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.
Best Answer
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 $n$, $$ \int \lvert f_n - 0\rvert = \int \lvert f_n\rvert = 1$$