1

LECTURE NOTES

IN MEASURE THEORY

Christer Borell Matematik Chalmers och Göteborgs universitet 412 96 Göteborg (Version: January 12) 2 PREFACE

These are lecture notes on integration theory for a eight-week course at the Chalmers University of Technology and the Göteborg University. The parts de…ning the course essentially lead to the same results as the …rst three chapters in the Folland book [F ] ; which is used as a text book on the course. The proofs in the lecture notes sometimes di¤er from those given in [F ] : Here is a brief description of the di¤erences to simplify for the reader. In Chapter 1 we introduce so called -systems and -additive classes, which are substitutes for monotone classes of sets [F ]. Besides we prefer to emphasize metric outer measures instead of so called premeasures. Through- out the course, a variety of important measures are obtained as image mea- sures of the linear measure on the real line. In Section 1.6 positive measures in R induced by increasing right continuous mappings are constructed in this way. Chapter 2 deals with integration and is very similar to [F ] and most other texts. Chapter 3 starts with some standard facts about metric spaces and relates the concepts to measure theory. For example Ulam’s Theorem is included. The existence of product measures is based on properties of -systems and -additive classes. Chapter 4 deals with di¤erent modes of convergence and is mostly close to [F ] : Here we include a section about orthogonality since many students have seen parts of this theory before. The Lebesgue Decomposition Theorem and Radon-Nikodym Theorem in Chapter 5 are proved using the von Neumann beautiful L2-proof. To illustrate the power of abstract integration these notes contain several sections, which do not belong to the course but may help the student to a better understanding of measure theory. The corresponding parts are set between the symbols

### and

""" respectively. 3

Finally I would like to express my deep gratitude to the students in my classes for suggesting a variety of improvements and a special thank to Jonatan Vasilis who has provided numerous comments and corrections in my original text.

Göteborg 2006 Christer Borell 4 CONTENT

1 Measures 1.1 -Algebras and Measures 1.2 Measure Determining Classes 1.3 Lebesgue Measure 1.4 Carathéodory’sTheorem 1.5 Existence of Linear Measure

2 Integration 2.1 Integration of Functions with Values in [0; ] 2.2 Integration of Functions with Arbitrary Sign1 2.3 Comparison of Riemann and Lebesgue Integrals

3 Further Construction Methods of Measures 3.1 Metric Spaces 3.2 Linear Functionals and Measures 3.3 q-Adic Expansions of Numbers in the Unit Interval 3.4 Product Measures 3.5 Change of Variables in Volume Integrals 3.6 Independence in Probability

4 Modes of Convergence 4.1 Convergence in Measure, in L1(); and in L2() 4.2 Orthogonality 4.3 The Haar Basis and Wiener Measure

5 Decomposition of Measures 5.1 Complex Measures 5.2 The Lebesgue Decomposition and the Radon-Nikodym Theorem 5.3 The Wiener Maximal Theorem and Lebesgue Di¤erentiation Theorem 5

5.4 Absolutely Continuous Functions and Functions of Bounded Variation 5.5 Conditional Expectation

6 Complex Integration 6.1 Complex Integrand 6.2 The Fourier Transform 6.3 Fourier Inversion 6.4 Non-Di¤erentiability of Brownian Paths

References 6 CHAPTER 1 MEASURES

Introduction

The Riemann integral, dealt with in calculus courses, is well suited for com- putations but less suited for dealing with limit processes. In this course we will introduce the so called Lebesgue integral, which keeps the advantages of the Riemann integral and eliminates its drawbacks. At the same time we will develop a general measure theory which serves as the basis of contemporary analysis and probability. In this introductory chapter we set forth some basic concepts of measure theory, which will open for abstract Lebesgue integration.

1.1. -Algebras and Measures

Throughout this course

N = 0; 1; 2; ::: (the set of natural numbers) Z = f:::; 2; 1g; 0; 1;; 2; ::: (the set of integers) Q =fthe set of rational numbersg R = the set of real numbers C = the set of complex numbers.

If A R;A+ is the set of all strictly positive elements in A: If f is a function from a set A into a set B; this means that to every x A there corresponds a point f(x) B and we write f : A B: A function2 is often called a map or a mapping.2 The function f is injective! if

(x = y) (f(x) = f(y)) 6 ) 6 7 and surjective if to each y B; there exists an x A such that f(x) = y: An injective and surjective2 function is said to be bijective.2 A set A is …nite if either A is empty or there exist an n N+ and a bijection f : 1; :::; n A: The empty set is denoted by : A2 set A is said f g ! to be denumerable if there exists a bijection f : N+ A: A subset of a denumerable set is said to be at most denumerable. !

Let X be a set. For any A X; the indicator function A of A relative to X is de…ned by the equation

1 if x A (x) = A 0 if x 2 Ac:  2 The indicator function A is sometimes written 1A: We have the following relations:

c = 1 A A A B = min( A; B) = A B \ and

A B = max( A; B) = A + B A B: [

De…nition 1.1.1. Let X be a set. a) A collection of subsets of X is said to be an algebra in X if has the following properties:A A

(i) X : (ii) A 2 A Ac ; where Ac is the complement of A relative to X: (iii) If2A; A B ) 2then A A B : 2 A [ 2 A

(b) A collection of subsets of X is said to be a -algebra in X if is an algebra with theM following property: M

If An for all n N+, then 1 An : 2 M 2 [n=1 2 M 8

If is a -algebra in X; (X; ) is called a measurable space and the membersM of are called measurableM sets. The so called power set (X), that is the collectionM of all subsets of X, is a -algebra in X: It is simpleP to prove that the intersection of any family of -algebras in X is a -algebra. It follows that if is any subset of (X); there is a unique smallest -algebra ( ) containingE ; namely the intersectionP of all -algebras containing : E E E The -algebra ( ) is called the -algebra generated by : The -algebra generated by all openE intervals in R is denoted by . It isE readily seen that the -algebra contains every subinterval of R. BeforeR we proceed, recall that a subset ERof R is open if to each x E there exists an open subinterval of R contained in E and containing x; the2 complement of an open set is said to be closed. We claim that contains every open subset U of R: To see this suppose x U and let x R ]a; b[ U; where < a < b < : Now pick r; s Q such2 that a < r <2 x < s < b: Then x 1]r; s[ U and it1 follows that U is2 the union of all bounded open intervals2 with rational boundary points contained in U: Since this family of intervals is at most denumberable we conclude that U : In addition, any closed set belongs to since its complements is open.2 It R is by no means simple to grasp the de…nitionR of at this stage but the reader will successively see that the -algebra hasR very nice properties. At the very end of Section 1.3, using the so calledR Axiom of Choice, we will exemplify a subset of the real line which does not belong to . In fact, an example of this type can be constructed without the Axiom ofR Choice (see Dudley’sbook [D]). In measure theory, inevitably one encounters : For example the real line has in…nite length. Below [0; ] = [0; [ 1: The inequalities x y and x < y have their usual meanings1 if x;1 y [[0 f1g; [. Furthermore, x  if x [0; ] and x < if x [0; [ : We de…ne2 1x + = + x =  1if x; y 2 [0; 1] ; and 1 2 1 1 1 1 2 1 0 if x = 0 x = x =
1 1
if 0 < x :  1  1 Sums and multiplications of real numbers are de…ned in the usual way.

If An X; n N+, and Ak An =  if k = n, the sequence (An)n N+ is called a disjoint denumerable2 collection.\ If (X;6 ) is a measurable space,2 the M collection is called a denumerable measurable partition of A if A = 1 An [n=1 and An for every n N+: Some authors call a denumerable collection of sets a2 countable M collection2 of sets. 9

De…nition 1.1.2. (a) Let be an algebra of subsets of X: A function  : [0; ] is called a contentA if A! 1

(i) () = 0 (ii) (A B) = (A) + (B) if A; B and A B = : [ 2 A \

(b) If (X; ) is a measurable space a content  de…ned on the -algebra is called aM positive measure if it has the following property: M

For any disjoint denumerable collection (A ) of members of n n N+ 2 M

( 1 An) = 1 (An): [n=1 n=1

If (X; ) is a measurable space and the function  : [0; ] is a positive measure,M (X; ; ) is called a positive measure space.M! The quantity1 (A) is called the -measureM of A or simply the measure of A if there is no ambiguity. Here (X; ; ) is called a probability space if (X) = 1; a …nite positive measure spaceM if (X) < ; and a -…nite positive measure space if X is a denumerable union of measurable1 sets with …nite -measure. The measure  is called a probability measure, …nite measure, and -…nite measure, if (X; ; ) is a probability space, a …nite positive measure space, and a -…nite positiveM measure space, respectively. A probability space is often denoted by ( ; ;P ): A member A of is called an event. As soon as we haveF a positive measure spaceF (X; ; ), it turns out to be a fairly simple task to de…ne a so called -integral M

f(x)d(x) ZX as will be seen in Chapter 2. 10

The class of all …nite unions of subintervals of R is an algebra which is denoted by 0: If A 0 we denote by l(A) the Riemann integral R 2 R 1 A(x)dx Z1 and it follows from courses in calculus that the function l : 0 [0; ] is a R ! 1 content. The algebra 0 is called the Riemann algebra and l the Riemann content. If I is a subintervalR of R, l(I) is called the length of I: Below we follow the convention that the empty set is an interval. If A (X), cX (A) equals the number of elements in A, when A is a 2 P …nite set, and cX (A) = otherwise. Clearly, cX is a positive measure. The 1 measure cX is called the counting measure on X: Given a X; the probability measure a de…ned by the equation a(A) = (a); if A 2 (X); is called the Dirac measure at the point a: Sometimes A 2 P we write a = X;a to emphasize the set X: If  and  are positive measures de…ned on the same -algebra , the sum  +  is a positive measure on : More generally,  +  is a positiveM measure for all real ; 0: Furthermore,M if E ; the function (A) = (A E);A ; is a positive measure. Below2 M this measure  will be denoted\ by E2and M we say that E is concentrated on E: If E ; the class 2 M E = A ; A E is a -algebra of subsets of E and the function M f 2 M  g (A) = (A), A E; is a positive measure. Below this measure  will be 2 M denoted by  E and is called the restriction of  to E: j M Let I1; :::; In be subintervals of the real line. The set

n I1 ::: In = (x1; :::; xn) R ; xk Ik; k = 1; :::; n   f 2 2 g n is called an n-cell in R ; its volume vol(I1 ::: In) is, by de…nition, equal to   n vol(I1 ::: In) =  l(Ik):   k=1 If I1; :::; In are open subintervals of the real line, the n-cell I1 ::: In is called an open n-cell. The -algebra generated by all open n-cells inRn is denoted by n: In particular, 1 = . A basic theorem in measure theory R R R states that there exists a unique positive measure vn de…ned on n such that R the measure of any n-cell is equal to its volume. The measure vn is called the n volume measure on n or the volume measure on R : Clearly, vn is -…nite. R 2 The measure v2 is called the area measure on R and v1 the linear measure on R: 11

Theorem 1.1.1. The volume measure on Rn exists.

Theorem 1.1.1 will be proved in Section 1.5 in the special case n = 1. The general case then follows from the existence of product measures in Section 3.4. An alternative proof of Theorem 1.1.1 will be given in Section 3.2. As soon as the existence of volume measure is established a variety of interesting measures can be introduced. Next we prove some results of general interest for positive measures.

Theorem 1.1.2. Let be an algebra of subsets of X and  a content de…ned on . Then, A (a)  isA …nitely additive, that is

(A1 ::: An) = (A1) + ::: + (An) [ [

if A1; :::; An are pairwise disjoint members of : (b) if A; B ; A 2 A (A) = (A B) + (A B): n \ Moreover, if (A B) < ; then \ 1 (A B) = (A) + (B) (A B) [ \ (c) A B implies (A) (B) if A; B : (d)  …nitely sub-additive, that is 2 A

(A1 ::: An) (A1) + ::: + (An) [ [ 

if A1; :::; An are members of : A

If (X; ; ) is a positive measure space M 12

(e) (An) (A) if A = n N+ An;An ; and ! [ 2 2 M

A1 A2 A3 ::: :   

(f) (An) (A) if A = n N+ An;An ; ! \ 2 2 M

A1 A2 A3 :::   

and (A1) < : (g)  is sub-additive,1 that is for any denumerable collection (A ) of n n N+ members of , 2 M ( 1 An) 1 (An): [n=1  n=1

PROOF (a) If A1; :::; An are pairwise disjoint members of ; A n n ( Ak) = (A1 ( Ak)) [k=1 [ [k=2 n = (A1) + ( Ak) [k=2 and, by induction, we conclude that  is …nitely additive.

(b) Recall that A B = A Bc: n \ Now A = (A B) (A B) and we get n [ \ (A) = (A B) + (A B): n \ Moreover, since A B = (A B) B; [ n [ (A B) = (A B) + (B) [ n and, if (A B) < ; we have \ 1 (A B) = (A) + (B) (A B). [ \

(c) Part (b) yields (B) = (B A) + (A B) = (B A) + (A); where the last member does not fall belown (A): \ n 13

n (d) If (Ai)i=1is a sequence of members of de…ne the so called disjunction n n A (Bk)k=1 of the sequence (Ai)i=1 as

k 1 B1 = A1 and Bk = Ak Ai for 2 k n: n [i=1   k k Then Bk Ak; i=1Ai = i=1Bi; k = 1; ::; n; and Bi Bj =  if i = j: Hence, by Parts (a) and[ (c), [ \ 6

n n n ( Ak) =  (Bk)  (Ak): [k=1 k=1  k=1

(e) Set B1 = A1 and Bn = An An 1 for n 2: Then An = B1 :::: Bn; n  [ [ Bi Bj =  if i = j and A = 1 Bk: Hence \ 6 [k=1 n (An) = k=1(Bk) and

(A) = k1=1(Bk): Now e) follows, by the de…nition of the sum of an in…nite series.

(f) Put Cn = A1 An; n 1: Then C1 C2 C3 :::; n    

A1 A = 1 Cn n [n=1 and (A) (An) (A1) < : Thus   1

(Cn) = (A1) (An) and Part (e) shows that

(A1) (A) = (A1 A) = lim (Cn) = (A1) lim (An): n n n !1 !1 This proves (f).

(g) The result follows from Parts d) and e). This completes the proof of Theorem 1.1.2. 14

The hypothesis ”(A1) < ”in Theorem 1.1.2 ( f) is not super‡uous. If 1 cN is the counting measure on N+ and An = n; n + 1; ::: ; then cN (An) = + f g + for all n but A1 A2 :::: and cN+ ( n1=1An) = 0 since n1=1An = : 1 If A; B X; the symmetric di¤erence\ A
B is de…ned by\ the equation 

A
B =def (A B) (B A): n [ n Note that = : A
B j A B j Moreover, we have A
B = Ac
Bc and ( 1 Ai)
( 1 Bi) 1 (Ai
Bi): [i=1 [i=1  [i=1

Example 1.1.1. Let  be a …nite positive measure on : We claim that to each set E and " > 0; there exists a set A; whichR is …nite union of 2 R intervals (that is, A belongs to the Riemann algebra 0), such that R (E
A) < ":

To see this let be the class of all sets E for which the conclusion S 2 R c is true. Clearly  and, moreover, 0 : If A 0, A 0 and c 2 S R  S 2 R 2 R therefore E if E : Now suppose Ei ; i N+: Then to each " > 0 2 S 2 S 2 S 2 i and i there is a set Ai 0 such that (Ei
Ai) < 2 ": If we set 2 R

E = 1 Ei [i=1 then (E
( 1 Ai)) 1 (Ei
Ai) < ": [i=1  i=1 Here c c E
( 1 Ai) = E ( 1 A ) E ( 1 Ai) [i=1 f \ \i=1 i g [ f \ [i=1 g and Theorem 1.1.2 (f) gives that

n c c ( E ( A ) (E ( 1 Ai) ) < " f \ \i=1 i g [ f \ [i=1 g if n is large enough (hint: i I (Di F ) = ( i I Di) F ): But then \ 2 [ \ 2 [ n n c c n (E
Ai) = ( E ( A ) E ( Ai) ) < " [i=1 f \ \i=1 i g [ f \ [i=1 g 15 if n is large enough we conclude that the set E : Thus is a -algebra 2 S S and since 0 it follows that = : R  S  R S R

Exercises

1. Prove that the sets N N = (i; j); i; j N and Q are denumerable.  f 2 g

2. Suppose is an algebra of subsets of X and  and  two contents on such that  A  and (X) = (X) < : Prove that  = : A  1

3. Suppose is an algebra of subsets of X and  a content on with (X) < : ShowA that A 1 (A B C) = (A) + (B) + (C) [ [ (A B) (A C) (B C) + (A B C): \ \ \ \ \

4. (a) A collection of subsets of X is an algebra with the following property: C If An ; n N+ and Ak An =  if k = n, then n1=1An . Prove2 that C 2is a -algebra.\ 6 [ 2 C C

(b) A collection of subsets of X is an algebra with the following property: C If En and En En+1; n N+; then 1 En . Prove2 that C is a -algebra. 2 [ 2 C C

5. Let (X; ) be a measurable space and (k)k1=1 a sequence of positive measures onM such that    ::: . Prove that the set function M 1  2  3 

(A) = lim k(A);A k !1 2 M is a positive measure. 16

6. Let (X; ; ) be a positive measure space. Show that M

n n n ( Ak)  (Ak) \k=1  k=1 q for all A1; :::; An : 2 M

7. Let (X; ; ) be a -…nite positive measure space with (X) = : Show that for anyMr [0; [ there is some A with r < (A) < : 1 2 1 2 M 1

8. Show that the symmetric di¤erence of sets is associative:

A
(B
C) = (A
B)
C:

9. (X; ; ) is a …nite positive measure space. Prove that M (A) (B) (A
B): j j

10. Let E = 2N: Prove that

cN(E
A) = 1 if A is a …nite union of intervals.

11. Suppose (X; (X); ) is a …nite positive measure space such that ( x ) > 0 for every x X:P Set f g 2 d(A; B) = (A
B); A; B (X): 2 P Prove that d(A; B) = 0 A = B; , d(A; B) = d(B;A) 17 and d(A; B) d(A; C) + d(C;B): 

12. Let (X; ; ) be a …nite positive measure space. Prove that M n n ( i=1Ai) i=1(Ai) 1 i

13. Let (X; ; ) be a probability space and suppose the sets A1; :::; An M n n 2 M satisfy the inequality (Ai) > n 1: Show that ( Ai) > 0: 1 \1 P

1.2. Measure Determining Classes

Suppose  and  are probability measures de…ned on the same -algebra , which is generated by a class : If  and  agree on ; is it then true thatM and  agree on ? The answerE is in general no. ToE show this, let M X = 1; 2; 3; 4 f g and = 1; 2 ; 1; 3 : E ff g f gg 1 Then ( ) = (X): If  = cX and E P 4 1 1 1 1  =  +  +  +  6 X;1 3 X;2 3 X;3 6 X;4 then  =  on and  = : In this sectionE we will6 prove a basic result on measure determining classes for -…nite measures. In this context we will introduce so called -systems and -additive classes, which will also be of great value later in connection with the construction of so called product measures in Chapter 3. 18

De…nition 1.2.1. A class of subsets of X is a -system if A B for all A; B : G \ 2 G 2 G

The class of all open n-cells in Rn is a -system.

De…nition 1.2.2. A class of subsets of X is called a -additive class if the following properties hold:D (a) X : (b) If A;2 D B and A B; then B A : 2 D  n 2 D (c) If (An)n N+ is a disjoint denumerable collection of members of the 2 class ; then 1 An : D [n=1 2 D

Theorem 1.2.1. If a -additive class is a -system, then is a - algebra. M M

PROOF. If A ; then Ac = X A since X and is a - 2 M n 2 M 2 M M additive class. Moreover, if (An)n N+ is a denumerable collection of members of ; 2 M c c c A1 ::: An = (A ::: A ) [ [ 1 \ \ n 2 M for each n; since is a -additive class and a -system. Let (Bn)1 be the M n=1 disjunction of (An)n1=1: Then (Bn)n N+ is a disjoint denumerable collection of 2 members of and De…nition 1.2.2(c) implies that 1 An = 1 Bn : M [n=1 [n=1 2 M

Theorem 1.2.2. Let be a -system and a -additive class such that : Then ( ) G: D G  D G  D

PROOF. Let be the intersection of all -additive classes containing : The class isM a -additive class and . In view of Theorem 1.2.1G is a -algebra,M if is a -systemG and  M in that  D case ( ) : Thus the Mtheorem follows if weM show that is a -system. G  M M Given C X; denote by C be the class of all D X such that D C .  D  \ 2 M 19

CLAIM 1. If C ; then C is a -additive class. 2 M D

PROOF OF CLAIM 1. First X C since X C = C : Moreover, if 2 D \ 2 M A; B C and A B; then A C;B C and 2 D  \ \ 2 M (B A) C = (B C) (A C) : n \ \ n \ 2 M

Accordingly from this, B A C : Finally, if (An)n N+ is a disjoint denumer- n 2 D 2 able collection of members of C , then (An C)n N+ is disjoint denumerable collection of members of andD \ 2 M

( n N+ An) C = n N+ (An C) : [ 2 \ [ 2 \ 2 M

Thus n N+ An C : [ 2 2 D

CLAIM 2. If A ; then : 2 G M  DA

PROOF OF CLAIM 2. If B ;A B : Thus B A: We have proved that and remembering2 G \ that2 G  Mis the intersection2 D of all G  DA M -additive classes containing Claim 2 follows since A is a -additive class. G D

To complete the proof of Theorem 1.2.2, observe that B A if and only 2 D if A B: By Claim 2, if A and B ; then B A that is A B: 2 D 2 G 2 M 2 D 2 D Thus B if B . Now the de…nition of implies that if G  D 2 M M M  DB B : The proof is almost …nished. In fact, if A; B then A B that2 is MA B : Theorem 1.2.2 now follows from Theorem2 M 1.2.1. 2 D \ 2 M

Theorem 1.2.3. Let  and  be positive measures on = ( ), where is a -system, and suppose (A) = (A) for every A M: G G (a) If  and  are probability measures, then  = :2 G (b) Suppose there exist En ; n N+; such that X = 1 En; 2 G 2 [n=1 20

E1 E2 :::; and  

(En) = (En) < ; all n N+: 1 2 Then  = :

PROOF. (a) Let = A ; (A) = (A) : D f 2 M g It is immediate that is a -additive class and Theorem 1.2.2 implies that = ( ) sinceD and is a -system. M G  D G  D G

(b) If (En) = (En) = 0 for all all n N+, then 2

(X) = lim (En) = 0 n !1 and, in a similar way, (X) = 0: Thus  = : If (En) = (En) > 0; set 1 1 n(A) = (A En) and n(A) = (A En) (En) \ (En) \ for each A : By Part (a)  = n and we get 2 M n

(A En) = (A En) \ \ for each A : Theorem 1.1.2(e) now proves that  = : 2 M

Theorem 1.2.3 implies that there is at most one positive measure de…ned n on n such that the measure of any open n-cell in R equals its volume. RNext suppose f : X Y and let A X and B Y: The image of A and the inverse image of!B are  

f(A) = y; y = f(x) for some x A f 2 g and 1 f (B) = x; f(x) B f 2 g 21 respectively. Note that 1 f (Y ) = X and 1 1 f (Y B) = X f (B): n n Moreover, if (Ai)i I is a collection of subsets of X and (Bi)i I is a collection of subsets of Y 2 2

f( i I Ai) = i I f(Ai) [ 2 [ 2 and 1 1 f ( i I Bi) = i I f (Bi): [ 2 [ 2 Given a class of subsets of Y; set E 1 1 f ( ) = f (B); B : E 2 E 1 If (Y; ) is a measurable space, it follows that the class f ( ) is a -algebra in X: NIf (X; ) is a measurable space N M 1 B (Y ); f (B) 2 P 2 M is a -algebra in Y . Thus, given a class of subsets of Y; E 1 1 (f ( )) = f (( )): E E

De…nition 1.2.3. Let (X; ) and (Y; ) be measurable spaces. The func- M N 1 tion f : X Y is said to be ( ; )-measurable if f ( ) . If we say that f :(X;! ) (Y; ) is measurableM N this means thatN f: X M Y is an ( ; )-measurableM ! function.N ! M N

Theorem 1.2.4. Let (X; ) and (Y; ) be measurable spaces and suppose generates : The functionM f : X YN is ( ; )-measurable if E N ! M N 1 f ( ) : E  M PROOF. The assumptions yield

1 (f ( )) : E  M 22

Since 1 1 1 (f ( )) = f (( )) = f ( ) E E N we are done.

Corollary 1.2.1. A function f : X R is ( ; )-measurable if and only 1 ! M R if the set f (] ; [) for all R: 1 2 M 2

If f : X Y is ( ; )-measurable and  is a positive measure on , the equation! M N M 1 (B) = (f (B)), B 2 N 1 de…nes a positive measure  on : We will write  = f ;  = f() or N  = f : The measure  is called the image measure of  under f and f is said to transport  to : Two ( ; )-measurable functions f : X Y and g : X Y are said to be -equimeasurableM N if f() = g(): ! As! an example, let a Rn and de…ne f(x) = x + a if x Rn: If B Rn; 2 2  1 f (B) = x; x + a B = B a: f 2 g 1 Thus f (B) is an open n-cell if B is, and Theorem 1.2.4 proves that f is ( n; n)-measurable. Now, granted the existence of volume measure vn; for R R every B n de…ne 2 R

(B) = f(vn)(B) = vn(B a):

Then (B) = vn(B) if B is an open n-cell and Theorem 1.2.3 implies that  = vn: We have thus proved the following

n Theorem 1.2.5. For any A n and x R 2 R 2

A + x n 2 R and vn(A + x) = vn(A): 23

Suppose ( ; ;P ) is a probability space. A measurable function  de…ned F on is called a random variable and the image measure P is called the probability law of : We sometimes write

() = P: L Here are two simple examples. If the range of a random variable  consists of n points S = s1; :::; sn 1 f g (n 1) and P = n cS;  is said to have a uniform distribution in S. Note that 1 P = n  :  n k=1 sk Suppose  > 0 is a constant. If a random variable  has its range in N and n   P = 1 e   n=0 n! n then  is said to have a Poisson distribution with parameter :

Exercises

1. Let f : X Y , A X; and B Y: Show that !   1 1 f(f (B)) B and f (f(A)) A:   2. Let (X; ) be a measurable space and suppose A X: Show that the function Mis ( ; )-measurable if and only if A : A M R 2 M

3. Suppose (X; ) is a measurable space and fn : X R; n N; a sequence of ( ; M)-measurable functions such that ! 2 M R

lim fn(x) exists and = f(x) R n !1 2 for each x X: Prove that f is ( ; )-measurable. 2 M R 24

4. Suppose f :(X; ) (Y; ) and g :(Y; ) (Z; ) are measurable. Prove that g f is (M ; !)-measurable.N N ! S  M S

n 5. Granted the existence of volume measure vn, show that vn(rA) = r vn(A) if r 0 and A n:  2 R

6. Let  be the counting measure on Z2 and f(x; y) = x; (x; y) Z2: The positive measure  is -…nite. Prove that the image measure f(2) is not a -…nite positive measure.

7. Let ;  : [0; ] be two positive measures such that (I) = (I) < for each openR! subinterval1 of R: Prove that  = : 1

n k 8. Let f : R R be continuous. Prove that f is ( n; k)-measurable. ! R R

9. Suppose  has a Poisson distribution with parameter : Show that P [2N] =  e cosh :

9. Find a -additive class which is not a -algebra.

1.3. Lebesgue Measure

Once the problem about the existence of volume measure is solved the exis- tence of the so called Lebesgue measure is simple to establish as will be seen in this section. We start with some concepts of general interest. If (X; ; ) is a positive measure space, the zero set  of  is, by M Z de…nition, the set at all A such that (A) = 0: An element of  is called a null set or -null set.2 M If Z

(A  and B A) B 2 Z  ) 2 M 25 the measure space (X; ; ) is said to be complete. In this case the measure  is also said to be complete.M The positive measure space (X; ; X ; ); f g where X = 0; 1 and  = 0; is not complete since X  and 0 = ; X : f g 2 Z f g 2 f g

Theorem 1.3.1 If (En)1 is a denumerable collection of members of  n=1 Z then 1 En : [n=1 2 Z

PROOF We have

0 ( 1 En) 1 (En) = 0  [n=1  n=1 which proves the result.

Granted the existence of linear measure v1 it follows from Theorem 1.3.1 that Q v1 since Q is countable and a v1 for each real number a. Suppose2 Z (X; ; ) is an arbitrary positivef g 2 Z measure space. It turns out that  is the restrictionM to of a complete measure. To see this suppose M is the class of all E X is such that there exist sets A; B such that M  2 M A E B and B A : It is obvious that X since : If   n 2 Z 2 M M  M E ; choose A; B such that A E B and B A : Then c2 Mc c c 2c M   n 2c Z B E A and A B = B A  and we conclude that E : If   n n 2 Z 2 M (Ei)1 is a denumerable collection of members of ; for each i there exist i=1 M sets Ai;Bi such that Ai E Bi and Bi Ai : But then 2 M   n 2 Z

1 Ai 1 Ei 1 Bi [i=1  [i=1  [i=1 where 1 Ai; 1 Bi . Moreover, ( 1 Bi) ( 1 Ai)  since [i=1 [i=1 2 M [i=1 n [i=1 2 Z

( 1 Bi) ( 1 Ai) 1 (Bi Ai): [i=1 n [i=1  [i=1 n

Thus 1 Ei and is a -algebra. [i=1 2 M M If E ; suppose Ai;Bi are such that Ai E Bi and Bi Ai 2 M 2 M   n 2  for i = 1; 2: Then for each i; (B1 B2) Ai  and Z \ n 2 Z

(B1 B2) = ((B1 B2) Ai) + (Ai) = (Ai): \ \ n

Thus the real numbers (A1) and (A2) are the same and we de…ne (E) to be equal to this common number. Note also that (B1) = (E): It is plain 26

that () = 0: If (Ei)i1=1 is a disjoint denumerable collection of members of ; for each i there exist sets Ai;Bi such that Ai Ei Bi and M 2 M   Bi Ai : From the above it follows that n 2 Z

( 1 Ei) = ( 1 Ai) = 1 (Ai) = 1 (Ei): [i=1 [i=1 n=1 n=1

We have proved that  is a positive measure on . If E  the M 2 Z de…nition of  shows that any set A E belongs to the -algebra : It follows that the measure  is complete and its restriction to equalsM: M The measure  is called the completion of  and is called the com- pletion of with respect to : M M

n De…nition 1.3.1 The completion of volume measure vn on R is called n Lebesgue measure on R and is denoted by mn: The completion of n with n R respect to vn is called the Lebesgue -algebra in R and is denoted by n: n R A member of the class n is called a Lebesgue measurable set in R or a Lebesgue set in Rn: A functionR f : Rn R is said to be Lebesgue measurable ! if it is ( ; )-measurable. Below, m1 is written m if this notation will not Rn R lead to misunderstanding. Furthermore, is written . R1 R

n Theorem 1.3.2. Suppose E and x R : Then E + x and 2 Rn 2 2 Rn mn(E + x) = mn(E):

PROOF. Choose A; B n such that A E B and B A v : Then, 2 R   n 2 Z n by Theorem 1.2.5, A + x; B + x n; vn(A + x) = vn(A) = mn(E); and 2 R (B + x) (A + x) = (B A) + x vn : Since A + x E + x B + x the theoremn is proved. n 2 Z  

The Lebesgue -algebra in Rn is very large and contains each set of interest in analysis and probability. In fact, in most cases, the -algebra n is su¢ ciently large but there are some exceptions. For example, if f : Rn RRn ! is continuous and A n, the image set f(A) need not belong to the class 2 R n (see e.g. the Dudley book [D]). To prove the existence of a subset of the realR line, which is not Lebesgue measurable we will use the so called Axiom of Choice. 27

Axiom of Choice. If (Ai)i I is a non-empty collection of non-empty sets, 2 there exists a function f : I i I Ai such that f(i) Ai for every i I: ![ 2 2 2

Let X and Y be sets. The set of all ordered pairs (x; y); where x X and y Y is denoted by X Y: An arbitrary subset R of X Y is called2 a 2   relation. If (x; y) R , we write x s y: A relation is said to be an equivalence relation on X if X2 = Y and

(i) x s x (re‡exivity) (ii) x s y y s x (symmetry) ) (iii) (x s y and y s z) x s z (transitivity) )

The equivalence class R(x) =def y; y s x : The de…nition of the equiv- f g alence relation s implies the following:

(a) x R(x) (b) R(2x) R(y) =  R(x) = R(y) \ 6 ) (c) x X R(x) = X: [ 2

An equivalence relation leads to a partition of X into a disjoint collection of subsets of X: Let X = 1 ; 1 and de…ne an equivalence relation for numbers x; y in X 2 2 by stating that x s y if x y is a rational number. By the Axiom of Choice it is possible to pick exactly one element from each equivalence class. Thus there exists a subset NL of X which contains exactly one element from each equivalence class.

