# Proving $f_n\to f$a.e implies $f_n\to f$ almost uniformly [duplicate]

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.

Question:

Can someone provide me any help?

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, metamorphyJan 14 at 9:47

• If $\lambda>0$ then $\mu(\{x:g(x)>\lambda\})<\infty$. – David C. Ullrich Jan 13 at 13:02
• @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. – Pedro Gomes Jan 13 at 15:13
• @PedroGomes no, it doesn't. It's exactly the answer to your question – Jakobian Jan 13 at 15:20
• @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 $\mu(\{x:g(x)>\lambda\})=0$. – Pedro Gomes Jan 13 at 15:21
• 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? – 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 $$|f_n|\le g$$.)
• 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? – Pedro Gomes Jan 13 at 17:50
• $E_k$ 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? – David C. Ullrich Jan 13 at 23:09