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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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    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.2012-01-17
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    Sorry I am new at this. How do I accept...?2012-01-17
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    This link http://math.stackexchange.com/questions/83424/omitting-the-hypotheses-of-finiteness-of-the-measure-in-egorov-theorem/84332#84332 may help you.2012-01-17
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    To the left of each answer is a check mark. Click on that check mark for an answer you feel is the best one and it will turn green, indicating that you have accepted that answer.2012-01-17
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    @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.2012-01-17
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    How are you defining "almost uniformly" in this context?2012-01-17
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    Alright, sorry guys, I'm not familiar with the etiquette here but comments are all appreciated. Edits coming2012-01-17
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    @Nate: "Almost uniform" usually means that for every $\epsilon\gt 0$ there exists $F\subseteq E$ such that $\mu(F)\lt\epsilon$ and the convergence is uniform in $E-F$. For example, $f_n(x)=x^n$ converges almost uniformly to $0$ on $[0,1]$.2012-01-17
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    @DavideGiraudo: I am having trouble understanding the pf you linked me to. I'm not saying the claim itself is wrong but consider that in the proof you only use the domination to establish $L^{1}$ convergence, and use this to prove the claim... consider a sequence of moving characteristic functions which have length $\frac{1}{n}$, these converge in $L^{1}$ but not almost uniformly since these functions will cover an infinite amount of length. In terms of where I think the proof itself is wrong, it only establishes the convergence for indices $n_{j}$; ...2012-01-17
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    ...though the use of Chebyshev guarantees that every index past the given $n_{j}$ will satisfy the Chebyshev condition, the set on which the condition is satisfied may be different, so the intersection of all the complements doesn't work for all indices. Convergence in the $L^{1}$ norm only implies almost uniform convergence if the speed of convergence is in an absolutely summable sense so that where you use Chebyshev we can control the tail of the series starting with $n_{j}$ to generate sets $A_{j}$ to form the final set. Let me know if you think this ar2012-01-17
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    ...gument is correct I am no authority but this relates to some of the other problems I did on Tao's website and think this is right2012-01-17
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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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    how did you obtain that $\frac{1}{\epsilon} \int (2g) < \epsilon$?2012-01-17
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    @RedRover: **Don't accept an answer until you are satisfied!** Accepting an answer discourages people from posting alternative answers (or even looking at the question), and signals to the person giving the answer that you understood it.2012-01-17
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    @RedRover: Sorry, that should be "$\lt \infty$", not $\lt\epsilon$.2012-01-17
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    Did you prove that condition just so you could use the continuity of measure for the limit?2012-01-17
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    @RedRover: I was looking at/copying from my old Measure Theory problem solutions, and I think that's right.2012-01-17
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    Alright thanks you're the man2012-01-17