# Bounded convergence theorem

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.

• 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

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 $n$, $$\int \lvert f_n - 0\rvert = \int \lvert f_n\rvert = 1$$
• 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
In order to bound the integral of a function, we need to bound either the measure of the domain of the integral, or the function itself. By using Egorov's Theorem we can ensure the existence of a measurable subset $F$ of the domain on which we have uniform convergence whose complement can be made to have arbitrarily small measure. On $F$ we can bound $|f_n-f|$ precisely because we have uniform convergence there. However if we don't have a uniform bound on $\{|f_n|\}_n$, and hence on $\{|f_n-f|\}_n$, $|f_n-f|$ may grow in such a way that nullifies the smallness of the measure of $F^c$. This would be a problem, since even though we can make the measure of $F^c$ arbitrarily small, we may be unable to make it zero (e.g. see this post).