Egerov's Theorem

Statement

Suppose $\{f_{n}\}$ is a sequence of measurable functions defined on a measurable set $E$ with $\mu (E)<\infty$ and $f_{n}\rightarrow f$ a.e. on E. /break Then: Given $\epsilon >0$ we may find a closed subset $A_{\epsilon }\subset E$ such that $\mu (E\setminus A_{\epsilon }\leq \epsilon$ and $f_{n}\rightarrow f$ uniformly on $A_{\epsilon }$

Proof

For any $n\in \mathbb {N}$, let $g_{n}=\inf _{k\geq n}f_{k}$.

By definition, $\liminf _{n\rightarrow +\infty }f_{n}=\lim _{n\rightarrow +\infty }(inf_{k\geq n}f_{k})=\lim _{n\rightarrow +\infty }g_{n}$. And $g_{n}\leq g_{n+1},\forall n\in \mathbb {N}$, so by Monotone Convergence Theorem, $\lim _{n\rightarrow +\infty }\int g_{n}=\int \lim _{n\rightarrow +\infty }g_{n}=\int \liminf _{n\rightarrow +\infty }f_{n}$.

Furthermore, by definition we have $g_{n}\leq f_{n}$, then $\int g_{n}\leq \int f_{n}$.

Since $\lim _{n\rightarrow +\infty }$ exists, taking $\liminf _{n\rightarrow +\infty }$ of both sides: $\int \liminf _{n\rightarrow +\infty }f_{n}=\lim _{n\rightarrow +\infty }=\liminf _{n\rightarrow +\infty }\int g_{n}\leq \liminf _{n\rightarrow +\infty }\int f_{n}$.