# Measure Theory/Measures on topological spaces

Definition (Borel σ-algebra):

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

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

where ${\displaystyle \tau }$ is the topology on ${\displaystyle X}$.

Definition (tight):

Let ${\displaystyle \Omega }$ be a topological space and let ${\displaystyle {\mathcal {F}}}$ be a ${\displaystyle \sigma }$-algebra on ${\displaystyle \Omega }$ that contains the Borel ${\displaystyle \sigma }$-algebra. A measure ${\displaystyle \mu :{\mathcal {F}}\to [0,\infty ]}$ is called tight iff for all sets ${\displaystyle A\in {\mathcal {F}}}$

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

The following proposition provides a class of tight measure spaces:

Proposition (Borel measure on Polish space is tight):

Let $\displaystyle {{definition|inner regular|Let [itex]\Omega$ be a topological space and let ${\displaystyle {\mathcal {F}}}$ be a ${\displaystyle \sigma }$-algebra on ${\displaystyle \Omega }$ that contains the Borel ${\displaystyle \sigma }$-algebra. A measure ${\displaystyle \mu :{\mathcal {F}}\to [0,\infty ]}$ is called inner regular iff for all sets ${\displaystyle A\in {\mathcal {F}}}$

${\displaystyle \mu (A)=\sup _{C\subseteq A \atop C{\text{ closed}}}\mu (C)}$.

Definition (outer regular):

Let ${\displaystyle \Omega }$ be a topological space and let ${\displaystyle {\mathcal {F}}}$ be a ${\displaystyle \sigma }$-algebra on ${\displaystyle \Omega }$ that contains the Borel ${\displaystyle \sigma }$-algebra. A measure ${\displaystyle \mu :{\mathcal {F}}\to [0,\infty ]}$ is called outer regular iff for all sets ${\displaystyle A\in {\mathcal {F}}}$

${\displaystyle \mu (A)=\inf _{O\supseteq A \atop O{\text{ open}}}\mu (O)}$.

Proposition (closed set with empty interior in σ-compact measure space is nullset):

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

Proof: Let

${\displaystyle \Omega =\bigcup _{n\in \mathbb {N} }K_{n}}$,

where the ${\displaystyle K_{n}}$ are compact. Then we have by countable subadditivity of measure

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

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