Problem
Given a locally compact Hausdorff space.
Consider a regular Borel measure in the sense: $$\mu:\mathcal{B}(\Omega)\to\overline{\mathbb{R}}_+:\quad\mu(E)=\inf_{E\subseteq U}\mu(U)\quad\mu(U)=\sup_{K\subseteq U}\mu(K)\quad(\mu(K)<\infty)$$
Then one can approximate the measure of opens by functions: $$\mu(U)=\sup\left\{\int h\mathrm{d}\mu:h\in\mathcal{C}_0(\Omega):\quad h\leq\chi_U\right\}$$ Does a similar approach work for compacta: $$\mu(K)=\inf\left\{\int h\mathrm{d}\mu:h\in\mathcal{C}_0(\Omega):\quad\chi_K\leq h\right\}$$ (I have the feeling this may fail.)
Attempt
For the former one has by regularity: $$m\leq\mu(U):\quad m\leq\mu(K)\leq\mu(U)\quad(K\subseteq U)$$ Applying Urysohn's lemma one finds: $$h\in\mathcal{C}_0(\Omega):\quad\mu(K)=\int\chi_K\mathrm{d}\mu\leq\int h\mathrm{d}\mu\quad(h\leq\chi_U)$$
For the latter one regularity is more twisted.
Fix a compact set $K$. As you point out, if $h \in \mathcal C_0(\Omega)$ and $\chi_K \le h$ then $$\mu(K) = \int \chi_K \, d\mu \le \int h \, d\mu$$ so that $$\mu(K) \le \inf \left\{ \int h \, d\mu :\ h \in \mathcal C_0(\Omega),\ \chi_K \le h\right\}.$$
On the other hand, let $U \supset K$ be open. Since there exists $h_0 \in \mathcal C_0(\Omega)$ satisfying $\chi_K \le h_0 \le \chi_U$ you have that $$ \inf \left\{ \int h \, d\mu :\ h \in \mathcal C_0(\Omega),\ \chi_K \le h\right\} \le \int h_0 \, d\mu \le \sup \left\{ \int h \, d\mu :\ h \in \mathcal C_0(\Omega),\ h \le \chi_U \right\},$$ i.e., $$\inf \left\{ \int h \, d\mu :\ h \in \mathcal C_0(\Omega),\ \chi_K \le h\right\} \le \mu(U).$$ Now take the infimum over all $U$ containing $K$: $$\inf \left\{ \int h \, d\mu :\ h \in \mathcal C_0(\Omega),\ \chi_K \le h\right\} \le \mu(K).$$