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let $X$ be a finite measure space and $\{f_n\}$ be a sequence of nonnegative integrable functions, $f_n \rightarrow f\ a.e.$ on $ X$. We know that $\lim_{n \rightarrow \infty}\int_X f_n d\mu=\int_X fd\mu$ and on any measurable $E_i \subset X$

I should apply Egoroff's theorem to conclude that $\lim_{n \rightarrow \infty}\int_X |f_n-f|d\mu=0$.

My attempt:

I broke the set $X$ to two sets: $F_\sigma$ on which $f_n \rightarrow f$ uniformly based on Egoroff's theorem and $X\backslash F_\sigma$ which is a very small set, i.e. $\mu\{X\backslash F_\sigma\}=\sigma$ and $f_n \nrightarrow f$

I want to show that on each of these sets, the integral is less than $\frac{\epsilon}{2}\ \forall \epsilon$ to finish. How can I show this for the set $X \backslash F_\sigma$?

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    I am missing something here. Take $f_n = n\cdot1_{[0,\frac{1}{n}]}$. Then $f_n(x) \to f(x) = 0$ a.e., but $\int f_n = 1$, and $\int f = 0$.2012-10-29
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    @copper.hat The "We know" is an assumption, I presume.2012-10-29
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    @AlexBecker: Thanks, I missed that interpretation completely...2012-10-29

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