This is problem 36 in Chapter 2 of the Folland's book.
If $\mu(E_n)<\infty$ for all $n\in\mathbb N$ and $\chi_{E_n}\to f$ in $L^1$, then $f$ is a.e. equal to the characteristic function on a measurable set.
My "proof":
Convergence in $L^1$ implies convergence a.e. of a subsequence. So say $\chi_{E_{n_k}} \to f$ except on a set $A$ with $\mu(A)=0$
There exists $N\in \mathbb N$ such that $|\chi_{E_{n_k}}(x)-f(x)|<1/2 $ for all $k\geq N$, for $x\in A^c$.
Now I think it is clear that if $k\geq N$ then ${E_{n_k}}$ and ${E_{n_{k+1}}}$ can differ at most by a null set. So taking $E=\bigcap _{k\geq N} {E_{n_k}}$, we are removing at most a null set $B$. Then I think $f=\chi_E$ on the complement of $A\cup B$.
That's my best attempt so far, but I did not use the assumption $\mu(E_n)<\infty$. I may have needed to use completeness of the Lebesgue measure to guarantee $B$ is measurable.
Best Answer
The proof is not correct, since $\mu(E_{n_k})$ is not necessarily equal to $\mu(E_{n_{k + 1}})$. As a particular example to see this, define a sequence of sets $E_k = [0, \frac{1}{k}]$, and note that we never have equality of the measure of two members of the sequence.
As a hint towards a full proof: You know that $f$ is measurable. Furthermore, $f(x) \in \{0, 1\}$ for almost every $x$. So try letting $E = \{x : f(x) = 1\}$ and go from there.
As a follow-up thought, where did we use that $\mu(E_n) < \infty$? (Hint: The convergence occurs in $L^1$).