Approximation of Opens by regular compacts

Question

Let $(X,\Sigma,\mu)$ be a Radon space and $X$ be a compact metric space. By definition of $\mu$, every Borel set $A \in \Sigma$ can be approximated from within by compact sets. However, if $A$ is itself a regular open subset of $X$, can it necessarily be approximated by regular compact subsets in the sense that $$ \mu(A)=\sup\{\mu(K):K\subseteq A,K \text{compact and regular}\,\}? $$ I expect that you can always add in the missing regular parts to any compact subset $K$ but I'm not sure if I'm missing something...

Answer 1

If $A$ is $any$ open subset of $X$. Let $d$ be the metric on $X$. For $n\in \Bbb N$ let $A_n=\{x\in A: B_d(x,1/n)\subset A\}.$ Let $C_n=\cup \{B_d(x,1/4n):x\in A_n\}.$ Let $D_n=Cl(int (Cl(C_n))).$

Any subset $Y$ of any topological space satisfies $V(V(Y))=V(Y)$ where $V(Y)=Cl(int(Y)).$

Therefore $$Cl(int(D_n))= Cl(int(Cl(int(Cl((C_n)=$$ $$=V(V(Cl(C_n)))=V(Cl(C_n))=Cl(int(Cl(Cn)))=D_n.$$

So $D_n$ is regular compact. We have $D_n\subset D_{n+1}\subset A.$ And $\cup_{n\in \Bbb N}D_n=A,$ so $\mu(A)=\mu(A_1)+\sum_{n=1}^{\infty}\mu (A_{n+1}\backslash A_n)<\infty.$

So $\lim_{n\to \infty} G(n)=0$ where $G(n)=\sum_{m\geq n}\mu(A_{m+1}\backslash A_n).$

So $\mu (A)=\lim_{n\to \infty}\mu(A_n)+G(n)=\lim_{n\to \infty}\mu(A_n)=\sup_{n\in \Bbb N}\mu(A_n).$

Remark: It is a fairly easy exercise to show that $D_n\subset \cup \{B_d(x,1/n): x\in A_n\}\subset A.$