Riesz-Markov-Kakutani Representation Theorem/Lemma 9

Lemma
$\MM_F = \set {E \in \MM: \map \mu E < \infty}$

Proof
By Lemma 2 and Lemma 8:
 * $E \in \MM_F \implies \paren {\forall K \text{ compact}: E \cap K \in \MM_F \implies E \in \MM}$

That is:
 * $\MM_F \subset \MM$

Conversely, suppose that:
 * $E \in \MM$
 * $\map \mu E < \infty$
 * $\epsilon \in \R{>0}$

By definition of $\mu$, there exists an open $V \supset E$ such that:
 * $\map \mu V < \infty$

By Lemma 5 and Lemma 6, there exists a compact $K \subset V$ such that:
 * $\map \mu {V \setminus K} < \epsilon$

By definition of $\MM$:
 * $E \cap K \in \MM_F$.

So by definition of $\MM_F$, there exists a compact $H \subset E \cap K$ such that:
 * $\map \mu {E \cap K} < \map \mu H + \epsilon$

Since $E \subset \paren {E \cap K} \cup \paren {V \setminus K}$, by Lemma 1:
 * $\map \mu E \le \map \mu {E \cap K} + \map \mu {V \setminus K} + 2 \epsilon$

Thus:
 * $E \in \MM_F$.

Therefore:
 * $\MM_F = \set {E \in \MM: \map \mu E < \infty}$