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.

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What have you tried? Additionally, you should consider accepting answers to some of your previous questions. This gives the answerers feedback and is a good way to reward people for answering your questions, making them more likely to do so in the future. – Alex Becker Jan 17 '12 at 19:51
Sorry I am new at this. How do I accept...? – Red Rover Jan 17 '12 at 19:55
@RedRover: Additionally, some consider it at least somewhat rude to have posts written solely in the imperative ("Show", "Prove", "Do", "Solve"), as if they were homework assignments to the group. It's also helpful if you include context (is this for a course? Which course? Is this an assignment? Self-study? What related material do you know?) and information about what you've tried already so that people don't spend their time telling you things you already know, or replying at an inappropriate level. – Arturo Magidin Jan 17 '12 at 20:03
How are you defining "almost uniformly" in this context? – Nate Eldredge Jan 17 '12 at 20:04

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.
how did you obtain that $\frac{1}{\epsilon} \int (2g) < \epsilon$? – Red Rover Jan 17 '12 at 21:09
@RedRover: Sorry, that should be "$\lt \infty$", not $\lt\epsilon$. – Arturo Magidin Jan 17 '12 at 21:14