# Egoroff's theorem, Real analysis

Let $(X,M,\mu)$ be a finite measure space, and E be a measurable set. Let $f_{n}$, f, g be $\mu$ measurable functions on E such that $f_{n}\to f$ a.e., and lim $\int_{E}|f_{n}-g|=0$. Prove that $f=g$ a.e.

What I tried: I am trying for sure to use Egoroff theorem since X has finite measure, so we have $f_{n}$ converges uniformly to f on the complement of a measurable set having measure less than $\epsilon$. But I am having hard time to put all these ideas together. Can someone help? I would truly appreciate! Thanks

• $f_{n}\to f$ a.s. implies $f_{n}\to f$ in measure. $\lim\int_{E}|f_{n}-g|=0$ implies $f_{n}\to g$ in measure, because $$\mu(\{x\in E: |f_n(x)-g(x)|\le\epsilon\})\le\frac{\int_{E}|f_n-g|}{\epsilon}.$$ So we must have $f=g$ a.s. – Professor Vector Dec 13 '17 at 14:27