Principles Of Real Analysis, third edition

PRINCIPLES OF REAL ANALYSIS Third Edition

CHARALAMBOS D. ALIPRANTIS Departments of Economics and Mathematics Purdue University and

OWEN BURKINSHAW Department of Mathematical Sciences Indiana University. Purdue University, Indianapolis

ACADEMIC PRESS San Diego London Boston New York Sydney Tokyo Toronto

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© 1998 by Academic Press

@

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Library of Congress Cataloging-in-Publication Data Aliprantis, Charalambos D. Principles of real analysis

I Charalambos

D. Aliprantis and Owen Burkinshaw.

p. em. Includes bibliographical references and index. ISBN 0-12-050257-7 (acid-free paper)

I.

Mathematical analysis.

QA300.A48

2. Functions of real variables.

I. Burkinshaw, Owen.

II. Title.

1998

515-dc21

98-3955 CIP

Printed in the United States of America 98 99 00 0 I

02 03 DS 9 8 7

6

5 4 3 2

I

CONTENTS Preface C HAPTER 1 .

lX

FUNDAMENTALS OF REAL ANALYSIS 1. 2. 3. 4. 5. 6. 7.

C HAPTER 2.

C HAPTER 3.

Elementary Set Theory Countable and Uncountable Sets The Real Numbers Sequences of Real Numbers The Extended Real Numbers Metric Spaces Compactness in Metric Spaces

1 9 14 22 29 34 48

TOPOLOGY AND CONTINUITY

57

8. 9. 10.

Topological Spaces Continuous Real-Valued Functions Separation Properties of Continuous Functions I I. The Stone-Weierstrass Approximation Theorem

57 66 80 87

THE THEORY OF MEASURE

93

12. 13. 14. 15 . 16. 17. 18. 19. 20.

Semirings and Algebras of Sets Measures on Semirings Outer Measures and Measurable Sets The Outer Measure Generated by a Measure Measurable Functions Simple and Step Functions The Lebesgue Measure Convergence in Measure Abstract Measurability

93 98 103 1 10 120 126 133 146 149

vii

CONTENTS

viii CHAPTER 4.

THE LEBESGUE INTEGRAL 2!." 22. 23. 24. 25. 2 6.

CHAPTER 5.

NORMED SPACES AND L p-SPACES 27. 28. 29. 30. 3!.

CHAPTER 6.

Normed Spaces and Banach Spaces Operators Between Banach Spaces Linear Functionals Banach Lattices Lp-Spaces

HILBERT SPACES 32. 33. 34. 35.

CHAPTER 7.

Upper Functions Integrable Functions The Riemann Integral as a Lebesgue Integral Applications of the Lebesgue Integral Approximating Integrable Functions Product Measures and Iterated Integrals

Inner Product Spaces Hilbert Spaces Orthonormal Bases Fourier Analysis

I6I I6I I66 I77 I 90 20 I 204 2I7 2I7 224 235 242 254 275 276 288 298 307

SPECIAL TOPICS IN INTEGRATION

325

Signed Measures Comparing Measures and the Radori-Nikodym Theorem 38. The Riesz Representation Theorem 39. Differentiation and Integration 40. The Change of Variables Formula

326

36. 37.

338 352

366 385

Bibliography

399

List of Symbols

401

Index

403

PREFACE This is the third edition of Principles of Real Analysis, first published in 1981. The aim of this edition is to accommodate the current needs for the traditional real analysis course that is usually taken by the senior undergraduate or by the first year graduate student in mathematics. This edition differs substantially from the second edition. Each chapter has been greatly improved by incorporating new material and by rearranging the old material. Moreover, a new chapter (Chapter 6) on Hilbert spaces and Fourier analysis has been added. The subject matter of the book focuses on measure theory and the Lebesgue integral as well as their applications to several functional analytic directions. As in the previous editions, the presentation of measure theory is built upon the notion of a semiring in connection with the classical Caratheodory extension procedure. We believe that this natural approach can be easily understood by the student. An extra bonus of the presentation of measure theory via the semiring approach is the fact that the product of semi rings is always a semiring while the product of a -algebras is a semiring but not a a -algebra. This simple but important fact demonstrates that the semiring approach is the natural setting for product measures and iterated integrals. The theory of integration is also studied in connection with partially ordered vector spaces and, in particular, in connection with the theory of vector lattices. The theory of vector lattices provides the natural framework for formalizing and interpreting the basic properties of measures and integrals (such as the Radon­ Nikodym theorem, the Lebesgue and Jordan decompositions of a measure, and the Riesz representation theorem) . The bibliography at the end of the book includes several books that the reader can consult for further reading and for different approaches to the presentation of measure theory and integration. In order to supplement the learning effort, we have added many problems (more than 150 for a total of 609) of varying degrees of difficulty. Students who solve a good percentage of these problems will certainly master the material of this book. To indicate to the reader that the development of real analysis was a collective effort by many great scientists from several countries and continents through the ages, we have included brief biographies of all contributors to the subject mentioned in this book. ix

X

PREFACE

We take this opportunity to thank colleagues and students from all over the world who sent us numerous comments and corrections to the first two editions. Special thanks are due to our scientific collaborator, Professor Yuri Abramovich, for his comments and constructive criticism during his reading of the manuscript of this edition. The help provided by Professors Achille Basile and Vinchenco Aversa of Universita Federico II, Napoli, Italy, during the collection of the bio­ graphical data is greatly appreciated. Finally, we thank our students (Anastassia Bax.evani, Vladimir Fokin, Hank Hernandez, Igor Kuznetsov, Stavros Muronidis, Mohammad Rahman, and Martin Schlam) of the 1 997-98 IUPUI graduate real analysis class who read parts of the manuscript and made many corrections and improvements.

C. D. ALIPRANTIS and 0. BURKINSHAW West Lafayette, Indiana June, 1998

PRINCIPLES OF

REAL ANALYSIS Third Edition

C H A P T ER

!

_ _ _ _________ __

FUNDAMENTALS OF REAL ANALYSIS If you are reading this book for the purpose of learning the theory of integration, it is expected that you have a good background in the basic concepts of real analysis. The student who has come this far is assumed to be familiar with set theoretic terminology and the basic properties of real numbers, and to have a good understanding of the properties of continuous functions. The first section of this chapter covers the fundamentals of set theory. We have kept it to the "minimum amount" of set theory one needs for any modem course in mathematics. The following two sections deal with the real and extended real numbers. Since the basic properties of the real numbers are assumed to be known, the fundamental convergence theorems needed for this book are emphasized. Sim­ ilarly, the discussion on the extended real numbers is focused on the needed results. The last two sections present a comprehensive treatment of metric spaces.

1.

ELEMENTARY SET THEORY

Throughout this book the following commonly used mathematical symbols will be employed: V

means "for all" (or "for each"); 3 means "there exists" (or "there is"); ==> means "implies that" (or simply "implies"); means "if and only if." ¢:=:::> The basic notions of set theory will be briefly discussed in the first section of this chapter. It is expected that the reader is familiar in one way or another with these concepts. No attempt will be made, however, to develop an axiomatic foundation of set theory. The interested reader can find detailed treatments on the foundation of set theory in references [8], [13], [17], and [20] in the bibliography at the end of this book. 1

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS

2

The concept of a set plays an important role in every branch of modem math­ ematics. Although it seems easy, and natural, to define a set as a collection of objects, it has been shown in the past that this definition leads to contradictions. For this reason, in the foundation of set theory the notion of a set is left undefined (like the points and lines in geometry), and is described simply by its properties. this book we shall mainly work with a number of specific "small" sets (like the Euclidean spaces IR" and their subsets), and we shall avoid making use of the "big" sets that lead to paradoxes. Therefore, a set is considered to be a collection of objects, viewed as a single entity. Sets will be denoted by capital letters. The objects of a set A are called the elements (or the members or the points) of A. To designate that an object x belongs to a set A, the membership symbol E is used, that is, we write x E A and read it: x belongs to (or is a member of) A. Similarly, the symbolism x ¢ A means that the element x does not belong to A. Braces are also used to denote sets. For instance, the set whose elements are a, b, and c is written as {a, b, c}. A set having only one element is called Two sets A and B are said to be equal, in symbols A = B, if A and B have precisely the same elements. A set A is called a subset of (or that it is included in) a set B, in symbols A c B , if every element of A is also a member of B . Clearly, A = B if and only if A c B and B c A both hold. If A B and B # A, then A is called a proper subset of B . The set without any elements is called the empty (or the void) set and is denoted by 0. The empty set is a subset of every set. If A and B are two sets, then we define the union A U B of A and B to be the set AU B {x: x E A or x E B}; the intersection An B of A and B to be the set {x: x E A and x E B }; AnB the set difference A \ B of B from A to be the set A \ B = {x: x E A and x ¢ B}. The set A \ B is sometimes called the complement of B relative to A. Two sets A and B are called disjoint if An B 0. A number of useful relationships among sets are listed below, and the reader is expected to be able to prove them: (AU B) n C = (An u (B n C); (An B) U =(AU C)n (B U C); In

a singleton.

c

1.

=

n.

=

m.

=

1. 2.

C)

C

Section 1: ELEMENTARY SET THEORY

3. 4.

3

(A\ C) U (8\C); . (A U 8)\C (A n 8)\C =(A\C) n (8\C). =

The identities ( 1) and (2) between unions and intersections are referred to as the distributive laws. We remind the reader how one goes about proving the preceding identities by showing ( 1 ). Note that an equality between two sets has to be established, and this shall be done by verifying that the two sets contain the same elements. Thus, the argument for (1) goes as follows: x E (AU 8) n C ¢=:? ¢=:?

X

E AU 8 and x E C ¢=:? (x E A or x E 8 ) and X E C

x E A n C or x

E

8 n C ¢=:? x E (An C)U (8 n C).

Another useful concept is the symmetric difference of two sets. If A and 8 are sets, then their symmetric difference is defined to be the set A6.8 =(A\8) U (8\A). The concepts of union and intersection of two sets can be generalized to unions and intersections of arbitrary families of sets. A family of sets is a nonempty ·set :F whose members are sets by themselves. There is a standard way for denoting a family of sets. If for each element i of a nonempty set a subset A; of a fixed set X is assigned. then {A; };et (or {A;: i E I } or simply {A;}) denotes the family whose members are the sets A;. The nonempty set is called the index set of the family, and its members are known as indices. Conversely, if :F is a family of sets, then by letting I = :F and A; = i for each i E I , we can express :F in the form {A;};et· If {A;};et is a family of sets, then the union of the family is defined to be the set

I,

I

U A;= {x:

ie/

3 i !E

I

such thatx E A;},

and the intersection of the family by

n A;= {x:

iE/

X

E A; for each

i E

/}.

Occasionally. Uie/ A; will be denoted by u A; and nie/ A; by n A;. Also, if = IN = { 1 . 2.. . . } (the set of natural numbers), then the union and intersection of the family will be denoted by A, and n:, A11, respectively. The dummy index n can be replaced, of course, by any other letter.

I

U::,

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS

4

The distributive laws for general families of sets now take the form

(U ) iel

A; nB = UCA; n B) iel

and

(n ) iel

A; U B = n (A; UB). iel

is called pairwise disjoint if for each pair i and j of distinct indices, the sets A; and A j are disjoint, i.e., A; n A j = (/). The set of all subsets of a set A is called the power set of A, and is denoted by P(A). Note that 0 and A are members of P(A). For most of our work in this book,

A family of sets

{A;};eJ

subsets of a fixed set X will be considered (the set X can be thought of as a frame of reference), and all discussions will be considered with respect to the basic set X.

Now, let X be a fixed set. If P(x) is a property (i.e., a well-defined "log­ ical" sentence) involving the elements x of X , then the set of all x for which

P(x) is true wiU be denoted by {x EX: P(x)}. For instance, if X = {1, 2, ... } and P(x) represents the statement "The number x E X is divisible by 2," then {x EX: P(x)} = {2, 4, 6, . . . }. If A is a subset of X, then its complement A e (relative to X) is the set A e = X\ A = {x E X: x ¢. A}. It should be obvious that (Ae)e = A, A n Ae =(/),and A U A e = X. Some other properties of the complement operation are stated next (where A and B are assumed to be subsets of X): A \ B = A n Be; 6. A C B if and only if Be C Ae; 7. (A U B)e = AenBe; 8. (A n B)e = Ae U Be.

5.

The identities (7) and (8) are referred to as De Morgan's1 laws. The generalized De Morgan's laws are going to be very useful, and for this reason we state them as a theorem,

Theorem 1.1 (De Morgan's Laws). X,

the following idemities hold:

For a family {A; };e1 of subsets of a set

and

1

(n ) IE/

A;

e

= U Af. ie/

Augustin De Morgan ( 1806-1871 ), a Britishmathematician. He is well known for his contributions

to mathematical logic.

5

Section 1: ELEMENTARY SET THEORY

Proof. We establish the validity of the first formula only, and we leave the

verification of the other for the reader. Note that

xE

( ) U A; iEI

c

{=:::=}

x rt U A; {=::=}X rt A; for all i E I iEI {=:::=} x E A� for all i E I {=::=}X E n A�, iEI

and this establishes the first identity.

•

By a function f from a set A to a set B, in symbols f: A --+ B (or A�B or even x 1-+ f(x)), we mean a specific "rule" that assigns to each element x of A a unique element y in B. The element y is called the value of the function f at x (or f(x). The element the image of x under f) and is denoted by f(x), that is, y y f(x) is also called the output of the function when the input is x. The set A f(x)} is called is called the domain of f , and the set {y E B: 3 x E A with y the range of f. It is tacitly understood that the sets A and B are nonempty. Two functions f: A --+ B and g: A --+ B are said to be equal, in symbols f g, if f(x) g(x) holds true for each x E A. A function f: A --+ B is called onto (or surjective) if the range of f is all of B; that is, if for every y E B there f(x). The function f: A --+ B is called exists (at least one) x E A such that y one-to-one (or injective) if x1 =I= x2 implies f(x1) =I= j(x2). Now, let f: X --+ Y be a function. If A is a subset of X, then the image /(A) of A under f is the subset of Y defined by

=

=

=

=

=

=

f(A)

= {y

E Y:

3

xEA

such that

y

=

= f(x)}.

Similarly, if B is a subset of Y, then the inverse image f-1(B) of B under f is the subset of X defined by f- 1(B) {x E X: f(x) E B}. Regarding images and inverse images of sets, the following relationships hold (we assume that {A;};e1 is a family of subsets of X and {B;};e/ a family of subsets of f):

=

9. /(U;e/A;) U;e/ f(A;); 10. f(n;e/A;) C nielf(A;) ; 11. /-I cuiE/ B;) Uie/f-l(B;); 12. f-l(nfe!B;) nielf-l(B;); e 13. /-1(B ) (/-1(B))c . Given two functions f: X --+ Y and g: Y --+ Z, their composition go f is the function go f: X --+ Z defined by (go f)(x) g(f(x)) for each x E X. ' If a function f: X --+ Y is one-to-one and onto, then for every y E Y there f(x); the unique element x is denoted by exists a unique x E X such that y f-1(y). Thus, in this case, a function f-1: Y --+ X can be defined by f-1(y) x, whenever f(x) y. The function f-1 is called the inverse of f. Note that (f o f-1)(y) y for ally E Y and (/-1 o f)(x) x for all x E X . The latter

== =

= =

=

=

=

=

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS relations are often written as f o f- 1 = Iy and f- 1 of = Ix , where Ix : X� X and /y(y) = y and ly: Y � Y denote the identity functions; that is, Ix (x) =

x

for all x E X and y E Y. Any function IN"�

x: X, where IN" = { 1 , 2, . . . } is the set of natural numbers, is called a sequence of X. The standard way to denote the value x (n) is by X11 (called the n'h term of the sequence). We shall denote the sequence x by {xn } . and we shall consider it both as a function and as a subset of X. A subsequence of a sequence {x11 } is a sequence {Yn} for which there exists a strictly increasing sequence {k11} of natural numbers (that is, 1 =::: k 1 < k2 < k3 < · · ) such that Y11 = Xkn holds for each n. If now {A; };e1 is a family of sets, then the Cartesian2 product TI;e1 A; (or TIA;) is defined to be the set consisting of all functions f: I -+ U;e/ A; such that x; = f(i) E A; for eachi E I. Such a function is called (for obvious reasons) a choice function and quite often is denoted by (x;);e/ or simply by (x;). If a family of sets consists of two sets, say A and B, then the Cartesian product of the sets A and B is designated by A x B. The members of A x B are denoted as ordered pairs, that is, ·

Ax B = {(a, b): a E A

and

b E B}.

Clearly, (a, b)

= (a 1 , b1) if and only if a = a 1 and b = b 1 • Similarly, the Cartesian , An} is written as A 1 x product of a finite family of sets {A 1 , x A, and its •

•

· · ·

•

members are denoted as n-tuples, that is,

A1 x · · ·x A11 = {(at, . . . , a11): a; E A;

for each i

=

1, . . . , n}.

(at , . . . , an ) = (bt, . . . , b11) if and only if a; b; fori = 1, . . . , n. If A 1 = A2 = · = A, = A, then it i s standard to write A 1 x · x A, a s A". Similarly, if the family o f sets {A; };e/ satisfies A; = A for each i E I, then TI;e/ A; 1 is written as A 1 , that is, A = { f I f: I -+ A}.

Here, again

=

·

•

·

· ·

When is the Cartesian product ofa family of sets {A; };e/ nonempty?

Clearly, if the Cartesian product is nonempty, then each

A; must be nonempty.

The following question may, therefore, be asked: •

If each A; is nonempty, is then the Cartesian product TIA; nonempty?

Although the answer seems to be affirmative, it is unfortunate that such a state­ ment cannot be proven with the usual axioms of set theory. The �ffirmative answer

2Rene Descartes o r Cartesius (1596-1650), an i nflu ential F ren ch philosophe r and mathe matician. He is t he foun der o fanalytic geo metry.

7

Section 1: ELEMENTARY SET THEORY

to the last question is known as "the axiom. of choice." This axiom will be assumed throughout this book without further explanation. One of its forms is stated below.

I f {A;};e1 is a nonempty family of sets such that A; is nonempty for each i I, then nA; is nonempty. Axiom of Choice.

E

A useful equivalent formulation of the axiom of choice is the following: •

a nonemptyfamily ofpairwis e disjoint sets such that A; #(/)for Ieach f {A;i};eE/I,is then ther e exi s ts a set E c Uie/ A; such that En A; consists of precisely one elementfor each I. i

E

For a discussion of the axiom of choice and its history, the reader is referred to [8] and [13]. B.y a (binary) relation on a set X we simply mean a subset n of X x X. If (X, E 1?.., then X is said to be in the relation 1?.. with and this is denoted by Among the most interesting relations are the equivalence relations. A relation n on a set X is called an equivalence relation if it satisfies the following three properties:

y)x'R.y.

a. b. c.

y,

(refl e xi v i t y ). y'R.x (symmetry). xx'R.'R.yy, y'R.z, x'R.z (transitivity).

xnx for each X E X If If

then and

then

Let n be an equivalence relation on a set X. Then the equivalence class determined by the element X E X is defined by [x] = E X: xny}. It is easy to observe that any two equivalence classes are either disjoint or else they coincide. Since X E [x] for each X E X' it follows that n partitions X. That is, there exists a family {A;}ie/ of pairwise disjoint sets (here, the family of equivalence classes) such that X = Uie/ A;. Conversely, if a family of pairwise disjoint sets {A;};e/ partitions X (i.e., X = Uie/ A;), then by letting

n = {(x,

y)

E X x X: 3 i

E

I

such that

{y

x

and

y

are in

A;},

an equivalence relation is defined on X whose equivalence classes are precisely the sets A;. Thus, the equivalence relations on a set correspond precisely in one-to-one fashion with its partitions. Another important type of relation is an order relation. A relation, denoted by <, on a set X is said to be a partial order for X (or that X is partially ordered by <) if it satisfies the following three properties: a.

{3.

y.

x < x< y If x If x

t y). x (rye flexivi (a! l li s ymmetr y ). x z (transitivity).

holds for every E X andy < x, then x = < y and y < z, then <

8

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS

An alternative notation for x < y is y > x. A set equipped with an order relation is called a partially ordered set. Now, let X be a partially ordered set. A subset Y of X is said to be a chain if for every pair x, y E Y, either x < y or else y < x; a chain is also referred to as a totally ordered set. If Y is a subset of X such that x < u holds for all x E Y and some u E X, then u is called an upper bound of Y. An element m E X is called a maximal element of X whenever the relation m < x implies x = m. (Warning: A partially ordered set may contain more than one maximal element.) The next statement guarantees the existence of maximal elements in certain partially ordered sets. It is known as Zorn 's3 lemma, and it is a powerful tool in analysis.

Zorn's Lemma. If eve!)' chain in a partially ordered set X has an upper bound in X, then X has a maximal element. Zorn's lemma (as a statement) is equivalent to the axiom of choice. For a detailed discussion, see

[13].

EXERCISES 1. Prove statements 2 through 13 of this section.

For two sets A and B show that the following statements are equivalent.

2.

a. b. c.

A c B. AU B=B. An B=A.

3.

Show that (A�B)�C=A�(B�C) hold for every triplet of set s A, B, and C. 4. Give an example of a function f: X --+ Y and two subsets A and B of X such that f(A n B)# f(A)n f(B). 5. For a function f: X --+ Y show that the following three statements are equivalent. a

b. c.

f is one-to-one. f(An B)=f(A)n f(B) holds for all A, B E P(X). For every pair of subsets A, B of X satisfying A n B = 0. we have f(A) n f(B)= (/).

6. Let f: X --+ Y be a function. Show that f{f-1 (A)) c A for all A £ Y, and B c f-1 (f(B)) for all B c X.

7. Show that a function f: X B c Y.

8. 9.

--+ Y is onto if and only if f(f-1 (B)) = B holds for all

Assume X -4 Y ....4- Z. If A f Z, show that (go f)-1 (A)=f-1 (g-1 (A)). Show that the composition of functions satisfies the associative Jaw. That is, show that if X -4 Y ....4- Z � V, then (h o g) o f= h o (g o f).

Max Zorn ( 1906-1 993), a Gen nan- bom Ameri can mathematician. He wo rke d in set theory, to po l­ ogy, and alge br a. 3

9

Section 2: COUNTABLE AND UNCOUNTABLE SETS 10. 11.

2.

f: X � Y. Show thatthe relation R �:m X, defined by .tt

Let equivalence are is an

If X and Y 'P(X U Y).

relation.

sets, then show that

'P(X) n 'P(Y)

=

'R.:c2 if /(:rt) = f( x2).

'P(X n Y) and 'P(X) U 'P(Y) �

COUNTABLE AND UNCOUNTABLE SETS

In this section we shall deal with questions concerning the "size" of a set. Two sets are said to have the "same number of elements" if their elements can be put in one-to-one correspondence with each other. By a one-to-one correspondence between two sets A and B, we mean a function f: A B that is one-to-one and onto. Definition 2.1. Two sets A and B are said to be equivalent (in symbols A � B) ifthere exist� a function f: A B that is one-to-one and o11to. It is easy to verify that for sets the following properties hold: 1. A� A. 2. If A � B, the B � A. 3. IfA� B and B� C.then A� C. The dividing line for the sizes of sets is the set of natural numbers {1, 2, 3, ... Any subset of 1N of the form { . . . , n} is called a segment of and n is called the number of elements of the segment. Clearly, two segments {1, . . . , n} and {1 , . .. , m} are equivalent if and only if n m. This shows that a proper subset of a segment cannot be equivalent to the segment. A set that is equivalent to a segment is called a finite set. The empty set is also considered to be finite with zero elements. set that is not finite is called an �

�

}

1N =

1N,

1,

.

=

A

infinite set.

Definition 2.2.

A set A is called countable if it is equivalent to

lN,

there exists a one-to-one correspondence of 1N with the elements of

that is, if

A:· There is a standard notation for a countable set A. It is usually written as A {a 1 , a2, }, often called an enumeration of the set A, and it indicates the one-to-one correspondence of A with the set of natural numbers An infinite set that is not countable is called an uncountable set. Our first result compares the infinite sets with the countable ones. =

•

•

•

lN.

Every infinite set contains a countable subset. (/>. a1 E {a 1 }. 1 az E

Theorem 2.3.

Let A be an infinite set; clearly A -:/= Pick A, and consider the set A A \ Since A is infinite, A is nonempty. Pick A and consider Proof. 1 =

1,

10

Chapter

1: FUNDAMENTALS OF REAL ANALYSIS

the set A2\ {a 1 , a2}. By the same arguments, there exists an element a3 e A2• Proceeding in this way, a set {a1 , a2, a3, } is obtained tha� is clearly countable and by construction it is a subset of A. • •

•

•

The following two principles of the natural numbers are needed in order to establish more properties of the countable sets. The first property is known as the "well-ordering principle" of N. Keep in mind that a subsetS of N is said have a least (or a first) element if there exists k e S such that k < n for each n e S; clearly k is uniquely determined.

THE WELL-ORDERING PRINCIPLE. Every nonempty subset ofN has

a least element.

The second property of 1N that is needed is known as the "principle of mathe­ matical induction," and is a powerful tool in mathematics.

THE PRINCIPLE OF MATHEMATICAL INDUCTION. Ifa subset S of

N satisfies the properties a. l b. n

e S, and + l ES whenever n ES,

then S =lN. We now continue our discussion about countable sets.

Theorem 2.4. Every subset ofa countable set is eitherfinite or else countable. Proof. Let A be a subset of a countable set Assume that A is not finite; then

it will be shown that A is countable. An easy argument shows that one can assume without loss of generality that A is a subset of 1N. Now, define a function f: 1N --+ A inductively as follows: f(l) =the least ele­ ment of A (that exists by the well-ordering principle); and then if f(l), . . . , f(n) have been defined, let f(n + l) be the least element of A\{f(l), . . . , f(n)}. (The least element again exists by the well-ordering principle, and the fact that A is not finite.) It is now left to the reader to verify that f is one-to-one and onto. Thus, • A�N It is interesting to observe that (in contrast with the finite sets) an infinite set can be equivalent to some of its proper subsets. To see this, let X = {2, 4, 6, ...}, and note that X is a proper subset ofN. Now the function f:N --+ X, defined by f(n) = 2n for each n, is one-to-one and onto, so that X � N. Some useful characterizations to ensure that an infinite set is countable are presented next.

Section 2: COUNTABLE AND UNCOUNTABLE SETS

11

For an infinite set A, t�e following statements are equivalent: AThere is countable. exists a subset B of IN and a function f: B A that is onto. There exists afunction g: A IN that is one-to-one. Proof. A B IN. f: IN A = B IN B A 1 f - (a) = {n E 8: f(n) = a} f . a E A. g: A IN g(a) 1 f- (a); g g(a) = g(b) a = f(g(a)) = f(g(b))g: Ab IN gA g(A). A g(A) IN, 2. 4 , g{A) IN A IN Theorem 2.5. i. ii. iii.

-+

-+

(i) ==> (ii) Since is countable, there exists -+ that is one­ to-one and onto. Thus, (ii) holds with (ii) ==>(iii) Assume that is a subset of and that f: -+ is an onto function. Note that since is onto, is nonempty for each Now, define -+ as follows: =the least element of this natural number exists by the well-ordering principle. To complete the proof, it must be shown that is one-to-one. Indeed, if holds, then = also holds, which shows that is one-to-one. (iii)==> (i) Assume -+ to be one-to-one. Then � Since is is an infinite subset of and hence, by Theorem (by hypothesis) infinite, � holds. Therefore, � holds, and the proof is complete. • The next two results are consequences of the preceding theorem. The first one asserts that the countable union of countable sets is countable.

Let {A , A2, } be a countable fami l y of set s such that each 1 A; is a countable set. Then A U:1 A, is a countable set. Proof. A, = {a;' , a�, . . . } n = 1, 2, A = U:1 A,. k B = {2 3": k, n E IN}. B A 3") = af. f(2k B A, A 2.5. Theorem 2.6. Let

·

onto

maps

•

•

•

=

for .. . , and -+ Now define f: by and hence is a countable set by Theorem

Also, let Then f

B

·

The Cartesian product of a finite collection of countable sets is always countable.

Let {A1 A,} be a finite collection of sets such that each A; is countable. Then A A, is countable. Proof. A; = IN i, A = IN" . , k,) =n p f: IN" IN f (k1, Pi' Pi2 p�". f 2.5, IN" Theorem 2.7.

,

1

•

•

•

,

x ··· x

It suffices to assume that for each and hence, Pick distinct prime numbers, say 1, , p,, and define -+ by It should be clear from the fundamental theorem of arithmetic (ev­ ery natural number has a unique factorization into primes) that is one-to-one. Hence, by Theorem is countable. • •

•

•

•

•

•

•

•

•

Our discussion will close with some results concerning the "big" sets. One may ask the question: To answer this question a definition is needed. Let us write -< B whenever there exists a one-to-one function -+ B; in other words, � B if is equivalent to a subset of B. It should be clear that in this sense one can say that "B has at least as many elements as

How large can a set be? f: A A."

AA

A

12

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS The relation-< satisfies the following properties: 1. 2. 3.

A -< A for all sets A. If A -< B and B -< C, then A If A-< Band B-< A, then A

-< C. B.

�

Statement (3) is known as the Schroder-Bemstein4•5 theorem, and is a very im­ portant result; for a proof see [13, p. 88], or [28, p. 29]. The next result shows that the power set of a given set has more elements than the set; it is due to G. Cantor.6

Theorem 2.8 (Cantor). If A is a set, then A

-< 'P(A) and A � 'P(A) hold.

Proof. If A = (/), then the result is trivial. So, assume that A # (/). Define f: A -)o 'P(A) by f(x) = {x} for x E A, and note that f is one-to-one. Hence, A-< 'P(A). To show that A � 'P(A)� assume by way of contradiction that A 'P(A). So, there exists g: A -)o 'P(A) that is one-to-one and onto. Consid�r the subset B of A defined by B = {x E A: x ¢. g(x)}, and then pick a e A such that g (a) = B. Now,

�

notice that a E B if and only if a ¢. completes the proof of the theorem.

B, which is impossible.

This contradiction

•

Imagine that for every set A a symbol (playing the role of a number) can be assigned that designates the number of elements in the set. Without going into detail, this symbol is called the cardinal number of A, and is denoted card A. It should be clear that if A is a finite set, say A = {a 1 , , a,}, then card A = n. But what about IN? We use the symbol �0 to denote the cardinal number of IN. By saying that a set A has cardinal number �0 (in symbols card A = �0), we simply IN. In general, card A = card B means A� B. mean that A If a and b are cardinal numbers, then a < b means that there exist two sets A and B such that card A = a, card B = b, and A -< B. Similarly, a < b among cardinal numbers means that there exist two sets A and B such that card A = a, card B = b, A -< B, and A � B. Note that according to the Schroder-Bemstein theorem, a < b and b < a guarantee a = b. We mention also a few things about "cardinal arithmetic." By Theorem 2.8, we know that there exists a set with cardinal number greater than �0; this set is 'P(IN). where 1R is the It can be shown that 'P(IN) � 1R (see Exercise 6 of Section •

•

•

�

5),

4Emst SchrOder ( 1 841-1902}, a Ger man mat hematician. He wo rked in algebra and m at hem atical lo gic. 5Fel ix Bernstei n ( 1878-J 956), a Germ an mat hematici an. He wo rk ed in s et theory and co ntributed decisively to the dev elo pment o fthe i nherit ance po pul at io n genetics. 6Geo rg C anto r ( 1 845-191 8}, a German mat hemat ician . He was t he founder o f s et t heo ry. He also cont rib uted to the fi eld of classical analysis.

13

Section 2: COUNTABLE AND UNCOUNTABLE SETS

set of all real numbers. The cardinal num!Jer of JR. is denoted by c, and is called the cardinality of the continuum; thus, �0 < c. A cardinal number a satisfying �0 � a, is called an infinite cardinal. x It is easy to see that if 2 = {0, I}, then 2 � P(X) for every set X. For this X reason it is a custom to denote the cardinal number of P(X) by 2card . Thus, the following inequalities hold: 0< 1 <2<· · < ·

n

,c < · · < �0 < 2K11 = c < 2c < 2- < · · · ·

The outstanding question dealing with infinite cardinals is the following: •

If a 2 a?

is an infinite cardinal, is there a cardinal number such that b

a < b <

The continuum hypothesis assumes a negative answer to the above question when a = �0• and the generalized continuum hypothesis assumes a negative answer for every infinite cardinal. It has been shown that the continuum hypothesis is "independent" of the ordinary axioms of set theory. There are models of set theory that satisfy· the continuum hypothesis and there are models that do not. The interested reader is referred to [8], [13], and [20] for extensive treatments of the cardinal numbers.

EXERCISES 1. Show that the set of all rational numbers is countable. 2. 3.

4.

5.

Show that the set of all finite subsets of a countable set is countable. Show that the union of an at-most countable collection of sets, each of which is finite, is an at-most countable set. Let A be an uncountable set and B be a countable subset of A. Show that A is equivalent to A \ B. Assume that f: A � B is a surjective (onto) function between two sets. Establish the following: a.

6. 7. 8. 9.

10.

card B � card A.

b. If A is countable, then B is at most countable.

Show that two nonempty sets A and B are equivalent if and only if there exists a function from A onto Band a function from B onto A. n Show that if a finite set X has n elements, then its power set P(X) has 2 elements. Show that the set of all sequences with values 0 or 1 is uncountable. If2 = {0, 1}, then show that 2X � P(X) for every set X. Any complex number that is a root of a (nonzero) polynomial with integer coefficients is called an algebraic number. Show that the set of all algebraic numbers is countable.

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS

14

11. For an arbitrary function f: JR. -t JR., show that the set

{

A= a 12.

E JR.:

}� 0 f(x) exists and .J�a f(x) ¥= f(a) }

is at-most countable. Show that the set of real numbers is uncountable by proving the following: a. (0, 1) � 'P(JR); and b. (0, 1) is uncountable.

[HINT: If (0, 1) is countable, then let {x1, x2, . ..} be an enumeration of (0, 1 ). For each n write x11 = O.dnl d112 ···in its decimal expansion, where each dij is 0, 1, ... , 9. Now consider the real number y of (0, 1) whose decimal expansion y = O.y1 Y2 ·· · satisfies Yn = 1 if d1111 ¥= 1 and Yn = 2 if d1111 = 1. To obtain a contradiction, show that y ¥= x11 for each n.] 13. Using mathematical induction prove the following: a. b. c.

14. 15.

If a > -1, then ( 1 +a)" > 1 + na for n = 1, 2, .. . (Bernoulli's7 inequality). If 0 < a < 1, then I + 311 a > (I + a)11 for n = I, 2, . . . . cos(mr) = (-i)11 for n = i, 2, . . . .

Show that the well-ordering principle implies the principle of mathematical induction. Show that the principle of mathematical induction implies the well-ordering principle.

3. THE REAL NUMBERS Without any doubt, the most important set for this book will be the set of real numbers lR = ( -oo, oo) . The set of real numbers is also known as the real line. The reason is that by considering a straight line, one can put (in the usual way) the real numbers in one-to-one correspondence with the points of the line. The terms "real line" and "real numbers" will be viewed as identical. While it is not our purpose to give a complete axiomatic development of the real numbers, it is important to stop and consider exactly what axioms characterize the real numbers. They consist of the field axioms, the order axioms, and the completeness axiom. In algebraic terminology, the set of real numbers is referred to as the one and only "complete ordered field." The name comes from the axiomatic foundation of the real numbers outlined below.

The real numbers are the members of a nonempty set lR equipped with two operations, + and · from lR x lR into lR., called addition and multiplication, that satisfy the following axioms:

Field Axioms The letters x, y, and z denote arbitrary real numbers, unless otherwise stated.

7Jacob (Jacques ) Bernoulli (1654-1 705), a Swiss math ematician and one o f the most promin ent m em bers ofthe famous B ernou lli mathematica l family. His major works were in ca lculus and proba­ bi lity t heory.

15

Section 3: THE REAL NUMBERS

Axiom

1.

x + y = y + x and xy = yx (the commutative laws).

Axiom 2. x + (y + z) = (x + y) + z and x(yz) = (xy)z (the associative

laws).

Axiom 3. x(y + z) = xy + xz (the distributive law).

Axiom 4. There exists an element 0 E JR. such that x + 0 = x for all x E JR.

Axiom 5. For each x E JR. there exists an element in JR. (denoted by -x) such

that x + (-x) = 0.

Axiom 6. There exists an element l E JR. with l f. 0 satisfying l x = x for

all X E JR.

·

Axiom 7. For each x f. 0 there exists an element in JR. (denoted by x-1)

satisfying xx- 1 =

l.

4

It can be shown that the zero element of Axiom is uniquely determined. Also, it can be established that the element -x given by Axiom 5 is uniquely determined, and that -x = (-l )x holds. In a similar manner, it can be seen that the element x- 1 of Axiom 7 which satisfies xx- 1 = l (where, of course, x =f. 0) is uniquely determined. From the field axioms, one can derive the familiar properties of addition and multiplication. For instance, 0 · x = 0, -(-x) = x, (-x)(-y) = xy, x- y = X + (-y) = -(y - X), (x-l )-I = X. (The reader will find the details in the exercises at the end of this section.) The next requirement is that JR. must be not merely a field but also an "ordered field." This means that JR. is equipped with an order relation > compatible with the algebraic operations via the following axioms: Order Axioms

Axiom 8. For any x, y E JR., either x .::::, y or y .::::, x holds.

Axiom 9. lfx > y, then x + z > y + z holds for each z E JR.

Axiom 10. Ifx > y and z > 0, then xz.::::, yz.

An alternative notation for x > y is y < x. Any number x E JR. satisfying x > 0 (i.e., x > 0 and x =f. 0) is called a positive number (and likewise, any number x with x < 0 is called a negative number). From the order axioms, one can derive the ordinary inequality properties of the real numbers. Let us mention one very useful property dealing with inequalities: •

lfx + E > y holds for each E >

0,

then x.::::, y holds.

Chapter 1:

16

FUNDAMENTALS OF REAL ANALYSIS

Indeed, if the conclusion is not true, then y - X > 0. Let E = � (y - X) > 0, and note that our hypothesis implies 4 (x + y) = x + 4 (y - x) > y. This in tum implies y - x < 0, which is a contradiction. The usual way to define the absolute value of a real number a is as follows: Ia I = a if a > 0 and Ia I = -a if a < 0. If a v b denotes the larger of the numbers a and b (for instance, (-=-1) v 2 = 2 and 1 v 1 = I), then a moment's thought reveals that I a I = a v (- a) for all a e IR. In particular, it follows that I a I = 1- a I for each a e IR. The absolute value satisfies the properties: 1. 2. 3.

Ia I > 0 for each a e IR, and Ia I = 0 if and only if a = 0; labl = Ia I · lbl for all a, b E IR; and I a + bl < I a I + lbl for all a, b E lR (the triangle inequality).

The non-negative nu�ber I a - b I can be viewed geometrically as the distance between the numbers a and b. The least understood property of the real numbers is the completeness axiom� or the axiom of continuity. Before stating this axiom, let us recall a few things. Let A be a nonempty subset of IR. An upper bound of A is any real number a such that x < a for all x e A; similarly, b e lR is a lower bound for A if b < x for all x e A. If A has an upper (resp. lower) bound, then A is said to be bounded from above (resp. below). If A is both, bounded from above and below, then A is called a bounded set. A real number is called a least upper bound (or a supremum) of A if it is an upper bound for A, and it is less than or equal to every other upper bound of A. That is, x e lR is a least upper bound for A if i. A is bounded from above by x, and ii. if A is bounded from above by y, then x

< y.

It should be clear that a given set A can have at most one least upper bound, denoted by sup A. A similar definition is given for the greatest lower bound (or infimum) of a set A, denoted by inf A. The completeness axiom asserts that every nonempty set bounded from above has a least upper bound and is stated next. The Completeness Axiom

Every nonempty set of real numbers that is bounded from above has a least upper bound. Axiom 11.

From this axiom, it follows easily that every nonempty set of real numbers bounded from below has a greatest lower bound. (If A is nonempty and bounded from below, then the set B = { b e IR: b � x V x e A} is bounded from above, and so sup B exists. Note that sup B = inf A.) It should be clear also that if a set A has a maximum (resp. a minimum) element, then max A = sup A (resp. min A =

17

S ection 3: THE REAL NUMBERS

A).

A A,

A

On the other hand, if the supremum 9f a set exists and sup E then sup inf is the maximum element of In other words, the supremum of a set generalizes the concept of the maximum element of a set. For more about this axiomatic foundation of the real numbers, see and the exercises at the end of this section. Regarding the supremum of a set, the following approximation property holds:

A.

[3]

1R

for Assume that the supremum of a subset A of exists. Then every E > there exists some x A such that A - E < x � A. < A - E, Proof. A x A E A, x A A - E < x < A. Corollary 3.2. The set of natura/numbers is unbounded.
Theorem 3.1. 0,

E

sup

sup

sup If for every x E we have then sup is an upper bound of which is less than the least upper bound. But this is impossible. Thus, sup there exists some E such that sup • 1N

Assume by way of contradiction that holds for each 11 E 1N and some e JR. Then by the completeness axiom, = sup 1N exists, and by Theorem which 1 3.1 there exists some E 1N with s This implies • is impossible. The next useful property of the real numbers is known as the "Archimedean property." It was used extensively by Archimedes in geometrical proofs but, most likely, it was introduced first by Eudoxus. 8

l f x andy are two positive real numbers, then there exists some natural number n such that nx > y. n, 3.2n. < yI n. Proof. nx < y Theorem 3.3 (The Archimedean Property).

holds for each then If bounded from above, contrary to Corollary

x.

for each

That is, 1N is

•

An important density property of the rational numbers is described in the next theorem. Recall that a rational number is any real number that can be written as a quotient of two integers.

Theorem 3.4.

number.

Between any two distinct real numbers there exists a cational

8Eudoxus o f Cnidus (ca 400-347 BC), a Gr eek scho lar o f great eminen ce and in fluenct!. He co nt rib uted to astronomy, mat he mat ics, geog raphy, and philosophy and helped in the wr iting o f the laws in his home town o f Cn idus (Asia Mino r). He in troducedthe "methodo f exhaustio n·• for co mput ing are as and vo lume s and was the fir st to g ive a r igorous de finit io n o f a re al n umber.

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS

18

Proof. It is easy to see that we need consider only positive numbers. So, let

a , b E JR. be such that 0 < a < b.

Consider first the set A = {11 E IN: 11 > max{ h� , t}}. Since IN is not bounded a from above, A is nonempty. Fix an element q E A. Clearly, 0 < l < b - a and I < bq. Now, let B = {n E IN: n < bq}. Since I E B, we see th�t B # (/J, and clearly B is a finite set. Let p = max B; note that p E B and p + I ¢. B. To finish the proof, w e shall show that a < < b holds. To this end, note first we must that by construction < b holds. On the other hand, since b < have

�

�

a = b - (b - a) <

p

r;t,

p + I - -I = , q q q •

and we are done.

With the help of the completeness axiom, we can also establish the existence of "roots" of real numbers.

Theorem 3.5.

For a rea/ number a and any natural number n > 2, we have

the following: I. Ifa > 0 and n is even, there exists a unique b > 0 such that bn = a. 2. Ifa E JR. and n is odd, there exists a unique b E JR. such that bn = a.

Proof. · We shall establish both cases by assuming a

> 0 and leave the easy details for completing the proof to the reader. If a = 0, then clearly b = 0, and so we can also suppose that a > 0. We shall establish first the uniqueness of b. To see this, assume that x > 0 and y > 0 satisfy xn = y" = a. Then

and since x"- 1 + x" -2y + · · · + xy"-2 + y"- 1 > 0, we infer that x - y = 0, or X = y. For the existence, consider the setS = {s > 0: s n < a }. By the Archimedean property, there exists some E lN such that rna > I , or � < a. So, 0 < (�)n < �� < a, and thus, S is nonempty and contains positive numbers. On the other hand, if k E lN satisfies a < k, then s :::; k for each s E S; otherwise s > k implies s" > k > a, a contradiction. Now, by the completeness property, b = supS > 0 exists in JR. We shall complete the proof by proving that bn = a. This will be done by eliminating the possibilities b" < a and b" > a.

m

19

Section 3: THE REAL NUMBERS

k

So, assume first that E 1N we have

li' < a

is possible. By the binomial theorem, for each

t

�

t

" ('�)bn-i = b" + (b + �) = ('�)bn-i �i < b" + �. k k i =O l k k i I l

(*)

=

where r = "L;'= G')b"-i > 0. By the Archimedean property there exists some 1 k E 1N such that k(a - b") > r, or r/ k < a - h". But then, it follows from (*) that for this k we have (b + t )" < b" + (a - b") = a, which implies b + t E S, a contradiction. When b" > a, the situation is similar. As previously, we can use the binomial theorem to get l " t (b - - ) > b" - k k'

for some fixed t > 0 and all k E lN. If we choose again some k E 1N with f < b" - a and t < b, then it follows from (**) that (b - tY' > a . But if we pick some s E S with 0 < b - t < s (such an s is guaranteed by Theorem 3.1 ) , then we have s" > (b - t )" > a, a contradiction. Hence, b" = a holds. • When feasible, the unique solution b of the equation b" = a provided by Theo­ rem 3.5 is called the nth-root of a and is denoted fo or a f. .

EXERCISES 1.

a b = ma (a b} and a b = min {a, b}. then show that a b = 1
If

x

v

1\

.

1\

v

2. 3.

Show that the real numbers .J2 and .J2 -/3 are irrational numbers. Show that between any two distinct real numbers there is an irrational number.

+

4. 5. This exercise will introduce (by steps) the familiar process of subtraction in the frame­ work of the axiomatic foundation of real numbers.

i.e., show that if x +0* = x

a.

Show that the zero element 0 is uniquely determined,

b.

Show that the cancellation law of addition is valid, i.e., show that x

for all x E

implies c. d.

1R and some O*

E IR, then

a = b.

0*

= 0.

a=

+a = x + b

a

Use the cancellation law of addition to show that 0 · 0 for all E IR. Show that f or the real number -a is the unique real number

each real number a

that satisfies the equation of

a.)

a

+x

= 0. (The real number -a is called the negative

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS

20

Show that for any two given real numbers a and b, the equation a + x = b has a unique solution, namely x = b + (-a). The subtraction operation - of .R is now defined by a - b = a + (-b); the real number a - b is also called the difference of b from a. f. For any real numbers a and b show that -(-a) = a and -(a + b) = -a - b.

e.

6.

This exercise introduces (by steps) the familiar process of division in the framework of the axiomatic foundation of real numbers. a. b. c.

d.

e. f. 7.

Show that the element 1 is uniquely determined, i.e., show that if I * x = x for all x E .R and some 1 * E .R, then I * = 1 . Show that the cancellation law of multiplication is valid, i.e., show that x a x · b with x # 0 implies a = b. Show that for each real number a # 0 the real number a- 1 is the unique real number that satisfies the equation x · a = 1. The real number x = a- t is called the inverse (or the reciprocal) of a. Show that for any two given real numbers a and b with a # 0, the equation ax = b has a unique solution, namely x = a- 1 b. The division operation ...;- (or /) of .R is now defined by b ...;- a = a- 1 b; as usual, the real number b ...;- a is also denoted by bfa or � For any two nonzero a, b E .R show that (a- 1 )- 1 = a and (ab)- 1 = a- l b- 1 • Show that T = a for each a, � = 0 for each b # 0, and ; = 1 for each a # 0. ·

·

=

Establish the following familiar properties of real numbers using the axioms of the real numbers together with the properties established in the previous two exercises. 1. n.

m.

The zero product rule: ab = 0 if and only if either a = 0 or b = 0. The multiplication rule of signs: (-a )b = a(-b) = -(ab) and (-a)(-b) = ab for all a. b E IR. The multiplication rule for fractions: For b, d # 0 and arbitrary real numbers a, c we have

a ac c -.=bd b d # 0, then ( * )- 1 �·

In particular, if * tv. The cancellation law of division: If a # 0 and x # 0, then !:� = � for each b. v. The division rule for fractions: Division by a fraction is the same as multi­ plication by the reciprocal of the fraction, i.e., whenever the fraction * ...;- � is defined, we have =

a 8.

b

c ...;- d

b · -;; - be

a =

d

ad

This exercise establishes that there exists essentially one set of real numbers that satisfies the eleven axioms stated in this section. To see this, let .R be a set of real numbers (i.e., a collection of objects that satisfies all eleven axioms stated in this section). a. Show that 1 > 0. b. A real number a satisfies a = -a if and only if a

=

0.

21

Section 3: THE REAL NUMBERS

If n = 1 + 1 + · · · + 1 (where the sum has "n summands" all equal to 1), then show that these elements are all distinct; as usual, we shall call the set 1N of all these numbers the natural numbers of JR. d. Let Z consist of lN" together with their negative elements and zero; we shall call, of course, Z the set of integers of JR. Show that Z consists of distinct elements and that it is closed under addition and multiplication. e. Define the set Q of rational numbers by Q = { �:: m, 11 E Z and n =/; 0}. Show that Q satisfies itself axioms 1 through 1 0 and that c.

a = sup{r

f.

E

Q: r � a } = inf{s

E

Q: a � s }

holds for each a E JR. Now, let JR.' be another set of real numbers and let Q' denote its rational numbers. If 1 ' denotes the unit element of JR', then we write n ' = 1' + 1' + + 1' for the sum having "n-summands" all equal to 1 '. Now, define the function f: Q -+ Q' by · · ·

( ) f m

11

= m

'

' n

and extend it to all of JR via the formula

f(a) = sup{f(r):

r � a}.

Show that JR and JR' essentially coincide by establishing the following: 1.

ii. iii. 9.

11.

JR.'.

f(a + b) = f(a) + /(b) and f(ab) = f(a)f(b) for all a , b E JR.

Consider a two point set a. b. c.

10.

a � b holds in JR. if and only if f(a) � f(b) holds in f is one-to-one and onto. R =

{0, 1 } equipped with the following operations: =

Addition (+) : 0 + 0 = 0, 0 + 1 = 1 + 0 = 1 and 1 + 1 Multiplication (·): 0 · 1 = 1 · 0 = 0 and 1 · 1 = 1, and Ordering: 0 ::: 0, 1 ::: 1 and 1 ::: 0.

0,

Does R with the preceding operations satisfy all eleven axioms defining the real numbers? Explain your answer. Consider the set of rational numbers Q equipped with the usual operations of addition, multiplication, and ordering. Why doesn't Q coincide with the set of real numbers? This exercise establishes the familiar rules of "exponents" based on the axiomatic foundation of real numbers. To avoid unnecessary notation, we shall assume that all real numbers encountered here are positive-and so by Theorem 3.5 all non-negative real numbers have unique roots. As usual, the "integer" powers are defined by a" = a · a · · · a, '-v-"

a0 -

1

'

aI _ - a,

and

ti-facror·s

Extending this to rational numbers, for each m , 11 E a-n = m

,r-=

va111

and

a--n m

1N

1 a -II = ­ a"

we define

= -;;;- = --. II�

1

an

1

ya'"'

. �hapter 1: FUNDAMENTALS OF REAL ANALYSIS

22

Establish the following properties: a. b. c.

a; = (.ifQ)m for all

E lN. .

ni1 11

If m . 11 p I q E lN satisfy �: = �I then a� = a � . If r and s are rational numbers, then: 1. aras = ar+s and �: = ar-s 11. (ab)" = a"b'' and (�l = �;, m. (a")5 = a"·t. I

I

4.

SEQUENCES OF REAL NUMBERS

We start with the familiar definition of the convergence of a sequence.

to x E IR iffor eveJ)' E > 0 there exists a natura/ number no (depending on E) such that Definition

4.1. A sequence {x,} ofreal numbers is said to lx, - xl

<E

converge

for all n > no.

The real number x is called the limit ofthe sequence {x,}, and we write x, or x = lim,....00x,, or simply x = lim x,.

-+

x,

We shall say that the terms of a sequence {x,} of a set A satisfy a property (P) eventually, if there exists some natural number no such that x, satisfies the proP.,erty (P) for ali n > no. In this terminology, a sequence of real numbers {x,} converges to some real number x if and only if for each E > 0 the terms x, are eventually E-close to x. It should be clear that if lim x, = x, then lim y, = x for every subsequence {y,} of {x,}. ·

Theorem

4.2. A sequence ofreal numbers can have at-most one limit.

Assume that a sequence of real numbers {x,} satisfies x = lim x, and y = lim x,. Let E > 0. Then there exists a natural number no such that lx, -xI < E and lx, - yl < E for all n > no . Now, fix n > no. and use the triangle inequality to get Proof.

0 < lx - Y l for all E

> 0.

<

lx - x,l + lx, - Yl

This implies x =

< E + E = 2E

y, and the proof is finished.

•

Recall that a sequence of real numbers {x, } is said to be bounded if there exists a real number M > 0 such that l x,l < M for all n. A sequence {x,} of IR is said to be increasing if x, :S Xn + l for each n, and decreasing if Xn+ l < x, for all n. A monotone sequence is either an increasing or a decreasing sequence.

23

Section 4: SEQUENCES OF REAL NUMBERS

The symbolism x, t x means that {xn} is _increasing and x = sup{x, } . Similarly, x, t x means that {x,} is decreasing with x = inf{x, }. If a sequence {x,} satisfies x, = c for all n, then it is called a constant sequence.

Theorem 4.3. Every monotone bounded sequence of real numbers is conver­

gent.

Proof. Assume that {x, } is increasing and bounded. Since {x,} is bounded, it follows from the completeness axiom that x = sup{x, : n E IN} exists in JR. We claim that X = lim Xn. Indeed, if E > 0 is given, then by Theorem 3 . I there exists no such that x E < x,0 < x. Since (x, } is increasing, it follows that lx - x,l = x - x, < x - x,11 < E for all n > n0, and thus, lim x, = x. The proof -

•

for the decreasing case is similar.

Notice that the preceding theorem implies that an increasing sequence {x,} of real numbers satisfies x, t x if and only if x = lim x,. The basic convergence properties of real sequences are listed below.

I . Every convergent sequence is bounded. 2.

3.

If x, = c for each n, then lim x, = c. If the three sequences {x,}, { y,}, and { z,} of JR. satisfy x, < z11 < n, and lim x11 = limy, = x, then {z, } converges and limz, = x.

For the next properties assume that lim x11 = x and limy, =

y, for all

y.

4.

For each a , f3

5.

The sequence {x,y,} is convergent and lim(x11 y,) = xy. If l y, l > a > 0 holds for all n, and some 8 > 0, then (x,jy,} converges and limxnfy, = x fy. If X11 > Y11 holds for all n > no, then x > y .

6.

7.

E JR. the sequence (ax, + f3y,} converges and lim(ax, + f3y,) = ax + f3y.

A real number x is said to be a limit point (or a cluster point) of a sequence of real numbers {x,} if for every n E 1N and E > 0, there exists k > n (depending on E and n) such that lxk - x l < E . The limit points of a sequence are characterized as follows:

Theorem 4.4. Let {x11} be a sequence of real numbers. Then a real number

is a limit point for (x,} ifand only if there exists a subsequence (xk, } of {xn} such that lim Xkn = x.

x

Proof. Assume that x is a limit point of {x, }. Choose a natural number kt such that lxk1 x I < 1. Now, inductively, if k1, , k11 have been selected, then -

•

•

•

24

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS

choose k11+t > kn such that lxk,+1 - x l < 11�1 • Thus, we construct a sequence of natural numbers {kn} such that kt < k2 < · · · and lxk, - x l < � for all n. Clearly, {xk,} is a subsequence of {Xn} satisfying lim Xk, = x. For the converse, assume that a subsequence {xk, } of {xn} satisfies lim Xk,. x. Let m E W and E > 0 be fixed. It must be shown that there exists p > m such that lxp - x l < E. To see this, choose some no such that lxk, - xl < E for all n > no. Pick n > max{no, m}, and let p = k11 • Then p > m (since k11 > n) and • lxp - xI < E , and the proof is finished. =

Among the limit points of a sequence, the largest and the smallest ones are of some importance.

Let {xn} be a bounded sequence ofiR. Then the limit superior of {xn} is defined by Definition 4.5.

lim sup x11

=

qnd the limit inferior of {xn} by lim inf x11

=

[ ]

inf sup xk , II k ?c. ./1

]

supr infn xk · 11 Lk�

If we write supxk k�ll

00

=

Xk V k=ll

00

and kinf Xk = 1\ Xk, >ll k=ll

then the preceding formulas can be rewritten as follows:

Also, since V�n+ 1 Xk < V�n Xk and 1\�n Xk < 1\':::.n+ 1 Xk for each n, it follows that 00

00

Xk -1, lim sup X11 and 1\ Xk t lim infX11 • V k=n k=n

If {xn} is a bounded sequence, then lim infx11 and lim sup Xn are the smallest and largest limit points of {x11}. In particular,

Theorem 4.6.

lim inf X11 < lim supxn.

Section 4: SEQUENCES

OF REAL NUMBERS

Proof. Let { x, } be a bounded sequence of JR. Put s

25

= lim sup x,. We shall

show that s is the largest limit point of {x;, }. The other case can be shown in a similar manner. We show first that s is a limit point. To this end, let m E lN and E > 0. Since v�, Xk .!,, S, there exists ll > m such that s < V�, Xk < s + E. This implies the existence of some k > n > m such that s - E < Xk < s + E. Hence, s is a limit point of {x, } . To finish the proof, we show that s is the largest limit point. Let x be a limit point of { x,}, and let E > 0. Then for each n E lN, there exists m > n such that x - E < x, < x + E. It follows that x - E < Vf::, xk for each n, and so, x - E < /\� 1 V'f::, ;q = s for each E > 0. Thus, x < s .. and the proof is complete. • 0 The next result is known as the Bolzano-Weierstrass9· 1 theorem.

Corollary 4.7 (Bolzano-Weierstrass).

convergent subsequence.

Every bounded sequence of JR. has a

Proof. Let {x, } be a bounded sequence. By Theorem which by Theorem

4.4

4.6,

{x,} has a limit point is the limit of a convergent subsequence of {x, } . •

If lim x, = x , then x is the only limit point of {x, }, and hence, lim sup x, lim inf x, = x holds. The converse of this statement is also true, as the following theorem shows:

Theorem 4.8. A bounded sequence {x, } ofreal numbers converges ifand only

if lim inf x, = lim sup x, = x. In this case, lim x, = x.

We assume that lim infx, = lim sup x, = x and show that lim x, The inequalities

Proof.

X11 - X <

00

00

v Xk - 1\ Xk

k=11

k=n

and

X - X11 <

00

00

k=n

k=n

= x.

v Xk - 1\ Xk

9 Bemard Bolzano ( 1781-1 848), a Czech mathematician, philosopher. and theologian. He is well known as one of the early mathematicians who emphasized lhe need for a foundation of mathematics as well as for rigorous proofs. 1°Karl Theodor Wilhelm Weierstrass ( 1 8 15-1 897). a German mathematician and one of the mosl prominent mathematicians of the nineteenth century. He made numerous contributions to analysis and he is remembered for his belief that the "highest aim of science is to achieve general results."

26

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS

imply that lxll - X I < v�ll Xk - 1\�1/ Xk. Since

,1�� it easily follows that lim Xn

[V

k=ll

=

Xk -

A xk] = x - x = 0,

k=n

•

x.

A sequence {x,} in JR is said to be a Cauchy 1 1 sequence if for each E > 0 there exists no (depending on E) such that lxn - Xm I < E for all n , m > no. Clearly, a Cauchy sequence must necessarily be bounded. Also, it should be clear that every convergent sequence is a Cauchy sequence. The converse is also true, and it is expressed by saying that the real numbers form a complete metric space.

A sequence of real numbers converges if and only if it is a Cauchy sequence.

Theorem 4.9.

Proof. We have only to show that if {x11} is a Cauchy sequence, then {x,}

coQverges in JR. By Corollary 4.7 there exists a subsequence {xk, } of {x11 } such that lim xk, = x . Now, letE > 0. Choose no such that lxk, -xl < E and lx�� -xml < E for n, m > no. Now, if n > n0, then k, > n > n0, and so lx" - x l < lx" - .Yk, I

+ lxk,

- xl < E

+ E = 2E. •

Hence, lim X11 = x.

Now, let { /11} be a sequence of real-valued functions defined on a nonempty set X. Suppose that there exists a real-valued function g such that 1-",(x)l < g(x) for all x E X and all n. Then for each fixed x E X, the sequence of real numbers {/11(x)} is bounded. Thus, lim sup -",(x) and lim inf f,,(x) both exist in JR. Consequently, lim sup /11 and lim inf [,, of the sequence of functions { /11 } can be defined for each x E X as (lim sup [,, )(x) = lim sup /11 (x)

and

(lim inf /11 )(x)

=

lim inf fn (x).

1 1 Augustin Lou is Cau chy ( 1789-1 875). agreat French ma themat icia n. He was a hig hly crea tive and pro lific res earcher who made num ero us or iginal co ntributions to all fields o fmat hem atics o f his t im e and to ma them at ical p hys ics. His 1821 boo k Cours d'analyse is the firs t rigorous written trea tm ent o f ca lcu lus. Becaus e he was o n the co ns er va tive s ide dur ing the period o f the French revo lutio n he was denied, i n many o ccas io ns , the academ ic appo intm ents appro p riate for a s cientist o f his s ta ture.

27

Section 4: SEQUENCES OF REAL NUMBERS

EXERCISES 1. Show that if lx I < 1 , then lim x11 = 0. 2.

Show that lim :-.:11 = :c holds if and only if every subsequence of {:c11} has a subsequence that converges to :c. 3. Consider two sequences {kn} and {m11} of strictly increasing natural numbers such that {kt , k2, . . . } U {m 1 , m2, . . . } = IN. Show that a sequence of real numbers {x11} converges in JR. if and only if both subsequences {:ck, } and {xm ,. ) of {x11 } converge in JR. and they satisfy lim Xk. = lim x111,. (in which case the common limit is also the limit of the sequence). In particular, show that a sequence of real numbers {:en} converges in JR. if and only if the "even" and "odd" subsequences {.t211 } and {.tz11- t } both converge in JR. and they satisfy lim X211 = lim X211- t . 11 4. Find the lim sup and lim inf for the sequence { (- 1 ) } . 5. Find the lim sup and lim inf of the sequence {xu} defined by r

•

6.

1• -3 I-

·r2n - 31 ·r2n-l

•

and

for

n = 1 , 2. . . .

and

lim inf(-x11) = -lim supx11 •

If {x11} and {Yn} are two bounded sequences, then show that a. b.

8.

� + X211

Let {x11 } be a bounded sequence. Show that lim sup( -:c11) = -lim inf:c11

7.

X2n + l =

lim sup(x11 + Yu) :S lim sup .t11 + lim sup Yn , and lim in f.t11 + lim inf y11 • lim inf(.r11 + y11 )

>

Moreover, show that if one of the sequences converges, then equality holds in both (a) and (b). Prove that the lim sup and lim inf processes "preserve inequalities." That is, show that if two bounded sequences {xu} and {y11} of real numbers satisfy Xu :S Yn for all then

n > no,

lim infXu 9.

10.

< lim inf Y11

'<

lim sup y11 •

0). Show that lim ::fo = 1 (and conclude from this that lim .ifQ = 1 for each a 1 [HINT: Let � = 0. Using Bernoulli's inequality (see + x11, where x11 1 Exercise 13 of Section 2), we see that Jn = (1 + :c11) 1 2: 1 + x11 .t11 , and so 0 < .t11 < for each

>

)n

n.]

>

n >n

If {x11} is a sequence of strictly positive real numbers, then show that '-"n+ J . .mf ·-1 1m Xn

+l . f ..:t ��� ��� < 1'1m sup X --. < 1.1m m :c11 < 1'1m sup ..:/X11 n

Xn

Conclude from this that iflim x� , +n • exists in JR., then lim .::JX;; also exists and lim � = • Xn+l . lim :c,. • x:o+l ; · "0+2 · · /' for each [HINT: Note that X11 = Xno · no+ l - 11-1 · no The sequence of averages of a sequence of real numbers {:c11} is the sequence {a11 } . :c, defined by a11 = +x2t· . +.tn . If {x, } is a bounded sequence ·of real numbers, then ·

11.

lim sup x11

and

•

n > no .]

28

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS show that

12.

In particular, ifx, � x, then show that a, � x. Does the convergence of {a,} imply the convergence of {x,}? For a sequence of real numbers {x11} establish the following: a. b.

13.

14.

lim infx, :::: lim inf a, :S: lim sup a, < lim sup x11 •

If x,+l - Xn � x in JR., then x,jn � x. If {x,} is bounded and 2x, :S: x,+l + X11-l holds for all n Xn +l - Xn t 0.

=

2, 3, . . . , then

Consider the sequence {x, } defined by 0 < XI < I and x11+1 = I - JI n = I , 2, . . . . Show that x11 ,j, 0. Also, show that x:�.;1 � ! . Show that the sequence {x,} defined by

-

x, for

is a convergent sequence. Assume that a sequence {x,} satisfies lxn+l - x,l :::: alxn - Xn-1 1 for 17 = 2, 3, . . . and some fixed 0 < a < l . Show that {x,} is a convergent sequence. 16. Show that the sequence of real numbers {x,}, defined by

15.

XJ = I and x11+1 = 17. 18.

20.

22. 23.

for 17 = 1 , 2, . . . ,

�(

x11 +

�)

,

.

11 = I , 2, . . .

Show that {x11} converges and that lim x, = ..fi. Define the sequence x, = Lk= 1 t for n = l , 2, . . . . Show that {x,} does not converge in JR. [HINT: Show that x211 - x, � ! · ] Let -oo < a < b < oo and 0 < A < I . Define the sequence {xn} by XJ = a, x2 = b and

Xn+2 21.

3 + x,

converges and determine its limit. Consider the sequence {xn } of real numbers defined by x1 = l and x,+ 1 = I + 1j. ., x for n = 1 , 2, . . . . Show that {x,} is a convergent sequence and that lim x, = ../2. Define the sequence {x11 } by XJ = I and

Xn+J =

19.

I

=

AX11 + ( 1 - A)X11+1

for n = l , 2, . . .

.

Show that {xn} converges in JR. and find its limit. Let G be anonempty subset of JR. which is a group under addition (i.e., ifx, y E G, then x + y E G and x E G). Show that between any two distinct real numbers there exists an element of G or else there exists a.E JR. such that G = {17a: 17 = 0, ±I, ±2, . . . }. [HINT: If G i= {0}, let a = infG n (0, oo). ] Determine the limit points of the sequence {cos n}. [HINT: Considerthe set G = {n+2nm: n, m integers} and use the previous exercise.] For each n define f, : [- I , I ] � JR. by f, (x) = xn . Determine lim sup f, and lim inf J,,.

-

Section 5: THE EXTENDED REAL NUMBERS 24.

29

that every sequence of real numbers has a monotone subsequence. Use this conclusion to provide an alternate proof or the Bolzano-Weierstrass property of the real numbers: Every bo11nded seqLLence has a convergent subsequence. (See Corol­

Show

lary 4.7.)

5. THE EXTENDED REAL NUMBERS

The extended real numbers R* are the real numbers with two elements adjoint. The two extra elements are denoted by oo (or +oo) and -oo, read plus infinity and minus infinity. Thus, R* = R U { -oo, oo} or, as is customarily written R* = [-oo, oo]. The algebraic operations for the two infinities are defined as follows: I.

oo + oo = oo and (-oo) - oo = -oo; 2. (±oo) · oo = ±oo and (±oo) · ( -oo) = =foo; 3. x + oo = oo and x - oo = -oo for each x E R; 4. x · (±oo) = ±oo if x > 0 and x (±oo) = =fOO if x < 0. ·

The expressions oo - oo and -oo + oo are left (as usual) undefined. In this book we shall agree that 5.

0 . 00 = 0.

Also, R* is ordered, with oo the largest element, and -oo the smallest element. Moreover, 6. -oo < x < oo for each x E R. In topology, if R* is endowed with an appropriate topology, R* is referred to as the two-point compactification of R. It turns out that the usual tangent function tan: [-lf , I] � R*, where, of course, tan(-I) = -oo and tan(�) = oo, is a homeomorphism. One reason for introducing the extended real numbers is that in the theory of measure, one needs to consider sets with infinite measure. Another is that if {x11} is an unbounded sequence of real numbers, then by using Definition 4.5 one can see that lim supx11 and lim inf xll exist in R* (the values may be plus or minus infinity). Thus, every sequence of real numbers has a limit superior and limit inferior in R*. A sequence {x11 } of real numbers converges to oo (denoted lim X11 = oo) if for every M > 0, there exists no (depending on M) such that X11 > M for all n > no. Similarly, lim x11 = -oo means that for every real number !vi < 0, there exists n0 such that X11 < i'v/ for all n > no. Theorem 4.3 can now be formulated as follows; the proof is left for the reader. Theorem 5.1. Every increasing sequence of rea/ numbers either converges to

a real number or to plus infinity.

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS

30

Recall that for a sequence {x11} of real numbers, the series 2:;:1 x11 is said to be convergent, if the sequence of partial sums {L%=1 Xk } converges in JR. If the sequence of partial sums converges to infinity, then we write 2::1 x11 = oo and say that the sum of the series is infinite. The definition of 2::1 x11 = -oo is similar. Observe that by Theorem 5. I every series of non-negative real numbers converges in 1R*. A series 2::1 x, of real numbers is said to be rearrangement invariant if for every one-to-one and onto function cr: 1N � 1N (called a permutation of JN), the series 2::1 Xan converges, and moreover 2::1 X11 = 2::1 Xa, ·

If {x11} is a sequence of zwzz-l1egative real numbers, then the series 2::1 X11 is rearral1gemem invariant. Proof. Let cr: 1N � 1N be a permutation of JN. Set a = 2::1 x, and b = 2::1 Xan (note that both series converge in JR.*). To show that a = b, it is enough Theorem 5.2.

to

establish (in view of the symmctrj of the situation) that b � a. The latter is equivalent to showing that L� =l Xam S a holds for all zz. If n E JN, putk = max{crt , . . . , cr11 } and observe that I:�,= I Xam < L�= t x; < a . The proof is now complete. • In addition to series, double series with non-negative terms will appear from time to time in this book. If {a11,111} is a double sequence with 0 < a11.nr < oo for each pair zz, m , then for each fixed zz the series :L�= 1 a11•111 converges in 1R* (possible to +oo). The double series 2::1 L�=t a, , 111 is now defined by :x:>

( 00 ) �� L L

00

k

L L a11,m = k n=l m=l

11= 1

m=l

a,,m

·

Note that the limit to the right always exists in IR*. The following result on interchanging the order of summation holds: Theorem 5.3.

If 0 < a11,111 <

oofor all m and n, then

I: I: a, m = I: I: a,,/11 . 00

00

n=l m=l

00

.

00

m=l n=l

2::1 L�=t a11 , m .and b each k and p we have Proof. Set

k

p

a

=

p

k

L L a11.m = L L1 a ,m < 11= 1 m=l

m=l 11=

,

=

:L::=I 2::1 a,. m . Note that for

( ) L L p

00

m=l

11 = 1

oc

an.m L 1 11 m=l =

an m < L .

00

=

b,

Section 5: THE EXTENDED REAL NUMBERS

31

from which it easily follows that a < b. .A similar argument shows that b < a. Thus, a = b, and we are done. •

Theorem 5.4. Let 0 < an,m < oo for all m , n. If a : 1N -+ 1N x 1N is one-to­ one and onto, then

00

L au"

00

=

11=1

Proof. Let a = 2:: : , au" and b

(n; , m; ) . Then

kL Lk i=l

au; =

i=l

all;,lll; <

00

L L a11 ,m·

11=1 m=l

=

2::: , L�=• a,,111•

For each

kL ( Loo ) Loo ( Loo ) 111=1

i=l

all;.m

<

11=1

111=1

a,,/11

i,

let

=b

holds for each k, and so a < b. On the other hand, for each k and m, there exists n such that for each 1 < and 1 < j < m there exists 1 ::::: r < n with a,. = (i, j). Thus,

kL L Loo

a;

i


m

i= l j=l

ai,j <

r=l

au, < a,

which shows that b < a. Therefore, a = b, and the proof is finished.

•

The series 2::: , au" is usually called a rearrangement of the double series . . �oo . =• � m=l a11.m mto a smgIe senes.

�oo

���

EXERCISES

1. Let {x11 } be �sequence of JR.*.

2.

Define a limit point of {x11 } in JR.* to be any element x of JR.* for which there exists a subsequence of {x11 } that converges to x. Show that lim sup x11 and lim infx, (use Definition 4.5) are the largest and smallest limit points of {x11 } in JR.*. Let {x11 } be a sequence of positive real numbers such that e = lim x�:· exists in JR. Show that: a. b.

3.

if e < 1, then lim xll = 0, and if e > 1, then limxn = 00.

[HINT: See the hint of Exercise 10 of Section 4.] Let 0 � a11 .111 � oo for all m. n, and let a: JN x 1N' � JN x JN be one-to-one and onto. Show that 00

00

00

00

L L a11,111 = L L Gu(n,m)·

n=lm=l

n=l m=l

32

4.

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS Show that

1 ., = ""' ""' ., L- L- n- +m00

00

00.

1

n=l m=l

x

1). � [�, % 1. x

5. This exercise describes the p-adic representation of a real number in (0, ). We assume that p is a natural number such that p ::: 2 and E (0,

1)

[ 1 [�, k,;l )

...,

), Divide the interval [0, into the p closed-open intervals [0, ), e; , 1 ), and number them consecutively from 0 to p - Then belongs pre­ cisely to one of these intervals, say 1 (0 � 1 < p). Next divide the interval into p closed-open intervals (of the same length), number them con­ secutively from 0 to p - 1, and let be the subinterval to which belongs. Proceeding this way, we construct a sequence of non-negative integers such that 0 � < p for each 12. Show that

a.

k,

b.

k k k2 {k1 } � -k, · x = 11=L1 p---n 1] , ( r;l 1]. {m11} ""' m, . x L-

Apply the same process as in (a) by subdividing each interval now into p open­ closed intervals. For example, start with (0, and subdivide it into the open­ ], . . . , closed intervals (0, As in (a), construct a sequence of non-negative integers such that 0 � < p for each Show that

m

� ] , (�, %

n.

11

=

c.

x

00

n=l

p"

Show by an example that the two sequences constructed in (a) and (b) may be different.

x = O.k1 k2 · · · . \ f(A)

6.

In order to make the p-adic representation of a number unique, we shall agree to take the one determined by (a) above. As usual, it will be written as Show that P(IN) � JR. by establishing the following:

7.

is at most If is an infinite set, and � B is one-to-one such that B countable, then show that A � B. for which the dyadic (i.e., p = 2) ii. Show that the set of real numbers of (0, representation determined by (a) and (b) of the preceding exercise are different is a countable set. be the dyadic representation determined iii. For each E (0, ), let by part (a) of the preceding exercise; clearly, each is either 0 or 1 . Let 11 E IN: Show that (0, 1) � 'P(lN') is one-to-one such that � 'P(IN) . P(IN) is countable, and con�lude from part (i) that (0, . For a sequence of real numbers show that the following conditions are equivalent: i.

A

f: A

x

a. b. c. d. e.

1)

1 x = O.k 1 k2 · · · k11 = 1 }. f:

k;

{ f(x) =\ f((O, 1)) {xn} L:�1 x, L:�1 X L:�1 lxnl{s,} {-1, 1} s11xn L:�1 {xk,} {x11 } L:�1 Xk,

1)

The series is rearrangement invariant in JR. For every permutation a of IN the series converges in JR. The series converges in JR. For every sequence of the series converges in JR. of the series For every subsequence converges in JR. a"

33

Section 5: THE EXTENDED REAL NUMBERS f.

k

For every E > 0, there exists an integer (depending on E) such that for every finite subset s of IN with min s :::: k, we have I LueS x, I < E.

,L�1 x,

8.

(Any series satisfying any one of the above conditions is also referred to as an unconditionally convergent series.) A series of the form L�=• ( - I )"- 1 x,, where x, > 0 for each n, is called an alter­ nating series. Assume that a sequence (x,} of strictly positive real numbers satisfies � 0. Then establish the following:

x, a. b.

9.

,L�1

The alternating series (-I yr-l x, converges in JR. If ,L�1 = oo, then the alternating series ,L� 1 (- I )"- 1 x, is not rearrange­ ment invariant.

x,

This exercise describes the integral test for the convergence of series. Assume that [ I , oo) -r [0, oo) is a decreasing function. We define the sequences (a,} and (r,} by

f:

II

a, = L f (k ) k=l

and

r,

=

}{" f(x)dx. 1

Establish the following: a.

b. c.

10.

f(l)

for all n. the sequence {a, - r11 } is decreasing-and hence, convergent in JR. Show that the series 2::� 1 converges in JR if and only if the improper Riemann integral f100 = lim,_ 00 exists in JR.

f(k) f(x) dx

f[ f(x) dx

c.

�

= oo. Prove with (at least) three different ways that 2:: 1 If a computer starting at 12 midnight on December 3 1 . 1939, adds one million terms of the harmonic series every second, what was the value (within an error of I) of the sum at 1 2 midnight on December 31, 1997? (Assume that each year has 365 days.) "oo (-l)n- l I · ( Show that + , ) = In2. , = 11+ 1 + ,+2 + · · · 2

Ln=1

11�00

I

12

!;

I

a,

1 (Toeplitz) 2 Let (a, } be a sequence of positive real numbers (i.e., > 0 for each n) and put b, = = oo. If {x,} is a sequence of 1 a;. Assume that b, t ,L� 1 real numbers such that � in JR, then show that

,L�'=

x, x

a;

'La;x; = x. .

1 II lim n-oo b" I=

12.

1

Use the preceding exercise to show that the series 2::�1 ,P does not converge in JR for 0 < p :=:: I and converges in JR for all p > 1 . The following are exercises related to the harmonic series 2::�1 f,: a. b.

11.

0 < a11 - r, <

I

1 (Kronecker) 3 Assume that a sequence of positive real numbers {b,} satisfies 0 < b < b1 < b3 < · · · and b, t oo. If a series of real numbers

,L� 1 x11

1

0tto Toeplitz. ( 1881-1940). a Gennan mathematician. He made many contributions to analysis­

13 Leopold Kronecker ( 1823-1 891 ), a German mathematician.

especially to the theory of integral equations and operator theory. to unify (successfully) arithmetic, algebra, and analysis.

He devoted his entire life to attempting

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS

34

converges in IR, then show that

1

II

L b;x; 11-+ 00 bll 1=1 . lim

= 0.

In particular, show that if {Yn} is a sequence of real numbers such that the series 00 .lli " � 0· �n=1 11 converges in 1R' then Yl +··+v, n

[HINT: Put bo = 0, so = 0, and Sn = x1 + · · · + x11 for each and note that Lf=1 b; x; = 2::7=1 b;(s; - s;-1) = b11s11 - 2::7=2 s;-1 (b; - b;_J). Now use the preceding exercise.]

n > 1,

6. METRIC SPACES A metric

distance) d

on a nonempty set X is a function d: X x X satisfying the three properties : a.

b. c.

(or a

--+

JR.

d(x, y) > 0 for all x , y E X and d(x, y) = 0 <=::::> x = y; d(x, y) = d(y,x) for aii x , y E X; d(x, y) < d(x, z) + d(z, y) for all x , y, z E X (the triangle inequality).

The pair (X, d) is called a

metric space.

In a met ri c space (X, d) the inequality

ld(x, z) - d(y, z)l < d(x, y) holds for all po ints x, y, z E X . Indeed, from the triangle inequality it follows that d(x, z) < d(x, y)+d(y, z), and therefore d(x , z) - d(y, z) < d(x, y). Interchang­ ing x and y gives d(y, - d(x, z) < d(x, y), from which the inequality follows.

z)

Here are some examples of metric spaces. The reader should be able to veri fy

by himself that the exhibited functions satisfy the properties of a d i stan ce .

Example 6.1. The set of real numbers 1R equipped with the distance for all x, y E IR. Example 6.2.

The Euclidean 1 4 space IR" equipped with the distance

d(x, y) =

( ?= II

1= 1

)

(x; - y; )2

d(x, y) = !x - yl

•

�

for x = (x 1 , . . . , x,) and y = (Y1, . . . , y11 ) in 1R11 • This distance on 1R11 is called the • Euclidean distance.

Example 6.3. Let X be a nonempty set. }hen the function d defined by d(x, y) = I if x =I= y and d(x, x) = 0 is a distance on X. This distance is called the discrete distance on X, and X with this distance is called a discrete metric space. • 14Euclid (ca 365-300 BC), a famous Greek geome1er and lhe mos1 celebraled malhemalician of all lime. His name was a synonym for geometry umil lhe lwemielh cemury. Euclid's fame resls upon his classic work on geomelC)', The Elements, (wrinen in lhirteen books) lhal had a profound influence on lhe human mind and inquiry more lhan any other work excepl the Bible.

Section 6: METRIC SPACES Example 6.4.

Let X = (0, oo). Then

l �

35

d(x, y) = � - � for x, y E X is a distance on X.

X

y

•

If Y is a subset of a metric space (X. d), then Y equipped with the distance d also becomes a metric space. Now, let us fix a metric space (X. d). If x E X . then the open bal l at x with radius r > 0 is the set B(x, r) = {y E X: d(x. y) < r}. The open subsets of X can now be defined in the usual way. A subset A of X is called open if for every x E A, there exists some r > 0 such that B (x. r) c A. Every open ball B (x. r) is an open set. Indeed, if y E B (x. r ) . then the open ball B(y, r1). where r 1 = r - d(x. y) > 0, satisfies B(y , ri) c B(x, r). Reason: z E B(y, ri) implies d(x, z) < d(x, y) + d(y, < d(x, y) + r 1 = r, and so z E B(x, r).

z)

Theorem 6.5. t.

H.

ttt.

For a metric space (X. d) thefollowing statements hold:

and 0 are open sets. Arbitrary unions ofopen sets are open sets. Finite intersections of open sets are open sets. X

Proof. (i) Obvious. (H) Let

{A;};e/ be a family of open subsets of X . Let x E U A;. Then there exists some i E I such that x E A;. Since A; is open, there exists r > 0 with B(x .r) C A; C U A;. Hence, U A; is open. (iii) Let {A I . . . . . A,} be a finite collection of open sets. If X E n7= 1 A;. then for each 1 < i < n there exists r; 0 such that B (x. r;) c A;. Put r = min {r 1 , . . . . r,}, and note that B(.! . r) c n�'=1 A;. Hence, n�'= 1 A; is an >

•

open set.

A point x is called an interior point of a subset A if there exists an open ball B (x. r) such that B (x. r) c A. The set of all interior points of A is denoted by A0 and is called the interior of A; clearly, A0 c A. It is easy to see that A0 is the largest open subset of X included in A. Also, note that A is open if and only if

A = A0•

A subset A of a metric space (X, d) is called closed if its complement ( = X \ A) is an open set. The properties of closed sets are stated next.

Theorem 6.6. i. tt. iii.

For a metric space (X. d) the following statements hold:

and 0 are closed sets. Arbitrary intersections of closed sets are closed sets. Finite unions of closed sets are closed sets. X

Ac

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS

36

Proof. (i) The result follows from xe = 0. 0e = X , and Theorem 6.5(i).

(ii) Let {A; l ie/ be a family of closed sets. Then by Theorem 6.5 and De Morgan's law, we see that ( nie/ A; )e = Uie/ A f is open. Thus, nie/ A; is a closed set. • (ii) Combine ( u;'=1 A; )e = n;'=1 Aj with Theorem 6.5(iii). It should be observed that a set A is open if and only if Ae is closed; and similarly, A is closed if and only if A e is open. Observe that a set which is not open is not necessarily closed, and vice versa. A point x E X is called a closure point of a subset A of X if every open ball at x contains (at least) one element of A ; that is, B(x, r) n A i= 0 for all r > 0. The set of all_:losure points of A is denoted by A, and is called the closure of A; clearly, A c A. Theorem 6.7.

For el'ery subset A of a metric space, A is the smallest closed

set rhar inciudes A.

Proof. Let A be a subset of a metric space. We show that A is closed. Indeed,

if x ¢ A, then there exists an open ball B (x , r) such that B(x, r) n A = 0. If y E B(x , r), then [since· B(x, r) is an open set] there exists 8 > 0 such that B (y , 8) c B (x , r). Thus, B(y, 8) n A = 0. and so y ¢ A. Consequently, B(x, r) C (A)e, and so (A)e is open, which shows that A is closed. Now, if B is a closed subset such that A c B , then for every x E Be there exists an open ball B(x, r) c Be. Thus, B (x , r) n B = 0. and in particular, B(x, r) n A = 0. This shows that no element of Be is a closure point of A, and •

�refure A C B .

An immediate consequence of the preceding theorem is that a set A is closed if and only if A = A. Every set of the form A = {x E X : d(x, a) < r}, called the closed ball at a with radius r, is a closed set. Indeed, assumed(x , a) > r , andputr1 = d(x, a)-r > 0. If d(y, x) < r1, then

d(a, y)

>

d(a, x) - d(y, x) > d(a, x) - r1 = r,

which shows that A e is open, and thus, A is closed. Observe that in a discrete metric space, B(a, r) may be a proper subset of {x E X: d(x, a) < r}. However, in the Euclidean space IR11, the closure of every open ball of radius r is the closed ball of radius r. (Why?) Lemma 6.8.

If A is a subset ofa metric space, then

A0

=

(Ae )e .

Section 6: METRIC SPACES

37

Proof. Notice that x E A0

<==>

3r > O with B(x , r) C A <==> 3r > 0 with B(x. , r) n Ac ¢::=> X fj Ac ¢::=> X E (AC )c ,

=0

•

and the proof is finished.

A point x is called an accumulation point of a set A if every open ball B (x, r) contains an element of A distinct from x; that is, B(x, r) n (A \ {x}) :j:. 0 for each r > 0. Note thatx need not be an element of A. Clearly, every accumulation point of a set is automatically a closure point of that set. The set of accumulation points of A is called the derived set of A, and is denoted by A'. It should be clear that A = A U A'. In particular, it follows that a set is closed if and only if it contains its accumulation points. A sequence {x11 } of a metric space (X, d) is said to be convergent to x E X (in symbols, lim X11 = x, or X11 -+ x) if lim d(x11 , x) = 0. From the triangle inequality it easily follows that a sequence in a metric space can have at most one limit. (See the proof of Theorem 4.2.) The next theorem characterizes the closure points of a set in terms of sequences.

Theorem 6.9. Let A be a subset of a metric space (X, d). Then a point x E X

belongs to A ifand only ifthere exists a sequence {x11} ofA such that lim X11 = x. Moreover, ifx is an accumulation point of A, then there e;'Cists a seqllence of A with distinct terms that converges to x. Proof. Assume that x belongs to the closure of A. For each n pick X11 E A

such that d(x. x,) < � - Then {x.,} is a sequence of A such that limx, = x. On the other hand, if a sequence {x11} of A satisfies lim X11 = x, then for each r > 0 there exists some k such that d (x, X11) < r for n > k. Thus, B (x, r) n A =j:. 0 for each r > 0, and so x E A. Next, assume that x is an accumulation point of A. Start by choosing some x1 E A such that x1 =j:. x and d(x, x1) < 1. Now, inductively, if x 1 , . . . , X11 E (A\ {x}) have been chosen, pick X11 + 1 E A \ {x} such that d (x, x,+ 1 ) < min { 11 1 , d (x , X11 )}. Then {x11} is a sequence of A satisfying x11 :j:. X111 if n :j:. m and lim x, = x. •

�

A subset A of a metric space (X, d) is called dense in X if A = X . According to Theorem a set A is dense in X if and only if for every x E X there exists a sequence {x,} of A such that lim X1 = x. Also, notice that a set A is dense if and only if V n A =j:. 0 holds for each nonempty open set V .

6.9.

1

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS

38

A point x E X is called a boundary point of a set A if every open ball of x contains points from A and Ac; that is, if r) n A # 0 and B(x. r) n Ac # 0 for all r > 0. The set of all boundary points of a set A is denoted by a A and is called the boundary of A. By the symmetry of the definition, a A = aAc holds for every subset A of X. Also, a simple argument shows that

B(x,

We now introduce the concept of continuity.

p)

Definition 6.10. Aftmction f: (X, d) -+ (Y, between two metric spaces is said to be continuous at a point a E X iffor every E > 0 there exists o > 0

(depending 011 E) such that p(f(x), f(a)) < E whenever d(x, a) < 8. The function f is said to be continuous on X (or simply continuous) iff is continuous at every point of X.

The next theorem presents the most useful characterizations of continuous func­ tions.

p)

Forafunction f: (X, d) -+ (Y, bern•een two metric spaces, the following statements are equivalent: 1. f is continuous on X. f- 1 (0) is an open subset of X whenever 0 is an open subset of Y. n. If lim x holds in X, then lim f(x ) = f(x) holds in Y. tv. f(A) c f(A) holdsjor eve1y subset A of X. v. f- 1 (C) is a closed subset of X whenever C is a closed subset ofY. Proof. (i) ==::} (ii) Let 0 be an open subset of Y and a E f- 1 (0). Since f(a) E 0 and 0 is open, there exists some r > 0 such that B(f(a), r) c 0. Now by the continuity of f at a, there exists o > 0 such that d(x, a) < 8 implies p(f(x), f(a)) < r. But this shows that B(a, o) c f- 1 (0). Therefore, a is an interior point of f- 1 (0), and hence, f- 1 (0) is open. (ii) ==::} (iii) Assume limx, = x in X and r > 0. Let V = B(f(x), r). By our assumption f- 1 (V) is an open subset of X, and since x belongs to it, there exists some 8 > 0 such that B(x, o) c f- 1 (V). Pick some k such that x, e B(x, o) for all n > k. Then f(x,) E V for all n > k, which shows that lim f(x,) = f(x). (iii) ==::} (iv) Let A be a subset of X. If y E f(A), then there exists x E A such that y f(x). Since x E A, there exists (by Theorem 6.9) a sequence {x,} of A such that limx, = x. But then, {f(x,)} is a sequence of f(A), and by assumption lim f(x,J f(x) = y. By Theorem 6.9 it follows that y E /(A); that is, f(A) c f(A). (iv) ==::} (v) Let C be a closed subset of Y; clearly, C = C holds in Y. Applying our assumption to the set A = f- 1 (C), we get f(A) c f(A) c C = C, which Theorem 6.11.

x, =

m.

=

,

=

Section 6: METRIC SPACES

39

A c !-tee)!-=t A. A A A = A, Aa = ee)E > 0. e = [B(f(a), E) ]c = { p(f(a), > E} . By hypothesis, !- t ee) is a closed subset of Since a rf. f-1(e), there exists some 8 > 0 such that B(a, 8) c [f-1(e)]c. But then, if d(x, a) < 8, then p(f(x), f(a)) < E holds, so that f is continuous at a. Since a is arbitrary, f is continuous on The proof of the theorem is now complete. : • shows that Since C· is always true, it follows that which shows that is a closed subset of X . (v) =? (i) Let E X and Consider the closed set y)

y E Y:

X.

X.

From statement (iii) it should be clear that the composition of two continuous functions between metric spaces (whenever it makes sense) must be a continuous function. Two metric spaces (X, and (Y, are called homeomorphic if there exists are both a one-to-one onto function (X, --* (Y, p) such that and continuous. Two distances and on a set X are called equivalent if a sequence of X satisfies lim if and only if lim By the preceding theorem, for this to happen it is necessary and sufficient for the identity mapping / : (X, --* (X, to be a homeomorphism. Rephrasing this last statement, and p are equivalent if and only if and generate the same open sets. such A metric space (X, is called bounded if there exists a number M that y) M for all y E X. The diameter of a subset of a metric space (X, d) is defined by

d)

p) f: d) f f-1 d p {x,} d(x,, x) = 0 p(x,, x) = 0. d) p) d d p 0 d) > d(x, < x, A d(A) = sup{d(x , x, A}. Thus, d) is bounded if and only if the diameter of is finite. If d is a distance on a set then the function p defined by p(x, = d(x,d(x, is also a distance on Moreover, p) is bounded and p is equivalent to d. We now tum our attention to complete metric spaces. A sequence {x,} of a metric space d) is called a Cauchy sequence if for every E > 0, there exists (depending on E) such that d(x,, Clearly, every E for all n, m > < convergent sequence is a Cauchy sequence. However, in general, the converse is not true. As an example, let = (0, oo) with distance d(x, = lx and x, = � for each Then {xn} is a Cauchy sequence that does not converge in y):

y E

(X,

X

X,

y)

y)

X.

1+

y)

(X,

(X,

no

Xm)

X

no.

y)

-

yl,

n.

X.

If a metric space has the property that all of its Cauchy sequences converge (in the space), then the metric space is called a complete metric space. Examples

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS

40

of complete metric spaces are provided by the Euclidean spaces lR" with their Euclidean distances. Recall that according to Theorem 4.9 the real numbers form a complete metric space. Here is another important example of a complete metric space.

Let n be a nonempty set We shall denote by B (D.) the set of all real­ valued functions defined on n that are bounded. That is, a function f: n ---+ IR belongs to B(Q) if and only if there exists a n umber M > 0 (depending on f ) such that 1/(w)l !S M for all w E n. Now for each f, g E B(Q) define

Example 6.12.

D(f, g) = sup {lf(w) - g(w)l: w E Q}. Note that since both

f and g are bounded, D( f, g) is a real number. We claim that D is a

distance on B(Q) and that, in fact, B(Q) is complete (with this distance). To see that D is a distance, only the triangle inequality is verified; the other two properties are trivial. Indeed, if f, g, h E B(Q), then for each w E n we have

l f(w) - g(w)l .!S lf(w) - h(w)l + lh(w) - g(w)l .!S D(f, lz) + D(h, g), and so D(f, g) < D(f, h) + D(h, g).

Next, we shall establish the completeness of B(Q). To this end, let {f,,} be a Cauchy sequence of B(Q). Then, given E > 0 there exists no such that D(f,. fm ) < E for all n, m > no. In particular, note that for each w E n, the inequality l f,(w) - fm (w)l .!S D(f,,. fm ) implies that {f,(w)} is a Cauchy sequence of real n umbers. Thus, {f,(w)} converges in IR for each w E n; let f(w) = lim f,(w). lt easily follows from the inequality l f,(w) - fm (w)l < E for all n. m > no that IJ,, (w) - f(w)l < E for all n > llO and all w E D.. This last inequality in tum implies that f E B(Q) and D(f,,. f) < E for all u > no. Hence, lim f, = f, and so B(Q) is complete. •

Closed subsets of complete metric spaces are complete metric space in their own right.

A

Let (X, d) be a complete metric space. Then a subset of X is closed if and only if (with metric d) is a complete metric space in its own right. Theorem 6.13.

A

A

A,

Let be closed. If {x,} is a Cauchy sequence of then {x,} is a Cauchy sequence of X. Since X is. complete, there exists x E X such that limx, = x. But since is closed, x E A. Thus, d) is a complete metric space. Conversely, assume that d) is a complete metric space. If a sequence {x,} of satisfies lim x, = x in X, then {x,} is a Cauchy sequence of X. But then, {x,} Proof.

A

A

(A,

(A,

Section 6: METRIC SPACES

41

is a Cauchy sequence of A, and hence, it must converge to a unique element of A. This element must be x. Thus, x E A, so that A is a closed subset of X. • The following important result dealing with complete metric spaces is due to G. Cantor. (Keep in mind that the diameter of a set A is defined by d(A) = sup{d(x, y): x , y E A}.)

Theorem 6.14 (Cantor). Let (X, d) be a complete metric space and let {A,}

be a sequence of closed, nonempty subsets of X such that An+ I A11 for each n and lim d(A,) = 0. Then the intersection n;:1 A, consists precisely of one element.

c

Proof. If X . y E n: I A,, then

X.

y E A, for each 11, and hence, 0 <

d(x, y) < d(A11 ) for each n. Thus, d(x , y) = O, n :,1 A11 contains at most one element.

so that x = y. This shows that

To show that n: A11 =j:. 0. proceed as follows. For each n, choose some 1 x, E A,. Then, it is easy to see that d(x,+p• x,) < d(A11) holds for each n and p, and from this it follows that {x,} is a Cauchy sequence of X. Thus, there exists X E X such that lim x, = X. and we claim that X E n:: I A,. Indeed, since Xm E A, for m > n, we get x E A11 for each n. But since each A11 is closed, A11 = A, holds; therefore, x E A, for each n, and we are done. •

A subset A of a metric space (X, d) is said to be nowhere dense if its closure

has an empty interior; that is, if (A)0 = (/). Since = holds for every subset B, it is easy to see that a subset A is nowhere dense if �d only if (A)c is dense in X. A classical nowhere dense subset of the real line is the so-called Cantor set. Because we shall use it later, we pause for a while to describe this set and its properties.

B0 (Bc)c

Example 6.15 (The Cantor Set).

as follows:

The Cantor set is a subset of

[0, l ] and is constructed

Let Co =

Ct =

[0, 1]. Then trisect [0, 1] and remove the middle open interval
Next, trisect each closed interval of C 1

and remove from each one of them the middle

open interval. Let C2 be the set remaining from C

[0, �] U [3, !J U [�. �] U [&. 1]�

intervals.

=

1 after these removals.

That is, Cz

=

note that Cz is the union of 22 = 4 disjoint closed

The (inductive) process of constructing C,+l from C, should be clear now. Trisect each

of the 2" disjoint closed intervals of C,, and remove from each one of them the middle open interval. What is left from C, is then Cn+l· Note that C,+l is the union of2"+ 1 disjoint

closed intervals. The graphs of the

first few constructions

are shown in Figure

1.1.

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS

42

Co o CI

0 ... J_

c2 -. o l c3

. . .......-...........

-·- .. ..

9

2

T

. - .....................-........ .._........ 1 2 L 2 l � 9 3 3 9 9 3

.... .

..

-··-.........- ..

.................................

- .-........_._ .

FIGURE 1.1. The Construction of the Cantor Set

Clearly, C11 +1 c C11 holds for al1 11. The Cantor set of [0, I] is now defined by C = n� 1 Cn . Next, we mention the most interesting properties of the Cantor set C. C subset I. Clearly, C is closed as it is an intersection of closed sets. It should also be clear from the above co st .action that C does not contain any interval and t1-!us, C has empty interior, i.e., is a nowhere dense set. [0, I ] C 2. 1 11 To see this, note that at the n1h step we remove 2 - open intervals each of which has length 3 -II; therefore, a total of 211 - 1 3 -II. Thus, we remove altogether a total 1 ! "'00 ( � 11 length of L11= "'00 I 211- 1 3-11 = 2 L11= I 3 )

The set is a closed 1 owbere dense n

ofJR..

n

The totallengtb oftbe removed il11ervalsfrom

to get equals o11e.

•

C � JR.

•

3.

=

The set bas cardinalit y t b at is, Perhaps the simplest way of proving this is by showing that {0, I}. Since [the mapping a g(a) one-to-one from onto and (see Exercise C

c,

·

P(.IN) � 2IN = {a11 } 1-+ 2IN 'P(.IN)] 'P(.IN) � JR. will follow that C � JR. The details are included below.

where 2 = {n E .IN: a11 = I } is 6 of Section 5), it

C � 2N,

=

If = {xn} E 2IN (i.e., each x11 is either 0 or I), then let y11 = 2x11 for each n, and n = L:� 1 3- Y11 · Clearly, each )'11 is either 0 or 2. Note first that E C. define IndeecJ,· since Yl =I= I , we have (j-, j); similarly, since )'2 =1= I, we can see that �, �) U �, � . By induction, we can verify that does not belong to any of the removed open intervals. Thus, E C which shows that is a mapping from 2IN into C . Now, we claim that is one-to-one. Indeed, if = {a11 } =J: b = (b11 } holds, then let k = min{n E .IN: 11 =I= b11 }; assume that bk = I and = 0. Then, in view of 2 L�k+l 3-11 = 3-k, we have

xf(x)

f(x) rt, (

f(b)

f(x) rt, ( ) f(x) x 1-+a f(x)

=

2

00

L

11=1

b113 -n 2: 2

k 1 -

L

11=l

11 b113- + 2 · 3-k

f(x)x 1-+ f(x) a ak >

2

L an 3-11 = 00

11=1

f(x)

f(a).

Finally, it is not difficult to see that C consists precisely of all numbers of [0, 1] with a triadic representation (see Exercise 5 of Section 5) assuming the values 0 or 2. This implies that is onto, proving that C � JR. •

x f(x) 1-+

Section 6: METRIC SPACES

43

A subset Y of a metric space is said to �e meager (or of first category) if there exists a sequence of nowhere dense subsets such that Y = U�1 A 15 metric space is called a Baire space if every nonempty open set is not a meager set. The next result characterizes the Baire spaces.

{An}

A,.

Theorem 6.16. For a metric space X thefollowing statements are equivalent: I. X is a Baire space.

2. Eve1y countable intersection of open dense sets is also dense. 3. If X = U:. 1 F, and each F, is a closed set, then the open set U:1 (Fn)0 is dense. Proof. (1) => (2) We shall use the following property of open dense sets: If B is an open dense set, then its complement Be is nowhere dense. (This follows = 0.) from Lemma 6.8 by observing that (B )o = ( Bc y c = ( B ) = Assume now that X is a Baire space and let {A,} be a sequence of open dense subsets of X. To show that A = n: 1 A11 is dense in X, we need to show that An 0 # 0 for each nonempty open set 0 of X. Suppose, if possible, that A n 0 = = Ac U oc, and thus 0 for some nonempty open set 0. Then X = (A n

c

c xc

·-

O)c

But this shows that 0 is a meager set, and thus, it is empty by hypothesis, contrary to the choice of 0. Hence, is dense in X. and (2) ==> (3) Let be a sequence of closed sets satisfying X = U:1 consider the open set 0 = U�1 For each n, let E = and note \ that E, is a nowhere dense closed set. In particular, the set E = U�1 E11 is a meager set. Since En is a closed nowhere dense set, it easily follows that each (E1 c is an open dense set. So, by our hypothesis, Ec = E )C is also a dense set. Now notice that

{F,}

A

(F,)0•

n

Fn (Fn)0

1)

n:l ( I/

ac = X \ 0 = and so Ec desired.

c

00

00

u F/1 \ u (FI/)0 n=l II=I

00

c UrFI/ \ (FI/)0]

Fn

=

E,

0. Since Ec is dense, we easily conclude that 0 is also dense, as

15Rene Louis Baire ( 1 874- 1 932), a French mathematician. Among other topics, he studied exten­ sively the properties of the pointwise limits of sequences of continuous functions.

44

·

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS

(3) ==> (1) Let V be a nonempty open set. If V is a meager set, then V can be written as a countable union v = u:) A,, where (A,)0 = 0 for each n. Then

is a countable union of closed sets, and so by our hypothesis the open set

vc, we see that vc is also dense in X. In particular, is dense in X. From (Vc)o we have V n. vc # 0, which is impossible. Hence, V is not a meager set and so X is a Baire space. •

c

The next result is known as Baire 's category theorem, and it plays an important :n nnnl y.,.: s

��1 ...

I Vl \..o J.

UJ 4LI

131

•

Theorem 6.17 (Baire). Every complete metric space is a Baire space. Proof. Let {A,} be a sequence of open and dense subsets of X. According to

Theorem 6.16, we have to show that the set A = n: I A, is dense in X . Conse­ quently, if x E X and r > 0 are given, we have to establish that B(x, r) n A # 0. Let C(a, r) = {x E X : d(x, a) < r } . Since A 1 is open and dense in X, there exists x1 E X and 0 < r1 < 1 such that C(X] , I" J ) B(x , r) n A ) . Now, inductively, if X ] , , x, and I"] , . • • , r, have been selected, choose x,+1 E X and r,+1 < ,!1 such that C (x,+1 , r,+1) B(x, , r,) n A,+1• Thus, there exists a sequence {x, } of X and a sequence {r, } B(x,, r,) n A,+1 of real numbers such that 0 < r, < �. and C (x + 1 , r,+J ) for each n. Now put C, = C(x, , r,) for each n. Then each C, is nonempty and closed, and Cn + l c C, holds for each 11. Moreover, d(C,) .:S 2r, < � implies lim d (C, ) = 0. Thus, by Theorem 6.14, there exists y E n: 1 C,. But then, it easily follows that y E B (x, r) n A, and we are done. •

c

•

,

•

•

c

c

A special case of the preceding theorem with many useful applications is stated next.

Theorem 6.18. lf(X, d) is a complete metric space and X = U: 1 A,, then (A, )0 # 0 for some n. A function f: {X, d) -+ (Y, p) between two metric spaces is called uniformly continuous if for every E > 0 there exists some > 0 (depending on E) such that p(f(x), f(y)) < E whenever d(x, y) < 6. Clearly, every uniformly continuous function is continuous.

8

Section 6: METRIC SPACES

45

If X = (0, 1 ] and Y = JR., both with the distance d (x, y) = ! x y I, then the function f: X � Y, where f (x) = x 2 , is uniformly continuous, while g: X � Y, with g(x) = x-1 , is continuous but not uniformly continuous. For the proof of the next theorem the following simple result is needed: -

•

In a metric space (X, d), if lim x, = x and lim y,

=

y, then

lim d(x, , y,) = d(x , y).

11-+00

To see this, note first that by the triangle ineqmt.lity _ we have ! d(x, z) - d(z, y) ! d(x, y), and then use the chain of inequalities ! d(x,, y,) - d(x, y )I < ! d(x,, y, )

<

- d(x, y,)! + !d(x, y,) - d(x, y)!

< d(x11 , x) + d(y11, y),

to establish the validity of the statement. Theorem 6.19. Let A be a subset of a metric space (X, d) and let (Y, p) be a

complete metric space. If f: A � Y is a uniformly continuous function, then f bas a unique uniformly continuous extension to tlte closure A of A. Proof. The uniqueness of the extension should be clear. We only have to establish its existence. Let x E A. Then by Theorem 6.9 there exists a sequence {x11} of A such that lim x, = x. We shall show that { f(X11)} is a Cauchy sequence of Y. To this end, let E > 0. By the uniform continuity of f, there exists > 0 such that p(f(x), f(y)) < E whenever x , y E A satisfy d(x, y) < Pick no such that d(x,, Xm) < for all n, m > no. Thus, p(f(x11), f(x11,)) < E for all n , m > no; that is, {f(x11)} is a Cauchy sequence. By the completeness of Y, there exists y E Y such that lim f(xn) = y. If now {y11} is another sequence of A such that limy, = x, then by the above {f(y11)} converges in Y. Let lim f(y11) = Ll . Now, for each n define z2" = X11 and Z2n-l = y11• Note that {z,} is a sequence of A such that lim z, = x. Thus, lim f(z,) exists in Y. In particular, we have lim f(z2,) = lim f(z211-J); that is, y = u. Therefore, lim f (x,) is independent of the choice of the sequence {x11} of A (as long as it converges to x). Now, define f*: A � Y by f*(x) = lim f(x,,), where {x11} is a sequence of A such that lim x11 = x. Clearly, f*(x) = f(x) for all x E A. To finish the proof, we show that f* is uniformly continuous. If E > 0, choose > 0 such that p(f(x), f(y)) < E whenever x, y E A satisfy d(x, y) < Now if x, y E A satisfy d(x , y) < choose two sequences {x,} and {y,} of A such that lim x11 = x and lim y11 = y. By the discussion before the theorem, we have lim d(x,, y, ) = d(x, y). Pick no such that d(x , y,) < for

8.

8

8.

8

8

8,

,

8

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS

46

all n

> no. Then p(f(x,), f(Yn)) < E for n > no, and so, by the same remark • p(f*(x), /*(y)) < E , which shows that f* is uniformly continuous on A.

A function f: (X, d) -+

(Y, p) between two metric spaces is called an isometry if p(f(x), f(y)) = d(x, y) holds for all x, y E X. If in addition f is onto, then (X, d) and (Y, p) are called isometric. Note that two isometric metric spaces

are

necessarily homeomorphic. Also, observe that every isometry is a uniformly continuous function. A complete metric space (Y, p) is called a completion of a m�tric space (X, d) if there exists an isometry f: (X, d) -+ (Y, p) such that f (X) is dense in Y. If we think of X and /(X) as identical, then X can be considered as a subset of Y. Any two completions of a metric space (X, d) must be isometric. Indeed, if (Y1 , p1) and (Y2 , P2) are two completions of (X, d), then there exist two isometries 1 f: X -+ Y1 and g: X -+ Y2• Thus, g is an isometry from X onto g(X) and f is an isometry from f(X) onto X. Then h = g o f- 1 is an isometry from /(X) onto g(X) c Y2. Since f(X) is dense in Y1 , h is uniformly continuous and Y2 is complete, it follows from Theorem 6. 1 9 that there exists a uniformly continuous extension h* of h to all of Y1• It is now straightforward to show that h* is an isometry from Yt onto Y2. Observe that if f: (X, d) -+ (Y, p) is an isometry and (Y, p) is complete, then the closure /(X) is a completion of X (since, by Theorem 6.13, /(X) is a complete metric space). Next, we shall use this observation to establish that every metric space has a completion. Theorem 6.20.

tion.

Every metric space has a unique (up to an isometry) comple­

Proof. let

(X, d) be a metric space. Fix an element a E X. For each x E X, let f�:: X --7 IR be defined by f�:(Y) = d(x , y) - d(y, a) for each y E X. From the triangle inequality it is easy to see that l f\· (Y) I < d(x, a) for each y E X, and this shows that f\· is a bounded function for each x E X. That is, f�: E B(X) for each x E X; see Example 6.12. We have established, therefore, a function f: X -+ B(X) by x � f�:· We claim that f is an isometry. (Recall that the distance on B(X) is given by D(f, g) = sup{ l f(x) - g(x)l: x E X }.) Indeed, note first that if x, z E X, then I f�: ()') - f: (Y) I

=

ld(x , y) - d(y , a) - [d(z , y ) - d(y , a) ]l

= l d(x , y) .....:. d(z, y)l < d(x, z )

holds for all y E X. On the other hand, lf\. (z) - f:(z)l = d(x, z). Hence, D (fx ! f :) = sup { lf�: (Y) - f: (Y ) I: Y

E X} = d(x, z).

Section 6:

47

METRIC SPACES

Since (B(X), D) is a complete metric .space (see Example 6.12), we see that (/(X), D ) is a completion of (X, d). The uniqueness of the completion (up to an isometry) was established in the • discussion before the theorem. The proof is now complete.

EXERCISES 1.

For subsets A and 8 of a metric space (X, d), show that:

(A n 8)0 = A0 n 8°. b. A0 U 8° C (A U B)0• c. A U 8 = A U B . d. A n B c A n B . e. If 8 is open, then A n 8 c A n 8 . a.

2.

Show that in a Euclidean space JR." with the Euclidean distance, the closure of any open ball B(a, r) is the closed ball {x E JR.": d(x, a) < r}. Give an example of a complete metric space for which the corresponding statement is false.

3.

If A is a nonempty subset of JR., then show that the set

8 = {a E A : There exists E > 0 such that (a, a + E) n A = 0}

4. 5.

6.

is at most countable. Let

f: (X, d) � (Y, p)

be a function. Show that

f

is continuous if and only if

J-1(8°) c (f-1(8))0 for every subset 8 of Y. Show that the boundary of a closed or open set in a metric space is nowhere dense. Is this statement true for an arbitrary subset?

Show that the set of irrational numbers is not a countable union of closed subsets of

JR. 7.

