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