Chapter 5 Decomposition of Measures

1 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. 2

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: 3

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 [ 4 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 5

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 for every A : 2 R 6

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 exercises. 7

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 8

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 9 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 10 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 11

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 12 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 13 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. 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

2. Suppose is a real measure and = 1 2; where 1 and 2 are …nite + positive measures. Prove that 1 and 2 :

3. Let 1 and 2 be mutually singular complex measures on the same - algebra: Show that 1 2 : j j?j j

4. 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 14

5.3. The Wiener Maximal Theorem and the Lebesgue Di¤erentiation 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

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: 15

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 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 16

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) 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)): 17

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

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 18

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 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

Now, if we introduce

N (A) = H(xj)x (A);A 1 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 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 # 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 20

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 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 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: 21

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 variation. A bounded increasing function on R is of bounded variation. Therefore the di¤erence of two bounded increasing functions on R is of bounded variation. Interestingly enough, the converse is true. In the special case ] ; [ = R we write F BV if F is of bounded variation. 2

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) 22

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.

(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 23

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 In particular, the range of the map I equals the set of all real-valued absolutely continuous maps on [a; b] :

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 24

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 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 25

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:

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].

### 5.5. Conditional Expectation 26

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 conditional 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]): """