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Suppose that $f_n$ is a sequence of measurable functions that are all bounded by M, supported on a set E of finite measure, and $f_n(x)\to f(x)$ a.e. x as $n\to \infty$. Then f is measurable, bounded, supported on E for a.e. x, and $\int |f_n-f|\to 0 $ as $n\to\infty$

At here, I understand the restriction of E to be finite measure is because we need to use Egorov theroem. But what is the reason to assume $f_n$ to be bounded? I check the Egorov theorem, it does not require the squence that converge to $f$ to be bounded in order to find a smaller set differs from E by $\epsilon$ and converges uniformly to $f$ on that set.

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In my opinion, it is needed to assume that to ensure that $f$ is a bounded measurable function and hence we can conclude that $\int_{E}f$ exist. In fact, $|f(x)|\le M$ for all $x\in E$. –  juniven Dec 17 '12 at 4:26
    
That make sense! –  Zhixia Zhang Dec 17 '12 at 4:33

1 Answer 1

up vote 3 down vote accepted

Take $X = [0,1]$ with Lebesgue measure. Then let $$f_n = n 1_{[0,\frac{1}{n})}.$$ Then $f_n \rightarrow 0$ a.e. However for all $n$, $$ \int \lvert f_n - 0\rvert = \int \lvert f_n\rvert = 1$$

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Also note that if $f_n \leq M$ for all $n$ and $E$ is of finite measure, then $M1_{E}$ is a dominating function for $f_n$. –  Deven Ware Dec 17 '12 at 4:41
    
Nice post! I like it. –  juniven Dec 17 '12 at 5:09

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