[HINT: Use Baire 's theorem.} Let (X, d) be a ·metric space. Show that if {x,} and then

8.

{d(x,, y,)} converges in JR.

10.

are Cauchy sequences of X,

Show that in a metric space a Cauchy sequence converges if and only if it has a convergent subsequence.

9.

{y,}

Prove that the closed interval [0,

1]

is an uncountable set:

6.14, and

a.

by using Cantor's Theorem

b.

by using Theorem

Let

{r1 , r1, . . . } be an enumeration of all rational numbers in the interval [0, and E [0, 1} let Ax = {n E N: r, :5 x }. Define the function f: [0, 1} � 1R by

6.18.

1]

for each x

the formula

1 f(x) = � - . L...t 2" neA.r

f restricted to the set of irrational numbers of [0, 1] is continuous. 11. This exercise concerns connected metric spaces. A metric space (X, d) is said to be connected whenever 0 and X are the only subsets of X that are simultaneously open and closed. A subset A of a metric space (X, d) s i said to be connected whenever Show that

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS

48

(A, d) is itself a connected metric space. Establish the following properties regarding connected metric spaces and connected sets. a.

b. c.

d. e.

f.

g.

12.

13. 14. 15.

A metric space (X, d) is connected if and only if every continuous function f: X ---+ is constant, where the two point set 1 } is considered to be a metric space under the discrete metric. If in a metric space (X, d) we have B � A c X, then the set B is a connected subset of (A, d) if and only if B is a connected subset of (X, d). Iff: (X, d) ---+ (Y, p) is a continuous function and A is a connected subset of X, then f(A) is a connected subset of Y. If {A; }; e/ is a family of connected subsets of a metric space such that ni e/ A; =/= 0. then U; e/ A; is likewise a connected set. If A is a subset of a metric space and a E A, then there exists a largest (with respect to inclusion) connected subset Ca of A that contains a. (The connected set Ca is called the component of a with respect to A.) If a, b belo!lg to a subset A of a metric space and Ca and Cb are the components of a and b in A, then either Ca = Cb or else Ca n Cb = (/J. Hence, the identity A = Uae A Ca shows that A can be written as a disjoint union of connected sets. A nonempty subset of JR. with at least two elements is a connected set if and only if it is an interval. Use this and the conclusion of (f) to infer that every open subset of JR. can be written as an at-most countable union of disjoint open intervals.

{0, 1}

{0,

Show that JR.n with the Euclidean distance is a connected metric space. Use this conclusion to establish that, if the intersection of two open subsets of JR.n is a proper closed set, then the two open sets must be disjoint. Let C be a nonempty closed subset ofJR. Show that a function f: C ---+ JR. is continuous if and only if it can be extended to a continuous real-valued function on JR. Show that a metric space is a Baire space if and only if the complement of every meager set is dense. A subset of a metric space is called co-meager if its complement is a meager set. For a subset A of a Baire space show that: a. b.

A is co-meager if and only if it contains a dense G 0-set. A is meager if and only if it is contained in an Fa -set whose complement is dense.

7. COMPACTNESS IN METRIC SPACES We shall discuss in this section the basic properties of compact sets. A family {A;};e / of subsets of a set X is said to cover a subset A of X if A c U;e/ A;. If a subfamily of {A; };e/ also covers A , then it is called a subcover. If (X, d) is a metric space, then any cover of a set consisting of open sets will be called an open cover of the set. Open covers of subsets of the Euclidean spaces JR." can always be reduced to countable ones, as the next classical result of E. Lindel0f16 shows. 16Emst Leonhard Lin del of ( 1 870-1940), a Fin mathematician. He contributed to function theory and he is known for his theorems on the existence of solutions to differential equations.

Section 7: COMPACTNESS IN METRIC SPACES

49

Every open cover of a subset ofJR" can be reduced to an at-most countable subcove1: Theorem 7.1 (Lindelof).

Call a point a = (a 1 , , a11) E JR." rational if each component a; is a rational number. Let A be a subset of JR." and { CJ } ; e / be an infinite open cover of A, i.e., A C U;e/ CJ; . Now, for each x E A choose first an index ix E:: I such that x E 0; , and then pick a rational point ax E JR." and a rational positive number rx such that x E 8(ax, l"x) CJ; , . Then the collection {8(ax, rx): x E A } is an at most countable (why?) open cover of A . Since each 8(ax, rx) is a subset of some CJ;, it easily follows that there exists an at-most countable subcover of {_0; };e/ for A. • Proof.

•

•

•

..

C

And now we are ready to introduce the important notion of compactness. Definition 7.2. Let (X, d) be a metric space. A subset A of X is said to be compact if every open cover of A can be reduced to a finite subcover. If X is itself a compact set, then (X, d) is referred to as a compact metric space.

The concept of compactness is of fundamental importance in analysis and in mathematics in general. The next result characterizes the compact sets in metric spaces and gives an indication of the usefulness of the compact sets. Theorem 7.3. For a subset A ofa metric space (X, d) tbefollowing statements

are equivalellt: 1 . A is a compact set. 2. Every infinite subset of A has an accumulation point in A . 3. Every seqnence in A ltas a subsequence which converges to a point of A .

Proof. (1) => (2) Let S be an infinite subset of the compact set A. Assume

by way of contradiction that S has no accumulation point in A. Thus, for every x E A there exists some rx > 0 such that 8 (x, rx) n (S \ {x}) = 0. Note that B(x' r_.;) n s c {x }. Clearly, A c Ux eA 8 (x' fx) holds, and, in view of the compactness of A , there exist x 1 , , x11 E A such that U7= 1 8 (x; , rx)· But then, •

s = An

s

.

.

AC

II

c

u [8 (x;' rx;) n s] c {XI' . . . ' XII } ,

i=l

which shows that S must be a finite set, contradicting our hypothesis. (2) => (3) Let {x11} be a sequence of A. If the sequence {x,} assumes only a finite number of distinct values, then there is nothing to prove-since it must have a constant subsequence.

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS

50

So, assume that {xn} assumes an infinite number of distinct values. This implies the existence of a subsequence {Yn} of {xn } such that Yn =F Ym holds whenever n =F m. (To see this, set ki = 1 and proceed inductively as follows: If ki < < k, with Xkn+l =F Xk; for each k, have been selected, then choose some ku +I 1 < i < n. Such an integer kn+I must exist since {i,} assumes an infinite number of distinct values. If we let Yn = Xk, , then the subsequence {Yn} of {x11 } has the desired properties.) Now, the set {YI , y2, } is an infinite subset of A, and so by our hypothesis it has an accumulation point in A, say x. Note that we can assume that y, =F x for each n; if Yk = x for some k, then replace {Yn } by {Yk+nlNext, choose m i with d(Ym 1 , x) < L Now, inductively, if m i < < m, have been selected, choose m u+I such that

>

•

•

·

·

·

•

-

-

·

Clearly, m n+I > m 11 must hold. This shows that {YmJ is a subsequence of {Y11 }, and hence, a subsequence of {xn}. In view of d (Ym, , x) < �, it follows that lim Ymn = x, as required. (3) => (1) Let A c Uie/ 0; be an open cover of A.

>

Claim I; There exists some � 0 such that for eachx E A we have B(x, �) c 0; for at least one i E I. (Any such number � 0 is called a Lebesgue number of A for the open cover {0; li e / -) To see this, assume that our claim is false. This means that for each 11, there exists some X11 E A such that B (xn, �) n Of =F 0 holds for each i E I. Let x E A be the limit of some subsequence of {x11 }. Pick some i E I with x E 0;, and then choose some 0 with B(x, r) c 0;. Next, select some n so that � < � and

>

r>

r)

d(x,xn) < �- It follows that B(x11, �) c B(x, c 0;, contrary to the selection of Xn- This establishes the validity of our claim. Claim II.

For each

r>

r > 0,

there exist X I , . . . , Xn

AC

II

E A such that

U B(xj. r).

j=I

To see this, let 0, and assume that the claim is false. Fix some XI E A, and then choose x2 E A \ B(xi, r). In general, using induction, choose Xu+ I E A \ U7= I B(x;, r). Clearly, d (xn , Xm ) >< holds for 11 =F m. This implies that no subsequence of {x11} can converge, which contradicts our hypothesis, and our second claim has been established. To finish the proof, pick a Lebesgue number-� 0 of A for { O;};e/ , and then choose XI , . . . , Xn E A such that A c Uj= 1 B(x j, �)- Now, for each j pick some

r

>

51

Section 7: COMPACTNESS IN METRIC SPACES

ij E I such that B(xj . 8) c Oir Thus, A c U�=l B(xj . 8) This shows that A is a compact set, and the proof is finished.

C U�=l

Oii holds.

•

The compact subsets of the Euclidean spaces are precisely the closed and 1 bounded subsets. This result (known as the Heine-Borel theorem 17· 8 ) gave rise to the present definition of compactness. The details follow.

Theorem 7.4 (Heine-Borel). A subset of a Euclidean space is compact ifand

only if it is closed and bounded.

Proof. Let A be a subset of some Euclidean space

Euclidean distance of

lR.k;

that is,

lR.k . Let us denote by d th_e

Assume that A is a compact set. We shall show first that A is a bounded set. Since A c U...-eA B(x, 1), there exists a finite number of points x 1 , . . . , X11 of A such that A c u�l=l B(Xj , 1). Let M = max{d(Xj, Xj ) : i , j = 1, . . . ' n}. If x, y E A, then choose i and j such that x E B (xi , 1) and y E B (xj , 1 ). Therefore,

so that A is bounded. Next, we show that A is closed by establishing that A contains its closure points. Let x E A. By Theorem 6.9 there exists a sequence {x,1} of A such that x = lim x11 • Now, by Theorem 7.3, {x,1} has a a subsequence that converges to some point of A. Since every subsequence of {x11} must converge to x, it follows that x E A . That is, A c A holds, so that A i s closed. Pick For the converse, assume that A is a closed and bounded subset of some M > 0 so that d(x, y) < M holds for all x, y E A. Fix an element y E A. , ak) E A, then ad . < d(a, 0) < d(a, y) + d(y, 0) < M + d(y, 0) If a = (a 1 , holds for each 1 < i < k. Thus, the set of real numbers consisting of the i 1h coordinates of the elements of A is a bounded set. Now, let {x,1} be a sequence of A. Since the sequence of the first coordinates of {x,1} is a bounded sequence of real numbers, it follows from Corollary 4.7 that there exists a subsequence {x,� } of {x,1} such that its first coordinates form a sequence

lR.k .

•

•

•

l

17 Heinrich Eduard Heine ( 1821-1881 ), a German mathematician. He studied several classes of functions and he was the first to formulate and introduce the notion of uniform continuity. 18Emile Borel ( 1 871-1958), a French mathematician. He made many contributions to analysis and probability. His 1898 classic book, Lerons sur Ia Tlreorie des Fonctions. laid the basis of measure theory.

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS

52

that converges in 1R. Now, proceed as follows: Choose a subsequence {x; } of {x�} so that its sequence of second coordinates converges in 1R, and so on. After k steps, the sequence {x!} is a subsequence of {x, l. having the property that for each 1 < i < k the sequence of its ;th coordinates form a convergent subsequence. This implies that {x�} converges in 1Rk , and since A is closed, it converges to some point of A. Thus, every sequence of A has a convergent subsequence in A. By • Theorem 7.3, A is a compact set, and the proof is finished. In general, the Heine-Borel theorem is not valid for arbitrary metric spaces. As in the above proof, if a subset of a metric space is compact, then the set is closed and bounded. However, a closed and bounded subset of a metric space need not be compact. Here is a simple example. Let X be an infinite set, and let d be the discrete distance on X [that is, d (x, y) = 1 if x =f. y, and d(x , x) = 0]. Note that B(x, 1) = {x} for each x E X. Clearly, X is closed :n1d bounded (with respect to this distance), and X = t..Jx.: x B (x 1). But this open cover cannot be reduced to a finite subcover. In fact, if X is uncountable, it cannot even be reduced to a countable subcover. The next result infonns us that continuous images of compact sets are also compact sets. ·

.

.

p)

Theorem 7.5.

Let f: (X, d) � (Y, be a continuousfunction, and let A be a compact subset ofX . Then f(A) is a compact subset ofY. Proof. Let f(A) c U i e / 0; be an open cover. Then A C Ui e / /- 1 (0;), and by the continuity of f each f- 1 ( 0;) is an open subset of X. By the compactness of A, there exist indices it , . . . , i, such that A c U�=t f- 1 (0;",). Thus,

so that

f(A) is compact.

•

From the last result and Theorem 7 .4, it easily follows that if f: (X, d) � 1R is continuous, then f attains a maximum and a minimum value on every compact subset of X . Every closed subset of a compact metric space (X, d) is compact. Indeed, if U [ Uie/ 0;] is an C is closed and {O;};e1 is an open cover of C , then X = open cover of X. Since X is compact," there exist indices i 1, i, such that X = U 0;1 U · · · U 0;". But then C c 0;1 U · · · U 0;,. , and hence, C is compact. A function f: (X, d) � (Y, is called an open mapping if f(A) is open whenever A is open. Similarly, f is called a closed mapping if f(A) is closed whenever A is closed.

cc

cc

p)

•

•

•

,

Section 7: COMPACTNESS IN METRIC SPACES

53

Let (X, d) be a compact metric space and suppose that f: (X, d) --* (Y, p) is a continuousfunction. Then f is a closed mapping. In particular, iff is one-to-one and onto, then f is a homeom01phism.

Theorem 7.6.

Proof. Let C be a closed subset of X. Then C is a compact subset of X, and

7.5

by Theorem the set f(C) is a compact subset of Y. Hence, f(C) is closed, and so, f is a closed mapping. If now f is in addition one-to-one and onto, then the relation (f- 1 )- 1 (A) = f(A) holds for every subset A of X. But then, it follows from the first part and Theorem 6.l l(v) that f- 1 is also continuous, and hence, that f is a homeomor­ phism. • A continuous function need not be uniformly continuous.

However, a continu­ ous function whose domain is compact is always uniformly continuous.

Let f: (X, d) (Y, p) be a continuous function. If (X, d) is compact, then f is uniformly continuous. Proof. Let E > 0. By the continuity of f, for every x there exists r:c > 0 such that p(f(y), f(x)) < E holds whenever d(x, y) < 2r:c. Then the col­ lection of open balls B (x, r:c) covers X, and since X is compact, there exists a finite number of points Xt , . . . , Xn in X such that X = U'�=1 B(x;, r:c) · Let c5 = min{r.\"J rJ:n} > 0, Now, assume that x, y E X satisfy d(x, y) < c5. There exists an integer i (1 < i < n) such that d(x, x;) < rx; · Clearly, p(f(x), f(x;)) < E . By the triangle Theorem 7.7.

--*

1 • • • I

inequality

d(y, x;) < d(y, x) + d(x, x;)

< c5

+ rx;

<

2rx;

holds. Therefore,

p(f(;-:), f(y)) < p(f(x), f(x;)) + p(f(x;), f(y))

< E+E

This shows that f is uniformly continuous, and we are finished.

=

2E. •

space (X, d) is called totally bounded if for each r > 0 there exists a finite number of points Xt , . . . , X11 such that X = U7=t B (x;, r). A compact metric space is totally bounded, but a totally bounded metric space need not be compact; take, for instance, X = (0, 1 ) with the distance d(x, y) = lx - yl. The next result shows the connection between compactness and completeness. A metric

Theorem 7.8.

totally bounded.

A metric space is compact if and only if it is complete and

54

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS

Proof. Let (X, d) be a metric space. Assume first that (X, d) is compact.

Clearly, (X, d) is totally bounded. By Theorem 7.3, every sequence {x,} has a limit point x in X and hence, if {xn } is a Cauchy sequence, then lim Xn = x . Thus, (X, d) is also complete. For the converse, assume that (X, d) is complete and totally bounded. According to Theorem 7 .3, we must show that every infinite subset of. X has an accumulation point. To this end, let A be an infinite subset of X. Let us write (for this proof only) C(a, r) = {x E X: d(a, x) < r}. Since X is totally bounded, there exists a finite subset F of X such that X = U eF C(x, 1). But then, since A is an infinite x set, there exists some x1 E F such that A n C(x1, 1) is an infinite set. Now, by induction, if X), x, have been chosen such that the set A n C(XJ , 1) n . . . n C (xn , *) in an infinite set, then argue as previously shown to select Xn+ 1 such that the set • • • •

nJ

A n Cfx, , l ) n c {x, . �) n . . . n c (x... . .!.) n c (xn.�.1, .· \ -· '

·

2J

'

is infinite. Next, put EII = c (xI

·

1 n + 1

)

1) n . . . n c (Xn . t) for each n; clearly, each En is nonempty and closed. Also, £,+1 c En and d(En) < � hold for each n. By Theorem 6.14, there exists a E X such that a E En for each n. Observe that if y E A n C(XJ, 1 ) n . . . n C(xn. * ), then d(a, y) < d(a, Xn) + d(xn . y) < � •

holds, from .which it follows that a is an accumulation point of A . The proof of the theorem is now complete. •

EXERCISES

Let f: (X, d) --+ (Y, p) be a function. Show that f is continuous if and only if f restricted to the compact subsets of X is continuous. [HINT: If lim x, = x, then the set {xn : n E N} U {x} is compact.] 2. A metric space is said to be separable if it contains a countable subset that is dense in the space. Show that every compact space (X, d) is separable. {HINT: For every n choose a finite subset Fn of X such that X = U eFn B (x, * ). x Now show that F = U� 1 Fn is dense in X .] 3. Show that if (X, is a separable metric space (see the preceding exercise for the definition), then card X :s c. 4. Let (X 1 , dJ), . . . , (Xn, be metric spaces, and let X = X 1 x · · · x Xn . If x = • (XJ , . . , x,) and y = ()'J, , y,), define 1.

d)

d,)

. • •

a

b.

Show that D1 and D2 are distances on X. Show that Dt is equivalent to D2.

Section 7: COMPACTNESS IN METRIC SPACES c. d. 5.

Show that (X, D 1 ) is complete if and �>nly if each (X;, d;) is complete. Show that (X, D t ) is compact if and only if each (X;, d;) is compact.

Let {(Xu, du)} be a sequence of metric spaces, and let X x = {xu} and y = {Y ) in X define

n

a. b. c.

55

-

�

l du(Xu , Yu ) d(x, y) = L...J - · ----­ l + d11(X11 , y11 )

n� t

Xu . For each

n= t 2"

Show that d is a distance on X. Show that (X, d) is a complete metric space if and only if each (X11 , d11) is complete. Show that (X, d) is a compact metric space if and only if each (X11 , d,) is compact.

A family of sets :F is said to have the finite intersection property if every fi�ite intersection of sets of :F is nonempty. Show that a metric space is compact if and only if every family of closed sets with the finite intersection property has a nonempty intersection. 7. Let f: X -+ X be a function from a set X into itself. A point a E X is called a fixed point for f iff (a) = a. Assume that (X, d) is a compact metric space and f: X -+ X satisfies d(f (x), j(y)) < d(x, y) for x =f:. y. Show that f has a unique fixed point. [IDNT: Show that the function g(x) = d(x, j(x)) attains its minimum value, that it must be zero.] 8. Let (X, d) be a metric space. A function f: X -+ X is called a contraction if there exists some 0 < ex < l such that d(j(x), j(y)) � exd(x, y) for all x, y E X; ex is called a contraction constant. Show that every contraction f on a complete metric space (X, d) has a unique fixed point; that is, show that there exists a unique point x E X such that j(x) = x. [IDNT: Pick a E X, and define the sequence x 1 = a, and x11+ 1 = f (x11 ) for 11 ::: l . Show that {x11} is a Cauchy sequence and then that its limit is a fixed point of f . ] 9. A property of a metric space is called a topological property if it is preserved in a homeomorphic metric space. 6.

a. b.

10.

Show that compactness is a topological property. Show that completeness, boundedness, and total boundedness are not topological properties.

Let (X, d) be a metric space . Define the distance of two nonempty subsets A and B of X by

d(A, B ) = inf{d(x, y): x E A and y E B ) . a. b. 11.

Give an example of two closed sets A and B of some metric space with A n B = 0 and such that d(A, B) = 0. If A n B = (/), A is closed, and B is compact (and, of course, both are nonempty), then show that d(A, B) > 0.

Let (X, d) be a compact metric space and f: X -+ X an isometry; that is, d(j(x), j(y)) = d(x, y) holds for all x, y e X. Then show that f is onto. Does the conclusion remain true if X is not assumed to be compact?

56 12.

Chapter 1: FUNDAMENTALS OF REAL ANALYSIS Show that a metric space (X, d) is compact if and only if every continuous real-valued function on X attains a maximum value.

13.

14.

This exercise presents a converse of Theorem 7.7. Assume that

(X, d) is a metric

space such that every real-valued continuous function on X is uniformly continuous. a.

Show that X is a complete metric space.

b.

Give an example of a noncompact metric space with the above property.

c.

If X has a finite number of isolated points (an element a

e X is said to be an iso­ lated point whenever there exists some r > such that B (a, r) n (X \ {a}) = (/)), then show that X is a compact metric space.

0

Consider a function 1: (X, d) I is the subset of X

� (Y, p) between two metric spaces.

x Y defined by G

=

{(x, y)

eX

x

The graph G of

Y : y = l(x)}.

p) is a compact metric space, then show that I is continuous if and only if G is a closed subset of X x Y, where X x Y is considered to be a metric space under the rtist�11ce D((:),: , y), (u. 1.1)) = d(x, u) + p(y, v); see Exercise 4 of this section. Does the result hold true if (Y, p) is not assumed to be compact? 15. A cover {V; lie/ of a set X is said to be a pointwise finite cover whenever each x E X If (Y,

belongs at most to a finite number of the V; . Show that a metric space is compact if

and only if every pointwise finite open cover of the space contains a finite subcover.

CHAPTER

2

_ _ _ _ _ _ _ _ _ _ _ _ _ _

TOPOLOG Y AND CONTINUIT Y The role of open and closed sets in metric spaces has been discussed previously. Now, the fundamental notion of an open set will be generalized by introducing the concept of a topological space. The properties of open and closed sets will b� studied in this setting. This chapter is devoted entirely to topological and function spaces and emphasizes the results needed for this book. For a detailed study of general topology, the reader is referred to the classical book on the subject [ 18]. For a modem detailed treatment of general topology, you might consult the mono­ graph [22]. The material has been arranged into four sections. The first section discusses the theory of topological spaces. The second section deals with the properties of continuous real-valued functions, and introduces the concepts of vector lattices and function spaces. The notions of pointwise and uniform convergence of sequences are discussed, and their relations to one another are investigated. In the third section of this chapter, we investigate extension and separation properties of real-valued continuous functions. Finally, the fourth section culminates the discussion with a detailed presentation of the classical Stone-Weierstrass approximation theorem.

8. TOPOLOGICAL SPACES

In

the previous sections, the fundamental properties of metric spaces were dis­ cussed. The open and closed sets played a basic role in that study. In this section the concepts used in a metric space will be generalized b.y introducing the notion of a topological space. Properties such as closeness, convergence, and continuity will be studied in this setting. The starting point is the definition of open sets.

A

Definition 8.1. Let X be a nonempty set. collection -r of subsets of X is said to be a topology on X if-r satisfies the following properties:

1. X E -r and (/) E -r. 2. If U and V belong to -r, then U n V E -r. 3. If { V; };eJ is afamily ofmembers of-r, then U ie/ V; E -r. 57

Chapter 2: TOPOLOGY AND CONTINUITY

58

If r is a topology on a set X , then the pair (X, r) is called a

topological space.

r, sometimes for simplicity we shall write X instead of (X, r ) . The members of r are called the open sets of X .

If there is no ambiguity about the topology

Some examples of topological spaces are presented next. It is expected that the

reader will be able to verify that the exhibited collections are indeed topologies.

Example 8.2. Let X be a nonempty set. Then r = { ct;, X} is a topology on X, called the indiscrete topology. This topology is the smallest possible (with respect to inclusion) topology on X. •

Example 8.3. Let X be a nonempty set. Then r = P(X) is a topology on X. Here every subset of X is an open set. This topology is called the discrete topology, and it is the "largest" possible topology on X. • Let (X, d) be a metric space, with the set X uncountable. Let r denote the collection of all subsets 0 of X such that for each x E 0 there exist r > 0 and an at-most countable subset A of X (both depending on x) such that x rt A and B (x, r) \ A c 0. Then • r is a topology on X .

Example 8.4.

Example 8.5. Let (X, d) be a metric space. Then the collection of all open subsets of X satisfies the properties for a topology; see Theorem 65. Thus, every metric space is a •

topological space.

Example 8.6. Let X = 1R* be the set of extended real numbers. Let r be the collection of subsets of X defined as follows. A subset 0 of X belongs to r if it satisfies these properties:

1.

2. 3.

Foreach x E OniR thereexists r > O (depending onx) such that (x -r,x+r) c 0. If oo E 0, then there exists a E 1R such that (a, oo] c 0. If -oo E CJ, then there exists a E 1R such that [-oo, a) c 0.

•

Then r is a topology on IR*.

Example 8.7. Let (X, of subsets of Y by

r) be a topological space, and Y a subset of X. Define a collection ry

=

{V n Y: V E r}.

Then (Y, r y) is a topological space. The topology r y is called the topology induced by r on Y, or the relative topology of r on Y. •

Unless otherwise stated, for our discussion here, we shall assume that (X,

r) is

a fixed topological space. The closed sets are now defined as in the case of metric spaces. A subset A of X is said to be closed if its complement is open (i.e.,

Ac E r).

By taking complements, the following properties of closed sets follow

directly from Definition 8.1:

1.

2.

3.

Ac

0 and X are closed sets (as well as open sets). Finite unions of closed sets are closed sets. Arbitrary intersections of closed sets are closed sets.

·

59

Section 8: TOPOLOGICAL SPACES

The balls of a metric space are now replaced in this setting by neighborhoods. A neighborhood of a point X is any open set containing X. Thus, an open set is a neighborhood for all of its points. In particular, note that a subset A of X. is open if and only if every point of x E A has a neighborhood Vr such that Vx c A. (Indeed, if A has this property, then A = U eA V.r is an open set, since it is a union x of open sets.) The interior A0 of a subset A of X is defined by

A0 = {x E X:

3 a neighborhood

V of x such that V

c A}.

The members of A0 are called the interior points of A. Clearly, A0 is an open set, and A0 c A. Moreover, A0 is the largest open set that is included in A. It should be also clear that A' is open if and only if A = A0• The closure and accumulation points of a set are define.d as in metric spaces. A point x E X is said to be a closure point for a set A if every neighborhood V of x contains (at least) one point from A (that is, if V n A =j:. (/) for every neighborhood V of x). The set of all closure points of A is denoted by A and is called the closure of A. Clearly, A c A. As in the proof of Theorem 6. 7, it can be shown that A is the smallest (with respect to inclusion) closed set that contains A. It follows from this that a set A is closed if and only if A = A. Some other useful properties that the closure operator satisfies are the following:

� B, then A c B. A U B = A U B for all subsets A and B. A0 = (Ac)c for every subset A. A subset A of X is said to be dense in X if A = X. A point x E X is said to be an accumulation point of a subset A if every neighborhood of x contains a point of A different from x; that is, if (V \ {x}) n A =j:. (/) holds for each neighborhood V of The set of all accumulation points of A is denoted by A' and is called the derived set of A. Clearly, A' c A. Also, it easily follows that A = A U A' and from this that a set A is closed if and only if A' c A. A point x E X is said to be a boundary point of a set A if V n A =j:. (/) and V n Ac =j:. (/) hold for every neighborhood V of x. The set of all boundary point of A is denoted by a A and is called the boundary of A. Observe that a. b. c.

If A

x.

aA = A n Ac, and hence, a A is always a closed set. Also, it is easily seen that a A = a A c holds for every subset A of X. Moreover, A = A U a A also holds. A subset A of X is said to be nowhere dense if (A)0 = (/). A set is called a meager set (or a set of first category) if there exists a sequence {A, } of nowhere dense sets such that A = A,. Observe that any subset of a meager set is itself necessarily a meager set.

U:.1

A

Chapter 2: TOPOLOGY AND CONTINUITY

60

A sequence {xn } in a topological space (X, r) is said to converge to x (denoted by

lim Xn = x) if for every neighborhood V of x there exists an integer k (depending on V) such that Xn E V for all n 2: k. Our first observation is that, in contrast with a metric space, a sequence can have more than one limit. For instance, every sequence in the topological space of Example 8.2 converges to every point of X. However, in a Hausdorff1 topological space, every sequence has at most one limit. A topological space (X, r) is said to be a Hausdorff space if for every pair x, y E X with x -:/= y there exist neighborhoods V of x and U of y such that V n U = (/J. That is, a Hausdorff space is a topological space in which any two distinct points can be separated by disjoint neighborhoods. Unless otherwise stated, all topological spaces encountered in this book will be Hausdorff. Our attention now turns to continuous functions. A function f: (X, r ) -+ (Y, r t ) between two topological spaces is said to be continuous at a point a E X if for every neighborhood V of f (a) there exists a neighborhood W of a such that f (x ) E V whenever x E W. If f is continuous at every point of X, then f is called a continuous function. The next result characterizes the continuous functions and is the parallel of Theorem 6. 1 1 .

For a function f: (X, r) -+ (Y, r1) benveen two topological spaces, the following statements are equivalent: l . f is a continuous function. 2. �-�(0) is open whenever 0 is an open subset ofY. 3. f(A ) c f(A) holds for every subset A of X. 4. f- 1 (C) is a closed subset of X whenever C is a closed subset ofY. Theorem 8.8.

( l ) ==> (2) Let 0 be an open subset of Y, and let a E f - l ( 0). Then there exists a neighborhood V of a such that f(x) E 0 for all x E V. Thus, V c f- 1 (0), and so a is an interior point of f- 1 (0). Since a e f- 1 (0) is arbitrary, it follows that f - I ( 0) is open. (2) ==> (3) Let A c X and put B = f(A). Since B is closed, Be is open, and thus, by our hypothesis f- 1 (Bc) = [f- 1 (B)]c is also open. This implies that the set f- 1 (B) is closed. Now, by virtue of A c f- 1 (B ) we obtain A c f- 1 (B). Therefore, f(A) c B JCA), as desired. C3)==> C4) Let C be a closed subset of Y. Put A = f- 1 CC). Then, f(A ) c JCA) c C = C holds, which shows that A c f- 1 CC) = A. Therefore, A = A, and so A = f- 1 CC) is a closed set. (4)==> ( l ) Leta E X, and let V be a neighborhood of f(a). Since vc is closed, f-1(Vc) = [f-1(V)]c is also closed by our assumption, and so W = f-1(V) is an

Proof.

=

1 Felix Hausdorff (1 868-1942), a Gennan mathematician. His main work was in topology and set theory. He is the founder of general topology and the theory of metric spaces.

61

Section 8: TOPOLOGICAL SPACES

open set. Observe now that a e W , and thus, W is a neighborhood of a. Clearly, x e W implies f(x) E V, so that f is contiimous at a. Since a is arbitrary, f is a continuous function. • A countable union of closed sets is not necessarily a closed set, and a countable intersection of open sets need not be an open set. Such sets are nevertheless of importance. A set is called an Fa-set if it is the union of countably many closed sets. Similarly, a set is said to be a G rset if it is the intersection of countably many open sets. The next theorem tells us that the prior classes of sets are in a dual relation.

A set is an Fa-set if and only if its complement is a G6-set. Similarly, a set is a Gs-set if and only if its complement is an Fa-set. Proof. Let be an Fa-set, so that A = where each A, is closed. Then A c = is a countable intersection of open sets, and thus, is a G s-set. The proof that the complement of a G 8-set is an Fa-set is similar. • Theorem 8.9.

U:1 An,

A n: 1 A�

The union of a countable collection of Fa-sets is again an Fa-set and the count­ able intersection of G8-sets is a G8-set. Also, it can be ·seen easily that a finite union or intersection of Fa-sets (resp. G 8-sets) is again an Fa -set (resp. a G 0-set). Now, consider a real-valued function f defined on a topological space (X, r); that is, f: X � JR. It is possible for f to be discontinuous everywhere. For instance, f: JR � JR with the value 1 on each irrational and 0 on each rational is discontinuous at every point. It is instructive to examine the set of point where a function is continuous, or the set of points where a function is discontinuous. To do this, we need some preliminary discussion. Let us denote by N." the collection of all neighborhoods of the point x. The oscillation w1(x) of f at the point x is the non-negative extended real number defined by

w1(x) =

inf

{

sup

V e.N.r :,yeV

l

l f(z) - f(y)l ·

A straightforward verification shows that f is continuous at the point x if and only if w1(x) = Rephrasing this last statement, one can see that f is discontin­ uous at the point x if and only if Wf(X) > 0. Thus, if = {x E X: Wf(x) > and denotes the set of all points of discontinuity of f, then = holds.

0.

D

Dn

D U: 1 Dn

�}

Let (X, r) be a topological space, and let f: X � JR. Then the set of all points of discontinuity of f is an Fa-set. In particular, the set ofpoints of continuity off is a G8-set.

Theorem 8.10.

D

62

Chapter 2: TOPOLOGY AND CONTINUITY

Proof. According to the discussion preceding the theorem, we have D u:l D,, where D, = {x E X: Wj(X) > � } . It suffices to show that each Dn is a closed set. To this end, let x � D,. Then w1 (x) < It follows from the defin ition of w1(x) that there exists a neighborhood V ofx such that sup:, el' l f(z)-f(y)l < y Since V is a neighborhood for each of its members, we easily get that

�.

�·

1

12 :::,ye V for each a e V . Thus, V c D�, which shows that x is an interior point of D�. Since x is arbitrary, D� is open, and hence, D, is closed. The proof of the theorem • is now complete.

WJ(a) < sup lf(z) - f(y)l

<-

We continue our discussion with the introduction of compact sets. Their defi­ nition is an abstract version of the one given for metric spaces. Definition 8.11. A subset A ofa topological space (X, r ) is said to be compact if every open cover of A can be reduced to a finite subcover. In particular, if X itself is a compact set, then (X, r) is called a compact topological space. It shoul� be obvious from the preceding definition that every finite subset of a topological space is automatically compact. Also, it should be clear that a finite union of compact sets must be a compact set. More properties of the compact sets are included in the next result. Theorem 8.12.

For a Hausdorfftopological space (X, r ) thefollowing state­

ments hold: 1 . Every compact subset of X is closed. 2. If B is a closed subset of a compact set A, then B is compact. Proof. (1) Let A be a compact subset of X. We must show that Ac is open. To this end, let x e A Then for each y e A there exists a neighborhood Vy of y and a neighborhood Uy of x such that Vy n Uy (/.>. Clearly, A c UyeA Vy . Since A is compact, there exist Y I ' . . . ' Yn E A such that A c u�l= l VYm ' Let

c.

=

() = n�l=l uYm . Then () is a neighborhood of X such that () n A = (/). Hence, () c A c, and so x . is an interior point of A c. Thus, A c is open, which means that A is closed. (2) Let {O;}; e1 bean open cover of B . Then the family of sets {Bc}u {O;: i e / } is an open cover for A. Choose indices it , . . . , in such that A c BcU0;1 U· · · U O;, . It follows that B c 0;1 U · · · U O;n holds, which shows that B is a compact set. (Notice that the proof of part (2) does not require X to be Hausdorff.) •

Section 8: TOPOLOGICAL SPACES

63

The arguments of the proof of part ( I ) �f the preceding theorem yield also the following separation result.

8.13. Suppose that A is a compact subset ofa Hausdorfftopological space and x ¢: A. Then there exist open sets V and W such that x E V , A c W, and V n W = 0 (and hence, also x ¢ W). Theorem

Continuous functions map compact sets to compact sets. The details follow.

Theorem 8.14.

If f: (X, i) � (Y, it) is a continuous function and A is a compact subset of X, then f(A) is a compact subset ofY. In particular, every continuous real-valuedfunction on a topological space (X, i ) will attain a maximum and a minimum value on every compact subset ofX. Proof. For the proof of the first part, repeat the proof of Theorem 7.5. For the second part, let A be a compact subset of X, and let f: X � .JR. be continuous. By the preceding, f (A) is a compact subset of .JR., and hence, by Theorem 7 .4, f (A) is closed and bounded. Thus, if a = sup{f(x): x E A} and b = inf{/(x): x E A}, then since a, b E f(A), there exist two points x, y E A such that a = f(x) and b = f (y). The proof of the theorem is now complete. • Two topological spaces (X, i) and (Y, it) are called homeomorphic if there t exists a one-to-one onto function f: (X, i) � (Y, it) such that f and f- are both continuous. Any such function is called a homeomorphism between (X, i) and (Y, it).

Theorem 8.15.

A one-to-one continuousfunctionfrom a compact topological space onto a Hausdorff topological space is a homeomorphism. Proof. Let f: (X, i) � (Y, it) be a one-to-one surjective continuous func­ tion, where (X, i) is compact and (Y, it) is Hausdorff. Assume that C is a closed subset of X . By Theorem 8.12(2), C is a compact subset of X , and a glance at Theorem 8.14 guarantees that /(C) is a compact subset of Y. Now, Theo­ t

rem 8.12(1) implies that (/- 1 )- (C) = f(C) is a closed set, and this (in view of Theorem 8.8(4)) shows that f- 1 is also continuous. • EXERCISES

1. For any subset A of a topological space show the following: a.

b.

c.

Ao = (A c)c . BA = A \ A0 • (A \ A 0)0 = (/).

64

Chapter 2: TOPOLOGY AND CONTINUITY

2. If A and B are two subsets of a topological space, then show the following: a. b.

A U B = A U B. (A U B)' = A' U B'.

3. If A is an arbitrary subset of a Hausdorff topological space, then show that its derived 4.

set A' is a closed set. Let X = IR, and let r be the topology on X defined in Example 8. In other words, A E r if and only if for each x E A there exist € > and an at-most countable set B (both depending on x) such that (x - €, x + E) \ B C A.

0

a. b. c.

Show that r is a topology on X. Verify that E (0, I). Show that there is no sequence {xn} of (0, l) with lim x11 = 0.

0

5.

If A is a dense subset of a topological space, then show that 0 £ A n 0 holds for every open set 0. Generalize this conclusion as follows: If A is open, then A n B � A n B for each set B. 6. If { V; }; e/ is
8.

a. b.

Every finite subset of X is closed. Every sequence of X converges to at-most one point.

a. b.

If r is the discrete topology, then f is continuous. If r is the indiscrete topology and r1 is a Hausdorff topology, then f is continuous if and only if f is a constant function.

For a function f: (X, r) ooo4 (Y, r1) show the following:

Let f and g be two continuous functions from (X, r) into a Hausdorff topological space (Y, r1 ). Assume that there exists a dense subset A of X such that f(x) = g(x) for all x E A. Show that f(x) = g(x) holds for all x E X. 10. Let f: (X, r) ooo4 (Y, rt) be a function. Show that f is continuous if and only if f- 1 (B 0) c [f- 1 (B ))0 holds for every subset B of Y. ll. If f: (X, r) ooo4 (Y, rt) and g: (Y, rt) ooo4 (Z, r2) are continuous functions, show that their composition g a f: (X, r) ooo4 (Z, r2) is also continuous. 12. Show that a function f: X ooo4 IR, where X is a topological space, is continuous at some a E X if and only if its oscillation at a is zero, i.e., wf(a) = 13. Show that a finite union of nowhere dense sets is again a nowhere dense set. Is this statement true for a countable union of nowhere dense sets? Show that the boundary of an open or closed set is nowhere dense. 14. 15. Let f: (X, r) ooo4 IR, and let D be the set. of all points of X where f is discontinuous. If De is dense in X, then show that D is a meager set. 16. Show that there is no function f: 1R ooo4 1R having the irrational numbers as the set of its discontinuities. [HINT: Us.e Exercise 6 of Section 6 and Theorem 8.10.] 9.

0.

Section 8: TOPOLOGICAL SPACES 17. 18.

65

Show that every closed subset of a metri� space is a Gc5-set and every open set is an F,.-set. Let B be a collection of open sets in a topological space (X, r ) If for each x in an arbitrary open set V there exists some B E B with x E B c V , then B is called a base for r. In general, a collection B of subsets of a nonempty set X is said to be a base if .

i. ii.

UneB B = X, and

for every pair A , B E B and x E A n B, there exists some C E B with x E C c

A n B.

Show that if B is a base for a set X, then the collection

r

19.

20.

{ V C X:

Vx

E V there exists B E B with x E B C V }

is a topology on X having B as a base. Let (X, r) be a topological space, and let B be a base for the topology r (see the preceding exercise for the definition). Show that there exists a dense subset A of X such that card A � card B. Let f: X � Y be a function. If r is a topology on X , then the quotient topology r determined by f on Y is defined by r1 = {0 � Y: f - 1 (0) E r}.

1

a. b. c.

21.

=

1)

1 1)

Show that r is indeed a topology on Y and that f: (X, r) � (Y, r is continuous. If g: (Y, r � (Z, r1) is a function, then show that the composition g o (X, r) � (Z, r1) is continuous if and only if g is continuous. Assume that f: X � Y is onto and that r* is a topology on Y such that (X, r) � (Y, r*) is an open mapping (i.e., it carries open sets of X onto open sets of Y) and continuous. Show that r * = r1.

f:

This exercise presents an example of a compact set whose closure is not compact. Start by considering the interval [0, I ] with the topology r generated by the metric d(x, y) = lx - y!. It should be clear that ([0, I ] , r) is a compact topological space. Next put X = [0, 1 ] U lN = · [O, I ] U {2, 3, 4, . }, and define .

r* a. b. c. d. e. 22.

f:

=

r U { [0, l ]

.

u A: A � lN }.

Show that r* is a non-Hausdorff topology on X and that r* induces r on [0, l ]. Show that (X, r*) is not a compact topological space. Show that [0, l ] is a compact subset of (X, r*). Show that [0, I ] is dense in X (and hence, its closure is not compact). Why doesn't this contradict Theorem 8.12( 1 )?

A topological space (X, r) is said to be connected if a subset of X that is simultane­ ously closed and open (called a clopen set) is either empty or else equal to X. a. b.

Show that (X, r) is connected if and only if the only continuous functions from (X, r) into {0. 1} (with the discrete topology) are the constant ones. Let f: (X. r) � (Y, r*) be opto and continuous. If (X, r) is connected, then show that (Y, r*) is also connected.

66

9.

Chapter 2: TOPOLOGY AND CONTINUITY CONTINUOUS REAL-VALUED FUNCTIONS

The properties of topological spaces are the tools for the study of continuous real-valued functions. In this section many important properties of continuous functions will be discussed, and function spaces will be considered. Also, the behavior of the limit of a sequence of continuous functions will be discussed in some detail. Let X be a topological space. The collection of all continuous real-valued functions on X will be denoted by C(X). The set C(X) is closed under addition and scalar multiplication. That is, if f and g are members of C(X), then the functions f + g and af defined by

(f + g)(x) = f(x) + g(x)

and

(af)(x) =

af(x)

for eachx E X and a E IR belong to C(X). Now it should be clear that C(X) is a vector space. Also, C(X) has a natural partial ordering defined by f > g whenever f(x) > g(x) holds for all x E X. Moreover, C(X) is also a vector lattice. This means that for every pair f, g E C (X) the least upper bound f v g, as well as the greatest lower bound f A g, both exist in C(X). They are given by the fomm­ las:

(f v �)(x) = max{f(x), g(x)}

and

(f A g)(x) = min{f(x), g(x)}

for each x E X. The absolute value lfl of a function f E C(X) is defined by If I = f v (-f). That is, lfl(x) = lf(x)l holds for each x E X; clearly, If I be­ longs to C(X). Note also that f v g and f A g satisfy the identities:

f v g = 4
and

f A g = 4
Observe that the above identities show that f v g and f A g are continuous func­ tions because they can be expressed as sums of continuous functions. Since most of the spaces of functions that one encounters are vector lattices, it is appropriate to stop and consider them more closely. Recall that a relation > on a nonempty set X is called an order relation if it satisfies the following properties: l.

2. 3.

u > u for all u E X (reflexivity). If u > v and v � u, then u = v (arztisymmetly). If u > v and v > w, then u > w (transitivity).

The symbolism v < u is an alternate notation for u > v. Also, u > v (or v < u) means u > v and u # v. An ordered vector space is a real vector space E equipped with an order relation satisfying the following two conditions:

Section 9: CONTINUOUS REAL-VALUED FUNCTIONS 4.

5.

+

+

If u > v, then u w > v w for all w E If u > v, then au > av for all a >

0. 0

67

E.

A vector LL in E is called positive if u > holds. The set of all positive vectors is denoted by £+. A vector lattice E is an ordered vector space with the additional property that for every two vectors u , v E E, the supremum u v v and the infimum u 1\ v exist in E. We remind the reader that two vectors u, v E E have a supremum w in E if w > u and w > v hold, and whenever z is an upper bound of { u, v}, then z > w holds. Clearly, w = u v v is uniquely determined. In other words, u v v is the smallest upper bound of the set { u, v}. The definition of u 1\ v is similar. If u, v, and w are vectors in a vector lattice E, then the following identities hold: a. b. c. d.

u v v = - [(- u) 1\ (- v)]; u v v w = (LL w) v (v + w); w) 1\ (v w); u 1\ v w = (u a(u v v) = (au) v (av) for each a >

+ +

++

+

0.

To indicate how one proves identities in a vector lattice, we shall establish (b). Put f = u v v + w and g = (u w) v (v + w). It suffices to show that f > g and g > f both hold. Note first that f = u v v w implies f - w = u v v , and so, u < f - w and v < f - w. Thus, u + w < f and v + w < f hold, so that f > (u + w) v (v + w) = g . On the other hand, g = (u w) v (v w) implies u + w < g and v + w < g. Hence, u < g - w and v < g - w, from which it follows that u v v < g - w. Therefore, f = u v v w < g also holds. If E is a vector lattice and u E E, then we define u+

=

uv

+ 0,

+ +

+

+

u- = ( -u) v 0 ,

and

lu i

=

u v ( -u).

The element u + is called the positive part, u- the negative part, and I u I the absolute value of u.

If u is a vector in a vector lattice, then thefollowing identities

Theorem 9.1.

hold: 1. u u+ - u-, 2. I tt l = u+ + u-, and 3 . u+ 1\ tL0. Proof. (1) Applying identity (b) above, we get =

=

+

tL- u from which u

=

=

( -u) v

u+ - u- follows.

0+u 0 =

vu

=

u+,

Chapter 2: TOPOLOGY AND CONTINUITY

68

(2) By using (b) and (d) we obtain lu i = =

u v (-u) = (2u) v 0 - u = 2(u v 0) - u = 2u+ - u 2u+ - (u+ - u-) = u + + u-.

(3) Using (c) and (a), we get

u + 1\ uand the proof is finished.

=

=

(u+ - u-) 1\ 0 + u- = u 1\ 0 + u - u) v 0] + u- = - u- + u- = 0,

-[(

•

Typical examples of vector lattices are provided by function spaces. A function space L is a vector space of real-valued functions defined on some nonempty set X such that the functions j v g and j 1\ g beiong to L for every pair f, g E L, where

(I v g)(x) = max{ l (x), g(x)} and (I 1\ g)(x) = min{l(x), g(x)} , hold for each x E X. Note that for every function I I I of L satisfy

I in a function space L , the elements I+ , 1-, and

l+(x) = max {l(x) , 0}, 1-(x) = max{-l(x), 0}, and ll l(x) = ll(x)l for each x E X. Here are some example of function spaces.

1. 2.

3.

4.

The vector space m_X of all real-valued functions defined on a set X. The vector space B(X) of all bounded real-valued functions defined on X. The vector space C(X) of all continuous real-valued functions on X (pro­ vided, of course, that X is a topological space). The vector space Cb(X) of all bounded continuous real-valued functions on a topological space X.

Consider a sequence {In} of real-valued functions defined on a set X such that lim In (x) exists in IR. for each x E X. Then a new function I can be defined by I (x) = lim In (x) for each x E X. If this happens, then the sequence {In} is said to converge pointwise to I (or that I is the pointwise limit of {In }) and is written symbolically as In ---+ I. In other words, In ---+ I if for each € > 0 and each x E X there exists some n0 (depending upon both E and x) such that lln(X) - l(x)l < € for all n > no.

69

Section 9: CONTINUOUS REAL-VALUED FUNCTIONS

A stronger concept of convergence of a· sequence of real-valued functions is that of uniform convergence. A sequence {f,,} of real-valued functions is said to converge uniformly on to a function f if for each E 0 there exists some n0 (dependingonlyuponE)suchthat l f,,(x)-f(x)l < E foralln noandallx E X. It should be clear that uniform convergence implies pointwise convergence. We are interested in determining what properties are possessed by a function that is the "limit" of a sequence of continuous functions. Note first that the pointwise limit of a sequence of continuous functions need not be a continuous function. For an example" take = [0, l], and let {f,,} be the sequence of functions defined by j,,(x) = x for each x E [0, l]. Then each j,, is a continuous function, and -+ f holds for the function f defined by f (x) = 0 if x E [0, l) and f ( l) l. fnClearly, f is not continuous. Also, it is easy to see that the convergence is not uniform. The uniform limit of a sequence of continuous functions is always a continuous function. The details follow. Theorem 9.2. Let be a topological space and let {j�, } be a sequence of If {f,,} converges uniformly to f on then f is a continuousfunction. Proof. We need to show that f is continuous at every point of Therefore, let a and E 0. Since { fn} converges uniformly to f on there exists k such that lfk(x) - f(x)l < E for all x E On the other hand, since fk is a continuous function, there exists a neighborhood V of a such that l fk(x) - fk(a)l < E for all E V. Now note that if x E V, then l f(x) - f(a)l < lf(x) - fk(x)l l fk(x) - fk(a)l + l fk(a) - f(a)l < E E E = 3E, and this shows that f is continuous at a, as desired. Let be a nonempty set, and let denote the collection of all bounded real­ valued functions defined on Clearly, is a function space. The uniform (or sup) norm of a function f E is defined by l flloo = sup lf(x)l . The uniform norm satisfies the three characteristic properties of a norm on a vector space, namely: l. llfllco 0 for each f E and l fl l oo 0 if and only if f = 0. 2. l afllco = I a I · ll f l oo for each f E 3. l f g l < llfll co l g l oo for all f, g Eand all a E lR. X

>

>

X

=

X

C(X).

X,

X.

EX

X,

>

X.

x

+

+

+

•

X

B(X)

X.

B(X)

B(X)

.r eX

B(X)

>

B(X)

+

co

+

=

B(X).

70

Chapter 2: TOPOLOGY AND CONTINUITY

D(f,

I / - gll 00

If we set for J, g E B(X), then D is a distance on g) = B(X) called the uniform distance (or the uniform metric) such that (B(X), D) is a complete metric space; see Example 6.12. Moreover, l;>y a straightforward of B(X) converges to some E verification one can show that a sequence B(X) with respect to D [i.e., lim D , 0] if and only if converges uniformly to on X. This justifies the name "uniform distance." Consider now a compact topological space X. Then by Theorem 8.14, every function E C(X) is bounded, and hence, C(X) c B(X). Therefore, C(X) equipped with the uniform distance is a metric space, which is actually complete, as the next result will show.

{/,} (f, /) =

f

f

{f,,}

f

If X is a compact topological space, then C(X) is a complete metric space (with the uniform distance). Proof. Let {f,,} be a Cauchy sequence of C(X). Note that the inequality lfn(X)- fm(X) I !I f,, - fm l oo E X. implies that {fn(x)} is a Cauchy sequence of real numbers for each Put f(x) lim f,, (x) for each x E X. We claim that f E C (X) and that lim l f,, - /I I � = To this end, let > Choose no such that 1 /, - fm l oo for n, m > no. It holds for all n , m > no and all x E X. follows from (*) that lf,,(x) - fm(x)l holds for all n > no and all x E X. That is, {fn l But then l f, ,(x) - f(x)l converges uniformly to f on X. Therefore, by Theorem 9.2, f E C(X), and clearly, lim l f,, - /lloo = A sequence of real-valued functions {f,} on a set X is said to increasing if fn f,+1 holds for all n (and, of course, is called decreasing if f,,+1 f,, holds Theorem 9.3.

<

x

=

0.

E

0.

<E

< E

< E

•

0.

be

<

<

for all n). An increasing or a decreasing sequence of functions is referred to as a monotone sequence of functions. It was observed that pointwise convergence need not imply uniform conver­ gence. The next useful result gives a condition under which pointwise convergence implies uniform convergence, which is a classical result known as Dini 's2 theorem.

If

Theorem 9.4 (Dint). Let X be a compact topological space. a monotone sequence of C(X) converges pointwise to a continuous function, then it also

converges uniformly.

{f,, } of C(X) is increasing and convergent f,,(x) t f(x) holds for each x E X.

Proof. Suppose that the sequence pointwise to some E C(X); that is,

f

2Ulisse Dini (1845-1 9 1 8), an Italian mathematician. He became Professor of mathematics at the University ofP isa at the age of 2 I . His main contributions were in the the theory of real functions and partial differential equations.

71

Section 9: CONTINUOUS REAL-VALUED FUNCTIONS =

Now, let E 0. For each n define 011 {x f(x) - �1(x) < E}. Clearly, it each On is open. Also, note that since �� (x) < fn+ (x) holds for all follows that On 011+1 holds for all n. Moreover, since �1(x) t f(x) holds for each we easily get X = u: On. In view of the compactness of there exists some k such that U� 0; = 011 for n k. But this merely says that 0 < f(x) - �1(x) < E Okt and so for all n k and all x In other words, { ��} converges uniform!y to f on If {�1 } is a decreasing sequence, then apply the above arguments to the sequence {- � } . Now, consider a sequence { f�1} of real-valued functions defined on a set For each n let Sn L.; /;. We say that the series L.: j;1 (x) converges uniformly on if the sequence of partial sums {sn } converges uniformly to a real-valued function defined on X. This is, of course, equivalent to saying that the sequence {s11} is a uniformly Cauchy sequence in the sense that for each E 0 there exists some no such that lsn (x) - s111 )I < E holds for all n, m no and all We shall present two useful criteria of uniform convergence of series of func­ tions. The first one is due to Weierstrass and is known as the Weierstrass >

X

c

E X,

>

E X:

1

x E X,

l

X,

X =

X

>

=

=l

E X.

X.

.

�

X

=

X.

1

1 =1

>

(x

X

>

E X.

K.

M-test.

9.5 (Weierstrass' M-test). Let {f11} be a sequence of real-valued functions defined on a set X and assume that for each n there exists some positive rea/number a11 such that I /11 I < a11 holdsfor all x If L.: all converges, then the series L-: 1 �1(x) converges uniformly on X. Proof. LetE 0. Picksomen0suchthat'L.Z�11 ak < Eholdsforalln, m n0. Now, notice that if m no and x then 111 111 m ls11(x) - Sm(x)l L fk (x) < L lfk (x)l < I: ak < E. This shows that the sequence {s11 } of partial sums is uniformly Cauchy, and hence, a uniformly convergent sequence. • The second criterion of uniform convergence of series is due to J. Dirichlet.3 Theorem 9.6 (Dirichlet's test). Let {fn} and {g11} be two sequences of real­ Theorem

E X.

(x)

1

>

>

n,

E X,

>

=

k=ll

k=ll

k=ll

valuedfunctions defined on a set X satisfying thefollowing properties:

3Johann Peter Gustav Lejeune Dirichlel ( 1 805-1859), a Gennan malhemaiician. He worked in number Iheory, analysis, and mechanics. The modem definition of a function was proposed by him in 1837. Because of his fundamenlal papers on lhe convergence of trigonomelric series, he is considered lo be one of the founders of Fourier analysis. He was Ihe Ieacher of Riemann.

72

Chapter 2: TOPOLOGY AND CONTINUITY

There exists a constant C 0 such that 1:�=7=1 .fi(x)l < C holdsfor all n and all x E X, i.e., the seque1ice ofpartial sums of the series L:: 1 /11 (x) is uniformly bounded. 2. For each n we have 811+ 1 (x) < g11(X ) for all x E X and {g11} converges uniformly to zero on X. Then the series L::1 f,1(x)g11(x) converges uniformly on X. Proof. fn} {g11} conditions. 1 tn(x) = L; = 1 /;(x). Clearly, Sn(x) = L:?= l Jj(x)g;(x) 1.

>

Assume that { and and

let

satisfy the stated

For each

11,

S11 (x) = L t;(x)[g;(x) - 8i+l(x)] + 811+ l (x)t11 (x). i=l II

Next, fix E 0 and then choose some n0 such that g11 (x) < E holds for all tn-:Jt , n-

...._. n ::::.

u

wnd "'

>

"'11

"

li..JO".,.' 1-..

. � V ., ii::: .�a. .

lsn(X) - Sm(x)l

II

-

L

i=m+ l


II

L

nnt-=• •v ....,

..........

�f ...._ n� "> J1n and)." .... n , , .... , , . :.... ·u ••

.

t= -

Y thP.n ...... ....,..

. ..

,

t;(x)[g; (x) - 8i+ t (x)] + g11 +l (x)t11 (x) - g,+l (x)t,(x) [g;(x) - gi+l(:r)] + Cg11+ 1 (X) + Cgm+l(x)

i=nt+l 2Cg m+l(x) < 2CE.

This shows that {s11} is a uniformly Cauchy sequence, and hence. the series of fu t s 2::1 f,1(x)g11(x) converges on X. • We now illustrate the preceding theorems with some examples. Example ��l sin��x sin11·��x I < J, � x nc ion

unifonnly

9.7.

2:�1 : 11

Consider the series of functions

11 -

.

11·

Since 1

oo. it follows from the Weierstrass· M-test that the series verges uniformly on IR.. consider the series of functions

Now.

<

f

II=I

cos(2n

11

1

- )x

and

� sin(2n �

11= l

-

11

2:�1 s1�1�1

:- 11-

and

con-

1 )x

over the interval (0. rr). We claim that both series converge uniformly on every closed subinterval [a. b] of (0. rr ). Notice first that the nth term each series is dominated by . However. since L�= 1 * = oo. we cannot apply Weierstrass· M-test to conclude that the series converge uniformly [a. b]. However. we shall reach this conclusion by applying Dirichlet's test. To do this. let = cos(2n - l)x g11(x) = *· Clearly, the sequence {g11} converges monotonically

of

/11

and

*

on

Section 9: CONTINUOUS REAL-VALUED FUNCTIONS

73

and uniformly to zero. On the other hand, from -elementary trigonometry, we know that

11 L cos(2k -

k=l

I )x

sin 2nx

. = -

2smx

2

nx L.:sm(2k - l)x = sinsmx . . 11

and

•

k=l

For instance, to establish the first identity note that

2 sin.t

II

L k=l

cos(2k -

l)x

II

L [sin((2k - l)x + x) - sin((2k - I ).t - x)] k=l II = L [sin(2k.t) - sin(2(k - l )x)] = sin 2nx. k= l =

Now, if we choose some 8 > 0 such that sin x

>

8 for all x E [a , b], then

II

L cos(2k - I )x k= l

I

< -

<

- 2 sinx - 28

holds for all n and all x E [a, b]. Hence, Dirichlet's test applies and shows that the series "oo .r ) ) "oo cos<2n- l )x 1m ·11 ar1 y, th e senes converges um.orm1 y on [a, b] (.\ gil(.\· = x 2 I::_1 srn< :-J> converges uniformly on every closed subinterval of (0, rr). •

. s· Assume now that a sequence {f,,} of real-valued continuous functions converges pointwise to some function f. Although f is not necessarily a continuous function, nevertheless, something can be said about the points of discontinuity of f. We shall show that the set of all points of discontinuity of f is a meager set. Recall that a set is called meager set if it can be written as a countable union of nowhere dense sets. To do this, we need a lemma. �n= l J!f

���=l

·

11

·'"

·

a

Lemma 9.8. Let X be a topological space. Assume that the sequence {f,} of C(X) converges pointwise to some function f. For E > and n E 1N, let

0

V,(E)

=

{x E X : lf,(x) - f(x)l .::: E}.

Also, pat O(E) = U::1 [V11(E)]0 • Then the set of all points of continuity off is the Gs-set n�1 0( � ). Proof. a E X E f lfk (a) - f(a)l E. f��(a) = f(a), k a U fk f lf x E U. x E lf(a) - f(x)j E k(x) - fk(a) l < E

Let be a point at which is continuous, and let 0. Since lim there exists some such that Also, by the continuity of and at the point there exists a neighborhood of such that and for all Then for we >

<

a

<

U

Chapter 2: TOPOLOGY AND CONTINUITY

74 have

l f(x) - fk (x)l < l f(x) - f(a) l + l f(a) - fk (a) l + l fk(a) - fk(x) l < E + E + E = 3E. This implies U C a E n:1 O(f, ).

[Vk(3E)]0, and so a

E

0(3E) for each E

>

0,

which shows that

For the reverse inclusion, assume that a E n: 1 0(�). To finish the proof, we have to show that f is continuous at a. So, let ·E > Pick n such that "l < E. From a e 0( "l) it follows that there exists some k such that a e [Vk( l )]0 • This r in tum means that there exists a neighborhood U of a with U c \'k(;; ); that is, l fk (x) - f(x)l < � < E holds for all x E U. Since fk is continuous at a, there exists a neighborhood W of a such that l fk (x)- fk (a) l < E for each x E W. Thus, if X E U n W, then

0.

lf(x) - f(a)l < l f(x) - fk (x)l + l fk(x) - fk (a) l + l fk (a) - f(a)l < E + E + E = 3E, which shows that

f is continuous at a.

•

We are ·now ready to prove that the set of discontinuities of a function that is the pointwise limit of a sequence of continuous functions is a meager set.

Theorem 9.9.

Let X be a topological space, and let {fn } be a sequence of C(X) . If {fn } converges pointrvise to a real-valued function f, then the set D of all poi11ts ofdisco12tinuity off is a meager set. Proof. According to Lemma 9.8, we have D U: 1 [0(� )]c. To complete the proof, it suffices to show that every set of the fonn [O(E)]c is a meager set; it =

will then follow that D is a meager set, since it will be a countable union of meager sets. So, let E > For each m define

0.

Fm(E) = n {x E X : lj;ll(x) - fm+i(x) l :5 E}, 00

i=l

and note that each Fm(E) is a closed set. Also, since for each x E X we have Fm(E). lim j;,(x) = f(x), it is easy to see that X = Using once more the fact that lim fn(x) = f(x) for each x E X , it is easy to see that Fm(E) C Vm(E) holds. Therefore, [Fm(E)]° C [Vm{E)]° C O(E) holds for

U�=l

Section 9: CONTINUOUS REAL-VALUED FUNCTIONS

75

each m, and so, u�= l [Fm(E)]0 c O(E) also holds. Now, observe that [O(E)]c

00

00

X

u Fm(E) \ u [Fm(E)t m=l m= I 00 00 c UrFm(E)\[Fm(E)]0] u oFm(E), m =l m=l =

X

\ O(E) C X \ u (Fm(E)]0

=

=

where the last equality holds since each Fm(E) is a closed set. Now observe that since each F111 (E) is closed, its boundary oF111 (E) is a nowhere dense set, and thus, according to the last inclusion [O(E)]c is a meager set, and the proof is finished.

•

Our next objective is to characterize the (uniform) compact subsets of To do this we need a definition. Let be a topological space, and let S be a subset of Then the set S is said to be equicontinuous at some x E if for each E there exists a neighborhood V of x such that E V implies l f(y) - f(x) l < E for every f E S. If S is equicontinuous at every point of then S is called an equicontinuous set. As we have seen before, in a metric space a closed and bounded set need not be compact. However, if is a compact topological space, then a closed and bounded (with respect to the uniform metric) subset of is compact if and only if it is equicontinuous. This result is known as the Ascoli-ArzeUt4· theorem and is stated next. C(X).

X

y

X

C(X). > 0

X,

X

C(X)

5

Let X be a compact topological space, and let S be a subset of C (X). Then thefollowing statements are equivalent: 1. S is a compact subset ofthe metric space C(X) (equipped, ofcourse, with the uniform metric). 2. S is closed, bounded, and equicontinuous.

Theorem 9.10 (Ascoli-Arzela).

Proof. (1) => (2)

We already know that a compact set is closed and bounded. What remains to be shown is that S is equicontinuous. To this end, let E 0. Choose f1 , . , J,, S such that S c U�'=1 B(J;, E). If x then pick a neighborhood V.r of x such that l fi(y) - Ji (x) l < E holds for all and all i . . , n. Now, let E Vx and f E S. Choose some with >

E X, y E V.r

=

E

. .

1,

.

y

i

4Guido Ascoli ( 1 887-1957), an llalian mathematician. He contributed to the theory of real functions and differential equations. 5 Cesare Arzela ( 1 846-1912), an llalian mathematician. He studied the convergence of sequences of real functions.

Chapter 2: TOPOLOGY AND CONTINUITY

76

f E B ( f; ,

E), and note that

l f (y) - f(x )l

< l f(y) - f;(y)l + l f;(y) - J;(x)l + 1/;(.r) - f(x)l < + +E =

E E

3E.

This shows that S is equicontinuous at x, and since x is arbitrary, S is an equicon­ tinuous set of functions. (2) ==> (1) Let {/11 } be a sequence of S. According to Theorem it suffices to show that u�a has a convergent subsequence. To this end, choose some M > satisfying I f(x) I < M for all x E X and f E S. Using the equicontinuity of S and the compactness of X , it is easy to see that for each k there exists a finite subset Fk of X and neighborhoods {\',.: y E Fd such that X = UyeF, Vy and l f (x) - f (y) l < t whenever x E Vy and f E S. Let F = U:, F;. Clearly, F is at most countable; assume F countable and iei F = {x1, x1, . . . } be an enumeration of F. since l f,,(xdl < /'.1 holds for all n , there exists a subsequence {g,�} of {/11} such that lim g,� (x 1 ) exists in JR. Similarly, there exists a subsequence {g� } of {g,�} so that lim g� (x�.) exists in JR. Continuing this way, we can choose (inductively) sequences {g;,} (i = I . 2, . . . ) such that

7.3,

0

Now,

a. b. c.

{g,�} is a subsequence of {/11}, {g;/ 1 } is a subsequence of {g;,} for each i = 1 , 2, . . . , and limu-Hx) g;, (x; ) exists in JR. for each i = 1 , 2, . . . .

Now, consider the diagonal sequence h, = g;; , and note that { h 11} is a subse­ quence of {/11} such that lim11_00 h11(x; ) exists in JR. for each i. Moreover, we claim that { h 1 } is a Cauchy sequence of C(X). To see this, fix k and then choose some no so that lh,(y) - h111(y)l < t holds for all n, m > no and all y E Fk . if x E X , then pick some y E Fk such that x E Vy, and note that 1

Now,

lh"(x) - hm(x)l

holds for alJ n, m

>

< lhu(X) - h, (y) l + lh"(y) - hm(Y) I + lhm(Y) - hm(x )l < t+f+t=f n0• That is,

llh" - hmlloo = sup { lh ,;(x ) - hm(x)l: X

E X} < f

holds for all n , m > n0, so that {h,} is a Cauchy sequence of C(X). By Theorem 9.3, { h 11 } converges to some h E C(X). Since S is closed, and the proof of the theorem is complete.

h

E

S,

•

77

Section 9: CONTINUOUS REAL-VALUED FUNCTIONS

Let S be a bounded equicontinuous subset of some C(X)-space with X compact. It is not difficult to establish that the (uniform) closure S of S is likewise bounded and equicontinuous, and thus by the Ascoli-Arzela theorem, S is a compact subset of C(X). This implies that every sequence of S has a subsequence that converges uniformly. In particular, every bounded equicontinuous sequence has a uniformly convergent subsequence. This last observation is very useful in establishing the existence of solutions to differential equations.

EXERCISES 1.

If u, v, and w are vectors in a vector lattice, then establish the following identities: a. b. c. d. e. f. g.

2.

u v v + u 1\ v = u + v; u - v v w = (u - v) 1\ (u - w); u - v 1\ w = (u - v) v (u - w); a(u 1\ v) = (au) 1\ (au) if a 2::. 0; l u - vi = u v v - u 1\ v; u v v = �(u + v + l u - vi); ll A v = cu + v - !u - vi).

If u and

a. b.

!

v are elements in a vector lattice, then show that:

Itt + vi v l u - v i = lui + l v l , and Ill + v i 1\ ill - v i = l l u l - lull.

[HINT:

lu + vi

that ILL I

v

lu - vi = (u + v) v (-u - v) v (u - v) v (-u + v) = (lui + v) v (l u i - v) = lu i + lvl. ]

3.

Show

4.

Show that the vector space consisting of all polynomials (with real coefficients) on

1\

!vi = 0 holds if and only if lu + vi = l u - vi holds.

R the vector space of all real-valued

is not a function space. Prove a similar result for differentiable functions on R. 5. Let X be a topological space. Consider the collection L on X defined by L = {f E

Rx :

3

of all real-valued functions

{/n} £ C(X) such that lim f, (x) = f(x)

VxE

X }.

that L is a function space. 6. Let L be a vector space of real-valued functions defined on a set X. If for every function f e L the function 1/1 [defined by 1/ l(x) = 1/(x) l for each x e X] belongs to L, then show that L is a function space. 7. Consider each rational number written in the form � , where n > 0, and m and n are integers without any common factors other than ± 1. Clearly, such a representation is unique. Now, define f: R � R by f(x) = 0 if x is irrational and j(x) = t if x = ':: as above. Show that f is continuous at every irrational number and discontinuous at every rational number. Show

Chapter 2: TOPOLOGY AND CONTINUITY

78

IR

Let f: [a, b] -+ be increasing [i.e., x < y implies f(x) � f(y)]. Show that the set of points where f is discontinuous is at-most countable. [HINT: If f is discontinuous at c, with a < c < b, then choose a rational number r such that limx tc f(x) < r < limxk f(x).] 9. Give an example of a strictly increasing function f: [0, 1] -+ IR which is continuous at every irrational number and discontinuous at every rational number. mapping if f(V) is open 10. Recall that a function f: (X, r) -+ (Y, r1 ) is called an is a continuous open mapping, then f whenever V is open. Prove that if f: IR -+ is a strictly monotone function-and hence, a homeomorphism. ge let 11. Let X be a nonempty set, and for any two functions 8.

open

IR

J,

d(f g) '

=

-���

l f(x) - g(x) l 1 + l f(x) - g(x)l

IRX

Establish the following: a. b. 12.

13.

(IRX, d) is a metric space. .t.. sequence satisfies d(_(11, f) -+ 0 for some f e {/11 } converges uniformly to f.

(Jr.} 5; lR.x

IRx if and only if

Let f, ft , /2 . . . . be real-valued functions defined on a compact metric space (X, d) such that Xn -+ x in X implies j11(Xn) -+ f(x) in If f is continuous, then show that the sequence of functions {J,,} converges uniformly to f. For a sequence { fn} of real-valued functions defined on a topological space X that converges uniformly to a real function f on X, establish the following:

IR.

a. · If x11 -+ x and f is continuous at x, then J,, (x11 ) -+ f (x ). b. If each /11 is continuous at some point xo e X, then f is also continuous at the point xo and lim

X-+Xo

14. 15.

16.

IR

lim /11(x)

11 -+00

18.

lim lim J,,(x) n-+00 X-+Xo

=

f(xo).

Let f, : [0, 1] -+ be defined by j11(x) = x" for x e [0, 1]. Show that (/11 } converges pointwise and find its limit function. Is the convergence uniform? Let g: [0, 1] -+ be a continuous function with g( l ) = 0. Show that the sequence of functions (/11 } defined by fn(x) = x"g(x) for x e [0, 1], converges uniformly to the constant zero function. Let (/11 } be a sequence of continuous real-valued functions defined on [a, b], and let {a, } and {b,} be two sequences of [a, b] such that lim a11 = a and lim b11 = b. If {/11} converges uniformly to f on [a, b], then show that

IR

nlim -+oo 17.

=

1h, a,

fn(x)dx

=

1h

f(x)dx.

a

Let (/11 } be a sequence of continuous real-valued functions on a metric space X such that { fn } converges uniformly to some function f on every compact subset of X. Show that f is a continuous function. Let (/11 } and {g,} be two uniformly bounded sequences of real-valued functions on a set X. If both {/n } and {g,} converge uniformly on X, then show that {fng11} also converges uniformly on X.

Section 9: CONTINUOUS REAL-VALUED FUNCTIONS 19.

20.

21.

22.

23.

79

Suppose that {.ft,} is a sequence of monotene real-valued functions defined on [a. b] and not necessarily all increasing or decreasing. Show that if {.ft, } converges point­ wise to a continuous function f on [a. b], then {j�,} converges uniformly to f on [a, b]. [HINT: Use the fact that f must be uniformly continuous on [a, b]. ] Let X be a topological space and let {!11 } be a sequence of real-valued continuous functions defined on X. Suppose that there is a function f: X --+ IR such that f(x) = lim .ft,(x) holds for all x E X. Show that f is continuous at a point a if and only if for each € > 0 and each m there exist a neighborhood V of a and some k > m such that lf(x) - fk(x)l < E holds for all x E V . Let {f, } be a uniformly bounded sequence of continuous real-valued functions on a closed interval [a, b]. Show that the sequence of functions {t/>11} , defined by t/>11 (x) = f�r .ft,(t} dt foreachx E [a, b} , contains a uniformly convergent subsequence on [a, b]. For each n, let f11 : IR --+ IR be a monotone (either increasing or decreasing) function. If there exists a dense subset A of IR such that lim j,(x} exists in IR for each x E A, then show that lim f, (x) exists in IR at most for all but countably many x. Consider a continuous function f: [0, oo) --+ JR. For each n, define the continuous function .ft,: [0, oo) --+ IR by f (x) = f(x."). Show that the set of continuous functions {ft , fz, . . . } is equicontinuous at x = I if and only if f is a constant function. Let (X, d) be a compact metric space and let A be an equicontinuous subset of C(X). Show that A is uniformly equicontinuous, i.e., show that for each E > 0 there exists some 8 > 0 such that x. , y E X and d(x, y) < 8 imply lf(x) - f(y}l < E for all ,

24.

f E A.

25.

26.

27.

28.

22

Let X be a connected topological space (see Exercise of Section 8 for the definition) and let A be an equicontinuous subset of C(X). If for some xo E X, the set of real numbers {f(xo): f E A} is bounded, then show that {f(x): f E A} is also bounded for each x E X . Let {j, } be an equicontinuous sequence in C(X), where X is not necessarily compact. If for some function f: X --+ IR we have lim f, (x) = f(x) for each x E X, then show that f E C(X). Let X be a compact topological space, and let {.ft,} be an equicontinuous sequence of C(X). Assume that there exists some f E C(X} and some dense subset A of X such that lim f, (x) = f (x) holds for each x E A. Then show that {j,} converges uniformly to f. Show that for any fixed integer n > the set of functions f E C[O. 1 ] such that there is some x E [0, 1 - �] for which

1

l f(x + h) - f(x)l :S nlz

whenever 0 < lz

<

�.

is nowhere dense in C[O, 1] (with the uniform metric). Use the preceding conclusion and Baire's theorem to prove that there exists a continuous real-valued function defined on [0, 1] that is not differentiable at any point of [O, 1]. 29. Establish the following result regarding differentiability and uniform convergence. Let {.ft,} be a sequence of differentiable real-valued functions defined on a bounded

Chapter 2: TOPOLOGY AND CONTINUITY

80

open interval (a, b) such that:

for some xo E (a, b) the sequence of real numbers {/11 (xo)} converges in IR., and b. the sequence of derivatives { f,�} converges uniformly to a function g: (a, b) --+ IR.. a.

Then the sequence { f11} converges uniformly to a function f: (a. b) differentiable at xo and satisfies f'(xo) = g (xo) .

10.

--+

IR that is

SEPARATION PROPERTIES OF CONTINUOUS FUNCTIONS

It is possible that the only real-valued continuous functions on a topological space are the constant ones. For instance, any indiscrete topological space has this property. Such spaces will be of little interest to us here. In this section, we shall describe a large class of topological spaces with an abundance of continuous real-valued functions. We start our discussion with two notions of separation.

Definiiion iO.i. a.

Two disjoiiit subsets A and B ofa topological space X are:

separated by open sets,

if !here exisl two disjoin! opel/ se/s V aud

salisfyiug A c V and B c W . a11d

b.

separated by a continuous function, f: X

--+ [0, 1]

b E B.

Lemma 10.2.

such !hal f(a)

=

W

if !here exisls a colllinuous funcJioll

Ofor each a E A and f(b)

= 1 for each

If two disjoin! subse/s of a Jopological space are separaled by

a conJinuous function , !hen they are also .fiepara!ed by open se/s.

Proof. Let A and B be two disjoint subsets of a topological space X and let

f: X

f(b)

--+ [0. I ] be a continuous function such that f(a) = 0 for each a E A and = 1 foreach b E B. IfV = {x E X : f(x) < � } and W = {x E X : f(x) > 41.

then V and

W are two disjoint open sets satisfying A c

V

and B C

W.

•

Topological spaces whose disjoint closed sets can be separated by open sets play an important role in mathematical analysis and they are referred to as normal spaces.

Definition

10.3.

A lopological space is said to be

normal

if every pair of

disjoin! closed se/s can be separated by open se/s.

Here are two classes of normal topological spaces.

Lemma 10.4. normal spaces.

Metric spaces and compact Hausdorff topological spaces are

81

Section 10: SEPARATION PROPERTIES OF CONTINUOUS FUNCTIONS

Let (X, d) be a metric space and let A be a nonempty subset of X. Define the distance function d(·, A): X 1R of A by d(x, A) = inf(d(x, y): y E A} . The non-negative number d(x, A) is called the distance of x from A . It is easy to see that ld(.r. A) - d(y, A)l < d(x, y) for all x, y E X, and so, d(·, A) is a uniformly continuous function on X. Now, if A and B are two nonempty closed disjoint subsets of X, consider the continuous function f: X --+ 1R. defined by f(x) d(x, A)-d(x, B). Clearly, the open sets V f 1 (( - , 0)) and W f 1 ((0, oo)) are disjoint sets satisfying A c V and B c W. Next, let X be a Hausdorff compact topological space and let A and B be two closed disjoint sets. By Theorem 8.12, both A and B are compact sets. Fix E A. Then for each h E B there exist neighborhoods Vh of a and Wh of h such that vh wh 0. From B c Uhe B wh and the compactness of B, there exist Wa. If Vu n;'=l vhi' then Vu is a h i , . . . , h, E B such that B c U�'= l wh; neighborhood of a satisfying Vu Wu (/). Now from A C UueA Vu and the compactness of A, we see that there exist a . . . , ak E A such that A c u;:::; Vu;. If V U7= 1 Va; and W n;=l Wai' then V and W are two disjoint open sets satisfying A c V and B c W. • The next6 result characterizes the normal topological spaces and is known as Uryson 's lemma. Proof.

--+

=

a

n

-

-

=

oo

=

=

=

n

=

=

=

Theorem 10.5 (Uryson's Lemma).

=

I,

I

For a topological space X thefollowing

statements are equivalent. 1. X is a normal space. 2. If A is a closed subset and V is an open subset of X satisfying A c V, then there exists an open set such that A c c c V. 3. Eve1y pair ofdisjoint closed sets can be separated by a contimzousftmction. 4. If C is a closed subset of X and f: C --+ [0, 1] is a continuous function, then there exists a continuous extension of f to all of X with values in [0, 1]. Proof. ( 1) ==> (2) A V

W

W W

Assume that is a closed set and is an open set such that A c V. Put C vc and note that A and C are two disjoint closed sets. So, there exist two disjoint open sets W and satisfying A c W and C c From we easily infer that W (/), and thus W c uc c cc V. W =

n U = 0.

U nU =

U. =

6Pavel Samuilovich Uryson ( 1898- 1924), a Russian mathematician. Although his scientific activity lasted for only five years-he drowned off the coast of Brittany (France) at the age of 26 while on vacation-he made several important contributions to general topology.

Chapter 2: TOPOLOGY AND CONTINUITY

82

(2) ==::?

(3) Let A and

B be two disjoint closed sets. Put V = Be and note that A c \1. Next, put r0 = 0, r1 = I , and let {r2, r3, . . . } be an enumeration of the rational numbers in the open interval (0, 1 ) . Our hypothesis implies the existence of two open sets V,.0 and V,.1 such that A C V,.1 C V ,.1 C V,.0 C V ,.0 C V . Now, proceed inductively. Assume that the open sets V,.0 , , V,., have been chosen so that r; < ri implies A c V,.j c V,.j c V,.1 • Observe that there •

•

•

are precisely two rational numbers ri and ri among ro, r1, , r11 such that r; < , r11 lies in the r11 + 1 < ri holds, and so no other rational number among ro, r1, open interval (ri, ri ). Clearly, r; = max{rk: rk < �"n+ l and 0 < k < n}. Similarly, rj = min{rk: �"n+ l < rk and 0 < k < }. By our hy.pothesis, there exists an open set V,.,+1 such that •

•

•

•

•

•

n

c u,.t�+l c c v,.11+ 1 v,.J. - v,.I . Thus, if Q denotes the set of all rational numbers, we can construct a collection of open sets r E Q n [0, i ] } with the foiiowing two propenies: a. b.

{ V,.:

A c V,. c V for each r E Q n [0, 1]. If r, s E Q n [0, 1 ] satisfy s > r, then Vs c

V,. holds.

Next, we define the function f: X -7 [0, 1 ] by

-I

f(x) -

sup{r: 0

x e V,. }

if X E Vo if x r{. Vo.

Clearly, f(x) = 1 for all x e A and f(x) = 0 for each x e v c = B . We claim that f is continuous. To see this, fix a e X and let E > 0. Assume first that 0 < f(a) < 1 holds. Choose two rational numbers s, t in [0, 1 ] such that f(a) < s < t < f(a) + E . From V, c Vs and f(a) < s it follows that a r{. � . If f(a) > 0, choose a rational number r e (0, 1 ) such that f(a) - E < r < f(a) and a e V,. . Put U = V,. \ V, ; clearly, U is a neighborhood of a. Also, note that if x E U, then r < f(x) < t holds. Thus, lf(x) - f(a)l < E holds for all x E U. If f(a) = 0, put U = X \ � . Then U is a neighborhood of a , and for x E U we have 0 < f(x) < t. Therefore, lf(x) - f(a)l < E holds for all x E U. Hence, in either case f is continuous at a. Finally, in case f(a) = 1, choose a rational number r E (0, 1) such that a E V,. and 1 - E < r. Clearly, V,. is a neighborhood of a.. Also, if x E V,., then r < f(x) < 1, from which it follows that lf(x) - f(a)l < E holds for all x E V,.. The above arguments show that f is continuous at each point a E X, and hence f is a continuous function separating A and B . ===:} (4 ) Let C be a (nonempty) closed subset of X and let f: C -7 [0, 1 ] be a continuous function. We shall consider f as a continuous function from C to [-1 , 1].

(3)

Section 10: SEPARATION PROPERTIES OF CONTINUOUS FUNCTIONS

83

Start by observing that if A and B are two disjoint closed subsets of X and .81 is an arbitrary closed interval, then there exists a continuous function ¢:X .81 with ¢ on A and ¢ .B on B. (Indeed, if 1/J: X I 1 is a continuous function such that 1/J 0 on A and 1/J on B, then the continuous function¢ has the desired properties.) The extension of the function f will be based upon the following property C 1 fc g: X lR 2 < 3,. for all -'" E X and lh(c) - g(c)l < 3 ,. for all c E C. To verify property we argue as follows: Let A -j]) and 1 B - ( J . The continuity of h guarantees that the disjoint sets A and B are closed in C. Since C is a closed subset of X. it follows that A and B are also closed subsets of X. So, there exists a continuous function g: X j . } 1 (i.e., < j for each E X) such that g -} on A and 1 on B. Now, let lcg(x) l E C. If c E then < h(c) < -1 g(c) and so lh(c) - < �r. If c E B, then g(c) 1 < < and hence, l h(c) - g(c) l < �,.. Finally, if g(c) l < �,. c � A U B, then - 1 < < j-, from which it follows that holds true in this case too. Now, we claim that there exists a sequence of continuous real-valued func­ tions on X such that for each we have 1 ( 2 ) 1/- l l g,(x) l < 3 3 for all E X. and ( ) for all c E C. /(c) - L gi(c) .:S 3 i=l The existence of the sequence {g11} can be established by induction as follows: For we apply property with and f. So, there exists a continuous functiong1:X --T lRsatisfying l g 1(x) l < ! foreachx E X and l/(c)-g,(c)I .:S � for each c E C. Now, for the induction step, assume that /1, f, have been selected to satisfy (*) and (**). Applying property (E) with (jY' and lz 1 f L�=1 g;, we see that there exists some continuous function g,+ 1 : X lR satisfying l gu+t(x)l < �(�Y' foreachx E X and 1 + 1 3 2 1 + 1 ( ) f(c) - L i=l g;(c) < for all c E C. [a,

--7

[a,

=

a

=

--7 [0,

= I

=

= (,8 - a)l/J + a

(E): If h: --7 [-r. r is a continuousftmction, then there exists a continuous un tion --7 satisfying lg(x)l

1

(E),

=

h

[

=

h - 1 ([-r,

r]).

--7 [-

x

-r

A.

=

g

=

h(c) h(c)

r,

=

g(c)l

=

lh(c) -

{g,}

n

x

2

II

11 =

1,

(E)

r=I

"

h=

•

r

•

•

,

=

=

--7

84

·

Chapter 2: TOPOLOGY AND CONTINUITY

1 From I:: 1 �(�)''- = 1 and (*}, we see that the series g(x) = I:: 1 g11(x) converges uniformly, and so (by Theorem 9.2) g defines a continuous function from X into [ - 1 , 1]. Now, a glance at (**) guarantees that g (c) = f(c) for all

c E C.

Next, consider the function lg I: X

extension of f to all of X.

�

[0, 1 ] and note that lg I is a continuous

B be two nonempty closed disjoint subsets of X. Then A U B is a closed set, and the function f: A U B � [0, 1], defined by f(a) = 0 for each a E A and f(b) = 1 for b E B, is continuous. Indeed, if a E A and € > 0, then Be is a neighborhood of a and lf(x) - f (a )l = 0 < € holds for each x e (A U B) n Be. This shows that f is continuous at a E A; similarly, f is continuous at every b e B. If g is a continuous extension of f to all of X with values in [0, 1], then g clearly separates A and B. By Lemma 10.2, A and B can (4) =? ( 1) Let A and

•

be separated by open sets, and so X is a normal space.

Our next result is the celebrated Tietze's7 extension theorem.

Let C be a closed subset ofa normal space X and let f: C � 1R be a continuous function. Then there exists a continuous extension off to all of X with values in lR. Proof. Let f: C � 1R be a continuous function, where C is a closed subset of a normal space X. Assume first that f(x) > 0 for each x E C. Put h = 1{1 and note that h: C � (0, 1) c [0, 1] is a continuous function. By Uryson's lemma (Theorem 10.5}, there exists a continuous extension h0: X � [0, 1] of h. 1 Next, let B = 170 ({I }) and note that B is closed. Also, since 0 < h0(x) < 1 Theorem 10.6 (Tietze's Extension Theorem).

for each x E C, we see that A n C = 0. Using Uryson 's lemma once more, we see that there exists some continuous function ¢: X � [0, 1] such that (c) = 1 =

¢

B. Now notice that the function g = ����;,,1 is a continuous extension of f to all of X with values in lR. For the general case, assume that f: C � 1R is continuous and write f = f+ - f-, where f + = f v 0: C � 1R and f- = (-f) v 0: C � 1R are two

for all c E C and ¢(b)

0 for all b E

continuous non-negative continuous functions. By the first part, there exist two notice that ¢

=

¢1

�

f+ and f-, respectively. - ¢ 1 : X � 1R is a continuous extension of f .

continuous functions ¢ 1 , ¢2: C

A topological space (X,

1R that extend

Now

•

T) is called locally compact if every point of X has a

neighborhood whose closure is a compact set. Clearly, every compact topological space is locally compact. By Theorem 7.4

a subset of lR" is compact if and only if it is closed and bounded. Thus, it follows

n

that a Euclidean space 1R is not compact but it is locally compact.

7 Heinrich Franz Friedrich Tietze ( 1880-1964 ), an Austrian mathematician. He worked in topology

and he is well known today for his famous "group transformations."

Section 10: SEPARATION PROPERTIES OF CONTINUOUS FUNCTIONS

85

For locally compact spaces we have the following separation property: Lemma 10.7. Let (X, r) be a Hausdorff locally compact topological space. Assume that V is an open set and A is a compact set such that A c_y . Then there exists an open set 0 with compact closure such that A c 0 c 0 c V .

Since each point of A has a neighborhood with compact closure, and since A can be covered by a finite number of these neighborhoods, it easily follows that there exists an open set W with compact closure such that A c W . Replacing W by W n V (if necessary), we can assume that A c W c V holds. If w n vc then 0 w satisfies A c 0 c 0 c v. Otherwise, if X W n vc, then ¢ A, and so by Theorem 8.13 there exists an open set Ux such that A c andx ¢ . Now,observethatthefamily {(V t : wnvc} isanopen cover of the compact set W n vc (its compactness follows from Theorem 8.12). Thus, there exists a finite subset F of w n vc such that w n vc c UxeF (U_ . ;)c. Note that ( n xeF ) n w n vc (/). Put 0 n xeF (Ux n W), and note that 0 is an open set such that A C 0 C W . It follows that 0 c W, and consequently, 0 n vc 0 n W n vc c ( n U.r ) n W n vc 0 . xeF • Hence, A c 0 c 0 c V holds, and the proof is finished. The following theorem is a "locally compact version" version ofUryson 's lemma. Proof.

= 0. x v...

v...

=

E

=

...

v...

X

E

.

=

=

=

Theorem 10.8 (Uryson). Let X be a Hausdorff locally compact topological space, and let A be a compact subset of X . If V is an open set such that A c V, then there exists a contimwusfunction f: X -+ [0, 1] such that f (x ) all x E A and f(x) = Ofor all x E

=

1 for

vc. Proof. The proof is identical to the proof of the implication (2) (3) of Theorem 10.5. The only difference in the proof is that instead of statement (2) we must invoke Lemma 10.7. • As an application of Theorem 10.8, we shall present a useful result dealing with "partitions of unity." Iff: lR is a function, then the closure of the set Y {x f(x) =f:. 0} is called the support of f and is denoted by Supp f. That is, Supp f Y. A function is said to have compact support if its support is a compact set. ==>

X -+

=

E X:

=

Theorem 10.9. Let X be a Hausdmff locally compact topological space, and V, are open sets such tlzat A c let A be a compact subset of X. If V1, •

•

•

,

Chapter 2: TOPOLOGY AND CONTINUITY

86

U;'=t V;, then there exist continuous real-valued functions /1

•

•

•

•

,

f,, on X

satisfying these properties: 1 . 0 < f;(x) < 1 holdsfor all x E X and each 1 < i < n. 2. Each J; has compact support, and Supp /; V;. 3. 2:::;'= 1 f;(x) = 1 holds for all x E A. Proof. Let x E A. Then there exists some i (1 < i < n) such that x E

C

V; . Since {x} is a compact set, it follows from Lemma 10.7 that there exists a neigh­ borhood Ux of x with compact closure such that Ux C V;. That is, every x E A has a neighborhood Ux with compact closure satisfying Ux c V; for some i. Let XI ' . . . ' Xm be a finite number of points of A such that A c u�:l Ux; . Next, for each i define 0; to be the union of all those Ux for which U.rj C F; holds (if no i such Uxi exists, then 0; = 0). Clearly, each 0; is an open set with compact closure

satisfying 0; c V;. Moreover, A c U;'= 1 0; holds. By Lemma 10.7, for each i there exists an open set B; with compact closure such that 0; c B; c B; c V;. By Theorem 10.8, for each i there exists a continuous function g;: X -+ [0 , 1] such that g;(x) = 1 for each x e 0; and g;(x) = 0 for each x ¢ B;. Also, by the same theorem, there exists a continuous function h: X -+ [0, l ] such that h (x ) = 1 for all x E A, and h(x) = O for all x E [ U7=1 O;]c. Put g = ( 1 - h ) + I:;'=1 g;, and note that g is a continuous function with g(x) > 0 for all x E X. Now let J; = g; Ig for i = 1, . . . , n . We leave it for the reader to verify that /1 , , f,, satisfy the desired properties. • •

•

•

Any collection of functions /1, , J,, that satisfies the properties of Theo­ rem 10.9 is referred to as a partition of unity for A subordinate to the open cover {V1 , , V, }. •

.

•

•

•

•

EXERCISES 1.

2.

3.

4.

Let (X, d) be a metric space and let A be a nonempty subset of X. The distance function of A is the function d(·, A): X ---? R defined by

d(x, A) = inf{d(x, a): a E A}. Show that d(x, A) = 0 if and only if x E A. Let (X, d) be a metric space, let A and B be two nonempty disjoint closed sets, and consider the function f: X ---+ [0, 1 ] defined by f(x) = d(x.�� ��l-.:.B>. Show that: a. f is a continuous function, b. f- 1 ({0}) = A and f- 1 ({ 1}) = B; and c. if inf{d (a , b): a E A and b E B) > 0, then f is uniformly continuous. Let A and B be two nonempty subsets of a metric space X such that A n B = A n B = (/). Show that there exist two open disjoint sets U and V such that A � U and B � V . Show that a closed set of a nonnal space is itself a normal space.

Section 11: THE STONE-WEIERSTRASS APPROXIMATION THEOREM

87

Let X be a normal space and let A and B be two disjoint closed subsets of X. Show that there exist open sets V and W such that A � V, B � W and V n W = (/). Show directly that a topological space is normal if and only if for each closed set A and each open set V with A c V , there exists an open set W such that A c W c W c V . For a closed subset A of a normal topological space X establish the following:

5. 6. 7.

a. There exists a continuous function f: X � [0, I ] satisfying f- 1 ({0}) = A if and only if A is a G0 -set. b. If A is a G 0-set and B is another closed set satisfying A n B (/), then there exists a continuous function g: X � [0, I] such that g- 1 ({0}) = A and g(b) I for e ach b E B . .

=

Show that a compact subset A qf a Hausdorff locally compact topological space is a G c5-set if and only if there exists a continuous function f: X � [0, l ] such that 1 A = j- ({0}). 9. A topological space X is said to be perfectly normal if for every pair of disjoint closed sets A and B there exists a continuous function f: X � [0, I] such that A = J-1 ({0}) and B = f- 1 ({ l }). (Part (b) of Exercise 2 above shows that every metric space is perfectly normal.) Show that a Hausdorff normal topological space is perfectly normal if and only if every closed set is a G0-set. 10. Show that a nonempty connected normal space is either a singleton or uncountable. 11. Let X be a normal space, let C be a closed subset of X, and let I be a nonempty interval. If f : C � I is a continuous function, then show that f has a continuous extension to all of X with values in I. 8.

11.

THE STONE-WEIERSTRASS APPROXIMATION THEOREM

In this section we shall present some conditions under which a linear subspace of

C(X), with X compact, is dense in C(X) with respect to the uniform metric. The

main result of this sort is known as the Stone-Weierstrass approximation theorem and is a classical result. ·Before stating and proving this theorem, we need some preliminary discussion.

L of real-valued functions defined on a set X is said to separate the points of X if for every pair of distinct points x and y of X there exists a function f E L such that f(x) =I= f(y).

A

collection

Our first result presents a property of vector spaces of functions that separate

the points. The constant function equals

1

is the function whose value at every point

1.

Let X be a nonempty set, and let L be a vector space of real­ valuedfunctions on X that separate the points of X and contains the constant function 1. Then given any two distinct points x and y of X and real numbers a and {3, there exists some f E L such that f(x) = a and f(y) = {3. Lemma 11.1.

88

Chapter 2: TOPOLOGY AND CONTINUITY

Proof. Since L separates the points of X there exists g

e

1 y - [(a

g(y); put y = g(x) - g(y). Then the function f = ag(y))l] belongs to L and satisfies f(.x) = a and f(y) = {3.

L such that g(x) # - f3)g + (f3g(.x) -

•

The next result presents a local approximation property that enables us to ap­ proximate a function from above at a given point.

Let X be a compact topological space, and let L be a function space ofcolltinuousfunctions that contains the constantfunction ! and sepaz-ates the poiuts of X. Then given a function g e C{X), a poiut a e X, and � > 0, thel·e exists a function f ill L such that Lemma 11.2.

f(a) = g(a) and f(.x) > g(x) - � for all x e X. Proof. For each x e X, there exists (by Lemma 1 1 . 1 ) a function f,. e L such that fx(a) = g(a) and fx(x) = g(x). Since f-: and g are continuous functions. there exists a neighborhood V:t of .x such that fJ.· (y) > g(y) - � for all y e Vx.

Since X = Ux eX v\' and X is compact, there exists a finite number of points , x, of X such that X = u�= l v\'m • Let f = !x i v . . . v fxn ; clearly, f e L Xt, and f(a) = g(a). Also, if x e X, then there exists some m such that .x e V"m' Thus, f(x) > f"nr(x) > g(x) - � holds, which shows that the function f satisfies the required properties. • •

•

•

The lattice version of the Stone-Weierstrass theorem8 is presented next.

Let X be a compact topological space, and let L be afunction space ofcontinuousfunctions separating the points ofX and containing the constant function 1. Then L is dense in C(X) with respect to the uniform metz-ic. Proof. Let g e C{X), and let � > 0. For each x e X, use Lemma 1 1 .2 to choose a continuous function fx e L such that f" > g - � and f"(x) = g(.x). From the inequality fx (x) = g (x) < g(x) + � and the continuity of f and g at x, it follows that there exists a neighborhood Vx of x such that fx (y) < g(y) + � for all y e Vx. Since X is compact, there are points x, , . . . , x, such that X = U�,=l Vxm . Let f = !xi 1\ • · 1\ fxn• and note that f e L. Also, since f > g - �. it easily follows that f > g - �. On the other hand, if X e X, then there exists some m such that X e v and so f (.x) < fxm (x) < g(x) + � holds. Thus, g(x) - � < f(x) < g(x) + � holds for all .x e X, and consequently, II/ - gll oo = sup { l f(x) - g(x)l: x e X} < �. The proof of the Theorem 11.3 (Stone-Weierstrass).

..

•

..nr

..nl ,

theorem is now complete.

•

Marshall Harvey Stone ( 1 903-1989), an American mathematician. He contributed to Boolean algebras, topology, the theory of functions, and functional analysis. R

Section 11: THE STONE-WEIERSTRASS APPROXIMATION THEOREM

89

Our next result deals with the uniform approximation of the square-root function by polynomials. Lemma 11.4.

There exists a sequence of polynomials that converges uni­ formly to .jX in tbe interval [0, 1 ].

Proof.

put

Start by defining P1(x)

0

=

for all x E [0, 1], and then, inductively,

for n > l. Clearly, P,} is a sequence of polynomials. We claim that 0 < P,(x) < .jX holds for each n and all x E [0, ]. The proof of the claim is by induction. For n 1 the claim is triv ial. Assume now that 0 < P,(x) < .jX holds for all E [0, 1 ] and some n. Clearly, 0 < P,+1(x) holds for all x E [0, Also, {

1

=

I].

x

-/X ·- P,+ l (x)

=

-/X - P,(x) - Hx - ( P,(x))2 ] [-/X 4c.JX +

P,(x)J[I P,(x))], and the two factors of the last product are nonnegative from our induction hypoth­ esis. Therefore, P,+ 1(x) < .jX forallx E [0, 1 ] . Now, from the definition of P,+ 1 and the fact that < x for each x E it follows that the sequence P,} is increasing and bounded on [0, 1 Thus, 0, { converges pointwise to some non-negative function f on [0, 1]. It easily follows that [f(x)]2 x for each x E [0, 1], and so, f(x) ,JX. Finally, since .jX is a continuous function and is increasing, Dini 's Theorem 9.4 shows that {P,} converges uniformly to .JX on [0, 1 ] . A vector space A of real-valued functions on a set X is called an algebra of functions whenever the product of any two functions in A is again in A. Thus, a x set A c JR. is an algebra if for every pair f, g E A and real numbers and {3 we have af + {Jg and fg in A; where, of course, (fg)(x) f(x)g(x) for each x E X . And now we are ready to state and prove the classical Stone-Weierstrass the­ orem. =

-

[P, (x)f

[

1 ], P, }

{

].

=

=

{ P,}

•

a

=

Theorem 11.5 (Stone-Weierstrass). Let X be a compact topological space,

and let A be an algebra of contimJOIIS real-valued functions on X separating the points of X and containing the constantfunction 1. Then A is dense in C(X) with respect to the uniform metric.

Let A denote the closure of A in C(X) with respect to the uniform metric. Then A is a closed algebra (why?) containing the constant function 1 and Proof.

90

Chapter 2: TOPOLOGY AND CONTINUITY

separating the points of X. We have to show that A = C(X). By Theorem 1 1.3, it suffices to show that A is a function space. To this end, let f E A with f :f= 0. Put a = llflloo = sup{lf(x)l: x E X} > 0. Let { P,} be the sequence of polynomials detennined by Lemma 1 1 .4 that converges unifonnly to .jX on [0, 1]. Since A is an algebra, the function g11 = P,( ) belongs to A for each n. Moreover, the sequence {g,} converges unifonnly on X to = Thus, 1 1 E A, and so, If I E A. Therefore, A contains the absolute value of every function of A. But then, since

/!J

�:

�

'-9.

f v g = �(f + g + If - g l)

and

f A g = �(f + g - I f - gl),

it follows that f v g and f A g belong to A for every pair f, g E A. In other words, A is a function space, and the proof of the theorem is complete. • Since the collection ofall polynomials on IRis an algebra ofcontinuous functions Lhal contains the constant function 1 and separates the points of JR., the follmving original result of K. Weierstrass follows immediately from the last theorem.

Any continuous real-va/uedftmction on a com­ pact subset A ofiR. is the uniform limit on A of a sequence ofpolynomials. Corollary 11.6 (Weierstrass).

EXERCISES 1.

Let X be a compact topological space. For a subset L of C(X), let L denote the uniform closure of L in C(X). Show the following: a. If L is a function space, then so is L . b. If L is an algebra, then so is L.

2.

[0, I].

Let L be the collection of all continuous piecewise linear functions defined on That is, E L if and only if E C[O, and there exists a finite number of points < x, = (depending on such that is linear on each interval = < XJ < Show that L is a function space but not an algebra. Moreover, show that L is dense in C[O, with respect to the uniform metric. ---+ R is the uniform limit of a sequence of 3. Show that a continuous function polynomials on if and only if it admits a continuous extension to [0, 4. If is a continuous function on [0, such that Jd = for n = 1, . . . , then show that = 0 for all x E I ). but fails to be 5. Show that the algebra generated by the set {1, x2 ) is dense in C[O, dense in C[ 6. Let us say that a polynomial is odd (resp. even) whenever it does not contain any monomial of even (resp. odd) degree. = if Show that a continuous function ---+ R vanishes at zero (i.e., and only if it is the uniform limit of a sequence of odd polynomials on

f 0 xo [Xm-1· x111 ]. f

· ·

·

1] (0, I) f(x) -I, I].

I

f

I]

f: (0, I) I] [0,

f: [0. I]

[0, I].

f)

f

1]. 0,

x" f(x) dx 0

1]

f(O) 0)

Section 11: THE STONE-WEIERSTRASS APPROXIMATION THEOREM

91

1 7. If f: [0, 1] -+ 1R is a continuous function such that f0 f ( 2l•+.y;) dx = 0 for 11 = 0, 1 , 2, . . . , then show that f(x) = 0 for all x E [0, 1]. Does the same conclusion hold true if the interval [0, 1 ] is replaced by the interval [-1, 1 ]? 8. Assume that a function f: [0, oo) -+ 1R is either a polynomial or else a continuous bounded function. Then show that f is identically equal to zero (i.e., show that f = 0) if and only if J;o f(x)e-nx dx = 0 for all n = 1 , 2, 3, . . . . 9. Show that a continuous bounded function f: [ 1 , oo) -+ 1R is identically equal to zero if and only if J;XJ x-11 f(x) dx = 0 for each n = 8, 9, 10, . . 10. Let A be an algebra of continuous real-valued functions defined on a compact topo­ logical space X which separates the points of X . Show that the closure A of A in C(X) with respect to the uniform metric is either all of C(X) or else that there exists some a E X such that A = {/ E C(X): f(a) = 0}. [HINT: A glance at the proof of Lemma 1 1 .4 shows that the polynomials P,(x) that approximate ../X uniformly on [0, I] have constant terms zero. This implies that 1/ 1 and Ji7T both belong to A for each f E A.] 2 11. Let A be the vector space generated by the functions I. sin x, sin x, sin3x, . . . def­ ined on [0. 1 ]. That is, f E A if and only if there is a nonnegative integer k and real numbers ao, a 1 , . . . , ak (all depending on f) such that f(x) = a11 sin"x for each x E [0, 1]. Show that A is an algebra and that A is.dense in C [O , 1] with respect to the uniform metric. Let X be a compact subset of lR. Show that C(X) is a separable metric space (with 12. respect to the uniform metric). 13. Generalize the previous exercise as follows: Show that if (X, d) is a compact metric space, then C(X) is a separable metric space. [HINT: Let {x,} be a countable dense subset of X , and let j,(x) = d(x, x,). If A is the algebra generated by the functions 1, !1 , fz, . . , show that A = C(X).] 14. Let X and Y be two compact metric spaces. Consider the Cartesian product X x Y equipped with the distance D1 given in Exercise 4 of Section 7, so that X x Y is a compact metric space. Show that if f E C(X x Y) and € > 0, then there exist functions {/J, . . . , J,,} c C(X) and {gJ, , g } c C(Y) such that .

.

L;,=O

.

. . .

,

,

f(x, y) - L f; (x)g; (y)

;= )

<€

holds for all (x, y) E X x Y. [HINT: Consider the algebra generated in C (X x Y) by the functions F(x, y) = f (x) and G(x. y) = g(y) for f E C(X) and g E C(Y).]

CHAPTER

3

__ __ __ __ __ __ __ __ __ __ __ __ _

THE THEORY OF MEASURE At the turn of the nineteenth century it was quite apparent to mathematicians that the properties of continuous functions and Riemann's theory of integration were not rich enough to solve many scientific problems. The inadequacies of the continuous functions led them to search for different classes of functions that would provide solutions to a variety of problems. Around the beginning of the twentieth century, the theory of measure was origi­ nated. At that time, it was realized that to get a better understanding of the structure of functions it was necessary to make a thorough study of the subsets of Euclidean spaces. To study these sets, it became clear that the classical notions of length, area, and volume needed to be_ generalized. The search for devising ways of assigning a concept of a "measure" to a given set of points has its roots in that period. E. Borel [4] in 1898 was the first to establish a measure theory on the subsets of the real numbers known today as Borel sets. Soon after (in 1902), H. Lebesgue [21] presented his pioneering work on Lebesgue measure, and a little later (around 1 9 1 8), C. Caratheodory introduced and studied the properties of outer measures. From then on a rapid development of the theory of measure, which included among its contributors the most prominent mathematicians of the first half of the twentieth century, followed. This chapter discusses in detail the theory of measure. We start our study by introducing the concept of a serniring of sets and then proceed by studying the properties of measures on sernirings. The important notion of the outer measure is introduced and studied next. It is followed by a detailed investigation of the mea­ surable sets and measurable functions. Our attention is then turned to the properties of simple and step functions and to the basic properties of the Lebesgue measure. The chapter culminates with an investigation of convergence in measure and a discussion on abstract measurability properties. 12. SEMIRINGS AND ALGEBRAS OF SETS

In this section the notion of a serniring of sets is introduced and its properties are studied. A serniring of sets is the simplest family of sets for which a measure 93

94

Chapter 3: THE THEORY OF MEASURE

theory can be built. It turns out that most "reasonable" collections of sets satisfy the semiring properties. Definition 12.1. Let X be a nonempty set. A collection S of subsets of X called a semiring if it satisfies tlle following properties:

is

Tlle empty set belongs to S; tbat is 0 E S. If A, B E S; then A n B E S; tbat is, S is closed underfmite imersections. 3. The set difference of any n-vo sets of S can be written as a finite union of painvise disjoint members of S. Tbat is, for every A, B E S; tbere exist C" . . . , Cn in S (depending on A and B) sucb tllat A \ B = U�'= 1 C; and C; n Cj = 0 ifi "# j. I. 2.

Now, let S be a semiring of subsets of X. A subset A of X is called a u-set with respect to S (or simply a u -set) if there exists a disjoint sequence {A,} of C' t: A A - ¥J rll : f ,. 1 ..J.. m J ' "'Uch th'll '-' \.1.�;;.. , n , t .d - Ui"= l •A•1. 'lllth 11 n nm - uoo ..... .4 1=1 .d•11• Tf A 1, • • • , An E S and A; n Aj = 0 for i "# j, then A is a u-set. To see this, put A; = 0 for i > n. It follows from Definition 12.1 that A \ B is a u-set for every pair A and B in S. Some basic properties of u -sets are included in the next theorem. "'

1

'

"

.:>

•

....

•�

l

•

• •

•·

·�·

Foz· a semiriug S, tbefollowiug statemems bold: 1. If A E S aud A" . . . , An E S, tbez1 A \ U�=l A; cau be writteu as afiuite uuiou ofdisjoiut sets of S (aud lleuce, it is a u-set). 2. For evezy sequeuce {A,} of S, tbe set A = U�1 A, is a u-set. 3. Coulltable Ulliolls audfi uite illtersectious ofu -sets are u -sets.

Theorem 12.2.

Proof. (I) We use induction on 11. For 11 = I , the statement is true from the def­

inition of the semiring. Now, assume the statement true for some n. Let A E S, and let A 1 , , A,, A,+l E S.By theinduction hypothesis, thereexistB1, • • • , Bk E S such that B = A \ U�'=l A; = u�=l B; and B; n Bj = 0 if i =I j. Consequently, • • •

k A \ U A; = B \ An+l = U(B; \ A,+ l)• =l 11+1 i

i =l

By property (3) of Definition 12.1, each B \ An+l can be written as a finite union 1 of disjoint sets of S. Since B; n B j = 0 if i # j, it easily follows that A \ U�',!1 A; can be written as a finite union of disjoint sets of S. This completes the induction and the proof of (I). (2) Let {A,} C S. Put A = U:, 1 A,, and then write A = U:1 B, with Bt = At and Bn+l = An + l \ U�'= l A; for n > I . Observe that B; n Bj = 0 if

Section 12: SEMIRINGS AND ALGEBRAS OF SETS

95

i =P j, and by statement ( 1) each B; is a ct -set. It now follows easily that A is itself a a-set. (3) The proof follows from (2), and property (2) of Definition

12. 1 .

•

The proof of part (2) of the preceding theorem also guarantees the validity of the following useful result:

If {A,} is a sequence of sets in a semiring S, then there exists a disjoint sequence { C,} of S such that U�, A11 = U:, C11 and for each n there exists some k with C11 c Ak . Lemma 12.3.

Some natural collections of sets happen to satisfy other properties that are stronger than mose of a semiring. The "algebra of sets" is such a collection, and its definition follows.

A nonempty collection S ofsubsets ofa set X which is closed underfmite intez-sections and complementation is called an algebra of sets (or simply an algebra). That is, S is an algebm whenever it satisfies the following properties: i. If A , B E S, then A n B E S. ii. If A E S, then Ac E S. Definition 12.4.

