Pointwise a.e. Convergence and Domination Imply Almost uniform convergence Background: I am in a course on measure theory prepping for an exam by doing problems off terence tao's website. here is a question copied from his notes that I am not able to solve. I have noted that the hypotheses imply $L^{1}$ convergence as well as convergence in measure. Almost uniformly is defined in the sense of the conclusion of Egorov's theorem. I appreciate any help. The level of my knowledge is having done a good amount of Royden, essentially I know measure theory on the real line and am not experienced with abstract spaces.
"${f_{n}}_{n \in \mathbb{N}}$ is a sequence so that $f : E \to \mathbb{R}$, each $f_{n}$ is measurable and for all $n$, $|f_{n}|\le g$ for $g$ absolutely integrable. $f$ is another measurable function, and $f_{n} \to f$ pointwise a.e. Show $f_{n} \to f$ almost uniformly."
Thanks to all for helpful comments. I also have just learned of the need to "Accept" answers so I will do that for good answers.
 A: This is Egorov's Dominated Convergence Theorem.
Let $N=\bigl\{x\in E\mid \{f_n(x)\}\text{ does not converge to }f(x)\bigr\}$. Since $|f|\leq g$ on $E-N$, we have
$$|f_n-f| \leq |f_n|+|f|\leq 2g$$
on $E-N$. Fix $\epsilon\gt 0$, and let
$$D_k(\epsilon) = \{x\in E\mid |f_k(x)-f(x)|\geq \epsilon\}.$$
Then
$$D_k(\epsilon)-N \subseteq \{x\in E\mid 2g(x)\geq \epsilon\}$$
hence
$$\bigcup_{k=1}^{\infty} D_k(\epsilon) - N \subseteq \{x\in E\mid 2g(x)\geq\epsilon\}.$$
By Chebyshev's Inequality,
$$\mu\left(\bigcup_{k=1}^{\infty} D_k(\epsilon)\right) \leq \mu\Bigl(\bigl\{ x\in E\mid 2g(x)\geq \epsilon\bigr\}\Bigr) \leq \frac{1}{\epsilon}\int (2g)d\mu\lt\infty$$
(since $g\in \mathcal{L}^1$).
On the other hand,
$$\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}D_k(\epsilon)$$
is a measurable set of measure zero, since $f_n$ converges to $f$ almost everywhere; hence $\lim\limits_{n\to\infty}\mu(\cup_{k=1}^{\infty}D_k(\epsilon))=0$. Now proceed as in Egorov's Theorem to conclude you have almost uniform convergence.