If we assume that NL we get a contradiction as follows. Let (ri)i1=1 be an enumeration of the2 rational R numbers in [ 1; 1]. Then

X 1 (ri + NL)  [i=1 and it follows from Theorem 1.3.1 that ri + NL = m for some i: Thus, by 2 Z Theorem 1.3.2, NL = m: 2 Z 28

Now assume (ri + NL) (rj + NL) = : Then there exist a0; a00 NL \ 6 2 such that ri + a0 = rj + a00 or a0 a00 = rj ri: Hence a0 s a00 and it follows that a0 and a00 belong to the same equivalence class. But then a0 = a00: Thus ri = rj and we conclude that (ri + NL)i N+ is a disjoint enumeration of Lebesgue sets. Now, since 2 3 3 1 (ri + NL) ; [i=1  2 2   it follows that 3 m( 1 (ri + NL)) = 1 m(NL):  [i=1 n=1 But then NL m; which is a contradiction. Thus NL = : 2 Z 2 R

In the early 1970’Solovay [S] proved that it is consistent with the usual axioms of Set Theory, excluding the Axiom of Choice, that every subset of R is Lebesgue measurable. From the above we conclude that the Axiom of Choice implies the exis- tence of a subset of the set of real numbers which does not belong to the class : Interestingly enough, such an example can be given without any use of theR Axiom of Choice and follows naturally from the theory of analytic sets. The interested reader may consult the Dudley book [D] :

Exercises

1. (X; ; ) is a positive measure space. Prove or disprove: If A E B and (AM) = (B) then E belongs to the domain of the completion: 

2. Prove or disprove: If A and B are not Lebesgue measurable subsets of R; then A B is not Lebesgue measurable. [

3. Let (X; ; ) be a complete positive measure space and suppose A; B , where MB A is a -null set. Prove that E if A E B (stated2 M n 2 M   otherwise = ). M M 29

4. Suppose E R and E = . Show there is an " > 0 such that  2 R m(B A) " n  for all A; B such that A E B: 2 R  

5. Suppose (X; ; ) is a positive measure space and (Y; ) a measurable space. Furthermore,M suppose f : X Y is ( ; )-measurableN and let 1 1 ! M N  = f ; that is (B) = (f (B));B : Show that f is ( ; )- 2 N M N measurable, where denotes the completion of with respect to  and M M the completion of with respect to : N N

1.4. Carathéodory’sTheorem

In these notes we exhibit two famous approaches to Lebesgue measure: One is based on the Carathéodory Theorem, which we present in this section, and the other one, due to F. Riesz, is a representation theorem of positive linear functionals on spaces of continuous functions in terms of positive measures. The latter approach, is presented in Chapter 3. Both methods depend on topological concepts such as compactness.

De…nition 1.4.1. A function  : (X) [0; ] is said to be an outer measure if the following properties areP satis…ed:! 1

(i) () = 0: (ii) (A) (B) if A B:   (iii) for any denumerable collection (An)n1=1 of subsets of X

( 1 An) 1 (An): [n=1  n=1 30

Since E = (E A) (E Ac) \ [ \ an outer measure  satis…es the inequality

(E) (E A) + (E Ac):  \ \ If  is an outer measure on X we de…ne () as the set of all A X such that M  (E) = (E A) + (E Ac) for all E X \ \  or, what amounts to the same thing,

(E) (E A) + (E Ac) for all E X:  \ \  The next theorem is one of the most important in measure theory.

Theorem 1.4.1. (Carathéodory’s Theorem) Suppose  is an outer measure. The class () is a -algebra and the restriction of  to () is a complete measure.M M

PROOF. Clearly,  () and Ac () if A (): Moreover, if A; B () and E 2X; M 2 M 2 M 2 M  (E) = (E A) + (E Ac) \ \ = (E A B) + (E A Bc) \ \ \ \ +(E Ac B) + (E Ac Bc): \ \ \ \ But A B = (A B) (A Bc) (Ac B) [ \ [ \ [ \ and Ac Bc = (A B)c \ [ and we get (E) (E (A B)) + (E (A B)c):  \ [ \ [ It follows that A B () and we have proved that the class () is an algebra. Now if A;[ B 2 M() are disjoint M 2 M (A B) = ((A B) A) + ((A B) Ac) = (A) + (B) [ [ \ [ \ 31 and therefore the restriction of  to () is a content. M Next we prove that () is a -algebra. Let (Ai)i1=1 be a disjoint denu- merable collection of membersM of () and set for each n N M 2

Bn = 1 i nAi and B = i1=1Ai [   [

(here B0 = ). Then for any E X;  c (E Bn) = (E Bn An) + (E Bn A ) \ \ \ \ \ n

= (E An) + (E Bn 1) \ \ and, by induction, n (E Bn) =  (E Ai): \ i=1 \ But then c (E) = (E Bn) + (E B ) \ \ n n c  (E Ai) + (E B )  i=1 \ \ and letting n ; ! 1 c (E) 1 (E Ai) + (E B )  i=1 \ \ c ( 1 (E Ai)) + (E B )  [i=1 \ \ = (E B) + (E Bc) (E): \ \  All the inequalities in the last calculation must be equalities and we conclude that B () and, choosing E = B; results in 2 M

(B) = i1=1(Ai):

Thus () is a -algebra and the restriction of  to () is a positive measure.M M Finally we prove that the the restriction of  to () is a complete measure. Suppose B A () and (A) = 0: If E MX;  2 M  (E) (E B) + (E Bc) (E Bc) (E)  \ \  \  and so B (): The theorem is proved. 2 M 32

Exercises

1. Suppose i : (X) [0; ] ; i = 1; 2; are outer measures. Prove that P ! 1  = max(1; 2) is an outer measure.

2. Suppose a; b R and a = b: Set  = max(a; b): Prove that 2 6 a ; b = (): f g f g 2 M

1.5. Existence of Linear Measure

The purpose of this section is to show the existence of linear measure on R using the Carathéodory Theorem and a minimum of topology. First let us recall the de…nition of in…mum and supremum of a non- empty subset of the extended real line. Suppose A is a non-empty subset of [ ; ] = R ; : We de…ne x and x for all x [ 1; ]1: An element[ f1b 1g[ ; ] is called1 a majorant of A 1if x b for all2 x1A1and a minorant if2x 1b for1 all x A: The Supremum Axiom states that2 A possesses a least majorant, which2 is denoted by sup A. From this follows that if A is non-empty, then A possesses a greatest minorant, which is denoted by inf A. (Actually, the Supremum Axiom is a theorem in courses where time is spent on the de…nition of real numbers.)

Theorem 1.5.1. (The Heine-Borel Theorem; weak form) Let [a; b] be a closed bounded interval and (Ui)i I a collection of open sets such that 2

i I Ui [a; b] : [ 2  Then i J Ui [a; b] [ 2  for some …nite subset J of I: 33

PROOF. Let A be the set of all x [a; b] such that 2

i J Ui [a; x] [ 2  for some …nite subset J of I: Clearly, a A since a Ui for some i: Let 2 2 c = sup A: There exists an i0 such that c Ui : Let c ]a0; b0[ Ui ; where 2 0 2  0 a0 < b0: Furthermore, by the very de…nition of least upper bound, there exists a …nite set J such that

i J Ui [a; (a0 + c)=2] : [ 2  Hence

i J i0 Uk [a; (c + b0)=2] [ 2 [f g  and it follows that c A and c = b. The lemma is proved. 2

A subset K of R is called compact if for every family of open subsets Ui; i I; with i I Ui K we have i J Ui K for some …nite subset J of I: The2 Heine-Borel[ 2 Theorem shows that[ 2 a closed bounded interval is compact. If x; y R and E;F R; let 2  d(x; y) = x y j j be the distance between x and y; let

d(x; E) = inf d(x; u) u E 2 be the distance from x to E; and let

d(E;F ) = inf d(u; v) u E;v F 2 2 be the distance between E and F (here the in…mum of the emty set equals ): Note that for any u E; 1 2 d(x; u) d(x; y) + d(y; u)  and, hence d(x; E) d(x; y) + d(y; u)  34 and d(x; E) d(x; y) + d(y; E):  By interchanging the roles of x and y and assuming that E = ; we get 6 d(x; E) d(y; E) d(x; y): j j Note that if F R is closed and x = F; then d(x; F ) > 0: An outer measure  : (R) [0; 2] is called a metric outer measure if P ! 1 (A B) = (A) + (B) [ for all A; B (R) such that d(A; B) > 0: 2 P

Theorem 1.5.2. If  : (R) [0; ] is a metric outer measure, then (): P ! 1 R  M

PROOF. Let F (R) be closed. It is enough to show that F (): To this end we choose2 PE X with (E) < and prove that 2 M  1 (E) (E F ) + (E F c):  \ \ Let n 1 be an integer and de…ne 

c 1 An = x E F ; d(x; F ) : 2 \  n  

Note that An An+1 and  c E F = 1 An: \ [n=1 Moreover, since  is a metric outer measure

(E) ((E F ) An) = (E F ) + (An)  \ [ \ and, hence, proving c (E F ) = lim (An) n \ !1 we are done. 35

c Let Bn = An+1 A : It is readily seen that \ n 1 d(Bn+1;An)  n(n + 1) since if x Bn+1 and 2 1 d(x; y) < n(n + 1) then 1 1 1 d(y; F ) d(y; x) + d(x; F ) < + = :  n(n + 1) n + 1 n Now (A2k+1) (B2k A2k 1) = (B2k) + (A2k 1)  [ k :::  (B2i)   i=1 and in a similar way k (A2k) i=1(B2i 1):  But (An) (E) < and we conclude that  1

1 (Bi) < : i=1 1 We now use that c E F = An ( 1 Bi) \ [ [i=n to obtain c (E F ) (An) + 1 (Bi): \  i=n c Now, since (E F ) (An), \  c (E F ) = lim (An) n \ !1 and the theorem is proved.

PROOF OF THEOREM 1.1.1 IN ONE DIMENSION. Suppose  > 0: If A R; de…ne 

(A) = inf k1=1l(Ik) the in…mum being taken over all open intervals Ik with l(Ik) <  such that

A 1 Ik:  [k=1 36

Obviously, () = 0 and (A) (B) if A B: Suppose (An)n1=1 is a denumerable collection of subsets ofR and let " > 0: For each n there exist open intervals Ikn; k N+; such that l(Ikn) < ; 2

An 1 Ikn  [k=1 and n 1 l(Ikn) (An) + "2 : k=1  Then A =def 1 An 1 Ikn [n=1  [k;n=1 and 1 l(Ikn) 1 (An) + ": k;n=1  n=1 Thus (A) 1 (An) + "  n=1 and, since " > 0 is arbitrary,

(A) 1 (An):  n=1

It follows that  is an outer measure. If I is an open interval it is simple to see that

(I) l(I):  To prove the reverse inequality, choose a closed bounded interval J I: Now, if  I 1 Ik  [k=1 where each Ik is an open interval of l(Ik) < ; it follows from the Heine-Borel Theorem that n J Ik  [k=1 for some n: Hence n l(J)  l(Ik) 1 l(Ik)  k=1  k=1 and it follows that l(J) (I)  and, accordingly from this, l(I) (I):  37

Thus, if I is an open interval, then

(I) = l(I).

Note that   if 0 < 1 2: We de…ne 1  2 

0(A) = lim (A) if A R:  0 ! 

It obvious that 0 is an outer measure such that 0(I) =l(I); if I is an open interval. To complete the proof we show that 0 is a metric outer measure. To this end let A; B R and d(A; B) > 0: Suppose 0 <  < d(A; B) and 

A B 1 Ik [  [k=1 where each Ik is an open interval with l(Ik) < : Let

= k; Ik A =  f \ 6 g and = k; Ik B =  : f \ 6 g Then = ; \ A k Ik  [ 2 and B k Ik  [ 2 and it follows that

k1=1l(Ik) k l(Ik) + k l(Ik)  2 2

(A) + (B):  Thus (A B) (A) + (B) [  and by letting  0 we have !

0(A B) 0(A) + 0(B) [  and 0(A B) = 0(A) + 0(B): [ 38

Finally by applying the Carathéodory Theorem and Theorem 1.5.2 it follows that the restriction of 0 to equals v1. R

We end this section with some additional results of great interest.

Theorem 1.5.3. For any  > 0;  = 0: Moreover, if A R 

0(A) = inf k1=1l(Ik) the in…mum being taken over all open intervals Ik; k N+; such that 2 1 Ik A: [k=1 

PROOF. It follows from the de…nition of 0 that  0: To prove the  reverse inequality let A R and choose open intervals Ik; k N+; such that  2 1 Ik A: Then [k=1 

0(A) 0( 1 Ik) 1 0(Ik)  [k=1  k=1

= k1=1l(Ik): Hence 0(A) inf 1 l(Ik)  k=1 the in…mum being taken over all open intervals Ik; k N+; such that 2 k1=1Ik A: Thus 0(A) (A); which completes the proof of Theorem 1.5.3.[  

Theorem 1.5.4. If A R; 

0(A) = inf 0(U): U A U open

Moreover, if A (0); 2 M

0(A) = sup 0(K): K A K closed bounded 39

PROOF. If A U, 0(A) 0(U): Hence  

0(A) inf 0(U):  U A U open

Next let " > 0 be …xed and choose open intervals Ik; k N+; such that 2 1 Ik A and [k=1  1 l(Ik) 0(A) + " k=1  (here observe that it may happen that 0(A) = ). Then the set U =def 1 1 Ik is open and [k=1

0(U) 1 0(Ik) = 1 l(Ik) 0(A) + ":  k=1 k=1  Thus inf 0(U) 0(A) U A  U open and we have proved that

0(A) = inf 0(U): U A U open

If K A; 0(K) 0(A) and, accordingly from this,  

sup 0(K) 0(A): K A  K closed bounded

To prove the reverse inequality we …rst assume that A (0) is bounded. Let " > 0 be …xed and suppose J is a closed bounded interval2 M containing A: Then we know from the …rst part of Theorem 1.5.4 already proved that there exists an open set U J r A such that 

0(U) < 0(J r A) + ":

But then

0(J) 0(J r U) + 0(U) < 0(J r U) + 0(J r A) + "  and it follows that 0(A) " < 0(J U): n 40

Since J r U is a closed bounded set contained in A we conclude that

0(A) sup 0(K):  K A K closed bounded

If A (0) let An = A [ n; n] ; n N+: Then given " > 0 and n 2 M \ 2 2 N+; let Kn be a closed bounded subset of An such that 0(Kn) > 0(An) ": Clearly, there is no loss of generality to assume that K1 K2 K3 ::: and by letting n tend to plus in…nity we get   

lim 0(Kn) 0(A) ": n !1  Hence 0(A) = sup 0(K): K A K compact and Theorem 1.5.4 is completely proved.

Theorem 1.5.5. Lebesgue measure m1 equals the restriction of 0 to (0): M

PROOF. Recall that linear measure v1 equals the restriction of 0 to and R m1 =v 1: First suppose E and choose A; B such that A E B 2 R 2 R   and BrA v1 : But then 0(ErA) = 0 and E = A (ErA) (0) since 2 Z [ 2 M the Carathéodory Theorem gives us a complete measure. Hence m1(E) = v1(A) = 0(E). Conversely suppose E (0): We will prove that E and m1(E) = 2 M 2 R 0(E). First assume that E is bounded. Then for each positive integer n there exist open Un E and closed bounded Kn E such that   n 0(Un) < 0(E) + 2 and

n 0(Kn) > 0(E) 2 : The de…nitions yield A = 1Kn;B = 1Un and [1 \1 2 R

0(E) = 0(A) = 0(B) = v1(A) = v1(B) = m1(E): 41

It follows that E and 0(E) = m1(E): 2 R In the general case set En = E [ n; n] ; n N+: Then from the above \ 2 En and 0(En) = m1(En) for each n and Theorem 1.5.5 follows by letting2 Rn go to in…nity.

The Carathéodory Theorem can be used to show the existence of volume measure on Rn but we do not go into this here since its existence follows by several other means below. By passing, let us note that the Carathéodory Theorem is very e¢ cient to prove the existence of so called Haussdor¤ mea- sures (see e.g. [F ]); which are of great interest in Geometric Measure Theory.

Exercises

1. Prove that a subset K of R is compact if and only if K is closed and bounded.

2. Suppose A and m(A) < : Set f(x) = m(A ] ; x]); x R: Prove that f is2 continuous. R 1 \ 1 2

3 3. Suppose A m and B = x ; x A : Prove that B m: 2 Z f 2 g 2 Z

4. Let A be the set of all real numbers x such that p 1 x j q j q3 for in…nitely many pairs of positive integers p and q: Prove that A m: 2 Z

5. Let I1; :::; In be open subintervals of R such that

n Q [0; 1] Ik: \  [k=1 n Prove that  m(Ik) 1: k=1  42

6. If E and m(E) > 0; for every ]0; 1[ there is an interval I such 2 R 2 that m(E I) > m(I). (Hint: m(E) = inf 1 m(Ik), where the in…mum \ k=1 is taken over all intervals such that 1 Ik E:) [k=1 

7. If E and m(E) > 0; then the set E E = x y; x; y E contains an open2 non-emptyR interval centred at 0:(Hint:f Take an interval2 g I with m(E I) 3 m(I): Set " = 1 m(I): If x "; then (E I) (x+(E I)) = :) \  4 2 j j \ \ \ 6

8. Let  be the restriction of the positive measure k1=1R; 1 to : Prove that k R inf (U) > (A) U A U open if A = 0 : f g

1.6. Positive Measures Induced by Increasing Right Continuous Functions

Suppose F : R [0; [ is a right continuous increasing function such that ! 1 lim F (x) = 0: x !1 Set L = lim F (x): x !1 We will prove that there exists a unique positive measure  : [0;L] such that R! (] ; x]) = F (x); x R: 1 2 This measure will often be denoted by F : The special case L = 0 is trivial so let us assume L > 0 and introduce

H(y) = inf x R; F (x) y ; 0 < y < L: f 2  g The de…nition implies that the function H increases. 43

Suppose a is a …xed real number. We claim that y ]0;L[; H(y) a = ]0;F (a)] ]0;L[ : f 2  g \ To prove this …rst suppose that y ]0;L[ and H(y) a: Then to each 2 n  positive integer n; there is an xn [H(y);H(y) + 2 [ such that F (xn) y: 2  Then xn H(y) as n and we obtain that F (H(y)) y since F is right continuous.! Thus, remembering! 1 that F increases, F (a)  y: On the other hand, if 0 < y < L and 0 < y F (a); then, by the very de…nition of H(y); H(y) a:  We now de…ne  = H(v1 ]0;L[) j and get (] ; x]) = F (x); x R: 1 2 The uniqueness follows at once from Theorem 1.2.3. Note that the measure  is a probability measure if L = 1:

Example 1.1.1. If 0 if x < 0 F (x) = 1 if x 0   then  is the Dirac measure at the point 0 restricted to . F R

Example 1.1.2. If x t2 dt F (x) = e 2 (a Riemann integral) p2 Z1 then F is called the standard Gaussian measure on R:

Exercises

1. Suppose F : R R is a right continuous increasing function. Prove that there is a unique positive! measure  on such that R (]a; x]) = F (x) F (a); if a; x R and a < x: 2 44

2. Suppose F : R R is an increasing function. Prove that the set of all discontinuity points! of F is at most denumerable. (Hint: Assume …rst that F is bounded and prove that the set of all points x R such that F (x+) F (x ) > " is …nite for every " > 0.) 2

3. Suppose  is a -…nite positive measure on : Prove that the set of all x R such that ( x ) > 0 is at most denumerable.R 2 f g

4. Suppose  is a -…nite positive measure on n: Prove that there is an at most denumerable set of hyperplanes of the typeR

xk = c (k = 1; :::; n; c R) 2 with positive -measure.

5. Construct an increasing function f : R R such that the set of discon- tinuity points of f equals Q. ! 45 CHAPTER 2 INTEGRATION

Introduction

In this chapter Lebesgue integration in abstract positive measure spaces is introduced. A series of famous theorems and lemmas will be proved.

2.1. Integration of Functions with Values in [0; ] 1

Recall that [0; ] = [0; [ : A subinterval of [0; ] is de…ned in the 1 1 [ f1g 1 natural way. We denote by 0; the -algebra generated by all subintervals of [0; ] : The class of all intervalsR 1 of the type ] ; ] ; 0 < ; (or of 1 1  1 the type [ ; ] ; 0 < ) generates the -algebra 0; and we get the following 1  1 R 1

Theorem 2.1.1. Let (X; ) be a measurable space and suppose f : X [0; ] : M ! 1 1 (a) The function f is ( ; 0; )-measurable if f (] ; ]) for every 0 < : M R 1 1 2 M  1 1 (b) The function f is ( ; 0; )-measurable if f ([ ; ]) for every 0 < : M R 1 1 2 M  1

Note that the set f > for all real if f is ( ; 0; )-measurable. f g 2 M M R 1 If f; g : X [0; ] are ( ; 0; )-measurable, then min(f; g); max(f; g), ! 1 M R 1 and f + g are ( ; 0; )-measurable, since, for each [0; [ ; M R 1 2 1 min(f; g) (f and g )  ,   max(f; g) (f or g )  ,   46 and f + g > = ( f > q g > q ): f g f g \ f g q Q [2 Given functions fn : X [0; ] ; n = 1; 2; :::; f = supn 1 fn is de…ned by the equation ! 1  f(x) = sup fn(x); n = 1; 2; ::: : f g Note that 1 1 f (] ; ]) = 1 f (] ; ]) 1 [n=1 n 1 for every real 0 and, accordingly from this, the function supn 1 fn is   ( ; 0; )-measurable if each fn is ( ; 0; )-measurable. Moreover, f = M R 1 M R 1 infn 1 fn is given by 

f(x) = inf fn(x); n = 1; 2; ::: : f g Since 1 1 f ([0; [) = 1 f ([0; [) [n=1 n for every real 0 we conclude that the function f = infn 1 fn is ( ; 0; )-   M R 1 measurable if each fn is ( ; 0; )-measurable. Below we write M R 1 fn f " if fn; n = 1; 2; :::; and f are functions from X into [0; ] such that fn fn+1 1  for each n and fn(x) f(x) for each x X as n : ! 2 ! 1 An ( ; 0; )-measurable function ' : X [0; ] is called a simple measurableM functionR 1 if '(X) is a …nite subset of![0; [1: If it is neccessary to be more precise, we say that ' is a simple -measurable1 function. M

Theorem 2.1.2. Let f : X [0; ] be ( ; 0; )-measurable. There exist ! 1 M R 1 simple measurable functions ' ; n N+; on X such that ' f : n 2 n "

PROOF. Given n N+, set 2

1 i 1 i Ein = f ( ; ); i N+ 2n 2n 2   47 and 1 i 1  = + 1 : n n Ein f ( ) 2 f1g i=1 1 X It is obvious that n f and that n n+1: Now set 'n = min(n; n) and we are done.  

Let (X; ; ) be a positive measure space and ' : X [0; [ a simple M ! 1 measurable function: If 1; :::; n are the distinct values of the simple function 1 ', and if Ei = ' ( i ); i = 1; :::; n; then f g n ' = i=1 i Ei : Furthermore, if A we de…ne 2 M

n n Ei (A) = 'd =  i(Ei A) =  i (A): i=1 \ k=1 ZA n Note that this formula still holds if (Ei)1 is a measurable partition of X and ' = i on Ei for each i = 1; :::; n: Clearly,  is a positive measure since each term in the right side is a positive measure as a function of A. Note that

'd = 'd if 0 <  1 ZA ZA and 'd = a(A) ZA if a [0; [ and ' is a simple measurable function such that ' = a on A: If2 is1 another simple measurable function and ' ;  'd d:  ZA ZA 1 To see this, let 1; :::; p be the distinct values of and Fj = ( j ); j = 1; :::; p: Now, putting Bij = Ei Fj; \ 

'd = ( ij(A Bij)) [ \ ZA

= ij(A Bij) = ij 'd = ij id \ A Bij A Bij Z \ Z \ 48

ij jd = d:  A Bij A Z \ Z In a similar way one proves that

(' + )d = 'd + d: ZA ZA ZA From the above it follows that

n ' Ad = i=1 i Ei Ad \ ZA ZA

n n = i=1 i Ei Ad = i=1 i(Ei A) \ \ ZA and

' Ad = 'd: ZA ZA If f : X [0; ] is an ( ; 0; )-measurable function and A , we de…ne ! 1 M R 1 2 M

fd = sup 'd; 0 ' f; ' simple measurable   ZA ZA 

= sup 'd; 0 ' f; ' simple measurable and ' = 0 on Ac :   ZA  The left member in this equation is called the Lebesgue integral of f over A with respect to the measure : Sometimes we also speek of the -integral of f over A: The two de…nitions of the -integral of a simple measurable function ' : X [0; [ over A agree. ! 1 From now on in this section, an ( ; 0; )-measurable function f : X [0; ] is simply called measurable. M R 1 ! 1The following properties are immediate consequences of the de…nitions. The functions and sets occurring in the equations are assumed to be mea- surable.

(a) If f; g 0 and f g on A; then fd gd:   A  A R R 49

(b) A fd = X Afd: R R (c) If f 0 and [0; [, then fd = fd:  2 1 A A R R

(d) fd = 0 if f = 0 and (A) = : A 1 R

(e) fd = 0 if f = and (A) = 0: A 1 R

If f : X [0; ] is measurable and 0 < < ; then f f 1([ ; ]) = ! 1 1  1 f and f  g

fd f d = f d:  f  g f  g ZX ZX ZX This proves the so called Markov Inequality

1 (f ) fd   ZX where we write (f ) instead of the more precise expression ( f ):  f  g

Example 2.1.1. Suppose f : X [0; ] is measurable and ! 1 fd < : 1 ZX We claim that 1 f = = f ( ) : f 1g f1g 2 Z To prove this we use the Markov Inequality and have

1 (f = ) (f ) fd 1    ZX 50 for each ]0; [ : Thus (f = ) = 0: 2 1 1

Example 2.1.2. Suppose f : X [0; ] is measurable and ! 1 fd = 0: ZX We claim that 1 f > 0 = f (]0; ]) : f g 1 2 Z To see this, note that

1 1 1 f (]0; ]) = 1 f ( ; ) 1 [n=1 n 1  

Furthermore, for every …xed n N+; the Markov Inequality yields 2 1 (f > ) n fd = 0 n  ZX and we get f > 0  since a countable union of null sets is a null set. f g 2 Z

We now come to one of the most important results in the theory.

Theorem 2.1.3. (Monotone Convergence Theorem) Let fn : X [0; ] , n = 1; 2; 3; ::::; be a sequence of measurable functions and suppose! 1 that fn f; that is 0 f1 f2 ::: and "   

fn(x) f(x) as n , for every x X: ! ! 1 2 Then f is measurable and

fnd fd as n : ! ! 1 ZX ZX

PROOF. The function f is measurable since f = supn 1 fn:  51

The inequalities fn fn+1 f yield X fnd X fn+1d X fd and we conclude that there exists an [0; ] such that  2 1R R R

fnd as n ! ! 1 ZX and fd:  ZX To prove the reverse inequality, let ' be any simple measurable function such that 0 ' f, let 0 <  < 1 be a constant, and de…ne, for …xed   n N+; 2 An = x X; fn(x) '(x) : f 2  g If 1; :::; p are the distinct values of ';

p An = ( x X; fn(x)  k ' = k ) [k=1 f 2  g \ f g and it follows that An is measurable. Clearly, A1 A2 ::: . Moreover, if   f(x) = 0; then x A1 and if f(x) > 0; then '(x) < f(x) and x An for 2 2 all su¢ ciently large n. Thus 1 An = X: Now [n=1

fnd  'd   ZAn ZAn and we get  'd  ZX since the map A 'd is a positive measure on : By letting  1, ! A M " R 'd  ZX and, hence fd:  ZX The theorem follows.

Theorem 2.1.4. (a) Let f; g : X [0; ] be measurable functions. Then ! 1 (f + g)d = fd + gd: ZX ZX ZX 52

(b) (Beppo Levi’s Theorem) If fk : X [0; ] , k = 1; 2; ::: are mea- surable, ! 1

k1=1fkd = k1=1 fkd ZX ZX

PROOF. (a) Let ('n)n1=1 and ( n)n1=1 be sequences of simple and measurable functions such that 0 ' f and 0 g: We proved above that  n "  n "

('n + n)d = 'nd + nd ZX ZX ZX and, by letting n ; Part (a) follows from the Monotone Convergence Theorem. ! 1

(b) Part (a) and induction imply that

n n k=1fkd = k=1 fkd ZX ZX and the result follows from monotone convergence.

