Chapter 4

Signed Measures

4.1 Signed Measures

(X, ) is throughout this section a measurable space. M Definition 4.1.1. A map µ : [ , ] is called a signed measure if M! 1 1 1. µ( ) = 0, ; 2. µ assumes at most one of the values ,i.e., ±1 either µ : [ , ), or µ : ( , ]. M! 1 1 M! 1 1 3. If (A ) are pairwise disjoint, then n ⇢M

µ An = µ(An), n N ⇣ [ ⌘ X2

where the right side n N µ(An) converges absolutely if µ An is finite. 2 P S Remark. Assume µ : [ , ] is a signed measure. W.l.o.g. we M! 1 1 can assume (by property (2)) that µ : ( , ]. Let (A ) be M! 1 1 n ⇢M pairwise disjoint. Define

+ N = n N : µ(An) 0 and N = n N : µ(An) < 0 . { 2 } { 2 } It follows that

µ A = µ(A )= µ(A ) > . n n n 1 n N n N n N ⇣ 2[ ⌘ X2 X2 69 70 CHAPTER 4. SIGNED MEASURES and either µ A = n 1 n N + ⇣ 2[ ⌘ then µ A = , n 1 or ⇣ [ ⌘ µ A < n 1 n N + ⇣ 2[ ⌘ then <µ A < , and µ(A ) < . 1 n 1 n 1 n N ⇣ [ ⌘ X2 Example 4.1.2. Let µ be a positive measure and f : X R¯ measurable ! on (X, ), so that M + f dµ < or f dµ < . 1 1 Z Z Define for A 2M + ⌫(A)= f dµ f dµ. Z Z Then ⌫ is a signed mesure. We call in this case the function f an extended µ-integrable function. Proposition 4.1.3. (Continuity from below and above) Let µ be a signed measure on (X, ). M If (A ) is an increasing sequence, then j ⇢M

µ An =limµ An . n !1 ⇣ [ ⌘ If (Aj) is a decreasing sequence and µ(A 1) R then ⇢M 2

µ An =limµ An . n !1 Proof. Homework ⇣ \ ⌘ Definition 4.1.4. Let µ : [ , ] be a signed measure. M! 1 1 We call A a a positive set for µ (or negative set for µ) if for all 2M B A, B we have µ(B) 0, (respectively if for all B A, B ⇢ 2M ⇢ 2M we have µ(B) 0).  We say that A is a null set for µ if for all measurable subsets B A, 2M ⇢ µ(B) = 0. Note: For A , being a null sets, does not only mean that µ(A)=0but 2M also that all measurable subsets have measure 0. 4.1. SIGNED MEASURES 71

Lemma 4.1.5. Let µ be a signed measure on (X, ). Then any mea- M surable subset of a positive (negative) set is positive (respectively negative), The union of countably many positive (negative) sets is positive (respectively negative).

Proof. The first claim is clear. To show the second claim let (An)bea sequence of positive sets in and let B An, B . Then for every n M1 ⇢ 2M n N, Bn = B An An has non negative measure. 2 \ \ j=1 S Thus S 1 1 µ(B)=µ B = µ(B ) 0. n n n=1 n=1 ⇣ [ ⌘ X

Theorem 4.1.6. (Hahn Decomposition Theorem) Let µ be a signed measure on (X, ). Then there exist P and N with the property that P M 2M 2M is positive and N is negative, P N = ,andP N = X. \ ; [ We call X = P N a Hahn Decomposition of µ. [ Moreover, if P is a positive and N a negative set, with P N = and 0 0 0 \ 0 ; P N = X, then P P and N N are null sets. 0 [ 0 4 0 4 0 Proof. Without loss of generality assume that µ is [ , )-valued. Put 1 1 m =sup µ(A):A is positive , 2M and let (Pj) be a sequence of positive sets for which µ(Pj) increases to m. We can assume that Pj Pj+1, for j N.WeletP = 1 Pj. It follows ⇢ 2 j=1 from Proposition 4.1.3 that µ(P )=m and from Lemma 4.1.5 that P is S positive. It also follows that µ(P )=m< . 1 We put N = X P . We claim that N. Assume that N is not negative, \ and thus that there is an A N, for which µ(A ) > 0. 0 ⇢ 0 We claim

