology on X consisting of all sets U of the form U = Z ∨ U = Z ∪ {x0 } ∪ (Fi1 ∩ Fi2 ∩ ... ∩ Fin ), where Z ⊂ Y , n < ω and {i1 , i2 , ..., in } ⊂ I. It can readily be verified that T is indeed a topology on X. Now, in the product space X × Y consider the closed set A = cl({(y, y) : y ∈ Y }). Observe that Y ⊂ pr1 (A) and that x0 is an accumulation point for Y , hence for pr1 (A) as well. Consequently, we must have x0 ∈ pr1 (A) from which it follows that there exists a point y0 ∈ Y satisfying the relation (x0 , y0 ) ∈ A. Further, for any i ∈ I and for any neighborhood U(y0 ) of y0 , the product set ({x0 } ∪ Fi ) × U(y0 ) is a neighborhood of (x0 , y0 ), so / {(y, y) : y ∈ Y } ∩ (({x0 } ∪ Fi) × U(y0)) = 0. This fact implies at once that Fi ∩U(y0 ) = 0/ and, by virtue of the closedness of Fi , we get y0 ∈ Fi . Finally, we come to the relation y0 ∈ ∩{Fi : i ∈ I} / which contradicts the assumption that ∩{Fi : i ∈ I} = 0. This result is converse to the Kuratowski theorem on closed projections. It was obtained by Mr´owka [180]. 4∗ . Let C[0, 1] denote the Banach space of all real-valued continuous functions on the unit segment [0, 1] and let D be the family of all those functions from C[0, 1] which are differentiable at least at one point of [0, 1]. Prove that D is a first category subset of C[0, 1]. For this purpose, fix a natural number n and consider the set Pn = {( f , x) ∈ C[0, 1] × [0, 1] : (∀h ∈ [−x, 1 − x])(| f (x + h) − f (x)|  n|h|)}.

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Verify that Pn is a closed subset of the product space C[0, 1] × [0, 1] and that D ⊂ ∪{pr1 (Pn ) : n < ω }. Applying Kuratowski’s theorem on closed projections, check that every set pr1 (Pn ) (n < ω ) is closed and, simultaneously, nowhere dense in C[0, 1], which yields the required result. Conclude from the said above that the family of all nowhere differentiable functions from C[0, 1] is residual (co-meager) in C[0, 1]. Consequently, there are many nowhere differentiable functions in C[0, 1]. This important statement was independently established by Banach [9] and Mazurkiewicz [171]. 5. Consider two mappings f1 : R → R and f2 : R → R such that f1 (x) = x, f2 (x) = 0 (x ∈ R). Observe that both of them are closed and continuous. The product mapping ( f 1 , f 2 ) : R2 → R2 is defined by the formula ( f1 , f2 )(x, y) = ( f1 (x), f2 (y)) = (x, 0) ((x, y) ∈ R2 ) and, in fact, coincides with the canonical projection pr1 : R × R → R. Check that ( f1 , f2 ) is not a closed mapping. 6. Generalize Theorem 4 to the case of an arbitrary family { fi : i ∈ I} of mappings with closed graphs. Namely, show that if all fi (i ∈ I) have closed graphs, then the product mapping ∏{ fi : i ∈ I} also has closed graph. 7∗ . Let X and Y be two topological spaces and let f : X → Y be a continuous mapping such that: (a) f is surjective and closed; (b) f −1 (y) is quasicompact for any point y ∈ Y . Prove that the inequality w(Y )  w(X) + ω holds true, where w(X) (respectively, w(Y )) denotes the topological weight of X (respectively, of Y ). For this purpose, choose an open base {Ui : i ∈ I} of X such that card(I) = w(X). Assume, without loss of generality, that I is an infinite set. Let J denote the family of all finite subsets of I. Clearly, card(J ) = card(I). Verify that the family of all sets of the form Y \ f (X \ ∪{Ui : i ∈ J}) (J ∈ J )

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is an open base of Y . To see this circumstance, take an arbitrary point y ∈ Y and consider its arbitrary open neighborhood W (y). According to (b), the set f −1 (y) is quasicompact and is contained in the open set f −1 (W (y)). Consequently, there exists a set J ∈ J for which we have f −1 (y) ⊂ ∪{Ui : i ∈ J} ⊂ f −1 (W (y)), thus it follows that y ∈ Y \ f (X \ ∪{Ui : i ∈ J}) and the set Y \ f (X \ ∪{Ui : i ∈ J}) is open in Y . Also, verify that Y \ W (y) ⊂ f (X \ f −1 (W (y))) ⊂ f (X \ ∪{Ui : i ∈ J}), which implies that Y \ f (X \ ∪{Ui : i ∈ J}) ⊂ W (y). This inclusion at once yields the required inequality w(Y )  w(X) + ω . 8. Deduce from the previous exercise that if X is a quasicompact topological space with w(X)  ω and Y is a Hausdorff continuous image of X, then Y is a compact metrizable space. In particular, a continuous image of a metrizable compact space is metrizable if and only if it is Hausdorff. 9. Let E be a Polish topological space and let R be an equivalence relation in E such that for any closed set F ⊂ E, the set R(F) is also closed in E. Show that the partition E/R canonically associated with R admits a Borel selector of type Fσ δ . 10. Let E be a compact metric space and let R be an equivalence relation in E whose graph is a closed subset of the product space E × E. Deduce directly from the theorem of Kuratowski and Ryll-Nardzewski (see [151] or Appendix 2) that there exists a Borel selector of the partition E/R. For this purpose, take into account the fact that the quotient space E/R is compact and metrizable (see Exercise 8) and consider the canonical surjection

φ : E → E/R. Associate with this surjection a set-valued mapping Φ : E/R → P(E)

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defined by the formula Φ(z) = φ −1 (z)

(z ∈ E/R).

Check that Φ satisfies the assumptions of the theorem of Kuratowski and Ryll-Nardzewski (where E/R is equipped with its Borel σ -algebra). Let f : E/R → E be a Borel selector of Φ. Keeping in mind that f is injective and applying the theorem that an injective metrizable Borel image of a Borel subset of a Polish space is also a Borel set (see [99], [148], [160], or Appendix 6), conclude that ran( f ) is the required Borel selector of the partition E/R. 11. Let E be an arbitrary set. Show that the A-operation over any class L ⊂ P(E) generalizes the standard operations of taking unions and intersections of countable families of sets from L . 12∗ . Equip the set ω with its discrete topology and consider the product space ω ω . As usual, ω ω is called the canonical Baire space of topological weight ω . Prove that any nonempty Polish topological space E is a continuous image of ω ω . Moreover, apply the fact that any nonempty Suslin set in E is a continuous image of ω ω (see Appendix 6) and infer from this fact that any Suslin set in E may be regarded as the second projection of some closed subset of the Polish product space ω ω × E. 13∗ . Let E1 be a topological space satisfying the following two conditions: (a) each point of E1 has a countable local base; (b) every closed subset of E1 is of type Gδ in E1 . Let E2 be a complete metric space and let f : E1 → E2 be a partial continuous mapping. Show that there exists a partial continuous mapping f ∗ : E1 → E2 extending f and defined on a Gδ -subset of E1 (Lavrentiev’s theorem [157] on extensions of partial continuous functions). Argue in the following manner. For any point x ∈ cl(dom( f )), let B(x) denote a local base of x and let Ω f (x) = inf{diam( f (U)) : U ∈ B(x)}, where diam( f (U)) denotes the diameter of f (U). The real number Ω f (x)  0 is usually called the oscillation of f at x. Check that for every point x ∈ dom( f ), the equality Ω f (x) = 0 holds true. Further, put X = {x ∈ cl(dom( f )) : Ω f (x) = 0}

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and show that the set X is of type Gδ in E1 . Now, let x ∈ X and let {xn : n < ω } ⊂ dom( f ) be a sequence converging to x. Verify that { f (xn ) : n < ω } is a Cauchy sequence in E2 and hence converges to some point y ∈ E2 . Putting f ∗ (x) = y, prove that the function f ∗ : X → E2 is well defined in this manner, which extends f and is continuous on X. 14∗ . Let E1 and E2 be two complete metric spaces and let f : E1 → E2 be a partial homeomorphism. By using the result of the previous exercise, show that there exists a partial homeomorphism f ∗ : E1 → E2 extending f and such that both sets dom( f ∗ ) and ran( f ∗ ) are of type Gδ in E1 and in E2 respectively (Lavrentiev’s theorem [157] on extensions of partial homeomorphisms). For this purpose, denote g = f −1 . In view of Exercise 13, there are two partial continuous mappings f ∗ : E1 → E2 , g ∗ : E2 → E1 extending f and g, respectively, and defined on some sets of type Gδ . In the product space E1 × E2 consider the set Z = {(x, y) : y = f ∗ (x)} ∩ {(x, y) : x = g∗ (y)}. Verify that dom( f ) ⊂ pr1 (Z) and ran( f ) ⊂ pr2 (Z). Moreover, check that both sets pr1 (Z) and pr2 (Z) are of type Gδ and the mapping f ∗ |pr1 (Z) establishes a homeomorphism between pr1 (Z) and pr2 (Z). 15. Let E1 be a metric space, E2 be a complete metrizable space and let f : E1 → E2 be a partial homeomorphism. By using the previous exercise, show that there exists a partial homeomorphism f ∗ : E1 → E2 extending f and defined on some Gδ -subset of E1 . Note that, in general, we cannot assert that the set ran( f ∗ ) is of type Gδ in E2 . 16∗ . Let E1 be a metric space, E2 be a Polish topological space and let f : E1 → E2 be a partial Borel mapping. Prove that there exists a Borel mapping f ∗ : E1 → E2 extending f . For this purpose, start with Lavrentiev’s theorem on extensions of partial continuous functions (see Exercise 13), take into account the Baire classification of Borel mappings (see Example 6 from Chapter 1 and Appendix 6) and then use the method of transfinite induction on α < ω1 . 17. Let E1 and E2 be two Polish topological spaces and let f : E1 → E2 be a partial Borel isomorphism. Prove that there exists a partial Borel isomorphism f ∗ : E1 → E2 extending f and such that dom( f ∗ ) is a Borel subset of E1 and ran( f ∗ ) is a Borel subset of E2 .

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For this purpose, apply the result of Exercise 16 and use an argument similar to that of Exercise 14. 18∗ . Let E be an infinite set and let Φ be a family of partial injections acting from E into itself, such that card(Φ) = card(E), (∀φ ∈ Φ)(card(dom(φ )) = card(E)). Prove that there exists a family {Xi : i ∈ I} of subsets of E satisfying the following conditions: (a) card(I) = 2card(E) ; (b) for any two distinct indices i ∈ I and j ∈ I and for each function φ ∈ Φ, we have

φ (Xi ) \ X j = 0. / In order to show this fact, denote by α the least ordinal number of cardinality card(E). Let {xξ : ξ < α } be an enumeration of all points in E and let {φξ : ξ < α } be an enumeration of all those partial functions which either belong to Φ or are inverse to a member of Φ. For every ξ < α , define by transfinite recursion a point pξ ∈ E \ ({pζ : ζ < ξ } ∪ {φζ (pη ) : ζ < ξ , η < ξ }) and put P = {pξ : ξ < α }. Further, verify that there exists a family {Xi : i ∈ I} of subsets of P satisfying condition (a) and such that card(Xi \ X j ) = card(E) (i ∈ I, j ∈ I, i = j). Check that this {Xi : i ∈ I} is the required family. 19. Let E be an uncountable Polish topological space and let X and Y be two subspaces of E. We shall say that these subspaces are incomparable in the descriptive sense if there exists no Borel isomorphism of X onto some subset of Y and there exists no Borel isomorphism of Y onto some subset of X. By using the results of Exercise 17 and Exercise 18, prove that there are 2c subspaces of E which are pairwise incomparable in the descriptive sense. In particular, conclude that there are 2c Borel types of subspaces of E. 20. Give an example of a partial homeomorphism g : R → R such that both sets dom(g) and ran(g) are homeomorphic to the Cantor space {0, 1}ω but there exists no homeomorphism g∗ : R → R extending g. 21∗ . Let f be a function acting from R into R and let x be a point of R.

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We shall say that x is a pleasant point with respect to f (or, briefly, f -pleasant point) if for each real number ε > 0, there exists a neighborhood V (x) = V (x, ε ) of x such that the set {y ∈ R : | f (y) − f (x)| < ε } is categorically dense in V (x), i.e., the intersection of this set with any nonempty open interval contained in V (x) is of second category in that interval. In accordance with the definition above, we shall also say that a point x ∈ R is unpleasant with respect to f (or, briefly, f -unpleasant point) if x is not f -pleasant. Show that, for any function f : R → R, the set of all f -unpleasant points is of first category in R; consequently, the set of all f -pleasant points is co-meager in every nonempty open subinterval of R. For this purpose, suppose otherwise, i.e., suppose that the set A = {x ∈ R : x is unpleasant with respect to f } is not of first category. For each point x ∈ A, there exists a strictly positive number ε (x) such that for any neighborhood V of x, the set {y ∈ R : | f (y) − f (x)| < ε (x)} is not categorically dense in V . Pick two rational numbers r(x) and s(x) satisfying the inequalities f (x) − ε (x)/2 < r(x) < f (x) < s(x) < f (x) + ε (x)/2. Further, for any pair (r, s) of rational numbers, put Ar,s = {x ∈ A : r(x) = r, s(x) = s}. Evidently, the equality A = ∪(r,s)∈Q×Q Ar,s holds true. By the assumption, A is not of first category, so there exists a pair (r0 , s0 ) ∈ Q × Q such that the set Ar0 ,s0 is not of first category, either. Consequently, there exists a nonempty open interval ]a, b[ ⊂ R such that the set Ar0 ,s0 is categorically dense in ]a, b[. Choose any point x0 ∈ ]a, b[ ∩ Ar0 ,s0 . For this point x0 , we may write f (x0 ) − ε (x0 )/2 < r0 < f (x0 ) < s0 < f (x0 ) + ε (x0 )/2.

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Analogously, for each point y ∈ Ar0 ,s0 , we have f (y) − ε (y)/2 < r0 < f (y) < s0 < f (y) + ε (y)/2. Therefore, | f (y) − f (x0 )| < s0 − r0 < ε (x0 ). In other words, the inclusion Ar0 ,s0 ⊂ {y ∈ R : | f (y) − f (x0 )| < ε (x0 )} is valid, which immediately implies that the set {y ∈ R : | f (y) − f (x0 )| < ε (x0 )} is categorically dense in ]a, b[ contradicting the definition of ε (x0 ). The obtained contradiction yields the desired result. 22∗ . Let f be an arbitrary function acting from R into R. Prove that there exists an everywhere dense subset X of R such that the function f |X is continuous (Blumberg’s theorem [15]). Argue in the following manner. Since R is homeomorphic to the open unit interval ]0, 1[, it suffices to establish Blumberg’s statement for any function f : ]0, 1[ → ]0, 1[. Let Gr( f ) denote the graph of f . Taking into account Exercise 21, recursively construct two sequences {Zn : n ∈ N}, {Xn : n ∈ N}, satisfying the following conditions: (a) for each natural number n, the set Zn can be represented in the form Zn = ∪{]ai , bi [×]ci , di [ : i ∈ I(n)}, where I(n) is a countable set, all intervals of the family {]ai , bi [ : i ∈ I(n)} are contained in ]0, 1[ and are pairwise disjoint, the set pr1 (Zn ∩ Gr( f )) is categorically dense in ]0, 1[ and for any i ∈ I(n), the length of ]ci , di [ is strictly less than 1/(n + 1); (b) for each natural number n, the set Xn is a finite (1/(n + 1))-net of ]0, 1[, i.e., for any point t from ]0, 1[, there exists a point x ∈ Xn such that |t − x| < 1/(n + 1); (c) the sequence of sets {Zn : n ∈ N} is decreasing by inclusion; (d) the sequence of sets {Xn : n ∈ N} is increasing by inclusion;

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(e) for each natural number n, the relation Xn ⊂ pr1 (Zn ∩ Gr( f )) holds and any point from Xn is f -pleasant. Afterwards, put X = ∪{Xn : n ∈ N} and verify, by using the conditions (a), (c), (d), (e), that X is everywhere dense in R and the restriction of f to X is continuous. 23. We say that a topological space E is a Blumberg space if for any function g : E → R, there exists an everywhere dense subset X of E such that the restriction g|X is continuous. Check that any Blumberg space is necessarily a Baire topological space. A useful survey of results concerning Blumberg’s theorem, its generalizations, and Blumberg spaces is presented in paper [28]. See also [27], [29], and [241]. 24∗ . We shall say that a metrizable topological space E is topologically complete if there exists a complete metric space homeomorphic to E. Show that any open subset U of a topologically complete metric space (E, d) is also topologically complete. For this purpose, assume that U = E and consider in the complete product space E × R the set K = {(x,t) : td(x, E \ U) = 1}. Verify that this K satisfies the following relations: (a) K is a closed subset of E × R and hence K is topologically complete; (b) K is homeomorphic to U. Conclude from the relations (a) and (b) the topological completeness of U. Further, suppose that {Vn : n < ω } is a countable family of topologically complete subspaces of a metrizable topological space E. Show that the space ∩{Vn : n < ω } is also topologically complete. For this purpose, consider the product space V = ∏{Vn : n < ω } which is topologically complete and take in this space the diagonal Δ = V ∩ {z ∈ E ω : (∀n < ω )(∀m < ω )(prn (z) = prm (z))}. Verify that Δ is a closed subset of V and the space ∩{Vn : n < ω } is homeomorphic to Δ. Taking into account the above-mentioned facts and Lavrentiev’s theorem on extensions of homeomorphisms, prove that for a metrizable space E, the following two assertions are equivalent:

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(c) E is topologically complete; (d) E is homeomorphic to a Gδ -subset of some complete metric space. Finally, by using (d) and the topological universality of [0, 1]ω in the class of all separable metrizable spaces, show that Polish spaces can be topologically characterized as Gδ -subsets of [0, 1]ω . In this connection, recall that the equivalence between (c) and (d) was first established by Alexandrov (see, for instance, [58] and [148]). 25∗ . Suppose that the inequality 2ω < 2ω1 is satisfied (for example, this inequality is fulfilled if the Continuum Hypothesis holds true). Let E be a separable metric space, E  be a metrizable space, and let g : E → E  be a Borel mapping. Demonstrate that g(E) is a separable subspace of E  . For this purpose, apply the fact that every nonseparable metrizable space contains an uncountable discrete subspace. Let now P be a Polish topological space or, more generally, a Suslin subset of a Polish topological space and let f : P → E  be a Borel mapping. Demonstrate within ZFC theory that f (P) is a separable subspace of E  . For this purpose, utilize the result of Exercise 12 and take into account the fact that any uncountable Suslin set is of cardinality continuum (see [10], [99], [148], [160], or Appendix 6).

Chapter 9

Weakly metrically transitive measures and nonmeasurable sets

In the previous chapters we were concerned with various extensions of σ -finite measures, which are obtained by using certain types of nonmeasurable sets or nonmeasurable functions. It was also mentioned that sometimes absolutely nonmeasurable sets can occur for concrete classes of measures (e.g., Vitali sets for the class of all translation-invariant extensions of the Lebesgue measure λ = λ1 on the real line R = R1 ). Obviously, the latter sets turn out to be useless from the point of view of the measure extension problem. In the present chapter, we continue our consideration of absolutely nonmeasurable sets (for nonzero σ -finite invariant measures) and show their connections with the notion of a weakly metrically transitive measure. The precise definition of weakly metrically transitive measures will be introduced later. In 1896 Minkowski published his famous work [173] in which he suggested a geometric approach to classical problems of number theory and extensively developed beautiful geometric methods in this theory. The main role in his methods was played by convex subsets of Euclidean spaces. Among many other important results, Minkowski proved in [173] the following fundamental statement. Minkowski’s Theorem. If C is a compact convex body in the Euclidean space Rn , symmetric with respect to the origin of Rn , and having volume greater than or equal to 2n , then C contains at least two nonzero points of the lattice Zn where Z denotes the set of all integers. This statement was then generalized in various directions (see, e.g., [76], [77] and the references therein). The argument used by Minkowski in the proof of his theorem is purely group-theoretical and measure-theoretical. Actually, by applying a similar argument, it can be established that there exists a subset of Rn nonmeasurable in the Lebesgue sense (see Theorem 2 beA.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7_9, © 2009 Atlantis Press/World Scientific

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low). In this context, it is interesting to notice that the rigorous concept of the Lebesgue measure was introduced some years later after Minkowski’s theorem was stated, namely, in the beginning of the twentieth century (see [158]). Moreover, it should be recalled that the existence of subsets of Rn which are nonmeasurable in the Lebesgue sense was first proved by Vitali [239] only in 1905. An analogous astonishing situation can be observed in connection with the famous Poincar´e theorem on recurrent points for an abstract space E equipped with a finite measure μ and with a single transformation g : E → E which preserves μ , that is we have g−1 (X) ∈ dom(μ ) & μ (g−1 (X)) = μ (X) for all sets X ∈ dom(μ ). The above-mentioned theorem was proved by Poincar´e before the basic concepts of measure theory (e.g., the countable additivity of measures) were introduced in real analysis (cf. [192]). The precise formulation of the Poincar´e theorem is given in Exercise 2 and its proof is also outlined therein. The main goal of this chapter is to demonstrate the role of Minkowski’s method in establishing the existence of nonmeasurable subsets of Euclidean spaces and, more generally, in establishing the existence of nonmeasurable sets in an abstract space equipped with a transformation group and with some measure which is invariant under that group. In the sequel, we will introduce the concept of a weakly metrically transitive invariant measure and will show how such a measure produces absolutely nonmeasurable sets with respect to the class of all its invariant extensions. We begin with several auxiliary notions concerning spaces endowed with transformation groups. Let E be a nonempty set, G be a group of transformations of E and let μ be a complete measure defined on a G-invariant σ -algebra of subsets of E and invariant under all transformations from G. In this case, we briefly say that μ is a G-invariant measure (see Chapter 2). Let X be some subset of E. We shall say that X is G-thick with respect to μ if there exists a countable subset (equivalently, countable subgroup) H of G such that

μ (E \ ∪{h(X) : h ∈ H}) = 0. We shall say that X is G-thin with respect to μ if (∀g ∈ G)(∀h ∈ G)(g = h ⇒ μ (g(X) ∩ h(X)) = 0).

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In particular, if a group G acts freely in a space E, then every singleton is G-thin in E. Example 1. If E = R2 and G = {0} × R, then the graph Gr( f ) of any partial function f :R→R is a G-thin subset of E. On the other hand, it also may happen that Gr( f ) is R2 -thick with respect to some translation-invariant extension μ of the two-dimensional Lebesgue measure λ2 (in this connection, see Chapter 19). Moreover, it may happen that there exists a countable family {gn : n < ω } of isometric transformations (i.e., motions) of R2 such that R2 = ∪{gn (Gr( f )) : n < ω }. For more details concerning the latter intriguing fact, see [49], [51], [222], and Exercise 8 of Appendix 1. Example 2. Suppose that μ is a σ -finite G-invariant measure on E metrically transitive (ergodic) with respect to G, i.e., for any μ -measurable set Z, the relation (∀g ∈ G)(μ (g(Z)Z) = 0) implies the disjunction

μ (Z) = 0 ∨ μ (E \ Z) = 0. Then it can easily be checked that every μ -measurable set X with μ (X) > 0 is G-thick in E and this property of μ is equivalent to its metrical transitivity with respect to the group G. In this context, let us recall that the left (right) Haar measure on a σ -compact locally compact topological group (H, ·) is metrically transitive with respect to the group of all left (right) translations of H (see Exercise 17 for Chapter 3). Example 3. Suppose that a group G of transformations of a given space E is countable and suppose, in addition, that G acts almost freely in E with respect to some G-invariant measure μ on E, i.e., for any two distinct transformations g ∈ G and h ∈ G, we have

μ ∗ ({x ∈ E : g(x) = h(x)}) = 0, where μ ∗ stands for the outer measure associated with μ . Consider the partition of E into all G-orbits of the points of E. Let X be an arbitrary selector of this partition (in short, G-selector). Then it is not difficult to verify that X is simultaneously G-thick and G-thin in E. The following statement is essentially due to Minkowski (cf. [173]).

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Theorem 1. Let (E, G) be a space equipped with a transformation group and let μ be a

σ -finite G-invariant measure on E. Let X be a G-thick subset of E and let Y be a G-thin subset of E. If both of these sets are measurable with respect to μ , then μ (Y )  μ (X). Proof. Only two cases are possible. (a) The group G is uncountable. In this case there is nothing to prove because the μ measurable G-thin subset Y of E is necessarily of μ -measure zero in view of the σ finiteness and G-invariance of μ and the almost disjointness of the family {g(Y ) : g ∈ G}. (b) The group G is countable. In this case, we obviously have

μ (E \ ∪{g(X) : g ∈ G}) = 0 and, consequently,

μ (Y ) = μ (Y ∩ (∪{g(X) : g ∈ G})) 

∑ μ (g(X) ∩Y ) =

g∈G

∑ μ (X ∩ g(Y ))  μ (X),

g∈G

which yields the required result. Theorem 1 has thus been proved. Let (E, G) be a space equipped with a transformation group and let μ be a σ -finite Ginvariant (or, more generally, G-quasi-invariant) measure on E. We shall say that μ is weakly metrically transitive (or weakly ergodic) with respect to G if for any ε > 0, there exist a μ -measurable set X with μ (X) < ε and a countable subset (equivalently, countable subgroup) H of G such that

μ (E \ ∪{h(X) : h ∈ H}) = 0. It is not hard to verify that any σ -finite non-atomic G-invariant metrically transitive measure μ on E is weakly metrically transitive. The converse assertion is not true in general. Indeed, it directly follows from the above definition that every G-invariant extension of a G-invariant weakly metrically transitive measure is also weakly metrically transitive. An analogous statement fails to be true for G-invariant metrically transitive measures. Example 4. Let us consider the concrete particular case, where: E = the real line R, G = the group of all translations of R,

μ = the Lebesgue measure on R (i.e., μ = λ ). Obviously, we may identify G with the additive group R. It is well known that μ is metrically transitive with respect to the group Q ⊂ R of all rational numbers.

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By applying the standard transfinite methods (see, e.g., [83], [85], [95], [108], [115], [176], [192], [198], or Exercise 4 from Chapter 6), it can be proved that there exist two subsets A and B of R satisfying the following conditions: (1) A ∩ B = 0/ and A ∪ B = R; (2) card(A) = card(B) = c, where c denotes the cardinality of the continuum; (3) μ∗ (A) = μ∗ (B) = 0; (4) both sets A and B are almost R-invariant, that is for all h ∈ R, we have card((h + A)A) < c, card((h + B)B) < c. By using these two sets, the original measure μ can be extended to an R-invariant measure

μ  which fails to be metrically transitive with respect to the group R (hence, with respect to the group Q ⊂ R). For this purpose, it suffices to define on the σ -algebra S = σ ({A, B} ∪ dom(μ )) the functional μ  by the formula

μ  ((A ∩ X) ∪ (B ∩Y)) = (1/2)(μ (X) + μ (Y)), where X and Y are arbitrary elements from dom(μ ). This definition of μ  is correct and μ  is a measure extending μ (cf. Chapter 2). Applying Marczewski’s method described in Chapter 2, we can also suppose that all subsets of R, whose cardinalities are strictly less than c, belong to dom(μ  ). Now, it is clear that

μ  is R-invariant and, simultaneously, is not metrically transitive in view of the almost invariance of the μ  -measurable sets A and B which both are of strictly positive μ  -measure; moreover, we have

μ  (A) = μ  (B) = +∞. At the same time, μ  as an extension of μ is weakly metrically transitive. Example 5. Suppose that a σ -finite G-invariant metrically transitive measure μ on E has at least one atom. Then it is not difficult to describe the structure of μ . Namely, in this case the space E admits a representation E = ∪{Ai : i ∈ I} for which the following relations are satisfied: (1) the set I is at most countable and the family {Ai : i ∈ I} is disjoint; (2) each set Ai (i ∈ I) is an atom of μ ;

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(3) for any i ∈ I and j ∈ J, there exists a transformation g ∈ G such that

μ (g(Ai )A j ) = 0; (4) for any transformation h ∈ G, the family {h(Ai ) : i ∈ I} almost coincides with the family {Ai : i ∈ I}, i.e., each member of the first family almost coincides with some member of the second family, and conversely. In other words, we obtain a certain G-invariant lattice of atoms of our measure μ , similar to the standard lattice {z + [0, 1]n : z ∈ Zn } of the Euclidean space Rn . Let (E, G) be a space equipped with a transformation group and let M be some class of G-invariant (G-quasi-invariant) measures on E. We would like to stress that, in general, the measures from M are defined on different σ -algebras of subsets of E. Recall (see Chapter 5) that a set X ⊂ E is absolutely nonmeasurable with respect to M if there exists no measure from M for which X is measurable. Now, let μ be a G-invariant (G-quasi-invariant) measure on E. We shall say that a set X ⊂ E is G-absolutely nonmeasurable with respect to μ if X is absolutely nonmeasurable with respect to the class of all G-invariant (G-quasi-invariant) measures on E extending μ . The next example highlights close relationships between convex sets and absolutely nonmeasurable sets. Example 6. Let E be an infinite-dimensional separable Banach space, G be the group of all translations of E, and let B be a bounded convex body in E (i.e., B is a closed bounded convex set in E with nonempty interior). It can be proved that B is absolutely nonmeasurable with respect to the class of all nonzero σ -finite G-invariant (more generally, G-quasi-invariant) measures on E (see [109] and Exercise 10 for this chapter). The above fact implies, in particular, that E does not admit a nonzero σ -finite G-quasi-invariant Borel measure (cf. Chapters 3 and 4). By using an algebraic isomorphism between the additive groups E and R, it can also be shown that there exists a subset of R which is absolutely nonmeasurable with respect to the class of all nonzero σ -finite R-invariant (more generally, R-quasi-invariant) measures on R. Example 7. As in Example 4, let us put: E = the real line R, G = the group of all translations of R,

μ = the Lebesgue measure on R (i.e., μ = λ ). We remember that any selector of R/Q is called a Vitali subset of R (see [10], [40], [72], [90], [143], [148], [150], [176], [183], [192], [239], [240], or Chapter 5). It is well

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known that all Vitali sets are R-absolutely nonmeasurable with respect to μ (see the abovementioned works and Exercise 2 from Chapter 5). Theorem 2. Let G be a countable group of transformations of E and let μ be a nonzero

σ -finite G-invariant weakly metrically transitive measure on E. Suppose also that G acts almost freely in E with respect to μ . Then every G-selector is G-absolutely nonmeasurable with respect to μ . Proof. Take an arbitrary G-selector X in E. We should verify that for any G-invariant measure μ  on E extending the original measure μ , the set X is not μ  -measurable. Suppose otherwise, i.e., X ∈ dom(μ  ) for some G-invariant extension μ  of μ . Then, in view of the countability of G and of the equality E = ∪{g(X) : g ∈ G}, we must have μ  (X) > 0. Let us denote μ  (X) = ε . According to our assumption, there exist a μ -measurable set Y with μ (Y ) < ε and a countable family H ⊂ G such that

μ (E \ ∪{h(Y ) : h ∈ H}) = 0. In other words, the set Y turns out to be a G-thick subset of E with respect to μ . Since μ  extends μ , the same Y is G-thick with respect to μ  as well. On the other hand, the selector X is a G-thin subset of E with respect to μ and hence with respect to μ  (see Example 3). By virtue of Theorem 1, the relation

ε = μ  (X)  μ  (Y ) = μ (Y ) < ε should be valid, which yields a contradiction. This contradiction completes the proof of Theorem 2. Remark 1. Clearly, Theorem 2 may be regarded as a generalization of Vitali’s theorem [239] stating the existence of subsets of R nonmeasurable in the Lebesgue sense. Indeed, put in Theorem 2: E = the real line R, G = the group Q of all rational numbers,

μ = the Lebesgue measure on R (i.e., μ = λ ). Then we directly come to Vitali’s classical result on the existence of λ -nonmeasurable sets in R. Therefore, Theorem 2 may be treated as an abstract version of the above-mentioned classical result. Some other abstract versions of Vitali’s theorem are presented in [115], [119], [195], [229], and [230] (cf. also [251]).

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In order to demonstrate once more that the notion of the weak metrical transitivity is closely connected with the existence of absolutely nonmeasurable sets, we need two auxiliary statements. The first of them deals with Borel selectors of a certain partition of a Polish topological space (see, for instance, [24] and Chapter 8). Lemma 1. Let E be a Polish topological space and let R(x, y) (x ∈ E, y ∈ E) be an equivalence relation on E satisfying the following conditions: (1) each equivalence class with respect to R(x, y) is a closed subset of E; (2) for any closed set F ⊂ E, the set R(F) = {y ∈ E : (∃x)(x ∈ F & (x, y) ∈ R)} is a Borel subset of E. Let E/R denote the quotient set associated with the equivalence relation R. Then there exists a Borel selector of E/R. A detailed proof of Lemma 1 was given in Chapter 8. Lemma 2. Let (G, ·) be a σ -compact complete metrizable topological group and let H be a closed subgroup of G. Then the family of all right (left) translates of H admits a Borel selector. Proof. Notice first that the given metrizable group G is separable as a union of countably many compact and hence separable subspaces. Moreover, applying the classical Baire theorem, we readily infer that G is locally compact. Denote by the symbol G/H the family of all right translates of H. Since H is a closed subgroup of G, the elements of G/H are also closed in G. In view of Lemma 1, it remains to verify that for any closed set F ⊂ G, the set ∪{Hx : x ∈ G, F ∩ Hx = 0} / = H ·F is Borel in G. Indeed, the sets H and F are closed subsets of a σ -compact space, so they are σ -compact as well. Define a continuous mapping

φ : G×G → G by putting

φ (x, y) = x · y (x ∈ G, y ∈ G).

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Obviously, the set H · F coincides with the image of H × F with respect to the mapping

φ , which immediately implies that H · F is representable as the union of countably many compact subsets of G. Therefore, H · F is Borel in G. Remark 2. As mentioned above, any σ -compact complete metrizable topological group (G, ·) is a locally compact Polish group (equivalently, is a locally compact group with a countable base). Also, it is well known that if a Polish group (G, ·) admits a nonzero

σ -finite Borel measure μ invariant under all left (right) translations of G, then G is σ compact and locally compact, and μ coincides with the left (right) Haar measure on G (in this connection, see Chapters 3 and 4). Now, we are ready to establish the following statement. Theorem 3. Let (G, ·) be an uncountable locally compact Polish group, let μ denote the left Haar measure on G, and let H be a countable subgroup of G. We consider H as a group of transformations of G. Namely, each h ∈ H is identified with the mapping h:G→G (we preserve the same notation h) defined by the standard formula h(x) = h · x (x ∈ G). In this way, we come to the partition of G into all left H-orbits. The following three assertions are equivalent: (a) μ is weakly metrically transitive with respect to H; (b) every H-selector in G is absolutely nonmeasurable with respect to μ ; (c) H is non-discrete in G. Proof. The implication (a) ⇒ (b) is a straightforward consequence of Theorem 2. Let us prove the implication (b) ⇒ (c). Suppose that a subgroup H of the given group G is discrete. Then it can easily be verified that H is countable and closed in G. Applying Lemma 2, we infer that there exists a Borel H-selector in G. This selector is μ -measurable and, therefore, it cannot be absolutely nonmeasurable with respect to our μ . We thus have shown that the implication ¬ (c) ⇒ ¬ (b) is valid. Consequently, we also have (b) ⇒ (c). Finally, let us establish the implication (c) ⇒ (a). Suppose that (c) is fulfilled and put P = cl(H), where cl(·) as usual denotes the closure operator. Evidently, P is an uncountable closed subgroup of G. Again, according to Lemma 2, there exists a Borel selector X of the family

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G/P of all right translates of P. Since there are uncountably many pairwise disjoint left translates of X, we must have μ (X) = 0. Also, it is clear that P · X = G. Fix ε > 0. Since μ (X) = 0, there exists an open set U ⊂ G such that X ⊂ U and μ (U) < ε . Obviously, P ·U = G. Taking into account that U is open and H is everywhere dense in P, we readily deduce that ∪{h(U) : h ∈ H} = H ·U = G, which yields the weak metrical transitivity of μ with respect to H, i.e., assertion (a) holds true. Thus, we have shown that the implication (c) ⇒ (a) is valid, which finishes the proof of the theorem. Remark 3. It is not difficult to verify that the implication (c) ⇒ (b) remains true for an arbitrary uncountable σ -compact locally compact topological group (G, ·) endowed with the left Haar measure μ . For the sake of completeness, we will present an easy proof of this fact, applying the classical argument of Vitali [239] to G. Let H be a countable non-discrete subgroup of G. Such subgroups always exist in G and a proof of this simple auxiliary assertion is left to the reader. Pick any H-selector X in G. Suppose to the contrary that there exists a left H-invariant extension μ  of μ such that X ∈ dom(μ  ). Taking into account the equality ∪{hX : h ∈ H} = G, we must have μ  (X) > 0. Further, since G is σ -compact, there is a compact set K ⊂ G satisfying the relation 0 < μ  (X ∩ K)  μ  (K) = μ (K) < +∞. For the sake of brevity, denote Y = X ∩ K. Let V be a compact neighborhood of the neutral element of G. Since H is non-discrete, the set H ∩V is infinite. Let us represent the last set as an injective sequence {hn : n < ω }. Clearly, the following two relations are satisfied: (1) hnY ∩ hmY = 0/ whenever n < ω , m < ω and n = m; (2) ∪{hnY : n < ω } ⊂ V · K.

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Relation (1) directly implies that

μ  (∪{hnY : n < ω }) =

∑ μ (hnY ) = +∞.

n<ω

On the other hand, by virtue of the compactness of V · K, relation (2) implies

μ  (∪{hnY : n < ω })  μ  (V · K) = μ (V · K) < +∞, which yields an obvious contradiction. The obtained contradiction gives us the required result. In a similar way Theorem 2 can be applied to the n-dimensional Lebesgue measure λn on the Euclidean space Rn (n  1). Indeed, we may consider λn as an invariant measure with respect to an arbitrary countable group H of isometric transformations (i.e., motions) of Rn . Notice that some deep properties of λn treated as an H-invariant measure on Rn are investigated in paper [142]. In the same paper various H-invariant extensions of λn are also studied. For our purpose, we need several well-known facts from the geometry of Euclidean spaces.

(i) Let x be a point of Rn and let G be a family of motions of Rn . If the set G(x) = {g(x) : g ∈ G} is relatively compact in Rn , then the family G is relatively compact in the topological group of all motions of Rn . Consequently, if G is a closed subset of the group of all motions of Rn , then the set G(x) is closed in Rn . (ii) Let F be a closed subset of Rn and let G be a closed subset of the group of all motions of Rn . Then the set G(F) = ∪{g(x) : g ∈ G, x ∈ F} admits a representation in the form of a countable union of compact subsets of Rn . Therefore, G(F) is of type Fσ in Rn and hence is Borel in Rn . (iii) The group of all motions of Rn acts almost freely in Rn with respect to λn . Indeed, for any two distinct motions g and h of Rn , the set {x ∈ Rn : g(x) = h(x)} is contained in some affine hyperplane of Rn and, consequently, this set is of λn -measure zero.

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(iv) Let G be an uncountable group of motions of Rn and let X be a G-selector. Then either X is nonmeasurable with respect to λn or λn (X) = 0. Notice that (iv) readily follows from (iii). More generally, if an uncountable group G of transformations of a base set E acts almost freely in E with respect to a σ -finite G-quasiinvariant measure μ on E, then any G-selector is either μ -nonmeasurable or has μ -measure zero. Taking these facts into account, we can formulate for λn a direct analogue of Theorem 3. Namely, we have Theorem 4. Let H be a countable group of motions of the Euclidean space Rn . The following three assertions are equivalent: (a) λn is weakly metrically transitive with respect to H; (b) every H-selector in Rn is absolutely nonmeasurable with respect to λn ; (c) H is non-discrete in the group of all motions of Rn . The proof of this statement is similar to the proof of Theorem 3, and all corresponding details are left to the reader. For further information closely connected with the topic of this chapter, see [39], [40], [46], [60], [83], [84], [104], [107], [115], [119], [164], [176], [229], [230], [249], and [251]. EXERCISES 1∗ . Prove Minkowski’s theorem on convex bodies, formulated in the beginning of the chapter. For this purpose, consider the set C/2 where C is a given compact convex body in the Euclidean space Rn , whose volume λn (C) is strictly greater than 2n , and verify that C/2 is not Zn -thin in Rn . Then reduce the case of a compact convex body C ⊂ Rn with λn (C) = 2n to the previous case by using an appropriate limit process. 2∗ . Let E be a set equipped with a finite measure μ and let g : E → E be a transformation of E preserving μ , i.e., for any set X ∈ dom(μ ), we have g−1 (X) ∈ dom(μ ), μ (g−1 (X)) = μ (X). Let k be a natural number and let A be any μ -measurable set. We denote Rk (A) = {x ∈ A : (∃n ∈ [k, ω [)( f n (x) ∈ A)}.

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Prove that

μ ∗ (A \ Rk (A)) = 0. Deduce from this fact that

μ ∗ (A \ ∩{Rk (A) : k < ω }) = 0. Conclude that μ -almost all points of A infinitely many times return to A via appropriate iterations of g. This result is the classical Poincar´e theorem on recurrence. 3. Let (E1 , G1 ) and (E2 , G2 ) be two spaces equipped with groups of transformations G1 and G2 respectively. Let μ1 be a σ -finite G1 -quasi-invariant measure on E1 and let μ2 be a

σ -finite G2 -quasi-invariant measure on E2 . Suppose that μ1 is weakly metrically transitive with respect to G1 and μ2 is weakly metrically transitive with respect to G2 . Show that the product measure μ1 ⊗ μ2 is weakly metrically transitive with respect to the product group G1 × G2 . 4. Complete the details of Example 4. 5. Complete the details of Example 5. 6∗ . Let E be a Polish topological space and let {Zi : i ∈ I} be a countable family of Borel subsets of E. Prove that the following two assertions are equivalent: (a) {Zi : i ∈ I} separates the points in E, i.e., for any two distinct points x ∈ E and y ∈ E, there exists an index i ∈ I such that card(Zi ∩ {x, y}) = 1; (b) the σ -algebra generated by {Zi : i ∈ I} coincides with the Borel σ -algebra B(E) of E. For establishing this equivalence, use Marczewski’s characteristic function of the given family {Zi : i ∈ I} and the classical result from descriptive set theory stating that if f : E → E  is a Borel injection into a Polish space E  , then f (X) is Borel in E  whenever X is a Borel subset of E (see [99], [148], and Appendix 6). 7. Let (E, || · ||) be a real Banach space, X be a bounded convex body in E and let G be a countable everywhere dense subgroup of the additive group E. Applying the result of the preceding exercise, show that the countable family {g(X) : g ∈ G} generates the Borel

σ -algebra B(E). 8. Let (E, ||·||) be an infinite-dimensional Banach space and let K be a σ -compact subset of E. Prove that there exists a nonzero vector e ∈ E such that every straight line in E parallel

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to e intersects K in at most one point (in other words, K is uniform in direction e). Deduce from this fact that K is an E-absolutely negligible subset of E. 9. Let (E, || · ||) be an infinite-dimensional separable Banach space. Using the previous exercise, infer that there is no nonzero σ -finite E-quasi-invariant Borel measure on E. For this purpose, apply the fact that any σ -finite Borel measure on E is Radon (cf. also Chapter 4 where a more general result is presented). 10∗ . Let E be an infinite-dimensional separable Banach space. Prove that any bounded convex body in E is an E-absolutely nonmeasurable subset of E. Prove the same fact for an arbitrary open bounded convex subset of E. For this purpose, take into account the results of Exercises 7 and 9. 11. Give detailed proofs of the statements (i)-(iv) of this chapter. 12. By using the statements (i)-(iv), show the validity of Theorem 4. 13. Let G be a metrizable topological group. Prove that the following three assertions are equivalent: (a) G is topologically complete and σ -compact; (b) G is separable and locally compact; (c) G is Polish and locally compact. Check that the topological vector space R(N) of all eventually finite real-valued sequences is σ -compact and metrizable, but the same vector space R(N) is not complete with respect to any metrizable topology compatible with the vector structure of this space. For this purpose, first observe that R(N) is representable as the union of countably many finite-dimensional vector subspaces, all of which are nowhere dense in R(N) . Then take into account the Baire theorem on category for complete metrizable topological spaces.

Chapter 10

Nonmeasurable subgroups of uncountable solvable groups

It is well known that basic ideas and concepts of modern theory of dynamical systems come from the classical theory of ordinary differential equations. Indeed, for a given autonomous system of first-order ordinary differential equations in a finite-dimensional phase space (e.g., in Rn ), the famous Liouville theorem establishes necessary and sufficient conditions, under which the corresponding phase flow preserves the Lebesgue measure (see, for instance, [196] and Exercise 2 of Chapter 3). Those conditions are trivially fulfilled for a Hamiltonian system in the same phase space. Consequently, we obtain a natural example of a dynamical system which is systematically used in studies of various physical (primarily, evolutional) processes. As is known, an abstract dynamical system can be treated as a triplet (E, G, μ ), where E is a nonempty phase space, G is some group of transformations of E, and μ is a nonzero finite (or σ -finite) G-invariant measure defined on some G-invariant σ -algebra of subsets of E. Many deep and important results were established in terms of (E, G, μ ). Among them, let us especially recall the Poincar´e recurrence theorem (see Exercise 2 for Chapter 9), the Birkhoff ergodic theorem (see [47] and [81]), the Krylov-Bogoliubov theorem on the existence of a dynamical system for a one-parameter group G of homeomorphisms of a nonempty compact metric space E (see Chapter 3). The last result was essentially generalized by Markov and Kakutani to the case of a solvable group G of homeomorphisms of a nonempty compact space E (for more details, see again Chapter 3). This generalization shows, in particular, the remarkable role of solvable transformation groups in the theory of dynamical systems and, equivalently, in the theory of invariant measures. Additionally, the classical results of Banach and von Neumann should be mentioned, which state that if G is a solvable group of isometric transformations of the Euclidean space Rn , then there always exists a nonnegative finitely additive G-invariant functional defined on the family P(Rn ) A.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7_10, © 2009 Atlantis Press/World Scientific

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of all subsets of Rn and extending the Lebesgue measure λn (see [22], [57], [83], [155], and [240]). The main goal of the present chapter is to discuss the measure extension problem for uncountable solvable groups equipped with nonzero σ -finite left invariant (left quasiinvariant) measures. Namely, it will be demonstrated below that special algebraic properties of an uncountable solvable group (G, ·) enable us to extend a given nonzero σ -finite left invariant (left quasi-invariant) measure μ on G by using some subgroups of G which are nonmeasurable with respect to μ . Let (G, ·) be an arbitrary group. As usual, in our further considerations, we will identify G with the group of all left translations of G. We have already mentioned in Chapter 2 that if μ is a nonzero σ -finite left G-invariant (left G-quasi-invariant) measure on an uncountable group (G, ·), then the domain of μ differs from the family of all subsets of G (see, for instance, [60], [104], and Exercise 11 of Appendix 1). In this connection, the following natural problem arises. Does there exist a left G-invariant (left G-quasi-invariant) measure μ  on G strictly extending μ ? In the classical case, when G coincides with the n-dimensional Euclidean space Rn (n  1) and μ is a translation-invariant extension of the standard Lebesgue measure on Rn , this problem was originally posed by Sierpi´nski (see [234] and [235]). Numerous works were devoted to related problems concerning more general situations (see [41], [45], [85], [105], [108], [115], [119], [135], [142], [195], [198], [242], [247], [249], and Exercise 20 from Chapter 3). There are various aspects of the above-mentioned Sierpi´nski problem and of its generalized version for an arbitrary uncountable group (G, ·). In the present chapter, we are going to discuss a certain aspect of this problem which is closely connected with the concept of absolutely negligible sets. We will try to demonstrate that the notion of an absolutely negligible set is of undoubted interest from the measure-theoretical and group-theoretical points of view and plays a significant role in questions concerning extensions of invariant (quasi-invariant) measures (see [105], [108], [115], and [198]). Let X be a subset of (G, ·). We recall (cf. Chapter 2) that X is G-absolutely negligible (in G) if for any σ -finite left G-invariant (respectively, left G-quasi-invariant) measure μ on G, there exists a left G-invariant (respectively, left G-quasi-invariant) measure μ  on G extending μ and satisfying the relation μ  (X) = 0. This definition directly implies that:

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1) if X is a G-absolutely negligible set in G and X belongs to the domain of a given σ -finite left G-invariant (left G-quasi-invariant) measure μ , then μ (X) = 0; 2) if X is a G-absolutely negligible set in G and X does not belong to the domain of a σ finite left G-invariant (left G-quasi-invariant) measure μ on G, then μ is strictly extendable, by using this X, to some left G-invariant (left G-quasi-invariant) measure μ  such that

μ  (X) = 0. The properties 1) and 2) naturally lead to the following method of extending μ . Denote by

ω the first infinite cardinal and suppose that a given group G admits a countable covering {Xn : n < ω } such that all sets Xn (n < ω ) are G-absolutely negligible in G. If our measure

μ is not identically equal to zero, then there always exists at least one index n < ω for which the set Xn does not belong to dom(μ ). Consequently, our μ can be strictly extended with the aid of this Xn (see Chapter 2). So we come to the following problem essentially motivated by the generalized version of Sierpi´nski’s problem mentioned above. Problem. Give a characterization of all those uncountable groups (G, ·) which admit a countable covering by G-absolutely negligible sets. This problem still remains open. The main goal of the present chapter is to demonstrate that any uncountable solvable group (G, ·) can be covered by a countable family of G-absolutely negligible sets. Moreover, a much stronger result will be obtained below in terms of certain subgroups of (G, ·). For this purpose, we need several auxiliary propositions the first of which was already stated earlier in Chapter 2. Lemma 1. Let (G, ·) be an arbitrary group and let X be a subset of G. The following two assertions are equivalent: 1) X is G-absolutely negligible; 2) for any countable family {gi : i ∈ I} ⊂ G, there exists a countable family { f j : j ∈ J} ⊂ G such that

 j∈J



f j(

gi X) = 0. /

i∈I

A detailed proof of Lemma 1 was given in Chapter 2 where even a more general result was presented concerning an analogous characterization of absolutely negligible sets in an abstract space E equipped with some group of its transformations. Obviously, Lemma 1 yields a purely algebraic characterization of all G-absolutely negligible sets in (G, ·). In particular, it immediately follows from this lemma that if X is a

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G-absolutely negligible subset of G and μ is an arbitrary σ -finite left G-quasi-invariant measure on G, then

μ∗ (∪{gi X : i ∈ I}) = 0 for every countable family {gi : i ∈ I} of elements of G (cf. Exercise 13 from Chapter 2). Here the symbol μ∗ as usual denotes the inner measure associated with μ . Lemma 2. Let G and H be two groups and let φ : G → H be an epimorphism (i.e., surjective homomorphism). Suppose that Y is an H-absolutely negligible subset of H. Then the set X = φ −1 (Y ) is G-absolutely negligible in G. Proof. Take any countable family {gi : i ∈ I} of elements of G and denote hi = φ (gi ) for each i ∈ I. Consider the set Y  = ∪{hiY : i ∈ I}. Since Y is H-absolutely negligible, there exists a countable family {hj : j ∈ J} ⊂ H such that ∩{hjY  : j ∈ J} = 0. / Further, since φ is a surjection, we can find a countable family {gj : j ∈ J} ⊂ G such that

φ (gj ) = hj for all j ∈ J. Now, a straightforward verification shows that the equality 

j∈J



gj (

gi X) = 0/

i∈I

holds true. We leave the corresponding details to the reader. Finally, in view of Lemma 1, we can conclude that the set X is G-absolutely negligible in G which ends the proof of Lemma 2. Remark 1. There are several notions in the theory of invariant and quasi-invariant measures, which are connected with the concept of a G-absolutely negligible set. Among them let us especially mention the notion of a G-negligible set and the notion of a G-absolutely nonmeasurable set (for precise definitions and more details, see [108], [115], [119], and Chapters 2 and 5). We would like to stress at this point that direct analogues of Lemma 2 are valid for those sets, too. Namely, let G and H be again two groups, φ : G → H be an epimorphism and let Y be a subset of H. Then the following two assertions are valid: a) if Y is H-negligible, then φ −1 (Y ) is G-negligible;

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b) if Y is H-absolutely nonmeasurable, then φ −1 (Y ) is G-absolutely nonmeasurable. These facts are useful in some constructions of G-negligible and G-absolutely nonmeasurable sets (cf. [108], [115], and [119]). Lemma 3. Let (G, ·) be a group and let H be a normal subgroup of G such that card(G/H)  ω . Fix a selector {gk : k ∈ K} of the countable disjoint family G/H. Suppose that a set Y ⊂ H is H-absolutely negligible in H. Then the set X = ∪{gkY : k ∈ K} turns out to be G-absolutely negligible in G. Proof. We may assume without loss of generality that if gk ∈ H, then gk coincides with the neutral element of G. Consider an arbitrary countable family {gi : i ∈ I} of elements from G. Obviously, for each index i ∈ I, we can write gi = gk(i) hi , where hi ∈ H. Further, since H is a normal subgroup of G, we have the inclusion ∪{gi X : i ∈ I} ⊂ ∪{gk Z : k ∈ K}, where Z = ∪{hi Y : i ∈ I} for some countable family {hi : i ∈ I} of elements of H. Keeping in mind that Y is Habsolutely negligible, we derive that there exists a countable subgroup F of H for which ∩{ f Z : f ∈ F} = 0. / Now, consider the family U = {gk f : k ∈ K, f ∈ F} of elements of G. This family is countable and we assert that the equality 

u(

u∈U



gk Z) = 0/

k∈K

is fulfilled. Indeed, suppose otherwise, i.e., suppose that there exists an element t∈



u(

u∈U



gk Z).

k∈K

Clearly, we can represent t in the form t = gk0 h0 for some elements k0 ∈ K and h0 ∈ H. Consequently, we have the relation t = gk0 h0 ∈

 f ∈F

gk0 f (



k∈K

gk Z)

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from which it readily follows that h0 ∈



f(

f ∈F



gk Z)

k∈K

or, equivalently, Fh0 ⊂ ∪{gk Z : k ∈ K}. Notice now that if gk does not belong to H, then gk Z ∩ H = 0/ and, therefore, Fh0 ∩ gk Z = 0. / This circumstance implies at once the inclusion Fh0 ⊂ Z and hence the relation h0 ∈ ∩{ f Z : f ∈ F}, which contradicts the above-mentioned equality ∩{ f Z : f ∈ F} = 0. / The obtained contradiction shows that  u∈U



u(

gk Z) = 0/

k∈K

and, consequently,  u∈U



u(

gi X) = 0. /

i∈I

The last equality establishes the G-absolute negligibility of X. Lemma 3 has thus been proved. Remark 2. All G-absolutely negligible sets may be regarded as small subsets of a given uncountable group (G, ·) because they form an ideal in the power set P(G) (see Exercise 14 from Chapter 2). However, the group operation applied to such subsets sometimes yields a non-small set. More precisely, it can be demonstrated that for every uncountable solvable group (G, ·), there exist two G-absolutely negligible sets A and B satisfying the equality A · B = G (in this connection, see Exercise 20 for Chapter 11). In what follows the symbol (G, ·) will denote an arbitrary group and (G, +) will denote an arbitrary commutative group. We do not suppose that G is equipped with a topology compatible with its algebraic structure or, equivalently, we may assume that G is endowed with its discrete topology. Let (G, ·) be an uncountable group. As previously shown, if μ is a nonzero σ -finite left G-quasi-invariant measure on G, then there exists a subset of G nonmeasurable with respect

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to μ (see [60], [104], and Exercise 11 for Appendix 1). This result is based on the classical theorem of Ulam stating that the first uncountable cardinal number ω1 is not real-valued measurable (see [238]). For an uncountable commutative group (G, +) equipped with a nonzero σ -finite G-quasiinvariant measure μ , a much stronger result can be established. Namely, it can be demonstrated the existence of a subgroup of (G, +) which is nonmeasurable with respect to μ . In general, the latter strong result does not hold for an uncountable non-commutative group (H, ·) endowed with a nonzero σ -finite left H-quasi-invariant measure. In this respect, commutative groups have some advantage because for any uncountable commutative group (G, +) equipped with a nonzero σ -finite G-quasi-invariant measure μ , there are uncountably many subgroups of (G, +) nonmeasurable with respect to μ (see [116] and [119]). Here we are interested in analogous statements closely connected with the existence of those nonmeasurable subgroups of (G, +) which can be used for obtaining G-invariant (G-quasi-invariant) extensions of initial nonzero σ -finite G-invariant (G-quasi-invariant) measures on G. In order to obtain such subgroups, we need some helpful facts from the general theory of commutative groups. The algebraic structure of an arbitrary commutative group is investigated more or less thoroughly. The following statement yields a description of this structure and will be crucial in further constructions. Kulikov’s Theorem. Any commutative group (G, +) admits a representation in the form G = ∪{Gn : n < ω }, where {Gn : n < ω } is an increasing (by inclusion) sequence of subgroups of G and, for each n < ω , the group Gn is a direct sum of cyclic groups (finite or infinite). For the proof of this important statement, see [70] and [152] (cf. also Appendix 5). Suppose now that our group (G, +) is uncountable. Then we may assume, without loss of generality, that all Gn in the above-mentioned representation are also uncountable. Let us express Gn in the form of a direct sum Gn =

∑ Gi ,

i∈In

where all Gi are cyclic groups (finite or infinite) and card(Gi )  2

(i ∈ I).

According to our assumption, we have card(In ) > ω . Further, let us represent every set In in the form In = ∪{In,k : k < ω },

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where {In,k : k < ω } is a partition of In and card(In,k ) > ω for each k < ω . Finally, let us put Gn,k =

∑

Gi

i∈In,0 ∪In,1 ∪...∪In,k

and consider the countable family {Gn,k : n < ω , k < ω } of subgroups of G. It is not difficult to verify the validity of the following two relations: (1) card(Gn /Gn,k ) > ω for any n < ω and k < ω ; (2) ∪{Gn,k : k < ω } = Gn for every n < ω . Relation (1) immediately implies that (∀n < ω )(∀k < ω )(card(G/Gn,k ) > ω ). Relation (2) implies at once that G = ∪{Gn,k : n < ω , k < ω }. Summarizing these two consequences, we obtain the following auxiliary assertion. Lemma 4. Any uncountable commutative group (G, +) admits a representation G = ∪{G j : j ∈ J}, where J is a countable set, G j is a subgroup of G for each j ∈ J, and the inequality card(G/G j ) > ω holds true. We have already mentioned Serpi´nski’s problem which deals with various translationinvariant extensions of the Lebesgue measure λn on the Euclidean space Rn . Lemma 4 enables us to give a solution of generalized Sierpi´nski’s problem concerning extensions of nonzero σ -finite translation-invariant (respectively, translation-quasi-invariant) measures on uncountable commutative groups by using the existence of certain nonmeasurable subgroups. For this purpose, we need the next simple auxiliary proposition. Lemma 5. Let (G, +) be an uncountable commutative group, μ be a σ -finite G-invariant (G-quasi-invariant) measure on G and let G0 be a subgroup of G such that card(G/G0 ) >

ω . Then there exists a G-invariant (G-quasi-invariant) measure μ  on G which extends μ and satisfies the equality μ  (G0 ) = 0. Proof. We apply a fairly standard argument based on Marczewski’s method (see Chapter 2). Namely, let us introduce the following family of sets in G: I = {X : X can be covered by countably many translates o f G0 }.

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Observe that I is a G-invariant σ -ideal of subsets of G. Moreover, if X ∈ I , then for some countable family {gi : i ∈ I} ⊂ G, we have X ⊂ ∪{gi + G0 : i ∈ I}. Therefore, X ∩ (h + X) = 0/ for each element h ∈ G not belonging to the set ∪{gi − g j + G0 : i ∈ I, j ∈ I}. This fact implies by easy transfinite induction that there exists an uncountable family {hξ :

ξ < ω1 } of elements of G such that the family {hξ + X : ξ < ω1 } consists of pairwise disjoint sets. We thus conclude that μ∗ (X) = 0 in view of the σ finiteness and G-quasi-invariance of μ . Now, we can utilize the method of extending μ described in Chapter 2. Namely, we suppose without loss of generality that μ is complete and introduce the σ -algebra S  = {Y X : Y ∈ dom(μ ), X ∈ I }, where  denotes the symmetric difference of sets. Further, we put

μ  (Y X) = μ (Y ) for any set Y X ∈ S  (here Y ∈ dom(μ ) and X ∈ I ). A simple verification shows that

μ  is well defined and is a complete G-invariant (G-quasi-invariant) measure extending μ . According to the definition, μ  (X) = 0 for all members X of the σ -ideal I . In particular, we have μ  (G0 ) = 0 which ends the proof of the lemma. Remark 3. In fact, the preceding argument establishes that G0 is a G-absolutely negligible subset of G. Observe that this argument essentially exploits the commutativity of a given group G. The reader can check it himself (herself). From the above lemmas we readily obtain the following result. Theorem 1. Let (G, +) be an uncountable commutative group. There exists a countable family {G j : j ∈ J} of subgroups of G such that: 1) G = ∪{G j : j ∈ J}; 2) for any nonzero σ -finite G-invariant (G-quasi-invariant) measure μ on G, at least one subgroup G j is nonmeasurable with respect to μ ; 3) if G j is nonmeasurable with respect to μ , then there exists a G-invariant (G-quasiinvariant) measure μ  on G extending μ and satisfying the relation μ  (G j ) = 0.

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Proof. Let {G j : j ∈ J} be as in Lemma 4 and let μ be an arbitrary nonzero σ -finite G-invariant (G-quasi-invariant) measure on G. By virtue of the equality G = ∪{G j : j ∈ J}, we may write 0 < μ (G) 

∑ μ ∗(G j ) j∈J

where

μ∗

denotes the outer measure associated with μ . Consequently, there exists an index

j ∈ J such that μ ∗ (G j ) > 0. In view of the relation card(G/G j ) > ω and of the G-quasi-invariance of μ , we easily infer that G j must be nonmeasurable with respect to μ . Finally, applying Lemma 5, we conclude that for the same G j , there exists a G-invariant (G-quasi-invariant) measure μ  on G extending μ and satisfying the equality

μ  (G j ) = 0 which completes the proof. Theorem 1 may be regarded as the first step in solving (within ZFC theory) the measure extension problem for nonzero σ -finite invariant (quasi-invariant) measures. By using a more complicated argument, this theorem can be generalized to the class of all uncountable solvable groups. For obtaining the desired result, we need one more notion. Let (G, ·) be an arbitrary group. We shall say that a family {G j : j ∈ J} of subgroups of G is admissible if: (i) card(J)  ω ; (ii) there exists a countable family {gk : k ∈ K} of elements of G such that G = ∪{gk (∪{G j : j ∈ J}) : k ∈ K}; (iii) every group G j ( j ∈ J) is G-absolutely negligible in G. It can easily be seen that if {G j : j ∈ J} is an admissible family for G and μ is an arbitrary nonzero σ -finite left G-invariant (left G-quasi-invariant) measure on G, then at least one group G j is nonmeasurable with respect to μ . Therefore, μ can be extended to a left Ginvariant (left G-quasi-invariant) measure μ  on G such that μ  (G j ) = 0 (cf. Lemma 5). Recall that a group (G, ·) is solvable if there exists a finite sequence (Γ0 , Γ1 , ..., Γn ) of subgroups of G such that:

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(a) Γ0 = G and card(Γn ) = 1; (b) for each integer k ∈ [0, n − 1], we have Γk+1 ⊂ Γk , the group Γk+1 is normal in Γk and the quotient group Γk /Γk+1 is commutative. Lemma 6. For every uncountable solvable group (G, ·), there exists an admissible family of its subgroups. Proof. Let (Γ0 , Γ1 , ..., Γn ) be a finite family of subgroups of G satisfying the abovementioned conditions (a) and (b). We use induction on n. Let

φ : Γ0 → Γ0 /Γ1 denote the canonical epimorphism. Only two cases are possible. 1. card(Γ0 /Γ1 )  ω . In this case, we obviously have card(Γ1 ) > ω . By inductive assumption, there exists an admissible family {G j : j ∈ J} of subgroups of Γ1 . Taking into account Lemma 3 and the inequality card(Γ0 /Γ1 )  ω , we derive that the same family {G j : j ∈ J} is admissible for Γ0 = G. 2. card(Γ0 /Γ1 ) > ω . In view of Theorem 1, there exists an admissible family {G j : j ∈ J} of subgroups of the uncountable commutative group Γ0 /Γ1 . Now, it is not difficult to verify that the family of groups {φ −1 (G j ) : j ∈ J} is admissible for the original group Γ0 = G. The details of this verification are left to the reader. The two preceding lemmas immediately imply an analogue of Theorem 1 for an arbitrary uncountable solvable group (G, ·), in terms of nonzero σ -finite left G-invariant (left Gquasi-invariant) measures on G. Namely, we have the next statement. Theorem 2. Let (G, ·) be an uncountable solvable group. There exists a family {G j : j ∈ J} of subgroups of G satisfying the following relations: 1) {G j : j ∈ J} is admissible; 2) for any nonzero σ -finite left G-invariant (left G-quasi-invariant) measure μ on G, at least one subgroup G j is nonmeasurable with respect to μ ; 3) if G j is nonmeasurable with respect to μ , then there exists a left G-invariant (left Gquasi-invariant) measure μ  on G extending μ and such that μ  (G j ) = 0. In connection with Theorem 2, we would like to formulate the following open problem. Problem. Give a characterization of all those uncountable groups (G, ·) for which there exists at least one admissible family of subgroups of G. Remark 4. Let (G, ·) be a connected σ -compact locally compact topological group and let D be a subgroup of G. Denote by ν the completion of the left Haar measure on G. Then,

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by using the Steinhaus property (see Chapters 3 and 4), it can be shown that

ν (D) = 0 ∨ ν∗ (G \ D) = 0. In the case of the equality ν∗ (G \ D) = 0, the group D is usually called thick with respect to

ν (or, briefly, ν -thick). Evidently, a ν -thick subgroup is everywhere dense in the original group G. Moreover, if a ν -thick subgroup D is proper (i.e., G = D), then D is necessarily nonmeasurable with respect to ν . Indeed, supposing for a while that a proper ν -thick subgroup D is ν -measurable (hence, is of full ν -measure) and taking any element g ∈ G\ D, we obviously get gD ∩ D = 0, / thus it follows that 0 = ν (G \ D)  ν (gD), 0 = ν (gD) = ν (D) > 0, and we come to a contradiction. Analogously, if D is an everywhere dense subgroup of an arbitrary σ -compact locally compact topological group (G, ·) equipped with the completion ν of the left Haar measure, then either ν (D) = 0 or D is ν -thick in G. Notice that there exists an uncountable commutative non-discrete locally compact topological group (Γ, +) which does not admit proper everywhere dense subgroups (see [46], [92], [103], [203], and Exercise 6 for Appendix 5). This fact also implies that for the same (Γ, +), there are no proper subgroups thick with respect to the completion μ of the Haar measure on Γ, i.e., there are no thick subgroups nonmeasurable with respect to μ . On the other hand, the existence of a μ -nonmeasurable subgroup of Γ follows directly from Theorem 2 and it must be underlined that no topological technique is needed for establishing this result. It should also be mentioned that every uncountable non-discrete locally compact group G is a resolvable topological space, i.e., G can be represented in the form G = A ∪ B, where A and B are some two disjoint everywhere dense subsets of G. More generally, the well-known theorem due to Hewitt states that every locally compact topological space E without isolated points is resolvable (see Exercise 7 for this chapter). In various natural situations the topological structure of a given topological group does not admit a nonzero σ -finite quasi-invariant measure. For instance, it can be proved that if G is a proper uncountable analytic (or co-analytic) subgroup of R, then there exists no nonzero σ -finite G-quasi-invariant Borel measure on G (see [115] and [117]).

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Similarly, if E is an arbitrary infinite-dimensional Banach space, then there does not exist a nonzero σ -finite E-quasi-invariant Borel measure on E (see, for instance, Chapter 4 of this book; cf. also [226] where the case of an infinite-dimensional Hilbert space E is considered in detail). Taking into account examples of this kind, it makes sense sometimes to ignore the topological concepts and pose the following general question. Let (G, ·) be an uncountable solvable group. Does there exist a nonzero σ -finite measure

μ defined on a sufficiently large σ -algebra of subsets of G and invariant under the group of all left translations of G? Theorem 2 shows that any such measure μ is defined on a proper subclass of the family of all subsets of G and that μ can be strictly extended by using some μ -nonmeasurable subgroup of G. However, the method of extending measures described in Theorems 1 and 2 does not essentially change the class of measurable sets. For instance, if the original measure μ on G is separable (see Exercise 11 for Chapter 1), then the extended measure

μ  produced by this method remains separable because all μ  -measurable sets are obtained with the aid of members of a certain G-invariant σ -ideal of subsets of G which does not change the metrical structure of μ . Therefore, it is reasonable to ask whether G admits a nonseparable σ -finite left G-invariant measure. It turns out that the answer to the latter question is positive under an additional set-theoretic assumption. In fact, the following statement can be proved within ZFC theory. (*) Let (G, ·) be a solvable group whose cardinality is greater than or equal to the cardinality of the continuum c. Then there exists a non-atomic nonseparable σ -finite left G-invariant measure on G. As an immediate consequence of (*), we have the next statement. (**) Let (G, ·) be an uncountable solvable group. Assuming the Continuum Hypothesis, there always exists a non-atomic nonseparable σ -finite left G-invariant measure on G. We do not know whether CH is necessary for the validity of (**). A detailed proof of the statements (*) and (**) is given in Chapter 17. Notice that an essential role in the proof is played by the following purely algebraic fact. If (H, +) is an uncountable commutative group, then there exists a homomorphic image (H1 , +) of (H, +) such that card(H1 ) = ω1 . Simple examples show that this fact fails to be true for uncountable non-commutative groups.