Theorem 2.1.5. Suppose w : X [0; ] is a measurable function and de…ne ! 1 (A) = wd; A : 2 M ZA Then  is a positive measure and

fd = fwd; A 2 M ZA ZA for every measurable function f : X [0; ] : ! 1

PROOF. Clearly, () = 0. Suppose (Ek)k1=1 is a disjoint denumerable col- lection of members of and set E = 1 Ek: Then M [k=1

( 1 Ek) = wd = wd = 1 wd [k=1 E k=1 Ek ZE ZX ZX 53 where, by the Beppo Levi Theorem, the right member equals

k1=1 Ek wd = k1=1 wd = k1=1(Ek): ZX ZEk This proves that  is a positive measure. Let A . To prove the last part in Theorem 2.1.5 we introduce the class of all2 M measurable functions f : X [0; ] such that C ! 1 fd = fwd: ZA ZA The indicator function of a measurable set belongs to and from this we conclude that every simple measurable function belongs toC : Furthermore, if C fn ; n N; and fn f ; the Monotone Convergence Theorem proves that f 2 C: Thus2 in view of" Theorem 2.1.2 the class contains every measurable function2 C f : X [0; ] : This completes the proofC of Theorem 2.1.5. ! 1

The measure  in Theorem 2.1.5 is written

 = w or d = wd:

Let ( n)1 be a sequence in [ ; ] : First put = inf k; k+1; k+2; ::: n=1 1 1 k f g and = sup 1; 2; 3; :: = limn n: We call the lower limit of ( n)n1=1 and write f g !1 = lim inf n: n !1 Note that = lim n n !1 if the limit exists. Now put = sup k; k+1; k+2; ::: and = inf ; ; ; :: = k f g f 1 2 3 g limn n: We call the upper limit of ( n)n1=1 and write !1

= lim sup n: n !1 Note that = lim n n !1 54 if the limit exists. Given measurable functions fn : X [0; ] ; n = 1; 2; :::; the function ! 1 lim infn fn is measurable. In particular, if !1

f(x) = lim fn(x) n !1 exists for every x X; then f is measurable. 2

Theorem 2.1.6. (Fatou’sLemma) If fn : X [0; ] ; n = 1; 2; :::; are measurable ! 1

lim inf fnd lim inf fnd: n  n ZX !1 !1 ZX PROOF. Introduce gk = inf fn: n k  The de…nition gives that gk lim infn fn and, moreover, " !1

gkd fnd; n k   ZX ZX and

gkd inf fnd:  n k ZX  ZX The Fatou Lemma now follows by monotone convergence.

Below we often write f(x)d(x) ZE instead of fd: ZE

Example 2.1.3. Suppose a R and f :(R; ) ([0; ] ; 0; ) is measurable. We claim that 2 R ! 1 R 1

f(x + a)dm(x) = f(x)dm(x): ZR ZR 55

First if f = ; where A , A 2 R

f(x + a)dm(x) = A a(x)dm(x) = m(A a) = ZR ZR

m(A) = f(x)dm(x): ZR Next it is clear that the relation we want to prove is true for simple mea- surable functions and …nally, we use the Monotone Convergence Theorem to deduce the general case.

Example 2.1.3, Suppose 1 an is a positive convergent series and let E be the set of all x [0; 1] such that 2 p a min x < n p 0;:::;n j n j n 2f g for in…nitely many n N+: We claim that E is a Lebesgue null set. 2 To prove this claim for …xed n N+; let En be the set of all x [0; 1] such that 2 2 p a min x < n : p N+ n n 2 j j Then if B(x; r) = ]x r; x + r[ ; x [0; 1] ; r > 0; we have 2 n p a E B( ; n ) n n n  p=0 [ and 2an m(En) (n + 1) 4an:  n  Hence 1 m(En) < 1 1 X and by the Beppo Levi theorem

1 1

En dm < : 0 1 1 Z X 56

Accordingly from this the set

1

F = x [0; 1] ; En (x) < ( 2 1 1) X is of Lebesgue measure 1. Since E [0; 1] F we have m(E) = 0:  n

Exercises

1. Suppose fn : X [0; ] ; n = 1; 2; :::; are measurable and ! 1

1 (fn > 1) < : n=1 1 Prove that

lim sup fn > 1  . n 2 Z  !1 

2 2. Set fn = n 0; 1 ; n N+: Prove that [ n ] 2

lim inf fndm = 0 < = lim inf fndm n 1 n ZR !1 !1 ZR (the inequality in the Fatou Lemma may be strict).

3. Suppose f :(R; ) ([0; ] ; 0; ) is measurable and set R ! 1 R 1

g(x) = 1 f(x + k); x R: k=1 2 Show that

gdm < if and only if f > 0 m: 1 f g 2 Z ZR

4. Let (X; ; ) be a positive measure space and f : X [0; ] an M ! 1 ( ; 0; )-measurable function such that M R 1 f(X) N  57 and fd < : 1 ZX For every t 0, set  F (t) = (f > t) and G(t) = (f t):  Prove that

fd = n1=0F (n) = n1=1G(n): ZX

2.2. Integration of Functions with Arbitrary Sign

As usual suppose (X; ; ) is a positive measure space. In this section when we speak of a measurableM function f : X R it is understood that f is an ( ; )-measurable function, if not otherwise! stated. If f; g : X R are measurable,M R the sum f + g is measurable since !

f + g > = ( f > q g > q ) f g f g \ f g q Q [2 for each real : Besides the function f and the di¤erence f g are mea- surable. It follows that a function f :X R is measurable if and only if + ! the functions f = max(0; f) and f = max(0; f) are measurable since + f = f f : We write f 1() if f : X R is measurable and 2 L !

f d < j j 1 ZX and in this case we de…ne

+ fd = f d f d: ZX ZX ZX Note that fd f d j j j j ZX ZX 58

+ since f = f + f : Moreover, if E we de…ne j j 2 M

+ fd = f d f d ZE ZE ZE and it follows that

fd = Efd: ZE ZX Note that fd = 0 if (E) = 0: ZE Sometimes we write f(x)d(x) ZE instead of fd: ZE If f; g 1(); setting h = f + g; 2 L h d f d + g d < j j  j j j j 1 ZX ZX ZX and it follows that h + g 1(): Moreover, 2 L + + + h h = f f + g g and the equation + + + h + f + g = f + g + h gives

+ + + h d + f d + gd = f d + g d + hd: ZX ZX ZX ZX ZX ZX Thus hd = fd + gd: ZX ZX ZX Moreover, fd = fd ZX ZX 59 for each real : The case 0 follows from (c) in Section 2.1. The case  + + = 1 is also simple since ( f) = f and ( f) = f :

Theorem 2.2.1. (Lebesgue’s Dominated Convergence Theorem) Suppose fn : X R; n = 1; 2; :::; are measurable and !

f(x) = lim fn(x) n !1 exists for every x X: Moreover, suppose there exists a function g 1() such that 2 2 L fn(x) g(x); all x X and n N+: j j 2 2 Then f 1(), 2 L lim fn f d = 0 n j j !1 ZX and

lim fnd = fd n !1 ZX ZX Proof. Since f g, the function f is real-valued and measurable since + j j f and f are measurable. Note here that

f (x) = lim fn(x); all x X: n !1 2

We now apply the Fatous Lemma to the functions 2g fn f ; n = 1; 2; :::; and have j j

2gd lim inf (2g fn f )d  n j j ZX !1 ZX

= 2gd lim sup fn f d: X n X j j Z !1 Z But X 2gd is …nite and we get R lim fn f d = 0: n j j !1 ZX Since

fnd fd = (f fn)d f fn d j j j j j j ZX ZX ZX ZX 60 the last part in Theorem 2.2.1 follows from the …rst part. The theorem is proved.

Example 2.2.1. Suppose f :]a; b[ X R is a function such that f(t; ) 1() for each t ]a; b[ and, moreover, assume! @f exists and
2 L 2 @t @f (t; x) g(x) for all (t; x) ]a; b[ X j @t j 2  where g 1(). Set 2 L F (t) = f(t; x)d(x) if t ]a; b[ : 2 ZX We claim that F is di¤erentiable and @f F 0(t) = (t; x)d(x): @t ZX

To see this let t ]a; b[ be …xed and choose a sequence (tn)n1=1 in ]a; b[ t which converges 2 to t : De…ne n f g 

f(tn; x) f(t ; x) hn(x) =  if x X: tn t 2  Here each hn is measurable and @f lim hn(x) = (t ; x) for all x X: n @t  !1 2

Furthermore, for each …xed n and x there is a  n;x ]tn; t [ such that hn(x) = @f 2  ( n;x; x) and we conclude that hn(x) g(x) for every x X: Since @t j j 2

F (tn) F (t )  = hn(x)d(x) tn t X  Z the claim above now follows from the Lebesgue Dominated Convergence The- orem.

Suppose S(x) is a statement, which depends on x X: We will say that 2 S(x) holds almost (or -almost) everywhere if there exists an N  such 2 Z 61 that S(x) holds at every point of X N: In this case we write ”S holds a.e. ” or ”S holds a.e. []”. Sometimesn we prefer to write ”S(x) holds a.e.” or ”S(x) holds a.e. []”: If the underlying measure space is a probability space, we often say ”almost surely”instead of almost everywhere. The term ”almost surely”is abbreviated a.s. Suppose f : X R; is an ( ; )-measurable functions and g : X R: ! M R ! If f = g a.e. [] there exists an N  such that f(x) = g(x) for every 2 Z x X N: We claim that g is ( ; )-measurable. To see this let R and2 usen that M R 2

g > = [ f > (X N)] [ g > N] : f g f g \ n [ f g \ Now if we de…ne A = f > (X N) f g \ n the set A and 2 M A g > A N:  f g  [ Accordingly from this g > and g is ( ; )-measurable since is an arbitrary real number.f g 2 M M R Next suppose fn : X R; n N+; is a sequence of ( ; )-measurable functions and f : X R!a function.2 Recall if M R !

lim fn(x) = f(x); all x X n !1 2 then f is ( ; )-measurable since M R 1 f > = k;l N+ n k fn > + l ; all R: f g [ 2 \  2 If we only assume that 

lim fn(x) = f(x); a.e. [] n !1 then f need not be ( ; )-measurable but f is ( ; )-measurable. To M R M R see this suppose N  and 2 Z

lim fn(x) = f(x); all x X N: n !1 2 n Then

lim X N (x)fn(x) = X N (x)f(x) n n n !1 62

and it follows that the function X N f is ( ; )-measurable. Since f = n M R X N f a.e. [] it follows that f is ( ; )-measurable. The next example showsn that f need not be ( ; )-measurable.M R M R

Example 2.2.2. Let X = 0; 1; 2 ; = ; 0 ; 1; 2 ;X ; and (A) = f g M f f g f g g A(0);A : Set fn = 1;2 ; n N+; and f(x) = x; x X: Then each 2 M f g 2 2 fn is ( ; )-measurable and M R

lim fn(x) = f(x) a.e. [] n !1 since x X; lim fn(x) = f(x) = 0; 1 n 2 !1 f g and N = 1; 2 is an-null set. The function f ois not ( ; )-measurable. f g M R

Suppose f; g 1(): The functions f and g are equal almost everywhere 2 L with respect to  if and only if f = g : This is easily seen to be an equivalence relation and the set off all6 equivalenceg 2 Z classes is denoted by L1(): Moreover, if f = g a.e. [] ; then

fd = gd ZX ZX since

fd = fd + fd = fd = gd X f=g f=g f=g f=g Z Zf g Zf 6 g Zf g Zf g and, in a similar way, gd = gd: X f=g Z Zf g Below we consider the elements of L1() members of 1() and two members of L1() are identi…ed if they are equal a.e. [] : FromL this convention it is straight-forward to de…ne f + g and f for all f; g L1() and R: Moreover, we get 2 2

(f + g)d = fd + gd if f; g L1() 2 ZX ZX ZX 63 and fd = fd if f L1() and R: 2 2 ZX ZX Next we give two theorems where exceptional null sets enter. The …rst one is a mild variant of Theorem 2.2.1 and needs no proof.

Theorem 2.2.2. Suppose (X; ; ) is a positive complete measure space M and let fn : X R; n N+; be measurable functions such that ! 2

sup fn(x) g(x) a.e. [] n N+ j j 2 where g L1(): Moreover, suppose f : X R is a function and 2 !

f(x) = lim fn(x) a.e. [] : n !1 Then, f L1(), 2 lim fn f d = 0 n j j !1 ZX and

lim fnd = fd: n !1 ZX ZX

Theorem 2.2.3. Suppose (X; ; ) is a positive measure space. M (a) If f :(X; ) ([0; ] ; 0; ) is measurable there exists a measur- M ! 1 R 1 able function g :(X; ) ([0; ] ; 0; ) such that f = g a.e. [] : M ! 1 R 1 (b) If f :(X; ) (R; ) is measurable there exists a measurable function g :(X; )M (R!; ) suchR that f = g a.e. [] : M ! R

+ PROOF. Since f = f f it is enough to prove Part (a): There exist simple -measurable functions ' ; n N+; such that 0 ' f: For each …xed M n 2  n " n suppose 1n; :::; knn are the distinct values of 'n and choose for each …xed 1 1 i = 1; :::; kn a set Ain ' ( in ) such that Ain and ' ( in) Ain  n f g 2 M n n . Set 2 Z kn n = i=1 in Ain : 64

kn Clearly n(x) f(x) if x E =def n1=1( i=1Ain) and (X E) = 0: We now de…ne g(x") = f(x); if2x E; and\ g(x[) = 0 if x X E:nThe theorem is proved. 2 2 n

Exercises

1. Suppose f and g are real-valued measurable functions. Prove that f 2 and fg are measurable functions.

2. Suppose f L1(): Prove that 2 lim f d = 0: f j j !1 Zj j

(Here f means f .) j j fj j g R R 3. Suppose f L1(): Prove that to each " > 0 there exists a  > 0 such that 2 f d < " j j ZE whenever (E) < :

4. Let (fn)1 be a sequence of ( ; )-measurable functions. Prove that n=1 M R the set of all x R such that the sequence (fn(x))n1=1 converges to a real limit belongs to 2 : M

5. Let (X; ; ) be a positive measure space such that (A) = 0 or for every A M: RShow that f L1() if and only if f(x) = 0 a.e. [] : 1 2 M 2

6. Let (X; ; ) be a positive measure space and suppose f and g are non-negativeM measurable functions such that

fd = gd; all A : 2 M ZA ZA 65

(a) Prove that f = g a.e. [] if  is -…nite. (b) Prove that the conclusion in Part (a) may fail if  is not -…nite.

7. Let (X; ; ) be a …nite positive measure space and suppose the functions M fn : X R; n = 1; 2; :::; are measurable. Show that there is a sequence ! ( n)n1=1 of positive real numbers such that

lim nfn = 0 a.e. [] : n !1

8. Let (X; ; ) be a positive measure space and let fn : X R; n = 1; 2; :::; be a sequenceM in L1() which converges to f a.e. [] as n! : Suppose f L1() and ! 1 2 lim fn d = f d: n j j j j !1 ZX ZX Show that

lim fn f d = 0: n j j !1 ZX

9. Let (X; ; ) be a …nite positive measure space and suppose f L1() is a boundedM function such that 2

f 2d = f 3d = f 4d: ZX ZX ZX Prove that f = for an appropriate A : A 2 M

10. Let (X; ; ) be a …nite positive measure space and f : X R a measurable function.M Prove that f L1() if and only if ! 2

1 ( f k) < : k=1 j j 1

1 11. Suppose f L (m): Prove that the series k1= f(x + k) converges for m-almost all x:2 1 66

12. a) Suppose f : R [0; [ is Lebesgue measurable and R fdm < : Prove that ! 1 1 lim m(f ) = 0: R

!1  b) Find a Lebesgue measurable function f : R [0; [ such that f = L1(m); m(f > 0) < ; and ! 1 2 1 lim m(f ) = 0:

!1 

13. (a) Suppose is an -algebra of subsets of X and  a positive measure M on with (X) < : Let A1; :::; An : Show that M 1 2 M

A1 A2 ::: An = 1 (1 A1 ) ::: (1 An ) [ [ [

and conclude that

(A1 A2 ::: An) = (Ai) (Ai Ai ) [ [ [ 1 \ 2 1 i n 1 i1bijections) x : 1; 2; :::; n 1 f g ! 1; 2; :::; n and let  = cX . A random variable  : X has the uniform f g n! ! distribution in X or, stated otherwise, the image measure P equals : Find the probability that  has a …xed point, that is …nd

P [(i) = i for some i 1; 2; :::; n ] : 2 f g

(Hint: Set Ai = x X; x(i) = i ; i = 1; :::; n; and note that the probability f 2 g in question equals (A1 A2 ::: An).) [ [ [

14. Let (X; ; ) be a positive measure space and f : X R an ( , )- measurable function.M Moreover, for each t > 1; let ! M R

1 a(t) = tn(tn f < tn+1): n= j j X1 67

Show that lim a(t) = f d: t 1+ j j ! ZX

15. Let (X; ; ) be a positive measure space and fn:X R; n N+; a sequence of measurableM functions such that ! 2

2 2 lim sup n ( fn n ) < : n j j 1 !1

Prove that the series 1 fn(x) converges for -almost all x X: n=1 2 P

16. Let (X; ; ) be a positive measure space and f:X R a measurable function. Furthermore,M suppose there are strictly positive! constants B and C such that 2 af a C e d Be 2 if a R:  2 ZX Prove that t2 ( f t) 2Be 2C if t > 0: j j 

2.3 Comparison of Riemann and Lebesgue Integrals

In this section we will show that the Lebesgue integral is a natural general- ization of the Riemann integral. For short, the discussion is restricted to a closed and bounded interval. Let [a; b] be a closed and bounded interval and suppose f :[a; b] R is a bounded function. For any partition !

: a = x0 < x1 < ::: < xn = b of [a; b] de…ne n S
f = i=1( sup f)(xi xi 1) ]xi 1;xi] and n s
f = i=1( inf f)(xi xi 1): ]xi 1;xi] 68

The function f is Riemann integrable if

inf S
f = sup s
f

b and the Riemann integral a f(x)dx is, by de…nition, equal to this common value. R Below an (( )[a;b]; )-measurable function is simply called Lebesgue R R measurable. Furthermore, we write m instead of m [a;b]: j

Theorem 2.3.1. A bounded function f :[a; b] R is Riemann integrable if and only if the set of discontinuity points of!f is a Lebesgue null set. Moreover, if the set of discontinuity points of f is a Lebesgue null set, then f is Lebesgue measurable and

b f(x)dx = fdm: Za Z[a;b]

PROOF. A partition
0 : a = x0 < x0 < ::: < x0 = b is a re…nement of a 0 1 n0 partition
: a = x0 < x1 < ::: < xn = b if each xk is equal to some xl0 and in this case we write

0: The de…nitions give S
f S
f and s
f s
f   0  0 if

0: We de…ne, mesh(
) = max1 i n(xi xi 1): First suppose f is Riemann integrable.  For each partition
let

n G
= f(a) a + i=1( sup f) ]xi 1;xi] f g ]xi 1;xi] and n g
= f(a) a + i=1( inf f) ]xi 1;xi] f g ]xi 1;xi] and note that

G
dm = S
f Z[a;b] and

g
dm = s
f: Z[a;b] 69

Suppose
k; k = 1; 2; :::; is a sequence of partitions such that
k
k+1,  b S
f f(x)dx k # Za and b s
f f(x)dx k " Za as k : Let G = limk G
k and g = limk g
k : Then G and g are ! 1 !1 !1 ( [a;b]; )-measurable, g f G; and by dominated convergence R R   b Gdm = gdm = f(x)dx: Z[a;b] Z[a;b] Za But then (G g)dm = 0 Z[a;b] so that G = g a.e. [m] and therefore G = f a.e. [m] : In particular, f is Lebesgue measurable and

b f(x)dx = fdm: Za Z[a;b] Set N = x; g(x) < f(x) or f(x) < G(x) : f g We proved above that m(N) = 0: Let M be the union of all those points which belong to some partition
k: Clearly, m(M) = 0 since M is denumerable. We claim that f is continuous o¤ N M: If f is not continuous at a point [ c = N M, there is an " > 0 and a sequence (cn)n1=1 converging to c such that2 [ f(cn) f(c) " all n: j j Since c = M, c is an interior point to exactly one interval of each partition 2
k and we get G
(c) g
(c) " k k  and in the limit G(c) g(c) ":  But then c N which is a contradiction. 2 70

Conversely, suppose the set of discontinuity points of f is a Lebesgue null set and let (
k)k1=1 is an arbitrary sequence of partitions of [a; b] such that
k
k+1 and mesh(
k) 0 as k : By assumption,  ! ! 1

lim G
(x) = lim g
(x) = f(x) k k k k !1 !1 at each point x of continuity of f. Therefore f is Lebesgue measurable and dominated convergence yields

lim G
dm = fdm k k !1 Z[a;b] Z[a;b] and

lim g
dm = fdm: k k !1 Z[a;b] Z[a;b] Thus f is Riemann integrable and

b f(x)dx = fdm: Za Z[a;b]

In the following we sometimes write

f(x)dx (A ) 2 R ZA instead of

fdm (A ): 2 R ZA In a similar way we often prefer to write

f(x)dx (A ) 2 Rn ZA instead of

fdmn (A ): 2 Rn ZA b Furthermore, a fdm means [a;b] fdm: Here, however, a warning is moti- vated. It is simple to …nd a real-valued function f on [0; [, which is bounded R R 1 71 on each bounded subinterval of [0; [ ; such that the generalized Riemann integral 1 1 f(x)dx Z0 is convergent, that is b lim f(x)dx b !1 Z0 exists and the limit is a real number, while the Riemann integral

1 f(x) dx j j Z0 sin x is divergent (take e.g. f(x) = x ): In this case the function f does not belong to 1 with respect to Lebesgue measure on [0; [ since L 1 b f dm = lim f(x) dx = : [0; [ j j b 0 j j 1 Z 1 !1 Z

Example 2.3.1. To compute n (1 x )n lim n dx n px !1 Z0

t suppose n N+ and use the inequality 1 + t e ; t R; to get 2  2 x n x (x)(1 ) e if x 0: [0;n] n   From this x n x (1 n ) e fn(x) =def (x) ; x 0 [0;n] px  px  and, in addition, x e lim fn(x) = : n x !1 p x x e 1 e Here L (m1 on [0; [) since 0 and px 2 1 px  x 1 e 1 x2 dx = 2 e dx = p: px Z0 Z0 72

Moreover fn 0 for every n N+ and by using dominated convergence we get  2 n x n (1 n ) 1 lim dx = lim fn(x)dx = n px n !1 Z0 !1 Z0 x 1 1 e lim fn(x)dx = dx = p: n px Z0 !1 Z0 Exercises

1. Let fn : [0; 1] [0; 1], n N; be a sequence of Riemann integrable functions such that! 2

lim fn(x) exists = f(x) all x [0; 1] : n !1 2 Show by giving an example that f need not be Riemann integrable.

2 n x 2. Suppose fn(x) = n x e j j; x R; n N+: Compute limn fn and j j 2 2 !1 limn fndm. !1 R R 3. Compute the following limits and justify the calculations: (a) x 1 sin(e ) lim dx: n 1 + nx2 !1 Z0 (b) n x n lim (1 + ) cos xdx: n n !1 Z0 (c) n x n 2x lim (1 + ) e dx: n n !1 Z0 (d) 1 x x lim (1 + )n exp( (1 + )n)dx: n n n !1 Z0 (e) x 1 sin( ) lim n n dx: n x(1 + x2) !1 Z0 73

(f) n x 1 + nx lim (1 )n cos xdx n n n + x !1 Z0 (g) 1 x n2 nx lim (1 + ) e dx: n n !1 Z0 (h) 1 1 + nx2 lim dx: n (1 + x2)n !1 Z0 (i) 1 lim pn (1 t2)n(1 + pn sin t )dt: n 1 j j !1 Z

4. Let (rn)n1=1 be an enumeration of Q and de…ne

n f(x) = 1 2 '(x rn) n=1 1 where '(x) = x 2 if 0 < x < 1 and '(x) = 0 if x 0 or x 1: Show that a)   1 f(x)dx = 2: Z1 b) b f 2(x)dx = if a < b: 1 Za c) f < a.s. [m] : 1 d) sup f(x) = + if a < b: a

x x2 n 5. Let n N+ and de…ne fn(x) = e (1 ) ; x R: Compute 2 2n 2 p2n lim fn(x)dx: n p2n !1 Z 74

p p 1 n 6. Suppose p N+ and de…ne fn(x) = n x (1 x) ; 0 x 1; for every 2   n N+: Show that 2 1 lim fn(x)dx = (p 1)!: n !1 Z0 7. Suppose f:[0; 1] R is a continuous function. Find ! 1 (n min(x;1 x))2 lim n f(x)e dx: n !1 Z0

2.4. Expectation

Suppose ( ; ;P ) is a probability space and  : ( ; ) (S; ) a random variable. RecallF that the probability law  of  is givenF by! the imageS measure P: By de…nition,

Bd = B()dP ZS Z for every B ; and, hence 2 S 'd = '()dP ZS Z for each simple -measurable function ' on S (we sometimes write f g = f(g)): By monotoneS convergence, we get 

fd = f()dP ZS Z for every measurable f : S [0; ] : Thus if f : S R is measurable, f L1() if and only if f() !L1(P1) and in this case ! 2 2 fd = f()dP: ZS Z In the special case when  is real-valued and  L1(P ); 2 xd(x) = dP: ZR Z The integral in the right-hand side is called the expectation of  and is denoted by E [] : 75 CHAPTER 3 Further Construction Methods of Measures

Introduction

In the …rst section of this chapter we collect some basic results on metric spaces, which every mathematician must know about. Section 3.2 gives a version of the Riesz Representation Theorem, which leads to another and perhaps simpler approach to Lebesgue measure than the Carathéodory The- orem. A reader can skip Section 3.2 without losing the continuity in this paper. The chapter also treats so called product measures and Stieltjes in- tegrals.

3.1. Metric Spaces

The construction of our most important measures requires topological con- cepts. For our purpose it will be enough to restrict ourselves to so called metric spaces. A metric d on a set X is a mapping d : X X [0; [ such that  ! 1

(a) d(x; y) = 0 if and only if x = y (b) d(x; y) = d(y; x) (symmetry) (c) d(x; y) d(x; z) + d(z; y) (triangle inequality). 

Here recall, if A1; :::; An are sets,

A1 ::: An = (x1; :::; xn); xi Ai for all i = 1; :::; n   f 2 g A set X equipped with a metric d is called a metric space. Sometimes we write X = (X; d) to emphasize the metric d: If E is a subset of the metric 76

space (X; d); the function d E E(x; y) = d(x; y); if x; y E; is a metric on j  2 E: Thus (E; d E E) is a metric space. The functionj '(t) = min(1; t); t 0; satis…es the inequality  '(s + t) '(s) + '(t):  Therefore, if d is a metric on X, min(1; d) is a metric on X: The metric min(1; d) is a bounded metric. The set R equipped with the metric d1(x; y) = x y is a metric space. More generally, Rn equipped with the metric j j

dn(x; y) = dn((x1; :::; xn); (y1; :::; yn)) = max xk yk 1 k n   j j is a metric space. If not otherwise stated, it will always be assumed that Rn is equipped with this metric. Let C [0;T ] denote the vector space of all real-valued continuous functions on the interval [0;T ] ; where T > 0: Then

d (x; y) = max x(t) y(t) 1 0 t T   j j is a metric on C [0;T ] : If (Xk; ek); k = 1; :::; n, are metric spaces,

d(x; y) = max ek(xk; yk); x = (x1; :::; xn) ; y = (y1; :::; yn) 1 k n   is a metric on X1 ::: Xn: The metric d is called the product metric on   X1 ::: Xn: If X = (X; d) is a metric space and x X and r > 0; the open ball with centre at x and radius r is the set B(x; r)2 = y X; d(y; x) < r : If E X and E is contained in an appropriate open ballf in2X it is said tog be bounded. The diameter of E is, by de…nition,

diam E = sup d(x; y) x;y E 2 and it follows that E is bounded if and only if diam E < . A subset of X which is a union of open balls in X is called open. In particular,1 an open ball is an open set. The empty set is open since the union of an empty family of sets is empty. An arbitrary union of open sets is open. The class of all 77 open subsets of X is called the topology of X: The metrics d and min(1; d) determine the same topology. A subset E of X is said to be closed if its complement Ec relative to X is open. An intersection of closed subsets of X is closed. If E X, E denotes the largest open set contained in E and  E (or E) the smallest closed set containing E:E is the interior of E and E its closure. The -algebra generated by the open sets in X is called the Borel -algebra in X and is denoted by (X). A positive measure on (X) is called a positive Borel measure. B B A sequence (xn)1 in X converges to x X if n=1 2 lim d(xn; x) = 0: n !1 If, in addition, the sequence (xn)1 converges to y X; the inequalities n=1 2 0 d(x; y) d(xn; x) + d(xn; y)   imply that y = x and the limit point x is unique. If E X and x X; the following properties are equivalent:  2

(i) x E: (ii) B2(x; r) E = ; all r > 0: \ 6 (iii) There is a sequence (xn)n1=1 in E which converges to x:

If B(x; r) E = , then B(x; r)c is a closed set containing E but not x: \ c Thus x = E: This proves that (i) (ii). Conversely, if x = E; since E is open there2 exists an open ball B(y;) s) such that x B(y; s)2 Ec Ec: Now choose r = s d(x; y) > 0 so that B(x; r) B(y;2 s): Then B(x; r) E = : This proves (ii) (i).  \ ) 1 If (ii) holds choose for each n N+ a point xn E with d(xn; x) < n and (iii) follows. If there exists an2 r > 0 such that2B(x; r) E = ; then (iii) cannot hold. Thus (iii) (ii). \ ) If E X, the set E E is called the boundary of E and is denoted by @E:  n A set A X is said to be dense in X if A = X: The metric space X is called separable if there is an at most denumerable dense subset of X: For example, Qn is a dense subset of Rn: The space Rn is separable. 78

n Theorem 3.1.1. (R ) = n: B R

n PROOF. The -algebra n is generated by the open n-cells in R and an R n n open n-cell is an open subset of R : Hence n (R ): Let U be an open n Rn  B n subset in R and note that an open ball in R = (R ; dn) is an open n-cell. If x U there exist an a Qn U and a rational number r > 0 such that x 2B(a; r) U: Thus U 2is an\ at most denumerable union of open n-cells 2  n and it follows that U n: Thus (R ) n and the theorem is proved. 2 R B  R

Let X = (X; d) and Y = (Y; e) be two metric spaces. A mapping f : X Y (or f :(X; d) (Y; e) to emphasize the underlying metrics) is said to! be continuous at the! point a X if for every " > 0 there exists a  > 0 such that 2 x B(a; ) f(x) B(f(a);"): 2 ) 2 Equivalently this means that for any sequence (xn)n1=1 in X which converges to a in X; the sequence (f(xn))n1=1 converges to f(a) in Y: If f is continuous at each point of X, the mapping f is called continuous. Stated otherwise this means that 1 f (V ) is open if V is open or 1 f (F ) is closed if F is closed. The mapping f is said to be Borel measurable if

1 f (B) (X) if B (Y ) 2 B 2 B or, what amounts to the same thing,

1 f (V ) (X) if V is open. 2 B A Borel measurable function is sometimes called a Borel function. A continuous function is a Borel function.