( ) For all A N,withµ(A)>0thereisaB A, so that µ(B)>µ(A). ⇤ ⇢ ⇢ Indeed assume that A N, is measurable with µ(A) > 0. A cannot be ⇢ positive (which would mean that all measurable A A have non negative 0 ⇢ measure), because otherwise A P would be positive and and µ(P A) > [ [ µ(P ). Then there exists a measurable C A, so that µ(C) < 0. But this ⇢ implies for B = A C, that µ(B) >µ(A). \ ( ) leads to finding recursively a sequence (A ) and a sequence of ⇤ j ⇢M natural numbers (nj) so that the following holds: 72 CHAPTER 4. SIGNED MEASURES

a) n1 is the smallest positive integers for which there is a measurable B N so that µ(B) > 1/n , and A N is such a set. ⇢ 1 1 ⇢ b) Assuming that Aj 1 N has been defined, then nj N, is the smallest ⇢ 2 positive integer so that there is a measurable B Aj 1 so that ⇢

µ(B) >µ(Aj 1)+1/nj,

and let Aj be such a set.

Let A = j1=1 Aj.Then T 1 1 >µ(A)= lim µ(Aj) > n , 1 j j !1 Xj=1 and thus n , as j . j !1 !1 Again there is an n N and a measurable B A with µ(B) >µ(A)+ 1 . 2 ⇢ n Since for some j N it follows that nj >n, this represents a contradiction 2 to the definition of nj. Now let (P 0,N0) be another partition of X into a positive set P 0 and a negative set N . It follows that P P is at the same time a subset of P as 0 \ 0 it is a subset of N 0. Thus it is positive as well as negative, which means, that for all measurable A P P it follows that µ(A) = 0. ⇢ \ 0 Definition 4.1.7. Let µ and ⌫ be two signed measures on a measurable space (X, ). We say that µ and ⌫ are mutually singular if there are two M disjoint sets E and F which are disjoint, with X = E F , so that E is a [ null set for µ and F is a null set for ⌫. We write in that case µ ⌫. ? Theorem 4.1.8. Let µ be a signed measure on a measurable space (X, ). M Let P and N as in the Hahn Decomposition Theorem 4.1.6 and define for A 2M + µ (A)=µ(A P ) and µ(A)= µ(A N). \ \ Then µ+ and µ are the unique measures on , which are mutually M singular, so that µ = µ+ µ . Proof. It follows from the definition that µ+ µ and µ = µ+ µ . ? Now assume µ = ⌫+ ⌫ is another decomposition into a di↵erence of two mutually singular (positive) measures. Let E,F be disjoint, with 2M X = E F , so that E is a ⌫+-null set and F is a ⌫ -null set. It follows that [ 4.2. THE LEBESGUE- AND THE RADON-NIKODYM THEOREM 73

E F is another Hahn decomposition, and it follows from Theorem 4.1.6 [ that P F is a µ- null set. Thus for any A 4 2M µ+(A)=µ+(A P )=µ(A P )=µ(A F )=⌫+(A F )=⌫+(A). \ \ \ \

Definition 4.1.9. For a signed measure µ we call the uniquely (positive) measures µ+ and µ which are mutually singular and for which µ = µ+ µ , the Jordan Decomposition of µ. We call the (positive) measure

+ µ = µ + µ | | the total variation of µ.

Example 4.1.10. so that

+ f dµ < or f dµ < . 1 1 Z Z Then define for A 2M + ⌫(A)= f dµ f dµ. Z Z Then

+ + ⌫ (A)= f dµ, ⌫(A)= f dµ, and ⌫ (A)= f dµ. | | | | ZA ZA ZA 4.2 The Lebesgue- and the Radon-Nikodym The- orem

Definition 4.2.1. Suppose that ⌫ is a signed measure, and µ a positive measure on (X, ). We say ⌫ is absolutely continuous with respect to µ, M and we write ⌫<<µ,if⌫(A) = 0 for all A for which µ(A) = 0. 2M Proposition 4.2.2. For a signed measure ⌫ and positive measure µ on (X, ) the following are equivalent: M 1. ⌫<<µ,

+ 2. ⌫ << µ and ⌫ << µ, 3. ⌫ << µ. | | 74 CHAPTER 4. SIGNED MEASURES

Proof. Let X = P N be a partition of X into a set P which is positive for [ ⌫, and a set N which is negative for ⌫. By Theorem 4.2.7 ⌫+(A)=⌫(A P ), \ and ⌫ (A)=⌫(A N). \ If ⌫<<µ, then it follows that for all A , not only ⌫(A) = 0 also for 2M all measurable B A,wehaveµ(B) =, and thus ⌫(B) = 0, and thus A is ⇢ null for ⌫ in the sense of Definition 4.1.4. From these observations the claimed equivalences follow easily.

