Proof of Closed–Open Regularity of a Finite Borel Measure on a Metric Space

Throughout, open and closed subsets of $X$ are Borel, all sets formed below from them are Borel because $\mathcal{B}(X)$ is a $\sigma$-algebra, and the values of $\mu$ are real numbers because $\mu$ is finite.

Let $\mathcal{A}$ be the set of those $B\in\mathcal{B}(X)$ such that for every real $\varepsilon>0$ there are a closed $F$ and an open $V$ with $F\subseteq B\subseteq V$ and $\mu(V\setminus F)<\varepsilon$. We show that $\mathcal{A}$ is a $\sigma$-algebra on $X$ containing every open set. Since $\mathcal{B}(X)$ is the $\sigma$-algebra generated by the open sets (Borel Sigma-Algebra of a Metric Space (i)), it then follows that $\mathcal{B}(X)\subseteq\mathcal{A}$, which is the claim (i).

Open sets. Let $V\subseteq X$ be open and let $\varepsilon>0$. If $V=X$, the closed set $F=X$ works, since $\mu(V\setminus F)=\mu(\emptyset)=0$. Otherwise $A=X\setminus V$ is nonempty and closed. For $n\in\mathbb{N}$ let

$$F_{n}=\{x\in X:\operatorname{dist}(x,A)\ge 1/n\}.$$

Each $F_{n}$ is closed, its complement $\{x\in X:\operatorname{dist}(x,A)<1/n\}$ being open by Compact Subsets of a Metric Space: Covers by Open Sets, Finite Unions and Increasing Covers; Sublevel Sets of the Distance and Closed Balls (v). Since $\operatorname{dist}(x,A)=0$ for $x\in A$, we have $F_{n}\subseteq V$; clearly $F_{n}\subseteq F_{n+1}$. Every $x\in V$ lies outside the closed set $A$, so $\operatorname{dist}(x,A)>0$, and hence $\operatorname{dist}(x,A)\ge1/n$ for some $n$; thus $\bigcup_{n}F_{n}=V$. By continuity from below, $\mu(V)=\sup_{n}\mu(F_{n})$, and since $\mu(V)$ is finite there is $n$ with $\mu(F_{n})>\mu(V)-\varepsilon$. Then

$$\mu(V\setminus F_{n})=\mu(V)-\mu(F_{n})<\varepsilon,$$

so $V\in\mathcal{A}$ with the closed set $F_{n}$ and the open set $V$. In particular $X\in\mathcal{A}$.

Complements. Let $B\in\mathcal{A}$ and $\varepsilon>0$, and take a closed $F$ and an open $V$ with $F\subseteq B\subseteq V$ and $\mu(V\setminus F)<\varepsilon$. Then $X\setminus V$ is closed, $X\setminus F$ is open,

$$X\setminus V\subseteq X\setminus B\subseteq X\setminus F,\qquad (X\setminus F)\setminus(X\setminus V)=V\setminus F,$$

so $X\setminus B\in\mathcal{A}$.

Countable unions. Let $(B_{m})$ be a sequence in $\mathcal{A}$, let $B=\bigcup_{m}B_{m}$ and let $\varepsilon>0$. For each $m\in\mathbb{N}$ choose a closed $F_{m}$ and an open $V_{m}$ with

$$F_{m}\subseteq B_{m}\subseteq V_{m},\qquad \mu(V_{m}\setminus F_{m})<\varepsilon/2^{m+1}$$

(the whole sequence $(F_{m},V_{m})_{m}$ is chosen by the axiom of dependent choice). The set $V=\bigcup_{m}V_{m}$ is open and contains $B$. Let $E=\bigcup_{m}F_{m}\subseteq B$. A point of $V\setminus E$ lies in some $V_{m}$ but not in $F_{m}$, so $V\setminus E\subseteq\bigcup_{m}(V_{m}\setminus F_{m})$, and countable subadditivity gives

$$\mu(V\setminus E)\le\sum_{m\in\mathbb{N}}\varepsilon/2^{m+1}=\varepsilon/2 .$$

The closed sets $G_{N}=F_{1}\cup\dots\cup F_{N}$ increase with union $E$, so by continuity from below and finiteness of $\mu(E)$ there is $N$ with $\mu(E\setminus G_{N})=\mu(E)-\mu(G_{N})<\varepsilon/2$. Now $G_{N}\subseteq B\subseteq V$ and $V\setminus G_{N}\subseteq(V\setminus E)\cup(E\setminus G_{N})$, so

$$\mu(V\setminus G_{N})\le\mu(V\setminus E)+\mu(E\setminus G_{N})<\varepsilon .$$

Hence $B\in\mathcal{A}$.

Conclusion. $\mathcal{A}$ is a subset of $\mathcal{P}(X)$ that contains $X$ and is closed under complements and countable unions, so it is a $\sigma$-algebra on $X$; it contains every open set, and therefore $\mathcal{B}(X)\subseteq\mathcal{A}$.