Three basic properties of an algebra are included in the next theorem.

Theorem 12.5.

For an algebra ofsets S, the following statements hold:

0, X E S. 2. The algebra S is closed wzder finite unions and intersections. 3. The algebra S is a semiring. Proof. (1) Since S is nonempty there exists some A E S. Now, by hypothesis Ac E S, and so, 0 = A n Ac E s. Moreover, X = 0c E s. (2) Let A, B E S. Then A U B = (Ac n Bc)c E S, and the rest of the proof can 1.

be completed easily by induction. (3) We have to verify only property (3) of Definition in view of the identity A \ B = A n Be.

12.1. But this is obvious

•

We continue by illustrating the notions of semiring and algebra of sets with examples. Example 12.6.

For every nonempty set

X, the

collection S =

{0, X}

sets. This is the "smallest" (with respect to inclusion) possible algebra.

is

an

algebra of

•

Chapter 3: THE THEORY OF MEASURE

96

Example 12.7. For every nonempty set X, its power set 'P(X) (i.e., the collection of all subsets of X) forms an algebra. This is the "largest" possible algebra •

Example 12.8. Let :F be a nonempty pairwise disjoint family of subsets of a set X. Then S = :F U {(/)} is a semiring of subsets of X. To see this, note first that (/) E S. Now, if A, B E S, then A n B is either empty or equal to A. Likewise, A \ B is either empty or equal to A. Thus, A, B E S implies that A n B and A \ B both belong to S, and so S is a

.

��.

Example 12.9. If a, b E IR, let us write [a, b) = (/) if a ::: b and (as usual) [a, b) = {x E IR: a < x < b } if a < b. Then the collection S = {[a, b): a, b E IR} is a semiring of subsets ofiR., which is not an algebra (for instance, notice that [0, I) U [2, 3) ¢. S). •

The semiring of the previous example is very important because of its many applications. Its analogue in higher dimensions is p resented next.

Example 12.10. Let S denote the collection of all subsets A of IR" for which there exist A [aJ, b1) x · · · x [a,, bn ). {If a; > b; holds intervals [aJ , b1 ), . . . , [a,, bn ) for some i, then [a;, b;) = (/), and so A = (/).) Then S is a semiring of subsets of IR". To see this, note first that only the third property of the semiring definition needs verification; the other two are trivial. The proof is based upon the following identity among sets A, B, C, and D:

such that

A

x

B \c

x

=

D = [(A \ C) x B] u [(A n C) x (B \ D)],

where the sets of the union on the right-hand side are disjoint. For the proof, use induction on n. For n = I the result is straightforward. Assume it now true for some n. We have to show .that any set of the form

can be written as a finite union of disjoint sets from the (n + I )-dimensional collection S. But this can be easily shown by letting

A = [aJ, bi) x · · · X [an, bn ), C = [CJ , di) X . . . X [cn, dn), in (*) and using the induction hypothesis.

B = [an+l• bn+ J), D = [cn+l• du+d •

An intennediate notion between semirings and algebras is that of a ring of sets. A ring of sets (or simp ly a ring)

is a nonempty collection of subsets R of a set X

satisfying these properties: a. If A , B E R, then A U B E R. b. If A , B E R, then A \ B E R.

Section 12: SEMIRINGS AND ALGEBRAS OF SETS

97

Every ring n contains the empty set. Indeed, since n is nonempty, there exists A E 7?., and so 0 = A \ A E n·. Clearly, every algebra of sets is a ring of sets. Also, a ring 1?. is necessarily a semiring. Indeed, if A , B E 7?., then the relation A n B = A \ (A \ B) shows that A n B E 7?.. · Another useful concept is that of a a-algebra of sets.

An algebra S ofsubsets ofsome set X is called a a -algebra if every union of a countable collection of members of S is again in S. That is, in addition to S being an algebra, U:, A, belongs to S for every sequence {A,} ofS. Definition

12.11.

By virtue of n�, A, = ( U:, A�)c, it easily follows that every a-algebra of sets is also closed under countable intersections. Every collection of subsets :F of a nonempty set X is contained in a smallest a -algebra (with respect to the inclusion relation). This a-algebra is the intersection of all a -algebras that contain :F (notice that P(X) is one of them), is called the a -algebra generated by :F. An important a -algebra of sets is the a -algebra of all Borel sets of a topological space. Its definition is given next.

The Borel sets of a topological space (X, r) are the mem­ bers of the a-algebra generated by tlze open sets. The a-algebra of all Borel sets of(X, r) will be denoted by B. Definition 12.12.

EXERCISES 1.

If X is a topological space, then show that the collection S=

{C n 0: C closed and 0 open} = {C1 \ C2: C1• Cz closed sets} is a semiring of subsets of X.

Let S be a semiring of subsets of a set X, and let Y c X. Show that Sy = {Y n A: A E S} is a semiring of Y (called the restriction semlring of S to Y). 3. Let S be the collection of all subsets of [0, I) that can be written as finite unions of subsets of [0. I) of the fonn [a, b). Show that S is an algebra of sets but not a u -algebra. 4. Prove that the u -sets of the semiring 2.

S = { [a, b): a, b E 1R and a ::: b }

fonn a topology for the real numbers. 5. Let S be a semiring of subsets of a nonempty set X. What additional requi rements must be satisfied for S in order to be a base for a topology on X? (For the definition of a base, � Exercise [ 18] of Section [8].) Prove that if such is the case, then each member of S is both open and closed in this topology.

98

Chapter 3: THE THEORY OF MEASURE

6. Let A be a fixed subset of a set X. Determine the two a-algebras of subsets of X generated by a. b. 7.

{A}. and { B: A � B � X}.

Let X be an uncountable set, and let

S = {£ � X: E or

Ec

is at most countable}.

Show that S is the a-algebra generated by the one-point subsets of X. 8. Characterize the metric spaces whose open sets form a a -algebra. 9. Determine the a-algebra generated by the nowhere dense subsets of a topological space. 10. Let X be a nonempty set, and let F be an uncountable collection of subsets of X. Show that any element of the a-algebra generated by F belongs to the a-algebra generated by some countable subcollection of F. 11. Show that every Fer and every G0-subset of a topological space is a Borel set. 12. Show that every infinite a-aigebra of seis has uncountably many sets. 13. Let (X, r) be a topological space, let B be the a -algebra of its Borel sets, and let Y be an arbitrary subset of X. If Y is considered equipped with the induced topology and By denotes the a-algebra of Borel sets of (Y, r ) , then show that

By

= {A

n

Y:

A E B}.

Let A J , , A, be sets in some semiring S. Show that there exists a finite number of pairwise disjoint sets 8 1 , . . . , 8111 of S such that each A; can be written as a union of sets from the B J , , B,. [HINT: Use induction on n and Theorem 12.2(1 ).]

14.

.

.

.

•

13.

.

.

MEASURES ON SEMIRINGS

The semirings would not be of importance to us if it were not for the purpose of defining measures on them. The concept of a measure can be thought of as a generalization of the concepts of length and area, and its · definition is given next.

A

real-valued function defined on a family of sets is referred to as a

set function.

A An

Let S be a semiring of subsets of a set X. set function J-L: S � [0, oo] is called a measure on S ifit satisfies the following properties:

Definition 13.1. 1.

2.

{An}

J-L(0) = 0, and whenever is a disjoint sequence of s satisfying

?L( Q A.) � ?L(A.)

holds; that is, J-L is a-additive.

=

u�,

E s. then

Section 13: MEASURES ON SEMIRINGS

99

A triplet (X. S, J.L), where X is a nonetnpty set, S is a semiring of subsets of X ,

and J.L is a measure on S is called a measure space.

Theorem 13.2. For a measure space (X, S, J.L), thefollowing statements hold: 1 . If A I · . . . , An E S are pairwise disjoint and U�'= 1 A; E S, then J.L(U ;'= 1 A; ) = I:�'= I J.L(A; ) . That is, J.L is finitely additive. 2. If A , B E S satisfy A c B , then J.L(A) < J.L(B) holds. That is, J.L is monotone.

Proof. ( 1 ) If A 1 , . . . , A11 E S are pairwise disjoint sets such that U�'= 1 A; E S, let A; = 0 for i > n. Then {A;} is a disjoint sequence of S satisfying U: 1 A; = U:'=1 A ; E S. Thus, by the a-additivity of J.L we have /.1.

(� ) (� ) II

A;

= J.L

00

A;

=

t; J.L(A;) = t; J.L(A;), OC

II

where the last equality holds by virtue of J.L(0) = 0. (2) Let A, B E S satisfy A c B. Choose a finite collection of disjoint sets ' C 1 , . . . , C11 of S such that B \ A = U� = 1 C; . Then B = A U ( B \ A ) = A U C 1 U U C, is a finite union of disjoint sets of S. Thus, from part ( 1) we have · · ·

md the proof of the theorem is complete.

•

We continue with some examples of measure spaces. Example 13.3 (The Counting Measure). Let X be a set, and let S = P(X). Define J.L:S -* [0. oc] by J.L(A) = oo if A is an infinite subset of X and J.L(A) = the number of

elements of A if A is a finite set. The reader can verify easily that (X, S, J.L) is a measure

��

The measure of the next example is known as a Dirac 1 measure.

.

Example 13.4 (The Dirac Measure). Let X be a nonempty set, and let S = P(X). Fix an element a E X, and define J.L: S -* [0, oo) by J.L(A) = 0 if a ¢. A and J.L(A) = 1 if • a E A. It is easy to see that (X, S, p.) is a measure space. Example

13.5. Let :F be a nonempty pairwise disjoint family of subsets of a set X,

and let S = :F U {(/)}. In Example 12 we verified that S is a semiring. Now, for each nonempty set A E :F fix some m A E [0, oo]. Then the set function J.L: S -* [0, oo), defined 1 Paul Adrien �Iaurice Dirac ( 1902-1984 ), a famous English theoretical physicist. He won the Nobel prize in Physics at the age of3 1 for his pioneering work in quantum theory. This measure was introduced by him in the comext of "della functions."

100

Chapter 3: THE THEORY OF MEASURE

by J.L(0) = 0 and J.L( A) = rn A if A e S is nonempty, is a measure. To see this, note that if A e S can be written as a disjoint union A = U�1 A, with {A,} s;;; S, then it must be the case that A = Ak for some k and An = 0 for n =ft k. This easily implies J.L(A) J.L(Ak) = L�I J.L(A,) and so J.L is a-additive. • =

Example 13.6. Assume that the function f: IR. --+ IR. is nondecreasing and left continuous; that is, limx ta f(x) = f(a) holds for each a e IR.. Consider the semiring S = {[a, b): a, b e IR. and a < b}; see Example 12. Now define J.L: S --+ [0, oo) by J.L([a, b)) = f(b)- f(a) if a < b and J.L(0) = 0. We claim that the set function J.L is a measure. To see that J.L is a -additive, let a < b and let [a, b) = U�1 [an, bn) with the sequence {[a,, b,)} disjoint; we can assume that an < bn for each n. Let 00

Rearranging

s = L J.L([an, bn )). l�l [at, bt)

• . . .

, [a,�: , b;J.

Since f is increasing, Ef

we c<�n

suppose that

1 [f(b; )-f(a; )] .:S f(bk )- f(ai) � f(b)-f(a), which implies s < f(b) - f(a) = J.L([a, b)).

For the reverse inequality, let 8 > 0 and 0 < E < b- a. For each n choose some c, < a, n satisfying f (a,) - f(x) < - 8 whenever Cn < X � a,. Since [a. b - E] c u�l <en. bn) and [a, b - E) is compact, it follows that [a, b - E) C u;t=l (Cn, bn) must hold for some Assume a1 = a. If b1 < b - E , then by rearranging, we can assume b1 e (c2. b2). Since (a1, b1) n (a2, b2) = 0. it follows that c2 < b1 .:S a2. Continuing this process, we obtain 2, Ci+l < b; < a; +l for 1 l � i � m - 1; note that E7�� [f(a;+;) - f(b;)] .:S 8� Consequently,

2

k.

m s ?: L [f(b; ) - f(a; )] i=l

m-l = f(bm ) - f(at) - L [f(a;+I ) - f(b; )] i=l > f(b - E) - f(a) - 8. Since 8

>

0 and

0 < E < b - a are arbitrary, the left continuity off at b implies s > f(b) - f(a).

The latter combined with (*) shows that s =

f(b) - f(a), so J.L is cr-additive.

•

Section 13: MEASURES ON SEMIRINGS

101

An important special case of the preceding example is the case when f(x) = x for all e The resulting measure is called the Lebesgue measure on S, and it will be denoted that is, J.([a, b)) = b - a. Later, the domain of this measure will be extended to include all the open and closed sets. IR.

x

by .A;

Example 13.7.

Consider the semiring S of Example 12.10. That is, the semiring S con­ sists of all subsets of JR." of the form [a,, b,) x · · · x [a , b11 ) with a < b; for l � i � n together with the empty set. Define A.: S � [0, oo) by A. 0) = 0 and

n , i ( n A.((a , , b,) X · · · X [au, bu)) = n(b; - a;). i=l

Then A. is called the Lebesgue measure on S. We postpone the proof of the a-additivity of A. until Section 18. A proof that requires some additional background will be presented in Theorem 18.1. •

The following theorem characterizes the set functions on semirings that are measures: Theorem 13.8. Let S be a semiring, and let J.L: S --7 [0, oo] be a setfunction. Then I.

J.L is a measure on S ifand only if J.L satisfies the following conditions:

0.

JJ-(0) = If E S and i =f. j, then S and If holds; that is,

A 1 , , A11 E S satisfy U�'=t A; c A and A; n A j = 0 for "L?=• J.L(A;) < J.L(A) holds. Ae {A11} C S satisfy A C U�. An, then J.L(A) < L�• J.L(A11) 3. J.L is a-subadditive. Proof. Assume that J.L is a measure on S. Then by definition, JJ-(0) = 0. For (2) assume that A E S, and that the disjoint sets A., . . . , An of S satisfy U�'=• A; c A. By Theorem 12.2(1), there exist disjoint sets B,, . . . , B111 of S such that< A \< u;'=• Ai = U�'�. B;. Put c. = A,, . . . ' Cu = Au. and Cu+i = B; for I m. Then the sets C , , . . . , Cn+m are disjoint and A = U7:." C;. By the i finite additivity property of J.L (see Theorem 13.2), we get m+n 11 J.L(A) = L J.L(C;) L J.L(A;). i=l i=l

2.

A

•

.

•

>

For the a-subadditivity of J.L, assume that A C U�, A11 holds with A e S and Put B, = A, and Bu+ l Au+ l \ U�'= • A; for > I. Then u�. Bn {Au} U�. A11 and B11 A, for each n. Also, the sequence {B11} is disjoint, and by Theorem 1 2.2( for each > 2 there exist pairwise disjoint sets C�', . . , c;; in such that Bu u�:. Cf'. Note that by (2) and U�:. Cj' c Au for each �. it follows that L;·:, J.L(Cj') J.L(A11). (For n 1 , we put k, = 1 and c: = A,.) c

s.

I).

s

c

k=

=

.

11

<

=

11

=

Chapter 3: THE THEORY OF MEASURE

102

A=

U: 1 (B11 n Now, observe that So, by the a-additivity of we get 00

kn

J-L

A) = U:1 u:::: 1 (Cj' n A), is a disjoint union. 00

k,

00

J-L(A) = L L J-L (Cf' n A ) < L L J-L(ci') < L J-L (A "). n=l i= l

11=l i = l

11=l

Conversely, if the set function J,L: S � [0, oo] satisfies the above three con­ ditions, then is a-additive by combining (2) and (3). Hence, J1 is a measure.

J-L

•

We close the section with the definition of a finitely additive measure. A set function JJ-: S � [0, oo], where S is a semiring, is called a finitely additive measure on S if it satisfies these properties: a. b.

J,L(C/>)

If

AJ,

= .

.

0. •

I

All

E s

are disjoini and u;'= I

A; E s' then

It easily follows that every finitely additive measure J-L is monotone; that is, if , B e S satisfy A c B, then J,L(A) < J,L(B) holds. By Theorem 13.2, every measure is a finitely additive measure, but the converse is not true. See Exercise 7 of this section.

A

EXERCISES 1.

Let {an} be a sequence of nonnegative real numbers. Set J,L(0) = 0, and for every nonempty subset A of 1N put" J,L(A) = Lne A a11 • Show that J.L: 'P(IN) � [0, oo] is a measure. 2. Let S be a semiring, and let J,L: S � [0, oo] be a set function such that J,L(A) < oo for some A E S. If J.L is a-additive, then show that J,L is a measure. 3. Let X be an uncountable set, and let the a-algebra S = {£ c X: E or Ec is at most countable};

see also Exercise 7 of Section 12. Show that J,L: S � [0, oo), defined by J,L(E) = 0 if E is at most countable and J,L(E) = 1 if Ec is at most countable, is a measure on S. 4. Let X be a nonempty set, and let f: X � [0, oo] be a function. Define J,L: 'P(X) � [0, oo] by J.L(A) = Lx eA f(x) if A # . 0 and is at most countable, J.L(A) = oo if A is uncountable, and = 0. Show that J,L is a measure. 5. Let S be a semiring, and let J,L: S � [0, oo] be a finitely additive measure. Show that if J,L is a-subadditive, then J,L is a measure. 6. Let {Jl-11} be an increasing sequence of measures on a semiring S; that is, J,L11(A) :=:: Jl-11 +1 (A) holds for all A E S and all 11. Define J,L: S -+ [0, oo] by J.L(A) = sup{J,Ln(A)} for each A E S. Show that J.L is a measure.

J-L(0)

Section 14: OUTER MEASURES AND MEASURABLE SETS 7.

8. 9.

103

Consider the semi ring S = { A c IR: A . is at most countable}. and define the set function J-1.: S --+ [0, oo] by J.L(A) = 0 if A is finite and J.L (A) = oo if A is countable. Show that J-1. is a finitely additive measure that is not a measure. Show that every finitely additive measure is monotone. Consider the set function J.L defined in Example 13. That is, consider a nondecreasing and left-continuous function f: IR --+ IR and then define the set function J-1.: S --+ [0, oo) by JJ.([a, b)) = j(b) - f(a), where S is the semiring S = {[a, b): -oo < a =:: b < oo}. Prove alternately the fact that J-1. is a measure.

14. OUTER MEASURES AND MEASURABLE SETS

The theory of outer measures will be presented in this section. The concept of an outer measure is due to C. Caratheodory2 and is defined as follows.

A set function J-1. : P(X) --+ [0, oo] defined on the power set P(X) of some set X is called an outer measure if it satisfies Definition 14.1 (Caratheodory). these properties:

JJ.(0) = 0. 2. JJ.(A) < JJ.(B) if A c B ; that is, J-1. is monotone. 3. JJ. < U:1 A,) < 2::1 JJ.(A,) holds for every sequence {A , } of subsets of X; that is, J.l is a-subadditive. I.

An outer measure J-1. need not be a-additive on P(X). However, as we shall see, there always exists a a-algebra of subsets (called the measurable sets) on which J-1. is a-additive. The details will be explained below. Throughout the rest of this section, J-1. will denote a fixed outer measure. The next definition describes the measurable sets and is also due to C. Caratheodory.

(Caratheodory). A subset E ofX is called measurable (more precisely, JJ.-measurable) whenever Definition 14.2

holds for all

A c X.

Since the a-subadditivity of J-1. implies

JJ.( A) = Jl((A n E ) U (A n Ec)) < J.t(A n E ) + J.l(A n Ec)

2Constantin Cnratheodory ( 1 873-1 950), a distinguished Greek mathematician. He made many significant contributions to pure and applied mathematics.

104

Chapter 3: THE THEORY OF MEASURE

for all subsets A and E, it easily follows that a subset E is measurable if and only if

holds for each subset A of X. The collection of all measurable sets will be denoted by A . That is, A =

{E

c X:

J.L(A) = J.L(A n E) + J.L(A n Ec) for all A c X}.

If clarity requires J.L to be indicated, then we shall write A 1� instead of A. The simplest measurable sets are the sets having outer measure zero. Before verifying this, we name these sets. Definition 14.3.

A set E is called a null set if 11(E) = 0.

It should be clear from the a-subadditivity property of J.L that a countable union of null sets is again a null set. The null sets will play an important role in the theory of integration .

Every null set is measurable. Proof. Let E c X with J.L(E) = 0. Then the monotonicity of 11 implies J.L(A n E) = 0 for each A c X. Consequently, for each subset A of X we have Theorem 14.4.

where the first inequality holds by virtue of the a-subadditivity of J.L. Thus, E is measurable. • For more properties of the measurable sets, we need the following: Lemma 14.5.

Let the sets E 1, J.L

( hl II

•

.

•

, E11 be disjoint and measurable. Then

)

(A n E;)

t; J.L(A n E;) II

=

holdsfor every subset A of X. Proof. The proof is by induction on n. Obviously, the result is true for n

Assume it now true for some n, and let the sets E 1 ,

•

•

•

=

1.

, E11, En + 1 be disjoint and

Section 14: OUTER MEASURES AND MEASURABLE SETS

lOS

measurable. If A c X, then

[ ·g ] [g ]

An An

E ; n En+ I = A n En+ t

E; n (En + t )' = A n

[� J E

Therefore, using the measurability of En+ 1 , we see that

�

(Q

) ( [Q ]) ( ([Q ] ) ( [Q ] 0 [� ]) �

(A n E;) = � A n =

� An

E;

E; n E n + t + � A n

= �(A n En+tl + �

n

E

=

)

E n (En+tl'

�(A n E; ) ,

where the last equality holds by the induction hypothesis. The induction is now complete, and the proof is finished. • We are now ready to establish that the collection of all measurable sets is a a-algebra.

Theorem 14.6.

E

The collection 1\. of all measurable sets is a a-algebra.

Proof. It should be clear from the definition of the measurable sets that if E /\., then

Ec E /\.; that is, 1\. is closed under complementation. Since JJ-(0) = 0,

we have 0 E 1\.; therefore, X E 1\.. Next, we show that if E 1, E2 E 1\., then E = E 1 U E2 E /\.. Indeed, note first that E = E 1 U (E'J n E2), and then that for every subset A of X the relations

JJ,(A)

< < = =

J..L ( A n E ) + JJ,(A n Ec) [JJ-(A n E1) + ll((A n ED n E2)]+JJ-((A n ED n ED JJ-(A n Ed + [ JJ- ((A n ED n E2) + JJ-((A n ED n ED] JJ,(A n E J ) + J..L (A n Ef} = JJ,(A)

imply E1 U E2 E /\.. It now follows easily that 1\. is closed under finite unions and finite intersections. Also, if E , , E2 E /\., then E 1 \ E2 = E 1 n E� E A. Thus, 1\. is an algebra of sets.

106

Chapter 3: THE THEORY OF MEASURE

To finish the proof, it remains to be shown that A is a cr-algebra of sets. To this end, let { E, } c A . Put E = u:, E,, and define G 1 = E 1 and G,+ t = E,+l \ U�'= t E; for n > 1 . Then (Gn} c A , G, n Gm = 0 if n =j:. m, and E = U: 1 G,. Let F, = U�'= 1 G; for n > 1 , and note that each F11 is a measurable set such that u:, F11 = E. Now if A c X, then

p,(A)

=

p,(A n Fn) + p,( A n F,� )

>

p,(A n Fn) + p,(A n Ec)

=

[ t.

]

p.(A n G;) + p.(A n E')

holds for each n, where the last equality holds by virtue of Lemma 14.5. Hence,

p,(A) >

[ r: 1 =1

]

p,(A n Gi) + p,(A n E c) > p,(A. n E) + p,(A n Ec) 2: J.L(A ),

and so E E A. Therefore, A is a cr-algebra. Remarkably, the outer measure tL restricted to A is a measure.

Let tL be an outer measure on X. Then (X, A , tL) is a measure space� that is, tL is a-additive on A .

Theorem 14.7.

Proof. Let {E11} be a disjoint sequence of A . Put E

subadditivity property of tL we have

p,(E) <

u:, E11• By the cr­

X

L p,(E,). II=

On

=

I

the other hand, Lemma 14.5 shows that

t p.(E,) t p.(E n E ) ( E n [ 0. E., ] ) < p.(E) =

holds for every k. Hence, L::, the proof is finished.

.,

= p.

p,(E11) < p,(E), so that p,(E) = L::, p,(E11), and •

When tL is restricted to A it is often referred to as the measure induced by the outer measure p,. It is useful to know that tL is subtractive on the measurable sets of finite measure.

Section 14: OUTER MEASURES AND MEASURABLE SETS

Theorem 14.8.

107

A and B be mea�urab/e sets such that A c B, with f.L(B) < oo. Then f.L(B \ A) = f.L(B) - f.L(A) holds. Proof. B = A U (B \ A), f.L(B) = f.L f.L(B \ A) = JJ.(B)-JJ.(A). • f.L(B) < oo, f.!( A)+ f.L(B \ A ). Let

Write

and then use the additivity of to get Since it follows that We are now ready to describe a process of constructing outer measures. Let F be a collection of subsets of a set X containing the empty set. Also, let Jl: F [0, oo] be a set function such that JJ-(0) 0. For every subset A of X we define I'' (A) inf { t, !L(A,. ): {A,} is a sequence ofF with A c Q A,. } . If there is no sequence {A,} ofF such that A c U�1 A,, then we let Jl*(A) That is, we adhere to the convention inf 0 --+

=

=

= oo.

= oo.

Theorem 14.9 (Caratheodory). The setfimction Jl*: P(X) --+ [0, oo] is an outer measure (called the outer measure generated by the setfunction Jl: F --+ [0, oo]) satisfying

Jl*(A) � f.L(A)

F. Proof. Clearly, Jl *(A) 0 holds for every A c X. If A, 0 for all n, then 0 < JJ- *(0) < L�1 f.L(0) 0, so that JJ-*(0) 0. For the monotonicity of f.L* assume A c B. If B c U� 1 A, with {A,} c F, then A c u:, A, , and so Jl*(A) :L:, f.L(A11). (If there is no sequence {A11 } ofF that covers B, then Jl*(B) and f.L*(A) f.L*(B) is obvious.) Thus, for each

A

E

>

=

=

=

<

=

a

oo,

<

For the -subadditivity of Jl *, let { A11} be an arbitrary sequence of subsets of X. If :L:, JJ-*(.-\11) oo, then clearly, f.!*( u:, A11) < :L:, Jl*(A11). Therefore, assume :L:, JJ-"'(A,) Let E 0. For each i, choose a sequence {A�,} ofF such that Ai C U:, A;,, and L�=l f.L(A�,) < Jl*(Ai) + 2-i E. Then A�, F for each i and n, and u:, A, C U:, U: , A;,. Hence, =

< oo.

>

E

108

Chapter 3: THE THEORY OF MEASURE

Since A = A U 0 U 0 U 0 U · · ·, it easily follows that JJ-*(A) < for each A E :F.

J.L(A) holds true •

It is easy to construct examples where JJ-*(A) < JJ-(A) is valid for some A E :F, i.e., JJ-* need not be (in general) an extension of JJ-. For instance, if X = { 1 , 2, 3}, :F = {0, { 1 }, { 1, 2}} and JJ-: :F -+ [O, oo] is defined by JJ-(0) = 0, JJ-({ 1}) = 2 and JJ-({ 1, 2}) = 1 , then JJ-*({ 1 }) = 1 . However, as we shall see in the next section, when JJ- is a measure, the outer measure J-L* is always an extension of JJ-. It turns out that the outer measure generated by JJ-* coincides with JJ-*, i.e., (JJ-*)* = JJ-*. The details follow.

Let :F be a collectio11 of subsets of a set X coutaiuiug the empty set and let JJ-: :F -+ [0, oo] be a setfuuctiou satis.fyiug JJ-(0) = 0. Assume is also a collection of subsets of X with :F c and that v: -+ [0, oo] denotes the restriction ofiL* to ¢, i.e., v(A) = J.l*(A)for each A E <1>. Then the outer measure generated by v coincides with JJ-*, i.e., v*(A) = J.l*(A) holdsfor each subset A of X. Proof. Let A c X. We first claim that Theorem 14.10.

JJ-* (A) < v* (A). If v*(A) = oo, then the inequality is obvious. So, assume v*(A) < oo. In this case, note that if a sequence {A,} of satisfies A c U: 1 A,, then the a -subadditivity of JJ-* implies JJ-*(A) < :L: 1 JJ-*(A,) = :L:, v(A,). Therefore,

IL'(A)

< inf

{ t,

v(A.,): {A.,]

c and

Ac

.Q A., }

=

v '(A).

Next, we shall show that

v* (A) < JJ-*(A) is also true. Again, if JJ-*(A) = oo, then the inequality is obvious. So, assume JJ-*(A) < oo and let E > 0. Then there exists a sequence {A11 } C :F C sat­ isfying A c U:, A, and :L: , JJ-(A,) < JJ-*(A) + E. Now, taking into ac­ count (from Theorem 14.9) that JJ-*(A,) < JJ-(A,) holds for each n, we see that

v* (A)

<

00

00

00

11=1

11=1

11=1

L v(A,) = L JJ-* (A,) < L JJ-(A,) < J.l"'(A) + E.

Section 14: OUTER MEASURES AND MEASURABLE SETS

109

Since E > 0 is arbitrary, it follows that v�(A) < J.L*(A). Thus, v*(A) holds for all subsets A of X.

=

J.L*(A)

•

Finally, we leave it as an exercise for the reader to prove the following result asserting that the only outer measures are the ones generated by set functions.

Theorem 14.11. Let v: P(X)

[0, oo] he a set function on the power set of a set X . Then v is an outer measure if and only if there exist a family F of subsets of X and a set fimction J.L: F --+ [0, oo], where 0 E F and J.L(0) = 0, whose generating outer measure coincides with v, i.e., v = J.L*. --+

EXERCISES

Unless otherwise stated, in the exercises below J.L is assumed to be an outer measure on some set X. l. 2.

Show that a countable union of null sets is again a null set. If A is a null set, then show that

3.

holds for every subset B of X. If a sequence { A u } of subsets of X satisfies :L::1 Jl(A,) < oo, then show that the set

J,L(B) = Jl( A U B) = J,L(B \ A)

(x E

X:

x belongs to Au for infin itely many

n

}

is a null set. 4. If E is a measurable subset of X, then show that for every subset A of X the following equality holds:

J,L(E U A) + J,L(E n A) = J,L(E) + J,L(A).

If A is a nonmeasurable subset of X and E is a measurable set such that A £; E, then show that J,L(E \ A) > 0. 6. Let A be a subset of X and let {£,1 } be a disjoint sequence of measurable sets. Show that 5.

I

7. Let {A,} be a sequence of subsets of X. Assume that there exists a disjoint sequence (811} of measurable sets such that A11 £; Bu holds for each n. Show that

8.

Show that a subset E of X is measurable if and only if for each E measurable set F such that F c E and J.L( E \ F) < E. 1

>

0 there exists a

Chapter 3: THE THEORY OF MEASURE

110

Assume that a subset E of X has the property that for each € > 0, there exists a measurable set F such that J-L(E AF) < €. Show that E is a measurable set. Let X = { I , 2, 3}, :F = {0, { I } , { I , 2} } and consider the set function 11-: :F --+ [0, oo] defined by JJ-(0) = 0, J.L({ 1 }) = 2 and J.L( { 1 , 2}) = I .

9. 10.

a. b.

Describe the outer measure J.L* generated by the set function J-L. Describe the a-algebra of all J-L*-measurable subsets of X (and conclude that the set { 1 } E :F is not a measurable set).

Prove Theorem I4. l l . Let A be the collection of all measurable subsets of X of finite measure. That is, A = {A E A: J-L(A) < oo}.

11. 12.

a. Show that A is a semiring. b. Define a relation :::::: on A by A :::::: B if J.L(AAB) = 0. Show that :::::: is an equivalence relation on A. c. Let D denote the set of all equivalence classes of A. For A E A let A denote the equivalence class of A in D. Now for A, i3 E D define d(A. B) = J-L(AAB). Show that d is \VeH defined and that (D, d) is a complete metric space. Wor this part, see also Exercise [3] of Section [31].)

15.

THE OUTER MEASURE GENERATED

BY A MEASURE

Throughout this section, (X, S, J-L) will be a fixed measure space. Our main objec­ tive here is to study the remarkable properties ofthe outer measure 11"' generated by J-L. Among other results, we shall establish that the a -algebra of all f.1 *-measurable subsets of X contains the members of the semiring S. It will then become appar­ ent that the semirings are the smallest collections of sets on which a "reasonable measure theory" can be built. Recall from the previous section that the outer measure J-L *: P(X) � [0, oo] generated by J-L is defined by

IL'(A)

=

inf

{ f;

IL(A,.): {A,.} is a sequence of S with A

C

Q }. An

If there is no sequence { A , } of S such that A C U: 1 A,, then we let J-L*(A) = oo. That is, we adhere to the convention inf 0 = oo. Recall also that (as we saw in Theorem 14.9) the set function J-L* is indeed an outer measure.

In general, J-L * need not be a measure, since it may fail to be a-additive on P(X). However, if J-L * is restricted to the a -algebra of all measurable sets, then we already know (by Theorem 14.7) that it is a-additive. The next theorem shows that the outer measure J-L * is in fact an extension of J-L from S to P(X ).

Section 15: THE OUTER MEASURE GENERATED BY A MEASURE

111

Theorem 15.1. The outer measure f.L"' {san extension of f.L. That is, tf A E S, then J.L*(A) = J.L(A) holds. know that J.L*(A) < J.L(A). Proof. Let A E S. From Theorem 14.9, we already . Now, let {A,} c S with A c u:,, A,. By the cr-subadditivity of f.L (Theorem 13.8), J.L(A) < :L:,, J.L(A,) holds, from which it follows immediately that J.L(A) < J.L*(A). Hence, J.L*(A) = J.L(A). • f.L* Caratheodory extension f.L. A"

f.L, f.L*

Since the outer measure generated by a measure is an extension of is of From now on the expression "the also called the outer measure of a set will also be referred to as "the measure of The measurable sets of an outer measure generated by a measure have a number of useful characterizations. Some of them are included in the next theorem.

Theorem 15.2.

A."

J.L)

f.L*

Let (X, S, be a measw·e space, and let be the outer measure generated by For a subset E of X tlte following statements are equivalet1t: 1. 2. 3. 4.

f.L.

E is measurable with respect to f.L*.

J.L(A) = J.L*(A n £) + J.L*(A n £c) holds for all A E S with J.L(A) < oo. J.L(A) > J.L*(A n £) + J.L*(A n £ c ) holds for all A E S with J.L(A) < oo. J.L*(A) > J.L*(A n £) + J.L*(A n £ c) lwldsfor all A C X. Proof. ( 1 ) :=::} (2) and (2) :=::} (3) are obvious. (3) ==> (4) Let A c X. If J.L*(A) = oo, then (4) holds trivially. Hence, assume J.L*(A) < oo. Let E > 0. Choose a sequence {A,} of S with A c U:,1 A11 and :L:,1 J.L(A11) < J.L*(A) + E. It follows that J.L(A,) < oo for each n, and so J.L*(A, n £) + J.L*(A, n Ec) < J.L(A,) holds for each n, by our hypothesis. Therefore,

00

00

L J.L*(A/1 n £) + L J.L*(A, n Ec) n=l n=l 00 = L[J.L*(A, n £) + J.L*(A n n Ec)] II= I 00 < L J.L(A,) < J.L*(A) + E 11=1 <

for all E > 0, so that

f.L*(A) > J.L*(A n £) + f.L*(A n £c).

112

Chapter 3: THE THEORY OF MEASURE

(4) ==> (I) By the a -subadditivity property of J.L *, we have hence J.L *(A) = J.L*(A n E) + J.L*(A n Ec) for all A

c X. Thus, E is measurable. •

A subset E of a measure space (X,' S, J.L) will be called measurable (more precisely J.L-measurable) if E is measurable with respect to the outer measure J.L * generated by J.L.

Theorem 15.3.

Every member of S is measurable. That is, S c A.

Let E E S. We have to show that E is a measurable set. If A E S, then there exist pairwise disjoint sets B1 , , B11 of S such that A n Ec = A \ E = U?=1 B;. Note that A n E, B1, . . . , B11 is a disjoint collection of Im!mbers of S such that A = (A n E) U B 1 U U Bn. By the a-subadditivity property of J.L* , ..... nbtain

Proof.

•

·

y�"" v

·

•

•

·

...

J.L*(A n E) + J.L*(A n Ec )

<

"

J.L*(A n E) + L J.L *(B ;) i=l "

= J.L ( A n E) + L J.L(B;) = J.L (A), i=l where the last two equalities hold true by virtue of Theorem 15.1 and the a­ • additivity of J.L on S. By Theorem 15.2, E is measurable, and so, S c A. Let us write A, t A to mean that the sequence {A,} of subsets of X satisfies A, c A n + I for each n and A = U: 1 A,. Similarly, A n ,J, A means Au+ I c A, for each n and A = n: 1 A n.

For a measure space (X, S, J.L) and a sequence ofmeasurable sets {En}, thefollowing statements hold: I. If En t £ , then J.L*(E,) t J.L*(£). 2. If E, ,j, E and J.L*(Et) < oo holds for some k, then 11*(£,) ,j, Jl*(E). Theorem 15.4.

Proof. (1) Let B1 = £1

and B, measurable, and B; n B j = 0 if i # Hence, by Theorem 14.7 we have

En \ £,_1 for n > 2. Then each B, is j. Also, E, = U�'=l B; and E = u�l B;. =

oo

n "' J.L*(B;) = lim "' J.L*(B; ). J.L *(E) = � n-oo � i=l i= l But

p.*(En) = 2:�'= 1 J.L*(B;), and thus, J.L *(E,) t J.L*(£) holds.

Section 15: THE OUTER MEASURE GENERATED BY A MEASURE

113

( 2) Observe that without loss of generaljty we can assume f.L*(£ I ) oo. Now, Et \ E, t E t \ E , and so, by part (1), limJL*(Et \ E") = JL*(El \ E). Applying Theorem 14.8, we get lim[JL*(£ 1 ) JL*(E11)] JL*(E I) JL*(E), from which it follows that f.L*(£11) -1- f.L*(E). •

<

=

-

-

As an application of the results obtained so far, we shall show that the Lebesgue outer measure on IR generalizes the ordinary concept of length. Remember that the Lebesgue measure A is themeasuredefined on the semiring S = {[a, b): a, b E IR } by A([a, b)) = b - a for all a .:;:: b. It is a custom to call (and we shall do so) the outer measure A"' generated by the Lebesgue measure on IR. A subset I of IR is called an interval if for every x, y E I with x y, we have [x, y] c / . If I has any one of the fonns [a, b), [a, b], (a, b], or (a, b) with oo, then I is called a bounded interval, and its length is defined -oo a by = b If I is unbounded, then its length is said to be infinite, which is written = oc .

A

<

<
£\·et)' imerva/ 1 of JR. is Lebesgue measurable and

besg e measurable.

111.

shows that

e

We shall more cases and leave the est first case is a bounded interval the ]. Let E11 ) each n . en each E is Lebesgue measurable, and £1 t lds Hence, is Lebesgue measurable, and by we get 11

1

.

>..*(/) = lim >..* ([a, h+_!.)) = (b+ � - a) = b-a = lll. of form I = [a, oo). = for note a I holds. Thus, I Theorem 15.4(1) n-oo

11

The second case is an interval 11, and th t F, t we have

n

the Put F, [a, a + n) is Lebesgue measurable, and by

>..*

A*(/) = lim ([a, rr-oo

lim

n-+oo

a + n))

each

n

= lim = oo =Ill, u.-.. co

•

as desired. Two important classes of measure spaces are introduced next

Definition 15.6. A measure space (X, S, f.L) is said to be

1. 2.

and exists a seqttence {X11} of subsets of X such that X JL* (X,) oo for each rz.

finite, if J.l*(X) a-finite. if there

<

U: 1 X u

and

oo ,

<

-

114

Chapter 3: THE THEORY OF MEASURE

Clearly, every finite measure space is a-finite. We also have the following:

Lemma 15.7. A measure space (X, S, J.L) is a-finite if and only if there exists

a disjoint sequence {Y,} ofS such that X = U:1 Y11 and J.L(Y,) < ooforeach n.

To see the "only if" part, assume that (X, S, J.L) is a-finite. Choose subsets X 1 , X2, . . . of X such that X = . U:1 X; and J.L*( X;) < oo for. each 00 i. Now, for each i, choose a sequence { A � } of S such that X; c U,=1 A� and 2::: 1 J.L(A�,) < J.L*(X;) + 1 . Then A�, E S, J,L(A;,) < oo for each n and i, and X = U:1 U: 1 A�,. Now, by applying Theorem 12.2 and Lemma 12.3, our claim • can be established.

Proof.

,

The measurable sets of a finite measure space have a simple characterization.

Let (X, S, J.L) be afinite measure space, and let E be a subset of X. Then E is measurable (fand only if

Theorem 15.8.

J.L*(E) + J.L*(Ec) = J.L*(X). Clearly, if E is measurable, then J.L* (E) + J.L* (Ec ) = J.L* (X) holds. For the converse, assume that J.L* (E) + J.L* (Ec) = J.L* (X) holds. Let A E S. The measurability of A applied to E and Ec gives

Proof.

J.L*(E) = J.L*(E n A) + J.L*(E n A c), J.L*(Ec) = J.L*(Ec n A) + J.L*(Ec n A c). By adding the last two equalities, we get

J.L*(X) = J.L*(E) + J.L*(Ec) = J.L*(E n A) + J.L*(Ec n A) + J.L*(E n A c) + Jl*(Ec n A c) > J.L* (A) + J.L*(Ac) > J.L*(A U A c) = J.L*(X). Therefore,

In view of J.L *(A c ) gives

< J.L*(E n Ac) + J.L* (Ec n Ac) and J.L *(X) < oo, the last equality

This last inequality is the one required by Theorem 15.2 for the measurability of • E , and the proof of the theorem is complete.

llS

Section 15: THE OUTER MEASURE GENERATED BY A MEASURE

Notice that Theorem 15.3 combined with Theorem I 2.2 shows that in the defi­ nition of the outer measure f.l we can consider only disjoint sequences of S. That is, we have the foiiowing useful result: *,

Lemma 15.9. I' *(A)

= inf

lf (X,

{ t,

S, J.l) is a measure space, then !'(A,.): I A,. I C

S, I A,. I

disjoint, and

AC

= inf(J.l*(B): B is a a-set and A c B ) holds tmefor el'ery suhset

A

of X (witlt

,Q } A,.

inf0 = oo).

The next result describes the uniqueness of the extension of J.l to the a-algebra of the measurable sets.

S, J.l) he a a-fil1ite measure space, :E a semiring of sets such that S c :E C /\ , and v a measure 011 :E . lf v = J.l 011 S, then v = J.l*

Theorem 15.10. on

Let (X,

:E .

In this case,

J.l *

is the one atld only extel1sion of J.L to a measure on

./\.

Proof.

Let v* be the outer measure generated by (X, :E, v ). Now, let A c X, and let (A11} be a sequence of S such that A C U:: 1 A11• Since J.l and v agree on S, we have

v*(A) <

00

00

n=l

n=l

L v(A,) = L J.l (A11).

This implies v*(A) < J.l*(A) for all A c X . Now, assume that A E :E satisfies J.l*(A) < oo. We shaH show that J.l*(A) < v(A) holds, and therefore, v(A) = J.l*(A) wiii be established in this case. To this end, let E > 0. Choose a disjoint sequence (A11 } of S such that A c U:1 A11 and 2:::, J.l(A11) < J.l *(A) + E . Put B = U: 1 A11, and note that B E .1\ J.L , B E /\11, and J.l*(B) < J.l*(A) + E. Now, observe that v*(B \ A) � J.L*(B \ A) = J.l*(B) - J.l*(A) < E holds. Therefore,

J.l '" (A) < J.l*(B) =

00

00

11=1

11=1

L J.l(A11) = L v(A11) = v*(B)

= v*(A) + v*(B \ A) < v(A) + E

for a]] € > 0. Hence, J.l*(A) < v(A).

116

Chapter 3: THE THEORY OF MEASURE

For the general case, let {Xn } be a disjoint sequence of S covering X such that JJ-*(X11) < oo for all n. Now, if A E :E, then by the above v(X, n A) = 11- *(X11 n A) holds for all n, and so

!L'(A) = IL'

( Q[X., )

= t. v(Xn

and the proof is finished.

n A]

n A)

= v*

= t. IL*(X.,

n A)

( Q[x., )

n A] = v(A),

•

A subset A of a measure space (X, S, J.L) is said to be a -finite if there exists a sequence { A11} of S such that A c U:1 A, and JJ-(A11) < oo for each n. Ciearly, the sequence { A11} can be chosen to be disjoint. Repeating verbatim the proof of the last theorem we can establish the following. •

Let (X, S, JJ-) be a measure space, :E be a semiring of subsets of X such that S C :E c A, and v be a measure on :E . If v = J.L on S, then v(A) = JJ-*(A) lwldsfor every a-finite set of :E .

It should be noted that the hypothesis of a-finiteness in Theorem 1 5 . 1 0 cannot be dropped. As an example, take X = IR, S = {[a, b): a, b E IR}, and define JJ-: S � [O, oo] by JJ-(0) = 0 and J.L([a, b)) = oo if a < b. Then (X, S, J.L) is a measure space that is not a -finite. Moreover, A = 'P(X) and 11 *(A) = oo for every nonempty subset A of X. It is easy to see now that the counting measure (see Example 13) is an extension of 11- to A that is different from 11- *. We continue with an approximation property of an arbitrary set by a measur­ able set.

Let (X, S, JJ-) be a measure space. /f A is a subset ofX, then there exists a measurable set E such that A C E and JJ-*(E) = /.L*(A). In particular, if S is a a-algebra, then for each A c X there exists some E E S with A c E and 11-*(E) = JJ-*(A).

Theorem 15.11.

Let A c X. If 11- *(A) = oo, then the measurable set E = X satisfies JJ-*(E) = 11- *(A). So, assume JJ-*(A) < oc. For each i , choose a sequence {A:, } ofthesemiringS such that A c u:, A:, and 'L:1 JJ-(A:, ) < JJ-*(A) + t• and let E; = u:, A�,. Then each E; is a measurable set such that A c E;. Put E = n:1 £11• Then A c E, and E is measurable (and,

Proof.

117

Section 15: THE OUTER MEASURE GENERATED BY A MEASURE of course,

E E S if S is a a-algebra). Alsq,

for each i implies

Jl*(E) = Jl*(A).

•

If A is a subset of X, then any measurable set

E satisfying A C J.L*(A) is called by many authors a measurable cover of A.

E and Jl*(E)

=

Is every set a measurable set? Of course, the answer is no. It is easy to construct

examples where not every set is measurable. Here is a simple example.

Example 15.12. Let X = { I , 2, 3} and S = {0. {3}. { I , 2}. X}; clearly S is a a-algebra. It should be easy to check that the set function J.L: S ---+ [0, oo), defined by

J.L(0) = 0, J.L({3}) = J.L({ I . 2}) = I .

and

J.L(X) = 2

is a measure. We claim that the set E = { I } is not measurable. To see this, note first that J.L*((2}) = J.L*(( I . 2}) = I. Now if A = { I . 2 } . then

J.L*(A n £) + J.L*(A n Ec) = J.L*(( 1 }) + J1,*({2)) = 1 + 1

>

J.L*(( 1}) =

I = J.L*(A), •

and this shows that E is not measurable.

It is considerably harder to demonstrate the existence of a non-Lebesgue measur­ able subset of the real line. The following classical example of a non-measurable subset of the real is due to G. Vitali.3

Example 15.13 (Vitali). Let A.* be the outer measure generated on 1R by the Lebesgue measure A.. We define a relation � on [0, 1] by saying that x � y whenever x - y is a rational number. An easy verification shows that � is an equivalence relation on [0, 1]. Thus, � partitions [0, 1 ] into equivalence classes. Let E be a subset of [0, 1 ] intersecting each equivalence class precisely at one point. By the axiom of choice such a set exists. We claim that E is not Lebesgue measurable. To see this, assume by way ofcontradiction that E is Lebesgue measurable. Let r1 . r2 be an enumeration of the rational numbers of [ - I , 1 ] . For each n put •

.

.

.

Eu = {r" + x : x E £ } = r11 + £, and notethateach £11 is Lebesgue measurable. Now, it is notdifficult to see that E11 nEm = 0 holds if n ::j:. m and that (since A.* is translation invariant) A.*(£11) = A.*(£) for each n. Moreover, [0, 1 ] � U� 1 £11 � [ - 1 , 2]. By the a-additivity of A.* we get

A.*

( ::0 )

00

U E" = 'L: A.*(E11) = 11�� [ nA.*(£) ] ::; A.*([- 1 , 2]) = 3. 11 =1 11=1

3 Giuseppe Vilali ( 1875 1932). an Italian mathematician. He contributed to the theory of real func­ tions and measure theory.

Chapter 3: THE THEORY OF MEASURE

118

This implies A*(£) = 0, and hence, A*( U� 1 £,) = 0. On the other hand, [0, l ] C U�1 £, shows that 1 � A*( U�1 £,), which is impossible. Therefore, E cannot be a • Lebesgue measurable set.

EXERCISES 1. 2.

Let (X, S, p.,) be a measure space, and let E be a measurable subset of X. Put SE = { E n A: A E S}, the restriction of S to E. Show that (£, SE , p.,*) is a measure space. Let (X, S, p.,) be a measure space. Show that p.,*(A)

3. 4.

5.

=

inf{p.,*(B): B is a cr-set such that A � B }

holds for every subset A of X. Complete the details of Example 15. Show that every countable subset of 1R has Lebesgue measure zero. For asubset A of1R. andrea1numbers a and b, definethe setaA+b = {ax+b: Show that a. b.

x

E A}.

A*(a A + b) = Ia lA *(A), and if A is Lebesgue measurable, then so is a A + b.

6. Let S be a semiring of subsets of a set X, and let p.,: S � [0, oo] be a finitely additive measure that is not a measure. For each A � X define (as usual) p.,*(A)

7.

=

inf

{ f:

p.,(A,): {A, ) � S and A c

n=l

n=l

A11

!·

Show by a counterexample that it is possible to have p., =f:. p.,* on S. Why doesn't this contradict Theorem 15 . I ? [HINT: Use Exercise 7 of Section 13.] Let E be an arbitrary measurable subset of a measure space (X, S. J.L) and consider the measure space (£, S£, v), where SE = (E n A: A E S) and v(E n A) = p.,*(E nA) (see Exercise l ofthis section). Establish the following properties regarding the measure space (£, S£, v): a. b.

The outer measure v* is the restriction of p.,* on E, i.e., v*(B) = 11*(B) for each B � E. The v-measurable sets of the measure space (£, SE, v) are precisely the sets of the form E n A where A is a p.,-measurable subset of X, i.e., A v = { F � E: F E

8.

0

All ).

Show that a subset E of a measure space (X, S, p.,) is measurable if and only if for each E > 0 there exist a measurable set AE and two subsets BE and Cf satisfying E = (AE U BE) \ CE , p.,*(BE) < E, and p.,*(Cf) < E.

9.

Let (X, S, p.,) be a measure space, and let A be a subset of X. Show that if there exists a measurable subset E of X such that A c E, J.L*(E) < oc, and p.,*(£) = p.,*(A) + p.,*(£ \ A), then A is measurable.

Section 15: THE OUTER MEASURE GENERATED BY A MEASURE 10. 11.

119

Let A be a subset of JR. with >..*(A) > 0. Sh
< L J.L* ( E, ) . II=

12.

13.

I

[HINT: Use the sequence {E,} described in Example 1 5 . 1 3.] Let (X. S, J.L) be a measure space, and let {A,} be a sequence of subsets of X such that A, � An + l holds for all n. If A = U�1 A,, then show that J.L *(A n ) t J.L*(A). [HINT: Use Theorems 15. 1 1 and 15.4(1).] For subsets of a measure space (X, S, J.L) let us define the following almost everywhere (a.e.) relations: a. b. c.

A c B a.e. if J.L*(A \ B) = 0; A = B a.e. if J.L*(Al:l.B) = 0; An t A a.e. if A, C An+ 1 a.e. for all n and A, � A a.e. is similar.)

A

= U� 1

A, a.e. (The meaning of

Generalize Theorem 15.4 by establishing the following properties for a sequence {En) of measurable sets:

i. If E11 t E a.e., then J.L*(En) t J.L*(E).

ii.

14.

If En

� E a.e. and J.L*(Ek) < oo for some k, then J.L*(E11) � J.L*(E).

Is (i) true without assuming measurability for the sets En? Give an example of a sequence {En) of measurable sets of some measure space (X, S, J.L) such that En+ I C En holds for all n and

n-+00 lim

15.

For a sequence

J.L*(E,)

> J.L*

( noo ) n=l

{An} of subsets of a set X, defihe

lim inf A, =

00

00

U n A;

tr=l i=n

lim sup An =

and

Now, let (X. S. J.L) be a measure space and let sets. Show the following: a. b.

16.

E, .

00

00

n U A;.

n=l i=n

{E,} be the sequence of measurable

J.L*(lim inf £,) _:s lim inf J.L*(E,). If J.L*CU�1 En) < oo, then J.L*{lim sup £11) > limsupJ.L*(En).

Give an example of a sequence {An} of subsets of some measure space (X. S, J.L) such that An+ 1 A, for each 11, J.L*(A < co, and

�

1)

lim J.L* (A,) > J.L* n-+00

( noo ) n= l

An

.

120

Chapter 3: THE THEORY OF MEASURE [HINT: If

u�, E; . J

{E,} is the sequence of sets described in Example

15,

then let A11

=

Let (X' sl' J.L I) and (X' s2. J.L2) be two measure spaces. Show that J.L I and Jl-2 generate the same outer measure on X if and only if J.L I = J.Li on S1 and /1- 2 = J.Lf on S2 both hold. Let (X, S, J.L) be a measure space. A measurable set A is called an atom if J.L*(A) > 0 and for every measurable subset E of A we have either J.L*(E) 0 or !J.*(A \ E) = 0. If (X, S, J.L) does not have any atoms, then it is called a nonatomic measure

17.

18.

=

space.

a. Find the atoms of the measure spaces of Examples 13.3 and 13.4. b. Show that the real line with the Lebesgue measure is a nonatomic measure space. This exercise presents an example of a measure that has infinitely many extensions to a measure on the a-algebra generated by S. Fix a proper nonempty subset A of a set · X (i.e., A ::j:. X) and consider the collection of subsets S {0. A } . a. Show that S is a semiring. b. Show that the set function ,tL: S -+ [0, oo] defined by J-1.((/J) = 0 and f..L(A) = 1 is a measure. c. Describe the Caratheodory extension J.L* of JL d. Determine the a-algebra of measurable sets A11• e. Show that p. has uncountably many extensions to a measure on the a-algebra generated by S. Why doesn't this contradict Theorem 1 5.1 0?

19.

16.

MEASURABLE FUNCTIONS

An important role in the theory of integration is played by the so-called almost everywhere relations. If J1. is an outer measure of X , then a relation involving the elements of X is said to hold almost everywhere (or that it holds for almost all x) if the set A of all points for which the relation fails to hold is a nu l l set (i.e., Ji. (A) = 0). For instance, if f and g are real-valued functions defined on X , then f < g al most everywhere means that Jl.({x E X: j(x) > g(x)}) = 0. Similarly, {/11 } converges to f almost everywhere if there is a set A of measure zero such that lim f,(x) = j(x) holds for all x ¢ A. If (X, S, Ji.) is a measure space, then by saying that a relation holds almost everywhere (abbreviated a.e.) we mean that the relation holds almost everywhere with respect to the outer measure Jl. * generated by Jl.. The basic almost everywhere relations that will be used in this book are summarized below. We assume that ( X , S, Jl.) is a given measure space and that f and g denote real-valued functions defined on X.

I. f = g a.e. if Jl.*( {x E X : j(x) # g(x)}) = 0.

f > g a.e. i f Ji. *({x E X : j(x) < g(x)}) = 0. 3. f, � f a.e. if Jl.*({x E X : f,,(x) fr f(x)}) = 0.

2.

Section 16: MEASURABLE FUNCTIONS 4.

5.

ffn t f n ,J, f

a.e. if a.e. if

fnfn fn+ l +l fn <

Throughout this section,

<

a.e. for all. a.e. for all

nn fn � f fn � f and and

121 a.e. a.e.

(X, S, f.L) will denote a fixed measure space.

Let f: X � lR be afunction. If f- 1 (0) is a measurable set for every open snhset 0 oflR., then f is called a measurable function. f (x) c x EX -1 -1 JR., 0 f (0) c ¢. 0 f (0) X c E 0. Definition 16.1.

Every constant function is measurable. Indeed, if = for all and if is an open subset of then 0 if and Recall that the Borel sets of a topological space are the members of the a­ algebra generated by the open sets. The first theorem gives a number of useful characterizations of the measurable functions. (This result is a special case of a more general result regarding measurability of functions; see Theorem 20.6.) =

=

For a function f: X � JR., thefollowing statements are equiv­ alent: ff-1((a, is measurable. b)) i s measurable for each bounded open inter v al (a, h) of JR. 3. f- 1 (C) is measurable for each closed subset C of f-1 ( [a, oo)) is measurable for each a E 5. f-1((-oo, a] ) is measurable for each a E 6. f- 1 (8) is measurable for each Borel subset B of Proof. ( => 1 (C) [f -1 (Cc)]c => (3) fcc (3) 1 ((-oo, a)) (f-1 ([a, oo))]c => (5) fa E JR. Theorem 16.2. l. 2.

JR.

4.

JR. JR.

JR.

(2) Obvious. l) It follows from the relation (2) and the fact that can be written as a countable union of bounded open intervals. . => (4) Obvious. (4) Note first that is measurable for each The result now follows from the identity =

=

<(-oo a)) = 0 r' ( (-oo. a + D) (5) => (6) A = c JR.: f- 1 A lR ( -oo, a] A a E lR f-1 (B) B (6) => r'

.

Let {A (A) is measurable}. An easy verification shows that is a a-algebra of subsets of such that belongs to A for contains the open subsets of each by hypothesis. It easily follows that JR., and hence, it also contains the Borel sets. Thus, is measurable for each Borel subset of JR. ( 1 ) Obvious. • Two functions that are equal almost everywhere are either both measurable or else both nonmeasurable. The details follow.

122

Chapter 3: THE THEORY OF MEASURE

Theorem 16.3. Iff is a measurablefunction and g: X

---7

IR satisfies

f=g

a.e., then g is a measurablefunctiou. Proof. If A = {x E X: f(x) # g(x)}, then from our hypothesis J.L*(A) = 0, and so, A is measurable. Now, let 0 be an open subset ofiR. Since f is a measurable function, f - 1 (0) is measurable, and hence, Ac ng - 1(0) = Ac n f-1(0) is a mea­ surable set. Also, since Ang - 1 (0) has outer measure zero, it is measurable. Hence, is a measurable set, so that g is a measurable function.

•

We continue with more properties of measurable functions.

Theorem 16.4. Iff and g are measurablefuuctions, then the thz·ee sets

{x E X: f(x) > g(x)}, {x E X : f(x) > g(x)}, and {x E X: f(x) = g(x)} are all measurable. Proof. (a) If r1 , r2, is an enumeration of the rational numbers of IR., then

a. b. c.

•

•

•

00

{x E X: f(x) > g(x)} = u ({x E X : f(x) > r11} n {x E X: g(x) < rn }], n=l

which is measurable; since it is a countable union of measurable sets. (b) Note that {x E X: f(x) > g(x)} = {x E X: g(x) > f(x)}c, which is mea­ surable by (a). (c) Observe that

{x E X : f(x) = g(x)} = {x E X : f(x) > g(x)} n {x E X : g(x) > f(x)}, which is measurable by (b).

•

The next result informs us that the usual algebraic combinations of measurable functions again produce measurable functions.

Theorem 16.5. For measurable functions f and g the following statements hold: I . f + g is a measurable function. 2. fg is a measurable function. 3 . l f l. f+, and f- are measurable functions. 4. f v g and f A g are measurable functions.

Section 16: MEASURABLE FUNCTIONS

123

Proof. ( 1 ) Note first that i f c is a const�nt number, then c - g is a measurable

function. [Reason: If a E IR, then {x E X : c - g(x) i s a measurable set.] Now, i f a E IR, then the set E X:

(f + g)- 1([a, oo)) = {x

> a} = {x E X : g(x) < c - a}

f(x) + g(x) > a} = {x

E X:

f(x) > a - g(x)}

is measurable by the above observation and Theorem 1 6.4. Thus, by Theorem 16.2, f + g is a measurable function. (2) Note first that f2 is a measurable function. Indeed, if a E IR, then we have {x E X : f2(x) < a} = 0 if a < O and {x E X : f2(x) < a} = f- 1 ( [-Ja, Ja ] ) if a > 0. Thus, f2 is a measurable function by Theorem 16.2. Also, if c is a constant number, then cf is measurable. [Reason: If A = {x E X : cf(x) > a}, then A = {x E X: f(x) > afc} for c > 0 and A = {x E X : f(x) < afc} for c < 0.] The result now follows from the above ·observations combined with ( 1 ) and the relation

I .

fg = 2 [(j + g)- - f - - g-]. .,

.,

.,

(3) The measurability of I f I follows from the relations {x E X : l f(x) l < a} = 0

if a

< 0,

and

{x E X : lf(x)l < a} = {x

+

E X:

For the measurability of f and

�

f(x) < a} n {x

E X:

f(x) > -a}

if

a > 0.

f- use the identities

f+ = (l f l + f)

I

and

f - = -(l f l - f). 2

and

f 1\ g = 2 (f + g - I f - gl)

(4) The identities

1

f v g = 2 (f + g + If - g l) show that f

1

v g and f 1\ g are measurable functions.

•

The next result states that the almost everywhere limit of a sequence of measur­ able functions is again a measurable function.

Theorem 16.6.

For a seqllellce {f11 } of measurable functions, the following

statements lzold: 1 . If fn � f a.e., then f is a measllrable function.

Chapter 3: THE THEORY OF MEASURE

124

is a bounded sequencefor each x , then lim sup f, and lim inf f,, are both measurable functions. Proof. (l) Let A = {x E X: lim fn(x) = f(x)}. Since f, -+ f a.e., it follows 2.

If {fn (X)}

that J.L*(Ac) = 0. Thus, Ac is measurable, and hence, A is also a measurable set. Now, let a E JR. Observe that the equality

1

and the measurability of each f; show that A n f- ((a , oo)) is a measurable set. Also, Ac n f- 1 ((a, oo)) is a measurable set since it is a subset of a set of measure zero. Thus,

is a measurable set. It now easiiy foiiows from Tneorem 16.2 Lhai f is a measurable function. (2) Let {f,(x)} be bounded for each x. We shall show that lim sup f, is a measurable function. The measurability of lim inf f, will then follow from the identity lim inf f, = - lim sup(-fn ). Note first that lim sup f,, = A: Vf:, fk. Fix a natural number n. Then the function h ' = J,, v f,+ 1 v · · · v J,, +m is measurable for each m [by Theo­ " rem 1 6.5(4}], and since hm t v�, j,, = g, (everywhere), it follows from part (1} that each gn is a measurable function. Since g, ,!, lim sup J,, (everywhere}, by ( l ) again it follows that lim sup J,, is a measurable function, and the proof is complete. •

1

The last two theorems show that the collection of all measurable functions fonns a function space containing the limits of its sequences that converge almost everywhere. Moreover, this collection is an algebra under the pointwise product. In other words, the usual operations applied to measurable functions produce measurable functions again. Therefore, it is natural to ask whether there is a function that fails to be measur­ able. Clearly, if every subset of X is measurable [i.e., A = P(X)], then every real­ valued function on X is measurable. On the other hand, if there are nonmeasurable sets, then there are also nonmeasurable functions. Indeed, if E is a nonmeasurable set, then the function f defined by f(x) = l if x E E and f (x) = 0 if x ¢. E is not measurable simply because the sef ({I}) = E is not measurable. We close this section with a basic theorem connecting almost everywhere con­ vergence and unifonn convergence. The result is due to D. F. Egorov.4

f-1

4 Dimilry Fedorovich Egorov ( 1 869-1931 ), a Russian malhemalician. He is crediled wilh lhe crealion

of Lhe famous Moscow school dealing wiLh Lhe Lheory of funclions of one variable.

Section 16: MEASURABLE FUNCTIONS

125

Let {f,,} be a sequence ofmeasurablefunctions such ·able subset of X such that J.L*(E) that J,, f a. e ., and let E be a measw Thenfor eve1y E there exists a measurable subset F ofE with J.L*(F) and with {f, } converging uniformly to f on E\ F. Proof. X, x E X. E X J.L*(E) n k E,.k {x E E: lf,,(x) - f(x)l 2-" m k}. k X. addition, E,,k E,,k+ l E, . k j;,(x) f(x) x E X, E,.k k E each fixed n,

Theorem 16.7 (Egorov). �

< oo. < E,

> 0,

First note that by removing a null set from we can assume without loss of general i ty that lim f,,(x) = f(x) holds for each Fix a measurable subset of with < oo. For each pair of positive integers and let <

=

for all

i s a measurable subset of In Clearly, each for all and n. S ince lim = for each

t

>

holds C it easily follows that

for

and so, by Theorem 15 .4, we see that

J.L*(E,,k) ft J.L*(E) n. J.L*(E) k, JJ-*(£ E,, t ,) J.L*(E) J.L*(E,, t J 2-"E. F U:1 (E \ E,,t,). \ F F E, I:: E,, t ,) J.L*(E \ J. L *(F) , > k11• m f(x)l x E E \ F n:, E,,t,. f E \ F, { �,} for each fixed

Now, let E > 0. Since < oo, for each n there exists a natural number such that = < Let = < Then is measurable, c and < E. Also, if This = then l f,11(X) < 2-" holds for each shows that j converges uniformly to on and the proof of the theorem is complete. •

EXERCISES 1.

Let (X S. J1) be a measure space. For a function f: X -+ IR show that the following statements are equivalent: I

a. b. c.

f is a measurable function.

/- 1 (( -oo� a)) is measurable for each a E IR. f- 1 ((a. oo)) is measurable for each a E IR.

Let (X. S. Jl.) be a measure space, and let A be a dense subset of IR. Show that a function f: X -+ IR is measurable if and only if (x E X: f(x) � a} is measurable for each a E A. 3. Give an example of a non-Lebesgue measurable function f: IR -+ IR such that 1/1 is a measurable function and f 1 ((a}) is a measurable set for each a E JR. 4. Show that if f: IR -+ IR is continuous a.e., then f is a Lebesgue measurable function. 5. Let f: IR IR be a differentiable function. Show that f' is Lebesgue measurable. 6. Let (X S. J1) be a measure space and let f: X -+ IR be a measurable function. Show that: 2.

-

-+

I

Chapter 3: THE THEORY OF MEASURE

126

l f !P is a measurable function for all p ::::. 0, and b. if f (x) :/= 0 for each x E X , then 1 If is a measurable function. a.

7. Let {/11} be a sequence of real-valued measurable functions on a measure space (X, S, J.L). Then show that the sets

A = {x E X:

f,(x) � oo}, {x E X: fn (x) � -oo}, and b. B c. C = {x E X : lim f, (x) exists in JR.} a.

=

are all measurable. Let (X, S, J.L) be a measure space. Assume that f: X � lR is a measurable function and g: lR � lR is a continuous function. Show that g o f is a measurable function. Let :F be a nonempty family of continuous real-valued functions defined on JR. Assume that there exists a function g: lR � lR such that f(x) ::=:: g(x) for each x E lR and all f E :F. Show that the supremum function h: lR � JR., defined by h(x) = sup{f(x): f E :F}, is (Lebesgue) measurable. Show that if f: X � lR is a measurable function, then either f is constant almost everywhere or else {exdusive!y) there exists a constant c such that

8.

9.

10.

J.L *({x E X: f(x) > c}) > 0 and J.L*({x E X: f(x) < c}) > 0.

17. SIMPLE AND STEP FUNCTIONS From this section on, the properties of the characteristic functions are needed. For this reason their properties are listed below, and the reader is expected to be able to verify them. If A is a subset of a set X, then the XA of A is the real-valued function defined on X by XA (x) = I if x E A and XA (x) = 0 if x � A. (The characteristic function XA is also called by many authors the of A and is also denoted by 1 A .) For subsets A and B of a set X, the following relations hold:

characteristic function

indicator

function

xx =

l. 2. 3.

X¢ = 0 and

5. 6.

XA \ B = XA - XAnB·

4.

1.

A C B if and only .if XA < XB · XAnB = XA · XB = XA 1\ XB · XAuB = XA + XB - XAnB = XA V XB·

If A =

U:1 A, and {A11} is a disjoint sequence of subsets of X, then 00

XA = 7.

L XA,·

n=l

XA x B = XA · XB · (Here the set B can be taken to be a subset of another

set Y .)

Again, throughout this section (X, S, J.L) is assumed to be a fixed measure space. If a measurable function lj>: X -+ JR. assumes only a finite number of v al ues, then 4>

Section 17: SIMPLE AND STEP FUNCTIONS

127

simple function.

is called a Clearly, finite sums, finite products, and finite suprema and infima of simple functions are again 'simple functions. In other words, the collection of all simple functions is a function space, which is in addition an algebra of functions. If ¢ is a simple function assuming the distinct nonzero values a 1 , , a11 , then the sets A; = {x E X: ¢(x) = ad are all measurable and pairwise disjoint, and ¢ = L�'= 1 a; XA, holds. This expression is called the of ¢. Let us agree also that the standard representation of the constant zero function is X¢ · In general, a simple function ¢ may be represented in more than one way in the form ¢ = L�'= 1 hjXB;· where the bj values are real numbers and the sets Bj are all measurable. A simple function ¢ is called a if ¢ has a representation of the form ¢ = L��� bj XB;, where each Bj is a measurable set of finite measure, i.e., J.L*(B j) < oo. Clearly, a simple function is a step function if and only if it vanishes outside of a set of finite measure.5 For this reason, many authors call a step function a simple function with finite support. It should also be clear that the collection of all step functions fonns a function space that is in addition an algebra of functions. The step functions will be the "building blocks" for the theory of integration. •

•

.

standard representation

step function

Definition 17 .1. Let ¢ be a step function with standard representation ¢ = L�= 1 a; XA;; that is, a 1 , a11 are the distinct nonzero values f ¢ and A; = {x E X: ¢(x) = a; } . Then the Lebesgue integral of¢ (or simply, the integral of¢) is defined by •

.

•

,

/ (¢) =

o

,

II

L: a;J.L*(A;). i= l

dJ.L,

dJ.L.

The integral of ¢ is usually denoted by }� ¢ For the or simply by J ¢ time being we shall use /(¢) to denote the integral of ¢ , and later on we shall switch to the conventional notation. Our first theorem describes the linearity property of the integral for step func­ tions. The difficulty of the proof lies in the fact that the standard representation of the sum of two simple functions is not necessarily the sum of their standard representations.

Theorem 17.2 (The Linearity of the Integral).

If ¢ and ljl are step func­

tions, then /(a¢ + {3ljl) = a / (¢) + {3/ (ljl) holds for all a. {3

E JR.

5 Notice that the definition of a step function is more general than the familiar one associated with

the Riemann integral. There. a function ¢: [a. b) ---+ IR is called a step function if there exists some partition P = {xu . .\·1 x, I of [a. h] such that ¢ is constant on each open subinterval (.\";-I , x; ). • • . . •

Chapter 3: THE THEORY OF MEASURE

128

aE

a/(¢) holds for each step function ¢ and JR. are step functions, then What needs to be shown, therefore, is that if ¢ and /(¢ 1/1) = /(¢) holds-the additivity property of the integral. Let ¢ = and = be the standard representations of ¢ and 1/1. Let = }, and note that U < oo. Next put and = = 0. Then the are pairwise U�'=1 . . " . . . . . 1 are pairwise dISJOmt, and U 1 dISJOmt, th e = U"' = L�=l in its standard representation. Since ¢ Now, write ¢ c 1/l(x) =I= Oimplies either ¢(x) =I= Oor1/f(x) =I= O,iteasilyfollows that U�=' = 0. Then the Put = U�=l and are pairwise disjoint, and

Proof. Clearly, /(cup)

1/1

+ /(1/1) 1/1 'Lj':1biXBj 'L�'= 1a;XA; E
+

Ao = E

=

\

J.l*(E)

\

Co E

\

Ck

A; i=O Bi = E .

Ck

co

n

(x) + Ck E.

m

ck = U U
is a disjoint union for 0 < k � r . Thus, = In addition, note that n hoids for all i, j, and k. Indeed, if = 0, then the equality holds trivially, and if then the equality holds by virtue of Putting the above together and using the (finite) additivity of 11- on A, we get

x E Ck nA; n Bi• r

I (¢ + 1/1)

=

=

=

r

11

111

L ckJ.l*(Ck) = L ckJ.l*(Ck) = L L L ckJ.l*(Ck n A; n Bj ) k=l k=O k=O i=O j=O 11

111

11

m

r

L L L (a; + bi)Jl*(Ck n A; n Bi) i=O j=O k=O L :L
=

r

m

m

11

:La; L J.l*(A; n Bj) + L bi L J.l*(A; n Bj) j=O i=O i=O j=O 11

m

:La;J.l*(A;) + L biJ.l*(Bj) j=O i=O = /(¢) + 1(1/1),

=

and the proof is finished.

•

From the preceding theorem, it should be clear that if¢ is a step function having with a representation ¢ = measurable with E IR and each

'Lj'= 1biX Bi'

bi

Bj

129

Section 17: SIMPLE AND STEP FUNCTIONS

finite measure. then I(¢) = }'=1 j J-L*(Bj) holds. In other words, the integral of a step function does not depend upon its particular representation. We continue with the monotone property of the integral.

"L

b

Theorem 17.3 (The Monotonicity of the Integral).

For stepfimctions ¢ and

1/1, the following statements hold: I. 2.

�f¢ > O a.e., then l (¢) > O. ln particn/ar, if¢ > 1/J a.e., then l(¢) > 1 (1/1). If¢ = 0 a.e., then I (¢) = O. ln particular, if¢ 1/1 a.e., then I (¢) = I (1/1). =

Proof.

( 1 ) Let ¢ = "L;'=1 a; XA; be the standard representation of ¢ . Since

¢ > 0 a.e., note that if a; < 0 holds for some i , then necessarily J-L*(A;) = 0 for that Thus, 1(¢) "L�'=1 a;�..t *(A;) > 0. Now, if ¢ > 1/J a.e., then ¢ - 1/1 > 0 a.e., and so 1 (¢) - 1 ( 1/1) = 1(¢ - 1/1) > 0. Hence, 1 (¢) > 1 (1/J). (2) If ¢ = 0 a. e., then ¢ > 0 a.e. and -¢ > 0 a.e. both hold. Thus, by part (I), 1(¢) > 0 and - 1 (¢) > 0 both hold. Therefore, 1(¢) = 0. Now, if ¢ = 1/J a.e., then ¢ - 1/1 = 0 a.e., and hence, by the preceding case, we • 1 (1/J). obtain 1(¢) - 1 (1/1) = 1 (¢ - 1/1) = 0. That is, 1 (¢)

i.

=

=

The next theorem describes a basic continuity property of the integral.

Theorem 17.4 (The Order Continuity of the Integral).

Let {¢,} be a sequ­

ence ofstepfnnctions. lf¢, ,!. O a.e., then 1(¢,) ,!. 0. In particular, 1j ¢ is a step fimction and ¢, t ¢ a.e., then I(¢,)

Proof.

t

I (¢).

Assume ¢, ,!. 0 a.e. Put

A, = {x E X: ¢,+1 > ¢,(x)}

and

Ao = {x E X: ¢,(x) -fr 0}.

By assumption. Jl*(A,) 0 for n = 0, 1, 2, . . . . Let A = U�o A,. Then J-L*(A) = 0 (use the a-subadditivity of J.L* to see this), and note that ¢, (x) ,!. 0 and ¢,+ 1 (x) < ¢, (x) hold for each x E Ac . Also, if 1/J, = ¢, · XAc, then 1/J, is a step function and 1/J, = ¢, a.e. for each n, 1/J, (x) ,!. 0 for each x E X , and by Theorem 17.3, we have I (1/J,) = I (¢,) for each n. Thus, replacing {¢,} by { Y,, } if necessary, we can assume without loss of generality th[!t ¢, (x) ,!. 0 holds for each x E X. Now, let E > 0. Put M = max{¢1 (x): x E X } and B = {x E X: ¢1 (x) > 0}; clearly, JJ-*(B) < oo. For each n , define E, = {x · E X: ¢,(x) > E }. Then J-L *(£1) < oo, each En is measurable, and £, ,!. 0 by virtue of ¢,(x) ,!. 0 for each x E X. By Theorem 1 5.4, we have JJ-*(£,) ,!. 0. Pick an integer k such that J-L*(Ek) < E. Now, if n > k, then taking into account that ¢k(x) < E for each =

Chapter 3: THE THEORY OF MEASURE

130 x

E B\

b

E

we see that

from which it follows that 0 < I (¢11)

E

� M J.L* ( d + EJ.L*(B) < E[M + J.L* (B)]

for all n > k. Therefore, lim /(¢11) = 0. Now, if¢ is a step function and ¢11 t ¢ a.e. holds, then ¢ - ¢11 ,!, 0 a.e. also holds, and so, by the previous case, I(¢) - I (¢11) = I (¢ - ¢11) ,!, 0. Hence, I (¢11) t I (¢), and the proof is finished. • As an application of the last theorem, we have the following result:

Theorem 1 i 5

Assume iziai rn•o sequences of siep fuzu:iiozzs { ¢11} and { lfrn } satisfy ¢11 t f a.e. azzd 1/111 t f a.e., where f: X � JR* is a givel1 fuzzction. Thezz .

.

lim I ( ¢,) = lim I ( 1/111)

/1�00

/1-t-00

holds, with the above limits possibly being izzfizzite. Proof. Note first that for each fixed m, we have lPm 1\ 1/111

t II

lPm 1\ f =

lPm

a. e.

Therefore, by Theorem 1 7 .4, lim11�oo / (¢111 1\ 1/111) = / (¢m). Now, by virtue of ¢m 1\ 1/111 < l/J, a.e. for all m and zz and the monotonicity of the integral, we get I (¢111)

=

lim I (¢111

II -to 00

1\

1/111) < lim I (1/ln) II-to 00

for all m . Hence, lim I (¢,) < lim I ( l/J,) holds. By the symmetry of the situation, • lim 1(1/111 ) < lim /(¢,), and so, lim / (¢11) = lim /(1/111 ). The next result describes another useful property of step functions.

Theorem 17.6. Let { ¢n} be a sequence of step functions. If A is a subset of

such that ¢, t holds. X

XA

a.e., then A is a measurable set, and lim /(¢n)

=

J.L*(A)

Section 17: SIMPLE AND STEP FUNCTIONS

131

I).

Proof. The measurability of A follows from Theorem 16.6( We can assume (how?) that ¢11 (x) holds for all E X. For each the set A11 = {x E X: ¢11{:r) > 0} is measurable with finite mea­ sure, and A11 A holds. It follows that and thus, by Theorem 17.5, we

t XA (x) n t

have

x XA, t XA.

where the last equality holds by virtue of Theorem 15.4.

•

We close this section with an important approximation property of measurable functions.

Let f: X --+ be a nzeasurableju11ctio11 satisfying f(x) 0 for all tx. f(x) Thenholds thereforexistsall x sequence {¢11 } of simple jimctions such that 0 E X. ¢,.(x) 11 Proof. A:, {x E X: (i i2- } 1)2-" f(x) f I,A;,2, . . . , n2'1 , A!. Aft 11 L/;. 1 2-"(i -l)XA:, , 0 ¢11(x) ¢,+1 (x) f(x) 0 f(x)- ¢11 (x) 2-" n. ¢11 (x) t f(x) x, Theorem 17.7.

>

IR

<

a

For each

11

let

=

< = 0 if i =I= j. Since -

for i = < is measurable, all the

and note that n are measurable sets. I '>" Now, for each define¢, = and note that {¢,} is a sequence < of simple functions. Also, an easy verification shows that < < < holds for all x and all n. Moreover, if x is fixed, then < holds for all sufficiently large Thus, holds for all and the proof of the theorem is finished. •

f:

f

If X --+ IR is a measurable function such that (x) > 0 for almost all x, then there exists a sequence {¢,1 } of positive simple functions such that ¢11 t a.e. holds. To see this, let = {x E X : > 0}, and then apply Theorem 17.7 to the measurable function

E fXE.

f(x)

f

EXERCISES Verify the identities about the characteristic functions at the beginning of Section 17. 2. Let ¢ be a step function and 1/J a simple function such that 0 ::: 1/J ::: ¢ a.e. Show that 1/1 is a step function. 3. Show that if (X. S, J1.) is a finite measure space, then every simple function is a step function. 4. Give an alternate proof of the linearity of the integral (Theorem 1 7.2) based on Exer­ cise 14 of Section 12. 5. Show that l/(¢)1 ::;: I (1¢1) holds for every step function ¢. 6. Let ¢ be a step function such that /(1¢1) = 0. Show that ¢ = 0 a.e. holds. 1.

Chapter 3: THE THEORY OF MEASURE

132

7.

Let ¢ be a step function. Let A = (x E X: ¢(x) # 0} and M = max(l¢(x)l: x E X } . Show that II(¢)1 < MJ.L*(A). 8. Let ( ¢,} be a sequence of step functions. Show that if ¢ is a step function and ¢, J, ¢ a.e. holds, then I(¢,) J, I(¢) also holds. 9. Let (¢11 } be a sequence of step functions and ¢ a simple function such that 0 < ¢, t ¢ a.e. holds. Show that if lim I (¢,) < oo, then ¢ is a step function. 10. Let (X, S, J.L) be a measure space, and let f: X ---+ IR be a function. Show that f is a measurable function if and only if there exists a sequence ( ¢,} of simple functions such that lim ¢, (x) = f (x) holds for all x E X . 11. Let (X, S, J.L) be a a-finite measure space, and let f: X ---+ IR be a measurable function such that f (x) 2: 0 for all x E X. Show that there exists a sequence ( ¢,} of step functions such that 0 :::; ¢, t f (x) holds for all x E X. 12. Give a proof of the order continuity of the integral (Theorem 1 7.4) based on Egorov ·s Theorem 16.7. 13. Let (X, S, J.L) be a measure space, and let f: X ---+ [0, oo) be a function. Show that f is a measurable function if and only if there exist non-negative constants c,, c2 . . . . and measurable sets E , , £2 , . . . such that f(x) =

00

L: c,xE, (x) 11=.1

14.

15.

holds for each x E X. Let (X, S, J.L) be a measure space satisfying J.L*(X) = 1 , and let £ , . E1 . . . . . £10 be ten measurable sets such that J.L * (E; ) = � holds for each Show that four of these sets have an intersection with positive measure. Is the conclusion true for nine measurable sets instead of ten? If f : X ---+ [0, 1 ] is a measurable function, then show that either f = XA a.e. for some measurable set A or else (exclusively) there exists a constant 0 < c < i such that

i.

J.L*((x E X: c < f(x) < 1 16.

-

c}) > 0.

Let (X, S. J.L) be a measure space, and let ¢: X ---+ IR be a .simple function having the standard representation ¢ = "£�'=1a; X A;. If ¢ > 0 a.e., then the sum "£'/= 1 a; J.L*(A;) makes sense as an extended real number (it may be infinite). Call this extended real number the Lebesgue integral of r/J, and write I(¢) = "f:.'/- 1 a; J.L*(A;). If ¢ and 1/1 are simple functions such that ¢ > 0 a.e. and 1/1 > 0 a.e., then show that I(¢ + 1/1) = I(¢) + /(1/J). b. If ¢ and 1/1 are simple functions such that 0 < ¢ :::; 1/1 a.e., then show that I(¢) ::: I(l/1). c. Show that if (¢,} and ( 1/1n } are two sequences of simple functions and f: X ---+ IR" such that 0 .::: ¢11 t f a.e. and 0 ::: 1/J, t f a.e., then lim I (¢11) = lim I (1/111 ) holds (with the limits possibly being infinite). d. Assume that (¢,} is a sequence of simple functions such that 0 :::; ¢, t x A a.e. holds. Show that lim I (¢11) = J.L *(A). e. Give an example of a sequence {¢,} of simple functions on some measure space such that ¢11 -1- 0 (everywhere) and lim I (¢, ) # 0. a.

Section 18: THE LEBESGUE MEASURE

133

17. Let (X, :E, /.L) be a measure space with :E being a a -algebra. Let us say that a function f: X -+ IR i s :E-measurahle if f- 1 (A) 'E I: for each open subset A of IR. Also, let lvt I; denote the collection of all I: measurable functions. Establish the foll owing: -

a. b. c.

M I; is a function space and an algebra of functions. ;'vt I; is closed under sequential pointwise limits. If /.L is a-finite and f: X -+ IR is a measurable function, then there exists a :£-measurable function g: X -+ IR such that f = g a.e.

18. THE LEBESGUE MEASURE The notion of the Lebesgue measure is the natural extension of the concepts of length, area, and volume. In particular, the Lebesgue measure of any geometric figure i n IR? turns out to be its area, while the Lebesgue measure of any geometric solid in IR3 is its volume. Throughout this section, d(x, y) will denote the Eu­ clidean distance of the vectors x1 = (x 1 , . . . , X11 ) and y = (YI · . . . , y11) of IRn. That 2 ! is, d(x . y) = [ L�1=1(x; - y;) ) . The objective of this section is to present in some detail the properties of 1 the Lebesgue measure on IR 1 • For this discussion, S will denote the semi ring consisting of the empty set and all sets of the form A = n��=l [a; . b; ). where -oo < a; < b; < oo for each i. This semi ring was discussed in Example 12. In Ex­ ample 1 3 we have promised the reader to show that the set function A.: S � [0, oo) 1 1 definedby A.(0) = O and A.( TI; = 1 [a;, b;)) = n; =1 (b; - a;) is a-additive. We shall do this next.

Theorem 18.1. The set function A.: S called the Lebesgue measure on S.

�

[0, oo) defined above is a measure,

Proof. The proof is by induction on the dimension of IR11 • Let us denote by

Sn the semiring S on IRn. and the corresponding set function by A · For n = 1 the n result has been established in Example 1 3 . Now. assume that A11 is a measure for some n. We have to show that An+ 1 is a-additive on Sn+ 1. To this end. let A X [a, b) = u�I[A; X [a; , b; ) ] , with A E Sn . A; E sll for each i, and the sequence {A; x [a; , b;)} being pairwise disjoint. It easily follows that XAx)u.b) = 2:: 1 XA; x)u;.b;)· Now for X = (x1 . . . . , X11) E IR11 and t E IR, the last equality yields

XA (X ) · XIa.b) (t ) =

co

L XA; (X) · XIa1 ,b1)(t). i=l

"'"'k n Fix x = (x 1 . . . . . Xn) E IR , and let ¢k(t) = L i =l XA1 (x) · X[a;,h; ) (t). Then each ¢k is a step function [for (IR, S1 , AI)] such that (/Jk(t) t XA (x) X [a.b)(f) for each "'"'k t E IR. By Theorem 17.4 we get Li=1( b; - a;)XA1(x ) t (b - a)XA(x ) for each ·

Chapter 3: THE THEORY OF MEASURE

134

, X11)

x

E IR". Since by our induction hypothesis (IR", = (x1 , . . . measure space, by applying Theorem 17.4 once more, we get

Sn, A11)

IS

a

k

(b; - a; )A.,(A;) t (b - a)A.,(A). L i=l That is, I

I

00

Au+ (A; X [a;, b; )) = Au+ (A X [a, b)). L i=l

•

Thus, A.11+ 1 is a -additive and the proof is finished.

An interval of IR.n is any set of the form 117= 1 /;, where each I; is an interval of IR. If each /; is in addition a bounded open interval of IR., then 11�'=1 I; is called a bounded open interval of IR". As in Example 15.5, we can show that each interval ofiR" is Lebesgue measurable, and moreover, A.*( 11�'= 1 I;) = 11�'= 1 I/; I, where I /; I denotes the length of I; . A useful formula for the outer Lebesgue measure is the following: A*(A) = inf

{ [,>

*(/; ) :

Each I; is a bounded open interval and A

c

Q /; }

for every subset A of IR". To prove the formula, note first that if A = 11;'=1 [a;, b;) (i.e., A e S), then for each E > 0 the bounded open interval If = Jl�'= 1 (a; - E, b;) satisfies A c I€ , and

A.*(/f ) - A.(A) = n(b; - a; + E) - n(b; - a;) i=l i=l II

II

�

0

as

E --+

0.

If A.*(A) = oo, then the formula holds trivially. If A.*(A) < oo, then given E > 0, choose a sequence {A;} of S such that 2::1 A.(A;) < A.*(A) + E. By the preceding, for each i there exists a bounded open interval I; such that A; c I; and A.*(/;) - A.(A;) < 2-iE. Then A c U �1 /; and

A.*(A) <

00

00

i=l

i=l

:L A.*(/;) < L [A.(A;) + 2-i E] < A.*(A) + E ,

which establishes the validity of the desired formula. •

From now on,for simplicity, A.* will be denoted by A. and it will be called the Lebesgue measure on IR." .

Section 18: THE LEBESGUE MEASURE

135

It should be clear that every finite sub �et of 1R" has Lebesgue measure zero. Therefore, by the a-subadditivity of A., every countable subset of 1R" likewise has Lebesgue measure zero. Our next result deals with an approximation property of the Lebesgue measur­ able sets by open sets.

Theorem 18.2. A subset E of 1R" is Lebesgue measurable if and only iffor each E > 0, there exists an open set 0 such that E c 0, and A.(O \ E) < E. Proof. Assume E to be Lebesgue measurable. First, consider the case when of bounded open intervals such A.( E) < oo. Given E > 0, choose a sequence that E c U�t and :L::t A.(/;) < A.(£) + E. Then 0 = u:l is an open set such that E c 0 andA.(O) < :L:: 1 A.(/;) < A.(E)+E hold. Since E is measurable, we get A.(O \ E) = A.(O) - A.(E) < E. Now, assume A.( E) = oo. For each i let B; = {x E 1R": d(x, 0) < i} and E; = E n B;. Then E = U:1 E;, and each E; is a Lebesgue measurable set such that A.(£;) < oo. By what was shown above, for each i there exists an open set 0; such that E; C 0; and A.(O; \ E;) < 2-; E. Put = U:t Clearly, 0 is an open set, E c and in view of 0 \ E c U:1 (0; \ E;), we get A.(O \ E) < :L::t A.(O; \ E;) < E. For the converse, assume that for each E > 0 there exists an open set 0 such that E c 0 and A.(0 \ E) < E. For each i choose an open set 0; such that E c 0; and A.(O; \ E) < i- 1 • Put G = n:t 0;. T�en G is a Lebesgue measurable set such that E c G . Also, A.(G \ E) < A.(0; \ E) < i - t for each i implies that A.(G \ E) = 0, and so G \ E is a Lebesgue measurable set. The measurability of E now follows from the identity E = G \ (G \ E), and the proof is. finished. •

{I;}

I;

I;

0

0,

0;.

The next result informs us that every Borel set is Lebesgue measurable. The collection of all Lebesgue measurable sets is denoted (as usual) by A .

Theorem 18.3. Every Borel subset of1R" is Lebesgue measurable.

Proof. Let -oo < a; < b; < oo for i = 1 , . . . , 11. Choose m such that a; + ; < b; for each 1 < i < 11. The relation ,

" (a

oo

[ [

J]

b)= a + b D ; , ; kld D ; k ' ,..} "

l

and the fact that A is a a-algebra imply that 0�=1(a , b;) E A. But then, since ; every open set can be written as a countable union of such sets, it follows that A contains every open set. Therefore, A must contain every Borel set since the Borel sets are the members of the a -algebra generated by the open sets. •

Chapter 3: THE THEORY OF MEASURE

136

It will become apparent in the course of this book that the link connecting measure theory, topology, and analysis is the concept of a regular Borel measure. Its definition follows.

Definition 18.4. Let (X, -r) be a Hausdorfftopological space, and let B be tile· a-algebra ofits Borel sets.A measure J.L: B -+ [0, oo] is called a regular Borel measure if it satisfies tbe following properties: J.L(K) < oo for eve1y compact set K. 2. If B is a Borel subset of X, then 1.

J.L( B ) = inf{J.L(O): 0

open and B

c 0} .

3. If 0 is an open subset of X, then JJ-(0)

=

sup{J.L(K): K

compact and K

c 0}.

measure J.L on the Borel sets of a topological space that satisfies J.L(K) < oo for each compact set K is referred to as a Borel measure. The next result informs us that property (3) of the definition of a regular Borel measure is true for all Borel sets. A

Lemma 18.5. If J.L is a regular Borel measure on a topological space, then J.L(B) = sup{J.L(K): K

compact and K

c

B}

bolds for eacb Borel set B with J.L(B) < oo. Proof. Let B be a Borel set with J.L(B) < oo, and let E 0. Pick an open set V with B c V, and J.L(V) < J.L(B) + E . Similarly, choose an open set W such that V \ B c W c V and >

J.L(W) < J.L (V \ B) + E = J.L (V ) - J.L(B) + E < 2E. Next, choose a compact set C such that C c V and J.L(V) < J1(C) K = C n we, and note that K is a compact subset of B. Moreover, 0 <

+ E. Put

J.L (B) - J.L(K) = J.L(B \ K) < J.L(V \ K) = J.L((V \ C) U W )

< [J.L(V) - J.L(C)] + J.L(W) < 3E holds, and the proof is finished.

•

Section 18: THE LEBESGUE MEASURE

137

It is important to realize that the Lebesgue measure is a regular Borel measure. The details follow.

Theorem 18.6.

The Lebesgue measure on IR" is a regular Borel measure.

Proof. ( l ) Let K be a compact subset of IR". Then K is bounded, and so there

exists some A = [a;, b;) with K c A. Therefore, A.(K) < 'A(A) < oo. (2) Let B be a Borel set, and let E > 0. By Theorem 18.2 there exists an open set V such that B c V and A.(V \ B) < E . Therefore,

n�'= 1

A(B)

< inf(A.(O): 0 is open and

<

B c 0} 'A(V) = 'A(V \ B) + 'A(B) < 'A(B) + E

holds for all E > 0, from which the desired equality follows. (3) Let 0 be an open subset of IR". Pick a sequence (K;} of compact sets with } be an enumeration of all closed balls 0 = u: 1 K;; for instance, let ( K 1 , K2 , with "rational" centers and rational radii that are included in 0. Now, for each K111 , and note that each C; is compact and C; t 0. It follows that let C; = 'A(C;) t 'A(O), and so A.( O) = sup('A(K): K compact and K c 0} holds. The proof of the theorem is now complete. • •

•

i

•

U�n=l

Later we shall prove a general result (see Theorem [38.4]) that will guarantee that every Borel measure on IR" is necessarily a regular B orel measure. If A is a subset of IR" and a E IR", then the set a + A = (a + x: x E A } is called the translate of A by the vector a. An alternate notation for a + A is A + a. It is easy to see that A.( A) = 'A(a + A) holds for every subset A of IR" and every a. Also, by using the identities

E n (a + A) = a + (E - a) n A

and

E n (a + A)c = a + (E - a) n Ac,

it can be shown easily that a subset A of IR" i s Lebesgue measurable i f and only if a + A is Lebesgue measurable for each a E IR". Let B denote the a-algebra of all Borel .subsets of IR". Then a + A E B for each A E B and each a. Indeed, if A = {A E B: a + A E B for all a E IR"}, then A is a a-algebra of sets containing the open subsets of IR"; hence, A = B. A measure 11-: B -+ [0, oo] is said to be translation invariant if /1-(A) = J.L(a + A) holds for all A E B and all a E IR". The following question is a natural one: •

Wltat ace tlte tcanslation. invaciant Borel measures on IR."?

138

Chapter 3: THE THEORY OF MEASURE

In answering this question, we shall show that besides a multiplication factor the only translation invariant Borel measure on .IR" is the Lebesgue measure. To establish this, the following result dealing with additive functions is needed. Recall that a function f: .IR � .IR is said to be additive if f (x + y) = f(x) + f (y) holds for all x, y E .IR. Lemma 18.7. Let f: .IR � .IR be an additive function. If f is continuous at

zero, then there exists a constant c such that f(x) = ex holds for all x E .IR. In particular, if f: .IR" � .IR is continuous at zero for each variable and additive for each variable separately, then there exists a constant c such that f(x,, . . . , X11) = ex, · · · X11 holdsfor all (x,, . . . , X11) E .IR". Proof. If f: .IR � .IR is additive, then f satisfies the following properties:

f(O) = 0 [Reason: f(O) = f(O + 0) = f(O) + f(O)], 2. f(-x) = -f(.x) for each .x E .1R [Reason: f(x) + f(-x) = f(.x - .x) = f(O) = 0], and 3. f(rx) = rf(.x) for all rational numbers r E .IR and all x E .IR. 1.

The proof of (3) goes by steps. If k is a positive integer, then

f(kx) = f(.x + · · · + x) = f(x) + · · · + f(x) = kf(x). So, nf U·II ) = f(�) = f(x), which shows that f( IIlx) = IIlf(x). Thus, if m and n II are positive integers, then

(m )

( )

1 m f -x = mf -x = -f(x). n n n It follows that f(r .x) = rf(x) holds for all rational numbers r and all x E .IR. Since f is continuous at zero, the identity f(.x - y) = f(x) - f(y) shows that f is continuous at every point of .IR. Now, if .x E .IR, then choose a sequence of rational numbers {r11} such that lim r, = .x. Consequently, lim rnfO) = .xf(l) = ex f(.x) = lim f(r,) = 11-+00 n-+oo

where c = f(l) is a constant, and the result has been proven for the one-dimensional case. The general case can be established now by induction. We leave the details as • an exercise for the reader.

139

Section 18: THE LEBESGUE MEASURE

We are now ready to prove that, aside from a multiplication factor, the Lebesgue measure is the one and only translation invariant Borel measure on IR".

Let J.L be a translation invariant Borel measure on IR". Then there exists a constant c such that J.,L(A) = cA.(A) holds for every Borel set A ofiR11• Proof. For each x E IR let lx = 0 if x = 0, lx = [0, x) if x > 0, and lx = [x, 0) if x < 0. Also, for every x = (xt , . , X11) E IR11 let Sgn x = 1 if n�=l X; > 0 and Sgnx = -1 if n;·= l X; < 0. Assume now that J.L is a translation invariant nonzero Borel measure on IR". IR by Define f: IR11 Theorem 18.8.

.

--+

f(xt

•

. . .

, x,)

= Sgnx J.L ·

.

( (I ) 1=1

lx; •

where x = (x 1 • • • • , x,) E IR". Then f is additive in each variable separately. For instance, to see that f is additive in the first variable, consider the case when a > 0, b < 0 with a + b > 0. Fix x2, . . . , X11 and let s = Sgn (x2, . . . , x,) and I = Clearly, [0, a ) \ [a + b, a) = [0, a + b) and [a + b, a) X I (a, 0, 0, . . . , 0) + [b, 0) x I. Since J.L is translation invariant,

n;'=2 1X; ·

J.L ([b, 0) x I) = J.L([a + b, a) x 1).

Consequently,

f(a, X2 ,

•

•

.

, X11) +

f(b, X2,

.

.

•

, X11)

= S( J.,L((O, a) X I) - J.L ((b, 0) I) = s[ J.,L([O, a) x I) - J.L([a + b, a) x = SJ.L(([O, a) \ [a + b, a)) x I) = SJ.,L((O, a + b) X = f(a + b, X2 , . , Xn). X

.

.

]

/)

/)]

The other cases can be proven easily in a similar manner. Next, note that f is continuous at zero for each variable separately. Indeed, if ak t 0 and I = n�·=2 1X;' then [ak . 0) X I -!.- (/;. Hence, By Theorem 15.4, J.L([ak . 0) x I) ,!.. 0, which implies limk--+-oo f(ak , x2 , x,) = 0. That is, f is left continuous at zero for the first variable. On the other hand, the identity f(b, x2, . . . , x,) = - f( -b, x2, . . . , x,) implies that f is also right continuous for the first variable. Therefore, f is continuous for each variable separately. By Lemma 18.7, there exists a constant c such that f(x 1 , , x,) = cx1 • x,. Since J.L "# 0, it follows that c > 0. • . . .

•

•

•

•

•

140

Chapter 3: THE THEORY OF MEASURE

Now, let A = fl;'= 1 [a;, b; ), where -oo < A = (a 1 , . . . . a,) + n;'=1 [0, b; - a;), we get JL (A) = Jl = c

(n II

[0, b; - a;)

•=I

;

)

=

a;

<

b;

< oo

for each i. Since

j(b1 - a h . . . , b, - a,)

n (b; - a ) = cA(A). i= l

1 1 Thus, A = c- Jl on S. But then Theorem 15.10 shows that A = c- Jl holds on B. • That is, J.L(A) = cA(A) holds for each A e B. A

natural question one may ask at this point is whether every Lebesgue mea­ surable set is a Borel set. The answer is negative. That is, there exist Lebesgue measurable sets that are not Borel sets. One way of establishing this is by a cardi­ nality argument. It can be shown that the a-algebra of ali Borei sets has cardinaiity c, while that of the Lebesgue measurable sets is 2'. Hence, the two a -algebras do not coincide; see [15, p. 1 33]. Next, we shall present a different proof of the existence of a Lebesgue measurable set that is not a Borel set, the steps of which have some independent interest. For simplicity, the arguments will be given for the one-dimensional case. In fact, the example will be a subset of [0, I ] . To do this, we need some preliminary discussion. The Cantor set C has been described in Example 6. 15. It was observed there that the total length of the open intervals removed from [0, I] to get C is equal to one. It follows from this that A(C ) 0. Thus, every subset of C has Lebesgue measure zero, and hence, every subset of C is Lebesgue measurable. Next, we shall describe the E -Cantor set. The process for constructing it is similar to the one described in Example 6.15. Fix 0 < E < I , and let o = 1 - E. Start with 1 Ao = [0, 1] and remove from the center of Ao an open interval of length 2- o. Let A 1 be the remaining set; that is,

=

A1 =

-

1

[o � - �4] [� U

'2

2

+

]

� 1 . 4'

Note that A 1 consists of 2 = 2 disjoint closed intervals of the same length and that A(A 1) = 1 2- I o. In the next step, remove from the center of each of the two disjoint closed intervals of A an open interval of length 2-3 o. Thus, the "total length" removed from A 1 is 2-28. Let A2 be the union of the remaining 22 disjoint closed intervals of equal length. Note also that A(A2 ) = A(A 1) - 2-2 8 = 1 (2- 1 + 2-2)8. For the general step, suppose that the set A, has been constructed consisting of the union of 2" disjoint closed intervals all of the same length and satisfying

-

1

Section 18: THE LEBESGUE MEASURE

-

141

A(A,) = 1 (2- 1 + + 2-")8. From ·the center of each of these 2" closed intervals, delete an open interval of length 2- 2"- 1 o. Let A,+1 be the union of the remaining 211+ 1 disjoint closed intervals of equal length. Since the "total length" removed from A, is 2-"-18, it follows that · · ·

Clearly, A,+1 c A11 holds for each 11, and the E-Cantor set is now defined to be the set CE = n�l A,. Clearly, c€ is a closed set, and nowhere dense in [0, I ] . Also, /1-+-00

A(CE) = lim A(A,) = 1 -

(� ) �

1 2- 1

o

II= I

=

l - o = E.

Lemma 18.9. For every 0 < E < I there exists a real-valued continuous func­ tion f: [0, I ]

I. 2.

3.

�

[0, I] such that:

f is onto, f is strictly increasing (and hence, f is one-to-one), and f maps the E-Cantor set C€ onto the Cantor set C .

..

Proof. Let { 1 1 , h, . } be the open intervals removed from [0, 1 ] in the con­

struction of C (counted from left to right, following the construction process in­ ductively). Similarly, let (11 , h . . . . } be the open intervals removed from [0, 1] in order to get the Cantor set C (counted again from left to right). Clearly, c€ = [0, 1 ] \ u::l /II and c = [0, 1 ] \ u:l J, . . Let /, = (a11 , b,) and 111 = (c11, d,) for each n . Now define f: [0, 1 ] """"* [0, 1] by steps as follows: �:

1. 2.

/(0) = 0. For x E /, = (an. b11 ) define f(x ) =

3.

dn - c,

b, - a,

(x - a, )

+ c,.

Note that f maps In onto J,. If x =j:. 0 and x E CE, define f(x) = sup

{

f(y): y < x and y E

U /, } .

II= I

A part of the graph for a typical function f is shown in Figure 3.1.

Chapter 3: THE THEORY OF MEASURE

142

. •

./

./

0

-

I�

FIGURE 3.1.

We leave the details as an exercise for the reader to verify that properties stated in the lemma.

f satisfies the

•

Since thesetC has cardinality c (see Example [6. 15], it follows from Lemma 1 8.9 that every E-Cantor set also has cardinality c. In view of Example 15, we know that there exist subsets of [0 , 1 ] that are not Lebesgue measurable. The next result describes an interesting property of the non-Lebesgue measurable subsets of [0, 1 ].

Let A be a non-Lebesgue measurable subset of [0, 1 ] . Then there exists some 0 < E < 1 such that for any Lebesgue measurable subset E of [0, 1 ] with A(£) > E, the set A n E is not Lebesgue measurable.

Lemma 18.10.

Proof. Assume by way of contradiction that the conclusion is false. Then for

each 0

E€ of [0, 1] such that A(£€) > E and A n E€ is Lebesgue measurable. Choose a sequence {E11 } of (0, 1 ) <E

< 1 there exists a Lebesgue measurable subset

such that lim E11 = l . Put E = U: 1 Ef, , and note that E is a Lebesgue measuraple subset of [0, 1 ] . Moreover, En < A(E€n) < A(£) < 1 holds for all n , and thus, A(£) = l . Now, since A(A n Ec) < A([O, 1 ] \ £) = 0, we have A(A n Ec) = 0, and so, A n Ec is

Section 18: THE LEBESGUE MEASURE

143

Lebesgue measurable. But then

A = (A n E) U (A n Ec) =

[Q

]

(A n £,,) U (A n Ec)

holds and shows that A is a Lebesgue measurable set. However, this is a contra­ diction, and the proof is complete. • It was observed before that every subset of the Cantor set is Lebesgue measurable (since it has Lebesgue measure zero). We are now ready to establish that there exists a Lebesgue measurable set that is not a Borel set.

Theorem 18.11. The Cantor set has a subset that is not a Borel set. Proof. Choose a subset A of [0, I ] that is not Lebesgue measurable.

By Lemma 18.10 there exists 0 < E < I such that A n E is not Lebesgue mea­ surable for all Lebesgue measurable subsets E of [0, I] with A.(E) > E. Next, consider the continuous function f of Lemma 18.9 that carries the E­ Cantor set C" onto the Cantor set C. By the continuity of f, we get that f is Lebesgue measurable (where, of course, it is assumed that the Lebesgue measure is restricted to [0, 1]). Since A.(C") = E, the set B = A n C" is not Lebesgue measurable. Therefore, the subset f(B) of the Cantor set, which is Lebesgue mea­ surable, cannot be a Borel set-since B = f-1 (f(B)) is not Lebesgue measurable. The proof of the theorem is now complete. • A useful result (due to H. Steinhaus6) regarding Lebesgue measurable sets will

be presented next. It will be shown that if A is a Lebesgue measurable subset of JR" such that A.( A) > 0, then zero is an interior point of A - A . Recall that if A and B are two subsets of JRn , then A - B = {a - b: a E A and b E B }. The set of B from A. A B is called the In order to establish this result, we need a lemma.

algebraic difference

-

Lemma 18.12. Let E be a subset oflR" such that 'A( E) > 0. Then for each

0<E <

A.(E n 1).

1,

..._

there exists a bounded open interval I of JR" such that E'A(/) <

Proof.

Fix 0 < E < 1 , and let E be a subset of JR" such that A.(E) > 0. Without loss of generality, we can assume that A.( E) < oo. Indeed, if Bk = {x E JR": d(O, x) < k}, then E = U:1(E n Bk ), and so A.(E n B,.:) > 0 must hold for some k; since E n Bk c E, we can replace E by E n Bk . 6Hugo Steinhaus ( 1 887-1972), a Polish malhematician. He made many curious and interesting contributions to functional analysis.

Chapter 3: THE THEORY OF MEASURE

144

Since 0 < E < 1, there exists a sequence {/;} of bounded open intervals such that E c u� l /; and E I:: I J....(/;) < J....( E). To complete the proof, we show that there exists some i such that E A(/;) < J....(E n /;). Indeed, if this is not the case, then J.... ( E n /;) < EJ.... (/; ) holds for all i, and so

( k} )

J....(E) = J....

E n /;

8 00

00

<

8 00

J....( E n /;)

<E

J....(l;)

< J....(E),

which is impossible. This completes the proof of the lemma.

is a Lebesgue measurable subset of IR" 0, then the zero vector is an interior point of E - E . If E

Theorem 18.13 (Steinhaus).

such that J....( E)

>

•

Proof. Assume that E is measurable with J....( E)

.

>

0. By Lemma 18.12 there

exists a bounded open interval / = 07=1 (a;, b;) such that � J.... (I) < J.... (E n /). 3 <' -t • fi-- ' t !' 'leX , cnoose o > u suc..:'n t 'nat 1� = 1rTII '" � 1 1i=l (a;. - o, u; + OJ Si1 LS tc::, ,..('�' < 2/\.\.1 " and then consider theopen interval J = (-o, o) x · . · X (-o, o) c IR". To complete the proof, we shall show that J c E - E. To see this, let x E J . Then "

'

r.

.

r

I.

<'\

7 \

' I 7\

(E n I ) u (x + E n I ) c I u (x + I ) c Is

�

holds. Therefore, J....( (E n I) U (x + E n /)) < J....(/8) < J....( I), which implies that ( E n / ) n(x +En/) # 0. Indeed, if(En/)n(x+En/) = 0, then by the additivity of J...., we get J....((E n I ) U (x + E n / )) = 2J.... ( E n I ) > J....( I) (here we use the measurability of E), which is impossible. Pick a vector y E (E n I ) n (x + E n I). Then y E E, and there exists z E E such that x +z = y. That is, x = y- z E E - E, • so that J c E - E , and the proof is complete.

�

EXERCISES I.

2.

3. 4. 5.

6.

=

n:,'=1 ni'=l

Let I = /; be an interval of JR.". Show that I is Lebesgue measurable and that 1/;1, where 1/;1 denotes the length of the interval /;. A.(/ ) Let CJ be an open subset of JR. Show that there exists an at-most countable collection {Ia: a E A} of pairwise disjoint open intervals such that 0 = UaeA Ia. Also, show that A.(CJ) = LaeA llal· Show that the Borel sets of IR.n are precisely the members of the a -algebra generated · by the compact sets. Show that a subset E of JR." is Lebesgue measurable if and only if for each E > 0 there exists a closed subset F of JR." s uch that F c E and A.(£ \ F) < E. Show that if a subset E of [0, l] satisfies A.(£) = l, then E is dense in [0, 1]. 0. If E c JR." satisfies A.(E) = 0, then show that £0

=

Section 18: THE LEBESGUE MEASURE 7.

145

Show that if E is a Lebesgue measurable subset o f IR", then there exist an Fu -set A and a G0-set B such that A � E � B and A.(B \ A) = 0.

