Measure Theory/Measures on topological spaces

Definition (Borel σ-algebra):

Let $X$ be a topological space. The Borel $\sigma$-algebra ${\mathcal {B}}(X)$ on $X$ is the $\sigma$-algebra generated by all open subsets of $X$, ie.

${\mathcal {B}}(X)=\sigma (\tau )$,

where $\tau$ is the topology on $X$.

Definition (tight):

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 tight iff for all sets $A\in {\mathcal {F}}$

$\mu (A)=\sup _{K\subseteq A \atop K{\text{ compact}}}\mu (K)$.

The following proposition provides a class of tight measure spaces:

Proposition (Borel measure on Polish space is tight):

Let Failed to parse (syntax error): {\displaystyle {{definition|inner regular|Let <math>\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

$\mu (A)=\mu (\Omega \cap A)=\mu \left(\bigcup _{n\in \mathbb {N} }(K_{n}\cap A)\right)\leq \sum _{n\in \mathbb {N} }\mu (A\cap K_{n})$.

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 $\Omega$. $\Box$