Example 3.1.1. Let f :(R;d1) (R;d1) be a continuous strictly increasing function and set (x; y) = f(x)!f(y) ; x; y R: Then  is a metric on R. j j 2 79

De…ne j(x) = x; x R: The mapping j :(R;d1) (R;) is continuous. We 2 ! claim that the map j :(R;) (R;d1) is continuous: To see this, let a R ! 2 and suppose the sequence (xn)n1=1 converges to a in the metric space (R;); that is f(xn) f(a) 0 as n : Let " > 0: Then j j! ! 1

f(xn) f(a) f(a + ") f(a) > 0 if xn a + "   and f(a) f(xn) f(a) f(a ") > 0 if xn a ":   Thus xn ]a "; a + "[ if n is su¢ ciently large. This proves that he map 2 j :(R;) (R;d1) is continuous. ! The metrics d1 and  determine the same topology and Borel subsets of R:

A mapping f :(X; d) (Y; e) is said to be uniformly continuous if for each " > 0 there exists a! > 0 such that e(f(x); f(y)) < " as soon as d(x; y) < : If x X and E;F X; let 2  d(x; E) = inf d(x; u) u E 2 be the distance from x to E and let

d(E;F ) = inf d(u; v) u E;v F 2 2 be the distance between E and F: Note that d(x; E) = 0 if and only if x E: If x; y X and u E; 2 2 2 d(x; u) d(x; y) + d(y; u)  and, hence d(x; E) d(x; y) + d(y; u)  and d(x; E) d(x; y) + d(y; E):  Next suppose E = : Then by interchanging the roles of x and y; we get 6 d(x; E) d(y; E) d(x; y) j j 80 and conclude that the distance function d(x; E); x X; is continuous. In fact, it is uniformly continuous. If x X and r > 0; the2 so called closed ball B(x; r) = y X; d(y; x) r is a2 closed set since the map y d(y; x); y X; is continuous.f 2  g ! 2If F X is closed and " > 0, the continuous function  1 X = max(0; 1 d( ;F )) F;" "
ful…ls 0 X 1 and X = 1 on F: Furthermore, X (a) > 0 if and only  F;"  F;" F;" if a F" =def x X; d(x; F ) < " : Thus 2 f 2 g X : F  F;"  F"

Let X = (X; d) be a metric space. A sequence (xn)n1=1 in X is called a Cauchy sequence if to each " > 0 there exists a positive integer p such that d(xn; xm) < " for all n; m p: If a Cauchy sequence (xn)1 contains a  n=1 convergent subsequence (xnk )k1=1 it must be convergent. To prove this claim, suppose the subsequence (xn )1 converges to a point x X: Then k k=1 2 d(xm; x) d(xm; xn ) + d(xn ; x)  k k can be made arbitrarily small for all su¢ ciently large m by choosing k su¢ - ciently large. Thus (xn)n1=1 converges to x: A subset E of X is said to be complete if every Cauchy sequence in E converges to a point in E: If E X is closed and X is complete it is clear that E is complete. Conversely, if X is a metric space and a subset E of X is complete, then E is closed: It is important to know that R is complete equipped with its standard metric. To see this let (xn)n1=1 be a Cauchy sequence. There exists a positive integer such that xn xm < 1 if n; m p: Therefore j j  xn xn xp + xp 1+ xp j jj j j j j j for all n p: We have proved that the sequence (xn)n1=1 is bounded (the reader can check that every Cauchy sequence in a metric space has this property). Now de…ne

a = sup x R; there are only …nitely many n with xn x : f 2  g

The de…nition implies that there exists a subsequence (xnk )k1=1; which con- verges to a (since for any r > 0; xn B(a; r) for in…nitely many n). The 2 81 original sequence is therefore convergent and we conclude that R is complete (equipped with its standard metric d1): It is simple to prove that the product of n complete spaces is complete and we conclude that Rn is complete. Let E X: A family (Vi)i I of subsets of X is said to be a cover of E  2 if i I Vi E and E is said to be covered by the Vi0s: The cover (Vi)i I is [ 2  2 said to be an open cover if each member Vi is open. The set E is said to be totally bounded if, for every " > 0;E can be covered by …nitely many open balls of radius ": A subset of a totally bounded set is totally bounded. The following de…nition is especially important.

De…nition 3.1.1. A subset E of a metric space X is said to be compact if to every open cover (Vi)i I of E, there is a …nite subcover of E, which means 2 there is a …nite subset J of I such that (Vi)i J is a cover of E: 2

If K is closed, K E; and E is compact, then K is compact. To see this,  let (Vi)i I be an open cover of K: This cover, augmented by the set X K is an open2 cover of E and has a …nite subcover since E is compact. Notingn that K (X K) = ; the assertion follows. \ n

Theorem 3.1.2. The following conditions are equivalent: (a) E is complete and totally bounded. (b) Every sequence in E contains a subsequence which converges to a point of E: (c) E is compact.

PROOF. (a) (b). Suppose (xn)n1=1 is a sequence in E: The set E can be ) 1 covered by …nitely many open balls of radius 2 and at least one of them 1 must contain xn for in…nitely many n N+: Suppose xn B(a1; 2 ) if 2 2 1 n N1 N0 =def N+; where N1 is in…nite. Next E B(a1; 2 ) can be 2  2 \ covered by …nitely many balls of radius 2 and at least one of them must 1 contain xn for in…nitely many n N1: Suppose xn B(a2; 2 ) if n N2; 2 2 j2 where N2 N1 is in…nite. By induction, we get open balls B(aj; 2 ) and  j in…nite sets Nj Nj 1 such that xn B(aj; 2 ) for all n Nj and j 1:  2 2  82

Let n1 < n2 < :::, where nk Nk; k = 1; 2; ::: . The sequence (xnk )k1=1 is a Cauchy sequence, and since E2 is complete it converges to a point of E .

(b) (a). If E is not complete there is a Cauchy sequence in E with no limit) in E: Therefore no subsequence can converge in E; which contradicts (b). On the other hand if E is not totally bounded, there is an " > 0 such that E cannot be covered by …nitely many balls of radius ": Let x1 E n 1 2 be arbitrary. Having chosen x1; :::; xn 1; pick xn E i=1 B(xi;"); and 2 n [ so on. The sequence (xn)n1=1 cannot contain any convergent subsequence as d(xn; xm) " if n = m; which contradicts (b).  6

(a) and (b) (c). Let (Vi)i I be an open cover of E: Since E is totally fbounded it isg )enough to show2 that there is an " > 0 such that any open ball of radius " which intersects E is contained in some Vi: Suppose on the n contrary that for every n N+ there is an open ball Bn of radius 2 2  which intersects E and is contained in no Vi: Choose xn Bn E and 2 \ assume without loss of generality that (xn)n1=1 converges to some point x in E by eventually going to a subsequence. Suppose x Vi and choose r > 0 2 0 such that B(x; r) Vi : But then Bn B(x; r) Vi for large n, which  0   0 contradicts the assumption on Bn:

(c) (b). If (xn)1 is a sequence in E with no convergent subsequence in ) n=1 E, then for every x E there is an open ball B(x; rx) which contains xn for 2 only …nitely many n: Then (B(x; rx))x E is an open cover of E without a …nite subcover. 2

Corollary 3.1.1. A subset of Rn is compact if and only if it is closed and bounded.

PROOF. Suppose K is compact. If xn K and xn = B(0; n) for every 2 2 n N+; the sequence (xn)n1=1 cannot contain a convergent subsequence. Thus2 K is bounded. Since K is complete it is closed. 83

Conversely, suppose K is closed and bounded. Since Rn is complete and K is closed, K is complete. We next prove that a bounded set is totally bounded. It is enough to prove that any n-cell in Rn is a union of …nitely many n-cells I1 ::: In where each interval I1; :::; In has a prescribed positive length. This is clear and the theorem is proved.

Corollary 3.1.2. Suppose f : X R is continuous and X compact: ! (a) There exists an a X such that maxX f = f(a) and a b X 2 2 such that minX f = f(b): (b) The function f is uniformly continuous:

PROOF. (a) For each a X; let Va = x X : f(x) < 1 + f(a) : The open 2 f 2 g cover (Va)a K of X has a …nite subcover and it follows that f is bounded. Let 2 (xn)1 be a sequence in X such that f(xn) sup f as n : Since X is n=1 ! K ! 1 compact there is a subsequence (xn )1 which converges to a point a X: k k=1 2 Thus, by the continuity of f; f(xnk ) f(a) as k : The existence of a minimum is proved! in a similar! 1 way.

(b) If f is not uniformly continuous there exist " > 0 and sequences n (xn)1 and (yn)1 such that f(xn) f(yn) " and xn yn < 2 n=1 n=1 j j j j for every n 1: Since X is compact there exists a subsequence (xn )1 of  k k=1 (xn)n1=1 which converges to a point a X: Clearly the sequence (ynk )k1=1 converges to a and therefore 2

f(xn ) f(yn ) f(xn ) f(a) + f(a) f(yn ) 0 j k k jj k j j k j! as k since f is continuous. But f(xnk ) f(ynk ) " and we have got a contradiction.! 1 The corollary is proved.j j

Example 3.1.2. Suppose X = ]0; 1] and de…ne 1(x; y) = d1(x; y) and  (x; y) = 1 1 ; x; y X: As in Example 3.1.1 we conclude that the metrics 2 j x y j 2 1 and 2 determine the same topology of subsets of X: The space (X; 1) 84

totally bounded but not complete. However, the space (X; 2) is not totally bounded but it is complete. To see this, let (xn)n1=1be a Cauchy sequence in (X; 2): As a Cauchy sequence it must be bounded and therefore there exists an " ]0; 1] such that xn ["; 1] for all n: But then, by Corollary 3.1.1, 2 2 (xn)n1=1 contains a convergent subsequence in (X; 1) and, accordingly from this, the same property holds in (X; 2): The space (X; 2) is not compact, since (X; 1) is not compact, and we conclude from Theorem 3.1.2 that the space (X; 2) cannot be totally bounded.

Example 3.1.3. Set Rˆ=R ; and [ f1 1g d^(x; y) = arctan x arctan y j j if x; y Rˆ: Here 2   arctan = and arctan = : 1 2 1 2 ^ Example 3.1.1 shows that the standard metric d1 and the metric d R R determine the same topology. j  We next prove that the metric space Rˆ is compact. To this end, consider a sequence (xn)1 in R.ˆ If there exists a real number M such that xn M n=1 j j for in…nitely many n; the sequence (xn)n1=1 contains a convergent subsequence since the interval [ M;M] is compact. In the opposite case, for each positive real number M, either xn M for in…nitely many n or xn M for   in…nitely many n: Suppose xn M for in…nitely many n for every M ^   2 N+: Then d(xn ; ) = arctan xn 0 as k for an appropriate k 1 j k 2 j! ! 1 subsequence (xnk )k1=1: The space Rˆ= (R,ˆ d^) is called a two-point compacti…cation of R. It is an immediate consequence of Theorem 3.1.2 that the product of …nitely many compact metric spaces is compact. Thus R^ n equipped with the product metric is compact. We will …nish this section with several useful approximation theorems.

Theorem 3.1.3. Suppose X is a metric space and  positive Borel measure in X: Moreover, suppose there is a sequence (Un)n1=1of open subsets of X such that X = 1 Un [n=1 85 and (Un) < ; all n N+: 1 2 Then for each A (X) and " > 0; there are a closed set F A and an open set V A such2 B that   (V F ) < ": n In particular, for every A (X); 2 B (A) = inf (V ) V A V open and (A) = sup (F ) F A F closed

If X = R and (A) = n1=1 1 (A) ;A ; then ( 0 ) = 0 and (V ) = n 2 R f g for every open set containing 0 : The hypothesis that the sets Un; n 1 f g 2 N+; are open (and not merely Borel sets) is very important in Theorem 3.1.3.

PROOF. First suppose that  is a …nite positive measure. Let be the class of all Borel sets A in X such that for every " > 0 there existA a closed F A and an open V A such that (V F ) < ": If F   1 n is a closed subset of X and Vn = x; d(x; F ) < n ; then Vn is open and, by Theorem 1.1.2 (f), (Vn) (F ) as n . Thus F and we conclude that contains all closed# subsets of X:! 1 2 A NowA suppose A : We will prove that Ac : To this end, we choose " > 0 and a closed set2 AF A and an open set V 2 AA such that (V F ) < ": Then V c Ac F c and, moreover, (F c V c) < " since n   n V F = F c V c: n n If we note that V c is closed and F c open it follows that Ac : 2 A Next let (Ai)1 be a denumerable collection of members of . Choose i=1 A " > 0: By de…nition, for each i N+ there exist a closed Fi Ai and an 2 i  open Vi Ai such that (Vi Fi) < 2 ": Set  n

V = 1 Vi: [i=1 86

Then (V ( 1 Fi)) ( 1 (Vi Fi)) n [i=1  [i=1 n 1 (Vi Fi) < ":  i=1 n But n V ( 1 Fi) = 1 V ( Fi) n [i=1 \n=1 f n [i=1 g and since  is a …nite positive measure

n (V ( i1=1Fi)) = lim (V ( i=1Fi)): n n [ !1 n [ Accordingly, from these equations

n (V ( Fi)) < " n [i=1 if n is large enough. Since a union of open sets is open and a …nite union of closed sets is closed, we conclude that i1=1Ai : This proves that is a -algebra. Since contains each closed[ subset2 of AX; = (X): A A A B We now prove the general case. Suppose A (X): Since Un is a …nite 2 B positive measure the previous theorem gives us an open set Vn A Un such Un n  \ that  (Vn (A Un)) < "2 : By eventually replacing Vn by Vn Un we can n \ Un \ n assume that Vn Un: But then (Vn (A Un)) =  (Vn (A Un)) < "2 :  n \ n \ Set V = 1 Vn and note that V is open. Moreover, [n=1

V A 1 (Vn (A Un)) n  [n=1 n \ and we get (V A) 1 (Vn (A Un)) < ": n  n=1 n \ By applying the result already proved to the complement Ac we conclude there exists an open set W Ac such that  (A W c) = (W Ac) < ": n n c Thus if F =def W it follows that F A V and (V F ) < 2": The theorem is proved.   n

If X is a metric space C(X) denotes the vector space of all real-valued continuous functions f : X R: If f C(X); the closure of the set of ! 2 87 all x where f(x) = 0 is called the support of f and is denoted by suppf: The vector space of6 all all real-valued continuous functions f : X R with ! compact support is denoted by Cc(X):

Corollary 3.1.3. Suppose  and  are positive Borel measures in Rn such that (K) < and (K) < 1 1 for every compact subset K of Rn: If

n f(x)d(x) = f(x)d(x); all f Cc(R ) n n 2 ZR ZR then  = :

PROOF. Let F be closed. Clearly (B(0; i)) < and (B(0; i)) < for every positive integer i: Hence, by Theorem 3.1.31 it is enough to show1 that (F ) = (F ): Now …x a positive integer i and set K = B(0; i) F: It is enough to show that (K) = (K): But \

Rn Rn  j (x)d(x) =  j (x)d(x) K;2 K;2 n n ZR ZR for each positive integer j and letting j we are done. ! 1

A metric space X is called a standard space if it is separable and com- plete. Standard spaces have a series of very nice properties related to measure theory; an example is furnished by the following

Theorem 3.1.4. (Ulam’s Theorem) Let X be a standard space and suppose  is a …nite positive Borel measure on X: Then to each A (X) and " > 0 there exist a compact K A and an open V A such2 B that (V K) < ":   n 88

PROOF. Let " > 0: We …rst prove that there is a compact subset K of X such that (K) > (X) ": To this end, let A be a dense denumerable subset of X and let (ai)i1=1 be an enumeration of A: Now for each positive integer j j; 1 B(ai; 2 ") = X; and therefore there is a positive integer nj such that [i=1

nj j j ( B(ai; 2 ")) > (X) 2 ": [i=1 Set nj j Fj = B(ai; 2 ") [i=1 and L = 1 Fj: \j=1 The set L is totally bounded. Since X is complete and L closed, L is complete. Therefore, the set L is compact and, moreover

c c (K) = (X) (L ) = (X) ( 1 F ) [j=1 j c (X) 1 (F ) = (X) 1 ((X) (Fj))  j=1 j j=1 j (X) 1 2 " = (X) ":  j=1 Depending on Theorem 3.1.3 to each A (X) there exists a closed F A and an open V A such that (V F2) < B ": But   n V (F L) = (V F ) (F L) n \ n [ n and we get

(V (F L)) (V F ) + (X K) < 2": n \  n n Since the set F L is compact Theorem 3.1.4 is proved. \

Two Borel sets in Rn are said to be almost disjoint if their intersection has volume measure zero.

Theorem 3.1.5. Every open set U in Rn is the union of an at most denu- merable collection of mutually almost disjoint cubes. 89

Before the proof observe that a cube in Rn is the same as a closed ball n in R equipped with the metric dn.

k PROOF. For each, k N+; let kbe the class of all cubes of side length 2 2 Q k whose vertices have coordinates of the form i2 ; i Z: Let F1 be the union 2 of those cubes in 1 which are contained in U: Inductively, for k 1; let Q  Fk be the union of those cubes in k which are contained in U and whose k 1 Q n interiors are disjoint from Fj: Since d(x;R U) > 0 for every x U it [j=1 n 2 follows that U = 1 Fj: [j=1

Exercises

d n 1. Suppose f :(X; ) (R ; d) and g :(X; ) (R ; n) are measur- able. Set h(x) = (fM(x)!; g(x)) RRd+n if x XM. Prove! thatRh :(X; ) d+n 2 2 M ! (R ; d+n) is measurable. R

2. Suppose f :(X; ) (R; ) and g :(X; ) (R; ) are measurable. Prove that fg is ( M; !)-measurable.R M ! R M R

3. The function f : R R is a Borel function. Set g(x; y) = f(x); (x; y) R2: Prove that g : R2 !R is a Borel function. 2 !

4. Suppose f : [0; 1] R is a continuous function and g : [0; 1] [0; 1] a Borel function. Compute! the limit !

1 lim f(g(x)n)dx: n !1 Z0

5. Suppose X and Y are metric spaces and f : X Y a continuous mapping. Show that f(E) is compact if E is a compact subset! of X. 90

6. Suppose X and Y are metric spaces and f : X Y a continuous bijection. 1 ! Show that the inverse mapping f is continuous if X is compact.

7. Construct an open bounded subset V of R such that m(@V ) > 0:

8. The function f : [0; 1] R has a continuous derivative. Prove that the !1 set f(K) m if K = (f 0) ( 0 ): 2 Z f g

9. Let P denote the class of all Borel probability measures on [0; 1] and L the class of all functions f : [0; 1] [ 1; 1] such that ! f(x) f(y) x y ; x; y [0; 1] : j jj j 2 For any ;  P; de…ne 2 (; ) = sup fd fd : f L j [0;1] [0;1] j 2 Z Z (a) Show that (P; ) is a metric space. (b) Compute (; ) if  is linear 1 n 1 measure on [0; 1] and  = k=0  k ; where n N+ (linear measure on [0; 1] n n 2 is Lebesgue measure on [0; 1] restricted to the Borel sets in [0; 1]).

n 10. Suppose  is a …nite positive Borel measure on R : (a) Let (Vi)i I be a n 2 family of open subsets of R and V = i I Vi. Prove that [ 2

(V ) = sup (Vi ::: Vi ): 1 [ [ k i1;:::;ik I k N+2 2 n (b) Let (Fi)i I be a family of closed subsets of R and F = i I Fi. Prove that 2 \ 2

(F ) = inf (Fi1 ::: Fik ): i1;:::;ik I \ \ k N+2 2 91

### 3.2. Linear Functionals and Measures

Let X be a metric space. A mapping T : Cc(X) R is said to be a linear ! functional on Cc(X) if

T (f + g) = T f + T g; all f; g Cc(X) 2 and

T ( f) = T f; all R; f Cc(X): 2 2 If in addition T f 0 for all f 0;T is called a positive linear functional   on Cc(X): In this case T f T g if f g since g f 0 and T g T f =    T (g f) 0: Note that Cc(X) = C(X) if X is compact. The main result in this section is the following

Theorem 3.2.1. (The Riesz Representation Theorem) Suppose X is a compact metric space and let T be a positive linear functional on C(X): Then there exists a unique …nite positive Borel measure  in X with the following properties: (a) T f = fd; f C(X): 2 ZX (b) For every E (X) 2 B (E) = sup (K): K E K compact

(c) For every E (X) 2 B (E) = inf (V ): V E V open 92

The property (c) is a consequence of (b), since for each E (X) and " > 0 there is a compact K X E such that 2 B  n (X E) < (K) + ": n But then (X K) < (E) + " n and X K is open and contains E: In a similar way, (b) follows from (c) since Xnis compact. The proof of the Riesz Representation Theorem depends on properties of continuous functions of independent interest. Suppose K X is compact and V X is open. If f : X [0; 1] is a continuous function such that  ! f and suppf V  V  we write f V  and if f and suppf V K   V  we write K f V:  

Theorem 3.2.2. Let K be compact subset X. (a) Suppose K V where V is open. There exists a function f on X such that  K f V:  

(b) Suppose X is compact and K V1 ::: Vn; where K is compact and  [ [ V1; :::; Vn are open. There exist functions h1; :::; hn on X such that

hi Vi; i = 1; :::; n  and

h1 + ::: + hn = 1 on K: 93

1 c PROOF. (a) Suppose " = 2 minK d( ;V ): By Corollary 3.1.2, " > 0: The X
continuous function f =  satis…es f ; that is K f K": K;" K   K"   Part (a) follows if we note that the closure (K") of K" is contained in V:

(b) For each x K there exists an rx > 0 such that B(x; rx) Vi for some 2 1  i. Let Ux = B(x; rx): It is important to note that (Ux) Vi and (Ux) 2  is compact since X is compact. There exist points x1; :::; xm K such that m 2 Ux K: If 1 i n; let Fi denote the union of those (Ux ) which are [j=1 i    j contained in Vi: By Part (a), there exist continuous functions fi such that Fi fi Vi; i = 1; :::; n: De…ne  

h1 = f1

h2 = (1 f1)f2 ::::

hn = (1 f1):::(1 fn 1)fn:

Clearly, hi Vi; i = 1; :::; n: Moreover, by induction, we get 

h1 + ::: + hn = 1 (1 f1):::(1 fn 1)(1 fn): n Since Fi K we are done. [i=1 

The uniqueness in Theorem 3.2.1 is simple to prove. Suppose 1 and 2 are two measures for which the theorem holds. Fix " > 0 and compact K X and choose an open set V so that  (V )  (K) + ": If K f V;  2  2    (K) = d fd = T f 1 K 1  1 ZX ZX = fd d =  (V )  (K) + ": 2  V 2 2  2 ZX ZX Thus 1(K) 2(K): If we interchange the roles of the two measures, the opposite inequality is obtained, and the uniqueness of  follows. To prove the existence of the measure  in Theorem 3.2.1; de…ne for every open V in X, (V ) = sup T f: f V  94

Here () = 0 since the supremum over the empty set, by convention, equals 0: Note also that (X) = T 1: Moreover, (V1) (V2) if V1 and V2 are open  and V1 V2: Now set  (E) = inf (V ) if E (X): V E 2 B V open

Clearly, (E1) (E2); if E1 E2 and E1;E2 (X): We therefore say that  is increasing.  2 B

Lemma 3.2.1. (a) If V1; :::; Vn are open,

n n ( Vi)  (Vi): [i=1  i=1

(b) If E1;E2; ::: (X); 2 B

( 1 Ei) 1 (Ei): [i=1  i=1

(c) If K1; :::; Kn are compact and pairwise disjoint,

n n ( Ki) =  (Ki): [i=1 i=1

PROOF. (a) It is enough to prove (a) for n = 2: To this end …rst choose g V1 V2 and then hi Vi, i = 1; 2; such that h1 + h2 = 1 on supp g: Then  [ 

g = h1g + h2g and it follows that

T g = T (h1g) + T (h2g) (V1) + (V2):  Thus

(V1 V2) (V1) + (V2): [  95

(b) Choose " > 0 and for each i N+, choose an open Vi Ei such (Vi) < i 2  (Ei) + 2 ": Set V = 1 Vi and choose f V: Since suppf is compact, [i=1  f V1 ::: Vn for some n: Thus, by Part (a),  [ [ n T f (V1 ::: Vn)  (Vi) 1 (Ei) + "  [ [  i=1  i=1 and we get

(V ) 1 (Ei)  i=1 since " > 0 is arbitrary. But 1 Ei V and it follows that [i=1 

( 1 Ei) 1 (Ei): [i=1  i=1

(c) It is enough to treat the special case n = 2: Choose " > 0: Set  = d(K1;K2) and V1 = (K1)=2 and V2 = (K2)=2: There is an open set U  K1 K2 such that (U) < (K1 K2)+" and there are functions fi U Vi [ [  \ such that T fi > (U Vi) " for i = 1; 2: Now, using that  increases \

(K1) + (K2) (U V1) + (U V2)  \ \

T f1 + T f2 + 2" = T (f1 + f2) + 2":  Since f1 + f2 U; 

(K1) + (K2) (U) + 2" (K1 K2) + 3"   [ and, by letting " 0; !

(K1) + (K2) (K1 K2):  [ The reverse inequality follows from Part (b). The lemma is proved.

Next we introduce the class

= E (X); (E) = sup (K) M 8 2 B K E 9 < K compact = : ; 96

Since  is increasing contains every compact set. Recall that a closed set in X is compact, sinceMX is compact. Especially, note that  and X . 2 M

COMPLETION OF THE PROOF OF THEOREM 3.2.1:

CLAIM 1. contains every open set. M

PROOF OF CLAIM 1. Let V be open and suppose < (V ): There exists an f V such that < T f: If U is open and U K =def suppf; then f U; and hence T f (U): But then T f (K): Thus < (K) and Claim 1 follows since Kis compact and K V: 

CLAIM 2. Let (Ei)i1=1 be a disjoint denumerable collection of members of and put E = 1 Ei: Then M [i=1

(E) = i1=1(Ei) and E : 2 M

PROOF OF CLAIM 2. Choose " > 0 and for each i N+, choose a compact i 2 Ki Ei such that (Ki) > (Ei) 2 ": Set Hn = K1 ::: Kn: Then, by Lemma 3.2.1 (c), [ [

n n (E) (Hn) =  (Ki) >  (Ei) "  i=1 i=1 and we get (E) 1 (Ei):  i=1 Thus, by Lemma 3.2.1 (b), (E) = i1=1(Ei). To prove that E ; let " be as in the very …rst part of the proof and choose n such that 2 M

n (E)  (Ei) + ":  i=1 97

Then (E) < (Hn) + 2" and this shows that E : 2 M

CLAIM 3. Suppose E and " > 0: Then there exist a compact K and an open V such that K2 ME V and (V K) < ":   n

PROOF OF CLAIM 3. The de…nitions show that there exist a compact K and an open V such that " " (V ) < (E) < (K) + : 2 2 The set V K is open and V K by Claim 1. Thus Claim 2 implies that n n 2 M (K) + (V K) = (V ) < (K) + " n and we get (V K) < ": n

CLAIM 4. If A ; then X A : 2 M n 2 M

PROOF OF CLAIM 4. Choose " > 0: Furthermore, choose compact K A and open V A such that (V K) < ": Then   n X A (V K) (X V ): n  n [ n Now, by Lemma 3.2.1 (b),

(X A) " + (X V ): n  n Since X V is a compact subset of X A; we conclude that X A : n n n 2 M

Claims 1, 2 and 4 prove that is a -algebra which contains all Borel sets. Thus = (X): M M B 98

We …nally prove (a). It is enough to show that

T f fd  ZX for each f C(X): For once this is known 2 T f = T ( f) fd fd   ZX ZX and (a) follows. Choose " > 0: Set f(X) = [a; b] and choose y0 < y1 < ::: < yn such that y1 = a, yn 1 = b; and yi yi 1 < ": The sets 1 Ei = f ([yi 1; yi[); i = 1; :::; n constitute a disjoint collection of Borel sets with the union X: Now, for each i; " 1 pick an open set Vi Ei such that (Vi) (Ei)+ n and Vi f (] ; yi[):    1n By Theorem 3.2.2 there are functions hi Vi; i = 1; :::; n; such that  hi =  i=1 1 on suppf and hif yihi for all i: From this we get  n n n T f =  T (hif)  yiT hi  yi(Vi) i=1  i=1  i=1 n n "  yi(Ei) +  yi  i=1 i=1 n n  (yi ")(Ei) + "(X) + (b + ")"  i=1 n fd + "(X) + (b + ")"  i=1 ZEi = fd + "(X) + (b + ")": ZX Since " > 0 is arbitrary, we get

T f fd:  ZX This proves Theorem 3.2.1.

It is now simple to show the existence of volume measure in Rn: For pedagogical reasons we …rst discuss the so called volume measure in the unit cube Q = [0; 1]n in Rn: 99

The Riemann integral f(x)dx; ZQ is a positive linear functional as a function of f C(Q): Moreover, T 1 = 1 and the Riesz Representation Theorem gives us a2 Borel probability measure  in Q such that f(x)dx = fd: ZQ ZQ Suppose A Q is a closed n-cell and i N+: Then  2

Q vol(A)  i (x)dx vol(A2 i )  A;2  ZQ and Q  i (x) A(x) as i A;2 ! ! 1 for every x Rn: Thus 2 (A) = vol(A): The measure  is called the volume measure in the unit cube. In the special case n = 2 it is called the area measure in the unit square and if n = 1 it is called the linear measure in the unit interval.

PROOF OF THEOREM 1.1.1. Let Rˆ=R ; be the two-point com- pacti…cation of R introduced in Example 3.1.3[ f1 and1g let R^ n denote the product of n copies of the metric space R^ : Clearly,

(Rn) = A Rn; A (R^ n) : B \ 2 B n o

Moreover, let w : Rn ]0; [ be a continuous map such that ! 1 w(x)dx = 1: n ZR Now we de…ne T f = f(x)w(x)dx; f C(R^ n): n 2 ZR 100

Note that T 1 = 1. The function T is a positive linear functional on C(R^ n) and the Riesz Representation Theorem gives us a Borel probability measure  on R^ n such that

f(x)w(x)dx = fd; f C(R^ n): n ^ n 2 ZR ZR As above we get w(x)dx = (A) ZA for each compact n-cell in Rn: Thus

(Rn) = lim w(x)dx = 1 i [ i;i]n !1 Z n and we conclude that  is concentrated on R : Set 0(A) = (A);A (Rn); and 2 B 1 dm = d : n w 0 n Then, if f Cc(R ); 2

f(x)w(x)dx = fd0 n n ZR ZR and by replacing f by f=w;

f(x)dx = fdmn: n n ZR ZR

From this mn(A) =vol(A) for every compact n-cell A and it follows that mn is the volume measure on Rn. Theorem 1.1.1 is proved.

