# proof of Egorov’s theorem

Let $E_{i,j}=\{x\in E:|f_{j}(x)-f(x)|<1/i\}.$ Since $f_{n}\to f$ almost everywhere, there is a set $S$ with $\mu(S)=0$ such that, given $i\in\mathbb{N}$ and $x\in E-S$, there is $m\in\mathbb{N}$ such that $j>m$ implies $|f_{j}(x)-f(x)|<1/i$. This can be expressed by

 $E-S\subset\bigcup_{m\in\mathbb{N}}\bigcap_{j>m}E_{i,j},$

or, in other words,

 $\bigcap_{m\in\mathbb{N}}\bigcup_{j>m}(E-E_{i,j})\subset S.$

Since $\{\bigcup_{j>m}(E-E_{i,j})\}_{m\in\mathbb{N}}$ is a decreasing nested sequence of sets, each of which has finite measure, and such that its intersection has measure $0$, by continuity from above (http://planetmath.org/PropertiesForMeasure) we know that

 $\mu(\bigcup_{j>m}(E-E_{i,j}))\xrightarrow[m\to\infty]{}0.$

Therefore, for each $i\in\mathbb{N}$, we can choose $m_{i}$ such that

 $\mu(\bigcup_{j>m_{i}}(E-E_{i,j}))<\frac{\delta}{2^{i}}.$

Let

 $E_{\delta}=\bigcup_{i\in\mathbb{N}}\bigcup_{j>m_{i}}(E-E_{i,j}).$

Then

 $\mu(E_{\delta})\leq\sum_{i=1}^{\infty}\mu(\bigcup_{j>m_{i}}(E-E_{i,j}))<\sum_{% i=1}^{\infty}\frac{\delta}{2^{i}}=\delta.$

We claim that $f_{n}\to f$ uniformly on $E-E_{\delta}$. In fact, given $\varepsilon>0$, choose $n$ such that $1/n<\varepsilon$. If $x\in E-E_{\delta}$, we have

 $x\in\bigcap_{i\in\mathbb{N}}\bigcap_{j>m_{i}}E_{i,j},$

which in particular implies that, if $j>m_{n}$, $x\in E_{n,j}$; that is, $|f_{j}(x)-f(x)|<1/n<\varepsilon$. Hence, for each $\varepsilon>0$ there is $N$ (which is given by $m_{n}$ above) such that $j>N$ implies $|f_{j}(x)-f(x)|<\varepsilon$ for each $x\in E-E_{\delta}$, as required. This the proof.

Title proof of Egorov’s theorem ProofOfEgorovsTheorem 2013-03-22 13:47:59 2013-03-22 13:47:59 Koro (127) Koro (127) 7 Koro (127) Proof msc 28A20