8.

Let ( £ , } be a sequence of nonempty (Lebesgue) measurable subsets of [0, l ] satisfying lim A.(£11) = a.

Show that for each 0 < E < l there exists a subsequence ( Ek, } of ( £11} such that

;..e n�, Ek, ) > €. Show that n�" Ek = 0

b.

9.

I.

is possible for each n = l , 2, . . . .

Assume that a function f: I --+ IR defined on a subinterval of IR satisfies a Lipschitz7 condition. That is, assume that there exists a constant C > 0 such that lf(x) - f(y)l � Clx - y l holds for all x, y E I. Show that f carries (Lebesgue) null sets to null sets.

In particular, if a function f:

I --+ IR

defined on a subinterval of IR has a contin­

uous derivative, then show that f carries null sets to null sets. [HINT: For the second part note that if J is a closed subinterval of holds for each

M l.r - y I

t E J , then by the Mean Value Theorem we have for all x , y E J . ]

10.

Show that the Lebesgue measure of a triangle in

11.

If fJ. is a translation invariant Borel measure on

the Lebesgue measure of a disk in

12.

IR2

1 / '(t)l �M 1/(x) - �

I and

f(y ) l

IR.2 equals its area. Also, determine

•

IR", then show that there exists some c 2: 0 such that f.J.*(A) = cA.( A) for all subsets A of IR". Show that an arbitrary collection of pairwise disjoint measurable subsets of IR, each

of which has positive measure, is at most countable.

13.

Let G be a proper additive subgroup of x

-

IR" (i.e., x , y E G imply x + y E G and E G). If G is a measurable set, then show that A.( G) = 0.

[HINT: Use Theorem 1 8 . 1 3 . ] 14.

Let f:

IR

--+

IR be additive (i.e., f(x + y)

Lebesgue measurable . Show that

=

f(x) + f(y) for all x, y

E IR) and

f is continuous (and hence, of the form f(x)

= ex).

[HINT: Use the properties of f shown during the proof of Lemma 18.7 to establish that f-1 ((0, E)) has positive Lebesgue measure for each E to show that f is continuous at zero . ]

>

0, and then use Theorem 18.13

15.

Show that an arbitrary union of proper intervals of IR is a Lebesgue measurable set.

16.

Let C be a closed nowhere dense subset of characteristic function

IR" such that A.(C)

>

0. Show that the

xc cannot be continuous on the complement of any Lebesgue

null set ofiR". Also, show that xc will be continuous on the complement of a properly chosen open set whose Lebesgue measure can be made arbitrarily small.

17.

Let

f: IR" --+ IR be a continuous function. Show that the graph G = {(XI

I

•

•

•

' XII , f(x l

I

• •

•

I

.r/1)): (XJ

I

•

•

•

I

.r/1)

E IR"}

of f has (n + 1 )-dimensional Lebesgue measure zero. [HINT: IfGk = ( (.r 1 ,

. . . , .r" , f(x l , . . . , .r")): l.ril �

A.( G k ) = 0 and then note that G k t G . ]

k

for i = l . . . . , n}, show that

7RudolfLipschitz ( 1832-1 903), a Gennan mathematician. His name is associated with the existence and uniqueness of solutions to differential equations.

Chapter 3: THE THEORY OF MEASURE

146

18.

L et X be a Hausdorff topological space, and let JL be a regular Borel measure on X. Show the following:

a.

If A

is an arbitrary subset of X, then

b.

If A

is a measurable subset of X with JL*(A) < oo, then

c.

If JL is a-finite and A is a measurable subset of X, then

JL*(A) = inf{JL(O): 0 open and A c 0}.

JL*(A) = sup{JL(K): K compact and K � A}. JL*(A) = sup{JL(K): K compact and K � A }.

=

is a measurable subset of 1R of positive measure and 0 < 8 < ).(A), then show that there exists a measurabl e subset B of A satisfying A.(B) 8. Let E be a Lebesgue measurable subset of 1R of finite Lebesgue measure. Show that the function fE : 1R � lR, defined by

19. If A 20.

...

f ( ,.\

J G \""• I

- l (f;' II. ( •• ..L f;' \) - .... I - I , •"'

is uniformly continuous.

19. CONVERGENCE IN MEASURE Let (X, S, J.L) be a measure space again. The collection of all real-valued measurable functions defined on X will be denoted by M. That is, M =

{/ E IR.x : f

J,L­

is J,L-measurable}.

It should be clear from Theorem 16.5 that M is closed under the ordinary algebraic and lattice operations. That is, M is a function space and an algebra. In the space M, a useful concept of convergence of sequences is defined as follows:

{[,1} of measurable functions com·erges in mea­ sure to a function f E M , in symbols [,1 � f, iffor every E > 0 we have Definition 19.1. A sequence lim

11�00

J.L* ({x

E X:

IJ,�(x) - f(x)l

2:

E)) = 0.

The next result summarizes the basic properties of convergence in measure.

Let {[,1 } and {g11 } be nvo sequences ofmeasurable functions, and let f, g E M. Then the following statements hold: 1. If fn � f and 8n � g, then afn + �g�� � af + �g for all a, � E JR. 2. If fn � f and J,� � g. then f = g a.e. Theorem 19.2.

Section 19: CONVERGENCE IN MEASURE

Proof.

147

(1) We can assume a, f3 =I= 0. from

lafn(x) + fJgn (x) - [af(x) + {Jg(x)] I < laii /J,(x) - f(x) l + lfJllgn (x) - g(x)l , it follows that

{x E X : lafn(X) + fJgn (X) - [af(x) + {Jg(x)] I

{

c x

e X:

lf,,(x) - f(x)l >

� } U { x e X : lg,(x) - g(x)l > 2�1 }·

(2) Let /J, � f and j� � g. Given E E X:

C {x E X :

>

0, the triangle inequality implies

lf(x) - g(x)l > 2E} l fn(X) - f(x)l > E } U {x E X : lf,,(x) - g(x)l > E},

from which it easily follows that t.t*({x E X : E > 0. Therefore,

iJ.'((x e X: f(x) # g(x)}) so that

f

=

g

E}

2 1

The conclusion now follows easily.

{x

>

= iJ.' (Q {

l f(x) - g(x) l

> 2E })

x E X: l f(x) - g(x)l >

=

0 for all

�}) =

a.e. holds.

0, •

The lattice operations are also continuous with respect to convergence in measure.

Theorem 19.3.

then

Ifa sequence { /J,} ofmeasurable functions satisfies /J, � f, + !,( � f , !,-; � f- , and lfnl � lfl .

Proof. The statements easily follow from the standard lattice inequalities

lfn+ - f+ l < If,, - fl , lfn- - f- 1 < 1 /J, - fl , and 1 1 /J, l - lfll < lfn - fl.

•

The convergence in measure is related to the pointwise convergence as follows.

Theorem 19.4.

�

for some f E fk, f a.e.

Ifa sequence {/J,} ofmeasurable functions satisfies f,, � f M, then there exists a subsequence {fk,. } of {/J,} such that

Chapter 3: THE THEORY OF MEASURE

148

Proof. Let f, �f. A simple inductive argument shows that there exists a

strictly increasing sequence {k,} of natural numbers such that !L'

( {x

E X:

lf,(x) - /(x}l

for all k > /.:.,. Let £, = {x E X : measurable set E = U:m

n�= l

?:_

l fkn (x) - f(x) l

E,. Then

�})

>

< 2-"

� } for each n, and define the

holds for all m, so that !.L*E) = 0. Also, if x ¢ E, then there exists some m such that X ¢ u:m E,, and so, lfk, (x) - f(x) l < � holds for each 12 2: 117. Therefore, lim fkn (x) = f(x) for each x E Ec, and so fkn -+ f a. e. holds. • Pointwise convergence does not imply convergence in measure. For instance, let X = IR with the Lebesgue measure A. , and define f, = X[11.n+ l 1 for each n. Clearly, lim f,(x) = 0 holds for each x e JR . Since

A.{{x E

X:

l fn(x) l

>

1})

=

A. {[n, n + 11)

=

I

holds for all n, it follows that { f, } does not converge to zero in measure. However, if the measure space is finite, then pointwise convergence implies convergence in measure.

Theorem 19.5. Assume !.L *(X) < oo. If a sequence {f,, } of measurable func­

tions satisfies fn

f a.e., then f,� f also holds. > 0 and for each n let £, = {x e X : l f,,(x) - f(x) l -+

2: E}. Now Proof. Fix E let o > 0. Since /.L*(X) < oo, it follows from Egorov's Theorem 16.7 that there exists a measurable set A such that /.L*(A) < o and { fn l converges uniformly to f

on Ac. Choose k such that l f,, (x) - f(x) l < E holds for all x e Ac and all n > k. Then E, c A holds for all n > k, and so !.L*(E,) < /.L*(A) < o for all n > k. Thus, lim /1-*(£,) = 0, and the proof is finished. • It is interesting to observe that there .exist sequences of measurable functions that converge in measure, but fail to converge at any points. An example of this type is presented next. Consider [0, 1] equipped with the Lebesgue measure A.. For each into the n subintervals [0, -];1, �], . . . 1].

Example 19.6. divide [0,

I]

[�,

, [ n�l ,

n sub­

Section 20: ABSTRACT MEASURABILITY

149

Enumerate all these intervals as follows:

[0, �], [�. ] [0, �]. [�. �]. [ �. ] [ �]. [�. �]. [ �. �]. [�. ] [0, �]. 1

I . 0,

.

I .

· · ·

.

Let f11 be the characteristic function of the nth interval of the above sequence. It is easy to see that /11 �0 holds and that {f11(.r)} does not converge to zero for any x E [0, 1 ] . B

EXERCISES 1.

Let { J,,} be a sequence of measurable functions and let f: X � IR. Assume that lim J..L *( {.r E X: jf,,(x) - f(x)l � E}) = 0 holds for every E > 0. Show that f is a measurable function.

{ f,} c M satisfies j�, t and f,, �f. Show that i, t f a.e. holds. 3. If {/11 } 5; }A satisfies f,� f and f, > 0 a.e. for each n, then show that f � 0 a.e. 2.

Assume that

holds.

Let { J,,} 5; ;\;f and {g,} 5; M satisfy f, ...1!:...:.. f, g, ...1!:...:.. g. and f, = g, a.e. for each n. Show that f g a.e. holds. 5. Let (X, S, J..L ) be a finite measure space. Assume that two sequences {j;,} and {g,} of M satisfy f, ...l!:...:.. f and g,...l!:...:.. g. Show that f,g,� fg. Is this statement true if J..L* (X) oo? 6. Show that a sequence of measurable functions { f,, } on a finite measure space converges to f in measure if and only if every subsequence of { f,} has in tum a subsequence which converges to f a.e. 7. Define a sequence {f,, } of ;\If to be J,.L-Cauchy whenever for each E > 0 and � > 0 there exists some k (depending on E and 8) such that J..L* ({.r E X: lf,,(.r) - fm (x) l � E}) < 8 holds for all n. m � k. Show that a sequence {/,} of ;\If is a ,u-Cauchy sequence if and only if there exists a measurable function f such that j�, �f.

4.

=

=

20. ABSTRACT MEASURABILITY This section can be skipped in the first reading. It deals with collections of sets with stronger properties than those of a semi ring. Most of these families were introduced briefly in Section 12. Let us start by recalling the notion of a ring of sets.

Definition 20.1. A nonempty collection n of subsets of a set X is said to be a ring if A. B E n imply A U B E n and A \ B E 'R. It should be obvious that every ring is closed under finite unions. Also, since an arbitrary ring is nonempty, it contains a set, say A E 'R, and so 0 = A \ A E 'R.

'R.

ISO

Chapter 3: THE THEORY OF MEASURE

That is, every ring contains the empty set. Also, from the identities

A�B = (A \ B) U (B \ A)

and

A n B = A \ (A \ B),

it easily follows that every ring is closed under symmetric differences and finite intersections. Recall that a nonempty collection of subsets of a set X is said to be an of sets if it is closed under finite intersections and complementation. From the identities

algebra

it should be obvious that every algebra is a ring. In the converse direction, we have the following result (whose easy proof is left for the reader):

Lemma 20.2.

A ring ·R, is an aigebra if and only if X E "R.

if it is closed under countable unions, i.e., if {A11} c 'R A ring 'R is called a a �mplies U�1 A11 E 'R. Notice that every a-ring is automatically closed under countable intersections. To see this, let { A11} be a sequence in a a -ring 'R and let A = n: 1 An. Then At \ A = U: 1 (At \ An) E 'R, and so A = At \ (At \ A) E 'R. The straightforward proof of the next result is also left for the reader.

-ring

Lemma 20.3.

A a-ring 'R is a a-algebra if and only if X E R.

The following scheme indicates the relationships between the various families of sets we have introduced so far: cr-ring a-algebra

�

�

algebra

FIGURE 3.2.

�

ring

�

=====>

semiring

The Implication Scheme

In general, no other implication is valid; see Exercise

3 at the end of this section.

Arbitrary intersections of a-algebras ·of subsets of a fixed set X are also a­ algebras. Thus, for every nonempty family :F of subsets of a set X there exists a smallest a-algebra (with respect to inclusion) that includes :F. This smallest by :F and is denoted by a(:F). It is a-algebra is called the the intersection of all a -algebras that contain :F (notice that P(X) is one of them).

a-algebra generated

Section 20: ABSTRACT MEASURABILITY

151

That is,

a(F) = n { :E :

:E is a a-algebra and :F c :E } .

If :E is a a-algebra and a family of sets :F of :E (i.e., :F c :E ) satisfies a(:F) = :E , then :F is called a family of generators for :E. The reader should also notice that for an arbitrary collection of subsets :F of a set X there always exists an algebra, a ring, and a a-ring generated by :F. In this book, we shall make use only of the a-algebras generated by families of sets. We now tum our attention to abstract measurability properties of functions. Notice that if f : X -+ Y is a function and A is either an algebra or a a-algebra (or a ring, or a a-ring, or a semiring), then

is likewise an algebra or a a-algebra (or a ring, or a a-ring, or a semiring).

Lemma 20.4. Iff : X -+ Y is afunction and :F is a nonemptyfamily ofsubsets

ofY, then

a(f- 1 (F)) = f- 1 (a(F)). Proof. Notice first that the identities f-1 (Ac) = [f- 1 (A)]c,

guarantee that f- 1 (:E) = {f-1 (A): A E :E ) is a a-algebra whenever :E is a a-algebra. So, f- 1 (a(F)) is a a-algebra satisfying f-1 (F) c f- 1 (a(F)). This implies

a(f- 1 (F)) C f- 1 (a(F)). For the reverse inclusion, let

Clearly, A is a a-algebra satisfying :F c A c a(F). Hence, A = implies f- 1 (a(F)) c a(f-1 (F)), and so f- 1 (a(F)) = a(f- 1 (F)) ,

a(F). This •

now a pair (X, :E), where X is a set and :E is a a-algebra of subsets of X. The members of :E are called the measurable sets of X . In an abstract setting, a measurable space is

152

Chapter 3: THE THEORY OF MEASURE

Definition 20.5. A function f: (X 1 , :E t) � (X2, :E2) between two measur­ able spaces is said to be measurable (more precisely ( :E 1 , 'L2 )-measurable) if

f-1(A) E :E t holdsfor each A E :E2, i.e., if f-1 C:E2 ) c :E t . A function f: X � Y between llvo topological spaces is said to be Borel measurable iff: (X, Bx ) � (Y, By) is measurable, i.e., if f-1(B) is a Borel subset of X for each Borel subset B ofY.

When (X, :E) is a measurable space and Y is a topological space, then a function f: X � Y is called measurable if f: (X, :E) � (Y, By) is measurable.

Let f: (X t , :E t ) � CX2 . :E2) be a function between two mea­ surable spaces and let :F be a family of generators for :E2, i.e . , a(:F) = :E2. Then f is measurable if and only if f-1 (A) E :E 1 for each A E :F. Proof. If f is measurable, then clearly f-1(A) E :E1 holds for each A E F. On the other hand, if f-1 (A) E :E1 for all A E :F, then it follows from Lemma 20.4 Theorem 20.6.

that

which shows that f is a measurable function.

•

We now continue with another important family of sets.

Definition 20.7. A family of subsets V of a set X is said to be a Dynkin8 system if it satisfies the following three ptoperties:

X E V, 2. A, B E V and A C B imply B \ A E V, and 3. wheuever a sequeuce { An } c V satisfies A11 t A, theu A E V. 1.

The reader should stop and verify that a Dynkin system is a a -algebra if and only if it is closed under finite intersections. The Dynkin systems play an important role because of the following result-known as Dynkin 's lemma.

Lemma 20.8 (Dynkin's Lemma). Let V be a Dyukiu system aud let :F be a family of sets such that :F c V. If :F is closed w1dec finite imersectimzs, thel1

the a-algebt·a genecated by F is inclqded in V, i.e., a(:F)

C 'D.

8Eugene Borisovich Dynkin ( 1 924-), a Russian mathematician. His main research interests are in probability theory. In 1977 he was appointed Professor of mathematics at Cornell University, Ithaca, New York.

153

Section 20: ABSTRACT MEASURABILITY

Proof. Assume that F c V is nonempty . and closed under finite intersections. Let us denote by V0 the smallest Dynk in system including F (i.e., V0 is the intersection of all Dynkin systems that include F). Clearly, F c V0 c V. Now let

V,

=

{A

Vo: A n C

E

E V0 for all C E

F}.

An easy verification shows that V1 is itself a Dynkin system satisfying A E V1 for all A E F (to establish this we must use the assumption that F is closed under finite intersections). Thus, F c V1 c Vo, and so by the definition ofV0, we infer that V1 Vo. Next, consider the family

=

V2

=

=

{A

E

Vo: A n B

E

Vo

for all

B

E

Vo}.

Again, it is easy to check that V2 is a Dynkin system such that F c V2 c V0. Thus, V0 ·v2, and therefore, V0 is a Dynkin system which is closed under finite intersections. This easily implies that V0 is a a-algebra, and from this we see that a (F) V0 c V, as desired. •

=

Here are a few striking consequences of Dynkin 's lemma: Corollary 20.9. Assume tltat V is a Dyukiu system aud L: is a a-algebra suclt titat V C :E . If V contaius a family of geuerators of L: wlticlt is closed uuder

fwite iutersectious, tlteu V

=

L:.

Let :E be a a-algebra of subsets of a set X and Jet F be a family of geuerators for :E (i.e., a(F) :E) whiclt is closed wtder fwite iutersectious. Tlteu two fwite measures J.L and v ou :E coiucide (i.e., J.L = v) if aud ouly if 1 . J,L(X) = v(X), aud 2. JJ.(F) v(F) for eaclt F E F. Corollary 20.10.

Proof.

=

Assume that two finite measures J.L and V

Then X

=

=

{A

E :E:

JJ.(A)

=

v on :E satisfy ( 1 ) and (2). Let

v(A)} .

E V and if A, B E V satisfy A c B , then

J,L(B \ A)

=

JJ.(B ) - JJ.(A)

=

v(B) - v(A)

=

v(B \ A),

154

Chapter 3: THE THEORY OF MEASURE

i.e.,

B \A

E V. Moreover, if a sequence

{A11} c V satisfies A11 t A, then

lim J.L(A11) = lim v(A,J = v(A), J.L(A) = n-+oo u-+oo

so that A E V. Thus, V is a Dynkin system and :F V = :E, and so J.L = v.

c V holds. By Corollary 20.9, •

Corollary 20.11. Twofinite measures defined on the Borel sets ofa topological

space are equal if and only if they coincide on the open sets. We now turn our attention to product semirings. If St and S2 are two semirings of subsets of X and Y respectively, then the product semiring St ® S2 is defined by

Using the identities

(A x B) n CAt x Bt) = (A n At) x (B n Bt) (A x B) \ (A t x Bt) = [(A \ A d x B ] u [(A n A t ) x (B \ Bt) L it is easy to verify that St ® S2 is indeed a semiring of X x Y. It should be noticed that St ® S2 need not be an algebra even when St and S2 are both a-algebras. The product a-algebra generated by St and S2 is the a-algebra a(St ® S2) generated

by St ® S2. If A is a subset of X x Y and x

of Y defined by

E X, then the x-section

Ax = { y E Y: Similarly, if y E Y, then the y-section

AY = {x

(x , y) E

Ax of A is the subset

A}.

AY of A is the subset of X defined by

E X : (x, y) E

A }.

The geometric meanings of the x- and y-sections are shown in Figure 3.3. Regarding sections, we have the following identities. (The families of sets below are assumed to be families of subsets of X x Y .)

' = nie/ (A ; )Y. 3. (Ac)x = (A.r )c and (Ac)Y = (AJY. 1.

The easy verification of these identities are left for the reader.

155

Section 20: ABSTRACT MEASURABILITY

y

X FIGURE 3.3. The Sections of a Set

I:

1 ) and (Y, :E2) be measurable spaces. !! A belongs to the product a-algebra, i.e., A E a(:E1 ® :E2), then A has "measurable sections." That is, Ax E "E2for each x E X and A>" E :E1 for all y E Y .

Lemma 20.12. Let (X,

Proof. Let A E a ( :E 1 ® :E2). By the symmetry of the situation, it suffices to

show that Ax E :E2 for each x E X. To this end, fix x E X and let

From (0 x 0).\. = 0 and (X x Y)x Y , we see that 0. X x Y E :E . Also, from the identities listed preceding the lemma, we see that :E is a a -algebra. Moreover, if A x B E 'Et x :E2, then (A x B)x = B E :E2 ifx E A and (A x B):c = 0 E :E2 if x ¢ A, and so :E 1 ® 2:2 c :E . This implies :E = a(:E1 ® :E2), and the proof is finished. • =

We now come to the notion of joint measurability for functions. Let (X, 2: 1 ), (Y, :E2), and (Z, :E3) be three measurable spaces, and let f: X x Y -+ Z be a function. We shall say that:

1. f is jointly measurable, if f : (X x Y, a ( :E 1 ® I;2)) 2.

3.

-+ (Z, :E3) is measur­

able, f is measurable in the variable x (resp. in y) if for each y E Y (resp. for each x E X) the function f( · , y): (X, :E 1 ) -+ (Z, :E3) (resp. the function f(x, · ): (Y, :E2) -+ (Z, :E3)) is measurable, and f is separately measurable if f is measurable in each variable.

Joint measurability always guarantees separate measurability.

Theorem 20.13. Let (X, :E 1 ), (Y, :E2) and (Z, :E3) be three measurable spaces.

If a function f: X measurable.

x Y -+ Z

is jointly measurable, then it is separately

156

Chapter 3: THE THEORY OF MEASURE x Y 1 = f- (A) E

Proof. Assume f: X

y E Y. Also, let B

f- 1 (A, y) = {x E X :

�

Z is jointly measurable, let A a(:E 1 ® :E2 ). Note that

f(x, y)

E A} = {x E X : (x, y) E

E :E3 and fix

B } = BY .

��

Now, a glance at Lemma 20.12 guarantees that BY E :E�t and consequently, the {Z, :E3) is measurable. Similarly, itcanbe shown that function f(·, y): (X, :E 1 ) the function f(x, · ) : {Y, :E2) • (Z, :E3) is measurable for each x E X .

Does separate measurability imply joint measurability? The answer is in general

negative. W. Sierpinski9 has constructed in [27] a non-Lebesgue measurable subset A of IR2 whose intersection with each line consists of at-most two points; see also [10 p. 130] and [9]. From this property it easily follows that the characteristic function XA is separately measurable but not jointly measurable. However, under certain conditions, separate measurability suffices for the joint measurability. In order to state such a result, we need the notion of a Caratheodory function.

� � �

Definition 20.14. Let (X, :E) be a measurable space, and let Y and Z be two topological spaces. A function f: X x Y Z is said to be a Caratheodory function if I . for each x E X the function f(x, · ) : Y Z is continuous, and (Z, Bz) (where Bz is tbe 2. for each y E Y , the function f(·, y): (X, :E)

a-algebra of all Borel sets of Z) is measurable.

Briefly, a function f(x, y) of two variables is a Caratheodory function if f is continuous in y and measurable in x. For the next result, keep in mind that every topological space is considered a measurable space equipped with the a-algebra of its Borel sets.

�

Theorem 20.15. Let (X, :E) be a measurable space, Y a separable metric

space, and Z a metric space. Then eve1y Caratheod01yfunction f: X x Y is jointly measurable.

Z

Proof. Let d denote the metric of Y and p the metric of Z. Also, let {y 1 , Y2 , . . . }

be a dense subset of Y . Since the closed sets form a family of generators for Bz, it suffices to show that f- 1 (C) E a( :E ® Br ). So, let C be a closed subset of Z. Start by observing that (since f(x' .r i s continuous for each X E X), f(x' y) belongs to the closed set C if and only if for each n there exists some Ym with 9Waclaw Sierpinski (1 882-1969), a Polish mathematician. He made many important contributions to set theory, topology, and number theory.

Section 20: ABSTRACT MEASURABILITY

157

and d(y, y,) < l II

p(f(x, y, ), C) = in f{ p ( / Ct , Y111 ), c): c E C }

<

�.

n

This implies

r' (C) =

0 .Q [

(x E X : j (x , y, ) E B � (C)j

( �) ] .

x 8 y,,

(*)

where B 1 (C) = {: E Z : p(: , C ) < � } . The continuity ofthefunction : � p(z, C) guarante'�s that B! (C) is an open set. Since f is measurable in x and B 1 (C) is open (and hence a Borel set), it follows that {x E X : f(x, y111) E B1 (C)} E :E for each m and n . Now, a glance at ( *) shows that f- 1 (C) E a(:E ® s;'), and thus, f is jointly measurable. • II

II

EXERCISES 1. Let 'R be a nonempty collection of subsets of a set X. Show that 'R is a ring if and

only if n is closed under symmetric differences and finite intersections. [HINT: Note that A D:.. B = (A \ B ) U (B \ A), A n B = A \ (A \ 8), A \ B = A D:.. (A n B), and A U B = (A D:.. B )D:.. ( A n B). ] 2. If 'R is a ring, then show that the collection A = [A E 'R: Either 3.

A or Ac belongs to 'R}

is an algebra of sets. In the implication scheme of Figure 3.4 show that no other implication is true by verifying the following regarding an uncountable set X : The collection of all singleton subsets of X together with the empty set is a semirino but not a rino. b. The collection of all finite subsets of X is a ring but is neither an algebra nor a a-ring. c. The collection of all subsets of X that are either finite or have finite complement is an algebra but is neither a a-algebra nor a a-ring. d. The collection of all at-most countable subsets of X is a a-ring but not an algebra. a.

0

0

cr-rino cr-aloebra ;=

�

0

� ring

� algebra �

===:;:.

FIGURE 3.4. The Implication Scheme

semiring

158

Chapter 3: THE THEORY OF MEASURE e.

4. 5.

6.

The collection of all subsets of X that are either at most countable or have at-most a countable complement is a a-algebra (which is, in fact, the a-algebra generated by the singletons).

Show that a Dynkin system is a a -algebra if and only if it is closed under finite intersections. Give an example of a Dynkin system which is not an algebra. A monotone class of sets is a family M of subsets of a set X such that if a sequence {Au} of M satisfies Au t A or A, � A, then A E M. Establish the following properties regarding monotone classes: a.

We have the following implication scheme: a-algebra ==> Dynkin system ==> monotone class

Give examples to show that no other implication in the above scheme is true. b. An algebra is a monotone class if and only if it is a a-algebra. c. The a-algebra a(A) generated by an algebra A is the smallest monotone class containing A. 7. 8.

9.

Show that if X and Y are two separable metric spaces, then Bx xY = Bx ® Br. Show that the composition function of two measurable functions is measurable. If (X, 'E) is a measurable space, then show that a. b.

the collection of all real-valued measurable functions defined on X is a function space and an algebra of functions, and any real-valued function on X which is the pointwise limit of a sequence of ('E, B)-measurable real-valued functions is itself ('E, B)-measurable.

Let (X, 'E) be a measurable space. A 'E-simple function is any measurable function ¢: X � IR which has a finite range, i.e, if ¢ has finite r.mge and its standard repre­ sentation ¢ = :L?= 1 ai xA; satisfies Ai E 'E for each Show that a function f: X � [0, oo) is measurable if and only if there exists a sequence {¢u} of 'E-simple functions such that l/>u(x ) t f(x) holds for each x e X. 11. Use Corollary 20.10 to show that if a measure J1. is a-finite, then Jl.* is the one and only extension of J1. to a measure on a(S). Give an example of a measure space such that Jl..* is not the only extension of J1. to a measure on a(S). 12. Show that the uniform limit of a sequence of measurable functions from a measurable space into a metric space is measurdble. 13. Let f, g: X � IR be two functions and let B denote the a-algebrd of all Borel sets of JR. Show that there exists a Borel measurdble function h: IR � IR satisfying f = h o g if and only if f- 1 (B) c g- 1 (B) holds. 14. Let (X, 'E) be a measurable space, Y, Z 1 and Z2 be separable metric spaces and Ill a topological space. Assume also that the functions fi : X X y -+ zi' = 2), are Caratheodory functions and g : Z 1 x Z2 � Ill is Borel measurable. Show that the function h: X x Y � Ill , defined by 10.

i.

(i 1,

h (x, y) = g (fl (x, y) , f2 (x , y)), is jointly measurable.

Section 20: ABSTRACT MEASURABILITY 15. 16.

159

Let (X, :E) be a measurable space and (l:'. d) a separable metric space. Show that a function f: X � Y is measurable if and only if the function x 1--+ d(y, f(x)), from X into JR., is measurable for each fixed y E Y. Let (X, S, f.J.) be a a-finite measure space, where S is a a-algebra. If f: X � JR. is a Att-measurable function, then show that there exists a S-measurable function g: X � JR. such that f = g a.e.

CHAPTER

4

__ _ _ _ _ _ _ _ _ _ __ _

THE LEBESGUE I NTEGRAL By definition a Riemann integrable function is bounded, and its domain is a closed interval. These two restrictions make the Riemann integral inadequate to fulfill the requirements of many scientific problems. H. Lebesgue 1 in his classical work [2 I ] introduced a concept of an integral (called today the Lebesgue integral) based on measure theory that generalizes the Riemann integral. It has the advantage of treating at the same time both bounded and unbounded functions and allows their domains to be more general sets. Also, it gives more powerful and useful convergence theorems than the Riemann integral. In this chapter we study the Lebesgue integral. The concepts of an upper function and its Lebesgue integral are introduced first. Consequently, the Lebesgue integrable functions are introduced as differences of two upper functions, and their properties are studied. The Lebesgue dominated convergence theorem that (under certain conditions) allows us to interchange the processes of limit and integration is proved, and various applications of this powerful result are presented. Next, it is shown that every Riemann integrable function is Lebesgue integrable, and that in this case the two integrals (the Riemann and the Lebesgue) coincide. Furthermore, the relationship between an improper Riemann integral and the Lebesgue integral is obtained. Finally, the chapter culminates with a study of product measures and iterated integrals. For this chapter, (X, S, J.l) will be a fixed measure space, and unless otherwise specified, all properties of the functions will be tacitly referred to in this measure space.

21. UPPER FUNCTIONS It was mentioned before that the step functions will be the "building blocks" for the Lebesgue integral. Recall that a function ¢ is a step function if and only if there exist a finite collection {A 1 , , A11} of measurable sets with J.l*(A;) < oo for i = I , . . . . n. and real numbers a 1 , , a11 such that ¢ = :L�'=1 a; XA; holds. •

.

•

•

•

•

1 Henri Leon Lebesgue ( 1 875-1 941 ), a prominent French mathematician. The founder ofthe modem theory of integration.

161

Chapter 4: THE LEBESGUE INTEGRAL

162

The real number /(¢) = L�= taip,*(Ai) is called the Lebesgue integral of ¢, and we have already seen (in Section 17) that it is independent of the particular representation of ¢. From now on, the Lebesgue integral of ¢ will be denoted by its conventional symbol J ¢ d p,, or fx ¢ d JJ.. If clarity requires the variable to be emphasized, then the notation J ¢(x) dp,(x) will be used. Thus, the integral of the step function is

The collection of all step functions has both the structure of an algebra and that of a function space.

The collection ofall stepfunctions under tile pointwise oper­ ations is a function space and an algebra.

Theorem 21.1.

Proof. The proof that the step functions form an algebra is straightforward.

To see that this collection is also a function space, note that if the step function ¢ has the standard representation ¢ = l:�'= taiXA;> then ¢+ = l:�'= 1 max{ai, O}XA; holds. Thus, ¢+ is a step function, and the conclusion follows. • The basic properties of the Lebesgue integral for step functions were discussed in Section 17. Here we shall deal with the almost everywhere limits of increasing sequences of step functions. Such limits are used to define upper functions.

A function f : X -+ IR is called an upper function if there exists a sequence {¢,} ofstepfunctions such that I . ¢n t f a .e., and 2. lim J ¢, dp, < oo.

Definition 21.2.

Any sequence of step functions that satisfies conditions (I) and (2) of the pre­ ceding definition will be referred to as a generating sequence for f. Note, in particular, that by Theorem 1 6.6 every upper function is a measurable function. The collection of all upper functions will be denoted by U. Clearly, •

every stepfunction is an upperfunction.

Also, observe that an upper function need not be a positive function. If f is an upper function with a generating sequence {¢,}, and if { 1/J,} is another sequence of step functions such that 1/J, t f a.e., then it follows from Theorem 17.5 that lim J ¢, dp, = lim J 1/J, d p, holds. Thus, { 1/Jn } is also a generating sequence for f, and therefore, the following definition is well justified.

Section 21: UPPER FUNCTIONS

163

Definition 21.3. Let f be an upper fttnction, and let {4>n} be a sequence of step functions such that f/>11 t f a.e. holds. Then the Lebesgue integral (or simply the integral) of f is defined by

I f df-L = J f/>11 df-L. 11-+ 00

lim

Again, we stress the fact that the value of the Lebesgue integral of an upper function is independent of the generating sequence of step functions. Also, it should be clear that if f is an upper function and g is another function such that g = f a.e., then g is also an upper function and J g d f-L = J f d f-L holds. The rest of this section is devoted to the properties of upper functions.

Theorem 21.4. For upperfunctions f and g, the following statements hold:

f + g is an upperfunction and f 0 and j (af) df-L = a ff df-L. 3. f v g and f 1\ g are upperfunctions. Proof. Choose two generating sequences {4>n} and { 1/111 } for f and g, respec­ 1.

tively. ( 1 ) Clearly, {f/1, + 1/111} is a sequence of step functions, and f/>11 a.e. holds. The result now follows by observing that

+ 1/111 t f + g

(2) Straightforward. (3) Note that both {f/>11 v 1/111} and { f/1, 1\ 1/J, } are sequences of step functions. Now, f/>11 1\ 1/ln t f 1\ g a.e. and lim J f/>11 1\ 1/111 df-L < lim J f/>11 df-L < oo, show that f 1\ g is an upper function. To see that f v g is an upper function, observe first that f/>11 v 1/111 t f v g a.e.

holds, and then use the identity

f/>u V 1/J, to obtain Jf/>11 V

=

f/>11 + 1/111 - f/>u

1/111 df-L = Jf/>11 d f-L + J 1/111 d f-L

-

1\

1/111

Jf/>11 1\ 1/111 df-L. This implies

This finishes the proof of the theorem. The next theorem states that the integral is a monotone function on U.

•

Chapter 4: THE LEBESGUE INTEGRAL

164

Theorem 21.5. If f and

are upper functions such that f f g dJ.L holds. In particular, if f E U satisfies f g

J f dJ.L > f f dJ.L > 0.

2:: g >

a.e., then 0 a.e., then

Proof. Let {,} and { l/f,} be generating sequences for f and g, respective]y.

Then 11 A 1/1, t g a. e. holds, and so { , A 1/1n } is also a generating sequence for g. By Theorem 17.3, we have J , d J.L > J
and the proof is finished.

•

It should be noted that if f is an upper function such that f 2:. 0 a.e., then there exists a sequence of step functions { 1/1n } satisfying l/f, > 0 a.e. for each n and + l/f, t f a.e. To see this, notice that if
Theorem 21.6. Let f : X

� IR. be a function. If there exists a sequence {f,,}

of upper functions such that fn t f a.e. and lim J fn d J.L < oo, then f is an upperfunction and J f dJ.L = lim f f,, dJ.L. Proof. For each i choose a sequence { i, it follows that J /; dJ.L = lim,_oc J
J fn dJ.L = J 1/ln dJ.L J f dJ.L,

and the proof is complete.

lim 11-+-0C

=

II

The integral satisfies an important convergence property for decreasing se­ quences. It is our familiar order continuity property of the integral.

Theorem 21.7. If {fn } is a sequence of upperfunctions such that f, .J, 0 a.e.,

then lim J f,, dJ.L = 0 holds.

For each n choose a step function , is an

Proof. Let E

a.e. and

j(f,,

>

0.

Section 21: UPPER FUNCTIONS

165

=

upper function). Let 1/111 /\�'= 1 k. Now, the almost everywhere inequalities

0 < ft,

- = 1/111

ft, -

II

=

=

II

II

II

V([t, -
i=l

<

imply

Thus,

for all n

> k, which shows that f ft, dJ.L ,{, 0

•

holds.

Finally, we mention that U is not in general a vector space since it fails to be closed under multiplication by negative real numbers. An example of this type is presented in Exercise 2 of this section.

EXERCISES 1. Let L be the collection of all step functions ¢ such that there exist a finite number of sets A 1 , , A, in S all of finite measure and real numbers a 1 , , a11 such that 2:7=• a; XA;. Show that L is a function space. Is L an algebra of functions? ¢ [HINT: Use Exercise 14 of Section 12.] 2. Consider the function f : 1R ---+ 1R defined by f(x) = 0 if x � (0, l], and f(x) .,fo if x E (11!1 , �] for some n. Show that f is an upper function and that - f is not an upper function. [HINT: A step function is necessarily bounded.] 3. Compute J f d).. for the upper function f of the preceding exercise. 4. Verify that every continuous function f : [a, b] ---+ IR is an upper function-with respect to the Lebesgue measure on [a, b]. 5. Let A be a measurable set, and let f be an upper function. If XA � f a e , then show that ,u*(A) < oo. 6. Let f be an upper function, and let A be a measurable set of finite measure such that a � f(x) � b holds for each x E A. Then show that

=

• • •

• • •

=

..

Chapter 4: THE LEBESGUE INTEGRAL

166

a. f XA is an upper function, and b. 7.

a 11 *(A) � j f XA d11 � hJ-L*(A).

Let (X, S, J-L) be a finite measure space. and let f be a positive measurable function. Show that f is an upper function if and only if there exists a real number M such that J ¢ dJ-L � M holds for every step function ¢ with ¢ � f a.e. Also, show that if this is the case, then

If d

J.L

8.

=

sup

{I

¢ dJ.L: ¢ is a step function with ¢ �

Show that every monotone function to the Lebesgue measure on [a, b] .

f

:

[a. b]

-+

f a.e.

J

.

1R is an upper function-with respect

22. INTEGRABLE FUNCTIONS It was

observed before that the collection U of all upper functions is not a vector

space. However, if we consider the collection of aii functions that can be written as

an almost

everywhere difference of two upper functions, then this set is a function space. The members of this collection are the Lebesgue integrable functions. The details will be explained in the following.

Definition 22.1. A function f : X � IR is called Lebesgue integrable (or simply integrable) ifthere exist two upperfunctions u and v such that f = u - v a.e. holds. The Lebesgue integral (or simply the integral) of f is defined by

It should be noted that the value of the integral is independent of the representa­ tion of f as a difference of two upper functions. Indeed, if f = u - v = u I - VI a.e. with u, LII , v, and vI all upper functions, then u + vI = u I + v a.e. hoi ds, and by Theorem 2 1 .4( 1) we have f u df.L + f VI dJL = f u 1 dJL + J v dJ-L. Therefore,

j U dJ.L - j V df.L

=

j UI dJ-L - j VI dJ.L.

An integrable function is necessarily measurable, and every upper function is Lebesgue integrable. Also, it is readily seen that if a function f . is Lebesgue integrable and g is another function such that f = g a.e., then g is also Lebesgue integrable and J g dJ-L = J f df.L holds.

Historical Note: The above introduction. of the Lebesgue integral is a modification of a method due to P. J. Daniell [5]. Daniell's 2 general approach to integration starts with a function space L on some nonempty set X , together with an "integral" : L -+ 1R is said to be an ime�:ral if

I

I

on L . The function

2P. J. Daniell (1 889-1946), a Briiish mathematician. He worked in functional analysis and the theory of integration.

167

Section 22: INTEGRABLE FUNCTIONS

1. 2. 3.

l (a¢ + {31/1) = a/ (¢) + {31(1/1) for all fX, f3 E IR and ¢, 1/1 E L, /(¢) > 0 whenever ¢ :;:: 0, and whenever {¢11} C L satisfies ¢11 (x) .l, 0 for each x E X, then I (¢11 ) .l,

0.

A function u : X IR is called an upper function if there exists a sequence { c/J11} C L with ¢11(x) t u(x) for all x E X and lim /(¢11) < oo. As in the proof of Theorem 17 .5, we can show that lim /(c/J11) is independent of the "generating" sequence {¢11 }. The real number I (u) = lim I (¢11) is the integral of u. Finally, f : X -+ IR is said to be integrable i f there exist two upper functions u and v with f = u v. The integral of f is then defined by �

-

l(f) = l(u) - l(v).

Here our approach to the Lebesgue integral can be considered as a "measure theoretical

Daniell method." The set of integrable functions has all the expected nice properties.

Theorem 22.2. The collection of all Lebesgue integrable functions is a func­

tion space.

Proof. Let f and g be two integrable functions with representations f = u - v a. e. and

g

=

u 1 - v1

a. e. Then the almost everywhere identities

J+g ct.f af

j+

=

(u + u 1 ) - (v + vJ),

= Cf.ll - Cf.V if =

=

ct. > 0,

[(-a)v] - [(-a)u] if + (u - v) = u v v - v

a <

0,

and

express the above functions as differences of two upper functions. This shows that the collection of all integrable functions is a function space. • Thus, if f is integrable, then If I is also an integrable function. In particular, it follows from the last theorem that a function f is Lebesgue integrable if and only if and f- are both integrable. The next result describes the linearity property of the integral. Its easy proof follows directly from Definition 22.1 and is left as an exercise for the reader.

j+

Theorem 22.3. Iff and g are two integrable functions, then

1 (ct./ + f3g) dJL 11 dJL + f31 g dp. = ct.

holdsfor all a. f3 E IR.

Every positive integrable function is necessarily an upper function.

Chapter 4: THE LEBESGUE INTEGRAL

168

Theorem 22.4. If an integrable function f satisfies f > 0 a.e.. then f is an upperfunction.

=u

v a.e. holds. Since each u and v is the almost everywhere limit of a sequence of step functions, there exists a sequence { 1{1,} of step functions such that 1{1n � f a.e. Since f > 0 � f a.e. also holds. a.e., it follows that By Theorem 17.7, there exists a sequence {s,} of simple functions satisfying 0 < s, t f a.e. Now, for each n let ¢, = s, 1\ Then {¢,} is a sequence of step functions such that 0 < ¢, t f a.e. holds. To complete the proof, we show that {j ¢, d J.L} is bounded. Indeed, from ¢, + v < f + v < u a.e. and Theorem 21.5, it follows that J ¢, dJ.L + J v dJ.L < J u d /1-. and therefore, J ¢, d J.L < J u dJ.L J v dJ.L < oo holds for all n. The proof of the theorem is • now complete.

Proof. Choose two upper functions u and v such that f

-

t:


-

If f is an integrable function, then by Theorem 22.4, t + and .r- are both upper functions, and so, f = f+ - f- is a decomposition of f as a difference of two positive upper functions. In particular,

(This formula is usually the one used by many authors to define the Lebesgue integral.) As an application of the preceding theorem, we also have the following useful result.

Theorem 22.5. If f is an integrable function, then for evet)' E > 0 the mea­

) > E} hasfinite measure.

surable set {x E X : l f(x l

Proof. Fix E > 0, let A

=

{x E X : lf(x)l > E}, and note that EXA < l f l holds. Now, � I f I is an integrable function, in fact, by Theorem 22.4 an upper function. Let {¢,} be a sequence of step functions such that ¢, t � I f I a.e. Then {¢, 1\ XA } is a sequence of step functions such that ¢, 1\ XA t XA a.e. holds. Thus, by Theorem 17 .6,

and the proof is finished.

•

Measurable functions "sandwiched" between integrable functions are integrable.

Section 22: INTEGRABLE FUNCTIONS

169

Theorem 22.6. Let f be a measurablefunction. If there exist two integrable

functions h and g such that h < f < g a.e., then f is also an integrable function. Proof. Writing the given inequality in the form 0 :::: f - h :::: g - h a.e., we see that we can assume without loss of generality that 0 < f :::: g a.e. holds. By Theorem 22.4, g is an upper function. Pick a sequence { ¢n} of step functions such that 0 < ¢, t g a. e. By Theorem 17.7 there exists a sequence { 1/r,} of simple functions such that 0 < 1/r, t f a.e. holds. But then {¢, 1\ 1/r, } is a sequence of step functions such that ¢, 1\ 1/r, t f a.e., and J ¢k 1\ Vrk dJJ- :::: lim J ¢, d!J- = J g dJJ- < oo for all k. Hence, f E U, and so, f is an integrable function. • More properties of the integral are included in the next theorem:

Theorem 22.7. For integrable functions f and g we have the following:

1 . J Ill d!J- = 0 if and only if f = 0 a.e. 2. If f > g a.e., then J f d!J- > J g d!J- . 3. If f dJJ-1 < J Ill dJJ-. Proof. ( 1) Clearly, if f = 0 a.e., then J Ill dJJ- = 0 holds. On the other hand, assume that J Ill dJJ- = 0. Since, by Theorem 22.4, 1 /1 is an upper function, there exists a sequence {¢,} of step functions such that 0 < ¢, t If I a.e. holds. By Theorem 21.5, it follows that J ¢n d!J0 for each n, and so, ¢, = 0 a.e. for each n. Thus, 1/1 = 0 a.e., and so f = 0 a.e. (2) Since f-g > 0 a. e., it follows from Theorem 22.4 that f-g is an upper func­ tion. But then, Theorem 2 1 .5 implies that J f dJJ- - J g dJJ- = j(J - g) dJJ- > 0, so that J f d!J- > J g dJJ- holds. • (3) The conclusion follows from (2) and the inequality -1/1 < f < 1 /1 . =

The reader has probably noticed that the functions we have considered so far are real-valued. It is a custom, however, to allow a function to assume infinite values, provided that the set of all points where the function equals -oo or oo is a null set. The reason for this is that neither the integrability character nor the value of the integral of a function changes by altering its values on a null set. Moreover, assigning any value to the sum of two functions at the points where the form oo - oo occurs does not affect the integrability and the value of the integral of the sum function (as long as the set of points of all such encounters has measure zero). If one does not want to deal with functions assuming infinite values (up, of course to null sets), then one may change the infinite values to finite ones (for instance, change all the infinite values to zero) without loosing anything regarding integrability. When an extended real-valued function f is said to define an inte­ grable function, it will be meant that f assumes the infinite values (or it is even

Chapter 4: THE LEBESGUE INTEGRAL

170

undefined) on a null set, and that if finite values are assigned to these points, then f becomes an integrable function. To summarize the preceding: •

Functions that are almost everywhere equal have the same integrability properties and can be considered as identical.

We continue with a theorem of B. Levi3 describing a basic monotone property of the integral.

Theorem 22.8 (Levi). Assume that a sequence {fn} of integrable functions satisfies /11 < f11+ 1 a.e. for each n and lim J /11 dJL < oo. Then there exists an illtegrablefunction f such that !11 t f a.e. (and hence, J fn d JJ. t J f dJL

holds).

Proof. Replacing { /11} by { f, - f1 } if necessary, we can assume without loss of generality that j,, > 0 a.e. hoids for each n. Aiso, an easy argument shows that we can assume that 0 < f, (x) t holds for all x E X . Let I = lim J fn dJL < oo. For each x E X let g(x) = lim .fn (x) E IR*, and consider the set E=

{x

E X : g(x) =

oo}.

Clearly, E = n:1 [U:dx e X : f,(x) > i}] holds, and so E is a measurable set. Next, we shall show that JL * (E) = 0. By Theorem 22.4, each /11 is an upper function. Thus, for each i there exists a sequence {¢J, } of step functions such that 0 < ¢� t n f; a.e. holds. For each n let '1/1n = V;'= 1 ¢j,, and note that { l/J, } is a sequence of step functions such that '1/111 t g a.e. and lim J l/J, dJL = lim J /11 d J.L = I . In particular, for each k the sequence of step functions { l/J, A f.: XE} satisfies '1/111 A k XE t f.: XE a. e. From Theorem 17 .6, it follows that JJ.*(E) < oo and kJL* (E) < lim J l/Jn dJL = I < oo for each k. Hence, J.L*(E) = 0. Now, define f : X -+ IR by f(x) = g(x) if x fl. E and f(x) = 0 if x E E. Then f11 t f a.e. holds, and the result follows from Theorem 21.6. II The series analogue of the preceding theorem is presented next.

Theorem 22.9. Let {/11} be a sequence of non-negative integrable functions 3 Beppo Levi (1 875-1961}, an halian malhemaiician. His main comribUiions were in algebraic 1opology, malhemaiical logic, and analysis.

Section 22: INTEGRABLE FUNCTIONS

171

such that L.: 1 J fn dJ.L < oo. Then Lr:: J f,, defines an integrable function and

Proof. For each n let g, .L�'= 1 f;, and note that each g, is an integrable function such that g, t L: J f; a.e. holds. Now, by Levi's theorem, L: J f,, =

defines an integrable function, and

holds.

•

The next result is known in the theory of integration as Fatou 's4 lemma.

Theorem 22.10 (Fatou's Lemma). Let {f:,} he a sequence ofintegrablefunc­ tions such that J,, > Oa.e. for each n and lim inf J f, dJ.L < oo. Then lim inf f,

defines an integrable function, ·and

I

lim inf f,, d 11-

I

< lim inf f, d J.t.

Proof. Without loss of generality, we can suppose that f,(x) > 0 holds for all x E X and all n. Given n, define g,(x) inf{f;(x) : i > n} for each x E X. Then each g, is a measurable function, and 0 < g, < f, holds for all 11. Thus, by Theorem 22.6, each g, is an integrable function. Now, observe that g, t and lim J g, dJ.t =s lim inf J f,, df..L < oo holds. Thus, by Theorem 22.8, there exists an integrable function g such that g11 t g a.e. holds. It follows that g lim inf f,, a.e., and therefore, lim inf fn defines an integrable function. Moreover, =

=

I

lim inf f,, dJ1

and the proof is finished.

=

Ig dJ.t

I

I

�� g, dJ.L < lim inf j, dJ.t,

= ��

•

We are now in the position to state and prove the Lebesgue dominated conver­ gence theorem-the cornerstone of the theory of integration. 4 Pierre Joseph Louis Fatou ( 1878-1929), a French mathematician. Besides his work in analysis, he also studied the motion of planets in astronomy.

Chapter 4: THE LEBESGUE INTEGRAL

172

Let { fn} be a sequence of integrable functions satisfying IfnI g a.e. for all nintegrable and somefunction fixed i12and tegrable function g. If f, f a.e., then f defines an lim jfn dj.L J lim fn dj.L jJ dj.L. Proof. Clearly, 1!1 < g a.e. holds, and the integrability of f follows from Theorem 22.6. Observe next that the sequence {g - fn} satisfies the hypotheses of Fatou 's lemma and moreover, lim inf(g - fn) - f a.e. Thus, Jg dJ-L - J f dJ.L = J(g - f) dJ.L Jlim inf(g - fn) dJ.L < lim inf J(g - fn) dj.L J g dj.L � lim sup J fn dj.L, Theorem 22.1 1 (The Lebesgue Dominated Convergence Theorem). < �

=

11-HXJ

=

II-+ 00

=

g

=

=

and hence,

J fn dj.L < J f dJ-L. Similarly, Fatou 's lemma applied to the sequence {g fn} yields Jg dJ.L + J f dJ-L J(g + f) dJ-L J lim inf(g + J,, ) dJ.L < lim inf J(g + fn) dJ-L J g dJ.L + lim inf J fn dj.L , lim sup

+

=

=

=

and so,

J f dJ.L < lim inf J fn dJ-L. Therefore, lim J fn dJ.L exists in and lim Jfn dJ.L J f d J.L holds. IR.,

=

•

The next theorem characterizes the Lebesgue integrable functions in terms of some given property. It is usually employed to prove that all Lebesgue integrable functions possess a given property.

Let S, J.L) be a measure space and let (P) be a property which may or may not be possessed by an integrable function. Assume that:

Theorem 22.12.

(X,

Section 22: INTEGRABLE FUNCTIONS

173

1. Iff and g are integrable functions with property (P ) , then f + g and af

for each a E IR also have property (P). 2. If f is an integrable function such that for each E > 0 there exists an integrable function g with property (P) satisfying J If - g l d�J- < E, then f has property (P ). 3. For each A E S with �J-(A) < oo, the characteristic function XA has property

(P).

Then every integrable function has property (P). 11- * (A) < oo. So, there exists a disjoint sequence {A,} of S such that A = U:1 A11• Put 8, = U;=l Ak for each n and note that 8, t A. From X B" = L�=l XAt• (3) and (1), we see that x8" has property (P) for each n. Since f IXA - x8, l d�J- = IJ-*(A) - �J-(8,,) -7 0, it follows from (2) that XA likewise has property (P). Next, assume that A is an arbitrary measurable set of finite measure and let E > 0. Then there exists a a -set 8 of finite measure such that A c 8 and IJ-*(8) < JL* (A) + E. This implies f IXA - XB I d�J- = 11-*(8) - �J-*(A) < E . From the above discussion and (2), we infer that XA satisfies property (P). Proof.

Assume first that

Now, from

A

is a a-set with

( 1 ) we see that every step function satisfies property (P).

then, it follows from

(2) that every upper function satisfies property (P).

But

Since

every integrable function is the difference of two upper functions, invoking

(1)

once more, we infer that indeed every integrable function satisfies property (P).

•

It is easy to see that for every subset

E of X, the collection SE

of subsets of E (called the restriction ofthe semiring S to

=

{EnA : A e S }

E) is a semiring of subsets

E is a measurable subset of X, then 11-* restricted to SE is also That is, (E, S£ , 11-*) is a measure space for every measurable subset

of E. If, in addition, a measure.

E of X. Also, a straightforward verification shows that the measurable subsets of (E, S£, 11-*) are precisely the subsets of E that are measurable subsets of X; see Exercise If

E

7 of Section 15.

is a measurable subset of

integrable over

X , then a function

f:E

-7

IR is said to be

E if f is integrable with respect to the measure space (£, SE, 11-* ).

f can be extended to all of X by assigning the values f (x) = 0 if x ¢ E. Then f so defined is an integrable function over X, and in this case fx f d�J- = JE f dJL holds. A function f : X -7 IR is said to be integrable over a measurable subset E of X if the function f XE is integrable over X, or equivalently, if f restricted to E is integrable with respect to the measure space (E, S£, JL*). In this case, we shall write J fXE d�J- = J£ f dJL. Of course, the domain of

The simple proof of the next result is left as an exercise for the reader.

174

Chapter 4: THE LEBESGUE INTEGRAL

Theorem 22.13. Every integrable function f is integrable over every mea­

surable subset of X. Moreover,

holds for every measurable subset E of X . The rest of this section deals with an observation concerning infinite Lebesgue integrals. If ¢ = :L7= 1 a; XA; is the standard representation of a positive simple function ¢, then the sum :L7= 1 a;p,* (A;) makes sense as an extended non-negative real number. If :L;'=t a; p,* (A;) = oo, then it is a custom to write J ¢ dp, = oo and say that the Lebesgue integral of¢ is infinite. Assume now that f : X -+ IR� is a function where there exists a sequence {¢11} of positive simple functions such that ¢11 t f a.e. hol4s. Then lim J ¢11 dp, exists as an extended reai number, and it can be seen easily that lim J ¢n d;.,;, is independent from the chosen sequence {¢11 }. In the case that lim J
In this manner, we can assign an "integral" to a much larger class of measurable extended real-valued functions. One advantage ofthe above extension ofthe integral is that a number of theorems can be phrased without Lebesgue integrability assumptions on the functions. For instance, Fatou's lemma can be stated as follows: •

If {f,,} is a sequence of measurable functions satisfying fn > 0 a.e. for each n, then

J

lim inf f,, d J1 < lim inf

J fn d

J..L

holds-where, of course, one or both sides of the inequality may be infinite.

Section 22: INTEGRABLE FUNCTIONS

175

EXERCISES 1. Show by a counterexample that the integrable functions do not fonn an algebra. 2. Let X be a nonempty set, and let 8 be the Dirac measure on X with respect to the point a (see Example 13.4). Show that every function f : X -+ IR is integrable and that J f d8 = j(a). 3. Let J.L be the counting measure on IN (see Example 13.3). Show that a function f : IN -+ IR is integrable if and only if L::,, lf(n)l < oo. Also, show that in this case J f dJ.L = L:,t f(n). 4. Show that a measurable function f is integrable if and only if I f I is integrable. Give an example of a nonintegrable function whose absolute value is integrable. 5. Let f be an integrable function, and let IE 11} be a sequence of disjoint measurable subsets of X. If E = u:,, £,, then show that

1£ f dJ.L 11=f1 1E,f dJ.L . =

6. 7. 8.

Let f be an integrable function. Show that for each € > 0 there exists some 8 > 0 (de­ pending on €) such that If£ f dJ.LI < € holds for all measurable sets with J.L*(E) < 8. [HINT: Note that I/ l A n t lfll Show that for every integrable function f the set (x E X : f(x) =I= 0} can be written as a countable union of measurable sets of finite measure-referred to as a a-finite set. Let f : IR -+ IR be integrable with respect to the Lebesgue measure. Show that the function g : [0, oo) -+ IR defined by g( t )

9.

10.

= sup

{ j l f(x

+ y)

- f(x) I d>..(x) : IYI

:::

t}

for t =:: 0 is continuous at t = 0. [HINT: Use Theorem 22 . 12.] Let g be an integrable function and let If�,} be a sequence of integrable functions such that If, I < g a.e. holds for all n. Show that if f, 4 f, then f is an integrable function and lim J I f, - !I dJ.L = 0 holds. [HINT: Combine Theorem 19.4 with the Lebesgue dominated convergence theorem.] Establish the following generalization of Theorem 22.9: If (/11 } is a sequence of inte­ grable functions such that L::,1 J If, I d J.L < oo, then L::, 1 f, defines an integrable function and

[HINT: By Theorem 22.9, the series g = L::,1 If,,! defines an integrable func­ tion and IL�= I f, l ::; g a.e . holds for each k. Now, use the Lebesgue dominated convergence theorem.] 11. Let f be a positive (a.e.) measurable function, and let ; i l e; = J.L*
Chapter 4: THE LEBESGUE INTEGRAL

176

Let ( fn) be a sequence of integrable functions such that 0 ::::; f,+ 1 ::::; fn a.e. holds for each n. Then show that fn ..J, 0 a.e. holds if and only if J /11 dJ.L ..J, 0. 13. Let f be an integrable function such that f(x) > 0 holds for almost all x. If is a measurable set such that fA f dJ.L = 0, then show that J.L*(A) = 0. 14. Let (X, S, J.L) be a finite measure space and let f : X � 1R be an integrable function satisfying f(x) > 0 for almost all x. If 0 < e ::::; J.L*(X), then show that

12.

A

inf 15.

A

J.L*(E) :::: e

I

> 0.

v(A) = fA! dJ.L

(X, A, v) is a measure space.

If A 11 denotes the a -algebra of all v-measurable subsets of X, then A 5; A 11 • Give an example for which A ::j: A 11• If J.L*((x E X : /(x) = 0)) = 0, then A = A11• If g is an integrable function with respect to the measure space (X, A . v), then fg is integrable with respect to the measure space (X, S, J.L) and

c. d.

17.

E All and

Let f be a positive integrable function. Define v : A � [0, oo) by for each E A . Show that: a. b.

16.

It f dJ.L : E

1g dv 1gf dJ.L . =

(A Change of Variable Fonnula) Let be an interval ofiR., and let f : -+ 1R be an in­ tegrable function with respect to the Lebesgue measure. For a pair of real numbers a and b with a ::j: 0, let J = ((x -b)/a : x E I). Show that the function g : J � 1R de­ fined by g(x) = f(ax +b) for x E J is integrabie and that f1 f d>.. = Ia I f1 g d).. holds. [HINT: Use Theorem 22.12.] Let (X, S, J.L) be a finite measure space. For every pair of measurable functions f and g let

I

I

- I 1 +I fI f- -g lgl dJ.L.

d(f' g) -

Show that (M, d) is a metric space. Show that a sequence Un) of measurable functions (i.e., (/n) 5; M) satisfies /11 � f if and only if lim d(f,, /) = 0. Show that (M, d) is a complete metric space. That is, show that if a sequence (/n) of measurable functions satisfies d(/11, fm) -+ 0 as n, m � oo, then there exists a measurable function f such that lim d(fn, /) = 0.

a. b. c.

18.

!I

f

=

Let

:

1R � 1R be a Lebesgue integrable function. For each finite interval f1 f d>.. and E1 = {x E I : f(x) > /J ). Show that

>.ln

{ (f - /J)d>.. . 11{ I f - /J i dJ.. = 2 1£,

19.

Let f : [0, oo) � 1R be a Lebesgue integrable function such that for each t :::: 0. Show that f(x) = 0 holds for almost all x.

I

J� f(x)dJ..(x)

let

=

0

Section 23: THE RIEMANN INTEGRAL AS A LEBESGUE INTEGRAL 20.

177

Let (X, S, J.l) be a measure space and let f, !1, h, . . . be non-negative integrable functions satisfying j,, -+ f a.e. and lim J /11 df.l = J f df.l. If E is a measurable set, then show that

n-+limoo }E{ !" df.l = }£{ f df.l.

21.

22.

If a Lebesgue integrable function f : [0, 1] -+ IR satisfies Jd x2" j(x) d>..(x) = 0 for each n = 0, 1 , 2, . . . , then show that f = 0 a.e. [HINT: Since the algebra generated by {1. x2} is uniformly dense in C[O, 1], we have

Jd g(x)f(x)dx = 0 for each g E C[O, 1].] 11 11 1 For each 11 consider the partition {0, 2-1 , 2 · 2- , 3 2- , , (211 - 1) 2-11 , 1 } the interval [0, 1 ] and define the function r11 : [0, 1 ] -+ IR by r11 (1) = - 1, and r11(x) = ( - 1 )"- I for (k - 1 )2 _, :=: x < k2 -n and each k = 1 , 2, . . . , 211 • ·

a. b.

lim

of

lot r11(x)j(x)d>..(x)

=

0.

[HINT: For (b) use Theorem 22.12.] Let { E11} be a sequence of real numbers such that 0 < E11 < 1 for each n. Also, let us say that a sequence { A11) of Lebesgue measurable subsets of [0, 1 ] is consistent with the sequence { E,} if I..(A11) = E11 for each n. Establish the following properties of !En}: a.

b. 24.

·

Draw the graphs of r 1 and r2. Show that if f : [0, 1 ] -+ IR is a Lebesgue integrable function, then

n-+oo

23.

• • •

The sequence {E, ) converges to zero if and only if there exists a consistent sequence {A,) of measurable subsets of [0, 1 ] such that L::,.1 XA, (x) < oo for almost allx. The series I:: 1 E11 converges in IR if and only if for each consistent sequence {A,) of measurable subsets of [0, I ] we have L::,.1 XA, (x) < oo for almost all x .

Let (X, S. p.) be a finite measure space and let f : X -+ IR be a measurable function. a. b.

Show that if f" is integrable for each n and lim J f" dJ.l exists in IR, then 1/(x)l � 1 holds for almost all x. 1 11 If / 1 is integrable for each 11, then show that J / dJ.l = c (a constant) for 11 = 1, 2, . . . if and only if f = XA for some measurable subset A of X.

23. THE RIEI\'IANN INTEGRAL AS A LEBESGUE INTEGRAL It will be shown in this section that the Lebesgue integral is a generalization of the Riemann5 integral. We start by reviewing the definition of the Riemann integral. 5Georg Friedrich Bernhard Riemann (1 826-1866), a German math�matician, one of the greatest mathematicians of all time. Although his life was short, his contributions left the impact of a real genius. He made path-breaking contributions to the theory of complex functions. space geometry, and mathematical physics.

Chapter 4: THE LEBESGUE INTEGRAL

178

For simplicity, the details will be given for functions of one variable, and at the end it will be indicated how to carry out the same results for functions of several variables. Unless otherwise specified, throughout our discussion, f will be a fixed bounded real-valued function on a closed interval [a, b] of JR. A collection of points P {x0, x1 , , x,} is called a partition of [a, b] if

= = Xo < X1 < · · < = b •

a

holds. Every partition P tervals

= {x0, x1,

•

•

X11

·

•

•

•

, x,} divides [a, b] into the n closed subin­

=

=

The length of the largest of these subintervals is called the mesh of P and is denoted by I P I ; that is. I P I max{x; - x; _ 1 : i 1 , . . . , n } . A partition P is said to be finer than another partition Q if Q c P holds. If P and Q are partitions, then P U Q is also a partition that is finer than both P and Q . For a partition P {xo, x1 , , x,} of [a, b], let

=

m; = inf{j(x) : x

=

•

E

•

•

M;

[x; _ 1 , x;]} and

= sup{f(x) : x

E

[x;_ 1 , x; ]}

for each i 1 , . . . , n. Then the lower sum S.(f, P) of f corresponding to the partition P is defined by S*(f, P)

n

= Li= l m (x - X;-1), ;

;

and similarly, the upper sum S* (f, P) of f by S* (f, P )

= Li=l "

M;(X;

-

Xi-1 ) .

Clearly, S*(f, P) < S*(f, P) holds for every partition P of [a, b]. Lemma 23.1.

then

Ifa partition P isfmer than anotherpartition Q (i.e., Q c P),

S* (f, Q) < S* (f, P) and S*(f, P) < S* (f, Q). Proof. We show that S* (f, Q) < S (f, P) holds. The other inequality can be * proven in a similar manner.

Section 23: THE RIEMANN INTEGRAL AS A LEBESGUE INTEGRAL

179

To establish the inequality, it is enough to assume that P has only one more point , x,} and P = Q U {t }. Then there exists than Q, say t. So, let Q (x0, x 1 , some k ( 1 < k < n) such that Xk-1 < t < x�.;, and thus, P = {xo, x 1 , . . . , Xk- l , t, XJ.:, , x11 } . Let c 1 = inf{f(x ) : x E [xk_1 , t] } and c2 = inf{f(x) : x E [t, xk]} . Observe that both m�.: < c1 and m �.: < c2 hold. Therefore, =

•

•

•

•

•

•

S* (f, Q) = <

II

l: m; (x; - x; - 1 ) = l: m; (x; - x;-J .) + mdxk - x�.: -1) i=l

i�k

L m;(x; - X;- t) + c1 (t - x�.:-d + c2(x�.: - t) i#

holds, and the proof is finished.

Lemma 23.2.

•

For every pair ofpartitions P and Q, we have

Proof. From Lemma 23.1, it follows that * S*(f, P) < S*(f, P U Q) < S (f, P U Q) < S* (f, Q) ,

as claimed.

•

The preceding lemma states that every upper sum is an upper bound for the collection of all lower sums of f, and similarly, every lower sum is a lower bound for the collection of all upper sums. Thus, if the lower Riemann integral of f is defined by /* (f) = sup{ S*(f, P) : P is a partition of [a , h]} .

and the upper Riemann integral of f by ! * (f) = inf{S*(f, P ) : P is a partition of [a, b] } ,

Chapter 4: THE LEBESGUE INTEGRAL

180

then

S.(J, P)

<

holds for every pair of partitions

I.(/ ) < / * (f) < S* (f, Q ) P and Q of [a, b].

Definition 23.3. A boundedfunction f : [a, b] -+ IR is called Riemann inte­ grable if I.(/) = I * (f). In this case, the common value is called the Riemann integral off and is denoted by the classical symbol f(x)dx. Historical Note:

Riemann's definition of the integral is

J:

a generalization of Eudoxus'

method of exhaustion as was used by Archimedes6 in his computation of the area of a circle. Interestingly, the value of the limit of the areas of the inscribed (or circumscribed) polygons that were employed by Archimedes to compute the area of the circle, was also called by him the integral

(ro rrav).

It should be historically correct to call the Riemann integral the

Eudoxus-Archimedes integral or the Eudoxus-Archimedes-Riemann integral (or even the

Archimedes-Riemann integral).

A characterization for the Riemann integrability of a function, known as Rie­ mann's criterion, is presented next.

Theorem 23.4 (Riemann's Criterion). A bounded function f : [a, b]

IR is Riemann integrable if and only iffor every E > 0 there exists a partition P of [a, b] such that S* (f, P) - S.(J, P) < E holds. -+

Riemann integrable and let E > 0. Then let I = J: j (x) dx. Then there exist t wo partitions P1 and P2 of [a, b] such that I - s.cJ, PI ) < E and S* (f, P2) - I < E . Then (by Lemma 23.1), the partition P = P1 U P2 satisfies

Proof. Assume

that

f

is

S* (J, P) - S* (f, P)

< =

<

S* (f, P2) - S* (f, P1) [S * (J, P2) - I ] + [I - S* (f, P1 )] E + E = 2E.

6Archimedes of Syracuse (287-212 BC), a Greek mathematician and inventor. He was the mo;t celebrated mathematician of antiquity and perhaps the best mathematician of all times. He used EudOxus' method of exhaustioo to compute areas and volumes very successfully. In his classic work, TheMeasurement of the Circ:/e, he established that a circle has the same area as a right triangle having ooe leg equal to the radius Of the circle and the other equal to the circumference of the circle, and that the VOlume of a sphere is four times the VOlume Of a right cooe with radius and height equal tO the radius of the sphere.

Section 23: THE RIEMANN INTEGRAL AS A LEBESGUE INTEGRAL

181

Conversely, if the condition is satisfied, then since 0

< /* (f) - I*(f) < S* (f, P) - S* (f, P)

holds for every partition P of [a, b], we have O < l*(f) - l* (f) < E for all E > 0. 0, or l (f) Hence, l * (f) - l* (f) l*(f), which shows that f is Riemann * integrable. • =

T

for i

{ t1 ,

=

•

1,

, x,} be a partition of [a, b].

A collection of points , t,} is said to be a selection of points for P if x; _ 1 < t; < x; holds . . . , n. We write

Now, let P =

{x0, x1 ,

=

•

=

•

•

•

•

Rt
;

f(t; )(x; - x - J ), L i= l II

=

and call it (as usual) a Riemann sum associated with the partition P. The following theorem ofJ.-C. Darboux7 presents some powerful approximation formulas for the Riemann integral. The theorem can be viewed as an abstract formulation of Eudoxus' exhaustion method.

Theorem 23.5 (Darboux). Let f : [a, b]

-r

JR. be Riemann integrable, and

let ( P11 } be a sequence ofpartitions of [a, b] such that lim

11--+oo

S* (f, P, )

=

lim

11-+-00

S*(f,

P,)

=

lim I P, l

lim Rt(P11 , T11 )

=

0.

Then

1bf(x)dx. a

In particular, if a sequence ofpartitions { P,} satisfies is a selection ofpoints for P11 , then

n--+oo

=

lim I P, l

0

=

and T,,

1bf(x)dx. a

l f(x) l < c holds for all x e [a, b]. Let E > 0. By Theorem 23.4, there exists a partition P {xo, x 1 , , Xm} of [a, b] such that S*(f, P) - S* (f, P) < E. Choose n0 such that Proof. Choose a constant c

> 0 such that

=

I Pn l

f

< 2cm --

7 Jean-Gastin

and

IP111

•

•

•

< min{XJ - xo, X2 - XJ , . . . , Xm - Xm- J }

Darboux ( 1842-1917), a French mathematician. He was a geometer who used his

geometric intuition to solve various problems in analysis and differential equations.

Chapter 4: THE LEBESGUE INTEGRAL

182

for all n

> n0. Fix n > n0, and let P, = { t0, I J ,

MJ = s p{ j(x) M; = s p{ j(x) u

u

:x :

x

• • .

E [XJ-J . Xj]}

for

[t;_1, t;]}

for

E

, tk}.

Put

j = 1, 2, . . . m, i = l , . . . , k. .

The definitions of mr and m; are analogous (replace the sups by infs). Then

0<

=

f." f(x) dx - S,(f, P,) < S*(f, P,) - S,(f, P,) k

(M; - m;)(t; - t;_1) = V + W, L l i= (M;

where V is the sum of the terms -m; )(t; - 1;_1) for which [t;_1, t; ] lies entirely in some subint�".'al of the partition P , a.Tld W is the sum of the remaining terms. The sums V and W are estimated separately. We estimate V first. Note that V = 'E1 + · · · + 'Em, where each 'E1 is the sum of the terms m;)(t; - 1;_1) for which [1;_1, t;] c [x1_h x1] holds. But if m{ holds. Also, the sum of the [1;_1, t;] c [x1_ h x1 ], then M; - m; < lengths of those subintervals of the partition P, that lie inside [xJ-l, x1] never exceeds Xj - Xj-l · Thus, "£ j < r - mr)(Xj - Xj-t) holds, and hence

(M; -

MJ -

(M

Ill

v < L (Mf - mr )<xj - Xj- l ) = S*(j. P) - S* (f, P) < E . j=l

Now, we estimate W. Let (M; - m;)(ti - 1;_1) be a term of the sum W. Since !Pn l < min{x1 - Xj-l : j = 1 , . . . . m}, there exists exactly one j with 1 < j < n such that x1_1 < t;-t < x1 < t; < XJ+l · Thus, the sum W has at most m terms, and since E

E

=(M; - m;)(t; - t;_1) < 2c !P,! < 2c · 2em m it follows that

W < m(Ejm) = E . Thus,

0< for all n

f." j(x) dx

- S,(f,

!'•J < V + W < < + < = 2£

> no. That is, lim s (/, P,) = I: f(x) dX.

* h 0 ,:::: S*(j, P, ) - Ia j(x) dx < V + W < 2E holds for all n > no, and so, lim S*(f, P,) = I: j(x) dx. The last part follows immediately from the Similarly,

Section 23: THE RIEMANN INTEGRAL AS A LEBESGUE INTEGRAL

183

inequalities

S* (f, P11) < Rf(P,. T,, ) < S* (J, P,). •

The proof of the theorem is now complete.

We are now ready to establish that the Lebesgue integral is a generalization of the Riemann integral. Here, " f is Lebesgue integrable" means that f is integrable with respect to the Lebesgue measure.

EveryRiemann integrableftmction f : [a, b] integrable, and in this case the two integrals coincide. That is,

Theorem 23.6.

---+

Proof. For each n let

I f dA = {f(x)dx. P11 = {x0, x 1 , . . . , x2, } (b - a )2-1 ;

2"

=

¢, = L m iX[x;_1,x;)

i= l

is Lebesgue

=

be the partition that divides [a, b] that is, x; a+ i (b-a)2-11 •

1 into 2 subintervals all of the same length Let

JR.

and

2"

1/J, = Li=l M;X[x;_1,x;)• M; = {f(x) x

sup : E [xi 1 • x;]}. where m ; inf {f (x) : x E [Xi - 1 • X;]} and Clearly. {¢11 } and { 1/111} are two sequences of step functions satisfying the propetries cf>��(x) t< f(x) < 1/111(x) ,J, for each x E [a, b). Now, if ¢,(x) t g(x) and 1/J,(x) ,J, h(x), then by Theorem 22.6, both functions g and h are Lebesgue integrable and g(x) < f(x) < h(x) holds for all x E [a, b). Also, by definition, J c/J11 d>... S*(f, P11) and J 1/ln d>... S* (f, P11 ). Therefore, since 1/J,(x) - cj>11(x) ,J, h(x) - g(x) > 0, it follows that

= o < Jch - g) d>... = lcl/111 II = 11-+00 S*(f, P11) - 11-+00 S*(J. P11) .• h=g=f f lim

-HXJ

lim

= II-+00 11/J,

cfJ,)d>... = lim

lim

= 0,

l

c/J, d>... II-+00

d>... - lim

=

where the last equality holds true by virtue of Theorem 23.5. This implies h - g 0 a.e and hence, a. e. holds. In particular, c/J, t f a.e. and 1/111 ,J, f a.e. hold, which show that is Lebesgue integrable-in fact, an upper function. Finally,

I f d>... = l lim

11-+oo

c/>11 d>... = lim

n-oo

and the proof of the theorem is finished.

S* (f, P, )

= 1"

f(x) dx.

u

•

Chapter 4: THE LEBESGUE INTEGRAL

184

The next theorem, due to H. Lebesgue and G. Vitali, characterizes the Riemann integrable functions in terms of their discontinuities. (The almost everywhere relations are considered with respect to the Lebesgue measure.)

Theorem 23.7 (Lebesgue-Vitali). A bounded functioll f : [a, b]

--+

IR is

Riemanll il7tegrab/e if and only if it is colltinuous almost everywhere. Proof. For each ll, let Pn. ¢n. and 1/f, be as they were introduced in the proof of Theorem 23. 6. Assume first that f is Riemann integrable. Then a glance at the proof of Theorem 23.6 guarantees the existence of a (Lebesgue) null subset A of [a , b] such that ¢,(x) t f(x) and 1/f,(x) -!- f(x) for all x ¢ A. Clearly, D = A U ( U: 1 P,) has Lebesgue measure zero, and we claim that f is continuous on [a, b] \ D. To see this, let s e [a , b] \ D and E > 0. Pick some n with f(s) - ¢n(s) < E and 1/f,(s) - f(s) < E . Then there exists some subinterval [x;_ 1 , X;] of the partition P;; such that. \' e (x; -1 , Xi). Clearly, ¢,(s) = m; and 1/f,(s) = M;. Therefore, if x e (x;-lt x;), then - E < m; -

f(s)

<

f(x) - f(s)

<

M; - f(s)

< E.

Since (x;_ 1 , x;) is a neighborhood of s, the last inequality shows that f is contin­ uous at s. This establishes that f is continuous almost everywhere. For the converse, assume that f is continuous almost everywhere. Let s # b be a point of continuity of f. If E > 0 is given, then choose some 8 > 0 such that

f(s) - E

<

f (x)

<

f (s) + E

for all x e [a, b] with lx - sl < 8. Pick some k so that IPkl < 8. Then for some subinterval [x;-1 , X;] of Pkt we must have s e [x;_1 , x; ). In particular, lx - s l < 8 must hold for all x e [x;_1 , x; ], and so, from (•) we get

j(s) - E

<

m; = ¢k(s)

<

j(s) + E.

Since ¢,(s) t, it easily follows that ¢,(s) t j(s). Similarly, 1/f,(s) -1- f(s) holds. Since f is continuous almost everywhere, we conclude that ¢, t f a.e. and 1/f, -1- f a.e. both hold. This shows that f is Lebesgue integrable. Moreover, by the Lebesgue dominated convergence tfieorem we have

S*(f, Pn) =

J¢, d/, t J f d).

and

S*(f, P,) =

J1/J, d), -1- J f d)..

Section 23: THE RIEMANN INTEGRAL AS A LEBESGUE INTEGRAL

185

Thus, lim[S*(f, Pn) - S* (f, P,)] 0, and-so, by Theorem 23.3, the function f is Riemann integrable. The proof of the theorem is now complete. • =

An immediate consequence of Theorem 23.7 is the following:

The collection ofall Riemann integrablefunctions on a closed interval is a fwzction space and an algebra offtmctions.

Theorem 23.8.

It is easy to present examples of bounded Lebesgue integrable functions that are not Riemann integrable. Here is an example:

Example 23.9. Let f : [0, 1] � IR be defined by f(x) = 0 if :c is a rational number and f (x) = 1 if x is irrational (in other words, f is the characteristic function of the irrationals of [0, 1]). Then f is discontinuous at every point of [0, 1], and thus, by T heorem 23.7, f is not Riemann integrable. On the other hand, f = 1 a.e. holds (since the set of rational numbers has Lebesgue measure zero), and so, f is Lebesgue integrable. Also, note that • f f d>.. = 1 holds. It follows from Theorem 23.7 that if a function f : [a, b] � JR. is Riemann integrable, then f restricted to any closed subinterval of [a, b] is also Riemann integrable there. Moreover, by the same theorem, if two functions f and g are Riemann integrable on [a, c] and [c, b], then the function h : [a, b] � JR., defined by h(x) = f(x) ifx E [a, c] , and h(x) = g(x) ifx E (c, b], is Riemann integrable. Clearly, by Theorem 23.7, every continuous function on a closed interval is Riemann integrable. To compute the Riemann (and hence, the Lebesgue) integral of a continuous function, one usually uses the fundamental theorem of calculus, one form of which is stated next. Since any "reasonable" calculus book contains a proof of this important result, its proof is omitted. (See also Exercise 6 at the end of the section.) The fundamental theorem of calculus is due to I. Newton8 and independently to G. Leibniz.9

Theorem 23.10 (The Fundamental Theorem of Calculus). For a continu­

ousftmction f : [a, b] � JR. we have the following:

8Isaac Newton ( 1642-1727). a great British mathematician, physicist, astronomer, and philoso­ pher. He discovered the law of gravity and was one of the founders of calculus. His pioneering original contributions to mathematics and science revolutionized the modem scientific app­ roach. 9Gottfried Wilhelm Leibniz ( 1646-1716), a prominent Gennan mathematician and philosopher. He was a person with "universal" scientific interests. Besides his philosophical and metaphysical contributions, he contributed substantially to mathematics, mathematical logic, and physics. Together with Newton, he is considered the founder of calculus.

Chapter 4: THE LEBESGUE INTEGRAL

186

If A : [a, b] --+- IR is an area function off (i.e., A(x) = I;rf(t)dt holds for all x E [a, b] and some fixed c E [a, b]), then A is an antiderivative of f. That is, A'(x) = f(x) holdsfor each x E [a, b]. 2. If F : [a, b] --+- IR is an amiderivative off, i.e., F '(x) = f(x) holds for each x e [a, b], then 1.

{f(x)dx = F(b) - F(a).

In a conventional way, the integral I: f(x)dx is defined to be - I: f(x)dx; that is, I: f(x)dx = - I <x)dx. Also, I: f(x)dx is defined to be zero. By doing so, the useful identity

!J

idf(x)dx = f .f(x)dx rd f(x)dx c

·

c

+

Jc·

Jc

holds regardless of the ordering between the points c, d, and e of [a, b]. We now indicate how to extend the above results to functions ofseveral variables. In the general case, the interval [a, b] is replaced by a cell J = [a 1 , b1] x · · · x [an, b,], and its Lebesgue measure is A.(J) = n7=1 (b; - a;). A partition P of 1 is a set of points of the form P = P1 x · · · x Pn, where P; is a partition of [a;, b;] for each i = 1, . . . , 12. Clearly, any partition P divides J into a finite number of subcells. Now, if f : J --+- IR is a bounded function and the partition P divides J into the subcells 11 , , lt. then we define again the numbers •

•

•

m; = M; =

. . . , x,) E J; }, sup { f(xt , . . . , Xn) : (X t , . . . , x,) E J; }. inf{ f(xt, . . . , Xn ) : (xt,

The lower and upper sums corresponding to the partition by the formulas

k

S* (f, P) = L m;A.(J;) i=l

and

P are defined as before k

S *(f, P ) = L M;A.(J;), i=l

respectively. The lower Riemann integral of f is defined (as before) by

I*(f) = sup( S*(f, P ) : P

is a partition of

J},

Section 23: THE RIEMANN INTEGRAL AS A LEBESGUE INTEGRAL

187

and the upper Riemann integral of f by. /*(f)

=

inf{S*(f,

P) : P

is a partition of J}.

As in the one-dimensional case, -00 < / (/) < / * ( f) < 00 *

=

holds. The function f is called Riemann integrable if / (f) /*(f). This * common number is called the Riemann integral of f and is denoted by

All the results given in this section are valid in this general setting. Their proofs parallel the ones presented here, and for this reason we leave them as an exercise for the reader.

EXERCISES 1. Let

f : [a, b]

� JR. be Riemann integrable. Show

1tl

1! !d 1 = f(x) dx + f(x) dx

every closed subinterval of [a, b ]. Also, show that ,

2.

c:

I!

holds for every three pon i ts c, d, and e of [a, b]. Let f : [a. b] � JR. be Riemann integrable. Then show that

1b f(x)dx U

3.

f(x) dx

-- L. f (a + - a ) . f" b =1 1 11-+00 = lim

u-Hx:>

b-a n

l=l

f,,(x)dx

u

5. 6.

i (b

11

)

n

Let (f,, ) be a sequence of Riemann integrable functions on [a, b] such that {/11 } converges uniformly to a function f. Show that is Riemann integrable and that lim

4.

that f is Riemann integrable on

For each n . let j�, : [0, 1] � JR. be defined by

f(x)dx.

u

/11 (x) =

"j�:.' for all x E [0, 1 ]. Then

show that lim f f,,(x)dx = � · [HINT: Use integration by parts.] Let f : [a. b] � JR. be an increasing function. Show that f is Riemann integrable. [HINT: Verify that f satisfies Riemann's criterion.] (The Fuudamental Theorem of Calculus) If f : [a , b] � JR. is a Riemann integrable function, define its area function A : [a, b] � JR. by A(x) = J�\- f(t)dt for each x E [a, b]. Show that

d

Chapter 4: THE LEBESGUE INTEGRAL

188

a. A is a uniformly continuous function. b. If f is continuous at some point of [a, b], then A is differentiable at and 1 A ( ) = j{c) holds. c. Give an example of a Riemann integrable function f whose area function A is differentiable and satisfies N =F f.

c

c

c

[HINT: For part (c) use the function defined in Exercise 7 of Section 9.] 7. (Arzela) Let {/11} be a sequence of Riemann integrable functions on [a, b] such that lim f,,(x) = f(x) holds for each x E [a, b] and f is Riemann integrable. Also, assume that there exists a constant M such that lf,,(x)l < M holds for all x E [a, b] and all Show that

n.

11-HXJ

lim

f(x) dx. J{b a fn(x) dx = J{b a

Determine the lower and upper Riemann integrals for the function of Example 23.9. 9. Let C be the Cantor set (see Example 6.15). Show that xc is Riemann integrable over [0, 1], and that Jd xc dx = 0. 10. Let 0 < E < 1 , and consider the E-Cantor set CE oi iO, ij. Show thai xcf is not Riemann integrable over [0. 1]. Also, determine I.(xcf ) and / *(xcf ). [HINT: Show that the set of all discontinuities of xcf is CE.] 11. Give a proof of the Riemann integrability of a continuous function based upon its uniform continuity (Theorem 7.7). 12. Establish the familiar change of variable formula for the Riemann integral of contin­ uous functions: If [a, bl---4 [c, d] ....L. IR are g differentiable ( e , g ha.li a derivative), 8.

i

ously

.

.

iab

continuousftmctions with continu­ continuous then

f(g(x))l(x)dx =

lg(b) g(a)

f(u)du.

13. Let f : [0, oo) � IR be a continuous function such that limx-oo f(x) = 8. Show that limn-oc J; f( x) dx = a8 for each a > 0. 14. Let f : [0, oo) � IR be a continuous function such that f(x + J) = f(x) for all x :::. 0. If g : [0, 1] � IR is an arbitrary continuous function, then show that

n

of {0, J]. Let f : [0, 1] � [0, oo) be Riemann integrable on every closed subinterval 1 Show that f is Lebesgue integrable over [0, J] if and only if limE�O fe f(x) dx exists in IR. Also, show that if this is the case, then J f d)... = limE�O JE1 f(x) dx. 16. As an application of the preceding exercise, show that the function f : [0, J] � IR defined by j(x) = xP if x E (0, J] and f(O) = 0 is Lebesgue integrable if and only if p > J. Also, show that if f is Lebesgue integrable, then 15.

-

f

j d)... =

l +�-. p

-

17. Let f : [0, l ] � IR be a function and define g : [0, J] � IR by g(x) = ef<x>.

Section 23: THE RIEMANN INTEGRAL AS A LEBESGUE INTEGRAL

f

189

g.

a. Show that if is measurable (or Borel measurable), then so is b. If f is Lebesgue integrable, is then 8 necessarily Lebesgue integrable? c. Give an example of an essentially unbounded function which is continuous on 2, . . . . (A function 1] such that /" is Lebesgue integrable for each n is "essentially unbounded," if for each M > the set {x e l] : f(x)l > M} has positive measure.)

(0,

18.

19.

f = 1,

0

f

[0,

[HINT: For consider the function f(x) = x-! .] Let f : [ l] -+ IR be Lebesgue integrable. Assume that f is differentiable at x 3 and Show that the function IR defined by for = x-1 x e and is Lebesgue integrable. ] x [c. -+ IR be a continuous function. Show that the Riemann integral Let off can be computed with two iterated integrations. That is, show that

(b) 0, f(O) (0, 1]= 0. g(O) = 0 f : [a, b d]

t 1d

f(x ,

y)dxdy

I

g: [0, 1] -+

=

g(x)

=0 f(x)

t [1d f(x, y) dy J dx = 1" [{f(x, y)dx J dy.

Generalize this to a continuous function of n variables. ] -+ IR are two continuous functions such ] -+ IR and g : 20. Assume that < < :5 for E b]. Let that E b] and A= E IR2

[a, b

f : [a, b g(x) x [a, f(x)

g(x)} .

a.

{(x, y)

:x [a,

f(x) y

Show that A is a closed set-and hence, a measurable subset of IR2. b. If lz : A IR is a continuous function, then show that h is Lebesgue integrable over A and that

21.

-+

1 hd'J.. = 1ab [ f g(x)lz(x, y)dy] dx. A

f(x)

Let -+ IR be a differentiable function-with one-sided derivatives at the end-points. If the derivative !' is uniformly bounded on then show that f' is Lebesgue integrable and that

f : [a, b]

22. Let /, 8 :

[a, b],

r

J[a.b]

!'

d'J... = f(b)- f(a).

[a. b] -+ IR be two Lebesgue integrable functions satisfying lx f(t)d'J.. (t) :::= lx g(t)d'J.. (t) for each E [a, b]. If ¢ : [a, b] -+ IR is a non-negative decreasing function, then show that the functions ¢f and cpg are both Lebesgue integrable over [a, b] and that x

they satisfy

lx cp(t)f(t) d'J.. (t) 1x cp(t)g(t) d'J..(t) <

[a, b]. [IDNT: ao a1

for all x E Prove it first for a decreasing function of the form ¢ 'Lf=l c; X[a;-�oa;)• where < < · · · < is a partition of b], and then use the fact that there

ak

[a,

=

190

Chapter 4: THE LEBESGUE INTEGRAL exists a sequence {¢n} of such step functions satisfying ¢11 (1) t ¢(t) for almost all t in [a, b]; see Exercise 8 of Section 2 1 .]

24. APPLICATIONS OF THE LEBESGUE INTEGRAL

If a function oo) --+ 1R (where, of course, E 1R) is Riemann integrable on every closed subinterval of [a, oo then its improper Riemann integral is defined by

f: [a,

)

a

,

}a

r)O lim rr f(x) dx' Ja f(x) dx = r-HX>

provided that the limit on the right-hand side exists in IR. The existence of the prior limit is also expressed by saying that the (improper Riemann) integral exists. Similarly, if - oo, --+ 1R is Riemann integrable on every closed subintervai of -oo, then f�00 [(x) dx is defined by

f : ( a] ( a], la f(x)dx ,._!�! 00 1° f(x)dx

J:O f(x) dx

=

-oo

whenever the limit exists in

Jboo f(x) dx

r

IR.

also exists for each

It should be clear that if

b>a

and

fa00

f(x) dx

exists, then

['"f(x)dx J."f(x)dx + {"f(x)dx. =

on eve1y Assume that f: i s Riemann integr a ble closed subinterval of [a, oo). Then faxf(x) dx exists if and only iffor eve1fory > 0 ther e exi s ts some M > 0 (depending on such that I J; f(x) dxl all s, t > M. Proof. I Jcr: f(x) dx M>0 r > M. s, t M, I/ -J�-f(x)dxl rf(x) dx J>(x) dx - t f(x) dx I - }[' f (x) d I - f (x) d j.fa a Theorem 24.1.

[a. oo) --+ 1R

<E

E)

E

Assume that

=

exists. Pick a real number

< E holds for all

that

If

2:.

such

then

= <

X +

X < 2E.

Conversely, assume that the condition is satisfied. If {a11} is a sequence of [a, oo) such that lim x} is = oo, then it is easy to see that the sequence

an

{j:n f(x) d

191

Section 24: APPLICATIONS OF THE LEBESGUE INTEGRAL

A f(x) dx IR. {b,} = [a, = B = J:" f(x) dx. b, !. a, f. Al - Bl < A - a f(x)dx a, f(x)dx B - f.h,a f(x)dx , A - BI < E E > A = B, fa00 f(x) dx

Cauchy. Thus, lim J:" of oo) with limb, oo; let

exists in lim

Now, let be another sequence From the inequality

+

+

it is easy to see that I holds for all 0. Thus, limit is independent of the chosen sequence. This shows that and the proof is finished.

and so the exists,

•

f(x) J: l f (x)l dx dxl s t, Lemma 24.2. If a function f: a, oo) -+ IR is Riemann itttegcable otz every closed subitzterval of[a, oo) and face 1/(x)l dx exists thetz J;� f(x) dx also exists and 1fa00 f(x) dxl faoo l f(x)l dx. f fa00 f (x) dx fa00 I f (x) I dx f(x) = si�I x = oo). J000 si�x dx = I }�00 si-�xl dx. From the preceding theorem, and the inequality < we have the following:

11:�

<

for

[

<

The converse of this lemma is false. That is, there are functions exist but for which improper Riemann integrals to exist. A well-known example is provided by the function I ) over [0, We shall see later that /(0) does not exist. To see the latter, note that

whose fails (with

but

2 1br -I x.l dx >- - 1 lsmxldx. . = krr X I

sin

br

(k-l)rc

holds for each

k

rc

­

(k-l)rr

k. Therefore,

J."rc -I x.l dx - L 1kir .xl d.'l. > -2 :L - . k X X 7C 0

sin

_

!sin

"

k=l

(k-l)rr

"

I

k= l

J000 lsi�- xl dx

which shows that does not exist in IR. If the improper Riemann integral of a function exists, then it is natural to ask whether the function is, in fact, Lebesgue integrable. In general, this is not the case. However, if the improper Riemann integral of the absolute value of the function exists, then the function is Lebesgue integrable. The details follow.

Let f: [a, oo) IR be Riemantz itztegcable otz every closed subinterval of [a, oo). Thetz f is Lebesgue itUegrahle ifand only ifthe improper Theorem 24.3.

-+

Chapter 4: THE LEBESGUE INTEGRAL·

192

Riemann integral J000 lf(x)l dx exists. Moreover, in this case

f is Lebesgue integrable over [a, oo). Then j+ is also Lebesgue integrable over [a, oo). Let {rn} be a sequence of [a, oo) such that lim rn = oo. For each n, let f,,(x) = j+(x) if x E [a, rn] and fn(X) = 0 if x rn . Then lim f,,(x) = j+(x) and O < f,,(x) < j+(x) hold for all x E [a, oo). Proof. Assume that

>

Moreover, by Theorem 23.6, {/11 } is a sequence of Lebesgue integrable functions such that J f,, dJ... = J�·" j+(x) dx. Thus, by the Lebesgue dominated convergence theorem

I f+ dJ... = 11-00 j f,, dJ... = n-oo 1,." j+(x)dx. lim

lim

0

This shows that J000 j+(x)dx exists and J000 j+(x)dx = J j+ dJ.... Similarly, j000 f-(x)dx exists and j000 f-(x)dx = J f- dJ... . But then, it follows from f = j+ f- and If I = j+ + f- that both improper Riemann integrals J000 f(x) dx and fa00 1 /(x)l dx exist. Moreover, -

{' J(x)dx = J f d

A

and

["' i f(x)l dx = j fl d . l

For the converse, assume that the improper Riemann integral

A

J000l/(x)l dx

n exists. Clearly, lim J;+ l f (x) l dx = j000l/(x)l dx. Define f,,(x) = 1/(x)l if x E [a, a+n] and fn(x) = Oifx a+ for each n . Note thatO < /11 (x) t 1/(x)l holds for all x a. Since j,, is Riemann integrable on [a, a + ], fn is an upper n + function on [a, oo) satisfying J f,, dJ... = J; fn (x)dx = faa+n lf(x)l dx. Thus, >

>

n

n

by Theorem 21.6, 1/1 is an upper function (and hence, Lebesgue integrable) such that f i J I dJ... = foc l f (x) l dx. The Lebesgue integrability of f now follows from a Theorem 22.6 by observing that f is a measurable function, since it is a measurable function on every closed subinterval of [a, oo). The proof of the theorem is now • complete. The next result deals with interchanging the processes of limit and integration.

Let (X, S, J,L) be a measure space, let J be a subinterval of IR, a d let f : X x J � IR be a function such that f(·, t) is a measurable function for each t E J. Assume also that there exists an integrable function g Theorem 24.4. n

193

Section 24: APPLICATIONS OF THE LEBESGUE INTEGRAL

such that for each t E J we have lf(x, 01 < g(x) for almost all x. Iffor some accumulation point to (including possibly ±oo) of J there exists a function h such that lim,_,0 f(x, t) = h(x) exists in JR.for almost all x, then h defines an integrable function, and lim

,_,0

j f(x, t) d�J-(x) = J lim f(x, t) d�J-(.x) = Jh d!J-. (--1' (1)

� JR. satisfies the hypotheses of the theorem. Let {t,} be a sequence of J such that lim t11 = to. Put h,(x) = f(x, t,) for x E X, and note that l h, I < g a.e. holds for each n and h, � ll a. e. By f:X

Proof. Assume that the function

x J

Theorem 22.6, each It, is integrable. Moreover, by the Lebesgue dominated convergence theorem, h defines an integrable function and ,_oc

lim

j f(x, t,) d!J-(x)

11-oo

= lim

J h, d!J- Jh d!J=

holds, from which our conclusion follows.

•

If f : X x (a, b) � JR. is a function and to E function at to is defined by

D,11 (x, t )

(a, b), then its difference quotient

- f(x, t) - f(x, to) t - to

-

-----­

for all x E X and all t E (a, b) with t #- to. To make the function D,0 defined everywhere, we let D,0(x, to) = for all x E X. As usual, lim,--..,0 D,0(x, t), whenever the limit exists, is called the partial derivative of f with respect to t at the point (x, t0) and is denoted by ex, t0). That is,

0

¥-

. D,0(x, t) = ltm . f(x, t) - f(x, to) . of to) = ltm -(x, at t - to ,_,0

r-ra

The next result deals with differentiation of a function defined by an integral.

Let (X, S, /.L) be a measure space and let f : X x (a, b) � IT·� be afimction snell that f(·, t) is Lebesgue integrablefor eve1y t E (a, b). Assume that for some to E (a, b) the partial derivative 1, (x, to) exists for almost all x. Suppose also that there exists an integrable function g and a neighborhood V of to such thatfor each t E V we have ID,0(x, t)l < g(x)for almost all E X. Then 1 . r, (·, to) defines an integrable fimction, and Theorem 24.5.

x

Chapter 4: THE LEBESGUE INTEGRAL

194

2.

thefunction F : (a, b) --+ IR defined by F(t) = J f(x, t) dJ.L(X) is differ­ entiable at to and

, faatt (x, to)dJ.L(x).

F (to) =

r,(x, to) 0 at each point x where the partial derivative does not exist. Then limr r Dr0(X, t) fr (x, to) holds for almost all x. Thus, by Theorem 24.4, �� (·, to) defines an integrable function and Proof. Put

=

-+

o

F(t) - F(to) t - to

=

=

=

J f(x, t)t -- tof(x, to) dJ.L(X) J Dr0(X, t) dJ.L(X) J ir (x, to)dJ.L(X) �

as t -+ to. This shows that F is differentiable at to and that F'(to) (x) holds.

=

frr (x, to) dJ1.

•

There is a criterion for testing the boundedness of the difference quotient function D111(x, t) by an integrable function that is very useful for applications. It requires the existence of a neighborhood V of to satisfying the following two properties:

1.

2.

the partial derivative (x, t) exists for all x and all t e V , and there exists a non-negative integrable function g such that for each t E V, we have I <x, t) l < g(x) for almost all x.

%f

o/i

Indeed, if the previous two conditions hold, then by the mean value theorem, it easily follows that for each t E V we have ID10(x, t)l < g(x) for almost all x. Next, as applications of the last two theorems we shall compute a number of classical improper Riemann integrals.

Theorem 24.6 (Euler).l0

We have J000e-x� dx

=

�.

Proof. The existence of the improper Riemann integral follows from the in­

equalities 0 < e-x2 < e-x for x > 1 and J;:x:� e-x dx = e- 1 • Also, since e-x2 for all x , the improper Riemann integral is a Lebesgue integral over [0, oo)

.

>0

10Leonhard Euler ( 1 707-1 783), a great Swiss mathematician. He was one of the most prolific writers throughout the history of science. He is considered (together with Gauss and Riemann) as one of the three greatest mathematicians of modem times.

Section 24: APPLICATIONS OF THE LEBESGUE INTEGRAL

For the computation of the integral, c�nsider the real-valued functions on defined by

195

IR

The derivatives of the previous functions will be determined separately. For f we use the fundamental theorem of calculus and the chain rule. Thus, f'(t) =

,

2e-t·

For g observe that

it 0

,

e-x· dx.

holds for every x E [0, I] and every t E IR in any bounded neighborhood of some fixed point to. The constant M depends, of course, upon the choice of the neighborhood of t0 . By Theorem 24.5, and the fact that the Lebesgue integrals are Riemann integrals, we get

for all t

E IR.

Substituting u = xt (for t :f: 0), we obtain g'(t) =

.,

-2e_1_

11 0

e

-11,

du =

for each t E IR. Thus, J'(t)+ g'(t) = 0 holds for each t holds for all t . In particular, c = f(O) + g(O) =

-2e_1• ,

11 0

e-x- dx .,

E IR, and so, j(t)+ g(t) = c (a constant)

11 1 0

dx

., + x-

- -4

,

and so j(t) + g(t) = :f for each t . Now, observe that for each x and t we have 2 e -t (

I +.r2)

1 + x2

<

1

+ x2 '

---

1

Chapter 4: THE LEBESGUE INTEGRAL

196

and clearly lim 1 -+oo

Thus, by Theorem 24.4, lim1_. 7C

4

-

00

e-12(l+x2) I + x2

g(t) =

0.

=

Consequently,

= lim f(t) + lim g(t) = l-+ l-400 00

2

from which it follows that J000 e-x dx =

Theorem 24.7. For each t 100

e

0.

.;;

(

1 00 0

e-x- dx ,

)2

,

•

.

JR. we have

..fiC e-··,.2 cos(2xt) dx = _ e-1 2 . � 0 2 F(t) = [000 e-.r cos(2xt) dx for all t e JR.

2

Since le-x cos(2xt)l < e-x- holds for all x and t, it follows that the improper Riemann integral F(t) exists and, moreover, is a Lebesgue integral over [0, oo). Now,

Pro.?f· Let

� [e-x2 cos(2xt) ] ar

=

2 < 2x e-x = g(x) l -2xe-x2 sin(2xt)l -

holds for all x > 0 and t . Hence, since the function g is positive on [0, oo) and the improper Riemann integral [000 g(x) dx exists (its value is I), g is Lebesgue integrable over [0, oo). Therefore, by Theorem 24.5 (and the remarks after it), it follows that

ar

2

2

F'(t) = roo � [e-x cos(2xt)] dx = -2 roo xe-x sin(2xt)dx. lo lo

Integrating by parts, we obtain

2

-2 rocxe -x sin(2xt) dx = lo

- 2t e-x2 sin(2x t) 100 0

2

2

roo e -x cos(2xt) dx lo

= -2t roo e -x cos(2xt) dx. · lo

Thus, F'(t) = -2t F(t) holds for all t . Solving the differential equation, we get e-12 , as and so, F(t) = F(t) = F(O)e-12 • By Theorem 24.6, F(O) = claimed. •

./{,

./{

Section 24: APPLICATIONS OF THE LEBESGUE INTEGRAL For the next result, the value of

si�x

at zero will be assumed to be 1 .

Theorem 24.8. If t > 0, then

1oc · 0

holds.

197

smx . t - e-.\ dx X

=

- - arctan t

rr

2

Proof. Fix some to > 0. The cases to > 0 and to

=

0 will be treated separately.

Case I. Assume to > 0. For each fixed t > 0, note that le-x' si�x I :5 e--� t holds for all x > 0. Thus, the improper Riemann integral exists as a Lebesgue integral over [0, oo). Let e-xt si�x dx for t > 0. Then F satisfies the following properties: F(t) =

J000

lim F(t)

1-+00

and

F

I

(t) = -

1

1 + t2

=

0,

for each t > 0.

To see (*), note that if g(x) = 1 for x E [0, 1 ] and g(x) = e-x for x > 1, then g is Lebesgue integrable over [0, oo) for each t > 0 and le-x' si-�x I < g(x) holds for all x > 0 and all t > I. On the other hand, lim,-+oo e-x t si�x = 0 holds for all

x > 0, and so, by Theorem To establish (**)

24.4, lim,-+oc F(t) note first that ..! a [e-x' sinx] x

=

0.

= -e-x' sin x holds for all x > 0 t and that for each fixed a > 0, the inequality le-x' si�x I < e-ax holds for each •

-

•

e-xt sinx dx for all t > a t 2:, a and all x > 0. By Theorem 24.5, F'(t) = (and all a > 0), where the last equality holds since the improper Riemann integral is a Lebesgue integral. Thus, F'(t) = e-xt sinx dx holds for all t > 0. Since from elementary calculus we have

J;

J000

lor0 -·

. x dx e .\ t sm

=

-

e-rt(t sinr + cosr) + -1 + t-., 1 + t2 '

by letting r ----* · we get F'(t) = Integrating (**) from t to to yields

1,!

F(t0) - F(t)

and by letting t

----*

oo

f 1 dx ., = tn

= -

for t > 0.

1

+ x-

it follows that F(t0)

=

arctan t - arctan t0,

� - arctan t0•

198

Chapter 4: THE LEBESGUE INTEGRAL

Case II.

Assume to = 0.

s�i x s�i J000 .Y

In this case,

is not Lebesgue integrable over [0, oo) However, the improper

Riemann integral parts yields

[' sinx dx =

ls

_

X

1

X

_

t cosx dx = cos s _ cos t

s ls X2

1 sin x 1 I -- dx < - + - + s t s x

oo sin x x

_

t

S

[

By Theorem 24. 1 • ro

dx = 0.

dx exists. Indeed, if 0 < s < t, then an integration by

cos x

and thus,

J

.

t cos x dx,

ls X2

[

1 dx 2 - = -. s x2 s

n-+oo Jruu+ l sinx

dx exists. In particular note that lim

e-.n si�.;·

'

,

_ .

Now, foreachn define f.1(t � = J�, dx for t > 0, andnotethatthereiation 1 lfn(n)I .:S f� e-.m dx = t-��-��- < * implies lim fn (n) = 0. By Theorem 24.5, we have

_ _, 1 1 11 (t ) _

1u o

e

-.n

. . sm x d.\

.

-

_

· n + cos n ) - 1 e- 111 (t sm , I + t,

and so, lim11-.. 00 J.:(t) = - 1�1 1 holds for all t > 0. Also,

Jn

I + (1 + t)e-1 1 -fl(t)l < 1 + t2 -

holds for each t > 0, and the dominating function for the sequence { f1: } is Lebesgue integrable over [0, oo). Let g11 = /1:Xro.11], and note that {g11} is a sequence of Lebesgue integrable for 1 functions over [0, oo). Also, since lg11 1 < 11.: and lim11_,.00 g11(t) = t > 0, the Lebesgue dominated convergence theorem yields lim

u-+oo

10n

J.:(t)dt = lim 11 -+oo

I

10x

g11(t ) dt = -

1

- �1

10oo

Since f� J.:(t)dt = j.1(n) - J.,(O) and lim J., (n) = Finally, letting n � oo in the identi�

n10 +l X

sinx

-- dx =

we easily get

1

si .,-

11 sinx - dx +

0 X

J000 � dx = I ·

1 "

0,

dt 7r 2 = -- . 2 I +t

we get lim /.1 (0)

11+ 1 sinx -- dx = /11 (0) +

X

u1 +l n

2

-

I·

sin x

-- dx, X

•

Section 24: APPLICATIONS OF THE LEBESGUE INTEGRAL

199

EXERCISES 1.

Show that Ia� _2, .\

0

holds for

2. 3. 4.

0. 1x. 2, . . . . e-1 2 ! .[if

11 =

Show that J0� Show that f(x) = Show that

dx

=

(2n)! e-xl = 2-1111! dx

for each

Let f : [0, oo)

---+

tion. Show that

!au( +

6. 7. 8.

1

> 0.

1

r :.... ll

) e-2x "

if and only if lim

(x

dA.

=

l.

(x)

=0

for each t > O.

Show that the improper Riemann integrals, J000 cos(x2)dx and J000sin(x2)dx (which as the Fresnel

11

integrals) both exist. Also, show that cos(x2) and sin(x2) are not Lebesgue integrable over (0, oo). are known

roo 5'�:x f(·, t

Show that JO Let

'

dx =

!f.

t

(X, S, J-L) be a measure space, T a metric space, and f : X t) is a measurable function for each

that for each

function F :

T

F(t) = is a continuous function. Show that

0,

holds for each t >

0.

fx

T ---+ IR a function.

E T and f(x,

let F(t) = f000

----

X

·

) is a continuous

t(x , t)d J-L(x),

laoo e-x _ e-xt 0

For each t >

x

X. Assume also that there exists an integrable function g such E T we have lf(x, t)l ::::; g(x) for almost all x E X. Show that the ---+ IR, defined by

function for each x E

10.

f

dx = l .

fj + t) x-+oo

1

Assume that

9.

fo 2

-

(0, oo) be a continuous, decreasing, and Lebesgue integrable func­

x-+oo ) 1xf(s) x ds = 0 lim

t

·

� is Lebesgue integrable over [ 1 , oo) and that J lim 11 -+00 0

5.

? -

dx = ln t

t���1 dx.

a.

Show that the integral exists as an improper Riemann integral and as a Lebesgue

b.

Show that F has a second-order derivative and that F"(t) + F(t) =

integraL

each t >

0.

1 1 Augustin Jean Fresnel (

f holds for

1788-1827) was French physicist who worked extensively in the field of optics. Using the integrals that bear his name, he was the first to demonstrate the wave conception of light.

a

200

11.

12.

Chapter 4: THE LEBESGUE INTEGRAL

cosx)dx,

Show that the improper Riemann integral, J02 ln(t and that it is also a Lebesgue integral. Also, show that lt

0.

holds for all t > Show that for each t >

0,

the improper Riemann integral

100

Lebesgue integral and that

The Gamma function for

J: x�;����1>

dx

0

exists as a

x d = -(l -e-1). x(l + x2) t>0 = la00x1 - le-x dx.

o

13.

exists for each t >

sin t

--=-

-

x

rr

2

is defined by an integral as follows:

f(t)

a.

b. c. d.

e.