"""

3.3 q-Adic Expansions of Numbers in the Unit Interval

To begin with in this section we will discuss so called q-adic expansions of real numbers and give some interesting consequences. As an example of an 101 application, we construct a one-to-one real-valued Borel map f de…ned on a proper interval such that the range of f is a Lebesgue null set. Another example exhibits an increasing continuous function G on the unit interval with the range equal to the unit interval such that the derivative of G is equal to zero almost everywhere with respect to Lebesgue measure. In the next section we will give more applications of q-adic expansions in connection with in…nite product measures. To simplify notation let ( ;P; ) = ([0; 1[ ; ([0; 1[); v1 [0;1[). Furthermore, let q 2 be an integer and de…neF a functionBh : R j 0; 1; 2; :::; q 1 of period one such that ! f g k k + 1 h(x) = k; x < ; k = 0; :::; q 1: q  q

Furthermore, set for each n N+; 2 n 1  (!) = h(q !); 0 ! < 1: n  Then 1 P [ = k] = ; k = 0; :::; q 1: n q

Moreover, if k1; :::; kn 0; 1; 2; :::; q 1 ; it becomes obvious on drawing a …gure that 2 f g

q 1 P 1 = k1; :::; n 1 = kn 1 = i=0 P 1 = k1; :::; n 1 = kn 1; n = i where each term in the sum in the right-hand side has the same value. Thus

P 1 = k1; :::; n 1 = kn 1 = qP 1 = k1; :::; n 1 = kn 1; n = kn and    

P 1 = k1; :::; n 1 = kn 1; n = kn = P 1 = k1; :::; n 1 = kn 1 P [n = kn] : By repetition,   

n P 1 = k1; :::n 1 = kn 1; n = kn = i=1P [i = ki] : From this we get 

n P 1 A1; :::n 1 An 1; n An = i=1P [i Ai] 2 2 2 2   102

for all A1; :::; An 0; 1; 2; :::; q 1 : Note that each! f [0; 1[ has a sog called q-adic expansion 2

i(!) ! = 1 : i=1 qi

(q) If necessary, we write n = n to indicate q explicitly. Let k0 0; 1; 2; :::; q 1 be …xed and consider the event A that a num- 2 f g ber in [0; 1[ does not have k0 in its q-adic expansion. The probability of A equals

P [A] = P [i = k0; i = 1; 2; :::] = lim P [i = k0; i = 1; 2; :::; n] n 6 !1 6

n q 1 n = lim i=1P [i = k0] = lim ( ) = 0: n n q !1 6 !1 In particular, if

(3) Dn = ! [0; 1[ ;  = 1; i = 1; :::; n : 2 i 6 n o then, D = n1=1Dn is a P -zero set. Set \ (2) 2i (!) f(!) = 1 ; 0 ! < 1: i=1 3i  We claim that f is one-to-one. If 0 !;!0 < 1 and ! = !0 let n be the (2) (2)  6 (2) least i such that i (!) = i (!0); we may assume that n (!) = 0 and (2) 6 n (!0) = 1: Then

(2) (2) n 2i (!0) n 1 2i (!0) 2 f(!0)  =  +  i=1 3i i=1 3i 3n (2) (2) n 1 2i (!) 4 2i (!) =  + 1 > 1 = f(!): i=1 3i i=n+1 3i i=1 3i Thus f is one-to-one. We next prove that f( ) = D: To this end choose (3) y D: If i (y) = 2 for all i N+; then y = 1 which is a contradiction. If 2 (3) 2 (3) k 1 is …xed and k (y) = 0 and i (y) = 2; i k + 1; then it is readily  (3)  seen that k (y) = 1 which is a contradiction. Now de…ne

1 (3) 2 i (y) ! = 1 i=1 2i 103 and we have f(!) = y:

Let Cn = Dn; n N+: The set C = n1=1Cn; is called the Cantor set. The Cantor set is a2 compact Lebesgue zero\ set. The construction of the Cantor set may alternatively be described as follows. First C0 = [0; 1]. Then 1 2 1 2 trisect C0 and remove the middle interval 3 ; 3 to obtain C1 = C0 3 ; 3 = 0; 1 2 ; 1 : At the second stage subdivide each of the closedn intervals 3 [ 3     of C1 into thirds and remove from each one the middle open thirds. Then    1 2 7 8 C2 = C1 ( 9 ; 9 9 ; 9 ): What is left from Cn 1 is Cn de…ned above. The n [ n n n set [0; 1] Cn is the union of 2 1 intervals numbered Ik ; k = 1; :::; 2 1; n   n  n where the interval Ik is situated to the left of the interval Il if k < l: Suppose n is …xed and let Gn : [0; 1] [0; 1] be the unique monotone in- ! creasing continuous function, which satis…es Gn(0) = 0;Gn(1) = 1;Gn(x) = n n k2 for x Ik and which is a¢ ne on each interval of Cn: It is clear that 2 n n Gn = Gn+1 on each interval Ik , k = 1; :::; 2 1: Moreover, Gn Gn+1 n 1 j j 2 and thus n+k n Gn Gn+k  Gk Gk+1 2 : j j k=n j j Let G(x) = limn Gn(x); 0 x 1: The continuous and increasing func- !1   tion G is constant on each removed interval and it follows that G0 = 0 a.e. with respect to linear measure in the unit interval.The function G is called the Cantor function or Cantor-Lebesgue function. Next we introduce the following convention, which is standard in Lebesgue integration. Let (X; ; ) be a positive measure space and suppose A and (Ac) = 0: If twoM functions g; h 1() agree on A; 2 M 2 L gd = hd: ZX ZX If a function f : A R is the restriction to A of a function g 1() we de…ne ! 2 L fd = gd: ZX ZX Now suppose F : R R is a right continuous increasing function and let  be the unique positive! Borel such that (]a; x]) = F (x) F (a) if a; x R and a < x: 2 If h L1() and E ; the so called Stieltjes integral 2 2 R h(x)dF (x) ZE 104 is by de…nition equal to hd: ZE If a; b R, a < b; and F is continuous at the points a and b; we de…ne 2 b h(x)dF (x) = hd Za ZI where I is any interval with boundary points a and b: The reader should note that the integral

h(x)dF (x) ZR in general is di¤erent from the integral

h(x)F 0(x)dx: ZR For example, if G is the Cantor function and G is extended so that G(x) = 0 for negative x and G(x) = 1 for x larger than 1, clearly

h(x)G0(x)dx = 0 ZR since G0(x) = 0 a.e. [m] : On the other hand, if we choose h = [0;1];

h(x)dG(x) = 1: ZR

3.4. Product Measures

Suppose (X; ) and (Y; ) are two measurable spaces. If A and B ; the setMA B is calledN a measurable rectangle in X Y: The2 product M -algebra2 N is, by de…nition, the -algebra generated by all measurable rectanglesM N in X Y: If we introduce the projections 

X (x; y) = x; (x; y) X Y 2  and Y (x; y) = y; (x; y) X Y; 2  105 the product -algebra is the least -algebra of subsets of X Y , M N S  which makes the maps X :(X Y; ) (X; ) and Y :(X Y; )  S !1 M 1  S ! (Y; ) measurable, that is = (X ( ) Y ( )):. NSuppose generates ;Mwhere NX ; andM [generatesN ; where Y . We claim thatE the class M 2 E F N 2 F

= E F ; E and F E  F f  2 E 2 Fg generates the -algebra : First it is clear that M N ( ) : E  F  M N Moreover, the class

1 E ; E Y ( ) = E X;  (E) ( ) f 2 M  2 E  F g M\  X 2 E  F is a -algebra, which contains and therefore equals . Thus A Y ( ) for all A and, inE a similar way, X BM ( )for all2 E  F 2 M  2 E  F B and we conclude that A B = (A Y ) (X B) (  ) for all2A N and all B : This proves that \  2 E F 2 M 2 N ( ) M N  E  F and it follows that ( ) = : E  F M N Thus ( ) = ( ) ( ) if X and Y : E  F E F 2 E 2 F n Since the -algebra n is generated by all open n-cells in R , we conclude that R k+n = k n: R R R Given E X Y; de…ne  

Ex = y;(x; y) E if x X f 2 g 2 and Ey = x;(x; y) E if y Y: f 2 g 2 If f : X Y Z is a function and x X; y Y , let  ! 2 2

fx(y) = f(x; y); if y Y 2 106 and f y(x) = f(x; y); if x X: 2

y Theorem 3.4.1 (a) If E ; then Ex and E for every x X and y Y: 2 M N 2 N 2 M 2 2 (b) If f :(X Y; ) (Z; ) is measurable, then fx is ( ; )- measurable for each xMX and N !f y is (O ; )-measurable for each y NY:O 2 M O 2

Proof. (a) Choose y Y and de…ne ' : X X Y by '(x) = (x; y): Then 2 !  1 1 1 = (' ( )) = ' (( )) = ' ( ) M M  N M  N M N y and it follows that E : In a similar way Ex for every x X: (b) For any set V 2 M; 2 N 2 2 O 1 1 (f (V ))x = (fx) (V ) and 1 y y 1 (f (V )) = (f ) (V ): Part (b) now follows from (a).

Below an ( ; 0; )-measurable or ( ; )-measurable function is simply called -measurable.M R 1 M R M

Theorem 3.4.2. Suppose (X; ; ) and (Y; ; ) are positive -…nite measurable spaces and suppose EM . If N 2 M N y f(x) = (Ex) and g(y) = (E ) for every x X and y Y; then f is -measurable, g is -measurable, and 2 2 M N fd = gd: ZX ZY

Proof. We …rst assume that (X; ; ) and (Y; ; ) are …nite positive measure spaces. M N 107

Let be the class of all sets E for which the conclusion of the theoremD holds. It is clear that the2 M class N of all measurable rectangles in X Y is a subset of and is a -system.G Furthermore, the Beppo Levi Theorem shows thatD is a G-additive class. Therefore, using Theorem 1.2.2, = ( ) Dand it follows that = : M N G  D D M N In the general case, choose a denumerable disjoint collection (Xk)k1=1of members of and a denumerable disjoint collection (Yn)n1=1of members of such thatM N 1 Xk = X and 1 Yn = Y: [k=1 [n=1 Set

k = Xk , k = 1; 2; ::: and

n = Yn , n = 1; 2; ::: . Then, by the Beppo Levi Theorem, the function

f(x) = n1=1 E(x; y) Yn (y)d(y) ZY

= n1=1 E(x; y) Yn (y)d(y) = n1=1n(Ex) ZY is -measurable. Again, by the Beppo Levi Theorem, M

fd = k1=1 fdk ZX ZX and

fd = k1=1(n1=1 n(Ex)dk(x)) = k;n1 =1 n(Ex)dk(x): ZX ZX ZX In a similar way, the function g is -measurable and N

y y gd = n1=1(k1=1 k(E )dn(y)) = k;n1 =1 k(E )dn(y): ZY ZY ZY Since the theorem is true for …nite positive measure spaces, the general case follows. 108

De…nition 3.4.1. If (X; ; ) and (Y; ; ) are positive -…nite measur- able spaces and E M, de…ne N 2 M N

y ( )(E) = (Ex)d(x) = (E )d(y):  ZX ZY The function   is called the product of the measures  and : 

Note that Beppo Levi’sTheorem ensures that   is a positive measure. Before the next theorem we recall the following convention. Let (X; ; ) be a positive measure space and suppose A and (Ac) = 0: IfM two functions g; h 1() agree on A; 2 M 2 L gd = hd: ZX ZX If a function f : A R is the restriction to A of a function g 1() we de…ne ! 2 L fd = gd: ZX ZX

Theorem 3.4.3. Let (X; ; ) and (Y; ; ) be positive -…nite measur- able spaces. M N

(a) (Tonelli’s Theorem) If h : X Y [0; ] is ( )-measurable and  ! 1 M N f(x) = h(x; y)d(y) and g(y) = h(x; y)d(x) ZY ZX for every x X and y Y; then f is -measurable, g is -measurable, and 2 2 M N fd = hd( ) = gd X X Y  Y Z Z  Z (b) (Fubini’sTheorem) (i) If h : X Y R is ( )-measurable and  ! M N ( h(x; y) d(y))d(x) < j j 1 ZX ZY 109 then h L1( ): Moreover, 2 

( h(x; y)d(y))d(x) = hd( ) = ( h(x; y)d(x))d(y) X Y X Y  Y X Z Z Z  Z Z

1 1 (ii) If h L (( )); then hx L () for -almost all x and 2  2

hd( ) = ( h(x; y)d(y))d(x) X Y  X Y Z  Z Z 1 y 1 (iii) If h L (( )); then h L () for -almost all y and 2  2

hd( ) = ( h(x; y)d(x))d(y) X Y  Y X Z  Z Z PROOF. (a) The special case when h is a non-negative ( )-measurable simple function follows from Theorem 3.4.2. RememberingM N that any non- negative measurable function is the pointwise limit of an increasing sequence of simple measurable functions, the Lebesgue Monotone Convergence Theo- rem implies the Tonelli Theorem.

(b) PART (i): By Part (a)

> ( h+(x; y)d(y))d(x) = h+d( ) 1 X Y X Y  Z Z Z 

= ( h+(x; y)d(x))d(y) ZY ZX and

> ( h(x; y)d(y))d(x) = hd( ) 1 X Y X Y  Z Z Z 

= ( h(x; y)d(x))d(y): ZY ZX + 1 In particular, h = h h L ( ): Let 2  + 1 A = x X;(h )x; (h)x L () : 2 2  110

Then Ac is a -null set and we get

( h+(x; y)d(y))d(x) = h+d( ) A Y X Y  Z Z Z  and

( h(x; y)d(y))d(x) = hd( ): A Y X Y  Z Z Z  Thus ( h(x; y)d(y))d(x) = hd( ) A Y X Y  Z Z Z  and, hence,

( h(x; y)d(y))d(x) = hd( ): X Y X Y  Z Z Z  The other case can be treated in a similar way. The theorem is proved.

PART (ii): We …rst use Theorem 2.2.3 and write h = ' + where ' 1 2 L ( ); is ( )-measurable and = 0 a.e. [ ] : Set  M N  + 1 A = x X;(' )x; (')x L () : 2 2 Furthermore, suppose E (x; y); (x; y) = 0 ;E and  f 6 g 2 M N ( )(E) = 0:  Then, by Tonelli’sTheorem

0 = (Ex)d(x): ZX

Let B = x X; (Ex) = 0 and note that B : Moreover (B) = 0 f 2 6 g 2 M and if x = B, then = 0 a.e. [] that is hx = ' a.e. [] : Now, by Part (i) 2 x x

hd( ) = 'd( ) = ( '(x; y)d(y))d(x) X Y  X Y  A Y Z  Z  Z Z = ( '(x; y)d(y))d(x) = ( h(x; y)d(y))d(x) A Bc Y A Bc Y Z \ Z Z \ Z 111

= ( h(x; y)d(y))d(x): ZX ZY

Part (iii) is proved in the same manner as Part (ii): This concludes the proof of the theorem.

If (Xi; i); i = 1; :::; n; are measurable spaces, the product -algebra M 1 ::: n is, by de…nition, the -algebra generated by all sets of the formM M A1 ::: An   where Ai i; i = 1; :::; n: Now assume (Xi; i;  ); i = 1; :::; n; are -…nite 2 M M i positive measure spaces. By induction, we de…ne 1 = 1 and k = k 1 k;  k = 1; 2; :::; n: The measure, n is called the product of the measures 1; :::; n and is denoted by  :::  : It is readily seen that 1   n

n = 1 ::: 1 (n factors) R R R and vn = v1 ::: v1 (n factors):   Moreover,

n ( ) =def ::: (n factors): Rn  R1 R1 R1

If A (R) ; by the Tonelli Theorem, the set A 0; :::; 0 (n 1 2 P n R1  f g zeros) is an mn-null set, which, in view of Theorem 3.4.1, cannot belong to n the -algebra ( ) : Thus the Axiom of Choice implies that R1 n = ( ) : Rn 6 R1

Clearly, the completion of the measure m1 ::: m1 (n factors) equals   mn: Sometimes we prefer to write

f(x1; :::; xn)dx1:::dxn A1 ::: An Z   instead of

f(x)dmn(x) A1 ::: An Z   112 or f(x)dx: A1 ::: An Z   Moreover, the integral

::: f(x1; :::; xn)dx1:::dxn ZAn ZA1 is the same as

f(x1; :::; xn)dx1:::dxn: A1 ::: An Z   De…nition 3.4.2. (a) The measure

x2 dx (A) = e 2 ;A 1 p2 2 R ZA is called the standard Gauss measure in R:

(b) The measure = ::: (n factors) n 1   1 is called the standard Gauss measure in Rn: Thus, if

2 2 n x = x + ::: + x ; x = (x1; :::; xn) R j j 1 n 2 q we have x 2 j j dx (A) = e 2 n ;A n: n p2 2 R ZA (c) A Borel measure  in R is said to be a centred Gaussian measure if  = f( ) for some linear map f : R R: 1 !

(d) A real-valued random variable  is said to be a centred Gaussian random variable if its probability law is a centred Gaussian measure in R: Stated otherwise,  is a real-valued centred Gaussian random variable if either

() = 0 (abbreviated  N(0; 0)) L 2 113 or there exists a  > 0 such that  ( ) = (abbreviated  N(0; )). L  1 2

(e) A family (t)t T of real-valued random variables is said to be a centred 2 real-valued Gaussian process if for all t1; :::; tn T; 1; :::; n R and every 2 2 n N+; the sum 2 n  = k=1 ktk is a centred Gaussian random variable:

2 2 n Example 3.4.1 Suppose x = x1 + ::: + xn if x = (x1; :::; xn) R : We claim that j j 2 n p 2 xi + x n n i k 16 lim (1 + ) d n(x) = p2 e k n 4k !1 R i=1 Z Y To prove this claim we …rst use that et 1 + t for every real t and have for each …xed i 1; :::; n ;  2 f g 2 x +x2 xi + xi i i 1 + e 4k : 4k 

Moreover, if k N+; then 2 2 xi + xi 1 1 2 1 1 + = ((xi + ) + 4k ) 0 4k 4k 2 4  and we conclude that 2 x +x2 xi + xi k i i (1 + ) e 4 : 4k  Thus, for any k N+; 2 n n 2 x +x2 xi + xi k i i 0 f (x) = (1 + ) e 4 = g(x) k def 4k def  i=1  i=1 Y Y where g 1( ) since 2 L n n x x2 i i dx 4 g(x)d n(x) = e n = Tonelli’sTheorem = n n p R R i=1 2 f g Z Z Y 114

n x x2 i i dxi n n e 4 = p2 e 16 : p i=1 R 2 Y Z Moreover, lim fk(x) = g(x) k !1 and by dominated convergence we get

n 2 xi + x n n i k 16 lim (1 + ) d n(x) = g(x)d n(x) = p2 e : k n 4k n !1 R i=1 R Z Y Z

Exercises

1. Let (X; ; ) and (Y; ; ) be two -…nite positive measure spaces. Let f L1()Mand g L1(N) and de…ne h(x; y) = f(x)g(y); (x; y) X Y: Prove2 that h L1(2 ) and 2  2  hd( ) = fd gd: X Y  X Y Z  Z Z

2. Let (X; ; ) be a -…nite positive measure space and f : X [0; [ a measurableM function. Prove that ! 1

fd = ( m)( (x; y); 0 < y < f(x); x X ):  f 2 g ZX 3. Let (X; ; ) be a -…nite positive measure space and f : X R a measurableM function. Prove that ( m)( (x; f(x)); x X ) = 0: !  f 2 g

1 4. Let E 2 and E [0; 1] [0; 1] : Suppose m(Ex) 2 for m-almost all x [0; 1] :2Show R that    2 1 m( y [0; 1] ; m(Ey) = 1 ) : f 2 g  2 115

5. Let c be the counting measure on R restricted to and R D = (x; x); x R : f 2 g De…ne for every A ( ) D ; 2 R  R [ f g

(A) = ( A(x; y)dv1(x))dc(y) ZR ZR and

(A) = ( A(x; y)dc(y))dv1(x): ZR ZR (a) Prove that  and  agree on  : (b) Prove that (D) = (D): R R 6

6. Let I = ]0; 1[ and

x2 y2 h(x; y) = ; (x; y) I I: (x2 + y2)2 2  Prove that  ( h(x; y)dy)dx = ; 4 ZI ZI  ( h(x; y)dx)dy = 4 ZI ZI and h(x; y) dxdy = : I I j j 1 Z  7. For t > 0 and x R let 2 1 x2 g(t; x) = e 2t p2t and @g h(t; x) = : @t Given a > 0; prove that

1 1 ( h(t; x)dt)dx = 1 a Z1 Z 116 and 1 1 ( h(t; x)dx)dt = 0 a Z Z1 and conclude that h(t; x) dtdx = : [a; [ R j j 1 Z 1  (Hint: First prove that 1 g(t; x)dx = 1 Z1 and @g 1 @2g = :) @t 2 @x2

8. Given f L1(m); let 2 1 x+1 g(x) = f(t)dt; x R: 2 x 1 2 Z Prove that g(x) dx f(x) dx: j j  j j ZR ZR 9. Let I = [0; 1] and suppose f : I R is a Lebesgue measurable function such that ! f(x) f(y) dxdy < : I I j j 1 Z  Prove that f(x) dx < : j j 1 ZI 1 10. Suppose A and f L (m): Set 2 R 2 d(y; A)f(y) g(x) = dy; x R: x y 2 2 ZR j j Prove that g(x) dx < : j j 1 ZA 117

11. Suppose that the functions f; g : R [0; [ are Lebesgue measurable and introduce  = fm and  = gm: Prove! that1 the measures  and  are -…nite and

( )(E) = f(x)g(y)dxdy if E :  2 R R ZE

12. Suppose  is a …nite positive Borel measure on Rn and f : Rn R a Borel measurable function. Set g(x; y) = f(x) f(y); x; y Rn: Prove! that f L1() if and only if g L1( ): 2 2 2 

13. A random variable  is non-negative and possesses the distribution func- tion F (x) = P [ x] : Prove thatE [] = 1(1 F (x))dx:  0 R

14. Let (X; d) be a metric space and suppose Y (X): Then Y equipped 2 B with the metric d Y Y is a metric space. Prove that j  (Y ) = A Y ; A (X) : B f \ 2 B g

15. The continuous bijection f :(X; d) (Y; e) has a continuous inverse. Prove that f(A) (Y ) if A (X) ! 2 B 2 B

16. A real-valued function f(x; y); x; y R; is a Borel function of x for every …xed y and a continuous function of y2for every …xed x: Prove that f is a Borel function. Is the same conclusion true if we only assume that f(x; y) is a real-valued Borel function in each variable separately?

17. Suppose a > 0 and 1 n a a  = e  a n! n n=0 X 118

where n(A) = (n) if n N = 0; 1; 2; ::: and A N: Prove that A 2 f g  1 (  )s =  a  b a+b for all a; b > 0; if s(x; y) = x + y; x; y N: 2

18. Suppose x2 1 xe f(t) = dx; t > 0: x2 + t2 Z0 Compute 1 lim f(t) and f(t)dt: t 0+ ! Z0 Finally, prove that f is di¤erentiable.

19. Suppose 1 tx ln(1 + x) f(t) = e dx; t > 0: 1 + x Z0 1 a) Show that 0 f(t)dt < : b) Show that f is in…nitely1 many times di¤erentiable. R

20. Suppose f is Lebesgue integrabel on ]0; 1[ : (a) Show that the function 1 1 1 g(x) = x t f(t)dt; 0 < x < 1; is continuous. (b) Prove that 0 g(x)dx = 1 f(x)dx: 0 R R R 3.5 Change of Variables in Volume Integrals

If T is a non-singular n by n matrix with real entries, we claim that 1 T (v ) = v n det T n j j 119

(here T is viewed as a linear map of Rn into Rn). Remembering Corollary 3.1.3 this means that the following linear change of variables formula holds, viz. 1 n f(T x)dx = f(x)dx all f Cc(R ): n det T n 2 ZR j j ZR The case n = 1 is obvious. Moreover, by Fubini’sTheorem the linear change of variables formula is true for arbitrary n in the following cases: (a) T x = (x(1); :::; x(n)); where  is a permutation of the numbers 1; :::; n: (b) T x = ( x1; x2; :::; xn); where is a non-zero real number. (c) T x = (x1 + x2; x2; :::; xn): Recall from linear algebra that every non-singular n by n matrix T can be row-reduced to the identity matrix, that is T can by written as the product of …nitely many transformations of the types in (a),(b), and (c). This proves the above linear change of variables formula. Our main objective in this section is to prove a more general change of variable formula. To this end let and be open subsets of Rn and 1 G : a C di¤eomorphism, that is G = (g1; :::; gn) is a bijective ! @gi continuously di¤erentiable map such that the matrix G0(x) = ( (x))1 i;j n @xj   is non-singular for each x : The inverse function theorem implies that 1 1 2 G : is a C di¤eomorphism [DI] : !

Theorem 3.5.1. If f is a non-negative Borel function in ; then

f(x)dx = f(G(x)) det G0(x) dx: j j Z Z

The proof of Theorem 3.5.1 is based on several lemmas. Throughout, Rn is equipped with the metric

dn(x; y) = max xk yk : 1 k n   j j Let K be a compact convex subset of : Then if x; y K and 1 i n; 2   1 d gi(x) gi(y) = gi(y + t(x y))dt dt Z0 1 n @gi =  (y + t(x y))(xk yk)dt k=1 @x Z0 k 120 and we get dn(G(x);G(y)) M(G; K)dn(x; y)  where n @gi M(G; K) = max k=1 max (z) : 1 i n z K @x   2 j k j Thus if B(a; r) is a closed ball contained in K;

G(B(a; r)) B(G(a); M(G; K)r): 

Lemma 3.5.1. Let (Qk)k1=1 be a sequence of closed balls contained in such that Qk+1 Qk  and diam Qk 0 as k : ! ! 1 Then, there is a unique point a belonging to each Qk and

vn(G(Qk)) lim sup det G0(a) : n vn(Qk) j j !1

PROOF. The existence of a point a belonging to each Qk is an immediate consequence of Theorem 3.1.2. The uniqueness is also obvious since the 1 diameter of Qk converges to 0 as k : Set T = G0(a) and F = T G: ! 1 Then, if Qk = B(xk; rk);

1 1 vn(G(Qk)) = vn(T (T G(Qk))) = det T vn(T G(B(xk; rk))) j j 1 1 1 n det T vn(B(T G(xk); M(T G; Qk)rk) = det T M(T G; Qk) vn(Qk): j j j j Since 1 lim M(T G; Qk) = 1 k !1 the lemma follows at once.

Lemma 3.5.2. Let Q be a closed ball contained in : Then

vn(G(Q)) det G0(x) dx:  j j ZQ 121

PROOF. Suppose there is a closed ball Q contained in such that

vn(G(Q)) > det G0(x) dx: j j ZQ This will lead us to a contradiction as follows. Choose " > 0 such that

vn(G(Q)) (1 + ") det G0(x) dx:  j j ZQ 2n Let Q = 1 Bk where B1; :::; B2n are mutually almost disjoint closed balls with the same[ volume. If

n vn(G(Bk)) < (1 + ") det G0(x) dx; k = 1; :::; 2 j j ZBk we get 2n vn(G(Q))  vn(G(Bk))  k=1 2n <  (1 + ") det G0(x) dx = (1 + ") det G0(x) dx k=1 j j j j ZBk ZQ which is a contradiction. Thus

vn(G(Bk)) (1 + ") det G0(x) dx  j j ZBk for some k: By induction we obtain a sequence (Qk)k1=1 of closed balls con- tained in such that

Qk+1 Qk;  diam Qk 0 as k ! ! 1 and

vn(G(Qk)) (1 + ") det G0(x) dx:  j j ZQk But applying Lemma 3.5.1 we get a contradiction. 122

PROOF OF THEOREM 3.5.1. Let U be open and write U = 1 Qi  [i=1 where the Qi0 s are almost disjoint cubes as in Theorem 3.1.5. Then

vn(G(U)) 1 vn(G(Qi)) 1 det G0(x) dx  i=1  i=1 j j ZQi

= det G0(x) dx: j j ZU Using Theorem 3.1.3 we now have that

vn(G(E)) det G0(x) dx  j j ZE for each Borel set E : But then 

f(x)dx f(G(x)) det G0(x) dx  j j Z Z for each simple Borel measurable function f 0 and, accordingly from this and monotone convergence, the same inequality holds for each non-negative Borel function f: But the same line of reasoning applies to G replaced by 1 G and f replaced by f(G) det G0 , so that j j

1 1 f(G(x)) det G0(x) dx f(x) det G0(G (x)) det(G )0(x) dx j j  j jj j Z Z

= f(x)dx: Z This proves the theorem.

