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Math Help - Lebesgue Integral

  1. #1
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    Lebesgue Integral

    Let f be a non-negative integrable function over a set E.Prove that for any \epsilon > 0,there exists a set X \subset E such that m(X)< \infty and \int_E f dm \le \int_X f dm + \epsilon
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  2. #2
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    Quote Originally Posted by problem View Post
    Let f be a non-negative integrable function over a set E.Prove that for any \epsilon > 0,there exists a set X \subset E such that m(X)< \infty and \int_E f dm \le \int_X f dm + \epsilon
    Let f_n:=f1_{X_n} where X_n= E \cap B_m(0,n) where B_m(0,n)=\{ x \in \mathbb{R} ^m : \Vert x \Vert \leq n \} that is f_n(x)=f(x) if x \in X_n and f_n(x)=0 otherwise. For all n \in \mathbb{N} \mu (X_n) < \infty, 0 \leq f_n \leq f for all n and f_n(x) \rightarrow f(x) a.e. in E then by the dominated convergence theorem we have \int_{E} \ f = \lim_{n \rightarrow \infty} \int_{E} \ f_n then for any \epsilon >0 there exists an N such that if n>N then (because f_n \leq f_{n+1} for all n and the monotony of the integral) \int_{E} \ f - \int_{E} \ f_n < \epsilon.
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