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Let $(X,S,\mu)$ be a measure space, and let $f,f_1,f_2,\dots:X\to [0,+\infty]$ be $\mu$-integrable such that $\lim\limits_{n\to\infty}f_n=f$ almost everywhere. Show that: $\lim\limits_{n\to\infty}\displaystyle\int|f_n-f|=0 \Leftrightarrow \lim\limits_{n\to\infty}\displaystyle\int|f_n|=\displaystyle\int|f|$ in which case $\lim\limits_{n\to\infty}\displaystyle\int f_n=\displaystyle\int f$.

"$\Rightarrow$" is not hard to show. But how to show "$\Leftarrow$"?

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up vote 1 down vote accepted

Let $g_{n}$ = $|f| - |f_{n}|$. Since $f_{n}$ converges to $f$ a.e., both $g_{n}^{+}$ and $g_{n}^{-}$ go to zero a.e. Further, $0 \leq g_{n}^{+} \leq |f|$ and $f$ is $\mu$-integrable. By the Dominated Convergence Theorem, we have $\int g_{n}^{+} d\mu$ goes to zero. By hypothesis, $\int g_{n} d\mu$ goes to zero. So, $\int g_{n}^{-} d\mu$ = $\int g_{n}^{+} d\mu$ - $\int g_{n} d\mu$ goes to zero and so $\int |g_{n}| d\mu$ = $\int g_{n}^{+} d\mu$ + $\int g_{n}^{-} d\mu$ goes to zero. Since all the $f$'s are non-negative, $|f-f_{n}| = ||f| - |f_{n}|| = |g_{n}|$ and since $\int |g_{n}|d\mu $ goes to zero, you have your answer.

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Thanks a lot. I have worked it out. – Frank Lu Oct 19 '12 at 1:24

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