View source for Egerov's Theorem/Bounded Convergence Theorem

==Statement==
<strong> Egerov's Theorem </strong>:Suppose <math> \mu </math> is a locally finite Borel measure and <math>\{f_n\}</math> is a sequence of measurable functions defined on a measurable set <math> E </math> with <math> \mu(E)<\infty </math> and <math> f_n \rightarrow f  </math> a.e. on E.

Then:
Given <math> \epsilon>0  </math> we may find a closed subset <math> A_\epsilon \subset E  </math> such that <math> \mu(E\setminus A_\epsilon) \leq \epsilon </math> and <math> f_n \rightarrow f  </math> uniformly on <math> A_\epsilon </math> <ref name="SS"> Stein & Shakarchi, ''Real Analysis: Measure Theory, Integration, and Hilbert Spaces'', Chapter 1 §4.3 </ref>

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

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

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

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


==Corollary==
<strong> Bounded Convergence Theorem </strong>: Let <math> f_n </math> be a seqeunce of measurable functions bounded by <math> M </math>, supported on a set <math> E </math> and <math> f_n \to f </math> a.e. Then

<math> \lim_{n \to +\infty}\int f_n=\int\lim_{n \to +\infty} f_n=\int f </math> <ref name="SS2"> Stein & Shakarchi, ''Real Analysis: Measure Theory, Integration, and Hilbert Spaces'', Chapter 2 § 1 </ref>

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

<math>\int |f_n-f|=\int_E |f_n-f|=\int_{A_\epsilon} |f_n-f|+\int_{E\setminus A_\epsilon}\leq \epsilon \mu(E)+2M \mu(E\setminus A_\epsilon)=\epsilon(\mu(E)+2M) </math>

Since <math> \epsilon </math> was arbitrary and <math> \mu(E)+2M </math> is finite by assumption we are done.

==References==