proof of Egorov’s theorem

Let 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 . 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

$$\mu (\bigcup _{j>m}(E-E_{i,j}))\underset{m\to \mathrm{\infty }}{\overset{}{\to }}0.$$

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

Let

$$E_{\delta }=\bigcup _{i\in \mathbb{N}}\bigcup _{j>m_i}(E-E_{i,j}).$$

Then

We claim that $f_n\to f$ uniformly on $E-E_{\delta }$. In fact, given $\epsilon >0$, choose $n$ such that . 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, .Hence, for each $\epsilon >0$ there is $N$ (which is given by $m_n$ above) such that $j>N$ implies for each $x\in E-E_{\delta }$, as required. This the proof.