Show that the integral

exists as an improper Riemann integral (and hence, as a Lebesgue integral). Show that r( �) = .fii. Show that f(t + 1 ) = t f(t) holds for all t > and use this conclusion to for n = 1 , 2, . . . . establish f(n + = Show that r is differentiable at every t > and that

1) n!

0,

0 f'(t) = la00 x1- 1 e-·'· lnxdx

holds. Show that r has derivatives of

all

r<" > (t) = holds for 14.

Let

b.

=

1,

fooo

2, . . . and all t >

f: [0, 1] � IR F : [0, 1] �

function

a.

n

order and that x ' - l e-x (lnx)n dx

0.

be a Lebesgue integrable function and define the real-valued IR by

F(t)

=

Jd f(x) sin(xt) d>.(x).

Show that the integral defining· F exists and that F is a unifonnly continuous function. Show that F has derivatives of all orders and that

201

Section 25: APPROXIMATING INTEGRABLE FUNCTIONS and

Fl2 •- 1 l ( t)

for n = 1 , 2, . . . and each c. Show that F = 0 (i.e.,

=

(- I )"

fo

i .r2n - l f(.r) cos(.
t E [0, 1]. F(t) 0 for all t E [0, 1]) if and only if f =

=

0 a.e.

25. APPROXIIVIATING INTEGRABLE FUNCTIONS The following type of approximation problem is commonly encountered analysis. •

m

Given an integrable function f, a family ofF of integrable functions, and E > 0, determine whether there exists a function g in the collection F such that f lf - gl d/1- < E .

By the definition of an integrable function, the following result is immediate.

Theorem 25.1. Let f be an integrable function and E a

step function ¢ such that

J If - ¢1 d/1- < E.

>

0. Then there exists

Let us denote by L the collection of all step functions ¢ for which there exist sets A1, A, E S all of finite measure, and real numbers a 1 , , a, (all depending on ¢) such that ¢ = E�'=1 ai X �; holds. Then L is a function space; see Exercise 1 of Section 21. . . . •

•

•

•

Theorem 25.2. Let f be an integrable function, and let E

>

0.

Then there

exists a function ¢ E L such that J If - ¢ 1 dJ1. < E . Proof. Let F denote the collection of all integrable functions f such that for each E > 0, there exists some function ¢ L such that J If -¢1 d11- < E. It should be clear that F is a vector space such that XA E F holds true for each A E S. Now, a glance at Theorem 22.1 2 guarantees that F consists of all integrable functions .

E

•

The next result deals with the approximation of integrable functions by continu­ ous functions. Recall that if X is a topological space and f : X -7 lR is a function, then the closure of the set {x E X : f(x) # 0} is called the support of f (denoted by Supp f). If the support of f happens to be a compact set, then f is said to have

compact support.

Theorem 25.3. Let X be a Hausdorff locally compact topological space, and

let f.L be a regular Borel measure on X. Assume that f is an integrable function

202

Chapter 4: THE LEBESGUE INTEGRAL

with respect to the measure space (X, B, Jl). Then given E > 0, there exists a continuousfunction g : X -+ IR with compact support suclz that J If -g I d f..L < E.

Proof. By Theorem 25.2, we can assume without loss of generality that f is

XA

the characteristic function of some Borel set of finite measure. So, assume f = for some Borel set A with J.L(A) < oo. Since Jl is a regular Borel measure, there exist a compact set K such that K c A and Jl(A) - J.L(K) < E (see Lemma 18.5) and an open set V such that A c V and J.L(V) - J.L(A) < E. By Theorem 10.8, there exists a continuous function g : X -+ [0, l ] (and hence, g is Borel measurable) with compact support such that g(x) = l for each x E K and Supp g c V. Clearly, g is integrable and IX A - g l < holds. Therefore,

Xv - XK J IXA - g

l d J.L <

Jcxv -

XK ) dJ.l = J.L( V) - Jl(K) < 2E ,

and the proof of the theorem is finished.

•

The following theorem describes an important property of the Lebesgue in­ tegrable functions on JR. It is usually referred to as the Riemann-Lebesgue

Lemma.

Theorem 25.4 (Riemann-Lebesgue). Iff : IR -+ IR is Lebesgue integrable,

then

lim

11-HX>

j

t(x) cos(nx) d}..(x)

= 11-+ lim

00

j

t(x) sin(nx) d).. (x)

= 0.

Proof. Note first that the inequality lf (x) cos(nx) l < lf(x) l for all x , com­

bined with Theorem 22.6, shows that the function f(x) cos(nx) is a Lebesgue integrable function for each n. Also, by Theorem 25.2, in order to establish Thus, let the theorem, it suffices to consider functions of the form f = In this case, the Lebesgue integrals are Riemann integrals, and f=

f

Xla.h)·

f(x) cos(nx) d)..(x)

=

as n -+ oo. Similarly, Iim

1h "

Xla.h)·

cos(nx) dx

=

l

n

-

l sin(n b)

f f(x) sin{nx) d)..(x) = 0.

-

sin(n a) l <

2

n

-+

0

•

A sequence of integrable functions { fn } is said to converge in the mean to some function f if lim f l fn - fl dJ.l = 0 holds.

Section 25: APPROXIMATING INTEGRABLE FUNCTIONS

203

Let { f,,} be a sequence.ofintegrablefunctions. Iff is an inte­ grablefunction such that lim Jlf, - !I df..L = 0, then there exists a subsequence {fkn } of {/,} such that [k, � f a.e. holds. Proof. Let E > 0. For each n let En = {x E X : lf,(x) - f (x ) l > E } . and note that each £11 is a measurable set of finite measure (see Theorem 22.5). Now, since EX£, < If,, - !I holds for each n, it follows that EJ..L* (£11) < flfn - !I df..L also holds for each n. Thus, lim J.L.*(£11) = 0, and so, j,, .J!:....:,. f. The conclusion Theorem 25.5.

•

now follows immediately from Theorem 1 9.4.

A sequence of integrable functions that converges in the mean to some function need not converge pointwise to that function. An example of this situation is provided by the sequence { /11 } of Example 1 9.6. EXERCISES 1.

Let f : 1R -4 1R be a Lebesgue integrable function. Show that lim 1 �00

lim j t(:c)sin(xt) dA.(x) = J t(:c)cos(xt)dA.(x) = 1-+00

0.

2. A function f : 0 -+ 1R (where 0 is a nonempty open subset of 1R") is said to be a C00-function if f has continuous partial derivatives of all orders. a. Consider the function p : 1R -+ 1R defined by p(x) = exp[xLt ] if lxl < I and = 0 if lxl > 1. Show that p is a C00-function such that Supp p = [-1, 1]. (Induction and L' Hopital's rule are needed here.) x b. ForE > and a E 1R show thatthe function f(x) = p( �a ) is also aC00-function with Supp f = [a - €, a + €].

p(x)

12

O

3.

Let [a. b] be an interval, E > 0 such that a + € < b - €, and p as in the previous exercise. Define lz : 1R -+ 1R by h (x ) = I: p( '-;x ) dt for all x E 1R. Then show that

a.

Supph C [a - E, b + E],

b. h(x) = c (a constant function) for all :c E [a + E, b - €], c. lz is a C00-function and h(l ) (x) = I: l,�, p('-;X ) dt holds for all .r E 1R, and t/z satisfies 0 < f(x) < for all X E 1R, f(x) = 1 for all d. the C00-function f xE €, b - E], and I IX[a ,l') - /1 dA. < 4€. l

[a +

=

1

4. Let f : 1R -4 1R be an integrable function with respect to the Lebesgue measure, and let € > 0. Show that there exists a C00-function g such that Ill - gl dA. < E. [HINT: Use Theorem 25.2 and the preceding exercise.] 12 Guillaume-Franc;ois-Antoine de L'H6pital

(1661-1704), a French mathematician. He is mainly

remembered for the familiar rule of computing the limit of a fraction whose numerator and denominator tend simultaneously to zero (or to ±oo).

Chapter 4: THE LEBESGUE INTEGRAL

204 5.

The purpose of this exercise is to establish the following general result: Iff : IR" � IR E> ! If for i = I , . . . , n , and put I = Choose € > 0 such that a. Let E - E for each Use Exercise 3 to select for each a C00-function f; : IR � IR such that 0 � f;(x) � 1 for all x, f;(x) = 1 if E E, - E], and Supp - E, E]. Now define II : IR" � IR by h(xl, . , Xn ) = 0�'= 1 f;(x;). Then show that h is a C00-function on IR" and that

is an illtegrablefunction (with respect to tbe Lebesgue measure) aud 0, tben there exists a C00-function g such that gl dA < €. a; < b; Of_1(a;, b;). a; + < b; i. x i [a+ b; f; � [a; b; + .. jlxt -hldA < [D(b; -a; + 2e) - D(b; - a;)] . 2

b. Let f : IR" � IR be Lebesgue integrable, and let E > 0. Then use part (a) to show that there exists a C00-function with compact support such that

g jlf - g j d), < €. a regular Borel measure on IR.n , f : IR" IR a J-L-integrable Junction, and 6. ELet J-L0. beShow that there exists a C00-function g : IRn IR such that fl f - g I dJ-L < E. let E 0. Show that there Let f [a, b] IR be a Lebesgue integrable function, exists a polynomial p such that f l f - pi d). < E, where the integral is considered, of course, over [a, b]. [HINT: Use the Stone-Weierstrass approximation theorem.] � �

>

7.

:

26.

and

�

>

PRODUCT MEASURES AND ITERATED INTEGRALS

Throughout this section, (X, S, J..L) and (Y, :E , v) will be two fixed measure spaces. The product semiring S ® :E of subsets of X x Y is defined by S ® :E

= {A x B : A E S

and

B

E :E}.

The members of S ® :E are called rectangles. The above collection S ® :E is indeed a semiring of subsets of X follows immediately from the identities

x Y.

This

(A x B) n (A1 x B1 ) = (A n A1 ) x (B n B1 ), and 2. (A x B) \ (A1 x Bd = [(A \ A1 > x B] u [(A n A J ) x (B \ B1 )], and the fact that A \ A 1 and B \ B 1 can be written as finite disjoint unions 1.

of members of S and :E, respectively. The product se miring was discussed in s ome detail in Section 20. Now, define the set function J..L x v : S ® :E � [0, oo] by J..L x v(A x B) = J..L(A) · v(B) for each A x B E S ® :E (keep in mind that 0 oo = 0). This set function is a measure on the product semiring S ® :E, called the product measure of J..L and v. The details follow. ·

·

Section 26: PRODUCT MEASURES AND ITERATED INTEGRALS

Theorem 26.1.

The setfimction J.t x J.t x v(A

x B)

v.: S

® :E

= J.t(A)

--*

[0,

205

oo] defined by

· v(B)

for each A x B E S ® :E is a measure. (This measure can be thought of as the generalization of the familiarformula "Length x Width" for computing the area ofa rectangle in elementary geometry!) Proof. Clearly, J.t x v(0) 0. For the a -additivity of J.t x v, let A x B E S ® :E and let {A, x B, } be a sequence of mutually disjoint sets of S ® :E such that A

X B

= u:l A,

X

=

B,. It must be established that

J.t(A) · v(B)

= L J.t(A,) 0:::

II= I

· v ( B,) .

Obviously, (*) holds true if either A or B has measure zero. Thus, we can assume that J.t(A) > 0 and v(B) > 0. Since = 2::1 we see that

XA,x B, •

XAx B

00

XA(x) XB(Y) = LXA,(x) · XB,(y) ·

n=l

holds true for all x and y. For each fixed y E B let K = {i E IN : y E B;} and note that L;e K for each x E X. Since the collection {A; : i E K } LieK J.t(A; ). Therefore, is pairwise disjoint (why?), the latter implies J.t(A)

XA(x )

=

XA; (x )

=

= L J.t(A; )XB/Y) = LJ.t(A ) XB,(Y) J.t(A,) = n. J.t(A,)

j.t(A) . XB(Y)

00

iE K

,

.

(**)

11=1

0 does not alter the sum in (*) or holds for all y E Y. Since a term with (**), we can assume that > 0 for all Now, if both A and B have finite measures, then by using Theorem 22.9 and integrating (**) term by term, we see that (*) is valid. On the other hand, if either A or B has infinite measure, then I::1J.t(A,) · v ( B,) = oo must hold. Indeed, if the last sum is finite, then, by Theorem 22.9, J.t(A)x8(y) defines an integrable function, which is impossible. Thus, in this case (*) holds with both sides infinite, and the proof is finished. • The next few results will unravel the basic properties of the product measure J.t x v . As usual (J.t x v)* denotes the outer measure generated by the measure space (X x Y, S ® :E, J.t x v) on X x Y.

Chapter 4: THE LEBESGUE INTEGRAL

206

Theorem 26.2.

tben

(Jl

x

If

A and B Y are measurable sets offmite measure, v)*(A B) = p.* v*(A B) p.*(A) v*(B). :E {A1 1 B1 1 } A B u:)AII BlAl. B 26.1, p.* v* cX

c

x

x

x

·

=

Proof. Clearly, S ® C Ap. ® Av holds. Now, let x be a sequence X c of s ® :E such that X Since, by Theorem X x E Ap. ® Av. it follows from is a measure on the semiring Ap. ® A v and Theorem 1 3.8 that 00

X

00

p.* v*(A B) 1L=1 p.* v*(A11 B11) 1L=1 v(A11 B11), <

x

x

and so,

x

x

x

=

x

< (p.

J1. x

J.l* v*(A B) v)*(A B). E 0 {An} L:: {B,}L::,v(B,J A v*U: p(A,) p.*(A)+E 1 1 A11, B U:1 B11, (B) +E. A B u:, u�=I AII Bm (Jl v)*(A B) n=lL m=lLJl v(A, B11 ) Ll =l m=LJ1(A,) v(B111) [�JL(A., )] · [� v(B,)] lJL*(A)+ <] · [v•(B) + <]

On the other hand, if c :E, with C and < so X

X

is given, choose two sequences c such that Then, X c X

>

X

00

:Xl

x

x

<

X

=

0.

That is,

x

=

(p. v)*(A B) p.*(A) v*(B) (p. v)*(A B) v*(A B) x

Therefore,

x

holds, and

00

·

I

<

=

for ali E >

00

<

c S and

x

<

=

p.* x

·

x

p.* x

v*(A B). x

holds, as required.

•

It is expected that the members of Ap. ® Av are J1. x v-measurable subsets of X x Y; that is, Ap. ® Av c A11-xv holds. The next theorem shows that this is actually the case.

If is a Jl.-measurable subset of X and a v-measurab/e subset of Y, then x is a J1. x v-measurable subset of X x Y. Theorem 26.3.

Proof. Assume

f.L(C)

·

v(D)

A A B A B

B

E Av. and fix C x D E S® :E with p. x E All-, < oo. In order to establish the J1. x v-measurability of

v(CA B,

x D) = x by

Section 26: PRODUCT MEASURES AND ITERATED INTEGRALS

207

Theorem 15.2, it suffices to show that

(J-L x v)*((C x D) n (A x B)) + (J-L x v)*((C

x D) n (A x B/) < J-L x v(C x D).

If f.1. x v(C x D) = 0, then the previous inequality is obvious (both sides are zero). So, we can assume JJ-(C) < oo and v(D) < oo. Clearly,

(C (C

x

X

D) n (A D) n (A

x

B) X B)c

= (C n A)

x

= [(C n A e) u [(C

(D n B), and

X

n Ae)

(D n B)] u [(C n A)

X

(D n Be)]

X

(D n Be)]

hold with every member of the previous union having finite measure. Now, the subadditivity of (J.L x v)* , combined with Theorem 26.2, yields

(J-L

X

v)*((C

X

D) n (A

X

B)) + (J-L

X

v)*((C

X

D) n (A

< J.L*(C n A) · v*(D n B ) + J.L*(C n Ae) v*(D n B)

X

B)e)

·

+ J.L*(C n A) · v *(D n Be) + J-L*(C n A e) · v*(D n Be)

= [J.L*(C n A) + J.L*(C n A e)] · [v*(D n B ) + v*(D n Be)] = J.L(C) · v(D)

=

J.L x v(C

x

D),

as required.

•

In general, it is not true that the measure JJ-* x v * is the only extension of J.L x v from S ® :E to a measure on AJ.L ® A v . However, if both (X, S, J.L) and (Y, :E, v) are a-finite measure spaces, then (X x Y, S ® I;, J-L x v) is likewise a a-finite measure space, and therefore, by Theorem 15.10, f.l* x v * is the only extension of J.L x v to a measure on the semiring AJ.L ® Av. Moreover, in view of AJ.L ® A v C AJ.Lxv and the fact that (J.L x v)* is a measure on AJ.Lxv, it follows in this case that (J.L x v)* = J-L* x v * holds on AJ.L ® A v . We now tum our attention to measurability properties of arbitrary subsets of X x Y. Let us recall a few definitions from Section 20. If A is a subset of X x Y Ax of A is the subset of Y defined by and x E X, then the

x-section

Ax = {y E Y : (x, Similarly, i f y E Y , then

y) E A } .

the y-section A Y o f A is the subset of X defined by AY = {x E X : (x, y) E A } .

The geometrical meanings of the x- and y-sections are shown i n Figure 4. 1 . Regarding sections of sets, we have the following identities.

Chapter 4: THE LEBESGUE INTEGRAL

208 )'

X

X

FIGURE 4.1.

The proofs of the previous identities are straightforward, and they are left as an exercise for the reader. The next theorem demonstrates the relationship between the J.L x v-measurable subsets of X x Y and the measurable subsets of X and Y, and it is a key result for this section. Recall that an extended real-valued function f that is undefined on a set of measure zero is said to define an integrable function if there exists an integrable function g such that f = g almost everywhere. That is, if arbitrary values are assigned to f on the points where it is undefined or attains an infinite value, then f becomes an integrable function. (The value of the integral does not depend, of course, upon the choices of these values.)

Let E be a J.L x v-measurable subset of X x Y satisfying (J.L x v)*(E) < oo. Then for J.,L-almost all x the set Ex is a v-measurable subset of Y, and the function x 1--7 v*(Ex) defines an integrable function on X such that

Theorem 26.4.

(J.L x

v)*(E) =

fx v*(Ex) dJ.,L(X).

Similarly,for v-almost all y, the set EY is a J.,L-measurable subset of X , and the function y J.L *(E Y) defines an integrable function on Y such that 1--7

(J.L x v) *(E) =

l J.L*(EY) dv(y).

209

Section 26: PRODUCT MEASURES AND ITERATED INTEGRALS

By the symmetry of the situatiQn, it suffices to establish the first for­ mula. The proof goes by steps.

Proof.

= A B E S :E. = B x xE EX,A =

Step I. Assume

x

E

if Clearly, Ex subset of Y for each

® and Ex and

(j) if x fj; A. Thus, Ex is a v-measurable

holds for all x E X. Since (!J- x v)"'(E) = (!J- x v)(A x B) = !J-(A) · v(B) < oo, two possibilities arise:

n

a.

Both A a d B have finite measure.

In this case, (*) shows that x 1--+ v*(Ex) is an integrable function (actually, it is a step function) satisfying

b.

l

v*(Ex) dfJ-(X) =

I

v(B)XA df-L = fJ-(A)

·

v(B) = (/-L x v)*(E).

Either A or B has infinite measure.

In this case, the other set must have measure zero, and so, (*) guarantees Thus, 1--+ v*(Ex) defines the zero function, v(E.. J = 0 for !J--almost all and hence

l

x.

x

== 0

v*(E.J dfJ-(X)

(/-L x v)*(E).

Step II. Assume that E is a a-set of S ® :E. Choose a disjoint sequence {E11} of S ® :E such that E U: 1 En. In view of Ex U:1 (E11).,. and the preceding step, it follows that Ex is a measurable subset of Y for each x E X. v *(Ex) and :L�'= t v((E;)x) for each x E X and all Now, define f(x)

=

n . By Step I, each

= J,,

If,, = t;II jr df-L

=

J,,(x) =

defines an integrable function and

X

= t;f-L II

v((E;)_,J d!J-(X)

X

v(E;)

t

(/-L

=

X

v)*(E) < oo.

Since {(E,).� } is a disjoint sequence of :E, we have v*(Ex) I:: 1 v((En)x). and so, fn(x) t f(x) holds for each x E X. Thus, by Levi's theorem (Theorem 22.8), f d�fines an integrabl� _f�nctio�. and

{ v*(Ex) dfJ-(X) = I f df-L =

lx

lim

ll-+oo

If,,

df-L =

� fJ8

X

v(E;)

=

(/-L

X

v)*(E).

210

Chapter 4: THE LEBESGUE INTEGRAL

Step l/1. Assume that E is a countable intersection of a-sets offmite measure. Choose a sequence {En} of -sets such that E n: l En, (Jl X \) )*(E I ) < 00, and En+ l C E11 for all n.

=

a

=

For eachn, letg11(X) = O if v*((En)...- ) = oo and g11(X) v*((En)x ifv*((E11)x < oo. By Step II, each g11 is an integrable function over X such that g11 d J1 = (Jl X v)*(E,) holds. In view of Ex = n:.l (En)x. it follows that E_,. is a \)­ measurable set for each x E X . Also, since v*((E1)x) < oo holds for jl-almost all x, it follows from Theorem 15.4 that gn(x) v*((E,Jx) {. v*(Ex) holds for J.L-almost all x. Thus, x H- v*(Ex) defines an integrable function and

J

=

}r v*(Ex) dJl{X) = nlim -oo X

! gil dJ1 = lim (Jl n-oo

X

v)*(En)

= (Jl

X

v)*(E),

where the last equality holds again by virtue of Theorem 1 5.4. Step N: Assume that E is a null set, i.e., (Jl x v)*(E) = 0. Arguing as in the proof of Theorem 15. 1 1 , we see that U"iere exists a measurable set G, which is a countable intersection of a-sets of finite measure, such that E c G and (Jl x v)*(G) = 0. By Step Ill, fx v*(Gx)dJ1(x) = (Jl x v)*(G) = 0, and so, by Theorem 22.7( 1), v*(Gx) 0 holds for jl-almost all x . In view of E.,. c G.,. for all x , we must have v*(Ex) 0 for jl-almost all x. Therefore, Ex is v-measurable for J.L-almost all x and x H- v*(Ex) defines the zero function. Thus,

= =

i v*(Ex) dJl(X) =

0 = (Jl

x

v)*(E).

e a Choose a v-measurable set F that is a countable intersection of -sets all of finite measure such that E F and (Jl v)*(F) = (Jl v)*(E). Let G = F \ E. Then G is a null set, and thus, by Step v*(G ) = 0 holds for jl-almost all x. Therefore, Ex is v-measurable and v*(Ex) = v*(Fx) holds for Jl-almost all x. By x H- v*(Fx) defines an integrable function, and so, x Step v*(Ex) defines Step V. The g ne l case. J1 x

r

c

x

IV,

x

ITI,

H-

an integrable function and (J.L x

v)*(E)

= (Jl

x

v)*(F)

= i v*(Fx) dJ.L(X) = i v*(Ex) djl(X)

holds. The proof of the theorem is now complete.

�

x

a

•

Now, let f : X x Y IR be a function. Then for each fixed x E X, the IR defined by f-:(Y) = f (x , symbol f-: will denote the function fx : Y for all y E Y. Similarly, for each y E Y the notation fY denotes the function IR defined by fY(x) = f(x, y) for all x E X. JY : X

�

�

y)

Section 26: PRODUCT MEASURES AND ITERATED INTEGRALS

Definition 26.5. Let f : X x Y

-+ JR. be afunction.

211

Then the iterated integral

JJf df.J.dv is said to exist l/ 1 . fY is an integrable function avec X for v-almost ally, at1d 2. the function g (y)

= f f>' dJ.L = l f(x, y) djJ.(X)

defines all imegl·able function over Y. The value of the itemted integml JJf df.J.d v is computed by starting with the innermost integration and by continuing with the second as follows:

holding y fixed while tile integmtioll over X is pelformed. The meaning of the iterated integral

JJf dvdf.J.

is analogous. That is,

If E is a J.L x v-measurable subset of X x Y with (J.L x v )*(£) < oo, then it is readily seen from Theorem 26.4 that both iterated integrals JJX E df.J.dv and JJ XE dvdf.J. exist, and that

ff

X£

dj.LdV

= ff

XE

dvdjJ.

=f

XE

d(J.L

X

v)

= (f.J.

X

v)*(£).

Since every J.L x v-step function is a linear combination of characteristic functions of fJ. x v-measurable sets of finite measure, it follows from the prior observation that if ¢ is a fJ. x v-step function, then both iterated integrals JJ¢ dJJ.d v and JJ¢ d vdfJ. exist and, moreover,

JJ¢ dJ.Ldv = JJ¢ dvdf.J. = J¢ d(J.L x

v).

The previous identities regarding iterated integrals are special cases of a more general result known as Fubini 's 1 3 theorem. 13Guido Fubini (

1879-1943),

an Italian mathematician. He made important contributions in analysis,

geometry, and mathematical physics.

212

Chapter 4: THE LEBESGUE INTEGRAL

Let f : X x Y -+ Then both iterated integrals exist and Theorem 26.6 (Fubini).

Jf

d(p, x

v)

IR be a

p, x

v-integrablefunction.

= JJ f dp,dv = JJf dvdp,.

Proof. Without loss of generality we can assume that f (x, y) > 0 holds for all

x and y.

Choose a sequence {t/>n } of step functions such thatO < holds for all x and y. Thus,

t/>11(x, y) t f(x , y)

By Theorem 26.4, for each n the function

8n(X) = J(t/>n).r dv = i

t/>11 (X , y) dv(y)

defines an integrable function over X; and clearly, 11 (x ) t holds for p,-almost all x. But then, by Levi's Theorem 22.8, there exists a p,-integrable function g : X -+ IR such that g11 (x ) t g(x) p,-a.e. holds. That is, there exists a p,-null subset A of X such that dv t g(x) < oo holds for all x ¢. A. Since (t/>n).\. t fx holds for each x, it follows that fr is v-integrable for all x ¢. A and

g

f{¢n}r

g"(x) = J(t/>ll)x dv = i t/>"(x,

y) dv(y) t

i fx dv

holds for all x ¢. A. Now, (*) combined with Theorem 21.6 implies that the function defines an integrable function such that

Similarly, J f d(p,

X

v)

= JJfdp,dv,

x � frfx dv

and the proof is complete.

•

The existence of the iterated integrals is by no means enough to ensure that the Y [0, 1], function is integrable over the product space. For instance, let X 2 2 2 I. (the Lebesgue measure), and f(x, y ) p, v (x - y )j(x + y2)2 if

= =

=

= =

Section 26: PRODUCT MEASURES AND ITERATED INTEGRALS

(x, y) =I= (0, 0) and f (0, 0)

=

213

0. Then, it is �asy to see that

JJ f djJ.dv = : -

and

JJ f dvdjJ. = :

.

Fubini 's theorem shows, of course, that f is not integrable over [0, I ] x [0, I]. However, there is a converse to Fubini 's theorem according to which the exis­ tence of one of the iterated integrals is sufficient for the integrability of the function over the product space. The theorem is known as Tonelli's theorem, and this result is frequently used in applications.

14

Theorem 26.7 (Tonelli). Let (X, S, J.l.) and (Y, :E , v) be two a-finite measure

spaces, and let f : X x Y -+ IR be a J.1. x v-measurable function. If one of the iterated integrals JJ Ill djJ.dv or JJ If I dvdJJ. exists, then the function f is JJ. x v -integrable--and hence, the other iterated integral exists and

J f d(J.l. x v) = JJ f djJ.dv = JJ f dvdjJ..

Proof. We can assume without loss of generality that f(x, y) > 0 holds for

all x and y. S ince (X, S, J.l.) and (Y, :E, v) are a-finite measure spaces, it is easy to see that the product measure space is also a a-finite measure space. Choose a sequence {A,} of JJ. x v-measurable sets such that (JJ. x v) * (A,) < oo for each n and A, t X x Y. By Theorem 17.7, there exists a sequence {1/t,} of J.1. x v­ simple functions such that 0 < 1/t,(x, y) t f(x, y) holds for all x and y. Let ¢, 1/t, · XA" for each n. Then {¢,} is a sequence of J.1. x v-step functions such that 0 < cp,(x, y) t f(x, y) holds for all x and y. Now, assume that JJ f djJ.dv exists. This means that for v-almost all y, the integral f(x, y) djJ.(x) exists and defines a v-integrable function. From ¢, (x. y) t f(x, y) it follows that ¢"(x, y) djJ.(x) t J f(x, y) dJJ.(x) holds for v-almost all y. But then, by applying the Lebesgue dominated convergence theo­ rem, we get

=

J

J

J,. d(p. v) = i [i
< oo.

v-upper function and that This shows that is a holds. The rest of the proof now follows immediately from Fubini 's theorem.

The Fubini and Tonelli theorems are usually referred to as "the method of computing a double integral by changing the order of integration." 14 Leonida Tonelli ( 1885-1946), an IIalian maihemaiician. He contribuled Io measure Iheory, Ihe Iheory of iniegraiion. and Io calculus of variaiions.

Chapter 4: THE LEBESGUE INTEGRAL

214

In general, it is a difficult problem to determine whether or not a given function IR is J.1 x v-measurable. However, in a number of applications the f:X x Y J.1 x v-measurability off can be established from topological considerations. For instance, if X Y = IR and J.1 = v =the Lebesgue measure, then it should be the product measure J.1 x v on IR2 is precisely the Lebesgue measure clear that 2 on IR . Therefore, every continuous real-valued function on IR.2 is necessarily J.1 x v-measurable. For more about the joint measurability of functions, see also Section 20.

�

=

EXERCISES 1.

Let (X, S, J.L) and (Y, �. v) be two measure spaces, and let A x B E

AJJ. ® Av.

a. Show that J.L*(A) · v*(B) ::: (J.L x v)*(A x B). b. Show that if J.L*(A) · v*(B) ::p 0, then (J.L x v)*(A x B) = J.L*(A) · v*(B). c. Give an example for which (J.L x v)*(A x B) ::p f.L*(A) · v*(B).

2. Let (X,.S, J.L) and (Y, �. v) be two cr-finite measure spaces. Then show that

(J.L x v)*(A

x

B) = J.L*(A) · v*(B)

holds for each A x B E AJJ. ® Av. 3. Let (X, S, J.L) and (Y, �. v) be two measure spaces. Assume that A and B are subsets of X and Y, respectively, such that 0 < J.L*(A) < oo, and 0 < t.•*(B) < oo. Then show that A x B is f.L x 11-measurable if and only if both A and B are measurable in their corresponding spaces. Is the above conclusion true if either A or B has measure zero? 4. Let (X, S, J.L) and (Y, �. v) be two cr-finite measure spaces, and let f : X x Y � IR be a Jl x v-measurable function. Show that for J.L-almost all x , the function f'" is a v-measurable function. Similarly, show that for v-almost all y, the function JY is J.L-measurable. 5. Show that if f(x, y) = (x2 - y2) j(x 2 + y2)2 , with /(0, 0) = 0, then

6. 7.

fo' [ fo'

]

/(x , y) dx dy = -

and

fo' [ fo 1

f(x ,

}

y) d)

x=

:

Let X = Y = IN, S = � = the collection of all subsets of IN, and f.L = v = the counting measure. Give an interpretation of Fubini 's theorem in this case. Establish the following result, known as Cavalieri's 1 5 principle. Let (X, S, J.L) and (Y, �. v) be two measure spaces, and let E and F be two J.L x v-measurable subsets of X x Y of finite measure. If v*(Ex) = v*(F.t) holds for J.L-almost all x. then (J.L

8.

:

x

v)*(E) = (J.L

x

v)*(F).

For this exercise, 'A denotes the Lebesgue measure on IR. Let (X, S, J.L) be a cr-finite measure space, and let f : X � IR be a measurable function such that j(x) > 0

15Bonaventura Cavalieri ( 1 589-1647). an Italian mathematician. He was a geometer who worked on problems regarding volumes of solids and wrote several monographs on this subject.

Section 26: PRODUCT MEASURES AND ITERATED INTEGRALS holds for all

E X. Then show that

x IR: 0 < y f(x)}, J1. x x IR. = ((x, y) x IR: 0 < f(x)} x x IR (J.l x x = ((x, f(x)) :x } J1. x x IR. x f dJ.l x =0 Y -+ IR g -+ IR f x Y -+ IR f(x, = )h f J1. x j fd(J.l x v) = (LgdJJ.) ([ h dv) .

a.

b.

c.

d. e.

9.

x

215

f

The set A = ((.Y, y) E X � called the ordinate set of , is a A.-measurable subset of X The set B EX is a J1. A.-measurable subset of � y X and A.)*(A) = (JJ. A.)*(B) holds. E X , is a The graph of f, i.e., the set G A.-measurable subset of X holds. If f is JJ.-integrable, then (JJ. A.)*(A) = J If f is JJ.-integrable, then (JJ. A.)*(G) holds.

be a JJ.-integrable function, and let Jz : be a v-integrable Let : X function. Define : X by y) g(x (y) for each x and y. Show that is v-integrable and that ·

10.

1r

rl oo (1r e-.\y. . x dx ) dy = d o --;- x 0 < E < r. E -+ o+ r -+ 00 loco --dx = -. f(x, ye-( l +x2>.v2

Use Tonelli's theorem to verify that sinx

sm

f

f

holds for each By letting steps) give another proof of the fonnula

11.

Show that if

0

y) =

and

sinx X

(and justifying your

rr

2

for each x and y, then

Use the previous equality to give an alternate proof of the fonnula

roo .J7i oJ e dx = --z · -x

12.

Show that

Iaoo (fo r > 0.

r

holds for all

e-xy

2

2

.

dx) dy = Ia (Ia e-xy2 dy) dx r -+ oo roo :c = � d 2 . lo ../X f000 cJ-jr d:c = J2ii ,

sinx

By letting

oo

sin x

show that

sinx

In a similar manner show that 13.

/2.

Using the conclusions of the preceding exercise (and an appropriate change of vari­ able), show that the values of the Fresnel integrals (see Exercise 6 of Section 24) are

� loco sm(x-)dx = fooo cos(x )dx = --. 0

.

?

0

2

4

Chapter 4: THE LEBESGUE INTEGRAL

216

J..L = the Lebesgue measure on [0, 1], and v = the counting measure on [0, 1]. Consider the "diagonal" fl. = {{x, x) : x E X} of X x Y. Show that:

14. Let X = Y = [0, 1],

a. b. c.

fl. is a J..L x v-measurable subset of X x Y , and hence, Xt. is a non-negative J..L x v­ measurable function. Both iterated integrals JJ Xt. dJ..Ldv and JJ Xt. dvdJ..L exist. The function x t. is not J..L x v-integrable. Why doesn't this contradict Tonelli's theorem?

15. Let f : IR � IR be Borel measurable. Then show that the functions f(x + y) and f(x - y) are both >.. x >..-measurable. [HINT: Consider first f = x v for some open set V .]

C HAPTER

5

__ _ _ _ _ _ _ _ _ _ _ _ _

NORMED SPACES AND Lp-SPACES The algebraic theory of vector spaces has been an integral part of modem mathe­ matics for some time. In analysis, one studies vector spaces from the topological point of view, taking into consideration the already existing algebraic structure. The most fruitful study comes when one attaches to every vector a real number, called the

norm of the vector. The norm can be thought of as a generalization of

the concept of length. A normed space that is complete (in the metric induced by its norm) is called a Banach 1 space.

A variety of diverse problems from different branches of mathematics (and science in general) can be translated to the framework of Banach space theory and solved by applying its powerlul techniques. For this reason the theory of Banach spaces is in the frontier of modem mathematical researc h. This chapter presents a brief introduction to the theory of normed and Banach spaces. After developing the basic properties of normed spaces, the three comer­ stones of functional analysis are proved-the principle of uniform boundedness, the open mapping theorem, and the Hahn-Banach theorem.2 Then a section is devoted to the study of Banach lattices; that is, Banach spaces whose norms are compatible with the lattice structure of the spaces. As we shall see, many Banach

lattices are actually old friends. Finally, the classical, Lp-spaces are investigated,

and the theory of integration is placed in its appropriate perspective.

27. NORMED SPACES AND BANACH SPACES A real-valued function

11·11

defined on a vector space

satisfies the following three properties:

X

is called a

norm

if it

1 Stefan Banach ( 1892-1945), a prominent Polish mathematician. He is the founder of the contem­

porary field of functional analysis. 2Hans Hahn (1879-1934), an Austrian mathematician and philosopher. He contributed decisively to functional analysis, general topology, and the foundations of mathematics.

217

Chapter 5: NORMED SPACES AND Lp·SPACES

218 I.

2. 3.

=

0 if and only if x ll x ll > 0 for each x e X , and ll x ll Ia ! · ll x ll for all x E X and a e JR.; ll ax ll llx + Yll < ll x ll + II YII for all x, y E X.

=

=

0;

Property (3) is referred to as the triangle inequality, and it is equivalent to the inequality ll x - Yll .:S ll x - zll + liz - Yll

for all x, y, and z in X . A vector space X equipped with a norm 11·11 is called a normed ''ector space, or simply a normed space. To avoid trivialities, the vector spaces will be tacitly as­ sumed to be different from {0}. Also, they will be assumed to be real vector spaces. On a normed space X, a metric is defined in terms of the norm 11·11 via the function d (x, y) ll x - y 11. From the properties of the norm, it is a routine matter to verify that d ( ·, ·) is indeed a metric on X . We shaH caH this metric on X t.'ie metric induced by the norm. A sequence {x,} in X is said to converge in norm to x if lim llx - x, II 0; that is, if {x,} converges to x with respect to the metric induced by the norm. In view of the triangle inequality,

=

=

l ll x II - ll y Il l

<

ll x - Yll

holds for all x and y in X. This readily implies that the norm considered as a function x r+ llx II for X into 1R is uniformly continuous. Also, by the triangle inequality it is easy to prove, but important to observe, that if x, x and y, -+ y in X and a, -+ a in JR., then x, + y, -+ x + y and a,x, -+ ax hold. A subset A of a normed space is said to be norm bounded (or simply bounded) if there exists some M > 0 such that ll x II < M holds for each x e A . Every Cauchy sequence {x,} of a nonned space is bounded. Indeed, to see this choose k such that ll x, - Xm II < I for all n, m > k, and let M max { I + ll x; ll : I < i < k}. Then the inequality ll x1111 < I + ll xk II for n > k implies ll x, II < M for all n. A nonned space X that is complete with respect to the metric induced by its norm is called a Banach space. In other words, X is a Banach space if for every Cauchy sequence {x,} of X there exists an element x e X such that lim ll x, - x II 0. Thus, Banach spaces special examples of complete metric spaces. Some examples of Banach spaces presented next.

�

=

are

are

=

=

o=�l= l Xf)� for each X = Example 27.1. The vector space R" with the norm llxll (x t , . . . , x11) E R" is a Banach space. This nonn is called the Euclidean norm, and it gives

the Euclidean metric. Example 27.2.

=

•

Let X be a nonempty set, and let B(X) be the vector space of all bounded real-valued functions defined on X. Then llfii"Xl sup{lf(x)l: x E X} for each f E B(X)

219

Section 27: NORMED SPACES AND BANACH SPACES

defines a norm on B (X), called the sup norm .. A glance at Example 6.12 guarantees that • the vector space B(X) with the sup norm is a Banach space.

. . .) such Example 27 3 Let e 1 denote the collection of all real sequences x = (..q , that :L� 1 lx, I < oo. It is easy to see that e 1 is a vector space under the algebraic operations .

x2 ,

.

x + y = (..q

+ Yl ,

x2 + Y2· . . . )

and

ax = (c:u 1 , ax2, . . ) . .

Moreover, if for every X E e I we define llx Il l = L� I lx, I' then 11·11 I is a norm on e I . The verification of the norm properties are left for the reader. We show next that f. 1 is actually a Banach space. To this end, let {x, J be a Cauchy sequence of e 1 . That is, for every E > 0 there exists some k such that llx, Xm Il l < E holds for all n, m > k. It follows that there exists some M > 0 such that Ux, II 1 � M for each n. Let x, = x2, . . . ) for each n. The relation

-

Cxl' ,

j x}' - xj" I � L jxj' - xj'j = llx, - x,ll 1 i=l 00

<E

for 1z, m > k implies that for each fixed i, the sequence of real numbers {xj' } is a Cauchy sequence. Let x; = lim,_.., 00 xj' for each i. Since for all n > k and each p we have p

p

p

p

i=l

i=l

i=l

i=l

L lx; I < L jx; - x}' I + L jxj' I � L lx; -xj' I + llx, II 1

(

� 1 1-+00 1=1 � lx;n - xj' I

= 1 lim

)

+

llx, II 1

< E + M < oo ,

= (.q ' X2, . . . ) E e I . Also, since :Lf= I lx;' - xj' l � llx, - Xm II I < E for all n , m > k, we have 'Lf=1 lx; - xjl = limm-+ oo C'Lf=l l xj - xjl) < E for all p and all n > k. Hence, llx - x11 11 1 < E holds for all n > k, so that {x11} converges to x in e 1 . That it follows that X

is, e I is a Banach space.

•

Example 27.4. Consider an interval [a, b] and a natural number k. Let Ck [a, b] denote the collection of all real-valued functions defined on [a, b] which have a continuous kth order derivative on [a, b] (with right and left derivatives at the endpoints). Clearly, c k [a, b] with the pointwise algebraic operations is a vector space. Moreover, if for each f E C k [a. b] we let

11!11 = llflloo + llf ' llco + · · · + llf(k)lloo. k then 11·11 defines a norm under which c [a, b] is a Banach space. k It is straightforward to verify that 11·11 is indeed a norm. To see that C [a, b] is complete k under 11·11, let {j;,} be a Cauchy sequence of C [a, b]. Then, it is easy to see that there exist continuous functions go, g 1 , . . . , 8k such that for each i = 0, l , . . . , k the sequence

Chapter 5: NORMED SPACES AND Lp·SPACES

220

of functions

{/�0} converges uniformly to

holds for eachx E from (*) that

[a,

gi.

Now, if 1 ::; i ::; k, then

b]. Thus, by the uniform convergence of the sequence {f,�0} it follows

gi- l (x) =gi-l (a) + 1xg;(t) dt holds for each x E (for 1

::;

=

[a, g;_ 1 (x) =

b]. By the fundamental theorem of calculus,

gi-l

is differentiable

g;(x) for each E b]. In particular, note i < k), and 1 , . . . , k. Therefore, go E b] and lim ll fn o ll = 0 holds, so that

for i is a Banach space.

x [a,

Ck[a,

-

g

that g; = gg> Ck[a,

b]

•

Two norms ll · ll t and 11·11 on a vector space X are said to be equivalent if there 2 exist two constants K > 0 and M > 0 such that

holds for each x E X. The reader should stop and verify that two norms are equivalent if and only if they generate the same open sets. In particular, note that if two norms on a vector space X are equivalent, then X is a Banach space with respect to one of them if and only if X is a Banach space with respect to the other. Also, observe that if II· II 1 is equivalent to II ·11 2 and II · 112 is equivalent to II · 113. then ll · l l t is equivalent to II· 113· In a normed vector space, all open balls "look alike." More precisely, any two open balls are homeomorphic. Indeed, if B(a, r) is an arbitrary open ball, then it is easy to see that the function x � a + rx is a homeomorphism from B(O, 1) onto B(a, r). For this reason, the open ball B(O, 1) = {x : llxll < 1 } plays an important role in the study of normed spaces and is called the open unit ball of the space ( and likewise {x: ll x II < 1 } is called the closed unit ball of the space).

Example 27.5.

It is instructive to consider various norms on 1R2. For v

define llvi!J = lxl llvll 2

+ IYI.

= ( ·2 + y2)! , x

llvlloo = max{lxl, jyl},

=

(x, y) E 1R2

Section 27: NORMED SPACES AND BANACH SPACES

l v II

I

..

221

=

FIGURE 5.1. The Unit Balls of Various Norms

where a and b are two, fixed, posi tive real numbers. The reader should stop and verify that the previous functions are norms and that they are, in fact, all equivalent. The geometric shape of the closed unit ball for each norm is illustrated in Figure 5.1. •

The fact that all norms in the preceding example were equivalent is not acci­ dental.

In afinite dimensional vector space all norms are equivaient. Proof. JR". l ·l z l x l 2 = (Xf · · x;) � . l ·llxz�= (x 1, x,) = x1e1 · x,e11 e11

Theorem 27.6.

Without loss of generality, we can assume that the finite dimensional denote its Euclidean norm; that is, + · + space is Let Also, let 11·1 1 be another norm on lR2 . It suffices to establish that 11·11 is equivalent to If + ··+ • • • , (where the are the standard basic unit vectors), then by the triangle inequality we have II

e X l xl = L ; ; = i l Thus, if M =

E�'=1

l edl.

then

Chapter 5: NORMED SPACES AND Lp-SPACES

222 holds for all x E the inequalities

IR". This establishes one-half of the desired inequality. In addition, l llx ii - IIYII I < llx - yll < M ll x - y lb

show that the function x r+ llx II from IR" with the Euclidean norm into IR is (uniformly) continuous. Let S = {x E IR": llx 112 = 1 } be the "unit sphere" for the Euclidean norm. Then S is closed and bounded, and hence, by the Heine-Borel Theorem 7 .4, S is compact for the Euclidean norm. In particular, the continuous function x r+ ll x II attains a minimum value on S, say at x0• Thus, llxll > llxoll holds for all x E S. Let K = llxoll· Since llxoll2 = 1, it follows that xo =I= 0, and so K = llxoll > 0. Now, if x E IRn is nonzero, then llxll/llxll2 = llxfllxlbll > K , and so

K llx ll2 < llx II holds for all x E IR". This establishes the other half of the desired inequality and !I completes the proof of the theorem. As an application of the last theorem, let us establish the following useful result:

Theorem 27.7.

is closed.

Eve1y finite dimensional vector subspace of a normed space

Proof. Let Y be a finite dimensional vector subspace of a normed space X.

Then Y can be identified (linearly isomorphically) with some IR". Hence, by Theo­ rem 27 .6, the norm of X restricted to Y must be equivalent to the Euclidean norm. In particular, Y is a complete metric space, and so (by Theorem 6 . 1 3) Y is closed. •

The structure of the open (or closed) unit ball reflects the topological and geo­ metrical properties of all balls. For example, a normed space is locally compact if and only if its closed unit ball is compact. (Recall that a topological space is locally compact if each point has a neighborhood whose closure is compact.) The next theorem tells us that the finite dimensional normed spaces are the only locally compact normed spaces.

Theorem 27.8. A

dimensional.

normed space is· locally compact if and only if it is finite

Proof. If the normed space is finite dimensional, then by Theorem 27 .6, its

norm is equivalent to the Euclidean norm. By Theorem 7.4, the closed unit ball is compact with respect to the Euclidean metric, and from this it follows (how?) that the space is locally compact.

Section 27: NORMED SPACES AND BANACH SPACES

223

For the converse, assume that X is a locally compact nonned space. Since all < 1} must E X: closed balls are homeomorphic, the closed unit ball V , X11 E V such that V C U�'= 1 B(x;, � and let Y be be compact. Choose XJ, X11 }. We shall show that X = Y, and this the linear subspace generated by will establish that X is fin ite dimensional. Assume by way of contradiction that X =F Y . Thus, there exists some EX the vector subspace Y is closed, it follows with x0 � Y . Since (by Theorem E Y} > 0 (also, see Exercise 1 of Section 10). that Y ) = inf x < Y). Then for some 1 < i < n we have Pick some E Y with belongs to Y , it follows that B(x; , � ). Since + •

•

•

{x1,

•

•

•

= {x

,

llxll ) .

x0

27.7)

d(x0,

{ll o yll: y llxo - yll 2d(xo, y (xo - y)fllxo - Yll E y llxo - yllx;

xo - y x; = ll xo - (y + llxo - yllx;)ll > d(xo, Y ) > -1 - llxo - Y ll 2 llxo - Yll llxo - Yll -

1 ->

---

2

which is impossible. This contradiction completes the proof of the theorem.

•

EXERCISES 1.

Let

X

be a normed space. Then show

sphere

2.

{.t =

E X: llx II =

d(x, y) llx - .vii).

X be

a.

Show that the mappings .t 1-+

a normed vector space. Fix

If A and B

3.

4.

X Let X

Let

a E X and a nonzero scalar a. .

a+

a. A + f3B

is an open set.

be a normed vector space, and let B

Show that B

= {x

E X: ll.tll � 1}.

a. and {3 are nonzero scalars,

E X: l .t II < be its open 1}

unit ball.

let {.t11 } be a sequence of X sue� that lim Xn = .t 1 holds. If y1 = n- (XJ + · · · + x11) for each then show that limy11 = x. (See also

of Section 4.) 5. Assume that two vectors x

n,

and y in a normed space

Then show that

7.

= {.t

isms.

be a normed space, and

Exercise 1 1

6.

morph

x and x 1-+ ax are both homeo

are two sets with either A or B open and

then show that

if its unit induced metric

is a Banach space if and only

1 } is a complete metric space (under the

Let

b.

that X

satisfy llx + yll = Ux ll + IIY II ·

ll a.x + /3yll = a.llxll + f311Y II holds for all scalars a. > 0 and {3 � 0. Let

X

be the vector space of all real-valued functions defined on [0, 1] having contin­

11/11 = 1/(0)1 + 11/'llco is a norm on X that is equivalent to the norm 11/lloo + 11/' ll oo· A series L:�1xn in a normed space is said to converge to x if limllx - E7=1x; ll = uous first-order derivatives. Show that

0. As usual, we write summable if L�=

x = L:� 1 x11• A series L:�1xn

1 llxnll < oo holds.

Show that a normed space

X is

is said to

be absolutely

a Banach space if and only if every absolutely

summable series is convergent.

8.

Show that a closed proper vector subspace of a normed vector sp�ce is nowhere dense.

Chapter 5: NORMED SPACES AND Lp·SPACES

224

9.

Assume that f: [0, 1] ---+- IR is a continuous function which is not a polynomial. By Corollary we know that there exists a sequence of polynomials {Pn } that converges uniformly to f. Show that the set of natural numbers

11.6,

{k E lN': k = degree of Pn for some n } is countable. [mNT: Show that the vector subspace of ] consisting of all polynomials of degree less than or equal to some m is closed.] This exercise describes some classes ofimportant subsets of a vector space. A nonempty subset of a vector space X is said to be:

C[O, 1

10.

A

=-A;

symmetric, if x e A implies -x e A, i.e., if A convex, if x, y E A implies /..x + - A.)y E A for all 0 =::: /.. =::: two vectors x, y E A the line segment joining x and y lies in A; c. circled (or balanced) if x E A implies /..x e A for each l A. I :::

a. b.

(1

1,

i.e., for every

1.

Establish the following: 1.

ii.

m.

iv.

11.

A circled set is symmetric. A convex and symmetric set containing zero is circled. Anonemptysubset B ofavectorspaceisconvex ifandonly ifaB+bB = (a+b)B holds for all scalars a ::;: 0 and b > 0. If is a convex subset of a normed space, then the closure A and the interior A 0 of A are also convex sets.

A

This exercise describes all norms on a vector space X that are equivalent to a given norm. So, let (X, 11·11) be a normed vector space. Let A be a bounded convex symmetric subset of X having zero as an interior point (relative to the topology generated by the norm 11 · 11). Define the function PA: X ---+- IR by PA (X)

=

inf(/.. > 0:

X E /..A } .

Establish the following: a. b. c. d.

The function p A is a well-defined not:m on X. The norm PA is equivalent to 11·11 , i.e., there exist two constants > 0 and K > 0 such that C llx II < pA (x) =::: K llx II holds true for each x E X. i.e., {x e X : pA (x) < 1 } = A. The closed unit ball of p A is the closure of Let 11 1 · 11 1 be a norm on X which is equivalent to 11·11 , and consider the norm bounded nonempty symmetric convex set B = (x e X: lllxlll ::: 1 } . Then zero is an interior point of B and lllx Il l = pB (x) holds for each x e X .

C

A,

28. OPERATORS BETWEEN BANACH SPACES In this section, we shall study linear operators (or transformations). Recall that a function T: X ---+ Y between two vector spaces is called a linear operator (or simply an operator) if T(ax + {Jy) = aT(x) + {JT(y) holds for all x, y E X, and all a, fJ E R. Observe that every linear operator T satisfies T(O) = 0.

Section 28: OPERATORS BETWEEN BANACH SPACES

225

If X and Y are two normed spaces, then th_e symbol 11·11 will be used (as usual) to denote the norm on both spaces.

Y be a linear operator between two normed Let T: spaces. Then the operator norm of T is defined by Definition 28.1.

X �

li T II = sup{IIT(x) ll: llxll = I } .

If

isthen finite,T isthencalled ians called a bounded operator (and, of course, unbounded operator). T

li T II liT II = oo,

if

Observe that

IIT(x)ll < II T II · IIxll holds for all x E X. To see this, note that if x f= 0, then the vector y = x/llxll satisfies II II = l, and so

y

IIT(x)ll llxll

_

T

( ) x

llxll

= IIT(y )ll < li T II

holds. In particular, it follows from the previous inequality that li T II = sup{II T (x) ll: llxll < 1 } . Next, some concrete examples of operators are presented.

X X.

X X 1,

Example 28.2. Let = C[O, 1] with the sup norm. Define T : � by T(f)(x) = xf (x) for each f E X and x E [0, 1]. Clearly, T is a linear operator such that IIT(/)1100 < II f II 00 for each f E This shows that II T II .5 1 (in fact, II T II = ) and so T is a bounded operator. •

The next example presents

be

an unbounded

operator.

Example 28.3. Let X the vector space of all real-valued functions on [0, 1] that have continuous derivatives with the sup norm. Also, let Y = C[O, 1] with the sup norm. Define D: � Y by D(f) = f' (the differential operator). It is easy to see that D is a linear operator. However, liD II = oo. Indeed if f, (x) = xn, then 11/11 lloo = 1 and II D(/11)1100 = sup{11x11- 1 : x E [0. 1]} = n hold for each n, implying II D II = oo. Hence, D is an unbounded • operator.

X

,

The next example is drawn from the important class of operators known as "integral operators."

Example 28.4. Let [a, b] be a (finite) closed interval, and let K : [a, b] x [a, b] � lR be a continuous function. Consider the vector space C [a, b] with the sup norm, and then define

226

Ch apter 5: NORMED SPACES AND Lp·SPACES

T: �[a, b] --+ C[a, b] by

T(j)(x)

=

ih

K(x, y)f(y)dy

for each f E C[a, b]. (The unifonn continuity of K on [a, b] x [a, b] can be invoked here to verify that indeed T(j) E C[a, b] for each f E C[a, b].) For obvious reasons, the operator T is called an integral operator and the function K is referred to as the kernel of T. Clearly, T is a linear operator. On the other hand, if

M

=

sup {IK(x , y)l: (x, y) E [a, b] x [a, b]} < oo,

then the estimate IT(j)(x)l ::: M(b - a)llfll oo shows that IIT(f)ll oo ::: M(b - a)llfll oo· Thus, II T II ::: M (b - a) < oo, and so is a bounded operator. It is worth verifying that T also satisfies the following important property:

T

•

If B = {/ E C[a, b]: 11/lloo < 1 } is the open unit ball of C[a, b], tlzeu T(B) is a compact subset ofC[a, b].

To see this, observe first that T(B) is closed and bounded. Thus, by the Ascoli-Arzela Theorem 9.10, we need only to show that T(B) is an equicontinuous subset of C[a, b]. To see this, fix xo E [a, b], and let E > 0. Since (by Theorem 7 .7) K is unifonnly continuous, there exists some 8 > 0 such that IK(x l , y) - K(x2, y)l < E holds whenever lxt - x2l < 8. Therefore, if x E [a, b] satisfies lx - xo I < 8 and f E B , then

IT(j)(x) - T(j)(xo)l

=

lib

I

[K(x . y) - K (xo . y)]f(y) dy ::: (b - a)E.

This shows that T(B) is equicontinuous at xo. Since xo is arbitrary, T(B) is equicontinuous everywhere. Thus, T(B) is compact. The preceding property is expressed simply by saying that T is a compact operator. In general, an operator T: X --+ Y between two Banach spaces is said to be a compact • operator if T(B) is a compact subset of Y (where B is the open unit ball of X).

Our next example is borrowed from the theory of differential equations. Example 28.5. Let Ck[a, h] be the Banach space of Example 27. Consider C[a, b] with the sup norm ll·lloo. and fix k functions po, Pl . . . . , Pk-l in C[a. b]. Now, define L : Ck[a, b] --+ C[a, b] by L(y) for each y E Ck[a, b).

=

k y + Pk - lYC -l ) + · · · + Pl y' + POY

Section 28: OPERATORS BETWEEN BANACH SPACES

227

Clearly, L is a linear (differential) operator. On the other hand, if

M=

l + IIPolloc +

IIP 1 lloo + · · · + IIPk - l lloo.

then it is easy to see that

IIL(y)ll oc � M[llylloo + 11/k - l >lloo + · · · + lllll oo + II Y II oo l = Mllyll holds. This shows that L is a bounded operator. It is also interesting to observe that L is onto. This follows from the standard existence theorem of solutions to an ordinary linear differential equation. • The bounded operators are precisely the continuous linear operators. The next theorem clarifies the situation.

For a linear operator T : X Y between two normed spaces, thefollowing statements are equivalent: 1 . T is a bounded operator. 2. There exists a real number M > 0 such that II T (x)ll < Mllxll holds for all

Theorem 28.6.

X

�

E X.

is continuous at zero. 4. is continuous. Proof. (1) => (2) We have seen before that IIT(x) ll 3.

T T

< IIT II · IIxll holds for all x E X. Thus, if T is a bounded operator, then (2) holds for any choice of the real number M with M > liT 11 . (2) => (3) Clearly, liml!x, II = 0 combined with the inequality IIT(x, ) ll < M ll x,ll implies lim!I T(x11)11 = 0. That is, T is continuous at zero. (3) => .(4) The (uniform) continuity of T follows immediately from the iden­ tity IIT(x) - T(y)ll = II T(x - y)ll and the simple fact that lim x, = x holds in a normed space if and only if lim(x, - x) = 0. (4) => ( 1 ) Assume by way of contradiction that liT II = oo. Then there exists a sequence {x,} of X with ll x11 11 = 1 and IIT(x11)11 > n for each n. Let y, = ·�· for each n, and note that l!y, ll = fr implies limy, = 0. But then, by the continuity of T, we must have lim T(y,) = 0, contrary to IIT(y,)ll = n 1 11 T(x, )ll > 1 for each n. Thus, II T II < oo holds, and the proof of the theorem is complete. • -

Let X and Y be two normed spaces. Then the collection of all bounded linear operators from X into Y will be denoted by L(X, Y). By Theorem 28.6, it follows that L(X, Y ) under the algebraic operations ( S + T)(x) = S (x) + T (x)

and

(aT)(x) = aT(x)

Chapter 5: NORMED SPACES AND Lp·SPACES

228

is a vector space. In fact, L(X, f) is a normed vector space under the operator norm II T II = sup{ I T (x) II: llx II = 1}. The details follow.

Let X and f be two normed spaces. Then L(X, f) is a normed vector space. Moreover, if f is a Banach space, then L(X, f) is likewise a Banach space. Proof. We verify first that T II T II = sup{ I T (x) II: llx II = I } is a norm on L(X, f). Clearly, from its definition liT II > 0 holds for all T E L(X, f). Also, the inequality IIT(x)ll < IIT II · IIxll shows that liT II = 0 if and only if T = 0. The proof of the identity llaT I = Ia I · liT I is straightforward. For the triangle inequality, let S, T E L(X, f), and let x E X with llxll = 1. Then Theorem 28.7.

r+

II(S + T)(x)ll = IIS(x) + T(x)ll

<

IIS(x)ll + IIT(x)ll

<

liS II + li T II

holds, which shows that liS + T II < liS II + liT 11. Thus, L(X, f) is a normed vector space. Now, assume that f is a Banach space. To complete the proof, we have to show that L(X, f) is a Banach space. To this end, let {Tn} be a Cauchy sequence of L(X, f). From the inequality IIT"(x) - Tm(x)ll < I Tn - Tm ll · llxll, it follows that for each x E X, the sequence {T (xn)} of f is Cauchy and thus convergent in f. Let T (x) = lim T11 (x) for each x E X, and note that T defines a linear operator from X into f. Since {Tn} is a Cauchy sequence, there exists some M > 0 such that iiTnll < M for all n. But then, the inequality IIT11(x)ll < IITnll · llxll < M IIxll, coupled with the continuity of the norm, shows that II T (x) II < M llx II holds for all x E X. Therefore, by Theorem 28.6, T E L(X, f). Finally, we show that lim T11 = T holds in L(X, f). Let E > 0. Choose k such that IIT11 - Tm II < E for all n, m > k. Now, the relation

for all n,

m > k implies

IIT(x) - Tn(x)ll

m-+oo IITm(X) - Tn(x)ll

= lim

for all n > k and x E X. That is, we have therefore, lim Tn = T holds in L(X, f):

liT - Tn ll

< <

Ellxll E for all n > k, and

A subset A of L(X, f) is said to be pointwise bounded if for every x subset {T(x): T E A} off is norm bounded.

•

E X the

Section 28: OPERATORS BETWEEN BANACH SPACES

A L(X, liT I < M

If a subset that

of Y) is norm bounded (i.e., if there exists some holds for each then i"n view of the relation

T E A,

T E A), I T(x)ll < I T I · I xl < Mll xll A

229

M

>

0 such

is· pointwise bounded. That is, every norm for each it follows that bounded set of operators is pointwise bounded. The converse of this statement is also true, provided that is a Banach space. The result is known as "the principle of uniform boundedeness," or as "the Banach-Steinhaus theorem," and it is a very powerful theorem.

X

Let X be a Ba­ be a normed space. Then a subset of L(X, Y) is norm space, and let nach bounded if and only if it is pointwise bounded. Proof. L(X, A c L (X, n = {x E X: II T (x)ll < n T E A} A X = u:, X 6.18, kE yE r 0 l x - Yll < r E X E A, = r, rx) x l ( y = l x l Yl l y rx E rii T(x)ll = II T (rx)ll = I T(y rx) - T(y)ll < I T (y rx)ll I T(y)ll < 2k. -1 = M E X < 2kt · x I T (x)l l l x l = A T E A. L(X, Y). li T I < M Theorem 28.8 (The Principle of Uniform Roundedness). Y

We already know that every norm bounded subset of Y) is point­ Y) be pointwise bounded. wise bounded. For the converse, let Start by observing that for each the set for all

E,

En is norm closed. Also, since is pointwise bounded, it follows that and Theorem there exists some holds. In view of the completeness of implies with (Ek)0 =f. 0. Choose such that Ek and > X Ek. with Now, let T and let l . Since it + follows that + Ek. and so

+

Thus,

holds for all

+

+

holds for all with l , and therefore, That is, is a norm bounded subset of

•

In applications, the next result is an often-utilized special case of the "Principle of Uniform Boundedness."

Let X be a Banach space, Y a normed space, and {T } a n E X, then T i s a sequence of L(X, I f T,(x) = T(x) holds for each x bounded operator. Corollary 28.9.

Y).

lim

Chapter 5: NORMED SPACES AND Lp-SPACES

230

The hypothesis that X is a Banach space in Theorem 28 . 8 is essential. The next example clarifies the situation.

Example 28.10.

Consider the vector space

terms are eventually zero. Then

X

consisting of all (real) sequences whose

X equipped with the sup norm is a normed space. On the other hand, the reader can verify easily that X is not a Banach space. Now, for each n define J,, : X -+ JR by II

fn (X) = L kXk k=l x = (XJ, x2 ) in X. From lf,,(x)l � (Lk=J k)llx !100, it follows that {f11} C L(X, JR). Moreover, lim f,,(x) = L:� 1 kxk holds for each x E X. (Observe that the series has in fact a finite number of nonzero terms.) Thus, {f,,} is pointwise bounded. However, it so that {fn} is not a norm bounded sequence. is easy to see that llfn II 2:. n hold for each for each

• . . .

n,

Hence, in general the Principie of Uniform Boundedness does not hoid true if X is not

•

assumed to be complete.

Another useful variation of the Principle of Uniform Roundedness (which is also due to S. Banach and H. Steinhaus) is known as the "Principle of Condensation of Singularities," and is stated next.

Theorem 28.11 (The Principle of Condensation of Singularities). For each

pair of natural numbers n and m , let Tn.m: X � Y be a bounded oper­ ator from a Banach space X to a normed space Y . If for each m we have lim sup11 IIT".m I = oo, then the set

lx

E X: lim sup i i

n-+oo

Tn,m (x) ll = oo for each

m

J

is a dense subset of X. Proof. For every n, m , and

Anmk

= {x

E X:

k,

let

IITp,m(X)II >

k

for some

p > n}.

Clearly, Anmk is an open set and we claim that it is also dense. We must show that if x e X and r > 0, then B(x, r) n Anmk # (/). So, let x e X and r > 0. such that By the Uniform Boundedness Principle there exists a vector

x0

lim sup ii

p-+00

Tp.m (Xo) ll = oo .

. Section 28: OPERATORS BETWEEN BANACH SPACES

Pick some E

>

0 such that E ll .toll <

231

r. From

it easily follows that either I!Tp.,(x)ll > k or else IITp,,(x + Exo) ll > k for some p > 11. That is, either X E B (x , r) n A nmk or X + EXo E B(x, r) n Anmk· Hence, 8 (x , r) n A,mk =/= 0, and so A,mk is an open dense set. Now, notice that the Baire Category Theorem 6. 17 guarantees that

n{Anmk:

II, m ,

k

=

I , 2, . . . }

is a dense set. Clearly, any point in this set satisfies lim sup11 IIT,_,(x) ll = oo for

�I

•

m.

Our next objective is to establish another important result of analysis known as the "open mapping theorem." It asserts that a surjective continuous linear operator between two Banach spaces is an open mapping (in the sense that it carries open sets onto open sets). To establish this result we need a lemma.

X

X -+

Let and Y be two Banach spaces, and let T : Y be an onto continuous linear operator. If zero is an interior point of a subset A of then zero is also an interior point of T(A). {rx : x E V} is Proof. Let V {x E llx II < 1}, and observe that r V the closed ball with center at zero and radius r. Since zero is assumed to be an interior point of A, there exists some r > 0 with r V c A. By the linearity of T, we must have T(r V) rT(V) c T(A). Hence, to establish the result it suffices Lemma 28.12.

=

X:

X,

=

=

to show that zero is an interior point of T(V). Clearly, U� 1 n V , and since T is an onto linear operator, Y U: 1 nT(V) also holds. By Theorem 6.18, there exists some k such that kT(V) has a nonempty interior. Since kT(V) kT(V), it follows that T(V) has an interior point. That is, there exist some y0 E T(V) and r > 0 such that B(y0, 2r) c T(V). Now, if y E Y satisfies IIY II < 2r , then y - Yo E T(V). Therefore,

X=

=

=

Y

= (y -

Yo) + Yo

E

T(V) + T(V)

c 2T(V) ,

where the last inclusion follows easily from the identity V + V {y E Y: IIYII < r } c T(V). By the linearity of T , it follows that

{y E Y:

IIYII < r 2-" }

c 2-"T(V)

=

=

2 V. That is,

T(2-"V)

holds for each n. Now, let y E Y be fixed such that IIYII < r 2- 1 . Since y e T(2 - 1 V), there exists some x 1 E 2- 1 V such that II Y- T (xJ) II < r 2-2 . Now proceed inductively. Assume that x, has been selected such that x, E 2-"v and IIY - :L;'=• T(x;)ll < r2-"- 1 .

Chapter 5: NORMED SPACES AND Lp·SPACES

232

Clearly, y - 2::�'- 1 T(x;) with IIY - 'L�';:{ T(x;)ll llx, ll < 2-" and y-

-- -- •

-

E T(2 " 1 V), and so there exists some Xn+1 E 2-" 1 V < r2 2 Thus, a sequence {x,} is selected such that "

II

L T (x; ) 1=1

hold for all n. Next, define

= II =

s,

ll sn+p - Sn

=

Y-

x1 +

· · ·

r( t ) 1= 1

+ x, for each n, and note that

II+P

L

i=n+1

x;

X;

.:S

ll+p

L llx; II < 2-"

i= n+1

=

for all n and p shows that {sn} is a Cauchy sequence. Let x lim s, in X. Then llxll < I:: : 1 llxn II < l (i.e., x E V), and by the continuity and linearity of T, we see that lim T(sn) T(x) = n-oo That is, y E T (V), and so {y now complete.

=

"

"" T(x1)

11-00 �

lim

i=1

=

y.

E Y: IIYII < �} c T (V ) . The proof of the lemma is

•

The open mapping theorem is stated next. The result is due to S. Banach.

Let X and Y be two Ba­ nach spaces, and let T: X Y be a bounded linear operator. If T is onto, then T is an open mapping (and hence, if in addition T is one-to-one, then it is a homeomorphism). Proof. Let 0 be an open subset of X, and let y e T(O). Pick a vector x E 0 T(x), and note that y T(O) T(x - 0) holds. Now observe such that y that zero is an interior point of x 0, and hence, by Lemma 28.12, zero is also an interior point of T(x - 0) y - T (0). This implies that y is an interior point of T (0). Since y is arbitrary, T ( 0) is an open set, and the proof of the theorem is Theorem 28.13 (The Open Mapping Theorem). -+

=

complete.

=

= -

•

=

Consider two normed spaces X and Y. Then a norm can be defined on X x Y by l!(x, y )ll llx II + lly ll. This norm is called the product norm. You should stop and check that indeed this function satisfies the properties of a norm. Some other (frequently used) norms equivalent to the product norm are

!l (x , y ) ll

=

? I

(ll x ll2 + llyll-)2

Section 28: OPERATORS BETWEEN BANACH SPACES and

ll (x , y) ll = max{llxll ,

233

II Y II}.

It should be noted that l im (x, , y,) = (x , y) holds in X x Y with respect to the product norm if and only if lim x, = x in X and lim y, = y in Y both hold. Moreover, if both X and Y are Banach spaces, then

x Y with the product nonn is likewise a Banach space. Unless otherwise specified, the Cartesian product of two normed spaces will be considered as a nonned space under its product nonn. The next example illustrates an application of the open mapping theorem to differential equations . Example 28.14.

X

Consider the differential equation

I Y(k) + Pk- l Y(k -J ) + · · · + P I Y + POY = q

·

(l)

where po, Pl· . . . , Pk-l and q are fixed continuous functions on a (finite) closed interval [a, b]. If y is a solution of (1) and c E [a, b], then the k real numbers y(c), y'(c), . . . , y
What we would like to demonstrate here (by using the open mapping theorem) is that the solutions of ( 1) depend continuously on their initial values. That is, we would like to show that "small perturbations" in the initial values cause "small perturbations" to the solutions. To do this, we need to translate the problem to the framework of Banach spaces. Consider the Banach space Ck [a, b] of Example 27. Clearly, its norm

controls the "sizes" of the functions and their derivatives. Also, consider the bounded linear operator L: C k [a. b] � C[a, b] of Example 28; that is,

L(y) = y + Pk- ly
[a, b] and define T: c k [a, b] � C[a, b] X IRk by T(y) = (L(y), y(c), y'(c), . . . , y
for each y E ck [a, b]. Then it is easy to check that T is linear and continuous. On the other hand, the existence and uniqueness of the solutions of (1) with prescribed initial values guarantees that T is one-to-one and onto. Thus, by the open mapping theorem, T - I (which exists and is linear) must be continuous.

Chapter 5: NORMED SPACES AND Lp·SPACES

234

r- 1

means that given E > 0 th ere exists some 8 > 0 such that 1 ur- 1 (/, x) - r (g , z)ll < E if (f, x) and (g. z) in C[a, b] X IRk satisfy II! - glloo + llx - zlloo < 8. Translating this statement back to the differential equation (1), we get the The continuity of

following: Suppose that

Yl and Y2 are two solutions of ( I ) satisfying the initial conditions

and

I Y2Cc) = fJ 1 , Y2(c) If

Ia; - {3; 1

<8

holds for each

i

=

=

fJ2 . . . . , y2(k-1)(c) = f3k ·

1, . . . , k, then

(k I ) (k I ) bi - Y2 II oo + II YI1 - Y2I II oo + · · · + II Y1 - - y2 - II oo < E. That is, the solutions of the differential equation values.

( I ) depend continuously on their initial •

Y is a function, then the subset G = {(x, T(x)): x E X} of X x Y is called the graph of T . Now, if X and Y are vector spaces and T is a linear operator, then G is a vector subspace of X x Y. Moreover, if T is a bounded operator, then it is a routine matter to verify that G is a closed subspace of X x Y . Recall that if T : X -+

The converse of this last statement is true for Banach spaces. The result is known as "the closed graph theorem," and it has numerous applications in analysis.

Let X and Y be llvo Banach spaces, and let T: X -+ Y be a linear operator. If the graph of T is a closed subspace of X x Y, then T is a bounded operator. Proof. Since the graph G of T is a closed subspace of the Banach space X x Y, it is a Banach space in its own right. The function (x, T(x)) x is a Theorem 28.15 (The Closed Graph Theorem).

1-+

linear operator from

G onto X that is clearly one-to-one and bounded. (It is bounded

ll(x, T(x))ll .) By the open mapping theorem, x is a homeomorphism. That is, x (x, T(x)) is a continuous (x, T(x)) function from X onto G. Now, observe that (x, T(x)) -+ T(x) is continuous. Therefore, x T(x) from X into Y is continuous since it is a composition of because

llxll

<

r+

ll xll + IIT(x)ll

=

r+

1-+

two continuous functions, and the proof of the theorem is complete.

•

EXERCISES 1.

Y be two Banach spaces and let T: X --+ Y be a bounded linear operator. Show that either T is onto or else T(X) is a meager set.

Let

X

and

Section 29: LINEAR FUNCTIONALS

235

2. Let X be a Banach space, T: X � X a bounded operator, and I the identity operator on X. If II T II < l , then show that I T is invertible. 1 [HINT: Show that (I - T)- = L:�0T".] -

3.

On C[O, I] consider the two norms llfll :o = sup{ l f(x)l: x E [O, l ] }

and

ll fll t =

fo

1

lf(x)l dx .

Then show that the identity operator / : (C[O, l] . ll·lloo) � (C[O, l]. l l · ll t ) is continu­ ous, onto, but not open. Why doesn't this contradict the open mapping theorem? 4. Let X be the vector space of all real-valued functions on [0, l ] that have continuous derivatives with the sup norm. Also, let Y = C [O, l ] with the sup norm. Define the mapping D : X � Y by D(f) = f' . Show that D is an unbounded linear operator. Show that D has a closed graph. Why doesn 't the conclusion in (b) contradict the closed graph theorem? 2 Consider the mapping T : C [O, I ] � C [O, I ] defined by Tf(x ) = x f(x) for all f E C[O, l ] and each x E [0, l]. a. b. c.

5.

6.

a. Show that T is a bounded linear operator. b. If / : C [O. l ] � C[O, l ] denotes the identity operator (i.e., !(f) = f for each f E C[O. l]), then show that II/ + Til = l + li T II.

Let X be a vector space which is complete in each of the two norms ll·ll t and 11 · 11 2· If there exists a real number M > 0 such that llx 11 1 .::: M llx 1!2 holds for all x E X, then show that the two norms must be equivalent. 7. Let X, Y, and Z be three Banach spaces. Assume that T: X � Y is a linear operator and S: Y � Z is a bounded, one-to-one linear operator. Show that T is a bounded operator if and only if the composite linear operator S o T (from X into Z) is bounded. [IDNT: Use the closed graph theorem.] 8. An operator P: V � V on a vector space is said to be a projection if ?2 = P holds. Also, a closed vector subspace Y of a Banach space is said to be complemented if there exists another closed subspace Z of X such that Y ffi Z = X. Show that a closed subspace of a Banach space is complemented if and only if it is the range of a continuous projection. [HINT: Use the closed graph theorem.]

29. LINEAR FUNCTIONALS

In this section, we shall discuss the basic properti es of continuous linear function­ als. Recall that a linear operator f: X � .IR, where X is a vector space, is called a linear functional on X. One of our objectives is to show that for every normed vector space X there are enough bounded linear functionals to separate the points of X . That is, for every p air of distinct vectors x and y of X, there exists a bounded linear functional on X such that f(x) i= f( y). This result (as well as many others) rests upon a classical

Chapter 5: NORMED SPACES AND Lp·SPACES

236

result known as the "Hahn-Banach theorem," which is one of the cornerstones of modern analysis. We shall discuss this theorem next. A mapping p: X -+ IR., where X is a vector space, is called a sublinear mapping if it satisfies the following two properties: a. b.

p(x + y) < p(x) + p(y) for all x, y e X; and p(ax) = ap(x) for all x e X and a > 0.

The crux of the proof of the Hahn-Banach theorem lies in the following lemma:

Let p be a sublinear mapping on a vector space X, Y a vector subspace of X, and xo rl. Y. If f is a linear functional on Y such that f (x) < p(x) boldsfor all .x e Y , then f can be extended to a linearfunctional g on tbe l'ector subspace Z generated by Y and x0 satisfying g(.x) < p(x)for all x e Z. Proof. Clearly, Z = {x + ax0: x e Y and a e IR}. Assume that g is a linear functional on Z that agrees with f on Y. Then g(x + axo) = f(.x) + ag(xo) holds for all x e Y and a e IR., and so g is determined uniquely by the value g(xo). Let c = g(x0). Thus, every real number c gives rise to a linear functional on Z that agrees with f on Y. Our objective is to show that there exists some value of c such that Lemma 29.1.

f(x) + ac < p(x + a.xo) holds for all x e Y and a e IR.. For a > 0 and x e Y , the inequality (*) is equiva­ lent to c < p(a- 1 x + xo) - f(a- 1 x) and to - f(a- 1 x) - p(-a- 1 .x - x0) < c for .x e Y and a < 0. Certainly, these inequalities [and hence, (*)] will be satisfied by a choice of c for which

- f(x) - p(-x - xo) < c < p(x + .xo) - f(x) holds for all .x e Y. If .x, y e Y, then the relations

f(y) - f(x) show that - f(x) Consequently, if

=

f(y - x) < p(y - x) = p(y + xo + (-.x - .xo)) < p(y + .xo) + p(-x - xo)

- p(-x - xo) < p(y + xo) - f(y)

s = sup{- f(x)- p(-x -xo): x

E Y}

and

holds for all

t = inf p( {

y+

x,

y e

Y.

.xo) - f(y): y E Y } ,

then s and t are both real numbers and s < t. But then any real number c such that s ::: c < t satisfies (**), and hence, (*). This completes the proof of the • lemma.

Section 29: LINEAR FUNCTIONALS

237

The classical Hahn-Banach theorem is stated next. This theorem is the heart of modern analysis, and it has far-reaching applications.

Theorem 29.2 (Hahn-Banach). Let p be a sublinear mapping on a vector

space X, and let Y be a vector subspace of X. If f is a linear functional on Y such that f(x ) < p(x) holds for all x E Y , then f can be extended to a linear functional g on all of X satisfying g(x) < p(x)for every x E X. Proof. Let C be the collection of all pairs (g, Z) such that Z is a vector subspace

of X containing Y and g is a linear functional on Z satisfying g(x) = f(x) for all x E Y and g(x) < p(x) for all x E Z. The collection C is nonempty since (f, Y) E C. Define an order relation on C as follows: (g2, Z2) > (g1 , Z 1 ) whenever Z 1 c Z2 and g2(x) = g1 (x) for all x E Z 1• It is easy to verify that > is indeed an order relation on C. Now, consider a chain {(g;. Z;): i E / } ofC (i.e., for every pair i, j E I either (g;, Z;) > (gj. Zj) or (gj, Zj) > (g;, Z;) holds). Let Z = U ; e 1 Z;, and note that Z is a vector subspace of X. Now, define g: Z � JR. by letting g(x) = g;(x) if x E Z;. Since {(g;, Z;): i E / } is a chain, the value of g(x) is independent of the chosen index i. Clearly, g is a linear functional. In addition, g(x) = f(x) holds for all x E Y, and g(x) < p(x) for each x E Z. Thus, (g, Z) E C, and clearly, (g, Z) > (g;, Z;) holds for all i E I. Therefore, every chain of C has an upper bound in C. By Zorn 's lemma, C has a maximal element, say (g, Z). To complete the proof, it suffices to show that Z = X. Indeed, if Z # X, then there exists some x0 E X with x0 ¢ Z. Let M be the vector subspace generated by Z and x0• By Lemma 29. I , there exists a linear functional h on M such that h(x) = g(x ) for allx E Z and h(x) < p(x ) forallx E M . But then (h, M) E C, and (h, M) > (g, Z) holds with (h, M) # (g, Z), contrary to the maximality property • of (g, Z). Hence, Z = X holds, and the proof of the theorem is complete. The next three theorems are applications of the Hahn-Banach theorem. The first result tells us that a continuous linear functional defined on a subspace can be extended to a continuous linear functional on the whole space with preservation of its original norm.

Theorem 29.3. Let Y be a vector subspace of a normed space X, and let f

be a continuous linearfunctional on Y . Then f can be extended to a continuous linear functional g on X such that ll g II = II f 11. = sup { l f(y ) l : y E

Y

and IIYII < 1 } < oo. Define p: X � JR. by p(x) = llfll · ll xll for each x E X. Note that p is a sublinear mapping on X such that f(x) < p(x) holds for all x E Y . By the Hahn-Banach theorem there exists a linear extension g of f to all of X such that g(x) < p(x) holds for all x E X . This implies lg(x)l < llfll· llxll for all x E X, and so llgll < llfll. On the

Proof. Let llf ll

Chapter 5: NORMED SPACES AND Lp-SPACES

238

other hand,

llf ll = sup {lf(y)l : y E Y and llyll < 1 } < sup { lg(x)l : x E X and llx II < 1 } = llgll also holds, so that II g II =

II f 11. Thus, g is a desired extension off to all of X.

•

For a nonned space X, the Banach space L(X, lR) is called the norm dual of X and is denoted by X * . That is, X * consists of all functions f: X � 1R that are both linear and continuous. The members of X * are called the continuous linear functionals on X. The norm dual X * plays an important role in the theory of Banach spaces, and part of this theory deals with the properties and structure of X * . The vectors of a nonned vector space are always separated by its continuous linear functionals.

Theorem 29.4. If x is a vector in some normed space X, then there exists a

continuous linear functional f on X such that llfll = 1 and f(x) = llxll. In particular, X * separates the points of X. Proof. If x = 0, then any f E X * with llfll = 1 satisfies f(x) = llxll. (When

X =P {0 }, Theorem 29.3 guarantees X * =P {0 } .) Thus, assume x # 0. Let Y = {ax: a E lR}, the vector subspace generated by x. Then, the formula g(ax) = a llxll defines a continu�ms linear functional on Y such that llgll = 1 and g(x) = llxll. By Theorem 29.3, there exists a continuous linear extension f of g to all of X with ll f ll = llgll = 1. Clearly, f(x) = llxll, as desired. To see that X * separates the points of X, let x , y E X with x =j:. y. By the preceding, there exists some f E X * such that f(x) - f(y) = f(x - y) • llx - y II # 0. Hence, f(x) =j:. f(y) holds, and the proof is complete. A vector always can be separated from a closed subspace that does not contain

it by a continuous linear functional. The details are included in the next result.

Theorem 29.5. Let Y be a vector subspace of a normed vector space X, and

let xo ¢ Y. Then there exists some f .E X * such that f(x) = 0 for all x E and f(xo) = 1 .

Y

Proof. Since xo ¢ Y, there exists some r > 0 such that llx - x0 II > r holds for all x E Y. Let Z = {x + axo : x E Y and a E lR}, the vector subspace generated

Section 29: LINEAR FUNCTIONALS by Y and xo. Now, define functional and

f: Z

� lR by

f(x

+ axo) =

239

a. Then f is a linear

holds for all x E Y and a E lR. It follows that f is a continuous linear functional Also, f(x) = 0 holds for each x E Y and on Z whose nonn does not exceed f(xo) = l . Now, apply Theorem 29.3 to extend f to a continuous l inear functional on X. •

r-1.

Let X be a nonned space. As we have seen in Theorem 28.7, the nonn dual X* [ = L(X, lR)] of X is always a Banach space. In particular, the nonn dual (X*)* of X* is likewise a Banach space. This Banach space is called the second dual of X and is denoted by X**. That is, X** = (X*)*. Every vector x E X gives rise to a continuous linear functional -� on X* via the fonnula

S:(f) = f(x) for all

f E X*. Indeed, clearly, S: is linear on X*, and the inequality IS:(f) l = l f(x) l

<

ll fll · llxll = ll.xll · ll f ll

shows that .t E X** and n.t-11 < llxll . On the other hand, by Theorem 29.4 there exists some f E X* with ll f ll = I and f(x) = ll.xll . Thus, llxll = f(x) = 1-t (f) l < 11 t ll and so 11-� 11 = llxll holds for all x E X. The mapping x H- -� (from X into X**) is called the natural embedding of X into its second dual X**. Summarizing the previous, we have the following theorem: -

,

The natural embedding x H- S: ofa normed space X into its second dual X** is a norm preserving linear operator (and hence, X can be considered as a subspace of X**).

Theorem 29.6.

-+

A linear operator T: X Y between nonned spaces that satisfies II T (x) II = llxll for all x E X is called a linear isometry. In this tenninology the preceding theorem can be phrased as follows: The natural embedding x H- -� is a linear

isometry.

In general, x H- S: is not surjective, and hence, X (when embedded in X**) is, in general, a proper subspace of X**. If X is not a Banach space, then x H- .t· cannot be onto, simply because X** is a Banach space. If the natural embedding of a Banach space X into its second dual X** is onto, then X is called a reflexive Banach space, and this is denoted by X** = X. The properties of the natural

Chapter 5: NORMED SPACES AND L,-SPACES

240

X**

X

embedding of into its second dual are utilized in many applications, some of which are illustrated below. then it is easy to see that its If Y is a vector subspace of a normed space closure Y is also a vector subspace (why?). In particular, if is a Banach space, then Y is the completion of Y. These observations will be used in the proof of the next theorem.

Theorem 29.7.

X,

X

The completion ofa normed space is a Banach space.

X be a normed space. Consider X embedded in X ** by its natural embedding. Then X ** is a Banach space and induces on X its original norm. Thus, X is the completion of X, which is likewise a Banach space. Proof. Let

•

The next result can be viewed as a"dual" of the Uniform Boundedness Principle.

Let A be a subset ofa normedspace X such that { f(x): x E A } is a bounded subset of JR. for each f E X*. Then A is a norm bounded subset of Theorem 29.8.

X.

X embedded in X ** . Then A as a subset of X ** is pointwise bounded on the Banach space X * . Thus, by the Principle of Uniform Boundedness (Theorem 28.8), A is norm bounded in X ** . It follows that A is also norm bounded X, and the proof is finished. Proof. Consider

•

in

EXERCISES I. Let I: X � JR. be a linear functional defined on a vector space X. The kernel of I is

the vector subspace

2.

Ker I = 1- 1 ((0}) = (x E X : l(x) = 0}. If X is a nonned space and I: X � JR. is nonzero linear functional, establish the following: a. I is continuous if and only if its kernel is a closed subspace of X. b. I is discontinuous if and only if its kernel is dense in X. Show that a linear functional I on a nonned space X is discontinuous if and only if for each a E X and each r > 0 we have I (B(a, r)) = (l(x): !Ia - xll < r} = JR.

3. Let 1. l1 , h, . . . , In be linear functionals defined on a common vector space X. Show

that there exist constants AI, , ).., satisfying I = 'L7 1 ')..;I; (i.e., I lies in the linear span of II . . . ' In ) if and only if nr= I Ker I; c Ker I. Prove the converse of Theorem 28.7. That is, show that if X and Y are (nontrivial) nonned spaces and L(X, Y) is a Banach space, then Y is a Banach space. .

4.

•

•

=

Section 29: LINEAR FUNCTIONALS

241

[HINT: Let {y,) be a Cauchy sequence c;>f Y. Pick f e X* with f # 0, and then consider the sequence {T, ) of L(X , Y) defined by T,(x) = f(.t)y,.] The Banach space B(JN) is denoted by 1!.':)0. That is, 1!.00 is the Banach space consisting of all bounded sequences with the sup nonn. Consider the collections of vectors

5.

co

=

{x

c = {x 6.

=

=

(.\: J , x2. X3, . . . ) E

f.oo: x, � 0 ). and

(.q , x2 • .t3, . . . ) e 1!. 00 :

lim x, exists in

JR. ).

Show that co and c are both closed vector subspaces of 1!.00• Let c denote the vector subspace of f.cx:. consisting of all convergent sequences (see Exercise 5 above). Define the limit functional L: c � JR. by L(x)

=

L(x1, x2 . . . . ) = lim 11-+00 x, ,

and p: eoo � JR. by p(X) = p(XJ, X2,

.

•

.

) = limsupx11•

Show that L is a continuous linear functional, where c is assumed equipped with the sup nonn. b. Show that p is sublinear and that L(x) = p(x) holds for each x e c. c. By the Hahn-Banach Theorem 29.2 there exists a linear extension of L to all of f.:o (which we shall denote by L again) satisfying L(x) � p(.\:) for all .t E 1!.00 • Establish the following properties of the extension L: 1. For each x E 1!.00, we have a.

lim infx, � L(x) � limsupx11 • 11-+00 /1-+00

n. m.

7.

L is a positive linear functional, i.e., x =:: 0 implies L(x) L is a continuous linear functional (and in fact II L II

=

1

:::,

0.

).

Generalize Exercise 6 above as follows. Show that there exists a linear functional £im: eoo � JR. satisfying the following properties: a. .Cim is a positive linear functional of nonn one. b. For each x (x,, x2, . . . ) E 1!.00, we have XJ + -t2 + · · · + x, 1 . x 1 + x2 + · · · + x, .c · tmmf . � 1m( x) � 1.tm sup 11-+oo n n 11-+oo In particular, .Cim is an extension of the limit functional L. ) e 1!. 00, we have c. For each x = (XJ, x2 =

.

• . . .

£im(.q , X2, X3 , . . ) = .Cim(X2, X3, X4, . . . ) . .

Any such linear functional .Cim is called a Banach-Mazur3 limit. [HINT: Define p : 1!.00 � JR. by p(x) = lim sup x1 +x,; and note that p is sublinear satisfying L(.r) = p(x) for all x e c.] 8. Let X be a nonned vector space. Show that if X* is separable (in the sense that it contains a countable dense subset), then X is also separable. .

3 Stanislaw

.

+x,

Mazur ( 1905-1981 ), a Polish mathematician and a close collaborator of Stefan Banach.

He made important contributions to functional analysis and probability theory.

Chapter 5: NORMED SPACES AND Lp·SPACES

242

[HINT: Let {ft , h, . . . } be a countable dense subset of X*. For each n choose x, E X with l lx,ll = 1 and !f,,(x,)l 2:: � !If, II, and let Y be the closed subspace generated by {xi, x2 }. Use Theorem 29.5 to show that Y = X.] 9. Show that a Banach space X is reflexive if and only if X* is reflexive. [HINT: If X =I= X**, then by Theorem 29.5 there exists a nonzero F E X*** such that F(x) = O for all x E X.] 10. This problem describes the adjoint of a bounded operator. If T: X .....:,. Y is a bounded operator between two normed spaces, then its adjoint is the operator T*: Y* � X* defined by (T* f)(x) = j(T x) for all f E Y* and all x E X. (Writing ll(x) = (x, the definition of the adjoint operator is written in "duality" notation as . . . .

·

h},

(Tx, j}

=

(x, T* f}

for all f E Y* and all x E X.) a. Show that T*: Y* � X* is a well-defined bounded linear operator whose norm coincides with that ofT , i.e., II T* II = II T 11 . b. Fix some g E X* and some u E Y and define S: X � Y by S(x) = g(x)u. Show that S is a bounded linear operator satisfying II S II = II g 11 · 11 u 11. (Any such operator S is called a rank-one operator.) c. Describe the adjoint of the operator S defined in the preceding part (b). d. Let A = [aij] be an m x n matrix with real entries. As usual, we consider the adjoint operator A* as a (bounded) linear operator from R" to Rm . De�cribe A*.

30. BANACH LATTICES As we have seen, both the continuous functions and the measurable functions have a natural ordering under which they are lattices. For this reason, it is important to consider Banach spaces that are also lattices. Let us begin by reviewing a few of the basic facts that will be needed about vector lattices from Section 9. equipped with an A partially ordered vector space is a real vector space order relation .:::. that is compatible with the algebraic structure as follows:

1. 2.

If x .:::. If x >

y, then x +

y, then ax

z

>

.:::.

X

holds for all z E holds for all a .:::. 0.

y+

ay

z

X.

and its members The set x+ = (x E x > 0) is called the positive cone of are called the positive vectors of Clearly, the sum of two positive vectors is again a positive vector. A partially ordered vector space is called a vector lattice (or a Riesz4 space) if for every pair of vectors x, y E both_ sup(x, y) and inf{x, y) exist As usual, sup{x, y) is denoted by x v y and inf{x, y) by x 1\ y. That is, x v y = sup{x, y) and x 1\ y = inf{x, y ) .

X:

X. X X

X,

4Frigyes (Fred�ric) Riesz (1880-1956), a dislinguished Hungarian maihemaiician. His name is closely associa1ed wilh Ihe developmem of funclional analysis in 1he tiTSI half of Ihe Iwentieth cemury.

Section 30: BANACH LATTICES

243

In a vector lattice, the positive part, the �egative part, and the absolute value of a vector x are defined by x+ = x V O, x- = (-x) v O,

and !x! = x V (-x),

respectively. By Theorem 9.1 the following identities hold: x = x+ - x-

and

!x! = x + + x -.

A number of useful inequalities are stated in the next theorem.

Theorem 30.1. If x, y, and z are vectors in a vector lattice, then the following

inequalities hold: l. 2. 3. 4. 5.

!x + y ! < l x l + l y l ; l lx l - l y l l < !x - y ! ; lx + - y+ l < !x - y ! ; !x v z - y v zl < !x - y!; !x 1\ z - y 1\ z ! < !x - y l .

<

lx l and y < I Y I . it follows that x + y < lx l + I Y I · Similarly, -(x + y) < lx l + ! y l holds, and thus

Proof. ( l ) From x

!

!x + y ! = (x + y) v [-(x + y) J < x l + I Y I (2) By ( l ) we seethat lx l = !x -y+y ! < !x-y l + ! yl. and so !x ! - l yl < lx -y !. Similarly, ly l - l x l < !x - y!. and hence, l lx i - I YII < lx - y !. (3) Note that x+ = ! <x + !x l ). Then to establish the required inequality, use ( 1) and (2) as follows: 1 1 1 ! x+ - y+ ! = l ) - (Y + l y l) = l(x - y) + ( !x ! - ! y l ) ! !x (x + z z l

1 I < 2 1x - Y l + 2 1 1xi - I Y I I < !x - y !. (4) Note that xv

z -

y v z = [(x - :) v 0 + z]

- [(y - z) v 0 + z] = (x - z)+ - (y - z)+

holds. (See the identities in Section 9 before Theorem !x v z - y v rl = !(x - z) +

9. 1.) Thus, by (3) we have

- (y - z) + l < ! (x - z) - (y - :) 1

=

!x - yl.

244

Chapter 5: NORMED SPACES AND Lp·SPACES

(5) Since x 1\ z (4) that

= -[(-x ) v (

-

z

)] and y 1\ z =

-

[(-y) v (-z)], it follows from

lx 1\ z - y 1\ z I = I (-y ) v (-z) - (-x) v (-z) I � I - y - (-x ) I = lx - y I, •

and the proof of the theorem is complete.

Let X be a vector lattice. A subset A of X is called order bounded if there exists an element y e X such that lx I < y holds for all x e A. A linear functional on X is said to be order bounded if it carries order bounded subsets of X onto order bounded subsets of JR. That is, a linear functional f: X � 1R is order bounded if for every y e x + there exists some M > 0 such that lf(x)l < M holds for all x E X with lxl � y. A linear functional f on X is called positive iff(x) > 0 holds for each x e X + . Clearly, every positive linear functional is order bounded. As we shall see in the L.,_ .. ._:••-- -- - d!U---- ,.., ­ rl l:no".- fU"""�:,..... nJ nrrle" hounrl nPVt t!hPQrPnl PUPI"1.1 aue \..,1\;;t vv 11Llc;;ta a.� a 1J1C1 Cll\,;C . -�J of two positive linear functionals. The collection of all order bounded linear functionals on a vector lattice X is denoted by x- and is called the order dual of X. Obviously, x- (under the usual algebraic operations) is a vector space. Moreover, if we define f > g whenever f(x) > g(x) holds for all x E x+, then it is easy to see that x- equipped with > is a partially ordered vector space. In actuality, x- is a vector lattice, as the next result of F. Riesz shows. -n.

·-

• -···,

""

V'a �

.& v

a

u ····�LU.

"'"'""

11\.,LaVIIQI \.f(UI

is a vector lattice, then its order dual x- is likewise a vector lattice. Moreover,

Theorem 30.2 (F. Riesz). If X

f + (x) = sup {f (y): 0 < y < x}, f- (x) = sup{- f (y): 0 < y < x}, and lf l(x) sup {f (y ) : lyl < x }

=

holdfor each f E X- and a/1 X E x + . Proof. In view of the identities

f v g = (f - g)+ + g and f 1\ g = -[(- f) v (-g)], we establish that x- is a vector lattice by proving that f+ exists for each f E x-. To this end, let f E x-. Define g: x+ � 1R by

g(x) = sup {f(y):

0
x}

Section 30: BANACH LATTICES

245

for each x E x+. The supremum i s finite since f is order bounded. Clearly, g(x) > 0, g(x) > f(x), and g(ax) = ag(x) hold for all x E x+ and a > 0. We claim that g(x + y) = g(x) + g(y) holds for all X ' y E x + . To see this, let X ' y E x+. If ll and u satisfy 0 < u < X and 0 < u < y ' then 0 < ll + u < X + y holds, and so f(u) + f(u) = f(u + u) < g(x + y); consequently, g(x) + g(y) < g(x + y). On the other hand, i f O < z < x + y, then let u = x A z, u = z - u , and note (by using the identities in Section 9 before Theorem 9. 1 ) that 0< u

=z-

x A z = z + (-x) v (-z) = (z - x) v 0 < y v 0 = y.

Therefore, 0 < ll < x and 0 < u < y. This implies f(:) = f(u + u) = f(u) + f(u) < g(x) + g(y), from which itfollows that g(x+y) < g(x)+g(y). That is,g(x+y) = g(x)+g(y). Now, for arbitrary x E X define g(x) = g(x+) - g(x-). Note that if x = u - u with u , u positive, thenx++u = u +x - , andso g(x+)+g(u) = g(u)+g(x-) holds by the additivity of g on x+. Therefore, g(x) = g(x+) - g(x-) = g(u) - g(v), and so the value of g(x) does not depend upon the particular representation of x as a difference of two positive vectors. The preceding observation, combined with ag(x) for all x , y E x+ and a > 0, implies g(x + y) = g(x) + g(y), and g(ax) that g is a linear functional on X. Finally, we show that g = J + holds in X"'. Indeed, if h is another positive linear functional satisfying f < h, then since f(y) < h(y) < h(x) holds for all 0 < y < X, it follows that g(x) < h(x) for each X E x + . Therefore, g is the least upper bound of f and zero. That is, g = J+ holds in X"'. This shows that X"' is a vector lattice and that J+(x) = sup{f(y): 0 < y < x} holds. The formula for f- follows from f- = (-f)+, and the formula for the absolute value follows from I f I = J+ + f-. II

=

The formula lfl(lxl) = sup{f(y): IYI < lx l } implies the following useful in­ equality lf(x)l < lfl(lxl), for all f E and all x E X. In view of the identity f v g = (f - g)+ + g and Theorem 30.2, the following identities hold:

X"'

f v g(x) = sup{f(y) + g(x - y): 0 < y < x}

Chapter 5: NORMED SPACES AND Lp-SPACES

246 and

f 1\ g(x) = inf{/(y) + g(x - y):

0

< y < x}

for all f, g E x- and x E x + . Sometimes it is important to know that the "dual" formulas of Theorem 30.2 are also true. More precisely we have the following:

X X

Let be a vector lattice, and let f be a positive linearfunc­ tional. Then for eveI)' x E the following identities hold: f(x+) = sup{g(x): g E x- and 0 < g < /}, f(x-) = sup {-g(x): g E x- and 0 < g < / } , and / ( lx l) = sup{g(x): g E x- and l gl < /}. Proof. We establish the formula for f(x+ ). Note first that ifO < g � f holds, then g(x) < g(x + ) < f(x+ ), and so sup{g(x): g E x- and 0 < g < /} < f(x + ). For the reverse inequality consider the function p: -+ IR defined by p(u) = f(u+). It is easy to see that p is a sublinear mapping on X such that p(u) > 0 holds for each u e Now, let Y = {ax: a e IR } , and define h: Y -+ IR by h(ax) = af(x+ ). Obviously, h(u) < p(u) holds for all u E Y, and so, by the Hahn-Banach Theorem 29.2, h can be extended linearly to all of so as to preserve the inequality h (u) < p(u) for all u e Next, observe that if u > 0, + then h(u) < p(u) = f(u ) = f(u) holds, and moreover, -h(u) = h( -u) < p(-u) = f((-u)+) = /(0) = 0. That is, 0 < h < f holds. Therefore, f(x+ ) = h(x) < sup{g(x): g E x- and 0 < g < /}, so that Theorem 30.3.

·

X

X.

X

X.

f(x+ ) = sup{g(x}:

g e x- and 0

< g < f}.

The other two formulas now follow easily from the relations f(x-) and /(l x l ) = f(x+ ) + f(x-).

X

= f (( -x)+ )

•

A norm 11 ·11 on a vector lattice is said to be a lattice norm whenever lx I < I y I in X implies ll x I I < II y 11- A normed vector lattice is a vector lattice equipped with a lattice norm. If a normed vector lattice is complete, then is referred to as a Banach lattice. Let be a normed vector lattice. Then it should be clear that llx II = ll lx I ll holds for all x E Also, in view of Theorem 30. 1 , the following inequalities are valid for all x, y E

X X. X:

llx+ - y+ ll < llx - y ll

X

and

X

ll lx i - IYI II < IIx - y ll .

Section 30: BANACH LATTICES

247

In particular, they imply that the mappings _x H- x+ and x H- lx I (both from X into X) are uniformly continuous. It is interesting to observe that most of the normed spaces one encounters in analysis are normed lattices or Banach lattices. Some examples follow.

Example 30.4. Let X be a nonempty set. The collection of all real-valued bounded func­ tions defined on X is denoted by B(X). Then B(X) is a vector lattice (in fact a function space) under the ordering f ::: g whenever j(x) ::: g(x) holds for all x e X. Also, under the sup norm 11/lloo sup{lf(.t)l: x e X}, the vector lattice B(X) is a Banach lattice. See Example 6.12. •

=

Example 30.5. Let X be a topological space. Denote (as usual) by Cc(X) the vector space of all continuous real-valued functions on X that have compact support. In other words, f e C(X) belongs to Cc(X) if and only if the set {.t e X : f(x) =I= 0} has compact closure. (When X is compact, Cc(X) = C(X).) Then Cc(X) under the pointwise ordering (i.e., f ::: g if f(x) ::: g(x) for all x e X), and the sup norm is a normed vector lattice. • Example 30.6. Consider C [0, 1] with the pointwise ordering, and define a norm by II f II = J� 1/(.t)l dx. Then C[O, I ] under this norm is a normed vector lattice, but not a Banach

.

��

Example 30.7. Let e 1 denote the vector space of all (real) sequences x = (.ta, x2, . . .) such that L�l lx/1 I < 00. With the pointwise algebraic and lattice operations e I is a function L� dx11 I. the vector lattice e 1 becomes a Banach lattice space. Under the norm llx II (see Example 27.3). •

=

A vector subspace Y of a vector lattice X is called a vector sublattice of X if for every pair x and y of Y, the elements x v y and x A y belong to Y . That is, Y is a vector sublattice of X if it is closed under the lattice operations of X. A vector subspace V of a vector lattice X is said to be an ideal whenever lxl < lyl and y E V imply x E V . In view of the identity x v y !<x + y + lx - yl), it follows that every ideal is a vector sublattice. Let X be a normed vector lattice and let A be an order-bounded subset of X. Pick some y E X such that lxl < y for each x E A, and consequently, llx II < llyll holds for all x E A. That is, every order-bounded set is norm bounded. It follows from this that every continuous linear functional on X carries order-bounded subsets of X onto bounded subsets of JR. In other words, the norm dual X* is a vector subspace of the order dual X"'. The next result shows that X* is, in actuality, an ideal of x-.

=

The norm dual X* ofa normed vector lattice X is an ideal of the order dual x- (and hence, X* is a \'ector lattice in its own right). Proof. Assume 1 / 1 < lgl holds in x- with g E X* and f E x-. We must

Theorem 30.8.

show that f E X*.

Chapter 5: NORMED SPACES AND Lp-SPACES

248

Let x E X satisfy ll x ll = < ll x ll = L Therefore,

IIYII

L Then for every y E X with

l f(x)l :S l fl(lx l) < lgl(lxl)

= sup {g(y) :

IYI

< lx l we have

IYI :S lx l } :S llgll ,

where the equality in the middle holds by virtue of Theorem 30.2. This shows that • f E X* and ll f ll < llgll . The preceding arguments also show that for a normed vectorlattice X the rehltion l f l < l g l in X* implies llfll :S llgll . That is, X* is a normed vector lattice. Since X* is in addition a Banach space (Theorem 28.7), the following result should be immediate.

The norm dual ofa normed vector lattice is a Banach lattice.

Theorem 30.9.

If X is a normed vector lattice, then Theorem 30.8 shows that X* c X"' holds. However, the norm dual of a Banach lattice always coincides with its order duaL

Theorem 30.10.

IfX is a Banach lattice, then X*

=

X"' holds.

Proof. Let f E X"'. Assume by way of contradiction that f is not continuous.

That is, assume that llf ll = sup {l f(x) l: ll x ll = 1 } = oo. Then there exists a sequence of vectors {x,} such that ll x, ll = 1 and l f(x11) 1 > 11 3 for all 1z. 2 Let = .L�= 1 k- 1xk!. Then

Yn

IIYn+p - Yn ll =

n+p L k-2 1xk l k=n+l

holds for all n and p, and hence, {y, } is a Cauchy sequence of X. Since X is complete, there exists some y E X with lim = y. Clearly, 0 < and we claim that 0 < y, < y holds for each n. < Indeed, note first that the inequalities

Y11

Yn

Yn+t.

- y)+ll < for all n and p. Therefore, 0 < (y, - y )+ = 0 holds for each n. But then, y, - y < implies 0 < < y for all n. Now, since n-2 1x, l < < y, it follows that

II (Yn

Yn

limp-+xiiY,+p - y ll = 0, and so (Yn - y) v 0 = (y11 - y)+ = 0

Yn

for each n, which is impossible. Thus,

f is continuous, and so X* = X"'.

•

Section 30: BANACH LATTICES

249

vector lattice, and let Y be a vector sublattice of X . We Let X be a normed . already know that Y is a vector subspace of X , and in view of the continuity of x r; x+, it is easy to see that Y is a vector sublattice of X . We also know that the natural embedding x r; -� of a normed space into its second dual is linear and norm preserving. If X is a normed vector lattice, then x r; -� is in addition lattice preserving. Indeed, for x E X and 0 < f e X*, Theorems 30.2 and 30.3 applied consecutively in connection with the fact that X * is an ideal of x- give

(.r)+(f)

= sup{.�(g): g

0 � g < f} = sup{g(x): g e X* and 0 < g < f} = sup{g(x): g e x- and 0 < g < f} e X * and

= f(x+ ) = ;+(f).

;+ holds for each x e X, and this shows that x r; .t preserves That is, (.t)+ the lattice operations. Thus, X "sits" in X** with its norm, algebraic, and lattice structures preserved. Since the closure of X in the Banach lattice X** is the com­ pletion of X, it follows from the preceding that the completion of X is a Banach lattice. Summarizing, we have the following result: =

Theorem 30.11.

The completion ofa normed vector lattice is a Banach lattice.

A linear isometry T: X """* Y between two normed vector lattices that satisfies T (x v y) = T (x) v T (y) for all x, y e X is called a lattice isometry. Two normed vector lattices X and Y are said to be isomorphic if there exists a lattice isometry from X onto Y . In other words, X and Y are isomorphic if there exists a mapping from X onto Y that preserves all structures: the norm, the algebraic, and the lattice. An operator T: X """* Y between two partially ordered vector spaces is called positive if T (x) > 0 holds for all x > 0. When X is a Banach lattice and Y a normed vector lattice, then every positive operator from X into Y is necessarily continuous. (To see this, repeat the arguments of the proof of Theorem 30.1 0.) Here is an example of a classical positive operator: Example 30.12 (The Laplace Transform). In this example, we shall denote by dt the Lebesgue measure on [0, oo). Consider the operator .C: L 1 ([0, oo)) -+ Ch([O, oo)) defined by £f (s =

)

fooo

e-sr f{t)dt, s ::: 0,

where Ch([O, oo)) denotes the Banach lattice of all (unifonnly) bounded continuous func­ tions on [0, oo). Since le-sr f{t)i � 1/(t)l holds for all s, t ::: 0 and for each s ::: 0 the function e-sr f(t) is measurable, it follows from Theorem 22.6 that the function e-sr f(t)

Chapter 5: NORMED SPACES AND L,-SPACES

250

is Lebesgue integrable over [0, oo) for each s ::: 0. Moreover, it should be clear that I.Cf(s)l =::: 11/111 for each s ::: 0 so that .Cf is a uniformly bounded function . In addition, f(t) holds for each so ::: 0, it follows from Theorem 24.4 since lim -+ o f(t) = that

s s e-st

lim s-+so

e-sot

.Cf(s) = =

s-Ho }0rooe-st fo00e-sot lim

j(t)d =

}0roos-+limsoe-st

f(t)d t

f(t)dt = .Cf(so),

which shows that .C f is indeed a bounded continuous function. (The preceding argument also shows that lims-+oo .Cf(s) = 0.) It is now a routine matter to verify that .C is a lin­ ear positive operator-and hence, a continuous operator. The operator .C is known as the Laplace5 transform and plays an important role in applications. It is useful to notice that the Laplace transform is also one-to-one. To see this, assume that some function f 1 ([0, oo)) satisfies .Cf(s) = f(t) dt = 0 for each s ::: 0. l Making thechangcofvariab!ex c:-1 , we get Jd xs - f(- lnx) dx = 0 foraii s ::;: 0. In par­ ticular, we have Jd xk f( - lnx)dx =0 for all k 0, 1 , 2, . . . . This implies /( - lnx)=O for almost all x E [0, 1] (see Exercise 21 of Section 22). Using the fact that the function In x carries null sets to null sets (see Exercise 9 of Section 18), we infer that f = 0 a.e. Thus, .C is one-to-one. •

J000e-st

EL

=

=

We shall close this section with a remarkable convergence property (due to P. P. Korovkin6) about sequences of positive operators on C[O, 1]. Korovkin's result demonstrates the usefulness of the order structures. If T: C [0, 1] � C [0 , 1] is an operator, then (for simplicity) we shall write Tf instead of T (/). For our discussion below, C [0, 1] will be considered equipped with the sup norm 11 · 1100• For instance, lim fn = f in C [0, 1] will mean limllf, - f lioc = 0, i.e., that {/,} converges uniformly to f. Also, 1, x, and x 2 will denote the three functions of C [0 , 1] defined by 1(t) = 1, x(t) = t, and x 2 {t) = t 2 for each t e [0, 1].

