Proposition 14.6.5. Let $X$ be a LCH space and $\mu: \cb_{X} \to [0, \infty]$ be a Borel measure such that:
For any $U \subset X$ open, $U$ is $\sigma$-compact.
For any $K \subset X$ compact, $\mu(K) < \infty$.
then $\mu$ is a regular measure on $X$.
Proof. By assumption (b), $f \mapsto \int f d\mu$ is a positive linear functional on $C_{c}(X; \real)$. By the Riesz Representation Theorem, there exists a Radon measure $\nu: \cb_{X} \to [0, \infty]$ such that for any $f \in C_{c}(X; \real)$, $\int f d\mu = \int f d\mu$.
Let $U \subset X$ be open, then by Proposition 14.6.2,
By assumption (a), there exists $\seq{K_n}\subset 2^{X}$ compact such that $K_{n} \upto U$. By Urysohn’s lemma, there exists $\seq{f_n}\subset C_{c}(X; [0, 1])$ such that $\one_{K_n}\le f_{n} \le \one_{U}$ for all $n \in \natp$, and $f_{n} \upto f$ pointwise. Using the Monotone Convergence Theorem,
Therefore $\mu(U) = \nu(U)$.
Now let $E \in \cb_{X}$ be arbitrary and $\eps > 0$. By Proposition 14.6.4, there exists $U \in \cn^{o}(E)$ and $V \subset E$ closed such that $\nu(U \setminus E), \nu(E \setminus V) < \eps$. In which case, since $U \setminus V$ is open,
so $\mu(U) = \mu(U \setminus E) + \mu(E) \le \mu(E) + \eps$. Therefore
and $\mu = \nu$.
Since $X$ is $\sigma$-compact, $\mu$ is $\sigma$-finite, so $\mu$ is regular by Proposition 14.6.3.$\square$