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Also, it is unknown whether any uncountable non-commutative group (G, ·) admits a non-atomic nonseparable σ -finite left G-invariant measure. This property is fulfilled if card(G)  2ω and, simultaneously, card(G) is not cofinal with ω . However, if these conditions do not hold for G, the question remains open. In this context, it should be pointed out that if (G, ·) is an uncountable σ -compact locally compact topological group, then the equality card(G) = (card(G))ω is always valid. Therefore, in this case the above-mentioned conditions are satisfied automatically. Finishing this chapter, let us recall that the first two (absolutely different) constructions of nonseparable translation-invariant extensions of the Lebesgue measure λ on R were done by Kakutani and Oxtoby [95], and Kodaira and Kakutani [141], respectively. Their constructions admit a straightforward generalization to a certain class of σ -compact locally compact groups (cf. [83]). In Chapter 16, we will consider one more method of obtaining nonseparable invariant extensions of nonzero σ -finite invariant measures, which essentially utilizes their nontrivial ergodic components. More information concerning extensions of measures invariant under various transformation groups can be found in [39], [41], [42], [45], [46], [69], [83], [85], [86], [95], [108], [115], [119], [141], [142], [195], [198], [234], [235], [242], [247], and [249]. EXERCISES 1. Prove the assertions a) and b) of Remark 1. 2. Verify that the group G of all isometric transformations of Rn , where n  2, is solvable. This algebraic fact implies that for n  2, there exists a functional

ν : P(Rn ) → [0, +∞[ which is finitely additive, G-invariant and extends the Lebesgue measure λn (see [22], [57], [83], [108], and [240]). On the other hand, such a ν does not exist for n  3 (see [155] and [240]). 3∗ . Recall that a topological space E is isodyne if for every nonempty open set U ⊂ E, the relation card(U) = card(E) is valid (equivalently, all nonempty open sets in E have one and the same cardinality).

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Suppose that the Generalized Continuum Hypothesis holds and let E be an isodyne Hausdorff topological space of second category. Prove that the relation (card(E))ω = card(E) is satisfied. For this purpose, use the fact that if X is an infinite subspace of E, then card(X)  22

card(Y )

,

where Y is any everywhere dense subset of X. Check also that this inequality is precise, i.e., for any infinite cardinal number a, there exist a compact Hausdorff space X and its everywhere dense subset Y such that a

card(Y ) = a, card(X) = 22 . On the other hand, take an arbitrary uncountable set F and equip F with the topology T = {X ⊂ F : card(F \ X)  ω }. Verify that (F, T ) is an isodyne T1 -space of second category. This circumstance shows that the assumption on E that it should be Hausdorff is essential for the validity of the equality (card(E))ω = card(E). 4. Let (G, ·) be a non-discrete σ -compact locally compact topological group. Infer from the result of the previous exercise that under the Generalized Continuum Hypothesis, the relation (card(G))ω = card(G) holds true. In connection with the last fact, it must be mentioned that for the same G, the equality card(G) = 2w(G) is satisfied, where the symbol w(G) denotes the topological weight of G (see, e.g., [46]). Actually, this result does not need any additional set-theoretical assumptions and trivially implies the equality (card(G))ω = card(G). 5∗ . Let E be a topological space satisfying the following conditions: (a) the smallest cardinality κ of nonempty open subsets of E is infinite; (b) there exists a base B of open sets in E such that card(B)  κ . Prove that E is a resolvable space. For this purpose, use the method of transfinite recursion and a Bernstein type construction (cf. [148] and [192]). 6. Let E be a topological space. Check that for every nonempty open set U ⊂ E, there exists a nonempty isodyne open set V ⊂ U. In other words, the family of all isodyne open subspaces of E forms a pseudo-base of open sets in E. Verify that if all members of some pseudo-base of open sets in E are resolvable, then E is also resolvable.

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For this purpose, consider a maximal (with respect to the inclusion relation) disjoint family of elements of this pseudo-base and show that the union of such a family is everywhere dense in E (cf. Exercise 12 for Appendix 3). 7∗ . Let E be an arbitrary locally compact topological space. Show that there exists a base of E whose cardinality does not exceed card(E), i.e., the inequality w(E)  card(E) holds true. By using the result of Exercise 5, infer from this fact that if a topological space Y is infinite, locally compact, and isodyne, then Y is resolvable. Finally, taking into account Exercise 6, conclude that if X is a locally compact topological space without isolated points, then X is resolvable (Hewitt’s theorem). In particular, any non-discrete locally compact topological group is resolvable. 8. Give an example of an uncountable group (G, ·) such that no homomorphic image of (G, ·) has cardinality ω1 . 9. Let (G, +) be an arbitrary uncountable commutative group. Show that there exists a countable family {G j : j ∈ J} of subgroups of G satisfying the following relations: (a) G = ∪{G j : j ∈ J}; (b) card(G/G j ) = card(G) for any j ∈ J. For this purpose, utilize an argument similar to that presented after the formulation of Kulikov’s theorem. 10∗ . Let (G, +) be an uncountable commutative σ -compact locally compact group and let

μ denote the completion of the Haar measure on G. Prove that there exists a countable family {H j : j ∈ J} of subgroups of G satisfying the following relations: (a) μ (G \ ∪{H j : j ∈ J}) = 0; (b) card(G/H j ) = card(G) for any j ∈ J; (c) each group H j ( j ∈ J) is μ -thick in G; (d) for every G-invariant extension μ  of μ , there exists an index j ∈ J such that H j is nonmeasurable with respect to μ  ; (e) if some H j is nonmeasurable with respect to μ  , then μ  can be extended to a G-invariant measure μ  such that μ  (H j ) = 0. In order to demonstrate the existence of {H j : j ∈ J} with properties (a)-(e), use the previous exercise and the remark made after Exercise 4.

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The result just formulated shows that in the case of an uncountable commutative σ -compact locally compact group (G, +) equipped with the completion μ of its Haar measure, the corresponding measure extension problem can be solved with the aid of a certain countable family of μ -thick subgroups of G.

Chapter 11

Algebraic sums of measure zero sets

This chapter and the next one are devoted to the following general problem concerning the behavior of small subsets of uncountable groups under the group-theoretical operation. Suppose that a group (G, ·) is given with an ideal I of subsets of G and, as a rule, this I is assumed to be left G-invariant. We will be interested in those situations, where there are two sets A ∈ I and B ∈ I such that A · B ∈ I . In other words, we will be dealing with those ideals of subsets of G, which are not preserved under the group-theoretical operation. Naturally, we will especially pay our attention to the case of a group (G, ·) equipped with a nonzero σ -finite complete G-invariant (G-quasi-invariant) measure. In this case, the role of I is played by the σ -ideal of all measure zero subsets of G. Analogously, if (G, ·) is a second category topological group, then the role of I is usually played by the σ -ideal of all first category subsets of G. This case is also of certain interest from a topological view-point. In addition, we especially want to examine the situations where the group-theoretical operation · is commutative. In such a case, it is more convenient to denote it by +. Let (G, +) be an arbitrary uncountable commutative group and let μ be a nonzero σ -finite complete G-invariant or, more generally, G-quasi-invariant measure on G. As usual, we denote by dom(μ ) the σ -algebra of all μ -measurable sets and by I (μ ) the σ -ideal of all

μ -measure zero sets in G. In particular, let G coincide with the real line R and let μ coincide with the standard Lebesgue measure on R, that is μ = λ . In connection with the said above, the well-known elementary fact should be mentioned, stating that there are two subsets A and B of R for which the relations

λ (A) = λ (B) = 0, A + B = R hold true. For more details, see Exercises 2, 3, and 4 of this chapter. Also, it must be noticed that by starting with the above-mentioned fact and applying the techique of Hamel A.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7_11, © 2009 Atlantis Press/World Scientific

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bases, Sierpi´nski [219] was able to establish the existence of two subsets A and B of R such that

λ (A ) = λ (B ) = 0, A + B ∈ dom(λ ). Analogous results are valid for the n-dimensional Euclidean space Rn (n  2) and for the Lebesgue measure λn on this space. Moreover, by using deep theorems on the structure of locally compact topological groups (see, for instance, [83], [177], [201], [202]) and the fundamental Mackey theorem [165], Sierpi´nski’s result can be extended to a wide class of uncountable topological groups equipped with σ -finite left invariant (left quasi-invariant) Borel measures (in this connection, cf. also [37], [44], [116], [118], [121], [131], [132], [137], and [236]). It is reasonable to ask whether similar statements hold true for more general situations when no topology is considered on a given group. Namely, it is natural to pose the following question. Let (G, +) be an uncountable commutative group equipped with a nonzero σ -finite complete G-invariant (G-quasi-invariant) measure μ . Do there exist two sets A ∈ I (μ ) and B ∈ I (μ ) whose algebraic sum A + B does not belong to dom(μ )? As already mentioned, the old theorem of Sierpi´nski [219] states that if G coincides with the real line R (or, more generally, with the Euclidean space Rn where n  1) and if μ coincides with the standard Lebesgue measure λ (λn ) on G, then the answer to this question is positive. Obviously, for an arbitrary uncountable commutative group (G, +) and for a nonzero σ finite complete G-invariant (G-quasi-invariant) measure μ on G, we do not have a direct analogue of this theorem because the σ -ideal I (μ ) can be very poor. For example, I (μ ) may coincide with the family of all countable subsets of G and, in this case, we readily get I (μ ) + I (μ ) ⊂ I (μ ) ⊂ dom(μ ). Various other and less trivial examples were constructed describing the situations, where these inclusions are valid. Therefore, the formulation of the question posed above should be replaced by another one. Here we suggest a slightly changed formulation in terms of extensions of a given measure, which seems to be sufficiently natural. Namely, the following problem is of interest from the measure-theoretical point of view. Problem. Let (G, ·) be an uncountable group and let μ be a nonzero σ -finite left Ginvariant (respectively, left G-quasi-invariant) measure on G. Does there exist a left Ginvariant (respectively, left G-quasi-invariant) complete measure μ  on G extending μ and

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such that for some sets A ∈ I (μ  ) and B ∈ I (μ  ), the relation A · B ∈ dom(μ  ) is satisfied?

Our aim in the present chapter is to establish that the answer to this question is positive for all uncountable commutative groups (G, +). Notice that for uncountable non-commutative groups, the situation is still unclear. It should also be pointed out that various aspects concerning the behavior of sets with a nice descriptive structure, under the operation of algebraic sum (i.e., Minkowski’s sum) in a vector space, have been discussed and investigated in many works (see, for instance, [62], [207], [227], [228], and Chapter 12). We begin our considerations with the case of an uncountable commutative group (G, +) and then we will briefly touch upon the case of uncountable non-commutative groups. Suppose first that (G, +) is representable in the form of a direct sum of cyclic groups, i.e., G = ∑{Gi : i ∈ I}, where each subgroup Gi (i ∈ I) is cyclic and is generated by a certain element gi ∈ Gi \ {0i} (consequently, card(Gi )  2). In this situation, we shall say that the set {gi : i ∈ I} is a basis for G. In some sense, {gi : i ∈ I} may be regarded as an analogue of a Hamel basis of any vector space over the field Q of all rational numbers. More precisely, every element g ∈ G admits a unique representation g = m1 gi1 + m2gi2 + ... + mk gik , where i1 , i2 , ..., ik are pairwise distinct elements of I and all summands on the right-hand side of this equality differ from zero (= the neutral element of G). Let us denote: X = {gi : i ∈ I}, p(g) = k

(g ∈ G),

X1 = Z · X = {mx : m ∈ Z, x ∈ X}, Xn+1 = Xn + X1

(n = 1, 2, ...).

In the above notation, the symbol Z as usual stands for the ring of all integers. Obviously, the following relations are valid: (a) p(x)  n for all elements x ∈ Xn ; (b) X1 ⊂ X2 ⊂ ... ⊂ Xn ⊂ ...;

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(c) G = ∪{Xn : 1  n < ω }. These simple relations will be used below. Let now (G, ·) be an arbitrary uncountable group. We remember from Chapter 2 that a set Y ⊂ G is G-absolutely negligible if for any σ -finite left G-invariant (respectively, left G-quasi-invariant) measure μ on G, there exists a left G-invariant (respectively, left Gquasi-invariant) measure μ  on G extending μ and satisfying the equality μ  (Y ) = 0. Various properties of absolutely negligible sets are considered in the monographs [108] and [115] (see also Chapters 2 and 10). Among those properties, let us especially emphasize a purely algebraic characterization of absolutely negligible sets which was mentioned several times earlier and which will be very useful in our further constructions. Lemma 1. Let (G, ·) be an arbitrary uncountable group and let Y be a subset of G. The following two assertions are equivalent: 1) Y is G-absolutely negligible in G; 2) for any countable family {gn : n < ω } of elements from G, there exists a countable family { fm : m < ω } of elements from G such that  m<ω

fm (



n<ω

(gnY )) = 0. /

Recall that a detailed proof of Lemma 1 was given in Chapter 2 (in fact, a more general proposition was proved in the same chapter). Lemma 2. Let (G, +) be a commutative group, H be a subgroup of G, and let Y be an H-absolutely negligible subset of H. Then Y is a G-absolutely negligible subset of G. Proof. Only two cases are possible. 1. card(G/H) > ω . In this case, by applying Lemma 1 it is not difficult to show that the subgroup H is a G-absolutely negligible set in G (see, for instance, Lemma 5 and Remark 3 from Chapter 10). Since Y ⊂ H, we easily infer that Y is G-absolutely negligible in G, too. Actually, here we do not need the assumption that Y is H-absolutely negligible in H because the inclusion Y ⊂ H directly yields the required result. 2. card(G/H)  ω . In this case, denote by {g j : j ∈ J} a selector of the countable family G/H and define Y  = ∪{g j + Y : j ∈ J}. Taking into account the fact that Y is H-absolutely negligible in H, it can be demonstrated that Y  is G-absolutely negligible in G (see Lemma 3 from Chapter 10). This circumstance

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immediately implies that each set g j +Y ( j ∈ J) is also G-absolutely negligible in G. Consequently, the set Y = (g j + Y ) − g j turns out to be G-absolutely negligible, too, which ends the proof of Lemma 2. We also need the following auxiliary proposition. Lemma 3. Let (G, +) be an uncountable commutative group representable in the form of a direct sum of cyclic groups and let X be a basis for G. Then X is a G-absolutely negligible subset of G. The proof of Lemma 3 can be carried out by the same method as in paper [106] where vector spaces over Q, instead of commutative groups, are under consideration. Actually, the argument presented in [106] works for uncountable commutative groups as well. Here we establish a more general statement containing Lemma 3 as a very particular case. Theorem 1. Let (G, +) be an uncountable commutative group representable in the form of a direct sum of cyclic groups and let X be a basis for G. Then all sets Xn (n = 1, 2, ...) defined earlier are G-absolutely negligible in G. Proof. Fix a natural number n  1 and consider the corresponding set Xn . Take any countable family { f j : j ∈ J} of elements from G. Since G is uncountable, its basis X is uncountable, too, so there exists an element f ∈ G such that p( fi − f j − f )  2n + 1 for all indices i ∈ J and j ∈ J. This circumstance implies that ( f + ∪{ f j + Xn : j ∈ J}) ∩ (∪{ f j + Xn : j ∈ J}) = 0. / Indeed, suppose to the contrary that ( f + ∪{ f j + Xn : j ∈ J}) ∩ (∪{ f j + Xn : j ∈ J}) = 0. / Then for some elements x ∈ Xn and x ∈ Xn and for some indices j ∈ J and i ∈ J, we must have x + f j + f = x + fi . Consequently, we get x − x = fi − f j − f , 2n  p(x − x ) = p( fi − f j − f )  2n + 1,

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which yields a contradiction. Now, in view of Lemma 1, we conclude that the set Xn is G-absolutely negligible in G. Remark 1. At present, it is known that every uncountable commutative group (G, +) can be covered by a countable family of G-absolutely negligible sets (see, e.g., Chapter 10). By virtue of Theorem 1, we are able to give another proof of this fact. Indeed, according to Kulikov’s theorem, we have a representation G = ∪{Gk : k < ω }, where all subgroups Gk (k < ω ) are direct sums of cyclic groups (see [70], [152], and Appendix 5). For any natural number k, denote by the symbol Xk a basis of the group Gk and for any natural number n > 0, denote by the symbol Xk,n the corresponding set in Gk . Then we may write the equality G \ {0} = ∪{Xk,n : k < ω , 0 < n < ω } from which, taking into account Lemma 2 and Theorem 1, the required result immediately follows, i.e., we come to a covering of G with countably many G-absolutely negligible subsets. For all uncountable non-commutative groups an analogous result is still unknown. In other words, we do not know whether any uncountable group (G, ·) admits a countable covering with G-absolutely negligible sets. This open question seems to be of interest from the view-point of the algebraic aspect of the general measure extension problem (cf. Chapter 2). The next auxiliary proposition plays the key role for obtaining the main result of this chapter. Lemma 4. Let (G, +) be a commutative group, μ be a σ -finite complete G-quasi-invariant measure on G and let H be a subgroup of G such that μ ∗ (H) > 0. Then H cannot be represented in the form H = A ∪Y, where μ (A) = 0 and Y is an H-absolutely negligible subset of H. Proof. Suppose to the contrary that H admits a representation in the above-mentioned form. Clearly, we may assume without loss of generality that A ∩Y = 0. /

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Further, since μ ∗ (H) > 0, we may consider the nonzero σ -finite complete measure ν on H which is produced by the original measure μ , i.e., ν is defined on the σ -algebra of sets S = {Z ∩ H : Z ∈ dom(μ )} and for any set Z ∩ H ∈ S , we have by definition

ν (Z ∩ H) = μ ∗ (Z ∩ H) = μ (Z ∩ H  ), where H  denotes a μ -measurable hull of H. It can easily be verified that ν is an H-quasiinvariant measure on H. In addition, we get A ∈ dom(ν ), ν (A) = 0. Therefore, the set Y = H \ A is ν -measurable and the relation

ν (Y ) = ν (H \ A) = ν (H) > 0 must be satisfied which contradicts the assumption that Y is an H-absolutely negligible subset of H. The contradiction obtained finishes the proof. Remark 2. It is essential in the formulation of the previous lemma that H is a subgroup of G. In this context, let us point out that if (G, +) is an arbitrary uncountable commutative group and μ is a nonzero σ -finite G-quasi-invariant measure on G, then there always exists a G-absolutely negligible subset Y of G with μ ∗ (Y ) > 0 (cf. Remark 1 above) and, in particular, Y turns out to be nonmeasurable with respect to μ . Moreover, if G coincides with the real line R and μ coincides with the standard Lebesgue measure λ on R, then there are even μ -thick G-absolutely negligible sets in G (see, for instance, Exercise 6). The existence of sets of this type can be established by the method of transfinite recursion using Lemma 1 and an argument similar to the classical Bernstein construction (cf. [107]). Now, we are ready to prove the following statement. Theorem 2. Let (G, +) be an uncountable commutative group and let μ be a nonzero

σ -finite G-invariant (respectively, G-quasi-invariant) measure on G. There exists a Ginvariant (respectively, G-quasi-invariant) complete measure μ  on G such that: 1) μ  extends μ ; 2) for some two sets A ∈ I (μ  ) and B ∈ I (μ  ), the relation A + B ∈ dom(μ  ) is satisfied. Proof. According to Kulikov’s theorem (see [70], [152], or Appendix 5), the given group G can be represented in the form G = ∪{Hk : k < ω },

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where {Hk : k < ω } is an increasing (by inclusion) sequence of subgroups of G and each Hk (k < ω ) is a direct sum of cyclic groups. Since μ is not identically equal to zero, there exists a natural number k such that μ ∗ (Hk ) > 0. Obviously, we may assume without loss of generality that Hk is uncountable. Let us put H = Hk and let X be a basis for H. By virtue of Theorem 1, the set X1 is H-absolutely negligible in H. In view of Lemma 2, the same X1 is a G-absolutely negligible subset of the original group G. Denote by μ  the minimal with respect to the inclusion relation, complete, G-invariant (G-quasi-invariant) extension of μ such that μ  (X1 ) = 0. By applying Lemma 4, it is not difficult to show that H remains to be a set of strictly positive outer measure with respect to μ  , i.e., H does not belong to the

σ -ideal I (μ  ). Further, since the equality H = ∪{Xn : 0 < n < ω } holds, there exists a smallest natural number n for which the set Xn is also of strictly positive outer μ  -measure. Notice that n > 1 because of the relation μ  (X1 ) = 0. Consequently, we can write Xn = Xn−1 + X1,

μ  (Xn−1 ) = μ  (X1 ) = 0. Now, using Theorem 1 once more, we see that Xn is H-absolutely negligible in H and, simultaneously, is not of μ  -measure zero. This fact immediately implies that Xn is not measurable with respect to μ  . Therefore, putting A = X1 , B = Xn−1 , we finally obtain A ∈ I (μ  ), B ∈ I (μ  ), A + B = Xn ∈ dom(μ  ), which yields the required result. Remark 3. In paper [118] several related questions are considered for an arbitrary uncountable vector space E over the field Q and for an arbitrary nonzero σ -finite E-invariant (E-quasi-invariant) measure μ given on E. In this case, some stronger results for μ are obtained by starting with the well-known fact that any vector subspace of E is a direct summand in E. The same property is true for any divisible subgroup of a commutative group (G, +). Remark 4. It would be interesting to generalize the main statement, that is Theorem 2 of the present chapter, to a maximally wide class of uncountable groups (G, ·) (not necessarily commutative). For some partial results in this direction, see [131], [132], and [139].

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Let (G, +) be again an uncountable commutative group and let μ be a nonzero σ -finite complete G-invariant or, more generally, G-quasi-invariant measure defined on some σ algebra of subsets of G. In general, we cannot assert that there are two subsets A and B of G satisfying the relations

μ (A) = μ (B) = 0,

A + B = G.

For instance, the σ -ideal I (μ ) can coincide with the family of all countable subsets of G and, consequently, I (μ ) can be closed under the operation of algebraic sum. In such a case, it follows that A + B = G for any sets A ∈ I (μ ) and B ∈ I (μ ). However, the question naturally arises whether there exists a G-invariant (G-quasiinvariant) measure μ  on G which extends μ and for which there are two sets A and B in G such that

μ  (A) = μ  (B) = 0,

A + B = G.

We are going to show that the answer to this question is always positive and, in fact, the required sets A and B can be chosen not depending on an initial measure μ . For this purpose, we need several auxiliary propositions. Note that G-absolutely negligible subsets of G play again a key role here. Remark 5. Let (G, ·) be an infinite group and let Z be a subset of G. We recall that Z is almost left G-invariant in G if for any g ∈ G, the relation card((gZ)ΔZ) < card(G) is valid. Suppose now that the cardinality of G is equal to ω1 , and let X be a subset of G. By using Lemma 1, it can be demonstrated that the following two assertions are equivalent: (a) X is not G-absolutely negligible in G; (b) there exists a countable family {gi : i ∈ I} ⊂ G such that the set Y = ∪{gi X : i ∈ I} contains an uncountable almost left G-invariant subset. Let (G, ·) be an uncountable group and let H be a subgroup of G. As usual, we denote by G/H the family of all left H-orbits in G. By using Lemma 1, we can prove the next auxiliary statement (cf. [121]). Lemma 5. Let (G, ·) be an uncountable group and suppose that a set Z ⊂ G has the following property:

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for any countable subgroup H of G, the relation card({T ∈ G/H : card(T ∩ Z)  2}) < card(G) is satisfied. Then Z is a G-absolutely negligible subset of G. Proof. Take any countable family { fi : i ∈ I} ⊂ G and denote by F the subgroup of G generated by this family. Since card(F)  ω , card(G) > ω , we can choose an element h ∈ G \ F. Further, denote by H the subgroup of G generated by {h} ∪ { fi : i ∈ I}. Obviously, card(H)  ω . According to our assumption, we have card({T ∈ G/H : card(T ∩ Z) > 1}) < card(G). Let us put P = ∪{T ∈ G/H : card(T ∩ Z)  1}, Z  = Z ∩ P. Then we can write card(Z \ Z  ) < card(G) and in view of Lemma 1, it is sufficient to demonstrate that 

g(



f Z  ) = 0. /

f ∈F

g∈H

Suppose to the contrary that there exists an element 

z∈

g∈H

Taking into account the definition of

Z,

g(



f Z  ).

f ∈F

we infer that there exists a unique element z ∈ Z 

for which the inclusion H · z ⊂ F · z is valid. Consequently, we have z ∈ F · z , F · z = F · z , H · z ⊂ F · z. The last inclusion implies at once that h·z = f ·z for some element f ∈ F. Therefore, we get h = f and h ∈ F, which contradicts the choice of h. The obtained contradiction ends the proof of Lemma 5. Lemma 6. Let (H, ·) be a group of cardinality ω1 . There exist two H-absolutely negligible sets A and B in H such that A · B = H.

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The proof of Lemma 3 is sketched in Exercise 9 for this chapter. Notice that a more general result is also valid. Namely, if card(H) is an uncountable regular cardinal, then there exist H-absolutely negligible sets A and B in H such that A · B = H (see Exercise 10). If (H, +) is an uncountable commutative group, then the assertion of Lemma 6 holds true without any assumption on card(H) (see Theorem 3 below). Lemma 7. Let (G1 , ·) and (G2 , ·) be two groups, φ : G1 → G2 be a surjective homomorphism and let Y be a G2 -absolutely negligible subset of G2 . Then the set X = ϕ −1 (Y ) is G1 -absolutely negligible in G1 . This auxiliary statement was proved in Chapter 10 (see Lemma 2 therein). Lemma 8. Let (G, +) be an arbitrary uncountable commutative group. There exists a commutative group (H, +) such that: 1) card(H) = ω1 ; 2) H is a homomorphic image of G. Proof. Since G is uncountable, it contains a subgroup G1 of cardinality ω1 . According to a well-known theorem from the theory of groups (see, e.g., [70], [152], and Appendix 5), there exists a divisible commutative group G2 containing G1 as a subgroup. We may assume, without loss of generality, that card(G2 ) = ω1 . Denote by ϕ0 the identity embedding of G1 into G2 . Since G2 is divisible, ϕ0 can be extended to a homomorphism

ϕ : G → G2 . Finally, putting H = ϕ (G), we see that the group H satisfies both relations 1) and 2) of the lemma. A natural generalization of Lemma 5 is given in Exercise 18 for this chapter. Now, we are ready to prove the following statement. Theorem 3. For any uncountable commutative group (G, +), there exist two G-absolutely negligible sets A and B in G such that A + B = G. Proof. Applying Lemma 8, we can find a commutative group (H, +) of cardinality ω1 which is a homomorphic image of G. Fix an epimorphism

ϕ : G → H.

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In view of Lemma 6, there are two H-absolutely negligible sets A and B in H for which A + B = H. Let us put A = ϕ −1 (A ), B = ϕ −1 (B ). According to Lemma 7, the sets A and B are G-absolutely negligible in G. Moreover, we can write A + B = ϕ −1 (A ) + ϕ −1 (B ) = ϕ −1 (A + B ) = ϕ −1 (H) = G, which ends the proof of Theorem 3. Remark 6. It immediately follows from Theorem 3 that if (G, +) is an arbitrary uncountable commutative group, then there exists a G-absolutely negligible subset C of G such that C + C = G. Indeed, take A and B as in Theorem 3 and put C = A ∪ B. Then the set C is G-absolutely negligible in G and we have C + C = (A ∪ B) + (A ∪ B) ⊃ A + B = G, which implies the required equality C + C = G. Let Z be a subset of a given uncountable group (G, ·). We recall (see Chapter 5) that Z is Gabsolutely nonmeasurable in G if for any nonzero σ -finite left G-quasi-invariant measure

μ on G, we have Z ∈ dom(μ ). At present, it is known that every uncountable commutative and, more generally, every uncountable solvable group (G, ·) contains a G-absolutely nonmeasurable subset (see [107] or [119]). In this connection, the following problem is of certain interest. Problem. Let (G, +) be an uncountable commutative group. Do there exist two Gabsolutely negligible sets A and B in G such that their algebraic sum A + B is a G-absolutely nonmeasurable subset of G? The problem just formulated remains open. The concrete particular case, where G = R, will be discussed later in the present chapter. In this context, we would like to recall once more a weaker result stating that for every uncountable commutative group (G, +) and for any nonzero σ -finite G-invariant (respectively, G-quasi-invariant) measure μ on G, there exists a G-invariant (respectively, G-quasi-invariant) measure μ  on G extending μ and such that

μ  (X) = μ  (Y ) = 0,

X + Y ∈ dom(μ  )

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for some G-absolutely negligible subsets X and Y of G (see Theorem 2 of this chapter). Remark 7. Suppose that for a given uncountable group (G, ·), there exist two G-absolutely negligible sets X ⊂ G and Y ⊂ G and a G-absolutely nonmeasurable set Z ⊂ G satisfying the relation X ·Y = Z. Then we can assert that there are also two G-absolutely negligible sets A and B in G for which A · B = G. Indeed, since Z is G-absolutely nonmeasurable, we have the equality ∪{gi Z : i ∈ I} = G for some countable family {gi : i ∈ I} ⊂ G (see Exercise 12). Putting A = ∪{gi X : i ∈ I}, B = Y, we get the required G-absolutely negligible sets A and B in G. Theorem 4. Let (G, +) be an uncountable commutative group and let μ be a σ -finite G-invariant (respectively, G-quasi-invariant) measure on G. There exists a G-invariant (respectively, G-quasi-invariant) extension μ  of μ such that

μ  (A) = μ  (B) = 0,

A+B = G

for some G-absolutely negligible subsets A and B of G which do not depend on an initial measure μ . Proof. The desired result easily follows from Theorem 3. Indeed, let A and B be as in Theorem 3. First, we extend our measure μ to a G-invariant (respectively, G-quasi-invariant) measure μ1 on G for which A ∈ dom(μ1 ), μ1 (A) = 0. Then we extend the measure μ1 to a G-invariant (respectively, G-quasi-invariant) measure

μ2 on G for which B ∈ dom(μ2 ), μ2 (B) = 0. Finally, we put μ  = μ2 and conclude that μ  is the required measure on G. It is natural to ask whether a result similar to Theorem 3 holds true for uncountable noncommutative groups. The following statement shows that a direct analogue of Theorem

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3 is valid for a wide class of such groups. However, to prove this analogue, we need an essentially different argument. Theorem 5. Let (G, ·) be an uncountable group such that (card(G))ω = card(G). Then there exist two G-absolutely negligible sets A and B in G for which the equality A·B = G holds true. Proof. Denote by α the least ordinal number whose cardinality is equal to card(G), and let {gξ : ξ < α } be an enumeration of all elements of G. Taking into account the equality (card(G))ω = card(G), we may introduce an α -sequence {Hξ : ξ < α } whose members are all countable subgroups of G. Now, we define recursively two α -sequences {aξ : ξ < α }, {bξ : ξ < α } of elements from G in the following manner. Suppose that for an ordinal ξ < α , the partial families {aζ : ζ < ξ } and {bζ : ζ < ξ } have already been defined. Then we denote: Gξ = the group generated by ∪{Hζ : ζ < ξ }; Cξ = {aζ : ζ < ξ } ∪ {bζ : ζ < ξ }. Obviously, we have the inequalities card(Gξ ·Cξ )  card(ξ ) + ω < card(G). Consequently, there exist two elements a and b from G such that (Gξ a) ∩ (Gξ ·Cξ ) = 0, /

(Gξ b) ∩ (Gξ ·Cξ ) = 0, /

ab = gξ .

We put aξ = a and bξ = b. Proceeding in this way, we are able to construct the α -sequences {aξ : ξ < α } and {bξ : ξ < α }. It immediately follows from our construction that the sets A = {aξ : ξ < α },

B = {bξ : ξ < α }

satisfy the relations card({T ∈ G/H : card(A ∩ T )  2}) < card(G), card({T ∈ G/H : card(B ∩ T )  2}) < card(G) for any countable subgroup H of G. In view of Lemma 5, both sets A and B turn out to be G-absolutely negligible in G. Finally, by virtue of the relation gξ = aξ · bξ ∈ A · B (ξ < α ),

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we obtain that G = A · B which completes the proof of Theorem 5. As soon as Theorem 5 is established, we may formulate a direct analogue of Theorem 4 for those uncountable groups (G, ·) which satisfy the equality (card(G))ω = card(G). We leave the formulation and proof of this analogue to the reader. Let us return to almost invariant subsets of groups (see Chapters 2, 3, and Remark 5). By applying some algebraic properties of these subsets, the next statement can be deduced. Theorem 6. Let (G, ·) be an arbitrary uncountable group whose cardinality is not cofinal with ω and let μ be an arbitrary σ -finite left G-invariant (respectively, left G-quasiinvariant) measure on G. Then there exist an almost left G-invariant set A in G and a left G-invariant (respectively, left G-quasi-invariant) measure μ  on G which extends μ and for which we have

μ  (A) = μ  (A−1 ) = 0, A · A = A · A−1 = G. Proof. By using an argument similar to that of the classical Sierpi´nski construction (cf. [176], [192], [222], or Exercise 19 for Chapter 3), it is not difficult to define a family {Ai : i ∈ I} of subsets of G satisfying the following relations: (a) card(I) = card(G) and card(Ai ) = card(G) for all i ∈ I; (b) for each i ∈ I, the set Ai is almost left G-invariant and is almost invariant with respect to the symmetry of G; (c) Ai ∩ A j = 0/ for any two distinct indices i ∈ I and j ∈ I. Since G is an uncountable group and μ is a σ -finite measure, we can find an index i ∈ I for which μ∗ (Ai ) = 0, where μ∗ as usual denotes the inner measure associated with μ . Let us put A = Ai . Applying the standard method of extending invariant measures (see [234], [235], and Chapter 2), we can extend μ to a left G-invariant (respectively, left G-quasiinvariant) measure μ  on G for which the equalities

μ  (A) = μ  (A−1 ) = 0 are valid. Now, take any g ∈ G. Since A is almost left G-invariant, we have card((gA)A) < card(G). Keeping in mind that card(A) = card(G), we easily infer that gA ∩ A = 0, / which yields the relation g ∈ A · A−1 and, therefore, G = A · A−1 = A · A.

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The last formula ends the proof of Theorem 6. Now, we wish to turn our attention to those algebraic sums of two absolutely negligible sets which are extremely bad from the measure-theoretical point of view. Namely, we are interested in the situation where A and B are fixed small subsets of (G, +) whose algebraic sum A + B is a set with utterly bad properties from the view-point of the theory of invariant (quasi-invariant) measures. The precise notion of smallness which plays a significant role in our further considerations is the same concept of absolute negligibility discussed in Chapter 2. The precise notion of ultimate badness is the concept of absolute nonmeasurability which was introduced earlier in Chapter 5. Let us mention that the structure of absolutely nonmeasurable sets can be rather simple in some infinite-dimensional vector spaces considered as commutative groups. Namely, the following proposition is valid. Lemma 9. Let E be an infinite-dimensional separable Hilbert space (over R) and let K be an arbitrary open ball in E. Then K is an E-absolutely nonmeasurable subset of E. The proof of Lemma 9 can easily be deduced from the result of Exercise 10 of Chapter 9. The assertion of this lemma readily implies the well-known fact that E does not admit a nonzero σ -finite Borel measure quasi-invariant under the group of all translations of E (see, e.g., [226] and Chapter 4). Lemma 10. Suppose that (G1 , +) and (G2 , +) are two isomorphic commutative groups. Then the following relations are equivalent: 1) there exist G1 -absolutely negligible subsets X and Y of G1 whose algebraic sum X + Y is G1 -absolutely nonmeasurable in G1 ; 2) there exist G2 -absolutely negligible subsets A and B of G2 whose algebraic sum A + B is G2 -absolutely nonmeasurable in G2 . We omit the trivial proof of Lemma 10. Now, we are able to establish the following statement. Theorem 7. There exist two R-absolutely negligible subsets of R such that their algebraic sum is an R-absolutely nonmeasurable set in R. Proof. Fix an infinite-dimensional separable Hilbert space (E, || · ||) and denote K = {e ∈ E : ||e|| < 2}. By virtue of Lemma 9, the open ball K is an E-absolutely nonmeasurable subset of E. Taking into account Lemma 10 and the fact that E and R are isomorphic as commutative

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groups, it is sufficient to show that there exist two E-absolutely negligible sets X and Y in E for which the equality X +Y = K holds true. We are going to define the required sets X and Y by using the method of transfinite induction. Let α be the least ordinal number of cardinality continuum, {kξ : ξ < α } be an enumeration of all elements from K and let {Hξ : ξ < α } be an enumeration of all countable subgroups of the additive group E. For any ξ < α , denote by Gξ the subgroup of E generated by the set ∪{Hζ : ζ < ξ }. Now, construct by transfinite recursion two α -sequences {xξ : ξ < α }, {yξ : ξ < α } of elements from E satisfying the following conditions: (1) ||xξ || < 1 and ||yξ || < 1 for each ξ < α ; (2) xξ + yξ = kξ for each ξ < α ; (3) (Gξ + xξ ) ∩ (Gξ + {xζ : ζ < ξ }) = 0/ for any ξ < α ; (4) (Gξ + yξ ) ∩ (Gξ + {yζ : ζ < ξ }) = 0/ for any ξ < α . Suppose that for an ordinal ξ < α , the partial ξ -sequences {xζ : ζ < ξ } and {yζ : ζ < ξ } have already been constructed. Let us put Zξ = {xζ : ζ < ξ } ∪ {yζ : ζ < ξ }, Kξ = {e ∈ E : ||e − kξ || < 1}, D = {e ∈ E : ||e|| < 1}. Notice that Kξ = D + kξ , card(Gξ + Zξ )  card(ξ ) + ω < card(E), card(Kξ ∩ D) = card(E). Consequently, there are two points x ∈ D and y ∈ D such that (Gξ + x) ∩ (Gξ + Zξ ) = 0, / (Gξ + y) ∩ (Gξ + Zξ ) = 0, / x + y = kξ .

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Let us define xξ = x and yξ = y. Proceeding in this manner, we are able to construct the required α -sequences {xξ : ξ < α } and {yξ : ξ < α } with properties (1) - (4). Now, putting X = {xξ : ξ < α }, Y = {yξ : ξ < α }, we easily deduce that X + Y = K in view of (1) and (2). We also conclude that both X and Y are E-absolutely negligible subsets of E in view of (3), (4), and Lemma 5. Theorem 7 has thus been proved. Actually, the preceding argument yields a much stronger result. Namely, we can assert that there exists an E-absolutely negligible set C ⊂ E such that C + C = K. Indeed, it suffices to put C = X ∪Y, where X and Y are the above-mentioned E-absolutely negligible subsets of E. Therefore, taking into account an algebraic isomorphism between the additive groups E and R, we claim that there exists an R-absolutely negligible set Z ⊂ R such that the algebraic sum Z + Z is R-absolutely nonmeasurable. Theorem 8. There are two subsets A and B of R having the following property: for every nonzero σ -finite R-invariant (R-quasi-invariant) measure μ on R, there exists an R-invariant (R-quasi-invariant) measure μ  on R extending μ and such that

μ  (A) = μ  (B) = 0, A + B ∈ dom(μ  ).

Proof. Indeed, it suffices to take as A and B any two R-absolutely negligible subsets of R whose algebraic sum A + B is R-absolutely nonmeasurable in R. The existence of such subsets is stated by Theorem 7. Let (G1 , +) and (G2 , +) be commutative groups and let φ : G1 → G2 be a surjective homomorphism. We already know that: (a) if a set Y ⊂ G2 is G2 -absolutely negligible, then the set X = φ −1 (Y ) is G1 -absolutely negligible;

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(b) if a set Y ⊂ G2 is G2 -absolutely nonmeasurable, then the set X = φ −1 (Y ) is G1 absolutely nonmeasurable. From the relations (a), (b), and Theorem 7 we can easily derive (under CH) that in every uncountable vector space E over the field Q of all rational numbers there exist two Eabsolutely negligible sets whose algebraic sum is E-absolutely nonmeasurable (cf. Exercise 17 below). It would be interesting to extend this result to the case of all uncountable commutative groups. EXERCISES 1. Show that there exists a totally disconnected compact subset Z of the Euclidean plane R2 = R × R such that pr1 (Z) is a non-degenerate segment of the line R × {0}. Notice that the first example of a subset of R2 of this type was constructed by Baire. 2. Let C denote the Cantor set on the real line R. Show that the equalities C + C = [0, 2], C − C = C + (−C) = [−1, 1] hold true. Conclude from these facts that the ideal of all Lebesgue measure zero subsets of R and the ideal of all nowhere dense subsets of R are not closed under the operation of algebraic sum (vector sum). 3. Deduce from the previous exercise that there exists a set X ⊂ R such that: (a) X is of Lebesgue measure zero; (b) X is of first category; (c) X + X = X − X = R. 4∗ . By using Exercise 3, show that there exists a subset D of the Euclidean space Rn , where n  1, satisfying the following conditions: (a) D is of λn -measure zero; (b) D is of first category; (c) D + D = D − D = Rn . Starting with this result and applying the method of transfinite recursion, construct a Hamel basis in Rn which is of first category and of λn -measure zero. 5. Let (G, +) be a commutative group and let H be an arbitrary subgroup of G. Verify that there exists a nonzero σ -finite G-invariant measure μ on G such that H ∈ dom(μ ).

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Conclude from this fact that H is not absolutely nonmeasurable with respect to the class of all nonzero σ -finite G-invariant measures on G. 6. By using the method of transfinite recursion, construct an R-absolutely negligible subset X of R which is thick with respect to the Lebesgue measure λ on R, i.e., λ∗ (R \ X) = 0. 7. Let (G, ·) be a group and let (H, ·) be an arbitrary homomorphic image of (G, ·). Suppose that H admits a representation H = A · B where A and B are some H-absolutely negligible subsets of H. Show that G admits an analogous representation G = A · B where A and B are some G-absolutely negligible subsets of G. 8. Prove the equivalence of assertions (a) and (b) of Remark 5. 9∗ . Give a detailed proof of Lemma 6. For this purpose, take any group (H, ·) of cardinality ω1 and consider an ω1 -sequence {Hξ : ξ < ω1 } of subgroups of H such that: (a) card(Hξ )  ω for any ordinal ξ < ω1 ; (b) ξ  ζ < ω1 ⇒ Hξ ⊂ Hζ ; (c) Hξ \ ∪{Hζ : ζ < ξ } = 0/ for each ordinal ξ < ω1 ; (d) H = ∪{Hξ : ξ < ω1 }. Verify that any partial selector of the family {(Hξ \ ∪{Hζ : ζ < ξ }) : ξ < ω1 } is an H-absolutely negligible subset of H. Let {hξ : ξ < ω1 } be an enumeration of all elements of H. By applying the method of transfinite recursion, construct two families {aξ : ξ < ω1 }, {bξ : ξ < ω1 } of elements of H satisfying the following conditions: (e) {aξ : ξ < ω1 } and {bξ : ξ < ω1 } are partial selectors of the above-mentioned family {(Hξ \ ∪{Hζ : ζ < ξ }) : ξ < ω1 }; (f) aξ · bξ = hξ for any ordinal ξ < ω1 . Then, putting A = {aξ : ξ < ω1 } and B = {bξ : ξ < ω1 }, demonstrate that A and B are the required H-absolutely negligible sets in H. 10. Let (H, ·) be a group such that card(H) is an uncountable regular cardinal. Show that there exist two H-absolutely negligible sets A and B in H such that A · B = H. For this purpose, use an argument similar to the argument of Exercise 9. Try to establish an analogous result for all those uncountable groups (H, ·) which satisfy the equality (card(H))ω = card(H).

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11. Formulate and prove a direct analogue of Theorem 4 for all those uncountable groups (G, ·) which satisfy the equality (card(G))ω = card(G). 12. Let E be a set, G be a group of transformations of E, and let X be a G-absolutely nonmeasurable subset of E. Show that there exists a countable family {gi : i ∈ I} ⊂ G such that E = ∪{gi (X) : i ∈ I}. For this purpose, suppose otherwise and consider the G-invariant σ -ideal I of subsets of E generated by {X}. Since this I is proper, there exists a probability G-invariant measure

μ on E such that (∀Z ∈ I )(μ (Z) = 0). In particular, the relation μ (X) = 0 holds true which contradicts the G-absolute nonmeasurability of X. The contradiction obtained yields the desired result. 13. Let us preserve the notation of the proof of Theorem 6. Give a detailed construction of the family {Ai : i ∈ I} of subsets of G satisfying the relations (a), (b), and (c). 14. Let (E, || · ||) be a Banach space over R. Prove that: (a) no closed half-space in E is E-absolutely nonmeasurable; (b) if E is infinite-dimensional and separable, then for every nonzero σ -finite E-quasiinvariant measure μ on E, there exists a closed half-space in E which is nonmeasurable with respect to μ . 15. By using the technique of Hamel bases, show that R and any vector space E over R with card(E) = c are isomorphic as commutative groups. In particular, an infinite-dimensional separable Hilbert space H is isomorphic to R in the group-theoretical sense. However, any isomorphism f between H and R is very bad from the point of view of descriptive set theory and general topology. For instance, check that f does not possess the Baire property. 16. Prove that there is an R-absolutely negligible set C having the following property: for every σ -finite R-invariant (R-quasi-invariant) measure μ on R, there exists an Rinvariant (R-quasi-invariant) extension μ  of μ such that

μ  (C) = 0, C + C ∈ dom(μ  ).

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Moreover, generalize this result to the case of any vector space E over Q with card(E)  c.

17. Assuming the Continuum Hypothesis prove that for every uncountable vector space E over Q, there exists an E-absolutely negligible set Z such that Z + Z is E-absolutely nonmeasurable. 18∗ . Let (G, +) be an arbitrary infinite commutative group. Show that for any infinite cardinal a  card(G), there exists a homomorphic image (H, +) of (G, +) such that card(H) = a. Prove this fact by applying two essentially different methods. Namely, the first one is similar to the argument used in the proof of Lemma 8 of this chapter and the second one is based on the structural representation of any commutative group, i.e., exploits Kulikov’s theorem. 19. Suppose that every commutative group (H, +) with card(H) = ω1 contains an Habsolutely negligible set X such that X + X is H-absolutely nonmeasurable. In this case, show that any uncountable commutative group (G, +) contains a G-absolutely negligible set Y such that Y + Y is G-absolutely nonmeasurable. 20∗ . Let (G, ·) be an uncountable solvable group. Show that there exist two G-absolutely negligible sets A and B in G satisfying the equality A · B = G. For this purpose, consider a composition series {e} = Gn ⊂ Gn−1 ⊂ ... ⊂ G1 ⊂ G0 = G for G and use induction on n. Observe that if n = 1, then G is commutative, so Theorem 3 of this chapter is applicable to G.

Chapter 12

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As mentioned in the previous chapter, there are examples which show that when the operation of algebraic sum is applied to subsets of the real line R or, more generally, to subsets of a given topological vector space E, then the descriptive structure of the obtained sets sometimes essentially differs from the structure of summands. We shall see below further vivid realizations of this phenomenon. Actually, the main goal of this chapter is to demonstrate that the algebraic sum (frequently called Minkowski’s sum) of two very small subsets of a topological vector space E can be absolutely nonmeasurable with respect to the class of the completions of all nonzero σ -finite continuous Borel measures on E. In preceding considerations we have been concerned with some interesting statements on certain algebraic and topological properties of the classical Cantor set on the real line R (see Exercises 2 and 3 for Chapter 11). In this connection, we would like to recall once more a simple but beautiful fact, which quite often is cited in various text-books of mathematical analysis and elementary topology (see, e.g., [72] and [176]). Namely, if C denotes the Cantor set on R, then the algebraic sum C + C = {x + y : x ∈ C, y ∈ C} coincides with the closed interval [0, 2] and the difference set C − C = {x − y : x ∈ C, y ∈ C} coincides with the closed interval [−1, 1]. The proof of both these assertions is not difficult and may be found in many sources (see, for instance, [72] or [176]). Moreover, in [72] and [176] a nice geometrical interpretation of this fact is given (cf. Exercise 1 for Chapter 11). As is well known, C is nowhere dense in R and its Lebesgue measure equals zero. Consequently, we infer that the algebraic sum of two small subsets of R (in the sense of Lebesgue measure or Baire category) can have A.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7_12, © 2009 Atlantis Press/World Scientific

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nonempty interior, so can be a non-small subset of R. Furthermore, putting D = ∪{nC : n is integer}, we easily get D + D = D − D = R, where the set D is of first category and of Lebesgue measure zero. We thus conclude that the algebraic sum of two small sets in R can coincide with R. It should be emphasized that here the smallness of sets is meant both in the sense of Baire category and Lebesgue measure. These examples are the simplest and historically earliest which show that the operation of an algebraic sum applied to small subsets of the real line leads to big sets. In particular, we see that certain descriptive properties of summands are not preserved under this operation. Among many other examples of this kind, let us mention the following two facts closely connected with the assertions formulated above. Example 1. There exist two Borel subsets X and Y of R such that the set X +Y is not Borel in R. As far as we know, this statement was first established in [227] and [228]. Further results in this direction were obtained by various authors (see, for instance, [37], [62], and [207]). Notice that if A and B are any two Borel subsets of R, then the set A + B which is a continuous image of the Borel subset A × B of the plane R2 is necessarily an analytic (Suslin) set in R. Consequently, A + B has the Baire property and is measurable in the Lebesgue sense. Moreover, A + B is universally measurable with respect to the class of the completions of all σ -finite Borel measures on R (see Appendix 2). Example 2. There exist two Lebesgue measure zero subsets X and Y of R such that the set X + Y is not Lebesgue measurable. We have already pointed out that this remarkable result is due to Sierpi´nski [219]. An analogous result is valid in terms of Baire category, and the arguments in both cases are similar. More precisely, in both cases the proofs of these results use the technique of Hamel bases and, of course, rely on an uncountable form of the Axiom of Choice. The phenomenon described in Example 2 is not accidental. It remains true for a large class of measures on the real line R. To give a formulation of the corresponding assertion, let us first recall that an affine transformation of R is any mapping h:R→R

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representable in the form h(x) = ax + b

(x ∈ R),

where a and b are some fixed real numbers and, in addition, a = 0. The family of all affine transformations of R is a group with respect to the standard composition operation. Let H denote this group. As usual, H is called the affine group of R and is significantly larger than the group of all isometric transformations of R. We recall that a measure μ defined on some H-invariant σ -algebra of subsets of R is quasiinvariant under H (or, in short, H-quasi-invariant) if for each h ∈ H and for every μ -measure zero set X, we have μ (h(X)) = 0. Obviously, the standard Lebesgue measure λ on R is H-quasi-invariant but is not Hinvariant. There are various extensions of λ which also are quasi-invariant under H. In other words, the class of such measures is sufficiently large. Notice that there exist even nonseparable measures belonging to this class (cf. [115]). The following statement may be regarded as a generalization of Example 2. Theorem 1. Let μ be a nonzero σ -finite complete measure on R quasi-invariant under the group H. Suppose that there are two μ -measure zero sets A and B such that A + B = R. Then there exists a μ -measure zero set X such that X + X is not measurable with respect to

μ. For the proof, see [116]. The technique of Hamel bases is also useful for establishing Theorem 1 but essentially new ideas are needed, too. Observe that Theorem 1 significantly generalizes Example 2. The Cantor set C which is of Lebesgue measure zero is not an absolutely small subset of R. Indeed, C carries a Borel probability measure ν whose completion is isomorphic with the restriction of λ to the σ -algebra of all Lebesgue measurable subsets of the closed unit interval [0, 1]. Actually, ν coincides with the product measure ∏n∈N νn where for each natural number n, the measure νn is defined on the family of all subsets of {0, 1} and

νn ({0}) = νn ({1}) = 1/2. We thus see that from the stand-point of the definition of ν , the set C is big because

ν (C) = 1 = 0. However, it was repeatedly underlined in preceding sections that there exist uncountable subsets of R, which are extremely small from the measure-theoretical view-point.

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Let us recall the precise definition of small subsets of R in the sense of topological measure theory (see Chapters 2 and 5). Let T be a topological space such that all singletons in T are Borel subsets of T . A set X ⊂ T is said to be universal measure zero in T if for every σ -finite diffused Borel measure

μ on T , we have μ ∗ (X) = 0. As mentioned earlier, several constructions of uncountable universal measure zero subsets of uncountable Polish topological spaces are known in the literature (see, for instance, [69], [79], [113], [148], [172], and Appendix 1). More delicate constructions of universally small sets are presented in [197] and [250]. The existence of a universal measure zero set of cardinality continuum cannot be established within ZFC theory (see [172]). However, assuming the Continuum Hypothesis, we easily get the existence of such sets. For example, as well known, all Luzin subsets of R are universal measure zero. By assuming Martin’s Axiom, the same result can be obtained for all generalized Luzin sets in R. Notice that Martin’s Axiom is much weaker than CH and does not bound from above the cardinality of the continuum c. Let us recall the definition of generalized Luzin subsets of Polish topological spaces. Let T be an uncountable Polish topological space. A set X ⊂ T is a generalized Luzin subset of T if card(X) = c and for each first category set Z ⊂ T , we have card(X ∩ Z) < c. Under Martin’s Axiom, there are generalized Luzin sets in T and any such set is universal measure zero (see Exercise 14 for Appendix 1). In fact, the construction of a generalized Luzin set imitates the classical construction of Luzin or the construction of Mahlo, where the Continuum Hypothesis is assumed instead of Martin’s Axiom (see [143], [148], [159], [176], [192], and [222]). By modifying these constructions, the existence of generalized Luzin sets with additional algebraic properties can be shown. For example, several authors proved that under Martin’s Axiom there exists a generalized Luzin set on the real line, which is simultaneously a vector space over the field Q of all rational numbers (cf. [40], [59], [172], and Chapter 5). In this connection, it should be pointed out that a much stronger statement was proved by Erd¨os, Kunen and Mauldin in their joint paper [59]. Namely, they have established the following statement. Theorem 2. Suppose that Martin’s Axiom holds. Then there exist two sets X ⊂ R and Y ⊂ R such that: 1) both X and Y are generalized Luzin sets in R;

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2) both X and Y are vector spaces over the field Q; 3) X +Y = R and X ∩Y = {0}; in other words, R is a direct sum of its vector subspaces X and Y . Proof. Let us make several preliminary remarks. In fact, we are able to give a proof of Theorem 2 which also works in a more general case, namely, for a Polish topological vector space E = {0} instead of R (cf. Lemma 1 below). Notice that our argument differs slightly from the one presented in [59] and, in some technical details, is easier. First, let us recall that if E is a vector space over an arbitrary field and U, V are vector subspaces of E, then there exists a vector space V  ⊂ V such that U + V = U + V , U ∩V  = {0}. Indeed, U ∩ V is a vector subspace of V , so we can take as V  a vector subspace of V complemented to U ∩V . Consequently, it suffices to prove the existence of sets X and Y satisfying the relations 1), 2), and such that X + Y = R. Let α denote the least ordinal number of cardinality continuum and let {zξ : ξ < α } be an enumeration of all points of R. Further, let {Fξ : ξ < α } be an enumeration of all first category Fσ -subsets of R. By using transfinite recursion, we shall construct two α sequences {Xξ : ξ < α }, {Yξ : ξ < α } of subsets of R, fulfilling the following relations: a) for each ξ < α , we have card(Xξ ) = card(Yξ ) = card(ξ ) + ω ; b) if ζ < ξ < α , then Xζ ⊂ Xξ and Yζ ⊂ Yξ ; c) if ξ < α , then Xξ and Yξ are vector spaces over Q; d) if ξ < α , then zξ ∈ Xξ + Yξ ; e) if ζ < ξ < α , then Fζ ∩ Xζ = Fζ ∩ Xξ and Fζ ∩Yζ = Fζ ∩Yξ . Suppose that for an ordinal ξ < α , the partial ξ -sequences {Xζ : ζ < ξ } and {Yζ : ζ < ξ } have already been constructed. Let us denote: X  = ∪{Xζ : ζ < ξ }, Y  = ∪{Yζ : ζ < ξ }.

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Let z be a point of R defined as follows: z = zξ if zξ does not belong to X  +Y  , and z is some point of R \ (X  +Y  ) if zξ belongs to X  + Y . Keeping in mind the relations card(X  )  card(ξ ) + ω < c, card(Y  )  card(ξ ) + ω < c and taking into account the fact that ∪{Fζ : ζ < ξ } is a first category set in R (see Remark 1 below), it is not difficult to show that there exist two points x ∈ R \ X  and y ∈ R \Y  such that: x + y = z, x ∈ X  + Q(∪{Fζ : ζ < ξ }), y ∈ Y  + Q(∪{Fζ : ζ < ξ }). We then put Xξ = Qx + X , Yξ = Qy + Y  . Proceeding in this way, we get the α -sequences {Xξ : ξ < α } and {Yξ : ξ < α }. Finally, we define X = ∪{Xξ : ξ < α }, Y = ∪{Yξ : ξ < α }. It follows from the above construction that X and Y are vector spaces over Q, are also generalized Luzin sets, and X +Y = R which ends the proof of the theorem. We immediately obtain from Theorem 2 that under Martin’s Axiom, the algebraic sum of two universal measure zero sets can coincide with R. Moreover, putting A = X ∪Y, where X and Y are as in Theorem 2, we get a generalized Luzin set A (which is universal measure zero as well) such that A + A = R.

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In particular, we infer that the algebraic sum A + A, where A is universal measure zero in R, can coincide with the whole R in certain models of set theory. In this connection, it makes sense to consider the question whether the algebraic sum of two universal measure zero sets can be very bad from the measure-theoretical point of view. First, let us remember the precise definition of extremely bad sets in the sense of topological measure theory (see Chapter 5). Let T be a topological space such that all singletons in T are Borel subsets of T . We say that a set Y ⊂ T is absolutely nonmeasurable in T if for every nonzero σ -finite continuous Borel measure μ on T , we have Y ∈ dom(μ  ), where μ  denotes the completion of μ . It turns out that absolutely nonmeasurable sets in an uncountable Polish space T admit a purely topological characterization. For this purpose, we need one notion from general topology. Let Z be a subset of a topological space T . Recall that Z is a Bernstein set if for each nonempty perfect set P ⊂ T , both intersections P ∩ Z and P ∩ (T \ Z) are nonempty. This definition directly shows that Z ⊂ T is a Bernstein set if and only if T \ Z is a Bernstein set. The well-known argument from Bernstein which is based on the method of transfinite induction establishes the existence of a Bernstein set in an uncountable Polish topological space T (cf. [12], [148], [176], [192], and Exercise 5 for Chapter 5). Moreover, any construction of such a set needs an uncountable form of the Axiom of Choice. Various properties of Bernstein sets are discussed in [12], [40], [148], [176], and [192] with their applications in general topology, measure theory, and real analysis. We would like to mention a simple fact concerning the cardinality of Bernstein sets. Namely, as known, every Bernstein subset of R is necessarily of cardinality continuum. This fact immediately follows from the existence of a disjoint family {Pj : j ∈ J} of nonempty perfect subsets of R, where card(J) = c. To get such a family {Pj : j ∈ J}, it suffices to take some Peano type mapping f : [0, 1] → [0, 1]2 , which is a continuous surjection and consider the family of all pre-images f −1 ({x} × [0, 1]) (x ∈ [0, 1]). These pre-images are pairwise disjoint and each of them is an uncountable closed set in R. So each of them contains a nonempty perfect subset.

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The next statement shows that in uncountable Polish topological spaces Bernstein sets are identical with absolutely nonmeasurable sets (cf. Exercise 18 from Chapter 5). Theorem 3. Let E be an uncountable Polish space and let Z be a subset of E. The following two assertions are equivalent: 1) Z is absolutely nonmeasurable in E; 2) Z is a Bernstein set in E. Proof. 1) ⇒ 2). Assume that for a set Z ⊂ E, relation 1) holds and let us show that Z is a Bernstein set. Supposing to the contrary that Z is not a Bernstein set, we deduce that for some nonempty perfect set P ⊂ E, we have either P ∩ Z = 0/ or P ∩ (E \ Z) = 0. / We may assume, without loss of generality, that P ∩ Z = 0. / According to a well-known theorem of descriptive set theory (see, e.g., [40], [99], [148], or Appendix 6), the topological spaces P and [0, 1] are Borel isomorphic. Let λ0 denote the restriction of the Lebesgue measure

λ to the Borel σ -algebra of [0, 1]. Using any Borel isomorphism between [0, 1] and P, we can transfer λ0 to P. In this manner, we obtain a Borel diffused probability measure μ on E concentrated on P, i.e.,

μ (P) = 1, μ (E \ P) = 0 and hence μ ∗ (Z) = 0. Consequently, Z turns out to be measurable with respect to the completion of μ . This circumstance contradicts 1) and, therefore, establishes the implication 1) ⇒ 2). 2) ⇒ 1). Let Z be a Bernstein set and let μ be an arbitrary nonzero σ -finite diffused Borel measure on E. Denote by μ  the completion of μ . We have to verify that Z is not measurable with respect to μ  . Suppose for a while that Z ∈ dom(μ  ). Then either

μ  (Z) > 0 or μ  (E \ Z) > 0. We may assume, without loss of generality, that μ  (Z) > 0. Since μ  is a Radon measure, there exists a compact set K ⊂ Z such that μ  (K) > 0. Since

μ  is diffused, K must be uncountable. So K contains a nonempty perfect subset, which contradicts the assumption that Z is a Bernstein set. Theorem 3 has thus been proved. The following auxiliary statement is a direct analogue of Theorem 2. Lemma 1. Suppose that Martin’s Axiom holds. Let E = {0} be a Polish topological vector space. Then there exist two subsets L1 and L2 of E such that: 1) both L1 and L2 are generalized Luzin sets in E; 2) both L1 and L2 are vector spaces over Q; 3) L1 + L2 = E and L1 ∩ L2 = {0}.

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Consequently, under Martin’s Axiom, there exist two universal measure zero sets in E whose algebraic sum coincides with E. The proof of this lemma is completely analogous to the proof of Theorem 2. The corresponding details are left to the reader. Now, we are going to determine some connection of Bernstein sets with the fact described in Example 2. The following result can be treated as a certain stronger version of Example 2. However, the proof of this version needs an additional set-theoretical assumption. Theorem 4. Suppose again that Martin’s Axiom holds. Let E = {0} be a Polish topological vector space. Then there exist two generalized Luzin subsets of E such that their Minkowski’s sum is a Bernstein set in E. Consequently, under Martin’s Axiom, there exist two universal measure zero sets in E whose Minkowski’s sum is absolutely nonmeasurable. Proof. According to Lemma 1, we have a representation E = X +Y

(X ∩Y = {0}),

where X and Y are generalized Luzin sets and both of them are vector spaces over Q. Let

α denote again the least ordinal number of cardinality continuum and let {Pξ : ξ < α } denote the family of all nonempty perfect subsets of E. We may assume, without loss of generality, that each of the partial families {Pξ : ξ < α , ξ is an even ordinal}, {Pξ : ξ < α , ξ is an odd ordinal} also contains all nonempty perfect subsets of E. By using the method of transfinite recursion, we shall construct an injective α -sequence {yξ : ξ < α } of points of E such that: a) {yξ : ξ < α } ⊂ Y ; b) for every ordinal ξ < α , we have (X + yξ ) ∩ Pξ = 0. / Suppose that for an ordinal ξ < α , the partial ξ -sequence {yζ : ζ < ξ } has already been defined. Consider the set Pξ and equip this set with a Borel probability diffused measure μ (cf. the proof of Theorem 3). Since all sets X + yζ (ζ < ξ ) are universal measure zero, we get

μ ∗ (Pξ ∩ (X + yζ )) = 0 (ζ < ξ ). In view of Martin’s Axiom (see Remark 1 below), we come to the relation

μ ∗ (∪{Pξ ∩ (X + yζ ) : ζ < ξ }) = 0.

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Therefore, we can write / Pξ \ ∪{X + yζ : ζ < ξ } = 0. Let z be a point of Pξ \ ∪{X + yζ : ζ < ξ }. By virtue of the equality E = X + Y, there exists a point y ∈ Y such that z ∈ X + y. We put yξ = y. Proceeding in this manner, we will be able to construct the desired α -sequence of points {yξ : ξ < α }. Afterwards, we define L1 = X, L2 = {yξ : ξ < α , ξ is odd}, L3 = {yξ : ξ < α , ξ is even} and we assert that L1 and L2 are the required sets. Indeed, according to the definition, L1 = X is a generalized Luzin set. Notice that L2 which is a subset of Y with card(L2 ) = c is a generalized Luzin set, too. Further, for any odd ordinal ξ < α , we have the relation / Pξ ∩ (X + L2 ) ⊃ Pξ ∩ (X + yξ ) = 0. Analogously, for any even ordinal ξ < α , we may write / Pξ ∩ (X + L3 ) ⊃ Pξ ∩ (X + yξ ) = 0. At the same time, taking into account the equality X ∩Y = {0}, we easily obtain (X + L2 ) ∩ (X + L3 ) = 0. / Thus both X + L2 and X + L3 are Bernstein subsets of E which completes the proof of Theorem 4. By slightly changing the above argument, we can also prove that if Martin’s Axiom is valid, then there exists a generalized Luzin set L such that L + L is a Bernstein subset of E (in this context, cf. [44]). By applying a similar technique, it can be established that under Martin’s Axiom there exist two generalized Luzin sets L1 and L2 which are vector spaces over Q and whose algebraic sum L1 + L2 is a Bernstein set. In this case, we necessarily have L1 = L2 . Remark 1. It is easy to see that the proof of Theorem 2 (as well as the proof of Theorem 4) does not need the full power of Martin’s Axiom. Indeed, in the proof of Theorem 2 we only use the following consequence of this axiom:

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if E is a second category topological space with a countable base, then the σ -ideal of all first category subsets of E is c-additive. Analogously, in the proof of Theorem 4 we use the following consequence of Martin’s Axiom: if E is a topological space with a countable base, in which all closed subsets are of type Gδ , and μ is a nonzero σ -finite Borel measure on E, then the σ -ideal generated by all

μ -measure zero subsets of E is c-additive For more details, see e.g. [10], [40], [67], and [216]. Remark 2. It should be noticed that an additional set-theoretical assumption in the formulation of Theorem 4 cannot be omitted. Indeed, as mentioned before, there is a model of set theory in which the Continuum Hypothesis fails to be true (i.e., ω1 < c) and the cardinality of any universal measure zero subset of R does not exceed ω1 (see [172]). If U1 and U2 are any two universal measure zero subsets of R in such a model, then card(U1 + U2 )  ω1 · ω1 = ω1 and, consequently, U1 +U2 cannot be a Bernstein set because the cardinality of every Bernstein set is equal to c. Recall that one purely algebraic version of Theorem 4 was discussed in Chapter 11 (see Theorem 7 therein). Notice that Theorem 7 from Chapter 11 has some advantage over Theorem 4 of this chapter because it does not need any additional settheoretical hypotheses. Remark 3. It would be interesting to extend the preceding results to a more general case where E is an uncountable Borel subgroup (nontrivial Borel vector subspace) of a Polish topological commutative group (Polish topological vector space). Namely, in connection with Lemma 1, the following two questions naturally arise for E assuming of course the Continuum Hypothesis or Martin’s Axiom. (a) Do there exist two universal measure zero subgroups X and Y of E such that X +Y = E? (b) Do there exist two universal measure zero subsets X  and Y  of E such that X  + Y  is absolutely nonmeasurable? As far as we know, these questions remain open. Also, it is natural to ask whether Lemma 1 admits a generalization to the case of all complete metrizable topological vector spaces of cardinality c. A canonical nonseparable representative of this class is the Banach space l∞ = {(xn )n<ω ∈ Rω : supn<ω |xn | < +∞},

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which plays an important role in many questions of functional analysis and probability theory. It turns out that the corresponding generalization of Lemma 1 is possible under the Continuum Hypothesis. However, the argument should be essentially changed in order to obtain the desired result. We need the following auxiliary statement which is important in topological measure theory (see, e.g., [192] or Appendix 3). Lemma 2. Let T be a metrizable topological space whose topological weight is not a real-valued measurable cardinal. Let μ be an arbitrary σ -finite Borel measure on T . Then

μ possesses a separable support, i.e., there exists a closed separable set F ⊂ T such that μ (T \ F) = 0. A detailed proof of Lemma 2 is given in Appendix 3. Also, some applications of this lemma are presented therein. Remark 4. As known, Martin’s Axiom implies that the cardinality of the continuum c is not real-valued measurable (see, for instance, [10], [40], and [67]). This fact is a direct consequence of the existence of generalized Luzin sets on the real line R and of the circumstance that all generalized Luzin sets are universal measure zero (under Martin’s Axiom).

Lemma 3. Assume Martin’s Axiom. Let E be a complete metrizable topological vector space whose topological weight is equal to c. There exist two universal measure zero sets U1 ⊂ E and U2 ⊂ E such that each of them is a vector space over Q and U1 + U2 is absolutely nonmeasurable in E. Proof. We only sketch the proof leaving all corresponding technical details to the reader. The argument is based on the fact that the family of all separable subspaces of E generates a c-additive ideal of subsets of E. Therefore, in view of Lemma 2, any Luzin type set for this ideal (cf. Exercise 14 for Appendix 1) turns out to be universal measure zero. Further, by using the method of transfinite induction, it can be proved that there exist two universal measure zero sets U ⊂ E and V ⊂ E which simultaneously are vector spaces over Q and satisfy the relations U ∩V = {0}, U + V = E. Applying again the method of transfinite induction, we can show that there exists a vector subspace V  of V such that U + V  is a Bernstein set in E (cf. the proof of Theorem 4). Notice that V  is a subset of a universal measure zero set and is also universal measure zero. Finally, it follows from Theorem 3 and Lemma 2 that any Bernstein subset of E is

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absolutely nonmeasurable in E, so we may put U1 = U, U2 = V  , which completes the proof. The following statement is a straightforward consequence of Lemma 1 and Lemma 3 formulated above. Theorem 5. Suppose that the Continuum Hypothesis holds. Let E be an arbitrary complete metrizable topological vector space with card(E) = c. There exist two universal measure zero sets U1 ⊂ E and U2 ⊂ E which are vector spaces over Q and whose Minkowski’s sum U1 + U2 is absolutely nonmeasurable in E. Remark 5. We do not know whether the assertion of Theorem 5 remains valid under Martin’s Axiom. Some relatively recent works were devoted to algebraic sums of small sets (see, for instance, [37], [44], [131], and [132]). Let us point out that in papers [131] and [132] similar questions are considered for an uncountable commutative group (G, +) which is not assumed to be endowed with a topology but is equipped with a nonzero σ -finite complete G-invariant or, more generally, G-quasiinvariant measure μ (cf. also Chapter 11). Obviously, for this μ , the question can be posed whether there are two μ -measure zero sets in G whose algebraic sum is nonmeasurable with respect to μ . In this context, the following open problem seems to be of interest (cf. the formulation of Theorem 1). Problem. Let (G, +) be an uncountable commutative group and let μ be a nonzero σ -finite complete G-invariant (G-quasi-invariant) measure on G such that A + B = G for some two

μ -measure zero sets A and B. Do there exist two μ -measure zero sets X and Y such that X + Y is nonmeasurable with respect to μ ? In Chapter 11 we were also concerned with a related problem which was formulated in terms of G-absolutely negligible sets and G-absolutely nonmeasurable sets. In view of the said above, it is reasonable to recall its formulation once again. Problem. Let (G, +) be an uncountable commutative group. Do there exist two Gabsolutely negligible sets A and B in G such that their algebraic sum A + B is G-absolutely nonmeasurable?

