# Egerov's Theorem/Bounded Convergence Theorem

## Statement

Egerov's Theorem : Suppose $\mu$ is a locally finite Borel measure and $\{f_{n}\}$ 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 $f_{n}\nrightarrow f$ is a null set. Fix $\epsilon >0$ and for $n,k\in \mathbb {N}$ we define $E_{k}^{n}=\{x\in E:|f_{j}(x)-f(x)|<{\frac {1}{n}}{\text{ for all }}j>k\}$ . Since $f_{n},f$ are measurable so is their difference. Then since the absolute value of a measurable function is measurable each $E_{k}^{n}$ is measurable.

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 $\mu (E\setminus {\tilde {A}}_{\epsilon })\leq \sum _{n=N}^{\infty }\mu (E-E_{k_{n}}^{n})<{\frac {\epsilon }{2}}$ .

Fix any $\delta >0$ . We choose $n\geq N$ such that ${\frac {1}{n}}\leq \delta$ . Since $n\geq N$ if $x\in {\tilde {A}}_{\epsilon }$ then $x\in E_{k_{n}}^{n}$ . And by definition if $x\in E_{k_{n}}^{n}$ then $|f_{j}(x)-f(x)|<{\frac {1}{n}}<\delta$ whenever $j>k_{n}$ . Hence $f_{n}\rightarrow f$ uniformly on ${\tilde {A}}_{\epsilon }$ .

Finally, since ${\tilde {A}}_{\epsilon }$ is measurable, using HW5 problem 6 there exists a closed set $A_{\epsilon }\subset {\tilde {A}}_{\epsilon }$ such that $\mu ({\tilde {A}}_{\epsilon }\setminus A_{\epsilon })<{\frac {\epsilon }{2}}$ . Therefore $\mu (E\setminus A_{\epsilon })<\epsilon$ and $f_{n}\rightarrow f$ on $A_{\epsilon }$ ## Corollary

Bounded Convergence Theorem : Let $f_{n}$ be a seqeunce of measurable functions bounded by $M$ , supported on a set $E$ with finite measure and $f_{n}\to f$ a.e. Then

$\lim _{n\to +\infty }\int f_{n}=\int \lim _{n\to +\infty }f_{n}=\int f$ ## Proof

By assumptions on $f_{n}$ , $f$ is measurable, bounded, supported on $E$ for a.e. $x$ . Fix $\epsilon >0$ , then by Egerov we may find a measurable subset $A_{\epsilon }$ of $E$ such that $\mu (E\setminus A_{\epsilon })<\epsilon$ and $f_{n}\to f$ uniformly on $A_{\epsilon }$ . Therefore, for sufficiently large $n$ we have that $|f_{n}(x)-f(x)|<\epsilon$ for all $x\in A_{\epsilon }$ . Putting this together yields

$\int |f_{n}-f|=\int _{E}|f_{n}-f|=\int _{A_{\epsilon }}|f_{n}-f|+\int _{E\setminus A_{\epsilon }}|f_{n}-f|\leq \epsilon \mu (E)+2M\mu (E\setminus A_{\epsilon })=\epsilon (\mu (E)+2M)$ Since $\epsilon$ was arbitrary and $\mu (E)+2M$ is finite by assumption we are done.

Note this theorem could also be proved in one line using dominated convergence theorem with dominating function $g=M\cdot \mathbf {1} _{E}$ , but traditionally one proves bounded convergence before dominating convergence.