Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 "$\Leftarrow$"?

share|improve this question
add comment

1 Answer 1

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.

share|improve this answer
    
Thanks a lot. I have worked it out. –  Frank Oct 19 '12 at 1:24
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.