### 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 ${m(E-A_{\epsilon})<\epsilon}$. $\Box$