2 Example 3.5.1. If f : R [0; ] is ( 2; 0; )-measurable and 0 < " < R < , the substitution ! 1 R R 1 1 G(r; ) = (r cos ; r sin ) yields

R 2 f(x1; x2)dx1dx2 = lim f(r cos ; r sin )rddr + "< x2+x2

R 2 = f(r cos ; r sin )rddr Z" Z0 and by letting " 0 and R ; we have ! ! 1 2 1 f(x1; x2)dx1dx2 = f(r cos ; r sin )rddr: 2 ZR Z0 Z0 The purpose of the example is to show an analogue formula for volume measure in Rn: n 1 n n Let S = x R ; x = 1 be the unit sphere in R : We will de…ne a f 2 j j g n 1 so called surface area Borel measure n 1 on S such that

1 n 1 f(x)dx = f(r!)r dn 1(!)dr n n 1 ZR Z0 ZS n for any ( n; 0; )-measurable function f : R [0; ] : To this end de…ne n R R 1 n 1 ! 1 G : R 0 ]0; [ S by setting G(x) = (r; !); where n f g ! 1  x r = x and ! = : j j x j j 1 n 1 n Note that G : ]0; [ S R 0 is given by the equation 1  ! n f g 1 G (r; !) = r!:

Moreover,

1 1 n 1 G (]0; a] E) = aG (]0; 1] E) if a > 0 and E S :    n 1 If E (S ) we therefore have that 2 B 1 n 1 vn(G (]0; a] E)) = a vn(G (]0; 1] E)):   We now de…ne

1 n 1 n 1(E) = nvn(G (]0; 1] E)) if E (S )  2 B and n 1 (A) = r dr if A (]0; [): 2 B 1 ZA 124

Below, by abuse of language, we write vn Rn 0 = vn: Then, if 0 < a n 1 j nf g  b < and E (S ); 1 2 B

G(vn)(]0; a] E) = (]0; a])n 1(E)  and G(vn)(]a; b] E) = (]a; b])n 1(E):  Thus, by Theorem 1.2.3, G(vn) =  n 1  and the claim above is immediate. To check the normalization constant in the de…nition of n 1; …rst note that R n n 1 R n 1 vn( x < R) = r d(!)dr = n 1(S ) j j n 1 n Z0 ZS and we get d n 1 n 1 vn( x < R) = R n 1(S ): dR j j

Exercises

1. Extend Theorem 3.5.1 to Lebesgue measurable functions.

2. The function f : R [0; [ is Lebesgue measurable and R fdm = 1: Determine all non-zero! real1 numbers such that hdm < ; where R R 1 n h(x) = 1 f( x + n); x RR: n=0 2 3. Suppose n x n n (A) = x ej j dx; A (R ); j j 2 B ZA 2 2 n where x = x1 + ::: + xn if x = (x1; :::; xn) R : Compute the limit j j 2 p n n lim  ln ( x R ; x  ):  !1 f 2 j j g 125

4. Compute the n-dimensional Lebesgue integral

ln(1 x )dx x <1 j j Zj j where x denotes the Euclidean norm of the vector x Rn: (Hint: n 1j j 2n=2 2 (S ) = (n=2) :)

1 5. Suppose f L (m2): Show that limn f(nx) = 0 for m2-almost all x R2: 2 !1 2

### 3.6. Independence in Probability

Suppose ( ; ;P ) is a probability space. The random variables  : ( ;P ) F k ! (Sk; k); k = 1; :::; n are said to be independent if S n P( ;:::; ) = P : 1 n k=1 k

A family (i)i I of random variables is said to be independent if i ; :::; i 2 1 n are independent for any i1; :::in I with ik = il if k = l: A family of 2 6 6 events (Ai)i I is said to be independent if ( A )i I is a family of independent 2 i 2 random variables. Finally a family ( i)i I of sub--algebras of is said to A 2 F be independent if, for any Ai i; i I; the family (Ai)i I is a family of independent events. 2 A 2 2

Example 3.6.1. Let q 2 be an integer. A real number ! [0; 1[ has a q-adic expansion  2 (q) k ! = 1 : k=1 qk (q) The construction of the Cantor set shows that (k )k1=1 is a sequence of independent random variables based on the probability space

([0; 1[ ; v1 [0;1[; ([0; 1[)): j B 126

Example 3.6.2. Let (X; ; ) be a positive measure space and let Ai ; M 2 M i N+; be such that 2 1 (Ai) < : i=1 1 The …rst Borel-Cantelli Lemma asserts that -almost all x X lie in Ai for at most …nitely many i: This result is an immediate consequence2 of the Beppo Levi Theorem since

1 d = 1 d < i=1 Ai i=1 Ai 1 ZX ZX implies that 1 < a.e. [] : i=1 Ai 1 Suppose ( ; ;P ) is a probability space and let (Ai)i1=1 be independent events such thatF 1 P [Ai] = : i=1 1 The second Borel-Cantelli Lemma asserts that almost surely Ai happens for in…nitely many i: To prove this, we use the inequality

1 + x ex; x R  2 to obtain P k+nAc = k+nP [Ac] \i=k i i=k i k+n k+n k+n P [Ai]  P [Ai] =  (1 P[Ai])  e = e i=k : i=k  i=k By letting n ; ! 1 c P [ 1 A ] = 0 \i=k i or

P [ 1 Ai] = 1: [i=k But then P [ 1 1 Ai] = 1 \k=1 [i=k and the second Borel-Cantelli Lemma is proved. 127

Theorem 3.6.1. Suppose 1; :::; n are independent random variables and  N(0; 1); k = 1; :::; n: If 1; :::; n R; then k 2 2 n n 2  k N(0;  ) k=1 k 2 k=1 k

PROOF. The case 1; :::; n = 0 is trivial so assume k = 0 for some k: We have for each open interval A; 6

n P [k=1 kk A] = d 1(x1):::d 1(xn) n 2  kxk A Z k=1 2

1 1 2 2 (x1+:::+xn) n e 2 dx1:::dxn: n  kxk A p2 Z k=1 2 2 2 Set  = 1 + ::: + n and let y = Gx be an orthogonal transformation such that p 1 y = ( x + ::: + x ): 1  1 1 n n Then, since det G = 1;

1 1 2 2 2 (y1 +:::+yn) P [k1=1 kk A] = n e dy1:::dyn 2 y1 A p2 Z 2

1 1 y2 = e 2 1 dy1 y1 A p2 Z 2 where we used Fubini’stheorem in the last step. The theorem is proved.

Finally, in this section, we prove a basic result about the existence of in…nite product measures. Let k; k N+ be Borel probability measures N+ 2 in R. The space R is, by de…nition, the set of all sequences x = (xk)k1=1 N+ of real numbers. For each k N+; set k(x) = xk: The -algebra 2 R is the least -algebra of subsets of RN+ which makes all the projections N+ S k :(R ; ) (R; ); k N+; measurable. Below, (1; :::; n) denotes S ! R 2 the mapping of RN+ into Rn de…ned by the equation

(1; :::; n)(x) = (1(x); :::; n(x)): 128

Theorem 3.6.1. There is a unique probability measure  on N+ such that R  =  :::  (1;:::;n) 1   n for every n N+: 2

The measure  in Theorem 3.6.1 is called the product of the measures  ; k N+; and is often denoted by k 2 1  : k=1 k

PROOF OF THEOREM 3.6.1. Let ( ;P; ) = ([0; 1[ ; v1 [0;1[; ([0; 1[) and set F j B (2) k (!) (!) = 1 ;! : k=1 2k 2

We already know that P = P: Now suppose (ki)i1=1 is a strictly increasing sequence of positive integers and introduce

(2)(!) ki 0 = 1 ;! : i=1 2i 2

n (2) (2) Note that for each …xed positive integer n; the R -valued maps (1 ; :::; n ) and ((2); :::; (2)) are P -equimeasurable. Thus, if f : R is continuous, k1 kn ! (2) n k (!) f()dP = lim f(k=1 )dP (!) n 2k Z !1 Z (2)(!) n ki = lim f(i=1 )dP (!) = f(0)dP n 2i !1 Z Z and it follows that P0 = P = P: By induction, we de…ne for each k N+ an in…nite subset Nk of the set k 1 k 2 N+ i=1 Ni such that the set N+ i=1Ni contains in…nitely many elements andn[ de…ne n[ (2) (!) nik  = 1 k i=1 2i where (nik)i1=1 is an enumeration of Nk: The map

(!) = (k(!))k1=1 129 is a measurable map of ( ; ) into (RN+ ; N+ ) and F R

P = 1 i k=1 where i = P for each i N+: 2 For each i N+ there exists a measurable map ' of ( ; ) into (R; ) 2 i F R such that P'i = i (see Section 1.6). The map

(x) = ('i(xi))i1=1

N+ N+ is a measurable map of (R ; ) into itself and we get  = (P ). This completes the proof of TheoremR 3.6.1.

""" 130 CHAPTER 4 MODES OF CONVERGENCE

Introduction

In this chapter we will treat a variety of di¤erent sorts of convergence notions in measure theory. So called L2-convergence is of particular importance.

4.1. Convergence in Measure, in L1(); and in L2()

Let (X; ; ) be a positive measure space and denote by (X) the class of measurableM functions f :(X; ) (R; ). For any f F(X); set M ! R 2 F

f 1= f(x) d(x) k k j j ZX and

2 f 2= f (x)d(x): k k sZX The Cauchy-Schwarz inequality states that

fg d f 2 g 2 if f; g (X): j j k k k k 2 F ZX To prove this, without loss of generality, it can be assumed that

0 < f 2< and 0 < g 2< : k k 1 k k 1 We now use the inequality 1 ( 2 + 2); ; R  2 2 131 to obtain f g 1 f 2 g2 j j j j d ( 2 + 2 )d = 1 f 2 g 2  2 f g ZX k k k k Z k k2 k k2 and the Cauchy-Schwarz inequality is immediate. If not otherwise stated, in this section p is a number equal to 1 or 2: If it is important to emphasize the underlying measure f p is written f p; : We now de…ne k k k k

p () = f (X); f p< : L f 2 F k k 1g The special case p = 1 has been introduced earlier. We claim that the following so called triangle inequality holds, viz.

p f + g p f p + g p if f; g (): k k k k k k 2 L The case p = 1; follows by -integration of the relation

f + g f + g : j jj j j j To prove the case p = 2; we use the Cauchy-Schwarz inequality and have

f + g 2 f + g 2 k k2kj j j jk2 = f 2 +2 fg d+ g 2 k k2 j j k k2 ZX 2 2 2 f +2 f 2 g 2 + g = ( f 2 + g 2) k k2 k k k k k k2 k k k k and the triangle inequality is immediate. Suppose f; g p(): The functions f and g are equal almost everywhere 2 L with respect to  if f = g : This is easily seen to be an equivalence relation and the setf of6 allg equivalence 2 Z classes is denoted by Lp(): Below we consider the elements of Lp() as members of p() and two members of Lp() are identi…ed if they are equal a.e. [] : FromL this convention it is straight-forward to de…ne f + g and f for all f; g Lp() and R and (p) p2 2 the function d (f; g) = f g p is a metric on L (): Convergence in the metric space Lp() = (Lkp(),d(p)k) is called convergence in Lp(): A sequence (fk)1 in (X) converges in measure to a function f (X) if k=1 F 2 F

lim ( fk f > ") = 0 all " > 0: k !1 j j 132

If the sequence (fk)k1=1 in (X) converges in measure to a function f (X) as well as to a functionFg (X); then f = g a.e. [] since 2 F 2 F " " f g > " f fk > fk g > fj j g  j j 2 [ j j 2 n o n o and " " ( f g > ") ( f fk > ) + ( fk g > ) j j  j j 2 j j 2 for every " > 0 and positive integer k: A sequence (fk)k1=1 in (X) is said to be Cauchy in measure if for every " > 0; F

( fk fn > ") 0 as k; n : j j ! ! 1 By the Markov inequality, a Cauchy sequence in Lp() is Cauchy in measure.

p Example 4.1.1. (a) If fk = k 0; 1 ; k N+; then [ k ] 2 1 fk 2;m= 1 and fk 1;m= : k k k k pk

1 2 Thus fk 0 in L (m) as k but fk 9 0 in L (m) as k : ! ! 1 ! 1

(b) L1(m) * L2(m) since

1 2 1 [1; [(x) L (m) L (m) 1 x 2 n j j and L2(m) * L1(m) since 1 (x) L1(m) L2(m): ]0;1] x 2 n j j p

Theorem 4.1.1. Suppose p = 1 or 2: (a) Convergence in Lp() implies convergence in measure: 133

(b) If (X) < ; then L2() L1() and convergence in L2() implies convergence in L1(1): 

p Proof. (a) Suppose the sequence (fn)n1=1 converges to f in L () and let " > 0: Then, by the Markov inequality,

1 p 1 p ( fn f ") fn f d = fn f j j  "p j j "p k kp ZX and (a) follows at once.

(b) The Cauchy-Schwarz inequality gives for any f (X); 2 F ( f 1d)2 f 2d 1d j j
 ZX ZX ZX or f 1 f 2 (X) k k k k and Part (b) is immediate. p

Theorem 4.1.2. Suppose fn (X); n N+: 2 F 2 (a) If (fn)n1=1 is Cauchy in measure, there is a measurable function f : X R such that fn f in measure as n and a strictly increasing ! ! ! 1 sequence of positive integers (nj)1 such that fn f a.e. [] as j . j=1 j ! ! 1 (b) If  is a …nite positive measure and fn f (X) a.e. [] as ! 2 F n ; then fn f in measure. ! 1 ! (c) (Egoro¤’s Theorem) If  is a …nite positive measure and fn f (X) a.e. [] as n ; then for every " > 0 there exists E such! that2 F(E) < " and ! 1 2 M

sup fk(x) f(x) 0 as n : k n j j! ! 1 x Ec 2

PROOF. (a) For each positive integer j; there is a positive integer nj such that j j ( fk fl > 2 ) < 2 ; all k; l nj: j j  134

There is no loss of generality to assume that n1 < n2 < ::: : Set

j Ej = fn fn > 2 j j j+1 j and  Fk = 1 Ej: [j=k If x F c and i j k 2 k  

fn (x) fn (x) fn (x) fn (x) j i j j j l+1 l j j l

1 c j k+1 (G ) (Fk) < 2 = 2 :  j=k X c Thus G is a -null set. We now de…ne f(x) = limj fnj (x) if x G and f(x) = 0 if x = G: !1 2 2 We next prove that the sequence (fn)n1=1 converges to f in measure. If x F c and j k we get 2 k  j+1 f(x) fn (x) 2 : j j j Thus, if j k  j+1 k+1 ( f fn > 2 ) (Fk) < 2 : j j j  Since " " ( fn f > ") ( fn fn > ) + ( fn f > ) j j  j j j 2 j j j 2 if " > 0; Part (a) follows at once.

(b) For each " > 0;

( fn f > ") = ]"; [( fn f )d j j 1 j j ZX 135 and Part (c) follows from the Lebesgue Dominated Convergence Theorem.

(c) Set for …xed k; n N+; 2 1 Ekn = 1 fj f > : [j=n j j k   We have 1 Ekn Z \n=1 2 and since  is a …nite measure

(Ekn) 0 as n : ! ! 1 k Given " > 0 pick nk N+ such that (Ekn ) < "2 : Then, if E = 1 Ekn , 2 k [k=1 k (E) < ". Moreover, if x = E and j nk 2  1 fj(x) f(x) : j j k The theorem is proved.

Corollary 4.1.1. The spaces L1() and L2() are complete.

p PROOF. Suppose p = 1 or 2 and let (fn)n1=1 be a Cauchy sequence in L ():

We know from the previous theorem that there exists a subsequence (fnj )j1=1 which converges pointwise to a function f (X) a.e. [] : Thus, by Fatou’s Lemma, 2 F p p f fk d lim inf fn fk d j j  j j j j ZX !1 ZX p p and it follows that f fk L () and, hence f = (f fk) + fk L (): 2 2 Moreover, we have that f fk p 0 as k : This concludes the proof of the theorem. k k ! ! 1

2 2 Corollary 4.1.2. Suppose n N(0; n); n N+; and n  in L (P ) as n : Then  is a centred Gaussian2 random2 variable. ! ! 1 136

2 PROOF. We have that  2= E  = n and  2  2=def  k n k n k n k !k k as n : q Suppose! 1 f is a bounded continuous function on R. Then, by dominated convergence,

E [f( )] = f(nx)d (x) f(x)d (x) n 1 ! 1 ZR ZR as n . Moreover, there exists a subsequence (nk )k1=1 which converges to  a.s.! 1 Hence, by dominated convergence

E f( ) E [f()] nk ! as k and it follows that  ! 1

E [f()] = f(x)d 1(x): ZR By using Corollary 3.1.3 the theorem follows at once.

Theorem 4.1.3. Suppose X is a standard space and  a positive -…nite Borel measure on X. Then the spaces L1() and L2() are separable.

PROOF. Let (Ek)k1=1 be a denumerable collection of Borel sets with …nite

-measures and such that Ek Ek+1 and k1=1Ek = X: Set k = Ek  and  [ p …rst suppose that the set Dk is at most denumerable and dense in L (k) for every k N+: Without loss of generality it can be assumed that each 2 member of Dk vanishes o¤ Ek: By monotone convergence

fd = lim fdk, f 0 measurable, k  ZX !1 ZX p and it follows that the set k1=1Dk is at most denumerable and dense in L (): From now on we can[ assume that  is a …nite positive measure. Let A be an at most denumerable dense subset of X and and suppose the subset rn; n N+; of ]0; [ is dense in ]0; [ : Furthermore, denote by the f 2 g 1 1 U 137

class of all open sets which are …nite unions of open balls of the type B(a; rn); a A; n N+. If U is any open subset of X 2 2 U = [V : V U and V ] [  2 U and, hence, by the Ulam Theorem

(U) = sup (V ); V and V U : f 2 U  g Let be the class of all functions which are …nite sums of functions of K the type  U ; where  is a positive rational number and U . It follows that is at most denumerable. 2 U SupposeK " > 0 and that f Lp() is non-negative. There exists a 2 sequence of simple measurable functions ('i)i1=1 such that

0 ' f a.e. [] :  i " p p Since f 'i f ; the Lebesgue Dominated Convergence Theorem shows j j  " that f ' p< for an appropriate k: Let 1; :::; l be the distinct k k k 2 positive values of 'k and set

l C = 1 + k=1 k:

Now for each …xed j 1; :::; l we use Theorem 3.1.3 to get an open 1 2 f g " Uj ' ( j ) such that 1 p< and from the above we k Uj 'k ( j ) 4C  f g k f g k " get a Vj such that Vj Uj and p< : Thus 2 U  k Uj Vj k 4C " 1 < Vj ' ( j ) p k k f g k 2C and l f  j p< " k k=1 Vj k Now it is simple to …nd a such that f p< ": From this we deduce that the set 2 K k k = g h; g; h K K f 2 Kg is at most denumerable and dense in Lp(): 138

The set of all real-valued and in…nitely many times di¤erentiable functions n ( ) n de…ned on R is denoted by C 1 (R ) and

( ) n ( ) n C 1 (R ) = f C 1 (R ); suppf compact : c 2 Recall that the support suppf of a real-valued continuous function f de…ned on Rn is the closure of the set of all x where f(x) = 0: If 6 n n f(x) = '(1 + xk)'(1 xk) ; x = (x1; :::; xn) R f g 2 k=1 Y 1 n where '(t) = exp( t ); if t > 0; and '(t) = 0; if t 0; then f Cc1(R ) . The proof of the previous theorem also gives Part (a) of the2 following

Theorem 4.1.4. Suppose  is a positive Borel measure in Rn such that (K) < for every compact subset K of Rn: The following sets are dense in L1()1; and L2(): (a) the linear span of the functions

n I ;I open bounded n-cell in R ;

( ) n (b) Cc1 (R ):

PROOF. a) The proof is almost the same as the proof of Theorem 4.1.3. First the Ek:s can be chosen to be open balls with their centres at the origin since each bounded set in Rn has …nite -measure. Moreover, as in the proof of Theorem 4.1.3 we can assume that  is a …nite measure. Now let A be an at most denumerable dense subset of Rn and for each a A let 2

R(a) = r > 0; ( x X; xk ak = r ) > 0 for some k = 1; :::; n : f f 2 j j g g

Then a AR(a) is at most denumerable and there is a subset rn; n N+ [ 2 f 2 g of ]0; [ a AR(a) which is dense in ]0; [ : Finally, let denote the class 1 n [ 2 1 U of all open sets which are …nite unions of open balls of the type B(a; rn); a A; n N+; and proceed as in the proof of Theorem 4.1.3. The result follows2 by2 observing that the characteristic function of any member of equals a …nite sum of characteristic functions of open bounded n-cells a.e.U [] : 139

Part (b) in Theorem 4.1.4 follows from Part (a) and the following

Lemma 4.1.1. Suppose K U Rn, where K is compact and U is open.  ( ) n Then there exists a function f Cc1 (R ) such that 2 K f U   that is f and suppf U: K   U 

n PROOF. Suppose  C1(R ) is non-negative, supp  B(0; 1); and 2 c 

dmn = 1: n ZR 1 Moreover, let " > 0 be …xed. For any g L (vn) we de…ne 2

n 1 f"(x) = " g(y)(" (x y))dy: n ZR Since k1+:::+kn @  1 g max L (vn); all k1; :::; kn N Rn k1 kn j j j @x1 :::@xn j2 2 n the Lebesgue Dominated Convergent Theorem shows that f" C1(R ): n n 2 Here f" C1(R ) if g vanishes o¤ a bounded subset of R : In fact, 2 c

supp f" (supp g)":  Now choose a positive number " 1 d(K;U c) and de…ne g = : Since  2 K"

f"(x) = g(x "y)(y)dy n ZR we also have that f"(x) = 1 if x K: The lemma is proved. 2 140

1 n Example 4.1.2. Suppose f L (mn) and let g : R R be a bounded Lebesgue measurable function.2 Set !

h(x) = f(x y)g(y)dy; x Rn: n 2 ZR We claim that h is continuous. To see this …rst note that

h(x +
x) h(x) = (f(x +
x y) f(x y))g(y)dy n ZR and

h(x +
x) h(x) K f(x +
x y) f(x y) dy j j n j j ZR = K f(
x + y) f(y) dy n j j ZR n n if g(x) K for every x R : Now …rst choose " > 0 and then ' Cc(R ) suchj thatj 2 2 f ' 1< ": k k Using the triangle inequality, we get

h(x +
x) h(x) K(2 f ' 1 + '(
x + y) '(y) dy) j j k k n j j ZR K(2" + '(
x + y) '(y) dy)  n j j ZR where the right hand side is smaller than 3K" if
x is su¢ ciently small. This proves that h is continuous. j j

Example 4.1.3. Suppose A and mn(A) > 0: We claim that the set 2 Rn A A = x x; x A f 2 g contains a neighbourhood of the origin. To show this there is no loss of generality to assume that mn(A) < : Set 1 n f(x) = mn(A (A + x)); x R : \ 2 141

Note that

f(x) = A(y) A(y x)dy n ZR and Example 4.1.2 proves that f is continuous. Since f(0) > 0 there exists a  > 0 such that f(x) > 0 if x < : In particular, A (A+x) =  if x < ; which proves that j j \ 6 j j B(0; ) A A: 

The following three examples are based on the Axiom of Choice.

Example 4.1.4. Let NL be the non-Lebesgue measurable set constructed in Section 1.3. Furthermore, assume A R is Lebesgue measurable and A NL: We claim that m(A) = 0: If not, there exists a  > 0 such that B(0; ) A A NL NL: If 0 < r <  and r Q, there exist a; b NL such that  2 2 a = b + r: But then a = b and at the same time a and b belong to the same equivalence class, which6 is a contradiction. Accordingly from this, m(A) = 0:

1 1 Example 4.1.5. Suppose A 2 ; 2 is Lebesgue measurable and m(A) > 0: We claim there exists a non-Lebesgue measurable subset of A: To see this note that   A = 1 ((ri + NL) A) [i=1 \ where (ri)1 is an enumeration of the rational numbers in the interval [ 1; 1] : i=1 If each set (ri + NL) A; is Lebesgue measurable \ m(A) = 1 m((ri + NL) A) i=1 \ and we conclude that m((ri + NL) A) > 0 for an appropriate i: But then \ m(NL (A ri)) > 0 and NL (A ri) NL; which contradicts Example \ \  4.1.4. Hence (ri + NL) A is non-Lebesgue measurable for an appropriate i: If A is a Lebesgue measurable\ subset of the real line of positive Lebesgue measure, we conclude that A contains a non-Lebesgue measurable subset. 142

Example 4.1.6. Set I = [0; 1] : We claim there exist a continuous function 1 f : I I and a Lebesgue measurable set L I such that f (L) is not Lebesgue! measurable.  First recall from Section 3.3 the construction of the Cantor set C and the Cantor function G. First C0 = [0; 1]. Then trisect C0 and remove the middle 1 2 1 2 1 2 interval ; to obtain C1 = C0 ; = 0; ; 1 : At the second 3 3 n 3 3 3 [ 3 stage subdivide each of the closed intervals of C1 into thirds and remove        1 2 7 8 from each one the middle open thirds. Then C2 = C1 ( ; ; ): We n 9 9 [ 9 9 repeat the process and what is left from Cn 1 is Cn. The set [0; 1] Cn is the n n n   n  n union of 2 1 intervals numbered Ik ; k = 1; :::; 2 1; where the interval Ik n is situated to the left of the interval I if k < l: The Cantor set C = 1 Cn: l \n=1 Suppose n is …xed and let Gn : [0; 1] [0; 1] be the unique the monotone ! increasing continuous function, which satis…es Gn(0) = 0;Gn(1) = 1;Gn(x) = n n k2 for x Ik and which is linear on each interval of Cn: It is clear that 2 n n Gn = Gn+1 on each interval I , k = 1; :::; 2 1: The Cantor function is k de…ned by the limit G(x) = limn Gn(x); 0 x 1: Now de…ne !1   1 h(x) = (x + G(x)); x I 2 2 where G is the Cantor function. Since h : I I is a strictly increasing and ! 1 continuous bijection, the inverse function f = h is a continuous bijection from I onto I: Set

A = h(I C) n and B = h(C): Recall from the de…nition of G that G is constant on each removed interval n Ik and that h takes each removed interval onto an interval of half its length. 1 1 Thus m(A) = 2 and m(B) = 1 m(A) = 2 : By the previous example there exists a non-Lebesgue measurable subset 1 M of B: Put L = h (M): The set L is Lebesgue measurable since L C 1  and C is a Lebesgue null set. However, the set M = f (L) is not Lebesgue measurable.

Exercises 143

1. Let (X; ; ) be a …nite positive measure space and suppose '(t) = M min(t; 1); t 0: Prove that fn f in measure if and only if '( fn f ) 0 in L1():  ! j j !

2. Let  = m [0;1]: Find measurable functions fn : [0; 1] [0; 1] ; n N+; j 2 ! 2 such that fn 0 in L () as n ; ! ! 1

lim inf fn(x) = 0 all x [0; 1] n !1 2 and lim sup fn(x) = 1 all x [0; 1] : n 2 !1

3. If f (X) set 2 F

f 0= inf [0; ]; ( f > ) : k k f 2 1 j j  g Let 0 L () = f (X); f 0< f 2 F k k 1g and identify functions in L0() which agree a.e. [] : (0) 0 (a) Prove that d = f g 0 is a metric on L () and that the corre- sponding metric space isk complete. k

(b) Show that (X) = L0() if  is a …nite positive measure. F

p p 4. Suppose L (X; ; ) is separable, where p = 1 or 2: Show that L (X; ; ) is separable. M M

5. Suppose g is a real-valued, Lebesgue measurable, and bounded function of period one. Prove that

1 1 1 lim f(x)g(nx)dx = f(x)dx g(x)dx n 0 !1 Z1 Z1 Z 144 for every f L1(m): 2

6. Let hn(t) = 2 + sin nt; 0 t 1; and n N+: Find real constants and such that   2 1 1 lim f(t)hn(t)dt = f(t)dt n !1 Z0 Z0 and 1 f(t) 1 lim dt = f(t)dt n h (t) !1 Z0 n Z0 for every real-valued Lebesgue integrable function f on [0; 1] :

n n n 7. If k = (k1; :::; kn) N+; set ek(x) = i=1 sin kixi; x = (x1; :::; xn) R ; n 2 1 2 2 and k = ( k ) 2 : Prove that j j i=1 i

lim fekdmn = 0 k n j j!1 ZR 1 for every f L (mn): 2

1 n 8. Suppose f L (mn); where mn denotes Lebesgue measure on R . Com- pute the following2 limit and justify the calculations:

lim f(x + h) f(x) dx: h n j j j j!1 ZR

9. The set A R has positive Lebesgue measure and  A + Q = x + y; x A and y Q f 2 2 g where Q stands for the set of all rational numbers. Show that the set

R (A + Q) n is a Lebesgue null set. (Hint: The function f(x) = m(A
(A x)); x R; is continuous.) 2 145

4.2 Orthogonality

Suppose (X; ; ) is a positive measure space. If f; g L2(); let M 2

f; g =def fgd h i ZX be the so called scalar product of f and g: The Cauchy-Schwarz inequality

f; g f 2 g 2 j h i jk k k k shows that the map f f; g of L2() into R is continuous. Observe that ! h i f + g 2= f 2 +2 f; g + g 2 k k2 k k2 h i k k2 and from this we get the so called Parallelogram Law

f + g 2 + f g 2= 2( f 2 + g 2): k k2 k k2 k k2 k k2 We will say that f and g are orthogonal (abbr. f g) if f; g = 0: Note that ? h i f + g 2= f 2 + g 2 if and only if f g: k k2 k k2 k k2 ? Since f g implies g f; the relation is symmetric. Moreover, if ? ? ? f h and g h then ( f + g) h for all ; R. Thus h? =def f ? L2(); f ?h is a subspace of L?2(); which is closed2 since the map f 2 f; h ; f ?L2(g) is continuous. If M is a subspace of L2(); the set ! h i 2

M ? =def h M h? \ 2 is a closed subspace of L2(): The function f = 0 if and only if f f: If M is a subspace of L2() and f L2() there exists at most? one point 2 g M such that f g M ?: To see this, let g0; g1 M be such that 2 2 2 f gk M ?; k = 0; 1: Then g1 g0 = (f g0) (f g1) M ? and hence 2 2 g1 g0 g1 g0 that is g0 = g1: ?

Theorem 4.2.1. Let M be a closed subspace in L2() and suppose f L2(): Then there exists a unique point g M such that 2 2

f g 2 f h 2 all h M: k k k k 2 146

Moreover, f g M ?: 2

The function g in Theorem 4.2.1 is called the projection of f on M and is denoted by ProjM f:

PROOF OF THEOREM 4.2.1. Set

(2) d =def d (f; M) = inf f g 2 : g M 2 k k and let (gn)n1=1 be a sequence in M such that

d = lim f gn 2 : n !1 k k Then, by the Parallelogram Law

2 2 2 2 (f gk)+(f gn) + (f gk) (f gn) = 2( f gk + f gn ) k k2 k k2 k k2 k k2 that is

1 2 2 2 2 4 f (gk + gn) + gn gk = 2( f gk + f gn ) k 2 k2 k k2 k k2 k k2 1 and, since (gk + gn) M; we get 2 2 2 2 2 2 4d + gn gk 2( f gk + f gn ): k k2 k k2 k k2 Here the right hand converges to 4d2 as k and n go to in…nity and we conclude 2 that (gn)n1=1 is a Cauchy sequence. Since L () is complete and M closed there exists a g M such that gn g as n : Moreover, 2 ! ! 1

d = f g 2 : k k

We claim that f g M ?: To prove this choose h M and > 0 arbitrarily and use the inequality2 2

(f g) + h 2 f g 2 k k2k k2 to obtain f g 2 +2 f g; h + 2 h 2 f g 2 k k2 h i k k2k k2 147 and 2 f g; h + h 2 0: h i k k2 By letting 0; f g; h 0 and replacing h by h; f g; h 0: Thus ! h i  h i  f g h? and it follows that f g M ?: The2 uniqueness in Theorem 4.2.1 2 follows from the remark just before the formulation of Theorem 4.2.1. The theorem is proved.

A linear mapping T : L2() R is called a linear functional on L2(): If h L2(); the map h f;! h of L2() into R is a continuous linear functional2 on L2(): It is a! very h importanti fact that every continuous linear functional on L2() is of this type.

Theorem 4.2.2. Suppose T is a continuous linear functional on L2(): Then there exists a unique w L2() such that 2 T f = f; w all f L2(): h i 2

2 2 PROOF. Uniqueness: If w; w0 L () and f; w = f; w0 for all f L (); 2 2 h i h i 2 then f; w w0 = 0 for all f L (): By choosing f = w w0 we get f f h i 2 ? that is w = w0:

1 Existence: The set M =def T ( 0 ) is closed since T is continuous and M is a linear subspace of L2() sincef g T is linear. If M = L2() we choose w = 0: Otherwise, pick a g L2() M: Without loss of generality it can be assumed that T g = 1 by eventually2 n multiplying g by a scalar. The previous theorem gives us a vector h M such that u =def g h M ?: Note that 2 2 2 0 < u 2= u; g h = u; g : Tok concludek h the proof,i h leti …xed f L2() be …xed; and use that (T f)g f M to obtain 2 2 (T f)g f; u = 0 h i or (T f) g; u = f; u : h i h i 148

By setting 1 w = u u 2 k k2 we are done.

