A problem in Folland's Real Analysis... 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.
 A: 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 $L^1$).
A: You need $\mu(E_n) < \infty$ for the functions $\chi_{E_n}$ to be in $L^1$. Now once you know that a subsequence $\chi_{E_{n_k}} \to f$ a.e., then $f$ can only take the values $1$ and $0$ a.e. Hence, $f$ must be a characteristic function of set a.e. (which is measurable since $f$ is)
A: Let $\mu(E_n) < \infty$ for $n\in\mathbb{N}$ and $\chi_{E_n}\to f $ in $L^1$. Then by proposition 2.29, we have $\chi_{E_n}\to f$ in measure. Then by theorem 2.30 there is a subsequent $\{\chi_{E_{n_j}}\}$ of $\{\chi_{E_n}\}$ that converges to $f$ a.e. That is there exists a measurable function $F\in M$ such that $\mu(F) = 0$. So for $x\in F^c$ we have $\chi_{E_{n_j}}\chi_{F^{c}}\to f\chi_{F^c}$, so we have for all $x\in X$, $f\chi_{F^c}(x) = 0$ or $f\chi_{F^c}(x) = 1$. So $\chi_{F^c}$ 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 measurable set $A$.
