The following problem is from Carothers' Real Analysis:
Suppose $f_n$ is a measurable sequence of functions such that $|f_n|\leq g\in L_1$ for all $n$. Prove that $f_n\to f$ almost everywhere implies $f_n\to f$ almost uniformly.
Thoughts/ Attempt:
At first glance, this looks like a mix of the Dominated Convergence Theorem and Egorov's Theorem.
A sequence of functions $(f_n)$ is said to converge almost uniformly on $D$ if:
$$\forall\epsilon>0\exists N\in\mathbb{N}:\forall n\geq N\implies \sup_{x\in D\setminus E}|f_n(x)-f(x)|<\epsilon \\\text{ (where $E\subset D$ is measurable and $m(E)<\epsilon$) }$$
It is clear that $f\in L_1$.
I tried a few things that didn't lead anywhere such as:
$f_n\to f \text{ a.e.} \implies \int f_n\to \int f \text{ a.e. }\implies |\int f_n-\int f|<\epsilon \text{ a.e. }\implies \int |f_n-f|<\epsilon \text{ a.e. }$
But doesn't lead anywhere about showing almost uniform convergence.
I think I have an idea on how to start now by somewhat imitating the proof of Egorov's Theorem, which defines a measurable set $$E(n,k)=\bigcup_{m=n}^{\infty}\left\{x\in D: |f_m(x)-f(x)|\geq \frac {1}{k}\right\}$$
and shows the measure of this set decreases to $0$ and then a subsequence $(n_k)$ is chosen such that $m(E(n_k,k))<\epsilon/2^k$. Then it is shown that $f_n\to f$ uniformly on $D\setminus E$, where $E=\cup_{k=1}^{\infty}E(n_k,k)$.
However I'm having trouble deciding how to define my set $E(n,k)$ given the hypotheses of the Dominated Convergence Theorem.
Any ideas on how to proceed would be appreciated. Hints are preferred over full solutions. Thanks!