Solution verification on Folland Real Analysis, Problem 2.36 Problem If $\mu(E_n) < \infty$ for every $n \in \Bbb{N}$ and $\chi_{E_n} \to f$ a.e., then $f$ is (a.e. equal to) the characteristic function of a measurable set.
Solution Let $L = \limsup E_n$ and $l = \liminf E_n$. I will first show that if $\chi_{E_n} \to g$, then $g = \chi_{L} = \chi_{l}$.
It is trivial that $g$, $\chi_L$, and $\chi_l$ is equal to either $0$ or $1$. Let $g(c) = 1$. Then there exists an $N \in \mathbb{N}$ such that $c \in E_n$ for every $n > N$. So $c \in \limsup E_n$ and $c \in \liminf E_n$, and the similar argument can be applied to show the case when $g(c) = 0$. It also follows that if $\chi_E$ converges to some $g$, then $\limsup E_n = \liminf E_n$.
Now let $E_{n_k}$ be the subsequence of $E_n$ such that $\chi_{E_{n_k}} \to f$ a.e. (the existence of these subsequence is given as theorem in the book.). Since $\limsup_{n \to \infty} E_n = \liminf_{n \to \infty} E_n$, it is also true that $$\limsup_{k \to \infty} E_{n_k} = \liminf_{k \to \infty} E_{n_k} = \lim_{n \to \infty} E_n,$$
and $f = \chi_{\lim E_n}$ a.e. The measurability of $\lim E_n$ is trivial.

How does it look? Is there any critical errors?
The reason I am not confident about my solution is that I cannot find the usage of the finiteness of $E_n$ given in this problem. Is it just to well-define the convergence of $L^1$, or is there some reasoning that I could not find out?
Thanks in advance.
 A: It looks good to me. Here is my proof for comparison.
Proof:
Let $\mu(E_n) < \infty$ for $n\in\mathbb{N}$ and $\chi_{E_n}\rightarrow f$ in $L^1$. Then by proposition 2.29, we have $\chi_{E_n}\rightarrow f$ in measure. Thus, by theorem 2.30 there is a subsequence $\{\chi_{E_{n_j}}\}$ of $\{\chi_{E_n}\}$ that converges to $f$ a.e. That is there exists a measurable set $F\in M$ such that $\mu(F) = 0$. So for $x\in F^c$ we have $\chi_{E_{n_j}}\chi_{F^c}\rightarrow f\chi_{F^c}$, so we have that for all $x\in X$,
$f\chi_{F^c}(x) =0$ or $f\chi_{F^c}(x) =1$. So $f\chi_{F^c}$ is the characteristic function of a set $A$. Since, for all $n$, $\chi_{E_{n_j}}\chi_{F^c}$ is a measurable function, we have that $f\chi_{F^c}$ is a measurable function. Since $f\chi_{F^c}$ is the characteristic function of the set $A$, we have that $A$ is a measurable set. So, $f\chi_{F^c}$ is the characteristic function of a mensaurable set (the set $A$).
Since, $f=f\chi_{F^c}$ a.e., we have that $f$ is (a.e. equal to) the characteristic function of a measurable set.
(Remark: this result does not need the assumption: $\mu(E_n) < \infty$ for $n\in\mathbb{N}$).
