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Exercise: Let $(f)_{n\in\mathbb{N}}$ be a sequence of functions such that $f_n\to f$a.e(almost everyehere) and there exists $g$ integrable such that $|f_n|\leqslant g$a.e for all $n\in\mathbb{N}$. Prove that $f_n\to f$ almost uniformly.

I think I can apply the following theorem:

Ergoroff Theorem: Consider $E\in\mathscr{F}$(sigma-algebra), and $E\in\Omega$ defined on a measure space $(\Omega,\mathscr{F},\mu)$. Suppose $\mu(E)<\infty$, and $\{f_n\}$ is a sequence of measurable functions on $E\to\mathbb{R}$ which are finite almost everywhere and converge almost everywhere to a function $f:E\to\mathbb{R}$ which is also finite almost everywhere. Then $f_n\to f$ almost uniformly in $E$.

I now that $f_n\to f$ a.e so $\lim_{n\to\infty}f_n(x)=f(x)\forall x\in E$, But to apply Ergoroff theorem I need to prove that $\mu(E)<\infty$ or $\mu(\Omega)<\infty$. I know by the Dominated convergence theorem that $\lim_{n\to\infty}\int |f_n-f| d\mu=0$ but I cannot see how shall I prove from there that $\mu(E)$ or $\mu(\Omega)$ are limited.


Can someone provide me any help?

Thanks in advance!

Note:$f_n$ does not necessarily converge to $f$ uniformly. So the question is not a duplicate.


marked as duplicate by Xander Henderson, mrtaurho, zz20s, Cesareo, metamorphy Jan 14 at 9:47

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  • $\begingroup$ If $\lambda>0$ then $\mu(\{x:g(x)>\lambda\})<\infty$. $\endgroup$ – David C. Ullrich Jan 13 at 13:02
  • $\begingroup$ @Jakobian My question is not a duplicate. If you read carefully the exercise you find out $f_n$ does not necessarily converge to $f$ uniformly. However the answer you provide assumes $f_n$ to converge uniformly so it is not answering this question. $\endgroup$ – Pedro Gomes Jan 13 at 15:13
  • $\begingroup$ @PedroGomes no, it doesn't. It's exactly the answer to your question $\endgroup$ – Jakobian Jan 13 at 15:20
  • $\begingroup$ @DavidC.Ullrich In order to apply Ergoroff I need to prove the measure of the domain where the function converges is finite. It is already assumed in the question when $f_n\to f$ a.e that $\mu(\{x:g(x)>\lambda\})=0$. $\endgroup$ – Pedro Gomes Jan 13 at 15:21
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    $\begingroup$ How do I know that you are pointing me in the right direction if you do not prove you are? Does it not sound like an authority fallacy? $\endgroup$ – Pedro Gomes Jan 13 at 15:32

Ok, a bigger hint. Let $$E_k=\{x:g(x)>1/k\}.$$Since $\mu(E_k)<\infty$, Egoroff shows that there exists $S_k\subset E_k$ such that $f_n\to f$ uniformly on $E_k\setminus S_k$ and $$\mu(S_k)<\epsilon/2^k.$$

So if $S=\bigcup_{k=1}^\infty S_k$ then $\mu(S)<\epsilon$. And it's possible to prove that $f_n\to f$ uniformly on $X\setminus S$. (There's still something to be proved in that last sentence, it's not quite just trivial by definition. Hint: So far we haven't used the fact that $|f_n|\le g$.)

  • $\begingroup$ Is it not $E_k=\{x:g(x)<1/k\}$. $S_k$ is subset of the set where $f_n$ converges to $f$? Or is there something I am missing? $\endgroup$ – Pedro Gomes Jan 13 at 17:50
  • $\begingroup$ $E_k$ is not what you say it is. So I don't know whether you're asking about $E_k$ or about $\{x:g(x)<1/k\}$. But in either case: So what? $\endgroup$ – David C. Ullrich Jan 13 at 23:09

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