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We know that if G is a vector space over Q with card(G)  c, then the latter problem has a positive solution. In particular, under CH, this problem is positively solved for all uncountable vector spaces over Q and among them are Rn (n  1), Rω , c0 , l1 , l2 , and l∞ . EXERCISES 1∗ . Let E = {0} be a normed vector space or, more generally, locally convex Hausdorff topological vector space. Prove that there exist two first category subsets X and Y of E such that X + Y does not have the Baire property in E. For this purpose, use the Hahn-Banach theorem and reduce the argument to the particular case E = R. 2. Let E = {0} be a Polish topological vector space. Show that if Martin’s Axiom is valid, then there exists a generalized Luzin set L ⊂ E such that L + L is a Bernstein subset of E. 3. Let E = {0} be a Polish topological vector space. Assuming Martin’s Axiom, demonstrate that there exist two generalized Luzin sets L1 and L2 which are vector spaces over Q and whose algebraic sum L1 + L2 is a Bernstein set. In this case, we obviously have L1 = L2 . 4. Verify the validity of the assertion of Remark 1 concerning the fact that the results presented in this chapter do not need the full power of Martin’s Axiom. 5∗ . Let E be a complete metric space of cardinality continuum. Suppose that E does not contain isolated points. Show that there exists at least one Bernstein subset of E. For this purpose, take into account the circumstance that the family of all closed separable subspaces of E is also of cardinality continuum and that every nonempty perfect subset of E contains a topological copy of the Cantor space {0, 1}ω which trivially is separable. Moreover, show that the following two assertions are equivalent: (a) c is not real-valued measurable; (b) if E is any complete metric space of cardinality c, then the class of all Bernstein subsets of E coincides with the class of all absolutely nonmeasurable subsets of E. In order to establish the equivalence (a) ⇔ (b), apply the special case of Lemma 2 stating that if c is not measurable in the Ulam sense, then for every metric space E with card(E) = c and for any σ -finite Borel measure μ on E, there exists a separable support of μ . 6. Complete the details of the proof of Lemma 3. 7. Let X1 and X2 be any two universal measure zero topological spaces. Show that their topological product X1 × X2 is universal measure zero, too. Infer from this circumstance

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that the topological product of a finite family of universal measure zero spaces is also universal measure zero. On the other hand, give an example of a countable family of universal measure zero spaces whose topological product is not universal measure zero. 8∗ . Assuming Martin’s Axiom, prove that there exists a universal measure zero subset of R whose image under some Borel mapping acting from R into R is not universal measure zero. For this purpose, use the previous exercise and take into account the existence of two generalized Luzin subsets L1 and L2 of R such that L1 + L2 = R. 9∗ . Let X be a universal measure zero subset of a Polish topological space E and let f : E → E be a Borel mapping acting from E into a Polish topological space E  . Suppose that (∀y ∈ E  )(card( f −1 (y))  ω ). Show that the set f (X) is universal measure zero. 10∗ . Demonstrate that an injective continuous Hausdorff image of a universal measure zero topological space is not, in general, universal measure zero. For this purpose, take into account the fact that the first uncountable cardinal ω1 equipped with the discrete topology is universal measure zero in view of the Ulam theorem [238] but the same ω1 equipped with its order topology is not universal measure zero (cf. Exercise 8 from Appendix 3).

Chapter 13

Absolutely nonmeasurable additive ´ Sierpinski-Zygmund functions

Here we will focus our consideration on functions acting from R into R and having bad structural properties (cf. Chapters 1 and 5). First of all, we mean those properties of functions which are bad from the measure-theoretical point of view. Notice that from the stand-point of general topology, bad functions are those which do not possess the Baire property. Notice also that, in many respects, the Baire property can be treated as a natural topological analogue of measurability (see, for instance, [40], [143], [148], [176], and [192]). Let E be a nonempty set and let M be a class of measures on E. In general, the domains of measures from M may be different σ -algebras of subsets of E. Let f : E → R be a function. We recall that f is absolutely nonmeasurable with respect to M if f is nonmeasurable with respect to any measure from M (see Chapter 5). In connection with this notion, it makes sense to remember the following two typical examples. Example 1. Let E = R = R1 and let M denote the class of all translation-invariant extensions of the Lebesgue measure λ = λ1 on R. Let X be an arbitrary Vitali subset of R. Then the characteristic function f = χX of X is absolutely nonmeasurable with respect to M (see Exercise 2 of Chapter 5, where a more general result is formulated). Example 2. Let E again coincide with R and let M be the class of the completions of all nonzero σ -finite diffused Borel measures on R. Denote by X a Bernstein subset of R and let g = χX be the characteristic function of X. As we know from Theorem 3 of Chapter 12, the function g is absolutely nonmeasurable with respect to M. Below, another interesting and important example of a function f : R → R will be given which also turns out to be absolutely nonmeasurable with respect to the same class M. Now, denote by ME the class of all nonzero σ -finite diffused measures on E. Let us emA.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7_13, © 2009 Atlantis Press/World Scientific

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phasize once more that the domains of measures from this class are various σ -algebras of subsets of E. We recall that a function f : E → R is absolutely nonmeasurable if f is absolutely nonmeasurable with respect to ME . Also, remember that a set X ⊂ R is universal measure zero if for every σ -finite diffused Borel measure μ on R, the equality μ ∗ (X) = 0 is satisfied, where μ ∗ as usual denotes the outer measure associated with μ . It was repeatedly mentioned in preceding sections that there exist uncountable universal measure zero subsets of R (see, e.g., [69], [79], [148], [172], and Appendix 1). A much stronger version of this result is contained in [197] and [250] where delicate examples of uncountable universally small subsets of R are constructed. Let us recall a characterization of absolutely nonmeasurable functions which was obtained in terms of universal measure zero sets and pre-images of singletons (see Chapter 5). Namely, for a function f : E → R, the following two relations are equivalent: (1) f is absolutely nonmeasurable; (2) the range of f is a universal measure zero subset of R and card( f −1 (t))  ω for each point t ∈ R. A detailed proof of the equivalence (1) ⇔ (2) was given in Chapter 5. In the same chapter, the question was considered whether Vitali type functions are absolutely nonmeasurable with respect to certain classes of extensions of λ . In this context, let us remember several notions. Let Q denote the set of all rational numbers. A function f : R → R is of Vitali type if the following two conditions hold: a) f (x) − x ∈ Q for every x ∈ R; b) ran( f ) is a selector of the partition R/Q. It was shown in Chapter 5 that if f : R → R is an arbitrary Vitali type function, then the Lebesgue measure λ can be extended to a measure μ on R such that f turns out to be measurable with respect to μ . In other words, f is relatively measurable with respect to the class of all extensions of λ . Let c as usual denote the cardinality of the continuum. We already know that the following three assertions are direct consequences of the equivalence between the relations (1) and (2) and of the existence of an uncountable universal measure zero set in R. I. If card(E) > c, then there are no absolutely nonmeasurable functions f with dom( f ) = E.

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II. There exists an injective absolutely nonmeasurable function f : E → R, where E is an arbitrary set with card(E) = ω1 . III. If the cardinality of any universal measure zero subset of R is strictly less than c, then there are no absolutely nonmeasurable functions f such that dom( f ) = R. In particular, the latter consequence shows that absolutely nonmeasurable functions f : R → R exist if and only if there exists a universal measure zero set X ⊂ R with card(X) = c. More generally, absolutely nonmeasurable functions f :E →R exist if and only if there exists a universal measure zero subset X of R with card(X) = card(E). This circumstance also shows that the existence of an absolutely nonmeasurable function acting from R into R cannot be established within the theory ZFC. Indeed, there are models of set theory in which the negation of the Continuum Hypothesis holds and the cardinality of any universal measure zero subset of R does not exceed the first uncountable cardinal ω1 (see, for example, [172]). Clearly, in such a model we do not have an absolutely nonmeasurable real-valued function defined on R. In this context, notice that the nonexistence of absolutely nonmeasurable functions acting from R into R can be deduced by using some other set-theoretical assumptions. For instance, it is easy to see that if c is a real-valued measurable cardinal, then no function acting from R into R is absolutely nonmeasurable because in this case R carries a probability continuous measure μ with dom(μ ) = P(R) and, therefore, all functions g : R → R turn out to be measurable with respect to μ . Recall that a function f : R → R is additive (or satisfies the Cauchy functional equation) if f (x + y) = f (x) + f (y)

(x ∈ R, y ∈ R).

In other words, f is a solution of the Cauchy functional equation if and only if f is a homomorphism of the additive group R into itself. There are many works devoted to the Cauchy functional equation (see, e.g., [143] and references therein). Obviously, all continuous solutions of the Cauchy functional equation are representable in the standard linear form x → ax

(x ∈ R),

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where a ∈ R. Any such function is called a trivial solution of the Cauchy functional equation. By using the technique of Hamel bases, we can get many nontrivial solutions of this equation. Actually, the family of all such solutions has cardinality 2c and all of them are nonmeasurable in the Lebesgue sense and do not possess the Baire property (for more details, see [40], [108], [143], [176], or Chapters 1 and 5). It should be pointed out that nontrivial solutions of the Cauchy functional equation have found interesting applications in the classical theory of equidecomposability of polyhedra (see, e.g., [18]). As mentioned earlier, within ZFC it is impossible to establish the existence of absolutely nonmeasurable solutions of the Cauchy functional equation. On the other hand, by assuming Martin’s Axiom and applying some properties of generalized Luzin subsets of R, it can be proved that there are additive absolutely nonmeasurable functions acting from R into R. In other words, under Martin’s Axiom, there exist absolutely nonmeasurable solutions of the Cauchy functional equation as seen in Chapter 5. Our main goal here is to strengthen this result in a certain direction. Namely, assuming the same Martin’s Axiom, we intend to show the existence of an additive absolutely nonmeasurable function acting from R into R, which is also utterly bad from the topological point of view. Sierpi´nski and Zygmund constructed in their remarkable work [225] a function f :R→R having the following property: for each subset Y of R with card(Y ) = c, the restriction f |Y is not continuous on Y . This classical result of Sierpi´nski and Zygmund was essentially motivated by a theorem of Blumberg [15] stating that for any function g : R → R, there exists an everywhere dense subset D of R such that the restriction g|D is continuous on D (see Exercise 22 from Chapter 8). Obviously, the set D is infinite because it is everywhere dense in R. Consequently, the existence of a Sierpi´nski-Zygmund function f : R → R shows that within ZFC theory we cannot assert the uncountability of D. Let us briefly sketch the standard construction of Sierpi´nski-Zygmund type functions. The construction is not effective, i.e., it appeals to the Axiom of Choice which is a necessary component in constructions of this sort. Let α be the least ordinal number of cardinality continuum, let {xξ : ξ < α } denote an injective family of all points of R, and let {gξ : ξ < α } be the family of all partial continuous functions acting from R into R whose domains are uncountable Gδ -subsets of R. By using the method of transfinite recursion, we define a function f : R → R.

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Suppose that for an ordinal ξ < α , all values f (xζ ) (ζ < ξ ) have already been defined and consider the point xξ . Since the set ∪{gζ (xξ ) : ζ < ξ , xξ ∈ dom(gζ )} is of cardinality strictly less than c, we may pick a point y ∈ R \ ∪{gζ (xξ ) : ζ < ξ , xξ ∈ dom(gζ )} and then put f (xξ ) = y. Proceeding in this manner, the desired function f will be completely determined. It immediately follows from the above definition of f that card({x ∈ R : f (x) = g(x)}) < c for any partial continuous function g : R → R whose domain is an uncountable Gδ -subset of R. But the last circumstance implies that we also have card({x ∈ R : f (x) = h(x)}) < c for any partial continuous mapping h : R → R whose domain is of cardinality continuum. Indeed, according to Lavrentiev’s theorem (see Exercise 13 from Chapter 8), there exists a partial continuous mapping h∗ : R → R extending h and defined on a Gδ -subset of R. Since card(dom(h)) = c, we claim that card(dom(h∗ )) = c. Consequently, card({x ∈ R : f (x) = h∗ (x)}) < c, which immediately yields card({x ∈ R : f (x) = h(x)}) < c. From the said above, we readily conclude that for any subset X of R with card(X) = c, the restriction f |X is not a continuous mapping. Thus, f is a Sierpi´nski-Zygmund function. Various works are devoted to Sierpi´nski-Zygmund functions (see, e.g., [7], [42], [123], [124], [185], and [199]). In those works different constructions are presented which yield further examples of Sierpi´nski-Zygmund functions with additional properties important from the view-point of real analysis. Let MR denote the class of the completions of all nonzero σ -finite diffused Borel measures on the real line R. It is not difficult to prove that every Sierpi´nski-Zygmund function f : R → R turns out to be absolutely nonmeasurable with respect to MR (see Exercise 2 of the present chapter). In particular, f is nonmeasurable with respect to the Lebesgue measure λ on R. At the same time, as mentioned above, we cannot assert that f is absolutely

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nonmeasurable. Moreover, it may happen that the graph Gr( f ) of f is a (λ ⊗ λ )-thick subset of the plane R2 and, in this case, f becomes measurable with respect to a suitable extension of λ (cf. Chapter 7). A much stronger result will be obtained in the next chapter. Namely, it will be demonstrated that there exists a Sierpi´nski-Zygmund function measurable with respect to some translation-invariant extension of λ and a certain transfinite construction of such a function will be done within ZFC theory. The following statement shows that under an appropriate set-theoretical assumption, there are absolutely nonmeasurable additive functions acting from R into R which are not Sierpi´nski-Zygmund functions. Theorem 1. Assume Martin’s Axiom. There exists a function f :R→R satisfying the following conditions: 1) f is injective and additive; 2) f is absolutely nonmeasurable; 3) the restriction of f to some set K ⊂ R with card(K) = c is the identical transformation of K and, consequently, f is not a Sierpi´nski-Zygmund function. Proof. We recall that a set L ⊂ R is a generalized Luzin subset of R if L is of cardinality continuum and for every first category set P ⊂ R, the inequality card(P ∩ L) < c holds true. It is well known that under Martin’s Axiom, there exists a generalized Luzin subset L of the real line, which is simultaneously a vector space over the field Q of all rational numbers (see, e.g., [40], [59], [172], or Chapter 12 of this book). Let HL denote a Hamel basis for L. Clearly, we have card(HL ) = c. Extend HL to a Hamel basis H of R. Let f0 : H → HL be a bijection such that for some set K ⊂ HL with card(K) = c, the restriction of f0 to K is the identical transformation of K. The bijection f0 can be extended to an isomorphism f (linear over Q) between R and L. Obviously, we may consider f as an injective group homomorphism from R into R whose range coincides with L. Since L is a universal measure zero subset of R (see Exercises 1 and 8 of Chapter 5 or Exercise 14 of Appendix 1) and since the function f is injective, we infer that f is absolutely nonmeasurable. At the same time, the restriction f |K coincides with the restriction f0 |K. Consequently, f |K is the

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identical transformation of K. This circumstance immediately implies that our f cannot be a Sierpi´nski-Zygmund function, which completes the proof of Theorem 1. The next statement establishes within ZFC that for any G ⊂ R which is a vector space over Q and has cardinality c, there is an injective additive Sierpi´nski-Zygmund function acting from R into G. Theorem 2. Let G be as above. There exists a function f :R→R satisfying the following conditions: 1) f is an injection; 2) f is a nontrivial solution of the Cauchy functional equation; 3) ran( f ) ⊂ G; 4) f is a Sierpi´nski-Zygmund function. Proof. We shall construct the required function f by using the method of transfinite recursion. Consider c as an initial ordinal number α equipped with its natural well-ordering which will be denoted by . Of course, this notation means that for any ordinal ξ < α = c, the relation card(ξ ) < c is valid. Let  be a well-ordering of R isomorphic to . Let {hξ : ξ < α } be an enumeration of all Borel mappings acting from uncountable Borel subsets of R into R. Under this notation, we are going to define three α -sequences {xξ : ξ < α }, {Vξ : ξ < α }, { fξ : ξ < α } satisfying the following relations: (a) {xξ : ξ < α } is a Hamel basis of R; (b) for each ordinal ξ < α , the set Vξ coincides with the vector subspace of R, over the field Q of all rational numbers, generated by {xζ : ζ  ξ }; (c) for each ordinal ξ < α , the function fξ is a group monomorphism acting from Vξ into G; (d) if ζ < ξ < α , then fξ extends fζ ; (e) if ξ < α , then we have fξ (qxξ + v) = hζ (qxξ + v) for all

ζ < ξ , q ∈ Q \ {0}, v ∈ ∪{Vη : η < ξ }

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such that qxξ + v ∈ dom(hζ ). Suppose that for an ordinal ξ < α , the partial ξ -sequences {xζ : ζ < ξ }, {Vζ : ζ < ξ }, { fζ : ζ < ξ } have already been constructed. Let us put: V  = ∪{Vζ : ζ < ξ }, f  = ∪{ fζ : ζ < ξ }. Applying (c) and (d), we claim that f  is a group monomorphism acting from V  into G and card(V  )  card(ξ ) + ω < c. Let x denote the least element (with respect to ) of the set R \ V  . We put xξ = x and denote by Vξ the vector space over Q generated by V  ∪ {x}. Further, we define D = {(1/q)(hζ (qx + v) − f (v)) : q ∈ Q \ {0}, ζ < ξ , v ∈ V  , qx + v ∈ dom(hζ )}. Since card(D) < c and card(ran( f  )) < c, we may choose an element y ∈ G \ (D ∪ ran( f  )). Clearly, there exists a unique group monomorphism fξ : Vξ → G extending f  and such that fξ (x) = y. Proceeding in this manner, we get the required families {xξ : ξ < α }, {Vξ : ξ < α }, { fξ : ξ < α }. Finally, denote f = ∪{ fξ : ξ < α }. It easily follows from our construction that {xξ : ξ < α } is a Hamel basis of R and, therefore, dom( f ) = ∪{Vξ : ξ < α } = R. Since all fξ (ξ < α ) are group monomorphisms, f is an injective homomorphism acting from R into G. Also, it can readily be verified that for any Borel function hξ (ξ < α ), we have card({z ∈ dom(hξ ) : f (z) = hξ (z)})  card(ξ ) + ω < c.

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This relation shows that f is a Sierpi´nski-Zygmund function (cf. [148] and [225]) and completes the proof. From Theorem 2 we easily obtain the following result. Theorem 3. Assuming Martin’s Axiom, there exists an injective additive absolutely nonmeasurable Sierpi´nski-Zygmund function. Proof. Let L denote again a generalized Luzin subset of R which simultaneously is a vector space over Q (see Chapter 5 or Chapter 12). Let in Theorem 2 the role of G be played by L. Then we get a group monomorphism f :R→L such that f is also a Sierpi´nski-Zygmund function. Now, keeping in mind the fact that L is universal measure zero and taking into account a characterization of absolutely nonmeasurable functions, we see that f is an absolutely nonmeasurable Sierpi´nski-Zygmund function. Theorem 3 has thus been proved. From the said above we can conclude that there are Sierpi´nski-Zygmund functions with ultimately bad measure-theoretical properties. On the other hand, in the next chapter the existence of a Sierpi´nski-Zygmund function will be established which is relatively measurable with respect to the class of all translation-invariant extensions of the Lebesgue measure

λ on R. Finally, it is reasonable to recall in this place that under Martin’s Axiom, there exist injective additive absolutely nonmeasurable functions f : R → R, which are not Sierpi´nskiZygmund functions (see Theorem 1 of the present chapter). EXERCISES 1. Show that no Sierpi´nski-Zygmund function has the Baire property. 2. Let MR denote the class of the completions of all nonzero σ -finite diffused Borel measures on R. Prove that every Sierpi´nski-Zygmund function is absolutely nonmeasurable with respect to MR . For this purpose, consider any Sierpi´nski-Zygmund function f : R → R and fix a nonzero

σ -finite diffused Borel measure μ on R. Suppose that f is measurable with respect to the completion μ  of μ . Since R is a Radon space, μ  is a Radon measure. Consequently, there exists a compact set K ⊂ R with μ  (K) = μ (K) > 0 such that the restriction f |K is continuous. Take into account the diffusedness of μ and infer that card(K) > ω from which

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it follows that card(K) = c. So a contradiction is obtained with the assumption that f is a Sierpi´nski-Zygmund function. 3. A function f : R → R is called weakly continuous at a point x ∈ R if there exist two sequences {tn : n < ω } and {rn : n < ω } of strictly positive real numbers tending to zero, such that limn→∞ f (x − tn ) = limn→∞ f (x + rn ) = f (x). Verify that every function f acting from R into R is weakly continuous at all points of some co-countable subset D of R. Also, check that D essentially depends on f . In particular, any Sierpi´nski-Zygmund function is weakly continuous on a co-countable subset of R. 4. By using the method of transfinite recursion, construct a Sierpi´nski-Zygmund function f : R → R whose graph Gr( f ) is λ2 -thick in R2 . Formulate and prove an analogous result in terms of the Baire property. 5. Show that there exists a Sierpi´nski-Zygmund function f : R → R whose graph Gr( f ) is a λ2 -measure zero subset of the plane R2 . 6. Show that there exists a Sierpi´nski-Zygmund function f : R → R whose graph Gr( f ) is a first category subset of R2 . 7∗ . Prove that there exists a family of functions fi : R → R

(i ∈ I)

satisfying the following three relations: (a) card(I) > c; (b) every fi (i ∈ I) is a Sierpi´nski-Zygmund function; (c) this family of functions is almost disjoint, i.e., the inequality card(Gr( fi ) ∩ Gr( f j )) < c holds true for any two distinct indices i ∈ I and j ∈ I. 8∗ . Assuming Martin’s Axiom, give an example of a Sierpi´nski-Zygmund function f : R → R such that for some set X ⊂ R of cardinality c, the restriction f |X is a semicontinuous function on X. 9. Let E and E  be two infinite sets and let Φ be a family of partial functions acting from E into E  . Suppose that card(E) = card(E  ) = card(Φ).

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Show that there exists a mapping f : E → E such that for any set X ⊂ E with card(X) = card(E) and for any partial function φ ∈ Φ, the relation f |X = φ |X is satisfied. In particular, take: E = a separable metric space of cardinality c; E  = the real line R; Φ = the family of all partial continuous functions φ : E → R such that dom(φ ) is a Gδ -set in E of cardinality c. Then the above-mentioned mapping f : E → R turns out to be a Sierpi´nski-Zygmund type function on E. 10. By using the result of the previous exercise and of Exercise 16 from Chapter 8, prove that there exists a function f :R→R having the following property: for every set X ⊂ R with card(X) = c, the restriction of f to X is not a Borel mapping. Conclude that any such f turns out to be a Sierpi´nski-Zygmund function but the converse assertion is not true in general (see Exercise 8 above). 11∗ . A set D ⊂ R is called nowhere Lebesgue measure zero if for each non-degenerate interval J ⊂ R, the intersection J ∩ D has strictly positive outer Lebesgue measure, i.e.,

λ ∗ (J ∩ D) > 0. Show that there exists a function f : R → R such that for every set D ⊂ R which is nowhere Lebesgue measure zero, the restriction f |D is discontinuous at all points of D. In order to demonstrate this fact, consider a partition {Xn : n < ω } of R such that X0 is an everywhere dense Gδ -set in R of λ -measure zero and all sets Xn (0 < n < ω ) are nowhere dense in R. Further, define a function f : R → R as follows: f (x) = n if and only if x ∈ Xn . Verify that f is the required one. For this purpose, take an arbitrary nowhere Lebesgue measure zero subset D of R and pick an arbitrary point x in R. Only two cases are possible. (a) There exists a neighborhood U(x) of x which intersects only finitely many sets from the family {Xn : n < ω }. In this case, utilizing the fact that all sets Xn (0 < n < ω ) are nowhere dense in R, we can find a nonempty open interval J ⊂ U(x) such that J ⊂ X0 .

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Since λ (X0 ) = 0, the interval J must also have Lebesgue measure zero. But this yields an obvious contradiction, so we claim that the situation just described is impossible. (b) Any neighborhood U(x) of x intersects infinitely many sets from the family {Xn : n <

ω }. In this case, it follows from the definitions of f and D that if x ∈ D, then f |D is unbounded on every neighborhood of x. The last circumstance directly implies the discontinuity of f |D at all points of D. The above-mentioned result is due to Brown [27]. It can easily be seen that this result does not need uncountable forms of the Axiom of Choice which are necessary in various constructions of Sierpi´nski-Zygmund type functions on R. 12. Let E be a set of cardinality c and let {Ti : i ∈ I} be a family of topologies on E satisfying the following two conditions: (a) card(I)  c; (b) each topology Ti (i ∈ I) has a countable base. Prove that there exists a function f : E → R such that for any topology Ti (i ∈ I), the corresponding mapping f : (E, Ti ) → R is a Sierpi´nski-Zygmund type function for the space (E, Ti ).

Chapter 14

´ Relatively measurable Sierpinski-Zygmund functions

In this chapter, we again will be dealing with Sierpi´nski-Zygmund functions. Our main goal remains the same. Namely, we consider structural properties of Sierpi´nski-Zygmund type functions from the measure-theoretical point of view and associate those properties with the measure extension problem (see Chapter 2 where three different aspects of this problem are discussed). The reader may think that the title of the present chapter appears misleading because all Sierpi´nski-Zygmund functions are nonmeasurable with respect to the Lebesgue measure λ on the real line R (in this connection, see Exercise 2 from Chapter 13 where a more general result is formulated). But here we are going to prove that there exist some Sierpi´nskiZygmund functions which are measurable with respect to certain translation-invariant extensions of λ . Let us remember several facts concerning absolutely nonmeasurable and relatively measurable real-valued functions. Let E be a nonempty set and let f : E → R be a function. We recall that f is absolutely nonmeasurable if f is nonmeasurable with respect to any nonzero σ -finite diffused (i.e., continuous) measure μ defined on a σ -algebra of subsets of E. Recall also that a set X ⊂ R is universal measure zero if for every σ -finite diffused Borel measure μ on R, the equality μ ∗ (X) = 0 is satisfied, where μ ∗ as usual denotes the outer measure associated with μ . As already said in preceding parts of the book, there exist uncountable universal measure zero subsets of R (see [69], [79], [148], [172], and Appendix 1). In addition, we have a characterization of absolutely nonmeasurable functions in terms of universal measure zero sets and pre-images of singletons. Namely, as shown in Chapter 5, A.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7_14, © 2009 Atlantis Press/World Scientific

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for a given function f : E → R, the following two assertions are equivalent: (1) f is absolutely nonmeasurable; (2) the range of f is a universal measure zero subset of R and the inequality card( f −1 (t)) 

ω holds for each point t ∈ R. For the proof of the equivalence (1) ⇔ (2), see again Chapter 5. Let c denote the cardinality of the continuum. We would like to mention the following two straightforward consequences of the equivalence between the assertions (1) and (2). I. If card(E) > c, then there are no absolutely nonmeasurable functions f : E → R; II. If the cardinality of any universal measure zero subset of R is strictly less than c, then there are no absolutely nonmeasurable functions f : R → R. The latter consequence shows us that absolutely nonmeasurable functions f : R → R exist if and only if there exists a universal measure zero set X ⊂ R with card(X) = c. More generally, absolutely nonmeasurable functions f :E →R do exist if and only if there exists a universal measure zero subset X of R such that card(X) = card(E). This circumstance also implies that the existence of an absolutely nonmeasurable function acting from R into R cannot be established within the theory ZFC. But there exists an injective absolutely nonmeasurable function f : E → R, where E is an arbitrary set of cardinality ω1 , and the latter fact holds true within ZFC theory (see Appendix 1). In this context it makes sense to recall that by assuming Martin’s Axiom and applying some properties of generalized Luzin subsets of R, it can be proved that there are injective additive absolutely nonmeasurable functions acting from R into R. In other words, under Martin’s Axiom, there exist absolutely nonmeasurable solutions of the Cauchy functional equation (see Chapters 5 and 12). Some further extensions of this result are closely connected with Sierpi´nski-Zygmund type functions (cf. Theorem 3 of the previous chapter).

As pointed out in Chapters 1 and 13, Sierpi´nski and Zygmund constructed in their classical work [225] a function f :R→R

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having the following property: for each subset Y of R with card(Y ) = c, the restriction f |Y is not continuous on Y . It was also mentioned in the same chapters that the result of Sierpi´nski and Zygmund may be regarded as a counter-version of Blumberg’s remarkable theorem [15] stating that for any function g : R → R, there exists an everywhere dense subset D of R such that the restriction g|D is continuous on D (see Exercise 22 from Chapter 8). Obviously, the set D must be infinite. On the other hand, the existence of a Sierpi´nski-Zygmund function shows that we cannot assert, in general, the uncountability of D. There are many works devoted to Sierpi´nski-Zygmund functions. In those works various constructions are presented, which yield further interesting examples of Sierpi´nskiZygmund functions with additional properties important from the view-point of real analysis, measure theory, and general topology (see [42] and references therein). It is known that every Sierpi´nski-Zygmund function f turns out to be nonmeasurable with respect to the completion of any nonzero σ -finite diffused Borel measure on R (see Exercise 2 for Chapter 13). In other words, every Sierpi´nski-Zygmund function is bad from the view-point of topological measure theory. At the same time, we cannot assert that any Sierpi´nski-Zygmund function f is absolutely nonmeasurable, because it may happen (e.g., in those models of set theory where the Continuum Hypothesis fails to be true) that the inequalities c > card( f −1 (t)) > ω hold for some point t ∈ R. In this case, by the equivalence of assertions (1) and (2), f must be measurable with respect to a certain nonzero σ -finite diffused measure on R. The present chapter is devoted to some much stronger measurability properties of certain Sierpi´nskiZygmund functions. As already said in the beginning of this chapter, the main goal is to demonstrate that there exist Sierpi´nski-Zygmund functions which are measurable with respect to appropriate translation-invariant extensions of the standard Lebesgue measure λ on R. For this purpose, we need several auxiliary propositions. Lemma 1. Let f : R → R be a function satisfying the following condition: for each subset Y of R with card(Y ) = c, the restriction f |Y is not a Borel mapping. Then f is a Sierpi´nski-Zygmund function.

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This lemma is trivial; however, it plays an essential role below. It is reasonable to point out that in Exercise 9 from Chapter 13 a general statement is formulated concerning the existence of Sierpi´nski-Zygmund type functions. It readily yields the existence of a function f : R → R which satisfies the condition of Lemma 1. Notice also that under Martin’s Axiom, there are Sierpi´nski-Zygmund functions for which this condition is not fulfilled (see Exercise 8 of the same Chapter 13). As usual, we denote by T (= S1 ) the one-dimensional unit torus in the Euclidean plane R2 . Recall that T is a commutative compact topological group canonically isomorphic to the group of all rotations of the plane R2 = R × R about its origin (0, 0). Lemma 2. Let φ : R → T be the canonical surjective continuous group homomorphism defined by

φ (x) = (cos(x), sin(x))

(x ∈ R).

There exists a mapping ψ : T → R such that: 1) φ ◦ ψ coincides with the identity transformation of T; 2) ψ is discontinuous only at one point of T (hence, ψ is a Borel mapping). We omit an easy proof of Lemma 2. Actually, in the sequel we only need the fact that ψ is a Borel mapping. So we may treat this simple lemma as a direct consequence of the theorem of Kuratowski and Ryll-Nardzewski [151] on measurable selectors (see Appendix 2). The torus T regarded as a compact topological group carries the Haar probability measure

ν . Obviously, the completion of ν coincides with the Lebesgue probability measure on the torus T, which is invariant under the group all translations of T. As usual, we denote by λ ⊗ ν the product measure whose multipliers are λ and ν . For our further purposes, it will be convenient to denote by the same symbol λ ⊗ ν the completion of the above-mentioned product measure. We recall that a function f : R → T has thick graph with respect to λ ⊗ ν if every (λ ⊗ ν )measurable set Z with (λ ⊗ ν )(Z) > 0 intersects the graph Gr( f ) of f ; in other words, f has thick graph if (λ ⊗ ν )∗ ((R × T) \ Gr( f )) = 0. We need the following lemma (cf. Chapters 2 and 7). Lemma 3. Let f : R → T be a group homomorphism whose graph Gr( f ) is thick with respect to λ ⊗ ν . For any set Z ∈ dom(λ ⊗ ν ), let us put Z  = {x ∈ R : (x, f (x)) ∈ Z}

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and introduce the class of sets S  = {Z  : Z ∈ dom(λ ⊗ ν )}. Finally, define a functional μ on S  by the formula

μ (Z  ) = (λ ⊗ ν )(Z)

(Z  ∈ S  ).

Then the following relations hold: 1) S  is a σ -algebra of subsets of R containing dom(λ ) and invariant under the group of all isometric transformations of R; 2) the functional μ is well defined; 3) μ is a measure on S  extending λ and invariant under the same group of all isometries of R; 4) the original homomorphism f is measurable with respect to μ . Proof. We use an argument similar to that given in the paper [141] (cf. also [115] or Chapters 2 and 7). If Z ∈ dom(λ ⊗ ν ) and Zi ∈ dom(λ ⊗ ν ) for each index i from a countable set I, then we can write R \ Z  = {x ∈ R : (x, f (x)) ∈ (R × T) \ Z}, ∪{Zi : i ∈ I} = {x ∈ R : (x, f (x)) ∈ ∪{Zi : i ∈ I}}. Now, taking into account the relations (R × T) \ Z ∈ dom(λ ⊗ ν ), ∪ {Zi : i ∈ I} ∈ dom(λ ⊗ ν ), we derive that S  is a σ -algebra of subsets of R. The invariance of S  under the group of all isometries of R follows directly from the invariance of dom(λ ⊗ ν ) under all translations of the product group R × T and under the symmetry of this group. If for a set Z  ∈ S  , we have simultaneously Z  = {x ∈ R : (x, f (x)) ∈ Z1 } and Z  = {x ∈ R : (x, f (x)) ∈ Z2 }, where Z1 and Z2 are some (λ ⊗ ν )-measurable sets, then {x ∈ R : (x, f (x)) ∈ Z1 Z2 } = 0. /

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In view of the (λ ⊗ ν )-thickness of the graph Gr( f ) of f , we infer that (λ ⊗ ν )(Z1 Z2 ) = 0, which also implies that (λ ⊗ ν )(Z1 ) = (λ ⊗ ν )(Z2 ). This equality establishes the correctness of the definition of μ . In a similar way, we prove the countable additivity of μ and its invariance under the group of all isometries of R. Further, if X is an arbitrary λ -measurable set, then we can write X = {x ∈ R : (x, f (x)) ∈ X × T}, X ∈ S ,

μ (X) = (λ ⊗ ν )(X × T) = λ (X) · ν (T) = λ (X) and, therefore, the measure μ extends λ . Finally, for any Borel set B ⊂ T, we have f −1 (B) = {x ∈ R : (x, f (x)) ∈ R × B}, R × B ∈ dom(λ ⊗ ν ), f −1 (B) ∈ S  , which shows that f is measurable with respect to μ and completes the proof of Lemma 3. Lemma 4. Let (G, +) be an infinite commutative divisible group such that, for any nonzero natural number n and for any element g ∈ G, the equation nx = g has at most countably many solutions in G. Let A be an arbitrary subset of G. There exists a subgroup [A] of G satisfying the following conditions: 1) A ⊂ [A]; 2) card([A]) = card(A) + ω ; 3) for each nonzero natural number n and for each element g ∈ [A], all solutions of the equation nx = g belong to [A] and, in particular, [A] is a divisible subgroup of G. Proof. Let B be any countably infinite subset of G. Construct by recursion an infinite sequence (G0 , G1 , ..., Gk , ...)

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of subgroups of G. First of all, put: G0 = the group generated by A ∪ B. Suppose now that the subgroup Gk has already been defined satisfying the following equality: card(Gk ) = card(A) + ω . Denote by Xk the set of all solutions of the equations nx = g

(n = 1, 2, ..., g ∈ Gk ).

Taking into account the assumption of the lemma, we get card(Xk )  card(Gk ) + ω = card(A) + ω . Let us put: Gk+1 = the group generated by the set Gk ∪ Xk . Proceeding in this way, we obtain the increasing sequence G0 ⊂ G1 ⊂ ... ⊂ Gk ⊂ ... of subgroups of G. Finally, denote [A] = ∪{Gk : k < ω }. A straightforward verification shows that [A] is the required group. Remark 1. Lemma 4 has a direct analogue for infinite non-commutative divisible groups (G, ·) and the proof of that analogue can be carried out similarly to the proof of Lemma 4. Lemma 5. There exists a group homomorphism f :R→T possessing the following properties: 1) the graph of f is thick with respect to the product measure λ ⊗ ν ; 2) for any uncountable Borel subset Y of R and for any Borel mapping h : Y → T, we have the inequality card({x ∈ R : f (x) = h(x)}) < c. Proof. We shall construct the required homomorphism f by using the method of transfinite recursion (cf. the proof of Theorem 2 from Chapter 13). Let α denote the least ordinal number of cardinality c and let  be an arbitrary wellordering of R isomorphic to α .

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Let {hξ : ξ < α } be an enumeration of all Borel mappings acting from uncountable Borel subsets of R into T. Let {Zξ : ξ < α } be an enumeration of all those Borel subsets of R × T whose (λ ⊗ ν )measure is strictly positive. We may assume, without loss of generality, that the range of the partial family {Zξ : ξ < α , ξ is an odd ordinal} coincides with the range of the entire family {Zξ : ξ < α }. Under this notation, we are going to define four α -sequences {xξ : ξ < α }, {yξ : ξ < α }, {Vξ : ξ < α }, { fξ : ξ < α } satisfying the following relations: (a) {xξ : ξ < α } is a Hamel basis of R; (b) for each ordinal ξ < α , the set Vξ is the vector subspace of R, over the field Q of all rational numbers, generated by {xζ : ζ  ξ }; (c) for each ordinal ξ < α , we have the group homomorphism fξ : Vξ → T such that fξ (xξ ) = yξ ; (d) if ζ < ξ < α , then fξ extends fζ ; (e) for each odd ordinal ξ < α , we have (xξ , yξ ) ∈ Zξ ; (f) if ξ < α , then we have fξ (qxξ + v) = hζ (qxξ + v) for all

ζ < ξ , q ∈ Q \ {0}, v ∈ ∪{Vη : η < ξ } such that qxξ + v ∈ dom(hζ ). Suppose that for an ordinal ξ < α , the partial ξ -sequences {xζ : ζ < ξ }, {yζ : ζ < ξ }, {Vζ : ζ < ξ }, { fζ : ζ < ξ }

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have already been constructed. Let us put: V  = ∪{Vζ : ζ < ξ }, f  = ∪{ fζ : ζ < ξ }. Applying (d), we may assert that f  is a group homomorphism acting from V  into T. Now, consider two possible cases. 1. The ordinal ξ is even. In this case, let x be the least element (with respect to ) of R \V  and let A = {hζ (qx + v) : ζ < ξ , q ∈ Q \ {0}, v ∈ V  , qx + v ∈ dom(hζ )} + f  (V  ). According to Lemma 4, we have card([A]) = card(A) + ω  card(ξ ) + ω < c. Choose an element y ∈ T \ [A] and put xξ = x, yξ = y. Taking into account the fact that T is a divisible group, we can extend f  to a group homomorphism fξ : Vξ → T satisfying the condition fξ (xξ ) = yξ . 2. The ordinal ξ is odd. In this case, we apply the classical Fubini theorem to the set Zξ and choose an element x ∈ R \ V  such that

ν ({t ∈ T : (x,t) ∈ Zξ }) > 0. Further, we choose an element y ∈ {t ∈ T : (x,t) ∈ Zξ } \ [A], where A is determined as in the previous case, and put again xξ = x, yξ = y. Obviously, we have (xξ , yξ ) ∈ Zξ . As above, using the divisibility of T, we define a group homomorphism fξ : Vξ → T extending f  and satisfying the condition fξ (xξ ) = yξ .

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Proceeding in this manner, we will be able to construct the required four α -sequences. Then we put f = ∪{ fξ : ξ < α }. In view of (a), this f is a group homomorphism acting from R into T. In view of (e), the graph of f is thick in the product space R× T (with respect to the product measure λ ⊗ ν ). Finally, relation (f) implies that the inequality card({x ∈ R : f (x) = hξ (x)}) < c is satisfied for any ordinal ξ < α . This circumstance also yields that for any set X ⊂ R with card(X) = c, the restriction of f to X is not a Borel mapping. To show the last fact, suppose otherwise, i.e., suppose that f |X is Borel. Then, according to a well-known statement from classical descriptive set theory, the function f |X can be extended to a T-valued Borel function f ∗ defined on a Borel subset of R (see, e.g., [58], [99], [148], or Exercise 16 from Chapter 8). Since the relation card(dom( f ∗ )) = c holds true, we conclude that f ∗ = hξ for some ordinal ξ < α . But this circumstance is impossible in view of the above-mentioned inequality. Lemma 5 has thus been proved. Theorem 1. There exists a Sierpi´nski-Zygmund function

χ :R→R measurable with respect to some extension μ of λ which is invariant under all isometric transformations of R. Proof. Let f be as in Lemma 5. By virtue of Lemma 3, f is measurable with respect to a certain measure μ on R which extends λ and is invariant under the group of all isometries of R. Let us put

χ = ψ ◦ f, where ψ : T → R denotes the Borel mapping of Lemma 2. We assert that χ is the required Sierpi´nski-Zygmund function. Indeed, χ is a μ -measurable function as the composition of two functions, first of which is

μ -measurable and the second one is Borel.

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Let X be a subset of R with card(X) = c. It suffices to show that the restriction χ |X is not a Borel mapping. Suppose otherwise, i.e., suppose that χ |X is Borel. Let φ : R → T denote the canonical epimorphism defined by

φ (x) = (cos(x), sin(x)) (x ∈ R). By virtue of Lemma 2, the composition φ ◦ ψ coincides with the identity transformation of T. Therefore, the mapping

φ ◦ χ |X = φ ◦ ψ ◦ f |X = f |X must be Borel, which is impossible in view of Lemma 5. The contradiction obtained ends the proof of our theorem. Remark 2. We cannot assert that the function χ of Theorem 1 is additive. In this connection, the following question seems to be interesting. Does there exist an additive Sierpi´nski-Zygmund function acting from R into R and measurable with respect to some translation-invariant extension of λ ? We do not know an answer to the posed question. In this context, notice that by applying a method analogous to the described above, it is possible to prove that there exists an additive Sierpi´nski-Zygmund function acting from R into R and measurable with respect to some translation-quasi-invariant extension of λ (see Exercise 3 below). Replacing R by a Polish topological vector space equipped with a nonzero σ -finite diffused Borel measure, we come to a certain analogue of Theorem 1. Theorem 2. Let E = {0} be a Polish topological vector space and let μ be a σ -finite diffused Borel measure on E. There exist a measure μ  on E extending μ and a Sierpi´nskiZygmund function χ : E → R measurable with respect to μ  . Moreover, if μ is invariant (quasi-invariant) under a group G ⊂ E, then μ  can also be chosen invariant (quasi-invariant) under the same G. The proof of this statement is similar to the proof of Theorem 1 and is left to the reader. Remark 3. It is well known that if E is an infinite-dimensional Polish topological vector space, then there exists no nonzero σ -finite Borel measure on E quasi-invariant under the group of all translations of E (see Chapter 4). On the other hand, it should be mentioned that, for a certain E, it is possible to construct a nonzero σ -finite Borel measure on E invariant under an everywhere dense vector subspace of E (see, e.g., [110], [115], or Chapter 3).

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We would like to finish this chapter with a statement which once again underlines bad behavior of all Sierpi´nski-Zygmund functions from the points of view of the Baire property and Lebesgue measurability. Theorem 3. Let h : R → R be an arbitrary Sierpi´nski-Zygmund function. The following two assertions are valid: 1) if every set in R of cardinality strictly less than c is of first category, then for any second category set X ⊂ R, the restriction h|X cannot be extended to a function h∗ : R → R possessing the Baire property; 2) if every set in R of cardinality strictly less than c is of Lebesgue measure zero, then for any set Y ⊂ R of strictly positive outer Lebesgue measure, the restriction h|Y cannot be extended to a function h∗ : R → R measurable in the Lebesgue sense. The proof of this statement is left to the reader (cf. Examples 7 and 8 from Chapter 1). EXERCISES 1. Deduce a weaker form of Lemma 2 (requiring only the Borel measurability of ψ ) directly from the theorem of Kuratowski and Ryll-Nardzewski stating the existence of measurable selectors of multi-valued functions (see Appendix 2). 2. Prove the assertion of Remark 1, that is establish an analogue of Lemma 4 for infinite non-commutative divisible groups (G, ·). 3. Show that there exists a function f :R→R satisfying the following conditions: (a) f is additive; (b) f is a Sierpi´nski-Zygmund function; (c) f is measurable with respect to some extension μ of λ which is quasi-invariant under the group of all isometries of R. For this purpose, equip R with a probability measure ν equivalent to λ and define f by the scheme described in the proof of Lemma 5. Further, for checking the R-quasi-invariance of μ , apply to the product measure λ ⊗ ν Theorem 2 from Chapter 4. 4. Assuming Martin’s Axiom, show that there exists a function g:R→R

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satisfying the following conditions: (a) g is a Sierpi´nski-Zygmund function; (b) for each set X ⊂ R with card(X) = c, the restriction g|X is relatively measurable with respect to the class of all nonzero σ -finite diffused measures on R; (c) g is measurable with respect to some extension μ of λ which is invariant under the group of all isometries of R. For this purpose, use an argument similar to the proof of Theorem 1 of this chapter and try to construct the required g in such a way that ran(g) would be a generalized Sierpi´nski subset of R (see Exercise 14 of Appendix 1). 5. Give a proof of Theorem 2 of this chapter by applying an argument analogous to the proof of Theorem 1 and taking into account Theorem 2 from Chapter 4. 6. Give a proof of Theorem 3. In connection with the last exercise, notice once more that according to a result of Novikov [188], for any second category set X ⊂ R, there exists a function f : X → R which does not admit an extension f ∗ : R → R having the Baire property and with dom( f ∗ ) = R. The assertion of Theorem 3 is much stronger and shows that, under Martin’s Axiom, there exists a universal function h : R → R of Sierpi´nski-Zygmund type such that for any second category set X ⊂ R, the restriction h|X cannot be extended to a function acting from R into R and having the Baire property. However, certain additional set-theoretical assumptions are needed for the validity of Theorem 3.

Chapter 15

A nonseparable extension of the Lebesgue measure without new null-sets

Let E be a set, S be a σ -algebra of subsets of E containing all singletons in E, and let μ be a nonzero σ -finite continuous, i.e., vanishing at the singletons measure on S . We recall from Chapter 2 that the general measure extension problem is to extend μ to a maximally large class of subsets of E. According to Ulam’s theorem (see, e.g., [69], [91], [192], [238], or Appendix 1), it is consistent with the axioms of contemporary set theory that the domain of any extension μ  of μ cannot coincide with the power set P(E) of E. For instance, this circumstance holds if card(E) is strictly smaller than the first inaccessible cardinal number. Much stronger results of purely set-theoretical flavor can be found in [10], [69], [91], [150], and [231]. Consequently, there always exists a set X ⊂ E such that X ∈ dom(μ  ). Then an easy argument shows that μ  can be extended to a measure μ  so that X becomes μ  -measurable (see Chapter 2 or Example 2 below). Thus, assuming that there are no large uncountable cardinals, we can conclude that there are no maximal extensions of the original measure μ . In this context, a result on extensions of measures should be recalled, which states that if {X j : j ∈ J} is an arbitrary partition of E, then there exists a measure ν on E extending μ and satisfying the relation {X j : j ∈ J} ⊂ dom(ν ). For the proof of this result, see [1], [13], or Chapter 2. Notice that the measures under consideration in [1] and [13] are assumed to be probability ones but the same argument works for arbitrary σ -finite measures (see the proof of Theorem 2 from Chapter 2). Actually, we do not need the above result in our further considerations but would like to mention one of its immediate corollaries. Namely, if a finite family of subsets of E is A.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7_15, © 2009 Atlantis Press/World Scientific

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given, then we can always extend μ to a measure which measures all these subsets. Indeed, it suffices to consider the finite partition of E generated by all members of the given family. An analogous assertion fails to be true for arbitrary countable families of subsets of E, i.e., if we have a countable family {Xn : n ∈ N} of subsets of E, then we cannot assert, in general, that there exists a measure ν on E extending μ and satisfying the inclusion {Xn : n ∈ N} ⊂ dom(ν ). Moreover, the existence of a Luzin set L on the real line R having the cardinality of the continuum (see Chapters 5 and 12) implies that there is a countably generated σ -algebra of subsets of R containing all singletons and not admitting any nonzero σ -finite continuous measure. Indeed, it is not difficult to show that every σ -finite continuous Borel measure given on a separable metric space is concentrated on some first category subset of the space (see Lemma 2 of Appendix 3). Since all first category subsets of L are at most countable, it follows that L does not admit a nonzero σ -finite continuous measure on the countably generated Borel σ -algebra of L. Now, using a one-to-one correspondence between L and R, we obtain the required σ -algebra of subsets of R. We know that the same L is a small subset of R because it has outer measure zero with respect to any σ -finite continuous Borel measure on R. In other words, L turns out to be a universal measure zero subspace of R. Recall that the existence of Luzin sets needs some additional set-theoretical axioms. For instance, the Continuum Hypothesis readily implies that there are Luzin subsets of R. Similarly to this circumstance, Martin’s Axiom implies that there are so-called generalized Luzin subsets of R (see Chapter 12 and Exercise 14 of Appendix 1). In this connection, it should be underlined that the existence of analogous small subsets of R having cardinality ω1 , where ω1 is the least uncountable cardinal, can be established within the theory ZFC (see [69], [79], [148], [172], [197], [250], or Appendix 1). This fact implies that there exists a certain countably generated σ -algebra of subsets of ω1 separating all points in ω1 and not admitting a nonzero σ -finite diffused measure on it. Once again, we would like to stress that the latter important result is valid within ZFC theory (see Theorem 5 of Appendix 1). If E is a nonempty set, S is a σ -algebra of subsets of E and μ is a measure on E with dom(μ ) = S , then we usually say that the triplet (E, S , μ ) is a measure space. All measures considered below are assumed to be nonzero, σ -finite, and continuous. If the need arises and without loss of generality, we can additionally suppose that a measure under consideration is also complete. Indeed, if necessary we may always replace a given

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measure by its completion. For any nonzero complete measure μ on a set E, we denote by I (μ ) the σ -ideal of all

μ -measure zero sets. For the sake of brevity, we shall say in the sequel that all members of I (μ ) are μ -null-sets in E. The symbol μ ∗ (respectively, μ∗ ) denotes the outer (respectively, inner) measure associated with μ . The symbol λ (= λ1 ) stands for the Lebesgue measure on R (= R1 ). Recall that λ is a complete measure; however, sometimes it is convenient to denote by the same symbol λ the restriction of the Lebesgue measure to the Borel σ -algebra B(R) of R. As usual, the symbols ω (= N) and c denote, respectively, the least infinite cardinal and the cardinality of the continuum. In this chapter we will be dealing with certain nonseparable extensions of the Lebesgue measure λ . For the definition of separable and nonseparable measure spaces, see Exercise 12 of Chapter 1. First of all, let us briefly discuss several typical examples concerning extensions of various

σ -finite measures. Example 1. A well-known method of extending a given σ -finite complete measure μ is based on adding to I (μ ) some new sets which are nonmeasurable with respect to μ and whose inner μ -measure is equal to zero (cf. [234], [235], and Chapter 2). Proceeding in this way, we come to a larger σ -ideal I  which properly contains I (μ ) and all of whose elements are of inner μ -measure zero. Then we consider the σ -algebra S  generated by dom(μ ) and I  . As seen in Chapter 2, any element U of this σ -algebra admits a representation U = (X ∪Y ) \ Z, where X ∈ dom(μ ), Y ∈ I  and Z ∈ I  . We define a functional μ  on S  by the formula

μ  (U) = μ  ((X ∪Y ) \ Z) = μ (X). A straightforward verification shows that the functional μ  is well defined in the sense that the value μ  (U) does not depend on a representation of U in the above-mentioned form and

μ  is also a measure on S  extending the initial measure μ . In addition, all members of I  become sets of μ  -measure zero. Example 2. The method of extending measures described in Example 1 has a weak side. Indeed, from the view-point of the theory of Boolean algebras, the complete measure μ and its extension μ  obtained in the previous example are the same. Nevertheless, by slightly

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changing the above method, we can achieve some difference between a measure and its extension if both of them are considered on the corresponding quotient Boolean algebras. For this purpose, let us take any set T ⊂ E nonmeasurable with respect to a complete measure μ . Obviously, we must have

μ∗ (T ) < μ ∗ (T ). If T0 denotes a μ -measurable kernel of T and T1 stands for a μ -measurable hull of T , then

μ (T1 \ T0 ) > 0 and the set T \ T0 regarded as a subset of T1 \ T0 satisfies the equalities μ∗ (T \ T0 ) = μ∗ ((T1 \ T0 ) \ (T \ T0)) = 0. So we may assume (replacing, if necessary, T by T \ T0 and E by T1 \ T0) that

μ∗ (T ) = μ∗ (E \ T ) = 0. Let S  denote the σ -algebra of all those subsets U of E which admit a representation in the form U = (X ∩ T ) ∪ (Y ∩ (E \ T )), where X ∈ dom(μ ) and Y ∈ dom(μ ). Define a functional μ  on S  by the equality

μ  (U) = (1/2)(μ (X) + μ (Y)). As earlier, μ  is well defined and turns out to be a measure on S  extending μ . Also, since T is μ  -measurable, μ  strictly extends μ . Moreover, we see that the quotient Boolean algebra associated with μ is properly contained in the quotient Boolean algebra associated with μ  . At the same time, we have the relation I (μ  ) = I (μ ), i.e., the measure μ  does not produce new null-sets. Further, considering more thoroughly the extensions of μ described in Example 1 and Example 2, we readily observe that both of these constructions do not essentially change the metrical structure of μ . Indeed, we can see that the metric space associated with a measure μ and the metric space associated with its extension μ  obtained by using any of the two described constructions have the same topological weight. Therefore, if the original measure μ is separable (i.e., if its metric space is separable), then the extended measure μ  is separable, too. In this context, we would like to recall that Kakutani and Oxtoby [95] gave a construction of a certain nonseparable extension of the Lebesgue measure λ on the real line R. Another

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construction of this kind was presented by Kodaira and Kakutani [141]. It is remarkable that both those extensions of λ are invariant under the group of all isometries of R. It should be observed that the topological weight of the metric space canonically associated with the extension of λ obtained by Kakutani and Oxtoby in [95] is equal to 2c . This extension necessarily yields new null-sets. In other words, there appear null-sets which are not of Lebesgue measure zero because the method of Kakutani and Oxtoby uses an uncountable σ -independent family of subsets of R such that the intersection of any countable subfamily of this family is a Lebesgue nonmeasurable set. Actually, the above-mentioned intersection becomes a null-set with respect to the extended measure (for more details, see [83], [108], or Exercise 4 of this chapter). The topological weight of the metric space canonically associated with the extension of λ obtained by Kodaira and Kakutani in [141] is equal to c. By applying their method, we can also obtain new null-sets (see, for instance, Theorem 2 below). In this connection, the following question seems to be interesting. Does there exist a nonseparable extension of λ all null-sets of which are precisely the λ null-sets? The question is of interest in view of the following circumstance. In many topics of real analysis, measure theory, and probability theory the inner structure of a measure under consideration does not play any role and only the induced concept ”almost everywhere” is essential. The standard example of this type is the well-known Lebesgue theorem stating that a bounded function f : [a, b] → R is integrable in the Riemann sense if and only if f is continuous almost everywhere on [a, b]. In such a case, f is also λ -measurable because Luzin’s C-property automatically holds for f . The just mentioned Lebesgue theorem does not need the notion of the Lebesgue measure. For its proof, it completely suffices to apply the notion of a null-set in the Lebesgue sense. Of course, numerous other examples of this kind can be pointed out. In this chapter, our goal is to demonstrate that under the Continuum Hypothesis there exists a nonseparable extension of λ which yields no new null-sets. It should be noticed that our argument may be regarded as a certain combination of the method of Kodaira and Kakutani [141] with the method of Luzin [159] by means of which he proved the existence of his set on the real line R (see, e.g., Chapter 12 or Exercise 14 from Appendix 1).

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We begin with some preliminary considerations which were already utilized in preceding parts of the book. Let (E1 , S1 , μ1 ) and (E2 , S2 , μ2 ) be two measure spaces such that μ1 is σ -finite and μ2 is a probability measure. Let f : E1 → E2 be a mapping and let Gr( f ) denote the graph of f . Suppose that Gr( f ) is (μ1 ⊗ μ2 )-thick in the product set E1 × E2, i.e., suppose that (μ1 ⊗ μ2 )∗ ((E1 × E2) \ Gr( f )) = 0. As in preceding sections, for any set Z ∈ dom(μ1 ⊗ μ2 ), let us put Z  = {x ∈ E1 : (x, f (x)) ∈ Z}. Further, introduce the class of sets S1 = {Z  : Z ∈ dom(μ1 ⊗ μ2)} and define the functional

μ1 (Z  ) = (μ1 ⊗ μ2 )(Z) (Z ∈ dom(μ1 ⊗ μ2 )). Then S1 is a σ -algebra of subsets of E1 and the functional μ1 is a measure on S1 extending

μ1 (cf. Chapter 2). In addition to this circumstance, the original mapping f turns out to be measurable with respect to the σ -algebras S1 and S2 , i.e., we have (∀Y ∈ S2 )( f −1 (Y ) ∈ S1 ). The remarks just made are rather simple and, in fact, are familiar to the reader from Chapters 2 and 5. Actually, these remarks are a starting point for our further constructions.

Lemma 1. Under the previous notation, if a measure μ1 is nonzero and a measure μ2 is nonseparable, then the measure μ1 is nonseparable, too. Proof. Take any set X ∈ dom(μ1 ) with 0 < μ1 (X) < +∞. The nonseparability of μ2 implies that for some ε > 0, there exists an uncountable family {Yi : i ∈ I} of μ2 -measurable sets, such that

μ2 (Yi Y j ) > ε

(i ∈ I, j ∈ I, i = j).

We leave the checking this fact to the reader (cf. also Exercises 1 and 2 from Chapter 17). Further, we may write

μ1 ((X × Yi ) (X × Y j ) ) = (μ1 ⊗ μ2)(X × (Yi Y j )) = μ1 (X) · μ2 (Yi Y j ) > μ1 (X) · ε

(i ∈ I, j ∈ I, i = j),

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thus it follows that μ1 is nonseparable which ends the proof of Lemma 1. Let J be a fixed set with card(J) = c. For any index j ∈ J, denote by ν j the restriction of the Lebesgue measure λ to the Borel

σ -algebra of the closed unit interval [0, 1]. Thus, ν j is a Borel diffused probability measure on [0, 1]. Let ν stand for the product of the family of measures {ν j : j ∈ J}, i.e.,

ν = ⊗ j∈J ν j . The following two auxiliary propositions are straightforward but, for the sake of completeness, we provide the short proofs here. Lemma 2. The cardinality of dom(ν ) is equal to c and the topological weight of the metric space canonically associated with ν is also equal to c. In particular, ν is a nonseparable measure. Proof. First, let us remark that any set Z ∈ dom(ν ) can be represented in the form B × [0, 1]J\J0 , where J0 is a countable subset of J (certainly, depending on Z) and B is a Borel subset of [0, 1]J0 . Taking into account the equality cω = c, this remark readily implies that card(dom(ν )) = c. Further, for each index j ∈ J, let us denote Z j = [0, 1/2] j × [0, 1]J\{ j} . Then we have

ν (Z j Zk ) = 1/2 ( j ∈ J, k ∈ J, j = k), thus it follows that the topological weight of the metric space canonically associated with

ν is equal to c which ends the proof of Lemma 2. Lemma 3. Let a be an infinite cardinal number satisfying the relation aω = a, and let (E1 , S1 ) and (E2 , S2 ) be two measurable spaces such that card(S1 )  a, card(S2 )  a. Then the cardinality of the product σ -algebra S1 ⊗ S2 does not exceed a either.

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Proof. The product σ -algebra S1 ⊗ S2 is generated by the family T of all sets Z having the form Z = X × Y , where X ∈ S1 and Y ∈ S2 . The cardinality of this family does not exceed a · a = a. Consequently, card(S1 ⊗ S2 ) = card(σ (T ))  aω . Taking into account the equality aω = a, we come to the required result. The next lemma plays a key role in our further consideration. Lemma 4. Assume the Continuum Hypothesis. Let J be a set whose cardinality is equal to c and let

ν = ⊗{ν j : j ∈ J}, where ν j ( j ∈ J) is the restriction of λ to the Borel σ -algebra of [0, 1]. There exists a mapping f : R → [0, 1]J satisfying the following relations: (1) the graph Gr( f ) of f is thick with respect to the product measure λ ⊗ ν ; (2) for any (λ ⊗ ν )-measure zero set Z, the set Z  = {x : (x, f (x)) ∈ Z} is of λ -measure zero. Proof. The required mapping f will be constructed by transfinite recursion. In what follows the symbol λ0 stands for the restriction of the Lebesgue measure on R to the Borel σ -algebra of R. Let α denote the least ordinal number of cardinality c. According to our assumption,

α = ω1 and hence card(ξ )  ω for each ordinal ξ < α . Let  denote a well-ordering of R which is isomorphic to α . Applying Lemmas 2 and 3, we easily derive the equality card(dom(λ0 ⊗ ν )) = c. Let {Zξ : ξ < α } be the family of all those (λ0 ⊗ ν )-measurable sets whose measure is strictly positive. Without loss of generality, we may suppose that the range of this family coincides with the range of the family {Zξ : ξ < α , ξ is odd}.

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Let {Tξ : ξ < α } be an enumeration of all those (λ0 ⊗ ν )-measurable sets whose measure is equal to zero. We are going to construct an α -sequence {(xξ , yξ ) : ξ < α } of points of the product space R × [0, 1]J . Suppose that for an ordinal ξ < α , the partial ξ -sequence {(xζ , yζ ) : ζ < ξ } has already been constructed. Consider two possible cases. (a) The ordinal ξ is even. In this case, denote by x the -least element of the set R \ {xζ : ζ < ξ } and for each ordinal

ζ < ξ , define Tζ (x) = {y : (x, y) ∈ Tζ }. Also, introduce a set Ξ by the formula Ξ = {ζ : ζ < ξ & ν (Tζ (x)) = 0}. Obviously, we have the equality

ν (∪{Tζ (x) : ζ ∈ Ξ}) = 0. Consequently, [0, 1]J \ ∪{Tζ (x) : ζ ∈ Ξ} = 0. / Choose a point y ∈ [0, 1]J \ ∪{Tζ (x) : ζ ∈ Ξ} and put (xξ , yξ ) = (x, y). (b) The ordinal ξ is odd. In this case, for each ordinal ζ < ξ , define Tζ0 = {x ∈ R : ν (Tζ (x)) > 0}. Since (λ0 × ν )(Tζ ) = 0, we have λ (Tζ0 ) = 0. Consider the set Zξ . Taking into account the circumstance that (λ0 ⊗ ν )(Zξ ) > 0 and applying the Fubini theorem, we can find a point x ∈ R \ ∪{Tζ0 : ζ < ξ } satisfying the relation ν (Zξ (x)) > 0. Choose a point y ∈ Zξ (x) \ ∪{Tζ (x) : ζ < ξ } and put (x, y) = (xξ , yξ ). Thus, in both cases (a) and (b) we have determined the point (xξ , yξ ) in the product space R × [0, 1]J .

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Proceeding in this manner, we are able to construct the desired α -sequence {(xξ , yξ ) : ξ <

α } of points of R × [0, 1]J . Now, let us put f (xξ ) = yξ

(ξ < α ).

So, a certain partial function f : R → [0, 1]J is defined. We assert that f is the required mapping. Indeed, the equality R = {xξ : ξ < α } holds true because the well-ordering  is isomorphic to α . Therefore, the domain of f coincides with R. The thickness of Gr( f ) is a straightforward consequence of the relation (xξ , yξ ) ∈ Zξ

(ξ < α , ξ is odd).

Finally, let us show that the inclusion {x : (x, f (x)) ∈ Tξ } ⊂ Tξ0 ∪ {xζ : ζ  ξ } is valid for each ordinal ξ < α . Take any element (x, f (x)) ∈ Tξ . According to our construction, we have (x, f (x)) = (xη , yη ) for some ordinal number η < α . Consequently, (xη , yη ) ∈ Tξ , yη ∈ Tξ (xη ). If η  ξ , then there is nothing to prove. Suppose now that ξ < η and consider two cases. (i) ν (Tξ (xη )) = 0. If η is even, then our construction yields yη ∈ Tξ (xη ). If η is odd, then our construction also yields yη ∈ Tξ (xη ). Therefore, this case is impossible for ξ < η . (ii) ν (Tξ (xη )) > 0. This relation immediately implies xη ∈ Tξ0 which yields at once the desired result. The proof of Lemma 4 is thus completed. Theorem 1. Under the Continuum Hypothesis, there exists a nonseparable measure λ  on the real line R extending λ and such that I (λ  ) = I (λ ). Proof. In view of Lemma 4, there exists a mapping f : R → [0, 1]J

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such that the following two relations are satisfied: (1) the graph of f is thick in the product space R × [0, 1]J with respect to the product measure λ ⊗ ν ; (2) for each (λ ⊗ ν )-measure zero set Z, the set Z  = {x : (x, f (x)) ∈ Z} is of λ -measure zero. Now, for any Z ∈ dom(λ0 ⊗ ν ), we put Z  = {x : (x, f (x)) ∈ Z},

λ  (Z  ) = (λ0 ⊗ ν )(Z). As said earlier, the functional λ  is well defined and is a measure extending λ0 . Obviously, the completion of λ  extends the Lebesgue measure on R. We preserve the same notation for the completion of λ  . By virtue of Lemmas 1 and 2, λ  is a nonseparable measure. More precisely, we can assert that the topological weight of the metric space canonically associated with λ  is equal to c. Finally, relation (2) yields at once the desired equality I (λ  ) = I (λ ), which ends the proof of the theorem. Remark 1. Theorem 1 was established by assuming the Continuum Hypothesis. It must be noticed that the existence of a nonseparable extension of λ without new null-sets cannot be proved within ZFC theory (cf. Exercise 3 below). In this connection, it should also be underlined that within ZFC there is a proper separable extension of λ whose null-sets are identical with the λ -null-sets (see Example 2 of this chapter). Remark 2. As pointed out at the end of the proof of Theorem 1, the topological weight of the metric space canonically associated with λ  is equal to c. We do not know whether it is possible to show (at least, in some models of set theory) the existence of an extension λ  of λ such that the topological weight of the metric space canonically associated with λ  is strictly greater than c and I (λ  ) = I (λ ). Remark 3. Also, we do not know within ZFC theory whether there exists a translationinvariant proper extension θ of λ such that I (θ ) = I (λ ). Let (E, S , μ ) be a probability measure space. The argument presented before Lemma 1 shows that if a mapping g:R→E

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is given whose graph is (λ ⊗ μ )-thick in R × E, then this mapping produces the measure

λg which is an extension of the Lebesgue measure λ on R. Theorem 2. Suppose that μ is diffused and card(S )  c. Then there exists a mapping h:R→E satisfying the following conditions: (1) the graph Gr(h) of h is (λ ⊗ μ )-thick in the product space R × E; (2) some set of λh -measure zero is thick in R with respect to λ ; in particular, we have I (λh ) = I (λ ).

Proof. Let B be a Bernstein subset of R (see [12], [30], [40], [148], [176], [192], or Chapters 2 and 5). We recall that according to the definition of Bernstein sets, both B and R \ B are totally imperfect subsets of R and this circumstance readily implies the equalities card(B) = card(R \ B) = c,

λ∗ (B) = λ∗ (R \ B) = 0. In particular, the sets B and R \ B are not measurable in the Lebesgue sense and, moreover, both of them are nonmeasurable with respect to a wide class of nonzero σ -finite diffused measures on R (in this connection, see Theorem 3 from Chapter 12). Choose a point y ∈ E and take the constant mapping h0 : B → {y}. By using the method of transfinite recursion, a mapping h1 : R \ B → E \ {y} can easily be constructed such that the graph Gr(h1 ) is (λ ⊗ μ )-thick in the product space R × E (cf. the proof of Theorem 1). Let h stand for the common extension of h0 and h1 . Clearly, the graph of h is also (λ ⊗ μ )-thick in R × E. Consider the set R × {y}. Since the measure λ is σ -finite and μ ({y}) = 0, we get (λ ⊗ μ )(R × {y}) = 0. Consequently, we must have

λh ({x : (x, h(x)) ∈ R × {y}}) = 0.

