Egerov's Theorem: Difference between revisions

Latest revision as of 21:02, 7 December 2020

Statement

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 $A_{\epsilon }$ ^[1]

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 $E$ and $f_{n}\to f$ a.e. Then

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

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

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

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

References

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

[SS-1] Stein & Shakarchi, Real Analysis: Measure Theory, Integration, and Hilbert Spaces, Chapter 1 §4.3

[SS2-2] Stein & Shakarchi, Real Analysis: Measure Theory, Integration, and Hilbert Spaces, Chapter 2 § 1

[1]

[2]

@@ Line 1: / Line 1: @@
 ==Statement==
-Suppose <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.
+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>
+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 \rightarrow f</math> is a null set. Fix <math>\epsilon>0 </math> and for <math> n, k \in \N </math> we define
+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>
+<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>
+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==
-For any <math> n \in \mathbb{N} </math>, let <math> g_n=\inf_{k\geq n} f_k </math>.
+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
-By definition, <math> \liminf_{n\rightarrow +\infty} f_n= \lim_{n\rightarrow +\infty} (inf_{k\geq n}f_k)=\lim_{n\rightarrow +\infty} g_n</math>.
-And <math> g_n\leq g_{n+1}, \forall n \in \mathbb{N} </math>, so by Monotone Convergence Theorem,
-<math> \lim_{n\rightarrow +\infty} \int g_n=\int \lim_{n\rightarrow +\infty} g_n = \int \liminf_{n\rightarrow +\infty} f_n</math>.
-Furthermore, by definition we have <math> g_n\leq f_n </math>, then <math> \int g_n\leq \int f_n </math>.
+<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> \lim_{n\rightarrow +\infty} </math> exists, taking <math> \liminf_{n\rightarrow +\infty} </math> of both sides:
+Since <math> \epsilon </math> was arbitrary and <math> \mu(E)+2M </math> is finite by assumption we are done.
-<math> \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</math>.
 ==References==

Egerov's Theorem: Difference between revisions

Latest revision as of 21:02, 7 December 2020

Contents

Statement

Proof

Corollary

Proof

References

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Egerov's Theorem: Difference between revisions

Latest revision as of 21:02, 7 December 2020

Statement

Proof

Corollary

Proof

References

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