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.

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

WLOG assume $f_{n}\rightarrow f$ for all $x\in E$ since the set of points at which $f_{n}\rightarrow f$ is a null set. Fix $\epsilon >0$ and for $n,k\in \mathbb {N}$ we define $E_{k}^{n}=\{x\in E:|f_{j}(x)-f(x)|<{\frac {1}{n}}{\text{ for all }}j>k\}$ Now for fixed $n$ we have that $E_{k}^{n}\subset E_{k+1}^{n}$ and $E_{k}^{n}\nearrow E$. Therefore using continuity from below we may find a $k_{n}$ such that $\mu (E\setminus E_{k_{n}}^{n})<{\frac {1}{2^{n}}}$. Now choose $N$ so that $\sum _{n=N}^{\infty }2^{-n}<{\frac {\epsilon }{2}}$ and define ${\tilde {A}}_{\epsilon }=\bigcap _{n\geq N}E_{k_{n}}^{n}$. By countable subadditivity we have that $\mu (E\setminus {\tilde {A}}_{\epsilon })\leq \sum _{n=N}^{\infty }\mu (E-E_{k_{n}}^{n})<{\frac {\epsilon }{2}}$

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: $\int \liminf _{n\rightarrow +\infty }f_{n}=\lim _{n\rightarrow +\infty }=\liminf _{n\rightarrow +\infty }\int g_{n}\leq \liminf _{n\rightarrow +\infty }\int f_{n}$.