Compact-set formula and local finiteness of the RMK measure

Suppose ${\mathbf{1}}_K\le f\in C_c(X)$. For $0<\epsilon <1$, the open set $U_{\epsilon }=\{f>1-\epsilon \}$ contains $K$. Every $g\prec U_{\epsilon }$ satisfies $g\le f\mathrm{/}(1-\epsilon )$, so positivity gives $\rho (U_{\epsilon })\le \mathrm{\Lambda }(f)\mathrm{/}(1-\epsilon )$. Outer regularity therefore yields $\mu (K)\le \mathrm{\Lambda }(f)\mathrm{/}(1-\epsilon )$, and then $\mu (K)\le \mathrm{\Lambda }(f)$.

Conversely, for every open $U\supseteq K$, choose an open $V$ with [L1, choose] $K\subseteq V\subseteq \bar{V}\subseteq U$ and compact closure. [L1] applied to $K\subseteq V$ supplies $f\in C_c(X)$ with ${\mathbf{1}}_K\le f\le {\mathbf{1}}_V$. Since $f=0$ off $V$, $\mathrm{supp}f\subseteq \bar{V}\subseteq U$, and hence $f\prec U$. Thus $\mathrm{\Lambda }(f)\le \rho (U)$. Taking first the infimum over $f$, then over $U$, gives the reverse inequality. For $K=\mathrm{\varnothing }$, $f=0$ gives both sides zero.

Choosing one relatively compact open neighbourhood $U$ of $K$ and the [step 1.1, step 1.2, L1] cutoff produced in step 1.2 gives $\mu (K)\le \mathrm{\Lambda }(f)<\mathrm{\infty }$.