Let {T,} be a sequence of positive operators from C[O, I] into C[O, 1]. /f lim Tnf = f holds when f equals 1, x, and x 2 , then lim Tn f = f holds for all f E C [0, I)Proof. Let f e C [0, I]. Our objective is to show that given E > 0 there exist constants C 1 , C2 , and C3 (depending on E) such that Theorem 30.13 (Korovkin).

5 Pierre Simon Marquis de Laplace ( 1749-1827), a distinguished French mathematician, physicist,

and astronomer. He made fundamental contributions to many fields including electricity, magnetism, planetary motion, and the theory of probability, 6Pavel Petrovich Korovkin ( 191 3-1985), a Russian mathematician. He worked in approximation theory.

Section 30: BANACH LATTICES

251

If this is done, then our hypothesis implies that II T11 f - f ll oo < 2E must hold for all sufficiently large n. Therefore, this will establish that lim T11 f = f holds for all f E C[O, 1 ] . To this end, start by observing that for each t in the interval [0, 1 ] the function 0 < g, E C [O, I ] defined by g,(s) = (s - t i satisfies g, = x2 - 2tx + t 2 1. Since each T,, is positive, T,,g, is a positive function. In particular, we have ., - 2tT11X + r-T111)(t) ., 0 < T11 g,(t) = (T,x=

(T11x 2 - x2)(t) - 2t (T,x - x)(t) + t2(T111 - l )(t) .,

(1)

.,

< II T,,x- - x-lloo + 2IIT��x - x lloo + UT,,l - llloo for each t E [0, 1]. Now, let M = II/ 1100, and let E > 0. By the uniform continuity of f on [0, 1 ] , there exists some 8 > 0 such that -E < f(s) - f(t) < E holds whenever s, t E [0, 1] satisfy Is - t I < 8. Next, observe that 2M 2M ., -E - 02 (s - t)-., < f(s) - f(t) < E + 02 (s - t)-

(2)

holds for all s, t E [0, 1]. Indeed, if Is - t l < 8, then (2) follows from -E < f(s) - f(t) < E. On the other hand, if Is - t I > 8, then (2) follows from the inequalities 2M ., - 2M 02 (s - t )2 < -2M < f(s) - f(t) < 2M < 02 (s - !)-. Since each T,, is positive and linear, it follows from (2) that 2M 2M -ET11 1 - [ l T11g1 < Tn f - f(t)T, l < ET,,l + T,g,. / 02 Next, let K

=

2 M /82

(3)

and evaluate (3) at t to get

I [ T,, J - f(t) T,,l ](!)I < ET,l( t) + K T11g,(t) =

E + E[ (T, l - 1)(!) ) + KT,,g,(t)

< E + E liT,, I - l lloo + K T,g,(t).

In particular, note that

I(T, f - f)(t)l < I [T, f - f(t)T,l](t) l + lf(t) l · I(T, l - l)(t) l < E + KT,, g,(t) + (M + E)I!T,,l - l lloo

(4)

252

Chapter 5: NORMED SPACES AND Lp-SPACES

holds for each t E

[0, 1]. Thus, by taking into account (1), it follows from (4) that •

This completes the proof of the theorem.

EXERCISES 1.

Let X be a vector lattice, and let f: x+ � [0, oo) be an additive function (that is, f(x + y) = f(x) + f(y) holds for all X, y E X+). Then show that there exists a unique linear functional g on X such that g(x) = f(x) holds for all x E x+. [IDNT: Use the arguments oftheproofofLemma 18.7 to show first that f(rx) = rf(x) holds for each x E x+ and each rational number r ::::. 0. Then define g (x) = f (x+) ­ f(x-) for each x E X.] 2. A vector lattice is called order complete if every nonempty subset that is bounded from above has a least upper bound (also called the supremum of the set). Show that if X is a vector lattice, then its order dual x- is an order complete vector lattice. [HINT: If A is a nonempty set of positive linear functionals such that f .:s g holds in x- for each f E A, put h(x) = sup{(vj'=l/;)(x): /; E A } for each X E x+. Use the p-eceding exercise to show that h extends to a positive linear functional and that = sup A.] 3. Show that the collection of all bounded functions on [0, 1] is an ideal of IR[O, 11. Also, show that C [O, I] is a vector sublattice of IR[O, I ] but not an ideal. 4. Let X be a vector lattice. Show that a norm 11·11 on X is a lattice norm if and only if it satisfies the following two properties:

h

a. If 0 :=:: x :=:: y, then llx II :=:: II y II, and b. llx II = ll lx I ll holds for all x E X.

=

Show that in a normed vector lattice X, its positive cone x+ is a closed set. [HINT: x+ {x E X : x- = 0}.] 6. Let X be a normed vector lattice. Assume that {xn } is a sequence of X such that x, :=:: Xn 1 holds for all n. Show that if lim x, x holds in X, then the vector x is + the least upper bound of the sequence {xn } in X. In symbols, Xn t x holds. [HINT: Observe that Xu+p x, ::::. 0 for all n and p and use the conclusion of the preceding exercise.] 7. Assume that x, --+ x holds in a Banach lattice and let {En} be a sequence of strictly positive real numbers, i.e., En > 0 for each n. Show that there exists a subsequence {xkn } of {.x, } and some positive vector u such that lxkn - x i :=:: E,,u holds for each n. [HINT: Pick a subsequence {y,} of {xn} satisfying II Y11 - xll < E,2-" for each n and let u :L:,.1 (En)- 1 l y11 - xl. N'ow, use the preceding exercise to conclude that (En)- 1 1Yn - x l .:S u holds for each n.] 8. Let T : X � Y be a positive operator between two normed vector lattices. If X is a Banach lattice, then show that T is continuous. [HINT: If is not continuous, then there exist a sequence {x11} of X and some E > 0 satisfying x11 � 0 and II Tx11 1l :2:. E for each n. By the preceding exercise, the-:e exists 5.

=

-

=

T

Section 30: BANACH LATTICES

9. 10.

253

a subsequence {y11 } of {x11 } and some u E .x+ satisfying IY111 .::;: �u for each n. Now, note that I I TY11 ll .5 t II T u II holds for each n .] Show that any two complete lattice norms on a vector lattice must be equivalent. [HINT: Apply the previous exercise.] The averaging operator A: l00 -+ l00 is defined by

A (x)

=

(

x1,

x1 + x2 XI + x2 + x3 2

,

3

,···,

x1 + x2 + · · + x" ·

n

)

, · · ·

for each x = (x1. x2, . . . ) E l00 • Establish the following: a. A is a positive operator. b. A is a continuous operator. c. The vector space

{.

.. · ·�2· .. · · · ) E e00 •. V = -� = (.�1 is a closed subspace of f00• Is V

=

{ x, +x2+11 ·+x, } ..

converges in

IR.}

e:r:;?

11. This exercise shows that for a normed vector lattice X, its norm dual X* may be a proper ideal of its order dual x- . Let X be the collection of all sequences {x,} such thatx, = 0 for all but a finite numberofterms (depending on the sequence). Show that: a. b. c. 12.

13. 14.

Determine the norm completion of the normed vector lattice of the preceding exercise. Determine the norm completion of the normed vector lattice of Example 30 when X is a Hausdorff locally compact topological space. Let X and Y be two vector lattices, and let T: X -+ Y be a linear operator. Show that the following statements are equivalent: a. b. c. d.

15.

X is a function space. X equipped with the sup norm is a normed vector lattice, but not a Banach lattice. If f: X -+ IR is defined by f(x) = 2::�1 nx11 for each x = {xn} E X , then f is a positive linear functional on X that is not continuous.

T(x v y) = T(x) v T(y) holds for all x, y E X. T(x 1\ y) = T(x) 1\ T(y) holds for all x , y E X . T(x) 1\ T(y) = 0 holds in Y whenever x 1\ y = 0 holds in X. I T(x)l = T(lxD holds for all x E X.

(A linear operator T that satisfies the preceding equivalent statements is referred to as a lattice homomorphism.) Let f00 be the Banach lattice of all bounded real sequences, that is, f00 = B(JN), and let {r1, r2, . . . } be an enumeration of the rational numbers of [0, 1]. Show that the mapping T : C[O, 1] -+ f00 defined by T(f) = (f(r1), j(r2), . . ) is a lattice isometry that is not onto. Let X be a norrned vector lattice. Then show that an element x E X satisfies x 2: 0 if and only if f(x) 2: 0 holds for each continuous positive linear functional f on X. [HINT: For the "if'' part use the second formula of Theorem 30.3 to obtain f(x-) = 0 for each continuous positive linear functional f.] .

16.

Chapter 5: NORMED SPACES AND Lp-SPACES

254

Let X be a Banach lattice. If 0 =:: x E X, then show that

17.

ll x ll = sup{f(x): 0

:5

f E X* and

11/11 =

1).

Assume that cp: [0, 1 ] --+ 1R is a strictly monotone continuous function and that T: C[O, 1] --+ C[O, 1 ] is a continuous linear operator. If T(cpf) = cpT(f) holds for each f E C[O, 1] (where cpf denotes the pointwise product of cp and f). Show that there exists a unique function h E C[O, 1] satisfying T(f) = hf for all f E C[O, 1]. Iff E C[O, 1], then the polynomials

18.

19.

where ('/) is the binomial coefficient defined by (k ) = k!(,7�k)!, are known as the Bernstein7 polynomials of f. Show that if f E C[O, 1 ] , then the sequence {811) of Bemstein polynomials of f converges unifonniy to j. [HINT: Consider the sequence { T, ) of positive operators defined by T,f(x) =

20. 21.

31.

{; ( ) "

n

k

f

(k ) -;;

.

xk ( l - x)"-k .

and apply Korovkin 's theorem.] Let T: C[O, 1] --+ C[O, l ] be a positive operator. Show that if Tf = f holds true when f equals 1, x, and x2, then T is the identity operator (that is, Tf = f holds for each f E C[O, 1]). (Korovkin) Let {T, } be a sequence of positive operators from C[O, 1] into C[O, 1 ] satisfying T, 1 = 1 . If there exists some c E [0, 1] such tha1 lim T,g = 0 holds for the function g(t) = (t -c)2, thenshow that lim T, f = f(c) · 1 holdsforall f E C[O, 1].

Lp-SPACES

Our attention is now turned from the study of general normed spaces to function spaces. Many of the classical spaces in analysis consist of measurable functions, and most of the important norms on such spaces are defined by integrals. The theory of integration enables us to study the remarkable properties of these spaces. Here the classical Lp-spaces will be considered. As we shall see, they are special examples of Banach lattices. Throughout this section (X, S, 11-) '!Viii be a fixed measure space, and unless otherwise specified, all properties of functions will refer to this measure space. It Sergei Nalanovich Bemslein (1 880-1968), a Russian malhemalician. He comribu1ed proximalion of funclions and probabilily lheory 7

.

10

1he ap­

Section 31: Lp·SPACES

255

P If I p> Then the collection of all measurable Definition 31.1. Let 0 < p < functions f for which l fiP is integrable will be denoted by L p(JJ.). X L p(J..L ) Lp(X) Lp(JJ.) Lp(X, S, J..L). f E Lp(JJ.), ctf L E

is important to keep in mind that if measurable for each 0.

f

is a measurable function, the

is also

oo.

to be indicated, then will be If clarity requires the measure space denoted by or even by is a vector space. Indeed, if It is easy to see that then clearly, E p(J..L) holds for all ct JR. On the other hand, the elementary inequality among the real numbers

L f E L < f- < If I , lfl , f f < < lfl f+ + L p(J..L). f E Lp(J..L) Lp(J.L. ) I f li p (! IJIP dJl r > 0 l ctf l p f. flip E L L p-norm f p I I l i f p(J..L) ct LE IR. lctl I f I p Lemma 31.2. If 0 < ).. < then

shows that p(J..L ) is closed under addition.8 Moreover, if and 0 imply that inequalities 0 to In other words, is a vector lattice. For each let

p(J..L ), then the belong and

I

=

is called the of Obviously, The number and and hold for all To obtain additional properties of the p-norms, we need an inequality.

=

·

·

1,

holds for eve/}' pair of nonnegative real numbers a and b. Proof. a b a1 > 0 b >xJ...0. x 0. f'(x) f: [0, xJ...-IRI ) > f. f x 1. x ajb x > 0. f(1) 0 < + f..x

The inequality is trivial if either or equals zero. Hence, assume and Consider the function oo) ---* defined by f(x) = - ).. + )..x Then , and so x = 1 is the for = f..( l It follows that attains its minimum at = Thus, only critical point of Now, let = to obtain the - x;. holds for all = 1 - ).. • desired inequality. 8To verify this inequality. note that lal

Ia + hi

S

(laiP)ii =:: (la lfl I

=

+ lbl'')fi , and so I

la.J + I/JI =:: 2(1al�' + lhlfl) fi . I

Chapter 5: NORMED SPACES AND Lp·SPACES

256

An important inequality between L P -nonns, known as Holder's9 inequality, is stated next.

Let 1 < p < oo and 1 < q < oo be such that * + � = 1. If f E L p(J.L) and g E Lq (J.L), then fg E L 1 (J.L) and

Theorem 31.3 (Holder's Inequality).

Proof. If f = 0 a.e. or g = 0 a. e. holds, then the inequality is trivial. So, assume

f -:j:. O a.e. and g -:j:. O a.e. Then II f li p > 0 and llgllq > 0. Now, apply Lemma 3 1 .2

with

1

).. = - , a = (lf(x)l/ ll f llp)P , p

and

l f(x )g(x)l < 1 l f(x ) I P II f II II II - P (II f II P )P

---P· g q

By Theorem 22.6,

+

b = (lg(x ) lfllgllq )q

lg(x) l q . q (IIg II q )q 1

fg E L 1 (J.L), and by integrating, we get _

_

_;;,!_Ifg_l _dJ.L

< _!._ + � = 1 . ll f llp . llgllq - p q

That is,

•

J lfgl dJ.L < II f liP · llg llq . as claimed.

For the special case p = q = 2, Holder's inequality is known as the Cauchy­ Schwarz 1 0 inequality; see also Theorem 32.2. The triangle inequality of the func­ 1 tion 11·11 P is referred to as the Minkowski 1 inequality. The details follow.

Let 1 < p < pair f, g E L p(J.L) tlze following inequality holds: Theorem 31.4 (Minkowski's Inequality).

oo.

Then for every

90no Ludwig HOlder ( 1 859-1937), a German malhemalician. He worked in group 1heory and geome1ry. He also con1ribu1ed 10 philosophical mallers concerning 1he foundalion of ma1hemaiics. 1 0 Hermann Amandus Schwarz ( 1 843-1921 ): a German maihemaiician. He worked in complex anal­ ysis and made several con1ribu1ions 10 1he 1heory of minimal surfaces. 11 Hermann Minkowski ( 1864-1 909), a German ma1hemalician. He s1udied eXIensively 1he geome1ric properties of convex se1s. His ideas in ma1hemaiical physics coniribUied greally 10 1he crea1ion of 1he 1heory of re1a1ivi1y.

Section 31: Lp-SPACES

257

Proof. For p = 1 the inequality is clearly true. Thus, we can assume 1 <

p <

oo. Let 1 < q < oo be such that l + l = L We already know that if f and /belcing to L p (J.J- ), then f + g likewise belongs to L p (J.J- ) . Next, observe thatsince (p - l )q = p, itfollowsthat l f + g iP- I E Lq(J.J-) . Thus, by Theorem 3 1 .3 both functions p 1/1 · I f + gl - t

p l g l · If + gl - l

and

belong to L 1 (J.J-) and we have the inequalities ·

(j

j

If I · If + g

j

p lgl · If + gl -t dJ.J- < llgllp · < II ! + gllp) � .

iP- '

dJL < l l f ll "

)

I f + g i
= 11/llp · ( llf + g llp)" , 1!.

I

'

p So, from I f + g! P = If + g l l f + gl -l < C l / 1 + lgl)lf + gj P l , we get
J

j

If + g i P d).l. <

J

p 1[1 . If + gl - l dj.J.

p lgl · If + gl -t dJ.J-

f f < 11/llp ·
The proof of the theorem is now complete.

•

Summarizing the preceding discussion: If 1 < p < oo, then a. b. c.

11 / llp > 0. lla/llp = lal · 11/llp , and II! + g llp < 11/llp + llgllp

hold for all f, g E L p (J.J-) and ex E JR. Obviously, by Theorem 22.7, 11 / llp = 0 if and only if f = 0 a.e. holds. Thus, unfortunately, the function ll · llp on L p(J.J- ) fails to satisfy the norm requirement

Chapter 5: NORMED SPACES AND Lp-SPACES

258

11/llp =

f=

0 imply 0. To avoid this difficulty, it is customary to call two that functions of L equivalent if they are equal almost everywhere. Clearly, this ), and introduces an equivalence relation on becomes a norm on the P for 1 < p < oo, is a normed space equivalence classes. In other words, L if we do not distinguish between functions that are equal almost everywhere. This means that in reality consists of equivalence classes of functions, but this should not pose a problem. In actual practice, the equivalence classes are relegated to the background, and the elements of L are thought of as functions (where two functions are considered identical if they are equal almost everywhere). An important advantage of the identification of functions that are equal almost everywhere is the following: A function of can assume infinite values or even be left undefined on a null set-since by assigning finite values to these points, the function becomes equivalent to a real-valued function of satisfies Also, it should be clear that if is a measurable function and E L < < a.e., then E is holds. In other words, (J..L) and (u' .....- iTi 'T'ha. a.+" e , +" 1 _::: y · ., l .,Hico ·-· · r l· " L L·l · C ,-.. ) I� i1 11o l11C::U C::L:tU a e jJ ..._ \J\.1 ca�11 and, in fact, a Banach lattice, as the next result of F. Riesz and E. Fischer12 shows.

p(J..L)

L p(J..L p(J..L),

11·11

L p(J..L)

p(J..L)

L p(J..L)

l g l 1/1

u

uLu

., ,...,...... .., uvuu •

g

g Lp

... .. .o .o ..,.. ..,.ovt

... lVt

1

Theorem 31.5 (Riesz-Fischer).

lattice.

Proof. Let

llgll p 11/lLJplp ;

-�

L.

- - -

r

L p(J..L). f p(J.L. )

:_

-

_

_

_�

ll·ll p

If 1 < p < oo, then L p(J..L) is a Banach

(f,,}

be a Cauchy sequence. By passing to a subsequence if nec­ < 2-" essary, we can assume without loss of generality that holds for each n. We must establish the existence of some E such that 0. l im 0 and f,,_ J ! for n > 2. Then Let

l f g f,=, l p = g · · + If,, fd 1 / d + 1 /:?. 1 + · , = 0 < g, t and

l /,,+1 - f,,llp f Lp(J..L)

-

holds for all n . By Levi's Theorem 22.8, there exists some 0 < t a.e. From

g, g

1 1,,+k - 1, 1 =

l +k

I: ct; - .ti- 1_)

i=ll+l

n+k

< L i=ll+l

12Emst Sigismund Fischer ( 1 875-1954), an Austrian

g Lp(J..L) E

such that

IJi - fi - l l = gn+k - g, ,

mathema ticia n. He in Ge nnany and studie d o rthononnal se quences o f fun ctio ns.

spent his s cie ntific career

Section 31: Lp-SPACES

259

it follows that { f,, } converges pointwise a. e. to some function f. Since

I f, I = !1 +

II

L(/; - /; - d i=2

< g, < g a.e.,

it follows that I f I < g a. e. holds, and hence, f E L P (JJ-). Now, in view of If - f, l < 2g a.e. and liml f,, - f i P = 0, the Lebesgue dominated convergence theorem implies lim II f - f,, ll P = 0, and the proof is finished. • A glance at the preceding proof reveals also the following interesting property of convergent sequences in L p·Spaces.

!fa sequence { f,,} c L p(JJ-), where 1 < p < oo, satisfies lim II / ­ f,, II P = 0, then there exist a subsequence { !k, } of { f, } and some g E L p(JJ-) such that !k, ---* f a.e. and I lk, I < g a.e.for each n.

Lemma 31.6.

In general, it is not true that lim II f - f,, ll P = 0 implies f, ---* f a.e . For instance, the sequence {f,,} of Example 19.6 satisfies limii J,, II p = 0 (for each 1 < p < oo), but { f,, (x)} does not converge for any x E [0, l ] . Also, it is easy to construct an example of a sequence in an L P -space that converges pointwise to some function of the space, but fails to converge in the X L p-nonn. For instance, consider JR. with the Lebesgue measure and f,, for each n. Then j, (x) � 0 holds for each x E JR. and f,, E Lp(lR) for all n and all 1 < p < oo. On the other hand, II /, II P = 1 holds for each n and 1 < p < oo, and so {/, } does not converge to zero with respect to any Lp-norm. The next useful result gives a condition for pointwise convergence to imply norm convergence in L p-spaces. =

Theorem 31.7. Assume 1 < p < oo. Let f E L p(JJ-) and let {f,,} be a sequence

of L p(JJ-) such that J,,

�

f a.e. If lim II /n il p = II f II P' then lim II f - !n il p = 0.

Start by observing that (a + b)P < 2 P -1 (aP + b P ) holds for each pair of nonnegative real numbers a and b. Indeed, for p = 1 the inequality is trivial. On the other hand, if 1 < p < oo, then the convexity of the function g(x) = x P (x > 0) implies

Proof.

Chapter 5: NORMED SPACES AND Lp·SPACES

260

and hence, (a + b)P < 2P- 1 (a P + b P) holds. In particular, for each pair of real numbers a and b we have

Thus, 0 < 2r- I (I J,, l P + If I P) - I fn - f I P a.e., and by applying Fatou 's le�ma (Theorem 22.1 0) and using the assumption lim I [11 I P dJ-L = J If I P dJ-L, we get 2P

j

lfiP dJL =

f

J

lim [2P- l (lf, I P + lfiP) - I f,, - fiP] dJ,L n -+ oo

/ f lfiP

< lim inf 11-+00 = 2p -

l

[2P- I (If�� I P + l f iP) - lfn - fiP] dJL

r

dJL + 2p- l lim

11-+00

l JriJ�, - fi P dJ-L

+ iim inf _ 11-+00 = 2P

j

lfi P dJl - lim sup 11-+00

/

f J

l f,IP dJL

I t, - fiP dJL.

Now, since J lfiP dJ,L < oo, the last inequality yields lim sup J I f,, - fiP dJ,L < 0. Hence, lim sup J lfn - fiP dJ,L = lim inf I f,, - fiP dJL = 0, so that lim

tl-+00

/

J

If,, - / I dJl = 0. •

Therefore, lim II [11 - f II = 0 holds, as required. P

A real number M is said to be an essential bound for a function f whenever lf(x)l < M holds for almost all x. A function is called essentially bounded if it has an essential bound. Therefore, a function is essentially bounded if it is bounded except possibly on a set of measure zero. The essential supremum of a function f is defined by

11 / ll oo = inf{M > 0: lf(x)l < M holds for almost all x } . If f does not have any essential bound, then it is understood that II f lloo = oo. Observe that lf(x)l < 11 /ll oo holds for almost all x. The following properties are easily verified, and they are left as exercises for the reader. I.

2.

If f = g a.e., then 11 /ll oo = ll g ll x.· II f II ';)C > 0 for each function f, and II f II 00 =

0 if and only if f = 0 a.e.

Section 31: Lp-SPACES 3.

4. 5.

261

llaflloo = lal · 11/lloo for all a E IR. II/ + gllx < 11/lloo + llglloo · If 1/1 < lgl, then 11/lloo < llglloo·

The collection of all essentially bounded measurable func­ tions is denoted by L00(JJ-).

Definition 31.8.

Here again, two functions are considered identical if they are equal almost everywhere. It should be obvious that with the usual algebraic and lattice operations L00(JJ-) is a vector lattice. Moreover, according to the above listed properties, L00 (JJ-) equipped with 11· 11 00 is a normed vector lattice that is actually a Banach lattice.

L00(JJ-) is a Banach lattice. Proof. Let {f,,} be a Cauchy sequence of L00(JJ.). We have to show that there exists some f E Lx(JJ.) such that lim ll / - J,, ll 00 = 0. Since for each pair m and n we have l f,,(x)-f,,(x)l < II J,, - fnr Ileo for almost all x, it follows that there exists a null set A such that l f (x) - fm (x)l < ll f,, - fm lloo holds for all m and n and all x rt A. But then lim fn(x) = f(x) exists in lR for all x rt A, and moreover, f is measurable and essentially bounded. That is, f E Lco (JJ.). Now, let E > 0. Choose such that llf,, - fm ll oo < E for all n , .m > k. Since 1/(x) - f,,(.x)l = lim,4-oo lf,(x) - f,,(x)l < E holds for all x ¢. A and n > it follows that II/ - f,, II < E for each n > This shows that lim II/ - f,, ll oo = 0, Theorem 31.9.

,

k

co

and the proof of the theorem is complete.

k.

k,

•

It is easy to verify that each step function belongs to every L p-space. Moreover, the collection of all step functions forms a vector sublattice of every L p -space. In addition, by Theorem 25.1 this vector sublattice is norm dense in L 1 (JJ.). The next result tells us that actually the vector lattice of step functions is norm dense in every L p(JJ.) with 1 < p < oo.

For eve1y I < p < oo, the collection ofall stepfunctions is

Theorem 31.10.

norm dense in L p(JJ-). Proof. Let 0 < f E Lp(JJ.). By Theorem 17.7, there exists a sequence of simple functions such that 0 < c/J, t f a.e. Clearly, each c/J, is a step function and (/ - c/J,)P .!- 0 a.e. holds. By the Lebesgue dominated convergence theorem we .get II! - c/J, II (j If - c/J,IP dJJ.)fi .!- 0. P Since every function of L p(JJ-) can be written as a difference of two positive functions of L p(JJ-), it easily follows that the step functions are norm dense in • Lp(JJ-).

{c/J,}

I

=

Chapter 5: NORMED SPACES AND Lp·SPACES

262

In case the measure is a regular Borel measure, the continuous functions with compact support are also norm dense in each Lp(J.L) for 1 < p < oo. The details are included in the next theorem.

Let J.L be a regular Borel measure on a Hausdorff locally compact topological space X. Then the collection of all comiuuous functions with compact support is norm dense in Lp(J.L) for every 1 < p < oo. Theorem 31.11.

Proof. Clearly, every continuous function with compact support belongs to

each L p(J.L). Now, let 1 < p < oo, f E L p(J.L), and E > 0. We must show thatthere exists some continuous function g with compact support such that II f - g II < E. P By Theorem 3 1 . 1 0 it suffices to assume that f = XA , where A is a measurable set such that J.L*(A) < oo. By Theorem 25.3, there exists a continuous function g : X --+ [0, 1] with compact support suchthat f iXA -:- g l dJ.L < 2-PfP. (Notethat i XA - g l < 2 holds.)But then II XA - cllv =

(/ (/

< 2

r (/ ) I

lxA - g i P d jL

=

p l lxA - c i · IXA - g l - dJL

I

IXA - g l dJL

and the proof is finished.

;;

r I

< 2 · r ' · E = E. •

Consider JR. equipped with the measure J.L that assigns to every subset of JR. the value zero, that is, J.L = 0. Then J.L is a regular Borel measure, and obviously, any two functions on JR. are equal J.L-almost everywhere. Thus, in this case, all functions on JR. can be identified with the zero function, a situation that is not very useful. Therefore, it is desirable to deal with regular Borel measures for which distinct continuous functions are not equivalent. To do this, we need to know where the measure is "concentrated" in the space.

Let J.L be a regular Borel measure on a Hausdorff locally compact topological space X . Then there exists a unique closed subset E of X with thefollowing two properties: 1 . J.L(Ec) = 0, and 0. 2. if V is an open set such that E n V # (/), then J.L(E n V )

Theorem 31.12.

Proof. Let 0 = U { V : V is opeo and J.L(V)

=

>

0}. Clearly, 0 is an open set,

and we claim that J.L(O) = 0. To see this, let K be a compact subset of 0. From the definition of 0 it follows that there exist open sets V1 , , V,, all of measure zero, such that K c U?= 1 V; . •

•

•

Section 31: Lp-SPACES

263

Hence, 11-(K) 0. Our claim now follows from Jl-(0) = sup{,u(K): K compact and K C 0}; see Definition 1 8.4. Now, let E oc. Then E is a closed set and ,u(Ec) = /1-( 0) = 0. On the other hand, if V is an open set such that E n V =F (/), then 11-CE n V) > 0 must hold true. Otherwise, if 11-CE n V) 0 holds, then ,u(V) = 11-(E n V) + 11-CEc n V) = 0 also holds, implying V c 0 = Ec, contrary to E n V =F (/). For the uniqueness of E assume that another closed set F satisfies ( I ) and (2). From it follows at once that Fe c 0, and so E = oc C F holds. On the other hand, since 11-CO n 0, it follows from (2) that 0 n F (/). Hence, c oc = E, so that F = E, and the proof is finished. •

= =

(I)

F

=

F) =

=

The unique set E determined by Theorem 3 1 . 12 is called the support of ,u and is denoted by Supp fl-. That is, Supp 11E. If we think of the measure space as a set over which some material has been distributed, then Supp 11- represents the parts of the set at which the material has been placed. IR", then If for example, the support of the zero measure is the empty set, while the support of the Lebesgue measure A satisfies SuppA = IR". Let 11- be a regular Borel measure on a Hausdorff locally compact topological space with Supp ,u = Then two continuous real-valued functions and holds for all x E This satisfy a.e. if and only if (x ) = on follows immediately by observing that if holds for some a E then f(x) =F g(x) holds for all x in some nonempty open set V (a neighborhood of for a.e. to be true. In particular, it follows Since 11-(V) > 0, it is impossible I defines a lattice norm on the function space of that II (j all continuous real-valued functions on with compact support. In general, equipped with an p-norrn is not a Banach lattice. However, by Theorem 31.11 the following result should be immediate.

= How "large" can the support ofa regular Borel measure be? X = X

X

f=g

X.

f li P = lf lP dfl-)li L

fX. g X, a).

ff(a) =P g(a) g(x) f=g Cc:(X), X

C,(X)

Let be a regular Borel measure on a Hausdor ff locally compact topological space X with =X. Then for each 1 < < the completion ofC,(X) with the L p-norm is the Banach lattice L L X = (0, f(x) x � 0<x <1 f(x) 0 x 1 L (J1) , L2(J1 ). 1 1 g(x) 0 0 < x < g x L (J1-). L2(J1-), L (X, S, J.1.) Theorem 31.13.

11-

p

Supp 11-

oo,

p(Jl-).

In general, the p-spaces are not "comparable." As an example, let oo) with the Lebesgue measure. Then the function = if and = if > belongs to but it does not belong to On the other 1 and (x) = hand, the function = if if x > 1 belongs to but not to 1 Two comparison results of the p-spaces are presented next. The first one is for the case that is a finite measure space. -

Chapter 5: NORMED SPACES AND Lp-SPACES

264

Let (X, S, JL) be a finite measure space, and assume that 1 < p < q < oo. Then Lq(JL) c L p(JL) holds. Proof. Clearly, in this case L00(JL) c Lp(JL) holds for each 1 < p < oo. Thus, assume 1 < p < q < oo. Let r = q I p > 1 , and then choose s > 1 such that : + � = 1 . If f E Lq(JL), then clearly, lfiP E L,.(JL). Since the constant function 1 belongs to LsCJL), it follows from Theorem 3 1.3 that IJIP = IJIP · 1 E L1 (JL). That is, f E L p(JL), and Theorem 31.14.

•

the proof of the theorem is complete.

Lq(JL) c L p(JL) holds, then Lq(JL) is an ideal of

It should be observed that if the vector lattice

L P(JL).

Some important examples of

L p-spaces are provided by considering the count­

ing measure on IN. In this case, the functions on and integration is replaced by summation. These 1 . ....tho.T.-. VLII""I 0 p .&II A b" "'"'n"'oS .1-J p=.::� pu'-""" t ""A u..&IU the" Lll J are denot LeU J "\,..

IN are denoted as sequences, L p-spaces are called the little A.., :+ Q

• ._,....... Y VI U.:J1 .l.l

..... ......

..,. }J

.....

00

th.,....

x 1 (x 1 , x2, ) such that E: 1 lxn IP < oo, and in this case llxllp = cr::l lxn iP)r. Similarly, loo is the vector space of all bounded •

l P consists of all sequences

•

•

-.....

t

Lll""ll

•

sequences with the sup norm.

L

The lP -spaces, unlike the general p-spaces, are always comparable. Note the contrast between the next theorem and the preceding one.

Theorem 31.15.

inclusion is proper. Proof.

If

1

Observe that if


x

=

(x1 , x2,

oo,

•

•

•

then

lp

c lq holds. Moreover, the

) belongs to some lp-space with 1 <

p < oo, then {xn} must be a bounded sequence (actually, convergent to zero), and hence, x E l00• That is, l C l00 holds for all 1 < p < oo. P ) E lp· Since E:1 lx11 1P < oo, Thus, assume 1 < q < oo. Let x = (x1, x2, q there exists some k such that lxn I < 1 for all n > k. This implies lxn l < lxn IP q for all n > k, and this shows that E: 1 lxnl < oo. Therefore, x E lq. and hence, lp c lq. I For the last part note that if we let Xn = for all n, then x = (x 1 , x2, . . . ) E lq .

but x

n - -,

¢. lp.

•

.

•

p and q in [1, oo] are called conjugate exponents if l + l = 1. We adhere to the convention 1 oo = 0, so that 1 and oo are conjugate �xpo�ents. Let p and q be two conjugate exponents. If g E Lq(JL), then it follows from Theorem 31.3 that fg E L1(JL) for each f E Lp(JL). Therefore, for each fixed g E Lq(JL) a real-valued function Fg can be defined on L p(JL) by 1\vo numbers

I

Section 31: Lp-SPACES

265

Fg

is a linear functional and, in fact, as the next result for all f E L p(Jl). Clearly, shows, a bounded linear functional.

Let I < p < oo, let q be its conjugate exponent, and let g E Lq(Jl). Then the linear functional defined by

Theorem 31.16.

1

F.c:
l gl q·

I

for f E L p(Jl) is a bounded linearfunctional on L p(Jl) satisfying F.c: II = Proof. First, we consider the case p = oo and q 1. From IF.c:{/ )1 < 11 /ll oo for each f E Loo(JJ.), it follows that F.c: is a bounded linear func­ < g 1! 1 holds. On the other hand, let f = S gn g, where tional and that > 0 and Sgn Sgn g(x) = I if = -1 if g(x) < 0. Then f belongs Therefore, to L 00(J1 ) and satisfies = I and = J dJJ. = =

l gl 1 ·

I Fg I I g(x) I /lloo

g(x) Fg(/)

lgl

I FNow, .c: l = wel gllconsider l· I < p < oo. By Holder's inequality

l g Il l ·

I F.c: l

< holds for all f E Lp(Jl). Hence, F.c: is a bounded linear functional, and holds. Now, let f = Sgn g . Clearly, f is a measurable.function, and fP= = holds, so that f E L p(Jl). Since fg = it follows that

ll g l q l g iP
That is,

lglq- l

l g l q,

I Fgll llgllq· Thus, I Fgll l gl q holds, and the proofis complete. >

=

•

The preceding theorem shows that for each 1 < p < oo a linear isometry g � F.c: can be defined from Lq(Jl) into L ; (JJ.), the norm dual of Lp{Jl). Observe that this isometry is also lattice preserving. Indeed, if 0 < f E L p(Jl ) , then by Theorem 30.2

(F,)+ (f)

=

=

sup( Fg(h):

0
<

I fg+ dJ1 Fg (f) =

+

f) .

= sup

{ Ihg dw 0

< /! <

f

}

Chapter 5: NORMED SPACES AND Lp-SPACES

266

Therefore, isometry.

(F.�)+ = F,�+

holds, which implies that g

H-

F.�

is also a lattice

Is every bounded linear functional on L p(J.J-) representable, as in Theorem 31 .1 6, I < p < oo. by aful1ctiol1 ofLq (J.J-)?

This is a classical result The answer is yes if of F. Riesz. A proof of this theorem, as well as some of its applications, is deferred until Section 37. Therefore, L ; (J.J-) and L q (J.J-) can be considered (under the above isomorphism) as identical Banach lattices. This is usually expressed by saying that for 1 the norm dual of L p(J.J-) is Lq(J.J-); in symbols, L ;(J.J-) Lq(JJ-). When the lattice isometry g Hfrom L1 (p.) to L�(p.) is rarely onto. The following example will clarify the situation.

< pp<=oooo

=

Fg

Let (X, S, 11-) be a measure space such that there exists a disjoint sequence of measurable sets {E,} with /1-*(En ) > 0 for each n and X = £11• Let L be the

Example 31.17.

U�1

collection of all real-valued functions f defined on X that are constant on each E,, assuming

on each

En

the value

sublattice of Loo(/1-).

f(E11), and for which

lim f(E,) exists in

Now, define a linear funciional F on L

R. Clearly, L is a vector

by F(f) = lim f(E,) for each f E L. It is clear holds for all f E L, and so is a continuous linear functional. By � can be extended to Loo(/1-) with preservation of its original norm. Denote Theorem 29.3, this extension y again. We claim that cannot be represented by a function of L (/1-). To see this, assume by way of contradiction that there exists some E Ll (/1-) satisfying F(f) = d11- for all E Loo(/1-). Let Gil E;)C and f,, XG,. Then {f,,} is a sequence of L, and = CU�'=I · ---+ 0, it F(f,) = I holds for each On the other hand, because � and follows from the Lebesgue dominated convergence theorem that J j,,g d11- -+ 0, which is impossible. � Fg from L 1 (/1-) to L � (JJ.) is not onto. Therefore, the lattice isometry • that

f

IF(f)l 1F/lloo b F F

= n.

F

1

g

fg lf,,gl F(f,l,)gl fng =

g

J

Later, we shall see (Theorem 37. 10) that if the measure is cr-finite. then the norm dual of L (J.J-) coincides with L00(J1 ) . The representation theorem for the bounded linear functionals on the l r-spaces can be proved directly.

1

Let 1 < p < oo, and let f be a continuous lineaz'[uzzctimzal ozsuclzz thatThen thez·e exists a uzzique y = (yl, ) lq (wlzere * � = f(x) l: x,y, holds for evezy x = (x1, ) lp. Proof. n, e, nth x = (x1 , x2, ) lp, Theorem 31.18. lp.

Y2, . . .

E

+

1)

00

=

It= I

x2, . . . E

For each let be the sequence having the value one at the .. . E then coordinate and zero at every other. Clearly, if

267

Section 31: Lp-SPACES

lim l x - :L�'= 1 x;e; l r = 0. Thus, f(x) = Z:::: 1 x,f(e,) holds. Let = f(e,) (y1, y2, . . . iq. If for each To complete the proof, we have to show that y = , then ly,l = l f(e,)l 11/11. so that y E ioo. let a, = y, · l y ,l q -2 if y, # 0 and a, = 0 if y, = 0. Now, for Then l a ,l P = l y, l q = a,y1 holds for all Moreover, Yn

n.

p

I

<

I < p < oo,

)

=

1

E

n.

t IYdq t,a; y; = t,a;J(e;) = f ( t, a1e1) =

<

l f l · t,a;e;

=

p

I

l fll ( t. l a;IP r

<:L�'= 1 l y;lq) ,. = <:L�'=1 ly;lq );; < 11!11 holds for. each n. This y (yl, Y2· . . .) belongs to iq and f(x) = :L:1xny,

Thus, implies that

I

-I

< oo

I

•

=

For a given normed space, it is often useful to have a characterization of its compact subsets. The Ascoli-Arzela theorem provided such a criterion for the compact subsets of (X)-spaces. Next, we shall characterize the compact sub­ sets of the Banach spaces L p([O, 1]). To do this, we need some preliminary discussion. Every function E L p([O, I]) will be considered defined on all oflR by =0 d>... instead of if rt [0, 1 ] . Also, for simplicity, we shall write If I < p < oo, then for E Lp([O, I ] ) and > 0 we define

C

f

t

h

f

f,1(t)

t

J: f(x) dx

[ =2h f(x) dx I

f (t) �a.b)f

t+ll

t-Il

for each E [0, I]. Note that the integral exists, since by Theorem 3 1 . 14 we have Lp([O, 1]) c L 1 ([0, I]). = f X(t,-ll.t,+ll> � Each f1, is a continuous function. Indeed, if t, � then < the Lebesgue dominated conver­ fX
t, g,

lg,l If I,

,_. oo

lim j,,(t,)

[ [ =2h ,_.oo f(x) dx = 2h f(x) dx = 1

tn+ll

lim

1

t+l1

-

t,-lr

t-Il

j, (t).

Chapter 5: NORMED SPACES AND Lp·SPACES

268

In particular, note that since C[O, I ] for each h > 0.

c L p([O, I]), it follows that [11

L p([O, I])

Let I < p < oo, and let f E Lp([O, I ]) Thenfor each h > 0 the continuous function [11 satisfies I a. 1/J,(t )l < (2h)-;; I f li for all t E [0, I ], and P b. I /J, I < II f I p . Proof. If p > I , then choose I < q < oo with l p + ! = I and apply HOlder's .

Lemma 31.19.

.

E

.

.

p

mequa1 1ty to get

q

p ' +h r llfh(t)IP = (2h)P } 1 · f(x) dx ,_11 f'+h 1 dx ) i < � lf(x)IP dx \2rt )P \ / t-h

I

( lt-h

[t + h lf(x)IP dx. ,

=

I

·

j'+h

2h t-h

Therefore,

[

1 t+h dx 1/J,(t)IP < 2h t-h lf(x)IP

(1)

holds for each 1 < p < oo and all t E [0, 1]. Also, ( I ) is obviously true for p and thus, statement (a) follows immediately. On the other hand, it follows from ( I ) that

h {1 lfh(t)IP dt < 2h l{1o [ ] lo [y
= 1,

'+ f J,_h lf(x)IP dx dt

I _

=

(2)

Here we have used the substitution x = t + y; see Exercise 16 of Section 22. Since the function f(t + y) is a Lebesgue measurable function on 1R2 (see Exercise 15 of Section 26), it follows from Tonell.i's Theorem 26.7 that

[ [ [;, 1/(t + y)IP dy ] dt t [{ 1/(1 y)IP dt ] dy < 2h { lf(x)IP dx . =

+

Section 31: LrSPACES

Thus,

269

(2) implies

1 ' lfh (tli' dt 1 ' Jf(x)J P dx, <

so that

ll f11 lip

<

•

II f liP holds.

A. N. Kolmogorov 13 characterized the compact subsets of Lp([O, 1])-spaces as follows:

Let l < p < oo, and let A be a closed and bounded subset of L p([O, 1]). Then the following statements are equivalent: I . The set A is compact lfor the L "-norm). 2. For each E > 0 there exists some 8 > 0 such that II f - /J, II P < E holds for all f E A and all 0 < h < 8. Proof. ( l ) ==} (2) Let E > 0. Since (by Theorem 3 1 . 1 1 ) C [0, l] is dense (for the L p -nonn) in L p([O, 1]) and A is compact, it is easy to see that there exist continuous functions fJ , . . . I f,, such that A c u:.'= I B(f;, E). By the uniform continuity of each f; , there exists some 8 > 0 such that If; (t) ­ f;(x)l < E holds for each l < i < n whenever t, x E [0, l ] satisfy lx - t I < 8. In particular, if 0 < h < 8, then Theorem 31.20 (Kolmogorov).

lf;(t) - (f; )h (t)l =

I

2h

lr+h [J;(t) - f;(x)] dx t-h

<E

holds. Thus, (j;)h ll p < E for each l < i < n and all 0 < h < 8. Now, if f E A, then choose I < i < 11 with f E B(f; , E). By Lemma 3 1 . 1 9 we have f1r - (J;)" li p < f - f; liP < E. Therefore,

II /; -

I

I

holds for all

f E A and all 0 < h

< 8. (2) ==} (1) According to Theorem 7.8, it is enough to show that A is totally bounded (for the L p-nonn). To this end, let E > 0. Fix some h > 0 such that f - fh P < E holds for all f E A. Next, choose M > O with ll f ll " < M forall f E A . Thenby Lemma 3 1 . 1 9 i t follows that

II

II

13 Andrey Nikolayevich Kolmogorov (1903-1987), a prominent Russian mathematician. He is the founder of modem probability theory. In addition, he studied the theory of turbulent flows and worked on dynamical systems in relation to planetary motion.

270

Chapter S: NORMED SPACES AND Lp-SPACES

holds for all t E [0, 1 ] and all f E A. Let A11 = {/1111: f E A}, where

!hh( t) =

1 2h

i

t+h

t-Il

/11 (x) dx.

Clearly, l .fi,11(t)l < K holds for all t E [0, 1 ] and f E A, and hence, A11 is a uniformly bounded set. Next, we claim that the set of continuous functions A11 is equicontinuous. To see this, note that if f E A and t < s, then

-

< <

1

2 /l

-1 2h

h is+ Ji, (x) dx - i s -h.Ji,(x) dx 1-11 t +h rs-11 , r s+h i l .fi,(x) l dx + l f,(x)l dx 1-h t+h

l

J

K 1 [2K(s - t)] = h(s - t) 2h

J

holds, and this shows that A11 is an equicontinuous set. Now, by the Ascoli-Arzela theorem, A11 is a totally bounded subset of C [O, 1] (for the sup norm). Choose functions /1 , . . . , f,, E A such that for each f E A there exists some 1 < i < n with II j,, - (/; )11, lloo < E. In particular, note that

II f - J; II P � II f - Ji, II P + II J, - Ji,l, II P + II .fi," - J; II P < 2E + II .fi," - J; II P < 2E + ll .fi,, - (j; ),,ll p + II (/; )"" - ( J;)hll p + II(J;), - /; li p

< 5E.

Thus, A is totally bounded (for the L p-nonn), and the proof is finished.

•

EXERCISES 1.

Let f E

L p(J.L), and let f > 0. Show that J.L*({x E X:

lf(x)i ::: E)) ::; f- p

f

l fi P dj.L .

be a sequence of some L p (J.L)-space with 1 ::; p < oo. Show that if lim llf, - flip = 0 holds in Lp(J.L), then {f,) converges in measure to f. [HINT: Use the preceding exercise.] 3. Let (X, S, J.L) be a measure space and consider the set

2. Let { f,)

E

=

{x A : A E A11. with J.L*(A) < oo ) .

271

Section 31: Lp·SPACES

Show that E is a closed subset of L 1 (J.L) (and hence, a complete metric space in its own right). Use this conclusion and the identity /;.(A b. B) = J IXA - XB I dJ.L to provide an alternate sol ution to Exercise 12(c) of Section 14. 4. Show that equality holds in the inequality of Lemma 3 1 .2 if and only if a = b. Use this to show that if f E Lp(J.L) and g E Lq(J.L), where 1 < p < oo and + = I , then J lfgl dJ.L = 11/llp llgllq holds if and only if there exist two constants C 1 and C2 (not both zero) such that C1 lflP = C2 lglq holds. 5. Assume that J.L*(X) = I and 0 < p < q � oo. If f is in Lq (J.L), then show that 11/l lp � 11/ll q holds. [HINT: Use Holder's inequality.] 6. Let f E L 1 (J.L) n L00(J.L). Then show that:

� �

·

a. f E L p(J.L) for each I < p < oo, b. If J.L*(X) < oo, then limp-+oo 11/llp = llflloo holds. [HINT: For (b), let E > 0. Then E = {x E X: lf(x)l > 11/lloo - E } has positive measure, and (11/lloo - E) · X E � 1!1 holds.] 1 7. Let f E L 2 [0. I ] satisfy 11/112 = I and ]0 f(x) d)... (x) > a > 0. Also, for each f3 E R, let EtJ = {x E [0, I]: f(x) > {J}. If O < f3 < a , show that )... (£f3) > ({3 -a)2. 8. Show that for I < p < oo each e p is a separable Banach lattice. 9. Show that f.'X is not separable. [HINT: Consider the collection of all sequences having zeros and ones as their entries.] 10. Show that Lco([O, I ]) (with the Lebesgue measure) is not separable. [HINT: Consider the set {X[O.x}: 0 < x < I }.] 11. Let X be a Hausdorff locally compact topological space, and fix a point a E X. Let J.L be the measure on X defined on all subsets of X by J.L( A ) = l if a E A and J.L( A) = 0 if a ¢. A. In other words, J.L is the Dirac measure (see Example 13.4). Show that J.L is a regular Borel measure and that S upp J.L = {a}. 1 12. If g E C [a, b] and f E L1 [a, b], then a. show that the function F: [a, b] --+- R defined by F(x) tinuous, and b. establish the following "Integration by Parts" formula:

b1

g(x)f(x) d)... (x)

=

g(x)F(x)

b b -1

=

J�r f(t) d)...(t) is con­

g'(x)F(x)dx.

u

[HINT: For (a) use Exercise 6 of Section 22. For (b) use Theorem 25.3 in connection with Lemma 3 1 .6 and the Lebesgue dominated convergence theorem.] 13. Let J.L be a regular Borel measure on R". Then show that the collection of all real­ valued functions on R" that are infinitely many times differentiable is norm dense in Lp(J.L) for each 1 < p < oo. [HINT: See Exercise 5 of Section 25.] Let (X, S, J.l.) be a measure space with J.L*(X) = 1. Assume that a function f E L 1 (J.L) 14. satisfies f(x) 2: M > 0 for almost all x. Then show that ln(f) E L 1 (J.L) and that J In(/) dJ.L < ln{j f dJ.L) holds. [HINT: Let t = f(x)/11! 1 1 1 in the inequality l - f � ln t < t - 1, and integrate.] oo. 15. Show with an example that Theorem 3 1 .7 is false when p =

272

16.

Chapter 5: NORMED SPACES AND Lp·SPACES This exercise shows that Theorem 3 1 . 16 is false when p = 1 and presents a necessary and sufficient condition for the mapping g 1-+ Fg from L00(J.L) into L j(J.L) to an isometry.

be

Show that for each g E L 00 (J.L), the linear functional FK (f) = J fg dJ.L for f E L I (J.L) is a bounded linear functional on L I (J.L) such that II Fcll � llgll 00 holds. b. Consider a nonempty set X and J.L the measure defined on every subset of X by J.L((/)) = 0 and J.L(A) = oo if A =I= (/). Then show that L I (J.L) = {0} and L00(J.L) = B(X) [the bounded functions on X] and conclude from this that g E L00(J.L) satisfies I Fg II = llg lloo if and only if g = 0. c. Let us say that a measure space (X, S, J.L) has the finite subset property whenever every measurable set of infinite measure has a measurable subset of finite positive measure. (A measure with the finite subset property is also called a iocally finite measure.) Show that the linear mapping g 1-+ Fg from L00(J.L) into Lj(J.L) is a lattice isometry if and only if (X, S, J.L) has the finite subset property. a.

17. Let (X, S, J.L) be a measure space. Assume that there exist measurable sets £ 1 , . . . , En such that 0 < J.L(E;) < oo for 1 < i < n, X = Ui I E;, and each £; does not contain

=

any proper nonempty measurable set. Then show that L6o(J.L) = L1 (J.L); that is, show that g 1-+ Fg from L I (J.L) toL6o(J.L)is onto. [HINT: If F E L6o(J.L), let c; = F (XE; ), g = [ci/ J.L*(E; )] XEI ' and then show that F = Fg holds.] 18. Let (X, S, J.L) be a measure space, and let 0 < p < 1.

Lf=l

a. b.

19.

·

11·11

Show by a counterexample that p is no longer a norm on L p(J.L). For each f, g E Lp(J.L) let d(f, g) = fl f - g! P dJ.L = (II/ - gllp)P . Show that d is a metric on L p(J.L) and that L p(J.L) equipped with d is a complete metric space.

[HINT: For the triangle inequality, observe that (a + b)P � aP + bP holds for every pair of non-negative real numbers a and b.] Let (X, S, J.L) a finite measure space. Then show that Theorem 3 1 . 1 0 is true for p = oo. That is, show that the step functions are norm dense in Lx(J.L).

be

20. If K is a compact subset of a metric space X, then show that there exists a regular 21. 22.

Borel measure J.L on X such that Supp J.L = K . If {fn} is a norm bounded sequence of L2(J.L), then show that J,,jn ---+ 0 a.e. Let (X, S, J.L) a measure space such that J.L*(X) = 1 . If f, g E L 1 (J.L) are two positive functi�ns satisfying f(x)g(x) � 1 for almost all x, then show that

be

23. Consider a measure space (X, S, J.L) with J.L*(X) = 1, and let f, g E L2(J.L). If

ff dJ.L = 0, then show that

Section 31: Lp·SPACES

273

24. If two functions f. g E L3 (Jl.) satisfy IIIII3 = ilg 113 = J I2g dJL = l , then show that g = III a.e. 25. For a function I E L 1 (JL) n Lz(JL) establish the following properties: a. I E L p(JL) for each 1 ::: p ::: 2, and b. limp- !+ III lip = III II! ·

26.

Assume that the positive real numbers ct 1 , a, satisfy 0 < ct1 < 1 for each i and , J,, are positive integrable functions on some measure space, "Li'= l ct; = 1 . If II · then show that 1 a. I� I:2 I,�n E L 1 (JL), and 1 b. J I�' I:2 Ir�'' dJL ::: (III! II! )a (II hill )a! · · · (ll frr ll l )an · ,

• • •

. . .

•

•

•

•

•

•

27. Let (X, S. JL) be a measure space and let {A11} be a sequence of measurable sets

satisfying 0 < JL*(A11) < oo for each n and lim Jl.*(A,) = 0. Fix l < p < oo and let I g11 = [JL*(A11)]-;j XA, for each n, where + = l. Prove that lim J Ig11 dJL = 0 for each I E Lp(JL). 28. Let (X, S. JL) be a measure space such that Jl. *(X) = 1 . For each 1 < p < oo define the set

� i

{

Cp = I E LJ (JL): Show that for each 0 < € <

j

III dJL =

I and

J IJIP dJL = 2 ) .

I there exists some 8p > 0 such that

JL*({x E X : II(x)l > €}) � 8p

29.

for each I E Cp. Let (X, S, JL) be a measure space and let 1

:::

p < oo and 0 < 11 < p.

Show that the nonlinear function 1/f: Lp(JL) � L e. (JL), where 1/f(J) = IIII7, is norm contmuous. b. If f,, � I and g11 � g hold in Lp(JL), then show that a.

.

,,

1 IJIP-171g 1 17 dJL. I J,,IP- 7Ign 1 17 dJ1. = 1lim 1 -+00 Let T : L p(J.I. ) � Lp(JL) be a continuous operator, where l < p < oo, and let 0 .s 11 ::: p. Show that:

J

30.

a.

J

If I E Lp(JJ.), then IIIP-171TIll7 E L 1 (JJ.) and

j

p 1 II -17 1TII11 dJL < IITII17 ( 11 IIIp )P .

b. If for some I E Lp(JJ.) with IIIIIp ::: 1 we have f i iiP-171TII'� dJL = 1171117 , then ITII = IITIII II .

31. Let (X, S. J.l.) be a measure space and let I E Lp(JL) for some l ::: p < oo. Show that the function g : [0, oo) � [0, oo) defined by p g(t ) = pt - l JL*({ x E X : II(x) l � t})

Chapter 5: NORMED SPACES AND Lp-SPACES

274

is Lebesgue integrable over [0, oo) and that

J IJIP dj.t

=

r

lro.oo)

g(t)dA(t) = p { 00t p-l J-L*({x E X : lf(x)l ::: t})dt .

lo

Let (X, S, p.) be a measure space and let f: X � JR. be a measurable function. If J-L*({x E X : lf(x)l 2: t}) � e-1 for all t 2: 0, then show that f E L p (J-L) holds for each 1 � p < oo. 33. Consider the vector space of functions

32.

{ f d>. = 0}. }JR. n Show that for each 1 < p < oo the vector space E is dense in L p(IR"). Is E dense in E=

34.

{f: JR." � IRI f

is a C00-function with compact support and

L 1 (JR.")? Let (0, oo) be equipped with the Lebesgue measure, and let 1 < p < oo. For each f E L p (A) let T(f)(x) = x- 1 fX. for x > 0. Then show that T defines a one-to-one bounded linear operator from L p((O, oo)) into itself such that liT II = [HINT: Assume that 0 � f E Cc((O, oo)). Then integrate by parts, and use Holder's inequality to get

J

(IIT(f)llp )P

=

pS.

flo JO' (�l lorxf(t) dt ) p dx =

1

1_P

(

lo lo

100 f(x)[ T(f)(x) ]P-l dx ( < _ foo T(f)(x) ]q
=

p p-

--

1

0

I

[

·

- _L1 11!11r · ( I IT(f)llr) � . p-

Now, use Theorem 3 1 . 1 1 . For

j11 (x) .=. x(n-'-l )p-'

if O

)

{ 00 { .xf(t) dt p d(x1-P)

liT II

<x< 1

use the sequence of functions and j,, (x) = 0 if x 2: 1.]

{!11 } defined by

CHAPTER

6

_ _ _ __ _ _ _ _ _ _ _ _ _

HILBERT SPACES The notion of norm can be thought of as a generalization of Euclid's concept of length. So, normed or Banach spaces can be viewed as finite or infinite dimensional analogues of the classical two- or three-dimensional spaces of Euclidean geometry. In the two- or three-dimensional spaces of Euclidean geometry there is, however, one more important ingredient that describes the length of a vector: it can be obtained by means of the familiar inner product. That is, the length of an arbitrary vector x is given by

llxll = �. where x · y denotes the inner product of the vectors x and y. In general, the inner productx · y = 2:7=t X; y; of two vectors x, y E lR11 can be considered as a function of two variables that satisfies certain linearity and homogeneity properties. Besides using the inner product to evaluate the norm of a vector, · the inner product also allows us to compute angles between vectors and express projections of vectors along lines and planes. For this reason, although all norms on lR.n are equivalent (Theorem 28.8), the Euclidean norm is the one that is more suitable for physical applications. Normed spaces whose norms are obtained (as above) by means of an inner product are called innerproduct spaces. If they are also complete, they are referred to as Hilbert 1 spaces. Naturally, these spaces inherit the most important properties of Euclidean spaces, and for physical applications they are the most suitable infinite dimensional analogues of Euclidean spaces. This chapter presents a brief introduction to Hilbert spaces. In the first section we study the basic properties of an inner product, and in the second section we · introduce Hilbert spaces. We discuss orthonomal bases in the third section. The final (fourth) section deals with some natural applications of Hilbert spaces to Fourier analysis. In particular, it discusses a few classical convergence properties 1 David Hilbert ( 1862-1943), a Gennan mathematician. He was a towering influential mathematical figure of modem times. In his 1900 address at the International Congress of Mathematicians. he proposed a list of 23 mathematical problems that have since stimulated and shaped the direction of present mathematical research.

275

276

Chapter 6: HILBERT SPACES

of Fourier series. Fourier analysis is today one of the basic tools of every engineer and applied scientist working in linear systems, antennas, mechanical vibrations, optics, bioengineering, various random processes, boundary value problems, and the design of filters used in the field of image and signal processing. Since the natural setting of Hilbert spaces is within the framework of complex vector spaces, in this chapter we shall deal mainly with complex vector spaces. The complex conjugate of a complex number ex = a + bz will be denoted by ex, i.e., ex = a - bz.

32. INNER PRODUCT SPACES An inner product on a real vector space X is a real-valued function of two variables

(·, ·): X

x

X � JR. such that:

(·, ·) is linear in tbefirst variable, i.e., (o:x + f3y, z) = o:(x, z) + f3(y, z)for ail x, y, z E X and aU reai numbers ex and {3; 2. (·, ·) is symmetric, i.e., (x, y) = (y, x)for all x, y E X; and 3. (x, x) > Ofor each x E X and (x, x) 0 if and only ifx 0. 1.

=

=

From (2) and ( l ), we see that every inner product on a real vector space is also linear in the second variable. A real inner product space is a real vector space equipped with an inner product. In case X is a complex vector space, a complex-valued function of two variables (·, ·): X x X � C is an inner product on X if a. b. c.

=

( ·) is linear in thefirst variable, i.e., (o:.x + f3y, z) o:(x, z) + f3(y, z)for all x, y, z E X and all complex numbers ex and {3; (x, y) (y, .x )for all x, y E X; and 0 ifand only ifx 0. (x, x) > Ofor each x E X and (x, ·,

=

.x) =

=

A complex inner product space is a complex vector space equipped with an inner product. It is easy to see that the inner product of a complex inner product space is additive in the second variable and "conjugate homogeneous" in the second variable. That is, • •

(x, y + z) = (x, y) + (x, z)for all x, y, z E X, and (x, o:y) = o:(x, y)for all x, y E X and all ex E C.

Every real vector space X can be "extended" to a complex vector space by means of its "complexification." If we define formally

Xc = X EB zX

= {x zy: +

.x ,

y E X},

Section 32: INNER PRODUCT SPACES

277

and equip it with the algebraic operations

(X + lJ) + + lJt) = X +X t + l(y + yt), (a + zf3)(x + zy) = ax - f3y + z(f3x + ay),

and

(-� 1

then X is a complex vector space-called the complexification of X . The reader should stop and verify that X with these operations is indeed a complex vector space. Also, it should be clear that the vector space X can be identified with the E X } of Xc . vector subspace Now, if X is a real inner product space with complexification X then we can define a function X x X c � C via the formula c

c

{x + zO: x ( ·}: (x + zy, Xt + zyt } = (x, X t ) + (y, Yt) + z[(y, xt ) - (x, yt )]. c.

0,

c

It is a routine matter to verify that ( 0 } is indeed an inner product on X 0,

(x + zO, Xt + zO} = (x, Xt)o ( ·, 0}

c.

and that

( ·)

That is, the inner product is an extension of the inner product from X to its complexification X Since most interesting results regarding inner product spaces are associated with complex inner product spaces and since every real inner product space can be extended (as above) to a complex inner product space, from now on, unless otherwise stated, all inner product spaces will be assumed to be complex inner product spaces. In particular, if X is a real inner product space, then we shall denote for brevity its complexification X by X again. In case we consider a real inner product space, we shall refer to specifically as a "real inner product space." Here are two examples of inner product spaces: c·

0,

c

The classical Euclidean spaces lR" provide the simplest examples of real inner product spaces. The inner product in lRn is defined by (x, y) L:�'= 1 x; y; for all vectors x = (.q , . x,) and y y,). Its complexification is simply C" whose inner product is defined by Example 32.1.

0

0 ,

= (Yt,

=

0 0 0 ,

II

(x, y) = L X; Yi

i= l

for all x, y E C" . The second example of an inner product space is the vector space C [a. b] of all con­ tinuous real-valued functions on a closed interval [a, b]. The inner product is given by the formula (f, g) J: f (x )g(x) dx Its complexification is the vector space of all continuous

=

0

Chapter 6: HILBERT SPACES

278

complex-valued functions defined on [a, b] and its inner product is given by

•

for all continuous functions f, g: [a, b) -+ C. In an inner product space X ,

the norm of a vector

x E X is defined by

llxll = J(x , x). Our immediate objective is to show that the preceding formula indeed defines a norm on X. To do this, we need to establish the classical Cauchy-Schwarz inequality.

Theorem 32.2 (The Cauchy-Schwarz Inequality). lfx vec ors

tra!}' t

in an inner product space, then

andy are two arbi­

l(x, y)l < llxii ii Y II·

Proof. Let x and y be two vectors in an inner product space. If x

then the inequality is obvious. So, we can assume each E IR we have

A

0< =

=

0 or y = 0,

x # 0 and y # 0. Then, for

(x + AY, x + AY)

(x, x) + A(X, y) + A(X, y) + A2(y, y)

llxll2 + 2ARe(x, y) + A211yll2 < llxll2 + 2AI(x, y)l + A2II Y II2• A

Now, notice that the last expression is a quadratic in which is always non-negative. This means that its discriminant is non-positive. That is,

Hence,

•

l(x, y)l < llxll llyll, as claimed.

We are now ready to establish that the formula

norm.

Theorem 32.3.

If X

prod

llx II = J<x, x) defines indeed a

t

is an inner uc space, then theformula llxll = J<x. x )

defines a norm on (called the norm induced by the inner product). X

279

Section 32: INNER PRODUCT SPACES

llx II > 0. for each x E C and x E X , then

Proof. Start by observing that

and only if x

=

0. Now, if a

and so I axil = l ct. l inequality to get

llx II = 0 if

llxll. Finally, for the triangle inequality use the Cauchy-Schwarz llx + Yll 2

This implies

E X and that

y,

(x + x + y ) = (x, x) + (x , + x) + = (x , x) + 2 Re(x , + < llxll2 + 21(x, y )l + IIYII 2 < llxll 2 + 211xii iiYII + IIYII2 = (llxll + llyll)2. =

y) (y, (y, y) y) (y, y)

llx + II .:S llx II + lly 11. and the proof is complete.

•

y

The inner product is a jointly continuous function . .

The inner pr o duct is a jointly continuous function . with respect to the induced nprm. That is, llxu - xll -+ and llYn - Yll -+ ==> (xu. Yu ) -+ (x , y).

Lemma 32.4.

0

0

Proof. The conclusion follows from the inequality

l(xu. Yu ) - (x, y ) l = l(xu -X, Yu ) +(x, Yu - y ) l and the fact that if II Yu

-

y II -+

0, then

<

IIYull ll xu -xll + llxii iiYu - Yll.

{ II Yull} is a bounded sequence.

•

A well known theorem of Euclidean geometry states that the sum of the squares of the sides of a parallelogram equals the sum of the squares of its diagonals. This fact is also true in any inner product space and is known as the

law.

Theorem 32.5 (The Parallelogram Law).

inner product space, then

parallelogram

If x and y are two vectors in an

Chapter 6: HILBERT SPACES

280

Proof. If x and y are two vectors in an inner product space, then note that

llx + yll2 + llx - y ll2

=

(x + y, x + y) + (x - y, x - y) = [(x , x) + (x, y) + (y, x) + (y, y)] + [(x, x) - (x, y) - (y, x) + (y, y)] 2(x, x) + 2(y, y) 2( 11x ll 2 + ll y ll 2),

= =

as claimed.

•

For a given norm 11·11 on a vector space X there exists at most one inner product on X that induces the norm 11 ·11 . To see this, assume that two inner products (·, ·) and {·, ·} on X induce 11· 11 . That is, assume that (x, x) {x , x } llx ll2 for all x E X. In particular, we have

= =

(x + y, x + y)

and therefore

= (x, x) + (x, y) + (y, x) + (y, y)

=

=

{x + y, x + y} {x, x} + {x, y} + {y, x } + {y, )'} ,

(x, y) + (y, x) = {x, y} + {y, x} for all x and y. If the vector space X is real, then the latter easily implies

(x, y) = {x, y}

for all x and y. If X is a complex vector space, then replacing x by z x in (*), we get z (x, y) - r (y, x) = z {x, y } - z {y, x} or

x=

(x, y) - (y, )

{x, y} - {y, x}

=

=

2 {x, y}, or (x, y) {x , y} for for all x and y. Adding this to (*) yields 2(x, y) all x and y. This establishes that a given norm is the induced norm of at-most one inner product Which norms are induced by inner products? Remarkably, as we shall see next, the ones that satisfy the parallelogram law. norm 11· 11 on a vector space is induced by an inner product if and only if it satisfies the parallelogram law, i.e., zf and only

Theorem 32.6. A

if

Section 32: INNER PRODUCT SPACES

281

holdstheforcaseallofvectors x andy. Moreover, the inner product ( ·, ) that induces 11 · 11 a real vector space is given by in ·

and in the case of a complex vector space by (x, y) = 4
Proof. Assume that a real Banach space X satisfies the parallelogram law. We

shall show that the formula 1

(x, y) = 4 (llx + yll- - l x -Yll2), x (x, x) = 11 · 11 . l 1 2 >0 x = 0. (x, y) = (y, x) '

is an inner product which induces the norm Clearly, for each Also, it should be clear that 0 if and only if and for all and y. Next, we shall show that the function (·, ·) is additive in the first variable. To do this, note first that

x (x,x x) = 4(u + v, w) +4(u - v , w) = ( l u + v + wll2 - l u + v- wll2) 2 + (l l u - v + wll2 - l u -v - wll ) = (llu + w + vll2 + l u + w- vll2) -(llu - w + vll2 + l u - w- vll2) = (21lu + wll2 + 211vl 2)-(211u - w!l2 + 21 1 vll2) = 2(11u + wll2 - l u - wll2) = 8(u, w). all u, v, w (u + v , w) + (u - v, w) = 2(u, w). � (x+y), v = 4 <x-y) , (2u, w)=2(u, w). =u v w=z (x, z) + (y, :) = (u + v, z) + (u - v , z) = 2(u, z) = (2u, z) = (x + y, z), ( ) 18.7, (rx, y) = r(x,y) r x, y X. ( ), Thus, for

When and

E X, we have

Now, letting u =

(••) yields in (••), we get

which is the additivity of in the first variable. we can establish that Now, as in the proof of Lemma as defined above, holds for each rational number and all E Since ·, ·

·,

·

282

Chapter 6: HILBERT SPACES

is a jointly continuous function (relative to the norm 1 1 · 11 ), it easily follows that (ax, y) = a(x, y) for all a E JR. and all x, y E X. This completes the proof of the real case. We leave the complex case as an exercise for the reader. • In an inner product space the notion of orthogonality is introduced by means of the inner product. Two vectors x and y are said to be orthogonal, in symbols x l. y, if (x, y) = 0. A set of vectors S in an inner product space is said to be an orthogonal set if 0 ¢ S and any two distinct vectors of S are orthogonal. An orthogonal set consisting of unit vectors is called an orthonormal set. Notice that if S is an orthogonal set, then the set of vectors {xI llx II: x E S} is automatically an orthonormal set. In the case of a real inner product space, we can also define the angle between two vectors. Indeed, if x and y are two non-zero vectors in a real inner product space, then (in view of the Cauchy-Schwarz inequality) there exists a unique angle f} satisfying 0 < f} < 180° and cos f}

=

(x, y) _____ llxll llyll

The angle f} is called the angle between the vectors x and y. So, in the real case, two vectors are orthogonal if and only if the angle between the vectors is (we also define the angle between the zero vector and any other vector to be One of the olde�t and most famous theorems in mathematics is the Pythagorean2 theorem. It states that the square of the hypotenuse of a right triangle equals the sum of the squares of its two legs. This result also carries over to inner product spaces.

goo goo).

If in an inner product space the vectors x1, X2 , . . . , x, are pailwise orthogonal, then Theorem 32.7 (The Pythagorean Theorem).

2Pylhagoras of Samos (ca 58�500 BC), a greal Greek philosopher and malhemalician. Born on lhe Greek island of Samos, he is crediled with many discoveries in malhemalics and science and was lhe firsllo associale number lheory wilh musical sounds. Pylhagoras moved lo lhe Greek cily of Crolona in soulhem llaly (a well known cily in ancienllimes for ils medical school and for many greal aihleles who won in the ancienl olympic games) and crealed lh� famous Pylhagorean school, which was in essence a secrel brolherhood. Following Pylhagoras' maxim "everylhing was nol lo be lold lo everybody," whal was done and 1aughl among lhe members of lhe brolherhood was kepi in profound secrecy from lhe ou1Side world. Due lo lhis secrecy, the exlend of knowledge and discoveries of the Pylhagorean school is fragmenled and highly speculalive.

Section 32: INNER PRODUCT SPACES

283

, x11 are pairwise orthogonal in an inner pro­ Proof. If the vectors x 1 , x2 , duct space, i.e., (X;, Xj) for -j:. j, then note that

=0 i ( II II ) II = = 8 II •

2

B

x;,

f;

xj

•

•

(x;, Xj) f;

II = B = �II (x;, x;)

as claimed.

2 llx; ll , •

An immediate consequence of the Pythagorean theorem is that non-zero orthog­ onal vectors are linearly independent.

In an inner product space any set of non zero pairwise or­ thogonal vectors is linearly independent. Proof. x 1 , x2, , x11 'L�'= 1 A;X; = 0. AIXI, A2X2, A2X11 'L7= 1 l A; 1 2 llx; 11 2 = A; = 0 i, x1 , x2 . . . , X11 'L7= 1 A;X; 11 2 = 0. Corollary 32.8.

•.. are non-zero pairwise orthogonal vectors , and let Since the vectors are also pair­ wise orthogonal, it follows from the Pythagorean theorem that This implies for each so that the vectors II • are linearly independent.

Assume that

•

.

.

Recall that if {A; };e/ is a family of non-negative real numbers, then the symbol L ; e/ A; denotes the supremum of the collection of all possible finite sums of the A;. That is, if F denotes the collection of all finite subsets of then

I,

where the supremum is taken in the extended real numbers.

Lemma 32.9. {Ai };eJ A; Lief A; < oo, L; e 1 A; Ai > �}, then Fn is a finite set (why?) and moreover, Proof. If F11 Ai > 0} U:1 F11 • Hence, A; > 0} is at-most countable. •

Ithen f # i0s holds a family of non-negative real numbers and for at most countably many indices i (in which case is either a finite sum or a series). = { i E I: = {i E I: {i E I:

One of the most basic properties regarding orthonormal sets is Bessel 's3 in­ equality. As we shall see later on, almost every aspect regarding the structure of orthonormal sets depends upon the next result. 3Friedrich Wilhelm Bessel ( 1784-1846), a notable Gennan astronomer and mathematician. By introducing new methods, he detennined the positions of several stars and planets quite accurately. He is also well known for a special class of functions which is today the indispensable tool of every scientist working in applied mathematics, physics, or engineering.

Chapter 6: HILBERT SPACES

284

Theorem 32.10 (Bessel's Inequality).

vectoz·s in an inner product space, then

If {x; };e/ is an oz·thozwl"malfamily of

L l(x, x;)l2 < llxll2 ie/

holdsfor each vector x. In particulaz; for each x all but an at most countable number of the (x, x;) vanish. Proof. Fix some vector x in an inner product space and let x 1 , x , . . . , Xn be 2

a finite orthonormal set. Then we have

II

II

n

II

- llxll2 - L(x, x;)(x;, x) - L(x, (x, Xj)Xj) + L L((x, x;)x;, (x, Xj)Xj) j=l i=l j=l i=l n n - llxf - L l(.x, x; )l2 - L l(.x, Xj )l2 + L l(x, x; )l2 i=l j=l i=l

1

II

llxll2 - L l(.x, Xi)l2. i=l Hence, L:�1 l(x, x;)l2 < ll.x ll2 . Now, if {x; }ie/ is an arbitrary orthonormal set, then the above conclusion easily implies L; e1 l(x, xi)l2 < llx 11 2 . In particular, • (x, x;) :f:. 0 must hold true for at most countably many indices i.

There is a standard procedure called the Gram4-Schmidt5 orthogonalization process for converting a countable (or .a finite) set of linearly independent vectors into an orthogonal set of vectors. As usual, let us use the symbol Span{z 1 , • • • to denote the vector subspace generated by the vectors 1 , . . . , Zn .

z

,

z11}

Let x1 , x , be a sequence of lineaz·ly independent vectors in an izmez· product space. If2we define the sequence ofvectoz·s y1 , y2 inductively by Theorem 32.11 (The Gram-Schmidt Orthogonalization Process). .

.

•

• . . .

� Yl = X) and Yn+l = Xn+l - L., i=l

(xll+l• y; ) y; for n = Ij y,. j1 2

1 , 2, . .

.

,

4Jorgen Pedersen Gram (1850-1916), a Danish mathematician. He taught at the University of Copenhagen and was the chairman of the Danish Insurance Council. 5 Erhard Schmidt (1 876-19�9), a German mathematician. He was a student of David Hilbert and worked on integral equations and the geometry of Hilbert spaces.

285

Section 32: INNER PRODUCT SPACES

then we have the following: 1 . The set {Yt , yz, . . . } is orthogonal, i.e., y, # Ofor each n and (y; , yj) = 0 for i # j, and 2. Span{Xt , Xz, . . . , Xu} = Span {Yt , yz, . . . , Ynl for n = 1 , 2, . . . . Moreover, the vectors Yt, Yz, . . . are (aside of scalar factors) uniquely determined in the sense that if Zt , zz, . . . is another sequence of orthogonal vectors satisfying Span{Xt , xz, . . . , X11} = Span{zt , Zz, . . . , Z11} for each n, then for each n there exists a 1wn-zero scalar A11 such that z11 = A11 Yn . Proof. First, we shall prove ( 1) and (2) by induction. For n = 1 , the claims ( 1 ) and (2) are obvious. So, for the induction step, assume that the vectors y1 , • • • , Yn

(as constructed by the above method) are all non-zero, pairwise orthogonal and satisfy

(t)

Span{Xt , Xz, . . . , Xn} = Span{yt , yz, . . . , Ynl· Now, de fine the vee tor I < j < n we have

Yn+ l = Xn+ l

-

"'II �i=l

(X,+J,J;)

IIYill2 y;

and note that 1or each &

are

so that the vectors Yt , . . . , y11, Yn+ l are pairwise orthogonal. Now, using (t), the • de�-mear1y m . l = Yn l + '\'II (Xu+ I 2 . 1 y Xn 1"dent"t + + �i=l IIYd.y;) l y; an d that X t , . . . , Xn+l pendent, we easily infer that Yn+ l # 0 and Span{Xt , xz, . . . , Xn, Xn+tl = Span{yt, yz, . . . , y11, Yn+tl- This completes the induction and the proof of ( 1 ) and (2). To establish the uniqueness of the sequence {y11}, assume that another sequence {z11} of pairwise orthogonal vectors satisfies .

Span{xt . X2, . . . , xn} = Span{yt , yz, . . . , yn} = Span{z t , Zz, . . . , zu} for each n. Clearly, z11 # 0 for each n. We shall prove by induction that for each n there exists a non-zero scalar A11 such that z11 = A11Y11 • For n = 1 , the conclusion is obvious. So, assume that there exist non-zero scalars A t , . . . , A11 such that z; = A;Y; for each i = 1 , . . , n. From .

Span{yt , yz, . . . , Yn. Yn + t l = Span{zt, Zz, . . . , Zn, Zn+d,

Chapter 6: HILBERT SPACES

286

1

we can write Z n+l = 'L-�':1 CiYi with c,+l # 0. Now from Zn+l - Cn+lYn +l = 'L�'= 1 ciYi and the fact that the vectors z,+l and Yn+l are both orthogonal to Span{y,, Y2· . . . , y, }, it easily follows that z,+l - c,+I Y,+ I l. :,71 - c, +lY,+ l · Thus, z,+l - Cn+l Yn+l = 0 or Zn+l = Cn+l Yn+l· This completes the induction and the proof of the theorem. •

EXERCISES 1. 2.

Let

c 1 , c2

, en ben (strictly) positive real numbers. Show that the function of two

variables(·, ·): IR" on •

.

.

.

IR.n .

x

IR.n � IR, defined by (x, y) = L:f=l CiXi Yi, is an inner product

Let (X, (· , ·)) be a real inner product vector space with complexification X c. Show that the function { · , ·): X c x Xc � C defined via the formula

{X + 1 )' , X I + 1 )'I ) = (X, .q ) + ()' , )'I ) + 1 [(y, X I ) - (X. )' J )].

A ,..,.. ..h,..... hot •h .., ,.. .....,. :.., . .-�..:1 h •I.e :-- - ---..l -� \I ' 7 ) " C• � a ro.Jv1 "'""" ,.." t""' L 'e l1V111l '''d U\..o\..U UJ L11 � '""e1 plUUu""L '"'"" ""' y v ' "0 v X h· on ';nm"'r "r"duc• ••

•

on X c is given by

llx + 'YII = Jcx , x) + (y , y) = Cllxll 2 + IIY if) ! .

3.

Let n be a Hausdorff compact topological space and let J1. be a regular Borel mea- ·

sure on n such that Supp J1.

defined by

= n. Show that the function ( ·): C(Q) X C(Q) � IR, · ,

(f, g) = £ tg dj1.,

is an inner product. Also, describe the complexification of C(Q) and the extension of

4.

the inner product to the complexification of C(Q).

Show that equality holds in the Cauchy-Schwarz inequality (i.e.,

if and only if

5. 6.

7.

8. 9.

x and y are linearly dependent vectors.

j(x. y)l = llxll llyll)

llxll = sup{ j(x. y)j: IIYII = 1}. holds if and only if llx + y 11 2 =

lfx is a vector inan innerproduct space, then.showthat

x ..L y llx ll2 + llyl! 2 • Does llx + yll 2 = llxll2 + IIYII 2 in a complex inner product space imply

Show that in a real inner product space

X j_ y?

{xn } in an inner product space satisfy l!xll . Show that x, � x.

Assume that a sequence

llxn II �

S be an orthogonal subset of an inner product complete orthogonal subset C such that S c C . Let

(x,.

x) � ll x 112

space. Show that there exists a

Show that the norms of the following Banach spaces cannot be induced by inner products: a. b. c.

10.

and

The norm

llxll = max{ lx t l , lx2 j, . . . , lxn l l on lR.n .

The sup norm on C[a, b].

The L p-norrn on any

Prove Theorem 32.6

L p(J.L)-space for each 1

for complex normed spaces.

� p � oo with p :;f 2.

Section 32: INNER PRODUCT SPACES

287

Let X be a complex inner product space and let T: X ---+ X be a linear operator. Show that T = 0 if and only if (T X '( ) = 0 for each X E X. Is this result true for real inner product spaces? 12. If {xn } is an orthonormal sequence in an inner product space, then show that Iim(x11 , y) = 0 for each vector y. 13. The orthogonal complement of a nonempty subset A of an inner product space X is defined by

11.

•

Al.

=

•

{.x E X: x .l.. y for all y E A } .

We shall denote (AJ..)l. by AJ..J.. . Establish the following properties regarding orthog­ onal complements:

Al. is a closed subspace of X, A c Al.l. and A n Al. = {0}. b. If A £ B , then B l. £ A l.. c. Al. A l. = [.C(A)]l. [.C(A)] l., where .C(A) denotes the vector subspace generated by A in X. d. If M and N are two vector subspaces of X, then Ml.l. + Nl.l. c (M + N)J.J.. e. If M is a finite dimensional subspace, then X = M E9 M l. . a.

=

14.

=

Let V be a vector subspace of a real inner product space X. A linear operator L: V is said to be symmetric if (Lx. y) = (x, Ly) holds for all x, y E V. a.

---+

X

Consider the real inner product space C [a, b] and let

V

=

.,

{ f E C -[a, b]: f(a) = f(b) = 0}.

Also, let p E C 1 [a, b] and q E C[a, b] be two fixed functions. Show that the linear operator L: V ---+ C[a, b], defined by

L (f)

=

(pj' )' + qf,

is a symmetric operator. Consider JR" equipped with its standard inner product and let A: JR" ---+ JR" be b. a linear operator. As usual, we identify the operator with the matrix A = [aij] representing it, where the jth column of the matrix A is the column vector Ae j . Show that A is a symmetric operator if and only if A is a symmetric matrix. (Recall that an n x n matrix B = [bij ] is said to be symmetric if bij = bji holds for all i and j.) c. Let L: V --+ X be a symmetric operator. Then L extends naturally to a linear operator L: Vc = {x + 1y: x, y E V } ---+ Xc via the formula L(x + 1y) = Lx + 1 Ly. Show that L also satisfies (Lu, v) = (u, Lv) for all u , v E Vc and that the eigenvalues of L are all real numbers. Show that eigenvectors of a symmetric operator corresponding to distinct eigen­ d. values are orthogonal. 15.

L:�'= 1

Xi Yi for all Let ( · , ·) denote the standard inner product on JR", i.e., (x, y) = x. y E JR". Recall that an n x n matrix A is said to be positive definite if (x, Ax) > 0 holds for all non-zero vectors x E JR". Show that a function of two variables ( ·, · } : JR" x JR" ---+ 1R is an inner product on JR" if and only if there exists a unique real symmetric positive definite matrix A such

Chapter 6: HILBERT SPACES

288 that

(x, y) = (x, Ay)

holds for all x, y E IR.n . (It is known that a symmetric matrix is positive definite if and only if its eigenvalues are all positive.)

33.

HILBERT SPACES

We start by introducing the notion of a Hilbert space.

Definition 33.1. A Hilbert space is an inner product space which is complete

under the norm induced by its inner product.

Again, we distinguish Hilbert spaces into real and complex. If space, then its complexification induced nonn is given by

1

He H EB H =

H

is areal Hilbert

is a complex Hilbert space whose

Here are two classical examples of Hilbert spaces:

Example 33.2.

The real Banach space f.2 is a real Hilbert space under the inner product

defined by

(x, y) = L: x, yn 00

n=l

.

for all x = (xt . x2 • . . . ) and y = ()'J , Y2· . . . ) in f.2. Its complexification, which we shall denote by f.2 again, consists of all square summable sequences of complex numbers. In this case, the inner product satisfies 00

(x, y) = L X,Yn· n=l

for all x =

(xt. x2

• . . .

) and y = (yt . Y2· . . .) in f.2.

•

Example 33.3.

The real Banach space L2(1J.) is a real Hilbert space. Its complexification, which we shall denote again by L2(1J.), consists of all complex valued measurable functions such that 1/1 2 dJ.L < oo holds (where, as usual, two functions are identical if they are equal almost everywhere). Then L2(J.L) is a complex Hilbert space with the inner product defined by

J

j

(J, g) = f(x)g(x) dJ.L. •

Section 33: HILBERT SPACES

J-L)

Example 33.4.

289

p:

Let (X, S, be a measure space and let X � (0, oo) be a measurable function-called a weight function. Then the collection of measurable functions

L2(p) L2(p) � by (f,g) fpfgdJ-L, then (·, ·) is a well-defined inner product. Indeed, since f E L2(p) is equivalent to having .JPf E L2(J-L), it follows from Holder's inequality that is clearly a real vector space. Now, if we define (·, ·):

JR.

x

=

2 pi ( ) pfg dJ-L Pifl2 l f g ( d dJ-L) I f J-LI f I

:::

2

2 I

<

00,

L2(p) L2(p) L2 p) ( L2(v)

is and so (-. ·) is well-defined. We leave it as an exercise for the reader to verify that a Hilbert space. As usual, we shall denote the complexification of by again. for the measure is exactly the Hilbert space The reader should also notice that 1\.IJ. � [0, oo] defined by

L2(p)

v:

v(A) ip(:r)dJ-L(:c) =

for each

AEA

•

w

The norm completion of an inner product space is a Hilbert space.

The norm completion X of an inner product space i s a of space. Moreover, Hilbert X and two sequences and satisfy and in X, then Theorem 33.5. X11 � x

X

if x, y e Yn 4 y

{x11 }

{y11}

X

(x , y) = lim (x,, y11 ). n--+oo

Proof. By Theorem 29.7 we know that X is

Banach space. We leave it as an exercise for the reader to verify that the formula ( · , · ) as defined above is a well-defined inner product that induces 11·11 on X. • a

Recall that in a normed space the distance from a vector x to a set A is defined by

d(x,

A ) = inf{llx - yll: y e A } .

Chapter 6: HILBERT SPACES

290

Our next important result asserts that a closed convex subset of a Hilbert space contains a unique vector of smallest distance from a given vector.

If A is a non-empty closed convex subset of a Hilbert space H, then for each x E H there exists a unique y E A such that

Theorem 33.6.

d(x, A) = llx - y ll . Proof. Let A be a non-empty closed convex subset of a Hilbert space H, let

x E H, and put d = d(x, A). Choose a sequence {y,} c A such that d = n-+oo lim llx - y, ll .

We claim that the sequence {y,} is a Cauchy sequence. To �ee this, note first that by the Parallelogram law, we have

llYn - Ym 11 2

ll(y, - X) - (Ym - x)ll 2 = 2 11Yn - x ll 2 + 211Ym - x ll 2 - ll(y, - x) + ( Ym - x)f y, Ym 2 = 211Yn - x ll 2 + 2 11Ym - x ll =

4

Since A is convex, it follows that

y,�y"'

xf.

1 ;

E A, and so d < II v,�y"' - x 11 . Therefore,

This shows that {Yn} is a Cauchy sequence. The completeness of H guarantees that {y,} converges to some vector y in H. Since A is closed, y E A and, clearly, d = d(x, A) = llx - y ll . For the uniqueness of y ' assume that another vector z E H satisfies nx - z II = d(x, A). Since A is a convex set, !CY + z) E A, and by the Parallelogram Law

d2

<

=

1

x - 2 (y + z)

2

1

4 11 cx - y) + (x - z)ll

=

2

� [211x - Yll 2 + 211x - zlf - ll(x - y) - (x - z)ll 2] 1

= -(2d2 + 2d2 - liz - yll 2 ) 4

.

1 = d2 - 4 ll z - y f < d2 • This implies

! II y - z 11 2 = 0 or y = z, and the proof is finished.

•

291

Section 33: HILBERT SPACES

Now, let X be an inner product space. If A is a non-empty subset of X, then the orthogonal complement A.l of A is the set of all vectors that are orthogonal to every vector of A. That is, A.l = {x E X: x y for all E A}. From tl)e linearity and continuity of the inner product it should be clear that A is always a closed subspace of X satisfying A.l = (A ).l and A n A.l = {0}; see also Exercise 13 of Section 32. Remarkably, when X is a Hilbert space and A is a closed subspace, then A together with A.l span the whole Hilbert space. Theorem 33.7. If M is a closed subspace of a Hilbert space H, then we have H = M EB M.i. Proof. Since M n M .l = {0}, it suffices to show that every vector x E H can written in the form x = y + z with y E M and z E M.i . So, letx E X. Since M is a closed and (as a vector subspace) convex set, there exists (by Theorem 33.6) a unique vector y E M such that d(x, M) ll x 11 . Let z = x - y and note that x y + z. We shall finish the proof9 by establishing that z M. To this end, let w E M and write (z , w) = r e' with 0. Then, for each real A we have j_

y

.l

be

=

-

=

y

.l

r >

liz II�.,

<

., ., liz + Ae 19 w ll- = li z II- + 2rA + A- ll w ll�. .,

.,

.:S

This implies 2rA + A2 11w ll 2 0 for each real A, and so 2r -A II w ll 2 for each A < 0. Hence, < 0, and therefore = 0. Consequently, (z, w) 0. In a Hilbert space the dense subspaces have a nice characterization in terms of an orthogonality condition. Corollary 33.8. A vector subspace M of a Hilbert space is dense if and only if the zero vector is the one and only vector orthogonal to M. Proof. Let M be a vector subspace of a Hilbert space H and let be its norm closure. Assume first that M = H and let x .l /vf. Pick a sequence {xn} C M such that x11 -+ x and note that 0 = (x11 , x) -+ (x, x) implies (x, x) = 0, or x = 0. For the converse, assume that M.l = {0}. Since (M).l = M.l, it follows from Theorem 33.7 that = M EB {0} = M EB(M).l = H, and soMis dense in H. Since the inner product is ajointly continuous function, it follows that every vec­ tor in an inner product space X defines a continuous linear functional f\': X C >

r

r

•

=

/vi

/vi

y

•

-+

Chapter 6: HaBERT'SPACES

292

via the formula

/y(x) = (x, y). X X

If is a Hilbert space, then (as we shall see next) all continuous linear functionals on are of this form.

If H is a Hilbert space and f: H C is a conti­ nuous linear functional, then there exists a unique vector y e H such that f(x) = (x, y) holdsfo all X E H. Moreover, we have f = y Proof. Let f: H C be a continuous linear functional on a Hilbert space and let M be its kernel, i.e., M = Ker f = f- 1 ({0}) = {x E H: f(x) = 0}. Since f is a continuous linear functional, it follows that M is a closed subspace. If M H, then y = 0 satisfies f(x) = (x, y) = 0 for each x e H. So, we0can assume that M is a proper closed subspace of H. Then, there exist some x e H with . . Now, notice f(x0) = and (by Theorem 33.7) some non-zero vector w E M.1 that ifx E H, then x - f(x )xo E M, and so(x- j(x)xo , w) = O or f(x)(xo, w) = (x, w). This implies f(w)(xo, w)=(w, w) 0, and so (xo , w) '# 0. Now, notice that the vector y = wj(w, xo ) satisfies f(x) = (x, y) for all x e H. Forthe uniqueness notethatif(x, y) = (x, y1)foreachx E H , then (x , y-y1) = 0 or 0 for each x E H , and so, by letting x = y - Y1, we get (y - Y1 , y - Y1) Theorem 33.9· (F. Riesz).

�

II

r

II II·

II

�

=

I

>

=

y = y1 • Finally, to establish the norm equality, note first that the Cauchy-Schwarz inequality

lf(x)l = ICx, y)l < l x li l Y

II

implies II II ::5 II y On the other hand, if y # 0, then letting = y I II y II, we see • = IIYII· that = ICY/IIYII . y)l = IIYII· Thus, = 1 and so >

l xl

f

1. 1 /1 lf(x)l

x

l fll

fy

If H is a Hilbert space, then Theorem 33.9 shows that a function y 1--7 where y) can be defined from H onto H * . In view of the properties

fy(x) = (x,

fy + fz = fy+z• afy = fay

and

l fyll =

IIYII,

it easily follows that this mapping is a "conjugate" linear isometry from H onto

Section 33: HILBERT SPACES

293

H * . By means of this isometry, we can establish that Hilbert spaces are reflexive Banach spaces.

Corollary 33.10. Proof. Let

Every Hilbert space is reflexive.

H be a Hilbert space and let F : H*

functional. Define ¢: H

� C be a continuous linear � C via the formula ¢(y) = F(/y). From

¢(y + z ) = F( /y :) = F(/y + f:) = F( /y ) + F(fz) = ¢(y) + ¢(z ) , + ¢(ay ) = F( /ay) = F(ii/y) = aF(/y) = a F( /y) = a cp(y) , and l¢(y ) l = IF( /y)l = I F(/y)l < II FII II /y ll = II FII IIyll, we see that ¢ E H*. So, by Theorem 33.9, there exists a unique x E X such that (y, x) = ¢(y) = F (/y) for all y E H. This implies .t(/y) = /y(x ) = (x, y ) = F(/y) for each y E H, and so F = .i. This shows that the natural embedding x � .i of H into its double dual H** is onto, and consequently H is a reflexive Banach space. • We shall close this section with a few more facts concerning orthogonal and orthonormal sets in Hilbert spaces. Let us say that an orthogonal set S of an inner product space is complete if x ..L s for each s e S implies x = 0. In view of Corollary 33.8, it should be noted that an orthogonal set S in a Hilbert space is complete if and only if the vector space generated by S is dense.

Every inner product space has a complete orthogonal set (and hence, it also has a complete orthonormal set).

Theorem 33.11.

Proof. The conclusion follows from Zorn's lemma by observing that the col­ lection of all orthogonal sets, ordered by inclusion, has a maximal element. To finish the proof, notice that an orthogonal set is maximal if and only if it is complete. • It is useful to note that if S is a complete orthogonal subset of a real Hilbert space H, then S is also a complete orthogonal set in the complexification Hc = H ffi z H of H. Indeed, if some x + z y satisfies x + z y ..L S, then (x + z y, s + z 0) = (x, s) + z (y, s) = 0 for all s E S, and so (x, s) = (y, s) = 0 for each s E S . Since S is complete, the latter implies x = y = 0, and so x + z y = 0. Now, assume that the vector subspace generated in a Hilbert space H by is the a countable collection of vectors X i , x2, . . . is dense in H . If y1, y2, orthogonal sequence of vectors produced by applying the Gram-Schmidt orthog­ onalization process to the sequence X i , x2 , then the orthogonal set {yi , y2, } is automatically complete. Indeed, if y ..L y, for each n, then y ..L x11 for each n,

M

,

•

•

•

•

•

•

•

.

.

Chapter 6: HILBERT SPACES

294 and so

H,

y _i M. Since M is dense in

we get y

_l M =

H,

and thus y

=

0. This

observation is used extensively in constructing complete orthogonal (or complete orthonormal) sequences of vectors. The next result illustrates this point.

[a, b] -+ (0, oo) is a measurable weight } c L2CP ). Also, let a sequence Po, PI, P2, . . . function such that of non-zero polynomials be such that a. each Pn is of degree n, and b. (Pn, Pm ) = J:p(x)Pn(x)-= P� m� (x�) dx = 0 for n ;f. m . Then the orthogonal sequence Po, PI , P2, . . . is complete and coincides (aside ofscalarfactors) with the sequence of orthogonalfunctions ofL�(p) that is ob­ tained by applying the Gram-Schmidt orthogonalizationprocess to the sequence of linearly independentfunctions { 1, x, x2, x3, } . Theorem 33.12.

Assume that

{1, x, x2 , x3,

.

p:

•

.

.

Proof.

•

•

We shall prove first that

holds for each n. For n = 0, the claim is obvious. So, assume that (*) is true for some n and let P be a polynomial of degree n + 1. Since Pn+l is a non­

zero polynomial of degree n + I , there exists some non-zero scalar Cn+l such that P -Gu+l Pn +l is a polynomial of degree less than or equal to n, i.e., P -Cn + l Pn+l E Span{1, x , x2 , . . . , x n }. So, from (*), we can write P - cu+ I Pn+l = L�=O ci Pi , or

P

=

E7:ci ci Pi. This shows that 2 Span {1, x , x ,

•

•

.

n I , x , xu + } C Span {Po , P,, P2, . . . , P, , Pn + d·

Since the reverse inclusion is obvious, the validity of (*) for n + I follows. Thus, (*) is true for each n. Now, a glance at Theorem 32. 1 1 guarantees that (aside of scalar factors) the sequence of polynomials P0 , P1 , P2, is the one •

•

•

obtained by applying the Gram-Schmidt orthogonalization process to the sequence 2 {1, x , x , }. •

•

•

Finally, in order establish that to show that the linear subspace

Po, P1 , P2 ,

M

•

•

•

is a complete sequence, it suffices

generated by

{ 1, x, x2 ,

•

.

.

} is dense in

L 2(p ).

M c C[a, b] c L2(p) and that M (by Corollary I 1 .6) is dense with respect to the sup norm in C[a, b]. This implies (how?) that C [a, b] To see this, note first that

L2(p). Next, notice that dense in L2(p), and so M is

is dense in the norm induced by the inner product in

Theorem 3 1 . 1 1 guarantees (how?) tliat dense in L 2(p). This shows that

C[a, b]

Po, PI , P2,

is . . . is a complete orthogonal sequence

in L2(p).

The following example illustrates the preceding theorem:

•

Section 33: HILBERT SPACES

295

Let Lz([ - l, l]) be the Hilbert. space of all square integrable functions on the closed interval [- 1 , 1] equipped with the Lebesgue measure. Then the functions 1, x, x2, are linearly independent. If the Gram-Schmidt orthogonalization process is applied to these polynomials, then the polynomials of the resulting orthogonal sequence are known as Legendre6 polynomials. We claim that, aside of scalar factors, the Legendre polynomials are given by the formula J.;'n (x2 - 1)2, the nth-order derivative of the polyno­ mial (x2 - l )". It is a custom to normalize the polynomials at l , in which case the Legendre polynomials are given by Example 33.13. .

.

•

l d" --1 (x2 - 1)11 • Pn(X) = -1 1 11! 2 dx 1 Clearly, each Legendre polynomial P11 is of degree 11 (since it is the nth-order derivative of a polynomial of degree 211) and satisfies the condition P11( 1 ) = 1. By Theorem 33. 12, in order to prove that the Legendre polynomials result from the Gram­ Schmidt orthogonalization process, we need only to show they are mutually orthogonal. Pick n and m such that n > m. To verify that P11 is orthogonal to P111, it suffices show that f�1 xk P11(x) dx = 0 for all k < n. The proof of this claim is based upon the observation that the function (x2 - 1)11 = (x - 1)" (x + 1)11 and all of its derivatives of order less than or equal to n - 1 vanish at the· points ±1. So, integrating by parts k times (where k < 11) and using the preceding observation, we get

1 1 xk P11(x)dx -I

-

d" 1-11 xk dx" (x2 - 1)" dx ( - l )kk! 1 d" -k 1 ----:- (x2 - lY' dx 2" n! - 1 dx 11-k

1 --

__

2"n!

I

(- l)kk! dll -k- 1 .., (x- - l)" 2"n! dxll-k- 1 -1

I

=

0.

Therefore P11 ..L P111 holds for all n =/= m. By Theorem 33.12, ( P11 } is a complete orthogonal sequence in the Hilbert space Lz([ -1, l ]). Finally, it is not difficult to see that the norms of the Legendre polynomials are given by Therefore, the sequence of polynomials Qo. Q 1 . Q2 , . . . , where II Pn l l =

J21�1 •

QII (x)

=

F[I I

n + I d" \"2 ( 2 dxII .

l_ _ 211 n.

_

I)1

1 .

is a complete orthonormal sequence of the Hilbert space L2([ - 1 , 6Adrien-Marie

r

I]).

•

r

Legendre ( 1752-1833), a F ench mathematician. His majo contribution was on

elliptic integrals that provided the analytical tools for modem mathematical physics. He spent over 40 years of his life attempting to prove Euclid's parallel postulate.

Chapter 6: HILBERT SPACES

296

EXERCISES 1. 2.

Verify that the inner product space L2(p) of Example 33 is a Hilbert space. Show that the Hilbert space L2[0, oo) is separable. 3. Let { 1/111 } be an orthonormal sequence of functions in the Hilbert space L2[a, b] which is also uniformly bounded. If {an} is a sequence of scalars such that an 1/111 -+ 0 a.e., then show that lim a11 = 0. 4. Let {¢11} be an orthonormal sequence of functions in the Hilbert space L2[- l , 1]. Show that the sequence of functions { 1/1n}, where

(x = Y,, )

( 2 )4 ( (x -2-)) ' b-a

¢n

2

b -a

-

b+a

is an orthonormal sequence in the Hilbert space L2[a, b]. 5. Show that the function (-, ·): X x X -+ C defined in Theorem 33.5 is a well-defined inner product on X that induces the norm of X. 6. Show that the closed unit ball of l2 is not a norm compact set. 7. Show that the Hilbert cube (the set of all x = (XJ , x2 , . . . ) e £2 such ihat lxn I ,::::: t holds for all n) is a compact subset of l2. 8. Show that every subspace M of a Hilbert space satisfies M = M .L.L . 9. For two arbitrary vector subspaces M and N of a Hilbert space, establish the following: a. b.

10. 11.

(M + N).L = M.L n N.L , and if M and N are both closed, then (M n N).L

= M.L + N.L .

Let X be an inner product space such that M = M .L.L holds for every closed subspace M. Show that X is a Hilbert space. Consider the linear operator V: L2[a, b] -+ L2[a, b] defined by

f(x ) = 1xf(t)dt.

V

12.

Show that the norm of the operator satisfies II V II =:: b - a. Let {x11 } be a norm bounded sequence of vectors in the Hilbert space l2, where x11 = (x�1 , x2_, x3, . . . ). If for each coordinate k we have lim11--+oc xJ: = 0, then show that lim (x , y) 11-+00 ,

13.

=0

holds for each vector y E l2. Let H be a Hilbert space and let {xn } be a sequence satisfying lim (x11, y) = (x, y) 11-+00

for each

y E H. Show that there exists a subsequence {xkn } of {xn } such that Jim n-+oo

}

II

X - - L.J �Xk·' = 0. n

i=l

Section 33: HILBERT SPACES 14.

Let p: [a, b] --+ (0, oo) be a measurable and �ssentially bounded function and for each n = 0, I , 2, . . . let P11 be a non-zero polynomial of degree n. Assume that

1 bp(x)P11(x)Pm(x) dx =

15.

297

0

for

n -:/:: m.

Show that each P, has n distinct real roots all lying in the open interval (a, b). In Example 33 we defined the sequence Po, P1, Pz, . . . of Legendre polynomials by the formulas

I dll P11(x ) = -- -- (x.2 l )II . 2" n ! dx " We also proved that these are (aside of scalar factors) the polynomials obtained by applying the Gram-Schmidt orthogonalization process to the sequence of linearly independent functions { 1, x, x2, . • . } in the Hilbert space L2([-1, 1]). Show that for each n we have -

P11 ( I) = 16.

17.

1

II P" II =

and

v�. z;;+I

Let {Ta }aeA be a family of linear continuous operators from a complex Hilbert space X into another complex Hilbert space Y. Assume that for each x E X and each y E Y, the set of complex numbers {(Ta(x), y): ex E A } is bounded. Show that the family of operators {Ta laeA is uniformly ·norm bounded� i.e., show that there exists some _ constant M > 0 satisfying II Ta II .::= M for all ex E A. Let {¢11} be an orthonormal sequence in a Hilbert space H and consider the operator T: H -+ H defined by 00

T(x) = L cx"(x, ¢n )¢", 11=1

18.

where {cx11} is a sequence of scalars satisfying lim cx11 = 0. Show that T is a compact operator. Assume that T, T*: H --+ H are two functions on a Hilbert space satisfying (Tx, y)

=

(x, T*y)

for all x, y E H. Show that T and T* are both bounded linear operators satisfying liT II = II T* II 19.

and

II TT* II = II T 11 2 •

Show that if T: H -+ H is a bounded linear operator on a Hilbert space, then there exists a unique bounded operator T*: H -+ H (called the adjoint operator of T) satisfying (Tx, y) = (x, T*y)

for all x, y E H. Moreover, show that II T II = liT* II.

Chapter 6: HILBERT SPACES

298

34. ORTHONORMAL BASES Recall that an orthogonal set S in an inner product space is said to be orthonormal if Us II = l holds for each s e S. A complete orthonormal set is a maximal orthonormal set. By Theorem 33.1 1, we know that every inner product space has a complete orthonormal set.

Definition 34.1. A complete orthonormal set in a Hilbert space is known as an orthonormal basis. An orthonormal basis will be denoted by {e; };e/ . That is, a family of vectors and only if l.

2.

{e; };e1 of a Hilbert space is an orthonormal basis if

(e;, ej) = �ii• where �ij is Kronecker's7 delta (defined by �ij = and �ij = 0 if i :f: j); and (x, e;) = 0 for all i implies x = 0.

l if i

=

j

The orthonormal sets that are bases are characterized as follows:

For an orthonormal family {e; };e1 of vectors in a Hilbert space the following statements are equivalent. l . The family {e; };e / is an orthonormal basis. 2. If x .l e; for each i e I , then x = 0. 3. For each vector x we have (x, e;) :f: 0for at-most countably many indices i and x = L;e1 (x, e;)e;, where the series converges in the norm. 4. For each pair of vectors x and y we have (x, e;) :f: 0 and (y, ei) :f: 0 for at-most countably many indices and (x, y) = L;e1(x, e;)(y, e;). 5. (Parseval's8 Identity) For each vector x we have llxll 2 = Li e/ l(x , e;) l 2 . Proof. ( l ) =? (2) This is an immediate consequence of the definition of the Theorem 34.2.

orthonormal basis. (2) =? (3) Fix some vector x . By Bessel's inequality (Theorem 32.1 0), we know that (x, e;) :f: 0 for at most countably many indices i. Let i 1 , i2 , be an enumeration of the indices for which (x, e;) :f: 0. (Here we consider the case where e;) =I= 0} is countable; the finite case can be dealt in a the set of indices {i e I : similar manner.) From Bessel's inequality, it also follows that 2::1 l(x, ei.t) 1 2 < •

•

•

(x,

llx 112 < oo. 7Leopold Kronecker ( 1823-1891 ), a German mathematician. He made imponant contributions to

the theory of equations and to the theory of algebraic numbers. Due to his belief that mathematics

should deal only with "finite numbers," he vigorously opposed Cantor's introduction of set theory. He is also famous for the quotation "God created the integers, all else is the work of man." 8Marc-Antoine Parseval des Chenes ( 1755-1 836), a French mathematician. He published very little

and he is only known for this identity. He was a royalist who was imprisoned in 1792 for publishing poetry against Napoleon's regime.

Section 34: ORTHONORMAL BASES

299

Now, the Pythagorean Theorem 32.7 guarpntees

2 m m L(x, e;t )eit = L j (x, eh)l 2 k=ll

k=ll

for all n and m. This shows that the sequence {2=�= t (x, eh)eh} is a Cauchy sequence, and thus the series 2:�1 (x, e;, )eh Li e/ (x, ei )e i is norm convergent. Let y = x - 2:�1(x, Since (x - L���(x, eit )eit , e;. ) = 0 for all m > I , 0 for all 11. it follows from the continuity of the inner product that (y, e;.) Consequently, (y, ei) = 0 for each i E I, and so, by our hypothesis, y = 0 or

=

eiJeh ·

x = L; e1(x, e;)e;. (3) ==> (4) Let x

=

=

Lie 1(x, e;)e; .and y = Lj e 1 (y , ej)ej. Since there are

only at most a countable number of non-zero terms in each sum and the two series converge in norm to x and y, respectively, it follows from the joint continuity of the inner product that

(x, y) =

e )

( Lie/

(x, e;)ei, L(y , j )ej

je/ = L L(x , ei)(y, ej )(ei, ej ) ie/ j eJ

= L(i x, e/

ei)(y, ej).

=

(4) ==> (5) The desired identity follows immediately by letting y x. (5) ==> ( 1 ) Suppose that {e;}ie / is not an orthonormal basis. This means that {ei };e / is not a maximal orthonorm�.I set. Thus, there exists a non-zero vector y that is orthogonal toeache; . But then, our hypothesis implies IIY f = L ; e 1 l (y , e;) l 2 0 contrary to the fact that y is a non-zero vector. Therefore, {e; };e1 is an orthonor­

mal basis, and the proof of the theorem is finished.

= •

If {e; };e/ is an orthonormal basis in a Hilbert space and x is an arbitrary vector, then thefamily ofscalars { (x, e; )Jiel is referred to as the family ofFourier9 coefficients ofx relative to the orthonormal basis {e;};e/· Definition 34.3.

The Hilbert spaces with countable orthonormal bases are precisely the infinite­ dimensional separable Hilbert spaces. 9Jean Baptiste Joseph Fourier ( 1768-1830), a French mathematician. He is the founder of Fourier analysis. In his 1807 memoir, On the Propagation of Heat in Solid Bodies, he introduced the idea of expanding functions as trigonometric series. Although this was a controversial revolutionary idea at that time, today it is the basis of modem "Fourier analysis."

Chapter 6: fiLBERT SPACES

300

An infinite-dimensional Hilbert space H is separable if and only if it has a countable orthonormal basis. Moreover, in this case, every orthonormal basis of H is countable. Theorem 34.4.

Proof. Let {e1, e2 , • • • } be a countable orthonormal basis in a Hilbert space H. Then the set of all finite linear combinations of the e, with rational coefficients (call a complex number rational if its real and imaginary parts are both rational) is a countable dense subset of H. Therefore, H is a separable Hilbert space. For the converse, assume that H is a separable Hilbert space and let {x 1 , x2, • • • } be a countable dense subset of H. We claim that there exists a strictly increasing sequence of natural numbers k1 < k2 < k3 · · · such that {xk, , Xk2, • • • } is a linearly independent sequence satisfying

for each n. We shall establish the existence of such a sequence by induction. Start by observing that we can assume x, # 0 for each n. Let k1 = I . For the induction step, assume that we selected natural numbers k1 < k2 < · · · < k, such that the vectors {xk, , Xk2, • • • , Xkn} are linearly independent and

If every vector x; with i > k, lies in the linear span of {xk. , Xk2 , • • • , xkJ, then the set {x 1, x2 , • • • } spans a finite dimensional vector subspace, and so it cannot be dense in the infinite dimensional Hilbert space H , a contradiction. Hence, there exists some vector x; with i > k, which does not lie in the linear span of {Xkp Xk2 , • • • , XkJ• Let

k,+l

=

min { i E lN:

X;

is not in the linear span of {xk, , Xk2 , • • • , Xkn } } •

Clearly,

This completes the induction and the proof of the existence of the sequence {xkn } . Since {x1, x2, • • • } is dense in it follows that the linear span of {xk, } is also dense in H. So, if we apply the Gram-Schmidt orthogonalization process to the sequence {xkJ, we get a complete orthogonal sequence. Normalizing this orthogonal sequence yields a countable orhononnal basis for H.

