Problem 2.36 in Folland. Is my solution correct? 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. 
 A: I don't understand the sentence 

So, we know convergence in $L^1$ is equivalent to: $\mu(E_n) < \int |f| < \infty.$

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.
A: 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 $n\in\mathbb{N}$).
