For the proof of Bounded Convergence Theorem, I see how to get most all the information, but I don't see exactly why $f$ is measurable. I assume I am missing something completely obvious. Here is the theorem as I understand it so far.
Theorem: Let $\{ f_n \}$ be a sequence of functions bounded by $M$, supported on a set $E$ of finite measure, and $f_n(x) \to f(x)$ a.e. $x$. Then $f$ is bounded measurable function supported on $E$ for a.e. $x$ and $\int |f_n(x) - f(x)|dx \to 0$ as $n \to 0$.
Proof: Almost everywhere we have, $|f(x)| - |f_n(x)| \le |f(x) - f_n(x)| < \epsilon.$ Therefore, $|f(x)| \le M$ almost everywhere, and $f(x) = 0$ for $x \in E$ almost everywhere.
Let $\epsilon > 0$. By Egorov's theorem we can find a measurable set $A \subset E$ such that $f_n \to f$ uniformly on $A$ and $m(E - A) < \epsilon$. Thus $|f_n(x) - f(x)| < \epsilon$ for large $n$ for all $x \in A$.
$\int |f_n(x) - f(x)|dx \le \int_A |f_n(x) - f(x)| dx + \int_{E - A} |f_n(x) - f(x)| dx \le \epsilon m(A) + 2M \epsilon.$
As $\epsilon$ is arbitrary, this completes the proof.
Is measurability something that I just get automatically?