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Suppose that $f_{n}$ is a sequence of measurable functions, in a finite measure space, $f_{n}\to f $ in $m$-measure and that there exists $g$ in $L^1$ such that $\vert f_n\vert \le g$. Prove that $$ \lim_{n\to +\infty}\Vert f_n-f\Vert_{L^1}=0. $$

What I obviously thought of doing was splitting the difference $|f_n-f|$ to the less than and greater than $\epsilon$ and bound the greater part by $2g$. I am stuck right there, I can show it is finite but can not show it is less than epsilon.

Next I thought of using the R. Fisher's argument of getting the subsequence of $f_n$ which converges a.e, and finiteness of space give you a. uniform by Egoroff). But that way I can only show result will be good for the case of subsequence. I am not sure if I can conclude from there though( by arguing that original sequence and its subsequence goes to the same limit). I am sure I am missing something here. I would love to get out of this confusion. Help please.

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    The subsequence argument can be put to work. You could try proving that any subsequence of $(f_n)$ has a subsequence which converges to $f$ in $L^1$, then apply [this lemma](http://math.stackexchange.com/a/70411/8157).2012-12-25

1 Answers 1