# Egerov's Theorem/Bounded Convergence Theorem

## Statement

Egerov's Theorem : Suppose ${\displaystyle \mu }$ is a locally finite Borel measure and ${\displaystyle \{f_{n}\}}$ is a sequence of measurable functions defined on a measurable set ${\displaystyle E}$ with ${\displaystyle \mu (E)<\infty }$ and ${\displaystyle f_{n}\rightarrow f}$ a.e. on E.

Then: Given ${\displaystyle \epsilon >0}$ we may find a closed subset ${\displaystyle A_{\epsilon }\subset E}$ such that ${\displaystyle \mu (E\setminus A_{\epsilon })\leq \epsilon }$ and ${\displaystyle f_{n}\rightarrow f}$ uniformly on ${\displaystyle A_{\epsilon }}$ [1]

## Proof

WLOG assume ${\displaystyle f_{n}\rightarrow f}$ for all ${\displaystyle x\in E}$ since the set of points at which ${\displaystyle f_{n}\nrightarrow f}$ is a null set. Fix ${\displaystyle \epsilon >0}$ and for ${\displaystyle n,k\in \mathbb {N} }$ we define ${\displaystyle E_{k}^{n}=\{x\in E:|f_{j}(x)-f(x)|<{\frac {1}{n}}{\text{ for all }}j>k\}}$. Since ${\displaystyle f_{n},f}$ are measurable so is their difference. Then since the absolute value of a measurable function is measurable each ${\displaystyle E_{k}^{n}}$ is measurable.

Now for fixed ${\displaystyle n}$ we have that ${\displaystyle E_{k}^{n}\subset E_{k+1}^{n}}$ and ${\displaystyle E_{k}^{n}\nearrow E}$. Therefore using continuity from below we may find a ${\displaystyle k_{n}}$ such that ${\displaystyle \mu (E\setminus E_{k_{n}}^{n})<{\frac {1}{2^{n}}}}$. Now choose ${\displaystyle N}$ so that ${\displaystyle \sum _{n=N}^{\infty }2^{-n}<{\frac {\epsilon }{2}}}$ and define ${\displaystyle {\tilde {A}}_{\epsilon }=\bigcap _{n\geq N}E_{k_{n}}^{n}}$. By countable subadditivity we have that ${\displaystyle \mu (E\setminus {\tilde {A}}_{\epsilon })\leq \sum _{n=N}^{\infty }\mu (E-E_{k_{n}}^{n})<{\frac {\epsilon }{2}}}$.

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

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

## Corollary

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

${\displaystyle \lim _{n\to +\infty }\int f_{n}=\int \lim _{n\to +\infty }f_{n}=\int f}$ [2]

## Proof

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

${\displaystyle \int |f_{n}-f|=\int _{E}|f_{n}-f|=\int _{A_{\epsilon }}|f_{n}-f|+\int _{E\setminus A_{\epsilon }}|f_{n}-f|\leq \epsilon \mu (E)+2M\mu (E\setminus A_{\epsilon })=\epsilon (\mu (E)+2M)}$

Since ${\displaystyle \epsilon }$ was arbitrary and ${\displaystyle \mu (E)+2M}$ is finite by assumption we are done.

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

## References

1. Stein & Shakarchi, Real Analysis: Measure Theory, Integration, and Hilbert Spaces, Chapter 1 §4.3
2. Stein & Shakarchi, Real Analysis: Measure Theory, Integration, and Hilbert Spaces, Chapter 2 § 1