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Complex Measures and the Radon-Nykodym Theorem

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Complex Measures and the Radon-Nykodym Theorem CHAPTER 5 Complex measures and the Radon-Nykodym theorem 5.1 Motivation. The Vitali-Cantor function. This chapter is the first step towards a differ- d entiation theory for (at the very least) the Lebesgue integral on R and R . Let us start with a simple example. If f is a continuous function on [0, 1] we may define its integral function by x F(x)= f (t) dt = f 1[0,x) dµ. Z0 Z respectively in Riemann and Lebesgue integral notation: Riemann and Lebesgue integral coincide for continuous functions. The first fundamental theorem of Calculus tells us that F(x + h) F(x) F 0(x) := lim = f (x) x (0, 1). h 0 h − ! 8 2 Conversely, if F is continuously differentiable on (0, 1), then F(x) F(0)= F 01[0,x) dµ. − Z 1 Both statements fail badly when the assumption of f continuous is replaced by f L (0, 1). 2 Here is a telling example. For n = 0, 1, . let En be the n-th iteration of the Cantor set, namely, 2n En = In,j j=1 [ n where the union is disjoint and In,j is a suitable closed interval of length 2− . Notice that In,j En+1 = In+1,2j 1 In+1,2j. Define the Riemann integrable function fn : [0, 1] C \ − [ ! 2n x 3 n+1 f 1 , F x f t dt. n = 2 In,j n( )= n( ) j=1 0 Å ã X Z It is easy to see that Fn is a continuous function, Fn(0)=0, Fn(1)=1 and Fn is constant on the complement of En. The easiest way to see that Fn converges uniformly on [0, 1] to a continuous function F on [0, 1] is the following. Notice that gn = fn+1 fn is zero unless x En, and there holds that − 2 n 1 2 + f x f x 1 x 1 1 x 1 x 2 1 x 3 n+1( ) n( ) In,j ( ) = 3 In+1,2j 1 ( )+ In+1,2j 1 ( ) 3 Jn,j ( )= − − − − where Jn,j is theÄ middle third of In,j. Inä particular,Ä since the above hasä mean zero on each In,j, Fn+1 = Fn outside En. Using the above display, one has 1 sup Fn 1 x Fn x + ( ) ( ) n+1 x [0,1] | − | 3 2 2 · 57 58 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM whence, by standard arguments, Fn is uniformly Cauchy and converges uniformly to a con- tinuous function F : [0, 1] R with F(0)=0, F(1)=1. Moreover, F is constant on the ! complement of the Cantor set E = En. It follows that F(x + h) F(x) F 0(x)=S lim = 0 h 0 h − ! for all x E (and therefore, F’=0 almost everywhere). Thus, the second fundamental theorem of62 Calculus fails for F. 1 Here, what happens is that the limit of fn should not be understood as an L function 1 (fn 0 pointwise a.e. and in L ), but rather as a measure which is concentrated on the Cantor! set, and thus is singular with respect to Lebesgue measure. The theory developed in the upcoming sections will allow us to correctly characterize the limit of fn. 5.2 Complex measures. Let be a σ-algebra on a set X . If E , and F 2F 1 E = Ej, Ej j, j = k = Ej Ek = j=1 2F8 6 ) \ ; [ then Ej : j N is called a measurable partition of the set E.Acomplex measure is a map { 2 } ⌫ : C with the property that, whenever E and Ej : j N is a measurable partitionF 7! of E, 2F { 2 } 1 (5.1) ⌫(E)= ⌫(Ej). j=1 X Notice that the convergence of the series (of complex numbers) is a part of the requirement, unlike the case of (positive) measure where ⌫ is allowed to take the value . 1 1 5.2.1 EXAMPLE. Let (X , , µ) be a measure space and f L (X , , µ). Let F 2 F ⌫f (E)= f dµ, A . ZE 2F Exercise 6 from Chapter 3 yields exactly that ⌫f is a complex measure on . We observe F for further use that the same statement yields ⌫ f is a finite positive measure and ⌫f (E) | | | | ⌫ f (E) for all E . | | 2F The second part of the above example suggests that, given a complex measure ⌫ on , one should look for a positive finite measure λ on such that ⌫(E) λ(E) for all A F . Such a measure would need to satisfy F | | 2F 1 1 λ(E)= λ(Ej) ⌫(Ej) j=1 ≥ j=1 | | X X whenever E and Ej : j N is a measurable partition of E. Thus, we define the map 2F { 2 } 1 (5.2) ⌫ : [0, ], ⌫ (E) := sup ⌫(Ej) E meas. part. of E | | F! 1 | | j j=1 | | { } X which is the least possible such measure λ and is termed the