Bounded convergence theorem

Question

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.

Answer 1

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

Answer 2

Deven Ware's answer is somewhat along the lines of saying "the reason for assuming uniform boundedness is that otherwise there are counterexamples" (which is a standard argument in mathematics). Here is another reason, which is rather philosophical (or heuristic), due to the proof of the Bounded Convergence Theorem using Egorov's Theorem:

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).