there exists a measurable set $A$ such that $f = \chi_{A}$ a.e.

Question

Hi need some hints with this question:

Prove that if $A_n$ is measurable, $\mu(A_n)< \infty$ for each $n$, and $\chi_{A_n}$ converges to $f$ in measure, then there exists a measurable set $A$ such that $f = \chi_{A}$ a.e.<

Answer 1

There is a subsequence $\chi _{A_{n_i}}$ which converges to $f$ almost everywhere. A sequence of $0$'s and $1$'s can only converge to $0$ or $1$. Hence $f(x) \in \{0,1\}$ for almost all $x$. Take $A=\{x : f(x)=1\}$ and verify that $f=\chi_A$ almost everywhere.