Proposition 4.2.3. For a signed measure ⌫ and positive measure µ on (X, ). M If ⌫<<µand ⌫ µ, then ⌫ =0. ? Proof. Since ⌫ µ, we can write X as X = E F ,whereE is µ-null and ? [ F is ⌫-null. But then, since ⌫<<µit also follows that E is ⌫-null.

Proposition 4.2.4. Let and µ be two (positive) finite measures on (X, ). M Then either µ or there is an ">0 and an E with µ(E) > 0,and ? 2M (A) "µ(A),forallA E, A . ⇢ 2M 1 Proof. For n N,letX = Pn Nn be the Hahn Decomposition of n µ, 2 [ 1 Then Pn P and Nn N.ThenN is negative for n µ, for all n N. % 1 & 2 Thus 0 (N) n µ(N), for all n N, and thus (N) = 0. If µ(P )=0   2 then µ.Ifµ(P ) > 0, then µ(Pn) > 0 for some n N, and since Pn is ? 2 a positive set for 1 µ, we conclude the claim by choosing vp = 1 , and n n E = P1,

Theorem 4.2.5. Let ⌫ be a finite signed measure and µ be a positive mea- sure on (X, ). M Then ⌫ is absolutely continuous with respect to µ if and only for all ">0 there is a >0 so that µ(A) <implies ⌫(A) <",forallA . | | 2M Proof. Since ⌫<<µif and only if ⌫ << µ, and since ⌫(A) ⌫ (A), | | | ||| we can assume without loss of generality that ⌫ is a positive measure. It is clear that the "- condition implies absolute continuity. To obtain, assume that ⌫ does not satisfy the "- condition, which means that there is an "> n so that for all n N there is an En ,withµ(En) < 2 and ⌫(En) " n 2 2M (taking =2 ) Take

1 1 F =limsupEn = Ej. n n=1 !1 \ j[=n 4.2. THE LEBESGUE- AND THE RADON-NIKODYM THEOREM 75

Thus for all n N 2 1 1 µ(F ) µ E µ(E ) 0as n .  j  j ! !1 ⇣ j[=n ⌘ Xj=n But by continuity from above, Theorem 2.3.3, (recall that we assumed that ⌫ is finite) it follows that

1 ⌫(F )= lim ⌫ Ej ", n !1 ⇣ j[=n ⌘ and thus ⌫ is not absolutely continuous with respect to µ.

Since for f L (µ), 2 1

⌫(A)= fdµ,A , 2M ZA defines a finite signed measure, which is absolutely continuous with respect to µ it follows that:

Corollary 4.2.6. Let f L (µ). Then for every ">0 there is a so that 2 1

">0 > A µ(A) < fdµ <". 8 9 8 2M ) A Z Assume that

Theorem 4.2.7. (Radon Nikodym Theorem) We are a given signed measure ⌫ and a positive measure µ on (X, ).Assumethat ⌫ and µ are -finite. M | | | | Then ⌫<<µif and only if there exists an measurable f : X R¯ which ! is an extended µ-integrable function so that for all A 2M ⌫(A)= f(x) dµ(x). ZA Moreover f is unique up to µ-a.e. equality.

Proof. First if there is an extended µ-integrable function so that for all A 2M ⌫(A)= f(x) dµ(x), ZA it easily follows that ⌫<<µ. 76 CHAPTER 4. SIGNED MEASURES

+ For the converse, after passing to ⌫ and ⌫, we can use Proposition 4.2.2, and assume that ⌫ is a positive measure. We first assume that ⌫ and µ are finite measures on (X, ), and define M by F = f : X [0, ]mble: A f(x) dµ(x) ⌫(A) F ! 1 8 2M A  n Z o We note the following: 1. = ,since0 , F6 ; 2F 2. if f,g , h = max(f,g). Indeed for every A 2F 2M hdµ= gdµ+ fdµ A A f g A gg )=⌫(A).  \{  } \{ } Put a =sup fdµ ⌫(X) < . f  1 2F Z and let gn , n N so that 2F 2

a =lim gn dµ. n !1 Z Putting fn := max(g1,g2,...gn) it follows from (b) that

a =lim gn dµ lim fn dµ a. n  n  !1 Z !1 Z Since the fn are increasing f =limfn =supfn exists pointwise and by the Monotone Convergence Theorem 3.2.4 we deduce that

