Sigma-compact open sets make locally finite Borel measures regular

If $U$ is open, write $U={\bigcup }_n{L}_n$ with $L_n$ compact. The compact sets $K_n={\bigcup }_{j\le n}{L}_j$ increase to $U$, so continuity from below [L1] gives $\mu (U)={\sup }_n\mu (K_n)$. Thus $\mu$ is inner regular on opens.

Since $X$ itself is sigma-compact, local compactness and finite [step 1.1, A1, choose] subcovers, chosen for each stage using [A1], give increasing relatively compact open sets $V_n$ with $X={\bigcup }_n{V}_n$. Fix one such $V=V_n$. Its total measure is finite because $\mu (V)\le \mu (\bar{V})<\mathrm{\infty }$. Every relatively open subset of $V$ is open in $X$, hence is compact-inner-regular by step 1.1. Every relatively closed subset of $V$ is sigma-compact: intersect it with a compact exhaustion of the sigma-compact open space $V$. It is therefore also compact-inner-regular by continuity from below.

Let ${\mathcal{R}}_V$ be the Borel subsets of $V$ that are both outer [step 2.1] regular in $V$ and inner regular by compact sets. The relatively open sets belong to ${\mathcal{R}}_V$: inner regularity is step 2.1 and outer regularity is immediate. A relatively closed set $F$ is inner regular by step 2.1; if $W=V\setminus F$, choose compact $K\subseteq W$ with $\mu (W\setminus K)<\epsilon$, and then the relatively open set $V\setminus K$ contains $F$ with excess below $\epsilon$. Thus closed sets also belong to ${\mathcal{R}}_V$. Because $\mu (V)<\mathrm{\infty }$, taking complements interchanges the two approximation properties. Countable unions preserve outer regularity by summable open errors chosen using [A1] and preserve inner regularity by first taking a finite partial union and then a finite union of compact cores. Hence ${\mathcal{R}}_V$ is a sigma-algebra containing the relatively open sets, so every Borel subset of $V$ is regular.

Let $E$ be Borel. Since $E\cap V_n\uparrow E$, the regularity just proved [step 3.1] on each $V_n$ gives $\mu (E)=\sup \{\mu (K):K\subseteq E,K\text{ compact}\}\mathrm{.}$ Indeed, when $\mu (E)<\mathrm{\infty }$, first choose $n$ so that $\mu (E\setminus V_n)$ is small and then take a compact core of $E\cap V_n$; when $\mu (E)=\mathrm{\infty }$, choose $n$ and then a compact core with arbitrarily large finite measure.

If $\mu (E)=\mathrm{\infty }$, outer regularity is automatic. Otherwise partition [step 3.1, step 4.1] $E$ into the Borel slices $E_n=E\cap (V_n\setminus V_{n-1})$, with $V_{-1}=\mathrm{\varnothing }$. Relative outer regularity in $V_n$ gives a relatively open, hence open-in-$X$, set $O_n\subseteq V_n$ containing $E_n$ with $\mu (O_n\setminus E_n)<\epsilon 2^{-n-1}$; use [A1] to choose all such $O_n$ at once. Then $O={\bigcup }_n{O}_n$ is open, contains $E$, and $\mu (O\setminus E)<\epsilon$. Thus every Borel set is outer regular as well as compact-inner-regular; in particular $\mu$ is Radon and is regular in the stronger convention.