If $\mu(E_n) < \infty$ for $n\in\mathbb{N}$ and $1_{E_n}\rightarrow f$ in $L^{1}$, then $f$ is (a.e. equal to) the characteristic function of a measurable set.
I am not sure how to define $E_n$ in order to prove this, any suggestions is greatly appreciated.
Since $1_{E_n}\rightarrow f$ in $L^{1}$ $$ \lim_{n\to\infty}\int_{E_n}|1_{E_n}-f|d\mu=0 $$ So for any $\epsilon>0$, there is a $n$ that $$ \int_{E_n}|1_{E_n}-f|d\mu<\frac{\epsilon}{2^n} $$ Let $F_k=\bigcup_{n=k}^{\infty}E_n$. Then $$ \int_{F_k}|1_{F_k}-f|d\mu\leqslant \sum_{n=k}^{\infty}\int_{E_n}|1_{E_n}-f|d\mu<\frac{\epsilon}{2^k} $$ Clearly $F_1\supset\cdots \supset F_k\cdots$. So let $F=\bigcap_{k=1}^{\infty}F_k$ and by monotone class theorem $$ \int_{F}|1_{F}-f|d\mu=\lim_{k\to\infty}\int_{F_k}|1_{F_k}-f|d\mu=0 $$
So $f=1_{F}$ a.e.