Measure Theory/Measures on topological spaces

The following proposition provides a class of tight measure spaces:

{{proposition|Borel measure on Polish space is tight|Let

{{definition|inner regular|Let $$\Omega$$ be a topological space and let $$\mathcal F$$ be a $$\sigma$$-algebra on $$\Omega$$ that contains the Borel $$\sigma$$-algebra. A measure $$\mu: \mathcal F \to [0,\infty]$$ is called inner regular iff for all sets $$A \in \mathcal F$$
 * $$\mu(A) = \sup_{C \subseteq A \atop C \text{ closed}} \mu(C)$$.}}

{{definition|outer regular|Let $$\Omega$$ be a topological space and let $$\mathcal F$$ be a $$\sigma$$-algebra on $$\Omega$$ that contains the Borel $$\sigma$$-algebra. A measure $$\mu: \mathcal F \to [0,\infty]$$ is called outer regular iff for all sets $$A \in \mathcal F$$
 * $$\mu(A) = \inf_{O \supseteq A \atop O \text{ open}} \mu(O)$$.}}

{{proposition|closed set with empty interior in σ-compact measure space is nullset|Let $$\Omega$$ be a topological space, let $$\mathcal F$$ be a $$\sigma$$-algebra on $$\Omega$$ that contains the Borel $$\sigma$$-algebra, and suppose that $$\mu$$ is a ... measure on $$(\Omega, \mathcal F)$$. Then every closed subset $$A \subseteq \Omega$$ that has empty interior is a nullset.}}

{{proof|Let
 * $$\Omega = \bigcup_{n \in \mathbb N} K_n$$,

where the $$K_n$$ are compact. Then we have by countable subadditivity of measure
 * $$\mu(A) = \mu(\Omega \cap A) = \mu \left( \bigcup_{n \in \mathbb N} (K_n \cap A) \right) \le \sum_{n \in \mathbb N} \mu(A \cap K_n)$$.

But closed subsets of compact sets are compact, and hence it suffices to prove that $$\mu(A) = 0$$ whenever $$A$$ is a closed, compact subset of $$\Omega$$.}}