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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 easily be proved using dominated convergence theorem with dominating function $g=M\cdot \mathbf {1} _{E}$, but traditionally one proves bounded convergence before dominating convergence.