MATH501_HW5 - Serge Ballif MATH 501 Homework 5 In all...

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Serge Ballif MATH 501 Homework 5 October 31, 2007 In all problems ( X, M , μ ) is a measure space 1. Let f : X × [ a, b ] where [ a, b ] is a finite interval. Assume that the function x 7→ f ( x, y ) is integrable for every y [ a, b ] . Assume also that y f ( x, y ) exists for every ( x, y ) and there exists a function g L 1 such that | y f ( x, y ) | ≤ g ( x ) for all ( x, y ) . Show that the function x 7→ y f ( x, y ) is measurable for every y and y Z f ( x, y ) ( x ) = Z y f ( x, y ) ( x ) . For a fixed y [ a, b ] define h n ( x, y ) := f ( x,y +1 /n ) - f ( x,y ) 1 /n for n . Then h n is measurable as a multiple of a difference of integrable functions. Moreover, lim n →∞ h n ( x ) exists and is equal to y f ( x, y ). The inequality | h n ( x ) | = | y f ( x, y ) | ≤ g ( x ) allows us to invoke the dominated convergence theorem as follows. y Z f ( x, y ) = lim h 0 R f ( x, y + h ) - R f ( x, y ) h = lim n →∞ Z h n = Z lim n →∞ h n = Z lim h 0 f ( x, y + h ) - f ( x, y ) h = Z y f ( x, y ) . 2. Prove: (a) k f n - f k 0 if and only if there exists E ∈ M such that μ ( E c ) = 0 and f n f uniformly on E . (b) ( L , k · k ) is a Banach space. (a) ( ) Suppose that k f n - f k 0. Then for any > 0, there exists N such that for n N , k f n - f k = inf { a 0 : μ ( { x X : | f n ( x ) - f ( x ) | > a } ) = 0 } < . Then for each m , there exists N such that whenever n > N the set E m := { x X : | f n ( x ) - f ( x ) | > 1 m } has measure 0. Define E c = S m =1 E m . Since E m % E c , we know by continuity from below that m ( E c ) = 0. To see that f n f uniformly on E we note that for x E , there exists N such that for n > N , | f n ( x ) - f ( x ) | < (else
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