Approximating Open Sets With Compact Sets in a $\sigma$-compact Polish Space

Question

Suppose that $X$ is a $\sigma$-compact Polish Space.

Suppose $\mu$ is a Borel measure on $X$ that is finite on compact sets.

How can we approximate any open set with compact sets so as to prove the inner regularity of $\mu$ on open sets?

Answer 1

Notation: $N=\mathbb N.$

Let $d$ be a metric for $X.$ Let $X=\cup_{n\in N}A_n$ where each $A_n$ is compact. Let $C_n=\cup_{j\leq n}A_j.$ Then $C_n$ is compact, with $ C_n\subset C_m$ when $n\leq m$. And $ \cup_{n\in N}C_n=X.$

Let $U$ be an open subset of $X.$ For $n\in N$ let $U_n=\{p\in U: B_d(p,1/n)\subset U\}$ and let $V(U,n)=$Cl$ (\;\cup_{p\in U_n}B_d(p,1/2n)\;).$

We have $\cup_{n\in N}V(U,n)=U,$ with $V(U,n)\subset V(U,m)$ when $n\leq m.$

For $n\in N$ let $ W(U,n)=C_n\cap V(U,n).$ Each $W(U,n)$ is compact and $W(U,n)\subset W(U,m)$ when $n\leq m.$

If $p\in U$ there exists $n(1)$ with $p\in C_{n(1)}$ and $n(2)$ with $p\in V(n(2),U)$, so $p\in W(\;U,\max (n(1),n(2))\;).$ Therefore $U=\cup_{n\in N}W(U,n).$

$$\text {Now }\quad \mu(U)=\mu(W(U),1)+\sum_{n=1}^{\infty}\mu(\;W(U,n+1)) \;\backslash \; W(U,n)\;)=$$ $$=\mu(W(U,1))+\sup_{n\in N}\sum_{j=1}^n\mu(\;W(U,n+1))\; \backslash \; W(U,n)\;)=$$ $$=\sup_{n\in N}\mu (W(U,n+1)).$$

Remark: I recall one article (essay) in which a Borel measure $\mu$ was called inner regular iff $\mu (T)=\sup \{\mu (W):W=\overline W\subset T\}$ (for all measurable $T$) and was called Radon regular (or Radon inner regular) iff $\mu(T)=\sup \{\mu(W):W\subset T \land W$ compact$\}$ (for all measurable $T$).