### 4.3. The Haar Basis and Wiener Measure

In this section we will show the existence of Brownian motion with continu- ous paths as a consequence of the existence of linear measure  in the unit interval. The so called Wiener measure is the probability law on C [0; 1] of real-valued Brownian motion in the time interval [0; 1] : The Brownian mo- tion process is named after the British botanist Robert Brown (1773-1858). It was suggested by Lous Bachelier in 1900 as a model of stock price ‡uctua- tions and later by Albert Einstein in 1905 as a model of the physical phenom- enon Brownian motion. The existence of the mathematical Brownian motion process was …rst established by Norbert Wiener in the twenties. Wiener also proved that the model can be chosen such that the path t W (t); 0 t 1; is continuous a.s. Today Brownian motion is a very important! concept  in probability, …nancial mathematics, partial di¤erential equations and in many other …elds in pure and applied mathematics. Suppose n is a non-negative integer and set In = 0; :::; n : A sequence 2 f g (ei)i In in L () is said to be orthonormal if ei ej for all i = j; i; j In 2 ? 6 2 2 and ei = 1 for each i In: If (ei)i In is orthonormal and f L (); k k 2 2 2

f i In f; ei ei ej all j I 2 h i ? 2 and Theorem 4.2.1 shows that

f i In f; ei ei 2 f i In iei 2 all real 1; :::; n: k 2 h i k k 2 k Moreover

2 2 2 f 2= f i In f; ei ei 2 + i In f; ei ei 2 k k k 2 h i k k 2 h i k and we get 2 2 i In f; ei f 2 : 2 h i k k 149

2 We say that (en)n In is an orthonormal basis in L () if it is orthonormal and 2 2 f = i In f; ei ei all f L (): 2 h i 2 2 n A sequence (ei)i1=0 in L () is said to be orthonormal if (ei)i=0 is ortho- normal for each non-negative integer n: In this case, for each f L2(); 2 2 2 1 f; ei f i=0h i k k2 and the series 1 f; ei ei i=0h i converges since the sequence

n ( f; ei ei)1 i=0h i n=0 2 of partial sums is a Cauchy sequence in L (): We say that (ei)i1=0 is an orthonormal basis in L2() if it is orthonormal and

2 f = 1 f; ei ei for all f L (): i=0h i 2

2 Theorem 4.3.1. An orthonormal sequence (ei)i1=0 in L () is a basis of L2() if ( f; ei = 0 all i N) f = 0 h i 2 )

Proof. Let f L2() and set 2

g = f 1 f; ei ei: i=0h i Then, for any j N; 2

g; ej = f 1 f; ei ei; ej h i h i=0h i i

= f; ej 1 f; ei ei; ej = f; ej f; ej = 0: h i i=0h ih i h i h i Thus g = 0 or f = 1 f; ei ei: i=0h i The theorem is proved. 150

As an example of an application of Theorem 4.3.1 we next construct an orthonormal basis of L2(), where  is linear measure in the unit interval. Set

H(t) = 0; 1 (t) 1 ;1 (t); t R [ 2 [ [ 2 ] 2 n 1 Moreover, de…ne h00(t) = 1; 0 t 1; and for each n 1 and j = 1; :::; 2 ,    n 1 n 1 hjn(t) = 2 2 H(2 t j + 1); 0 t 1:   n 1 Stated otherwise, we have for each n 1 and j = 1; :::; 2  1 n 1 j 1 j 2 2 2 ; n 1 t < n 1 ; 2  2

8 1 n 1 j j hjn(t) = > 2 2 > 2 ; n 1 t n 1 ; > 2   2 <> 0; elsewhere in [0; 1] : > > > n 1 It is simple to show that:> the sequence h00;hjn; j = 1; :::; 2 ; n 1; is orthonormal in L2(). We will prove that the same sequence constitute an orthonormal basis of L2(): Therefore, suppose f L2() is orthogonal to n 1 2 each of the functions h00;hjn; j = 1; :::; 2 ; n 1: Then for each n 1 and n 1   j = 1; :::; 2 j 1 2 j 2n 1 2n 1 fd = fd j 1 j 1 2 n 1 n 1 Z 2 Z 2 and, hence, j 1 2n 1 1 fd = fd = 0 j 1 2n 1 n 1 0 Z 2 Z since 1 1 fd = fh00d = 0: Z0 Z0 Thus k n 1 2 n 1 fd = 0; 1 j k 2 j n 1    Z 2 and we conclude that

1 b 1[a;b]fd = fd = 0; 0 a b 1:    Z0 Za 151

Accordingly from this, f = 0 and we are done. 2 The above basis (hk)k1=0 = (h00;h11; h12; h22; h13; h23; h33; h43; :::) of L () is called the Haar basis. Let 0 t 1 and de…ne for …xed k N   2 1 t

ak(t) = [0;t](x)hk(x)dx = hkd Z0 Z0 so that 2 [0;t] = k1=0ak(t)hk in L (): Then, if 0 s; t 1;   1 min(s; t) = (x) (x)dx = 1 ak(s)hk; [0;s] [0;t] h k=1 [0;t]i Z0

= 1 ak(s) hk; = 1 ak(s)ak(t): k=0 h [0;t]i k=0 Note that 2 t = k1=0ak(t):

If (Gk)k1=0 is a sequence of N(0; 1) distributed random variables based on a probability space ( ; ;P ) the series F

k1=0ak(t)Gk converges in L2(P ) and de…nes a Gaussian random variable which we denote by W (t): From the above it follows that (W (t))0 t 1 is a real-valued centred Gaussian stochastic process with the covariance  

E [W (s)W (t)] = min(s; t):

Such a process is called a real-valued Brownian motion in the time interval [0; 1] : Recall that

(h00;h11; h12; h22; h13; h23; h33; h43; :::) = (hk)k1=0: We de…ne

(a00;a11; a12; a22; a13; a23; a33; a43; :::) = (ak)k1=0 and

(G00;G11;G12;G22;G13;G23;G33;G43; :::) = (Gk)k1=0: 152

It is important to note that for …xed n;

t ajn(t) = (x)hjn(x)dx = 0 for at most one j: [0;t] 6 Z0 Set U0(t) = a00(t)G00 and n 1 2 Un(t) =  ajn(t)Gjn; n N+: j=1 2 We know that 2 W (t) = n1=0Un(t) in L (P ) for …xed t: The space C [0; 1] will from now on be equipped with the metric

d(x; y) = x y k k1 where x = max0 t 1 x(t) : Recall that every x C [0; 1] is uniformly continuous.k k1 From this,  rememberingj j that R is separable,2 it follows that the space C [0; 1] is separable. Since R is complete it is also simple to show that the metric space C [0; 1] is complete. Finally, if xn C [0; 1] ; n N; and 2 2

n1=0 xn < k k1 1 the series

n1=0xn converges since the partial sums

n sn =  xk; k N k=0 2 forms a Cauchy sequence. We now de…ne

 = ! ; n1=0 Un < : f 2 k k1 1g Here  since 2 F Un = sup Un(t) k k1 0 t 1 j j t Q 2 for each n: Next we prove that  is a null set. n 153

To this end let n 1 and note that  n n P Un > 2 4 P max ( ajn Gjn ) > 2 4 : n 1 k k1  1 j 2 k k1j j       But 1 ajn = n+1 k k1 2 2 and, hence,

n n 1 n + 1 P Un > 2 4 2 P G00 > 2 4 2 : k k1  j j h i Since   1 y2=2 dy x2=2 x 1 P [ G00 x] 2 ye e  ) j j  xp2  Zx we get n n 2n=2 P Un > 2 4 2 e k k1  and conclude that  

1 1 n E 1 n = P Un > 2 4 < : [ Un >2 4 ] 1 "n=0 k k1 # n=0 k k 1 X X   From this and the Beppo Levi Theorem (or the …rst Borel-Cantelli Lemma) P [] = 1: The trajectory t W (t; !); 0 t 1; is continuous for every ! : Without loss of generality,! from now on we can therefore assume that2 all trajectories of Brownian motion are continuous (by eventually replacing by ): Suppose 0 t1 < ::: < tn 1   and let I1; :::; In be open subintervals of the real line. The set

S(t1; :::; tn; I1; :::; In) = x C [0; 1] ; x(tk) Ik; k = 1; :::; n f 2 2 g is called an open n-cell in C [0; 1] : A set in C [0;T ] is called an open cell if there exists an n N+ such that it is an open n-cell. The -algebra generated by all open cells2 in C [0; 1] is denoted by : The construction above shows that the map C W : C [0; 1] ! 154 which maps ! to the trajectory

t W (t; !); 0 t 1 !   is ( ; )-measurable. The image measure PW is called Wiener measure in C [0F; 1]C: The Wiener measure is a Borel measure on the metric space C [0; 1] : We leave it as an excersice to prove that

= (C [0; 1]): C B

""" 155 CHAPTER 5 DECOMPOSITION OF MEASURES

Introduction

In this section a version of the fundamental theorem of calculus for Lebesgue integrals will be proved. Moreover, the concept of di¤erentiating a measure with respect to another measure will be developped. A very important result in this chapter is the so called Radon-Nikodym Theorem.

5:1: Complex Measures

Let (X; ) be a measurable space. Recall that if An X; n N+, and M  2 Ai Aj =  if i = j, the sequence (An)n N+ is called a disjoint denumerable \ 6 2 collection. The collection is called a measurable partition of A if A = 1 An [n=1 and An for every n N+: A complex2 M function 2on is called a complex measure if M

(A) = n1=1(An) for every A and measurable partition (An)n1=1 of A: Note that () = 0 if  is a complex2 M measure. A complex measure is said to be a real measure if it is a real function. The reader should note that a positive measure need not be a real measure since in…nity is not a real number. If  is a complex measure  = Re +iIm , where Re =Re  and Im =Im  are real measures. If (X; ; ) is a positive measure and f L1() it follows that M 2 (A) = fd; A 2 M ZA is a real measure and we write d = fd. 156

A function  : [ ; ] is called a signed measure measure if M! 1 1

(a)  : ] ; ] or  : [ ; [ (b) (M!) = 0 1 1 M! 1 1 and (c) for every A and measurable partition (An)1 of A; 2 M n=1

(A) = n1=1(An) where the latter sum converges absolutely if (A) R: 2

Here = and + x = if x R: The sum of a positive measure1 1 and a1 real measure1 and the di¤erence1 of2 a real measure and a positive measure are examples of signed measures and it can be proved that there are no other signed measures (see Folland [F ]). Below we concentrate on positive, real, and complex measures and will not say more about signed measures here. Suppose  is a complex measure on and de…ne for every A M 2 M

 (A) = sup 1 (An) ; j j n=1 j j where the supremum is taken over all measurable partitions (An)n1=1 of A: Note that  () = 0 and j j  (A) (B) if A; B and A B: j j j j 2 M  The set function  is called the total variation of  or the total variation measure of : It turnsj j out that  is a positive measure. In fact, as will shortly be seen,  is a …nite positivej j measure. j j

Theorem 5.1.1. The total variation  of a complex measure is a positive measure. j j

PROOF. Let (An)n1=1 be a measurable partition of A: 157

For each n; suppose an <  (An) and let (Ekn)1 be a measurable j j k=1 partition of An such that

an < 1 (Ekn) : k=1 j j

Since (Ekn)k;n1 =1 is a partition of A it follows that

1 an < 1 (Ekn)  (A): n=1 k;n=1 j jj j Thus 1  (An)  (A): n=1 j j j j To prove the opposite inequality, let (Ek)k1=1 be a measurable partition of A: Then, since (An Ek)1 is a measurable partition of Ek and (An Ek)1 \ n=1 \ k=1 a measurable partition of An;

1 (Ek) = 1 1 (An Ek) k=1 j j k=1 j n=1 \ j

1 (An Ek) 1  (An)  k;n=1 j \ j n=1 j j and we get  (A) 1  (An): j j  n=1 j j Thus  (A) = 1  (An): j j n=1 j j Since  () = 0, the theorem is proved. j j

Theorem 5.1.2. The total variation  of a complex measure  is a …nite positive measure. j j

PROOF. Since   +  j jj Re j j Im j there is no loss of generality to assume that  is a real measure. Suppose  (E) = for some E : We …rst prove that there exist disjoint sets jA; Bj such1 that 2 M 2 M A B = E [ 158 and (A) > 1 and  (B) = : j j j j 1 To this end let c = 2(1+ (E) ) and let (Ek)k1=1 be a measurable partition of E such that j j n  (Ek) > c k=1 j j for some su¢ ciently large n: There exists a subset N of 1; :::; n such that f g c k N (Ek) > : j 2 j 2 c Set A = k N Ek and B = E A: Then (A) > 1 and [ 2 n j j 2  (B) = (E) (A) j j j j c (A) (E) > (E) = 1: j j j j 2 j j Since =  (E) =  (A)+  (B) we have  (A) = or  1(B) =j j: If j (jB) < j wej interchange Aj andj B and1 have j (Aj) > 1 and1  (Bj ) =j : 1 j j j j 1 Suppose  (X) = : Set E0 = X and choose disjoint sets A0;B0 such that j j 1 2 M A0 B0 = E0 [ and (A0) > 1 and  (B0) = : j j j j 1 Set E1 = B0 and choose disjoint sets A1;B1 such that 2 M

A1 B1 = E1 [ and (A1) > 1 and  (B1) = : j j j j 1 By induction, we …nd a measurable partition (An)n1=0 of the set A =def n1=0An such that (An) > 1 for every n: Now, since  is a complex measure,[ j j

(A) = n1=0(An): But this series cannot converge, since the general term does not tend to zero as n : This contradiction shows that  is a …nite positive measure. ! 1 j j 159

If  is a real measure we de…ne 1 + = (  +) 2 j j and 1  = (  ): 2 j j + The measures  and  are …nite positive measures and are called the positive and negative variations of ; respectively . The representation

+  =   is called the Jordan decomposition of :

Exercises

1. Suppose (X; ; ) is a positive measure space and d = fd; where f L1(): ProveM that d  = f d: 2 j j j j

2. Suppose ; ; and  are real measures de…ned on the same -algebra and   and  : Prove that    min(; )  where 1 min(; ) = ( +    ): 2 j j

3. Suppose  : C is a complex measure and f; g : X R measurable functions. ShowM! that !

(f A) (g A)  (f = g) j 2 2 jj j 6 160 for every A : 2 R

5.2. The Lebesque Decomposition and the Radon-Nikodym Theo- rem

Let  be a positive measure on and  a positive or complex measure on : The measure  is said to beM absolutely continuous with respect to  (abbreviatedM  << ) if (A) = 0 for every A for which (A) = 0: If we de…ne 2 M  = A ; (A) = 0 Z f 2 M g it follows that  <<  if and only if

 : Z  Z

For example, n << vn and vn << n: The measure  is said to be concentrated on E if  = E , where E 2 M  (A) =def (E A) for every A : This is equivalent to the hypoth- \ 2 M esis that A  if A and A E = : Thus if E1;E2 , where 2 Z 2 M \ 2 M E1 E2; and  is concentrated on E1; then  is concentrated on E2: More-  over, if E1;E2 and  is concentrated on both E1 and E2; then  is 2 M concentrated on E1 E2: Two measures 1 and 2 are said to be mutually \ singular (abbreviated 1 2) if there exist disjoint measurable sets E1 and ? E2 such that 1 is concentrated on E1 and 2 is concentrated on E2:

Theorem 5.2.1. Let  be a positive measure and ; 1; and 2 complex measures. (i) If 1 <<  and 2 << ; then ( 11 + 22) <<  for all complex numbers 1 and 2: (ii) If 1  and 2 ; then ( 11 + 22)  for all complex ? ? ? numbers 1 and 2: (iii) If  <<  and  ; then  = 0: (iv) If  << ; then ? << : j j

PROOF. The properties (i) and (ii) are simple to prove and are left as exer- cises. 161

To prove (iii) suppose E is a -null set and  = E: If A , then (A) = (A E) and A E2 Mis a -null set. Since  <<  it follows2 M that \ \ A E Z and, hence, (A) = (A E) = 0: This proves (iii) \ 2 \ To prove (iv) suppose A and (A) = 0: If (An)1 is measurable 2 M n=1 partition of A; then (An) = 0 for every n: Since  << ; (An) = 0 for every n and we conclude that  (A) = 0: This proves (vi). j j

Theorem 5.2.2. Let  be a positive measure on and  a complex measure on : Then the following conditions are equivalent:M M(a)  << : (b) To every " > 0 there corresponds a  > 0 such that (E) < " for all E with (E) < : j j 2 M

If  is a positive measure, the implication (a) (b) in Theorem 5.2.2 is, ) in general, wrong. To see this take  = 1 and  = v1: Then  <<  and if we choose An = [n; [ ; n N+; then (An) 0 as n but (An) = for each n: 1 2 ! ! 1 1

PROOF. (a) (b). If (b) is wrong there exist an " > 0 and sets En , ) n 2 M n N+; such that (En) " and (En) < 2 : Set 2 j j

An = 1 Ek and A = 1 An: [k=n \n=1 n+1 Since An An+1 A and (An) < 2 , it follows that (A) = 0 and   using that  (An) (En) ; Theorem 1.1.2 (f) implies that j j j j

 (A) = lim  (An) ": n j j !1 j j  This contradicts that  << : j j

(b) (a). If E and (E) = 0 then to each " > 0; (E) < "; and we conclude) that 2(E M) = 0: The theorem is proved: j j 162

Theorem 5.2.3. Let  be a -…nite positive measure and  a real measure on . M(a) (The Lebesgue Decomposition of ) There exists a unique pair of real measures a and s on such that M  = a + s; a << ; and s : ? If  is a …nite positive measure, a and s are …nite positive measures. (b) (The Radon-Nikodym Theorem) There exits a unique g L1() such that 2 da = gd: If  is a …nite positive measure, g 0 a.e. [] : 

The proof of Theorem 5.2.3 is based on the following

Lemma 5.2.1. Let (X; M; ) be a …nite positive measure space and suppose f L1(): 2(a) If a R and 2 fd a(E); all E  2 M ZE then f a a.e. []. (b) If b R and 2 fd b(E); all E  2 M ZE then f b a.e. []. 

PROOF. (a) Set g = f a so that gd 0; all E :  2 M ZE Now choose E = g > 0 to obtain f g 0 gd = gd 0  E  ZE ZX 163 as g 0 a.e. [] : But then Example 2.1.2 yields g = 0 a.e. [] and we E  E get E Z: Thus g 0 a.e. [] or f a a.e. [] : 2   Part (b) follows in a similar way as Part (a) and the proof is omitted here.

(k) (k) PROOF. Uniqueness: (a) Suppose a and s are real measures on such that M  = (k) + (k); (k) << ; and (k)  a s a s ? for k = 1; 2: Then (1) (2) = (2) (1) a a s s and (1) (2) <<  and (1) (2) : a a a a ? (1) (2) (1) (2) Thus by applying Theorem 5.2.1, a a = 0 and a = a : From this (1) (2) we conclude that s = s . 1 (b) Suppose gk L (); k = 1; 2; and 2 da = g1d = g2d:

Then hd = 0 where h = g1 g2: But then hd = 0 h>0 Zf g and it follows that h 0 a.e. [] : In a similar way we prove that h 0 a.e. 1 1  []. Thus h = 0 in L (); that is g1 = g2 in L ():

Existence: The beautiful proof that follows is due to von Neumann. First suppose that  and  are …nite positive measures and set  = +: Clearly, L1() L1() L2(): Moreover, if f : X R is measurable   !

f d f d f 2d (X) X j j  X j j  s X Z Z Z p and from this we conclude that the map

f fd ! ZX 164 is a continuous linear functional on L2(): Therefore, in view of Theorem 4.2.2, there exists a g L2() such that 2 fd = fgd all f L2(): 2 ZX ZX Suppose E and put f = to obtain 2 M E 0 (E) = gd  ZE and, since  ;  0 gd (E):   ZE But then Lemma 5.2.1 implies that 0 g 1 a.e. [] : Therefore, without loss of generality we can assume that 0 g(x) 1 for all x X and, in addition, as above   2

fd = fgd all f L2() 2 ZX ZX that is f(1 g)d = fgd all f L2(): 2 ZX ZX A S Put A = 0 g < 1 , S = g = 1 ; a =  ; and s =  : Note that A S f  g f g  =  +  : The choice f = S gives (S) = 0 and hence s : Moreover, the choice ? n f = (1 + ::: + g ) E where E ; gives 2 M (1 gn+1)d = (1 + ::: + gn)gd: ZE ZE By letting n and using monotone convergence ! 1 (E A) = hd: \ ZE where h = lim (1 + ::: + gn)g: n !1 165

Since h is non-negative and

(A) = hd ZX it follows that h L1(): Moreover, the construction above shows that  = 2 a + s: In the next step we assume that  is a -…nite positive measure and  a …nite positive measure. Let (Xn)n1=1 be a measurable partition of X such that (Xn) < for every n: Let n be …xed and apply Part (a) to the pair Xn Xn 1 Xn Xn  and  to obtain …nite positive measures ( )a and ( )s such that

Xn Xn Xn Xn Xn Xn Xn  = ( )a + ( )s; ( )a <<  ; and ( )s  ? and Xn Xn Xn Xn d( )a = hnd (or ( )a = hn )

1 Xn where 0 hn L ( ): Without loss of generality we can assume that  2 Xn hn = 0 o¤ Xn and that ( )s is concentrated on An Xn where An : Xn  2 Z In particular, ( )a = hn: Now

Xn  = h + n1=1( )s where

h = n1=1hn and hd (X) < :  1 ZX 1 Xn Thus h L (): Moreover, s =def 1 ( )s is concentrated on 1 An 2 n=1 [n=1 2 : Hence s : Z Finally if ?is a real measure we apply what we have already proved to the positive and negative variations of  and we are done.

Example 5.2.1. Let  be Lebesgue measure in the unit interval and  the counting measure in the unit interval restricted to the class of all Lebesgue measurable subsets of the unit interval. Clearly,  << : Suppose there is an 166 h L1() such that d = hd. We can assume that h 0 and the Markov inequality2 implies that the set h " is …nite for every" > 0: But then f  g n (h ]0; 1]) = lim (h 2 ) = 0 n 2 !1  and it follows that 1 = (h = 0) = h=0 hd = 0; which is a contradiction. f g R

Corollary 5.2.1. Suppose  is a real measure. Then there exists

h L1(  ) 2 j j such that h(x) = 1 for all x X and j j 2 d = hd  : j j

PROOF. Since (A)  (A) for every A , the Radon-Nikodym Theorem impliesj that djj= hdj  for an appropriate2 M h L1(  ): But then d  = h d  (see Exercisej j 1 in Chapter 5.1): Thus2 j j j j j j j j  (E) = h d  ; all E j j j j j j 2 M ZE and Lemma 5.2.1 yields h = 1 a.e. [  ] : From this the theorem follows at once. j j

Theorem 5.2.4. (Hahn’s Decomposition Theorem) Suppose  is a real measure. There exists an A such that 2 M + A Ac  =  and  =  :

PROOF. Let d = hd  where h = 1: Note that hd = d  : Set A = h = 1 : Then j j j j j j f g 1 1 d+ = (d  +d) = (h + 1)d = d 2 j j 2 A 167 and + d = d d = ( 1)d = c d: A A The theorem is proved.

If a real measure  is absolutely continuous with respect to a -…nite positive measure , the Radon-Nikodym Theorem says that d = fd for an approprite f L1(): We sometimes write 2 d f = d and call f the Radon-Nikodym derivate of  with respect to :

Exercises

1. Let  be a -…nite positive measure on (X; ) and (fn)n N a sequence of measurable functions which converges in -measureM to a measurable2 function f: Moreover, suppose  is a …nite positive measure on (X; ) such that M  << : Prove that fn f in -measure. !

2. Suppose  and n; n N, are positive measures de…ned on the same 2 -algebra and set  = n1=0n. Prove that a)   if n ; all n N: ? ? 2 b)  <<  if n << ; all n N: 2

3. Suppose  is a real measure and  = 1 2; where 1 and 2 are …nite + positive measures. Prove that 1  and 2 :  

4. Let 1 and 2 be mutually singular complex measures on the same - algebra: Show that 1 2 : j j?j j 168

5. Let (X; ; ) be a -…nite positive measure space and suppose  and  are two probabilityM measures de…ned on the -algebra such that  <<  and  << : Prove that M 1 d d sup (A) (A) = d: A j j 2 X j d d j 2M Z 6. Let (X; ) be a measurable space and suppose ; : R and are real measures. ProveM that M! ( + )+ + + +: 

5.3. The Wiener Maximal Theorem and the Lebesgue Di¤erentia- tion Theorem

We say that a Lebesgue measurable function f in Rn is locally Lebesgue in- 1 1 tegrable and belongs to the class Lloc(mn) if f K L (mn) for each compact n 1 2 subset K of R : In a similar way f Lloc(vn) if f is a Borel function such 1 2 n 1 that f L (vn) for each compact subset K of R : If f L (mn); we K 2 2 loc de…ne the average Arf(x) of f on the open ball B(x; r) as 1 A f(x) = f(y)dy: r m (B(x; r)) n ZB(x;r)

It follows from dominated convergence that the map (x; r) Arf(x) of Rn ]0; [ into R is continuous. The Hardy-Littlewood maximal! function  1 f  is, by de…nition, f  = sup Ar f or, stated more explicitly, r>0 j j

1 n f (x) = sup f(y) dy; x R : r>0 m (B(x; r)) j j 2 n ZB(x;r) n n The function f  :(R ; (R )) ([0; ] ; 0; ) is measurable since B ! 1 R 1

f  = sup Ar f : r>0 j j r Q 2 169

Theorem 5.3.1. (Wiener’sMaximal Theorem) There exists a positive 1 constant C = Cn < such that for all f L (mn); 1 2 C mn(f  > ) f 1 if > 0:  k k

The proof of the Wiener Maximal Theorem is based on the following remarkable result.

n Lemma 5.3.1. Let be a collection of open balls in R and set V = B B: C [ 2C If c < mn(V ) there exist pairwise disjoint B1; :::; Bk such that 2 C k n i=1mn(Bi) > 3 c:

PROOF. Let K V be compact with mn(K) > c; and suppose A1; :::; Ap  2 C cover K: Let B1 be the largest of the Ai0 s (that is, B1 has maximal radius), let B2 be the largest of the Ai0 s which are disjoint from B1; let B3 be the largest of the Ai0 s which are disjoint from B1 B2; and so on until the process [ k stops after k steps. If Bi = B(xi; ri) put Bi = B(xi; 3ri): Then i=1Bi K and [  k n k c < i=1mn(Bi) = 3 i=1mn(Bi): The lemma is proved.

PROOF OF THEOREM 5.3.1. Set

E = f  > : f g

For each x E choose an rx > 0 such that Ar f (x) > : If c < mn(E ); 2 x j j by Lemma 5.3.1 there exist x1; :::; xk E such that the balls Bi = B(xi; rxi ); i = 1; :::; k; are mutually disjoint and2

k n i=1mn(Bi) > 3 c: But then n n n k 3 k 3 c < 3 i=1mn(Bi) < i=1 f(y) dy f(y) dy: j j  n j j ZBi ZR 170

The theorem is proved.

1 Theorem 5.3.2. If f L (mn); 2 loc 1 lim f(y)dy = f(x) a.e. [mn] : r 0 m (B(x; r)) ! n ZB(x;r)

1 PROOF. Clearly, there is no loss of generality to assume that f L (mn): n n 2 Suppose g Cc(R ) =def f C(R ); f(x) = 0 if x large enough . Then 2 f 2 j j g n lim Arg(x) = g(x) all x R : r 0 ! 2

Since Arf f = Ar(f g) (f g) + Arg g;

lim Arf f (f g)+ f g : r 0 ! j j j j Now, for …xed > 0; mn(lim Arf f > ) r 0 ! j j

mn((f g) > ) + mn( f g > )  2 j j 2 and the Wiener Maximal Theorem and the Markov Inequality give

mn(lim Arf f > ) r 0 ! j j 2C 2 ( + ) f g 1 :  k k n 1 Remembering that Cc(R ) is dense in L (mn); the theorem follows at once.

1 If f L (mn) we de…ne the so called Lebesgue set Lf to be 2 loc 1 Lf = x; lim f(y) f(x) dy = 0 : r 0 m (B(x; r)) j j  ! n ZB(x;r)  Note that if q is real and 1 Eq = x; lim f(y) q dy = f(x) q r 0 m (B(x; r)) j j j j  ! n ZB(x;r)  171

c then mn( q QEq ) = 0: If x q QEq; [ 2 2 \ 2 1 lim f(y) f(x) dy 2 f(x) q r 0 m (B(x; r)) j j  j j ! n ZB(x;r) c for all rational numbers q and it follows that mn(Lf ) = 0: n A family x = (Ex;r)r>0 of Borel sets in R is said to shrink nicely to a nE point x in R if Ex;r B(x; r) for each r and there is a positive constant ;  independent of r; such that mn(Ex;r) mn(B(x; r)): 

Theorem 5.3.3. (The Lebesgue Di¤erentiation Theorem) Suppose 1 f L (mn) and x Lf : Then 2 loc 2 1 lim f(y) f(x) dy = 0 r 0 m (E ) j j ! n x;r ZEx;r and 1 lim f(y)dy = f(x): r 0 m (E ) ! n x;r ZEx;r

PROOF. The result follows from the inequality 1 1 f(y) f(x) dy f(y) f(x) dy: m (E ) j j  m (B(x; r)) j j n x;r ZEx;r n ZB(x;r)

Theorem 5.3.4. Suppose  is a real or positive measure on n and suppose R  vn: If  is a positive measure it is assumed that (K) < for every compact? subset of Rn. Then 1

(Ex;r) lim = 0 a.e. [vn] r 0 v (E ) ! n x;r

If Ex;r = B(x; r) and  is the counting measure cQn restricted to n then Rn  vn but the limit in Theorem 5.3.4 equals plus in…nity for all x R : The hypothesis? "(K) < for every compact subset of Rn" in Theorem2 5.3.4 is not super‡ous. 1 172

PROOF. Since (E)  (E) if E n; there is no restriction to assume that  is a positivej measurejj j (cf. Theorem2 R 3.1.4). Moreover, since

(Ex;r) (B(x; r))

vn(Ex;r)  vn(B(x; r)) it can be assumed that Ex;r = B(x; r): Note that the function (B( ; r)) is Borel measurable for …xed r > 0 and (B(x; )) left continuous for …xed
x Rn:
2 A Suppose A  and vn = (vn) : Given  > 0; it is enough to prove that 2 Z F vn where 2 Z (B(x; r)) F = x A; lim >  2 r 0 m (B(x; r))  ! n  To this end let " > 0 and use Theorem 3.1.3 to get an open U A such that  (U) < ": For each x F there is an open ball Bx U such that 2 

(Bx) > vn(Bx):

If V = x F Bx and c < vn(V ) we use Lemma 5.3.1 to obtain x1; :::; xk such [ 2 that Bx1 ; :::; Bxk are pairwise disjoint and

n k n 1 k c < 3 i=1vn(Bxi ) < 3  i=1(Bxi )

n 1 n 1 3  (U) < 3  ":  n 1 Thus vn(V ) 3  ": Since V F n and " > 0 is arbitrary, vn(F ) = 0 and the theorem is proved.  2 R

Corollary 5.3.1. Suppose F : R R is an increasing function. Then F 0(x) exists for almost all x with respect! to linear measure.

