Proposition 14.6.2. Let $X$ be a LCH space and $\mu: \cb_{X} \to [0, \infty]$ be a Radon measure, then
For any $U \subset X$ open,
\[\mu(U) = \sup\bracs{\int f d\mu \bigg | f \in C_c(X; [0, 1]), \supp{f} \subset U}\]For any $K \subset X$ compact,
\[\mu(K) = \inf\bracs{\int f d\mu \bigg | f \in C_c(X; [0, 1]), f \ge \one_K}\]
Proof. (1): Let $U \subset X$ be open. By Urysohn’s lemma, for any $K \subset U$ compact, there exists $f_{K} \in C_{c}(X; [0, 1])$ such that $f_{K}|_{K} = 1$ and $\supp{f_K}\subset U$. In which case,
\[\mu(K) \le \int f_{K} d\mu \le \mu(U)\]
By (R3’),
\[\mu(U) = \sup_{K \subset U \text{ compact}}\mu(K) \le \sup_{K \subset U \text{ compact}}\int f_{K} d\mu \le \mu(U)\]
(2): Let $K \subset X$ be compact and $U \in \cn^{o}(K)$. By Urysohn’s lemma, there exists $f_{U} \in C_{c}(X; [0, 1])$ such that $f_{U}|_{K} = 1$ and $\supp{f_U}\subset U$. In which case,
\[\mu(K) \le \int f_{U} d\mu \le \mu(U)\]
By (R2),
\[\mu(K) = \inf_{U \in \cn^o(K)}\mu(U) \ge \inf_{U \in \cn^o(K)}\int f_{U} d\mu \ge \mu(K)\]
$\square$