Lusin's theorem for a Radon measure

First let $B$ be Borel of finite measure and let $\delta >0$. By [L1], choose open $U\supseteq B$ of finite measure and compact $L\subseteq U$ with $\mu (U\setminus L)<\delta \mathrm{/}2$. By outer regularity choose open $O\supseteq U\setminus B$ with $\mu (O)<\mu (U\setminus B)+\delta \mathrm{/}2$. Then $H=L\setminus O$ is compact and contained in $B$. Since $B$ is disjoint from $U\setminus B$, $\mu (B\cap O)<\delta \mathrm{/}2$, and hence $\mu (B\setminus H)\le \mu (U\setminus L)+\mu (B\cap O)<\delta$. This proves compact approximation for each finite-measure Borel set using only [L1].

Since $f$ is complex-valued and $\mu (E)<\mathrm{\infty }$, continuity from above gives an $M>0$ for which $B=\{x\in E:\mathrm{\mid }f(x)\mathrm{\mid }\le M\}$ satisfies $\mu (E\setminus B)<\epsilon \mathrm{/}4$. For each $m\ge 0$, partition the disk $\mathrm{\mid }z\mathrm{\mid }\le M$ into finitely many nonempty disjoint Borel cells $C_{m,j}$, $1\le j\le J_m$, each of diameter less than $2^{-m}$. Such cells can be obtained by intersecting the disk with a finite half-open square grid. Their preimages $B_{m,j}$ partition $B$ into finite-measure Borel sets.

Apply step 1.1 to choose compact $H_{m,j}\subseteq B_{m,j}$ with $\mu (B_{m,j}\setminus H_{m,j})<\epsilon 2^{-m-3}\mathrm{/}{J}_m$. Put $H_m={\bigcup }_{j=1}^{J_m}{H}_{m,j}$ and $K={\bigcap }_{m\ge 0}{H}_m$. Each $H_m$ is compact and thus closed in $X$; their intersection is a closed subset of the compact $H_0$, hence compact. Moreover $K\subseteq B\subseteq E$ and $\mu (E\setminus K)\le \mu (E\setminus B)+{\sum }_{m\ge 0}\mu (B\setminus H_m)<\epsilon \mathrm{/}4+{\sum }_{m\ge 0}\epsilon 2^{-m-3}=\epsilon \mathrm{/}2<\epsilon$.

For each $m$, the finitely many disjoint compact sets $K\cap H_{m,j}$ cover $K$. Each is relatively clopen, because its complement is a finite union of closed sets. On it, the values of $f$ lie in $C_{m,j}$ and have oscillation below $2^{-m}$. Given $x\in K$ and a positive tolerance, choose $m$ so $2^{-m}$ is smaller; its clopen cell piece is a neighbourhood witnessing continuity at $x$. The empty $K$ case is vacuous. Thus $f{\mathrm{\mid }}_K$ is continuous.