H,

301

Section 34: ORTHONORMAL BASES

For last part, assume that {b�t b2 , • • . } is a countable orthonormal basis of a _ For each n let Hilbert space and that {e; };e1 is another orthonormal basis of

H

1., =

H.

{i E /: l(e;, b.,)l �}· >

From Parseval's identity 1 = ll bu ii 2 = L; e; l ( bu , e; ) l 2 , t follows that 111 is a finite set. On the other hand, if i E I, then using Parseval 's identity once more, we see that 1 = lle; 11 2 = I:: 1 l(e;, bu) l 2, and so (e;, bn ) i= 0 for some n. Therefore, I = U: 1 111• This shows that I is at most countable-and hence, a countable • s�.

i

In Example 33.13, we saw that the sequence of normalized Legendre polyno­ mials Qo, Q 1 , Q 2, . . . , where

Q (r) · "

� n + d'' 2 2n n! 2 dx11 • ' 1

1

--

-- (r - 1 )11

is an orthonormal basis of the Hilbert space L 2 ([ - 1 , 1 ]). We now present two more examples of orthonormal bases in some classical Hilbert spaces. The first example exhibits an orthonormal basis in the Hilbert space L 2 ([0, oo)). Example 34.5. Let us denote the differential operator by D, i.e., Df = f' and ok f = f (k) , the kth derivative of f. The Laguerre 1 0 polynomials are defined by the fonnulas

for n =

0, 1, 2, . . . .

Clearly, each Laguerre polynomial L 11 (x) is of degree

define the Laguerre functions

¢o, 2, . . .

n (x) = We claim that ( o, 2, 4>3, . . . } is

by

�n. e-i Ln (x), an

n = 0, 1,

2,

11.

Next, we

....

orthononnal basis in the Hilbert space

L z([O, oo)) of

all square integrable functions on the interval [0, oo) equipped with the Lebesgue measure. This claim will be established by steps. STEP I: The sequence (o, 2, 4>3, . . . } is orthogonal.

Let m <

Then we need to show that

J000 Lm (x)L11(x)e-x dx = O. Since

polynomial of degree m, it suffices to verify that for all k < n we have 11.

L111(x) is a

10Edmond Nicolas Laguerre ( 1 834-1 886), a French mathematician. He worked in geometry and on approximation methods in analysis.

Chapter 6: HILBERT SPACES

302 To establish

() *

, integrate by parts k times to get

Therefore, the Laguerre functions are mutually orthogonal. STEP ll:

The sequence

(

.{¢o, ¢2, ¢3, . . . } is

orthonormal.

Taking into account *) and integrating by parts n times, we get

l e-2 Ln Il .. 1

2

n!

=

_

1

--2 (n!)

1-? -

1 looo 1oo Ln(x)Ln(x)e-.\. dx -(-l)nx" Ln (x)e-x dx 1oo (-l )nxn Dn(x"e-x. )dx 1 looox' e-.\. dx =

o

- ( ! 2 (n!)(n!) = n ) STEP ill:

The

f n

o

1.

sequence {¢o, ¢2 , ¢3, . . . } is

Assume that a function

2

= -2 (n!) (n!) o

(n!)- o

1

(n!)

complete.

2 0, oo)) satisfies f l_ ¢ for each 11, i.e.,

f E L ([

We must show = 0 a.e. To this end, let holds for all = 0, 1 , . . . This implies

. foooxng(x)dx

n

g (x) f(x)e-f =

and note

J;o g(x)xLn(x) dx = 0

= 0, for n = 0, 1 , 2, . . . .

(**)

g

f

To show that = 0 a.e., it suffices to prove that = 0 a.e. To see this, start by considering the complex valued function F(s) =

loooe-.ug(x) dx

define for all complex numbers s with real part Res > 0. Next, fix a complex number so with Reso > 0. Then there exist some neighborhood V of so and some a > 0 such that Res > a holds for all s E V. This implies that there exists some > 0 such that

M ae-sx g(x) '-xe-sx g(x) Mlg(x)l I I I I x E oo). as

=

<

holds for all s E V and all [0, Now, a glance at Theorem 24.5 (and the discussion after its proof) guarantees that F is differentiable at so and that F'(so) = -]000 In other words, F is an analytic function.

xe-sxg(x) dx.

303

Section 34: ORTHONORMAL BASES

Next, observe that for each n, Holder's inequality implies

Ia In particular, for each s Ia

00

lx11g(x)l dx

00

=

!f(x)lx 11 e-� dx

00

)!

x2n e -x dx 2

=

IIJIIJ(2n)!.

and each n, we have

>0

foo

< 11!11 ( Ia

oo s11-lx" g(x)l dx < llfii-J(2n)! s" n!

n!

=

a11.

<s <

1· Since limn �oo a�:1 = 2s, it follows that the series .L�o a11 converges for all 0 But then, Theorem 22.9 (see also Exercise 10 of Section 22) implies that the series L�o ( - 1 )11 (s;;r g(x) = e-sx g (x) defines an integrable function and that

oo s" oo x"g(x) dx e-sx g(x) dx L( -1 )11 F(s) Loo n. Lo o 11=0 holds for all 0 < s < � - Now, a glance at (**) yields F(s) Loo e-s.\ g(."C)dx for 0 < s < 2-I . Since F is an analytic function, the identity theorem for analytic functions implies F (s) 1

=

=

=

0

=

0,

.

=

0

for all complex numbers s with Res > 0; see for instance [2, Theorem 16.25, p. 462]. That is, the Laplace transform of g equals zero and so g = 0 a.e. (see Example 30.12), and the proofs of our claims are finished. • Example 34.6. Consider the weighted Hilbert space Lz(p) with the weight p(:r) = e-�.r• defined over the whole real line ( -oo, oo). The functions 1'

,.2

"' "" ...

"'

,.3 , . . .

' ·"

belong to L2(p) and they are clearly linearly independent. We claim that the linear span of this sequence i� dense in Lz(p). To see this, assume that a function f E L 2(p) satisfies foo f(x).�11 e_ l2·r- dx = 0 c10r each n = 0 , 1 , 2 , ·; . . Now, coosa'der the funct10n g(x) = � f(x)e- �x-. Since the functions lz(x) = f(x)e-lx- and B(x) = e-�x- belong to L 2 (1R), it follows that g = Jze E L 1 (1R) and

_00

·

!-00oo

x" g(x) dx

=

0, n = 0, 1, 2,

..

.

.

We must show that f = 0 a.e. or, equivalently, that g = 0 a.e. To see this, we introduce the complex-valued function G: C --+ G(z)

=

!00

j (x)e- � x2 ixz dx .

-oo

C by

Chapter 6: HILBERT SPACES

304

It should be clear that this (complex) integral is well defined and that (as in the previous example) G is differentiable at each point (and so G is an entire function). At z = 0, we have

for each

n = 0, l , 2, . . .

.

Therefore, G(z) = 0 for each z E C, and consequently =

100f(x)e-ix2e-ls.�: dx 1 00 g(x)e-1sx -oo

-oo

dx

=

0

for all -oo < s < oo. In other words, the "Fourier transfonn" of g is zero, and this guarantees that g = 0 a.e.; see, for instance [15, p. 408] or [26, p. 187]. We now claim that if we apply the Gram-Schmidt orthogonalization process to the sequence of functions 1, x, x2 , . . . , then the members of the resulting complete orthogonal sequence {H,) (aside of scalar factors) are the Hermite 1 1 polynomials which are defined by I ,.2 \ nI1 I .--:;.. n • •, (• · ) - (- '1 )II t:- 'iI ..,.2 u .... \c: - } ' ·

where D denotes the differentiation operator. Clearly, each H,(x) is a polynomial of degree n. Now, if k < n, then-integrating by parts k times-yields

This implies (Hm. H,) =

100 Hm(x)H,(x)e-�';

dx = 0

-oo

for n # m. Finally, a glance at Theorem 32. 1 1 guarantees that (aside of scalar factors) the sequence {H, } is indeed the one obtained by applying the Gram-Schmidt orthogonalization process to the sequence of linearly independent functions I, x, x2 , x3 , . . . . Nonnalizing the sequence of functions {Hn) we get to a complete orthononnal basis for L 2(P ) . •

Next, we shall use Theorem 34.2 to present a concrete realization of an arbitrary Hilbert space. To do this, we need some preliminary discussion. Recall that a linear operator T: -+ Y betwee n two norrned spaces is called norm preserving (or an isometry) if.IITxll = llxll holds for all x Similarly, a linear operator L: H1 -+ H2 between two Hilbert spaces is said to be inner product preserving if (Lx, Ly) = (x, y) holds true for all x, y e H1•

X

E X.

1 1 Charles Hermiie ( 1 822-1901 ), a French maihemaiician. He worked in Ihe Iheory of funciions. He was lhe firs1 1o prove (in 1873) 1ha1 e is a Iranscendemal number.

Section 34: ORTHONORMAL BASES

305

From Theorem 32.6 the following result should be immediate:

linear oper a tor L: H1 H2 between two Hilbert spaces i s norm preserviug if and only if it is inner product preserving. Lemma 34.7. A

�

Now, let Q be an arbitrary nonempty set. If x: Q C is an arbitrary complex­ valued function (which it will also be denoted as a family {x(q)}qeQ ). then its e2-norm is defined by �

llxl lz =

(L ) q eQ

l x(q) l 2

I

2

complex-valued function x: Q C is said to be square summable if llx112 < This is, of course, equivalent to saying that x(q) #- 0 for at most countably many q and i2f q1, q2, .. . is an enumeration of the set {q E Q: x(q) =f:. 0}, then z=:, jx(q11)1 It should be clear that the collection l2(Q) of all square summable functions with the pointwise algebraic operations and the l2-norm is a normed vector space. In actuality, it can be easily shown that l2( Q) is a Hilbert space.

A

�

oo.

< oo.

The vector space l2 ( Q) of all square summable complex-valued functions defil7ed on a nonempty set Q under the inner product (x, y) qL:eQ x(q)y(q) is a Hilbert space. Lemma 34.8.

=

We now ready to establish that every Hilbert space is linearly isometric to a concrete l2(Q) Hilbert space. are

Every Hilbert space H is linearly isometric to a Hilbert space ofthethelinear formoperator l (Q) . Specifically, if {e; };e/ is an orthonormal basis of H, then L: H lz(l), defil1ed by

Theorem 34.9. z

�

L(x) = {(x, e;)};e/.

is a sw]ective lil1ear isometry.

Clearly, L is linear and by Parseval 's Identity (Theorem 34.2(5)) it is also an isometry. \Ve leave it as an exercise for the reader to show that L is also a surjective linear operator. ProQf.

•

Chapter 6: Hll..BERT SPACES

306

Corollary 34.10.

An infinite dimensional Hilbert space H is separable ifand only if it is linearly isometric to

l2 •

EXERCISES 1. 2.

(ei li e/ and (h ljeJ be two orthonormal bases of a Hilbert space. Show that I and J have the same cardinality. Let (ei l ie/ be an orthonormal basis in a Hilbert space H. If D is a dense subset of H, Let

then show that the cardinality of D is at least as large as that of I. Use this conclusion to provide an alternate proof ofTheorem 34.4 by proving that for an infinite-dimensional Hilbert space

c.

H has a countable orthonormal basis. H is separable. H is linearly isometric to

Let

I be an arbitrary nonempty set, and for each i E

a. b.

3.

H the following statements are equivalent:

l2.

family of functions

4.

l x II

Let

(ei liei = 1.

elements.

5.

(ei lie/

i iet ei

=

X! i l· Show rhar the

is an orthonormal basis for the Hilbert space l2(I).

I: l(x,

k2

be an orthonormal basis in a Hilbert space and let x be a unit vector, i.e,

Show that for each k

E 1N the set (i E

e; )I

:::: t l has at most

M be a closed vector subspace of a Hilbert space H and let (e; l i e/ be an or­ thonormal basis of M; where M is now considered as a Hilbert space in its own right under the induced operations. If x E H, then show that the unique vector of M closest

Let

to x (which is guaranteed by Theorem

33.6) is the vector

y =

L; e1(x, e; )e;.

(fn

6. Let (e11 l be an orthonormal basis of a separable Hilbert space. For each n, let J,, =

7. 8. 9.

10.

en+ l - en . Show that the vector subspace generated by the sequence

Prove Lemma

34.7.

Prove Lemma 34.8. Complete the details of the proof of Theorem Let

34.9.

L2 [0, f L [0, a2, . . . 1 f{¢n L 2(JJ.) {aJ, f ¢n dJ.L.

(en l be an orthonormal sequence of vectors in

11.

(¢n { ¢n {¢J, ¢2, 2::�1 L,�1

12.

L2[0, f L,�1 (f, L2[0, ¢n)l2 2 11/11 2::� 11(/, . L2(JJ.), J.L f)

the Hilbert space

2rrl we have

pose that for each continuous function f in Show that

l is dense.

(en l is an orthonormal basis.

=

l be an orthonormal sequence of vectors in the Hilbert space 2rrl we have = pose that for eachcontinuous function in 2 Show that l is an orthonormal basis. Let . . . l be an orthonormal basis ofthe Hilbert space where Let

measure. Fix a function

E

coefficients relative to

1. i.e.,

an¢n a11¢n

and let

an = J

2rr ]. Sup­

en )en .

2rr]. Sup­

is a finite

be its sequence of Fourier

Show that (although the series

need not converge pointwise almost everywhere to

the Fourier series

can be integrated term-by-term in the sense that for every measurable set

E we have

1 f dj.L f 1 tPn dj.L. E

=

n=l

an

E

Section 35: FOURIER ANALYSIS [HINT: If s11

=

Lk=l a11r/J11, then the Cauchy-Schwarz inequality implies

I Lt L d!L -

13.

307

s, d!L

1

2

<

(L

If - Snl d!L

r

� II / - s,II'!L'(E). ]

Establish the following "perturbation" property of orthonormal bases. If orthonormal basis and u;· };e/ is an orthonormal family satisfying

{e; };e/

is an

L lie; - /; 112 < oo,

ie / then {f; };e/ is also an orthonormal basis. 35.

FOURIER ANALYSIS

In this section, we shall study some basic properties of periodic functions. Recall that a function f: ffi. -+ C is said to be periodic if there exists some p > 0 such that f(x + p) = f(x) for each x E ffi.. The positive number p > 0 is called a period of f. In essence, a periodic function is completely determined by its values on the closed interval [0, p]. Any function f: [0, p] -+ C that satisfies f(O) f(p) gives rise to a periodic function via the formula f(x + p) f(x) for each x E ffi.. Notice that if f: ffi. -+ is a periodic function, then the function g: ffi. -+ C, defined by g(x) = f({;x), is a periodic function with period 2rr. From now on, unless otherwise stated, all periodic functions encountered in this section will be assumed to have period 2rr . As a matter of fact, they will be defined only on [0, 2rr] (and they will be tacitly assumed defined on all of ffi. f(x)). That is, a periodic function is any function via the fonnula f(x + 2rr) f: [0, 2rr] -+ C satisfying /(0) = /(2rr). The closed interval [0, 2rr] will always be considered equipped with the Lebesgue measure: we shall write dx instead of d).. ( x). The Hilbert space L2[0, 2rr] is as­ sumed equipped with the inner product

=

=

C

=

(/, g)

=

["f(x)g(x) dx,

and so it induces the standard Lz-norrn II / 112

=

.j(f, /)

= ([

"

l f(x)l 2 dx

). I

'

A straightforward computation shows that

if if

nn =

=j:. m

m.

Chapter 6: HILBERT SPACES

308

This means that the countable collection of functions e"'x = cosnx + 1 sin nx, n = 0, ±1, ±2, ±3, . . . , is an orthogonal subset of L2[0, 2rr]. Therefore, the collection

I

1

e 11/X:. n - 0, ±1, ±2, ±3, . . . _

.Jirr

l

is an orthonormal set of functions in L2[0, 2rr ]. However, for historical reasons, these functions are normally used in their non-normalized form. We shall establish here that the orthogonal set

{ewx :

11 = 0, ±1, ±2, . . . }

{ Jk, e'n.\· :

is, in fact, a complete orthogonal set-and so n = 0, ±1, ±2, . . } is an orthonormal basis for the Hilbert space L2 [0, 2rr ]. A trigonometric polynomial is any periodic function P of the form P(x) =

1

m

2ao L(a11 +

II== I

.

cos nx + b sin nx)

n

ao, a1, .. . , au

where and b1 , . . . , bu are complex numbers. From 1 sm x it follows that sm nx =

elnX

_ e -11/X

and

-----

2z

COS 11X =

(1)

eu

= cos x +

-----

2

and so every trigonometric polynomial P can also be written in the form m

P(x) =

nx L Cue' n=-m

(2)

e'nx

for appropriate complex coefficients c,. Similarly, using the identities cos nx + 1 sin nx, cos( -nx) = cos nx, and sin( -nx) = -sin nx, we see that every expression P(x) as in (2) can be written as in (1). Thus, we have shown that every trigonometric polynomial P can be written in the following two ways: P(x) =

m 2ao n=lL(a,

1

+

.

m

cos nx + b, sin nx) =

L Cnewx, n=-m

where the coefficients are related by the identities

ao

= 2co,

a, Cn =

- c_n, b, = z(c, - c_n)

Section 35: FOURIER ANALYSIS

309

and taking the inner product of a trigonometric polynomial P with the function Now, kn e' . it follows from the orthogonality properties that 1 /. 2 ±m P(x)e dx, n 0, ±1, ±2 Cn =

11'

21l'

-

-IIIX

=

0

. . . •

.

Next, assume that f L 1 [0. 21l'] (nothing precludes f from being also complex­ valued). Then the inequality l f(x)e"rx I < lf(x) l coupled with Theorem 22.6 guarantees that f(x)e x E L 1 [0. 21l'] for each integer n. This observation will be used in the next definition. E

'

n

The Fourier coefficients of a function f E L [0, 21l'] are thetermsofthedouble sequence ofcomplexnumbers c-2 · c_ 1 , co. c 1 . c2 defined by Definition 35.1.

1

. . . •

Cn =

for each n = 0. ± 1 , ±2, is theformal series

.

..

.

• . . .

211' /. . f(x)e-"'·t dx 1

21l'

0

The Fourier series of a function f E L 1 [0. 21l'] 00

L Cn e'"x.

n=-oo

An important consequence of the Riemann-Lebesgue lemma (Theorem 25.4) is that the Fourier coefficients of an integrable function converge to zero. If f E L1 [0, 27r], then its Fourier coefficients satisfy

Theorem 35.2.

lim c, = lim c_, = 0. L 1 [0, 21l' ]. Then, by the Riemann-Lebesgue lemma (Theo­ n-oo

n-oo

Let f rem 25.4). we have 211' 211' /. /. lim f(x) sin nx dx = lim f(x) cos nx dx = 0. Proof.

E

,_ 00 0

,_ 00 0

Chapter 6: HILBERT SPACES

310

So, from the identity

c,

. 2 1 1 f21rf(x)e-mx dx = -- { JTf(x)[cos nx - 1 sin nx ] dx = -2n lo 2n lo 1 f(x)cos nx dx - -1 1 2JTf(x) sin nx dx, = -

12Jr

21Z' 0

21Z' 0

it easily follows that

•

Cn -+ 0 and c_, -+ 0.

Using that the set of functions { e' nx : n = 0, ± 1 , ±2, . . . } is orthogonal and the formulas cosnx = -----

2

and

sin nx =

e'"x - e-t nx 21

-----

we see that the functions

�, cos x , sin x, cos 2x, sin 2x, . . . , cos nx, sin nx, . . . are mutually orthogonal in L2[0, 2n]. With respect to this orthogonal set of func­ tions, the Fourier series of any function f e L [0, 2n] can also be written as

1

00

"'"""

� Cn e 11 =-oo where the coefficients

I IIX

=

1

2ao + �(a, cosnx + b11 smnx) . 00

"'"""

·

n= l

ao, a1, a2, . . . and b1, b2, . . . are given by

ao = 2co = n� lo{21rf(x) dx, 2 On = Cn + c_n = _!_ { "f(x) cos nx dx, and n lo bn = l (Cn - c_n ) = _!_ {2Jrf(x) sin nx dx. n lo

If we express the Fourier series of a function in terms of the cosine and sine functions, then we shall also refer to the coefficients and as the Fourier coefficients of f. We are now ready to establish a result that will guarantee that the set of orthog­ onal functions { e'nx : n = 0, ±1, ±2, . . . } is complete in L2[0, 2n].

ao, a1, a2,

.

.

.

b1, b2,

•

.

.

311

Section 35: FOURIER ANALYSIS

Theorem 35.3. A Lebesgue integrable function over

coefficients all vanish is almost everywhere zero. 1

[0, 2rr] whose Fourier

Let f E L [0, 2rr] be a function satisfying !.2rrf(x)e'1 x dx 0 (t) 0 for all n 0, ± 1, ±2, . . . . Assume at the beginning that f is also real valued. We shall establish that f 0 a.e. holds true by considering two cases. First, we shall assume that f is also continuous and then we shall consider f to be an arbitrary integrable function. CASE Assume that f is a continuous function satisfying (t)for all integers n. Assume by way of contradiction that f =I= 0. This implies that there exists some x0 (0, 2rr) satisfying f(xo) =I= 0. Replacing f by -f (if necessary), we can assume that f(x0) 0. Choose some constant C 0 and some 0 < < min{ 1, x0, 2rr - x0} such that f(x) C holds for all x E I (x0 - x0 Now, we need a trigonometric polynomial that is greater than one on the interval (xo- x0 and less than one on the complement Ic [0, 2rr] \I. The reader should verify that the trigonometric polynomial P(x) cos(x - xo) satisfies P(x) for all x E I and 0 < P(x) < 1 on Ic. From (t), it follows that 2 1 f(x)[P(x)]" dx 1 f(x)[P(x)]" dx /.0 rr f(x)[P(x)]" dx 0 for 0, 2, 3, . . . . Therefore, 1 f(x)[P(x)]" d - 1 f(x)[P(x)r dx holds for all n 0, 1, 2, 3, . . . . We claim that the right-hand side of equation (*) is bounded while the left-hand side is unbounded. To see this, first observe that since 0 < P(x) holds for each x E Ic, we have 2 1 f(x)[P(x)]" dx < 1 1/(x)[P(x)]"l dx < /.0 rr lf(x)l dx. Proof.

=

=

=

1:

E

>

>

E,

>

=

+ E)

E + E).

E,

=

=

1+

1 + COSE

.

> 1

I

11 =

+

IC

=

=

1,

I

x =

IC

=

< 1

IC

IC

312

Chapter 6: HaBERT SPACES

Thus, the right-hand side of (*) is bounded. On the other hand, let J be a closed subinterval of I . Since P is strictly greater than one on the open interval I , it follows that there exists some a > I satisfying P (x) > a > I for each x e J . Therefore, for each n we have

[

f(x)[P(x) ]n dx >

[

" f(x)[P(x)] dx > Can /..(1).

This shows that the left-hand side of equation (*) is unbounded, a contradiction. This contradiction establishes that f = 0. CASE II: Assume that f E

Lt [0, 2JT] satisfies (t)for all integers n .

Consider the function F : [0, 2rr] -+

F(x) =

IR defined by

[xf(t)dt .

... u

J02rr

Clearly, F is a continuous function and from F(O) = 0 and F(2rr) = f(t) n dt = 0, it also follows that F is a periodic function. We claim that F ..L e' x holds for all n # 0. To see this, apply the integration by parts formula (see Exercise 12 of Section 3 1 ) to get 0 =

{h

Jo

e-rnxf(x)dx = e-"'xF(x) l 6rr -

= F(O) - F(2rr) - zn =

-z n

{ 21r

Jo

Jo{ 2rr

{h

Jo

(-zn)e-"u:F(x)dx

F(x)e-wx dx

F(x)e-w.r dx.

Therefore,

{ 2rr

Jo

F(x)e-"'x dx

=

0 for

n =

± 1 , ±2, . . . .

Next, let C = 2� J02rr F(x)dx, and consider the continuous function G: [0, 2rr] -+ IR defined by G(x) = F(x) - C.

Since 1 ..L e"':r , it follows that G is perpendicular to er nx for all n -::J: 0. Any easy inspection also shows that G (x) ..L 1 holds. That is, G ..L e' "x holds for all integers

Section 35: FOURIER ANALYSIS

313

n = 0, ±1, ±2, ±3, . . . . But then, by CASE G(x) = 0 for all x E [0, 2rr]. This implies F(x) = for f(t) dt = C for all x E [0, 2rr ]. Since F(O) = 0, it follows that C 0, and so for f(t) dt = 0 for all x E [0, 2rr]. This implies that f = 0 a.e.

I,

=

II

(see Exercise 19 of Section 22), and the proof of CASE is complete. Finally, consider the complex case. That is, assume that f = g + l h is an inte­ grable function whose Fourier coefficients all vanish. Then, it easily follows that both g and h have their Fourier coefficients all equal to zero. Therefore, by the preceding conclusion, g = h = 0 a.e. This implies f = 0 a.e., and the proof of the theorem is complete. •

The two sets of orthogonalfunctions I. {e"1x: n = 0, ±1, ±2, . . . }, and 2. { �, cos sin x, cos 2x, sin 2x, . . . , cos nx, sin nx, . . . } are both complete in the Hilbert space L2[0, 2n]. Proof. By Theorem 34.2, it suffices to show that if f E L2[0, 2rr] is perpen­ dicular to e1 11x for each integer 11, then f = 0. So, assume that f E L2[0, 2rr] satisfies f l. e1 11x for each integer n , i.e., assume that the Fourier coefficients of f Corollary 35.4.

x,

are all zero. Since L2[0, 2rr] c L1 [0, 2rr] holds, it follows from Theorem 35.3 that f = 0. Therefore, the set of functions {e111x: n = 0, ±1, ±2, . . . } is a complete orthogonal subset of the Hilbert space L2[0, 2rr ], and the proof is complete. • Parseval's identity for the complete sequence of orthogonal functions {e -111x } takes the following form:

Iff E L2[0, 2rr], then its Fourier series I::-oo cue-1'1.-c is norm convergent to f in L2[0, 2rr] and Corollary 35.5.

I1

-

2rr 0

Proof.

271'

l f(x) l 2 dx =

00

L

n=-oo

lcu l 2 .

From Corollary 35.4, we know that the collection of functions

I .../2iI ie-IIIX.. n

- 0 , ± I , ±2 , . . .

l

is an orthononnal basis for the Hilbert space L2[0, 2rr]. The conclusions now follow from Theorem 34.2. • Corollary 35.6.

Iff belongs to L2[0, 2rr], then its Fourier series

. -111x = a ""'"' Cne 2o + ""'"' � (a11 cos nx + b11 sm nx) � 00

00

11=-00

11=1

Chapter 6: HILBERT SPACES

314

can be illtegrated term-by-term in any closed subinterval [u, v] of[O, 2rr] in the sense that

lv "

f(x)dx

1

= 2ao(v =

a; Let Sn (X) we also know that II f - Sn II then Proof.

1 "

11 f(x)dx

-

1 11 "

-

u)

v l +L oo

n=l "

(a, cos nx + b, sin nx)dx.

+I:Z=1 (a, coskx + b, sinkx). From Theorem 34.2,

� 0. Now, if [u, v] is a closed subinterval of [0, 2rr],

Sn (x)dx

v l � ({ ,

<

l f(x) - S (x)l dx n lf(x) - s.(x)l 2 dx

- il f - sn ll v v - u.

This implies

n 1 -ao(v - u) + 2 k= I

L

vl II

(a, cos nx

+ b, sin nx)dx =

and the desired conclusion follows.

1

" 11

I

I

)' ({ )'

s,(x)dx

! 2 dx

�

lv "

f(x)dx, •

The fact that the Fourier series of a function f E £ 2 [0, 2rr] is norm convergent in the Hilbert space £2 [0, 2rr] should not be interpreted to mean that the partial sums of the Fourier series converge pointwise to f almost everywhere. Pointwise convergence of the. Fourier series is a much more difficult and delicate problem. The study of the convergence of the Fourier series is the subject of inquiry of the field of Fourier analysis with special emphasis on the following two basic problems:

1. 2.

The Convergence Problem: When does the Fourier series converge at a

given point x (or at each point of a given set)? The Representation Problem: If the Fourier series converges at some point x, what is the relationship between the sum of the series and the value of the function f(x) at the point x?

We remark that it is possible for the Fourier series to diverge at every point and, moreover, there exist continuous functions whose Fourier series diverge on an uncountable set. It is even possible to construct examples of continuous peri­ odic functions whose Fourier series diverge at a countable number of points in the interval [0, 2rr]; see also Example 35.1 1 at the end of this section. The most

Section 35: FOURIER ANALYSIS

315

striking result concerning the pointwise con'{ergence of the Fourier series was es­ tablished by the Swedish mathematician Lennart Carleson in 1966. He proved that the Fourier series of any function in L2 [0, 2rr ] converges almost everywhere. 1 2 The unpleasant situation regarding the pointwise convergence of the Fourier series can be corrected by considering the sequence of arithmetic means of the partial sums of the Fourier series. Our final objective in this section is to prove that the sequence of arithmetic means of the partial sums of the Fourier series of a periodic function f converges to the value of the function at every point of continuity. To do this, we need some preliminary discussion. For the rest of our discussion, f: [0, 2rr] C is a fixed periodic integrable function which will also be assumed defined on all of via the formula f(x + 2rr) f(x). Its Fourier series is "" • -IIIX = a2"0 + "" � < a cosnx + b, smnx ) � . 11=-CO 11=1 The nth partial sum of the Fourier series is I 1 2ao + L (ak cos kx + bk sin kx) k= J 1 [ ?rrf(t) dt ] + L [-1 ?rrf(t) cos(kt) dt ] cos kx ? - rr Lo k=J rr Lo 2rr 1 [ + L rr f(t ) sin(kt) dt sin kx ] 0 k=l 11 1- 12rrf(t) [-1 + L(cos kx cos kt + sink."( sin kt) ] dt 2 k=J rr o ] 2 i!' 1 1 [ -rr o f(t ) 2- + L cos k(x - t) dt k=I - -rr 12rr f(t)DII(x - t) dt, o where Dll (t) = 21 + L cos kt. k=J --+

1R

=

c,e

:>C

s,(x)

00

,

=

1

-

II

1

11

1

II

1

II

12

0n the convergence and growth of partial sums of Fourier series, Acta Matlzematica 116 ( 1966).

1 35-157.

Chapter 6: HILBERT SPACES

316

D1

is called the Dirichlet kernel of order The trigonometric polynomial Dirichlet kernel can be expressed in a closed form.

The Dirichlet ket·nel satisfies D11(t) = ( . �)t

Lemma 35.7.

n.

The

sin n +

2 sm 21

•

Proof. Notice that I

II

= 21

L ek

r

= 21

e -'

2 n 1

1

L e'k1

Dn(t) = 2 k=lL kt k=-n k=O e-w1 I] ------ - e-"11 [e'<2n+l)1 e11 - 1 2e'�(e'� -e-'�) e-' e'-- 2(e'2 -e-'2) ( �)t 2 � -

cos

+

1

2

II

'

e'
_

(

I ·1 (11+2>

I

11+2J1 -

-

� � ,

sin

n

•

t

•

_

--.,... ,..

+

sin

as desired.

(x)}f.

•

We now consider the sequence {a, of the arithmetic means of the partial and sums of the Fourier series of the function That is, =

( ) - so(x) s,(.x ) s2(.x)

a, .x

·

+

_

+

n+l

+···+

so(.x) �ao s, (x) n - , &

, 10r

_

0

.

1,2.. .

2 2 [ 1 11'f(t)Do(x -t) dt - 1o 11'f(t)Dt t) dt · {21rf(t)D,(x - t) dt ]] 21 1rrr Jo[ -t) dt o2 f(t) n LDk(X k=O 1 11'f(t)K,(x - t) dt,

Using the Dirichlet kernel and some algebraic simplifications, we see that .

a,(x)

=

I

n+1

··· +

=

1 -

=

1 -

rr

1r

1 rr o

+

1

1

+1

0

11

•

1

rr

(x -

+ ··

317

Section 35: FOURIER ANALYSIS

where the expression K,(t) =

1

ll

+1

IIL k=O

Dk(t) =

2(11

1 + 1)

.Stn Lk=O sin (k + -2 )t "2

II

1

1

of order n. Thus, the sequence {a, } of means of is now called the the partial sums of the Fourier series of f are given by the formula Fejer 13 kernel

1 a,(x) = -

121Tf(t)K,(x - t) dt.

Jr 0

It is not difficult to express the Fejer kernel in a close form. Multiplying the numerator and the denominator of K, (t) by 2 sin � and using the trigonometric identity 2 sin(k + !t) sin !t = cos kt -cos(k + 1)t� we get [ sin(n + 1) � ] 2 1 - cos(n + l)t 1 K,(t) 4(n + 1) sin2 � 2(n + 1) sin � This formula shows that the Fejer kernel is a positive even function. If, as usual, we put Kn(2krr) = then K, is a continuous function on all of JR. A simple computation (see Exercise 2 at the end of this section) also shows that = ---

=

"!1 ,

2 -

1 K,(t) dt 1T

Jr 0

=

1.

Now, let us take another look at the nth arithmetic mean of the partial sums of the Fourier series. By the preceding discussion, we have a,(x)

=

flo2rrf(t)K,(x - t)dt .!_ 1xx-+rr f(x rr 1T - 1 f(x u)K,(u) du. -1T .!_

=

rr

1

+ u)K,(u)dtt

rr

+

Jr

Taking into account that K,(-u)

1° f(x - rr

=

K,(u),

+ u)K11(u)du

=

it easily follows that

frrlo f(x - u)Kn(u)du,

13Leopold Fejer ( 1 880-1959), a Hungarian mathematician. He worked mainly on Fourier series and their singularities.

Chapter 6: HILBERT SPACES

318

and consequently, an(X) =

�r rr r Jo

[f

(x

+ u) + f(x - u)

·

2

]

(**)

Kn(u)du.

Next, we introduce a new function a1 defined by the formula )

af (X =

1.

1m

f (x + u) + f (x - u)

11--+0+

2

.

The function a1 is, of course, defined at those points for which the preceding limit exists. For instance, if f is continuous at some point x, then a1(x) = f (x) and if f has a jump discontinuity atx0, then a1(xo) = f<xo+>;t<xo->. We are now ready to state and proof the following remarkable result of L. Fejer concerning the convergence of the sequence of arithmetic means {an }. Theorem 35.8 (Fejer) .

Let f: [0, 2rr] � C be a periodic integrablefunction. Then, the sequence {an} ofarithmetic means of the partial sums of the Fourier series of f satisfies .

hm an(x)

n --+oo

= af(x) =

.

f (x + u) + f (x - u)

hm ------2

u--+0+

at every point x for which a1 (x) exists. Moreover, if f is continuous on some closed subinterval [a, b] of [0, 2rr], then {an} converges uniformly to f on [a , b]. Proof. Assume that a1(x) exists at some point x E [0, 2rr]. Then there exists

some 0 < 8 < rr such that

f (x + u) + f (x - u) 2

- af (X ) < E

for all 0 < u < 8. Notice that the 8 depends upon the E and the point x. f f is con­ tinuous on [a , b ], then notice that (in view of the uniform continuity of f) we can select the 8 to be independent of x E [a, b]. Now, from (*) and (**), it follows that

I

lan(x) - af(X) I =

<

[

]

{rr f(x + u) + f(x - u) - af(x) Kn(u)du 2 . rr Jo 2 { o f(x + u) + f(x - u) - af(X) Kn(u)du rr Jo 2 2

2

+rr

1rr o

f(x + u) + f(x - u) 2

- af (X ) Kn (U ) d U . ·

319

Section 35: FOURIER ANALYSIS

Now, observe that

2 1.8 rr o

u) + f(x - u) - af(X) K1 (U) dU 8 2 2 < � { EK11 ( t)dt < rrEK11(t)dt =E. rr Jo lo f(x

+

rr

Moreover, since K11(t) < 2(n+ l�sin2 � for each 8 <

x

<

rr ,

it follows that

2 !.7T f(x + tt) + f(x - u) - at (;\: ) K ( ) u 2 rr 8 f.7T f(x + u) + f(x - u) - a1(x) du -< C , < n+ 2 - (11 + )rr sin- o constant C f �.1 2 J;rr l � a ( ) l du is independent ofx. where the Finally, notice that i e "1:hoose some no such that I�I < E, then the pre­ ceding estimates show that -

11 U d

·

1

.

2

., 8

1

f<x+n> f(x-u>

�

=

-

f

1

x

holds true for all 11 > no (and all x E [a, b] if f is continuous on [a, b]). This completes the proof of the theorem. The following consequence of the preceding theorem plays an important role in applications. •

Let f: [0, 2rr] C be a periodic integrable function. As­ off converges at some point x and that sume that the Fourier series exists, then . 1 1 ao """""' ( ) = """" �"' c"e - · = 2 + � (a1 cos nx + b1 smnx ) . Corollary 35.9.

---+

af(x)

00

at x

n=-oo

r

00

n= l

Inthenparticular, ifthe Fourier series converges at some point of continuity off, f(x) = L c"e-'1 x = a; + L(a, cos nx + b11 sin nx). Proof. Assume that s holds and that exists. Theorem Since is also convergent to s (see Exercise 1 1 we know that 1 of Section 4), it follows that s = a1(x), and we are done. x

oo

oo

n=-oo

n=l

(x) a (x) ---+ af(x). Sn

af(X)

---+

{a11(x)}

By

35.8, •

Chapter 6: HILBERT SPACES

320

Here is· an example that demonstrates the far-reaching implications of Corol­ lary 35.9. Example 35.10.

Consider the periodic function f: [0, 21r] � 1R defined by f(x)

=

{

if 0 < X < 1l' 0 if 1r ::: x < 21r 1 if X = 21l'. 1

A direct computation shows that the Fourier coefficients of f are given by

ao

=

an

-

.!. 1r

f 2"f(x)dx

lo

Jo

= .!. [" l dx 1r

=

1,

.!. {21rf(x)cosnxdx = .!_ ["cos 1l' h 1l' h 1

r rr

bn - ...:.. 1 2 1r

Jo

1 f(x)sinnx dx = TC

1o"

x dx = 0, and

n

sinnx dx =

1

- cos ll1l' nTC

Thus, the Fourier coefficients satisfy ao 1, a11 = 0 for each n, b, = 0 for n even. Therefore, the Fourier series of f is given by =

(

b11

1 2 sinx sin3x sin5x sin7x - + - - + -- + -- + -- + · · · 1 3 5 2 1C 7

1

-

=

)

1 - ( -1) 1 llTC

11;

for n odd and

.

This series converges at each x; see Example 9.7. By Corollary 35.9, this Fourier series converges to f(x) at the points x of continuity of f. More precisely, we have 1 2 -+2

1C

(

sin x

sin3x

sin5x

3

5

sin7x 7

- + -- + -- + -- + · · ·

1

) { ;:::

f(x) if x E (0, 1r) U (TC, 2TC),

!

This conclusion is typical and very powerful for applications.

if X = 0, TC, 2TC.

•

Finally, we close the section with an example that guarantees the existence of a continuous periodic function whose Fourier series diverges on a given countable set. Example 35.11. This example demonstrates the existence of a continuous periodic func­ tion whose Fourier series diverges on a countable set of points. Consider the Banach space C [0, 2TC] of all continuous real-valued functions with the sup norm. Let X be the subspace of C[O, 2TC] consisting of all continuous periodic real-valued functions. Clearly, X is a closed subspace of C[O, 2TC], and hence, it is a Banach space in its own right Next, fix a point x E [0, 2TC] and define the linear functional S11: X -+ 1R by the nth partial sum of the Fourier series of the given function evaluated at the point x. That is, for

Section 35: FOURIER ANALYSIS

each I E X we let Sn(l) =

L qe-,J:..� 11

•

•

k=-11

321

{ 2rr = - J l(t)D1 (x - t)dt. I

o

1f

1

0

It should be clear that this formula defines a bounded linear functional on X whose norm (according to Exercise 3 at the end of the section) is given by the formula liS" II =

I -

la2rr ID"(x - t)i dt.

1f 0

Next, we estimate this norm. Since the Dirichlet kernel D, is periodic, it follows that

{27T

Jo

ID"(x - t)i dt

=

{27T ID11(t)l dt.

Jo

So, using the inequality jsin u I ::: ju I and changing the variable, we get

la2rr j D11 (l)j dt = la2rr sin (n . o

o

>

+

2 sm

!orr jsin(2n

- o

+

t) t

r 2

l)uj

lu i

dt

l sin u j du lu i

2n

=

211

> This implies

I

+

+

dLL =

rr ) +l k 1( .l: k rr k=O [ 1(k+l)rr - L (k J:.rr k=O

!orr jsin(2n. I)u I du o lsm uj la <2"+t )rr jsinuj

=

I )rr

-du lu i

o

]

jsin uj du =

{xt , x2 ,

2u

L (k +2 l)rr .

k=O

and consequently limiiS1111 = oo independently of the point x. XJ, • • • } be a (possibly dense) countable subset of the interval [0, 2rr]. Now, let Let S11,111 be the bounded linear functional (defined as above) on X whose value at I E X is the nth partial sum of the Fourier series qf I evaluated at the point x111• From the above discussion, for each fixed m we have lim

1/-+00 IISu,m

II = 00.

Chapter 6: HILBERT SPACES

322 .

Therefore, by the Principle of Condensation of Singularities (Theorem 28.1 1 ), there exists some f E X (i.e., a continuous periodic real-valued function f on [0. 2rr]) satisfying

L Cke-l kxm

lim sup ISn,m(/)1 = lim sup

fl

k =-n

fl-+00

for each

m.

=

00

This implies that the Fourier series of f diverges at each point Xm

•

•

EXERCISES 1.

Show that sin11 x is a linear combination of {1, sinx, cosx, sin 2x, cos 2x, sin 3x, cos 3x, . . . , sin nx, cos nx }.

2.

Furthermore, show that the coefficients of the cosine terms are zero when n is an odd integer, and the coefficients of the sine terms are zero when n is an even integer. Show that the Dirichlet kernel Dn and the Fejer kernel Kn satisfy I

rr

ltr D11(t)dr = ltr K11(t)dr = I I

-

rr

-1r

.

-1r

3. Show that the norm of the linear functional Sn defined in Example 35 satisfies

2 IISull = ..!_ { 7r1Du(X - t)! dt. rr

4.

lo

Show that the sequence of functions

I ( ;) , { ; ) }

I

2

2

I

2

cosx ,

(rr) 2

I

2

cos2x ,

(;) 2

I

2

cos3x ,

(;) 2

I

2

cos 4x, . . .

I

is an orthonormal basis in L2[0, rr]. Also show that the preceding sequence is an orthogonal sequence of functions in L2[0, 2rr] which is not complete. S. Show that the sequence of functions

j \G)

sin x ,

! G)

sin2x,

G)!

sin 3x ,

! G)

sin4x ,

·

·

·

I

is an orthonormal basis of L2[0, rr]. Also prove that this set of functions is an orthog­ onal set of functions in L 2 [0, 27r] which is not complete. 6. The original Weierstrass approximation theorem showed that every continuous func­ tion of period 27r can be uniformly approximated by trigonometric polynomials. Establish this result. 7. Find the Fourier coefficients of the function / (x)

=

I

}

0

if 0 < X <

if

!

-

2

zr.

� x < 2rr.

Section 35: FOURIER ANALYSIS 8.

323

Find the Fourier series of the function s nx if 0 _::: X < 1r f (x) - smx if rr _::: x < 2rr.

=

I

�

9. Show that for each 0 < x < 2rr we have

n= l

10. Show that

holds for all 0 _::: x 11. Show that

00

-.

smnx x = rr - 2 � LJ 11 ·

_:::

00 cos nx rr 2 x-2 = rrx - - + 2 l: --=-2 n2 n=l

= 0 yields :L� 1 ,:� = �� .) 4 oo ( cos2nx rr sin nx ) = -rr+4 L

2rr. (The value x ?

.c

3

3

?

n=l

_

n

11

holds for each 0 < x < 2rr. 12. Consider the "integral" operator T: L2[0, rr] --+ Lz[O, rr] defined by Tf(x) where the kernel K: [0, rr]

x

= !orr K(x, t)f(t)dt,

[0, rr] --+ 1R is given by

�

K (x, t) = LJ n=l

[sin(n + l)x] sinnt 2 11

Show that the norm of the operator T satisfies II T II = rr /2. [HINT: Use the basis described in Exercise 5.]

•

7

CHAPTER

_ __ _ _ _ _ _ __ _ _ _ _

SPECIAL TOPICS IN INTEGRATION

The powerful techniques of measure and integration theory are utilized in many scientific contexts. However, in a number of applications, set functions that are not measures appear naturally. For this reason a study of more general set functions promises to be very fruitful. In this chapter the set functions known as "signed measures" will be investigated. Loosely speaking, a signed measure is an extended real-valued a-additive func­ tion on a a-algebra of sets. The first section of the chapter deals with the algebraic and lattice structures of signed measures, while the following adheres to compar­ ison properties of signed measures. The most important comparison properties are those of "absolute continuity" and "singularity." Regarding absolute continu­ ity, the far-reaching classical Radon-Nikodym theorem is proven here: If a finite signed measure v on a a-algebra is absolutely continuous with respect to a a-finite measure J..L , then there exists a (unique) J..L-integrable function f such that v(A) i f dJ..L This powerful theorem is used to show that L;CJ..L) Lq(J..L) holds for all holds for each 1 After comparing signed measures, our attention is turned to regular Borel mea­ sures on a Hausdorff locally compact topological space Another classical result, known as the "Riesz Representation Theorem," is proved: If F is a positive linear functional on then there exists a unique regular Borel measure J..L such that F(f) Jf dJ..L holds for all f E As an application of this theorem, the norm dual of will be characterized in terms of regular Borel measures. I;

=

A E I;.

=

< p < oo.

X.

Cc(X),

=

Cc(X).

Cc(X)

325

�hapter 7: SPECIAL TOPICS IN INTEGRATION

326

The last two sections of this book deal with differentiation and integration in Rn . First, the study focuses on the differentiation of Borel signed measures on R". It will be shown that every Borel signed measure is differentiable al­ most everywhere. This basic theorem will then be used effectively to derive the classical results about ordinary derivatives of functions of bounded variation. Finally, a detailed proof of the familiar "change of variables formula" will be presented.

36. SIGNED MEASURES

:E

X,

Throughout this section, will be a fixed a-algebra of subsets of a set and the measures considered will be assumed defined on :E . Since a variety of measures will be studied, it is a custom (for simplicity) to call the members of :E the measurable subsets of Let JL and v be two measures, and let a > 0. Then the two set functions JL + v �11d a f.l· defined by

X.

(/L + v)(A) (aJL)(A)

JL(A) + v(A), = a JL(A) =

for each A e are obviously measures. That is, the collection of all measures on :E is closed under addition and also by multiplication by nonnegative scalars. Clearly, this collection cannot be a vector space since multiplication by - 1 yields negative-valued s�t functions. An order relation < can be introduced among measures by letting JL < v whenever JL(A) < v(A) holds for each A e :E. The reader should stop and check that < is indeed an order relation. Remarkably, the collection of all measures under this ordering is a lattice. That is, for every pair JL and v of measures, the least upper bound JL v v and the greatest lower bound f..L 1\ v exist. The details are included in the next theorem.

:E

The collection of all measuze s ozz :Eforms a lattice, wherefor each pair ofmeasuz-es azzd v the lattice opez·arions are given by :E and A}, and JL v v(A) + v(A :E and A} JL 1\ v(A) + v(A for each A Moreover, Theorem 36.1.

e

f..L

= sup{JL(B)

\ B): B

= inf{J.L(B)

\ B) : B

e

e

:E.

/L

1\ v + J.l.. v v = /L + v.

B c

B c

327

Section 36: SIGNED MEASURES 'E .

Let and v be a pair of measur�s on For each A 'E define w(A) = sup{J.L(B) + v(A \ B): B E 'E and B c A } . First, we shall verify that w is a measure, and then, that it is the least upper bound of J.L and v. Clearly, w(A) 0 holds for each A E I: and w(0) = 0. It remains to be shown that w is a-additive. To this end, let { An } be a disjoint sequence of 'E and put A = u:I A" . If B E satisfies B c A, then Proof.

J.L

E

>

I:

00

= L[J.L(A/1 n n

=l

B) + v(A n \ (An n B))]

and so, w(A) E:1 w(A11) holds. For the reverse inequality, note that if w(A) = then w(A) = E: 1 w(A11) = is clearly true. Hence, assume w(A) and let E 0; clearly, w(A11) w(A) holds for all Thus, for each theren exists some B11 E with B11 C A11 and J.L(B11) + v(A11 \ B11) w(A11) - E2- . Obviously, {B n} is a disjoint sequence of 'E, and if B = U: 1 Bn c A, then U:1 (A11 \ B11) = A \ B holds. Moreover, �

oo,

oo

< oo,

n.

< oo

00

= L[ J.L (B11)

n=l

n

>

+

v (A11 \

00

>

>

'E

<

00

B11)] n=l L[w(A,) - E2-"]

= L w(A11) - E

holds for each E as required.

n=l

>

0.

That is, w(A)

>

2:::: 1 w(A11) and so w(A) = E:1 w(A,),

Chapter 7: SPECIAL TOPICS IN INTEGRATION

328

It should be clear that J.L < w and v < w both hold. Now, assume that another measure rr satisfies J.L < rr and v < rr . Let A e l:. If B e l: satisfies B c A, then

J1(B) + v(A \ B) < rr(B) + rr(A \ B)

=

rr(B

U (A \ B)) =

rr(A),

and so, w(A) < rr(A) holds for each A e l:. That is, w < rr, and therefore, w is the lest upper bound of J.L and v. That is, w = 11- v v. The proof of the infimum parallels the preceding one and is left for the reader. To see that J.L 1\ v + J.L v v = J.L + v holds, let A e l:. Then for every B e l: with B c A we have

�-t(B) + v(A \ B) + J.L v v(A)

>

J,L(B) + v(A \ B) + J.L(A \ B ) + v(B) = J.L(A) + v(A), 11- 1\ v(A ) + J,L(B) + v( A \ B) 5 J,L(A \ B) + v(B) + J,L(B) + v(A \ B) = J.L(A) + v(A). By taking the inf and sup over all measurable subsets B contained in A, the preceding inequalities yield J.L 1\ v(A) + J.L v v(A) = J,L(A) + v(A), and the proof is complete. • The formula J.L 1\ v + J.L v v = J.L + v is reminiscent of the familiar identity in vector lattices. The next result describes an order completeness property of the lattice of all measures.

/fa sequence {Jln} ofmeasures satisfiesby J.Lt (i.e., J.Ln < 1-Ln+ l foreachn), thenthesetfunction J.L: [0, oo] defined (A) limJ,L11(A) for each A i s a measur e . Mor e over, J.L11 t Jl holds; that is , J.L is the least upper bound of the sequence {J.Ln l· A I: Proof. 0 < J.L(A) < J.L(0) 0

. Theorem 36.2. e

J..l11

l: --+

l:

=

oo for each e and Clearly, = hold. Also, if A c B, then J.L(A) < J,L(B) is obviously true. For the a-additivity, let {A,} be a disjoint sequence of l:. Let A = A11 • In view of J.Lk(A) = .L::1 J.Lk(An) < �-t(A,) for each k, it follows that J,L(A) < .L:: 1 J,L(A,). On the other hand, for each k we have

U:1

.L:: 1

k

k

i=

i=

•

� J,L(A;) = lim � � J.Ln(A;) = lim J.Ln �l n-+oo 11-HXJ l

(U )

and so, L�t J,L(A;) < J,L(A) also holds. Thus, The verification of J.Ln t J.L is straightforward.

k

i=l

A;

J,L(A)

=

< lim J,L,(A) n-+oc

=

J,L(A),

.L::, J,L(An) , as required.

•

329

Section 36: SIGNED MEASURES

As mentioned before, multiplicationofa measure by - 1 yields a negative-valued set function. For this reason, it is desirable to· consider a -additive set functions that also assume extended negative values. However, if we do this, we run immediately into trouble. Suppose that a set function J..L: I: JR.* satisfies J..L(A) oo and with A n B 0. If J..L is to be additive, then J..L(A B) J..L(A) + J..L(B) J..L(B) must hold, and we face the problem of having to give a meaning to the expression The preceding difficulty can be avoided by excluding from the range of the set function at least one of the infinite values. If this is assumed, then the form does not appear, and the additivity property does not cause problems. To be distinguished from a measure, such a a-additive set function is usually referred to as a signed measure. Its precise definition follows. i s said to be a signed measure if Definition 36.3. A set function J..L : I: it satisfies the following properties: a. J..L assumes at most one of the values oo and -oo, b. J..L(0) 0, and c. J..L is a-additive, that is, if {A,} is a disjoint sequence of members of then /1( u:::l A,) L�I J,.L(A,) holds. IfJ..L is a signed measure, and a disjoint sequence {A,} of satisfies IJ..L( u:, A,)I then (since every permutation of {A,} has the same union as the original sequence) it follows from (c) of the preceding definition that :L: 1 J..L(A,) is re­ arrangement invariant. Thus, :L: 1 IJ..L(A11) l is convergent in JR. (see Exercise 7 of Section 5). Clearly, every measure is a signed measure. Also, (b) and (c) of Definition 36 together show that every signed measure is finitely additive. The next few results will reveal that signed measures behave in a manner similar to measures. The first one informs us that a signed measure is always subtractive. Theorem 36.4. Let J..L be a signed measure on I:, and let A E I: be such that IJ..L(A)I oo.If B E I: satisfies B c A, then IJ..L(B)I oo and J..L( A \B) J..L( A)- J..L( B). Proof. The identity A (A \ B) B combined with the additivity of J..L imply J..L(A) J..L( A \B) J..L(B). Since J..L( A) is a real number and J..L assumes at most one of the values oo and it follows that both J..L(A \B) and J..L(B) are real numbers. Now. the identity J..L( A \B) J..L(!\) - J..L( B) should be obvious. An immediate and useful conclusion of the preceding theorem is the following: If a set A E :E has a measurable subset of infinite signed measure, then A itself has infinite signed measure. --+

= -oo, = oo - oc

U

==

= =

oo - oo.

oo - oo

--+

JR.*

=

I:,

==

I:

< oo,

1

<

<

=

=

=

U

+

-oo,

=

•

Chapter 7: SPECIAL TOPICS IN INTEGRATION

330

The usual continuity properties of measures are inherited by signed measures. J.L

For a signed measure and a sequence {A,} of �, the fol­ lowing statements hold: IIff A,A t,!.. A,A and thenJ.L(AdJ.L(An) = J.L(A). is a real numberfor at least one then Theorem 36.5. a. b.

lim

1

1

k,

lim

11-00

J.L(An) = J.L(A).

Proof. Repeat the proof of Theorem 15.4 taking into consideration the con•

clusion of Theorem 36.4.

A measurable set A is said to be a positive set with respect to a signed measure J.L, in symbols A > 0, whenever J.L(E n A) > 0 holds for each E Equivalently, A set A with is a positive set whenever J.L(E) > 0 holds true for all E E C A. Obviously, the empty set is a positive set. Also, it should be clear that any measurable subset of a positive set is likewise a positive set. In addition, any countable union. of positive sets is a positive set. To see this, let {An } be a sequence of positive sets and let A = U:1 An. Now, let B 1 = A 1 , Bn+l = An+l \ U;'= 1 A; for > I , and note that { Bn} is a disjoint sequence such that A = U:1 B11 • Since B11 C An holds for all it follows that each Bn is a positive set. Therefore, for each measurable set E we have J.L(E n A) = J.L( U:1 E n B,) = :L:1 J.L(E n B11) > 0, so that A is a positive set. Similarly, a set A is called a negative set for a signed measure J.L, in symbols As before, measurable A < 0, whenever J.L(A n E) < 0 holds for each E subsets of negative sets are negative, and countable unions of negative sets are likewise negative sets. The next lemma is a basic result for this section and guarantees the existence of nonempty positive sets.

E :E. E �

E�

n

n,

E �.

Let be a signed measure on :E, and let E �with J.L(E) > 0. Then there exists a positive set A suclz that A and > 0.

Lemma 36.6.

J.L

CE

E J.L(A)

Proof. If for every measurable subset B of E we have J.L(B)

> 0, then E is

itself a positive set and there is nothing to prove. Thus, assume that there exists some B with B c E and J.L(B) < 0. By Zorn's lemma (how?) there exists a maximal collection C of mutually dis­ joint measurable subsets of E such that J.L(C) < 0 holds for each C C. We claim that C is at-most countable. To see this, note first that C = U:1 C,, where C: J.L(C) < II On the other hand, if some Cn is not finite, then it C11 = {C must contain a countable subset of C, say { C 1 , C2, . . } . But then, the measurable

E� E

_1. }.

E

.

Section 36: SIGNED MEASURES

331

set C = U: 1 C; satisfies C c E, and J-L(C) = 2::: 1 J-L(C;) = This implies J-L(E) = J-L(E \ C)+ J-L(C) = which is a contradiction. Thus, each Cu is finite, and so C is at most countable. Hence, the set D = Ucec C belongs to :E , and we claim that A = E \ D is a positive set satisfying J-L(A) > 0. Indeed, since 0 < J-L(E) = J-L(A) + J-L(D) and J-L(D) < 0, it is easy to see that J-L(A) > 0 holds. On the other hand, if J-L(F) 0 holds for some measurable subset F of A, then we can incorporate F into C and violate the maximality property of C. Thus, A is a positive set, and the proof is finished. If is a signed measure on :E and two disjoint sets A and B satisfy A > 0, B < 0, and A B = then the pair (A, B) is referred to as a Hahn decomposition of X with respect to J-L. Loosely speaking, a Hahn decomposition is a splitting of the space into two pieces, where is positive on one of the pieces and negative on the other. Such a splitting always exists and is "essentially" unique, the next theorem shows. -oo.

-oo,

<

11-

•

X,

U

11-

X

as

Let J-L be a signed measure on :E . Then has a Hahn decom­ position with respect to J-L. That is, there exist a positive set A and a negative setMoreover, B such that = A B and A n B = 0. if (A, B) and (Ah B1) are two Hahn decompositions of with respect to J-L, then a. J-L(A�AI) = J-L(B �B1) = 0, b. J-L(E n A) = J-L(E A 1 ), and Theorem 36.7.

X

X

U

X

n

c. J-L(E n B) = J-L(E n B1) holdfor each E E :E . Proof. We can assume without loss of generality that J-L(E) # holds for each E :E (otherwise, replace by -J-L). > Put a = sup{p.(E): E 0} and note that a > 0. Choose a sequence {Au} of positive sets such that lim J-L(An) = a. Then A = U: 1 Au is a positive set, and since J-L(Au) < J-L(A) < a, it follows that a = J-L(A) Now, we claim that B = \ A is a negative set for J-L. To see this, assume by way of contradiction that there exists some measurable subset C of B with J-L(C) 0. Then, by Lemma 36.5, there exists a positive set E with E c C and J-L(E) > 0. It follows that A U E > 0 and a + J-L(E) = J-L(A) + J-L(E) = J-L(A U E) a < which is impossible. Thus, (A, B) is a Hahn decomposition of with respect to J-L. For the "uniqueness" of the Hahn decomposition, let (A 1 , B 1) be another Hahn decomposition of Since A \ A 1 c A and A \ A 1 = A n A) = A n B 1 C B1, it oo

11-

E

X

< oo.

>

<

oo,

X

X.

Chapter 7: SPECIAL TOPICS IN INTEGRATION

332

follows that tL(A \ A 1 ) > 0 and tL(A \ A 1 ) < 0, that is, tL(A \ A 1 ) = 0. Similarly, tL(A 1 \ A) = 0, and so

tL(A � A J ) = tL((A \ A J ) U (A I \ A)) = tL(A \ A J) + tL(AI \ A) = 0. Now, if E

E

l;, then

tL(E n A) = tL(E n ( (A \ A 1) u (A n A 1) ]) = tL(E n (A \ A 1 )) + tL(E n A n A 1 ) = tL(E n A n A J ) = tL(E n (A1 \ A)) + tL(E n A n A 1 ) = tL(E n [ (A 1 \ A) u (A 1 n A) ]) = tL(E n A J).

By the symmetry of the situation, the corresponding formulas for also true, and the proof is finished.

B

and

B1

are •

If (A, B) is a Hahn decomposition of X with respect to a signed measure JL, then tL and tL restricted to A and B, respectively, are in actuality measures. As we shall see, these two measures determine the structure of JL. -

l;,

Let tL be a signed measure on and let (A, B) be a Hahn decomposition of with respect to JL. Then the three setfunctions

Definition 36.8.

X

tL+(E) = tL�E n A), tL-(E) = -tL(E n B), ltLI(E) = tL(E n A) - tL(E n B) = tL+(E) + tL-(E),

and

l;

forthe each E E are called the positive variation, the negative variation, and total variation ofJL, respectively.

A glance at Theorem 36.7 guarantees that the values of tL+, tL -, and ltL I are independent of the chosen Hahn decomposition. Also, it should be clear that tL+, tL- , and I J.l l are measures on l;. A signed measure tL is said to be a finite signed measure if JL(A) E IR holds for each A E l;. Note that in view of·Theorem 36.4, a signed measure tL is finite if and only if ltL(X)I < oo holds. Similarly the expression "tL is a finite measure" means that 0 < JL(A) < JL(X) < oo holds for all A E l; . Finally, let us say (as usual) that a signed measure is a-finite if there exists a disjoint sequence {An} of l; witb X = U:1 An and tL(An) E IR for each

n.

Section 36: SIGNED MEASURES

333

From Definition 36.8 it follows that holds. This identity is known as the Jordan 1 decomposition of the signed measure decomposition, then in view of the inequalities JJ-/1-+. Moreover, if (A,andB)JJ--is(E)a Hahn < -J.L(B) for each E E :E, it follows that at least (£) < JJ-(A) one of the measures J.L+ and J.L- is a finite measure. Thus, after all, the Jordan decomposition shows that each signed measure is the difference of two measures, at least one of which is a finite measure. Some other useful expressions for the different variations of a signed measure are presented next. Then for every E the Theorem 36.9. Let 11- be a signed measure on. following formulas hold: 1 . JJ- + (£) = sup{JJ-(F): F E :E and F £}. 2. - (E) = sup{ -J.L(F): F E :E and F £}. 3. I JJ - 1 ( £) = sup{:L IJJ-(F;)I: {F;} is finite and disjoim with U F; c £}. Proof. Let (A, B) be a Hahn decomposition of X with respect to JJ-, and let E :E. (1) Clearly, 11- + (£) = JJ-(E n A) < sup{JJ-(F): F E :E and F c £} holds. On the other hand, if F E :E satisfies F c E, then JJ-(F) = JJ-(F A) + JJ-(F n B) < JJ-(F n A) < JJ-(E n A) = JJ-+(E), so that sup{J.L(F): F E :E and F c £} < JJ-+(E). Therefore, the identity in (1) holds. (2) This follows from (1) by observing that 11- - = (-JJ-)+. (3) Let v(E) = sup { L IJJ-(F; )I: { F;} is a finite disjoint collection. of :E with U F; c E } . Then, IJ.LI(E) = JJ- +(E) + 11--(E) = JJ-(E n A) - J.L (E n B) = IJJ-(E n A) I + IJJ-(E n B)l < :E .

E :E

c

11-

c

c :E

E

n

v(E).

1Camille Jordan ( 1 838-1921 ), an eminent French mathematician. He was considered a "universal'' mathematician because he published papers in virtually all areas of mathematics of his time. His main contributions were in group theory.

Chapter 7: SPECIAL TOPICS IN INTEGRATION

334

Now, let { F1 , . . . , F,} be a finite disjoint collection of � such that U�1= 1 F; C E. Then II

L IJ.L( F;)I i=l

=

=

II

II

i=l

i=l

L IJ.L+(F;) - J.L-(F; ) I < L [J.L+( F; ) + J.L-(F;)] n

B IJ.LI( F;) = IJ.LI ( � F;) < IJ.L I(E). 11

This implies v(E) < IJ.LI (E). Thus, IJ.LI(E) = v(E) holds, and the proof of the theorem is complete. • It is easy to establish, but important to observe, the following inequality regard­ ing the total variation: IJ.L(A) I < IJ.L I (A)

for each A E � -

Summarizing, the different variations of a signed measure J.L satisfy the following identities: a. J.L = J.L+ - J.L- and IJ.L I = J.L+ + J.L-. b. J.L+ 1\ J.L- = 0. c. J.L- = ( -J.L)+. They are, of course, reminiscent of the usual identities of vector lattices. Fur­ thermore, they suggest that the set of all signed measures forms a vector lattice. Unfortunately, this is not the case. Although the (pointwise) sum of two measures is always defined, the sum of two signed measures may fail to exist for the very simple reason that by adding the signed measures pointwise one might encounter the expression oo - oo. However, if we restrict ourselves to the collection M(�) of all finite signed measures, then we obtain a vector lattice. Note first that a signed measure J.L satisfies J.L E M(�) if and only if IJ.LI(X) < oo. Indeed, if IJ.LI(X) < oo, then clearly, J.L E M(�). On the other hand, if J.L E M(�). then the identity J.L = J.L+ - J.L­ with either J.L+ or J.L- finite shows that J.L+ and J.L- are both finite, and so IJ.L I (X) = J.L+(X) + J.L-(X) < oo. (Because of the last statement, the finite signed measures are also known as the signed measures of finite total variation.) Clearly, if J.L, v E M(�). then J.L + tt and CXJ.L belong to M(�). where, of course, (J.L + v)(A) = J.L(A) + v(A) and (aJ.L)(A) = CXJ.L(A) for each A E � and a E JR. That is, M(�) is a vector space. Now, if J.L < v means J.L(A) < v(A) for each A E �, then < is an order relation under which M ( �) is a vector lattice. Its lattice

Section 36: SIGNED MEASURES

335

operations are given by J-L v v(A) = sup{,u(B) + v(A \ B): B E :E and B A}, and J-L v(A) = inf{J-L(B) + v(A \ B): B E :E and B A}. The preceding formulas, combined with Theorem 36.9, show that J-L+ = J-L v 0, ,u- = (-,u) v O, and 1 1-L I = J-L v (-,u) in M(:E) are precisely the positive, negative, and total variations of ,u as given by Definition 36.8. The next thing to observe is that l .ull = IJ-L!( X) defines a norm on M(:E). Indeed, (A )I < l.ui( A) < IJ-LI(X) = l .ull for each a. l.ui (X) > 0 holds, and since I J-L E :E, we have 1 1-L I = 0 if and only if J-L = 0. A b. l aJ-LII = la,ui (X) = Ia! l.ui (X) = lal · (IJ-LI(X)) = lal · I J-LI . c. 1 1-L +vii = l .u + vi(X) < (IJ-LI + lvi)(X) = IJ-LI(X) + lvi(X) = l .ull + I vii . ItI J-LisI also true that < lvl holds, then is a lattice norm on M(:E). Indeed, if 1 1-L 11 · 11 I = l.ui (X) < l v i(X) = I vii . Therefore, M(:E) is a nonned vector lattice which is actually a Banach lattice, the next result shows. c

1\

c

·

as

The collection of allfmite signed measures on a a-algebra is a Banach lattice.

Theorem 36.10.

According to the preceding discussion, it remains to be shown that M(:E) is a Banach space. To this end, let {J-L,} be a Cauchy sequence of M(:E). We have to show that there exists some ,u E M(:E) such that lim I J-L, - .u I = o. Let E > 0. Choose some k such that I J-L, - J-Lm I < E for all m, n > k. The inequalities Proof.

1/-+00

show that {,u,(A)} is a Cauchy sequence of real numbers for each A E :E. Let J-L(A) = lim J.t,(A), so that J-L is a real-valued set function. From ( ) it follows that ( ) IJ-L,(A )- J-L(A)I < E for each A E :E and n > k. *

**

336

Chapter 7: SPECIAL TOPICS IN INTEGRATION

Clearly, f.J-(0) = lim f.J-,(0) = 0. For the a-additivity of f.-L , let {A, } be a disjoint sequence of 'E, and let A = U:1 A,. Then p

p

L

[f.-Lk (A ;) - f.-Ln (A ; )]

i=l

<

L l f.-Lk(A;) - f.-Ln(A; )I < lf.-Lk - f.-L,I(A) i=l

< l lf.-Lk - f.-Ln II < E

'

holds for all p and ll > k. Hence, r�=r=l [f.-Lk (A;) - f.-L(A ; )] I � E holds for each p . Now, choose no such that lf.-Lk (A) - 2:f= 1 f.-Lk (A ; )I < E for all p > no, and note that p

p

i=l

i=l

f.-L(A) - L f.-L(A;) < lf.-L(A) - f.-Lk(A)I + f.-Lk (A) - L f.-Lk(A;) p

+ L [f.-Lk (A ;) - f.-L(A; )]

< 3E

i=l

holds for all p > no. Therefore, f.-L(A) = 2::1 f.-L(An), and so f.-L E Finally, combining (**) with Theorem 36.9, we get

M('E).

(f.-Ln - f.-L)+(X ) = sup{f.-Ln(A) - f.-L(A): A E 'E } < E and (f.-Ln

- f.-L)- (X) < E for all n

> k. Hence,

holds for all n > k. That is, lim II f.l-11

- f.-L II

=

•

0, as required.

EXERCISES 1.

Give an example of a signed measure and two Hahn decompositions (A, B) and (A 1 , B 1 ) of X with respect to the signed measure such that A =F A 1 and B =F B I · 2. If 11- is a signed measure, then show that 11-+ A 11-- = 0. 3. If 11- is a signed measure, then show that for each A E :E we have if.LI(A) = sup

{�

if.L(A n ll: I A n } is a disjoint sequence of E with

Q

An

=

}

A .

4. Verify that if 11- and v are two finite signed measures, then the least upper bound 11- v v and the greatest lower bound 11- A v in M(:E) are given by

11- v v(A) 11- A

v(A)

for each A E :E.

=

=

sup{�.t(B) + v(A \ B): B E :E and B c A}. and inf(�.t(B) + v(A \ B): B E :E and B c A }

337

Section 36: SIGNED MEASURES

Let A be the Lebesgue measure on the Lebesgue measurable subsets of 1R. If J.L is the Dirac measure, defined by J.L(A) = 0 if 0 rf. A and J.L(A) = l if 0 E A, describe A v J.L and A 1\ J.L. 6. Show that the collection of all a-finite measures fonns a distributive lattice. That is, show that if J.L, v, and w are three a-finite measures, then

5.

(J.L v v) 1\ w = (J.L 1\ w) v (v 1\ w) and (J.L 1\ v) v w = (J.L v w) 1\ (v v w). [HINT: Every vector lattice is a distributive lattice.] 7. If :E is a a-algebra of subsets of a set X and J.L: :E � 1R* is a signed measure, then show that AJ1.+ n AJ1.- = A IJJ. I· 8. Let J.L and v be two measures on a a-algebra :E with at least one of them finite. Assume also that S is a semiring such that S c :E , X E S, and that the a-algebra generated by S equals :E . Then show that J.L = v on :E if and only if J.L = v on S. [HINT: See Theorem 1 5 . 1 0; see also Exercise l l of Section 20.] 9. Let (X, S, J.L) be a measure space and let f E L 1 (J.L). Then show that v(A) = for each A E

All

v+(A) =

i f dJ.L

defines a finite signed measure on A ll. Also, show that

i t+ dJ.L,

v - (A) =

i !- dJ.L

and l v i(A) =

i Ill dJ.L

hold for each A E All. [HINT: Use Theorem 1 5 . 1 1.] 10. Let v be a signed measure on :E. A function f: X � 1R is said to be v-integrable iff is simultaneously v+- and v- -integrable (in this case, we write J f dv = J f dv+ Jf d v-). Show that a function f is v-integrable if and only if f E L 1 ( Ivi). 11. Show that the Jordan decomposition is unique in the following sense: If v is a signed measure, and J.LI and J.L2 are two measures such that v = J.LI - J.L2 and J.LI 1\ J.L2 = 0, then J.L 1 = v+ and J.L2 = v - . 12. In a vector lattice x11 -1- x means that x11+ 1 ::.: x, for each n and that x is the greatest lower bound of the sequence (x11}. A nonned vector lattice is said to have a-order continuous norm if x, -1- 0 implies lim llx11 ll = 0. a. Show that every L p(J.L) with 1 ::.: p < oo has a-order continuous nonn. b. Show that L00([0, 1]) does not have a-order continuous nonn. c. Let :E be a a -algebra of sets, and let (J.L,} be a sequence of M (:E) such that J.L11 -1- J.L. Show that lim J.L11(A) = J.L(A) holds for all A E :E . d. Show that the Banach lattice M(:E) has a-order continuous nonn. 13. Prove the following additivity property of the Banach lattice M(:E): If J.L, v E M(:E)

disjoint (i.e., I J.LI 1\ jv l = 0), then II J.L + vii = II J.L il + II vii holds. [HINT: In a vector lattice l.t l 1\ IYI = 0 implies j.t Yl = lxl IYI. Reason:

are

+

+

+

+

+

lx yl ?::: l lxl - lyl l = lxl v IYI - lxl 1\ IYI = lxl v IYI = lxl IYI ?::: lx y j.] 14. Let :E be a a-algebra of subsets of a set X and let (J.L11} be a disjoint sequence of MCE). If the sequence of signed measures (J.L11} is order bounded, then show that limiiJ.L,II = 0.

Chapter 7: SPECIAL TOPICS IN INTEGRATION

338

37. COMPARING MEASURES AND THE RADON-NIKODYM THEOREM Again, in this section, :E will be a fixed a-algebra of subsets of a nonempty set X and all set functions will be assumed defined on :E. Two important comparison notions for signed measures will be introduced here. Both notions will be defined in terms of measure properties. The first one is referred to as the "concept of absolute continuity."

sigz z e d measure is said to be absolutely continuous with respect to another signed measuz · e in symbols << or whenever E :E and 0 imply 0. Definition 37 .1. A

A

v J.l, v(A) =

IJ.LI(A) =

v

J.l >> v,

J.l

Some characterizations of absolute continuity in terms of variations are included in the next result.

For a pair of sigz z e d measures az z d the followiz z g state­ ments are equivalent: << << and << Theorem 37 .2. a.

b. c.

v v+ I vI

J.l

J.l. IJ.ll << I J.L ! .

v-

v,

I J.l l .

(a) => (b) Let A E :E satisfy IJ.L!(A) = 0. If B E I: satisfies B then IJ.LI(B) = 0, and so by our hypothesis v(B) = 0. Thus,

Proof.

v+(A) = sup{v(B): B

E :E

and B

c A}

=

c A,

0.

Similarly, v-(A) = 0, and so, both v+ << IJ.ll and v- << IJ.ll hold. (b) => (c) If IJ.LI(A) = 0, then by hypothesis v+(A) = v-(A) = I vi = v+ + v-, it follows that l v i(A) = 0, and thus I vi << IJ.ll. (c) => (a) It follows from the inequality lv(A)I < l v i (A).

0.

Since •

If a measure v is absolutely continuous with respect to another measure J.l, then it is natural to expect some relationship between the J.,L-measurable and v-measurable sets. The following theorem tells us the precise relationship:

For two measures and on a a-algebra we have the following: << and a subset of X satisfies 0, then 0. fevery<<J.,L-measurable and is a-function finite, thenis alsocv-measurable. holds. In particular, in this case Theorem 37.3. a. b.

lf v l v

J.l J.l

J.l

E

J.l

Ap.

v

J.l*(E) =

A 11

I:

v*(E) =

Section 37: COMPARING MEASURES AND THE RADON-NIKODYM THEOREM 339

Proof. (a) Assume JL*(E) = 0. By Theqrem 15.1 1 , there exists some A E :E

such that E c A and JL(A) = JL*(E). From v << JL it follows that v(A) = 0. Therefore, 0 < v*(E) < v(A) = 0, so that v*(E) = 0. (b) Assume E is JL-measurable such that JL*(E) < oo. By Theorem 15. 1 1 , there exists some A E I: with E c A and JL(A) = JL*(E) < oo. But then JL*(A \ E) = 0, and by (a) we have v*(A \ E) = 0. Therefore, A \ E is v-measurable. The v­ measurability of E now follows from the identity E = A \ (A \ E). Finally, the preceding conclusion, coupled with the a-finiteness of JL, easily implies that AJ.L c A v holds. • Now, let JL and v be two measures. If v < JL, then it should be clear that v << JL also holds. The converse of the latter is false. For instance, 2JL << JL holds for each measure JL, and 2JL < JL is true only if JL = 0. However, if v << JL holds, then (up to a multiplication factor) JL and v are "locally" comparable in 'the order sense. The exact details follow.

Let be a finite nonzero measure, and let be a a-finite measure.andIfsomeis absolutewith ly continuous with respect to then there exist some such that 0 0 v

Theorem 37.4. v

E >

A E :E

JL,

< JL(A) < oo

EJL(B)

holdsfor all

B E :E

with n.

JL

< v(B)

B C A.

Proof. Let {En} _be a disjoint sequence of I: such that X = U:1 E11 and

Since v is a nonzero measure, there exists some k with JL(E11) < oo for .each v(Ek ) > 0. Choose some E > 0 so that v(Ek ) - EJL(Ek ) = (v - EJL)(Ek ) >

0.

By Lemma 36.6 there exists a measurable subset A of Ek that is a positive set for the signed measure v - EJL satisfying (v - EJL)(A) > 0. Clearly, JL(A) < oo holds. Now observe that JL(A) > 0. Indeed, if JL(A) = 0 holds, then the last inequality yields v(A) > 0, contradicting the absolute continuity of v with respect to JL. On the other hand, the inequality (v - EJL)(B) > 0 for each B E I: with B c A implies E JL(B) < v(B ), and the proof of the theorem is finished. • The opposite notion to absolute continuity is that of singularity. Two signed measures JL and v are said to be singular (or orthogonal), in symbols JL j_ v, if there existtwodisjointsets A andB of:E with AUB = X and I JLI(A) = l vi(B) = 0. If JL is a signed measure, then Theorem 36.7 shows that JL+ and JL- are two singular measures. The singularity concept is characterized in terms of lattice properties as follows.

Chapter 7: SPECIAL TOPICS IN INTEGRATION

340

Two signed measures J.L and v are singular if and only if IJ.L I /\ I v i = o.

Theorem 37.5.

v. Choose two disjoint sets A and B of l; with A u B = X and IJ.LI(A) = lvi(B) = 0. Then Proof. Assume that J.L 0<

..l

IJ.LI A lvi(X) = IJ.L I A lvi(A) + IJ.L I A lvi(B) < I J.LI(A) + lvi(B) = 0,

so that IJ.LI /\ I vi = 0. Conversely, assume that IJ.LI /\ I vi = 0. By Theorem 36.1, for each n there exists some E, E � such that IJ.LI(En) + lvi(E�) < 2- " . Let A, = u:n E;' and then let A = n:1 A, and B = Ac. Note that

n,

holds for all and thus, IJ.LI(A) = 0. Now, obse�e that lvi(A�) = I vi( n�, Ej) < 2-i for all i > implies lvi(A�) = 0 for each n. Therefore, since B = Ac = U: 1 A�, it follows that lvi(B) = 0. Hence, X = A U B and IJ.LI(A) = lvi(B) = 0 hold, proving that J.L ..l v. •

n

The following list of statements presents a number of useful relationships be­ tween absolute continuity and singularity. The set functions J.L, v, and w are signed measures on a common a -algebra �.

1. If J.L << w and v << w, then IJ.LI + lvl << w. 2. If J.L ..l w and v ..l w, then IJ.LI + I vi ..l w.

3. 4. 5.

If J.L << w and I vi < IJ.LI. then v << w. If J.L ..l w and I vi < IJ.LI. then v ..l w. If v << J.L and v ..l J.L, then v = 0.

Their proofs are straightforward applications of the definitions and Theorems 37.2 and 37.5. For instance, the proof of (5) goes as follows: Since v ..l J.L, there exists some A e � such that lvi(A) = IJ.LI(Ac) = 0. From v << J.L and Theorem 37.2, lvi(Ac) = 0, and so, lvi(X) = lvi(A)+ lvi(Ac) = 0. That is, I vi = 0, so that v = 0. Now, let J.L be a a-finite measure. A classical result of H. Lebesgue asserts that any other a -finite measure v can be det:omposed as a sum of two measures v1 and v2 such that v1 << J.L and v2 ..l J.L. Its proof will be based upon the following simple property of vector lattices: Lemma 37.6. In

a vector lattice

x,

t x mpli

s

i e

Xn 1\

x

y t 1\ y for each y.

Section 37: COMPARING MEASURES AND THE RADON-NIKODYM THEOREM 341

Note first that X11 A y t < x A y h_olds. On the other hand, if x11 A y < z holds for each n and some z, then Theorem 30.1 (5) shows that Proof.

(X A y- .:)+ < (X A y -X11 A y)+ < !x A y -X11 A Yl < X -X11• Hence, X1 <+ x - (x A y - z)+ < x holds for each n. But then X11 t x implies is, x A y is the least upper bound of < z. That (x{x1 AAy soz) that= 0,X11andA so,t x AA y holds, y x y as claimed. 1 y}, •

-

We are now ready to state and prove the Lebesgue decomposition theorem.

Let J.L and v be two a-finite measures on I:. Then there exist llvo l nique measures v1 and v2 such that v = v1 + v2, where v1 J.L and v2 J.L. Proof. Assume first that both J.L and v are finite measures. Then the formula v1(A) = sup{(v A 11J.L)(A): n = 1, 2, ... < v(A) defines a finite measure on (Theorem 36.2). Moreover, if J.L(A) = 0, then clearly (v A nJ.L)(A) = 0 for each n. Thus, v1 (A) = 0, that is, v1 Now, let note that v2 is a finite measure such that v1 + v2 = v. Next, v2observe = v -thatv1,v and A l J .L t v1 implies (J.L + v) A (n + l )J.L = J.L + v A nJ.L t J.L + v1, and so, by Lemma 37.6, V A l )J.L = V A (J.L + v) A (n + )J.L t V A (J.L + VJ ). On the other hand, v A (n )J.L t v1 implies v1 = v A (J.L + v1 ). Thus, 0 < v2 A J.L = (v - vJ) A J.L = V A (J.L + VJ) - VJ = VJ - VJ = 0, and hence, by Theorem 37.5, v2 J.L. For the uniqueness of the decomposition, note first that v1 v2• Indeed, since sets A and B with X = A B v2and J.L, there exist twoThendisjoint measurable J.L implies v1 (B) = 0, so that v1 v2• Now, v2(A) = J.L(B) = 0. v1 assume that another pair of measures cv1 and cv2 satisfies cv1 J.L, cv2 J.L, and 2. Clearly, v1v2 - cvJ.L1 .=Itcvfollows 2 - v2 holds, and on one hand v1 - cv1 J.L, vand= oncv1 thecvother that v1 - cv1 = cv2 - v2 = 0, so that = theandgeneral V2 = VJ For case, choose a disjoint sequence { A11 } of with v(A11) < X = U:, A" . Let v" (A) = v(A n A") and J.L" (A) = J.L(A n A") J.Lfor(A1each 1)
<<

j_

}

I:

<< J.L.

(11 +

1

+1

j_

j_

j_

<<

+

cv2

CV2.

CVJ

oo,

2:.

<<

j_

I:

I: .

j_

U

j_

<<

oo,

Chapter 7: SPECIAL TOPICS IN INTEGRATION

342

n

for each there exists a unique pair of measures (v;', v2) such that v!' << J-L11 , v;' .l J-L11, and Vn = v!' + v2. Now, define 00

v1(A) = L v'; ( A) ll=l

and

00

v2(A) = L v�(A) ll=l

for each A e �. It is a routine matter to verify that v1 << Jl, v2 .l J-L, and v = v1 + v2. The uniqueness of v1 and v2 follows easily from the uniqueness of the decomposition of each v,. • The identity v = v1 + v2 provided by the preceding theorem (where v1 << J-L and v2 _L J-L), is called the Lebesgue decomposition of v with respect to J-L. Now, assume that J-L is a measure and f e L 1 (J-L). Then the formula

for each A e :E defines a (typical) finite signed measure that is absolutely con­ tinuous with respect to f-L. Sometimes formula (*) is referred to as the indefinite integral of f. A direct verification shows that

hold for each A e :E. . It is natural to ask whether the converse of the preceding statement is true. That is to say, whenever v is absolutely continuous with respect to J1 . does there exist some f e L 1 (J1) such that v is given by (*)? In general, the answer is negative; see Exercise 7 at the end of the section. On the other hand, whenever J-L is a -finite and v is finite, the answer is yes. This result is of great import�ce because of its wide range of applications. It is known as the Radon-Nikodym theorem. It was established first by J. Radon2 (in 1913) for the Euclidean spaces with the Lebesgue measure, and later (in 1930) 0. Nikodym3 extended it to the general case.

Let v be a finite signed measure which is absolutely continuous with respect to a a -finite measure Then there exists a Theorem 37.8 (Radon-Nikodym).

f-L.

Johann Radon (1 887-1956), an Aus1rian ma1hema1ician. He is well known for his con1ribu1ions 1o analysis and especially 10 lhe calculus of variations. 3 0non Martin Nikodym (1 889-1974), a Polish ma1hema1ician. He con1ribu1ed 10 Boolean algebras, measure 1heory, and ma1hematical physics. 2

Section 37: COMPARING MEASURES AND THE RADON-NIKODYM THEOREM 343

unique function f L 1 (J.L) such that = if dJ.L holdsfor all in

v(A)

A E I: .

Proof. (Uniqueness.) This part is straightforward and is left for the reader;

6 at the end of this section.

see Exercise J.L-a.e.)

(The function

f

is, of course, unique

(Existence.) Clearly, v+ and v- are finite measures, both are absolutely contin­

J.L,

uous with respect to and v = v+ - v- holds. This shows that we can assume without loss of generality that v is itself a finite measure. By Theorem 37.3, we have :E C AJ.L C A11 • Now, let C

= { g L 1 (p.): g 0 E

>

igdp.

p.-a.e. and

< v*(A) for all A E /1."

g g(x) f(x)}

Observe that C is closed under finite suprema. Indeed, if f, then the two sets E =

{x

E A:

2:

f(x) g(x)}

and

F

= {x

E A:

)·

E C and A E AJJ.,

>

are J.L-measurable (and hence, v-measurable) and disjoint, and E U F = A . Therefore,

i f v gdj.L = £ ! v gdj.L + f. ! v gdj.L = £! dj.L + lgdj.L < v*(E)

f0v g

so that Since

+

v*(F)

=

v*(E U F)

E C. E C, the set C is nonempty. Let

=

a = { Jg dp.: g ) {g,} J g, dJ.L = a.0 sup

EC

v*(A),

< v(X) < oo .

= g1 J gn dJ.L fX Lt(J.L)d J,,xA ! A· fJ,,xA J.L

Choose a sequence of C with lim Define J,, v ···v for each n , and note that {f,, } is a sequence of C with < f,, t and lim f,, = < oo. By Levi's theorem (Theorem 22.8) there exists some such E = t In view of 0 < and that f,, t < v*(A), and the Lebesgue dominated convergence theorem, it follows

a = a. f ffdJ.L JA !n dJ.L

Chapter 7: SPECIAL TOPICS IN INTEGRATION

344

that fA t dJ.L < v* (A) for each A E Ap.; that is, t E C. To finish the proof, we show next that v(A) = fA t d J.L holds for all A E :E. Clearly, w(A) = v*(A) - fA t df.J- for A E Ap. defines a finite measure on Ap. which is absolutely continuous with respect to J.L*. Assume by way of contradiction that w # 0. Then, by Theorem 37 .4, there exists some € > 0 and some B E Ap. such that 0 < J.L*(B) < oo and € J.L*(E) < w(E) for each E E Ap. with E c B. Moreover, the function g = t + € xs > 0 belongs to L 1 (J.L), and since a = Jt dJ.L < Jg dJ.L, the function g does not belong to C. On the other hand, if A E Ap., then

1g dJ.L = 1(t + =

€Xs) dJ.L =

1 t dJ.L + €J.L*(B n A) < 1 t dJ.L + w(B n A)

t df.l- + v * (B n A) - f t d J.L = f t dJ.L + v * (B n A) f 1BnA 1A 1A \ B

< v.-t (A \ B ) + v*(B n A) = v*(A), and therefore g oom��.

E C,

which is a contradiction. The proof of the theorem is now

•

The unique (J.L-a.e.)'function t determined by the Radon-Nikodym theorem is called the Radon-Nikodym derivative of v with respect to Jl., in symbols

dv = t or d v = t dJ.L. dj.L

-

The function t is also referr�d to as the density function of v with respect to J.L. In Section 31 it was shown that if ( X, S, J.L) is a measure space and I < p < oo, then for each g E Lq(f.l-), where * + � = I , the function (**) for t E L p(J.L) defines a continuous linear functional on L p(J.L) such that II Fg II = llg llq· Also, in the same section we proved that the mapping g � Fg from Lq(f.l-) to L;(11-) was a lattice isometry and promised the reader to establish that this lattice isometry is also onto. That is (according to F. Riesz), every member of L ;CJ.L) can be obtained from a function of Lq(J.L) as in (**). With the help of the Radon-Nikodym theorem, we are now in the position to establish this claim.

Let (X, S, f.J-) be a measure space, I < p < oo, and let F be a continuous /ineartunctional on L p(J.L). Then there exists a unique

Theorem 37.9 (F. Riesz).

Section 37: COMPARING MEASURES AND THE RADON-NIKODYM THEOREM 345 l+!

g L(/(J.L), where p

= 1 , such that F(f) = Ifg df.L holdsfor all f Lp(J.L). Moreover, I FII = l gl qProof. g g Step I. Assume that (X, S, J.L) is a finite measure space. v: JR = F(XA) AE E

I

(/

•

E

The uniqueness of istence" of has two steps.

has been discussed before. The proof of the "ex­

=

0. Define Ap. -+ by v(A) Ap.- Clearly, v(0) for each Also, if (Au} is a disjoint sequence of measurable sets and then

A = U:1 A11,

II

L XA; - XA i=l

Therefore, by the continuity of

F,

we have

v(A) A= 2::1 v(A11) J.L*(A)=

That is, holds, and hence, v is a finite signed measure on Ap.. Moreover, if E Ap. satisfies 0, then. 0; therefore, 0, so that v is absolutely continuous with respect to By the Radon-Nikodym theorem, there exists a unique function in L 1 such that holds for all E Aw Thus, for every E Ap. we have By linearity, it follows that for every step function Now let

F(O) = v(A) g df.L = J A F(XA) = JxAgdJ.L. l/J.

A = (x E X: g(x)

A

> 0}

v(A) = F(XA) = J.L*. g (J.L) A F(l/J) = J¢gdJ.L

XA =

and

x

B = { E X:

g(x) <

0},

f. l/J n t f = l/J"g+ t fg+ limlll/JnXA - fXAIIp = tPnXAg F Jl/J, g+ dJ.L = F (l/J1 XA) += F(f XA) fg- LI(J.L) F(fX B ) =jg+-Jfg-dJ.L. F(fXA) = jfjg dJ.L. Lp(J.L) fg LI(J.L ) F (f) = J fg dJ.L. g q (J.L). = l g lq- I XEnSgng. = l g(x)l f p(J.L)

and letO < E L p (J.L). Choose a sequence ofstep functions {¢11} with O < Then and 0. By the continuity lim of we have lim < oo, and so, by Levi's theorem Similarly, E and E L I (J.L) and Thus, for every we have and E E To complete the proof of this case, it remains to be shown that EL < n}, andthendefinef,, To this end, let£, {x E X : holds, and moreover, we have Since each /11 is essentially bounded, { , } c L

Chapter 7: SPECIAL TOPICS IN INTEGRATION

346

1 lglq XE dJL 1J.g dJL ..

=

=

( 1 lf,. IP dJL r I

= F( f,. )

< IIFII .

II F I I · (J lgl q XE.. dJL r . I

from which it follows that lg l q XEn d 11-F < II F II < oo for each 11. Levi's theorem now implies that g E Lq(J.L). Step II. The general case. For each A E AJ.L write Lp(A) = {/ E Lp(/1-): f = 0 on Ac}, and note that Lp(A) = {/XA: f E Lp(/1-)}. Also, let

(j

I

C = {A E AJ.I.: Ji.* (A) < oo}. By the preceding case, it should be obvious that for each A unique gA E Lq(A) such that

holds for all f

E

E C

there exists a

L p(/1-). Moreover,

holds. Let a. = sup{ llgAIIq: A E C} < IIFII < oo. Now, observe that if A, B E C satisfy A c B. then gA = gB on A; therefore, lgA I < lgB I. and so, llgAIIq < llgBIIq· It follows that there exists an increasing sequence {A,} of C such that II g A,,lq a < oo. The latter, combined with Levi's theorem, shows that the function g defined by g (x ) = lim gAu (x) for each x E X belongs to Lq(J.L). Clearly, g vanishes outside of the set A = U: 1 A,. Next, we claim that F vanishes on Lp(Ac). Indeed, if F is not zero on Lp(Ac), then in view of the norm denseness of the step functions there exists some B E C with B c Ac, so that F does not vanish on Lp(B). Thus, g8 :;{: 0, and since B n An = 0 for all n , it follows that for each n

t

·

holds. Since lim llgA"IIq = a , it follows that 11gB llq = 0, which is impossible. Hence, F vanishes on Lp(Ac).

Section 37: COMPARING MEASURES AND THE RADON-NIKODYM THEOREM 347

Now, let f E Lp(J..L ). Note first that sinc.e g E Lq(J..L), we have fg E L 1 (J..L ) . Thus, from fgAn 4 fg, IfgA, I < Ifg I , and the Lebesgue dominated con­ vergence theorem, it follows that lim J fgAn d f.l = J fg d f.l. Similarly, it fol­ lows that lim II/ XAn - f XA l i p = 0, and so, by the continuity of F we must have lim F(f XA,) = F(f XA). Therefore, F(f)

= =

F(fXA + fXA < ) lim F(fXA,)

n-oo

=

F(fXA) + F(f XA< )

=

j8

lim

t A

n-oo

,

j

=

F( /XA)

dJ..L = tg d f.l ,

as required. The equality I I F II = ll gllq was established in Theorem theorem is now complete.

31. 16. The proof of the •

The conjugate space of L 1 (J.L) will be considered next. If we assume that (X, S, J..L) is a a-finite measure space, then the norm dual of L 1 (J..L) is lattice iso­ morphic to L00(J.L). For a general measure space this may not be the case (see Exercise 16 of Section 31 ).

Theorem 37.10. Let (X, S, J..L) be a a-finite measure space, and let F be a

continuous linear ftmctional on L 1 (J..L) . Then there exists a unique g such that F(f) =

holdsfor all f

E

Lt(J.L. ). Moreover,

E

L00(J..L )

Jfg df.l

IIFII

L r (J..L) =

=

!l g!l00. /n other words,

L oo(J..L) .

Proof. First consider a finite measure space (X, S, J.L). Then the arguments of

Step I in the proof of Theorem 37.9 also apply here and show that there exists a unique function g E L 1 (J..L) such that F(f) = Jfg dJ..L holds for all f E L 1 (J..L ). Next, we show g E Loo(J..L ) and II F II = 11 8 11oo · Indeed, if E > 0, and A = {x E X: I I F II + E < lg(x)l } , then ( !IF II + E)J.L*(A)

<

i lgl d ig J..L

=

Sgn g df.l

=

F(XA Sgn g)

< II F II · II X ASgn g !I t = II F II · J..L* (A) holds. Thus, J..L * (A) = 0, and so, lg(x)l < II F II + E for almost all x. This shows that g E L00(J.L) and that llgll00 < IIFII + E . Since E > 0 is arbitrary, llgll00 < II F II holds. On the other hand, II F II < llglloo holds trivially, and thus, I I F II = llgll00 •

Chapter 7: SPECIAL TOPICS IN INTEGRATION

348

To extend the theorem to the a-finite case, let {X,} be a disjoint sequence of measurablesatisfying X = U: 1 Xn andJL*(X,) < oo foreach n . By thepreceding case, for each n there exists a unique e L00(J..L ) which vanishes off Xn, with and satisfying dJ..L for each E Lt(J..L). Let !lg,lloo < be the function that equals 8n on each X,. Clearly, is measurable, and since it follows that g E L00(J..L ) . 118lloo < We leave now the details for the reader to verify that dJL holds for all E L 1 (J..L) . •

g, F(fXxn) = Jfg,

I FI I FI .

f

g F(f) = Jfg

g

·

f

As an application of Theorem 37.9 we shall establish that every L p-space with 1 < p < oo is a reflexive Banach space. Recall that a Banach space is said to be reflexive if the natural embedding of the space to its second dual is onto.

Let (X, S, J..L) be a measure space. /f 1 < p < oo, then Lp(J..L) is a reflexive Banach lattice. Proof. Let 4> e L ;*(J..L) . Then for e Lq(J..L ) defines a bounded linear functional on Lq(J..L) , since Theorem 37.11.

lfr(g) = tj>(F�:) g

f l/f(g) = Jgf /(F.r: ) = F�:(f) = Jgf = l/f(g) . =f

By Theorem 37.9, there exists some E Lp(J..L) such that dJ..L holds dJ..L for all e Lq(J..L ). On the other hand, -..tJ> is, the na�ural embedding of L p(J..L) to its s�cond dual L ;*(J..L ) is onto, and hence, Lp(J..L) is a reflexive Banach lattice. •

g

g

EXERCISES 1.

Verify the following properties of signed measures: J.J.

« J.J.. b. v « J.J. and J.J. « w imply v « w. c. If 0 :::: v .:::: J.J., then v « J.J. . d. If J.J. « 0, then J.J. = 0. a.

2. 3.

(l)

Verify statements through (5) following Theorem 37.5. Le J.J. and v be two measures on a a-algebra I: . If v is a finite measure, then show

t

iv l

that the following statements are equ a ent: a.

b.

c.

v

« J.J. holds.

For each sequence {An } of I: with

lim J.J.(An) = 0, we have lim v(An) = 0.

0 there exists some 8

> 0 (depending on E) such that whenever A e I: satisfies J.J.(A ) < 8, then v(A) < E holds.

For each E >

Section 37: COMPARING MEASURES AND THE RADON-NIKODYM THEOREM 349

4.

[fllNT: If (a) does not imply (b), then there exists some E > 0 and a sequence {An} of 'E such that p.(A,) < z-" and v(A,) > €. Let A = n�1 U:::, A;, and note that p.(A) = 0, while v{A) ::: E.] Let p. be a finite measure and let { v,} be a sequence of finite measures (all on 'E) such that v, « p. holds for each n . Furthermore, assume that lim Vn (A) exists in R for each A E 'E. Then show that: a. b.

For each E > 0 there exists some c5 > 0 such that whenever A E :E satisfies p.(A) < 8, then v,(A) < E holds for each n. The set function v: 'E � [0, oo], defined by v(A) = lim Vn{A) for each A E 'E, is a measure such that v « p..

[HINT: Consider :E equipped with the distance d(A, B) = p.(AtlB), where A and B are identified if J.L(A 68) 0. Then {:E, d) is a complete metric space; see Exercise 3 of Section 31. Next, use v, « p. to verify that each v, is a well-defined continuous function on :E . Furthermore, if E > 0, then each C1.: {A E :E: lvn(A) - Vm(A)l ::S E for n, m ::: k} is a closed set, and :E = u�l ck holds. By Theorem 6.18, some ck has an interior point. Use this and the preceding exercise to establish (a). Statement (b) follows from (a).] 5. Let { v,} be a sequence of nonzero finite measures such that lim v,{A) exists in R for each A E 'E. Show that v(A) = lim Vn{A) for A E 'E is a finite measure. [HINT: Consider the finite measure p.{A) = L� l 2-"[vn(A)fvn(X)], and then apply the conclusion of the preceding exercise.] 6. Verify the uniqueness of the Radon-Nikodym derivative by proving the following statement: If (X. S, p.) is a measure space and t E L 1 (p.) satisfies fA t dp. = 0 for all A E S, then t = 0 a.e. 7. This exercise shows that the hypothesis of a-finiteness of p. in the Radon-Nikodym theorem cannot be omitted. Consider X = [0, 1], 'E the a-algebra of all Lebesgue measurable subsets of [0, 1 ], v the Lebesgue measure on 'E and p. the measure defined by p.(C/J) = 0 and p.{A) = oo if A =/= 0. (Incidentally, p. is the largest measure on :E.) Show that: =

=

a. v is a finite measure, p. is not a-finite, and v « p.. b. There is no function t E L 1 (p.) such that v(A) = fAt dp. holds for all A E 'E. 8.

Let p. be a finite signed measure on :E. Show that there exists a unique function t E L 1 (jp.l) such that

holds for all A E :E. 9. Assume that v is a finite measure and p. is a a-finite measure such that v « p.. Let g = dvjdp. E L I (J-L) be the Radon-Nikodym derivative of v with respect to p.. Then show that: a. Iff = {x E X: g(x) > O}, then YnA is a p.-measurablesetforeachv-measurable set A. b. If t E L 1 (v), then tg E L I (J-L) and ft dv = ftg dp. holds.

350

Chapter 7: SPECIAL TOPICS IN INTEGRATION

15.10)

[HINT: Note first that :E c AJl c A v holds, and then (by Theorem that v*(A) = A must hold for each A E Jl; hence, v*(Yc) = For (a) use Theorem 37.3 fA g dJL and the fact that :E is a a-algebra. Now, if¢ = 2::?=1 a; XA; is a v-step function, then

0.

n

II

i=l

i=l

J¢dv = La;v*(A;) = L a;v*(A; n Y) = L a; 1

10.

II

i=l

0

0

' w-a.e.,

holds. [HINT: Use the preceding exercise.] All measures considered here will be assumed defined on a fixed a-algebra :E. a. b. c.

Call two measures JL and v equivalent (in symbols, JL = v) if JL « v and v « JL both hold. Show that = is an equivalence relation among the measures on :E. If JL and v are two equivalent a-finite measures, then show that AJl = A,,. Show that if JL and v are two equivalent finite measures, then

dJL dv · = dv dJL

-

d.

e.

f.

12.

f

g dJL = ¢g dJL,

and so, ¢g E L t (JL). Now, if O � f E L1 (v), then pick a sequence {¢11} of v-step functions such that � ¢n (x) t f (x) for all x E X. To finish the proof, observe that � ¢11g t fg and f¢n8 dv = f¢, dv t ff dv < oo.] Establish the chain rule for Radon-Nikodym derivatives: If w is a a -finite measure and v and JL are two finite measures (all on :E) such that v « JL and JL « w, then v « w and

dv dv dJL = dw dJL · dw

11.

A;nY

-

1 a.e. holds.

If JL and v are two equivalent fin.ite measures, then show that f �---+ f · ��, from L 1 (JL) to L 1 (v), is an onto lattice isometry. Thus, under this identification, L 1 (JL) = L 1 (v) holds. Generalize (d) to equivalent a-finite measures. That is, if JL and v are two equiv­ alent a -finite measures, then show that the Banach lattices L 1 (J.L) and L 1 ( v) are lattice isometric. Show that if JL and v are two equivalent a-finite measures, then the Banach lattices L p(JL) and L p(v) are lattice isometric for each 1 � p � oo.

Let JL be a a-finite measure, and let AC(JL) be the collection of all finite signed measures that are absolutely continuous with respect to JL; that is,

AC(JL) = { v E M(:E): a. b.

v

« JL}.

Show that AC(JL) is a nonn closed ideal of M(:E) (and hence, AC(JL) with the norm II v ii = jvj(X), is a Banach lattice in its own right). For each f E L 1 (JL), let JL 1 be the finite signed measure defined by JLJ(A) = fA f d JL for each A E :E. Then show that f �---+ JLf is a lattice isometry from L 1 (JL) onto AC(JL).

13. Let :E be a a-algebra of subsets of a set X and JL a measure on :E. Assume also that :E* is a a-algebra of subsets of a set Y and that T: X � Y has the property that r- t (A) E :E for each A E :E*.

Section 37: COMPARING MEASURES AND THE RADON-NIKODYM THEOREM 351 a. b.

1 Show that v(A) = J.L(T - (A)) for each A e �* is a measure on �*. If E L 1 (v), then show that o T E L I (J.L) and

f

f £t dv fxt T dJ.L. =

w

o

c. If J.L is finite and is a a-finite measure on �* such that v « w, then show that there exists a function E L 1 (w) such that

g

fxt T dJ.L = £tg dw o

holds for each f E L 1 (v) .

14. Let (X, S, J.L) be a a -finite measure space, and let g be a measurable function. Show

that if for some 1 < p < oo we have fg E L I (J.L) for all f E Lp(J.L), then g E Lq(J.L), where + = 1 . Also, show by a counterexample that for 1 < p < oo, the a ­ finiteness of J.L cannot be dropped. We can assume that 2: 0 (why?). Then = J for E Lp( ) defines a positive linear functional on L p(J.L). By Theorem 30. 1 0, is continu­ ous. Now, by Theorems 37.9 and 37.10 there exists some h E Lq(J.L) such that Jfg dJ,L = J fh for each E Lp(J.L). Show that = h a.e. holds.] a measurable function, and 1 < p < 15. Let (X, S, J.L) be a a-finite measure space, oo. Assume that there exists some real number M > 0 such that ¢g E L 1 and J¢g � Mil¢ lip holds for every step function ¢. Then show that:

� �

[HINT:

g

dJ,L

a. b.

f

g

dJ,L g Lq(J.L), where � + � = and ffgdJ,L Mll/llq holds for all f

F(f) fgdJ,L f F g

J.L

(J.L)

1,

E

<

E Lp(J.L).

16. Let 11- be a Borel measure on JR.k and suppose that there exists a constant

c,

c.

c > 0 such

that whenever a Borel set E satisfies A( E) = then J,L(E) Show that J.L coincides with i.e., show that = Let J.L and v be two a-finite measures on a a-algebra � of subsets of a set X such that 17. v « and v # 0. Show that there exist a set E E � and an integer n such that

A, J.L

a. b. 18.

J.L A.

v(E) >

0; and A E � and A C

E imply

�J.L(A) < v(A) < nJ.L(A).

Let J.L be a finite Borel measure on a. b.

A,

=

[ 1, oo) such that

and J.L « J,L(B) = aJ.L(aB) for each a > 1 and each Borel subset B of aB = {ab: b E B}.

dJ.L/dA [dJ.L/dA](x) cjx2

[1, oo), where

If the Radon-Nikodym derivative is a continuous function, then show that > there exists a constant 0 such that = for each 19. Let be a finite Borel measure on (0, oo) such that

J.L

c

x :::. 1.

A,

a. J.L « and b. J.L(aB) = J,L(B) for each a > 0 and each Borel subset B of (0, oo). If the Radon-Nikodym derivative is a continuous function, then show that there exists = for each > 0. a constant > 0 such that

c

[dJ.L/dA](x) cfx

x

Chapter 7: SPECIAL TOPICS IN INTEGRATION

352

38. THE RIESZ REPRESENTATION THEOREM In this section, X will denote a Hausdorff locally compact topological space.

Recall that Cc(X) denotes the vector lattice of all real-valued continuous functions defined on X having compact support. That is, f: X -+ 1R belongs to Cc(X) if and only if f is continuous and vanishes off a compact set. The main purpose of our discussion here is to characterize the positive linear functionals on Cc(X). As usual, a linear functional F on Cc(X) is said to be positive if 0 < F(f) holds whenever 0 < f E Cc(X). Clearly, every continuous function on X is Borel measurable. Thus, if J.L is a Borel measure on X (i.e., J.L is a measure on the a-algebra B generated by the open sets and satisfies J.L(K) < oo for each compact set K), then the formula

f dJ.L, f

F(f) = f

E Cc(X),

defines a positive linear functional on Cc(X). In general, if F is a positive linear functional on Cc(X) and a Borel measure J.L on X satisfies F(f) = J f dJ.L for each f E Cc(X), then J.L is called a representing measure for F (or that J.L represents F). The following basic result (known as the Riesz representation theorem) will be obtained: Cc(X)

Every positive linear functional on i s repr e sented by a unique regular Borel measure. Recall that a Borel measure J.L on is called a regular Borel measure (see X

Definition 18.4) if J.L satisfies the following two extra properties:

B

B

a. J.L( ) = inf{J,L(V): V open and c V } for each Borel set B; and b. J.L(V) = sup{J.L(K): K compact and K c V } for each open set V. Of course, the preceding two conditions are enough to ensure that whenever a Borel set has finite measure, then

B

J.L(

B = sup{J.L(K): K compact and K B) c

)

holds; see Lemma 1 8.5. Let us start by introducing some standard notation. Let V be an open set, and let < I holds for each E X f E Cc(X). The symbol f -< V means that 0 < and Supp f C V. Similarly, for a compact set K and a function f E Cc(X), the I for notation K -< f means that 0 < f < I holds for all x E X, and f all x E K. The notation K -< f -< V simply means that K -< f and f -< V both hold. In this terminology, for instance, Theorem I 0.9 can be stated as follows: If K C U7= 1 V; K V; fJ , . . . , fn E Cc(X) f; -< V; 2:7= 1 f; = 1 K. Now, let F be a positive linear functional on Cc(X). For each open subset V of X , define

f(x)

x

(x) (x) = holds with compact �nd each open, then there exist functions such that for each i and on J.L(V) = sup{F(f): f

-<

V}.

353

Section 38: THE RlESZ REPRESENTATION THEOREM

Clearly, 0 < JL(V) < oo holds for each open set V . It is easily seen that JL(V) < JL(W) holds for every pair of ope·n sets with V c W (since f -< V implies -< W). This observation allows us to extend the set function JL from the open sets to all subsets A of by defining

f

X

.u(A) = inf{.u(V): V open and A

c

V}.

As expected, the set function .u is an outer measure.

Let be a positive linear functional on Cc(X). Then the set function (previously defined) is an outer measure on X. Proof. 0 X. 0, 0. c B, .u(B) Theorem 38.1.

F

.u

Clearly, < JL(A) < oo holds for every subset A of Since f -< 0 if and only if f = it follows that .u(0) = Also, it should be obvious that if A then .u(A) < holds. To complete the proof, it remains to be shown that .u is a-subadditive. To this end, let {A,} be a sequence of subsets of X . Let A = U:1 A,. If 2::1 .u(A,) = oo, then .u(A) < 2::1 ,u.(A,) is obvious. Hence, assume 2::1 choose an open set Vn such that A, c V, JL(A,) < oo. Let E > 0. For each and .u(V,) < JL(A,) + E2-". Put V = U: 1 Vn . Now, if f -< V holds, then K = Supp f c U:1 V,, and in view ofthe compactness of K there exists some such , E that K C U7,� 1 V, . By Theorem 10.9, there exist functions /1 such that f,, -< V, for n = I, . . . , m and 2:�'= 1 f,, = 1 on K . Clearly, < 2::7= 1 holds, and by the positivity of F it follows that

n,

• . . .

