Egorov's theorem

Let $\epsilon >0$, and let $N$ be a measurable null set outside which $f_n(x)\to f(x)$. For $m,k\ge 1$ put $E_{k,m}:={\bigcup }_{j\ge k}\{\mathrm{\mid }{f}_j-f\mathrm{\mid }>1\mathrm{/}m\}\mathrm{.}$ For fixed $m$ the sets $E_{k,m}$ decrease with $k$, each lies in $X$, and ${\bigcap }_{k=1}^{\mathrm{\infty }}{E}_{k,m}\subseteq N$ because outside $N$ only finitely many $j$ satisfy $\mathrm{\mid }{f}_j(x)-f(x)\mathrm{\mid }>1\mathrm{/}m$. Therefore [L3] and [L5] give $\mu (E_{k,m})\downarrow \mu \text{\hspace{0.17em}⁣}\left({\bigcap }_{k=1}^{\mathrm{\infty }}{E}_{k,m}\right)=0.$ So for each $m\ge 1$ there is a least index $k(m)$ with $\mu (E_{k(m),m})<\epsilon 2^{-m}$.

Put $E:={\bigcup }_{m=1}^{\mathrm{\infty }}{E}_{k(m),m}$. Then [L4] and step 1.1 give $\mu (E)\le {\sum }_{m=1}^{\mathrm{\infty }}\mu (E_{k(m),m})<{\sum }_{m=1}^{\mathrm{\infty }}\epsilon 2^{-m}=\epsilon \mathrm{.}$ If $x\in X\setminus E$, then for every $m\ge 1$ and every $j\ge k(m)$ one has $\mathrm{\mid }{f}_j(x)-f(x)\mathrm{\mid }\le 1\mathrm{/}m$. Hence for any $\eta >0$ one may choose $m$ with $1\mathrm{/}m<\eta$ and then $j\ge k(m)$ gives $\mathrm{\mid }{f}_j(x)-f(x)\mathrm{\mid }<\eta$. So $f_n\to f$ uniformly on $X\setminus E$.

Since $\epsilon >0$ was arbitrary, step 2.1 is exactly [L2]. Therefore $f_n\to f$ almost uniformly.