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Therefore, {x : (x, h(x)) ∈ R × {y}} ∈ I (λh ). But it is easy to check the validity of the relation B = {x : (x, h(x)) ∈ R × {y}} from which it follows that the set {x : (x, h(x)) ∈ R × {y}} is thick with respect to λ . Moreover, the above-mentioned set is a Bernstein subset of R and hence is thick with respect to any σ -finite continuous Borel measure on R, which completes the proof of Theorem 2. In exercises below we give some additional information about nonseparable extensions of

σ -finite measures. This topic will be continued in subsequent sections but from another point of view (see especially the next two chapters). EXERCISES 1. Let E be a metrizable topological space and let μ be a σ -finite Borel measure on E such that μ (U) > 0 for every nonempty open set U ⊂ E. Show that the following two assertions are equivalent: (a) E is a separable space; (b) μ is a separable measure. 2∗ . Investigate the question whether the assertion of Theorem 1 remains valid under Martin’s Axiom instead of the Continuum Hypothesis. 3∗ . Let (E, S , μ ) be a σ -finite complete measure space. We introduce the following notation (cf. [40]): add(I (μ )) = the smallest cardinal number of a subfamily of I (μ ) whose union does not belong to I (μ ); cov(I (μ )) = the smallest cardinal number of a covering of E with members of I (μ ). Let T (= S1 ) as usual denote the unit torus. Consider the compact topological group Tω1 equipped with the completion ν of its Haar probability measure. Prove that add(I (ν )) = ω1 . Deduce from this equality that the conjunction of Martin’s Axiom and the negation of the Continuum Hypothesis implies the relation add(I (ν )) < add(I (λ )) = c.

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It can be shown that the inequality cov(I (ν )) < cov(I (λ )) also does not contradict ZFC theory. The latter circumstance enables to demonstrate that the assertion of Theorem 1 of this chapter cannot be established within ZFC. 4∗ . Verify that the method of extending σ -finite invariant measures to nonseparable ones, due to Kakutani and Oxtoby (see [83], [95], or Exercise 18 for Chapter 3), necessarily produces new null-sets. 5. Suppose that E is a hereditarily Lindel¨of topological space, i.e., for any open set V ⊂ E, if V is represented in the form V = ∪{Vi : i ∈ I}, where all Vi (i ∈ I) are also open in E, then there exists a countable set I  ⊂ I such that V = ∪{Vi : i ∈ I  }. Further, suppose that μ is a finite Borel measure on E which is outer regular, i.e., for any Borel set X ⊂ E, we have

μ (X) = inf{μ (V ) : X ⊂ V, V is open in E}. Show that the topological weight of the metric space canonically associated with μ does not exceed w(E) + ω , where w(E) as usual denotes the topological weight of E. For this purpose, fix a base {V j : j ∈ J} of open sets in the given space E such that card(J) = w(E) and consider the family L of the unions of all finite subfamilies of {V j : j ∈ J}. Verify that the canonical image of L is everywhere dense in the metric space associated with μ . 6. Let E be a compact topological space with w(E)  ω and let μ be a finite Radon measure on the Borel σ -algebra B(E). Prove that the topological weight of the metric space canonically associated with μ does not exceed w(E). 7. Let E be a topological space and let D ⊂ E be an everywhere dense subset of E. Denote by Cb (E, R) the Banach space (with respect to the norm of uniform convergence) of all real-valued bounded continuous functions on E. Show that the inequality card(Cb (E, R))  2card(D)+ω holds true. Infer from this fact that card(Cb (E, R))  2w(E)+ω ,

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where w(E) as earlier stands for the topological weight of E. 8. Let E be a completely regular topological space and let Cb (E, R) be as in the previous exercise. Prove that w(E)  w(Cb (E, R)). Argue in the following manner. Take any family { f j : j ∈ J} ⊂ Cb (E, R) which is everywhere dense in Cb (E, R) and for which card(J) = w(Cb (E, R)). Check that this family separates the points and closed subsets of E. Deduce from this circumstance that E can be topologically embedded in the Tychonoff cube [0, 1]card(J) and conclude that w(E)  card(J) = w(Cb (E, R)). Infer from the said above that if E is a compact space, then the following two assertions are equivalent: (a) the Banach space C(E, R) of all real-valued continuous functions on E is separable; (b) E has a countable base (equivalently, E is metrizable). 9∗ . Let E be a nonempty topological space equipped with a nonzero finite Borel measure

μ such that μ (U) > 0 for every nonempty open set U ⊂ E. Suppose, in addition, that card(Cb (E, R)) > c. Prove that there exists a strictly positive integer n and an uncountable family { f j : j ∈ J} ⊂ Cb (E, R) such that

 E

( f j (x) − fi (x))2 d μ (x)  1/n

for any two distinct indices j ∈ J and i ∈ J. Deduce from this fact that the Hilbert space L2 (μ ) of all real-valued square integrable μ -measurable functions is nonseparable and, consequently, the measure μ is nonseparable, too. For this purpose, use the Erd¨os-Rado combinatorial theorem of Ramsey type (see [61] or [144]). 10. Let E be a completely regular topological space and let μ be a nonzero finite Borel measure on E such that μ (U) > 0 for every nonempty open set U ⊂ E. Show that if w(E) > c, then μ is a nonseparable measure. For this purpose, apply the result of the previous exercise. In particular, if (E, ·) is a compact topological group satisfying the relation w(E) > c, then the Haar probability measure on E is nonseparable. 11∗ . Let (E, S , μ ) be a probability measure space such that:

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(a) μ is universal, i.e., dom(μ ) = P(E); (b) μ is a non-atomic measure. Prove that μ cannot be a perfect measure. In order to establish this fact, utilize Exercise 17 for Appendix 1. In particular, conclude from the said above that no universal extension of the Lebesgue measure λ can be perfect. In connection with this result, a deep statement due to Gitik and Shelah should be mentioned, according to which the so-called Maharam type of any universal extension of λ is necessarily strictly greater than c (see [73]; cf. also [96]).

Chapter 16

Metrical transitivity and nonseparable extensions of invariant measures

In this chapter we continue our consideration of nonseparable extensions of σ -finite measures. However, here we will be dealing with those nonseparable extensions of σ -finite measures which are invariant with respect to a fixed group of transformations of a base set E, assuming that the cardinality of E is equal to the cardinality of the continuum. As is widely known, metrical transitivity (i.e., ergodicity) plays an extremely important role in various questions of the theory of invariant and quasi-invariant measures. Here we would like to confirm this circumstance by demonstrating that under the Continuum Hypothesis, any nonzero σ -finite metrically transitive left (right) invariant measure on a group of cardinality continuum admits a nonseparable left (right) invariant extension. An application of this result to the left (right) Haar measure on a σ -compact locally compact topological group of the same cardinality is also presented. Various left (right) invariant extensions of the left (right) Haar measure given on a Polish locally compact group are known in the literature. For instance, some nonseparable invariant extension of the Haar probability measure on an uncountable compact metrizable commutative group is thoroughly considered in the monograph by Hewitt and Ross [83]. The construction of this extension imitates the remarkable construction of Kakutani and Oxtoby [95] which was done in a particular case, namely, for the Lebesgue measure ν on the one-dimensional unit torus T = S1 . By using slight modifications, the same construction works for the Lebesgue measure λn on the n-dimensional Euclidean space Rn , where n  1, and, more generally, for the left Haar measure on an uncountable locally compact Polish topological group (cf. [108]). It seems at first sight that specific topological properties of the Haar (respectively, Lebesgue) measure play an important role in the above-mentioned constructions. However, we will demonstrate in the sequel that the situation is different. The main goal of this chapter is to show that topological concepts are inessential for constructing nonsepaA.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7_16, © 2009 Atlantis Press/World Scientific

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rable left (right) invariant extensions of a nonzero σ -finite left (right) invariant metrically transitive measure given on a group of cardinality continuum. In fact, we will show that a dominant role in constructions of such a kind is played by the metrical transitivity of this measure. The notion of metrical transitivity for measures invariant or, more generally, quasi-invariant under transformation groups was already introduced in preceding sections of this book. See, e.g., Chapter 3 or Chapter 9 where the related notion of weak metrical transitivity is also discussed. All measures considered in the present chapter are assumed to be diffused (continuous), i.e., all of them vanish at the singletons. This assumption is necessary for our further considerations (cf. Example 2 in the end of this chapter). Let E be a set, G be a group of transformations of E and let μ be a nonzero σ -finite G-invariant (G-quasi-invariant) measure on E. Recall that μ is metrically transitive (i.e., ergodic) if for every μ -measurable set X ⊂ E with μ (X) > 0, there exists a countable family {gi : i ∈ I} ⊂ G such that

μ (E \ ∪{gi(X) : i ∈ I}) = 0. It is well known that the metrical transitivity of an invariant measure μ is closely connected with its uniqueness property. Indeed, if μ has the uniqueness property, then μ is necessarily metrically transitive. Conversely, if G is uncountable and acts freely in E and μ is complete, then the metrical transitivity of μ turns out to be equivalent to the uniqueness property (for more details, see [105], [108], [115], and Exercise 1 of this chapter). Example 1. The left Haar measure μ on a σ -compact locally compact group (G, ·) is metrically transitive. More precisely, let H be an arbitrary everywhere dense subgroup of G. Then the same μ considered as a left H-invariant measure is also metrically transitive (see, for instance, [83] and [115]). Recall that a set Y ⊂ E is almost G-invariant with respect to μ if the equality

μ ∗ (g(Y )Y ) = 0 holds true for all transformations g ∈ G. Clearly, Y ⊂ E is almost G-invariant with respect to μ if and only if E \ Y is almost Ginvariant with respect to μ .

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Moreover, the union of a countable family of almost G-invariant sets with respect to μ is almost G-invariant with respect to μ , too. Therefore, the class of all almost G-invariant sets with respect to μ forms a G-invariant

σ -algebra of subsets of E. Recall also that a set Z ⊂ E is μ -thick in E if μ∗ (E \ Z) = 0. We need two auxiliary propositions. Lemma 1. Let E be a set, G be a group of transformations of E and let μ be a nonzero

σ -finite complete G-quasi-invariant metrically transitive measure on E. Suppose that a set Y ⊂ E is almost G-invariant with respect to μ . Then at least one of the following three relations is valid: 1) μ (Y ) = 0; 2) μ (E \ Y ) = 0; 3) both sets Y and E \ Y are thick with respect to μ and, consequently, both of them are nonmeasurable with respect to μ . Proof. Suppose that 3) does not hold, i.e., at least one of the sets Y and E \ Y is not thick with respect to μ . Without loss of generality, we may assume that Y is not thick with respect to μ . Then there exists a μ -measurable set T of strictly positive μ -measure, such that T ∩Y = 0. / Since μ is metrically transitive, we can find a countable family {gi : i ∈ I} ⊂ G satisfying the relation

μ (E \ ∪{gi (T ) : i ∈ I}) = 0. Since Y is almost G-invariant with respect to μ , we may write

μ (Y (∩{gi (Y ) : i ∈ I})) = 0, thus it follows that

μ (Y ∩ (∪{gi (T ) : i ∈ I})) = 0 and, consequently, μ (Y ) = 0 which ends the proof of Lemma 1. As usual, we denote by ω the first infinite cardinal, by ω1 the first uncountable cardinal, and by c the cardinality of the continuum. Moreover, ω (respectively, ω1 and c) can be identified with the least ordinal number whose cardinality is equal to ω (respectively, to ω1 and c).

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For our further purposes, we need the notion of Ulam’s matrix which is very important in combinatorial set theory, measure theory and set-theoretic topology (see [144], [148], [192], and [238]). Let U be a set of cardinality ω1 . Recall that a double family {Un,ξ : n < ω , ξ < ω1 } of subsets of U is an Ulam matrix over U if the following two conditions are satisfied: (*) for any n < ω , the family {Un,ξ : ξ < ω1 } is disjoint; (**) for any ξ < ω1 , the set U \ ∪{Un,ξ : n < ω } is at most countable. The existence of Ulam matrices was first proved by Ulam in his classical work [238] (see also [144], [192], and Exercise 18 for Appendix 1). Ulam’s result directly implies that ω1 is not a real-valued measurable cardinal. Some applications of Ulam matrices to the theory of invariant and quasi-invariant measures are presented in [108], [115], and [119]. The next lemma is crucial for our further considerations. Lemma 2. Let E be a set with card(E) = ω1 and let G be a group of transformations of E such that card(G)  ω1 . Suppose that μ is a nonzero σ -finite G-quasi-invariant metrically transitive measure on E. Then there exists a disjoint family {Yξ : ξ < ω1 } of subsets of E such that: 1) each set Yξ (ξ < ω1 ) is thick with respect to μ ; 2) for any Ξ ⊂ ω1 , the set ∪{Yξ : ξ ∈ Ξ} is almost G-invariant with respect to μ . Proof. We may assume, without loss of generality, that the given measure μ is complete. Consider the partition {Oi : i ∈ I} of E into G-orbits. Only two cases are possible. 1. All G-orbits are of μ -measure zero. In this case, we necessarily have the equality card(I) = ω1 because our measure μ is nonzero. Let {In,ξ : n < ω , ξ < ω1 } denote an Ulam matrix over I. We define the sets Zn,ξ = ∪{Oi : i ∈ In,ξ }

(n < ω , ξ < ω1 ).

Taking into account the conditions (*) and (**), we easily get the following relations: (a) for each n < ω , the family {Zn,ξ : ξ < ω1 } is disjoint; (b) for each ξ < ω1 , the set E \ ∪{Zn,ξ : n < ω } is of μ -measure zero.

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It follows from (a) and (b) that there exist a natural number n and an uncountable set Ξ ⊂ ω1 such that the inequality μ ∗ (Zn,ξ ) > 0 is satisfied for all ordinals ξ ∈ Ξ. Clearly, we also have Zn,ξ ∩ Zn,ζ = 0/ (ξ ∈ Ξ, ζ ∈ Ξ, ξ = ζ ). Further, since our measure μ is metrically transitive and each set Zn,ξ (ξ ∈ Ξ) is Ginvariant, we can conclude (in view of Lemma 1) that all sets Zn,ξ (ξ ∈ Ξ) are thick with respect to μ . Let

φ : ω1 → Ξ be any bijection acting from ω1 onto Ξ. Putting {Yξ : ξ < ω1 } = {Zn,φ (ξ ) : ξ < ω1 }, we come to the required family of subsets of E. 2. There exists at least one index i0 ∈ I for which we have μ ∗ (Oi0 ) > 0. In this case, we obviously obtain card(Oi0 ) = ω1 , card(G) = ω1 . Also, by Lemma 1, the orbit Oi0 is a thick set with respect to our measure μ . Let us denote V = Oi0 and let us fix a point x ∈ V . Clearly, the group G acts transitively on V , i.e., V = G(x) = {g(x) : g ∈ G}. Let us represent the group G in the form G = ∪{Gζ : ζ < ω1 }, where {Gζ : ζ < ω1 } is an increasing (by inclusion) family of subgroups of G satisfying the following relations: (c) Gζ = ∪{Gη : η < ζ } for all ζ < ω1 ; (d) card(Gζ )  ω for all ζ < ω1 . It is not difficult to show the existence of such a representation of G (cf. Exercise 19 from Chapter 3 or Exercise 9 from Chapter 11). Evidently, we may write V = G(x) = ∪{Gζ (x) : ζ < ω1 }. Now, putting Vζ = Gζ (x) \ ∪{Gη (x) : η < ζ }

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for any ζ < ω1 , we see that {Vζ : ζ < ω1 } is a disjoint covering of V by countable sets. In addition, for every subset Θ of ω1 , the set ∪{Vθ : θ ∈ Θ} is almost G-invariant with respect to the given measure μ . Further, consider an Ulam matrix {Un,ξ : n < ω , ξ < ω1 } over the set U = ω1 and define Zn,ξ = ∪{Vζ : ζ ∈ Un,ξ }

(n < ω , ξ < ω1 ).

Again, taking into account the conditions (*) and (**), we derive that: (e) for each n < ω , the family of sets {Zn,ξ : ξ < ω1 } is disjoint; (f) for each ξ < ω1 , the set V \ ∪{Zn,ξ : n < ω } is at most countable and, consequently, the set ∪{Zn,ξ : n < ω } is thick with respect to μ . It follows from the relations (e) and (f) that there exist a natural number n and an uncountable set Ξ ⊂ ω1 such that the inequality μ ∗ (Zn,ξ ) > 0 is valid for all ordinals ξ ∈ Ξ. Using once more the metrical transitivity of μ and applying Lemma 1, we infer that all sets Zn,ξ (ξ ∈ Ξ) are thick with respect to μ . Let

φ : ω1 → Ξ be any bijection acting from ω1 onto Ξ. Putting again {Yξ : ξ < ω1 } = {Zn,φ (ξ ) : ξ < ω1 }, we obtain the required family of subsets of E. Lemma 2 has thus been proved. The next auxiliary proposition is a direct consequence of the previous lemma. Lemma 3. Let α denote the least ordinal number of cardinality c and let E be an arbitrary set with card(E) = c. Let G be a group of transformations of E such that card(G)  c. Suppose that μ is a nonzero σ -finite G-invariant metrically transitive measure on E. Then assuming the Continuum Hypothesis, there exists a disjoint family {Yξ : ξ < α } of subsets of E such that: 1) each set Yξ (ξ < α ) is thick with respect to μ ; 2) for any Ξ ⊂ α , the set ∪{Yξ : ξ ∈ Ξ} is almost G-invariant with respect to μ . Now, we are able to establish the following statement. Theorem 1. Under the assumptions of Lemma 3, there exists a nonseparable G-invariant extension μ  of μ . More precisely, μ  can be chosen so that the Hilbert dimension of the space L2 (μ  ) is equal to 2c .

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Proof. We preserve the notation of Lemma 3. Take a family {Yξ : ξ < α } as in that lemma. According to a result of Tarski (see [83], [108], [150], or Exercise 3 for this chapter), there exists a family {Ξ j : j ∈ J} of subsets of α such that: (1) card(J) = 2c ; (2) for any injective sequence { jk : k < ω } ⊂ J, we have card(∩{Ξjk : k < ω }) = c, where Ξjk = Ξ jk or Ξjk = c \ Ξ jk for all indices jk (k < ω ). Now, for each index j ∈ J, let us define A j = ∪{Yξ : ξ ∈ Ξ j }. In this way we come to the family {A j : j ∈ J} of almost G-invariant subsets of E which are mutually σ -independent (in the generalized sense) with respect to μ . The last phrase means that for any injective sequence { jk : k < ω } ⊂ J, the equality

μ∗ (E \ ∩{Ajk : k < ω }) = 0 holds true, where Ajk = A jk or Ajk = E \ A jk . Now, applying the Kakutani-Oxtoby method of extending invariant measures (see [83], [95], or Exercise 18 from Chapter 3), we can extend our μ to a G-invariant measure μ  so that all sets from the family {A j : j ∈ J} become μ  -measurable. Moreover, the above-mentioned method guarantees that the Hilbert dimension of the space L2 (μ  ) is equal to 2c (see again [83] and [95]) which completes the proof of Theorem 1. Unfortunately, the obtained extension μ  is not metrically transitive because all sets A j ( j ∈ J) are almost G-invariant with respect to μ  . It would be interesting to find a method of extending measures by means of which a given invariant (respectively, quasi-invariant) metrically transitive measure can be extended to a nonseparable invariant (respectively, quasi-invariant) metrically transitive measure. More precisely, we would like to formulate the following question. Problem. Let E be a set with card(E) = c and let G be a group of transformations of E such that card(G)  c. Suppose that μ is a nonzero σ -finite G-invariant metrically transitive measure on E. Does there exist a nonseparable G-invariant metrically transitive measure

μ  on E extending μ ? We do not know an answer to this question. In this context, notice that the method described in the paper by Kodaira and Kakutani [141] preserves the property of metrical transitivity of the Lebesgue measure, but we do not know any purely measure-theoretical version of their method.

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Theorem 1 directly implies the following statement. Theorem 2. Assume the Continuum Hypothesis. Let G be a σ -compact locally compact group of cardinality c and let μ be a left invariant metrically transitive extension of the left Haar measure on G. Then there exists a left invariant extension μ  of μ such that the Hilbert dimension of the space L2 (μ  ) is equal to 2c . Remark 1. It should be mentioned that a measure μ in Theorem 2 can be nonseparable itself (in this connection, see [115]). This circumstance also indicates that Theorem 2 cannot be established only by using the methods developed in [83] and [95]. Remark 2. As pointed out in the beginning of this chapter, the construction of Kakutani and Oxtoby [95] as well as its generalization presented in the monograph [83] by Hewitt and Ross is based on special topological properties of the Lebesgue measure (respectively, Haar measure). On the other hand, Theorem 1 shows that replacing those properties by the metrical transitivity, we can get a much stronger assertion. However, the Continuum Hypothesis was essentially used in our argument and, at this moment, we do not know whether additional set-theoretical assumptions are necessary for obtaining the above-mentioned stronger result. Example 2. Let E be a set of cardinality c and let E  be an infinite countable subset of E. Define G as the group of all those transformations of E whose restrictions to the set E \ E  coincide with the identical transformation of E \ E  . Obviously, we have card(G) = card(E) = c. Further, define a measure ν on the power set P(E) of E by putting:

ν (Z) = card(Z ∩ E  ) if card(Z ∩ E  ) < ω ; ν (Z) = +∞ if card(Z ∩ E  )  ω . It can easily be verified that ν is a σ -finite, G-invariant, metrically transitive and separable measure. At the same time, ν admits no proper extensions because the domain of ν is the whole power set P(E). In particular, ν does not admit a nonseparable G-invariant extension. This simple example shows that the assumption of the diffusedness of a measure μ in the formulation of Theorem 1 is rather essential for the validity of this theorem. Example 3. Let E and G be as in Example 2. In a similar way we can define a probability measure ν1 on the power set P(E), which is G-quasi-invariant, metrically transitive, and

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separable. Analogously to the measure ν from the previous example, this ν1 has no proper extensions and, consequently, does not admit a nonseparable G-quasi-invariant extension. EXERCISES 1∗ . Let E be a set, G be a group of transformations of E and let μ be a σ -finite G-invariant complete measure on E. Suppose that G acts almost freely in E with respect to μ , i.e., for any two distinct transformations g ∈ G and h ∈ G, we have

μ ∗ ({x ∈ E : g(x) = h(x)}) = 0. Prove that the following two assertions are equivalent: (a) μ has the uniqueness property, i.e., for every σ -finite G-invariant measure ν with dom(ν ) = dom(μ ), there exists t ∈ [0, +∞[ such that ν = t μ ; (b) μ is metrically transitive (ergodic). For this purpose, use the fact that any μ -measurable set X ⊂ E with μ ∗ (X) > 0 contains a

μ -nonmeasurable subset (see Exercises 9, 10 and 11 of Appendix 1). Also, show that (a) always implies (b) (without the assumption of the completeness of μ ). 2. Deduce Lemma 3 from Lemma 2. 3∗ . Prove that for any infinite set T such that (card(T ))ω = card(T ), there exists a family {T j : j ∈ J} of subsets of T satisfying the following conditions: (a) card(J) = 2card(T ) ; (b) {T j : j ∈ J} is σ -independent in the set-theoretical sense, i.e., for every injective sequence { jn : n ∈ N} ⊂ J and for every function f : N → {0, 1}, we have f (0)

card(T j0 f (n)

where T jn

f (1)

∩ ... ∩ T jn

f (n)

= T \ T jn if f (n) = 1.

∩ T j1

= T jn if f (n) = 0 and T jn

f (n)

∩ ...) = card(T ),

4. Let (E, G) be a space equipped with a transformation group and let μ be a σ -finite G-invariant (G-quasi-invariant) measure on E. Let Y be a μ -measurable almost G-invariant subset of E with μ (Y ) > 0. Then Y determines a nonzero σ -finite G-invariant (G-quasi-invariant) measure μY by the formula

μY (X) = μ (X ∩Y ) (X ∈ dom(μ )). If μY is ergodic (i.e., metrically transitive), then we say that μY is an ergodic component of

μ. Obviously, μ is ergodic if and only if μE is an ergodic component of μ .

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Show that the nonseparable translation-invariant extension of λ obtained by the method of Kakutani and Oxtoby (see [83], [95], [108], or Exercise 18 from Chapter 3) has no ergodic components. 5. Let (E, G) be a space equipped with a transformation group and let μ be a σ -finite G-invariant (G-quasi-invariant) measure on E. We say that a family {Xi : i ∈ I} of nonempty subsets of E is admissible for μ if the following relations are satisfied: (a) all sets from {Xi : i ∈ I} are pairwise disjoint; (b) μ ∗ (∪{Xi : i ∈ I}) > 0; (c) for each index i ∈ I, we have μ (Xi ) = 0; (d) for any subset J of I, the set ∪{Xi : i ∈ J} is almost G-invariant with respect to μ . Supposing that there exists at least one admissible family for μ , it is convenient to introduce the notation: a(μ ) = inf{card(I) : there exists an admissible f amily {Xi : i ∈ I} f or μ }. Verify that a(μ ) is an uncountable regular cardinal and a(μ )  card(E). 6∗ . Prove that there exists a translation-invariant extension ν of the Lebesgue measure λ on R such that a(ν ) = ω1 . More generally, suppose that E is a vector space over the field Q of all rational numbers with card(E)  ω1 and μ is a nonzero σ -finite E-invariant (E-quasi-invariant) measure on E. Show that there exists an E-invariant (E-quasi-invariant) extension μ  of μ such that a(μ  ) = ω1 . 7. Suppose that (E, G) is a space equipped with a transformation group and satisfying the following two relations: (a) card(E) = card(G); (b) G acts transitively in E. Let μ be a nonzero σ -finite G-invariant (G-quasi-invariant) measure on E such that μ∗ (Z) = 0 for all sets Z ⊂ E with card(Z) < card(E). Check that if c f (card(E)) > ω , then μ can be extended to a G-invariant (G-quasi-invariant) measure on E for which there exists at least one admissible family of subsets of E. 8∗ . Let E be a set, G be a group of transformations of E and let μ be a nonzero σ -finite G-invariant measure on E. In addition, suppose that an uncountable disjoint family {Xt : t ∈ T } of μ -thick subsets of E is given such that for each T0 ⊂ T , the set ∪{Xt : t ∈ T0 } is almost G-invariant with respect to μ .

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Prove that there exists a G-invariant extension μ  of μ such that the Hilbert dimension of L2 (μ  ) is strictly greater than card(T ). For this purpose, apply the results of the preceding exercises. 9∗ . Assume that the Generalized Continuum Hypothesis holds. Let E be a set, G be a group of transformations of E, and let μ be a nonzero σ -finite G-invariant measure on E. Suppose that these five conditions are satisfied: (a) c f (card(E)) > ω ; (b) card(G) = card(E) and G acts transitively in E; (c) μ has at least one ergodic component; (d) a(μ ) is strictly less than the first uncountable inaccessible cardinal number; (e) μ∗ (Z) = 0 for all sets Z ⊂ E with card(Z) < card(E). Prove that there exists a G-invariant extension μ  of μ such that the Hilbert dimension of the space L2 (μ  ) is greater than or equal to 2a(μ ) . 10. Assume the Generalized Continuum Hypothesis. Let E be a set, G be a group of transformations of E and let μ be a nonzero σ -finite G-invariant measure on E. Suppose also that: (a) the cardinality of E is strictly less than the first uncountable inaccessible cardinal number and c f (card(E)) > ω ; (b) card(G) = card(E) and G acts transitively in E; (c) μ has at least one ergodic component; (d) μ∗ (Z) = 0 for all sets Z ⊂ E with card(Z) < card(E). Show that there exists a G-invariant extension μ  of μ such that the Hilbert dimension of L2 (μ  ) is greater than or equal to 2c and, in particular, μ  is a nonseparable extension of μ . 11. Prove that there exists a nonseparable translation-invariant extension of the Lebesgue measure λ on R, which does not have the Steinhaus property (for the definition of this property, see Chapters 3 and 4). 12. Let E be a set equipped with a nonseparable σ -finite measure μ . Show that there exists a μ -measurable set Y ⊂ E with μ (Y ) > 0 such that the restriction of μ to the σ -algebra SY = {Y ∩ X : X ∈ dom(μ )} of subsets of Y is a non-atomic measure. In order to demonstrate this assertion, take into account the fact that any maximal disjoint family of atoms of a σ -finite measure is at most countable.

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13∗ . Let T as usual denote the one-dimensional unit torus. Consider the product space G = Tc as a compact commutative group equipped with its Haar probability measure ν . Prove that there exists a subgroup H of G satisfying the following relations: (a) card(H) = c < 2c = card(G); (b) H is ν -thick in G. Infer from (a) and (b) that H is nonmeasurable with respect to the completion of ν . In connection with the last exercise, it should be mentioned that the existence of a subset of R which is nonmeasurable with respect to λ and whose cardinality is strictly less than card(R) = c cannot be established within ZFC theory. Indeed, as well known, if Martin’s Axiom holds true, then any subset X of R with card(X) < c is of λ -measure zero and, consequently, λ -measurable.

Chapter 17

Nonseparable left invariant measures on uncountable solvable groups

In the previous chapter we were concerned with certain nonseparable extensions of invariant measures given on a space E which is equipped with a transformation group G such that card(E) = card(G) = c, where c denotes the cardinality of the continuum. In connection with the result presented in that chapter, the following question arises. What can be said in the situation when card(E) = card(G) > c? In particular, a priori it is not clear whether there are σ -finite invariant (quasi-invariant) nonseparable measures in the above-mentioned more general case. However, under some purely set-theoretical assumptions and rather natural assumptions on a group G, the existence of such measures can be proved. Notice that among those assumptions, the Generalized Continuum Hypothesis was heavily exploited (see Exercises 9 and 10 for the previous chapter). Assuming the Continuum Hypothesis, we are going to establish that every uncountable solvable group admits a nonseparable non-atomic σ -finite left invariant measure. Several related questions concerning properties of such measures will be discussed, too. Let E be a nonempty set, S be a σ -algebra of subsets of E and let μ be a measure on E with dom(μ ) = S . As usual, the triple (E, S , μ ) is called a measure space. For any two sets X ∈ S and Y ∈ S having finite μ -measure, we may put d(X,Y ) = μ (XY ). As is well known, the function d is a quasi-metric and after the corresponding factorization it yields a complete metric space canonically associated with μ (cf. Exercise 12 of Chapter 1). A.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7_17, © 2009 Atlantis Press/World Scientific

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We recall that a measure μ is nonseparable if the above-mentioned metric space is nonseparable. It is also known that the following two assertions are equivalent: (1) μ is nonseparable; (2) there exist a real ε > 0 and an uncountable family {Yi : i ∈ I} of μ -measurable subsets of E such that d(Yi ,Y j ) > ε

(i ∈ I, j ∈ I, i = j).

Nonseparable probability measures naturally appear when we deal with uncountable products of probability measure spaces. For example, if T denotes the one-dimensional unit torus endowed with the standard Lebesgue probability measure, then the product group of uncountably many copies of T carries the product measure which is nonseparable and, in fact, coincides with the Haar probability measure on this product group. Uncountable direct sums of probability measure spaces also yield examples of nonseparable measures. However, in the latter case the obtained measures fail to be σ -finite, and this circumstance is a weak side of the operation of taking direct sums of probability measure spaces. Many years ago, a problem was formulated about the existence of nonseparable translationinvariant extensions of the standard Lebesgue measure λ on the real line R and about the existence of nonseparable translation-invariant extensions of the Lebesgue probability measure θ on the torus T (see [234] and [235]). This problem was intensively investigated by several authors and was solved positively in the classical works [95] and [141] by Kakutani, Kodaira and Oxtoby. Moreover, some generalized versions of the above-mentioned problem were considered for a Haar measure on a σ -compact locally compact topological group (see [83], [129], and the previous chapter). For further information on methods of extending invariant and, more generally, quasiinvariant measures which are given on various spaces equipped with their transformation groups, see [41], [42], [45], [46], [83], [85], [86], [95], [105], [108], [115], [119], [124], [129], [133], [141], [142], [193], [195], [198], [234], [235], [242], [247], [249], and Chapters 2 and 3 of this book. Let (G, ·) be an arbitrary uncountable group. In our consideration below we do not assume that G is equipped with a topology compatible with its algebraic structure. There are trivial examples of probability left-invariant measures defined on some σ algebras of subsets of G. For instance, we may introduce the probability measure μ0 defined

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on the family of all countable and co-countable subsets of G and vanishing at all singletons in G. Obviously, the measure μ0 is invariant under the group of all left (right) translations of G. In this context, the following natural question arises. Does there exist a nonseparable non-atomic σ -finite left-invariant measure on G? Using the Continuum Hypothesis, our goal is to show that if G is solvable, then the answer to this question is positive. Moreover, it turns out that if G is uncountable and commutative, then there always exists a nonseparable non-atomic invariant probability measure on G. In order to demonstrate this fact, we need several auxiliary propositions. The first lemma is purely algebraic in nature and is well known. Lemma 1. Let (G, +) be an uncountable commutative group. There exists a commutative group (H, +) such that: 1) card(H) = ω1 ; 2) H is a homomorphic image of G. A detailed proof of this statement was given in Chapter 11 (see Lemma 8 therein). Remark 1. One application of the above lemma has already been presented in the same Chapter 11 where algebraic sums of so-called absolutely negligible sets in uncountable commutative groups were discussed. Also, as demonstrated in preceding sections, these sets play an essential role in some questions concerning extensions of nonzero σ -finite left invariant (left quasi-invariant) measures on groups. Recall that if a commutative group (G, +) is given and ν is a left invariant measure on G, then ν automatically is a right invariant measure on G. In this case, we simply say that ν is an invariant (or G-invariant) measure on G. Starting with Lemma 1, it is not difficult to obtain the following statement. Lemma 2. Suppose that every commutative group of cardinality ω1 admits a nonseparable non-atomic σ -finite invariant measure. Then every uncountable commutative group admits a nonseparable non-atomic σ -finite invariant measure. Proof. Let (G, +) be an arbitrary uncountable commutative group. In view of Lemma 1, there exists a surjective homomorphism

φ : G → H, where H is some commutative group of cardinality ω1 . According to our assumption, H admits a nonseparable non-atomic σ -finite invariant measure μ . Consider the family of

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sets S = {φ −1 (Y ) : Y ∈ dom(μ )} and define a functional ν on S by putting

ν (φ −1 (Y )) = μ (Y )

(Y ∈ dom(μ )).

It can easily be verified that: (a) the definition of ν is correct because φ is a surjection; (b) S is a G-invariant σ -algebra of subsets of G; (c) ν is a non-atomic σ -finite measure on S . Let us check that ν is also nonseparable and G-invariant. Indeed, since the measure μ is nonseparable, there exists an uncountable family {Yi : i ∈ I} of μ -measurable subsets of H such that d(Yi ,Y j ) > ε

(i ∈ I, j ∈ I, i = j)

for some strictly positive real number ε , where d denotes the quasi-metric canonically associated with μ . Then we may write

ν (φ −1 (Yi )φ −1 (Y j )) > ε

(i ∈ I, j ∈ I, i = j),

which shows the nonseparability of ν . Take now an arbitrary set X ∈ S and an arbitrary element g ∈ G. Then X = φ −1 (Y ) for some set Y ∈ dom(μ ). Denoting h = φ (g), we easily come to the equality g + X = φ −1 (h + Y ) which yields

ν (g + X) = μ (h + Y ) = μ (Y ) = ν (φ −1 (Y )) = ν (X), i.e., ν turns out to be a G-invariant measure. Lemma 2 has thus been proved. Lemma 3. Suppose again that every commutative group of cardinality ω1 admits a nonseparable non-atomic σ -finite invariant measure. Then every uncountable solvable group admits a nonseparable non-atomic σ -finite left invariant measure. Proof. Let (G, ·) be an arbitrary uncountable solvable group. There exists a composition series {e} = G0 ⊂ G1 ⊂ ... ⊂ Gn−1 ⊂ Gn = G

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of subgroups of G, where e denotes the neutral element of G and for each natural number i ∈ [0, n−1], the group Gi is normal in Gi+1 and the quotient group Gi+1 /Gi is commutative. We must show that the group G = Gn admits a nonseparable non-atomic σ -finite left invariant measure on some σ -algebra of its subsets. For this purpose, we use the induction method on n. If n = 1, then the quotient group G = G1 /G0 is commutative and we may directly apply Lemma 2. Suppose now that our assertion has already been proved for uncountable solvable groups whose composition series contains at most n subgroups, and let us establish the validity of our assertion for those uncountable solvable groups G whose composition series consists of n + 1 subgroups, i.e., G = Gn . Let ψ : Gn → Gn /Gn−1 denote the canonical epimorphism. Consider two possible cases. 1. The commutative group Gn /Gn−1 is uncountable. In this case we apply again Lemma 2 and equip the group Gn /Gn−1 with a nonseparable non-atomic σ -finite left invariant measure μ . Further, as in the proof of Lemma 2, we define S = {ψ −1 (Y ) : Y ∈ dom(μ )},

ν (ψ −1 (Y )) = μ (Y ) (Y ∈ dom(μ )). In this manner we obtain the σ -algebra S of subsets of Gn = G and a nonseparable nonatomic σ -finite left invariant measure ν on S (cf. the proof of Lemma 2). Therefore, we conclude that our group G admits a nonseparable non-atomic σ -finite left invariant measure. 2. The commutative group Gn /Gn−1 is at most countable. In this case, taking into account the uncountability of our group Gn , we first derive that the subgroup Gn−1 is uncountable, too. According to the inductive assumption, Gn−1 admits some nonseparable non-atomic σ -finite left invariant measure μ . Denote by {gk : k ∈ K} a selector of the countable family Gn /Gn−1 . For each index k ∈ K, we have a canonical bijection between the sets Gn−1 and gk · Gn−1 . By using this bijection, we transfer the measure μ to the set gk · Gn−1 and denote the obtained measure by μk . Let us underline that one and only one of the measures μk (k ∈ K) coincides with the original measure μ . Now, let us put

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S  = the family of all subsets X of Gn such that (X ∩ gk · Gn−1 ) ∈ dom(μk ) for each k ∈ K. It can easily be seen that S  is a σ -algebra of sets in Gn . Further, for any set X ∈ S  , we define

ν (X) =

∑ μk (X ∩ gk · Gn−1).

k∈K

A direct verification shows that ν is a non-atomic σ -finite left invariant measure on the group Gn = G. Finally, ν is nonseparable since it contains as a direct summand the nonseparable measure

μ. We thus conclude that G admits a nonseparable non-atomic σ -finite left invariant measure which completes the proof of Lemma 3. Lemma 4. Assume the Continuum Hypothesis. Then every commutative group (H, +) with card(H) = ω1 admits a nonseparable non-atomic invariant probability measure. Proof. It is known (see, e.g., Exercise 19 from Chapter 3) that there exists a partition {Hξ : ξ < ω1 } of H satisfying the following relations: (a) card(Hξ )  ω for any ordinal ξ < ω1 ; (b) for each subset Ξ of ω1 , the set Z(Ξ) = ∪{Hξ : ξ ∈ Ξ} is almost H-invariant in H, i.e., we have (∀h ∈ H)(card((h + Z(Ξ))Z(Ξ))  ω ). Also, according to an old result on σ -independent families of sets due to Tarski, if E is a set of cardinality continuum, then there exists an uncountable family {Ei : i ∈ I} of subsets of E such that (c) for any injective sequence {in : n < ω } ⊂ I, we have card(∩{Ein : n < ω }) = c, where Ein = Ein ∨ Ein = E \ Ein

(n < ω ).

Notice that the existence of {Ei : i ∈ I} can be established directly by using the method of transfinite induction and without appealing to the above-mentioned result of Tarski (cf. Exercise 3 for Chapter 16). In view of our assumption c = ω1 , we immediately obtain an analogous uncountable family {Ξi : i ∈ I} of subsets of ω1 , i.e., the following relation holds:

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(d) for any injective sequence {in : n < ω } ⊂ I, we have card(∩{Ξin : n < ω }) = c, where Ξin = Ξin ∨ Ξin = ω1 \ Ξin

(n < ω ).

Now, for every index i ∈ I, let us put Ai = ∪{Hξ : ξ ∈ Ξi } and consider the family {Ai : i ∈ I} of subsets of H. This family is σ -independent, too, and each set Ai (i ∈ I) is almost H-invariant in H by virtue of relation (b). Let μ0 denote the probability H-invariant measure on H which is defined for all countable and co-countable subsets of H and vanishes at all singletons in H. Applying the KakutaniOxtoby method of extending invariant measures (cf. [83], [95], [108], or Exercise 18 from Chapter 3), we can extend μ0 to a non-atomic H-invariant measure μ on H such that {Ai : i ∈ I} ⊂ dom(μ ), d(Ai , A j ) = 1/2 (i ∈ I, j ∈ I, i = j). Since the set I is uncountable, the last relation yields that the measure μ is nonseparable. Moreover, we may take I so that card(I) = 2c and, consequently, the topological weight of the metric space canonically associated with μ will attain its maximum value. Lemma 4 has thus been proved. Remark 2. Unfortunately, the role of the Continuum Hypothesis in Lemma 4 is still unclear. In other words, we do not know whether it is possible to prove this lemma within the theory ZFC. Remark 3. Lemma 4 can be easily generalized to all those infinite groups (H, ·) which satisfy the equality (card(H))ω = card(H), but are not necessarily commutative or solvable. The proof can be carried out by the same method of σ -independent families of sets. However, this method does not work for those uncountable groups (H, ·) whose cardinality is cofinal with ω . Also, it is reasonable to point out in this place that under the Generalized Continuum Hypothesis, the following two assertions are equivalent for an infinite group H: (1) (card(H))ω = card(H);

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(2) card(H) is not cofinal with ω . Finally, let us recall that if G is an uncountable σ -compact locally compact topological group, then the equality (card(G))ω = card(G) is automatically valid (see, for instance, [46], [83], [94], or Exercise 4 from Chapter 10). Taking into account Lemmas 1 - 4, we come to the following statement. Theorem 1. If the Continuum Hypothesis holds, then every uncountable solvable group admits a nonseparable non-atomic σ -finite left invariant measure. For commutative groups, we also have a certain analogue of this theorem within the theory ZFC. Theorem 2. Any commutative group (G, +) with card(G)  c admits a nonseparable nonatomic G-invariant probability measure. The proof of Theorem 2 follows from the corresponding analogues of Lemmas 1, 2, and 4.

In connection with the results presented in this chapter, the following two open questions seem to be of interest. Problem. Is it true that every uncountable group (G, ·) admits a nonseparable non-atomic

σ -finite left invariant measure? More generally, is it true that any nonzero σ -finite left invariant measure given on G admits a nonseparable left invariant extension? Problem. Is it true that every uncountable solvable group (G, ·) admits a nonseparable nonatomic σ -finite left invariant measure having the uniqueness property? More generally, is it true that any nonzero σ -finite left invariant measure on G with the uniqueness property admits a nonseparable left invariant extension with the same property? Rich information about the uniqueness property for invariant measures can be found in [47], [80], [81], [83], [105], [108], [115], [182], [198], and [248]. Notice that a nonseparable invariant extension of the Lebesgue measure constructed by Kakutani and Oxtoby [95] does not possess this property. On the other hand, a nonseparable invariant extension of the Lebesgue measure constructed by Kodaira and Kakutani [141] has the uniqueness property (for more details, see [115]). EXERCISES

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1. Let μ be a σ -finite measure. Show that the following two assertions are equivalent: (a) μ is nonseparable; (b) there exist a real ε > 0 and an uncountable family {Yi : i ∈ I} of μ -measurable subsets of E such that d(Yi ,Y j ) > ε

(i ∈ I, j ∈ I, i = j),

where d denotes the quasi-metric canonically associated with μ . 2. Let (E1 , S1 , μ1 ) and (E2 , S2 , μ2 ) be two σ -finite measure spaces and suppose that a mapping f : E1 → E2 satisfies the relations (∀Y ∈ S2 )( f −1 (Y ) ∈ S1 ), (∀Y ∈ S2 )(μ1 ( f −1 (Y )) = μ2 (Y )), which simply means that μ2 is a homomorphic image of μ1 under f . Show that if μ2 is nonseparable, then μ1 is nonseparable, too. Formulate and prove a more general result in terms of the topological weights of the metric spaces canonically associated with μ1 and μ2 , respectively. 3. Let (G, +) and (H, +) be two commutative groups and let φ : G → H be a surjective homomorphism. Suppose that H is equipped with a σ -finite measure μ . Show that the pre-image φ −1 (μ ) has the same weight as μ . 4. Generalize Lemma 4 to the case of all those uncountable groups (H, ·) which satisfy the equality (card(H))ω = card(H). For this purpose, use Tarski’s result on σ -independent families of sets and Exercises 18 and 19 from Chapter 3. 5. Assuming the Generalized Continuum Hypothesis, show the equivalence of assertions (1) and (2) of Remark 3. 6. Prove that the nonseparable translation-invariant extension of λ obtained by the KodairaKakutani method [141] has the uniqueness property. 7. Let E be a set and let G be a group of transformations of E. Show that the following two assertions are equivalent:

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(a) there exists a finite (σ -finite) nonseparable non-atomic G-invariant measure on E; (b) there exists a finite (σ -finite) nonseparable G-invariant measure on E. 8. Let (E, S , μ ) be a σ -finite measure space and let ν be an arbitrary σ -finite measure equivalent to μ . Is it true that ν is separable if and only if μ is separable? More generally, is it true that the topological weight of the metric space canonically associated with ν is equal to the topological weight of the metric space canonically associated with μ ? 9∗ . Let (G, ·) be an uncountable solvable group and let P(G) as usual denote the family of all subsets of G. Prove that there exists a functional

ν : P(G) → [0, 1] satisfying the following conditions: (a) ν (G) = 1; (b) ν is finitely additive, i.e., ν (X ∪Y ) = ν (X) + ν (Y ) for any two disjoint sets X ⊂ G and Y ⊂ G; (c) ν is right invariant, i.e., ν (Xg) = ν (X) for each set X ⊂ G and for each element g ∈ G. The functional ν is called a right invariant mean for G (cf. [83] and [240]). In order to establish the existence of ν , consider the family F = F(G) of all real-valued bounded functions on G, equipped with the norm of uniform convergence, that is || f || = supg∈G | f (g)|

( f ∈ F).

For any function f : G → R and for any element g ∈ G, define the function fg : G → R by the formula fg (h) = f (h · g) (h ∈ G). Further, let Φ = F ∗ denote the conjugate space of F endowed with the weak topology

σ (F ∗ , F). For each φ ∈ Φ and for each g ∈ G, define φg ∈ Φ by the relation φg ( f ) = φ ( fg ) ( f ∈ F). In this manner, every g ∈ G produces the continuous affine mapping Tg : Φ → Φ, determined by the formula Tg (φ ) = φg

(φ ∈ Φ).

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Check that the obtained family {Tg : g ∈ G} satisfies the assumptions of Markov-Kakutani’s theorem from Chapter 3 and transforms the compact convex set K = {φ ∈ Φ : φ is positive, φ (χG ) = 1} into itself (here χG denotes the characteristic function of G, which is identically equal to 1). Conclude from these facts that there exists ψ ∈ K invariant under all mappings from {Tg : g ∈ G}, i.e., ψ is a linear positive normalized right invariant functional on F(G). Finally, for every set X ⊂ G, put

ν (X) = ψ (χX ) where χX denotes the characteristic function of X, and check that ν is the required functional. 10∗ . Let E be a set, S be an algebra of subsets of E and let

μ : S → [0, +∞] be a finitely additive functional on S , i.e.,

μ (A ∪ B) = μ (A) + μ (B) for any two disjoint sets A ∈ S and B ∈ S . Prove that there exists a functional

μ  : P(E) → [0, +∞], which extends μ and is finitely additive on P(E). For this purpose, use the Zorn lemma and an argument similar to the method of extending measures described in Chapter 2. 11∗ . Let E be a set, G be a group of transformations of E and let S be a G-invariant algebra of subsets of E. Suppose that a functional

μ : S → [0, +∞] is given such that: (a) μ is finitely additive, i.e., μ (A ∪ B) = μ (A) + μ (B) for any two disjoint sets A ∈ S and B∈S; (b) μ is G-invariant. Assuming that G is a solvable group, prove that there exists a functional

μ  : P(E) → [0, +∞]

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which extends μ and is also finitely additive and G-invariant. In order to demonstrate the existence of μ  , first extend μ to a functional

μ  : P(E) → [0, +∞] which is finitely additive (see Exercise 10). Then consider a linear positive normalized right invariant functional

ψ : F(G) → R for the given group G (see Exercise 9). Further, for any set A ⊂ E, define the function fA : G → R by putting fA (g) = μ  (g(A)) (g ∈ G). In general, fA is an unbounded real-valued function, so we cannot assert that fA ∈ F(G). However, for each natural number n, the relation min( fA , n) ∈ F(G) holds true. Now, define

μ  (A) = limn→+∞ ψ (min( fA , n)) and check that μ  is the required functional. For this purpose, observe that if A and B are any two disjoint subsets of E, then min( fA + fB , n)  min( fA , n) + min( fB , n)  min( fA + fB , 2n), whenever n < ω . This relation readily implies the finite additivity of μ  . Furthermore, if g ∈ G, A ⊂ E and n < ω , then

ψ (min( fg(A) , n)) = ψ (min(( fA )g , n)) = ψ ((min( fA , n))g ) = ψ (min( fA , n)) from which the G-invariance of μ  easily follows. In particular, infer from the stated above that there exists a nonnegative finitely additive functional which is defined on the family of all subsets of the plane R2 , is invariant under the group of all isometric transformations of R2 and extends the two-dimensional Lebesgue measure λ2 (Banach’s classical theorem). In order to show the validity of Banach’s theorem, take into account the fact that the group of all motions of R2 is solvable and apply the general result presented in this exercise.

Chapter 18

Universally measurable additive functionals

This chapter is devoted to those additive functionals on a real Hilbert space H which have good, in some generalized sense, measurability properties with respect to a certain class of measures on H but, simultaneously, are bad from the topological point of view, that is are discontinuous at all points of H. Primarily, additive functionals on infinite-dimensional vector spaces (over R) will be under consideration in this chapter. In particular, for an infinite-dimensional separable Hilbert space H, the problem of measurability of additive functionals f :H→R with respect to various extensions of σ -finite diffused Borel measures on H will be discussed (cf. Chapters 2 and 5). More precisely, it will be shown below that there exists a real-valued additive everywhere discontinuous functional f on H such that for any σ -finite diffused Borel measure μ on H, this f can be made measurable with respect to an appropriate extension of μ . Special attention will be paid to the case where μ is invariant or quasi-invariant under a certain subgroup of H (in this connection, see also Chapters 3 and 4). Let E be a topological space such that all singletons in E are Borel subsets of E. For example, if every singleton in E is closed (i.e., E is a T1 -space), then E automatically satisfies the above-mentioned condition. Recall that a Borel measure μ on E is diffused (or continuous) if μ ({x}) = 0 for each point x ∈ E. As in preceding sections, we denote by the symbol ME the class of the completions of all nonzero σ -finite Borel measures on E. A set X ⊂ E is called universally measurable (or absolutely measurable) with respect to ME if for any measure μ ∈ ME , we have X ∈ dom(μ ) (see Chapter 5). A.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7_18, © 2009 Atlantis Press/World Scientific

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It is well known that if E is a Polish space, then all analytic (also, co-analytic) subsets of E are universally measurable with respect to ME . For the proof, see, e.g., [24], [52], [99], [187], or Appendix 2. The family of all universally measurable sets in E forms a σ -algebra of subsets of E. A straightforward verification of this simple fact is left to the reader. Recall that a functional f : E → R is universally measurable with respect to the class ME if f is measurable with respect to any measure μ ∈ ME (cf. Chapter 5). Let H denote an infinite-dimensional separable Hilbert space (over the field R of all real numbers) and let f :H→R be a universally measurable additive functional with respect to ME . Then f turns out to be continuous (cf. [66], [97], [153], and [154]). In fact, f is continuous whenever it is universally measurable with respect to the class of the completions of all probability diffused Borel measures on H. To demonstrate this circumstance, take such an f and suppose to the contrary that f is not continuous. Then, for every natural number n, there exists an element hn ∈ H satisfying the relations ||hn || = 1, | f (hn )| > n · 4n. Let us put en = hn /4n . Obviously, we have ||en || = 1/4n , | f (en )| > n

(n < ω ).

Further, consider the Cantor space C = {0, 1}ω as a compact commutative group with respect to the product topology and addition operation (modulo 2). Equip C with the completion μ  of the Haar probability measure μ , which is isomorphic to the standard Lebesgue measure λ on the segment [0, 1] and, consequently, these two measures can be identified. Let us introduce a mapping

ψ :C →H by the formula

ψ ( χA ) =

∑ en

(A ⊂ ω ),

n∈A

where χA denotes the characteristic function of a set A ⊂ ω . It can easily be verified that: (i) ψ is injective and additive; (ii) ψ is continuous;

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(iii) |( f ◦ ψ )(χ{n})| = | f (en )| > n for each n < ω . Let θ denote the distribution in H of the random variable ψ . Clearly, θ is a Borel diffused probability measure on H and, according to our assumption, f is measurable with respect to the completion of θ . This circumstance implies that the composition f ◦ ψ is measurable with respect to μ  . Since this composition is also additive on the power set P(ω ), we conclude (see Exercise 1 for this chapter) that the set {( f ◦ ψ )(χ{n}) : n < ω } must be bounded in R which is impossible in view of relation (iii). The contradiction obtained yields the required result. Obviously, the same argument works for an arbitrary infinite-dimensional Banach space (E, || · ||) instead of H. We thus see that only continuous linear functionals on H can be universally measurable with respect to the class of the completions of all σ -finite (equivalently, probability) diffused Borel measures on H. For a finite-dimensional Hilbert space over R, that is for Euclidean space, we have a much stronger result which states that if an additive functional f : Rn → R is measurable with respect to the standard Lebesgue measure λn on Rn , then f is continuous (see, e.g., [108] and [143]). An analogous assertion is valid for any homomorphism f acting from a σ -compact locally compact topological group (Γ, ·) into the additive group R and measurable with respect to the completion of the Haar measure on Γ. Notice that the socalled Steinhaus property of the Haar measure plays a significant role in the proof (see [80], [83], or Exercise 3 of this chapter). Below, we shall introduce an essentially different notion of universal measurability of functionals. Also, it will be shown in the sequel that there exist universally measurable additive functionals on H (in the sense of our notion) which are everywhere discontinuous. Then we will demonstrate how such universally measurable additive functionals can be applied to some questions connected with the measure extension problem. Let E be an arbitrary nonempty set and let M be a class of measures on E. We do not assume, in general, that the measures from M are defined on one and the same σ -algebra of subsets of E. We recall once more from Chapter 5 that a functional f :E →R

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is universally (absolutely) measurable with respect to M if f is measurable with respect to all measures from M. But here we need a modified notion of universal measurability of functionals. We shall say that a functional f : E → R is universally measurable with respect to M (in the generalized sense) if for each measure μ ∈ M, there exists a measure μ  on E extending μ and such that f becomes measurable with respect to μ  . Evidently, this definition is more general than the definition given in the beginning of the chapter. Let us also remark that in both these definitions we may replace the real line R by an uncountable Polish space or, more generally, by an uncountable Borel subset of a Polish space since, according to the well-known result of descriptive set theory, any two uncountable Borel subsets of a Polish space are Borel isomorphic. Recall that a subset X of a topological space E is a Bernstein set in E if for each nonempty perfect set P ⊂ E, we have P ∩ X = 0/ & P ∩ (E \ X) = 0. / Various properties of Bernstein sets are discussed in [30], [40], [72], [143], [148], [172], [176], and [192]. In particular, any uncountable Polish space contains a Bernstein subset and the cardinality of such a subset is equal to the cardinality of the continuum which as usual is denoted by c. For our further purposes, we need two lemmas. Lemma 1. A separable Hilbert space H = {0} can be represented in the form H = X1 + X2

(X1 ∩ X2 = {0}),

where X1 and X2 are Bernstein subsets of H and, simultaneously, they are vector spaces over the field Q of all rational numbers. Proof. The argument is fairly standard (cf. Exercise 6 for Chapter 5). Denote by α the least ordinal number of cardinality continuum and let {Pξ : ξ < α } be an enumeration of all nonempty perfect subsets of H. By using the method of transfinite recursion, it is not difficult to construct two α -sequences {xξ : ξ < α } ⊂ H, {yξ : ξ < α } ⊂ H satisfying the following conditions: (a) the family {xξ : ξ < α } ∪ {yξ : ξ < α } is linearly independent over Q; (b) for any ordinal ξ < α , we have xξ ∈ Pξ and yξ ∈ Pξ .

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Now, let us take as X1 the vector space (over Q) generated by {xξ : ξ < α }. Further, let X2 denote a maximal vector subspace V of H (over Q) such that V ∩ X1 = {0},

{yξ : ξ < α } ⊂ V.

The existence of V is evident and it is also easy to see that X1 and X2 are the required Bernstein subsets of H. Remark 1. The proof given above works for an arbitrary Banach space (E, || · ||) of cardinality continuum, not necessarily separable (see Exercise 4 for this chapter). Lemma 2. Let H = {0} be a separable Hilbert space. There exists an additive functional f :H→R having the following property: for any σ -finite diffused Borel measure μ on H and for any σ -finite measure ν on R, the graph Gr( f ) of f is a (μ ⊗ ν )-thick subset of the product space H × R, that is this graph intersects every (μ ⊗ ν )-measurable set of strictly positive measure. Proof. First, applying Lemma 1, we represent H in the form H = X1 + X2

(X1 ∩ X2 = {0}),

where X1 and X2 are Bernstein subsets of H and, simultaneously, they are vector spaces over Q. Observe now that in view of the equalities card(X1 ) = c, card(R) = c, the vector spaces X1 and R (again over Q) are isomorphic to each other. Let

φ : X1 → R denote some isomorphism between these two spaces. We define a functional f as follows. Take any x ∈ H. This x admits a unique representation in the form x = x1 + x2 , where x1 ∈ X1 and x2 ∈ X2 . Let us put f (x) = f (x1 + x2 ) = φ (x1 ). Obviously, f is an additive functional on H. Further, fix a σ -finite diffused Borel measure

μ on H and a σ -finite measure ν on R. We assert that the graph of f is (μ ⊗ ν )-thick in H × R. Indeed, if Z is an arbitrary (μ ⊗ ν )-measurable set with (μ ⊗ ν )(Z) > 0, then, according to the Fubini theorem, there exists a point t ∈ R such that μ (Z(t)) > 0 where Z(t) = {x ∈ H : (x,t) ∈ Z}.

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Consider the point x1 = φ −1 (t) ∈ X1 . Since X2 is a Bernstein subset of H and Z(t) is an uncountable Borel subset of H, we have X2 ∩ (Z(t) − x1) = 0. / Choose a point x2 from the set X2 ∩ (Z(t) − x1) and define x = x1 + x2 . Then we get x = x1 + x2 ∈ Z(t), (x,t) ∈ Z, (x, f (x)) = (x,t), t = f (x), (x, f (x)) ∈ Z, which completes the proof of Lemma 2. Let f : H → R be as in Lemma 2. Notice that such an f is everywhere discontinuous on H because the graph of any measurable (in particular, continuous) functional is always of measure zero, so it cannot be thick in the product space H × R. Fix a Borel diffused probability measure ν on R. For instance, we may take as ν an arbitrary Borel probability measure equivalent to the standard Lebesgue measure λ on R assuming that the latter measure is restricted to the Borel σ -algebra of R. Now, let μ be any σ -finite diffused Borel measure on H. For each (μ ⊗ ν )-measurable set Z ⊂ H × R, we denote Z  = {x ∈ H : (x, f (x)) ∈ Z}. Further, we put S  = {Z  : Z ∈ dom(μ ⊗ ν )}. It can easily be verified that S  is a σ -algebra of subsets of H containing the Borel σ algebra of H. Finally, we define a functional μ  on S  by putting

μ  (Z  ) = (μ ⊗ ν )(Z)

(Z ∈ dom(μ ⊗ ν )).

A straightforward verification shows that the definition of μ  is correct in view of the (μ ⊗

ν )-thickness of the graph of f (see Chapter 2). Also, μ  turns out to be a measure on S  which extends the original measure μ . Keeping in mind all these facts, we are able to prove the following statement. Theorem 1. The additive functional f : H → R just described is universally measurable (in the generalized sense) with respect to the class of all σ -finite diffused Borel measures on H.

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Proof. It suffices to show that for any σ -finite diffused Borel measure μ on H, the functional f is measurable with respect to the measure μ  . For this purpose, take an arbitrary Borel subset B of R. Clearly, we may write f −1 (B) = {x ∈ H : f (x) ∈ B} = {x ∈ H : (x, f (x)) ∈ H × B} ∈ S  which establishes the measurability of f with respect to μ  and, therefore, finishes the proof. Let us indicate some other properties of f nice from the measure-theoretical point of view. It is well known that if H is an infinite-dimensional separable Hilbert space, then there exists no nonzero σ -finite Borel measure on H invariant (quasi-invariant) under the group of all translations of H (see, for instance, Chapter 4). At the same time, there exist various nonzero σ -finite Borel measures on H which are invariant under everywhere dense vector subspaces of H (see Chapter 3). Let μ be any such measure on H. We already know that f is measurable with respect to an appropriate extension μ  of μ . So it is natural to ask whether the measure μ  can be chosen to be invariant under the same everywhere dense vector subspace of H. It turns out that the answer to this question is always positive. In order to demonstrate this fact, let us slightly change the construction presented above (cf. [141] and Chapter 14). We replace the real line R by the one-dimensional unit torus T regarded as a compact commutative group. The torus T is equipped with the Haar probability measure which, actually, coincides with the Lebesgue probability measure on T invariant under all translations of T. We denote the latter measure by the same symbol ν . The following statement is true. Lemma 3. Let H = {0} be a separable Hilbert space. There exists a group homomorphism f :H →T having the property that for any Borel σ -finite diffused measure μ on H, the graph of f is a (μ ⊗ ν )-thick subset of the product space H × T. Proof. The argument given below is very similar to that of the proof of Lemma 2. Again, let us represent H in the form H = X1 + X2

(X1 ∩ X2 = {0}),

where both X1 and X2 are Bernstein subsets of H and, simultaneously, they are vector spaces over the field Q of all rational numbers. Taking into account the facts that X1 and R are isomorphic as vector spaces over Q and that there is a canonical surjective group homomorphism from R onto T, it is easy to define a surjective group homomorphism

φ : X1 → T.

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Now, if x is an arbitrary element of H, then x admits a unique representation in the form x = x1 + x2

(x1 ∈ X1 , x2 ∈ X2 ).

So, we may define f (x) = f (x1 + x2 ) = φ (x1 ). In this way, we obtain a group homomorphism f : H → T. Let μ be an arbitrary σ -finite diffused Borel measure on H. Take any set Z ∈ dom(μ ⊗ ν ) with (μ ⊗ ν )(Z) > 0. According to the Fubini theorem, μ (Z(t)) > 0 for some t ∈ T. Since φ is a surjection, we can find x1 ∈ X1 such that φ (x1 ) = t. Keeping in mind that X2 is a Bernstein subset of H, we get / X2 ∩ (Z(t) − x1) = 0. Choose an element x2 from the set X2 ∩ (Z(t) − x1 ) and put x = x1 + x2 . As in the proof of Lemma 2, we easily conclude that (x, f (x)) ∈ Z, which shows the (μ ⊗

ν )-thickness of the graph of f in the product space H × T. Starting with the homomorphism f : H → T described in the preceding lemma, we can extend any σ -finite diffused Borel measure μ on H to the measure μ  on the same H (by using the scheme indicated before). Theorem 2. Let H = {0} be a separable Hilbert space and let μ be a σ -finite Borel measure on H invariant (quasi-invariant) under some subgroup G of H. Then the measure

μ  is also invariant (quasi-invariant) under G. Proof. It suffices to consider the case of an invariant measure μ (for quasi-invariant measures the argument is absolutely analogous). Take any set Z  from the domain of μ  . According to the definition, we may write Z  = {x ∈ H : (x, f (x)) ∈ Z}, where Z belongs to the domain of μ ⊗ ν . For each element g ∈ G, we have Z  + g = {x ∈ H : (x − g, f (x − g)) ∈ Z} = {x ∈ H : (x, f (x)) ∈ Z + (g, f (g))}. Since the product measure μ ⊗ ν is (G × T)-invariant (see Exercise 9 from Chapter 3), we may write (μ ⊗ ν )(Z) = (μ ⊗ ν )(Z + (g, f (g)))

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and, consequently,

μ  (Z  ) = μ  (Z  + g) which completes the proof of the theorem. Obviously, for any nonzero σ -finite Borel measure μ on H, the measure μ  obtained by using the group homomorphism f : H → T is a proper extension of μ . But the σ -algebra dom(μ  ) is not significantly larger than the σ -algebra dom(μ ). Indeed, it can easily be seen that dom(μ  ) is a countably generated σ -algebra and, therefore, the measure μ  is separable. If we wish to extend μ to a nonseparable measure, then we must replace the torus T by the infinite-dimensional torus Tc which is the topological product of continuumly many copies of T. Let now ν denote the Haar probability measure on Tc . This measure is nonseparable. Moreover, there exists a family of sets {Zi : i ∈ I} ⊂ dom(ν ) having the following properties: (1) card(I) = c; (2) (∀i ∈ I)(ν (Zi ) > 0); (3) for every ν -measurable set Z with ν (Z) > 0, there is a set Zi contained in Z. The proof of the existence of {Zi : i ∈ I} with the above-mentioned properties can be found, e.g., in [83] (cf. also Exercise 11 from Chapter 2). Lemma 4. There exists a group homomorphism f : H → Tc such that for any σ -finite diffused Borel measure μ on H, the graph of f is a (μ ⊗ ν )-thick subset of H × Tc . Proof. Fix a family of sets {Zi : i ∈ I} ⊂ dom(ν ) satisfying the relations (1) - (3) formulated above. Without loss of generality, we may suppose that the set I of indices is well-ordered and its order type coincides with the least ordinal number α whose cardinality is equal to c. In other words, we may put {Zi : i ∈ I} = {Zξ : ξ < α }. As before, we represent our Hilbert space H in the form H = X1 + X2

(X1 ∩ X2 = {0}),

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where X1 and X2 are some Bernstein subsets of H and, simultaneously, they are vector spaces over Q. Now, it is not difficult to construct two injective α -sequences {yξ : ξ < α } ⊂ X1 , {zξ : ξ < α } ⊂ Tc satisfying the following conditions: (a) the family {yξ : ξ < α } is linearly independent over Q; (b) for each ξ < α , we have zξ ∈ Zξ . Further, we put

φ (yξ ) = zξ

(ξ < α ).

Taking into account condition (a) and the fact that Tc is a divisible group, we can extend the mapping φ to a group homomorphism

φ : X1 → Tc . In view of condition (b), the set φ (X1 ) is ν -thick in Tc . Now, take any x ∈ H. We have a unique representation x = x1 + x2 , where x1 ∈ X1 and x2 ∈ X2 . Define f (x) = f (x1 + x2 ) = φ (x1 ). In this manner, we get a group homomorphism f : H → Tc . Let us show that f is the required homomorphism. Consider any σ -finite diffused Borel measure μ on H. Let Z be an arbitrary (μ ⊗ ν )-measurable set such that (μ ⊗ ν )(Z) > 0. Applying again the Fubini theorem and the ν -thickness of φ (X1 ) in Tc , we see that

μ (Z(t)) > 0 for some t ∈ φ (X1 ). Choose an element x1 ∈ X1 such that φ (x1 ) = t. Since X2 is a Bernstein set in H, we have X2 ∩ (Z(t) − x1) = 0. / Consequently, we may pick some point x2 ∈ X2 ∩ (Z(t) − x1 ). Then we put x = x1 + x2 . An easy verification shows that (x, f (x)) ∈ Z (cf. the final part of the proof of Lemma 2). Thus, the graph of f is (μ ⊗ ν )-thick in the product space H × Tc and the proof is complete. Applying the previous lemma, we readily come to the following result. Theorem 3. Let H = {0} be a separable Hilbert space and let a group homomorphism f : H → Tc

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be as in Lemma 4. For every σ -finite diffused Borel measure μ , denote by μ  the extension of μ obtained by using this group homomorphism. Then we have: 1) μ  is a nonseparable measure; 2) if μ is invariant (quasi-invariant) under a group G ⊂ H, then μ  is invariant (quasiinvariant) under the same G; 3) f is measurable with respect to μ  . This statement also indicates that f is universally measurable in the generalized sense, i.e., for any σ -finite diffused Borel measure μ on H, there exists an extension μ  of μ such that f is μ  -measurable. Remark 2. Obviously, the preceding results are valid in the particular case where the ndimensional Euclidean space Rn (n  1) is taken as H. Thus, we may assert that there exists an additive functional f : Rn → R which is discontinuous at all points of Rn and is universally measurable (in the generalized sense) with respect to the class of all σ -finite diffused Borel measures on Rn . Remark 3. It should be noticed that if a vector space E (over the field Q) is of cardinality continuum, then, assuming Martin’s Axiom, there exists an injective additive functional g:E →R which is absolutely nonmeasurable with respect to the class of all nonzero σ -finite diffused measures on E. Indeed, the existence of such a g is implied by the fact that there exists a generalized Luzin subset of R which simultaneously is a vector space over Q (for more details, see [59] or Chapters 5 and 12). Consequently, under Martin’s Axiom, there are real-valued absolutely nonmeasurable additive functionals on a separable Hilbert space H = {0}. Remark 4. As has already been mentioned, the standard notion of universal measurability of functionals on a Hilbert space yields nothing new for the class of additive functionals because every universally measurable additive functional turns out to be continuous. Moreover, it was shown in [66] that even the mid-point convexity of a given functional with its universal measurability (in the standard sense) implies its continuity. This result can be regarded as an analogue of Sierpi´nski’s old theorem stating that any Lebesgue measurable mid-point convex function defined on a finite-dimensional Euclidean space is necessarily continuous (see [220]; cf. also [122] and [143]).

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In this context, we should also indicate paper [97] where the universal measurability (in the standard sense) of linear functionals, with respect to the class of all Gaussian measures on a Hilbert space H, is envisaged. EXERCISES 1∗ . Consider the standard Cantor space C = 2ω = {0, 1}ω . This C is a compact commutative group with respect to the addition operation (modulo 2). The same space may be regarded as the family P(ω ) of all subsets of ω in which the symmetric difference of sets is considered as a basic operation +, i.e., we have X + Y = XY = (X \ Y ) ∪ (Y \ X) (X ⊂ ω , Y ⊂ ω ). Actually, these two algebraic structures are canonically isomorphic, so they can be treated as identical to each other by identifying any subset X of ω with its characteristic function

χX . Let ν : P(ω ) → R be a function. We shall say that ν is additive if

ν (X + Y ) = ν (X) + ν (Y ) for any two disjoint subsets X and Y of ω . We shall say that ν : P(ω ) → R is measurable in the Luzin sense if it is measurable with respect to the completion μ  of the Haar probability measure μ on C. Recall that μ  is isomorphic to the Lebesgue measure on the closed unit interval [0, 1]. Prove that if ν is additive and measurable in the Luzin sense, then limn→+∞ ν ({n}) = 0. In particular, conclude that the sequence {ν ({n}) : n < ω } is necessarily bounded in R. Do it in the following manner. Suppose otherwise, i.e., suppose that the above-mentioned equality fails to be true. This fact means that there exist a neighborhood V of the neutral element in C and a strictly increasing sequence {nk : k < ω } of natural numbers such that

ν ({nk }) ∈ V

(k < ω ).

