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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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    @RedRover: So, to summarize the note on accepting: you *should* accept answers once you are satisfied; it is also a good idea to wait at least a couple of hours before accepting an answer, even if you are satisfied quickly (because questions without an accepted answer tend to attract more attention than questions with accepted answers, and so having no accepted answers in a recent question may encourage others to post alternative solutions); but don't rush into accepting an answer. Like a lot of things, it's a balancing game: try not to be too quick nor too slow in accepting answers.2012-01-17

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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.

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    Alright thanks you're the man2012-01-17