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?
Thanks in advance!
Note:$f_n$ does not necessarily converge to $f$ uniformly. So the question is not a duplicate.
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 $|f_n|\le g$.)