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Signed Measures 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 λ and µ be two (positive) finite measures on (X, ). M Then either µ λ or there is an ">0 and an E with µ(E) > 0,and ? 2M λ(A) "µ(A),forallA E, A .
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