Riesz-Markov-Kakutani Representation Theorem/Lemma 2

Lemma
Let $K \subset X$ be compact.

Then:
 * $K \in \MM_F$

and:
 * $\map \mu K = \inf \set {\Lambda f : K \prec f}$

Proof
Let $f \in \map {C_c} X: K \prec f$.

Let $\alpha \in \R \cap \openint 0 1$.

Define:
 * $V_\alpha = \map {f^{-1} } {\openint \alpha \to}$

Then $K \in V_\alpha$ and:
 * $\forall g: \map {C_c} X: g \prec V_\alpha: \alpha g \le f$

Thus:

Since $\alpha$ can be arbitrarily close to $1$:
 * $\map \mu K \le \inf \set {\Lambda f: K \prec f}$

By definition of $\mu$, for all $\epsilon \in \R_{>0}$, there exists a $V \in \tau: V \supset K$ such that:
 * $\map \mu V < \map \mu K + \epsilon$

By Urysohn's Lemma, there exists an Urysohn function $f$ for $K$ and $V$.

By definition of $\mu$:
 * $\Lambda f \le \map \mu V < \map \mu K + \epsilon$

Since $\epsilon$ was arbitrary:
 * $\map \mu K \ge \inf \set {\Lambda f: K \prec f}$

Therefore:
 * $\map \mu K = \inf \set {\Lambda f: K \prec f}$

By definition of positive linear functional, $\Lambda$ is finite.

Thus:
 * $\map \mu K < \infty$

So:
 * $K \in \MM_F$