## Month: April, 2014

### Littlewood’s three principles (3)

Finally the principle states that every convergent sequence nearly converges uniformly. Let’s figure out what the word “nearly” means in this context and this brings us to the Egorov’s theorem.

Theorem 1 (Egorov)Suppose ${\left\{f_{n}\right\}_{i=1}^{\infty}}$ is a sequence of measurable functions defined on a measurable set ${E}$ with ${m(E)<\infty}$, and assume ${f_{i}\rightarrow f}$ almost everywhere on ${E}$. Given ${\epsilon>0}$, we can find a closed set ${A_{\epsilon}\subseteq E}$ such that ${m(E-A_{\epsilon})\leq\epsilon}$ and ${f_{i}\rightarrow f}$ uniformly on ${E}$.

Proof: We may assume ${f_{i}\rightarrow f}$ for every ${x\in E}$ since we can always find a subset ${E'\subseteq E}$ which differs with ${E}$ by a measure zero set such that the convergence holds everywhere on ${E'}$, then we just rename ${E'}$ by ${E}$. Define

$\displaystyle E_{k}^{n}:=\left\{x\in E:\left|f_{j}(x)-f(x)\right|<\frac{1}{n}\;\forall j>k\right\}$

Note for fixed ${n}$ we have the relation ${E_{k}^{n}\subseteq E_{k+1}^{n}}$ and ${E_{k}^{n}\nearrow E}$ as ${k\rightarrow\infty}$. Then by the continuity of measures we know ${\lim_{k\to \infty}m(E_{k}^{n})=m(E)}$ which implies we can find an integer ${k_{n}\in\mathbb{N}}$ such that ${m(K-E_{k_{n}}^{n})<\frac{1}{2^{n}}}$. By construction, we have ${\left|f_{j}(x)-f(x)\right|<\frac{1}{n}}$ whenever ${j>k_{n}}$ and ${x\in E_{k}^{n}}$. Then we choose ${N\in\mathbb{N}}$ so that ${\sum_{n=N}^{\infty}<\frac{\epsilon}{2}}$ and let ${\hat{A}_{\epsilon}:=\bigcap_{n\geq N}E_{k}^{k_{n}}}$.
First we notice the relation ${m(E-\hat{A}_{\epsilon})=m(\bigcup_{n\geq N}E-E_{k}^{k_{n}})\leq\sum_{n=N}^{\infty}(E-E_{k}^{k_{n}}<\frac{\epsilon}{2})}$. Next for any ${\delta>0}$, we can choose ${N'\in\mathbb{N}}$ such that ${\frac{1}{n}<\delta}$ when ${n>N'}$. Now we indeed have completed our job as for any ${\delta>0}$ we can find an integer ${k_{n}}$ such that for all ${j>k_{n}}$ we \left|f_{j}(x)-f(x)\right|<\delta<\frac{1}{n} for all ${x\in\hat{A}_{\epsilon}}$.

Finally, as a technical requirement, we can find a closed subset ${A_{\epsilon}\subseteq\hat{A}_{\epsilon}}$ such that ${m(\hat{A}_{\epsilon}-A_{\epsilon})<\frac{\epsilon}{2}}$ which leads us to the desired property ${m(E-A_{\epsilon})<\epsilon}$.                                                                                                                                                                           $\Box$

### Littlewood’s three principles (2)

The second principle asserts that every function is nearly continuous. Now we investigate the meaning of “nearly” in terms of mathematics.

Theorem 2 (Lusin) Suppose f is measurable and finite valued on E with E of finite measure. Then for every ${\epsilon>0}$ there exists a closed set ${F_{\epsilon}}$ with

$\displaystyle F_{\epsilon}\subseteq E\;\mathrm{and}\;m(E-F_{\epsilon})\leq\epsilon$

and such that ${f|_{F_{\epsilon}}}$ is continuous.

