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So I had this question on a recent test and I wasn't able to do it.

Let $g(x)$ be integrable and let $\{f_n\}$ be a sequence of measurable functions such that $|f_n|\leq g(x)$ and $f_n\rightarrow f$ almost everywhere. I want to show that $\int |f_n-f|\rightarrow 0$.

I think the solution some how involves the Lebesgue convergence theorem, but I don't know how to apply it. I could use some help.


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Isn't this actually the statement of the Lebesgue Dominated Convergence Theorem? – Jonas Teuwen Oct 30 '11 at 18:50
Or at most a rephrasing. – Jonas Teuwen Oct 30 '11 at 19:06
up vote 2 down vote accepted

You can argue like that

$|f_n-f|\to 0$ a.e. by hypothesis. Since $|f_n(x)|\leq g(x)$ for all $n$ and $f_n$ converges to $f$, a.e. we have that $|f(x)|\leq g(x)$ a.e. By the triangular inequality we get the bound $|(f_n-f)(x)|\leq 2 g(x)$. Now we can use the Lebesgue Dominated Convergence Theorem.

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Huh! really straight forward, once you establish the bounds. Thanks – Kuku Oct 30 '11 at 19:17
You are welcome Kuku. – Leandro Oct 30 '11 at 19:23
@Leandro Sorry to bother, but how can we explicitly show that $|f(x)|\leq g(x)$ from those two assertions? It seems obvious, but I'm hoping for a formal reason? thanks :) – mathmath8128 Oct 30 '11 at 20:11
@aengle: consider each $x$ for which $\lim\limits_n f_n(x) = f(x)$. Then this assertion is obvious since it is a numerical limit – Ilya Oct 30 '11 at 20:23

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