Since ν is μ  -measurable and μ  is Radon, there exists a compact set P ⊂ C satisfying the relation μ  (P) > 1/2 and such that the restriction ν |P is continuous in view of Luzin’s classical theorem on the structure of real-valued Lebesgue measurable functions.

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The invariance of the Haar measure μ implies the relation P + P = C. This relation means that each subset Z of ω admits a representation Z = XY , where X ∈ P and Y ∈ P. Applying this fact to the sequence {nk : k < ω }, infer that {nk } = Ak Bk

(k < ω ),

where all sets Ak and Bk belong to P. Deduce from these relations that, without loss of generality, we may suppose Ak = {nk } ∪ Bk , nk ∈ Bk

(k < ω ).

Further, taking into account the compactness and metrizability of P and passing if necessary to an appropriate subsequence, assume that limk→+∞ Ak = A, limk→+∞ Bk = B for some A ∈ P and B ∈ P. Now, keeping in mind that Ak = Bk + {nk }

(k < ω )

and limk→+∞ {nk } = 0, / conclude that A = B. Since the restriction ν |P is continuous, we can write limk→+∞ ν (Ak ) = ν (A) = ν (B) = limk→+∞ ν (Bk ), from which it follows that limk→+∞ ν ({nk }) = limk→+∞ (ν (Ak ) − ν (Bk )) = 0. But this circumstance contradicts the assumption that ν ({nk }) ∈ V for all k < ω . The contradiction obtained yields the required result. 2∗ . Let H be an infinite-dimensional separable Hilbert space. By using the result of the previous exercise, prove the continuity or, equivalently, boundedness of an arbitrary universally measurable (in the usual sense) additive functional f : H → R. Also, consider the question whether an analogous result can be established for an arbitrary universally measurable (in the usual sense) additive functional g : H → T. 3. Let (Γ, ·) be a σ -compact locally compact topological group equipped with its Haar measure μ = μΓ . Suppose that an additive functional f :Γ→R

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is given which is measurable with respect to the completion of μ . Verify that f is continuous. For this purpose, apply the Steinhaus property of μ (see Chapter 3) and show that f is bounded on some neighborhood of the neutral element of G, from which the required result easily follows. 4∗ . Prove an analogue of Lemma 1 for an arbitrary Banach space (E, || · ||) of cardinality continuum. For this purpose, take into account the fact that the family of all separable perfect subsets of E is also of cardinality c. 5. Give a detailed proof of Theorem 3. 6∗ . A function f : Rn → R is called mid-point convex if the relation f ((x + y)/2)  ( f (x) + f (y))/2 holds true for all points x and y from Rn . Prove Sierpi´nski’s theorem stating that any mid-point convex Lebesgue measurable function f : Rn → R is continuous and, therefore, is convex in the usual sense. This theorem strengthens the well-known result of Fr´echet which states that if a solution of the Cauchy functional equation is measurable in the Lebesgue sense, then it is continuous (i.e., turns out to be a trivial solution of this equation). 7∗ . Recall that if E is a vector space, then the symbol dim(E) denotes the algebraic dimension of E, i.e., dim(E) stands for the cardinality of any basis of E (this cardinality does not depend on a choice of a basis). As usual, we say that a vector space E is finite-dimensional if dim(E) < ω . Otherwise, we say that E is infinite-dimensional. Let E be an infinite-dimensional topological vector space such that the neutral element 0 of E admits a fundamental system (i.e., local base) {Vi : i ∈ I} of its open neighborhoods, satisfying the inequality card(I)  dim(E). Show that there exists a linear functional f :E →R which is discontinuous at all points of E.

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For this purpose, denote by α the least ordinal number whose cardinality is equal to card(I). Let {Vξ : ξ < α } be an enumeration of all sets from the local base {Vi : i ∈ I}. By using transfinite induction on ξ < α , define an α -sequence {eξ : ξ < α } of linearly independent elements of E such that eξ ∈ Vξ for each ξ < α . Then put f (eξ ) = 1 (ξ < α ) and extend this partial function to a linear functional f : E → R. Verify that such an f is discontinuous at 0 and hence it is discontinuous at all points of E. Conclude from this result that if E is any infinite-dimensional metrizable topological vector space, then there exists a linear functional f : E → R everywhere discontinuous on E. On the other hand, if E is an arbitrary vector space and the symbol F denotes the family of all linear functionals on E, then the so-called weak topology σ (E, F) on E is defined in such a way that each functional f ∈ F becomes continuous on the topological vector space (E, σ (E, F)). This fact shows, in particular, that the cardinality of any local base of the neutral element of (E, σ (E, F)) is strictly greater than dim(E). 8. Let H be a Hilbert space over R (not necessarily separable) and let f :H→R be an additive functional universally measurable with respect to the class MH . Verify that f is continuous.

Chapter 19

Some subsets of the Euclidean plane

Here we would like to consider those subsets of the Euclidean plane R2 which possess rather strange geometrical and measure-theoretical properties. In particular, we will be dealing with thick (massive) subsets of R2 having small linear sections in all possible directions, and we will investigate these sets from the point of view of the measurability with respect to translation-invariant extensions of the classical Lebesgue measure λ2 on R2 . Many examples of paradoxical sets in a finite-dimensional Euclidean space that have small sections by certain affine hyperplanes in this space are known in the literature. Some of them have already been discussed in this book (cf. Chapters 2, 5, and 7). In this context, it is reasonable to remind that one of the earliest examples of this sort is due to Sierpi´nski who was able to construct a bijective function f : R→R such that its graph Gr( f ) is λ2 -thick in R2 . Obviously, the λ2 -thickness of Gr( f ) directly implies that Gr( f ) is nonmeasurable with respect to λ2 and hence f is not measurable in the Lebesgue sense. At the same time, any straight line in R2 parallel to the axis of abscissae or to the axis of ordinates meets the graph of f in exactly one point. Further, it should be mentioned in this context that Mazurkiewicz constructed a subset X of R2 having the property that for each straight line l in R2 , the set l ∩ X consists of precisely two points (see [170]). Notice that X can be chosen to be λ2 -thick and, consequently, nonmeasurable with respect to λ2 (see Exercise 1 for this chapter). Later, various examples of sets with small sections, but being large in a certain sense, were given by other authors (cf. [72], [105], [119], [176], and [192]). In the present chapter, we will consider some sets of this type in connection with the following natural question. How small are such sets from the point of view of translation-invariant extensions of the Lebesgue measure λ2 ? A.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7_19, © 2009 Atlantis Press/World Scientific

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Namely, we are going to demonstrate in our further considerations that there are translationinvariant extensions of λ2 concentrated on sets with small linear sections in all directions. Actually, a more stronger result will be obtained stating that the corresponding sets can have small sections by all analytic curves lying in the plane. Let λn denote the standard n-dimensional Lebesgue measure on the Euclidean space Rn , where n  1. Below, we need the following auxiliary statement, which also turns out to be rather useful in many other situations (cf. [108], [181], and [240]). Lemma 1. Let Z be a λn -measurable subset of the Euclidean space Rn with λn (Z) > 0 and let {Mi : i ∈ I} be a family of analytic manifolds in Rn , such that: 1) card(I) is strictly less than the cardinality of the continuum; 2) for each index i ∈ I, the dimension of Mi is strictly less than n. Then the relation

λn∗ (Z \ ∪{Mi : i ∈ I}) > 0 / is satisfied. In particular, we always have Z \ ∪{Mi : i ∈ I} = 0. The proof of this lemma is not difficult and can be carried out by induction on n (here the classical Fubini theorem plays a dominant role). For more details, see [108], [181], [240], and Exercise 3 of this chapter. Some applications of Lemma 1 to certain questions of the geometry of Euclidean spaces may be found in [108] and [240]. Notice also that this lemma does not hold longer true for topological manifolds in Rn . For example, there are models of set theory in which the Euclidean plane can be covered by a family of Jordan curves, whose cardinality is strictly less than the cardinality of the continuum (cf. Exercise 5 of this chapter). As usual, we denote by c the cardinality of the continuum. Starting with the abovementioned lemma, we are able to establish the following result. Theorem 1. Let G be the group of all analytic diffeomorphisms of Rn onto itself. Then there exists a subset X of Rn such that: 1) card(X) = c and X is almost G-invariant, i.e., (∀g ∈ G)(card(g(X)X) < c); 2) X is λ n -thick in Rn ; 3) for any analytic manifold M in Rn with dim(M) < n, we have card(M ∩ X) < c.

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Proof. We use the method first suggested and successfully applied by Sierpi´nski (see [222]; compare also [83], [95], [108], [115], [176], [192], [234], and [235]). This method has already been exploited in preceding sections of the present book (see, for instance, Chapters 3 and 16). Let α denote the least ordinal number of cardinality continuum. Since the equality card(G) = c holds, we may write G = ∪{Gξ : ξ < α }, where {Gξ : ξ < α } is some α -sequence of subgroups of G which satisfies these two conditions: (a) for each ordinal ξ < α , we have the inequality card(Gξ )  card(ξ ) + ω ; (b) the family {Gξ : ξ < α } is increasing by inclusion. Also, we may consider the family {Yξ : ξ < α } consisting of all those Borel subsets of Rn which are of strictly positive λn -measure. Finally, we denote by {Mξ : ξ < α } the family of all those analytic manifolds in Rn whose dimensions are strictly less than n. Now, by using the method of transfinite recursion, we define an injective α -sequence {xξ : ξ < α } of points in Rn satisfying the following relations: (i) for any ξ < α , the point xξ belongs to Yξ ; (ii) for any ξ < α , we have xξ ∈ Gξ (∪{Mζ : ζ  ξ } ∪ {xζ : ζ < ξ }). Notice that Lemma 1 guarantees, at each ξ -step of our recursion, the existence of a needed point xξ . So the recursion can be continued up to α . In this way, we will be able to construct the required α -sequence of points {xξ : ξ < α }. Finally, putting X = {Gξ (xξ ) : ξ < α }, we can readily check that this X is the desired set, which finishes the proof of Theorem 1. The next statement is a trivial consequence of the above theorem. Theorem 2. Let G denote the group of all isometric transformations (i.e., motions) of Rn . Then there exists a subset X of Rn such that: 1) card(X) = c and X is almost G-invariant; 2) X is λn -thick in Rn ;

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3) for each analytic manifold M in Rn with dim(M) < n, we have card(X ∩ M) < c. In particular, for any affine hyperplane L in Rn , the inequality card(X ∩ L) < c is true. Moreover, if the Continuum Hypothesis holds, then according to Theorem 2, for each analytic manifold M in Rn with dim(M) < n, we have card(X ∩ M)  ω . Let us point out one application of Theorem 2 to a particular case of the measure extension problem. Dealing with various invariant extensions of the Lebesgue measure λn , the following question naturally arises. Does there exist a measure μn on Rn extending λn , invariant under the group of all isometric transformations of Rn , and concentrated on some subset X of Rn with small sections by all affine hyperplanes? Of course, here the smallness of sections means that for any affine hyperplane L in Rn , the cardinality of L ∩ X is strictly less than the cardinality of the continuum. Theorem 2 immediately yields a positive answer to this question. Indeed, let G be again the group of all isometric transformations of Rn . Consider the G-invariant σ -ideal of subsets of Rn generated by Rn \ X, where X is the set indicated in Theorem 2. We denote this σ -ideal by J . Then for any set Z ∈ J , the inner λn measure of Z is equal to zero since X is almost G-invariant and λn -thick in Rn . Taking this fact into account and applying the standard methods of extending invariant measures (see, for instance, [234], [235], or Chapter 2), we derive that there exists a measure μn on Rn satisfying the following relations: 1) μn extends λn ; 2) μn is invariant under the group G; 3) J ⊂ dom(μn ); 4) for each set Z ∈ J , we have μn (Z) = 0. By virtue of relation 4), we also have

μn (Rn \ X) = 0, that is our measure μn is concentrated on X. At the same time, we know that X has small sections by all affine hyperplanes in Rn and, moreover, by all analytic manifolds in Rn whose dimensions are strictly less than n.

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In particular, for n = 2, we obtain that there exists a measure μ2 on the Euclidean plane R2 , such that: (1) μ2 is an extension of the two-dimensional Lebesgue measure λ2 ; (2) μ2 is invariant under the group of all isometric transformations of the plane R2 ; (3) μ2 is concentrated on some subset X of R2 having the property that all linear sections of X are small (i.e., they are of cardinality strictly less than the cardinality of the continuum). Notice that the question on the existence of a measure μ2 on the plane, satisfying conditions (1) - (3), was formulated by Mabry (oral communication). Moreover, we see that the abovementioned support X of μ2 has a stronger property, namely, for each analytic curve l in R2 , the cardinality of l ∩ X is strictly less than c. In this context, let us point out that in [105] an analogous question was considered for those extensions of λ2 which are invariant under the group of all translations of R2 . More precisely, it was demonstrated in [105] that there exists a translation-invariant extension of

λ2 concentrated on a subset of R2 all linear sections of which are at most countable. Remark 1. If the Continuum Hypothesis holds, then we also can conclude that the abovementioned subset X of the plane has the property that all its linear sections are at most countable and, obviously, the same is true for sections of X by all analytic curves lying in the plane. Thus, in the case of the validity of CH, the measure μ2 is concentrated on a set all linear sections of which are countable. In this connection, the following question seems to be natural. Does there exist a measure ν2 on the plane R2 , extending λ2 , invariant under the group of all isometric transformations of R2 , and concentrated on a set all linear sections of which are finite? It turns out that the answer to this question is negative (cf. Exercise 7). Remark 2. By assuming some additional set-theoretical axioms, a result much more stronger than Theorem 1 can be established. Namely, let us suppose that for any cardinal κ < c, the space Rn cannot be covered by a κ -sequence of λn -measure zero sets. In fact, it suffices to assume this property of Rn only for n = 1, i.e., that the real line R cannot be covered by a family of λ1 -measure zero sets, whose cardinality is strictly less than c. For example, it is well known that this hypothesis follows directly from Martin’s Axiom (see, e.g., [10], [40], [67], [145], and [216]). We say that a group G of transformations of Rn is admissible if each element g from G preserves the σ -ideal of all λn -measure zero subsets of Rn . For instance, if G coincides with the group of all affine transformations of Rn , then G is

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admissible. Let us fix an admissible group G with card(G) = c. Then, applying an argument similar to the proof of Theorem 1, we easily obtain the next statement which strengthens this theorem. There exists a subset X of Rn satisfying the following conditions: 1) card(X) = c and X is almost G-invariant; 2) X is λn -thick in Rn ; 3) for any λn -measure zero subset Z of Rn , we have card(Z ∩ X) < c. Condition 3) shows, in particular, that the above-mentioned set X is a generalized Sierpi´nski subset of Rn . For the definition and various properties of Sierpi´nski sets, see [40], [148], [176], [192], and [222] where the dual objects to the Sierpi´nski sets (i.e., Luzin sets) are discussed as well. The general definition of Luzin type sets for a given ideal of subsets of an abstract space E is formulated in Exercise 14 of Appendix 1. Recall that from the measure-theoretical view-point and, more exactly, from the point of view of topological measure theory, Luzin sets in R (respectively, in Rn ) are very small because they turn out to be universal measure zero subsets of R (respectively, of Rn ). Here we would like to give one more construction of a subset of R2 of Luzin type. This construction is in connection with a certain infinite version of a well-known problem from combinatorial geometry, first considered by Erd¨os and Szekeres. Let E be a vector space (over R) and let X be a subset of E. We recall that X is said to be convexly independent in E if for each point x ∈ X, the relation x ∈ conv(X \ {x}) is valid, where conv(·) as usual denotes the operation of taking convex hulls of subsets of E. In such a case, it is also said that the points of X are in convex position. In their joint paper [63] Erd¨os and Szekeres proved the following statement. Erd¨os-Szekeres Theorem. For any natural number m  3, there exists a smallest natural number N(m) possessing the following property: every set consisting of N(m) points in general position in the plane R2 contains a subset of m points which is convexly independent, i.e., those m points are in convex position and, consequently, they are the vertices of some convex polygon. In addition, Erd¨os and Szekeres posed the problem to determine the precise value of N(m) and in connection with this problem they conjectured that N(m) = 2m−2 + 1.

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The above-mentioned intriguing problem was then investigated by many mathematicians. However, it still remains unsolved. An extensive survey about this problem and closely related questions of combinatorial geometry can be found in [178] where a further list of references is also presented. As said above, in this chapter we would like to discuss some infinite variant of the Erd¨osSzekeres problem. The main attention will be paid to those infinite sets of points, which are either countable or are of cardinality continuum. Recall that the basic technical tool utilized by Erd¨os and Szekeres in [63] is the classical combinatorial theorem of Ramsey [204]. By applying this theorem, we easily get an upper bound for N(m). The same method successfully works for point-sets lying in Euclidean spaces of higher dimension (see, e.g., [178]). Here we only need a countably infinite version of the Ramsey theorem which is formulated as follows. Ramsey Theorem. Let i be a nonzero natural number, A be an arbitrary infinite set, Fi (A) denote the family of all i-element subsets of A and let {L1, L2 , ..., Lk } be a finite covering of Fi (A). Then there exist a nonzero natural number r  k and an infinite subset B of A such that Fi (B) ⊂ Lr . It is well known that from this infinite version of the Ramsey theorem its more popular finite version can be derived with the aid of the K¨onig lemma (see, for instance, [10]) or with the aid of the existence of a nontrivial ultrafilter on the set ω of all natural numbers. Further, by applying the above-mentioned infinite version of the Ramsey theorem, the following result can easily be obtained. Theorem 3. Let n  2 be a natural number and let A be an infinite subset of the Euclidean space Rn , such that card(A ∩ l) < ω for all straight lines l lying in Rn . Then there exists an infinite convexly independent set B ⊂ A. In particular, if card(A) = ω , then we have card(B) = card(A). Proof. The assumption of the theorem implies that there are affine linear manifolds L in Rn satisfying the relation card(A ∩ L)  ω . Let L be such a manifold whose dimension is minimal. Observe that dim(L ) > 1 and every affine hyperplane H in L has the property that card(H ∩ A) < ω . By using an easy induction, we can define an infinite set A ⊂ A ∩ L of points in general position in L . Now, applying to A the above-mentioned infinite version of the Ramsey theorem (namely, for i = dim(L ) + 2 and k = 2), we get the required infinite convexly independent set B ⊂ A ⊂ A which ends the proof.

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Remark 3. It should be noticed that the use of the Ramsey theorem for obtaining the assertion of Theorem 3 is not necessary. Indeed, let us first consider the case of the Euclidean plane R2 . Suppose that A is an infinite subset of R2 such that card(A ∩ l) < ω for any straight line l lying in R2 . Then A contains an infinite countable subset A of points in general position. Only three cases are possible: 1) there exist a ray p ⊂ R2 and an infinite set {an : n < ω } ⊂ A , such that {an : n <

ω } ∩ p = 0, / the sequence of points {an : n < ω } converges to the end-point o of p and the rays oan (n < ω ) converge to p; 2) there exist a ray p ⊂ R2 and an infinite set {an : n < ω } ⊂ A , such that {an : n < ω } ∩ p = 0, / limn→∞ ||an || = +∞, limn→∞ dist(an , p) = 0; 3) there exist a ray p ⊂ R2 and an infinite set {an : n < ω } ⊂ A , such that / limn→∞ ||an || = +∞, limn→∞ dist(an , p) = +∞ {an : n < ω } ∩ p = 0, and the rays oan (n < ω ) converge to p, where o is again the end-point of p. In each of these cases, it is not difficult to define (by induction) a polygonal convex curve P with infinitely many sides whose all vertices belong to the set {an : n < ω } and hence to the set A, as well. In the first case, P can be chosen in such a way that the lengths of its sides will be tending to zero. In the second and third cases, P can be chosen so that the lengths of its sides will be tending to infinity. Suppose now that A is an infinite subset of the Euclidean space Rn , where n > 2, and suppose that card(A ∩ l) < ω for all straight lines l lying in Rn . Again, there exists an infinite countable set A ⊂ A of points in general position in some affine linear submanifold of Rn . Without loss of generality, we may assume that the above-mentioned manifold coincides with Rn . Further, since A is countable, we can find a plane L in Rn such that the orthogonal projection of Rn onto L, restricted to A , is injective and transforms A to a set of points in general position in L. As we already know, the latter set contains an infinite convexly independent subset. Now, we can readily verify that the pre-image with respect to the above-mentioned projection of this subset yields the required infinite convexly independent subset of A and hence of A, as well. It would be interesting to investigate the question whether it is possible to derive the Erd¨os– Szekeres result directly from Theorem 3 without using the finite version of the Ramsey theorem. Now, let us discuss the case when uncountable point-sets in the plane R2 are under consideration and examine an analogue of the Erd¨os-Szekeres problem for such sets.

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Let A be an uncountable subset of the plane. Assuming that all points of A are in general position, the following natural question arises. Does there exist an uncountable convexly independent subset of A? We shall demonstrate below that, in general, the existence of such a subset of A cannot be guaranteed (within ZFC theory). Let α denote the least ordinal number of cardinality continuum. Consider the family of all convex curves in R2 . Obviously, we can represent this family in the form of a transfinite sequence {Cξ : ξ < α }. By using the method of transfinite recursion, let us construct a family {aξ : ξ < ω } of points in general position in the plane. Suppose that, for an ordinal

ξ < α , the partial family of points {aζ : ζ < ξ } has already been defined. Take the curves Cζ (ζ < ξ ). Since every curve Cζ is convex, the family of all those straight lines whose intersection with Cζ contains more than two elements is at most countable. Hence there exists a straight line l ⊂ R2 such that (∀ζ < ξ )(card(l ∩Cζ )  2). Clearly, we can find a point aξ ∈ l not belonging to the set ∪{Cζ : ζ < ξ } ∪ {aζ : ζ < ξ }. Moreover, by slightly modifying the above argument, we can always choose a point aξ ∈ l in such a way that all the points aζ (ζ  ξ ) will be in general position in the plane R2 . Proceeding in this manner, we are able to construct the required family of points {aξ : ξ <

α }. Finally, we put A = {aξ : ξ < α }. Then for A, the following statement is true. Theorem 4. The set A possesses these three properties: 1) card(A) = c; 2) the points of A are in general position in R2 ; 3) no subset of A of cardinality continuum is convexly independent. Proof. The properties 1) and 2) follow immediately from our construction. Let us check the validity of 3). Suppose to the contrary that some set B ⊂ A of cardinality continuum is convexly independent and put B = cl(conv(B)),

C = bd(B ),

where cl(·) and bd(·) denote the operations of taking the closure and boundary, respectively. Then B is a closed convex subset of R2 with nonempty interior and C as the boundary of B is a convex curve. Notice that all points of B must belong to C . This fact can

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easily be deduced, e.g., from the well-known Steinitz theorem which says that a point x belongs to the interior of the convex hull of a set X ⊂ Rn if and only if there exists a set Y ⊂ X with card(Y )  2n, such that x ∈ int(conv(Y )) (see, e.g., [76] and [77]). Consequently, we get card(A ∩C )  card(B) = card(α ) = c. On the other hand, there exists an ordinal ξ < α such that Cξ = C . Taking into account the definition of A, we may write A ∩Cξ ⊂ {aζ : ζ  ξ } and hence card(A ∩C ) = card(A ∩Cξ )  card(ξ ) + ω < c. The obtained contradiction finishes the proof. Remark 4. It should be observed that we were able to construct the set A of Theorem 4 because of a rather nice structure of convex curves lying in the plane R2 . More precisely, all convex curves in the plane turn out to be so thin in R2 that any family {Ci : i ∈ I} of such curves, with card(I) < c, does not form a covering of R2 . Indeed, this moment is crucial for the transfinite construction which was described before the formulation of Theorem 4. Notice that for more general classes of curves lying in the plane, the corresponding situation can be significantly different. Namely, there are models of set theory in which the existence of a family {Pi : i ∈ I} of homeomorphic images of [0, 1], satisfying the relations card(I) < c,

∪ {Pi : i ∈ I} = R2 ,

is possible (see, e.g., Exercises 5 and 6 below). In this situation, no construction analogous to the presented above can be carried out. Remark 5. Theorem 4 admits a direct generalization to the case of the Euclidean space Rn , where n  2. In other words, there exists a set A ⊂ Rn of cardinality continuum whose points are in general position in Rn but no subset of A of the same cardinality is convexly independent. The proof of this result is rather similar to the proof of Theorem 4. However, some insignificant technical details occur in the case of Rn . Remark 6. In various topics of real analysis, measure theory, and set-theoretical topology analogous constructions are known for obtaining those sets which almost avoid the members of a given family of sets (cf. [148], [159], [176], [192], and [225]). In most

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situations, the given family of sets forms an ideal of subsets of an original space. For example, we have the ideal of all measure zero sets in a nonzero complete measure space and the ideal of all first category sets in a nonempty Baire topological space. As a rule, the corresponding transfinite constructions need some additional set-theoretical axioms. In this connection, see [148], [176] or [192] where by assuming the Continuum Hypothesis the classical constructions of so-called Luzin sets and Sierpi´nski sets are discussed in detail (see also Exercise 14 for Appendix 1). EXERCISES 1∗ . We say that X ⊂ R2 is a Mazurkiewicz subset of R2 if for every straight line l ⊂ R2 , we have card(X ∩ l) = 2. Show that if a Mazurkiewicz set X is λ2 -measurable, then λ2 (X) = 0. Formulate and prove an analogous assertion in terms of the Baire property and Baire category. In the latter case, use the topological analogue of the Fubini theorem due to Kuratowski and Ulam (see [148] or [192]). Further, by applying the method of transfinite induction, prove that the following two statements are valid: (a) there exists a Mazurkiewicz subset of R2 which is thick with respect to λ2 ; (b) there exists a Mazurkiewicz subset of R2 which is of λ2 -measure zero and nowhere dense in R2 . 2. Let J be a set of cardinality continuum and let {l j : j ∈ J} be an injective family of all straight lines in R2 . Fix any family {n j : j ∈ J} of natural numbers such that n j  2 for each index j ∈ J. By applying the method of transfinite induction, show that there exists a set Y ⊂ R2 such that (∀ j ∈ J)(card(l j ∩Y ) = n j ). In particular, if n j = 2 for all j ∈ J, then Y is a Mazurkiewicz subset of R2 . 3. Give a proof of Lemma 1 by using the Fubini theorem. Moreover, generalize this lemma to the case of analytic submanifolds {Mi : i ∈ I} of a given analytic manifold M equipped with an appropriate analogue of the Lebesgue measure, where card(I) < c and dim(Mi ) < dim(M) for all i ∈ I. 4∗ . Prove the following statement due to Sierpi´nski. If n  3, then there exists a free group G of rotations of the space Rn about its origin, such that card(G) = c.

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For this purpose, keep in mind Lemma 1 and construct the required G by means of the method of transfinite recursion. 5∗ . As known, there is a model of set theory (e.g., Cohen’s original model) in which a certain subset of R having cardinality strictly less than c is of strictly positive outer λ1 measure and hence is not measurable with respect to λ1 . Show that in such a model there exists a covering {Xi : i ∈ I} of R by nowhere dense compact subsets Xi (i ∈ I), for which card(I) < c. Deduce from this fact that in the same model there is a covering {Zi : i ∈ I} of R2 by Jordan curves Zi (i ∈ I), where again card(I) < c. Recall that a Jordan curve is usually defined as a homeomorphic image of the onedimensional unit torus S1 (= T). 6∗ . Infer the main result of the previous exercise from the following set-theoretical hypothesis: c is measurable in the Ulam sense. In other words, prove that if c is a real-valued measurable cardinal, then there exists a covering {Xi : i ∈ I} of R such that card(I) < c and all Xi (i ∈ I) are nowhere dense compact sets in R. 7∗ . Show that if a set Z ⊂ R2 is finite in a direction e ∈ R2 , then Z is necessarily R2 negligible. For the definition of negligible sets, see Exercise 9 of Chapter 5. Derive from this fact that there exists no subset Z of R2 such that: (a) Z is finite in a fixed direction e ∈ R2 ; (b) μ (R2 \ Z) = 0 for some nonzero σ -finite translation-invariant measure μ on R2 . 8. Give a proof of the statement formulated in Remark 2. In other words, show the existence of a subset X of Rn satisfying conditions 1), 2) and 3) formulated therein. 9. Let us identify the plane R2 with the field C of all complex numbers and consider two mappings

φ : C → C, ψ : C → C defined as follows:

φ (z) = z + 1, ψ (z) = ei z

(z ∈ C),

where, as usual, i2 = −1, ei = cos(1) + isin(1).

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Obviously, φ is a translation of R2 and ψ is a rotation of R2 about its origin (0, 0). Denote by G the semigroup of motions of R2 generated by {φ , ψ } and put X = G((0, 0)). Taking into account the fact that ei is a transcendental number, show that

φ (X) ∩ ψ (X) = 0, / φ (X) ∪ ψ (X) = X. In other words, the countable set X ⊂ R2 admits a decomposition (i.e., partition) into two subsets such that each of them is G-congruent to X. This result is due to Mazurkiewicz and Sierpi´nski (cf. [240]) and does not need any form of the Axiom of Choice. Show also that if Y is a nonempty bounded subset of R2 , then there exists no partition {Y1 ,Y2 } of Y such that the sets Y , Y1 and Y2 are pairwise congruent. 10∗ . Let T2 denote the group of all translations of R2 and let g be a rotation of R2 such that all gn (1  n < ω ) are distinct from the identical transformation of R2 . Denote by G the group of transformations generated by T2 ∪ {g}. Prove that there exists a set X ⊂ R2 satisfying the following relations: (a) X is T2 -negligible; (b) X is G-absolutely nonmeasurable. In connection with the result presented in Exercise 10, let us point out that it is unknown whether there exists a set Y ⊂ R2 which is T2 -absolutely negligible and, simultaneously, M2 -absolutely nonmeasurable where M2 stands for the group of all isometric transformations (i.e., motions) of R2 . 11. Consider in detail the cases 1) - 3) of Remark 3 and, for each of them, construct the required polygonal convex curve P with infinitely many sides. 12. Generalize Theorem 4 to the case of the Euclidean space Rn , where n  2. In other words, prove that there exists a set X ⊂ Rn satisfying the following relations: (a) card(X) = c and all points of X are in general position; (b) no subset Y of X with card(Y ) = c is convexly independent. Conclude from this fact that, under the Continuum Hypothesis, there exists an uncountable set X ⊂ Rn of points in general position such that no uncountable subset of X is convexly independent. 13. Let k  3 be a natural number. By using the method of transfinite recursion, demonstrate that there exists a set Y ⊂ R2 satisfying the following conditions:

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(a) Y is λ2 -thick in R2 ; (b) every circumference in R2 has exactly k common points with Y . Condition (b) shows that Y may be regarded as a certain analogue of the Mazurkiewicz set for the family of all circumferences in the plane. 14∗ . Let k > 0 be a natural number and let {X j : j ∈ J} be a family of subsets of R2 . We shall say that this family is a k-homogeneous covering of R2 if every point in R2 belongs exactly to k sets from the family. Show that there exists no 1-homogeneous covering of R2 by Jordan curves. On the other hand, use the method of transfinite recursion and prove that for k  2, there exists a khomogeneous covering of R2 by pairwise congruent circumferences. Moreover, for any even k > 0, give a direct construction (within ZF theory) of a khomogeneous covering of R2 by pairwise congruent circumferences. Taking into account the result of Exercise 13, conclude that the following two dual statements are equivalent within ZF theory and are valid within ZFC theory: (a) there exists a subset X of R2 such that every circumference of radius 1 in R2 meets X in precisely three points; (b) there exists a family of circumferences of radius 1 in R2 such that every point of R2 belongs to exactly three circumferences of the family. Finally, give an example of a set X ⊂ R2 which has the property formulated in (a) but simultaneously is such that card(C ∩ X) = 4 for some circumference C ⊂ R2 . 15∗ . Let Z be a subset of R2 satisfying the following two conditions: (a) every straight line in R2 meets Z in at least two points; (b) Z contains a simple arc, that is Z contains a set homeomorphic to [0, 1]. Demonstrate that there exists a straight line l ⊂ R2 such that card(l ∩ Z)  3. Conclude from this fact that no Mazurkiewicz set (see Exercise 1) contains a simple arc. Show also that no Mazurkiewicz set can be an Fσ -subset of R2 . For this purpose, suppose otherwise and infer that some Mazurkiewicz set must contain a simple arc which contradicts the said above. 16∗ . Let Y be a subset of R2 . We shall say that Y is an ot-set if for any three pairwise distinct points y1 , y2 , y3 from Y , the triangle [y1 , y2 , y3 ] is non-degenerate and has an obtuse angle. Demonstrate the validity of the following assertions:

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(a) every ot-set is contained in a maximal (with respect to inclusion) ot-set; (b) no finite ot-set is maximal; (c) there exists a countable maximal ot-set. In addition to these facts, verify that {(cos(t), sin(t)) : 0  t < π } is a maximal ot-set in R2 whose cardinality is equal to c. 17∗ . Let f : R2 → R2 be a partial homeomorphism such that both sets dom( f ) and ran( f ) are homeomorphic to the Cantor space {0, 1}ω . Prove that there exists a homeomorphism f ∗ : R2 → R2 extending f (cf. Exercise 20 from Chapter 8). It should be noticed that the result just formulated cannot be extended to those partial homeomorphisms f : R3 → R3 whose domains and ranges are homeomorphic to the Cantor space. In order to verify this circumstance, we need to apply some properties of the so-called Antoine necklace (see, for instance, [19] where this necklace is thoroughly considered in connection with a ”wild” topological embedding of the sphere S2 into R3 ). 18∗ . Within ZF theory, show that there are two subsets X and Y of R2 satisfying the following relations: (a) card(X) = card(Y ) = c; (b) for any two isometric transformations f and g of R2 , the set f (X) ∩ g(Y ) contains exactly two points. Formulate and prove an analogous result replacing 2 by an even natural number 2k > 2. Applying the method of transfinite recursion, construct subsets X  and Y  of R2 such that: (a’) card(X  ) = card(Y  ) = c; (b’) for any two isometric transformations f and g of R2 , the set f (X  )∩g(Y  ) is a singleton. It should be mentioned that the existence of X  and Y  satisfying both relations (a’) and (b’) was first established by Sierpi´nski. 19∗ . Let E be a set and let {Xi : i ∈ I} be a family of subsets of E. Recall that {Xi : i ∈ I} is said to be independent (in the set-theoretical sense) if for each finite subset J of I and for every function f : J → {0, 1}, we have f ( j)

∩{X j f ( j)

where X j

f ( j)

= X j if f ( j) = 0, and X j

: j ∈ J} = 0/

= E \ X j if f ( j) = 1.

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It is well known that for any infinite set E, there exists an independent family {Xi : i ∈ I} of subsets of E such that card(I) = 2card(E) (in this connection, see [150]). The fact just indicated easily implies that the family of all nontrivial ultrafilters in E has cardinality 22

card(E)

.

Suppose now that E = R2 . Construct an infinite independent family of convex polygons in R2 . On the other hand, show that there exists no uncountable independent family of simple polygons in E. 20∗ . We say that a compact set P ⊂ R2 is a quasi-polygon in R2 if P has nonempty interior and the boundary of P can be represented as the union of countably many line segments. Observe that any polygon in R2 is a quasi-polygon and, obviously, the converse assertion fails to be true. Prove that there exists an uncountable independent family of convex quasi-polygons in R2 .

Some other results concerning combinatorial and descriptive properties of subsets of the Euclidean space Rn (in particular, of the plane R2 ) can be found in [2], [48], [49], [51], [54], [55], [76], [77], [105], [156], [169], [222], and [223].

Chapter 20

Restrictions of real-valued functions

This last chapter may be regarded as dual to Chapter 1 in which various extensions of partial functions were considered. In 1897 Poincar´e wrote that on the intuitive level every function f : R → R is closely related to some continuous function (see [200]). This phrase can be expressed in the following form: certain (of course, nontrivial) restrictions of f turn out to be continuous on their domains. Here we intend to consider those restrictions of real-valued functions which have rather good descriptive properties. Let us formulate the main problem which will be touched in the present chapter. Let f : R → R be a function. Does there exist a large (or, at least, non-small) subset X of R such that the restriction f |X possesses nice structural properties? For instance, we may require that f |X should be continuous (continuous at some points of X) or should belong to a given Baire class or should be measurable in the Lebesgue sense or should be with the Baire property (with the Baire property in the restricted sense). We will give only several results in this direction which seem to be interesting from the point of view of real analysis. To illustrate very briefly the said above, let us begin with the simplest fact of this type. Let g : R → R be a function. It is natural to ask whether there exists a set X ⊂ R with card(X) = card(R) = c such that the restriction g|X is bounded. The answer to this question is trivially positive. Indeed, consider the sets Xn = g−1 ([−n, n]) (n < ω ). Then, in view of the equality R = ∪{Xn : n < ω }, at least one set Xn0 is of cardinality continuum. Putting X = Xn0 , we see that (∀x ∈ X)(| f (x)|  n) A.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7_20, © 2009 Atlantis Press/World Scientific

313

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and so we come to the required result. Evidently, our argument was based on the relation c f (c) > ω which can be proved only with the aid of the Axiom of Choice. In a more general situation, if we have a set E with card(E)  ω , then we may assert that the following two statements are equivalent: 1) c f (card(E)) > ω ; 2) for any function f : E → R, there exists a set X ⊂ E such that card(X) = card(E) and the restriction f |X is bounded. We thus see that the simple question posed above which, at first sight, is of purely analytic flavor turns out to be closely connected with set-theoretical characteristics of the continuum. In the sequel, several other (less trivial) examples of this kind will be pointed out and discussed in detail. Now, let us recall the classical theorem of Blumberg [15] which is rather surprising because no assumptions on functions under consideration are required in its formulation (cf. Poincar´e’s prediction at the beginning of this chapter). Theorem 1. Let f : R → R be an arbitrary function. Then there exists a subset D of R satisfying the following conditions: (a) D is everywhere dense in R; (b) the restricted function f |D is continuous. The proof of this theorem was sketched in Exercises 21 and 22 for Chapter 8. There are many works devoted to various versions of Blumberg’s theorem (see [8], [27], [28], [29], [89], [206], and [241]). In view of the existence of Sierpi´nski-Zygmund type functions (see Chapters 1, 13 and 14), it is impossible to assert that such a D can be chosen to be uncountable. So the class of all functions acting from R into R should be replaced by a narrower class if we want to get good restrictions of functions to some non-small (e.g., in the sense of cardinality) subsets of R. It seems natural to treat all nonempty perfect sets in R as non-small subsets of R. We now turn our attention to those real-valued functions whose restrictions to nonempty perfect subsets of R are not totally discontinuous. Let E be a topological space and let f : E → R be a function. We shall say that f is of Baire zero class if f is continuous at all points of E, i.e., f is continuous on E. The family of all continuous functions acting from E into R is usually denoted by the symbol C(E, R). In accordance with the definition above, the notation Ba0 (E, R) will be used for the same family of functions. Thus, we have Ba0 (E, R) = C(E, R).

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The next important family of functions is the class of all Baire 1 functions. By the standard definition due to Baire (see [3], [4], and [5]), a function f :E →R is of Baire 1 class (or of first Baire class) if there exists a sequence { fn : n < ω } of functions from Ba0 (E, R) such that limn→+∞ fn (x) = f (x) for each point x ∈ E. In other words, f : E → R belongs to the first Baire class if and only if f can be represented as the pointwise limit of a sequence of functions belonging to Ba0 (E, R). It is well known that the functions of first Baire class play a significant role in various topics of real analysis. The following simple but important example emphasizes this circumstance. Example 1. Let E = R and let f : E → R be a derivative. This assumption means that there exists a continuous function g : E → R such that g (x) = f (x) for all x ∈ R. Define a sequence {gn : 1  n < ω } of real-valued functions on R by the formula gn (x) = n(g(x + 1/n) − g(x))

(x ∈ R, n = 1, 2, ...).

Obviously, we have limn→+∞ gn (x) = f (x)

(x ∈ R).

Since each gn (1  n < ω ) is a continuous function on R, we conclude that f belongs to the first Baire class. For any topological space E, the family of all functions f : E → R belonging to the first Baire class will be denoted by the symbol Ba1 (E, R). Notice that the other Baire classes Baξ (E, R) of real-valued functions on E are naturally introduced by iterating the limit process and using the method of transfinite recursion on

ξ < ω1 (in this connection, see [5], [99], [148], [162], [183], and Appendix 6). Here our main interest is concentrated on functions from the class Ba1 (E, R). The following simple properties are easily implied by the definition of Ba1 (E, R). (i) Ba1 (E, R) is a linear algebra over the field R; in other words, if we have f ∈ Ba1 (E, R), g ∈ Ba1 (E, R), a ∈ R and b ∈ R, then a f + bg ∈ Ba1 (E, R),

f · g ∈ Ba1 (E, R).

(ii) If f ∈ Ba1 (E, R), g ∈ Ba1 (E, R) and g(x) = 0 for all x ∈ E, then f /g ∈ Ba1 (E, R).

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(iii) If f ∈ Ba1 (E, R) and φ : ]a, b[ → R is a continuous function such that ran( f ) ⊂ ]a, b[, then

φ ◦ f ∈ Ba1 (E, R). Below, we will consider some other properties of the class Ba1 (E, R), which are not so trivial. Lemma 1. Let E be a topological space, {an : 1  n < ω } be a sequence of strictly positive real numbers such that

∑

1n<ω

an < +∞,

and let { fn : 1  n < ω } ⊂ Ba1 (E, R) be a sequence of functions such that | fn (x)| < an

(x ∈ E, 1  n < ω ).

Define a function f : E → R by the formula f (x) = f1 (x) + f2 (x) + ... + fn(x) + ...

(x ∈ E).

Then f also belongs to Ba1 (E, R). Proof. First of all, let us notice that the function f is well defined since the series f1 (x) + f2 (x) + ... + fn(x) + ...

(x ∈ E)

converges uniformly with respect to x ∈ E (in view of the relations | fn (x)| < an for any x ∈ E and n  1). Further, since fn ∈ Ba1 (E, R) for each natural number n  1, we can write fn (x) = limk→+∞ fn,k (x)

(x ∈ E),

where fn,k (k = 1, 2, ...) are some real-valued continuous functions on E. Without loss of generality, we may assume that | fn,k (x)|  an

(x ∈ E, n = 1, 2, ..., k = 1, 2, ...).

Now, let us put hk (x) = f1,k (x) + f2,k (x) + ... + fk,k (x)

(x ∈ E, k = 1, 2, ...).

Clearly, the functions hk are continuous on E. It suffices to show that f (x) = limk→+∞ hk (x)

(x ∈ E).

For this purpose, take an arbitrary real number ε > 0. There exists a natural number m such that am+1 + am+2 + ... + ai + ... < ε /3.

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Consequently, we have | fm+1 (x)| + | fm+2 (x)| + ... + | fi (x)| + ... < ε /3 | fm+1,k (x)| + | fm+2,k (x)| + ... + | fi,k (x)| + ...  ε /3

(x ∈ E),

(x ∈ E, k = 1, 2, ...).

For any x ∈ E and any k > m, we can write the inequalities | f (x) − hk (x)|  | f1 (x) − f1,k (x)| + | f2 (x) − f2,k (x)| + ... + | fm(x) − fm,k (x)|+ | fm+1 (x)| + | fm+2 (x)| + ... + | fi (x)| + ... + | fm+1,k (x)| + | fm+2,k (x)| + ... + | fk,k (x)|  | f1 (x) − f1,k (x)| + | f2 (x) − f2,k (x)| + ... + | fm (x) − fm,k (x)| + 2ε /3. If x ∈ E is fixed, then we can find k0 < ω so large that for all natural numbers k > k0 , the relation | f1 (x) − f1,k (x)| + | f2 (x) − f2,k (x)| + ... + | fm (x) − fm,k (x)| < ε /3 will be satisfied. But this relation immediately yields the inequality | f (x) − hk (x)| < ε for every integer k > k0 . Therefore, we get limk→+∞ hk (x) = f (x)

(x ∈ E),

which completes the proof of Lemma 1. We need this lemma in order to prove the following result due to Baire (cf. [5], [99], [148], [162], and [183]). Theorem 2. Let E be a topological space. Suppose that a sequence { fn : n < ω } ⊂ Ba1 (E, R) is given, uniformly converging to a function f : E → R. Then f also belongs to the class Ba1 (E, R). Proof. According to our assumption, for any natural number k, there exists a natural number nk such that | f (x) − fnk (x)| <

1 2k+1

(x ∈ E).

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Evidently, we may assume that n0 < n1 < ... < nk < ... . Let us consider the series of functions ( fn1 − fn0 ) + ( fn2 − fn1 ) + ... + ( fnk+1 − fnk ) + ... . Since the inequalities 1 1 1 + < 2k+2 2k+1 2k hold true for all x ∈ E, we can apply Lemma 1 to this series. So we obtain that the function | fnk+1 (x) − fnk (x)|  | fnk+1 (x) − f (x)| + | fnk (x) − f (x)| <

g = ( fn1 − fn0 ) + ( fn2 − fn1 ) + ... + ( fnk+1 − fnk ) + ... belongs to Ba1 (E, R). But it is easy to see that g = limk→+∞ fnk+1 − fn0 = f − fn0 . According to property (i), we finally get f = g + fn0 ∈ Ba1 (E, R). The theorem has thus been proved. Remark 1. Theorem 2 implies that for E = 0, / the family of all bounded functions from B1 (E, R) becomes a Banach space with respect to the norm of uniform convergence, i.e., with respect to the standard sup-norm || f || = supx∈E | f (x)|. Remark 2. Theorem 2 can be directly generalized to the case of the Baire class Baξ (E, R), where ξ is an arbitrary ordinal number strictly less than ω1 . The proof essentially remains the same as above (cf. [148]). Lemma 2. Let E be a topological space and let g ∈ Ba1 (E, R). Then for every t ∈ R, the pre-images g−1 (] − ∞,t[) = {x ∈ E : g(x) < t}, g−1 (]t, +∞[) = {x ∈ E : g(x) > t} are Fσ -subsets of E. Proof. Take any t ∈ R. Because of g ∈ Ba1 (E, R), there exists a sequence {gn : n < ω } ⊂ Ba0 (E, R) such that limn→+∞ gn (x) = g(x)

(x ∈ E).

It is not difficult to verify the following two relations: g(x) < t ⇔ (∃k < ω )(∃n < ω )(∀m ∈ [n, ω [)(gm (x)  t − 1/k),

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g(x) > t ⇔ (∃k < ω )(∃n < ω )(∀m ∈ [n, ω [)(gm (x)  t + 1/k). These relations yield at once that the above-mentioned sets {x ∈ E : g(x) < t}, {x ∈ E : g(x) > t} are of type Fσ in E, and the lemma is proved. Lemma 3. Let E be a normal topological space, g : E → R be a function and suppose that ran(g) = {t1 ,t2 , ...,tk }. If for any integer i ∈ [1, k], the set Xi = {x ∈ E : g(x) = ti } = g−1 (ti ) is an Fσ -subset of E, then g ∈ Ba1 (E, R). Proof. Obviously, we can write E = X1 ∪ X2 ∪ ... ∪ Xk , Xi = Fi,0 ∪ Fi,1 ∪ ... ∪ Fi,n ∪ ...

(i = 1, 2, ..., k),

where all Fi,n (1  i  k, n < ω ) are closed subsets of E and Fi,0 ⊂ Fi,1 ⊂ ... ⊂ Fi,n ⊂ ... . Let us put Fn = F1,n ∪ F2,n ∪ ... ∪ Fk,n

(n = 0, 1, 2, ...)

and define a function gn : Fn → R

(n = 0, 1, 2, ...)

by the formula: gn (x) = ti if and only if x ∈ Fi,n . Because the finite family of closed sets {F1,n , F2,n , ..., Fk,n } is disjoint, the function gn is continuous on the closed set Fn . By the Tietze-Urysohn theorem, gn admits a continuous extension g∗n : E → R. Now, it is easy to check that limn→+∞ g∗n (x) = g(x) for all points x ∈ E and, therefore, g ∈ Ba1 (E, R) which finishes the proof of the lemma. Lemma 4. Let E be a topological space in which every open set is of type Fσ (or, equivalently, in which every closed set is of type Gδ ). Let a set X ⊂ E be representable in the form X = A1 ∪ A2 ∪ ... ∪ Ak ,

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where all A j ( j = 1, 2, ..., k) are Fσ -subsets of E. Then X is representable in the form X = B1 ∪ B2 ∪ ... ∪ Bk , where B j ⊂ A j ( j = 1, 2, ..., k), all sets B j are also of type Fσ and, in addition, they are pairwise disjoint. Proof. Obviously, we have the equality X = F1 ∪ F2 ∪ ... ∪ Fi ∪ ..., where all sets Fi (1  i < ω ) are closed in E and each Fi is contained in some set A j(i) . Let us put C1 = F1 , C2 = F2 \ F1 , ..., Ci = Fi \ (F1 ∪ ... ∪ Fi−1), ... . The family of sets {Ci : 1  i < ω } is disjoint and in view of our assumption on E, all Ci are of type Fσ in E. Moreover, we have X = C1 ∪C2 ∪ ... ∪Ci ∪ ... . Now, for any natural number j ∈ [1, k], define the set B j by the formula B j = ∪{Ci : j is the least number such that Ci ⊂ A j }. Clearly, the family {B1 , B2 , ..., Bk } is disjoint, all sets B j ( j = 1, 2, ..., k) are of type Fσ in E and Bj ⊂ Aj

( j = 1, 2, ..., k),

X = B1 ∪ B2 ∪ ... ∪ Bk . Lemma 4 has thus been proved. Recall that a topological space E is perfectly normal if E is normal and each open set in E is of type Fσ . For such spaces, the next important statement due to Lebesgue is true (cf. [5], [148], and [183]). Theorem 3. Let E be a perfectly normal space and let f : E → R be a function. The following three assertions are equivalent: (1) f ∈ Ba1 (E, R); (2) for any t ∈ R, both sets {x ∈ E : f (x) < t} and {x ∈ E : f (x) > t} are of type Fσ in E; (3) for any open set U ⊂ R, the pre-image f −1 (U) is an Fσ -subset of E. Proof. The equivalence (2) ⇔ (3) is trivial. The implication (1) ⇒ (2) was established by Lemma 2 even for an arbitrary topological space E. Consequently, it remains to prove the

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implication (2) ⇒ (1). Suppose that (2) is valid and suppose first that ran( f ) ⊂ ]0, 1[ or, equivalently, 0 < f (x) < 1

(x ∈ E).

For any integer n  1, consider the sequence {t0 ,t1 , ...,tn } of points of R determined by the conditions t0 = 0, t j+1 − t j = 1/n

( j = 0, ..., n − 1).

In fact, we have t j = j/n for j = 0, ..., n. Further, introduce the following sets: A0 = {x ∈ E : f (x) < t1 }, An = {x ∈ E : f (x) > tn−1 }, A j = {x ∈ E : t j−1 < f (x) < t j+1 }

( j = 1, ..., n − 1).

Obviously, we have the equality E = A0 ∪ A1 ∪ ... ∪ An and all A j ( j = 0, 1, ..., n) are Fσ -subsets of E. Applying Lemma 4, we get another representation E = B0 ∪ B1 ∪ ... ∪ Bn , where all B j ( j = 0, 1, ..., n) are also Fσ -subsets of E, are pairwise disjoint, and Bj ⊂ Aj

( j = 0, 1, ..., n).

Further, define a function fn : E → R by putting: fn (x) = t j if and only if x ∈ B j . According to Lemma 3, the function fn belongs to Ba1 (E, R). Now, take an arbitrary x ∈ E. Then x ∈ B j for some integer j ∈ [0, n]. If j = 0, then we have t0 < f (x) < t1 , fn (x) = t0 , | f (x) − fn (x)| < 1/n. If j = n, then we have tn−1 < f (x) < tn ,

fn (x) = tn , | f (x) − fn (x)| < 1/n.

Finally, if 1  j  n − 1, then t j−1 < f (x) < t j+1 , fn (x) = t j , | f (x) − fn (x)| < 2/n. All these relations show that limn→+∞ fn (x) = f (x) uniformly with respect to x ∈ E. By virtue of Theorem 2, we obtain that f ∈ Ba1 (E, R).

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Suppose now that f : E → R is an arbitrary function satisfying (2). Fix any increasing homeomorphism φ : R → ]0, 1[ and consider the function φ ◦ f . This function also satisfies (2) and ran(φ ◦ f ) ⊂ ]0, 1[. As demonstrated above, φ ◦ f ∈ Ba1 (E, R). Consequently, in view of (iii), f = φ −1 ◦ (φ ◦ f ) ∈ Ba1 (E, R) which completes the proof of Theorem 3. Example 2. Let E be a perfectly normal space and let X be a subset of E. Denote by χX the characteristic function of X. It is easy to check that if X is closed in E, then relation (2) of Theorem 2 is satisfied for f = χX . Therefore, according to this theorem, we have

χX ∈ Ba1 (E, R). Now, if Y is an open subset of E, then taking into account the equality χY = 1 − χE\Y , we see that χY ∈ Ba1 (E, R), too (by virtue of property (i)). The above-mentioned facts also directly follow from Lemma 3. Example 3. Let E be a subinterval of R and let f : E → R be a monotone function. It is easy to verify that for any t ∈ R, both sets {x ∈ E : f (x) < t}, {x ∈ E : f (x) > t} are some subintervals of E. Because each interval in E is an Fσ -subset of E, we claim (in view of Theorem 3) that f ∈ Ba1 (E, R). The same conclusion is true for those f : E → R which are of finite variation on E. Indeed, such functions are representable in the form of the difference of two increasing functions on E (see, e.g., [183] or [210]) and it suffices to apply property (i) of the class Ba1 (E, R). Example 4. Let E be again a perfectly normal space and let f : E → R be an upper semicontinuous function. According to the definition of upper semicontinuous functions, for any t ∈ R, the set {x ∈ E : f (x) < t} is open in E and hence is an Fσ -subset of E. At the same time, the set {x ∈ E : f (x) > t} = ∪n<ω {x ∈ E : f (x)  t + 1/(n + 1)} is the union of countably many closed sets, i.e., is also an Fσ -subset of E. Using Theorem 3, we deduce that f ∈ Ba1 (E, R). From this fact it immediately follows again by virtue of property (i) that g ∈ Ba1 (E, R) for any lower semicontinuous function g : E → R.

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Actually, the characteristic function χX of a closed set X ⊂ E (see Example 2) is upper semicontinuous. Recall that a topological space E is Baire if no nonempty open subset of E is of first category in E. For such an E, an important result due to Baire is well known which yields an essential information about the structure of the set D( f ) of all discontinuity points of an arbitrary function f ∈ Ba1 (E, R). Theorem 4. Let E be a Baire space and let f ∈ Ba1 (E, R). Then the set D( f ) of all discontinuity points of f is of first category in E. Consequently, we have C( f ) ∩U = 0/ for any nonempty open set U ⊂ E (where C( f ) as usual denotes the set of all continuity points of f ). Proof. For each point x ∈ E, let us put Ω f (x) = inf{diam( f (V )) : V ∈ V (x)}, where V (x) is the filter of all neighborhoods of x and diam( f (V )) denotes the diameter of f (V ). The real number Ω f (x)  0 is usually called the oscillation of f at x (cf. Exercise 13 from Chapter 8). As is well known (see Exercise 3 for this chapter), the equality C( f ) =



{x ∈ E : Ω f (x) < 1/n}

1n<ω

is valid, where all sets {x ∈ E : Ω f (x) < 1/n} are open in E. So, it suffices to demonstrate that all these sets are everywhere dense in E. In other words, it suffices to show that for any real number ε > 0 and for any nonempty open set U ⊂ E, there exists a nonempty open set W ⊂ U such that (∀x ∈ W )(∀y ∈ W )(| f (x) − f (y)| < ε ). Taking into account the fact that f belongs to Ba1 (E, R), choose a sequence { fk : k < ω } ⊂ Ba0 (E, R) satisfying the relation f (x) = limk→+∞ fk (x)

(x ∈ E).

Further, for any natural number k, define the set Xk = {x ∈ E : (∀i  k)(∀ j  k)(| fi (x) − f j (x)|  ε /3)}. Obviously, all sets Xk (k < ω ) are closed in E and (∀k < ω )(Xk ⊂ Xk+1 ), E = ∪{Xk : k < ω }.

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Consequently, we have U = (U ∩ X0 ) ∪ (U ∩ X1 ) ∪ ... ∪ (U ∩ Xk ) ∪ ... . Since E is a Baire space, there exists a natural number n such that int(U ∩ Xn ) = 0. / Let V ⊂ U ∩ Xn be a nonempty open set. If x is an arbitrary point of V , then (∀i  n)(∀ j  n)(| fi (x) − f j (x)|  ε /3). Putting j = n and tending i to +∞, we get (∀x ∈ V )(| f (x) − fn (x)|  ε /3). Therefore, we can write | f (y) − f (x)|  | f (y) − fn (y)| + | fn (y) − fn (x)| + | fn (x) − f (x)|  2ε /3 + | fn(y) − fn (x)| for any two points x and y from V . Finally, because fn is a continuous function, there exists a nonempty open set W ⊂ V such that (∀x ∈ W )(∀y ∈ W )(| fn (y) − fn (x)| < ε /3). This relation implies at once that (∀x ∈ W )(∀y ∈ W )(| f (y) − f (x)| < ε ) which ends the proof of the Baire theorem. Remark 3. The proof presented above is fairly standard and, actually, is based on the classical argument originally due to Baire (cf., for instance, [183] and [192]). More general versions of the Baire theorem with further information about it can be found in [58] and [148]. In fact, the main part of the Baire theorem remains true for an arbitrary topological space E (see Exercise 6 of this chapter). The next classical result is widely known as the Cantor-Baire stationarity principle (see [58], [148], and [183]). Theorem 5. Let E be a hereditarily Lindel¨of topological space and let F0 ⊃ F1 ⊃ ... ⊃ Fξ ⊃ ... (ξ < ω1 )

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be a decreasing (with respect to the inclusion relation) ω1 -sequence of closed subsets of E. Then there exists an ordinal α < ω1 such that (∀ξ ∈ [α , ω1 [)(Fξ = Fα ). Proof. For each ordinal ξ < ω1 , let us denote Gξ = E \ Fξ and put G = ∪{Gξ : ξ < ω1 }. Then the family {Gξ : ξ < ω1 } is an open covering of an open set G ⊂ E. Since E is hereditarily Lindel¨of, there exists a countable set Ξ ⊂ [0, ω1 [ such that G = ∪{Gξ : ξ ∈ Ξ}. In view of the regularity of ω1 , the set Ξ is bounded from above in [0, ω1 [, i.e., there exists an ordinal α < ω1 for which we have sup(Ξ) < α . Now, it is easy to verify that Fξ = Fα for all ordinals ξ ∈ [α , ω1 [. Remark 4. A certain analogue of the Cantor-Baire stationarity principle was obtained by Luzin for decreasing ω1 -sequences of Fσ -subsets of R and this analogue was successfully applied to some deep problems of descriptive set theory (in this connection, see [160], [162], and [188]). The next example yields a good illustration of the usefulness of the Cantor-Baire stationarity principle. Example 5. Suppose again that E is a hereditarily Lindel¨of topological space. Take an arbitrary closed subset X of E and define by transfinite recursion an ω1 -sequence {Xξ : ξ <

ω1 } in the following manner: (a) X0 = X; (b) Xξ +1 = (Xξ ) for any ξ < ω1 , where (Xξ ) denotes the set of all accumulation points of Xξ ; (c) Xξ = ∩{Xζ : ζ < ξ } for any limit ordinal ξ < ω1 . Applying the Cantor-Baire stationarity principle to {Xξ : ξ < ω1 }, we infer that there exists a smallest ordinal ξ0 < ω1 for which we have Xξ = Xξ0

(ξ0  ξ < ω1 ).

Since E is hereditarily Lindel¨of, all sets Xξ \ Xξ +1 (ξ < ω1 ) are at most countable. Taking into account the equality X = (∪{Xξ \ Xξ +1 : ξ < ξ0 }) ∪ Xξ0 ,

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we conclude that X admits a representation in the form X =Y ∪Z

(Y ∩ Z = 0), /

where Y = Xξ0 is a perfect subset of E and Z is at most countable. Recall that this representation was first obtained by Cantor and Bendixson (the CantorBendixson theorem). Remark 5. According to the classical Baire theorem (see [5], [148], and [183]), for any function g : R → R, the following two assertions are equivalent: (1) g is of first Baire class; (2) for every nonempty closed set F ⊂ R, there are points x ∈ F at which the function g|F is continuous. In other words, (2) yields a characterization of all functions acting from R into R and belonging to the first Baire class. This characterization is given in terms of certain restrictions of functions. Unfortunately, we do not have an analogous nice characterization of the derivatives on R, which form an important proper subclass of Ba1 (R, R) (see Example 1). Notice that the proof of the equivalence (1) ⇔ (2) is outlined in Exercise 9 for this chapter. The Cantor-Baire stationarity principle is heavily exploited in that proof. According to the definition introduced by Luzin, a topological space X is always of first category (or X is perfectly meager) if each nonempty dense in itself subset of X is a first category space. Let E be a topological space and let X be a subspace of E such that, for every perfect set P ⊂ E, the set X ∩ P is of first category in P. Then it is not difficult to show that the space X is always of first category (see, e.g., [148]). Luzin proved that there exists an uncountable subspace X of R which is always of first category (see [148] or [162]). Other constructions of uncountable universally small sets can be found in [197] and [250] (cf. also Appendices 1 and 6). Here we wish to give one construction of such a kind. It needs Martin’s Axiom because a perfectly meager set obtained by means of this construction possesses deep additional properties. As usual, we denote by λ the standard Lebesgue measure on the real line R. Theorem 6. Suppose that Martin’s Axiom holds. Then there exists a set X ⊂ R satisfying the following conditions: (1) for every nonempty perfect set P ⊂ R, the intersection X ∩ P is of first category in P; (2) for each Lebesgue measurable set Y ⊂ R with λ (Y ) > 0, the intersection X ∩ Y is nonempty;

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(3) X is a generalized Sierpi´nski subset of R. Proof. Let α denote the least ordinal number whose cardinality is equal to c (actually, we may identify α with c). Let {Zξ : ξ < α } denote the family of all those Borel subsets of R which have strictly positive Lebesgue measure, i.e., {Zξ : ξ < α } = B(R) \ I (λ ), and let {Tξ : ξ < α } denote the family of all those Borel subsets of R which have Lebesgue measure zero, i.e., {Tξ : ξ < α } = B(R) ∩ I (λ ). For any ordinal ξ < α , we fix a partition {Zξ0 , Zξ1 } of Zξ such that Zξ0 is a Lebesgue measure

zero set and Zξ1 is a first category subset of Zξ . Notice that the existence of such a partition

follows directly from Lemma 2 of Appendix 3. Now, we define an injective α -sequence {xξ : ξ < α } of real numbers. Suppose that ξ < α and that the partial ξ -sequence {xζ :

ζ < ξ } has already been determined. Let us consider the set Dξ = (∪{Zζ0 : ζ  ξ }) ∪ {xζ : ζ < ξ } ∪ (∪{Tζ : ζ  ξ }). Martin’s Axiom implies that this Dξ is also of Lebesgue measure zero. Hence we have / So we can choose a point xξ from the above-mentioned nonempty difference Zξ \ Dξ = 0. of two sets. In this way, we are able to define the entire α -sequence {xξ : ξ < α } of points of R. Now, put X = {xξ : ξ < α }. We are going to show that X is the required set. Let P be any nonempty perfect subset of R. If λ (P) = 0, then for some ordinal ξ < α , we may write P = Tξ . Consequently, from the method of construction of the set X, we immediately obtain card(P ∩ X) < c. Applying Martin’s Axiom once more, we see that the intersection P ∩ X is of first category in P. Suppose now that λ (P) > 0. Then for some ordinal ξ < α , we can write P = Zξ . Therefore, P ∩ X ⊂ Zξ1 ∪ {xζ : ζ < ξ }. Taking account of the fact that the set P does not have isolated points, we obtain from the last inclusion that P ∩ X is again of first category in P. Hence condition (1) is satisfied for

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our set X. Furthermore, since xξ ∈ Zξ for each ordinal ξ < α , we conclude that condition (2) holds for the set X, too. The validity of condition (3) follows directly from our construction. Theorem 6 has thus been proved. Lemma 5. Let E be a hereditarily Lindel¨of T1 -space always of first category and let f : E → R be a function. Then for each subspace X of E, the set D( f |X) is of first category in X. Proof. The argument is very easy, so we only sketch it. Namely, we begin with establishing the fact that X admits a representation in the form X = Y ∪ Z, where Y is dense in itself, Z is at most countable, and Y ∩Z = 0/ (cf. Example 5 and Exercise 8). Then we verify the inclusion D( f |X) ⊂ Y ∪ (X  ∩ Z), where X  denotes the set of all accumulation points of X (in E). Finally, observe that both sets Y and X  ∩ Z are of first category in X which completes the proof. Theorem 7. Let E be a subspace of R satisfying the following relations: (a) card(E) = c; (b) E is always of first category. Then there exists a function f : E → R having the following properties: (c) f is not Borel (consequently, f does not belong to the class Ba1 (E, R)); (d) for any subspace Z of E, the set D( f |Z) is of first category in Z. Proof. Since the equality card(E) = c holds, the family B(E, R) of all Borel mappings acting from E into R is also of cardinality continuum. But the family of all mappings acting from E into R is of cardinality 2c . So we can choose a function f : E → R which is not Borel on E, i.e., such an f has property (c). In view of Lemma 5, the same f has property (d) which ends the proof. Remark 6. Recall that the existence of a subspace E of R satisfying the relations (a) and (b) was demonstrated by using Martin’s Axiom (see Theorem 6 of this chapter). The classical result formulated in Theorem 7 is due to Luzin (see [148] and [162]). In particular, it shows that property (d) of f does not imply the relation f ∈ Ba1 (E, R). Theorem 8. Let f : R → R be a Lebesgue measurable function (or let f : R → R have the Baire property). Then there exists a nonempty perfect set P ⊂ R such that the restriction f |P is monotone on P.

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Proof. The argument given below is fairly standard and is usually applied in many similar situations (cf. [122]). As is well known, there exists a nonempty perfect subset T of R such that the restriction f |T is continuous. This fact is true for all Lebesgue measurable functions and for all those functions which possess the Baire property. Of course, we may assume that diam(T ) < 1 where the symbol diam(T ) stands for the diameter of T . Let us denote g = f |T and suppose that for every nonempty perfect set Q ⊂ T , the restriction g|Q is not decreasing. Then by using ordinary induction, we can construct a dyadic system (Ti1 i2 ...ik )i1 ∈{0,1},i2 ∈{0,1},...,ik ∈{0,1}

(k < ω )

of nonempty perfect subsets of T satisfying the following conditions: (a) T0/ = T ; (b) Ti1 i2 ...ik ik+1 ⊂ Ti1 i2 ...ik ; (c) Ti1 i2 ...ik 0 ∩ Ti1 i2 ...ik 1 = 0; / (d) diam(Ti1 i2 ...ik ) < 1/2k ; (e) if (i1 , i2 , ..., ik ) ≺ ( j1 , j2 , ..., jk ), then x < y and g(x) < g(y) for all points x ∈ Ti1 i2 ...ik and y ∈ T j1 j2 ... jk , where  denotes the standard lexicographical ordering in the set of all k-sequences whose terms belong to {0, 1}. The details of the above-mentioned construction are left to the reader. Now, we define P=



(∪{Ti1 i2 ...ik : (i1 , i2 , ...ik ) ∈ {0, 1}k }).

k<ω

It follows from the construction of P that g|P = f |P is an increasing function on P which completes the proof of Theorem 8. The natural question arises whether the preceding theorem admits further generalizations. In particular, can we assert that it is possible to choose the above-mentioned nonempty perfect set P in such a way that λ (P) > 0? We will show that the answer to the posed question is negative. For this purpose, we need the existence of a continuous function f : R → R which is nowhere approximately differentiable. This fact is much deeper than the existence of a continuous nowhere differentiable function. The definition of an approximate derivative is given in Appendix 4 with all other auxiliary notions needed for our purposes. The following classical result was first obtained by Jarnik (see [88]). Theorem 9. There exist continuous bounded functions acting from R into R which are nowhere approximately differentiable.

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Actually, Jarnik proved that almost all (in the sense of the Baire category) functions from the Banach space C[0, 1] are nowhere approximately differentiable. Notice that this result significantly strengthens the corresponding result of Banach and Mazurkiewicz for the usual differentiability (see Exercise 4 of Chapter 8). Remark 7. In [122] the existence of a nowhere approximately differentiable function is applied to the question concerning some relationships between the notions of supmeasurability and weak sup-measurability of functions acting from R × R into R. Now, we are ready to prove the following statement. Theorem 10. There exists a continuous function f : R → R such that for every perfect set P ⊂ R with λ (P) > 0, the restriction f |P is not monotone on P. Proof. Let f : R → R be a continuous, bounded, and nowhere approximately differentiable function (see Theorem 9 above). We are going to show that f is the required function. Take any perfect set P ⊂ R whose Lebesgue measure is strictly positive. We assert that f |P cannot be monotone. Suppose otherwise, i.e., f |P is either increasing or decreasing. Without loss of generality, we may assume that f |P is increasing. Denote by f∗ : R → R some increasing function extending f |P (see Example 5 from Chapter 1). According to the classical result of Lebesgue, f ∗ is differentiable almost everywhere. Consequently, there exists x ∈ P such that x is a density point of P and f ∗ is differentiable at x. By virtue of the relation f ∗ |P = f |P, this circumstance immediately implies the fact that the original function f is approximately differentiable at x. But the latter contradicts the definition of f . The contradiction obtained finishes the proof. Remark 8. Actually, the function f of Theorem 10 is such that for every set X ⊂ R with

λ ∗ (X) > 0, the restriction f |X is not monotone; the proof remains the same as before. It should be underlined that this result holds true within ZFC theory. The question naturally arises whether in ZFC theory there exists a function g : R → R such that for every set X ⊂ R with λ ∗ (X) > 0, the restriction g|X is not continuous. As shown by Roslanowski and Shelah [208], the existence of g cannot be proved within ZFC. On the other hand, assuming Martin’s Axiom, any Sierpi´nski-Zygmund function can play the role of the abovementioned g. In view of Theorems 8 and 10, it makes sense to introduce the following definitions. Let f : R → R be a function and let L be a family of subsets of R.