Proof: Since by assumption, ${f}$ is measurable and finite valued on E, we can find a sequence of step functions, ${f_{n}}$ such that ${f_{n}\rightarrow f}$ almost everywhere. Note that each ${f_{n}}$ is a step function which means it is continuous almost everywhere as it contains only at most countably many discontinuities which have Lebesgue measure 0. So without loss of generality, say for each ${n}$ we can find sets ${E_{n}}$ with ${m(E_{n})\leq\frac{1}{2^{n}}}$ so that ${f_{n}}$ is continuous outside ${E_{n}}$. Now to proceed, we need something extra which is obtained from the third littlewood’s principle, i.e., Egrovff’s theorem. It enables us to find a set ${A_{\frac{\epsilon}{3}}}$ with ${m(E-A_{\frac{\epsilon}{3}})\leq\frac{\epsilon}{3}}$ such that ${f_{n}\rightarrow f}$ uniformly on ${A_{\frac{\epsilon}{3}}}$. Then for the integer N such that ${\sum_{n\geq N}\frac{1}{2^{n}}\leq\frac{\epsilon}{3}}$, we consider

$\displaystyle F':=A_{\frac{\epsilon}{3}}-\bigcup_{n\geq N} E_{n}$

For every ${n\geq N}$, the step function ${f_{n}}$ is continuous on ${F'}$; therefore, the uniform limit ${f}$ is also continous on ${F'}$. To complete the proof, notice that ${F'}$ is also measurable so we can approximate from below by a closed set ${F_{\epsilon}}$ such that ${m(F'-F_{\epsilon})\leq\frac{\epsilon}{3}}$ which gives us the desired ${m(E-F_{\epsilon})\leq\epsilon}$.

### A Mild Glimpse Into Measure Theory (2)

3. Measures and Outer measures

Let ${(E,\mathscr{M})}$ be a measurable space.

Definition: A measure on ${(E,\mathscr{M})}$ is a function ${\mu:\mathscr{M}\rightarrow [0,\infty]}$ such that:
(1) ${\mu(\emptyset)=0}$
(2) For all families ${\left\{A_{n}\right\}_{n\in\mathbb{N}}}$ of disjoint measurable sets,

$\displaystyle \mu(\bigcup_{n\in\mathbb{N}} A_{n})=\sum_{n\in\mathbb{N}}\mu(A_{n})$

Remarks:

• ${\mu(A)=+\infty}$ is allowed
• Property (2) is called ${\sigma}$-additivity
• We can also derive ${\mu(\bigcup_{n=1}^{k} A_{n})=\sum_{n=1}^{k}\mu(A_{n})}$

Proposition 6 : Let ${A,B\in\mathscr{M}}$, we have the following:
(1) If ${A\subseteq B}$, then ${\mu(A)\leq\mu(B)}$.
(2) If ${\mu(A)<+\infty}$ and ${A\subseteq B}$, then ${\mu(A\setminus B)=\mu(A)-\mu(B)}$
(3) ${\mu(A\cup B)=\mu(A)+\mu(B)-\mu(A\cap B)}$
(4) If ${A_{n}\in\mathscr{M}}$ for all ${n\in\mathbb{N}}$ and ${A_{n}\subseteq A_{n+1}}$ for all ${n\in\mathbb{N}}$, then ${\mu(\bigcup_{n\in\mathbb{N}}A_{n})=\lim_{n\rightarrow\infty}\mu(A_{n})}$.
(5) If ${B_{n}\in\mathscr{M}}$ for all ${n\in\mathbb{N}}$ and ${B_{n}\supseteq B_{n+1}}$ for all ${n\in\mathbb{N}}$, then ${\mu(\bigcap_{n\in\mathbb{N}}B_{n})=\lim_{n\rightarrow\infty}\mu(B_{n})}$.
(6) If ${A_{n}\in\mathscr{M}}$, then ${\mu(\bigcup_{n\in\mathbb{N}}A_{n})\leq\sum_{n\in\mathbb{N}}A_{n}}$.

To be updated.