Lemma 4.1.9468BD94D4D83

Proof. The sets $\{ |f|>N\} $ decrease to the $\mu $-null set $\{ |f|=\infty \} $ as $N\to \infty $, and $\mu $ is finite, so we can choose $N$ with $\mu (\{ |f|>N\} )<\varepsilon /3$. The truncation $f_ N=\max \{ -N,\min \{ f,N\} \} $ is bounded, hence belongs to $L^1(\mu )$. By proposition 2.3.5 there are continuous functions $g_ j$, $j\geq 1$, with $\int |g_ j-f_ N|\, \mathrm d\mu <\varepsilon 2^{-2j}/3$. By Chebyshev's inequality, the set $E_ j=\{ |g_ j-f_ N|>2^{-j}\} $ satisfies $\mu (E_ j)<\varepsilon 2^{-j}/3$, so $\mu (\bigcup _ jE_ j)<\varepsilon /3$. On the Borel set $F=\mathbb R^ d\setminus (\{ |f|>N\} \cup \bigcup _ jE_ j)$ we have $f=f_ N$ and $|g_ j-f|\leq 2^{-j}$ for every $j$, so $g_ j\to f$ uniformly on $F$. Since $\mu (\mathbb R^ d\setminus F)<2\varepsilon /3$, inner regularity gives a compact set $L\subset F$ with $\mu (F\setminus L)<\varepsilon /3$. Then $\mu (\mathbb R^ d\setminus L)<\varepsilon $, and $f|_ L$ is finite and continuous as a uniform limit of continuous functions. $\square$