(4.1) fdµ=lim fn dµ = a. n Z !1 Z Since each f we also deduce Monotone Convergence Theorem that for n 2F all A 2M

(4.2) fdµ=lim fn dµ ⌫(A). n  ZA !1 ZA Claim: ⌫(X)=a = f(x) dµ(x). Indeed, we put R (A)=⌫(A) f(x) dµ(x), for A . 2M Z 4.2. THE LEBESGUE- AND THE RADON-NIKODYM THEOREM 77

Since <<µ, it follows that by Proposition 4.2.4 either µ in which ? case by Proposition 4.2.3 = 0, and thus the claim follows, or there is an ">0 and E ,withµ(E) > 0 and (B) "µ(B) for all B ,with 2M 2M B E. But this would mean which that for all A ⇢ 2M ⌫(A)=⌫(A E)+⌫(A E) ("1 (x)+f(x)) dµ(x), \ \ E ZA which implies that "1 + f , and E 2F

("1E(x)+f(x)) dµ(x) >a, ZX which is a contradiction, and thus we proved the claim. We secondly claim that

fdµ= ⌫(A), for all A . 2M ZA Assume not, and assume that for some A we have 2M fdµ<⌫(A). ZA Then we deduce from (4.2) applied to Ac that

a = fdµ= fdµ+ fdµ<⌫(A)+⌫(Ac)=⌫(X), c ZX ZA ZA which contradicts (4.1). In the general case we can find a sequence of disjoint sets (A )in , n M which are pairwise disjoint, with X = An, and so that µ(An) as well as ⌫(An) are finite (first partition X into a sequence of sets (Bj) S ⇢M whose µ-measure is finite, and then partition each Bj into a sequence of sets (Bj,i) whose ⌫-measure is finite). ⇢M ¯ We now can find a nonnegative measurable function fn : X R so that !

fn dµ = ⌫(An A) for all A and n N. \ 2M 2 ZA and so that f 0 on Ac . Our claim follows then from the Corollary 3.2.6 n ⌘ n of the Monotone Convergence Theorem if we put f = n1=1 fn. In order to show uniqueness of f, assume that g : X [0, ] is measur- P! 1 able with the property that

gdµ= ⌫(A), for all A , 2M ZA 78 CHAPTER 4. SIGNED MEASURES and assume that µ( f>g ) > 0. Then { } ⌫( f>g )= fdµ> gdµ= ⌫( f>g ), { } f>g f>g { } Z{ } Z{ } which is a contradiction. Similarly we show that µ( f

Theorem 4.2.10. (Lebesgue Decomposition) Let µ be a -finite positive and ⌫ a -finite signed measure on (X, ). Then ⌫ is the sum of two signed M measures ⌫0 and ⌫1, where ⌫0 is singular with respect to µ and ⌫1 absolutely continuous with respect to µ. Moreover the measures ⌫0 and ⌫1 are unique. Proof. We can again assume that ⌫ is a positive measure, by considering ⌫+ and ⌫ separately. Define = µ + ⌫.Then is a -finite measure and it follows for positive measurable functions f that

fd= fdµ+ fd⌫. Z Z Z Moreover µ is absolutely continuous with respect to , and thus, by the Radon Nikodym Theorem 4.2.7 there is a measurable f : X [0, ]so ! 1 that fd= µ(A), for all A . 2M ZA We put

X = x X : f(x) > 0 and X = x X : f(x)=0 , + { 2 } 0 { 2 } which is a measurable decomposition of X into two disjoint measurable set. We define for A 2M ⌫ (A)=⌫(A X ) and ⌫ (A)=⌫(A X ). + \ + 0 \ 0 It follows

µ(X )= fd= 0 and ⌫ (X )=⌫(X X )=⌫( )=0, 0 0 + + \ 0 ; ZX0 which implies that µ and ⌫0 are mutually singular. It remains to show that ⌫1 is absolutely continuous with respect to µ. 80 CHAPTER 4. SIGNED MEASURES

If µ(A) = 0, then

0=µ(A)

= fd ZA = fd⌫+ fdµ ZA ZA =0 = fd⌫|+{z } fd⌫ X0 A X+ A Z \ Z \ =0

= | {z fd}⌫= 1X+ Af(x) d⌫1. \ X+ A Z \ Z On the set A X the function f is strictly positive. Thus it follows from \ + 1X+ Af(x) d⌫1 = 0 that ⌫1(A)=⌫(A +) = 0. \ \ R