Proposition 2.2.2D0882F84DCFC

Proof. Assume first that $\mu (\Omega )<\infty $, and let $F\subset \Omega $ be relatively closed; we may assume that $F\neq \varnothing $. The open sets $U_ j=\{ x\in \Omega :\operatorname {dist}(x,F)<j^{-1}\} $ decrease to $F$, so $\mu (U_ j)\to \mu (F)$ because $\mu $ is finite. Hence $F$ can be approximated in measure from outside by open sets. Passing to complements, every open set can be approximated from inside by relatively closed sets.

Let $\mathcal A$ be the class of Borel sets $E$ such that for every $\varepsilon >0$ there are an open set $U$ and a relatively closed set $F$ with $F\subset E\subset U$ and $\mu (U\setminus F)<\varepsilon $. We have just seen that $\mathcal A$ contains the relatively closed sets, and $\mathcal A$ is closed under complements because the condition is symmetric. To see that $\mathcal A$ is closed under countable unions, let $E=\bigcup _ jE_ j$ with $E_ j\in \mathcal A$, and choose $F_ j\subset E_ j\subset U_ j$ with $\mu (U_ j\setminus F_ j)<\varepsilon 2^{-j}$. The open set $U=\bigcup _ jU_ j$ satisfies $\mu (U\setminus E)<\varepsilon $. Since $\mu (\bigcup _{j\leq N}E_ j)\to \mu (E)$, there is $N$ such that the relatively closed set $F=\bigcup _{j\leq N}F_ j$ satisfies $\mu (E\setminus F)\leq \mu (E\setminus \bigcup _{j\leq N}E_ j)+\sum _{j\leq N}\mu (E_ j\setminus F_ j)<2\varepsilon $, and hence $\mu (U\setminus F)<3\varepsilon $. Thus $\mathcal A$ is a $\sigma $-algebra containing the open sets, so it contains every Borel set $E$. If $F\subset E$ is relatively closed and $(K_ j)$ is a compact exhaustion of $\Omega $, then the sets $F\cap K_ j$ are compact and $\mu (F\setminus K_ j)\to 0$. Hence every Borel set can be approximated in measure from outside by open sets and from inside by compact sets.

In general, let $(K_ j)$ be a compact exhaustion of $\Omega $, put $K_0=\varnothing $, and let $V_ j=\operatorname{Int}K_{j+1}$. The measures $\mu _ j(E)=\mu (E\cap V_ j)$ are finite, hence Radon by the first part. For inner regularity, let $E$ be a Borel set and $\varepsilon >0$. A compact set $K\subset E$ with $\mu _ j(E\setminus K)<\varepsilon $ satisfies $\mu (K)\geq \mu _ j(K)>\mu (E\cap V_ j)-\varepsilon $, and $\mu (E\cap V_ j)\to \mu (E)$ as $j\to \infty $. For outer regularity, note that the Borel sets $B_ j=K_ j\setminus K_{j-1}$ partition $\Omega $ and that $E\cap B_ j\subset V_ j$. By outer regularity of $\mu _ j$ there is an open set $O_ j\subset V_ j$ containing $E\cap B_ j$ with $\mu (O_ j\setminus (E\cap B_ j))<\varepsilon 2^{-j}$; it suffices to intersect an approximating open set with $V_ j$. Then $O=\bigcup _ jO_ j$ is an open set containing $E$ with $\mu (O\setminus E)<\varepsilon $. Finally, $\mu $ is $\sigma $-finite because $\Omega =\bigcup _ jK_ j$ and each $\mu (K_ j)$ is finite. $\square$