TITLE: Convergence in measure of characteristic functions
QUESTION [6 upvotes]: I was having trouble starting this problem.  I would appreciate some help.  Thanks in advance.
Let $E_1, E_2, \ldots$ be measurable sets.  Suppose that the functions $f_j = 1_{E_j}$ converge in measure to a limit function $f$.  Show that $f$ is $a.e.$ equal to $1_E$ for some measurable set $E$.

REPLY [3 votes]: A similar approach but I think phrased much more simply (using an important concept which is often used): Pick a subsequence $E_{n}$ (I do not re-label it) s.t. $\chi_{E_{n}} \to f$ pointwise a.e. This implies that $f = \limsup_{n \to \infty} \chi_{E_{n}}$. Recall that for measurable sets $\{E_{n}\}$
$$
E:= \limsup_{n \to \infty} E_{n} := \bigcap_{N \in \mathbb{N}} \bigcup_{n \geq N} E_{n}
$$
It is straightforward to check that $\limsup_{n \to \infty} \chi_{E_{n}} = \chi_{E}$, so $f = \chi_{E}$ a.e.