F(f) <

m

111

oo

m fm Cc(X) f, f

oo

L F(f,, ) < L1 JL(V,) < L ,U (Vn) < L JL(A11) + E . n= n=1 11= 1 11=1

Thus, since thelatterholdsforeach f -< V , it follows that .u(V) < 2::1 JL(A,)+E. Now, notice that A c V implies .u(A) < .u(V) < 2::, .u(A,) + E for all E > 0, from which it follows that JL(A) =

Jl

(Q ) � A..

The proof of the theorem is now complete.

<

JL (A ., ). •

The outer measure .u determined by the positive linear functional F is called the outer measure induced by F (or the outer measure associated with F). The next goal is to show that .u is a regular Borel measure on B.

Let be a positive linear functional on Cc(X), and let be set i s J L -measurable outer measure. Then ever y Borel (i. e . itsandinduced , restricted to is a regular Borel measure. Theorem 38.2. JL

F

B

.u B C AJL),

354

Chapter 7: SPECIAL TOPICS IN INTEGRATION

Proof. The proof goes by steps.

Step I. For each compact set K, we have J.L(K) < oo. Let K be a compact set. Choose an open set V with compact closure such that such that V g. K C V. By Theorem 10.9, there exists a function g E Now, if f V holds, then f < g, and so F(f) < F(g). Hence,

Cc(X )

-<

-<

J.L(K) < J.L(V) = sup{F(f): f -< V} < F(g) < oo. Step 11. IfK1 and K2 are disjoint compact sets, then J.L(KI UK2)=J.L(KJ)+J.L(K2)· In view of the a -subadditivity of J.L, it suffices to show JJ. (K 1 ) + J.L(K2) < J.L(K1UK2). Since K1 c K2 holds, there exists (by Lemma 10.7) an open set V1 with compact closure such that K1 C V1 C V1 C K2. Thus, K2 C (VJ)c holds. Let V2 = (VJ)c, and note that V1 and V2 are two disjoint open sets. Now, let E > 0. Choose ar1 open set V such that K1 U K2 c \l and J.L(V) < J.L(KJ UK2)+E; clearly, K1 C VnV1 and K2 C V nV2. Next, selecttwo continuous functions f and g such that f -< V n V1 , g -< V n V2 , J.L(V n \11 ) < F(f) + E, and J,L(V n V2) < F(g) + E. From VI n v2 = 0. it follows that f + g -< v Now, 0

note that

J.L(KJ ) + J.L(K2) < J.L(V n V1) + J.L(V n V2) < F(f) + F(g) + 2E = F(f + g) + 2E < J,L(V) + 2E < J,L(KJ U K2) + 3E, 0. Therefore, J.L(K 1) + J.L(K2) < J.L(K 1 U K2), as claimed. StepIll. Forevei)'Subset A ofX have J.L(A)= inf{J,L(V): V open and A C V}. This is precisely the definition of J.L. Step IV. For every open set V, J.L( V) = sup{J.L(K): K compact and K C V}. To see this, let V be an open set and let r E JR. satisfy r < J.L(V). Choose a continuous function f V with r < F(f), and let K Supp f. Now if W is an open set such that K c W, then f W holds, and so r < F(f) < J.L(W). holds for all E

>

we

-<

Therefore,

=

-<

J.L(V) > J,L(K) = inf{J,L(W): W open and K

c

W } :;:: F(f) > r

holds for every r E IR with r < J.L(V ) : and the desired identity follows.

Step V. Every Borel set is J.L-measurable, that is, B C Aw Note first that if K is a compact set and V is an open set disjoint from K, then J.L(K) + J.L(V) = J.L(K U V) holds. Indeed, by Step IT, for every compact subset K1

Section 38: THE RIESZ REPRESENTATION THEOREM

355

of V we have J.L(K) + J.L(K

I) =

J.L(K

u

K I) < J.L(K

u V)

< J.L(K) + J.L (V) ,

and the claim follows from the identity of Step IV. Since Ap.. is a
•

and the proof is finished.

Now, we come to the main result of this section. Namely: The regular Borel measure, induced by a positive linear functional on Cc(X), is the only regular Borel measure that represents F.

F

For ever y positive li­ there exists a unique regular Borel measure such

Theorem 38.3 (The Riesz Representation Theorem). Cc(X),

F on nearfu.nctional that

J.L

F(f) = Jf dJ.L holds for ever y f E Cc:(X). Moreover, the representing measure is the regular Borel measure induced F on X. v X Proof. (Uniqueness.) J f dJ.L = J f dv f Cc(X). = by

that

Let J.L and be two regular Borel measures on holds for each E We shall show that J.L(A)

such v(A )

Chapter 7: SPECIAL TOPICS IN INTEGRATION

356

holds for each Borel set A. In view of the regularity properties of J.l and v , it is v(K) holds for each compact set K. enough to establish that J.,L(K) To this end, let K be a compact set. Given E > 0, choose an open set V such that K c V and J.,L�V) J.,L(K) + E. By Theorem 10.9, there exists a function X\', we have E Cc(X) such that K -< -< V . But then, since XK

f

=



x

z) = (x JR."

4Alfred Haar ( 1 885-1933), a Hungarian mathematician. He studied orthogonal systems of functions and partial differential equations and is best remembered for his introduction of the integral that bears his name on Hausdorff locally compact topological groups.

Section 38: THE RIESZ REPRESENTATION THEOREM

363

and fa defined respectively by

fa (x) = f(ax)

and

fa (x) = f(xa)

for each x E G. Note that fa and fa both belong to Cc(G) and that fa = fa holds if G is commutative. A linear functional F: Cc(G) ---+ lR is said to be left­ invariant (resp. right-invariant) if F(f) = F(fa) (resp. F(f) = F(/0)) holds for all a E G. Here is the fundamental result regarding left-invariant linear functionals.

Theorem 38.8 (Haar). !fG isaHausdorfflocallycompact topological group,

then there exists a unique (aside of scalar factors) non-zero positive linear functional H: Cc(G) ---+ lR which is left-invariant. By the symmetry of the situation, there exists, of course, also a unique non-zero positive linear functional on Cc(G) which is right-invariant. The unique non-zero left-invariant positive linear functional Cc(G) ---+ lR is called the left-Haar integral of G. If G is commutative, then the left- and right-Haar integrals coincide and the common integral is called the Haar integral of G. Now, let G be a Hausdorff locally compact topological group and let H: Cc(G) ---+ lR be its unique left-Haar integral. By the Riesz Representation Theorem 38.3, there exists a unique (aside of scalar factors) regular Borel measure f..LG satisfying

H:

H(f) =

l f(g)dJ..LG(g).

It is easy to prove that the left-Haar measure satisfies f..LG (aA) = J..L G (A) for each Borel set A. The measure /1-G is called the left Haar measure of G. Here are two examples of the Haar integral and the Haar measure on two familiar Hausdorff locally compact topological groups.

1. 2.

Consider the additive Hausdorff locally compact topological group G = JR." . Its Haar integral coincides with the Lebesgue integral and its Haar measure is the Lebesgue measure. Consider G = (0, oo) as a commutative group under the ordinary multipli­ cation. Then the Haar integral H: G ---+ lR is given by H (f) = dx for for each f E Cc(G) and the Haar measure satisfies J.LG(A) = each Borel subset A of G.

J000 f�x> JA �r

For proofs and details regarding the Haar measure and Haar integral we refer the reader to [14, Chapter 6].

364

Chapter 7: SPECIAL TOPICS IN INTEGRATION

EXERCISES Unless otherwise specified, in the exercises below X will denote a Hausdorff locally compact topological space. 1.

2.

3.

If X is a compact topological space, then show that a continuous linear functional F

IIFII holds. Let X be a compact topological space, and let F and G be two positive linear func­ tionals on C(X). If F( 1) + G(1) � IIF - Gil, then show that F A G = 0. [HINT: Show that F v G(1) = !I F - G il.] Let co(X) = { f E C(X): V E > 0 3 K compact with lf(x)l < E V ¢ K}. Show on.C(X) is positive if and only if F(1) =

x

that:

a. b.

4.

5.

6.

7.

F be a positive linear functional on Cc(X), and let 11- be the outer measure induced by F on X. Show that if 11-* is the outer measure generated by the measure space Y R then " * hnldc fnr t J Let ( \•• '

"\

( A \ - ,_ ", (A \ ._,

'""'""'' "Uh" .. -• -• "' -'-'-


Let 11- and v be two regular Borel measures on X. Then show that 11- � v holds if and - ' ,_.,,

••

•• ,.-

\• •1

•

••-•

u

&-•

I '-

V.I.

.IL

ff dJJ, 2: f f dv

for each 0 � f E Cc(X). Fix a point x E X, and define F(f) = f(x) for each f E Cc(X). Show that F is a positive linear functional on Cc(X) and then describe the unique regular Borel measure 11- that satisfies F(f) = Jf dJJ, for each f E Cc(X). What is the support of 11-? Let X be a compact Hausdorff topological space. If 11- and v are regular Borel measures, then show that the regular Borel measures 11- v v and 11- A v satisfy only if

a.

b.

8.

co(X) equipped with the sup norm is a Banach lattice. The norm completion of Cc(X) is the Banach lattice co(X).

Supp(JJ- v v) = Supp 11Supp(JJ- " v) c Supp 11-

u Supp v, and n Supp v.

Use (b) to show that if Supp 11- n Supp v = 0, then 11- ...L v holds. Also, give an example for which Supp(JJ- " v) =f: Supp 11- n Supp v.

Characteriz� the positive linear functionals F on Cc(X) that are also lattice homomor­ phisms, that is, F(f v g) = max{F(/), F(g)} holds for each pair f, g E Cc(X).

· [HINT: Let 11- be the representing regular Borel measure for F. Show that Supp 11contains at most one point.]

9.

If X is an uncountable set, then show that a.

c;(X) is not separable, and

b.

C [O, 1] (with the sup norm) is not a reflexive Banach space.

x E X define the positive linear functional F.\·(f) = f(x). Show that II Fx - Fy II = 2 if x =f: y. For (b) combine (a) with Exercise 8 of Section 27 and [HINT: For every

Exercise 12 of Section 12.]

10.

For a finite signed measure 11- on B, show that the following statements are equivalent: a.

b. c.

J1 belongs to Mb(X).

11-+ and 11- - are both finite regular Borel measures. For each A E B and E > 0, there exist a compact set

K £ A £ V such that IJJ-(B)I

K and an open set V with B £ V \ K.

< E holds for all B E B with

Section 38: THE RIESZ REPRESENTATION THEOREM 11.

A sequence {xn} in a nonned space is said -to converge weakly to some vector x if lim f(x11 ) = f(x) holds for every continuous linear functional f. a. b.

12.

365

Show that a sequence in a nonned space can have at most one weak limit. Let X be a Hausdorff compact topological space. Then show that a sequence {J,,} of C(X) converges weakly to some function f E C(X) if and only if Un } is nonn bounded and lim /nCr) = f(x) holds for each x E X.

[HINT: For (b) use Theorem 28.8 and the Riesz representation theorem.] Let J-L be a regular Borel measure on X, and let f E L 1 (J-L). Show that the finite signed measure v. defined by

v(E) =

Lt

dJ-L

for each Borel set E. is a (finite) regular Borel signed measure. In other words, show that v E Mb(X). [HINT: Use the fact that {x E X : f(x) =f. 0} is a a-finite set.] 13. Generalize part (3) of Theorem 38.5 as follows: If J-L and v are two regular Borel mea­ sures on a Hausdorff locally compact topological space and one of them is a-finite, then show that J-L 1\ v is also a regular Borel measure. 14. Show that every finite Borel measure on a complete separable metric space is a regular Borel measure. Use this conclusion to present an alternate proof of the fact that the Lebesgue measure is a regular Borel measure. 15. Let X be a Hausdorff compact topological space. If ljJ: X � X is a continuous func­ tion, then show that there exists a regular Borel measure on X such that

holds for each f E C(X). [HINT: If Lim is a Banach-Mazur limit (see Exercise 7 of Section 29) and w E X is a fixed point, then consider the positive linear functional F: C ( X) � 1R defined by

2 F(f) = £im(f(l/J(w)). /(l/J (w)), f(ljJ3 (w)), . . ).] .

16.

This exercise gives an identification of the order dual c;(X) of Cc(X). Consider the collection M(X) of all fonnal expressions J-L 1 - JJ- 2 with J-L 1 and JJ-2 regular Borel measures. That is,

M (X) = {JJ- 1 - JJ-2: JJ- 1 and JJ-2 are regular Borel measures on X}.

a. b.

Define JJ-1 - JJ-2 = v1 - v2 in M (X) to mean J-L 1 (A) + v2( A) = v1 (A) + J-L2(A) for all A E B. Show that = is an equivalence relation. Denote the collection of all equivalence classes by M(X) again. That is, JJ- 1 - JJ-2 and v1 - v2 are considered to be identical if JJ-1 + v2 = v1 + JJ-2 holds. In M(X) define the algebraic operations

(JJ-1

- JJ-2) + ( v i

- v 2)

=

a ( J-L I

- JJ-2 )

=

(JJ- 1 + v t ) - (JJ-2 + V2).

{

aJJ-1 - aJJ-2 ( -a)JJ-2 - (-a )JJ- 1

if a ::: 0 if a < 0 .

366

Chapter 7: SPECIAL TOPICS IN INTEGRATION Show that these operations are well-defined (i.e., show that they depend only upon the equivalence classes) and that they make M(X) a vector space. c. Define an ordering in M(X) by J.q - J..L2 2: v1 - v2 whenever J..L I(A) + v2(A) > v1 (A) + J..L2(A) holds for each A e B. Show that > is well defined and that it is an order relation on M(X) under which M{X) is a vector lattice. d. Consider the mapping J..L = J..L l - J..L2 �---* Fp. , from M(X) to c;-(X), defined by Fp.(f) = Jf dJ..L 1 - J f dJ..L2 for each f E Cc(X). Show that Fp. is well-defined and that J..L �---* Fp. is a lattice isomorphism (Lemma 38.6 may be helpful here) from M(X) onto c;-(X). That is, show that c;-(X) = M(X) holds.

17. This exercise shows that for a noncompact space X, in general c;(X) is a proper ideal of c;-(X). Let X be a Hausdorff locally compact topological space having a sequence (On} of open sets such that On c On+ I and 011 :f:. On+ l for each n, and with X = U� 1 On. a. Show that if X is cr-compact but not a compact space, then X admits a sequence (0n } of open sets with the above properties. b. Choose XJ E 0: a.nd Xn E 0;; \ On-l for n � 2. Then shO\V that F (f) =

L f(xn ), 00

n=l

for f E Cc:(X),

defines a positive linear functional on Cc(X) that is not continuous. c. Determine the (unique) regular Borel measure J..L on X that represents F. What is the support of J..L?

39. DIFFERENTIATION AND INTEGRATION In this section the differentiability properties of a Borel signed measure on lR.k will be studied. The derivative of a measure will always be taken with respect to the Lebesgue measure. For a differentiation theory of arbitrary set functions in a more general context, the interested reader is referred to Chapter 8 of [34]. The results obtained about differentiation of signed measures will be used to derive the classical properties of functions of bounded variation. Throughout this section, the domain of all signed measures will be the a-algebra B of all Borel sets of lR.k. Also, unless otherwise specified, the expression "al­ most everywhere" will be synonymous with "almost everywhere with respect to the Lebesgue measure." The main Borel sets considered for the purposes of dif­ ferentiation will be the open balls. Recall that the open ball with center at x and radius r is the subset of lR.k defined by {y E lR.k: - y II < r}, where 2 Y ll = k Y; ) 1 2 • Occasionally, measures defined on the Borel subsets of an open set V of 1Rk appear. By assigning the value zero to the Borel subsets of vc, it is easy to see that these measures can be assumed defined on all Borel sets oflR.k . Therefore, theorems about Borel measures on lRJ: can be applied equally well to such a situation.

llx -

(Li=l (x; -

!

.

llx

Section 39: DIFFERENTIATION AND INTEGRATION

367

Let J.L be a signed measure on the a-algebra of all Borel sets of 1R.k. For each x E 1Rk and r > 0, we define the following two extended real numbers: L'l;(x) = sup t.,(x)

=

inf

l l ��;; : J.L(B) >..(B)

I, I·

: B is an open ball of radius < r and x E B -

B is an open ball of radius < r and x E B

and

Observe that x is not required to be the center of the open balls B where the sup and inf are taken. Clearly, -00 < L'lr (X) < L'l;(x) < 00 holds for each x E 1Rk and all r > 0. Also, it should be obvious that if 0 < r < then l:l.;(x) < L'l;(x), and fls(x) < l:l.r (x) both hold for each x. Therefore, the limits

s,

D* J.L(X) = lim L'l;(x) r�O

and

D*J.L(x) = lim L'lr(x) r�O

k exist in 1R* , and they satisfy -oo < D * J.L(X) < D * J.L(x) < oo for each x E 1R . The extended real numbers D*J.L(X) and D * J.L(x) are called the lower and upper derivatives of J.L (with respect to >... ) at the point x. If they are real and equal, then this common value is called the derivative of J.L at x.

Definition 39.1.

Let be a signed measure on Band J.L

-

k x E 1R .

oo < D*J.L(x) = D* J.L(X) < oo

If

holds, then i s said to be differentiable at the point The common value is called the derivative of at and is denoted by that is, J.L

J.L

x. DJ.L(x),

x

m i�:� m

Here is a rephrasing of the preceding definition: A real number satisfies = D J.L(X) if for each E > 0, there exists some 8 > 0 such that I - I <E holds for each open ball B with x E B and with a radius less than 8. An alternative definition of the derivative using sequences is this: A signed measure J.L on is differentiable at some point x E 1Rk if and only if there exists holds for every sequence (B11} of open a real number such that lim balls containing x whose radii tend to zero. (The number is, of course, the derivative DJ.L(X) ) The expression "D J.L(x) exists" is synonymous (as usual) with "J.L is differen­ tiable at x ." It is easy to see that if two signed measures J.L and v are differentiable

m

Bm

i�::�

.

=

m

m

368

Chapter 7: SPECIAL TOPICS IN INTEGRATION

at some point x and their sum J.L + v defines a signed measure, then J.L + v is also differentiable at x and D(J.L + v)(x) = DJ.L(X) + Dv(x) holds. Similarly, if J.L - v defines a signed measure, then D(J.L - v)(x) = DJ.L(X) - Dv(x). Our first objective is to show that a finite signed Borel measure J.L that is ab­ solutely continuous with respect to the Lebesgue measure is differentiable al­ most everywhere and that the derivative D J.L coincides with the Radon-Nikodym derivative dJ.L/dA.. This will be the key differentiation result for this section. To prove this theorem, we need some preliminary discussion. The reader can verify easily the following property of the Lebesgue measure: If A is a subset of JR.k and r > 0, ihen the sei r A = irx : x E A} satisfies A.(r A) = ,-k A.( A).

This, combined with the fact that >.. is translation invariant, shows that if B is an open ball with radius r and B * is another open ball with radius ar, then A.(B * ) = ak A.(B) holds. The latter will be used in the proof of the next lemma. Lemma 39.2. Let B 1 ,

•

•

, B, be open balls in IRk . Then there exist painvise

•

disjoint open balls Bk, , . . . , Bk"' among the B1 , . . . , Bn such that

Proof. Let r; be the radius of B;. Rearranging the balls, we can assume without loss of generality that r1 > r2 > · · · > �'n· Put k1 = 1, and let k2 be the smallest integer (if there is any) such that Bk2 is

disjoint from Bk , . Let k3 be the smallest integer such that Bk3 is disjoint from Bk, and , Bn } has been exhausted. This Bk2 • Continue this process until the finite set { B 1 , gives us the open balls Bk, , , Bk"' , and we claim that they satisfy the property of the lemma. To see this, let Ak; be the open ball with the same center as Bk; and with radius three times that of Bk; · Now, note that each B; must intersect some Bki · It follows that B; c Akj must hold for that j. �us, u�=l B; c U7=1 Akj' and so, .

.

as claimed.

•

•

•

.

•

Section 39: DIFFERENTIATION AND INTEGRATION

369

k. �:(x), JR. k a r > { x JR. : �:(x) > a} {r11 } rn oo oo l k {x E JR. : D*JJ,(X) > a} = u n X E JR.k : �;,,(x) > a + m1 1 e JR.k : D* JJ,(x) > a} a e JR. Lemma 39.3. Let be a Borel measure on JR.k , and let A be a Borel set such that JJ,(A) = Then DJJ,(x) = holdsfor almost all points in A. k, A Proof. JR. (x) x. DA({x JJ,(A ) = 0. *JJ,(x)e A :D*D*JJ,JJ,(x) 0}) = > 0. E A({x e A: D* JJ,(X) > E}) = 0 > E = {x e A: D*JJ,(x) > E}, E E > 0. K E V A c V. e K, D *JJ,(x) > E B K , x B c V JJ,(B) > EA(B). B , B11 K. Bk" . . . , Bkm B B11

Now, let JJ- be a signed measure on Ftom the definition of it is easy to see that for each e 1R and 0 the set e is an open set. Thus, if is a sequence of positive real numbers with .J,. 0, then the identity

m=l 11= 1

shows that {x is a Borel set for each will be used in the next key lemma.

This observation

JJ-

0.

0

x

Let J.l be a Borel measure on and let be a Borel set such that < Since JJ- is a measure, 0 < holds for each To establish the lemma it must be shown that 0. For this, for each it is enough to show that and note that is a To this end, fix Let Borel set. Let be an arbitrary compact subset of and an arbitrary open set such that Now, if x then holds, and so, there exists an open ball containing with and In view of the there exists a finite number of such open balls 1 , that compactness of cover Let that satisfy be the disjoint open balls among 1 , , Lemma 39.2. Then •

.

•

A

•

•

Since JJ- is also regular Borel measure (Theorem 38.4), it follows that But then the regularity of implies for every compact subset K of and we are done.

E.

•

A(A(£)K ) ==

0 0,

•

Now, we come to the main differentiation theorem regarding Borel measures.

k that is absolutely Ever y finite si g ned Borel measur e on JR. continuous with respect to the Lebesgue measure (i. e . , << A), is differentiable almost eve1ywhere. coincides (a. e . , of course) with the Radon­ Moreover, i t s derivative D Nikodym derivative dJJ,/dA; that is, DJJ, = dJJ,/dA holds. Theorem 39.4.

JJ-

JJ-

JJ-

Chapter 7: SPECIAL TOPICS IN INTEGRATION

370

f = djJ-jd). be the Radon-Nikodym derivative provided by Theorem That is, JJ-(E) JE ! d). holds for each Borel set E. Fix a real number r, and let A {x lR.k : f(x) r}. Clearly, A is Lebesgue measurable, and by Theorem ).(A) < there Also, by Theorem exists a Borel set such that A C F and ).(F) ).(A); clearly, \A) k Next, consider the Borel measure v on lR. defined by v(E) lFnE { [f(x) -r] d).(x) l{AnE [f(x) -r] d).(x) for each Borel set E. Note that if B is an open ball, then JJ-(B) -r).(B) la{ [f(x) -r] d).(x) < lan{ A [f(x) -r] d).(x) v(B) Proof. Let

E L 1 ().)

37.8.

=

=

>

E

22.5,

15. 1 1 ,

oo.

F

).(F

=

=

0.

=

=

=

=

JJ-(B) < r + v(B) ).(B) ).(B)

B, from which it follows that D* JJ-(X) � r + D * v(x) for each x lR.k . Since it follows from Lemma that D v(x) holds for almost all x. and consequently, (*) implies D* JJ-(X) < r for almost all x *D JJ-(X) < rThus� for almost all E A Since f (x) < r implies A it follows that the set Er {x lR.k : f(x) < r < D*JJ-(x)} satisfies Er c A Therefore, by the last conclusion, each E,. has Lebesgue mea­ sure zero. But then {x lR.k : D * JJ-(x) f(x)} C UreQ Er. where Q is the set of all rational numbers, shows that {x lR.k : D* JJ-(X) f(x)} has Lebesgue measure zero. Applying the preceding arguments to -JJ- , and taking into consideration that d(-JJ-)jd). -f and D*(-JJ-) -D*JJ-, we easily infer that the set

holds for each open ball

E

v(Fc) E pc.

=

39.3

0,

x

=

=

E pc, x E

c.

E

c.

E

>

E

=

=

has Lebesgue measure zero. In other words,

>

0

c,

Section 39: DIFFERENTIATION AND INTEGRATION

holds for almost all x E 1Rk . That is, a.e. holds, as desired.

371

DJ.l.(X) exists for almost all x and DJ.1. = f

•

As a first application of the preceding theorem, let us establish a classical result.

If E is a Lebesgue measurable subset of JR., then >..(En(x�E,x+E)) = 1 for almost all x in E, and limE-o+ >..(En(x�E ..t+E)) = 0 for almost all x in Ec. limE-o+

Theorem 39.5. 1. 2.

Proof. Without loss of generality, we can assume that )..(£)

the finite Borel measure J.l. on 1R defined by

J.l.(A) = )..( £ n A) =

< oo.

Consider

J. XE d).. .

Clearly, J.l. << ).. and dJ.1./d).. = XE · By Theorem 39.4, DJ.l. the formulas in ( 1 ) and (2) follow.

= XE a.e. holds, and •

A point x of a Lebesgue measurable subset E of 1R for which ( 1 ) in the pre­ ceding theorem holds is referred to as a density point of E. In this terminology, Theorem 39.5( 1 ) is usually stated as follows: If E is a Lebesgue measurable subset

of

JR., then almost every point of E is a density point.

Now, let J.l. be a signed Borel measure. If J.l. is singular with respect to the Lebesgue measure ).. , then we shall show that its derivative equals zero almost everywhere. This, coupled with Theorem 39.4, will yield the following important differentiation result.

Every signed Borel measure J.l. on JR.k is differentiable almost everywhere. Moreover, if J.1. = J.l.l + J.l-2 is the Lebesgue decomposition of J.1. (i.e., J.l- 1 << ).. and J.l-2 ..L >..), then Theorem 39.6.

DJ.l-2 = 0 a.e. and

DJ.l. = DJ.l-1

=

dj.l. a.e. d)..

-

Proof. Let J.l. be a Borel signed measure on JR.k . By considering separately the

positive and negative parts of J.l., we can assume without loss of generality that J.l. is a Borel measure. Clearly, ·J.l. is a-finite. Let J.l. = J.l. l + J.l-2 be the Lebesgue decomposition of J.1. (see Theorem 37.7), where J.l.l << >.. and J.l-2 ..L A. If B11 = {x E 1Rk : llx II < n}, then the set function defined by v,;(E) = J.l.l (E n 811) for each Borel set E is a finite Borel measure satisfying v11 << )... Thus, by Theorem 39:4, v11 is differentiable for almost all points of 811, and as a routine verification shows, D J.l.l (x) = D Vn (x) holds for almost all x E 811• Since 811 t 1Rk , it follows that J.l- 1 is differentiable almost everywhere.

Chapter 7: SPECIAL TOPICS IN INTEGRATION

372

On the other hand, JL2 ..L ).. implies the existence of a Borel set A with JL2(A) = A(Ac) = 0. But by Lemma 39.3, DJL2(x) = 0 holds for almost all points x of A, and hence (since A(Ac ) = 0), DJL2(x) = 0 for almost all points x ofJR.k . Using Theorem 39.4, we see that

djL DJL(x) = DJLt(X) + DJL2(x) = DJL1(x) = -(x) d).. holds for almost all x

E JR.k , and the proof is finished.

•

Having established the basic differentiation properties of measures, our attention is now turned to the relationship between measures and real valued functions defined on an interval. In the following discussion some of the deepest classical results of ordinary derivatives will be obtained. Recall that a function f: I -+ JR. (where I is an interval) that satisfies f(x) < f(y) whenever x < y is called an increasing function (�Tld if f(x) > f(y) whenever x < y, then f is called decreasing). An increasing or a decreasing function is referred to as a monotone function. Likewise, if x > y implies f (x) > f(y), then f is called a strictly increasing function (and iff(x) < f(y) whenever x > y, then f is called strictly decreasing). A strictly increasing or a strictly decreasing function is known as a strictly monotone function. Every monotone function f has the property that both limits rtx

f(x-) = lim f(t) =

r--..c

lim

f(t)

and

lim f(t) f(x+) = lim f(t) = r-x+ r.t.x

exist in JR. for each x. Moreover, the oscillation of f at any point x is given by

wf(x) = lf(x+) - f(x-)1 < oo. Therefore, the discontinuities of a monotone function are jump discontinuities, and as the next result shows, they are at-most countably many.

Theore� 39.7.

The set ofall discontinuities ofa monotonefunction is at-most

Proof. Let f: I

-+

countable.

JR. be monotone. Replacing f by - f (if necessary), we can assume that f is increasing. Since the interior of the interval can be written as a countable union of open finite intervals, it suffices to show that f has at-most countably many discontinuities in each finite open interval. To this end, let (a, b) be a finite open subinterval of I . If A is the set of all discontinuities of f in (a, b), then A = U: A,, where 1 A, = {x E (a , b): f(x+) - f(x-) > � }. Now, if a < XJ < · · · < Xk < b belong

Section 39: DIFFERENTIATION AND INTEGRATION

373

to some A,, then it is easy to see that

k

- <

n

k

L r= I .

[f(x;+) - f(x;-)] f(b) -f(a) <

< oo

holds, which implies that each A, is finite. Hence, A is at-most countable, and the proof is finished. • An immediate and very useful consequence of the preceding theorem is that a monotone function has a point of continuity in every open interval. In applications this observation is translated as follows: Iff is a monotone function, then for every point there exist sequences and such that x, -1, and with continuous at each and y,. Recall that if a function 1R � 1R is increasing and left continuous (i.e., satisfying limrtx for each , then a set function J.L 1 can be defined on < by the semiring S In Example 1 3 .6 we verified that 1 is indeed a measure. Clearly, every Borel set is J.Lrmeasurable; therefore, the outer measure j restricted to B is a Borel measure. For simplicity, the restriction of J.L j to B will be denoted by J.L 1 again, and it will be called the Borel measure induced by It is a routine matter to verify that On the other hand, for < holds for each open interval the relation

x

{x,} x, f: f(t) = f(x) = {[a, b): a b} J.L J.L f. f(b) - f(a+)

{y,} x, y, t x xl([a, ) b)) = f(b)- f(a). J.L

(a, b). f(b)- f(a) = J.LI ([a, b)) = J.LI ({a}) + J.LI((a, b)) f a f ({a}) = J.L

f

J.LI ((a,ab)) b,=

shows that is continuous at some E 1R if and only if J.L 1 0 holds. The first "differentiation" relation between and 1 is stated in the next impor­ tant theorem. Keep in mind that the open balls of 1R are precisely the finite open intervals.

Let f: 1R 1R be an incz·easing left continuous function, and Then the Boz·el measure J.L is difef rentiable at x0 ifand letonly.\:0ifbefa irsea/number. differentiable at x0• Moreover, in this case D!-LI(xo) = f'(xo). Proof. f x0 m = f'(x0). - f(xo) m + m - f(x)x -x0 lx -xol 8. (*)

Theorem 39.8.

--+

1

Assume first that is differentiable at E > 0, choose 8 > 0 such that E <

<

E

and let

whenever 0 <

Given

<

374

Chapter 7: SPECIAL TOPICS IN INTEGRATION

If an

open interval (a, b) satisfies x0 E (a, b) and b - a < 8, then it is easy to see from (*) that m - E < f;!a < m + E. This implies JL t ((a , b)) f(b) - f(a +) -m < E --::... -- - m = b-a A((a, b))

for each open interval (a, b) with x0 E (a, b) and b - a < 8. Hence, DJL 1(x0) = m holds. For the converse, assume that m = D JL1(x0) exists in 1R and let E > 0. Choose some 8 > 0 such that whenever x0 E (a, b) and b - a < 8, then JL1((a, b)) --..::... -- - m < E. b-a

Let b > xo satisfy b - xo < 8. Since f has at most countably many discontinu­ ities (Theorem 39.7). there exists a sequence {ar.} such that o,. < x0, lim a,. = x0 , b - an < 8, and f is continuous at each a,. Thus, JL !((an, b)) = f(b) - f(an+) = f(b) - f(an). and so f(b) - f(an) b - a,

-m < E

for each n . By the left continuity of f, it follows that f(b) - f(xo) <E ----- - m b - x0

for each b > xo with b - x0 < 8. This means that f is differentiable from the right at x0 and thus, it is right continuous at x0• Reason:

.

b.J.xu

hm[f(b) - f(xo)] =

. [

b.J.xu

hm

f(b) - f(xo) (b - xo) b - Xo ·

Now, by the symmetry of the situation, we have

]

= m 0 = 0. ·

f(a) - f(xo)

----- - m < E

a - xo

for all a < xo with xo - a < 8. Thus, .f'(xo) exists, and f'(xo) the theorem is now complete.

= m. The proof of •

By Theorem 39.7 we know that a monotone function has at-most countably many discontinuities. Now, we are ready to prove that a monotone function also has a derivative at almost every point.

Section 39: DIFFERENTIATION AND INTEGRATION

375

Every monetonefunction is differentiable almost everywhere. Proof. ff ff : f * f* (x) f(x-) f(t) f(t). f* (x) < f(x) x. f(x11* } f X11 x t f ; f x (x1 1 ) f(x f(x-) (x-) n) n * * f*(x).) x. w1.(x) w1(x) f(x+) f(x-) f f* f* . f 1. * J;(a) a. f a, * f* (a) = f(a). E - f(a) < m + E < !x - a! < m - E < j. (x)x -a x ! x -a! < (x11 } < + x, < !x11 -a! < 8, f Xn. - f(a) < f(x) - f(a) < f(x + ) - f(a) m -E < f*(x)x -a x -a x-a I. f(xn)- f(a) < m + E. n-oo a Theorem 39.9 (Lebesgue).

Let be a monotone function. Replacing by - (if necessary), we can assume that is increasing. Define a new function JR. --7- JR. by =

= lim tfx

= lim t-x+

Clearly, holds for all Moreover, is increasing and left continu­ ous. (Indeed, for each x, there exists a sequence and with such that = lim therefore, continuous at each = lim = = Also, a similar argument shows that = = holds for each In particular, this shows that and have precisely the same points of continuity. Now, let J.Lf. be the Borel measure induced by By Theorem 39.6, fJ. is differ­ entiable almost everywhere, and hence, by Theorem 39.8, is also differentiable almost everywhere. Assume that m = exists at some point Then is continuous at and so Given > 0, choose some 8 > 0 such that holds whenever 0

Now, fix some

Xn

with 0

0

8, and then choose a sequence such that and with continuous at each It follows that

-----

- tm

8.

,;,__ _ _ ___;_ _

x, -

Therefore,

f(x) - f(a) - m < E x-a < !x - a! < 8. f f'(a) * I ..;.___;,_,__

f'(a) J:(a) =m f. f

holds whenever 0 Hence, exists and is differentiable, so is holds. That is, at every point where differentiable almost everywhere.

= Therefore,

is

•

For the rest of the section, only real-valued functions defined on a (finite) closed interval will be considered. To simplify matters (when the circumstances

[a, b]

Chapter 7: SPECIAL TOPICS IN INTEGRATION

376

b] f(x) f(b) x > b. f(x) = f(a) x . . {t0, . , t1 1 } = b] to t11 b f: [a, b]

require it), a function defined on [a, will also be tacitly assumed defined on all of 1R by if < a and if = We start by reviewing some basic properties of functions of bounded variation. Recall that a collection of points P is a partition of an interval [a, if a = < t1 < < = holds. Let -4 1R be a function. Then the (total) variation V1 off over is defined to be · · ·

V1

= sup

{

t. l/(1;)- f(t; )I: P {to, tn} -1

=

[a, b]

a b] } .

is a partition of [ ,

. . . •

f

If V1 < oo holds, then is said to be a function of bounded variation. A function of bounded variation is necessarily a bounded function. [Reason: If < < b, then

a x (1/(x)l - lf(a)l) (1/(x)l - lf(b)l) < l f(x) - f(a)l + lf(b) - f(x)l < VI .] Also, it is a routine matter to verify that if f and are functions of bounded variation .(on [a , b]) and then f af, fg, and Ill are all functions of bounded variation. Thus, if B V [a, b] denotes the collection of all real-valued functions of bounded variation on [a , b], then the latter shows that B V [a, b] is a function space and an algebra of functions. Every monotone function f is of bounded variation, and V1 lf(b) - f(a)l holds. [Indeed, if f is increasing, and a = to < t1 _< < tn = b, then L lf(t;) - f(t;-J)I L [f(t;)- f(t;_J)] = f(b) - f(a).] +

g

+ g,

a E JR.,

=

· · ·

n

n

=

i=l

i=l

Therefore, every function that can be written as a difference of two monotone functions is of bounded variation. In actuality, these are the only types of functions that are of bounded variation. The following discussion will clarify the situation. -4 1R be a function of bounded variation. Then restricted Now, let to any closed subinterval [c, d] of [a, is of bounded variation there. As a matter of fact, if for any closed subinterval [c, d] of [ we denote the (total) variation of f over [c, d] by Var1(c, d), i.e.,

f: [a, b]

b]

f

a b] ,

Var1(c, d) = { II sup

8 1/(t;) - f(t;_J)I: P = {to, . . . , t,} is a partition of [c, d] } ,

Section 39: DIFFERENTIATION AND INTEGRATION

377

then it is easy to verify (and the reader should do so) that Varf(c, d) = Varf(c,

e) + Varf (e , d).

for all a < c < e < d < b. Therefore, we can define a new real function Vt(·) by Vf(a) = 0 and Vt(x) = the variation of f over [a, x] for a < x < b. Note that V1(b) = v1. The function V1(·) is called the variation function of f , and its basic properties are included in the next theorem.

For a function f: [a, b] � 1R of bounded variation, the following state_ments hold: l . The total variation function Vt(·) off is increasing. 2. l f(y) - f(x)l < Vt(Y ) - Vt(x) holds for all a < x < y < b. 3. The function g: [a, b] � 1R defined by g(x) = Vt(x) - f(x) is increasing. 4. The function f is continuous at some xo E [a, b] if and only if its total variation function V1 is continuous at x0• Proof. (1) Assume a < x < y < b. If a = to < t 1 < · · · < t11 = x < y Theorem 39.10.

holds, then

II

L l f(t;) - f(t; _, )l + l f(y) - f(x) l < Vf(y). i=l

Hence, Vt(x) < Vt(x) + l f(y) - f(x)l < Vt(Y) holds. (2) The preceding inequality immediately gives (2). (3) If a < x < y < b holds, then (2) implies

f(y) - f(x) < l f(y) - f(x)l < Vt(Y) - Vt(x), so that g(y) - g(x) = [Vt(Y ) - f(y)] - [Vf(x) - f(x)] > 0. (4) If V1 is continuous at xo, then it should be obvious from (2) that f is also continuous at x0. For the converse, assume that f is continuous at x0 E [a, b ]. We shall establish that v1 is right-continuous at xo, and we shall leave the identical arguments for the left continuity of Vf at xo for the reader. So, assume a < x0 < b and let E > 0. Choose some o > 0 such that x E [a, b] and l x - xol < o imply l f(x) - f(xo)l < E . Next, fix some xo < s < b such that s - x0 < o, and then select a partition P = {a = to < t1 < · · · < t, = s] of [a, s] such that

Vt(S) - E < L l f(t;) - f(t;- J)I . II

i=l

378

Chapter 7: SPECIAL TOPICS IN INTEGRATION

By adding the point x0 to the partition that

P, we can assume without loss of generality

P = {a = to < t1 < · · · < tj = xo < tj+ l < · · · < t, = s}. Now, assume that xo

< x < lj+ l · Then, we have

v,(x) + var,(x, s) = v,(s)

j

fl

< L lf(t; ) - f(t;- dl + E i=l

< L l f(t; ) - f(t;-dl + l f(x) - f(xo) l + l f(x) - f(tj+dl i= l n

+ L l f(t; ) - f(t;-dl + E i=j+2 < Vf (Xo) + if(x) - f(xo)l + Varf(X, s) + E < Vf(Xo) + Varf(X , s) + 2E. This implies 0 ,:5 Vf(X) - Vf(Xo) right-continuous at x0 , as desired.

< 2E

for all XQ

<

X

< lj+ l•

and

SO

Vf is •

An immediate application of the preceding theorem is the following:

Theorem 39.11. If f: [a, b] � R

is of bounded variation, then 1 .. f is the difference of two increasing functions (which, in addition, can be taken to be continuous iff is also continuous), and 2 . f is differentiable almost everywhere.

Proof. For ( 1 ) apply Theorem 39.10 to f(x) = Vf(x) - [V,(x) - f(x)], and

•

for (2) use Theorem 39.9.

Let f: [a, b] � R be an increasing function and consider it defined on all of R by f(x) = f(b) if x > b and f(x) = f(a) if x < a. Then the set function

J.l.f([c,d)) =

f(d-) - f(c-)

defines a measure on the semiring {[c, d): c, d E R and c < d}. Clearly, the measurable sets of J.l. 1 contain the Borel sets of R, and hence, J.L1 is a Borel measure. Also, J.l.f vanishes off [a, b]. That is, Supp Jl 1 c [a, b], and therefore, J.l. 1 is a finite Borel measure. Now, consider a function f: [a, b] � R of bounded variation. Again, f is considered defined on all of R by f(x) = f(b) if x > b and f(x) = f(a) if

Section 39: DIFFERENTIATION AND INTEGRATION

a. By Theorem 39. 1 1 , there exist two increasing functions g and h on [a, b] such that f = g - h on [a, b]. Then the set function

x <

379

defined

for each Borel subset E of 1R defines a finite Borel signed measure. It is easy to see (by Theorem 15. 10) that the value fJ-J(E ) does not depend upon the particular representation of f as a difference of two increasing functions. In addition, note that JJ-f vanishes off [a, b] and that

JJ-J( [c, d)) = f(d-) - f(c-)

MJ({x}) = f(x+) - f(x-).

and

The finite Borel signed measure JJ-1 is referred to as the Lebesgue-Stieltjes5 measure generated by the function of bounded variation f.

Which Lebesgue-Stieltjes measures are absolutely continuous with respect to the Lebesgue measure? The answer is provided by the next theorem. For a continuous function f: [a, b] � 1R ofbounded varia­ tion, the following statements are equivalent: 1 . fJ-f is absolutely continuous with respect to the Lebesgue measure. 2. For each E > 0, there exists some 8 > 0 such that if(a1, b 1 ), . . . , (an, b11) are disjoint open subintervals of [a, b] satisfying "L�'= 1 (b; - a; ) < 8, then "L�'= 1 l f(b;) - f(a;) l < E holds.

Theorem 39.12.

Proof. (1)

===>

(2) Assume (2) is false. Then there exists some E > 0 so that given any 8 there exist disjoint open subintervals (a 1 , b 1 ), , (a11, b11) of [a, b] such that •

i

=l

Note that the open set 0

and

L l f(b; ) - f(a; )l > E . i

=l

= U�'= 1 (a; , b;) satisfies A.(O) = "L�'= 1 (b; - a;) < 8 and 11 < L l f(b; ) - f(a; ) l = L IJJ-J ((a; , b;))l < L IMJI ((a; , b;)) = IMJI( O). i= l i=l i=l n

E

•

II

II

L(b; - a;) < 8

•

n

Thus, for each n there exists an open set 0, c (a, b) such that A.(011) < 2-u and IMJI(O,) > E. Let A = n�l u�ll 0; , and note that A is a Borel set. Since

5Thomas Jan Stieltjes ( 1 856-1894), a Dutch mathematician. He published extensively in all parts of analysis in his time. He is remembered today mainly for his generalization of the Riemann integral.

380

Chapter 7: SPECIAL TOPICS IN INTEGRATION

it follows that >..( A) = 0. On the other hand,

holds for each n, and so, IILJI(A) = limi/LJI( u�ll tJ;) > E, contrary to 1/LJI << A (recall that IILJI << >.. if and only if /Lf << .>..). Hence, (2) must be true. (2) ==> ( l ) LetA beaBorel setwith>.. ( A) = 0. Wecan supposethat A c (a, b). Let E > 0. Choose some fJ > 0 so that statement (2) is satisfied. Since /Lf is a finite regular Borel signed measure and >..(A) = 0 holds, there exists an open set tJ such that A c tJ c (a, b), 1/LJ(tJ) - /LJ(A)I < E, and .>..(0) < !J. Let tJ = U(a;, b;) be written as an (at tnost countable) union of disjoint open intervals. Then, it is easy to see that t"' VJ 111/(/ '1\1 -

� ' 'j(("" � /-A' ....,· , i

b·'' 1/1

-

-

"("·, )] -< L... �.., l f(J.. "/· )-f'n \-1·) l

J.. · )-J �.., [J.f(VI L... i

..

i

< e:

� �·

Hence, IJ.LJ(A)I < IlLJ(A ) - ILf(tJ) I + IlLf(tJ)I < 2E holds for each E > 0. That is, ILJ(A) = 0, so that IL1 << .>.. , and the proof of the theorem is complete. • Statement (2) of the preceding theorem is taken as the definition of absolute continuity of functions. Definition 39.13. A function f: [a, b] � 1R is said to be absolutely contin­ uous iffor every E > 0 there exists some fJ > 0 such that whenever (a 1 , b 1 ) , , •

•

•

(an, bn) are disjoint open subinten1als of [a, b], then 1/

1/

i=l

i=l

L(b; - a;) < fJ implies L l f(b; ) - f(a; ) I < E. It should be clear that an absolutely continuous function is necessarily contin­ uous. The converse is false; see Exercises 7 and 8 at the end of this section. Also, it is easy to verify that if f and g are absolutely continuous functions on [a, b] and a E lR, then f + g, af, fg, and If I are all absolutely continuous functions. Thus, the collection AC [a, b] of all absolutely continuous functions on [a, b] is a function space and an algebra of functions.

For an absolutely continuous function f: [a, b] � lR, the following statements hold: 1 . f is of bounded variation, and hence, AC [a, b] is a vector sublattice of B V [a, b]. Theorem 39.14.

Section 39: DIFFERENTIATION AND INTEGRATION 2.

381

The variationftmction V1(·) is absolutely continuous, and hence, difference of two increasing absolutely continuous functions.

f is the

Proof. (1) For any closed subinterval [c, d] of [a, b], we denote (as usual) by

Var1(c, d) the total variation off over [c, d]. Keep in mind that if c < e < d, then Var1(c, d) = Var1(c, e) + Var1(e, d).

Next, choose some 8 > 0 so that the definition of absolute continuity is satisfied with E = I . Now, fix some natural number n with h-n a < 8, and let a = to < t1 < b�a for each i . Clearly, · · · < tn = b be the partition of [a, b] with t; - t;-1 = Var1 (t; - t , t;) < I holds for each i, and thus, Vt = Varf(a, b) =

II

VartCti- t , t;) < n < oo. L = i l

, t11 } is a partition (2) If [c, d] is a closed subinterval of [a, b] and P = {t0, t1, of [c, d], then write v(c, d, P) = L:;'= t l f(t;) - f(t; 1)1. Note that •

•

•

-

Vt(d) - Vt(c) = Varf(c, d) = sup{v(c, d, P): P is a partition of [c, d]}.

Now, let E > 0. Choose some 8 > 0 for which the definition of absolute continuity of f is satisfied. Assume that (at, bt ), . . . , (an, bn) are pairwise disjoint open subintervals of [a, b] such that L�=1 (b; -a;) < 8. For each I < < n, let P; be an arbitrary partition of [a;, b;]. Each P; subdivides (a; , b;) into a finite number of open subintervals. Clearly, these open subintervals taken together are pairwise disjoint, and the sum of their lengths equals I:�= I (b; - a;) < o. The absolute continuity of f applied to the open subintervals yields

i

It follows that 11

L[Vf(b;) - V1(a;)] < E, i= l

so that V1(x) is absolutely continuous.

•

The final theorem of this section is a classical result. It asserts that for the Lebesgue integration, the fundamental theorem of calculus holds true precisely for the absolutely continuous functions. (As is customary, the Lebesgue measure on IR is denoted by dt .)

Chapter 7: SPECIAL TOPICS IN INTEGRATION

382

�heorem 39.15. A function only iff' E L.t ([a, b]) and

f: [a, b]

--+

f(x) - f(a) =

IR

is absolutely continuous zf and

[ f'(t) dt

holds for each x E [a, b]. Proof. Assume that f is absolutely continuous. By Theorem 39.14, we can suppose that f is increasing. Also, by Theorem 39.12, JL1 << ).. holds, and so, by d). holds for some E L 1([a, b]). the Radon-Nikodym theorem JLt(E) Now, combine Theorems 39.4 and 39.8 to obtain that f' a.e. To get the desired identity let E = [a, x). =

JE g

g

=

g

The converse is straightforward and is left as an exercise for the reader. See also Exercise 3 of Section 37. • ·

EXERCiSES 1.

If Jl is a Borel measure on

IRk , then show that Jl J.. ).. holds if and only if DJ.L(x) = 0

2.

Generalize Theorem

to Lebesgue measurable subsets of

for almost all

if

E

is

points.

3.

x.

39.5

JR.k . That is, show that

a Lebesgue measurable subset of IRk , then almost all points of E are density

Write Br(a) for the open ball with center

at a e JR.k and radius r . If I is a Lebesgue k integrable function on IR , then a point a e IRk is called a Lebesgue point for I if lim )..

\

{ ll(x) - l(a)l dJ..(x) = 0. (Br a )) JB,(a) Show that if 1 is a Lebesgue integrable function on JR.k , then almost all points of JR.k r-+0+

are

Lebesgue points.

[HINT: Let Q be the set of all rational numbers in JR. Apply Theorem 39.4 to conclude

that for each

a e Q there exists a null set Ea such that lim

holds for all

x

¢ Ea. Let E

Lebesgue point.] 4.

Let

\

r-+0+ A(Br X

))

al dJ..(t) ll(x) - al { JB,(x) ll(t) =

= UaeQ Ea, and then show that every point of Ec is a

I: JR. --+

JR. be an increasing, left continuous function. Show directly (i.e., with­ out using Theorem 38.4) that the Lebesgue-Stieltjes measure Jlf is a regular Borel

measure.

5.

(Fubini) Let

Un) be a sequence of increasing functions defined on [a, b] such that I:::1 ln(x) = l(x) converges in JR. for each x E [a, b]. Then show that I is differ­ entiable almost everywhere and that I' (x) = 2:�1 I� (x) holds for almost all x . [HINT: Replacing each In by In - ln(a), we can assume that In ::::_ 0 holds for each n. Let sn = It + · · · + 111 , and note that each sn is increasing and sn(x) t l(x) for

Section 39: DIFFERENTIATION AND INTEGRATION

6. 7.

383

each x. By Theorem 39.9, f and all the j;, and s11 are differentiable almost every­ where. Since sn+l - s11 = f"+ 1 (an increasing function), s:,+ 1 (x) ::: s:, (x) must hold for almost all x; similarly, /'(x) > s:,(x) for almost all x. Now, choose a subsequence {sk, } of {s"} such that :[:�1 [f(x) - Sk, (x)] :=: L� l [/(b) - Skn (b)] < oo. Observe that {f - Skn} is a sequence of increasing functions. Use the preceding arguments to obtain that sL � !' a.e., and then conclude that L�l J,; = f' a.e. holds.] Suppose { f, } is a sequence of increasing functions on [a, b] and that f is an increasing function on [a, b] such that J.1.. J,, t J.1.. 1 . Establish that f' (x) = lim 1,: (x) holds for almost all x. This exercise presents some basic properties of functions of bounded variation on an interval [a, b]. Iff is differentiable at every point and I!' (x) I � M < oo holds for all x E [a, b], then show that f is absolutely continuous (and hence, of bounded variation). b. Show that the function f: [0, 1 ] --+ .lR defined by /(0) = 0 and f(x) =x 2 cos(x -2) for 0 < x < 1 is differentiable at each x, but not of bounded variation (and hence, f is continuous but not absolutely continuous). c. If f is a function of bounded variation and 1/(x) l ::: M > 0 holds for each x E [a , b] , then show that j is a function of bounded variation. d. If a function f: [a, b] � .lR satisfies a Lipschitz condition (i.e., if there exists a constant M > 0 such that 1 /(x) - f(y)l < M ix - yl holds for all x, y E [a, b]), then show that f is absolutely continuous.

a.

8.

This exercise presents an example of a continuous increasing function (and hence, of bounded variation) that is not absolutely continuous. Consider the Cantor set C as constructed in Example 6.15. Recall that C was obtained from (0, 1] by removing certain open intervals by steps. In the first step we removed the open middle third interval. At the nth step there were 2"- 1 closed intervals, all of the same length, and we removed the open middle third interval from each one. of them. Let us denote by I;', . . . , I;,,_, (counted from left to right) the removed open intervals at the nth step. Now, define the function f: [0, 1] � [0, 1] as follows: i. ii. iii.

/(0) = 0; 1 if x E If' for some 1 :=: i :=: 2"- , then f(x) = (2i - 1)/2"; and if x E C with x ¥= 0, then f(x) = sup{f(t): t < x and t E [0, I ] \ C).

Part of the graph of f is shown in Figure 7.1. a. b. c. d. 9. 10.

Show that f is an increasing continuous function from [0, 1] to [0, 1 ] . Show that f'(x) = 0 for almost all x. Show that f is not absolutely continuous. Show that J.1.. 1 j_ A. holds.

Let f: [a, b] � .lR be an absolutely continuous function. Then show that f is a constant function if and only if f'(x) = 0 holds for almost all x. Let f and g be two left continuous functions (on .IR). Show that J.1..1 = J.l..g holds if and only if f - g is a constant function.

Chapter 7: SPECIAL TOPICS IN INTEGRATION

384

y

1

l r r

1.

4

r

..!.

-----e

2

i• �

..!.

4

i.

.1.. .1 .1.. 9 9 3

2 3

2 9

.! 1

X

9

FIGURE 7.1. 11. This exercise presents another characterization of the nann dual of C [a, b]. Start by

letting L denote the collection of all functions of bounded variation on [a, b] that are left continuous and vanish at a.

a. b.

c. d. 12.

13.

Show that L under the usual algebraic operations is a vector space, and that f H- /1-f• from L to Mb([a, b]), is linear, one-to-one, and onto. Define f � g to mean that f - g is an increasing function. (Note that f 2:. g does not imply f � g.) Show that L under � is a partially ordered vector space such that f � g holds in L if and only if /1-f 2:. 11-K in Mb([a. b]). Establish that L with the nann 11/11 = Vl fl is a Banach Janice. Show, with an appropriate interpretation, that C*[a, b] = L .

If f: [a, b] � IR is an increasing function, then show that f' E L;pa, b]) and that 1: f'(x)dx ::: f(b) - f(a) holds. Give an example for which f'(x) dx < f(b) - f(a) holds. [HINT: Let gn(X) = n[f(x + �)- f(x)] for each n and x E [a. b]. Note that gn � !' a.e. holds. Now, apply Fatou's lemma to the sequence {gn).] If f: [a, b] � IR is an absolutely continuous function, then show that

fa

Vt

= 1hl f'(x) l dx

holds. 14. For a continuously differentiable function f: [a, b] properties: a.

�

IR, establish the following

The signed measure 11- f is absolutely continuous with respect to the Lebesgue measure and d11-f fdA. = f' a.e.

Section 40: THE CHANGE OF VARIABLES FORMULA

385

b. If g: [a, b] -+ 1R is Riemann integrable. then gf' is also Riemann integrable and

JgdJkt = lhg(x)f'(x)dx.

15. For each n, consider the increasing continuous function j., (x) = If f: 1R -+ 1R is

16.

{ -:1) n(x

17.

1R -+ 1R defined by

if X > 0, if 1 - II! < \: < if X < - * .

+I

a continuous function, then show that

a. f is J1- f, -integrable for each n, and b. lim Jf dJk J,, = f(l). Let f: 1R -+ 1R be a bounded function and let E = {x E JR.:

/11:

1

•

1'

J\r) exists in JR.}.

If >..( £) = 0, then show that >..(f(E)) = 0. This exercise presents an example of a continuous function f: 1R -+ 1R which is nowhere differentiable. Consider the function ¢: [0, 2] -+ 1R defined by ¢ (."C) = ."C if 0 < x ,::: 1 and ¢(x) = 2 - x if < x .::: 2. Extend ¢ to all of 1R (periodically) so that ¢(x) = ¢(x + 2) holds for all x E JR. Now define the function f: 1R -+ 1R by

1

f(x) =

)1 1 ( 3 1L=0 00

4

¢(4nx).

Show that f is a continuous nowhere differentiable function. (Compare this conclusion with Exercise 28 of Section 9.) 40. THE CHANGE OF VARIABLES FORMULA The objective of this section is to establish the familiar formula known as the "change of variables formula." To do this, we need some preliminary discussion. For the rest of the section, V and W will be two fixed open sets of some Euclidean space JR.k . Also, 11·11 will always denote the Euclidean norm on JR.k. We remind the reader about the representation of a linear mapping on JR.k by a matrix. Let T : JR.k -+ JR.k be a linear mapping, and let {e 1 , , ek } be the standard k basis of JR. (that is, the ith coordinate of ej is 1 if i = j and 0 if i # j). For each j, there exist constants a1j, . . . , akj (uniquely determined) suoh that . L..Jki = J aije;. If we define the matrix T(ej ) = "\' •

A=

GJJ GJ2 a21 a22

G Jk a2k

•

•

386

Chapter 7: SPECIAL TOPICS IN INTEGRATION

then T(x) = Ax holds for each x e 1Rk. The matrix A is called the matrix representation of the linear mapping T (with respect to the standard basis). Conversely, any k x k matrix A = [aij] defines a linear operator T: 1Rk � 1Rk by the formula T(x) = Ax for 1 each x e 1Rk. Also, it is easy to see that if M = k ·max{ laij 1 : i, j = , . . . , k}, then II Ax II < M llx II holds so that each matrix defines a continuous linear operator. The determinant ofT is simply the determinant of the matrix A ; that is, det T = det A . A function T : V � 1Rk is said to be differentiable at some point0a of (the open set) V if there exists a linear operator A : 1Rk � 1Rk and some r > such that T(x) = T (a) + A(x - a) + o(x - a) holds for all x e V with ll x - a II V to 1Rk such that

< r. Here (as usual) o(x - a) is a function from

1.

un

x-+-a

o(x - a)

ll x - a ll

_ -

0

.

The linear operator A is denoted by T '(a) and is called the derivative ofT at the point a. It should be clear that if T is differentiable at a, then T must be continuous at a. A function T: V � 1Rk is said to be differentiable ifT'(x) exists at each point x e V. Moreover, if T = (T1 , T,) is differentiable, then it follows that each partial derivative •

8T·

1

ax;

•

•

,

h--+-0

(x) = lim

T · (x 1

+ he;) - T·(x) h 1

exists at every point x e V . In addition, the matrix representation of T'(x) is

determinant is called the This matrix is called the Jacobiall6 matrix, and Jacobian. The Jacobian determinant is denoted by lr(x); that is, lr(x) = det T'(x) = det[(8T;/8xj )(x)].

its

6Carl Gustav Jacob Jacobi ( 1 804-1851 }, a Gennan mathematician. He was a prolific writer who contributed to many diverse mathematical disciplines-including number theory, mathematical physics, mechanics, and the history of mathematics.

Section 40: THE CHANGE OF VARIABLES FORMULA

387

A function T: V --+ IRk is said to be C 1.-differentiable if T is differentiable and all of its partial derivatives are continuous functions on V . Observe that if T is C 1 -differentiable, then the Jacobian JT(·) is a continuous real valued function on V . It is important to know that C 1 -differentiable mappings carry null sets onto null sets. The details follow.

Lemma 40.1. Let V c IRk be open and let T: V --+ IRk be a C 1 -differentiable function. If A c V satisfies A.(A) = 0, then A.(T(A)) = 0 also holds. Proof. Assume first that there exists some M > 0 such that l(aTifaxj)(x)l < M holds for each x E V and i, j = 1 , . . . , k. Let C = kM + 1 . If a E V , then by the differentiability of T there exists some open ball Ba c V with center at a such that

II T(x) - T(a)ll < kMIIx - a ll + llx - a ll = C llx - all holds for each x E Ba. Let E > 0. Choose an open set 0 with A c 0 c V and A.(O) < E, and then select a sequence {K11} of compact sets such that 0 = U: 1 K11 • Since T is continuous, each T(K11) is compact, and the identity T(O) = U: 1 T(K11) shows that T ( 0) is a Borel set. Now, let K be an arbitrary compact subset of T(O). For each y E K fix some z E 0 with y = T(z). By the above, there exists an open ball By c 0 with center at z such that

II T(x) - Y ll < C ll x - zll holds for each x E By. Let B; be the open ball with center at y, and radius C times · , Yu E K that of By; clearly, T( By ) c B;. Since K is compact, there exist y 1 , such that K c U;'= 1 B;;. By Lemma 39.2, there exist pairwise disjoint balls Bj, . . . , B,�, among the B>�I , . . . , BY*II such that •

Note that the corresponding B 1 , balls. Therefore,

A(K) < A <

( Q B;)

�

•

•

•

, Bm

(3C)'

(3C)kA.(0) < (3C)kE

•

•

are necessarily pairwise disjoint open

t, }._(8

f) =

(3C)'A

(Q B ) 1

388

Chapter 7: SPECIAL TOPICS IN INTEGRATION

holds for each compact subset K ofT (0). Since T (0) is a Borel set, the regularity of ). implies A.(T(O)) < (3C)kE, and hence, A.(T(A)) < A.(T(O)) < (3C)kE for each E > 0. That is, A.(T(A)) = 0. For the general case, let B 1 , B2 , be an enumeration of the open balls with "rational" centers and rational radii whose closures lie entirely in V. Clearly, V = U: 1 Bn. Since each B, is compact and T is C 1 -differentiable, it follows easily that T : B n --+ IRk satisfies the hypotheses of the preceding case. Thus, A.(T(AnBn )) = 0 holds for each n, and consequently, we have A.(T(A)) = A.( U:1 T(A n Bn )) = 0, as required. • •

•

•

The notion of a diffeomorphism plays an important role for this section. Its definition follows.

Definition 40.2. Let V and W be two open sets ofiRk . Aftmction T: V --+ W is said to be a diffeomorphism if a. b. c. d.

T Is one-to-one and onto, T is C 1 -differentiable, lr(x) i= 0 holdsforall x E V, and T is a homeomorphism (from V onto W).

We remark that (a), (b), and (c) are enough to ensure that T is an open map­ ping and thus, a homeomorphism. However, property (d) has been added to the definition of a diffeomorphism in order to emphasize its importance. Also, it should be noted that the inverse mapping T - l : W --+ V is necessarily a diffeo­ morphism. For details about this and other properties of differentiable functions see [2]. The next theorem tells us that a diffeomorphism preserves the Borel and the Lebesgue measurable sets.

Let T : V --+ W be a diffeommplzism between two open sub­ sets of IRk , and let E be a subset of V. Then: a. T(E) is a Borel set if and only if E is a Borel set. b. T(£) is Lebesgue measurable if and only if E is Lebesgue measurable. Theorem 40.3.

Proof. (a) Let Bv and Bw denote the Borel sets of V and W, respectively.

Clearly, T(Bv) = {T(£): E E Bv } is a a-algebra of subsets of W containing the open sets of W; hence, Bw c T(Bv ). By the symmetry of the situation, B1, c T (Bw) holds, and from this· we get T (Bv) = Bw. The conclusion now follows easily from the last identity. (b) Let E be Lebesgue measurable. We can suppose A.(£) < oo (why?). Choose a Borel set B with E C B c V and A.(B \ £) = 0. By Lemma 40.1, ).(T (B \ £)) 0,

-l

=

Section 40: THE CHANGE OF VARIABLES FORMULA

so that T(B \ E) is Lebesgue measurable. set, and hence, the relation T(E)

=

389

�ow from (a) the set T(B) is a Borel

T(B) \ T(B \ E)

shows that T (E) is Lebesgue measurable. The converse should be obvious from the symmetry of the situation. • Now, assume that T : V ---+ W is a diffeomorphism. From the preceding theorem it is easy to see that the set function J.L(E)

=

A(T(E))

defined for each Lebesgue measurable subset E of V is a measure. On the other hand, it follows from Lemma 40.1 that J.L << A holds. The proof of the "change of variables formula" rests upon the fact that the Radon-Nikodym derivative dJ.L/dA satisfies (dJ.LfdA)(x) = llr(x)l for each x E V . To establish this identity, we need to know how a linear operator alters the volume of the unit cube of lR.k . Lemma 40.4.

matrix), then

If A : lR.k

---+

lR.k is a linear operator (which we identify with a

A(A(E))

=

I det A I · A( E)

holds for all Lebesgue measurable subsets E oflR.k . Proof. If A is not invertible, then A maps lR.k onto a linear. subspace oflower di­ mension, and hence, onto a set of measure zero (why?). Hence, A(A(E)) = ldet A I · A(E) = 0 holds trivially. Now, assume that A is invertible; clearly, A is a diffeomorphism. Let J.L(E) = A(A(E)) for each Borel set E . Then J.L is a Borel measure, and by Theorem 15.10 it is enough to establish J.L = ldet A I · A on the Borel sets. Start by observing that J.L is a translation invariant Borel measure on lR.k , and hence, by Lemma 18.7, there exists a constant C such that J.L = C A holds. In particular, if U = [0, 1 ] x · · · x [0, 1], then C = J.L(U). To complete the proof, we must show that C = ldet A I . To this end, note first that if A = A 1 o A2 and C , C 1 , and C2 are the constants associated with A, A t . and A2, respectively, then C

=

A(A(U))

=

A(A1 (Az(U)))

=

C J A(Az(U))

=

C1 C2A(U)

=

C 1 C2.

390

Chapter 7: SPECIAL TOPICS IN INTEGRATION

If we denote the matrix representing A by A again, then the matrix A can be written as a product of matrices A = A I A 2 An, where each A; is either · · ·

1. 2. 3.

a matrix obtained from the (k x k) identity matrix by multiplying one of its rows by a nonzero constant, a matrix obtained from the identity matrix by interchanging two of its rows, or a matrix obtained from the identity matrix by adding to one of its rows a nonzero multiple of some other row of the identity.

(The three types of matrices described previously are known as elementary ma­ trices.) Thus, since C = CIC2 C, [where C; = A.(A;(U))] and det A = det AI det A 2 det An, in order to complete the proof it suffices to verify the formula for the above three types of matrices. · · ·

·

· · ·

Type 1 .

Without loss of generality we can assume that the matrix A is obtained from the identity by mu!t!p!ying its first row by a nonzero number a . Note that det A = Ia I. Here U is mapped onto [0, a] x [0, 1 ] x x [0, I] if a > 0 and [a, 0] x [0, 1 ] x x [0, 1] if a < 0. In either case, C = A.(A(U)) = Ia I = ldet A I .

Type 2 .

matrix

· · ·

· · ·

In this case, det A = -1. It i s easy to see that U is mapped under A onto U . Hence, C = A.(A(U)) = A.(U) = 1 = ldet A I .

Type 3.

It is easy to see that this case is reduced to the matrix A obtained from the identity matrix by adding to its first row the second one. Clearly, det A = 1 . Now, observe that if x = (xi, x2, . . . , xk ), then Ax = (xi + x2, x2, X3, , xk). 2 In particular, when k = 2 the unit square in is mapped under A onto the parallelogram P as shown in Figure 7 .2. Obviously, A.(P) = 1 . On the other hand, if we consider the Lebesgue measure on JR.k (where k > 2) as the product measure of the Lebesgue measure on JR.2 and the Lebesgue measure on JR./.:-2 , then U is mapped under A onto the set P x Ut. where U1 is the (k - 2)-dimensional unit cube. By Theorem 26.2, A.(A(U)) = )�(P) A.(UI) = 1 = ldet AI, and the proof of • the lemma is complete.

JR.

·

0

XI

0

FIGURE 7.2. The Transformation of the Unit Square

.

•

.

391

Section 40: THE CHANGE OF VARIABLES FORMULA

Now, we are in the position to establish the identity

df.J./d'A = l iT I·

Theorem 40.5. Let T: V � W be a diffeomorphism between two open sub­ sets of lR.k . If J.J.(E) = 'A(T (E)) for each Borel set E of V, then J.1. is a Borel measure on V whose derivative satisfies

D J.J.(x) = lly(x) l for all x E V . Proof. Let a E V . Replacing T by S(x) = T(a + x) - T(a), we can assume without loss of generality that a = 0 and T (0) = 0. We shall prove first the result for the case when the Jacobian matrix equals the k

Jy(O) = 1 .

x

k identity matrix I, and so

Let E > 0. Fix 0 < a < i such that 1 - E < ( 1 - 2a )k < ( 1 and then select some o > 0 so that

II T(x) - x ll

<

+ 2ai

< 1 + E,

a llxll

holds for each x E V with llxll < o. (Since T(x) = T'(x)x + o (x) = x + o(x) such a o > 0 always exists.) Now, let B c V be an open ball containing zero with center at x0 and radius r < �. Let B 1 and B2 be the open balls both with center at xo and radii ( 1 - 2a)r and ( 1 + 2a )r, respective! y. We claim that

Indeed, note first that if x E B, then llx II < llx - xo II

IIT (x) - xoll

<

II T(x ) - x ll

+ llx - xo ll

<

+ llxo II

a llx ll

+r

<

<

2r

<

o, and so

(1 + 2a)r

for each x E B, which shows that T (B) c B2. Now, write

Bt = [Bt n T(B)] U [Bt \ T (B)] , and note that the two sets of the union are disjoint. Since T is a diffeomorphism, T(B) is an open set, and hence, B1 n T(B) is also open. Since

II T(xo ) - xoll

<

allxoll

<

2ar

<

( 1 - 2a)r,

it follows that T(x0) E B1 n T(B), and so B1 n T(B) # (/). Also, if llx - xoll = r holds, then llx II < 2r < o implies

r = llx - xoll

<

llx - T(x) ll

+ IIT(x) - xoll

<

2ar

+ II T(x) - xoll,

Chapter 7: SPECIAL TOPICS IN INTEGRATION

392

and so II T (x) - xo II > ( 1 - 2a )r. This shows that no boundary point of B is mapped under T into B 1 • In other words, B 1 \ T (B) = B 1 \ T (B) holds. Since T is continuous, T(B) is compact, and hence, B 1 \ T(B) is also an open set. But then, it follows that B 1 \ T(B) = (/) (see Exercise 1 at the end of this section), and so, B 1 = B 1 n T(B) c T(B) holds. Now, if c is the Lebesgue measure of the open ball with center at zero and radius one, then (1 - E)A.(B)

= c(l - E)rk < c( l

- 2a/rk = A.(B 1 ) <

J.L(B) < A.(B2) = c(l + 2a/rk < ( 1 + E)crk = ( 1 + E)A.(B).

=

A.(T(B))

Thus, 1 E < J.L(B)/A.(B) < 1 + E holds for all open balls B c V containing zero and having radius less than 8/2. 'That is, D J.L(O) = 1 holds. For the general case, let A be the Jacobian matrix at zero. Since det A = lr (O ) "# 0, the matrix A is invertible. Then S (x) = A -I (T (x)) defines a diffeomorphism between V and A -I (W) whose Jacobian matrix at zero equals I . By the preceding case, it follows that the Borel measure v ( E) = A.(S(E)) satisfies Dv(O) = 1 . Now, by Lemma 40.4, we have -

v(E) = ldet A -I I

·

A.(T(E)) = ldet Al-1 • J.L(E).

This implies DJ.L(O) = ldet AI = llr(O)I and the proof is finished. A particular case of the "change

•

of variables fonnula" is stated next.

Let T: V � W be a diffeomorphism between two open sub­ sets of JR.k such that A.(W) < oo. Then for each Lebesgue measurable subset E ofV we have Theorem 40.6.

A.(T (E)) =

L llr I dA..

A.(T(E)) for each Lebesgue measurable subset E of V. Then, by Lemma 40.1 , J.L << A. holds, and hence, by the Radon-Nikodym theorem J.L(E) = JE (dJ.L/dA.) dA. holds for each Lebesgue measurable set E. Now, com­ bine Theorem 39.4 and the preceding theorem to obtain dJ.LfdA. = D J.L = llrl Proof. Let J.L(E) =

393

Section 40: THE CHANGE OF VARIABLES FORMULA

Therefore, A.(T(E)) =

l llr I dA..

•

holds for each Lebesgue measurable subset of V. We now come to the main result of this section. E

Let T: V � W be a k diffeomorphism between two open sets of JR. . Then for every function f E L 1 (W), theftmctiorz (f T) · llr I belongs to L 1 (V) and

Theorem 40.7 (The Change of Variables Formula). o

fw f dA. = fv (f

holds.

o

T ) · llrl dA.

< oo.

Assume first that A.(W) Let f = X£, where E is a Lebesgue measurable subset of W. By Theorem we have Proof.

40.6,

1 [ f dA. = A.(£) = A.(T(T - (£))) = [

lw

=

fv (X£

o

T) · llrl dA. =

lr-1(£) llr I dA. (f o T ) · llr l dA..

fv

Thus, the conclusion is valid for characteristic functions of Lebesgue measurable subsets of W . It follows that the formula is true for step functions, and then by a simple continuity argument for each f L 1 (W). Now, if A.(W) = then let W, = {x E<W : llxll n} and Vn = r- 1 (W,). Clearly, A.(W,) for each n, and so, if f L 1 (W), then by the preceding case (and Levi's Theorem 22.8) [ f dA. = lim [ f dA. = lim 1 (f T ) · llr I dA. = 1 (f T) llr I dA. , lw lw, and the conclusion follows. The formula appearing in the preceding theorem is referred to as the change of variables formula and is commonly written as follows: E

oo,

< oo

<

0

,_,.oo v,

"-�'oo

o

E

v

o

·

•

{ lr(V)

f(y)dy =

1 f(T(x)) · llr(x)l dx. v

Chapter 7: SPECIAL TOPICS IN INTEGRATION

394

The following version of Theorem 40.7 is most often used in applications, and its proof follows immediately from the preceding theorem.

Let T: A B be a a mapping between nvo Lebesgue mea­ surable subsets of 1R.k. Assume that there exist two open sets V c A and W is a diffeomorphism, and 'A(A \ V) W c B such that T(V) W, T: V 'A(B \ W) 0. Thenfor each f E L 1 (B), the function (f T) · liT I (defined a.e. on A) belongs to L 1 (A) and Theorem 40.8.

-+

-+

=

=

=

o

f. f d). i(f =

holds.

0

T) . liT I d).

EXERCISES 1.

Show that an open ball in a Banach space is a connected set. That is, show that if B is an open ball in a Banach space such that B = 0 1 U 0 holds with both 01 and 0 2 2 open and disjoint, then either 01 = 0 or 0 = (/;. 2 [lllNT: If a E 01 and b E 02, then let a = inf{t E [0, I]: ta + (1 r)b E 01 }, and note that c = aa + (I a)b E B. To obtain a contradiction, show that c ¢. 0 1 and c ¢. 02.] 1 Let T: V -+ 1Rk be C -differentiable. Show that the mapping x � T' (x) from V into L(JRk, 1Rk) is a continuous function. Show that the Lebesgue measure on 1R2 is "rotation" invariant. (Polar Coordinates) Let E = {(r, 0) E JR2: r � 0 and 0 � 0 � 2rr}. The transfor­ mation T : E -+ 1R2 defined by T(r, 0) = (r cos O, r sinO), or as it is usually written -

-

2. 3.

4.

x = r cos 0

and

y = r sin 0,

is called the polar coordinate transformation on 1R2, shown graphically in Figure

7.3.

a. Show that /..(E \ E0) = 0. b. If A = {(x, 0): x � 0} , then show that A is a closed subset of 1R2 whose (two­ dimensional) Lebesgue measure is zero. c. Show that T: E0 -+ JR2 \ A is a diffeomorphism whose Jacobian determinant satisfies lT(r, 0) = r for each (r, 0) E E0• d. Show that ifG is a Lebesgue measurable subset of E with /..(G \ G0) = 0, then T(G) is a Lebesgue measurable subset of JR2. Moreover, show that if f E L1 (T(G)), then

[

lT holds.

f d/.. =

1·la[ f(r

cos 0, r sin O)r dr dO

Section 40: THE CHANGE OF VARIABLES FORMULA

(}

395

y

2rr

; . . 'E

.

:

(} .

'

.

.

. .. . . . . · cr-. e)· .. . . . :· .

.

T -------....

·

.

:

.. ..

;" � . .

r

.

. r

:· . . .

FIGURE

(x,y)

y

X

X

7.3. The Polar Coordinate Transformation

5. This exercise uses polar coordinates (introduced in the preceding exercise) to present an alternate proof of Euler's formula J000 e-x 2 dx = .J7i/2.

For each r > 0, let Cr = {(x, y) E JR2: x2 + y2 Sr = [0, r] X [0, r]. Show that Cr c Sr c c,..Ji· b. If f(x, y) = e-<x2+y2), then show that a.

1

C,

f dA. <

r

Js,

f dA. �

1

C, .fi

� r2,

.�.:

> 0, y > 0}

and

f dA.,

where >.. is the two-dimensional Lebesgue measure. c. Use the change of variables to polar coordinates and Fubini 's theorem to show that

L f dJ. f' { =

,-r' I dt dO

=

: (I - ,-'').

d. Use (b) to establish that

. and then let 6.

r

�

oo

to obtain the desired formula.

In JR4, "double" polar coordinates are defined by x = r cos(},

7.

y = r sin(},

z = p cos ¢,

w = p sin ¢.

State the change of variables formula for this transformation, and use it to show that the "volume" of the open ball in 1R4 with center at zero and radius a is �.rr2a4. (Cylindrical Coordinates) Let E = {(r, (}, z) E JR3: r :=:: 0, 0 � (} � 2rr, z E JR.}. The transformation T: E � 1R3 defined by T (r, (}, z) = (r cos 0, r sin(}, z) or as it is usually written X = r

COS(},

y

= r sin 0,

Z

= Z,

is called the cylindrical coordinate transformation, shown graphically in Figure 7.4.

Chapter 7: SPECIAL TOPICS IN INTEGRATION

396

z

z (r,

(x, y, z)

(), z)

l

2rr

()

FIGURE 7.4.

y

The Cylindrical Coordinate Transformation

a. Show that A.(£ \ £0) = 0. b. If A = {(x, 0, z) E 1R.3: x � 0, z E JR.}, then show that A is a closed subset of1R.3 whose (three-dimensional) Lebesgue measure is zero. c. Show that T: £0 -+ 1R.3 \ A is a diffeomorphism whose Jacobian detenninant satisfies Jr(r, () , z) = r for each (r, (), z) E £0• d. ShowthatifG isaLebesguemeasurable subsetof E with A.(G \ G0) = 0, then T(G) is a Lebesgue measurable subset of 1R.3. Moreover, show that if f E L 1 (T(G)), then

[

lr

f dA. =

holds. 8.

jj.laf

f(r cos(), r sin(), z) r dr d() dz

(Spherical Coordinates) Let

E = {(r, (), ¢) E 1R.3 : r � 0, 0 ,:5 () The transfonnation T : E

-+

,:5 2:rr,

0 ,:5 ¢

,:5 :rr } .

1R.3 defined by

T(r, (), ¢) = (r cos () sin ¢, r sin () sin ¢, r cos ¢), or as it is usually written x = r cos() sin¢,

y

= r sin() sin¢,

z = r cos ¢,

is called the spherical coordinate transformation, shown graphically in Figure

7.5.

a. Show that A.( £ \ £0) = 0. b. If A = { (x, 0, z): x � 0 and z E JR.}, then show that A is a closed subset of 1R.3 whose (three-dimensional) Lebesgue measure is zero. c. Show that T: £0 -+ 1R.3 \ A is a diffeomorphism whose Jacobian detenninant satisfies lr(r, (), ¢) = -r2 sin¢.

397

Section 40: THE CHANGE OF VARIABLES FORMULA

z c/>

(x, y, z) r

()

y

FIGURE 7.5. The Spherical Coordinate Transformation d.

Show that if G is a Lebesgue measurable subset of E with A.(G \ G0) = 0, then T(G) is a measurable subset of 1R.3. In addition, show that if f E L 1 (T(G)), then

{

}T(G)

holds.

f dA. =

!!Jc{

f(rcos () sin cp , r sin O sincp, r coscp)r2 sin cp dr dO d(jJ

BIB LIOGR APH Y I . C. D. Aliprantis and K. C. Border, Infinite Dimensional Analysis, Studies in Economic Theory, #4, Springer-Verlag, New York & Heidelberg, 1994. 2. T. M. Apostol, Mathematical Analysis, Second Edition, Addison-Wesley, Reading, MA, 1974. 3. G. Birkhoff and S. MacLane, A Survey ofModern Algebra, Third Edition, Macmillan, New York, 1965. 4. E. Borel, Le�ous sur Ia T!Leorie des Fonctions, Gauthier-Villars, Paris, 1898. (Third Edition, 1928.) 5. P. J. Daniell, A general form of integral, Annals of Matlzematics 19 (1917), 279-294. 6. M. M. Day, Normed Lineaz· Spaces, T hird Edition, Springer-Verlag, Heidelberg & New York, 1973. 7. 1. Dieudone, Foundations ofModern Analysis, Academic Press, New York & London, 1969. '8. A. A. Fraenkel, Abstract SetTlzeory, Fourth Edition, North-Holland, Amsterdam, 1976. 9. M. Frantz, On Sierpinski's nonmeasurable set, Fundamenta Mathematicae 139 (1991), 17-22. 10. B. R. Gelbaum and J. M. H. Olmsted, Theorems and Counterexamples in Mathematics, Springer-Verlag, Heidelberg & New York, 1990. 1 1 . C. Goffman, Real Functions, Prindle, Weber & Schmidt, New York, 1953. 12. C. Goffman and G. Pedrick, First Course in Functional A11alysis, Prentice-Hall, Englewood Cliffs, NJ, 1965. · 13. P. R. Halmos, Naive Set Theory, Springer-Verlag, Heidelberg & New York, 1974. 14. P. R. Halmos, Measure Theozy, Van Nostrand, New York, 1950. 15. E. Hewitt and K. Stromberg, Real and Abstract Analysis, Third Edition, Springer­ Verlag, Heidelberg & New York, 1975. ·16. T. Jech, Set Theory, Academic Press, New York, 1978. ·17. I. Kaplansky, Set Theozy and Metric Spaces, Allyn and Bacon, Boston, 1972. 18. 1. L. Kelley, General Topology, Graduate Texts in Mathematics, #27, Springer-Verlag, Heidelberg & New York, 1975. 19. A. N. Kolmogorov and S. V. Fomin, Measure, Lebesgue Integrals, and Hilbert Space, Academic Press, New York & London, 1961. ·20. K. Kuratowski and A. Mostowski, Set Theozy, North-Holland, Amsterdam, 1976. 21. H. Lebesgue, Integrale, longueur, aire, Annali Mat. Pura Appl., Ser. 3, 7 (1902), 23 1-359.

399

400

BIBLIOGRAPHY

22. J. R. Munkres, Topology: a Fh-st Course, Prentice-Hall, Englewood Cliffs, NJ, 1 975. 23. F. Riesz and B. Sz.-Nagy, Fuuctional Analysis, L. Boron translator, F. Ungar, New York, 1955. 24. H. L. Royden, Real Analysis, Third Edition, Macmillan, New York & London, 1 988. 25. W. Rudin, Principles of Mathematical Analysis, Third Edition, McGraw-Hill, New York, 1 976. 26. W. Rudin, Real and Complex Analysis, Third Edition, McGraw-Hill, New York, 1987. 27. W. Sierpinski, Sur un probleme concernant les ensembles mesurables superficiellement, Fundameuta Mathematicae 1 ( 1 920), 1 12-1 1 5 . 28. G. F. Simmons, lutroduction to Topology and Modern Analysis, McGraw-Hill, Macmil­ lan, New York, 1 963. 29. G. Strang, Linear Algebm and its Applications, Second Edition, Academic Press, New York, 1 980. 30. A. E. Taylor, General Theozy of Functions and Integration, Blaisdell, Waltham, MA, 1965. 3 1 . A. E. Taylor and D. C. Lay, Introduction to Functional Analysis, Second Edition, Wiley, l'le\V York, 1980. 32. A. Torchinsky, Real Variables, Addison Wesley, New York, 1988. 33. R. L. Wheeden and A. Zygmund, Measure and Integral, Marcel Dekker, Inc., New York & Basel, 1 977. 34. A. C. Zaanen, Integration, North-Holland, Amsterdam, 1967.

LIS T OF S YMBOLS (xi ); e / . 6 A1 , 6

Ac , 4

B(Q), 40 C(X), 66 Cb(X), 68 Fa -set, 6 1 Fa . G&, 61 G&-set, 61 Loo. 261

e2(Q}, 305 e p . 264 0. 2

L p (�J.), 255 A, Ap.. 104 IN, 6 n;eJ A; , 6 12 nie/ A ; , 3

�o .

Ui e/ A; , 3 XA. 126

C,

13

E, 2

A0 , 59

fJ,

*.

107 x-. 244

a A, 38 ±oo, 29 U, 162 a(:F), 150 d(x, A), 8 1 f: A -+ 8, 5 g 0 J, 5 Xn t X, 23

A, 59

401

I NDEX absolute value, 16 absolute value of vector, 67, 243 absolutely continuous function, 380 absolutely continuous measure, 338 absolutely summable series, 223 accumulation point, 37, 59 addition of real numbers, 14 additive function, 138 additivity of set functions (1'

98

finite, 99 adjoint of operator, 242 adjoint operator, 297

of continuity, 16 axiom of choice, 7 axiom of continuity, 16 axioms field, 14 order, 1 5 Baire space, 43

Baire 's category theorem, 44 Baire, Rene L. ( 1874-1932), 43 balanced set, 224 ball closed, 36

algebra of functions, 89

open, 35

algebra of sets, 95, 150 0'-, 97

unit, 220 Banach lattice, 246

algebraic difference of sets, 143 algebraic number, 13

Banach space, 218

almost everywhere relation, 120

Banach, Stefan ( 1892-1945), 217, 218,

reflexive, 239

alternating series, 33 angle between vectors, 282 antisymmetry, 7, 66 Archimedean property, 17 Archimedes (287-2 12 BC), 17, 180

229, 232, 237 Banach-Mazur limit, 241 Banach-Steinhaus theorem, 229 base, 65 for topology, 65

area function, 186, 187 Arzela, Cesare ( 1 846-1912), 75, 188

basis

Ascoli, Guido ( 1887-1957), 75 Ascoli-Arzela theorem, 75

Bernoulli's inequality, 14 Bernoulli, Jacob (1 654-1705), 14

associative laws, 15

Bernstein polynomials, 254

atom, 120

Bernstein, Felix ( 1 878-1956), 12

averaging operator, 253 axiom

Bernstein, Sergei N. ( 1880-1968), 254

completeness. 16

Bessel, Friedrich W. (1784-1846), 283

orthonormal, 298

Bessel's inequality, 284

403

404

INDEX

binary relation, 7 Balzano, Bernard (178 1-1848), 25, 29 Borel measurable function, 152 Borel measure, 136 induced by a monotone function, 373 Borel sets, 97 Borel signed measure, 362 Borel, Emile (187 1-1958), 5 1 bound essential, 260 lower, 16 upper, 16 boundary of set, 38, 59 boundary point, 38, 59 bounded from above set, 16 bounded from below set, 1 6 bounded metric space, 39 bounded operator, 225 bounded set, 1 6 C 1 -differentiable function, 387 C00-function, 203 cancellation law of addition, 1 9 of division, 20 of multiplication, 20 Cantor set, 41, 140 €-, 141 Cantor, Georg (1 845-1918), 12, 4 1 Caratheodory extension of measure, I l l Caratheodory function, 156 Caratheodory, Constantin ( 1873-1950), 103, 107' 1 1 1 ' 156 cardinal infinite, 13 cardinal number, 1 2 cardinality of the continuum, 1 3 Cartesian product, 6 Cartesius ( 1596-1650), 6 Cauchy sequence, 26, 39 in measure, 149 uniform, 71 Cauchy, Augustin L. (1789-1875), 26, 39, 278

Cauchy-Schwarz inequality, 278 Cavalieri's principle, 214 Cavalieri, Bonaventura (1589-1647), 214 cell, 186 chain, 8 chain rule, 350 change of variable, 176 change of variables formula, 393 characteristic function, 126 choice function, 6 circled set, 224 clopen set, 65 closed ball, 36 closed graph theorem, 234 closed mapping, 52 closed set, 35, 58 closed unit ball, 220 closure of set, 36, 59 closure point, 36, 59 cluster poipt, 23 co-meager set, 48 commutative group, 362 commutative laws, 1 5 compact metric space, 49 compact operator, 226 compact set, 49, 62 compact topological space, 62 complement orthogonal, 287, 291 complement of set, 4 complemented subspace, 235 complete metric space, 39 complete orthogonal set, 293 complete orthonormal set, 298 completeness axiom, 1 6 completion of a metric space, 46 complex inner product space, 276 complexification of a real vector space, 277 component of a point, 48 composition of functions, 5 condensation of singularities principle, 230 conjugate exponents, 264

INDEX

connected metric space, 47 connected set, 47 connected topological space, 65 constant sequence, 23 continuous function, 38, 60 at a point, 60 continuum, 13 continuum hypothesis, 13 generalized, 1 3 contraction, 55 convergence pointwise, 68 uniform, 69 convergence in measure, 146 convergent sequence, 22, 37, 60 in mean, 202 in measure, 146 weakly, 365 convergent series in a Banach space, 223 convex set, 224 countable set, 9 counting measure, 99 cover, 48 measurable of a set, 1 17 of set, 48 open, 48 pointwise finite, 56 cylindrical coordinate transformation, 395 cylindrical coordinates, 395 Daniell, P. J. (1889-1946), 166 Darboux, Jean-Gastin ( 1842-19 17), 181 De Morgan's laws, 4 De Morgan, Augustin ( 1806-1871 ), 4 decomposition Hahn, 331 Jordan, 333 Lebesgue, 342 decreasing function, 372 decreasing sequence, 22 decreasing sequence of functions, 70 dense subset, 37, 59 density function, 344

405

d�nsity point, 371 derivative, 386 lower of measure, 367 of measure, 367 Radon-Nikodym, 344 upper of measure, 367 derived set, 37, 59 Descartes, Rene (1596-1650), 6 determinant, 386 Jacobian, 386 diameter of a set, 39 diffeomorphism, 388 difference algebraic of sets, 143 set, 2 symmetric, 3 difference of two numbers, 20 differentiable function, 386 differentiable measure, 367 differential operator, 225, 227 Dini 's theorem, 70 Dini, Ulisse (1845-1918), 70 Dirac's measure, 99 Dirac, Paul A. M. (1902-1984), 99 Dirichlet kernel, 3 1 6 Dirichlet, Johann P. G. L. (1805-1859), 71, 316 discrete distance, 34 discrete metric space, 34 discrete topology, 58 disjoint sets, 2 distance, 34 discrete, 34 Euclidean, 34 , uniform, 70 distance function, 8 1 distance of a point from a set, 8 1 distributive law, 15 distributive laws, 3, 4 division, 20 division law for fractions, 20 domain of function, 5 double series, 30

406 dual order, 244

INDEX finer partition, 178 finite additivity, 99

Dynkin system, 152

finite intersection property, 55

Dynkin, Eugene B. (1924- ), 152

finite measure space, 1 1 3 finite set, 9

E -Cantor set, 141

finite signed measure, 332

Egorov, Dimitry F. (1 869-1931}, 124

finite subset property, 272

Egorov's theorem, 125 element

finitely additive measure, 102 first category set, 43, 59

first, 10

first element, 10

least, 10

Fischer, Ernst G. (1875-1954}, 258 Fisher, Ernst S. (1 875-1954), 258

element of a set, 2

fixed point, 55

elementary matrix, 390

Fourier coefficients, 299, 3 1 0

empty set, 2

Fourier coefficients of a function, 309

enumeration, 9

Fourier series of a function, 309

equality of sets, 2

Fourier, Jean B. J. (1768-1830), 299

equality of two functions, 5

Fresnel integrals, 199, 215

equicontinuous set, 75

Fresnel, Augustin J. ( 1788-1 827}, 199

equivalence class, 7

Fubini's theorem, 2 1 2

equivalence relation, 7

Fubini, Guido (1 879- 1943), 212, 382

equivalent distances, 39

function, 5 C 1 -differentiable, 387

maximal, 8

equivalent norms, 220 equivalent sets, 9

C00-, 203

essential bound, 260

absolutely continuous, 380

essential supremum, 260

additive, 138

essentially bounded function, 260

area, 186, 187

Euclid (ca 365-300 BC}, 34

Borel measurable, 152

Euclidean distance, 34

Caratheodory, 156

Euclidean norm, 218

characteristic, 126

Eudoxus of Cnidus (ca 400-347 BC), 17

choice, 6

even polynomial, 90

continuous, 38, 60

eventually true property, 22

continuous at a point, 60 contraction, 55

Fa -set, 6 1

decreasing, 372

family

density, 344

pairwise disjoint, 4

differentiable, 386

family of generators for u -algebra, 1 5 1

distance, 8 1

family of sets, 3

essentially bounded, 260

Fatou, Pierre J. L. (1 878-1929}, 171

Gamma, 200

Fejer kernel, 317

increasing, 372

Fejer, Leopold (1880-1959}, 317, 3 1 8

indicator, 126

field axioms, 14

injective, 5

INDEX

407

integrable, 166

greatest lower bound, 1 6

integrable over a set, 173 inverse, 5

group, 362 commutative, 362

isometry, 46

topological, 362

jointly measurable, 1 5 5 Lebesgue integrable, 166 measurable, 121, 152 measurable in a variable, 155 monotone, 372

Holder's inequality, 256

Holder, Otto Ludwig ( 1 859-1937), 256

Haar, Alfred ( 1885-1933), 362 Hahn decomposition, 331

one-to-one, 5 onto, 5

Hahn-Banach theorem, 237

periodic, 307

harmonic series, 33

Riemann integrable, 180, 187 separately measurable, 155 set, 98 simple, 127 square summable, 305 step, 127 strictly decreasing, 372 strictly increasing, 372 strictly monotone, 372 surjective, 5 uniformly continuous, 44

Hahn, Hans ( 1 879-1934), 217, 237, 33 1 Hausdorff, Felix ( 1 868-1 942), 60

Heine, Heinrich E. ( 1821-1881 ), 5 1 Heine-Borel theorem, 5 1 Hermite polynomial, 304 Hermite, Charles ( 1822-190 I), 304 Hilbert cube, 296 Hilbert space, 288 Hilbert, David (1 862-1943), 275 homeomorphic metric spaces, 39 homeomorphic topological spaces, 63 homeomorphism, 63

upper, 162 variation, 377 weight, 289 function measurable in a variable, 155

ideal, 247 identity

function of bounded variation, 376

Parseval, 298 image of a point, 5

function space, 68

image of set,

function with compact support, 85, 201 functional

improper Riemann integral, 190 increasing function, 372 increasing sequence, 22

left-invariant, 363 limit, 241

5

of functions, 70

right-invariant, 363

indefinite integral, 342 index set of family, 3

Gamma function, 200

indicator function, 126

generalized continuum hypothesis, 13 generating sequence for upper function,

indiscrete topology, 58 induced measure, 106

162 Gram, Jorgen P. ( 1850-1916), 284 Gram-Schmidt orthogonalization process, 284 graph of operator, 234

induced topology, 58 inequality Bernoulli's, 14 Bessel, 284 Cauchy-Schwarz, 278

408 (cont.) inequality Holder's, 256 Minkowski, 256 triangle, 16, 34, 2 1 8 infimum, 1 6 infinite cardinal, 1 3 infinite set, 9 infinity minus, 29 plus, 29 initial values, 233 injective function, 5 inner product, 276 inner product preserving operator, 304 inner product space complex, 276 inner, 276 input, 5 integrable function, 166 over a set, 173 Riemann, 180 integral, 127, 163, 166 Haar, 363 improper Riemann, 190 indefinite, 342 iterated, 2 1 1 Lebesgue, 1 27, 132, 163, 166 lower Riemann, 179, 186 of step function, 127 Riemann, 180, 187 upper Riemann, 179, 187 integral operator, 226 integral test, 33 interior of set, 35, 59 interior point, 35, 59 intersection, 2 intersection of sets, 3 interval, 134 inverse function, 5 inverse image of set, 5 inverse of real number, 20 isolated point, 56 isometric metric spaces, 46

INDEX isometry, 46 lattice, 249 linear, 239 isomorphic normed vector lattices, 249 iterated integral, 2 1 1 Jacobi, Carl G. J. ( 1 804-185 1), 386 Jacobian, 386 Jacobian matrix, 386 jointly measurable function, 155 Jordan dec�mposition of a signed measure,

333 Jordan, Camille (1838-1921), 333 kernel Dirichlet, 3 1 6 Fejer, 3 17

kernel of functional, 240 kernel of operator, 226 Kolmogorov, Andrey N. ( 1 903-1987), 269 Korovkin, Pavel P. ( 1 9 1 3-1985), 250, 254 Kronecker, Leopold ( 1 823-1891 ), 33, 298

L p-norm, 255 L p-space, 255 L'Hopital, Guillaume F. A. (1661-1704), 203 Laguerre functions, 30 I Laguerre, Nicolas E. ( 1834-1886), 301 Laplace transform, 250 Laplace, P. S. M. ( 1 749-1827), 250 lattice homomorphism, 253 lattice isometry, 249 lattice norm, 246 law distributive, 1 5 parallelogram, 279 laws associative, 15 commutative, 1 5 least element, 1 0 least upper bound, 1 6 Lebesgue decomposition, 342

INDEX

Lebesgue integrable function, 166 Lebesgue integral, 127, 132, 163, 166 Lebesgue integral of step function, 127 Lebesgue measure, 10 I. 1 1 3, 133, 134 Lebesgue number for open cover, 50 Lebesgue point, 382 Lebesgue, Henri L. ( 1875-1941 ), 50, 1 6 1 , 172, 1 84, 202, 341 , 375

Lebesgue-Stieltjes measure, 379 left Haar measure, 363 left translate of function, 362 left-Haar integral, 363 left-invariant functional, 363 Legendre polynomials, 295 Legendre, Adrien-Marie ( 1752-1833), 295

Leibniz, Gottfried W. ( 1646-1 7 1 6), 185 Levi, Beppo ( 1 875-1961 ), 170 limit Banach-Mazur, 241 pointwise, 68 limit functional, 241 limit inferior, 24 limit of sequence, 22 limit point of sequence, 23 limit superior, 24 LindelOf, Ernst L. ( 1870-1940), 48 linear functional, 235 order bounded, 244 positive, 244 linear isometry, 239 linear operator, 224 linearity of the integral, 127 Lipschitz condition, 145, 383 Lipschitz, Rudolf ( 1 832-1903), 1 45 little L p -spaces, 264 locally compact topological space, 84 locally finite measure, 272 lower bound, 1 6 greatest, 1 6 lower derivative of measure, 367 lower Riemann integral, 1 79, 186 lower sum, 178

rp apping closed, 52 sublinear, 236 mathematical induction, I 0 matrix elementary, 390 Jacobian, 386 positive definite, 287 symmetric, 287 matrix representation of operator, 386 maximal element, 8 Mazur, Stanislaw ( 1 905-1981 ), 241 meager set, 43, 59 measurable cover of set, 1 17 measurable function, 1 2 1 , 152 measurable set, I 03, 1 12, 1 5 1 measurable space, 1 5 1 measure, 98 absolutely continuous, 338 Borel, 136 counting, 99 differentiable, 367 Dirac, 99 finitely additive, I 02 Haar, 363 induced, I 06 Lebesgue, 1 0 1 , 1 13, 133, 134 Lebesgue-Stieltjes, 379 locally finite, 272 orthogonal, 339 outer, 103 outer generated by a measure, I 07 product, 204 regular Borel, 136, 352 representing a linear functional, 352 signed, 329 singular, 339 translation invariant, 137 measure space, 99 cr -finite, 1 1 3 finite, 1 1 3 nonatomic, 120 with the finite subset property, 272

410

member of a set, 2 membership symbol, 2 mesh of partition, 178 menic, 34 induced by a norm, 218 uniform, 70 menic space, 34 bounded, 39 compact, 49 complete, 39 connected, 47 discrete, 34 separable, 54 totally bounded, 53 menic spaces homeomorphic, 39 Minkowski inequality, 256 Minkowski, Hermann (1864--1909), 256 minus infinity, 29 monotone class, 158 monotone function, 372 monotone sequence, 22 monotone sequence of functions, 70 monotonicity of measure, 99 monotonici ty of the integral, 129 multiplication law for fractions, 20 multiplication of real numbers, 14 multiplication rule of signs, 20 natural embedding of normed space, 239 negative number, 15 negative of real number, 19 negative part of vector, 67, 243 negative set, 330 negative variation of a signed measure, 332 neighborhood of point, 59 Newton, Isaac (1642-1727), 185 Nikodym, Otton M. (1889-1974), 342 nonatomic measure space, 120 norm, 217 Lp. 255 e2, 305

INDEX

a-order continuous, 337 Euclidean, 218 in an inner product space, 278 lattice, 246 of operator, 225 sup, 69, 219 uniform, 69 norm bounded set, 218 norm dual of Banach space, 238 norm induced by an inner product, 278 norm preserving operator, 304 normal topological space, 80 normed vector lattice, 246 with a-order continuous norm, 337 normed vector space, 218 nowhere dense set, 41, 59 null set, 104 number algebraic, 13 cardinal, 12 Lebesgue, 50 negative, 15 positive, 15 odd polynomial, 90 one-to-one function, 5 onto function, 5 open mapping, 52 open ball, 35, 366 open cover, 48 open mapping, 52 open mapping theorem, 232 open set, 35, 58 open unit ball, 220 operator, 224 adjoint, 297 averaging, 253 bounded, 225 compact, 226 differential, 225, 227 inner product preserving, 304 integral, 226 Laplace, 250

INDEX linear, 224 norm preserving, 304 positive, 249 projection, 235 rank-one, 242 symmetric, 287 unbounded, 225 operator norm, 225 order partial, 7 order axioms, 1 5 order bounded linear functional, 244 order bounded set, 244 order complete vector lattice, 252 order continuity of the integral, 129, 164 order dual, 244 ordered pair, 6 ordered vector space, 66, 242 ordinate set of function, 2 1 5 orthogonal complement, 287, 291 orthogonal measures, 339 orthogonal set, 282 complete, 293 orthogonal vectors, 282 orthonormal basis, 298 orthonormal set, 282, 298 complete, 298 oscillation, 372 oscillation of function, 61 outer measure; 103 induced by a positive functional, 353 outer measure generated by a measure, 107 output, 5 p-adic representation of number, 32 pairwise disjoint family, 4 parallelogram law, 279 Parseval's identity, 298 Parseval, Marc-Antoine C. (1755-1836), 298 partial derivative, 193 partial order, 7 partially ordered set, 8

partially ordered vector space, 242 partition, 186 finer, 178 partition of interval, 178, 376 partition of set, 7 partition of unity, 86 perfectly normal space, 87 period of function, 307 periodic function, 307 permutation of lN, 30 plus infinity, 29 point accumulation, 37, 59 boundary, 38, 59 closure, 36, 59 cluster, 23 density, 371 fixed, 55 interior, 35, 59 isolated, 56 Lebesgue, 382 limit, 23 point of a set, 2 pointwise bounded set, 228 pointwise convergence, 68 pointwise finite cover, 56 pointwise limit, 68 polar coordinate transformation, 394 polar coordinates, 394 polynomial even, 90 Hermite, 304 odd, 90 positive cone, 242 positive definite matrix, 287 positive linear functional, 244 positive number, 15 positive operator, 249 positive part of vector, 67, 243 positive set, 330 positive variation of a signed measure, 332 positive vector, 67, 242 power set, 4

411

INDEX

412 principle

Riemann integral, 1 80, 187

condensation of singularities, 230

improper, 190

mathematical induction, 10

lower, 179, 186

uniform boundedness, 229

upper, 179, 187

well ordering, 1 0 product Cartesian, 6 . inner, 276

Riemann sum, 1 8 1 Riemann's criterion, 180 Riemann, Bernhard G. F. ( 1826-1 866), 177, 180, 202

product a -algebra, 154

Riesz representation theorem, 355

product measure, 204

Riesz space, 242

product semiring, 154, 204

Riesz, Frigyes (l 880-1956), 242, 244, 258, 292, 344, 355

projection, 235 proper subset, 2

right translate of function, 362

property

right-invariant functional, 363

Archimedean, 1 7 eventually L-uc, 22

ring cr ,

150

finite intersection, 55

ring of sets, 96, 149

topological, 55

root of real number, 19

Pythagoras of Samos (ca 580-500 BC), 282 Pythagorean theorem, 282

a -additivity, 98 a -algebra product, 154 a -algebra generated by a collection, 97

quotient topology, 65

a -algebra generated by a family, 150 a-algebra of sets, 97

Radon, Johann ( l 887-1956), 342

a-compact set, 357

Radon-Nikodym derivative, 344.

a -finite measure space, 1 1 3

Radon-Nikodym theorem, 342

a -finite set, 1 1 6, 175

range of function, 5

a-finite signed measure, 332

rank-one operator, 242

a-ring, 150

rearrangement invariant series, 30

a-set, 94

reciprocal of real number, 20

Schmidt, Erhard (1 876-1959), 284

rectangle, 204

Schroder, Ernst ( 1841-1 902), 12

reflexive Banach space, 239

Schroder-Bernstein theorem, 12

reflexivity, 7, 66

Schwarz, Hermann A. ( 1843-1921 ), 256,

regular Borel measure, 136, 352

278

regular Borel signed measure, 362

second dual of Banach space, 239

relation, 7

section of set, 154

a.e., 120

segment, 9

binary, 7

selection of points, 1 8 1

equivalence, 7

semiring

relative topology, 58

product, 154, 204

representing measure, 352

restriction, 97

restriction semiring, 97

semiring of sets, 94

Riemann integrable function, 180, 187

separable metric space, 54

INDEX separately measurable function, 155 separation by a continuous function, 80 by open sets, 80 separation of points by a family, 87 sequence, 6 Cauchy, 26, 39 constant, 23 convergent, 22, 37, 60 convergent in mean, 202 convergent to ±oo, 29 decreasing, 22 increasing, 22 monotone, 22 of averages, 27 sequence convergent, 22, 37 to ±oo, 29 sequence of averages, 27 sequence of functions decreasing, 70 increasing, 70 monotone, 70 series absolutely summable, 223 alternating, 33 convergent in Banach space, 223 double, 30 Fourier, 309 harmonic, 33 rearrangement invariant, 30 unconditionally convergent, 33 uniformly convergent, 71 set, 2 Fa . 61 Go. 6 1 €-Cantor, 141 CT, 94 cr-compact, 357 cr-finite, 1 16, 175 atom, 120 balanced, 224 Borel, 97 bounded, 16 bounded from above, 16

. bounded from below, 1 6 Cantor, 41, 140 circled, 224 clopen, 65 closed, 35, 58 co-meager, 48 compact, 49, 62 connected, 47 convex, 224 countable, 9 dense, 37, 59 derived, 37, 59 empty, 2 equality, 2 equicontinuous, 75 finite, 9 index, 3 infinite, 9 meager, 43, 59 measurable, 103, 1 12, 151 negative, 330 norm bounded, 218 nowhere dense, 41, 59 null, 104 of first category, 43 open, 35, 58 order bounded, 244 ordinate, 2 1 5 orthogonal, 282 orthonormal, 282, 298 partially ordered, 8 pointwise bounded, 228 positive, 330 power, 4 symmetric, 224 totally ordered, 8 uncountable, 9 void, 2 set difference, 2 set function, 98 cr-subadditive, 103 set inclusion, 2 set intersection, 2, 3 set of first category, 59

413

414

INDEX

set union, 2, 3 sets equivalent, 9 signed measure, 329 cr-finite, 332 Borel, 362 finite, 332 regular Borel, 362 simple function, 127 :E, 158

singleton, 2 singular measures, 339 space Loo(J.L), 261 Baire, 43 Banach, 218 function, 68 Hausdorff, 60 Hilbert, 288 measurable, 1 5 1 measure, 99 metric, 34 Riesz, 242 topological, 58 spherical coordinate transformation, 396 spherical coordinates, 396 square summable function, 305 standard representation of simple function, 127

Steinhaus, Hugo (1887-1972), 144, 229 step function, 127 Stieltjes, Thomas J. (1856-1894), 379 Stone, Marshall H. (1903-1989), 88, 89 Stone-Weierstrass theorem, 88, 89 strictly decreasing function, 372 strictly increasing function, 372 strictly monotone function, 372 subcover of set, 48 sublinear mapping, 236 subsequence, 6 subset proper, 2 subset of set, 2

subtraction, 20 sum Riemann, 1 8 1 sum of arbitrary numbers, 283 sup norm, 69, 219 support of function, 85, 201 support of measure, 263 supremum, 16 essential, 260 surjective function, 5 symmetric difference, 3 symmetric matrix, 287 symmetric operator, 287 symmetric set, 224 symmetry, 7 term sequence, 6 Tietze's extension theorem, 84 Tietze, Heinrich F. F. (1880-1964), 84 Toeplitz, Otto (1881-1940), 33 Tonelli, Leonida (1 885-1946), 213 topological group, 362 topological property, 55 topological space, 58 compact, 62 connected, 65 locally compact, 84 normal, 80 perfectly normal, 87 topology, 57 discrete, 58 indiscrete, 58 induced, 58 quotient, 65 relative, 58 total variation of a function, 376 total variation of a signed measure, 332 totally bounded metric space, 53 totally ordered set, 8 transformation cylindrical coordinate, 395 polar coordinate, 394 spherical coordinate, 396 transitivity, 7, 66

INDEX translate of function, 362 translation invariant measure, 137 translation of a set, 137 triangle inequality, 16, 34, 2 1 8 trigonometric polynomial, 308 unbounded operator, 225 unconditionally convergent series, 33 uncountable set, 9 unifonn boundedness principle, 229 unifonn convergence, 69 of series, 7 1 unifonn distance, 70 unifonn metric, 70 unifonn nonn, 69 unifonnly Cauchy sequence, 7 1 unifonnly continuous function, 44 union, 2 union of sets, 3 upper bound, 8, 16 least, 16 upper derivative of measure, 367 upper function, 1 62 upper Riemann integral, 179, 187 upper sum, 178 Uryson's lemma, 8 1 , 85 Uryson, Pavel S. ( 1898-1924), 8 1 , 85

variation function, 377 vector positive, 67, 242 vector lattice, 67, 242 nonned, 246 order complete, 252 vector space nonned, 218 ordered, 66 vector sublattice, 247 Vitali, Giuseppe ( 1 875-1932), 1 17, 184 void set, 2 weakly convergent sequence, 365 Weierstrass M-test, 7 1 Weierstrass, Karl T. W. ( 1 8 1 5-1897), 25, 29, 71, 88-90 weight function, 289 well ordering principle, 10 x-section of set, 207 y-section of set, 207 zero product rule, 20 Zorn's lemma, 8 Zorn, Max (1906-1993), 8

415