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Unformatted text preview: Math 318 HW #9 Solutions 1. Let A be a bounded, measurable set and let ( f n ) be a sequence of measurable functions on A converging to f . Suppose ϕ ∈ L ( A ) and that  f n ( x )  ≤ ϕ ( x ) for all x ∈ A and all n = 1 , 2 ,... . Show that lim n →∞ Z A f n g dm = Z A fg dm if g is measurable and essentially bounded on A , meaning that there exists M > 0 such that  g ( x )  ≤ M a.e. Proof. By assumption, there exists M > 0 such that { x ∈ A :  g ( x )  > M } has measure zero. Therefore,  f n ( x ) g ( x )  =  f n ( x )  g ( x )  ≤ Mϕ ( x ) a.e. Since ϕ ∈ L ( A ), Theorem 25.4(3) implies that Mϕ ∈ L ( A ). Also, since f n is measurable for all n = 1 , 2 ,... and g is measurable, Proposition 18.4 implies that f n g is measurable for all n = 1 , 2 ,... . Finally, lim n →∞ f n g = fg by the Algebraic Limit Theorem. Thus, the hypotheses of Corollary 28.10 are satisfied by ( f n g ), fg , and Mϕ , and so we see that lim n →∞ Z A f n g dm = Z A fg dm,...
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 Spring '08
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 Math, lim, Continuous function, Dominated convergence theorem

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