MATH501_HW9 - Serge Ballif MATH 501 Homework 9 1 Let and be...

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Serge Ballif MATH 501 Homework 9 December 12, 2007 1. Let μ and ν be measures on ( X, M ) such that ν μ . Set λ = μ + ν . Show that if f is the Radon-Nikodym derivative of ν with respect to λ , f = , then 0 f < 1 μ -a.e. and = f 1 - f . We first note that μ + ν λ implies that μ λ , so the Radon-Nikodym derivative f exists. Seeking a contradiction we suppose that f ( x ) 1 on a set E of positive measure. Then ν ( E ) = R E f dλ R E = λ ( E ). We cannot have ν ( E ) > λ ( E ) because λ = μ + ν . The only way to have equality is if μ ( E ) = 0. However, this would be contrary to the fact that ν μ . Thus, we have arrived at a contradiction, and we must therefore conclude that f < 1 λ -a.e. (and hence μ -a.e.). Similarly, f cannot be negative on a set E of positive measure, because μ ( E ) is not negative. Therefore, 0 f < 1 μ -a.e. We know ν ( E ) = R E 1 = R E f dλ = R E f dμ + R E f dν . Subtracting R E f dν from both sides yields R E (1 - f ) = R E f dμ . Now we use the fact that R kh = R kg implies that h = g a.e. (for k > 0) to see that R E 1 = R E f 1 - f . Therefore = f 1 - f . 2. Let ( X, M ) be a measurable space and let ( μ n ) be a sequence of complex measures on ( X, M ) such that n =1 k μ n k < . Prove: (a) For every E ∈ M , the series n =1 μ n ( E ) converges absolutely in and defines a measure μ . (b) For every E ∈ M , the series n =1 | μ n | ( E ) converges absolutely in + and defines a finite positive measure ν such that μ ν . (a) n =1 | μ n ( E ) | ≤ n =1 | μ n | ( E ) n =1 | μ n | ( X ) = n =1 k μ n k < , so the series converges absolutely. Clearly μ ( ) = 0. To see that μ is countable additive we consider the sequence ( E n ) ⊂ M of disjoint measurable sets. μ [ k =1 E k ! = X n =1 μ n [ k =1 E k !
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