If $\mu(E_n) < \infty$ for all $n \in \mathbb{N}$ and $\chi_{E_n} \to f$ in $L^1$, then is $f$ almost everywhere equal to the indicator of a measurable set.
So, we know convergence in $L^1$ is equivalent to:
$\mu(E_n) \to \int |f| < \infty.$ We also know that convergence in $L^1$ implies convergence in measure. Also, convergence in measure implies convergence in measure of any subsequence $\{ \chi_{E_{n_k}} \}$ to $f$. Now, by theorem 2.30 in Folland, any any of these (sub-)sequences converging in measure will have a (sub-)subsequence $\{ \chi \}_{E_{n_{k_j}}}$converging to $f$ a.e. We know that $\limsup_j \chi_{E_{n_{k_j}}} = \chi_{\limsup_j {E_{n_{k_j}}}}$ (same holds for $\liminf_j$). Since $\lim_j \chi_{{E_{n_{k_j}}}} = \limsup_j \chi_{E_{n_{k_j}}} = \liminf_j \chi_{E_{n_{k_j}}}$, we can define $f = \chi_E$ where $E = \{{ x \in X | \limsup_j \chi_{E_{n_{k_j}}} = \liminf_j \chi_{E_{n_{k_j}}}} \}$ which is measurable. Now recall for an arbitrary sequence, if we can for all subsequences of it, find a further subsequnce that is convergent, that limit is the limit of the original sequence. Using this fact, we can conclude that $\chi_{E_n} \to f$ almost everywhere.
Best Answer
I don't understand the sentence
Anyway the main idea is correct. Convergence in $L^1$ implies that a subsequence $\{ \chi_{E_{n_k}} \}$ converges to $f$ pointwise a.e.
The second part of the proof is a bit confusing. You cannot define $f=\chi_E$ and anyway your set $E$ is not correct.
Since $\{ \chi_{E_{n_k}} \}$ converges to $f$ pointwise a.e., take a point $x$ such that $\chi_{E_{n_k}}(x)\to f(x)$. Since $\chi_{E_{n_k}}(x)$ only takes values $0$ and $1$ and the limit exists, necessarily $\chi_{E_{n_k}}(x)=0$ for all $k$ large, in which case $\chi_{E_{n_k}}(x)=0\to 0$, or $\chi_{E_{n_k}}(x)=1$ for all $k$ large in which case $\chi_{E_{n_k}}(x)=1\to 1$. Hence, $f(x)$ can only be $0$ or $1$, which shows that $f$ is the characteristic function of a set.