# Clarification on bounded convergence theorem.

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?

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For any sequence of measurable functions $\{f_n\}$, both their lim sup and lim inf as $n\rightarrow \infty$ are measurable. Because the limit function $f$ here is equal to the lim sup (actually both) it is measurable.

Proof: I will use the fact that a function is measurable if the subset

$$f^{-1}((a, \infty])$$

is measurable.

First we prove that $\sup_n$ $f_n$ is measurable. Note that $\{\sup f_n>a\}=\cup_n\{f_n>a\}$, and as a countable union of measurable sets, it is measurable. As $\inf_n f_n$ is $-\sup (-f_n(x))$, we get the same result for $\inf f_n$.

Now note that $\lim$ $\sup$ as $n\rightarrow \infty$ of $f_n(x)$ is just

$$\inf_k\{\sup_{n\ge k} f_n\}$$

and you're done.

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