Proof of Finite Borel Measures on a Locally Compact Separable Metric Space Are Radon

Throughout, compact subsets of $X$ are closed and hence Borel, and the values of a finite Borel measure $\mu$ on $X$ are real numbers in $[0,\mu(X)]$.

Inner regularity by compact sets. Let $\mu$ be a finite Borel measure on $X$, let $B\in\mathcal{B}(X)$ and let $s=\sup\{\mu(K):K\subseteq X\text{ compact},\ K\subseteq B\}$. By monotonicity $\mu(K)\le\mu(B)$ for every such $K$, so $s\le\mu(B)$. For the reverse inequality let $\varepsilon>0$. By Closed–Open Regularity of a Finite Borel Measure on a Metric Space (i) there are a closed $F$ and an open $V$ with $F\subseteq B\subseteq V$ and $\mu(V\setminus F)<\varepsilon/2$; in particular $\mu(B\setminus F)<\varepsilon/2$. Let $(K_{m})$ be the compact sets of Exhaustion of a Locally Compact Separable Metric Space by Compact Sets, with Cut-Off Functions (i). Each $F\cap K_{m}$ is a closed subset of the compact set $K_{m}$, hence compact; the sets $F\cap K_{m}$ increase with $m$ and their union is $F\cap X=F$. By continuity from below $\mu(F)=\sup_{m}\mu(F\cap K_{m})$, so there is $m$ with $\mu(F)-\mu(F\cap K_{m})<\varepsilon/2$. Then

$$\mu(B)-\mu(F\cap K_{m})=\mu(B\setminus F)+\bigl(\mu(F)-\mu(F\cap K_{m})\bigr)<\varepsilon,$$

and $F\cap K_{m}$ is a compact subset of $B$, so $s>\mu(B)-\varepsilon$. As $\varepsilon>0$ was arbitrary, $s\ge\mu(B)$, which proves (i).

Radon. Let $\mu$ be a finite Borel measure on $X$. For compact $K$, $\mu(K)\le\mu(X)<\infty$. Inner regularity on open sets is the case $B=V$ of (i). For outer regularity let $B\in\mathcal{B}(X)$. Every open $V\supseteq B$ has $\mu(V)\ge\mu(B)$, so the infimum is at least $\mu(B)$. Given $\varepsilon>0$, Closed–Open Regularity of a Finite Borel Measure on a Metric Space (i) gives a closed $F$ and an open $V$ with $F\subseteq B\subseteq V$ and $\mu(V\setminus F)<\varepsilon$; then $\mu(V)-\mu(B)=\mu(V\setminus B)\le\mu(V\setminus F)<\varepsilon$. Hence the infimum equals $\mu(B)$, and $\mu$ is a Radon measure, which proves (ii).

Signed Radon. Let $\nu$ be a finite signed Borel measure on $X$. Its total variation $|\nu|$ is a finite Borel measure by Properties of the Total Variation: a Finite Measure, the Jordan Decomposition, Linear Combinations, the Total Variation Norm, and Integrability (i), so it is a Radon measure by (ii). By Finite Signed Radon Measure on a Metric Space (i), $\nu$ is a finite signed Radon measure, which proves (iii).