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We say that f is relatively monotone with respect to L if there exists at least one set X ∈ L for which the restriction f |X is monotone. We say that f is absolutely nonmonotone with respect to L if f is not relatively monotone with respect to L (i.e., there exists no set X ∈ L for which f |X is monotone). Example 6. Any function f : R → R turns out to be relatively monotone with respect to the class of all countably infinite subsets of R. This fact can easily be deduced from the infinite version of the Ramsey theorem (see Chapter 19) and also admits a direct simple proof. In some sense, the above-mentioned result cannot be strengthened because, under the Continuum Hypothesis, any Sierpi´nski-Zygmund function is absolutely nonmonotone with respect to the class of all uncountable subsets of R. In connection with Example 6, the following statement seems to be of interest. Theorem 11. Assuming the Continuum Hypothesis, there exists a function g : R → R which is not a Sierpi´nski-Zygmund function but is absolutely nonmonotone with respect to the family of all uncountable subsets of R. Proof. Let us take an arbitrary continuous nowhere differentiable function g1 : R → R and an arbitrary Sierpi´nski-Zygmund function g2 : R → R. Let L be a Luzin subset of R, the existence of which is implied by the Continuum Hypothesis (see Exercise 14 for Appendix 1). Define the function g : R → R as follows: g(x) = g1 (x) if x ∈ L and g(x) = g2 (x) if x ∈ R \ L. Let us check that g is the required function. First of all, g is not a Sierpi´nski-Zygmund function because g|L = g1 |L is continuous. Let now X be an uncountable subset of R. Only two cases are possible. 1. card(X ∩ L)  ω . In this case, we have card(X ∩ (R \ L)) > ω . Since g2 is a Sierpi´nskiZygmund function, the restriction g|(X ∩ (R \ L)) = g2 |(X ∩ (R \ L)) is not monotone (see Example 6 of this chapter or Example 8 of Chapter 1). 2. card(X ∩ L) > ω . Suppose for a while that g|(X ∩ L) = g1 |(X ∩ L) is monotone. Then the continuous function g1 should be monotone on the closure of X ∩ L. Further, since g1 is nowhere differentiable, the set cl(X ∩ L) is necessarily nowhere dense in R. Taking into account the fact that L is a Luzin set, we get card(L ∩ cl(X ∩ L))  ω ,

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which contradicts the inclusion X ∩ L ⊂ L ∩ cl(X ∩ L) and the uncountability of X ∩ L. The obtained contradiction finishes the proof. Some other results concerning restrictions of various functions to non-small subsets of the real line can be found in the works [8], [27], [28], [29], [31], [35], [42], [89], [148], [206], [208], and [214]. EXERCISES 1. Suppose that E is a separable topological space (i.e., E contains a countable everywhere dense subset). Show that card(Ba1 (E, R))  c. In fact, show that if E is nonempty and separable, then card(Ba1 (E, R)) = c. 2. Verify that the set Q ⊂ R of all rational numbers is not a Gδ -subset of R. Deduce from this circumstance that the characteristic function χQ (the so-called Dirichlet function) does not belong to Ba1 (R, R). Actually, χQ is of second Baire class. More generally, prove that if E is an uncountable Polish space without isolated points and X is a countable everywhere dense subset of E, then the characteristic function χX does not belong to Ba1 (E, R). 3. Let E be a topological space and let f : E → R be a function. As usual, denote by D( f ) the set of all discontinuity points of f and, respectively, denote by C( f ) the set of all continuity points of f . Verify that: (a) Ω f (x) = 0 if and only if f is continuous at x (equivalently, Ω f (x) > 0 if and only if x is a discontinuity point of f ); (b) for any t ∈ R, the set {x ∈ E : Ω f (x)  t} is closed in E; (c) the set D( f ) is representable in the form D( f ) = E1 ∪ E2 ∪ ... ∪ En ∪ ..., where En = {x ∈ E : Ω f (x)  1/n} for each integer n  1. Conclude from (b) and (c) that D( f ) is an Fσ -subset of E and therefore the set C( f ) of all continuity points of f is a Gδ -subset of E. Let B be a base of open sets in R and let F = {Y ⊂ R : R \ Y ∈ B}. Prove the equality D( f ) = ∪{cl( f −1 (Y )) \ f −1 (Y ) : Y ∈ F }.

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Generalize these results to an arbitrary function f : E → E  where E  is a given metric space. 4∗ . Let E be a nonempty resolvable topological space (see Chapter 10) and let X be an Fσ -subset of E. Show that there exists a function f : E → R such that X coincides with the set D( f ) of all discontinuity points of f . For this purpose, first represent X in the form X = F1 ∪ F2 ∪ ... ∪ Fn ∪ ..., where all sets Fn are closed in E and Fn ⊂ Fn+1 for each natural number n  1. Further, put F0 = 0/ and for any integer n  1, define a function fn : E → {0, 1} satisfying the relations: (a) fn is equal to zero at all points of the set E \ Fn ; (b) Ω fn (x) = 1 for each point x ∈ Fn . Now, take a sequence {an : n  1} of strictly positive real numbers, such that an+1 + an+2 + ... + ak + ... < an

(n = 1, 2, ...).

For example, it suffices to put an =

1 3n

(n  1).

Finally, consider the function f = a1 f1 + a2 f2 + ... + an fn + ... . This function is well defined because the series on the right-hand side of the above equality converges uniformly on E. Verify that: (c) f is continuous at all points of the set E \ X; (d) for any integer n  1 and for all points x ∈ Fn \ Fn−1, we have Ω f (x)  an − ∑ ak > 0. k>n

Conclude from (c) and (d) that D( f ) = X. Some further results closely connected with Exercise 4 can be found in [17] and [189]. 5. Let E be a perfectly normal topological space and let f : E → R be a function whose graph is closed in the product space E × R. Applying the Kuratowski theorem on closed projections (see Chapter 8), show that f is of first Baire class.

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Give an example of a function f : R → R satisfying the following two conditions: (a) the graph Gr( f ) is closed in R × R; (b) D( f ) is a nonempty perfect subset of R. 6. Let E be an arbitrary topological space and let f ∈ Ba1 (E, R). Demonstrate that the set D( f ) is of first category in E (for this purpose, use Exercise 3 and Lemma 2). Another way to show this fact is based on the Banach statement (see Exercise 12 for Appendix 3) which leads to a representation of E in the form E = E0 ∪ E1 , where E0 is an open Baire subspace of E and E1 is a first category closed subset of E. Applying Theorem 4 to the set D( f |E0 ), we get the required result. Deduce from this result that if f ∈ Ba1 (E, R) and X is an arbitrary subspace of E, then the set D( f |X) is of first category in X. In particular, conclude that if E is a complete metric space, f ∈ Ba1 (E, R) and X is a nonempty closed subspace of E, then there exist points in X at which f |X is continuous. 7. Give another proof of the Cantor-Bendixson theorem (see Example 5) that does not use the method of transfinite induction. For this purpose, take an arbitrary closed set X in a hereditarily Lindel¨of space E, consider the set of all condensation points of X and take it as Y . Then define the set Z by the equality Z = X \ Y and check that Z is at most countable. 8∗ . Let E be a topological space. We say that E is scattered if E does not contain a nonempty dense in itself subset. Demonstrate that the following two assertions are equivalent: (a) E is scattered; (b) E can be represented in the form of an injective α -sequence E = {eξ : ξ < α }, where α is some ordinal number and, for any ξ < α , the element eξ is an isolated point of the set {eζ : ξ  ζ < α }. Notice that the implication (b) ⇒ (a) is trivial. Supposing now that (a) is valid, use the method of transfinite recursion for obtaining the required representation of E. Finally, demonstrate that every topological space X can be expressed in the form X = Y ∪ Z where Y is a perfect subset of X, Z is a scattered subset of X, and Y ∩ Z = 0. / This classical result is due to Cantor.

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9∗ . Let E be a separable metric space and let g : E → R be a function such that for every nonempty closed set F ⊂ E, there exists a point x ∈ F at which the restricted function g|F is continuous. Prove that g is of first Baire class. This very important statement is due to Baire and yields a characterization of all functions belonging to Ba1 (E, R). The following argument enables to establish the above-mentioned result. Take any a ∈ R and b ∈ R such that a < b, and denote A = {x ∈ E : g(x) > a}, B = {x ∈ E : g(x) < b}. Clearly, we have the equality E = A ∪ B. Further, construct by transfinite recursion an

ω1 -sequence F0 ⊃ F1 ⊃ ... ⊃ Fξ ⊃ ... (ξ < ω1 ) of closed subsets of E. Put F0 = E. Suppose that for a given ξ < ω1 , the partial family {Fζ : ζ < ξ } has already been defined. If ξ is a limit ordinal, then we put Fξ = ∩{Fζ : ζ < ξ }. If ξ = η + 1, consider the set Fη . Only two cases are possible. / In this case, we define Fξ = Fη = 0. / (i) Fη = 0. (ii) Fη = 0. / In this case, there exists a point x ∈ Fη at which the function g|Fη is continuous. Consequently, there exists an open neighborhood V (x) of x such that Fη ∩V (x) ⊂ A ∨ Fη ∩V (x) ⊂ B. Then we define Fξ = Fη \ V (x). Proceeding in this way, we will be able to construct the sets Fξ (ξ < ω1 ). Observe now that for each ordinal ξ < ω1 , we have Fξ \ Fξ +1 ⊂ A ∨ Fξ \ Fξ +1 ⊂ B and according to the Cantor-Baire stationarity principle, for some α < ω1 , the equalities Fξ = 0/

(α  ξ < ω1 )

are valid. Deduce from these facts that there exist two sets A and B such that A ⊂ A, B ⊂ B, A ∪ B = E, A ∩ B = 0/ and both A and B are Fσ -subsets of E. Now, let {bn : n < ω } be a strictly decreasing sequence of real numbers satisfying the relation limn→+∞ bn = a. Put again A = {x ∈ E : g(x) > a}

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and for each n < ω , define Bn = {x ∈ E : g(x) < bn }. As above, show the existence of Fσ -sets An and Bn such that An ⊂ A, Bn ⊂ Bn , An ∪ Bn = E, An ∩ Bn = 0. / Finally, denote X = ∪{An : n < ω } and verify that X = {x ∈ E : g(x) > a} = A. Hence A is an Fσ -subset of E. By using a similar argument, demonstrate that B = {x ∈ E : g(x) < b} is an Fσ -subset of E, too. Conclude, in view of the Lebesgue theorem (i.e., Theorem 3 of this chapter) that the function g is of first Baire class. In connection with Exercise 9, we would like to mention that more general versions of the result presented above can be found in monograph [148] where a different argument is applied. Namely, it is proved there that if E is a complete metric space and g : E → R is a function, then the following two assertions are equivalent: (1) g ∈ Ba1 (E, R); (2) for any nonempty closed set F ⊂ E, there exists a point x ∈ F at which the restricted function g|F is continuous. The Baire theorem (i.e., Theorem 4 of this chapter) and the preceding exercise establish the equivalence (1) ⇔ (2) in the case of a Polish space E. For a nonseparable complete metric space E, the proof of (1) ⇔ (2) relies on properties of the Montgomery operation (see [148] and Exercise 4 from Appendix 3). Notice that this operation needs uncountable forms of the Axiom of Choice. In Luzin’s opinion, the Baire characterization of all functions of first class is a jewel of real analysis (see [162]). 10. Let E be a Polish topological space and let f : E → R be a function whose set of discontinuity points is at most countable. Show, by applying Exercise 9, that f ∈ Ba1 (E, R). Infer from this fact that any function g : [a, b] → R of finite variation on a segment [a, b] belongs to the first Baire class. Another way to establish this result was indicated in Example 3.

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11. Let E be a topological space and let X be a subset of E such that for every perfect set P ⊂ E, the intersection X ∩ P is of first category in P. Show that X is always of first category. 12∗ . A function f : R → R is called nowhere constant if for any non-degenerate subinterval U of R, the restriction f |U is not constant. Let F denote the family of all those continuous functions acting from R into itself which are nowhere constant. Prove under the Continuum Hypothesis that there exists a set X ⊂ R having the following property: if f ∈ F and g ∈ F are any functions from F such that f (X) = g(X), then f = g. Every set X with this property is called a magic set for the family F (cf. [11] and [42]). Argue by using the method of transfinite induction. Let { fξ : ξ < ω1 } be an injective enumeration of all functions from F. Suppose that for an ordinal number ξ < ω1 , the two families {xζ ,n : ζ < ξ , n < ω }, {yη ,ζ : η < ζ < ξ } of points of R are defined in such a way that for η < ζ < ξ , the relation yη ,ζ ∈ fζ (W ) \ fη (W ) holds, where W = {xζ ,n : ζ < ξ , n < ω }. Further, denoting W  = {yη ,ζ : η < ζ < ξ }, put U = ∪{ fζ (W ) : ζ  ξ }, V = (∪{ fζ−1 (U) : ζ  ξ }) ∪ (∪{ fζ−1 (W  ) : ζ  ξ }). Observe that the set U is countable and the set V is of first category in R. Then represent the family { fζ : ζ < ξ } in the form of an ω -sequence {gn : n < ω } and construct by ordinary recursion an injective ω -sequence {xξ ,n : n < ω } of points in R satisfying the following relations: (a) xξ ,n ∈ V for all n < ω ; (b) fξ (xξ ,n ) = gn (xξ ,m ) for all n < ω and m < ω . Afterwards, define an appropriate family {yη ,ξ : η < ξ } of points in R. Finally, put X = {xξ ,n : ξ < ω1 , n < ω }

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and check that X is a magic set for the family F. Obtain the same result under Martin’s Axiom (instead of the Continuum Hypothesis). A similar statement on the existence of a magic set for those Lebesgue measurable functions which are not constant on subsets of R of strictly positive λ -measure can be found in [35]. In connection with Exercise 12, the result of Ciesielski and Shelah should also be mentioned which states that there exists a model of ZFC where there are no magic sets for the family F (see [42]). 13∗ . Assuming Martin’s Axiom, prove that there exists a partial continuous function f : R → R satisfying the following two conditions: (a) the set dom( f ) is λ -thick in R; (b) for any set Y ⊂ dom( f ) with card(Y ) = c, the restriction f |Y is not monotone. For this purpose, consider a suitable restriction of a continuous nowhere approximately differentiable function whose existence is guaranteed by Theorem 9 of the present chapter (see also Appendix 4). 14. Let (E, ) be an infinite, Dedekind complete, dense, linearly ordered set endowed with its order topology and let f :E →R be a function having the Baire property. Formulate and prove for f an analogue of Theorem 8. Apply the obtained result to the particular case where (E, ) is a Suslin line (see Exercise 16 from Appendix 3).

Appendix 1

Some set-theoretical facts and constructions

The main goal of this Appendix is to present several purely set-theoretical statements and constructions which are useful in various fields of modern mathematics and, therefore, have a number of nontrivial and, quite often, unexpected applications. Primarily, we will be concerned with fundamental facts of infinite combinatorics. The reader will see in the sequel that some of them are closely connected with the measure extension problem (cf. Chapters 2, 16) and, actually, were essentially inspired by this problem. We begin with the standard definition of a tree which is a special (but extremely important) case of a partially ordered set. Let (E, ) be a partially ordered set. We say that (E, ) is a tree if the following conditions are satisfied: (a) (E, ) has a least (smallest) element; (b) for any e ∈ E, the set E(e) = {x ∈ E : x ≺ e} is well-ordered. It trivially follows from condition (a) that any tree is nonempty. The least element of a tree (E, ) is called its root. For each element e ∈ E, the ordinal type of the above-mentioned well-ordered set E(e) is called the height of e and is usually denoted by o(e). In particular, the height of the root is 0. If β is an ordinal, then the set Eβ of all those elements from E whose heights are equal to

β can be considered. This set is called the β -th level of E. In other words, we have Eβ = {e ∈ E : o(e) = β }. Notice that all elements of the β -th level are pairwise incomparable (because no wellordered set is isomorphic to its proper initial subinterval). A.B. Kharazishvili, Topics in Measure Theory and Real Analysis, Atlantis Studies in Mathematics 2, DOI 10.1007/978-94-91216-36-7, © 2009 Atlantis Press/World Scientific

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We also define o(E) = sup{o(e) + 1 : e ∈ E}. The ordinal o(E) is called the height of a given tree (E, ). It can readily be seen that the height of E coincides with the least element of the class of all those ordinals α for which the α -th level of (E, ) is empty. A tree (E, ) is called an α -tree if its height is equal to α . In particular, an ω -tree is a tree all whose n-th levels are nonempty, where n < ω , but its

ω -th level is empty. We say that a set X is a branch in a tree (E, ) if X is a maximal linearly ordered (by the induced ordering) subset of (E, ). Actually, any branch is a maximal well-ordered subset of (E, ) (this fact follows directly from the definition of trees). Let α be an ordinal number. We say that a set X ⊂ E is an α -branch in (E, ) if X is a branch with height equal to α . We say that (E, ) is a K¨onig tree if (E, ) is an ω -tree all levels of which are finite. The following important statement is due to K¨onig (cf. [91], [144], [145], and [150]). Theorem 1. Let (E, ) be a K¨onig tree. Then there exists an ω -branch in (E, ). Proof. For every natural number n, a subset X of E will be called a partial n-branch in (E, ) if X is linearly ordered by the induced ordering, has nonempty intersection with all k-th levels of (E, ) where k < n, but does not intersect the k-th levels of (E, ), where k  n. Obviously, every partial n-branch contains exactly n elements. An element e ∈ E will be called admissible if for any natural number n > 0, there exists a partial n-branch containing e. It can readily be proved (by ordinary induction) that there are admissible elements in every nonempty level of (E, ). In particular, the root of (E, ) is admissible. Moreover, suppose that a finite sequence (e0 , e1 , . . . , en ) of admissible elements is constructed such that the set {e0 , e1 , . . . , en } is linearly ordered and each ek belongs to the k-th level (i.e., this set is a partial (n + 1)-branch in (E, )). Then, by using the finiteness of the (n + 1)-th level of (E, ), it is not difficult to infer that there exists an admissible element en+1 from the (n + 1)-th level such that en ≺ en+1 . Continuing in this manner, we will obtain an ω -branch (e0 , e1 , . . . , en , . . .)

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of the given K¨onig tree (E, ). This completes the proof. As already mentioned, the assertion just proved is due to K¨onig and is known in the literature as the K¨onig lemma. It yields useful consequences in graph theory, combinatorics and many other areas of mathematics. For example, by applying this lemma it can be established that if any finite subgraph of a given countable graph is planar (i.e., has an isomorphic copy in the Euclidean plane R2 ), then the graph is planar, too. Notice also that the proof of the K¨onig lemma needs some weak form of the Axiom of Choice. In other words, Theorem 1 is not provable within ZF theory (see Exercise 2 below).

It is natural to ask whether certain analogues of the K¨onig lemma are valid for those trees whose height is uncountable but all levels have relatively small sizes in comparison with the height. First, let us introduce a relevant notion. We shall say that the branch property holds for a given tree (E, ) if there exists at least one branch in (E, ), whose height is equal to the height of (E, ). The K¨onig lemma states that all K¨onig trees have the branch property. So it is reasonable to ask whether an analogous result is true for ω1 -trees with countable levels. However, the answer to the question just posed is negative. The corresponding counterexample was originally constructed by Aronszajn (cf. [91], [144], and [145]). Theorem 2. There exists an ω1 -tree (E, ) such that: (1) all levels of (E, ) are countable; (2) (E, ) does not have the branch property. Proof. Recall that the symbol ω <ω1 denotes the family of all functions f such that dom( f ) is a proper initial subinterval of ω1 and ran( f ) is a subset of ω . Using this notation, let us put F = { f ∈ ω <ω1 : f is an injection}. For any two functions f ∈ F and g ∈ F, define f ∼ g if and only if dom( f ) = dom(g),

card({ξ ∈ dom( f ) : f (ξ ) = g(ξ )}) < ω .

Applying the method of transfinite recursion, we will construct an ω1 -sequence { f ξ : ξ < ω1 } ⊂ F

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satisfying the following three relations: (a) dom( fξ ) = [0, ξ [ for each ordinal ξ < ω1 ; (b) fα |[0, β [ ∼ fβ for any two ordinals α < ω1 and β < α ; (c) card(ω \ ran( fξ )) = ω for each ordinal ξ < ω1 . Notice first that if for some ξ < ω1 , the function fξ is already defined, then the role of fξ +1 can be played by any function from F which extends fξ and whose domain coincides with [0, ξ ] = [0, ξ + 1[. The case of limit countable ordinals needs a more delicate argument. Let ξ < ω1 be a limit ordinal and suppose that the partial family of functions { fζ : ζ < ξ } satisfying the relations (a), (b), (c) has already been defined. Take an arbitrary strictly increasing sequence of ordinal numbers {ξn : n < ω } such that lim ξn = ξ ,

n→+∞

and construct (by ordinary recursion) a sequence { fn : n < ω } ⊂ F with the following properties: (i) fn ∼ fξn for each n < ω ;  |[0, ξ [ = f  for each n < ω . (ii) fn+1 n n

Simple details of such a construction are left to the reader. Now, let fξ denote the unique function defined on [0, ξ [ and extending all functions fn (n <

ω ). Obviously, the existence of fξ follows directly from property (ii).

Also, it is easy to verify that fξ ∈ F and fξ |[0, ζ [ ∼ fζ for all ordinals ζ < ξ . Further, it is not hard to show that an injective mapping fξ : [0, ξ [ → ω

can be defined satisfying the next two conditions: (iii) fξ (ζ ) = fξ (ζ ) for any ordinal ζ ∈ [0, ξ [ \ {ξn : n < ω }; (iv) the set ω \ ran( fξ ) is infinite. A straightforward verification shows that fξ is the required function, i.e., the relations (a), (b), (c) are valid for the partial family { fζ : ζ  ξ }. Finally, having the ω1 -sequence of functions { f ξ : ξ < ω1 } ⊂ F with properties (a), (b), and (c), we put T = { f ∈ F : (∃ξ < ω1 )( f ∼ fξ )}

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and equip T with its standard ordering  (in other words, f  g if and only if g extends f ). Summarizing all the said above, we readily conclude that the partially ordered set (T, ) is an ω1 -tree, all levels of (T, ) are countable, and (T, ) does not possess the branch property because there exists no injective function acting from ω1 into ω . Theorem 2 has thus been proved. (T, ) is usually called an Aronszajn tree. At first sight, Aronszajn trees seem to be useless in measure theory and real analysis. But it turns out that such trees can successfully be applied to concrete problems in the abovementioned fields of mathematics. For example, in connection with the general measure extension problem Ryll-Nardzewski posed the following natural question. Let (E, S , μ ) be a probability measure space and let L be a family of subsets of E. Suppose that for every countable subfamily L  of L , there exists a measure μ  extending μ and defined on the σ -algebra generated by S ∪ L  . Does there exist a measure ν also extending μ and defined on the σ -algebra generated by S ∪ L ? It turns out that the answer to this question is negative. The corresponding counterexample was constructed by K.P.S. Bhaskara Rao with the aid of an Aronszajn tree. Theorem 2 shows that within ZFC theory the K¨onig lemma does not admit a natural generalization to all uncountable cardinals and we have seen that even the first uncountable cardinal provides us with a counterexample. Nevertheless, the branch property holds for some trees of large cardinality. For instance, cardinals which are real-valued measurable (i.e., measurable in the Ulam sense) may be treated as a certain kind of large cardinals. We recall that a cardinal a is real-valued measurable if there exists a nonzero σ -finite (equivalently, probability) diffused measure μ on a with dom(μ ) = P(a), where P(a) denotes the power set of a. The existence of real-valued measurable cardinals cannot be established within ZFC theory (see [91], [145], [238], or Exercise 18). For the least of such cardinals, the next statement is valid. Theorem 3. Let (E, ) be a tree satisfying the following two conditions: (a) o(E) is the first real-valued measurable cardinal; (b) all levels of E are of cardinality strictly less than o(E). Then (E, ) has the branch property. Proof. In fact, the argument given below shows that a more general result is true. Namely, if κ is a cardinal number with a κ -additive diffused probability measure on P(κ )

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and (E, ) is a κ -tree all levels of which are of cardinality strictly less than κ , then (E, ) has the branch property. Of course, in this formulation κ is identified with the least ordinal of cardinality κ and we recall that some approaches to the notion of a cardinal number automatically yield such an identification. To establish the above-mentioned generalized result, let us define E ∗ (x) = {y ∈ E : x  y} for each x ∈ E and let Eα be the α -th level of E for any ordinal α < κ . Denote by μ a

κ -additive diffused probability measure on P(E) and put t0 (x) = μ (E ∗ (x)), t1 (α ) = sup{t0 (x) : x ∈ Eα }. Since the equality

μ (∪{E ∗ (x) : x ∈ Eα }) = 1 holds for any α < κ , we infer in view of the κ -additivity of μ that (∀α < κ )(t1 (α ) > 0). Furthermore, if α  β < κ , then we have t1 (α )  t1 (β ). Therefore, there exist an ordinal γ < κ and a real number r > 0 such that (∀α ∈ [γ , κ [)(t1 (α ) = r). Now, we introduce the set S = {x : (∃α  γ )(x ∈ Eα & t0 (x) > (1/2)r)}. Obviously, this S satisfies the relation S ∩ Eα = 0/ for all those α  γ which are strictly smaller than κ . In particular, we have card(S) = κ . The set S possesses also the following property: (∀y ∈ S)(∃x ∈ S)(x ∈ Eγ & x  y).

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Indeed, to verify this fact, take any y ∈ S. Then y ∈ Eα for some α  γ and t0 (y) > (1/2)r. Pick an element x  y from the level Eγ . Clearly, the inclusion E ∗ (y) ⊂ E ∗ (x) holds from which it follows that t0 (x) > (1/2)r and hence x ∈ S, as well. Taking into account the circumstance that card(Eγ ) < κ , we deduce that card(S ∩ E ∗ (z)) = κ for some z ∈ Eγ . Finally, let us demonstrate that any two distinct elements u and v from S ∩ E ∗ (z) are comparable with respect to . Supposing to the contrary that u  v and v  u, we get / E ∗ (u) ∩ E ∗ (v) = 0, E ∗ (u) ∪ E ∗ (v) ⊂ E ∗ (z), t0 (u) > (1/2)r,

t0 (v) > (1/2)r

and, consequently, t0 (z)  t0 (u) + t0(v) > r, which is impossible in view of the relation t0 (z)  t1 (γ ) = r. The obtained contradiction shows that S ∩E ∗ (z) is a linearly ordered subset of E whose cardinality equals κ . Evidently, this subset can be included in some κ -branch of E. Theorem 3 has thus been proved. Combining Theorems 2 and 3, we readily obtain the following result which is due to Ulam but was established by him in absolutely another manner (see [148], [176], [192], and [238] where a transfinite matrix of a certain combinatorial type is essentially exploited; compare also with Exercise 18 for this Appendix). Theorem 4. ω1 is not a real-valued measurable cardinal. Proof. Indeed, suppose to the contrary that ω1 is real-valued measurable. The unique infinite cardinal strictly less than ω1 is ω , which obviously is not real-valued measurable. So we conclude that ω1 necessarily is the first real-valued measurable cardinal. According to Theorem 3, all ω1 -trees with countable levels should have the branch property. But this circumstance contradicts the existence of an Aronszajn tree (see Theorem 2).

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Theorem 4 is of special interest to us because it is closely connected with the general measure extension problem (see Chapter 2). It is also connected with one classical result of Sierpi´nski [218] (cf. [222], [223]), which will be proved below. To present this result with some related ones, we need several auxiliary propositions. Lemma 1. Let E be a set and let f : E → R be a partial function. There exists a countably generated σ -algebra S of subsets of E such that f is measurable with respect to B(R) and S , i.e., we have (∀X ∈ B(R))( f −1 (X) ∈ S ). Proof. Let {Vn : n < ω } be a countable topological base of R. For example, we may take the family of all open intervals in R with rational end-points. Putting S = σ ({ f −1 (Vn ) : n < ω }), we get the required countably generated σ -algebra of subsets of E. Lemma 2. Let E be a subset of R and let f : E → E be a partial function. Then there exists a countably generated σ -algebra S of subsets of E such that Gr( f ) ∈ S ⊗ S , where Gr( f ) as usual denotes the graph of f . This statement can readily be deduced from Lemma 1 (see Exercise 6). In addition to this fact, it should be noticed that we may assume without loss of generality that S contains all singletons in E. Indeed, it suffices to consider the σ -algebra σ (S ∪ B(E)), instead of S , and take into account the fact that if S is countably generated, then σ (S ∪ B(E)) is also countably generated. Lemma 3. Let Z be any subset of the product set ω1 × ω1 . There exists a countably generated σ -algebra S of subsets of ω1 such that Z ∈ S ⊗ S . Proof. In the product set ω1 × ω1 consider the following two subsets: A = {(ξ , ζ ) : ξ < ζ },

B = {(ξ , ζ ) : ζ  ξ }.

Obviously, we have the equalities A ∪ B = ω1 × ω1 , A ∩ B = 0. / In other words, {A, B} is a partition of the product set ω1 × ω1 , which is frequently called the Sierpi´nski partition of ω1 × ω1 . Additionally, the sets A and B satisfy the following two relations:

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(a) card(A ∩ (ω1 × {ζ }))  ω for every ζ < ω1 ; (b) card(B ∩ ({ξ } × ω1))  ω for every ξ < ω1 . Evidently, we can write the equality Z = (Z ∩ A) ∪ (Z ∩ B). This equality and the relations (a) and (b) show that Z admits a representation in the form Z = ∪{Zn : n < ω }, where for each n < ω , either Zn or Zn−1 is the graph of a partial function acting from ω1 into

ω1 . Since ω1 does not exceed the cardinality of the continuum, we can apply Lemma 2 to a set E ⊂ R with card(E) = ω1 and, in fact, here we identify E with ω1 . According to the above-mentioned lemma, every set Zn belongs to Sn ⊗ Sn , where Sn is a certain countably generated σ -algebra of subsets of ω1 . Now, it directly follows from this circumstance that Z belongs to S ⊗ S where the σ -algebra S = σ (∪{Sn : n < ω }) is countably generated, too. Lemma 3 has thus been proved. The next auxiliary proposition is a stronger version of Theorem 4 and is closely connected with the Sierpi´nski partition of ω1 × ω1 . Lemma 4. There exists a σ -algebra S of subsets of ω1 satisfying the following conditions: (1) S is countably generated; (2) S contains all singletons in ω1 ; (3) there is no nontrivial σ -finite diffused measure on S . Proof. In view of Lemma 3, there exists a countably generated σ -algebra S of subsets of

ω1 such that A ∈ S ⊗S, where A is the first member of the Sierpi´nski partition {A, B} of ω1 × ω1 . We may also assume that S contains all singletons in ω1 (cf. the remark made after Lemma 2). It remains to show that S does not admit a nonzero σ -finite diffused measure. Indeed, let μ be a σ -finite diffused measure with dom(μ ) = S . Consider the product measure

ν = μ ⊗ μ. Keeping in mind the relations (a) and (b) of the proof of Lemma 3 and using the Fubini theorem, we readily infer that

ν (A) = 0, ν (B) = 0.

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Consequently, the equalities (μ (ω1 ))2 = ν (ω1 × ω1 ) = ν (A ∪ B) = ν (A) + ν (B) = 0, μ (ω1 ) = 0 must be valid which completes the proof. Now, we are able to establish within ZFC theory the existence of an uncountable universal measure zero subset of the real line R. Taking into account the fact that all uncountable Polish topological spaces are Borel isomorphic (see Appendix 6), it suffices to show the existence of an uncountable universal measure zero set in some uncountable Polish topological space E. We will show the existence of such a set for the Cantor discontinuum, that is for E = 2ω = {0, 1}ω . Namely, we have the following classical statement (see, e.g., [69], [79], [113], [148], [162], and [172]). Theorem 5. There exists an uncountable universal measure zero subset of the Cantor space 2ω . Proof. Let {Ξn : n < ω } be a countable family of subsets of ω1 satisfying the following conditions: (a) {Ξn : n < ω } separates the points in ω1 , i.e., for any two distinct ordinals ξ < ω1 and

ζ < ω1 , there is Ξn such that card(Ξn ∩ {ξ , ζ }) = 1; (b) the σ -algebra σ ({Ξn : n < ω }) does not admit a nonzero σ -finite diffused measure. Notice that the existence of {Ξn : n < ω } is directly implied by Lemma 4. Now, we define a mapping

φ : ω1 → {0, 1}ω by putting

φ (ξ ) = (φn (ξ ))n<ω

(ξ < ω1 ),

where φn (ξ ) = 1 if ξ ∈ Ξn and φn (ξ ) = 0 if ξ ∈ Ξn . In other words, we associate with the family of sets {Ξn : n < ω } its characteristic function

φ which is frequently called Marczewski’s function (see [148] and [150]). Since {Ξn : n < ω } separates the points in ω1 , the mapping φ is injective. Denote Z = φ (ω1 ).

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Since φ is an injection, we have the equality card(Z) = ω1 . Further, from the definition of φ it immediately follows that Ξn = φ −1 (Z ∩ {t ∈ 2ω : tn = 1}) for all n < ω . Suppose for a moment that there exists a nonzero σ -finite diffused Borel measure μ on Z. Then in view of the preceding relation, we come to a nonzero σ -finite diffused measure ν defined on σ ({Ξn : n < ω }) by the formula

ν (φ −1 (Y )) = μ (Y ) (Y ∈ B(Z)). But this gives a contradiction with condition (b). Therefore, Z is a universal measure zero subspace of 2ω which completes the proof. There are several other constructions of uncountable universal measure zero subspaces of the real line, which use essentially different ideas. For more details, see [197] and [250]. Once again, we would like to underline that the existence of an uncountable universal measure zero subset of R directly implies the nonmeasurability (in the Ulam sense) of ω1 . The nonmeasurability in the Ulam sense of ω1 has a nontrivial application in the general theory of σ -finite invariant (quasi-invariant) measures (see Exercises 9, 10, and 11 below). There are also applications of Ulam’s transfinite matrix to some questions concerning the existence of those subsets of a given second category topological space, which do not possess the Baire property (see [148] and Exercise 19 for this Appendix). EXERCISES 1. Let (E, ) be a tree and let (X, ) be a branch in (E, ). Verify that, for any element x ∈ X, the set {y ∈ X : y  x} coincides with {y ∈ E : y  x}. 2∗ . Prove that the following two assertions are equivalent within ZF theory: (a) the K¨onig lemma; (b) for any family {Xn : n < ω } of nonempty finite sets, there exists its selector. 3∗ . Show that if a countable graph is such that all its finite subgraphs are realizable in the Euclidean plane R2 , then the entire graph is also realizable in the plane R2 . Give an example of a graph of cardinality ω1 , all countable subgraphs of which are realizable in R2 , but the graph cannot be realized in R2 .

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4. Verify that the K¨onig lemma is essentially used in the Kakutani-Oxtoby construction of a nonseparable translation-invariant extension of the Lebesgue measure λ on the real line R (see Exercise 18 for Chapter 3; cf. also Chapters 16 and 17). 5∗ . Prove an analogue of the K¨onig lemma for those infinite trees all levels of which are finite. For this purpose, keep in mind the fact that an arbitrary infinite set admits an appropriate nontrivial ultrafilter of its subsets and use an argument similar to the proof of Theorem 3. 6. Give a detailed proof of Lemma 2. For this purpose, consider any partial function f : E → E, where E is a subset of R, and introduce the following two sets associated with this partial function: Γ f = {(x, y) ∈ E × E : x ∈ dom( f ), y  f (x)}, Γ f = {(x, y) ∈ E × E : x ∈ dom( f ), y  f (x)}. Observe that Gr( f ) = Γ f ∩ Γ f . By virtue of Lemma 1, there exists a countably generated σ -algebra T of subsets of E such that f becomes measurable with respect to the σ -algebras B(E) and T . Then show that Γ f ∈ T ⊗ B(E), Γ f ∈ T ⊗ B(E) and, consequently, Gr( f ) ∈ T ⊗ B(E). To prove these relations, first consider the case where f is a partial step-function and then approximate an arbitrary partial function acting from E into itself by a sequence of partial step-functions. Conclude from the said above that S = σ (T ∪ B(E)) is the required countably generated

σ -algebra of subsets of E and, moreover, all singletons in E belong to S . 7. Let {Xξ : ξ < ω1 } be a family of subsets of ω1 . Show that there exists a countably generated σ -algebra S of subsets of ω1 such that {Xξ : ξ < ω1 } ⊂ S . For this purpose, apply Lemma 3 to the set Z = ∪{{ξ } × Xξ : ξ < ω1 }

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which is contained in the product set ω1 × ω1. 8∗ . Assuming the Continuum Hypothesis, prove that there exist two subsets Z1 and Z2 of the Euclidean plane R2 , satisfying the following conditions: (a) Z1 is uniform in the direction parallel to the line {0} × R; (b) Z2 is uniform in the direction parallel to the line R × {0}; (c) there is a countable family {hn : n < ω } of translations of R2 such that ∪{hn + (Z1 ∪ Z2 ) : n < ω } = R2 . Conversely, derive the Continuum Hypothesis from the assumption that there exist two subsets Z1 and Z2 of R2 satisfying the above-mentioned relations (a), (b), and (c). Observe that relation (c) implies the existence of a function f : R → R such that R2 = ∪{gn (Gr( f )) : n < ω } for some countable family {gn : n < ω } of isometric transformations of R2 . The results just formulated are essentially due to Sierpi´nski (cf. [105], [218], [222], and [223]). 9∗ . Let E be a set, G be a group of transformations of E and let μ be a nonzero σ -finite G-quasi-invariant measure on E. Assume that G is uncountable and acts almost freely in E with respect to μ , i.e., for any two distinct transformations g ∈ G and f ∈ G, we have

μ ∗ ({x ∈ E : g(x) = f (x)}) = 0. Prove that dom(μ ) = P(E); in other words, there exists a subset of E nonmeasurable with respect to μ . For this purpose, suppose otherwise, i.e., suppose that dom(μ ) = P(E) and assume, without loss of generality, that μ is a probability G-quasi-invariant measure on E. Denote by H a subgroup of G with card(H) = ω1 and let X be an arbitrary H-selector in E. First, check that: (a) ∪{h(X) : h ∈ H} = E; (b) μ (h(X) ∩ f (X)) = 0 for any two distinct transformations h ∈ H and f ∈ H. Then derive from relation (b) that (c) μ (h(X)) = 0 for each transformation h ∈ H. Taking into account both relations (a) and (c), construct by transfinite recursion an ω1 sequence {Xξ : ξ < ω1 } of subsets of E such that: (d) ∪{Xξ : ξ < ω1 } = E; (e) Xξ ∩ Xζ = 0/ whenever ξ < ω1 , ζ < ω1 and ξ = ζ ;

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(f) μ (Xξ ) = 0 for all ξ < ω1 . Finally, for each set Ξ ⊂ ω1 , put

ν (Ξ) = μ (∪{Xξ : ξ ∈ Ξ}) and verify that ν is a probability diffused measure on ω1 whose domain coincides with P(ω1 ). This circumstance yields a contradiction with the fact that the cardinal ω1 is not real-valued measurable (see Theorem 4). 10. Preserving the notation and assumptions of Exercise 9, strengthen the result obtained therein by proving that any set Y ⊂ E with μ ∗ (Y ) > 0 contains a subset nonmeasurable with respect to μ . Reduce this case to the situation of Exercise 9 by using the following argument. Suppose otherwise, i.e., P(Y ) ⊂ dom(μ ) and consider a set Y  ⊂ E which is almost G-invariant with respect to μ and is representable in the form Y  = ∪{gk (Y ) : k < ω }, where {gk : k < ω } is some countable family of transformations from G. The existence of Y  follows from Exercise 13 of Chapter 2. Check that P(Y  ) ⊂ dom(μ ). Further, put

ν (Z) = μ (Z ∩Y  ) (Z ⊂ E) and verify that ν is a nonzero σ -finite G-quasi-invariant measure on E with dom(ν ) = P(E) which contradicts the result of Exercise 9. 11. Deduce from the two preceding exercises that if E is a set and G is an uncountable group of transformations of E acting freely in E, then for every nonzero σ -finite G-quasiinvariant measure μ on E, the relation dom(μ ) = P(E) holds true. More generally, any set Y ⊂ E with μ ∗ (Y ) > 0 contains a subset nonmeasurable with respect to μ . 12∗ . Let X and Y be two sets and suppose that two injective functions f : X → Y, g : Y → X are given. Prove (within ZF theory) that there exist four sets A, B, A , B satisfying the following relations: (a) A ∩ A = 0/ and A ∪ A = X; (b) B ∩ B = 0/ and B ∪ B = Y ; (c) ran( f |A) = B; (d) ran(g|B ) = A .

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In order to show the existence of such sets, denote Z = X \ g(Y ), φ = g ◦ f and then put A = Z ∪ φ (Z) ∪ φ 2 (Z) ∪ ... ∪ φ n (Z) ∪ ..., A = X \ A, B = f (A), B = Y \ B. Finally, verify that g(B ) = A . This result is due to Banach (cf. [150]). It may be regarded as a strengthened form of the well-known Cantor-Bernstein theorem. Indeed, we can define a function h:X →Y by putting h(x) = f (x) if x ∈ A and h(x) = g−1 (x) if x ∈ A . Clearly, h is a bijection between the given sets X and Y . Moreover, if the original functions f and g have nice structural properties, then, in many cases, h also has the same properties. For instance, let X and Y be two topological spaces such that X contains a Borel subset which is Borel isomorphic to Y and, conversely, Y contains a Borel subset which is Borel isomorphic to X. Check that there exists a Borel isomorphism between X and Y . Analogously, let X and Y be two topological spaces such that X contains a topological copy of Y and Y contains a topological copy of X. Show that there exist two sets A ⊂ X and B ⊂ Y such that A is homeomorphic to B and X \ A is homeomorphic to Y \ B. 13. Infer the above-mentioned result of Banach from Tarski’s fixed-point theorem (see Exercise 10 for Chapter 1). 14∗ . Let E be a set with card(E)  ω and let I be an ideal of subsets of E such that {x} ∈ I for all x ∈ E. Obviously, this condition is equivalent to the relation E = ∪{X : X ∈ I }. Let us denote cov(I ) = the smallest cardinality of a covering of E by sets belonging to I . Recall that a base of I is any family B ⊂ I such that, for every set X ∈ I , there exists a set from B containing X. We denote co f (I ) = the smallest cardinality of a base of I . Suppose that the equalities cov(I ) = co f (I ) = card(E) are fulfilled. Prove that there exists a subset D of E such that:

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(a) card(D) = card(E); (b) card(X ∩ D) < card(E) for all sets X ∈ I . Argue in the following manner. Let α be the least ordinal number whose cardinality is equal to card(E). In view of the equality co f (I ) = card(E), there exists a base {Xξ : ξ < α } of I . Since cov(I ) = card(E), we have card(E \ ∪{Xζ : ζ < ξ }) = card(E) for each ξ < α . Define by transfinite recursion an α -sequence {xξ : ξ < α } of points of E such that xξ ∈ E \ ({xζ : ζ < ξ } ∪ (∪{Xζ : ζ < ξ })) for any ξ < α . Verify that D = {xξ : ξ < α } is the required set. In particular, take E = R and put: I1 = the σ -ideal of all first category subsets of R; I2 = the σ -ideal of all λ -measure zero subsets of R, where λ as usual denotes the Lebesgue measure on R. Under the Continuum Hypothesis, the equalities cov(I1 ) = co f (I1 ) = card(R) = ω1 , cov(I2 ) = co f (I2 ) = card(R) = ω1 are automatically satisfied. Then the set D1 for I1 is called a Luzin subset of R and the set D2 for I2 is called a Sierpi´nski subset of R. Under Martin’s Axiom, we have cov(I1 ) = co f (I1 ) = card(R) = c, cov(I2 ) = co f (I2 ) = card(R) = c. In this case, the set D1 for I1 is called a generalized Luzin subset of R and the set D2 for I2 is called a generalized Sierpi´nski subset of R. Suppose that the Continuum Hypothesis holds. Let L be a Luzin set and let S be a Sierpi´nski set. Show that: (a) L is universal measure zero; (b) the outer λ -measure of any uncountable subset of S is strictly positive (hence, any uncountable subset of S is nonmeasurable with respect to λ ). Suppose that Martin’s Axiom holds. Let L be a generalized Luzin set and let S be a generalized Sierpi´nski set. Show that:

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(a’) L is universal measure zero; (b’) the outer λ -measure of any subset X of S with card(X) = c is strictly positive and hence X is nonmeasurable with respect to λ . 15. Prove that under Martin’s Axiom there exists a countable family {Xn : n < ω } of subsets of R such that the Lebesgue measure λ on R cannot be extended to a measure defined on the σ -algebra σ (dom(λ ) ∪ {Xn : n < ω }). In connection with this result, let us notice that the following assertion is consistent with ZFC theory: for any countable family {Yn : n < ω } of subsets of R, there exists a measure extending λ and defined on σ (dom(λ ) ∪ {Yn : n < ω }). The consistency of this assertion with the standard axioms of set theory was first established by Carlson [36]. 16∗ . A result much stronger than Theorem 5 is known. Namely, there exists an uncountable universal measure zero subset X of R which simultaneously is a vector space over the field Q of all rational numbers (see, for example, [197] or Exercise 17 from Appendix 6). By using the existence of such an X, define an additive function f :R→R which is absolutely nonmeasurable with respect to the class of all nonzero σ -finite translation-quasi-invariant measures on R (cf. Chapter 5). 17∗ . Let E be an infinite set and let {Xi : i ∈ I} be a family of subsets of E such that card(I)  card(E), (∀i ∈ I)(card(Xi ) = card(E)). Prove that there exists a family {Y j : j ∈ J} of subsets of E satisfying the following relations: (a) card(J) > card(E); (b) card(Y j ∩Yk ) < card(E) for any two distinct indices j ∈ J and k ∈ J (this circumstance means that {Y j : j ∈ J} is an almost disjoint family of subsets of E); (c) card(Xi ∩Y j ) = card(E) for all indices i ∈ I and j ∈ J. For this purpose, construct the required family {Y j : j ∈ J} by using the method of transfinite recursion. Putting E = R and taking the family of all nonempty perfect subsets of R as {Xi : i ∈ I}, infer from the above result that there exists an almost disjoint family {Y j : j ∈ J} of Bernstein sets in R for which card(J) > c.

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18∗ . For every ordinal number ζ < ω1 , define an injective function fζ : [0, ζ [ → ω and introduce a double family of sets {Uξ ,n : ξ < ω1 , n < ω } where Uξ ,n = {ζ : ξ < ζ < ω1 , fζ (ξ ) = n}. Verify the following properties of this family: (a) Uξ ,n ∩Uη ,n = 0/ for any n < ω and for any two distinct ordinals ξ < ω1 and η < ω1 ; (b) ω1 \ ∪{Uξ ,n : n < ω } ⊂ [0, ξ ] for each ordinal ξ < ω1 . Any family {Uξ ,n : ξ < ω1 , n < ω } of subsets of ω1 , which satisfies the relations (a) and (b), is usually called Ulam’s (ω1 , ω )matrix (over ω1 ). Infer from the existence of Ulam’s matrix {Uξ ,n : ξ < ω1 , n < ω } that there is no nonzero

σ -finite diffused measure μ on ω1 such that {Uξ ,n : ξ < ω1 , n < ω } ⊂ dom(μ ) and, consequently, ω1 is not a real-valued measurable cardinal. In a similar way, construct Ulam’s (ωα +1 , ωα )-matrix of subsets of ωα +1 , where α is an arbitrary ordinal number, and show that if the cardinal ωα is not real-valued measurable, then ωα +1 is not real-valued measurable, either. Further, verify that if {ai : i ∈ I} is a family of cardinal numbers such that card(I) and all ai (i ∈ I) are not real-valued measurable, then the cardinal ∑{ai : i ∈ I} is not real-valued measurable, either. Derive from these facts that all those cardinal numbers which are strictly less than the first weakly inaccessible cardinal are not real-valued measurable. Under the Generalized Continuum Hypothesis, this result also yields that all those cardinal numbers which are strictly less than the first strongly inaccessible cardinal are not realvalued measurable. It is well known that the existence of weakly (strongly) inaccessible cardinals cannot be proved within ZFC theory (see, e.g., [91], [145], [150], and [215]). In view of the said above, the assumption that there are no real-valued measurable cardinal numbers does not contradict ZFC theory. All results of the previous exercise are due to Ulam (see [238]). They are closely connected with the general measure extension problem (see Chapter 2) and essentially inspired the development of combinatorial set theory and the theory of large cardinals.

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19∗ . Let E be a topological space satisfying the following conditions: (a) card(E) = ω1 ; (b) every singleton in E is of first category; (c) E is of second category; (d) E has the Suslin property, i.e., any disjoint family of nonempty open sets in E is at most countable. Prove that there exists a disjoint family {Xξ : ξ < ω1 } of subsets of E none of which has the Baire property. For this purpose, use Ulam’s transfinite matrix and Exercise 12 from Appendix 3. 20. Let λ2 as usual denote the two-dimensional Lebesgue measure on the Euclidean plane R2 . Show that if the cardinality of the continuum is real-valued measurable, then there exists a countably additive functional μ defined on dom(λ2 ) such that: (a) μ = λ2 ; (b) μ (T ) = 1 for every square T in R2 congruent to [0, 1[2 . For this purpose, utilize the fact that the real-valued measurability of c implies the existence of a measure ν on R extending λ , with dom(ν ) = P(R). By virtue of the classical Vitali theorem, ν cannot be invariant under all translations of R. Hence, there are a set X ∈ dom(ν ) and an element h ∈ R such that

ν (X + h) = ν (X). We may assume, without loss of generality, that X is a bounded subset of R and, consequently, X is contained in some closed interval [a, b] ⊂ R. Now, identifying the real line with the subset R × {0} of the plane R2 , define a functional μ on dom(λ2 ) by the formula

μ (Z) = λ2 (Z) + ν (Z ∩ [a, b]) − ν ((Z ∩ [a, b]) + h)

(Z ∈ dom(λ2 )).

Then verify that μ is countably additive,

λ2 (X) = 0, μ (X) = ν (X) − ν (X + h) = 0 and μ (T ) = 1 for all those squares T in the plane R2 which are congruent to the unit square of this plane. Derive from the said above that μ is the required functional (this result is due to Jacab and Laczkovich [87]). Conversely, it was proved that the existence of a countably additive functional μ on dom(λ2 ) with properties (a) and (b) implies the real-valued measurability of c (for more details, see [87]). Therefore, we can conclude that if c is not real-valued measurable, then

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every countably additive functional μ on dom(λ2 ) satisfying condition (b) necessarily coincides with λ2 . An easy induction on n yields that the same result remains true for the Euclidean space Rn , where n > 2. On the other hand, it is not difficult to show for R1 = R that there are many countably additive functionals μ on dom(λ1 ) satisfying the following relations: (a’) μ = λ1 ; (b’) μ (T ) = 1 for every interval T ⊂ R whose length is equal to 1. 21. Let L be a Luzin subset of R and let a function f : L → R have the Baire property. Prove that the set f (L) is universal measure zero. Derive from this fact that for any nonempty perfect set P ⊂ R, there exists a nonempty perfect set P ⊂ P such that P ∩ f (L) = 0; / in particular, f (L) is totally imperfect in R. 22. Let S be a Sierpi´nski subset of R and let f : R → R be a Lebesgue measurable function. Prove that for any nonempty perfect set P ⊂ R, there exists a nonempty perfect set P ⊂ P such that P ∩ f (S) = 0; / in particular, f (S) is totally imperfect in R.

Appendix 2

The Choquet theorem and measurable selectors

Let X and Y be any two sets and let F : X → P(Y ) be a multi-valued (set-valued) mapping, that is for any x ∈ X, we have F(x) ⊂ Y . In fact, this situation is equivalent to giving a binary relation GF ⊂ X × Y which is defined as follows: (∀x ∈ X)(∀y ∈ Y )((x, y) ∈ GF ⇔ y ∈ F(x)). This binary relation is usually called the graph of the multi-valued mapping F. A function f : X → Y is called a selector of F if f (x) ∈ F(x) for all x ∈ X. The Axiom of Choice states that if F(x) = 0/ for any x ∈ X, then there always exists a selector of F. However, only this fact is not satisfactory in those situations where some nice additional structural properties of selectors are needed, for example, continuity, semicontinuity, measurability, Baire property, certain algebraic properties and so on. In this Appendix, we will envisage the problem of the existence of nice selectors (in the above-mentioned sense) and will establish several theorems stating that under some natural assumptions, such selectors always exist. Recall that in Chapter 8 we were concerned with a rather particular case of this problem. Namely, we proved therein the existence of Borel selectors for certain partitions of a Polish topological space X. Of course, from the point of view of classical descriptive set theory, those selectors may be regarded as subsets of X with good structural properties (see Appendix 6). Since we are primarily interested in measurability properties of various sets and functions, we will restrict ourselves to the statements on the existence of measurable selectors. The statements of such a kind are extremely important in analysis, probability theory and stochastic processes, optimization theory, and the theory of differential equations. There are many works devoted to this topic (see, e.g., [38], [52], [68], [99], [149], [151], [162], [188], and especially the excellent book by J.E. Jayne and C.A. Rogers, Selectors, Princeton University Press, Princeton, 2002). 359

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We shall concentrate our attention on the following two classical results: Luzin-Jankov-von Neumann theorem (see, e.g., [99]) and the theorem of Kuratowski and Ryll-Nardzewski [151] which have a number of applications in different domains of modern mathematics. Notice that the first result can easily be deduced from the second one (for more details, see Theorems 8 and 11 below). But first of all, we wish to discuss the well-known Choquet theorem on capacities because this theorem is closely connected with measurable selectors. We shall formulate and prove a general version of the Choquet theorem. Then we will present its applications to some problems concerning measurability of various sets and functions (in particular, selectors). Let E be a base set and let L be some class of subsets of E. Suppose also that a function

ν : P(E) → R is given. We shall say that ν is a capacity on E for the class L (or with respect to L ) if the following three conditions hold: 1) if X ⊂ Y ⊂ E, then ν (X)  ν (Y ); 2) if a sequence {Xn : n < ω } of subsets of E is increasing by the inclusion relation, then

ν (∪{Xn : n < ω }) = lim ν (Xn ); n→+∞

3) if a sequence {Xn : n < ω } ⊂ L is decreasing by the inclusion relation, then

ν (∩{Xn : n < ω }) = lim ν (Xn ). n→+∞

The next example vividly shows that capacities can be frequently met in measure theory and related fields of mathematics, such as real analysis, abstract harmonic analysis, and probability theory. Example 1. Let (E, S , μ ) be a measure space equipped with a finite measure μ . We take as L any class of sets contained in the σ -algebra S . Let us define the function ν by the formula

ν (X) = μ ∗ (X) (X ⊂ E), where μ ∗ denotes the outer measure associated with μ . It is easy to check that the function

ν is a capacity with respect to the class L . As a rule, the class L is taken in such a way that S = σ (L ); in other words, it is usually assumed that L generates S . Let L be a class of subsets of E. In our further considerations, we denote by A (L ) the analytic class generated by L (see Chapter 8 or Appendix 6). The symbol Lδ will denote the class of the intersections of all countable families of sets from L .

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The following fundamental result is due to Choquet (cf. [24], [40], [52], and [187]). Theorem 1. Let E be a base set and let L be some class of subsets of E, closed under finite unions and finite intersections. Furthermore, let ν be any capacity on E with respect to L . Then for every set X from the analytic class A (L ), the following relation holds:

ν (X) = sup{ν (Y ) : Y ∈ Lδ & Y ⊂ X}. Proof. As usual, we denote by ω <ω the family of all finite sequences of natural numbers. Take any set X ∈ A (L ). Obviously, we can write X=

  t∈ω ω k<ω

Ft0 ...tk ,

where {Ft : t ∈ ω <ω } is some countable system of sets from the given class L (see Chapter 8). Since L is closed under finite intersections, we can assume (without loss of generality) that the above-mentioned system of sets is regular, i.e., Ft0 ...tk tk+1 ⊂ Ft0 ...tk

(k < ω ).

For any infinite sequence t = (t0 ,t1 , . . . ,tk , ...) ∈ ω ω , let us put Ft = Ft0 ∩ Ft0t1 ∩ ... ∩ Ft0t1 ...tk ∩ ... . It should be mentioned that Ft may not belong to L . At the same time, it is clear that the following equality holds: X=





(

Ft ).

n<ω {t:t0 n}

Now, taking into account the properties 1) and 2) of the capacity ν , for any real number

ε > 0, we can find a natural index n0 such that ν (X) − ε < ν (G0 ), where the set G0 is defined by the formula G0 =

 {t:t0 n0 }

Ft .

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Starting with this G0 , we continue our construction by ordinary recursion. Suppose that a finite sequence (n0 , n1 , . . . , nk ) of natural numbers and a finite sequence (G0 , G1 , . . . , Gk ) of sets are constructed in such a way that 

Gr =

(r = 0, . . . , k)

Ft

{t:t0 n0 ,...,tr nr }

and the relations

ν (X) − ε < ν (Gr )

(r = 0, . . . , k)

are satisfied. Let us consider the set Gk . Obviously, we can write Gk =





(

Ft ).

n<ω {t:t0 n0 ,...,tk nk ,tk+1 n}

From the inequality

ν (X) − ε < ν (Gk ) and the properties 1) and 2) of ν it follows again that there exists a natural index nk+1 such that for the set 

Gk+1 =

Ft ,

{t:t0 n0 ,...,tk nk ,tk+1 nk+1 }

the inequality

ν (X) − ε < ν (Gk+1 ) holds true, too. Proceeding in this way, we will be able to construct the required infinite sequences (n0 , n1 , . . . , nk , ...), (G0 , G1 , . . . , Gk , ...). Further, for every natural number k, we put 

Hk =

t0 n0 ,...,tk nk

Ft0 ...tk .

Obviously, in the above formula we use the union of finitely many sets. Therefore, Hk ∈ L . Moreover, it is easy to check that the sequence {Hk : k < ω } is decreasing with respect to the inclusion relation and Gk ⊂ Hk

(k = 0, 1, 2, ...).

Condition 3) for ν implies that

ν (X) − ε  inf{ν (Gk ) : k < ω }  lim ν (Hk ) = ν (∩{Hk : k < ω }). k→+∞

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Now, applying the regularity of the system {Ft : t ∈ ω <ω } and the K¨onig lemma on ω -trees with finite levels (see Appendix 1), we readily get the inclusion ∩{Hk : k < ω } ⊂ X. So, denoting H = ∩{Hk : k < ω }, we conclude that the set H belongs to the class Lδ and H ⊂ X, ν (X) − ε  ν (H). Since ε > 0 was taken arbitrarily, we obtain the desired result. The theorem just proved can be formulated as follows. All analytic sets over the original class L (i.e., all sets from A (L )) are capacitable with respect to any capacity for L . Example 2. Let (E, S , μ ) be a measure space with a finite (or, more generally, σ -finite) measure μ and let L be any class of subsets of E, which generates the σ -algebra S . According to Example 1, the function μ ∗ is a capacity for the class L . Therefore, by the Choquet theorem, any analytic set over L is capacitable with respect to μ ∗ . But, as it is easy to check, this fact means the following: any analytic set over L is measurable with respect to the completion μ  of the measure μ . It is also clear that if the original measure μ is complete, then we simply obtain the μ measurability of any analytic set over the class L . Now, let E be an arbitrary Polish topological space, μ be an arbitrary finite (or, more generally, σ -finite) Borel measure on E and let L be the family of all closed subsets of E. Applying the above considerations to this particular case, we see that every analytic (Suslin) subset of E is μ  -measurable. If E = R, then we immediately obtain the old classical result which says that all analytic subsets of R are Lebesgue measurable. Hence, all complements of the analytic subsets of R are also Lebesgue measurable. Furthermore, we readily derive that the sets belonging to the σ -algebra generated by the family of all analytic subsets of R are good from the measure-theoretical point of view. Notice that the same sets are good from the topological point of view because they have the Baire property and, moreover, they have the Baire property in the restricted sense (in this connection, see [99] or [148]). Let us give some other applications of the Choquet theorem. Theorem 2. Let (E, S , μ ) be a complete probability space (or, more generally, a complete measure space with a σ -finite measure μ ). Let K be a nonempty locally compact topological space with a countable base and let (K, B(K)) be the measurable space canonically

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associated with K (i.e., K is equipped with its Borel σ -algebra B(K)). Denote by pr1 : E × K → E the canonical projection of E × K onto E. Then for any set Z belonging to the product

σ -algebra S ⊗ B(K), the set pr1 (Z) belongs to the σ -algebra S . Proof. We may assume, without loss of generality, that μ is a probability measure. Let us consider the family Comp(K) of all compact subsets of the given topological space K. Evidently, the class Comp(K) is closed under finite unions and finite intersections. It is also clear that the Borel σ -algebra B(K) is generated by Comp(K). In the product set E × K consider the family of all sets of the form ∪{Xi × Yi : 1  i  m}, where m is an arbitrary natural number, Xi are elements of the σ -algebra S and Yi are members of Comp(K). We denote the class of all sets described in this manner by the symbol L . The class L is closed under finite unions and finite intersections, too. It is also obvious that the σ -algebra generated by L coincides with the product σ -algebra S ⊗ B(K). Let us define a real-valued function ν on the family of all subsets of E × K by the formula

ν (Z) = μ ∗ (pr1 (Z)) (Z ⊂ E × K). We are going to verify that the function ν is a capacity on E × K with respect to the class L . Indeed, it is clear that Z1 ⊂ Z2 ⇒ ν (Z1 )  ν (Z2 ) for any two sets Z1 ⊂ E × K and Z2 ⊂ E × K. Moreover, if {Zn : n < ω } is an increasing sequence of subsets of E × K and Z = ∪{Zn : n < ω }, then we have pr1 (Z) = ∪{pr1 (Zn ) : n < ω },

ν (Z) = μ ∗ (pr1 (Z)) = lim μ ∗ (pr1 (Zn )) = lim ν (Zn ). n→+∞

n→+∞

The latter formula immediately follows from standard properties of the outer measure μ ∗ associated with the given probability measure μ . Further, we must check that if {Zn : n < ω } is a decreasing sequence of sets from the class L , then

ν (∩{Zn : n < ω }) = lim ν (Zn ). n→+∞

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For this purpose, let us first prove the equality pr1 (∩{Zn : n < ω }) = ∩{pr1 (Zn ) : n < ω }

(∗).

In fact, it is sufficient to establish the inclusion ∩{pr1 (Zn ) : n < ω } ⊂ pr1 (∩{Zn : n < ω }) because the converse inclusion does always hold.

Let x be an arbitrary element of

∩{pr1 (Zn ) : n < ω }. This circumstance means that for any natural number n, there exists a point yn such that (x, yn ) ∈ Xn,i(n) × Yn,i(n) , where Xn,i(n) × Yn,i(n) is some component of the representation of the set Zn in the form Zn = ∪{Xn,i × Yn,i : 1  i  m(n)}. In particular, we have yn ∈ Yn,i(n) and hence Yn,i(n) = 0. / Moreover, since the sequence of sets {Zn : n < ω } is decreasing by the inclusion relation, we may assume without loss of generality that the families of sets {Xn+1,i : 1  i  m(n + 1)},

{Yn+1,i : 1  i  m(n + 1)}

are respectively inscribed in the families {Xn,i : 1  i  m(n)},

{Yn,i : 1  i  m(n)}.

Now, applying the K¨onig lemma on ω -trees with finite levels (see Appendix 1), we can find a decreasing sequence of sets {Xn,i(n) × Yn,i(n) : n < ω } such that all the sets Yn,i(n) are nonempty and the relation x ∈ Xn,i(n) holds true for every n < ω . Since all Yn,i(n) are also compact and decrease by inclusion, we obviously have ∩{Yn,i(n) : n < ω } = 0. / Let y be any element of ∩{Yn,i(n) : n < ω }. Then it is clear that (x, y) ∈ ∩{Zn : n < ω } and, consequently, x ∈ pr1 (∩{Zn : n < ω }). Thus, the required equality (∗) is proved.

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Now, taking into account the μ -measurability of all sets from the family {pr1 (Zn ) : n < ω }, we can write

ν (∩{Zn : n < ω }) = μ ∗ (pr1 (∩{Zn : n < ω })) = μ (∩{pr1 (Zn ) : n < ω }) = lim μ (pr1 (Zn )) = lim ν (Zn ).

n→+∞

n→+∞

In this way, we have checked that the function ν is a capacity with respect to the class L . The Choquet theorem implies that any set from the class A (L ) is capacitable with respect to ν . Since K is a locally compact topological space with a countable base, the inclusion S ⊗ B(K) ⊂ A (L ) holds (see Appendix 6). Consequently, for any set Z ∈ S ⊗ B(K), we have

ν (Z) = sup{ν (D) : D ∈ Lδ & D ⊂ Z}. In other words, we obtain the equality

μ ∗ (pr1 (Z)) = sup{μ (pr1 (D)) : D ∈ Lδ & D ⊂ Z}. This equality immediately yields the measurability of the set pr1 (Z) with respect to the original measure μ which ends the proof. The statement just established is sometimes called the theorem on measurable projection. Now, we are ready to consider an application of the Choquet theorem to certain questions about the existence of measurable selectors. First, we introduce one definition and prove an auxiliary assertion. Let (E, S , μ ) be a complete probability space or, more generally, a complete measure space with a σ -finite measure μ , let K be a locally compact space with a countable base, and let B(K) as usual denote the Borel σ -algebra of K. Let Z be a subset of the Cartesian product E × K. We say that Z is a measurable functional graph if the following two conditions are satisfied: (a) Z ∈ S ⊗ B(K); (b) for every e ∈ E, the corresponding section Z(e) = {k ∈ K : (e, k) ∈ Z} contains at most one point. The next result gives a characterization of measurable functional graphs in terms of measurable partial mappings acting from E into K (cf. [52]).

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Theorem 3. Let Z ⊂ E × K. The following two relations are equivalent: 1) Z is a measurable functional graph in E × K; 2) there exist a set X ∈ S and a measurable (with respect to the σ -algebras B(K) and S ) mapping g : X → K such that the equality Z = {(x, y) ∈ X × K : g(x) = y} holds true. Proof. First, let us prove the implication 2) ⇒ 1). Let relation 2) be valid. Consider a mapping

ψ : X ×K → K ×K defined by the formula

ψ (x, y) = (g(x), y) (x ∈ X, y ∈ K). Observe that ψ is measurable with respect to the product σ -algebras B(K) ⊗ B(K) = B(K × K), (S |X) ⊗ B(K), where S |X denotes the restriction of the σ -algebra S to the set X, i.e., S |X = {X ∩Y : Y ∈ S }. Analogously, the diagonal {(y, y) : y ∈ K} of the product space K × K is measurable with respect to the σ -algebra B(K) ⊗ B(K). Hence, the pre-image ψ −1 ({(y, y) : y ∈ K}) is measurable with respect to (S |X) ⊗ B(K). But it is obvious that

ψ −1 ({(y, y) : y ∈ K}) = {(x, y) ∈ X × K : g(x) = y} = Z. These equalities immediately yield that Z is a measurable graph in the product set E × K. Now, we are going to prove the implication 1) ⇒ 2). Suppose that relation 1) holds. Let us put X = pr1 (Z). The preceding theorem on measurable projection says that X is measurable with respect to the σ -algebra S . Let us define a mapping g : X → K in the following way: for every x ∈ X, the value g(x) is equal to the unique point from the intersection Z ∩ ({x} × K). Let us check that this mapping is measurable with respect to the σ -algebras B(K) and S |X. For this purpose, take any Borel subset Y of K. Then we can write g−1 (Y ) = pr1 (Z ∩ (E × Y )). But it is clear that the set Z ∩ (E × Y ) is measurable with respect to the product σ -algebra S ⊗ B(K). Using the theorem on measurable projection once more, we obtain that the set

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g−1 (Y ) is measurable with respect to the σ -algebra S , so the measurability of our mapping g is established which completes the proof of the theorem (cf. Exercise 6 from Appendix 1). Among various situations in which the statements just presented may be efficiently applied, the most important is the case when a given topological space K coincides either with the real line R or with the positive half-line R+ = {x ∈ R : x  0}. It should be noticed that the last case can be frequently met in the theory of stochastic processes (see, for example, [52], [187], and [205]). It will be convenient to formulate some results below in terms of the half-line R+ . Taking into account the existence of a Borel isomorphism between R+ and any uncountable Polish topological space K (see Appendix 6), it can easily be shown that analogous results remain valid for K, too. As usual, we assume that the half-line R+ is equipped with its Borel σ -algebra B(R+ ). Consider any complete probability space (E, S , μ ) or, more generally, a complete measure space with a σ -finite measure μ . Let Z be a subset of the Cartesian product E × R+ such that pr1 (Z) = E. In the sequel, we shall say that the function d Z : E → R+ defined by the formula dZ (e) = inf{t ∈ R+ : (e,t) ∈ Z}

(e ∈ E)

is the debut of Z. The following easy proposition is true. Theorem 4. Let (E, S , μ ) be a complete probability space (or, more generally, a complete measure space with a σ -finite measure μ ) and let a subset Z of E × R+ be measurable with respect to the product σ -algebra S ⊗ B(R+ ). Suppose also that pr1 (Z) = E. Then the debut dZ of Z is a real-valued function measurable with respect to the σ -algebras B(R+ ) and S . Proof. Indeed, for any real number t  0, the pre-image dZ−1 ([0,t[) = {e ∈ E : dZ (e) < t} coincides with the set pr1 (Z ∩ (E × [0,t[)). The set Z ∩ (E × [0,t[) is measurable with respect to the σ -algebra S ⊗ B(R+ ). By the theorem on measurable projection, the set

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dZ−1 ([0,t[) is measurable with respect to S . Hence, the function dZ is measurable with respect to B(R+ ) and S . The next result is a typical representative from a large group of the so-called uniformization theorems or, equivalently, theorems on measurable selectors (see [52]). Theorem 5. Let (E, S , μ ) be any complete probability space (or, more generally, a complete measure space with a σ -finite measure μ ) and let a set Z ⊂ E × R+ be measurable with respect to the product σ -algebra S ⊗ B(R+ ). Suppose also that pr1 (Z) = E. Then there exists a function g : E → R+ such that: 1) g is measurable with respect to the σ -algebras B(R+ ) and S ; 2) the graph Gr(g) of g is contained in the given set Z. Proof. It suffices to show that for every real number ε > 0, there exist a set Xε ∈ S and a measurable function gε : X → R+ such that the following two conditions hold: (a) μ (E \ Xε ) < ε ; (b) the graph of gε is contained in Z. Indeed, as soon as this fact is established, the next steps of the proof become evident: after finding a set Xε and a function gε , we move to the measure space (E \ Xε , S |(E \ Xε ), μ |(S |(E \ Xε ))) and to the set Z ∩ ((E \ Xε ) × R+ ), and continue analogous process for the real number ε /2, the new measure space and the new set. After countably many steps, we will fill the whole space E except, perhaps, a

μ -measure zero set with a disjoint countable family of the constructed μ -measurable sets. Then we obtain the required function g as a common extension of all functions of type gε and some function defined on a μ -measure zero subset of E. So, let ε be any strictly positive real number. Consider the class Comp(R+ ) of all compact subsets of the half-line R+ . Moreover, let us introduce the class L of all those sets Z ⊂ E × R+ which can be represented in the form Z = ∪{Xi × Yi : 1  i  m}, where m is an arbitrary natural number, Xi are members of S and Yi are members of Comp(R+ ). Earlier, we have already mentioned that the class L generates the product

σ -algebra S ⊗ B(R+ ) and that L is closed under finite unions and finite intersections. Further, it is easy to see that S ⊗ B(R+ ) ⊂ A (L ).