total variation of ⌫. It should be pretty obvious that if ⌫ is a complex measure taking nonnegative values, then ⌫ = ⌫. | | 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM 59 5.2.2 LEMMA. Let ⌫ be a complex measure on . Then the total variation ⌫ of (5.2) is a finite positive measure on . F | | F PROOF. Let us first prove that ⌫ is a measure. Let Ej , F` be any two measurable | | { } { } partitions of E. Then F` Ej : j N is a measurable partition of F` and, viceversa F` { \ 2 } { \ Ej : ` N is a measurable partition of Ej. Thus, with the first equality justified by the convergence2 } of the sums, there holds 1 1 1 1 1 1 1 1 ⌫(F`) = ⌫(F` Ej) ⌫(F` Ej) = ⌫(F` Ej) ⌫ (Ej). ` 1 | | ` 1 j 1 \ ` 1 j 1 | \ | j 1 ` 1 | \ | j 1 | | = = = = = = = = X X X X X X X X Taking the supremum over F` in the left hand side we obtain { } 1 (5.3) ⌫ (E) ⌫ (Ej). | | j=1 | | X For the opposite inequality, let Aj,` : ` N be a measurable partition of each Ej such that { 2 } 1 j ⌫ (Ej) "2− + ⌫(Aj,`) . | | `=1 | | X Clearly Aj,` : j, ` N is a measurable partition of E. Thus { 2 } 1 1 1 (5.4) ⌫(Ej) " + ⌫(Aj,`) " + ⌫ (E)+" j=1 | | j=1 `=1 | | | | X X X and " > 0 is arbitrary. Combining (5.3) with (5.4) yields countable additivity of ⌫ . Let us move to ⌫ being a finite measure. We will need the following Sublemma,| | whose proof is left as an exercise.| | 5.2.3 SUBLEMMA. Let z1,...,zN be complex numbers. Then there is a subset of indices S 1, . , N such that ⇢ { } N 1 z z . j ⇡ j j S ≥ j=1 | | 2 X X Suppose first that there exists a measurable set E with ⌫ (E)= . Then there must | | 1 exist a partition Ej : j T of E and a finite N such that { 2 } N ⌫(Ej) > ⇡(1 + ⌫(E) ). j=1 | | | | X Using the sublemma, choose a subcollection of indices S 1, . , N such that ⇢ { } (1 + ⌫(E) ) ⌫(Ej) = ⌫(A), A = Ej. | | j S | j S X2 2 c [ Let B = E A . It must be by finite additivity of ⌫ that \ ⌫(B) = ⌫(E) ⌫(A) ⌫(E) ⌫(A) 1 | | | − | ≥ | | − | | ≥ 60 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM So we have found two disjoint measurable sets A, B such that ⌫(A) , ⌫(B) 1 and A B = E, whence, without loss of generality ⌫ (A)= . | | | | ≥ [ We now show that if ⌫ (X )=| | we reach1 a contradiction. By repeating the above procedure, replacing E with| | A each1 time, we construct a sequence Bj of pairwise disjoint sets with ⌫(Bj) 1. But this contradicts the convergence of the series | | ≥ 1 1 ⌫ Bj = ⌫(Bj), Çj=1 å j=1 [ X which completes the proof. ⇤ 5.2.4 REMARK. It is easy to see, and we leave the proof as an exercise, that the space M(X , ) : ⌫ : C complex measures F { F! } is a linear space and ⌫ X = ⌫(X ) k k 1 is a norm on M(X , ). Furthermore, if µ is a positive measure on and f L (X , , µ), F F 2 F then ⌫f = ⌫ f and ⌫f = f 1. | | | | k k k k Let ⌫ be a complex measure on taking real values. We can decompose F ⌫ ⌫ ⌫ ⌫+ ⌫ , ⌫ : , = + − ± = | |±2 such that the measures ⌫± are finite positive measures on . This decomposition is referred to as the Jordan decomposition of ⌫. F 5.3 Absolute continuity. Let (X , , µ) be a measure space and ⌫ be an arbitrary (complex or positive) measure on . We sayF that ⌫ is absolutely continuous with respect to µ, and write ⌫ µ, if F Î (5.5) E , µ(E)=0 = ⌫(E)=0. 2F ) It is immediate to see that if ⌫ is a complex measure then ⌫ ⌫ . A measure ⌫ (positive or complex) on is said to be concentrated on the set A ifÎ | | F 2F ⌫(A B)=⌫(B) B . \ 8 2F If ⌫1, ⌫2 are two measures on such that ⌫j is concentrated on Aj and A1, A2 partition X , then we say that ⌫1, ⌫2 are mutuallyF singular and write ⌫1 ⌫2. It is not difficult to find examples for each property, thus? we leave these as an exercise, as well as the following properties. 5.3.1 PROPOSITION. Let µ be a positive measure, ⌫, ⌫1, ⌫2 be either positive or complex mea- sure on the same σ-algebra . Then if ⌫ is concentratedF on A, then ⌫ is concentrated on A as well; • if ⌫1 ⌫2 then ⌫1 ⌫2 ; | | • ? | | ? | | if ⌫1 µ, ⌫2 µ then ⌫1 + ⌫2 µ; • ? ? ? if ⌫1 µ, ⌫2 µ then ⌫1 + ⌫2 µ; • if ⌫ Î µ andÎ⌫ µ then ⌫ Î⌫ ; 1 Î 2 1 2 • ⌫ µ, ⌫ µ if and? only if ⌫ =?0.
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