PROOF. Let D be the set of all points of discontinuity of F: Suppose < 1 a < b < and " > 0: If a < x1 < ::: < xn < b; where x1; :::; xn D and 1 2

F (xk+) F (xk ) "; k = 1; :::; n  173 then n n"  (F (xk+) F (xk )) F (b) F (a):  k=1  Thus D [a; b] is at most denumerable and it follows that D is at most \ N denumerable. Set H(x) = F (x+) F (x); x R; and let (xj)j=0 be an enumeration of the members of the set H > 0 2: Moreover, for any a > 0; f g

H(xj) (F (xj+) F (xj ))  xj

N (A) =  H(xj)x (A);A j=0 j 2 R then  is a positive measure such that (K) < for each compact subset K of R. Furthermore, if h is a non-zero real number1 ; 1 1 1 (H(x + h) H(x) (H(x + h) + H(x)) 4 (B(x; 2 h ) j h j h  4 h j j j j j j and Theorem 5.3.4 implies that H0(x) = 0 a.e. [v1]. Therefore, without loss of generality it may be assumed that F is right continuous and, in addition, there is no restriction to assume that F (+ ) F ( ) < : By Section 1.6 F induces a …nite positive1 Borel 1 measure1 such that

(]x; y]) = F (y) F (x) if x < y: Now consider the Lebesgue decomposition

d = fdv1 + d

1 where f L (v1) and  v1: If x < y; 2 ? y F (y) F (x) = f(t)dt + (]x; y]) Zx and the previous two theorems imply that

F (y) F (x) lim = f(x) a.e. [v1] y x y x # 174

If y < x; x F (x) F (y) = f(t)dt + (]y; x]) Zy and we get F (y) F (x) lim = f(x) a.e. [v1] : y x y x " The theorem is proved.

Exercises

1 1. Suppose F : R R is increasing and let f L (v1) be such that ! 2 loc F 0(x) = f(x) a.e. [v1] : Prove that

y f(t)dt F (y) F (x) if < x y < :  1  1 Zx

5.4. Absolutely Continuous Functions and Functions of Bounded Variation

Throughout this section a and b are reals with a < b and to simplify notation 1 we set ma;b = m [a;b]: If f L (ma;b) we know from the previous section that the function j 2 x (If)(x) =def f(t)dt; a x b   Za has the derivative f(x) a.e. [ma;b] ; that is

d x f(t)dt = f(x) a.e. [m ] : dx a;b Za Our next main task will be to describe the range of the linear map I: A function F :[a; b] R is said to be absolutely continuous if to every " > 0 there exists a  > !0 such that

n n  bi ai <  implies  F (bi) F (ai) < " i=1 j j i=1 j j 175

whenever ]a1; b1[ ; :::; ]an; bn[ are disjoint open subintervals of [a; b]. It is ob- vious that an absolutely continuous function is continuous. It can be proved that the Cantor function is not absolutely continuous.1.

1 Theorem 5.4.1. If f L (ma;b); then If is absolutely continuous. 2

PROOF. There is no restriction to assume f 0: Set 

d = fdma;b:

By Theorem 5.2.2, to every " > 0 there exists a  > 0 such that (A) < " for each Lebesgue set A in [a; b] such that ma;b(A) < : Now restricting A to be a …nite disjoint union of open intervals, the theorem follows.

Suppose < and F :] ; [ R: For every x ] ; [ we de…ne 1   1 ! 2 n TF (x) = sup i=1 F (xi) F (xi 1) j j where the supremum is taken over all positive integers n and all choices n (xi)i=0 such that < x0 < x1 < ::: < xn = x < :

The function TF :] ; [ [0; ] is called the total variation of F: Note that ! 1 TF is increasing. If TF is a bounded function, F is said to be of bounded varia- tion. A bounded increasing function on R is of bounded variation. Therefore the di¤erence of two bounded increasing functions on R is of bounded vari- ation. Interestingly enough, the converse is true. In the special case ] ; [ = R we write F BV if F is of bounded variation. 2

Example 5.4.1. Let f:R R be a Lebesgue integrable function and de…ne !

1 y g(x) = ej jf(x y)dy if x R: 2 Z1 176

We claim that g is a continuous function of bounded variation. x To prov this claim put h(x) = ej j if x R so that 2 1 g(x) = h(x y)f(y)dy. Z1 We …rst prove that the function h is continuous. To this end suppose

(an)n N+ is a sequence of real numbers which converges to a real number a: Then2 h(an y)f(y)) f(y)) if n N+ and y R j j2j j 2 2 and since f 1(m) by dominated convergence, 2 L 1 lim g(an) = lim h(an y)f(y)dy = n n !1 Z1 !1 1 h(a y)f(y)dy = g(a) Z1 and it follows that g is continuous. We next prove that the function h is of bounded variation. Recall that the total variation function Th(x) of h at the point x is the supremum of all sums of the type n

h(xi) h(xi 1) i=1 j j X where < x0 < x1 < ::: < xn = x < : 1 1 We claim that h is the di¤erence of two bounded increasing functions. Setting

(x) = emin(0;x) and observing that h(x) = (x) + ( x) 1 the claim above is obvious and

C =def sup Th < : 1 Moreover, if < x0 < x1 < ::: < xn < ; 1 1 n

g(xi) g(xi 1) = i=1 j j X 177

n 1 1 h(xi y)f(y)dy h(xi 1 y)f(y)dy i=1 j j X Z1 Z1 n 1 h(xi y) h(xi 1 y) f(y) dy  i=1 j jj j X Z1 n 1 h(xi y) h(xi 1 y) f(y) dy i=1 j j! j j Z1 X 1 1 C f(y) dy = C f(y) dy < :  j j j j 1 Z1 Z1 Hence g is of bounded variation.

Theorem 5.4.2. Suppose F BV: 2 (a) The functions TF + F and TF F are increasing and 1 1 F = (TF + F ) (TF F ): 2 2 In particular, F is di¤erentiable almost everywhere with respect to linear measure. (b) If F is right continuous, then so is TF :

PROOF. (a) Let x < y and " > 0: Choose x0 < x1 < ::: < xn = x such that

n i=1 F (xi) F (xi 1) Tf (x) ": j j Then TF (y) + F (y) n i=1 F (xi) F (xi 1) + F (y) F (x) +(F (y) F (x)) + F (x)  j j j j TF (x) " + F (x)  and, since " > 0 is arbitrary, TF (y) + F (y) TF (x) + F (x): Hence TF + F is  increasing. Finally, replacing F by F it follows that the function TF F is increasing. 178

(b) If c R and x > c; 2 n Tf (x) = TF (c) + sup i=1 F (xi) F (xi 1) j j where the supremum is taken over all positive integers n and all choices n (xi)i=0 such that c = x0 < x1 < ::: < xn = x:

Suppose TF (c+) > TF (c) where c R: Then there is an " > 0 such that 2

TF (x) TF (c) > " for all x > c: Now, since F is right continuous at the point c, for …xed x > c there exists a partition

c < x11 < ::: < x1n1 = x such that n1 i=2 F (x1i) F (x1i 1) > ": j j But TF (x11) TF (c) > " and we get a partition

c < x21 < ::: < x2n2 = x11 such that n2 i=2 F (x2i) F (x2i 1) > ": j j Summing up we have got a partition of the interval [x21; x] with

n2 n1 i=2 F (x2i) F (x2i 1) +i=2 F (x1i) F (x1i 1) > 2": j j j j By repeating the process the total variation of F becomes in…nite, which is a contradiction. The theorem is proved.

Theorem 5.4.3. Suppose F :[a; b] R is absolutely continuous. Then 1 ! there exists a unique f L (ma;b) such that 2 x F (x) = F (a) + f(t)dt; a x b:   Za 179

In particular, the range of the map I equals the set of all real-valued absolutely continuous maps on [a; b] which vanish at the point a:

PROOF. Set F (x) = F (a) if x < a and F (x) = F (b) if x > b: There exists a  > 0 such that

n n  bi ai <  implies  F (bi) F (ai) < 1 i=1 j j i=1 j j whenever ]a1; b1[ ; :::; ]an; bn[ are disjoint subintervals of [a; b] : Let p be the least positive integer such that a + p b: Then TF p and F BV: Let 1  1  2 F = G H; where G = (TF + F ) and H = (TF F ): There exist …nite 2 2 positive Borel measures G and H such that

G(]x; y]) = G(y) G(x); x y  and H (]x; y]) = H(y) H(x); x y:  If we de…ne  = G H ; (]x; y]) = F (y) F (x); x y:  Clearly, (]x; y[) = F (y) F (x); x y  since F is continuous. Our next task will be to prove that  << v1. To this end, suppose A 2 R and v1(A) = 0: Now choose " > 0 and let  > 0 be as in the de…nition of the absolute continuity of F on [a; b] : For each k N+ there exists an open set 2 Vk A such that v1(Vk) <  and limk (Vk) = (A): But each …xed Vk is  !1 a disjoint union of open intervals (]ai; bi[)i1=1 and hence

n  bi ai <  i=1 j j for every n and, accordingly from this,

1 F (bi) F (ai) " i=1 j j and (Vk) 1 (]ai; bi[) ": j j i=1 j j 180

Thus (A) " and since " > 0 is arbitrary, (A) = 0: From this  << v1 and thej theoremj follows at once.

Suppose (X; ; ) is a positive measure space. From now on we write 1 M 1 c f L () if there exist a g () and an A such that A  and f(2x) = g(x) for all x A: Furthermore,2 L we de…ne2 M 2 Z 2

fd = gd ZX ZX (cf the discussion in Section 2). Note that f(x) need not be de…ned for every x X: 2

Corollary 5.4.1. A function f :[a; b] R is absolutely continuous if and only if the following conditions are true:! (i) f 0(x) exists for ma;b-almost all x [a; b] 1 2 (ii) f 0 L (ma;b) 2 x (iii) f(x) = f(a) + f 0(t)dt; all x [a; b] : a 2 R

Exercises

1. Suppose f : [0; 1] R satis…es f(0) = 0 and ! 1 f(x) = x2 sin if 0 < x 1: x2  Prove that f is di¤erentiable everywhere but f is not absolutely continuous.

2. Suppose is a positive real number and f a function on [0; 1] such that 1 f(0) = 0 and f(x) = x sin x ; 0 < x 1. Prove that f is absolutely continuous if and only if > 1:  181

3. Suppose f(x) = x cos(=x) if 0 < x < 2 and f(x) = 0 if x R ]0; 2[ : Prove that f is not of bounded variation on R. 2 n

4 A function f :[a; b] R is a Lipschitz function, that is there exists a positive real number C !such that

f(x) f(y) C x y j j j j for all x; y [a; b] : Show that f is absolutely continuous and f 0(x) C 2 j j a.e. [ma;b] :

5. Suppose f :[a; b] R is absolutely continuous. Prove that ! x Tg(x) = f 0(t) dt; a < x < b j j Za if g is the restriction of f to the open interval ]a; b[ :

6. Suppose f and g are real-valued absolutely continuous functions on the compact interval [a; b]. Show that the function h = max(f; g) is absolutely continuous and h0 max(f 0; g0) a.e. [ma;b]. 

7. Suppose (X; ; ) is a …nite positive measure space and f L1(): De…ne M 2 g(t) = f(x) t d(x); t R: j j 2 ZX Prove that g is absolutely continuous and

t g(t) = g(a) + ((f s) (f s))ds if a; t R:   2 Za

8. Let  and  be probability measures on (X; ) such that   (X) = 2: Show that  : M j j ? ### 182

5.5. Conditional Expectation

Let ( ; ;P ) be a probability space and suppose  L1(P ): Moreover, supposeF is a -algebra and set 2 G  F (A) = P [A] ;A 2 G and (A) = dP; A : 2 G ZA It is trivial that  = P  and the Radon-Nikodym Theorem shows there exists a uniqueZ Z \L G1( ) Zsuch that 2 (A) = d all A 2 G ZA or, what amounts to the same thing,

dP = dP all A : 2 G ZA ZA Note that  is ( ; )-measurable. The random variable  is called the con- ditional expectationG R of  given and it is standard to write  = E [ ] : G j G A sequence of -algebras ( n)1 is called a …ltration if F n=1

n n+1 : F  F  F

If ( n)n1=1 is a …ltration and (n)n1=1 is a sequence of real valued random variablesF such that for each n;

(a)  L1(P ) n 2 (b)  is ( n; )-measurable n F R (c) E  n =  n+1 j F n   then (n; n)n1=1 is called a martingale. There are very nice connections between martingalesF and the theory of di¤erentiation (see e.g Billingsley [B] and Malliavin [M]): """ 183 CHAPTER 6 COMPLEX INTEGRATION

Introduction

In this section, in order to illustrate the power of Lebesgue integration, we collect a few results, which often appear with uncomplete proofs at the un- dergraduate level.

6.1. Complex Integrand

So far we have only treated integration of functions with their values in R or [0; ] and it is the purpose of this section to discuss integration of complex valued1 functions. Suppose (X; ; ) is a positive measure. Let f; g L1(): We de…ne M 2

(f + ig)d = fd + i gd: ZX ZX ZX If and are real numbers,

( + i )(f + ig)d = (( f g) + i( g + f))d ZX ZX

= ( f g)d + i ( g + f)d ZX ZX = fd gd + i gd + i fd ZX ZX ZX ZX = ( + i )( fd + i gd) ZX ZX = ( + i ) (f + ig)d: ZX 184

We write f L1(; C) if Re f; Im f L1() and have, for every f L1(; C) and complex2 ; 2 2 fd = fd: ZX ZX Clearly, if f; g L1(; C); then 2 (f + g)d = fd + gd: ZX ZX ZX Now suppose  is a complex measure on : If M f L1(; C) = L1( ; C) L1( ; C) 2 def Re \ Im we de…ne

fd = fdRe + i fdIm: ZX ZX ZX It follows for every f; g L1(; C) and C that 2 2 fd = fd: ZX ZX and (f + g)d = fd + gd: ZX ZX ZX

### 6.2. The Fourier Transform

n Below, if x = (x1; :::; xn) and y = (y1; :::; yn) R ; we let 2 n x; y =  xkyk: h i k=1 and x = x; y : j j h i If  is a complex measure on n (orp n) the Fourier transform ^ of  is de…ned by R R i x;y n ^(y) = e h id(x); y R : n 2 ZR 185

Note that ^(0) = (Rn): 1 The Fourier transform of a function f L (mn; C) is de…ned by 2 ^ f(y) = ^(y) where d = fdmn:

n Theorem 6.2.1. The canonical Gaussian measure n in R has the Fourier transform y 2 j 2j ^n(y) = e :

PROOF. Since = ::: (n factors) n 1 1 it is enough to consider the special case n = 1: Set

1 x2 g(y) = ^ (y) = e 2 cos xydx: 1 p2 ZR Note that g(0) = 1: Since cos x(y + h) cos xy x j h jj j the Lebesgue Dominated Convergence Theorem yields

1 x2 g0(y) = xe 2 sin xydx p2 ZR (Exercise: Prove this by using Example 2.2.1). Now, by partial integration,

2 x= 2 1 x 1 y x g0(y) = e 2 sin xy e 2 cos xydx p2 x= p2 R h i 1 Z that is g0(y) + yg(y) = 0 and we get y2 g(y) = e 2 : 186

If  = ( ; :::;  ) is an Rn-valued random variable with  L1(P ); 1 n k 2 k = 1; :::; n; the characteristic function c of  is de…ned by

i ;y n c(y) = E e h i = P^( y); y R : 2 For example, if  N(0; ); then  = G; where G N(0; 1); and we get 2 2 i G;y c(y) = E e h i = ^ ( y) 1

 2y2 = e 2 : Choosing y = 1 results in

i 1 E[2] E e = e 2 if  N(0; ): 2 n   Thus if (k)k=1 is a centred real-valued Gaussian process

n 1 i ykk n 2 E e k=1 = exp( E ( yk ) 2 k=1 k     1 n 2 2 = exp( k=1ykE k 1 j

n Theorem 6.2.2. Let (k)k=1 be a centred real-valued Gaussian process with uncorrelated components, that is

E   = 0; j = k: j k 6   187

Then the random variables 1; :::; n are independent.

6.3 Fourier Inversion

1 ^ 1 Theorem 6.3.1. Suppose f L (mn): If f L (mn) and f is bounded and continuous 2 2 i y;x ^ dy n f(x) = e h if(y) n ; x R : d (2) 2 ZR

PROOF. Choose " > 0: We have

"2 2 dy "2 2 dy i y;x 2 y ^ i y;x u 2 y e h ie j j f(y) n = f(u) e h ie j j n du n (2) n n (2) ZR ZR ZR  where the right side equals

x u 1 2 dv du 1 2 du i v; v 2 u x f(u) e h " ie 2 j j n n = f(u)e 2" j j n n n p2 p2 "n n p2 "n ZR ZR  ZR

1 z 2 dz = f(x + "z)e 2 j j n : n p2 ZR Thus "2 2 dy 1 2 dz i y;x 2 y ^ 2 z e h ie j j f(y) n = f(x + "z)e j j n : n (2) n p2 ZR ZR By letting " 0 and using the Lebesgue Dominated Convergence Theorem, Theorem 6.3.1! follows at once.

n n Recall that Cc1(R ) denotes the class of all functions f : R R with compact support which are in…nitely many times di¤erentiable. If!f n ^ 1 2 Cc1(R ) then f L (mn). To see this, suppose yk = 0 and use partial integration to obtain2 6 1 f^(y) = e i x;y f(x)dx = e i x;y f (x)dx h i h i x0 k d iy d ZR k ZR and 1 i x;y (l) f^(y) = e h if (x)dx; l N: l xk (iy ) d 2 k ZR 188

Thus y l f^(y) f (l)(x) dx; l N k xk j j j j n j j 2 ZR and we conclude that

sup (1+ y )n+1 f^(y) < : y Rn j j j j 1 2 ^ 1 and, hence, f L (mn): 2

n ^ 1 Corollary 6.3.1. If f C1(R ); then f L (mn) and 2 c 2

i y;x ^ dy n f(x) = e h if(y) n ; x R : n (2) 2 ZR

Corollary 6.3.2 If  is a complex Borel measure in Rn and ^ = 0; then  = 0:

n f^(y) PROOF. Choose f Cc1(R ). We multiply the equation ^( y) = 0 by (2)n and integrate over R2n with respect to Lebesgue measure to obtain

f(x)d(x) = 0: n ZR n Since f C1(R ) is arbitrary it follows that  = 0: The theorem is proved. 2 c

6.4. Non-Di¤erentiability of Brownian Paths

Let ND denote the set of all real-valued continuous function de…ned on the unit interval which are not di¤erentiable at any point. It is well known that ND is non-empty. In fact, if  is Wiener measure on C [0; 1], x ND a.e. [] : The purpose of this section is to prove this important property2 of Brownian motion. 189

Let W = (W (t))0 t 1 be a real-valued Brownian motion in the time interval [0; 1] such that every path t W (t); 0 t 1 is continuous. Recall that !   E [W (t)] = 0 and E [W (s)W (t)] = min(s; t): If 0 t0 ::: tn 1     and 1 j < k n  

E [(W (tk) W (tk 1))(W (tj) W (tj 1)]

= E [(W (tk)W (tj)] E [W (tk)W (tj 1)] E [W (tk 1)W (tj)]+E [W (tk 1)W (tj 1)] = tj tj 1 tj + tj 1 = 0: From the previous section we now infer that the random variables

W (t1) W (t0); :::; W (tn) W (tn 1) are independent.

Theorem 7. The function t W (t); 0 t 1 is not di¤erentiable at any point t [0; 1] a.s. [P ] : !   2

PROOF. Without loss of generality we assume the underlying probability space is complete. Let c; " > 0 and denote by B(c; ") the set of all ! such that 2

W (t) W (s) < c t s if t [s "; s + "] [0; 1] j j j j 2 \ for some s [0; 1] : It is enough to prove that the set 2 1 1 1 B(j; ): k j=1 k=1 [ [ is of probability zero. From now on let c; " > 0 be …xed. It is enough to prove P [B(c; ")] = 0 : 190

Set j + 1 j Xn;k = max W ( ) W ( ) k j 3 and k 0; :::; n 3 : Let n > 3 be so large that2 f g 3 ": n  We claim that 6c B(c; ") min Xn;k :  0 k n 3  n     If ! B(c; ") there exists an s [0; 1] such that 2 2 W (t) W (s) c t s if t [s "; s + "] [0; 1] : j j j j 2 \ Now choose k 0; :::; n 3 such that 2 f g k k 3 s ; + : 2 n n n   If k j < k + 3;  j + 1 j j + 1 j W ( ) W ( ) W ( ) W (s) + W (s) W ( ) j n n jj n j j n j 6c  n 6c and, hence, Xn;k : Now  n 6c B(c; ") min Xn;k  0 k n 3  n     and it is enough to prove that

6c lim P min Xn;k = 0: n 0 k n 3  n !1     But n 3 6c 6c P min Xn;k P Xn;k 0 k n 3  n   n   k=0   X   191

6c 6c = (n 2)P Xn;0 nP Xn;0  n   n     1 6c 6c = n(P W ( ) )3 = n(P ( W (1) )3 j n j n j j pn   12c n( )3:  p2n where the right side converges to zero as n . The theorem is proved. ! 1

Recall that a function of bounded variation possesses a derivative a.e. with respect to Lebesgue measure. Therefore, with probability one, a Brown- ian path is not of bounded variation. In view of this an integral of the type

1 f(t)dW (t) Z0 cannot be interpreted as an ordinary Stieltjes integral. Nevertheless, such an integral can be de…ned by completely di¤erent means and is basic in, for example, …nancial mathematics.

""" 192 CHAPTER 6 COMPLEX INTEGRATION

Introduction

In this section, in order to illustrate the power of Lebesgue integration, we collect a few results, which often appear with uncomplete proofs at the un- dergraduate level.

6.1. Complex Integrand

So far we have only treated integration of functions with their values in R or [0; ] and it is the purpose of this section to discuss integration of complex valued1 functions. Suppose (X; ; ) is a positive measure. Let f; g L1(): We de…ne M 2

(f + ig)d = fd + i gd: ZX ZX ZX If and are real numbers,

( + i )(f + ig)d = (( f g) + i( g + f))d ZX ZX

= ( f g)d + i ( g + f)d ZX ZX = fd gd + i gd + i fd ZX ZX ZX ZX = ( + i )( fd + i gd) ZX ZX = ( + i ) (f + ig)d: ZX 193

We write f L1(; C) if Re f; Im f L1() and have, for every f L1(; C) and complex2 ; 2 2 fd = fd: ZX ZX Clearly, if f; g L1(; C); then 2 (f + g)d = fd + gd: ZX ZX ZX Now suppose  is a complex measure on : If M f L1(; C) = L1( ; C) L1( ; C) 2 def Re \ Im we de…ne

fd = fdRe + i fdIm: ZX ZX ZX It follows for every f; g L1(; C) and C that 2 2 fd = fd: ZX ZX and (f + g)d = fd + gd: ZX ZX ZX

### 6.2. The Fourier Transform

n Below, if x = (x1; :::; xn) and y = (y1; :::; yn) R ; we let 2 n x; y =  xkyk: h i k=1 and x = x; y : j j h i If  is a complex measure on n (orp n) the Fourier transform ^ of  is de…ned by R R i x;y n ^(y) = e h id(x); y R : n 2 ZR 194

Note that ^(0) = (Rn): 1 The Fourier transform of a function f L (mn; C) is de…ned by 2 ^ f(y) = ^(y) where d = fdmn:

n Theorem 6.2.1. The canonical Gaussian measure n in R has the Fourier transform y 2 j 2j ^n(y) = e :

PROOF. Since = ::: (n factors) n 1 1 it is enough to consider the special case n = 1: Set

1 x2 g(y) = ^ (y) = e 2 cos xydx: 1 p2 ZR Note that g(0) = 1: Since cos x(y + h) cos xy x j h jj j the Lebesgue Dominated Convergence Theorem yields

1 x2 g0(y) = xe 2 sin xydx p2 ZR (Exercise: Prove this by using Example 2.2.1). Now, by partial integration,

2 x= 2 1 x 1 y x g0(y) = e 2 sin xy e 2 cos xydx p2 x= p2 R h i 1 Z that is g0(y) + yg(y) = 0 and we get y2 g(y) = e 2 : 195

If  = ( ; :::;  ) is an Rn-valued random variable with  L1(P ); 1 n k 2 k = 1; :::; n; the characteristic function c of  is de…ned by

i ;y n c(y) = E e h i = P^( y); y R : 2 For example, if  N(0; ); then  = G; where G N(0; 1); and we get 2 2 i G;y c(y) = E e h i = ^ ( y) 1

 2y2 = e 2 : Choosing y = 1 results in

i 1 E[2] E e = e 2 if  N(0; ): 2 n   Thus if (k)k=1 is a centred real-valued Gaussian process

n 1 i ykk n 2 E e k=1 = exp( E ( yk ) 2 k=1 k     1 n 2 2 = exp( k=1ykE k 1 j

n Theorem 6.2.2. Let (k)k=1 be a centred real-valued Gaussian process with uncorrelated components, that is

E   = 0; j = k: j k 6   196

Then the random variables 1; :::; n are independent.

6.3 Fourier Inversion

1 ^ 1 Theorem 6.3.1. Suppose f L (mn): If f L (mn) and f is bounded and continuous 2 2 i y;x ^ dy n f(x) = e h if(y) n ; x R : d (2) 2 ZR

PROOF. Choose " > 0: We have

"2 2 dy "2 2 dy i y;x 2 y ^ i y;x u 2 y e h ie j j f(y) n = f(u) e h ie j j n du n (2) n n (2) ZR ZR ZR  where the right side equals

x u 1 2 dv du 1 2 du i v; v 2 u x f(u) e h " ie 2 j j n n = f(u)e 2" j j n n n p2 p2 "n n p2 "n ZR ZR  ZR

1 z 2 dz = f(x + "z)e 2 j j n : n p2 ZR Thus "2 2 dy 1 2 dz i y;x 2 y ^ 2 z e h ie j j f(y) n = f(x + "z)e j j n : n (2) n p2 ZR ZR By letting " 0 and using the Lebesgue Dominated Convergence Theorem, Theorem 6.3.1! follows at once.

n n Recall that Cc1(R ) denotes the class of all functions f : R R with compact support which are in…nitely many times di¤erentiable. If!f n ^ 1 2 Cc1(R ) then f L (mn). To see this, suppose yk = 0 and use partial integration to obtain2 6 1 f^(y) = e i x;y f(x)dx = e i x;y f (x)dx h i h i x0 k d iy d ZR k ZR and 1 i x;y (l) f^(y) = e h if (x)dx; l N: l xk (iy ) d 2 k ZR 197

Thus y l f^(y) f (l)(x) dx; l N k xk j j j j n j j 2 ZR and we conclude that

sup (1+ y )n+1 f^(y) < : y Rn j j j j 1 2 ^ 1 and, hence, f L (mn): 2

n ^ 1 Corollary 6.3.1. If f C1(R ); then f L (mn) and 2 c 2

i y;x ^ dy n f(x) = e h if(y) n ; x R : n (2) 2 ZR

Corollary 6.3.2 If  is a complex Borel measure in Rn and ^ = 0; then  = 0:

n f^(y) PROOF. Choose f Cc1(R ). We multiply the equation ^( y) = 0 by (2)n and integrate over R2n with respect to Lebesgue measure to obtain

f(x)d(x) = 0: n ZR n Since f C1(R ) is arbitrary it follows that  = 0: The theorem is proved. 2 c

6.4. Non-Di¤erentiability of Brownian Paths

Let ND denote the set of all real-valued continuous function de…ned on the unit interval which are not di¤erentiable at any point. It is well known that ND is non-empty. In fact, if  is Wiener measure on C [0; 1], x ND a.e. [] : The purpose of this section is to prove this important property2 of Brownian motion. 198

Let W = (W (t))0 t 1 be a real-valued Brownian motion in the time interval [0; 1] such that every path t W (t); 0 t 1 is continuous. Recall that !   E [W (t)] = 0 and E [W (s)W (t)] = min(s; t): If 0 t0 ::: tn 1     and 1 j < k n  

E [(W (tk) W (tk 1))(W (tj) W (tj 1)]

= E [(W (tk)W (tj)] E [W (tk)W (tj 1)] E [W (tk 1)W (tj)]+E [W (tk 1)W (tj 1)] = tj tj 1 tj + tj 1 = 0: From the previous section we now infer that the random variables

W (t1) W (t0); :::; W (tn) W (tn 1) are independent.

Theorem 7. The function t W (t); 0 t 1 is not di¤erentiable at any point t [0; 1] a.s. [P ] : !   2

PROOF. Without loss of generality we assume the underlying probability space is complete. Let c; " > 0 and denote by B(c; ") the set of all ! such that 2

W (t) W (s) < c t s if t [s "; s + "] [0; 1] j j j j 2 \ for some s [0; 1] : It is enough to prove that the set 2 1 1 1 B(j; ): k j=1 k=1 [ [ is of probability zero. From now on let c; " > 0 be …xed. It is enough to prove P [B(c; ")] = 0 : 199

Set j + 1 j Xn;k = max W ( ) W ( ) k j 3 and k 0; :::; n 3 : Let n > 3 be so large that2 f g 3 ": n  We claim that 6c B(c; ") min Xn;k :  0 k n 3  n     If ! B(c; ") there exists an s [0; 1] such that 2 2 W (t) W (s) c t s if t [s "; s + "] [0; 1] : j j j j 2 \ Now choose k 0; :::; n 3 such that 2 f g k k 3 s ; + : 2 n n n   If k j < k + 3;  j + 1 j j + 1 j W ( ) W ( ) W ( ) W (s) + W (s) W ( ) j n n jj n j j n j 6c  n 6c and, hence, Xn;k : Now  n 6c B(c; ") min Xn;k  0 k n 3  n     and it is enough to prove that

6c lim P min Xn;k = 0: n 0 k n 3  n !1     But n 3 6c 6c P min Xn;k P Xn;k 0 k n 3  n   n   k=0   X   200

6c 6c = (n 2)P Xn;0 nP Xn;0  n   n     1 6c 6c = n(P W ( ) )3 = n(P ( W (1) )3 j n j n j j pn   12c n( )3:  p2n where the right side converges to zero as n . The theorem is proved. ! 1

Recall that a function of bounded variation possesses a derivative a.e. with respect to Lebesgue measure. Therefore, with probability one, a Brown- ian path is not of bounded variation. In view of this an integral of the type

1 f(t)dW (t) Z0 cannot be interpreted as an ordinary Stieltjes integral. Nevertheless, such an integral can be de…ned by completely di¤erent means and is basic in, for example, …nancial mathematics.

""" 201

REFERENCES

[B] Billingsley, P. (1995) Probability and Measure. Wiley&Sons.

[BO] Bogachev, V. I. (2007) Measure Theory, Springer.

[D] Dudley, R. M. (1989) Real Analysis and Probability. Wadsworth&Brooks.

[DI] Diedonné, J. (1960) Foundations of Modern Analysis. Academic Press.

[F ] Folland, G. B. (1999) Real Analysis; Modern Techniques and Their Ap- plications. Second Edition. Wiley&Sons.

[M] Malliavin, P. (1995) Integration and Probability. Springer.

[R] Rudin, W. (1966) Real and Complex Analysis. McGraw-Hill.

[S] Solovay R. M. (1970) A model of set theory in which every set of reals is Lebesgue measurable. Annals of Mathematics 92, 1-56.