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Using the same argument as in the proof of the theorem on measurable projection, we can find for the given ε and Z, a set Z ∗ ⊂ Z belonging to the class Lδ and such that

μ (pr1 (Z) \ pr1 (Z ∗ )) = μ (E \ pr1 (Z ∗ )) < ε . For the set Z ∗ , we may consider its debut dZ ∗ : pr1 (Z ∗ ) → R+ which is defined on the set pr1 (Z ∗ ). Notice that for each element x ∈ pr1 (Z ∗ ), the section Z ∗ (x) = {t ∈ R+ : (x,t) ∈ Z ∗ } is a nonempty compact subset of R+ because it coincides with the intersection of some countable family of nonempty compact subsets of R+ . Consequently, we have dZ ∗ (x) = inf(Z ∗ (x)) ∈ Z ∗ (x). From this fact it immediately follows that the graph of the function dZ ∗ is contained in the set Z ∗ , so it is contained in the original set Z as well. It remains to take into account that, in view of Theorem 4, the function dZ ∗ is measurable with respect to the σ -algebras B(R+ ) and S . Therefore, we can put Xε = pr1 (Z ∗ ), gε = dZ ∗ , and Theorem 5 has thus been proved. It is obvious that the notion of the debut of a given set lying in the Cartesian product E × R+ can be generalized to those subsets of E × R+ whose first projections are not necessarily the whole space E. After this generalization it becomes clear that if a given set Z ⊂ E × R+ is measurable with respect to the product σ -algebra S ⊗ B(R+ ), then its debut is a measurable function acting from the set pr1 (Z) into R+ . This fact also indicates how we can generalize Theorems 4 and 5 to the case of any set Z ⊂ E × R+ measurable with respect to the product σ -algebra S ⊗ B(R+ ). Therefore, using the same argument as in the proof of Theorem 5, we obtain the following classical result. Theorem 6. Let E be a Polish topological space and let Z be an analytic subset of the topological product E × R+ . Suppose that a Borel finite (or, more generally, σ -finite) measure μ on E is given and let μ  denote the completion of μ . Then there exists a function g : pr1 (Z) → R+ such that: 1) g is measurable with respect to the σ -algebras B(R+ ) and dom(μ  ); 2) the graph of g is contained in the set Z.

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A detailed proof of this statement is left to the reader. Now, let P be an arbitrary uncountable Polish topological space. As is well known (see, e.g., [40], [99], [148], or Appendix 6), there exists a Borel isomorphism h : (R+ , B(R+ )) → (P, B(P)). It is not difficult to check that for this isomorphism, the equality h(A (R+ )) = A (P) holds true. Taking into account this trivial remark, we can immediately deduce from Theorem 6 the following result. Theorem 7. Let E1 and E2 be two Polish topological spaces and let Z be an analytic subset of the topological product E1 × E2 . Suppose also that a Borel finite (or σ -finite) measure

μ on E1 is given and let μ  denote the completion of μ . Then there exists a mapping g : pr1 (Z) → E2 such that: 1) g is measurable with respect to the σ -algebras B(E2 ) and dom(μ  ); 2) the graph of g is contained in the set Z. In many cases, the theorem formulated above is sufficient for various applications in measure theory and real analysis. But, sometimes, we need more subtle formulations of results of this type (see, for instance, the theorem of Kuratowski and Ryll-Nardzewski below). Now, let us turn our attention to set-valued mappings with those measurability properties which are not directly connected with the notion of measure. First, let us recall that the pair (X, S ) is an abstract measurable space if X is a base set and S is some σ -algebra of subsets of X. For instance, if E is an arbitrary topological space, then the pair (E, B(E)) is a canonical example of a measurable space. In this context, it should be mentioned that there are some uncountable subspaces E of R for which B(E) does not admit any nonzero σ -finite diffused measure (see Appendix 1). Suppose that (X, S ) is a measurable space. Let Y be a topological space and let F : X → P(Y ) be a set-valued mapping such that F(x) is a nonempty closed subset of Y for every point x ∈ X. In the sequel, only those set-valued functions F will be considered which satisfy this condition. We say that F is a weakly measurable set-valued mapping if for any open subset U of Y , we have {x ∈ X : F(x) ∩U = 0} / ∈S.

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It is not difficult to check the validity of the following auxiliary proposition. Lemma 1. Let (Y, d) be a separable metric space. For a set-valued mapping F : X → P(Y ), these two assertions are valid: 1) F is weakly measurable if and only if for each point y ∈ Y , the function gy : X → R defined by the formula gy (x) = d(F(x), y)

(x ∈ X)

is measurable with respect to the σ -algebras B(Y ) and S ; 2) if F is weakly measurable, then the graph of F is a measurable subset of the product space (X × Y, S ⊗ B(Y )). The proof of Lemma 1 is left to the reader. Let f be a mapping acting from X into Y . We recall that f is a selector of a given set-valued mapping F : X → P(Y ) if for any element x ∈ X, we have f (x) ∈ F(x). In view of the relation (∀x ∈ X)(F(x) = 0), / the Axiom of Choice immediately yields the existence of a selector of F. But if we want to get a measurable selector, then we must use a more delicate additional argument. For our further purposes, we also need the next simple auxiliary proposition. Lemma 2. Let (X, d) be an arbitrary metric space. Then there exist a Banach space (E, || · ||) and an isometric embedding

φ : (X, d) → (E, || · ||). In particular, if (X, d) is a complete metric space, then the image φ (X) is a closed subset of the space E. Moreover, the space E can be chosen in such a way that w(E)  w(X) + ω , where w(E) (respectively, w(X)) denotes the topological weight of E (respectively, of X). Proof. Without loss of generality, we may assume that X is a nonempty space. Let t be a fixed element of X. Further, for each element y ∈ X, let us consider a mapping fy : X → R defined by the formula fy (x) = d(x, y) − d(x,t)

(x ∈ X).

It is easy to see that | fy (x)|  d(y,t)

(x ∈ X).

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Hence the function fy is bounded. Now, we can consider the family E of all bounded functions f acting from X into R. Equip E with the standard norm || f || = supx∈X | f (x)|. Obviously, E is a Banach space with respect to this norm. Moreover, it is not difficult to check that || fy − fz || = d(y, z) for any two elements y and z from X. Consequently, a mapping φ : X → E defined by the formula

φ (y) = fy

(y ∈ X)

is an isometric embedding of X into E. A priori, the topological weight of E may be significantly greater than the topological weight of X. But if we take the closed vector subspace of E generated by the set φ (X), then the weight of this subspace does not exceed w(X) + ω . Lemma 2 is thus proved. Notice that the same argument leads to the following classical result: any metric space (X, d) can be isometrically embedded into some complete metric space E in such a way that X becomes everywhere dense in E. Let us recall that this result was first obtained by Hausdorff (see, e.g., [58], [101], and [148]). We also wish to remark that in the theory of metric spaces there are theorems much stronger than Lemma 2. For instance, if (X, d) is an arbitrary separable metric space, then there exists an isometric embedding of X into the separable Banach space C[0, 1] of all continuous real-valued functions defined on the unit segment [0, 1]. In other words, the separable Banach space C[0, 1] turns out to be universal for the class of all separable metric spaces. Recall that this important result is due to Banach and Mazur (see, e.g., [56]). However, for our further considerations, Lemma 2 is completely sufficient. Now, we can formulate and prove a famous theorem concerning the existence of measurable selectors. This important theorem was established by Kuratowski and Ryll-Nardzewski in [151]. Theorem 8. Let (X, S ) be a measurable space, let Y be a Polish space and let F : X → P(Y ) be a set-valued mapping satisfying the following conditions: 1) for each element x ∈ X, the set F(x) is nonempty and closed in Y ; 2) F is weakly measurable. Then there exists a mapping f : X → Y such that: a) f is a selector of F; b) f is a measurable mapping acting from (X, S ) into (Y, B(Y )).

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Proof. Taking into account Lemma 2, we may assume without loss of generality that Y is a separable Banach space. Let d be a complete metric on Y agreed with the norm of Y ; in other words, for any two points y and z from Y , we put d(y, z) = ||y − z||. Furthermore, denote by {yk : 1  k < ω } a countable family of points of Y which is everywhere dense in this space. Using the ordinary method of mathematical recursion, we shall construct a sequence { fn : 1  n < ω } of mappings acting from X into Y and satisfying the following relations: (1) ran( fn ) ⊂ {yk : 1  k < ω } for any natural number n  1; (2) d( fn (x), fn−1 (x)) < 1/2n−1 for each natural number n  2 and for each element x ∈ X; (3) d(F(x), fn (x)) < 1/2n−1 for any natural number n  1 and for any element x ∈ X; (4) fn is a measurable mapping acting from (X, S ) into (Y, B(Y )) for any natural number n  1. Let x be an arbitrary point of the space X. Then we put f1 (x) = yk where k is the smallest natural number for which d(F(x), yk ) < 1. Evidently, we obtain a certain mapping f1 : X → Y. Since F is a weakly measurable set-valued mapping, it is easy to check that f1 is a measurable mapping acting from (X, S ) into (Y, B(Y )). Suppose now that a partial finite sequence of functions { f1 , f2 , ... , fn } satisfying the above relations (1) - (4) has already been defined. Fix again an arbitrary element x ∈ X and consider the following two sets: V = {y ∈ Y : d(y, fn (x)) < 1/2n}, W = {y ∈ Y : d(y, F(x)) < 1/2n }. Clearly, V and W are open subsets of the space Y . Moreover, since d(F(x), fn (x)) < 1/2n−1, there exists a point z ∈ F(x) such that d(z, fn (x)) < 1/2n−1. Now, it can easily be seen that the point y = (z + fn (x))/2

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belongs to the set V ∩ W . Hence V ∩ W is a nonempty open subset of Y . Let k be the smallest natural number for which yk ∈ V ∩W . Let us put fn+1 (x) = yk . Taking into account the fact that x is an arbitrary element of X, we have a certain mapping fn+1 : X → Y. Since our set-valued mapping F is weakly measurable, it is not difficult to check that fn+1 is a measurable mapping acting from (X, S ) into (Y, B(Y )). Also, it is not hard to show that all relations (1) - (4) remain true for the partial sequence { f1 , f2 , ... , fn , fn+1 }. Proceeding in such a way, we are able to construct the required infinite sequence of functions { f1 , f2 , ... , fn , ... }. Finally, we put f (x) = lim fn (x) n→+∞

(x ∈ X).

Notice that this definition of f is correct because, in view of relation (2), the sequence of functions { fn : 1  n < ω } uniformly converges on X. Thus, we have a mapping f acting from X into Y . Now, using relation (3), we can readily conclude that f is a measurable selector of the original set-valued mapping F. The proof of Theorem 8 is completed. Some applications of Theorem 8 may be found in [99], [149], and [151] (in this connection, see also [38] and [68]). Here we wish to present only two statements which can be obtained with the aid of Theorem 8. The first of these statements will be formulated in terms of topological measure theory. The second one will be formulated in terms of Baire category.

Theorem 9. Let E1 and E2 be two Polish topological spaces and let A be an analytic subset of the product space E1 × E2 . Suppose that E1 is equipped with a σ -finite Borel measure μ and suppose that for almost all (with respect to μ ) points x ∈ pr1 (A), the section A(x) = {y ∈ E2 : (x, y) ∈ A} is an uncountable subset of E2 . Then there exists a set B ⊂ A such that: 1) B is a Borel subset of the space E1 × E2;

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2) for almost all (with respect to μ ) points x ∈ pr1 (A), the section B(x) = {y ∈ E2 : (x, y) ∈ B} is a nonempty perfect compact subset of E2 . Theorem 10. Let E1 and E2 be two Polish topological spaces and let A be an analytic subset of the product space E1 × E2 . Suppose that for almost all (in the Baire category sense) points x ∈ pr1 (A), the section A(x) is an uncountable subset of E2 . Then there exists a set B ⊂ A such that: 1) B is a Borel subset of the space E1 × E2; 2) for almost all (in the Baire category sense) points x ∈ pr1 (A), the section B(x) is a nonempty perfect compact subset of E2 . As mentioned above, the proofs of Theorems 9 and 10 are based on the theorem of Kuratowski and Ryll-Nardzewski (for more details, see Exercises 12 and 13 of this chapter). It is easy to check that both Theorems 9 and 10 may be treated as some parameterized versions of the classical result of Alexandrov and Hausdorff stating that any uncountable analytic set (in a Polish topological space) contains a nonempty perfect compact subset (see [99], [148], [160], [162], or Appendix 6). Furthermore, from the theorem of Kuratowski and Ryll-Nardzewski we can easily obtain another classical result concerning the uniformization of an analytic subset of the product of two Polish topological spaces. This result is due to Luzin, Jankov and von Neumann (cf. [99]). It essentially strengthens Theorem 7, which was obtained by using the Choquet theorem. Theorem 11. Let E1 and E2 be two Polish spaces, let A be an analytic subset of the product space E1 × E2 and let S be the σ -algebra of subsets of pr1 (A), generated by the family of all analytic subsets of pr1 (A). Then there exists a mapping f : pr1 (A) → E2 satisfying the following relations: 1) the graph Gr( f ) of f is contained in the set A; 2) f is a measurable mapping acting from (pr1 (A), S ) into (E2 , B(E2 )). In particular, f is measurable with respect to the completion of an arbitrary σ -finite Borel measure given on the Suslin space pr1 (A) and f has the Baire property in the restricted sense.

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Proof. Let us equip the set ω of all natural numbers with the discrete topology and take the product space ω ω which usually is called the canonical Baire space of countable topological weight. Since our set A is analytic, there exists a continuous function g : ω ω → E1 × E2 such that g(ω ω ) = A (see Exercise 12 from Chapter 8). Denote by G the graph of g, i.e., put G = {(t, x, y) ∈ ω ω × E1 × E2 : g(t) = (x, y)}. Obviously, G is a closed subset of the product space ω ω × E1 × E2 (see Theorem 2 from Chapter 8). Now, we define a set-valued mapping Φ : pr1 (A) → P(ω ω × E1 × E2 ) by the formula Φ(x) = G ∩ (ω ω × {x} × E2)

(x ∈ pr1 (A)).

It is not difficult to check that Φ satisfies the assumptions of the theorem of Kuratowski and Ryll-Nardzewski. According to this theorem, there exists a selector

φ : pr1 (A) → ω ω × E1 × E2 of Φ which is measurable with respect to the σ -algebras B(ω ω × E1 × E2 ) and S . It remains to put f = pr3 ◦ φ and to verify that f is the required mapping. The corresponding easy details are left to the reader. The assumption that Y is a Polish topological space is very essential in the formulation of the theorem of Kuratowski and Ryll–Nardzewski. However, in some particular cases, the result of this theorem remains true if we replace Y by a nonseparable complete metric space Y  . Moreover, sometimes we do not need even the completeness of the space Y  , for example, in those cases where all sets F(x) are compact and nonempty. Various situations closely connected with the theorem of Kuratowski and Ryll-Nardzewski are discussed in monograph [68] where it is shown that some additional set-theoretical axioms are necessary for appropriate generalizations of this theorem (see also [38]). EXERCISES

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1. Let E be any base set and let L be some class of subsets of E. We say that L is a quasicompact class (in the sense of Marczewski) if for every countable family {Zn : n < ω } ⊂ L , the relation ∩{Zn : n < ω } = 0/ implies that there exists a finite subfamily {Zn1 , Zn2 , . . . , Znk } of {Zn : n < ω } such that Zn1 ∩ Zn2 ∩ ... ∩ Znk = 0. / For example, if E is an arbitrary Hausdorff topological space and L is the class of all compact subsets of E, then L is a quasicompact class of sets. Now, let L be any quasicompact class of subsets of a base set E and let L ∗ denote the class of all finite unions of elements from L . Prove that L ∗ is also a quasicompact class. For this purpose, apply the K¨onig lemma on ω -trees with finite levels (see Appendix 1.) 2. Give generalizations of some results presented in this Appendix to the case when a locally compact topological space K with a countable base is replaced by a set F equipped with a quasicompact class L ⊂ P(F). 3. Give a detailed proof of Theorem 6 of this chapter and deduce Theorem 7 from it. 4∗ . Starting with the fact that there are two disjoint co-Suslin sets in R which cannot be separated by any two disjoint Borel subsets of R (see [99], [148], [160], and [162]), prove that there exists a Borel set Z ⊂ R2 such that: (a) pr1 (Z) = R; (b) there is no Borel function f : R → R whose graph is contained in Z. This deep result is due to Luzin and Novikov (cf. [162], [188]) and shows that for the class of all Borel subsets of the plane R2 , the uniformization problem has a negative solution if a selector is required to be Borel. This fact also explains why in Theorem 5 of this Appendix we should consider measurable (but not Borel) functions. As follows from the theorem of Kuratowski and Ryll-Nardzewski, we only have the guaranty that a selector can be chosen to be measurable with respect to the σ -algebras B(R) and S , where S is generated by the family of all analytic subsets of R. 5. Let (E, S ) be a measurable space and let X be a subset of E. We say that X is absolutely (or universally) measurable with respect to the σ -algebra S if for every σ -finite measure

μ defined on S , the relation X ∈ dom(μ  ) holds, where μ  denotes the completion of μ . Deduce from the Choquet theorem that all members of the analytic class A (S ) are absolutely measurable with respect to the σ -algebra S .

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In particular, if E is an arbitrary topological space, then all analytic sets in E (i.e., the members of A (F ) where F denotes the class of all closed subsets of E) are absolutely measurable with respect to the Borel σ -algebra B(E). Moreover, let us consider the particular case (E, S ) = (R, B(R)). In this case, give an example of a Lebesgue measurable subset of R which is not absolutely measurable with respect to the Borel σ -algebra B(R). 6. Let E be an arbitrary Hausdorff topological space and let X be a subset of E which is a Radon space with respect to the induced topology. Show that X is absolutely measurable with respect to the Borel σ -algebra B(E). Let E be an arbitrary Radon topological space and let X be a subset of E which is absolutely measurable with respect to the Borel σ -algebra B(E). Show that the set X equipped with the induced topology is a Radon topological space. Conclude that every analytic subset of a Polish topological space P is a Radon space. The same is true for the complement of any analytic subset of P and, more generally, for each member of the σ -algebra generated by the family of all analytic sets in P. 7. Assuming Martin’s Axiom, show that there exists a subset of the real line R, which is absolutely measurable with respect to the Borel σ -algebra B(R) but does not have the Baire property in R. Conversely, give an example (within ZFC theory) of a first category subset of R which is not absolutely measurable with respect to B(R). 8∗ . Let us consider the measure space (R, dom(λ ), λ ) where λ is the standard Lebesgue measure on R. Let P be an arbitrary projective subset of R and let B(P) denote the Borel

σ -algebra of the topological space P. Let us put S = dom(λ ) ⊗ B(P) and consider the canonical projection pr1 : R × P → R. Assuming that all projective subsets of R are Lebesgue measurable, prove that (∀X ∈ S )(pr1 (X) ∈ dom(λ )). Similarly, let us put L = Ba(R) ⊗ B(P),

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where Ba(R) is the σ -algebra of all those subsets of R which have the Baire property. Assuming that all projective subsets of R have the Baire property, prove that (∀X ∈ L )(pr1 (X) ∈ Ba(R)). Recall that the Lebesgue measurability and the Baire property of all projective subsets of R follow from the Axiom of Projective Determinacy (see, e.g., [10], [91], and [99]). 9∗ . Let X be an analytic subset of a Polish topological space, let E be a metric space equipped with a σ -finite Borel measure μ , and let f :X →E be a Borel mapping. Show that for every real number ε > 0, there exists a subset Y of X such that: (a) Y is a compact subspace of X; (b) the restriction of f to Y is a homeomorphism between the spaces Y and f (Y ); (c) μ  ( f (X) \ f (Y )) < ε , where μ  denotes the completion of μ . 10. Prove Lemma 1 of this chapter. 11∗ . Give a proof of Theorem 9 from this chapter. For this purpose, consider the family C = C (E2 ) of all nonempty compact subsets of E2 and endow C with the Vietoris topology (see [58], [101], [149]). Verify that the obtained topological space is Polish. Then define an appropriate analytic subset D of the product space E1 × C and apply to D the Luzin-Jankov-von Neumann theorem. 12∗ . Give a proof of Theorem 10 from this chapter. Use the same hint as in the previous exercise. 13∗ . This exercise deals with extensions of Carath´eodory type partial functions which are important in various questions of real analysis, measure theory, the theory of ordinary differential equations, optimization, and probability theory. Let X be a set equipped with a σ -finite complete measure μ and let Y be a compact metric space. Suppose that a partial mapping f0 : X × Y → R is measurable with respect to the σ -algebras B(R) and dom(μ ) ⊗ B(Y ). By using the theorem on measurable projection and the theorem of Kuratowski and RyllNardzewski, prove that the following two relations are equivalent: (a) for each x ∈ X, the partial mapping f0 (x, ·) : Y → R is uniformly continuous;

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(b) there exists a mapping f : X × Y → R extending f0 and satisfying Carath´eodory’s conditions, i.e., for each x ∈ X, the mapping f (x, ·) : Y → R is continuous and for each y ∈ Y , the mapping f (·, y) : X → R is μ -measurable. In order to establish the equivalence (a) ⇔ (b), consider the separable Banach space C(Y ) = C(Y, R) of all real-valued continuous functions on Y and define an appropriate set-valued weakly measurable mapping F : X → P(C(Y )) associated with the given mapping f0 . Then apply to F the theorem of Kuratowski and Ryll-Nardzewski. 14∗ . Let E be a Suslin space and let S be a countably generated subalgebra of the Borel

σ -algebra B(E). Suppose that μ is a probability measure with dom(μ ) = S . Demonstrate that there exists a Borel measure μ  on E extending μ . For this purpose, use an appropriate theorem on the existence of measurable selectors and Marczewski’s characteristic function (cf. the proof of Theorem 5 from Appendix 1). The result of Exercise 14 is due to Ershov (see [64]). It can be shown that, in general, the same result fails to be true for a co-Suslin subspace E of a Polish topological space. 15. Assuming Martin’s Axiom, prove that there exists a subset Z of the Euclidean plane R2 , satisfying the following relations: (a) for each point x ∈ R, the cardinality of the corresponding section Z(x) = {y ∈ R : (x, y) ∈ Z} is equal to c; (b) every selector of the set-valued mapping Φ : R → P(R) defined by Φ(x) = Z(x)

(x ∈ R)

is absolutely nonmeasurable with respect to the class of all nonzero σ -finite diffused measures on R. For this purpose, consider a disjoint family {Lx : x ∈ R} of generalized Luzin sets in R such that ∪{Lx : x ∈ R} is also a generalized Luzin subset of R, and put Z = ∪{{x} × Lx : x ∈ R}.

Appendix 3

Borel measures on metric spaces

This Appendix is devoted to some topological aspects of the theory of σ -finite measures which are given on the Borel σ -algebras of metric spaces. Namely, here we are going to describe the topological structure of supports of such measures. Since every nonzero σ finite measure is equivalent to a probability measure (defined on the same σ -algebra), we may restrict our further considerations to the case of probability measures. Let E be an arbitrary topological space and let μ be a σ -finite Borel measure on E. We recall that a Borel set X ⊂ E is a support of μ (or, in other words, μ is concentrated on X) if

μ (E \ X) = 0. Obviously, for a Borel probability measure μ on E, the following two assertions are equivalent: 1) X is a support of μ ; 2) μ (X) = μ (E) = 1. For our purposes, we need one important result concerning supports of σ -finite Borel measures given on a metric space (E, d) whose topological weight is not measurable in the Ulam sense, i.e., is not real-valued measurable. We recall that the topological weight of E is the smallest cardinality among the cardinalities of all open bases of E. The topological weight of E is usually denoted by the symbol w(E). For example, taking the classical topological vector spaces RN = Rω = {(xn )n<ω : (∀n < ω )(xn ∈ R)}, c0 = {(xn )n<ω ∈ Rω : lim xn = 0}, n→+∞

l∞ = {(xn )n<ω ∈ Rω : supn<ω |xn | < +∞}, 383

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we have respectively w(RN ) = w(c0 ) = card(N) = ω , w(l∞ ) = card(R) = c. Recall that an infinite cardinal number a is real-valued measurable or measurable in the Ulam sense if there exists a nonzero σ -finite measure defined on the σ -algebra of all subsets of a and vanishing at all one-element subsets of a. Obviously, in this definition we may consider any set A with card(A) = a, instead of a. Also, it is clear that a is real-valued measurable if and only if there exists a diffused probability measure μ on a with dom(μ ) = P(a). Extensive information about real-valued measurable cardinals and related topics of combinatorial set theory is presented in [69], [91], [144], [145], [150], and [231]. Here we only wish to recall the important fact that the existence of real-valued measurable cardinals cannot be established within ZFC theory. Consequently, the assumption that all cardinals are not real-valued measurable is consistent with ZFC (cf. Exercise 18 for Appendix 1). So, in order to manipulate with real-valued measurable cardinals, we are forced to appeal to additional set-theoretical axioms. Let us observe that if a cardinal a is measurable in the Ulam sense and b is another cardinal greater than a, then b is also measurable in the Ulam sense. This fact follows directly from the definition. Indeed, if A and B are any two sets such that card(A) = a, card(B) = b, A ⊂ B, and μ is a diffused probability measure with dom(μ ) = P(A), then a functional

ν : P(B) → [0, 1] defined by the formula

ν (X) = μ (X ∩ A) (X ⊂ B) is a diffused probability measure with dom(ν ) = P(B) and, in fact, this ν is concentrated on the set A. First, we would like to formulate and prove the following auxiliary statement from the theory of metric spaces (see [148], [174], and [192]). Lemma 1. Let (E, d) be a metric space, I be a set of indices linearly ordered by some relation , and let {Ai : i ∈ I}, {Bi : i ∈ I}, {Ci : i ∈ I}

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be three families of subsets of E such that: 1) all sets Ai (i ∈ I) are open in E; 2) for each index i ∈ I, we have Bi = Ai \ ∪{A j : j ≺ i}; 3) for each index i ∈ I, the set Ci is closed in E and Ci ⊂ Bi . Then the set C = ∪{Ci : i ∈ I} can be represented as the union of some countable family of closed subsets of E (in other words, C is of type Fσ in E). Proof. For any index i ∈ I and for any natural number n  1, let us put Cin = {z : z ∈ Ci , d(z, E \ Ai )  1/n}. Evidently, we have the equality ∪{Cin : 1  n < ω } = Ci . Furthermore, each set Cin is closed in E. Now, if j ≺ i, x ∈ Cin , y ∈ Cnj , then d(x, y)  1/n. Consequently, we get (∀n)(1  n < ω ⇒ d(Cin ,Cnj )  1/n) for any two distinct indices i and j from I (here we use the assumption that the set I is linearly ordered by ). Thus for each natural number n  1, the set C(n) = ∪{Cin : i ∈ I} turns out to be closed in E. It remains to observe that C = ∪{Ci : i ∈ I} = ∪{C(n) : 1  n < ω } which shows that C is an Fσ -set in E. Lemma 1 has thus been proved. Now, we are able to prove the following important statement (cf. [192]). Theorem 1. Let (E, d) be a metric space whose topological weight is not real-valued measurable, and let μ be an arbitrary σ -finite Borel measure on E. Then there exists a separable support of μ , i.e., there exists a closed separable subset F = Fμ of E such that

μ (E \ F) = 0.

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In particular, μ is a separable measure. Proof. Since every nonzero σ -finite measure is equivalent to a probability measure, we may assume, without loss of generality, that our μ is a Borel probability measure on E. Fix a natural number k and consider a covering {Uk,i : i ∈ I(k)} of E by open balls whose diameters are less than or equal to 1/(k + 1). Obviously, we can suppose that card(I(k))  w(E). Moreover, applying the Zermelo theorem, we can also suppose that the set I(k) is wellordered by some ordering relation . Further, for each index i ∈ I(k), we denote Zk,i = Uk,i \ ∪{Uk, j : j ≺ i}. Clearly, the family of sets {Zk,i : i ∈ I(k)} is disjoint and, simultaneously, is a covering of E. In addition to this circumstance, each set Zk,i (as a difference of two open sets) admits a representation in the form Zk,i = ∪{Zk,i,m : m < ω }, where all sets Zk,i,m are closed in E. Since μ is a probability measure, the set of all those indices i ∈ I(k) for which μ (Zk,i ) > 0, is at most countable. Let us denote this set by T (k) and put J(k) = I(k) \ T (k). We are going to demonstrate that

μ (∪{Zk,i : i ∈ J(k)}) = 0. Suppose otherwise, that is

μ (∪{Zk,i : i ∈ J(k)}) > 0. Then for any subset J of J(k), we may write ∪{Zk,i : i ∈ J} = ∪{Zk,i,m : i ∈ J, m < ω }. According to Lemma 1, the set Zk,m = ∪{Zk,i,m : i ∈ J}

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is Borel in E and hence the set ∪{Zk,i : i ∈ J} = ∪{Zk,m : m < ω } turns out to be Borel in E, too. Now, putting

ν (J) = μ (∪{Zk,i : i ∈ J}) for every set J ⊂ J(k), we come to the nonzero finite measure ν defined on the σ -algebra of all subsets of J(k) and vanishing at all one-element subsets of J(k). In other words, we derive that card(J(k)) is a real-valued measurable cardinal. But this is impossible because of the relation card(J(k))  card(I(k))  w(E) and in view of our assumption that w(E) is not real-valued measurable. The contradiction obtained shows that

μ (∪{Zk,i : i ∈ J(k)}) = 0 which implies, in particular, that our measure μ is concentrated on the set U(k) = ∪{Uk,i : i ∈ T (k)} for each natural number k. Now, it can easily be checked that the set X = ∩{U(k) : k < ω } is a separable subspace of E and μ is concentrated on X as well. Thus, it suffices to put F = cl(X) where cl(X) denotes the closure of X. The proof of Theorem 1 is complete. Some applications of this theorem will be given later (see especially Exercises 2 and 3). Remark 1. The assumption that a metric space E in Theorem 1 has non-real-valued measurable topological weight is very essential for the validity of this theorem. Indeed, suppose that a real-valued measurable cardinal exists and take such a cardinal a. Further, choose any set E with card(E) = a and equip E with the discrete topology (or, equivalently, with the discrete metric). Then E becomes a metrizable space whose topological weight is equal to a. Let μ be a nonzero σ -finite measure defined on the family of all subsets of E and vanishing at all one-element subsets of E. Then, obviously, μ is a Borel measure on E which does not possess a separable support because any separable subset of E is at most countable. In other words, the assertion of Theorem 1 fails to be true for such a μ . In addition, let us notice that for the validity of Theorem 1, the assumption of metrizability of E is essential, too (cf. Exercise 8 below).

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We thus see that under some natural assumption which does not contradict the axioms of set theory, any σ -finite Borel measure on a metric space E is concentrated on a separable subspace of E. Furthermore, if we deal with a Borel probability measure defined on a separable metric space, then the question arises whether there exists a support of this measure with more or less nice geometrical properties. Since every separable metric space is isometrically contained in a separable Banach space and, more precisely, according to the classical Banach-Mazur theorem, every separable metric space can be realized as an isometric copy of some subset of the function space C[0, 1], it is reasonable to ask whether each Borel probability measure on a separable Banach space admits a support with good geometrical properties. This and similar questions are studied in many works but we do not intend to discuss them here. We only touch upon the topological structure of supports of Borel probability measures. The next auxiliary proposition shows certain connections between σ -finite Borel measures on separable metric spaces and sets of first category in these spaces. Lemma 2.. Let E be a topological space satisfying the following conditions: 1) there exists a countable set X ⊂ E which is everywhere dense in E; 2) for any point x ∈ X, the one-element set {x} is a Gδ -subset of E. Let μ be a σ -finite Borel measure on E vanishing on X (equivalently, for each x ∈ X, we have μ ({x}) = 0). Then there exists a set P ⊂ E such that: (a) P is an Fσ -subset of E; (b) P is of first category in E; (c) P is a support of μ . In particular, if E is an arbitrary separable metric space, then any σ -finite diffused Borel measure on E is concentrated on some first category subset of E. Proof. If μ is identically equal to zero, then there is nothing to prove. So we may assume that μ = 0. Moreover, since every σ -finite measure is equivalent to a probability measure, we may also suppose that μ (E) = 1. Actually, in our further consideration we only use the assumption that μ is a finite measure. Fix an enumeration {xn : n < ω } of all points of X. For each natural index n < ω , let {Un,m : m < ω } be a countable family of open subsets of E such that ∩{Un,m : m < ω } = {xn }. Taking into account the upper semicontinuity of the finite measure μ , we infer that for any two natural numbers n and k, there exists a natural number m(n, k) satisfying the relation

μ (Un,m(n,k) ) < 1/2n+k+1.

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Let us define U(k) = ∪{Un,m(n,k) : n < ω }. Obviously, the set U(k) is open and everywhere dense in E. Also, the relation

μ (U(k))  ∑{1/2n+k+1 : n < ω } = 1/2k holds true. Consequently, Q = ∩{U(k) : k < ω } is a Gδ -subset of E and the equality μ (Q) = 0 is valid. Denoting P = E \ Q, we see that the set P satisfies the relations (a), (b), and (c) which ends the proof of the lemma. From Theorem 1 and Lemma 2 we easily obtain the following important statement (see [192]). Theorem 2. Let (E, d) be a metric space whose topological weight is not real-valued measurable and let μ be a σ -finite diffused Borel measure on E. Then there exists a disjoint covering {A, B} of E such that μ (A) = 0 and B is a first category set in E. Proof. In view of Theorem 1, the measure μ admits a closed separable support C. By virtue of Lemma 2, the set C can be represented in the form C = C ∪C

(C ∩C = 0), /

where C is of μ -measure zero and C is of first category in C (hence, C is of first category in E as well). Putting A = C ∪ (E \ C), B = C , we get the required result. EXERCISES 1. Verify the validity of the relations w(RN ) = w(c0 ) = ω , w(l∞ ) = c. 2. Let (E, d) be a complete metric space whose topological weight is not real-valued measurable, and let μ be a Borel probability measure on E. Show that for each Borel set X ⊂ E and for any real number ε > 0, there exists a compact set K ⊂ X satisfying the relation

μ (X \ K) < ε .

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Deduce from this fact that if ν is an arbitrary σ -finite Borel measure on E and X is any Borel set in E, then

ν (X) = sup{ν (K) : K ⊂ X, K is compact}. Thus, there exists a set X  ⊂ X such that: (a) ν (X \ X  ) = 0; (b) X  can be represented as the union of a countable family of compact subsets of E. Infer from these results that every σ -finite Borel measure on E is Radon and hence E is a Radon space. Finally, taking into account the said above, conclude that the following statement holds true: under the assumption that there are no cardinals measurable in the Ulam sense, every complete metric space is Radon. 3. Let (E, d) be a metric space whose topological weight is not real-valued measurable, and let μ be a σ -finite diffused Borel measure on E. Demonstrate that there exists a set X ⊂ E satisfying the following relations: (a) X is a separable Fσ -subset of E; (b) X is of first category in E; (c) μ is concentrated on X, i.e., X turns out to be a support of μ . 4. Let E be a topological space, α be an ordinal number and let {Uξ : ξ < α } be an

α -sequence of open subsets of E such that E = ∪{Uξ : ξ < α }. For each ordinal ξ < α , denote Mξ = Uξ \ ∪{Uζ : ζ < ξ }. Let {Xξ : ξ < α } be an arbitrary α -sequence of subsets of E. We put M({Xξ : ξ < α }) = ∪{Mξ ∩ Xξ : ξ < α } and we say that the set M({Xξ : ξ < α }) is obtained from {Xξ : ξ < α } by applying the Montgomery operation (in short, M-operation) whose basis is the given α -sequence {Uξ : ξ < α }. Verify that: (a) if for any ξ < α , we have Xξ = ∪{Xξ ,i : i ∈ I}, then M({Xξ : ξ < α }) = ∪{M({Xξ ,i : ξ < α }) : i ∈ I};

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(b) if for any ξ < α , we have Xξ = ∩{Xξ ,i : i ∈ I}, then M({Xξ : ξ < α }) = ∩{M({Xξ ,i : ξ < α }) : i ∈ I}; (c) if for any ξ < α , we have Xξ = Yξ \ Zξ , then M({Xξ : ξ < α }) = M({Yξ : ξ < α }) \ M({Zξ : ξ < α }). 5. Let E be a topological space. For each ordinal ξ < ω1 , define by transfinite recursion the family Fξ (E) of subsets of E. Namely, first put F0 (E) = the family of all closed subsets of E. If ξ is odd, then let Fξ (E) coincide with the family of all countable unions of members from ∪{Fζ (E) : ζ < ξ }. If ξ > 0 is even, then let Fξ (E) coincide with the family of all countable intersections of members from ∪{Fζ (E) : ζ < ξ }. Finally, put F (E) = ∪{Fξ (E) : ξ < ω1 }. Verify the validity of the inclusion F (E) ⊂ B(E), where B(E) as usual denotes the Borel σ -algebra of E. Show that if a topological space E is such that every open set in E is of type Fσ or, equivalently, every closed set in E is of type Gδ , then F (E) = B(E). In particular, this equality holds true for any metrizable topological space E. For an ordinal ξ < ω1 , we say that a set X ∈ F (E) is of order ξ if X ∈ Fξ (E) \ ∪{Fζ (E) : ζ < ξ }. The classical theorem due to Lebesgue states that if E is an uncountable Polish space, then for each ξ < ω1 , there are subsets of E of order ξ (see, e.g., [99], [148], [160], [162], [183], or Appendix 6). 6. Let E be a metric space, ξ be a nonzero countable ordinal number and let X be a subset of E. Suppose that X is locally Borel of order  ξ , i.e., for any point x ∈ E, there exists a neighborhood U(x) such that the set U(x) ∩ X is Borel of order  ξ . Show that X is Borel of order  ξ . For this purpose, starting with the result of Exercise 4 and Lemma 1, use the method of transfinite induction on ξ < ω1 .

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7. Give an example of a metric space E such that: (a) the topological weight of E is equal to ω1 ; (b) there exists a subset of E which is locally Borel but is not Borel in E. 8∗ . Equip the least uncountable ordinal ω1 with its order topology and consider the Borel

σ -algebra B(ω1 ). Prove that there exists a probability measure μ on B(ω1 ) such that: (a) μ is diffused and two-valued; (b) μ (F) = 1 for any uncountable closed subset F of ω1 . Taking into account the circumstance that every separable subspace of ω1 is at most countable, show that μ is not Radon and does not admit a separable support. Conclude from this fact that the assumption of metrizability of a space E in the formulation of Theorem 1 is essential. Show also that the same assumption of metrizability is essential for the validity of Theorem 2. The probability measure μ just described is usually called the Dieudonn´e measure on ω1 (see [53], [71], [80], and [101]). Verify that μ is invariant under the semigroup of all those mappings f : ω1 → ω1 which are representable in the form f (ξ ) = η + ξ

(ξ < ω1 ),

where η = η ( f ) < ω1 . In other words, f coincides with the left shift of ω1 corresponding to η . Notice that ω1 is a locally compact topological space, every point of which has a countable local base. Define an analogue of μ for the compact space ω1 + 1 and check that this analogue turns out to be a Borel probability measure which is not Radon and does not admit a separable support. 9∗ . Let E be a separable Hilbert space (over R) equipped with the inner product < ·, · >. We recall that a set H ⊂ E is a (closed) half-space in E if there exist a vector e ∈ E \ {0} and a real number t such that H = {x ∈ E : < e, x >  t}. Let μ be a Borel probability measure on E. Prove that μ is completely determined by its values on the half-spaces of E. This phrase means that if ν is any Borel probability measure on E such that μ (H) = ν (H) for all half-spaces H ⊂ E, then μ = ν .

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Deduce from this fact that any Borel probability measure on E is completely determined by its values on all balls of E. In connection with Exercise 9, it should be mentioned that an analogous result does not longer hold for more general separable metric spaces. The first example of a separable metric space E, in which some Borel probability measure is not determined by its values on all balls of E, was given by Davies (see [50]). 10. Let (E, S , μ ) be a space equipped with a nonzero complete measure. We shall say that a partition {Ei : i ∈ I} of E is admissible for μ if the following two relations are satisfied: (a) Ei ∈ dom(μ ) for all i ∈ I; (b) for every set X ∈ dom(μ ), there exists a countable set J ⊂ I such that

μ (X \ ∪{Ei : i ∈ J}) = 0. Assume the Continuum Hypothesis. Let μ be a nonzero complete measure such that there exists a family {Zi : i ∈ I} ⊂ dom(μ ) satisfying the following conditions: (c) card(I)  c; (d) for any Z ∈ dom(μ ) and for any real ε > 0, there is an index i ∈ I such that μ (Z \ Zi ) < ε . Prove that there exists an admissible partition for μ . 11. Let (E, d) be a metric space whose topological weight is not measurable in the Ulam sense, and let μ be a σ -finite Borel measure on E. Assuming Martin’s Axiom, show that the completion μ  of μ is c-additive, i.e., for every disjoint family {Xi : i ∈ I} ⊂ dom(μ  ) with card(I) < c, the equality

μ  (∪{Xi : i ∈ I}) = ∑{μ  (Xi ) : i ∈ I} holds true. 12∗ . Let E be a topological space and let {Ui : i ∈ I} be a family of open subsets of E such that each Ui (i ∈ I) is of first category in E. Prove that the open set U = ∪{Ui : i ∈ I} is also of first category in E. This important statement is due to Banach. In order to show the validity of Banach’s statement, consider a maximal (with respect to the inclusion relation) disjoint family {V j : j ∈ J} of open sets in E such that for any index j ∈ J, the set V j is contained in some set Ui , where i = i( j). Verify that: (a) the set V = ∪{V j : j ∈ J} is of first category in E; (b) U ⊂ cl(V ) = V ∪ bd(V ), where cl(V ) denotes the closure of V and bd(V ) stands for the boundary of V .

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Taking into account the fact that bd(V ) is nowhere dense in E, conclude that U is of first category in E. Infer from the said above that an arbitrary topological space E admits a representation in the form E = E1 ∪ E2

(E1 ∩ E2 = 0), /

where E1 is a closed first category subspace of E and E2 is an open Baire subspace of E. 13. Let E be a topological space and let X be a subset of E which locally is of first category in E, i.e., for any point e ∈ E, there exists a neighborhood U(e) such that the set U(e) ∩ X is of first category in E. Show that X is of first category in E. For this purpose, apply the result of Exercise 12. 14. Let (G, ·) be an arbitrary topological group. Prove that the disjunction of the following two assertions is valid: (a) G is a first category topological space; (b) G is a Baire topological space. For this purpose, apply again the result of Exercise 12. 15. Let (G, ·) be a topological group and let A and B be two second category subsets of G such that both of them have the Baire property. Show that the set A · B = {a · b : a ∈ A, b ∈ B} contains a nonempty open subset of G (the Banach-Kuratowski-Pettis theorem). 16∗ . A nonempty linearly ordered set (S, ) is called a Suslin line if it satisfies the following conditions: (a) S has neither least element, nor greatest element; (b) S is dense in itself, i.e., for any two elements x ∈ S and y ∈ S such that x < y, there exists z ∈ S such that x < z < y; (c) S is complete in the Dedekind sense, i.e., for every nonempty bounded from above set X ⊂ S, there exists sup(X) in S; (d) any disjoint family of nonempty open subintervals of S is at most countable. (e) (S, ) is not isomorphic to the real line R equipped with its standard order. The existence of a Suslin line does not contradict the axioms of ZFC theory. On the other hand, it follows from Martin’s Axiom and the negation of the Continuum Hypothesis that there is no Suslin line (for more details, see [40], [91], [144], and [145]).

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Verify that the next two relations (e’) and (e”) are equivalent to condition (e): (e’) S does not contain a countable subset having common elements with every nonempty open subinterval of S; (e”) S is not metrizable with respect to its order topology. Let us endow (S, ) with its order topology and let μ be an arbitrary σ -finite Borel measure on S. Prove that μ has a separable support in S, i.e., there exists a separable closed set F ⊂ S such that μ (S \ F) = 0. It is useful to compare the last exercise with Exercise 8.

Appendix 4

Continuous nowhere approximately differentiable functions

In Exercise 4 for Chapter 8, an application of Kuratowski’s theorem on closed projection to the question of the existence of nowhere differentiable functions in the Banach space C[0, 1] was given. It was also underlined that such pathological functions are typical, i.e., they constitute a residual (co-meager) set in C[0, 1]. Recall that this classical result is due to Banach [9] and Mazurkiewicz [171]. Analogous questions on the existence of continuous nowhere differentiable functions with respect to other concepts of derivative are much deeper and more complicated. Among those concepts the notion of an approximate derivative is of especial interest and importance (cf., for instance, [32], [162], [183], and [210]). As mentioned at the end of Chapter 20, the first example of a nowhere approximately differentiable function from the Banach space C[0, 1] is due to Jarnik (see [88]). Moreover, he showed that such functions also are typical in the space C[0, 1]. In this Appendix we present one precise construction of a continuous function acting from R into R, which is nowhere approximately differentiable. This construction is due to Mal´y [167]. It is not difficult and, at the same time, is rather vivid from the geometrical point of view. We begin with some preliminary notions and facts. Let λ denote the standard Lebesgue measure on R and let X be an arbitrary λ -measurable subset of R. We recall that x ∈ R is a density point for (of) X if lim

h→0, h>0

λ (X ∩ [x − h, x + h])/2h = 1.

According to the classical theorem of Lebesgue (see [32], [158], [161], [183], and [210]), almost all points of X are its density points. The notion of a density point turned out to be rather deep and fruitful not only for real analysis but also for general topology, probability theory and some other domains of math397

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ematics. For example, by using this notion the important concept of the density topology on R was introduced and examined. Then various properties of the density topology were extensively investigated by many authors (Pauc, Goffman, Waterman, Nishiura, Neugebauer, Tall, and others). This topology and its further generalizations were studied from different points of view (see [75], [147], [166], [194], [205], and [237]). We shall touch upon the density topology in several exercises of this Appendix. Now, let f : R → R be a function and let x ∈ R. We recall that f is approximately continuous at x if there exists a λ -measurable set X such that: (i) x is a density point of X; (ii) the function f |(X ∪ {x}) is continuous at x. Exercises 3 and 4 for this Appendix show that all Lebesgue measurable functions can be described in terms of approximate continuity. Let f : R → R be a function and let x ∈ R. We recall that f is approximately differentiable at x if there exists a Lebesgue measurable set Y ⊂ R for which x is a density point and for which there is a limit lim

y→x, y∈Y, y=x

This limit is denoted by

 (x) fap

f (y) − f (x) . y−x

and is usually called an approximate derivative of f at

 (x) is uniquely determined by f and x (in this the point x. It should be noticed that fap

connection, see Exercise 7). For our purposes below, we need two simple auxiliary propositions. Lemma 1. Let f : R → R be a Lebesgue measurable function, let x be a point of R and  (x), suppose that f is approximately differentiable at x. Then for any real number t1 > fap

we have lim λ ({y ∈ [x − h, x + h] \ {x} : ( f (y) − f (x))/(y − x)  t1 })/2h = 0.

h→0+

 (x), we have Similarly, for any real number t2 < fap

lim λ ({y ∈ [x − h, x + h] \ {x} : ( f (y) − f (x))/(y − x)  t2 })/2h = 0.

h→0+

Proof. Since the argument in both cases is completely analogous, we shall consider only  (x). There exists a λ -measurable set X such that x is a density point of the case of t1 > fap

X and lim

y→x,y=x,y∈X

 ( f (y) − f (x))/(y − x) = fap (x).

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Fix ε > 0 for which  (x) + ε < t1 . fap

Then there exists a real number δ > 0 such that for any strictly positive real h < δ , we have  (x) + ε ). (∀y ∈ X ∩ [x − h, x + h] \ {x})(( f (y) − f (x))/(y − x)  fap

But if δ > 0 is sufficiently small, then

λ (X ∩ [x − h, x + h])/2h  1 − ε for all strictly positive reals h < δ . So we obtain the relation

λ ({y ∈ [x − h, x + h] \ {x} : ( f (y) − f (x))/(y − x)  t1 })/2h  ε , and the lemma is proved. Actually, in our further considerations we need only the following auxiliary assertion which is an immediate consequence of Lemma 1. Lemma 2. Let f : R → R be a Lebesgue measurable function, let x be a point of R, and suppose that for every strictly positive real number t, the relation

λ ({y ∈ [x − h, x + h] \ {x} : | f (y) − f (x)|/|y − x|  t}) >0 2h holds true. Then f is not approximately differentiable at x. lim inf h→0+

In particular, suppose that two sequences {hk : k ∈ N}, {tk : k ∈ N} of real numbers are given and satisfy the following conditions: (1) hk > 0 and tk > 0 for all integers k  0; (2) limk→+∞ hk = 0 and limk→+∞ tk = +∞; (3) the lower limit lim inf λ ({y ∈ [x − hk , x + hk ] \ {x} : k→+∞

| f (y) − f (x)|/|y − x|  tk })/2hk is strictly positive. Then, in view of Lemma 2, we can assert that the function f is not approximately differentiable at x. After these simple preliminary remarks, we are able to begin the construction of a nowhere approximately differentiable function.

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First of all, let us put f1 (0/9) = 0, f1 (1/9) = 1/3, f1 (2/9) = 0, f1 (3/9) = 1/3, f1 (4/9) = 2/3, f1 (5/9) = 1/3, f1 (6/9) = 2/3, f1 (7/9) = 3/3, f1 (8/9) = 2/3, f1 (9/9) = 3/3 and extend (uniquely) this partial function to a continuous function f1 : [0, 1] → [0, 1] in such a way that f1 becomes affine on each segment [k/9, (k + 1)/9], where k = 0, 1, . . . , 8. We shall start with this function f1 . In our further construction, we also need an analogous function g acting from the segment [0, 9] into the segment [0, 3]. Namely, we put g(x) = 3 f1 (x/9)

(x ∈ [0, 9]).

Obviously, g is continuous and affine on each segment [k, k + 1], where k = 0, 1, . . . , 8. Also, another function similar to g will be useful in our construction. Namely, we denote by g∗ the function acting from [0, 9] into [0, 3] whose graph is symmetric to the graph of g, with respect to the straight line {(x, y) ∈ R × R : y = 3/2}. In other words, we put g∗ (x) = 3 − g(x) for all x ∈ [0, 9]. Suppose now that for a natural number n  1, the function fn : [0, 1] → [0, 1] has already been defined, such that: (a) fn is continuous; (b) for each segment of the form [k/9n , (k + 1)/9n], where k ∈ {0, 1, . . . , 9n − 1}, the function fn is affine on [k/9n , (k + 1)/9n] and the image of this segment with respect to fn is some segment of the form [ j/3n , ( j + 1)/3n ], where j ∈ {0, 1, . . . , 3n − 1}. Let us construct a function fn+1 : [0, 1] → [0, 1].

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For this purpose, it suffices to define fn+1 on any segment [k/9n , (k + 1)/9n ], where k ∈ {0, 1, . . . , 9n − 1}. Here only two cases are possible. 1. fn is increasing on [k/9n , (k + 1)/9n]. In this case, let us consider the following two sets of points of the plane: {(0, 0), (0, 3), (9, 3), (9, 0)}, {(k/9n , fn (k/9n )), (k/9n , fn ((k + 1)/9n)), ((k + 1)/9n, fn ((k + 1)/9n)), ((k + 1)/9n, fn (k/9n ))}. Because we have here the vertices of two rectangles, there exists a unique affine transformation h : R2 → R2 satisfying the conditions h(0, 0) = (k/9n , fn (k/9n )), h(0, 3) = (k/9n , fn ((k + 1)/9n)), h(9, 3) = ((k + 1)/9n, fn ((k + 1)/9n)), h(9, 0) = ((k + 1)/9n, fn (k/9n )). Let the graph of the restriction of fn+1 to the segment [k/9n , (k + 1)/9n] coincide with the image of Gr(g) under action of h. 2. fn is decreasing on [k/9n , (k + 1)/9n]. In this case, let us consider the following two sets of points of the plane: {(0, 0), (0, 3), (9, 3), (9, 0)}, {(k/9n , fn ((k + 1)/9n)), (k/9n , fn (k/9n )), ((k + 1)/9n, fn (k/9n )), ((k + 1)/9n, fn ((k + 1)/9n))}. Here we also have the vertices of two rectangles, so there exists a unique affine transformation h ∗ : R2 → R2 satisfying the relations h∗ (0, 0) = (k/9n , fn ((k + 1)/9n)), h∗ (0, 3) = (k/9n , fn (k/9n )), h∗ (9, 3) = ((k + 1)/9n, fn (k/9n )), h∗ (9, 0) = ((k + 1)/9n, fn ((k + 1)/9n)).

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Let the graph of the restriction of fn+1 to the segment [k/9n , (k + 1)/9n] coincide with the image of Gr(g∗ ) under action of h∗ . The function fn+1 has thus been determined. It immediately follows from the above construction that the corresponding analogues of the conditions (a) and (b) hold true for fn+1 , too. In other words, fn+1 is continuous and for each segment of the form [k/9n+1 , (k + 1)/9n+1], where k ∈ {0, 1, . . . , 9n+1 − 1}, the function fn+1 is affine on [k/9n+1 , (k + 1)/9n+1] and the image of this segment with respect to fn+1 is some segment of the form [ j/3n+1, ( j + 1)/3n+1], where j ∈ {0, 1, . . ., 3n+1 − 1}. Moreover, our construction directly yields the relation (∀x ∈ [0, 1])(| fn+1 (x) − fn (x)|  1/3n). In addition, let [u, v] = [k/9n , (k + 1)/9n] be an arbitrary segment on which fn is affine. Then it is not hard to check that fn+1 ([u, (2u + v)/3]) = fn ([u, (2u + v)/3]), fn+1 ([(2u + v)/3, (2v + u)/3]) = fn ([(2u + v)/3, (2v + u)/3]), fn+1 ([(u + 2v)/3, v]) = fn ([(u + 2v)/3, v]). Proceeding in this way, we come to the sequence of functions ( f1 , f2 , . . . , fn , . . . ) which are uniformly converging to some continuous function f . Obviously, f also acts from [0, 1] into [0, 1]. We assert that f is nowhere approximately differentiable on the segment [0, 1]. In order to demonstrate this fact, let us take an arbitrary point x ∈ [0, 1] and fix a natural number n  1. Clearly, there exists a number k ∈ {0, 1, . . . , 9n − 1} such that x ∈ [k/9n , (k + 1)/9n]. Therefore, we have fn (x) ∈ [ j/3n , ( j + 1)/3n]

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for some number j ∈ {0, 1, . . ., 3n − 1}. For the sake of simplicity, denote [u, v] = [k/9n , (k + 1)/9n], [p, q] = [ j/3n , ( j + 1)/3n ]. From the remarks made above it immediately follows that for all natural numbers m > n, we have fm (x) ∈ [p, q] and, consequently, f (x) ∈ [p, q], too. Further, we may assume without loss of generality that fn is increasing on [u, v] because the case when fn is decreasing on [u, v] can be considered completely analogously. Suppose first that f (x)  (p + q)/2 and put D1 = [(2v + u)/3, v]. Then for each point y ∈ D1 , we may write f (y) ∈ [(2q + p)/3, q]. Hence, we get ( f (y) − f (x))/(y − x)  ((2q + p)/3 − (p + q)/2)/(v − u) = (1/6)(3n). Suppose now that f (x)  (p + q)/2 and denote D2 = [u, (2u + v)/3]. In this case, for any point y ∈ D2 , we may write f (y) ∈ [p, (2p + q)/3]. Hence, we get ( f (x) − f (y))/(x − y)  ((p + q)/2 − (2p + q)/3)/(v − u) = (1/6)(3n). Thus, in both of these cases we have the inequality

λ ({y ∈ [x − 1/9n, x + 1/9n] \ {x} : | f (y) − f (x)|/|y − x|  (1/6)(3n )})  (1/3)(1/9n) or, equivalently,

λ ({y ∈ [x − 1/9n, x + 1/9n] \ {x} : | f (y) − f (x)|/|y − x|  (1/6)(3n )})  (1/6)λ ([x − 1/9n, x + 1/9n]).

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The latter relation immediately implies that our function f is not approximately differentiable at x (see Lemma 2 and the comments after it). Remark 1. The function f constructed above has a number of other interesting properties (in this connection, see [167]). Now, starting with an arbitrary continuous nowhere approximately differentiable function acting from [0, 1] into [0, 1], we can easily get an analogous function for R. We thus come to the following classical result (first obtained by Jarnik in 1934). Theorem 1. There exist continuous bounded functions acting from R into R, which are nowhere approximately differentiable. Remark 2. As already mentioned, Jarnik established a much stronger theorem. Let us remember once again the precise formulation. Namely, Jarnik proved that almost all (in the sense of the Baire category) functions from the Banach space C[0, 1] are nowhere approximately differentiable. Clearly, this result essentially strengthens the corresponding result of Banach and Mazurkiewicz obtained by them for the usual differentiability. Further investigations showed that analogous statements hold true for many kinds of generalized derivatives. In this connection, we refer the reader to the fundamental paper [33] where category analogues of Theorem 1 for generalized derivatives are discussed in detail (see also [34]). In Chapter 15 of [122], a nontrivial application of nowhere approximately differentiable functions is demonstrated to the question concerning some relationships between the supmeasurability and weak sup-measurability of functions acting from R × R into R. The concept of an approximate derivative relies essentially on the notion of a density point, so it is reasonable to touch upon the density topology Td on R and to consider several elementary properties of this topology. A precise definition of the density topology and some of its interesting features are given in Exercises 9, 10, 11 for this Appendix. One of the most important facts concerning the density topology states that (R, Td ) is a completely regular topological space (see, for instance, [192], [194], and [237]). This property of Td implies highly nontrivial consequences in real analysis. To illustrate the just said, we shall sketch a proof of the existence of an everywhere differentiable function

φ :R→R which is not monotone on any non-degenerate subinterval of R (notice that the construction presented below is due to Goffman [74]).

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Consider any two disjoint countable sets A = {an : n ∈ N} ⊂ R, B = {bn : n ∈ N} ⊂ R such that both of them are everywhere dense in R. Taking into account Exercise 10 and the fact that (R, Td ) is completely regular, we can find for each natural number n, an approximately continuous function fn : R → [0, 1] satisfying the relations 0  fn (x)  1

(x ∈ R),

fn (an ) = 1, (∀x ∈ B)( fn (x) = 0). Analogously, for each natural number n, there exists an approximately continuous function gn : R → [0, 1] such that 0  gn (x)  1

(x ∈ R),

gn (bn ) = 1, (∀x ∈ A)(gn (x) = 0). Now, define a function h : R→R by the formula h=

∑ (1/2n)( fn − gn).

n∈N

It is easy to check that: a) h is bounded and approximately continuous; b) h(a) > 0 for all a ∈ A; c) h(b) < 0 for all b ∈ B. Let φ denote an indefinite integral of h. Keeping in mind the above relations a), b) and c), it is not difficult to show that: i) φ is everywhere differentiable on R and φ  (x) = h(x) for each x ∈ R; ii) φ is not monotone on any nonempty open subinterval of R.

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We see that with the aid of the density topology Td , it is possible to prove this very nontrivial fact of mathematical analysis (another, more classical proof is presented in [98]). Remark 3. The density topology on R can be regarded as a particular case of the so-called von Neumann topology. Let (E, S , μ ) be a space with a complete probability measure or, more generally, with a complete nonzero σ -finite measure. Then, in accordance with a deep theorem of von Neumann and Maharam (see [166], [192], [194], and [205]), there exists a topology T = T (μ ) on E such that: (1) (E, T ) is a Baire space satisfying the Suslin condition; (2) the family of all those subsets of (E, T ) which have the Baire property in (E, T ) coincides with the σ -algebra S ; (3) a set X ⊂ E is of μ -measure zero if and only if X is of first category in (E, T ). We say that T = T (μ ) is a von Neumann topology associated with the given measure space (E, S , μ ). Notice that T is not unique in general. This circumstance is not so surprising because the proof of the existence of T , similarly to the proof of the well-known Hahn-Banach theorem, is essentially based on uncountable forms of the Axiom of Choice which are not needed in the case of Td . There are many nontrivial applications of a von Neumann topology in various branches of contemporary mathematics. For instance, some applications of this topology to the general theory of stochastic processes can be found in [205]. Remark 4. There exist translation-invariant extensions of the Lebesgue measure λ for which an analogue of the classical Lebesgue theorem on density points does not hold. For example, there are a measure μ on R and a μ -measurable set X ⊂ R, such that: (1) μ is an extension of λ ; (2) μ is invariant under the group of all isometric transformations of R; (3) there is only one μ -density point for X, i.e., there exists a unique point x ∈ R for which we have

μ (X ∩ [x − h, x + h]) = 1. 2h A more detailed account of the measure μ and its other extraordinary properties can be lim

h→0+

found in [115]. EXERCISES 1. Let (tn )n∈N be a sequence of strictly positive real numbers such that lim tn = 0,

n→+∞

lim tn /tn+1 = 1.

n→+∞

Appendix 4. Continuous nowhere approximately differentiable functions

407

Let X be a Lebesgue measurable subset of R and let x ∈ R. Prove that the following two assertions are equivalent: (a) x is a density point of X; (b) limn→+∞ λ (X ∩ [x − tn , x + tn])/2tn = 1. 2. Let X be a Lebesgue measurable subset of R and let x ∈ R. Show that the following two assertions are equivalent: (a) x is a density point of X; (b) limh→0+, k→0+ λ (X ∩ [x − h, x + k])/(h + k) = 1. 3∗ . Let g : R → R be a function, let ε be a fixed strictly positive real number and suppose that for any λ -measurable set X with λ (X) > 0, there exists a λ -measurable set Y ⊂ X with

λ (Y ) > 0, such that (∀x ∈ Y )(∀y ∈ Y )(|g(x) − g(y)| < ε ). Demonstrate that there exists a λ -measurable function h:R→R for which we have (∀x ∈ R)(|g(x) − h(x)| < ε ). Infer from this fact that if a function g : R → R satisfies the above condition for every ε > 0, then g is measurable in the Lebesgue sense. 4. Let f : R → R be a function. By applying the result of Exercise 3 and utilizing Luzin’s theorem on the structure of Lebesgue measurable functions (i.e., the so-called C-property), show that the following two assertions are equivalent: (a) the function f is measurable in the Lebesgue sense; (b) for almost all (with respect to λ ) points x ∈ R, the function f is approximately continuous at x. 5. Let f : R → R be a locally bounded Lebesgue measurable function and let F(x) =

 x 0

f (t)dt

(x ∈ R).

Prove that for any point x ∈ R at which the function f is approximately continuous, we have F  (x) = f (x). Check also that the local boundedness of f is essential for the validity of this assertion. 6. Verify that if a function f : R → R is approximately differentiable at x ∈ R, then f is also approximately continuous at x.

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7. Check that an approximate derivative of a function f : R→R at a point x ∈ R is uniquely determined, i.e., it does not depend on the choice of a Lebesgue measurable set Y for which x is a density point and for which the corresponding limit exists. Check also that the family of all functions acting from R into R and approximately differentiable at a point x ∈ R forms a vector space over R. 8. Demonstrate that if a function f : R → R is differentiable in the usual sense at a point  (x) = f  (x). x ∈ R, then f is approximately differentiable at x and fap

Give an example which shows that the converse assertion is not true. 9. For any Lebesgue measurable subset X of R, let us denote d(X) = {x ∈ R : x is a density point f or X}. Further, denote by Td the family of all those Lebesgue measurable sets Y ⊂ R for which Y ⊂ d(Y ). Verify that: (a) Td is a topology on R strictly extending the standard Euclidean topology of R; (b) (R, Td ) is a Baire topological space and satisfies the Suslin condition, i.e., no nonempty open set in (R, Td ) is of first category and any disjoint family of nonempty open sets in (R, Td ) is at most countable; (c) every first category set in (R, Td ) is nowhere dense and closed; consequently, the family of all those subsets of (R, Td ) which have the Baire property in (R, Td ), coincides with the Borel σ -algebra of (R, Td ); (d) a set X ⊂ R is Lebesgue measurable if and only if X has the Baire property in (R, Td ); (e) a set X ⊂ R is of Lebesgue measure zero if and only if X is a first category subset of (R, Td ); (f) the space (R, Td ) is not separable. The above-mentioned topology Td is called the density topology on R. In a similar way, the density topology can be introduced for the Euclidean space Rn (n  2) equipped with the n-dimensional Lebesgue measure λn . 10. Let f : R → R be a function and let x ∈ R. Prove that the following two assertions are equivalent: (a) f is approximately continuous at x; (b) f regarded as a mapping acting from (R, Td ) into R is continuous at x. 11∗ . By starting with the result of the previous exercise, show that the topological space (R, Td ) is connected.

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For this purpose, suppose to the contrary that there exists a partition {A, B} of R into two nonempty sets A ∈ Td and B ∈ Td . Then define a function f : R→R by putting f (x) = 1 for all x ∈ A, and f (x) = −1 for all x ∈ B. Obviously, f is a bounded continuous mapping acting from (R, Td ) into R and according to Exercise 10, f is approximately continuous at each point of R. Further, define F(x) =

 x 0

f (t)dt

(x ∈ R).

By applying Exercise 5 of this Appendix, demonstrate that the function F is differentiable everywhere on R and F  (x) = 1 ∨ F  (x) = −1 for each x ∈ R. This circumstance yields a contradiction with the Darboux property of any derivative (cf. [32] and [183]). 12∗ . Supposing that the Continuum Hypothesis holds, prove that there exists a function f : R → R satisfying the following relations: (a) for each uncountable set X ⊂ R, the restriction f |X is not monotone; (b) there are a set Y ⊂ R with card(Y ) = c and an everywhere differentiable function g : R → R such that f |Y = g|Y ; consequently, f is not a Sierpi´nski-Zygmund function. Formulate and prove an analogous result assuming Martin’s Axiom instead of the Continuum Hypothesis.

Appendix 5

Some facts from the theory of commutative groups

The fundamental role of group theory in many fields of contemporary mathematics, such as the theory of algebraic equations, geometry, topology, analysis, and the theory of differential equations, is widely known. In the class of all groups the subclass of all commutative (i.e., Abelian) groups is of special interest. For instance, as demonstrated in Chapters 3, 10 and 17, certain properties of commutative and solvable groups are important in studies of invariant and quasi-invariant measures. In this Appendix, we would like to recall some material from the general theory of infinite commutative groups. This theory is thoroughly considered, for example, in remarkable monographs [70], [83], and [152]. We shall present only those statements about commutative groups, which were essentially utilized in the preceding chapters of this book. For instance, the reader could see in Chapters 3, 10 and 17 how specific features of a given commutative transformation group G imply the corresponding properties of those measures which are invariant (or quasi-invariant) with respect to G. Here we do not touch upon deep relationships between commutative groups and modules. Obviously, any commutative group may be regarded as a module over the ring Z of all integers. Numerous facts of the theory of commutative groups can be treated from a more general view-point, as facts of the theory of modules. Such an approach is fruitful for both of these theories and very often leads to important results. Concerning finite commutative groups, we may say that there is no problem connected with their structural description. Indeed, the classical theorem says that any finitely generated (in particular, finite) commutative group G is representable as a direct sum of cyclic groups. The proof can be carried out by induction on card(X), where X is a finite set of generators of G (in this connection, see [70] or [152]). For infinite commutative groups, we do not have such a nice result and the corresponding 411

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theory becomes more complicated. Moreover, there are many infinite commutative groups which do not admit a nontrivial representation in the form of a direct sum (some simple examples will be given below). It is also known that there exist commutative groups with the above-mentioned property whose cardinalities are arbitrarily large. Let (G, +) be a commutative group and let 0 denote the neutral element of G. We recall that G is free if it contains at least one subset X satisfying the following condition: for any commutative group H and for any mapping

φ : X → H, there exists a unique homomorphism φ  : G → H extending φ . Actually, this notion is a particular case of the definition of a free object for a given abstract algebraic structure. A set X in the definition above is usually called a basis of the group G. A finite commutative group G is free if and only if G is trivial (in other words, G = {0}). The group (Z, +) of all integers is a natural example of an infinite free commutative group. The one-element set {1} is a basis of this group (as well as the one-element set {−1}). It can easily be verified that Z is not decomposable; this phrase means that Z does not admit a representation in the form of a direct sum of its two proper subgroups. Evidently, the direct sum of a family of free commutative groups is also a free commutative group. More precisely, if {Gi : i ∈ I} is a family of free commutative groups and Xi is a basis of Gi for each i ∈ I, then the set X = ∪{Xi : i ∈ I} is a basis of the free commutative group G = ∑ Gi . i∈I

Of course, we identify here each set X j ( j ∈ I) with its image under the canonical monomorphism from G j into ∑i∈I Gi . It immediately follows from the said above that any commutative group is a homomorphic image of some free commutative group. This fact is a particular case of the general statement from the theory of algebraic systems. We say that a commutative group (G, +) is projective if for any two commutative groups (H, +) and (H  , +) satisfying the relation (H, +) ⊂ (H  , +) and for an arbitrary homomorphism

φ : G → H  /H,

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there exists a homomorphism φ  : G → H  such that

φ = ψ ◦ φ , where ψ : H  → H  /H is the canonical epimorphism. The following statement yields a characterization of free commutative groups. Theorem 1. Let (G, +) be a commutative group. Then these three assertions are equivalent: 1) G is a free group; 2) G can be represented as a direct sum of a family of groups, all of which are isomorphic to Z; 3) G is a projective group. For the proof of this statement, see [70] or [152]. Thus, if (G, ·) is a free commutative group, then we may write G = ∑ Zi , i∈I

where I is some set of indices and, for each i ∈ I, the group Zi is isomorphic to Z. In other words, we have Zi = {kei : k ∈ Z}, where ei denotes a certain element of Zi generating Zi and such that kei = lei

(k ∈ Z, l ∈ Z, k = l).

