Proposition 14.6.3. Let $X$ be a LCH space and $\mu: \cb_{X} \to [0, \infty]$ be a Radon measure, then
$\mu$ is inner regular on all its $\sigma$-finite sets.
If $X$ is $\sigma$-compact or $\mu$ is $\sigma$-finite, then $\mu$ is regular.
Proof. (1): Let $E \in \cb_{X}$ with $\mu(E) < \infty$ and $\eps > 0$, then there exists
$U \in \cn^{o}(E)$ with $\mu(U \setminus E) < \eps/2$.
$K \subset U$ compact with $\mu(K) > \mu(U) - \eps/2$.
$V \in \cn^{o}(U \setminus E)$ with $\mu(V) < \eps/2$.
In which case, $K \setminus V \subset E$ is compact with
so $\mu$ is inner regular on $E$.
Now suppose that $E$ is $\sigma$-finite, then there exists $\seq{E_n}\subset \cb_{X}$ with $\mu(E_{n}) < \infty$ for each $n \in \natp$ and $E = \bigsqcup_{n \in \natp}E_{n} = E$. Let $N \in \natp$ and $\seqf{K_n}\subset 2^{X}$ compact with $K_{n} \subset E_{n}$ for each $n \in \natp$, then $\bigcup_{n = 1}^{N} K_{n} \subset E$ is compact. Hence
As the above holds for all $N \in \natp$,