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Unformatted text preview: Math 245A Homework 6 Brett Hemenway November 30, 2005 46. Let X = Y = [0 , 1], M = N = B [0 , 1] , = Lebesgue measure, and = counting measure. Let D = { ( x, x ) : x [0 , 1] } X Y . Then D x = { ( x, x ) } and D y = { ( y, y ) } , thus Z Z D dd = Z d = 0 and Z Z D dd = Z 1 d = 1 Let { A i B i } be a sequence of rectangles in M N such that D S i =1 A i B i . For each x [0 , 1], x A j B j for some j , then x A j B j . So [0 , 1] S i =1 ( A i B i ). Since ([0 , 1]) = 1, we have that ( A j B j ) > 0 for some j . Then ( A j ) > 0, and ( B j ) > 0, so ( B j ) = . Thus i =1 ( A i B i ) = for any covering set { A i B i } . Thus ( D ) = , so have Z D d ( ) = ( D ) = 49. a. Let E MN and ( E ) = 0. Then by Fubinis theorem 0 = ( E ) = Z E d = Z E x d = Z E y d So ( E x ) = ( E y ) = 0. b. Let f be L measurable and f = 0 a.e. Let A = { ( x, y ) X Y : f ( x, y ) 6 = 0 } 1 We have ( A ) = 0, so there exists a set E MN with A E and ( E ) = 0. Then by Fubinis theorem 0 = Z fd = Z f x d = Z f y d Thus ( E x ) = ( E y ) = 0. But A x E x and A y E y , so since and are complete, A x , A y are measurable and have measure zero. To prove Theorem 2.39, Let f be a Lmeasurable function. Then by Theorem 2.12, there is a measurable function g such that f = g a.e. If (i) f 0, or (ii) f L 1 ( ), then applying lemma (b) to f g we have that f x is Nmeasurable for a.e. x and f y is Mmeasurable for a.e. y . then by Fubinis theorem x 7 R g x d and y 7 R f y d are measurable, and by lemma (b)...
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This note was uploaded on 04/06/2010 for the course MATH various taught by Professor Tao/analysis during the Spring '10 term at UCLA.
 Spring '10
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