Clearly, the set {ei : i ∈ I} can be regarded as a basis of G = ∑i∈I Zi . Let us mention an important statement, according to which any subgroup of a free commutative group is free, too. This statement remains true for non-commutative groups as well (see, for instance, [152]). We say that a commutative group (G, +) is divisible if for every natural number n > 0 and for any element g ∈ G, the equation nx = g has at least one solution in G. Evidently, the product of any family of divisible commutative groups is a divisible commutative group; however, the standard proof of this simple fact is based on the Axiom of Choice. Example 1. The additive group (Q, +) of all rational numbers is divisible. All nonzero elements in Q are of infinite order. Similarly to Z, the group Q is not decomposable. The commutative group (S1 , ·) is divisible, too. Recall that S1 denotes the unit circumference

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in R2 equipped with the natural group operation which is the restriction to S1 of the usual multiplication operation of complex numbers (here R2 is identified with the field C of all complex numbers). Moreover, for every nonzero n ∈ N, there are elements in S1 of order n. Clearly, there are also continuumly many elements in S1 of infinite order. Example 2. Let p be an arbitrary prime natural number. Fix a natural number k and denote by Gk the subgroup of (S1 , ·) consisting of all those elements z ∈ S1 for which the equality k

zp = 1 is valid. In this way, we obtain a strictly increasing (by inclusion) sequence of finite commutative groups G0 ⊂ G1 ⊂ ... ⊂ Gk ⊂ ... . Let us define Γ p = ∪{Gk : k < ω }. Then Γ p is also a subgroup of S1 and card(Γ p ) = ω . Any group isomorphic to Γ p is usually called a quasi-cyclic group of type p∞ . Actually, the group Γ p may be considered as the inductive limit of the family of groups (Gk )k<ω with respect to their canonical embeddings

φk,k+1 : Gk → Gk+1

(k < ω ).

Let us verify that Γ p is divisible. First, let us observe that it suffices to establish the following fact: for any prime natural number q and for any element s ∈ Γ p , there exists an element z ∈ Γ p such that zq = s. Indeed, assume that this fact is true. If n is an arbitrary nonzero natural number, then we can represent n in the form n = q1 q2 · · · qk for some finite sequence (q1 , q2 , . . . , qk ) of prime numbers. According to our assumption, the equalities q

q

k−1 s = zq11 , z1 = zq22 , . . . , zk−2 = zk−1 , zk−1 = zk k

are satisfied for certain elements z1 , z2 , . . . , zk−1 , zk of Γ p , thus it immediately follows that q q2 ···qk

s = zk 1

= znk .

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415

Now, let q be a prime natural number and let s be an arbitrary element from Γ p . Then there exists a natural number k such that k

s p = 1. Only two cases are possible. 1. q = p. In this case, the numbers pk and q are co-prime. According to a well-known theorem from elementary number theory, we may write mpk + lq = 1 for some two integers m and l. Therefore, putting z = sl , we obtain k

zq = slq = s1−mp = s. Thus, z belongs to Γ p and is a solution of the equation xq = s. 2. q = p. In this case, consider an element z ∈ S1 such that z p = s. Since zp

k+1

k

= s p = 1,

we conclude that z belongs to Γ p and is a solution of the equation xq = s. Finally, let us point out the following interesting property of the group Γ p : every proper subgroup of Γ p is necessarily finite. We leave to the reader the checking this fact which also yields that Γ p is not decomposable. A commutative group (G, +) is called injective if for any two commutative groups (H, +) and (H  , +) satisfying the relation (H, +) ⊂ (H  , +) and for an arbitrary homomorphism

φ : H → G, there exists a homomorphism φ  : H  → G extending φ . The next statement yields a characterization of divisible commutative groups. Theorem 2. Let (G, +) be a commutative group. Then the following three assertions are equivalent: 1) G is a divisible group; 2) G can be represented as the direct sum of a family of groups, all of which are isomorphic either to Q or to a group of type p∞ , where p is an arbitrary prime number; 3) G is an injective group. The proof of this theorem may be found in [70] and [152]. From assertion 2) we infer at once that if (G, +) is a divisible commutative group whose all nonzero elements are of infinite order, then G is a vector space over Q.

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Notice that assertion 3) easily implies the next useful fact. If G is a divisible subgroup of a commutative group H, then G is a direct summand in H; in other words, we have a representation H = G + G

(G ∩ G = {0})

for some subgroup G of H. To see this circumstance, let us consider the identity mapping

φ0 : G → G which obviously is an isomorphism of G onto itself. Applying the injectivity of G, we claim that there exists a homomorphism

φ : H→G extending φ0 . Let us put G = ker(φ ) = φ −1 (0). Then G is a subgroup of H, and it can readily be verified that H is a direct sum of G and G . This property characterizes all divisible commutative groups. In other words, if (G, +) is a direct summand in every larger commutative group, then G is divisible. Let us mention, in addition to the said above, that any direct summand in a divisible commutative group is divisible itself. Indeed, suppose that (G, +) is a divisible commutative group and let G = H + H

(H ∩ H  = {0})

be a representation of G in the form of the direct sum of its two subgroups H and H  . Let us demonstrate that H is divisible, too. Take any natural number n > 0 and an arbitrary element h ∈ H. Since G is divisible, there exists an element x ∈ G such that nx = h. Clearly, we may write x = y + z, where y ∈ H and z ∈ H  . Further, we have nx = n(y + z) = ny + nz = h, h − ny = nz, h − ny ∈ H, nz ∈ H  . Consequently, h − ny = nz = 0 and h = ny which shows that H is divisible. Theorem 2 also implies that for any commutative group (G, +), there are sufficiently many homomorphisms acting from G into (S1 , ·). Indeed, take any nonzero element g ∈ G and consider the group [g] generated by g. In view of Example 1, the group S1 contains a subgroup T isomorphic to [g]. Let

φ : [g] → T

Appendix 5. Some facts from the theory of commutative groups

417

be an isomorphism between these two subgroups and let

φ  : G → S1 be a homomorphism extending φ . Obviously, we have φ  (g) = e, where e stands for the neutral element of S1 . It immediately follows from the said above that if x ∈ G, y ∈ G and x = y, then there always exists a homomorphism

φx,y : G → S1 such that φ (x) = φ (y). In other words, the family of all homomorphisms acting from G into S1 separates the elements of G. Keeping in mind this circumstance and applying the standard argument, we see that every commutative group G can be embedded in the corresponding product group Sκ1 where κ is some cardinal number. More precisely, we may take κ equal to card(G × G). From this fact we readily derive that for any commutative group (G, +), there exists a divisible commutative group (G , +) containing G as a subgroup. Moreover, we may assume without loss of generality that card(G )  card(G) + ω . Let us point out another important consequence of the above-mentioned fact. Namely, every infinite commutative group G can be endowed with a non-discrete Hausdorff topology compatible with the group operation in G. This result does not hold for infinite noncommutative groups. For instance, assuming CH, Shelah constructed a group of cardinality continuum, which does not admit a non-discrete Hausdorff group topology (see [213]). Later, Ol’shanskii gave an example (without using extra axioms) of an infinite group which also does not admit a non-discrete Hausdorff group topology. We have the following important statement. Theorem 3. Let (G, +) be a commutative group and let H be a subgroup of G. If G can be represented in the form of a direct sum of cyclic groups, then H can be represented in the same form. A detailed proof of Theorem 3 is given in [152]. It is useful to compare this theorem with the result mentioned earlier and stating that any subgroup of a free commutative group is free, too. Example 3. It can easily be verified that the additive group Q is not a direct sum of cyclic groups.

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However, there exists an increasing (by the inclusion relation) countable family {Hn : n <

ω } of subgroups of Q, such that each Hn is representable as a direct sum of cyclic groups and Q = ∪{Hn : n < ω }. Indeed, for n < ω , it suffices to put Hn = the subgroup of Q generated by {1/n!}. An analogous representation is trivially valid for any group of type p∞ where p is an arbitrary prime natural number (see Example 2). Finally, we are able to establish the following statement which is due to Kulikov and was essentially exploited in this book. Theorem 4. Let (G, +) be an arbitrary commutative group. Then there exists an increasing (by inclusion) countable family {Gn : n < ω } of subgroups of G such that: 1) ∪{Gn : n < ω } = G; 2) each group Gn (n < ω ) is a direct sum of cyclic groups. Proof. We know that G can be embedded in a divisible commutative group G . In view of Theorem 2, this G is the direct sum of a family of groups which are isomorphic either to Q or to a group of type p∞ , where p is a prime natural number. Taking into account Example 3, we see that G can be represented in the form G = ∪{Gn : n < ω }, where {Gn : n < ω } is an increasing (by inclusion) countable family of subgroups of G and each Gn is a direct sum of cyclic groups. Now, let us put Gn = G ∩ Gn

(n < ω ).

Clearly, Gn is a subgroup of G for any n < ω . Moreover, we have (∀n < ω )(Gn ⊂ Gn+1 ),

∪ {Gn : n < ω } = G.

Finally, since Gn is a subgroup of Gn , we may apply Theorem 3 and conclude that Gn is a direct sum of cyclic groups. Theorem 4 has thus been proved. The last theorem played a key role in our considerations concerning the existence of nonmeasurable sets and the measure extension problem (see, for instance, Chapter 10). This theorem also implies that the family of all subgroups of an uncountable commutative group is rather large from the measure-theoretical view-point. For example, it is known that if G

Appendix 5. Some facts from the theory of commutative groups

419

is a commutative group of cardinality ω1 , then there always exists a subgroup of G nonmeasurable with respect to a given nonzero σ -finite diffused measure on G (see [119] or Exercise 7 of this Appendix). For non-commutative groups of the same cardinality, an analogous fact does not longer hold because there are groups G with card(G) = ω1 but without proper uncountable subgroups (see [213]). Let us underline once more that many important statements concerning the structure of infinite commutative groups are presented in fundamental monographs [70], [83], [152]. Extensive information about locally compact commutative groups can be found in [46], [83], [177], and [202]. Notice that there are rather natural questions concerning infinite commutative groups, which are not decidable within ZFC theory. Of course, the famous Whitehead problem is of special interest among questions of such a kind. Since the topological formulation of the Whitehead problem looks much more attractive than its purely algebraic formulation, we will give below the topological version of this problem. Let κ be any cardinal number. Consider the topological product G = Sκ1 . This G may be regarded as a compact, commutative, and arcwise-connected group (the last property is easily verified). Whitehead’s question can be formulated as follows. Is any compact commutative arcwise-connected group isomorphic to the product of some copies of S1 ? As demonstrated by Shelah, this question is undecidable within ZFC theory (see his remarkable works [211] and [212]). EXERCISES 1∗ . Let p be a prime natural number. Show that the group Γ p has the property that any its proper subgroup is a cyclic group of order pk for some natural number k. In particular, Γ p does not contain proper infinite subgroups. For this purpose, consider a representation of Γ p in the form Γ p = G0 ∪ G1 ∪ G2 ∪ · · · ∪ Gk ∪ · · · , where each subgroup Gk is cyclic of order pk , and verify that Gk is a unique cyclic subgroup of Γ p whose order is pk . Let a0 = 1 and, consequently, G0 = [a0 ], and let Gk = [ak ] for k > 0. Suppose that G is a proper subgroup of Γ p . Then there exists a smallest natural number n such that an+1 ∈ G.

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Check that G = Gn . The inclusion Gn ⊂ G is trivial because of the relation an ∈ G. To show the reverse inclusion G ⊂ Gn , suppose otherwise, i.e., suppose that g ∈ G \ Gn for some element g ∈ Γ p . There exists a smallest k > n such that g ∈ Gk = [ak ] and, therefore, g ∈ Gk−1 = [ak−1 ]. The group [g] is finite cyclic and its order cannot be less than pk−1 because the only cyclic subgroup of Γ p with order pk−1 is Gk−1 . Consequently, the order of [g] is equal to pk and [g] must coincide with Gk . This fact immediately implies that ak ∈ G and hence an+1 ∈ G contradicting the definition of n. In connection with Exercise 1, let us also remark that the groups Γ p are unique (with exactness to an isomorphism) infinite commutative groups having the property that all their proper subgroups are finite (see [152]). 2∗ . Give a direct proof of the fact that if (G, +) is an arbitrary commutative group, (H, +) is a divisible commutative group and φ : G → H is a partial homomorphism, then φ admits an extension φ ∗ : G → H to a group homomorphism. For this purpose, consider any partial homomorphism φ  : G → H extending φ . Let g be an arbitrary element of G \ dom(φ  ). Show that φ  can be extended to a partial homomorphism

φ  : G → H such that g ∈ dom(φ  ). Indeed, only two cases are possible. 1. kg ∈ dom(φ  ) for some natural number k > 0. We may assume, without loss of generality, that k is the least natural number with this property. Since H is divisible, there exists h ∈ H satisfying the equality kh = φ  (kg). Put φ  (g) = h and prove that the latter equality correctly determines the partial homomorphism φ  : G → H whose domain coincides with the group generated by dom(φ  ) ∪ {g}. 2. kg ∈ dom(φ  ) for all natural numbers k > 0. In this case, pick an arbitrary element h ∈ H and put φ  (g) = h. Verify that this relation also correctly determines a partial homomorphism φ  : G → H whose domain coincides with the group generated by dom(φ  ) ∪ {g}. Finally, apply the Zorn lemma for obtaining the required homomorphism φ ∗ : G → H extending the initial partial homomorphism φ . 3. Show that if a commutative group (G, +) has the property that it is a direct summand in every commutative group containing G, then G is divisible. 4∗ . Check that any uncountable commutative group (G, +) contains a proper subgroup H with card(H) = card(G). In connection with this exercise, a deep result of Shelah [213] should be mentioned stating the existence of a group (Γ, ·) of cardinality ω1 , which does not contain proper uncountable subgroups. The existence of such a (Γ, ·) solves a famous problem posed by Kurosh [152]

Appendix 5. Some facts from the theory of commutative groups

421

many years ago. 5∗ . Let E be a compact topological group and let X be an infinite subgroup of E. Verify that the topology on X induced by the topology of E is not discrete. Deduce from this fact that any infinite commutative group (G, +) is isomorphic to an everywhere dense subgroup of some closed subgroup of Sκ1 , where κ is an appropriate cardinal number. Taking this circumstance into account, conclude that G admits a non-discrete Hausdorff topology which is totally bounded and compatible with the algebraic structure of G. The above result was first obtained in [102]. In connection with Exercise 5, it should also be noticed that any infinite commutative group admits a non-discrete metrization whose metric is invariant under all translations of the group (see [70]). 6∗ . Consider the two-element multiplicative discrete group {−1, 1} as a subgroup of the countable multiplicative group D = {cos(2π n/2m) + isin(2π n/2m) : n ∈ Z, m ∈ Z}. The commutative group D is 2-divisible, i.e., the equation x2 = d is solvable in D for any d ∈ D. Let us introduce the product group G = Dω which also is 2-divisible. Then H = {−1, 1}ω is a compact subgroup of G. By using H, we define a topology T on G. First of all, the set H remains in its product topology T1 and we equip any translate gH (g ∈ G) with the topology which is obtained from T1 via the left translation g. Further, a set U ⊂ G is open in the new topology T if and only if for every g ∈ G, the intersection U ∩ gH is open in gH. Verify that the pair (G, T ) is a commutative non-discrete locally compact topological group. Show that G does not contain proper everywhere dense subgroups. For this purpose, consider any everywhere dense subgroup G0 of G. Let g be an arbitrary element of G. Since the set gH is open, we must have / gH ∩ G0 = 0. So we may take x ∈ H such that gx ∈ G0 . Since G is 2-divisible, there exists y ∈ G such that y2 = x. Again, the set yH is open, so yH ∩ G0 = 0. / We may choose z ∈ H such that yz ∈ G0 . Observe now that the order of any element from H does not exceed 2. Therefore, denoting by e the neutral element of G, we can write x = y2 = y2 e = y2 z2 = (yz)2 ∈ (G0 )2 = G0 ,

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gx ∈ G0 , g ∈ G0 x−1 = G0 , which yields the required equality G0 = G. In connection with the previous exercise, see also [46], [92], [103], and [203]. 7∗ . Let (G, ·) be a commutative group of cardinality ω1 and let μ be a nonzero σ -finite diffused measure on G. Prove that there exists a subgroup of G nonmeasurable with respect to μ . For this purpose, use an appropriate modification of Ulam’s (ω × ω1 )-matrix and Kulikov’s theorem on the structure of commutative groups.

Appendix 6

Elements of descriptive set theory

As is widely known, general set theory is concerned with abstract sets and various relations between them. Descriptive set theory primarily deals with those subsets of the real line R or, more generally, of a Polish topological space E, which have rather good structure and can be described more or less effectively, e.g., without the aid of uncountable forms of the Axiom of Choice. Here we would like to recall the beginnings of this beautiful and important branch of mathematics. From the view-point of real analysis and general topology, the most interesting properties of a subset of R are its measurability in the Lebesgue sense and the so-called Baire property which may be regarded as a certain topological analogue of measurability (see [40], [148], [176], and [192]). Recall that a set X in a topological space E has the Baire property if it admits a representation in the form X = (U ∪Y ) \ Z where U is an open subset of E and Y and Z are some first category subsets of E. The family of all sets in E having the Baire property is denoted by Ba(E). In fact, Ba(E) coincides with the σ -algebra generated by the Borel σ -algebra B(E) and the family of all first category subsets of E. If E is of second category, then this family forms a σ -ideal of subsets of E. Let E  be a topological space and let g : E → E  be a mapping. We say that g has the Baire property if for each open set V ⊂ E  , the pre-image g−1 (V ) has the Baire property in E. Observe that if there exists a first category set X ⊂ E such that g|(E \ X) is continuous, then g necessarily has the Baire property. The converse assertion is also true under some assumption on E  . Namely, suppose that the topology of E  is countably generated and a mapping g : E → E  has the Baire property. Let {Vn : n < ω } denote a countable base of 423

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E  . For each n < ω , we may write g−1 (Vn ) = (Un ∪Yn ) \ Zn where Un is an open subset of E and Yn and Zn are some first category sets in E. Let us put X = ∪{Yn ∪ Zn : n < ω }. Obviously, X is a first category subset of E and it can readily be verified that the restriction g|(E \ X) is continuous. For extensive information about the Baire property and the Baire property in the restricted sense, see [40], [148], [176], and [192]. Recall that the Borel σ -algebra B(E) is generated by the family of all open (equivalently, by the family of all closed) subsets of E. The following simple auxiliary proposition turns out to be useful in many questions of descriptive set theory. Lemma 1. Let E be a topological space such that any open set in E is of type Fσ (or, equivalently, any closed set in E is of type Gδ ). Then B(E) coincides with the monotone class generated by the family of all open sets in E (equivalently, by the family of all closed sets in E). An easy proof is left to the reader (cf. Exercise 5 for Appendix 3). Let E be a topological space and let f : E → R be a function. We say (see Chapter 20) that f is of Baire zero class if f is continuous at all points of E, i.e., f is continuous on E. The family of all continuous functions acting from E into R is usually denoted by the symbol C(E, R) (or by C(E)). In accordance with the definition above, we use the notation Ba0 (E, R) for the same family of functions. Thus, we have Ba0 (E, R) = C(E, R). Suppose now that for an ordinal ξ < ω1 , the Baire classes Baζ (E, R) (ζ < ξ ) have already been determined. We say that a function f : E → R belongs to the class Baξ (E, R) if there exists a sequence of functions { fn : n < ω } ⊂ ∪{Baζ (E, R) : ζ < ξ } which pointwise converges to f , i.e., lim fn (x) = f (x)

n→+∞

(x ∈ E).

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By proceeding in this way, it becomes possible to define the classes Baξ (E, R) for all ordinals ξ < ω1 . Clearly, these classes increase by the inclusion relation. Further, putting Ba(E, R) = ∪{Baξ (E, R) : ξ < ω1 }, we obtain the class of all Baire functions acting from E into R (see [5], [99], [148], and [162]). In view of the regularity of ω1 , this class is closed with respect to the pointwise convergence of sequences of functions. We say that a function f ∈ Ba(E, R) is of Baire order ξ < ω1 if f ∈ Baξ (E, R) \ ∪{Baζ (E, R) : ζ < ξ }. In Chapter 20 we were concerned with some important properties of the first Baire class Ba1 (E, R). Recall that if E = R, then all monotone functions and all derivatives belong to Ba1 (R, R). The following three simple assertions are valid for Baξ (E, R): (1) Baξ (E, R) is a linear algebra over the field R; in other words, if we have f ∈ Baξ (E, R), g ∈ Baξ (E, R), a ∈ R and b ∈ R, then a f + bg ∈ Baξ (E, R), f · g ∈ Baξ (E, R); (2) if f ∈ Baξ (E, R), g ∈ Baξ (E, R) and g(x) = 0 for all x ∈ E, then f /g ∈ Baξ (E, R); (3) if f ∈ Baξ (E, R) and φ ∈ Baη (R, R), then φ ◦ f ∈ Baξ +η (E, R). In fact, the above-mentioned relations (1), (2), and (3) can readily be checked by the method of transfinite induction. Let us also formulate a less trivial property of the class Baξ (E, R). Namely, if { fn : n < ω } is a sequence of functions from this class, which uniformly converges to a function f : E → R, then f also belongs to Baξ (E, R) (cf. Theorem 2 from Chapter 20). Recall that the symbol B(E, R) denotes the family of all Borel functions acting from E into R. The following statement shows a close connection between Baire and Borel functions. Theorem 1. If E is a perfectly normal topological space, then the equality Ba(E, R) = B(E, R) holds true. In particular, this equality is satisfied for any metric space E.

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Keeping in mind Lemma 1, the Tietze-Urysohn theorem on extensions of real-valued continuous functions and the relation Ba(E, R) = ∪{Baξ (E, R) : ξ < ω1 }, we can prove Theorem 1 by using the method of transfinite induction on ξ < ω1 . We leave the corresponding technical details to the reader (see Exercise 3). Recall that ω ω as usual denotes the canonical Baire space of countable topological weight. We also recall that the symbol ω <ω denotes the family of all finite sequences of natural numbers. In the sequel, we do not assume that this family is equipped with a topology. For any s ∈ ω <ω , we put lh(s) = card(dom(s)). Actually, lh(s) coincides with the length of s. Further, if s = {s0 , s1 , ..., sk } ∈ ω <ω and n < ω , then the symbol s ∗ n denotes the extended sequence {s0 , s1 , ..., sk , sk+1 } ∈ ω <ω where sk+1 = n. In our further considerations, it will be convenient to identify each natural number k with the set {0, 1, ..., k − 1}. In fact, this assumption is redundant since in modern set theory the natural numbers and, more generally, the ordinal numbers are usually introduced in the von Neumann sense (see [10], [91], [145], and [150]). In Chapter 8, the notion of A-operation was briefly considered over appropriately indexed countable families of subsets of a base set E. We would like to recall that if a countable family of sets {Xs : s ∈ ω <ω } ⊂ P(E) is given, then the set A({Xs : s ∈ ω <ω }) =

 φ ∈ω ω

(∩{Xφ |k : k < ω })

is the result of A-operation applied to this family (see [99], [148], [150], [160], and [162]). In many cases we may assume, without loss of generality, that an initial family {Xs : s ∈

ω <ω } is regular, i.e., (∀n < ω )(∀s ∈ ω <ω )(Xs∗n ⊂ Xs ). Indeed, if necessary, we may replace this family by {Ys : s ∈ ω <ω } where Yn0 n1 ...nk = Xn0 ∩ Xn0 n1 ∩ ... ∩ Xn0n1 ...nk

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for any finite sequence s = (n0 , n1 , ..., nk ) of natural numbers. Clearly, we have A({Xs : s ∈ ω <ω }) = A({Ys : s ∈ ω <ω }). Also, suppose for a while that an initial regular family {Xs : s ∈ ω <ω } satisfies the following condition: / (∀s ∈ ω <ω )(∀t ∈ ω <ω )((s = t & lh(s) = lh(t)) ⇒ Xs ∩ Xt = 0). Then we come to the equality A({Xs : s ∈ ω <ω }) =



(



Xs ),

k<ω s∈ω <ω , lh(s)=k

which is easily verified. Consequently, in this case, the A-operation is reduced to the operations of countable unions and countable intersections. If a class L of subsets of E is given, then the symbol A (L ) stands for the class of all those sets which can be represented in the form A({Xs : s ∈ ω <ω }), where {Xs : s ∈ ω <ω } ⊂ L . The members from A (L ) are usually called analytic sets over the original class L . It readily follows from the definition of the class A (L ) that if L is closed under finite unions and finite intersections, then the same property holds true for A (L ). Moreover, under the assumption of the closedness of L with respect to finite unions and finite intersections, we can assert that A (L ) is closed under countable unions and countable intersections. Indeed, consider an arbitrary countable family of sets {Wi : i < ω } ⊂ A (L ). Then for each i < ω , we have Wi = A({Xsi : s ∈ ω <ω }) where {Xsi : s ∈ ω <ω } ⊂ L is a regular family of sets. Let us define Y0/ = E, Yin0 n1 ...nk = Xni 0 n1 ...nk for any i < ω and for any (n0 , n1 , ..., nk ) ∈ ω k+1 . Applying the A-operation to the obtained family {Ys : s ∈ ω <ω } ⊂ L , we easily infer that ∪{Wi : i < ω } ∈ A (L ). Further, let us define Zn0 n1 ...nk =

 ik

(∪{Xmi 0 m1 ...mk : m0 + m1 + ... + mk  n0 + n1 + ... + nk})

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for any (n0 , n1 , ..., nk ) ∈ ω k+1 . Applying the A-operation to the obtained family {Zs : s ∈

ω <ω } ⊂ L and taking into account the K¨onig lemma on ω -trees with finite levels (see Appendix 1), we derive that ∩{Wi : i < ω } ∈ A (L ). From the point of view of numerous applications, the most important case is when the role of E is played by an uncountable Polish topological space and the A-operation is applied to appropriately indexed countable families of closed subsets of E, i.e., the role of L is played by the class of all closed subsets of E. In this way we come to the Suslin subsets of E. The next statement shows that they admit a nice topological description (see [99], [148], [160], and [162]). Theorem 2. Let E be a Polish space and let X be a nonempty Suslin subset of E. Then there exists a continuous surjection f : ω ω → X. Conversely, any continuous image (in E) of the space ω ω is a nonempty Suslin subset of E. Proof. For every s ∈ ω <ω , denote U(s) = {z ∈ ω ω : (∀i)(i < lh(s) ⇒ z(i) = s(i))}. It is obvious that the family of sets {U(s) : s ∈ ω <ω } forms a base of ω ω consisting of closed-open sets. Since the given Polish space E is nonempty, it is not difficult to construct a regular system {Fs : s ∈ ω <ω } of nonempty closed subsets of E such that: (1) limlh(s)→+∞ diam(Fs ) = 0; (2) E = A({Fs : s ∈ ω <ω }). Actually, relation (2) shows that E is a continuous image of the Baire space ω ω . Indeed, for any z ∈ ω ω , we may put {φ (z)} = Fz0 ∩ Fz0 z1 ∩ ... ∩ Fz0z1 ...zk ∩ ... . It is easy to verify that this formula determines the continuous surjection

φ : ω ω → E. Since the given nonempty set X ⊂ E is Suslin, we can write X = A({Xs : s ∈ ω <ω })

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where {Xs : s ∈ ω <ω } is also a regular system of closed subsets of E. Taking into account the above relations (1) and (2), we may assume without loss of generality that lim

lh(s)→+∞

diam(Xs ) = 0.

Now, let us put / Z = {z ∈ ω ω : ∩{Xz|k : k < ω } = 0}. It can readily be checked that Z is a nonempty closed subset of ω ω . Consequently, Z is a nonempty Polish space as well. Let us define a mapping g:Z→E in the following way: for each z ∈ Z, the value g(z) is equal to the unique point from ∩{Xz|k : k < ω }. An easy verification shows that g is continuous and g(Z) = X. Keeping in mind the same relations (1) and (2) and applying their analogues to Z, we infer that there exists a continuous surjection

ψ : ω ω → Z. In this manner we come to the required continuous surjection f = g ◦ ψ of ω ω onto X. Conversely, let Y ⊂ E be such that there exists a continuous surjection h : ω ω → Y. Then for each z ∈ ω ω , we may write {h(z)} ⊂ ∩{h(U(z|k)) : k < ω } ⊂ ∩{cl(h(U(z|k))) : k < ω }. By virtue of the continuity of h, we get lim diam(cl(h(U(z|k)))) = 0

k→+∞

from which it immediately follows that {h(z)} = ∩{cl(h(U(z|k))) : k < ω }. Finally, denoting Ps = cl(h(U(s)))

(s ∈ ω <ω ),

we conclude that for the given set Y , the equality Y = A({Ps : s ∈ ω <ω }) holds true, which completes the proof of Theorem 2.

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By using Theorem 2, it can readily be shown that a metrizable continuous image of a Suslin space is also Suslin and the topological product of a countable family of Suslin spaces is Suslin, too. If E is an arbitrary Polish or, more generally, perfectly normal topological space, then we obviously have the inclusion B(E) ⊂ A (E). Therefore, Theorem 2 directly yields the fact that any nonempty Borel subset of a Polish space E can be considered as a continuous image of the Baire space ω ω . In this context, it should be noticed that for Borel subsets of Polish spaces, a more precise result can be established. Namely, we have the next important statement essentially due to Luzin (see [99], [148], [160], and [162]). Theorem 3. Let E be a Polish space. Every Borel subset of E may be regarded as a bijective continuous image of some Polish space. Proof. Denote by L the class of all those subsets of E which are bijective continuous images of Polish spaces. Since any open subset U of E is a Polish space (see Exercise 24 for Chapter 8), we obtain at once that U belongs to L . Let {Xi : i ∈ I} be an arbitrary disjoint countable family of subsets of E belonging to the class L . For each index i ∈ I, denote by Pi a Polish space such that there exists a continuous bijection fi : Pi → Xi . Let P be the topological sum of the family of spaces {Pi : i ∈ I}. Clearly, P is a Polish space, too. Without loss of generality, we may treat each Pi (i ∈ I) as a subset of P. Let f : P → ∪{Xi : i ∈ I} denote the common extension of all mappings fi (i ∈ I). It can easily be seen that f is a continuous bijection, thus it immediately follows that ∪{Xi : i ∈ I} ∈ L . Let now {Yi : i ∈ I} be an arbitrary countable family of subsets of E belonging to the class L . Again, for each index i ∈ I, denote by Qi a Polish space such that there exists a continuous bijection hi : Qi → Yi . Both topological products ∏{Qi : i ∈ I} and E I are Polish spaces. Define a continuous mapping h : ∏{Qi : i ∈ I} → E I by putting h(q) = (hi (qi ))i∈I

(q = (qi )i∈I ∈ ∏{Qi : i ∈ I}).

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Obviously, h is injective and continuous. Further, denote Δ = {z ∈ E I : (∀i ∈ I)(∀ j ∈ I)(pri (z) = pr j (z))}, Q = h−1 (Δ). Since Q is a closed subset of ∏{Qi : i ∈ I}, we see that Q is a Polish space as well. Additionally, the restricted mapping h|Q : Q → (∏{Yi : i ∈ I}) ∩ Δ is a continuous bijection. Taking into account the simple fact that the set ∏{Yi : i ∈ I} ∩ Δ is homeomorphic to the set ∩{Yi : i ∈ I}, we infer that ∩{Yi : i ∈ I} belongs to the class L . All the said above enables us to conclude that L contains the Borel σ -algebra B(E) (cf. Exercise 1) which completes the proof. In connection with Theorem 3, see also Exercise 9 of this Appendix. The following statement is a direct consequence of Theorem 3: any uncountable Borel subset of a Polish space has cardinality continuum. It is remarkable that the same statement is valid for all uncountable Suslin sets. In order to show this fact, we need one auxiliary proposition. Lemma 2. Let E be an arbitrary separable metric space and let g be a continuous mapping acting from E onto some uncountable metric space. Then there exist two open balls U ⊂ E and V ⊂ E such that: (1) both sets g(cl(U)) and g(cl(V )) are uncountable; (2) g(cl(U)) ∩ g(cl(V )) = 0. / The proof of this lemma is quite simple and is left to the reader. Now, we are able to formulate and prove the famous Alexandrov-Hausdorff theorem concerning the perfect subset property of Suslin sets in a Polish space (see [99], [148], [160], and [162]). Theorem 4. Let E be a Polish topological space, Y be an uncountable Suslin subset of E and let g : E → Y be a continuous surjection. Then there exists a set X ⊂ E homeomorphic to the Cantor space {0, 1}ω such that the restriction g|X : X → g(X) is a bijection (hence, in view of the compactness of X, this restriction is also a homeomorphism between X and g(X)).

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Proof. By using Lemma 2 and ordinary recursion, it is not difficult to construct a dyadic system {Fs : s ∈ {0, 1}<ω } of closed balls in E such that for any s ∈ {0, 1}<ω , the following relations are satisfied: (a) Fs∗0 ∪ Fs∗1 ⊂ Fs ; (b) g(Fs∗0) ∩ g(Fs∗1) = 0; / (c) g(Fs ) is an uncountable set; (d) diam(Fs ) < 1/(lh(s) + 1). Having this system of balls, we may put X(k) = ∪{Fs : s ∈ {0, 1}<ω , lh(s) = k}

(k < ω ),

X = ∩{X(k) : k < ω }. A direct verification shows that the set X is homeomorphic to the Cantor space by virtue of the relations (a), (b), and (d). In addition to this circumstance, relation (b) implies that g|X is an injection. Keeping in mind the compactness of X, we conclude that g|X is a homeomorphism between the sets X and g(X). Remark 1. As an immediate consequence of Theorem 4, we obtain that the cardinality of any uncountable Suslin (hence Borel) subset of a Polish space is equal to c. We thus see that Suslin sets in some sense realize the Continuum Hypothesis: they are either countable or they are of cardinality continuum. Unfortunately, the same statement fails to be true for co-Suslin sets (in this connection, see [10], [91], [99], [150], [162], and [188]; cf. also Exercise 14). Let E be an arbitrary topological set and let X and Y be two subsets of E. According to Luzin’s classical definition (see [160], [162]), we say that X and Y can be separated by Borel sets if there exist two Borel sets X  ⊂ E and Y  ⊂ E such that / X ⊂ X  , Y ⊂ Y  , X  ∩Y  = 0. The following auxiliary statement is valid. Lemma 3. Let E be a topological space, let X and Y be two subsets of E and suppose that X = ∪{Xn : n < ω }, Y = ∪{Yn : n < ω }. If the given sets X and Y cannot be separated by Borel sets, then there exists a pair (m, n) of natural numbers such that the sets Xm and Yn also cannot be separated by Borel sets.

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Proof. The argument is quite easy. Indeed, suppose to the contrary that for any pair (m, n) of natural numbers, the sets Xm and Yn can be separated by some Borel sets. This assumption means that there exists a Borel set Pmn ⊂ E satisfying the relation Xm ⊂ Pmn ⊂ E \ Yn. Now, putting Bm = ∩{Pmn : n < ω }

(m < ω ),

we see that X ⊂ ∪{Bm : m < ω }, Y ∩ (∪{Bm : m < ω }) = 0, / which contradicts the condition that X and Y cannot be separated by Borel sets. The obtained contradiction completes the proof. The next classical result is due to Luzin and is usually called the separation principle for analytic (Suslin) sets (see [160] and [162]). Theorem 5. Let E be a Polish topological space and let X and Y be two disjoint Suslin subsets of E. Then X and Y can be separated by Borel sets. Proof. Suppose to the contrary that X and Y cannot be separated by Borel sets. Let f : ω ω → X, g : ω ω → Y be two continuous surjections (see Theorem 2). Using the notation from the proof of Theorem 2, we may write X = ∪{ f (U(n)) : n < ω }, Y = ∪{g(U(m)) : m < ω }. By virtue of Lemma 3, there exist natural numbers n0 and m0 such that the sets f (U(n0 )) and g(U(m0 )) cannot be separated by Borel sets. Proceeding by ordinary recursion, we will be able to construct two infinite sequences a = (n0 , n1 , ..., nk , ...) ∈ ω ω , b = (m0 , m1 , ..., mk , ...) ∈ ω ω such that for every k < ω , the corresponding sets f (U(a|k)) and g(U(b|k)) cannot be separated by Borel sets. On the other hand, keeping in mind the continuity of both mappings f and g, we readily infer that for a sufficiently large natural index k, the sets f (U(a|k)) and g(U(b|k)) are contained in some disjoint open neighborhoods of the two distinct points f (a) ∈ X and g(b) ∈ Y . We thus come to a contradiction, which ends the proof of Theorem 5.

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Remark 2. Let E be an arbitrary uncountable Polish space. As shown by Luzin and Novikov (see, e.g., [160], [162]), there are two disjoint co-Suslin sets X and Y in E, which cannot be separated by Borel sets. This important fact immediately yields that there exists a Suslin subset of E, which is not Borel (see also Exercise 13). The separation principle for analytic sets implies many useful consequences. For instance, let E be a Polish space and let X ⊂ E be an analytic set such that its complement E \ X is also analytic. Then we can assert that X is a Borel subset of E. This classical result is due to Suslin and follows directly from Luzin’s separation principle for analytic sets. Let us give another example of an application of this principle. Suppose that E and F are any two Polish spaces and h is a mapping acting from E into F such that the graph Gr(h) of h is a Suslin subset of the product space E × F. Then we can assert that h is a Borel mapping and, therefore, Gr(h) is a Borel subset of E × F. To see this circumstance, take any Borel set B ⊂ F. Clearly, we may write h−1 (B) = pr1 (Gr(h) ∩ (E × B)), h−1 (F \ B) = pr1 (Gr(h) ∩ (E × (F \ B))), which implies that h−1 (B) and h−1 (F \ B) are Suslin sets in E. In addition, we have h−1 (B) ∩ h−1(F \ B) = 0, / h−1 (B) ∪ h−1 (F \ B) = E. By virtue of the separation principle, both sets h−1 (B) and h−1 (F \ B) must be Borel, which shows that h is a Borel mapping. The next auxiliary proposition may be regarded as a natural generalization of Theorem 5 to the case of a countable disjoint family of Suslin sets. Lemma 4. Let E be a Polish space and let {Xn : n < ω } be a countable disjoint family of Suslin sets in E. Then there exists a countable disjoint family {Yn : n < ω } of Borel sets in E such that Xn ⊂ Yn for all n < ω . Proof. According to Theorem 5, for any pair (n, m) of distinct natural numbers, there exists a Borel set Bnm such that Xn ⊂ Bnm ⊂ E \ Xm . Now, we put Y0 = ∩{B0m : 1  m < ω }

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and for n ∈ ω \ {0}, define by recursion Yn = (∩{Bnm : m < ω , m = n}) \ (Y0 ∪Y1 ∪ ... ∪Yn−1 ). It is easy to check that {Yn : n < ω } is the required family of Borel sets. We also have the following important statement (see [99], [148], and [160]). Theorem 6. Let E1 and E2 be two Polish spaces, X be a Borel subset of E1 , and let a mapping g : X → E2 be injective and continuous. Then the image g(X) is a Borel subset of E2 . Proof. In view of Exercises 8 and 9, it suffices to show that for every Polish space E and for any injective continuous mapping g : ω ω → E, the set ran(g) is Borel in E. In order to establish this fact, fix a natural number k and consider the corresponding disjoint countable family of Suslin sets (s ∈ ω k ).

g(U(s))

It follows from the separation principle formulated in Lemma 4 that there are pairwise disjoint Borel sets Ys (s ∈ ω k ) in E for which the relations (s ∈ ω k )

g(U(s)) ⊂ Ys

are satisfied. Now, we define by recursion the following Borel sets: Yn∗ = Yn ∩ cl(g(U(n)))

(n < ω ), (s ∈ ω <ω , n < ω ).

∗ = Ys∗n ∩ cl(g(U(s ∗ n))) ∩Ys∗ Ys∗n

Using the method of induction on lh(s), it is not difficult to check the inclusions Ys∗ ⊂ Ys , g(U(s)) ⊂ Ys∗ ⊂ cl(g(U(s)))

(s ∈ ω <ω ).

Consequently, for any z ∈ ω <ω , we get the equality ∗ : k < ω }, {g(z)} = ∩{Yz|k

which directly implies that g(ω ω ) =



(



z∈ω ω k<ω

∗ Yz|k ).

Keeping in mind the circumstance that for each k < ω , the family {Ys∗ : s ∈ ω k } of Borel sets in E is disjoint, we come to the relation g(ω ω ) =

 k<ω

(



s∈ω <ω ,

lh(s)=k

Ys∗ )

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from which it follows that g(ω ω ) is a Borel subset of E. Theorem 6 has thus been proved. Theorems 3 and 6 yield that the Borel subsets of Polish topological spaces coincide with the bijective continuous images of Gδ -subsets of the canonical Baire space ω ω (cf. Exercise 9). In addition to this result, it directly follows from Theorem 6 that if X and Y are two Borel subsets of Polish spaces and f :X →Y is a continuous bijection, then f −1 : Y → X is a Borel bijection and, consequently, f turns out to be a Borel isomorphism between X and Y . Remark 3. The assertion of Theorem 6 remains valid in the case of an injective Borel mapping g : X → E2 (see Exercise 10). In this context, it should be also mentioned that a much stronger result holds true. Namely, let E and E  be two Polish spaces and let f : E → E  be a partial Borel mapping defined on a Borel subset of E and such that card( f −1 (y))  ω for every point y ∈ E  . Then for any Borel set B ⊂ E, the image f (B) is a Borel subset of E  . Moreover, the set dom( f ) admits a representation dom( f ) = ∪{Xn : n < ω }, where all Xn (n < ω ) are pairwise disjoint Borel subsets of E and for each index n < ω , the restriction f |Xn is a Borel isomorphism between Xn and f (Xn ) (see [99], [148], [160], and [162]). So far, we were concerned with various properties of Borel and analytic subsets of a Polish space. As said earlier, these classes of sets are important from the point of view of numerous applications. However, descriptive set theory also deals with more general classes of sets, which are of paramount interest from the point of view of logical foundations of contemporary mathematics. For example, one such class is formed by the so-called projective subsets of the real line, which were introduced and thoroughly studied by Luzin and Sierpi´nski (see [10], [91], [99], [148], [150], [160], [162], and [188]). Here we do not intend to discuss deep properties of projective sets. Notice that some of them (e.g., the measurability in the Lebesgue sense) turn out to be independent of ZFC theory. However, they become provable under the assumption of the existence of large cardinals of a certain type. Below, we only present the classical definition of projective sets (cf. [148] and [150]).

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Let E be an arbitrary Polish space. We define the classes of sets Pr0 (E), Pr1 (E), . . . , Prn (E), . . . by ordinary recursion. First, let us put Pr0 (E) = B(E). Suppose now that for a natural number n  1, the classes Prk (E), where k < n, have already been determined. If n is odd, then we define Prn (E) as the class of all continuous images (in E) of the sets from Prn−1 (E). If n is even, then we define Prn (E) as the class of all complements of the sets from Prn−1 (E). Finally, we put Pr(E) = ∪{Prn (E) : n < ω }. By definition, any member of Pr(E) is called a projective subset of E. We thus see that Pr0 (E) = B(E), Pr1 (E) = A (E), i.e., Borel and analytic sets are only the first two steps in the classical construction of the Luzin-Sierpi´nski projective hierarchy. EXERCISES 1. Give a detailed proof of Lemma 1. Actually, deduce this lemma from a more general fact stating that if R is an algebra of subsets of a base set E, then the σ -algebra generated by R coincides with the monotone class generated by R. Further, let L be a family of subsets of E such that: (a) R ⊂ L ; (b) L is closed under the unions of all countable disjoint families of its members; (c) L is closed under the intersections of all countable families of its members. Verify that L contains the σ -algebra generated by R. 2. Let ξ < ω1 . Show that the Baire class Baξ (E, R) is closed with respect to the uniform convergence of sequences of its elements (cf. Theorem 2 from Chapter 20). 3∗ . Let E be an arbitrary topological space. By using the method of transfinite induction, prove that: (a) Ba(E, R) ⊂ B(E, R);

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(b) if E is perfectly normal, then Ba(E, R) = B(E, R). To show the validity of (b), argue in the following manner. First, reduce the argument to the case of the characteristic function of any Borel subset X of E, i.e., prove that the function

χX belongs to Ba(E, R). For this purpose, use the Tietze-Urysohn theorem, Lemma 1 and the method of transfinite induction on the order of X (cf. Exercise 5 from Appendix 3). 4. Let E and E  be two Polish spaces and let f : E → E  be a Borel mapping. Show that for any Suslin set X ⊂ E, its image f (X) is a Suslin subset of E  . Obviously, the result just formulated generalizes Theorem 2 of this Appendix. 5. Give a detailed proof of Lemma 2. 6. Let E be a nonempty complete metric space without isolated points. Show that there exists a subset X of E which is homeomorphic to the Cantor space {0, 1}ω . For this purpose, construct a dyadic system of closed balls in E similar to the system of balls indicated in the proof of Theorem 4. 7. Let E be a zero-dimensional Polish topological space and let Z be a subset of E such that: (a) Z is of type Gδ ; (b) Z is everywhere dense in E; (c) E \ Z is everywhere dense in E. Prove that Z is homeomorphic to the canonical Baire space ω ω . For this purpose, show that Z can be expressed as the result of A-operation applied to some regular system {Fs : s ∈ ω <ω } of nonempty closed-open sets in E satisfying the relations: (i) limlh(s)→+∞ diam(Fs ) = 0; (ii) Fs ∩ Ft = 0/ for any two distinct s ∈ ω <ω and t ∈ ω <ω such that lh(s) = lh(t). 8. Prove that an arbitrary uncountable zero-dimensional Polish space E can be represented in the form E = Z ∪D

(Z ∩ D = 0), /

where Z is homeomorphic to ω ω and D is at most countable. In order to show this fact, denote by P the set of all condensation points of E and observe that P is an uncountable perfect subset of E whose complement is countable. Let D0 be a countable everywhere dense subset of P. Put Z = P \ D0 , D = (E \ P) ∪ D0

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and, keeping in mind the result of Exercise 7, check that these Z and D yield the required decomposition of E. 9. Demonstrate that any Borel subset of a Polish space can be regarded as a bijective continuous image of some Polish subspace of the Baire space ω ω . For this purpose, take into account the existence of a canonical continuous bijection acting from the space ω ω onto the interval ]0, 1], which implies the existence of a continuous bijection of ω ω = ω ω ×ω onto the product space ]0, 1]ω . Further, use the fact that ]0, 1]ω contains topological copies of all Polish spaces and apply Theorem 3. 10. Let E1 and E2 be two Polish spaces, X be a Borel subset of E1 and let g : X → E2 be an injective Borel mapping. Prove that g(X) is a Borel subset of E2 . Argue in the following manner. In the product set X × E2 consider the graph of g, i.e., consider the set G = {(x, y) ∈ X × E2 : g(x) = y}. First, show that G is a Borel subset of X × E2 . For this purpose, denote by d any metric on E2 compatible with the topology of E2 and introduce a function

φ : X × E2 → [0, +∞[ defined by the formula

φ (x, y) = d(g(x), y)

(x ∈ X, y ∈ E2 ).

Check that φ is a Borel mapping and G = φ −1 (0), which yields that G is Borel. Finally, consider the injective continuous mapping pr2 |G : G → E2 and apply to it Theorem 6 of this Appendix. 11∗ . Let X and X  be any two uncountable Borel subsets of Polish spaces E and E  respectively. Demonstrate that there exists a Borel isomorphism f : X → X . For this purpose, first show that the Cantor space {0, 1}ω is Borel isomorphic to the Hilbert cube [0, 1]ω . Then, applying the topological universality of [0, 1]ω for the class of all separable metrizable spaces, the Alexandrov-Hausdorff theorem from this Appendix and Banach’s theorem on two injections from Exercise 12 of Appendix 1, obtain the required result.

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12∗ . In this exercise, we will be dealing with the extended real line [−∞, +∞] = R ∪ {−∞, +∞} which is equipped with the standard ordering and topology. So [−∞, +∞] becomes isomorphic to the closed interval [−1, 1]. Let E be an uncountable Polish space. Prove the classical Lebesgue theorem stating that for any ordinal ξ < ω1 , there exists a mapping Φξ : E × [0, 1] → [−∞, +∞] satisfying the following conditions: (a) Φξ is Borel; (b) for each function f ∈ Baξ (E, R), there is a point t ∈ [0, 1] such that Φξ (·,t) = f . In order to demonstrate this important statement, first observe that in view of Exercise 11, it suffices to reduce the argument to the case E = [0, 1]. Consider any Peano type mapping

φ = (φk )k<ω : [0, 1] → [0, 1]ω . This phrase simply means that φ is a continuous surjection. Let {Qk : k < ω } denote the sequence of all polynomials on [0, 1] with rational coefficients. Show that we can take Φ0 (x,t) = lim sup Qk (x)φk (t) (x ∈ [0, 1],t ∈ [0, 1]). k→+∞

If for ξ < ω1 , the mapping Φξ is already defined, then we can put Φξ +1 (x,t) = lim sup Φξ (x, φk (t)) k→+∞

(x ∈ [0, 1],t ∈ [0, 1]).

Finally, if ξ < ω1 is a limit ordinal, then take a strictly increasing sequence {ξk : k < ω } of ordinals such that limk→+∞ ξk = ξ and put Φξ (x,t) = lim sup Φξk (x, φk (t)) (x ∈ [0, 1],t ∈ [0, 1]). k→+∞

This procedure enables us to define the required mapping Φξ for any ordinal number ξ <

ω1 . By starting with the existence of Φξ for each ξ < ω1 , derive that Baξ ([0, 1], R) \ ∪{Baζ ([0, 1], R) : ζ < ξ } = 0/ and that the same holds true for an arbitrary uncountable Polish space E instead of [0, 1]. Argue in the following manner. Suppose to the contrary that Baξ ([0, 1], R) \ ∪{Baζ ([0, 1], R) : ζ < ξ } = 0/

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and deduce from this assumption that Ba([0, 1], R) = Baξ ([0, 1], R). Then, by using the mapping Φξ , construct a Borel mapping Φ : [0, 1] × [0, 1] → [0, 1] such that for any Borel function f : [0, 1] → [0, 1], there exists a point t ∈ [0, 1] for which Φ(·,t) = f . Further, define a Borel mapping Ψ by the formula Ψ(x,t) = lim

n→+∞

nΦ(x,t) 1 + nΦ(x,t)

(x ∈ [0, 1],t ∈ [0, 1])

and check that: (a) ran(Ψ) = {0, 1}; (b) for any Borel function g : [0, 1] → {0, 1}, there exists a point t ∈ [0, 1] such that Ψ(·,t) = g. Finally, put h(x) = 1 − Ψ(x, x) (x ∈ [0, 1]) and verify that: (c) the function h is Borel; (d) ran(h) ⊂ {0, 1}. Now, keeping in mind relation (b), take a point t ∈ [0, 1] for which h = Ψ(·,t) and infer the equalities h(t) = 1 − Ψ(t,t) = Ψ(t,t), Ψ(t,t) = 1/2, contradicting relation (a). Thus, the obtained contradiction yields the existence of Baire functions of any order ξ < ω1 and, consequently, the existence of Borel sets in [0, 1] of the same order ξ . 13∗ . Let E be an arbitrary uncountable Polish space. Prove the classical Suslin theorem (see [99], [148], [160], and [162]) stating that there exists a Suslin subset of E which is not Borel (consequently, there exists a co-analytic subset of E which is not analytic). Argue in the following manner. By virtue of Exercise 11, it suffices to establish this fact for some concrete Polish space E, e.g., in the case where E = {0, 1}ω .

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Consider the product space K = {0, 1}ω × [0, 1] and denote by Comp(K) the family of all compact subsets of K. Endowed with the standard Hausdorff metric, this family becomes a compact metric space (see [58], [101]). Consequently, there exists a Peano type mapping g : {0, 1}ω → Comp(K), i.e., g is a continuous surjection. Let I denote the set of all irrational numbers from [0, 1]. As widely known, this set is homeomorphic to the Baire space ω ω . Define a set-valued mapping G : {0, 1}ω → P({0, 1}ω ) by the formula G(t) = pr1 (({0, 1}ω × I) ∩ g(t))

(t ∈ {0, 1}ω ).

Verify the validity of the following relation: ran(G) = A ({0, 1}ω ). Further, in the Cantor space {0, 1}ω consider the set X = {t ∈ {0, 1}ω : t ∈ G(t)}. Suppose for a while that X is a Suslin set in {0, 1}ω . Then for some t0 ∈ {0, 1}ω , we must have G(t0 ) = X. It easily follows from this equality that t0 ∈ X ⇔ t0 ∈ X, which yields a contradiction. Therefore, X cannot be a Suslin subset of {0, 1}ω . On the other hand, show that the set {0, 1}ω \ X = {t ∈ {0, 1}ω : t ∈ G(t)} is Suslin. For this purpose, check the validity of the relation {0, 1}ω \ X = pr1 (D), where the Borel set D ⊂ K is defined as follows: D = {(t, r) ∈ K : (t, r) ∈ g(t) & r ∈ I}. Infer from the said above the required result. 14∗ . Let E be a base set, L be a class of subsets of E and let {Xs : s ∈ ω <ω } be a countable system of sets from L . By using the method of transfinite recursion on ξ < ω1 , define the following sets:

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(a) Xs0 = Xs for all s ∈ ω <ω ; ξ +1

(b) Xs

ξ

ξ

ξ

= Xs ∩ (∪{Xs∗n : n < ω }) for any ξ < ω1 and for all s ∈ ω <ω ; ζ

(c) Xs = ∩{Xs : ζ < ξ } for a limit ordinal ξ < ω and for each s ∈ ω <ω . Check (by using transfinite induction) that if ζ  ξ < ω1 , then Xsξ ⊂ Xsζ

(s ∈ ω <ω ).

Further, for any ξ < ω1 , put: Yξ = ∪{Xnξ : n < ω }; Tξ = ∪{(Xsξ \ Xsξ +1) : s ∈ ω <ω }; Zξ = Yξ \ Tξ . Verify that all the sets Xsξ , Yξ , Tξ , Zξ

(ξ < ω1 , s ∈ ω <ω )

belong to the σ -ring generated by the given class L . Prove the following important equalities due to Sierpi´nski: A({Xs : s ∈ ω <ω }) = ∩{Yξ : ξ < ω1 } = ∪{Zξ : ξ < ω1 }. In order to establish these equalities, argue in the following manner. First, show the validity of the inclusion (i) ∪{Zξ : ξ < ω1 } ⊂ A({Xs : s ∈ ω < ω }). Take an arbitrary z ∈ ∪{Zξ : ξ < ω1 }. Then for some ordinal ξ < ω1 , we must have z ∈ Yξ and z ∈ Tξ . Consequently, for some natural number n0 , the relations z ∈ Xnξ0 , z ∈ Xnξ0 \ Xnξ0+1 ξ +1

are satisfied, thus it follows that z ∈ Xn0 . Keeping in mind the inclusion Xnξ0+1 ⊂ ∪{Xnξ0 n : n < ω }, ξ

obtain that z ∈ Xn0 n1 for some natural number n1 . Continuing this process by ordinary recursion, construct an infinite sequence

φ = (n0 , n1 , n2 , ...) ∈ ω ω such that ξ

z ∈ ∩{Xφ |k : k < ω } ⊂ ∩{Xφ |k : k < ω } and, therefore, z ∈ A({Xs : s ∈ ω <ω }). This result obviously yields (i).

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Now, check the validity of the inclusion (ii) A({Xs : s ∈ ω <ω }) ⊂ ∩{Yξ : ξ < ω1 }. For this purpose, show that if ξ < ω1 , r < ω and φ ∈ ω ω , then the relation ξ

∩{Xφ |k : k < ω } ⊂ Xφ |r holds true (use the method of transfinite induction on ξ < ω1 ). Consequently, ξ

∩{Xφ |k : k < ω } ⊂ Xφ (0) ⊂ ∪{Xnξ : n < ω } = Yξ , which immediately yields (ii). Finally, verify the validity of the relation (iii) ∩{Tξ : ξ < ω1 } = 0. / To show this fact, suppose otherwise, i.e., suppose that there exists an element x ∈ ∩{Tξ :

ξ < ω1 }. Then for any ordinal ξ < ω1 , we can find a sequence s = s(ξ ) ∈ ω <ω such that x ∈ Xsξ \ Xsξ +1. Since ω <ω is countable and ω1 is uncountable, there are two distinct ordinals ξ < ω1 and

ζ < ω1 for which s = s(ξ ) = s(ζ ). We may assume, without loss of generality, that ξ < ζ . ζ

ξ +1

But in this case the relation x ∈ Xs ⊂ Xs

must be true, thus a contradiction follows.

Taking into account the relations (i), (ii) and (iii) established above, derive the validity of the Sierpi´nski equalities. Infer from these equalities that any analytic set over the class L can be represented as the intersection of an ω1 -sequence of elements from the σ -ring generated by L , and as the union of an ω1 -sequence of elements from the same σ -ring. Conclude that if D is a co-Suslin subset of a Polish topological space, then card(D) ∈ {0, 1, ..., n, ..., ω , ω1 , c}. The same relation remains valid if D is any Borel image (contained in a Polish space) of a co-Suslin set. 15. We preserve the notation of Exercise 14. Suppose that a base set E is equipped with a topology satisfying the Suslin condition. Show that if all sets from {Xs : s ∈ ω <ω } have the Baire property, then the set X = A({Xs : s ∈

ω <ω }) has the Baire property, too. For this purpose, take s ∈ ω <ω and consider the sets Xsξ \ Xsξ +1

(ξ < ω1 ),

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which are pairwise disjoint. Assuming without loss of generality that E is a Baire space, ξ

ξ +1

derive that there exists an ordinal α (s) < ω1 such that all sets Xs \ Xs

are of first category

in E whenever α (s)  ξ < ω1 . Further, put

α = sup{α (s) : s ∈ ω <ω } and conclude that the set Tα is of first category in E. Taking into account the inclusions X \ Zα ⊂ Yα \ Zα ⊂ Tα , obtain the desired result. Suppose now that E is equipped with a σ -finite complete measure μ . Show that if all sets from a family {Xs : s ∈ ω <ω } are μ -measurable, then the set X = A({Xs : s ∈ ω <ω }) is

μ -measurable, too. For this purpose, apply a method analogous to the described above. In fact, the second part of Exercise 15 directly follows from its first part. In order to establish this implication, it suffices to consider a von Neumann topology associated with a given σ -finite complete measure μ . In addition, the invariance of the Baire property under the A-operation remains valid for an arbitrary topological space E (see [148]). 16∗ . Prove that there exists a nonempty perfect subset P of R which is linearly independent over the field Q of all rational numbers. By using the Zorn lemma, extend P to a Hamel basis of R and claim that there are Hamel bases in R which contain nonempty perfect subsets. On the other hand, show that no Hamel basis in R can be a Suslin subset of R (Sierpi´nski’s theorem). In connection with the result of Exercise 16, let us remark that von Neumann [186] was able to construct a nonempty perfect set of algebraically independent real numbers. 17∗ . Let E and E  be two Polish spaces and let f : E → E  be a Borel mapping such that (∀y ∈ E  )(card( f −1 (y))  ω ). If X ⊂ E is a universal measure zero set, then its image f (X) is universal measure zero, too (see Exercise 9 from Chapter 12). Taking into account the above-mentioned fact and using Remark 3 of this Appendix, prove that there exists an uncountable universal measure zero subset of R which is simultaneously a vector space over the field Q.

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For this purpose, start with the existence of a nonempty perfect set P ⊂ R linearly independent over Q and pick an uncountable universal measure zero set X ⊂ P. Denote by Y the linear hull of X (over the same Q) and check that Y is a universal measure zero subset of R. 18∗ . Show that there exists a Borel isomorphism f : [0, 1] → [0, 1] which does not preserve the Baire property of subsets of [0, 1]. On the other hand, let E1 and E2 be two uncountable Polish spaces without isolated points. Prove that there exists a Borel isomorphism g : E1 → E2 satisfying the following condition: a set X ⊂ E1 is of first category in E1 if and only if its image g(X) is of first category in E2 . Derive from this condition that both mappings g and g−1 preserve the Baire property. In order to establish the existence of the required Borel isomorphism g, use Exercise 8 of this Appendix. 19. Let X ⊂ R be an analytic non-Borel set (such a set exists in view of the result of Exercise 13). According to one of Sierpi´nski’s equalities, we have X = ∪{Zξ : ξ < ω1 } where all Zξ (ξ < ω1 ) are some Borel subsets of R (see Exercise 14). Verify that Zξ \ ∪{Zζ : ζ < ξ } = 0/ for uncountably many ordinals ξ < ω1 . By starting with this fact and taking into account the argument presented in Exercise 15, give one more construction of an uncountable universal measure zero set in R.

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Subject Index

Absolutely measurable function, 89

Aronszajn tree, 343

absolutely negligible set,

autonomous system of ordinary

31

absolutely nonmeasurable function, 11

differential equations, 54

absolutely nonmeasurable set,

Axiom of Choice, 4

80

A.D. Alexandrov’s theorem, 26

Baire classes of functions, 315

admissible family of sets for a measure,

Baire order, 7

266

Baire property, 3

admissible family of subgroups, 168

Banach-Kuratowski-Pettis theorem, 49,

admissible partition for a measure, 393

394

admissible vector subspace, 78

Banach-Mazur theorem, 373

Alexandrov-Hausdorff theorem, 431

Banach theorem on two injections, 352,

algebraic aspect of the measure extension

353

problem, 19 algebraic sum of sets,

Bernstein construction, 86, 91

177

Bernstein set,

almost continuous function, 111 almost disjoint family of sets, almost invariant set,

Blumberg’s theorem, 11, 141

355

Borel measure, 40

25

Borel partial function, 7

almost measurable function, 111

Borel selector, 129, 131

analytic manifold, 298 analytic set,

34, 88, 91, 205

branch of a tree, 340

132

branch property, 341

A-operation, 132

Brower’s theorem, 16

approximate derivative, 398 approximately continuous function, 398

Canonical Baire space, 377, 426

approximately differentiable function,

Cantor-Baire stationarity principle, 325

398

Cantor-Bendixson theorem, 326, 334 457

458

Topics in Measure Theory and Real Analysis

Cantor set on the real line, 195 capacity in the Choquet sense,

domain of a partial function, 2 14

dynamical system,

159

Caratheodory conditions, 12 Caratheodory’s theorem, 19

Erdos-Szekeres theorem, 302

cardinal measurable in the Ulam sense,

ergodic measure, 147

61

Family of nonempty sets,

Cauchy functional equation, 9

4

fixed-point property, 16

Choquet theorem, 359, 361 closed equivalence relation, 128

free commutative group, 412

closed mapping, 125

Generalized Luzin set,

co-analytic set,

generalized Sierpinski set,

96

commutative group, 10

84, 202 327

global solution of an autonomous system,

complete partially ordered set,

17

55

completely regular topological space,

G-negligible set,

255

Godel’s incompleteness theorem, 2

continuous nowhere approximately

G-thick set,

differentiable function, 397

G-thin set,

91

146 146

continuous nowhere differentiable function, 126, 135

Haar measure, 41

Continuum Hypothesis, 8

Hahn-Banach theorem, 1

convex body, 145

Hamel basis,

convex polygon, 312 convexly independent set,

92

Hamiltonian system, 56 302

countably continuous partial function, 9

height of a tree, 339 Henry’s theorem, 27 Hewitt’s theorem, 170, 174

Darboux property, 123 debut of a set,

homogeneous covering, 310

368

decreasing partial function, 7

Increasing partial function, 6

density point, 397

independent family of sets,

density topology, 404

inductive limit of a family of groups, 414

Dieudonne measure, 392

inductive limit of a family of invariant

diffused measure, 9

measures, 47

Dini right derived number, 16

injective commutative group, 415

Dirichlet function, 122

intermediate value property, 123

divisible commutative group, 10

invariant measure, 24

311

Subject Index

459

isodyne topological space,

93, 172

Markov-Kakutani theorem, 40, 51 Martin’s Axiom, 8

Jarnik’s theorem, 329, 404

Mazurkiewicz set,

Jordan curve, 308

measurable functional graph, 366

Konig lemma, 60, 340 Konig tree,

340

Kronecker’s theorem, 56 Krylov-Bogoliubov theorem, 40 Kulikov’s theorem, 165 Kuratowski’s theorem on closed projection, 125

307

measurable selector, 13, 359 measure extension problem, 19 metric space canonically associated with a measure, 18 metrically transitive measure, 147 mid-point convex function, 294 Minkowski’s sum of sets,

199

Minkowski’s theorem, 145

Lavrentiev’s theorem, 137, 138

monothetic topological group, 42

Lebesgue integral, 19

monotone class of sets,

Lebesgue measurable function, 8

monotone partial function, 7

Lebesgue measure, 14, 41

Montgomery operation, 390

Lebesgue theorem on density points, 58 level of a tree,

339

Liouville theorem, 39 Lipschitz condition, 16 locally compact topological group, 41 local solution of an autonomous system, 55 locally finite measure, 41 lower semicontinuous partial function, 5 Luzin-Jankov-von Neumann theorem, 112, 376

Negligible set,

57

91

nonmeasurable set,

14

nonseparable extension of an invariant measure, 257 nonseparable measure, 18 normal topological space, 1 normed vector space, 1 nowhere constant function, 337 nowhere Lebesgue measure zero set,

225

null-set, 241

Luzin’s C-property, 37

One-dimensional unit torus, 42

Luzin set,

one-parameter group of transformations,

11

Mackey’s theorem, 66 Marczewski’s characteristic function, 157, 348

55 oscillation of a function at a point, 137 ot-set, 310

Marczewski’s method of extending

Partial function, 1

measures, 20

partial isometry, 17

460

Topics in Measure Theory and Real Analysis

partial selector of a family of sets, partially ordered set,

4

17

resolvable topological space, 170 Riemann integral, 19

partially recursive function, 2

Riesz theorem, 49

Peano type mapping, 205

right invariant mean, 278

perfect measure, 37

root of a tree, 339

perfectly meager topological space, 326 perfectly normal topological space, 320

Scattered topological space, 334

pleasant point with respect to a function,

selector of a family of sets,

140

separable measure, 18

Poincare theorem on recurrence, 157

separation principle for analytic sets,

points in general position, 302

set-function, 13

Polish locally compact group, 42

set-theoretical aspect of the measure

problem of extending partial functions, 1

extension problem, 19

product of a family of groups, 44

set-valued mapping, 16

projective commutative group, 412

Sierpinski partition, 15, 346

projective set,

Sierpinski set,

437

4 8

100

Sierpinski-Zygmund function, 9, 215 Quasi-compact class of sets,

378

solvable group, 159, 168

quasicompact topological space, 4

Sorgenfrey topology, 94

quasi-cyclic group, 414

Steinhaus property, 48

quasi-invariant measure, 24

step-function, 22

quasi-polygon, 312

step-function associated with a countable

quotient Boolean algebra, 244

partition, 97

quotient space, 128

support of a measure, 383 Suslin line, 338, 394

Radon measure, 26

Suslin set,

Radon-Nikodym derivative, 64

Suslin theorem, 441

Radon-Nikodym theorem, 64

symmetric group, 23

89

Radon topological space, 37 Ramsey theorem, 303

Tarski’s fixed-point theorem, 17

range of a partial function, 3

theorem of Kuratowski and

real-valued measurable cardinal, 61

Ryll-Nardzewski, 13, 373

recursive function, 2

theorem on measurable projection, 366

relatively measurable function, 79

thick set,

relatively measurable set,

Tietze-Urysohn theorem, 1

80

29

Subject Index

461

topological aspect of the measure

Vietoris topology, 380

extension problem, 19

Vitali equivalence relation, 84

topologically complete metrizable space,

Vitali partition, 85

142

Vitali set,

totally imperfect set,

34

83

Vitali type function, 84

transformation group, 23

von Neumann topology, 406

tree, 339 Weak product of a family of groups, 44

Tychonoff theorem, 4

weakly continuous function, 224 Ulam’s theorem, 21

weakly measurable set-valued mapping,

Ulam transfinite matrix, 260

371

universal measure zero set,

weakly metrically transitive measure,

11

universally measurable function, 89

145, 148

universally measurable functional in the

Whitehead problem, 419

generalized sense,

Wiener measure, 13

284

unpleasant point with respect to a

Wiener stochastic process, 13

function, 140 upper semicontinuous partial function, 5

Zermelo-Fraenkel set theory, 4 Zermelo’s theorem, 4

Vector field,

54

Zorn lemma, 4