Outer Measures

Definition: Outer Measure

Let $X$ be a set.

An outer measure on $X$ is a function ${\mu }^{\ast }:\mathcal{P}(X)\to {\mathbb{R}}_{\ge 0}\cup \{\infty \}$ from the powerset of $X$ to the non-negative extended real numbers with the following properties:

• ${\mu }^{\ast }(\varnothing )=0$
• For all subsets $A$ and $B$ of $X$, we have

$$A\subseteq B⟹{\mu }^{\ast }(A)\le {\mu }^{\ast }(B)$$

• For all countable subcollections $\{S_i{\}}_{i=1}^{\infty }$ of $\mathcal{P}(X)$, we have

$${\mu }^{\ast }\left(\bigcup _{i=1}^{\infty }{S}_i\right)\le \sum _{i=1}^{\infty }{\mu }^{\ast }(S_i)$$

NOTATION

We usually write ${\mathbb{R}}_{\ge 0}\cup \{\infty \}$ as $[0,\infty ]$.

WARNING

An outer measure is not necessarily a measure on $(X,\mathcal{P}(X))$.

Theorem: Alternative Definition

Let $X$ be a set.

A function ${\mu }^{\ast }:\mathcal{P}(X)\to {\mathbb{R}}_{\ge 0}\cup \{\infty \}$ from the powerset of $X$ to the non-negative extended real numbers is an outer measure on $X$ if and only if ${\mu }^{\ast }$ has the following properties:

• ${\mu }^{\ast }(\varnothing )=0$
• For all subsets $A\subseteq X$ and all countable subcollections $\{S_i{\}}_{i=1}^{\infty }$ of $\mathcal{P}(X)$ we have

$$A\subseteq \bigcup _{i=1}^{\infty }{S}_i⟹{\mu }^{\ast }(A)\le \sum _{i=1}^{\infty }{\mu }^{\ast }(S_i)$$

PROOF

Proof of (1):

TODO

Proof of (2):

TODO

Definition: Carathéodory Measurability

Let $X$ be a set with an outer measure ${\mu }^{\ast }$.

A subset $E\subseteq X$ is Carathéodory-measurable relative to ${\mu }^{\ast }$ or ${\mu }^{\ast }$-measurable if

$${\mu }^{\ast }(S)={\mu }^{\ast }(S\cap E)+{\mu }^{\ast }(S\setminus E)$$

for every subset $S\subseteq X$.

Carathéodory's Extension Theorem

If $X$ is a set with an outer measure ${\mu }^{\ast }$, then:

• The collection of all ${\mu }^{\ast }$-measurable subsets of $X$ is a σ-algebra $\Sigma$ on $X$.

• The restriction ${\mu }^{\ast }{|}_{\Sigma }$ of ${\mu }^{\ast }$ on $\Sigma$ is a measure $\mu$ on $(X,\Sigma )$.

• The measure space $(X,\Sigma ,\mu )$ is complete.

PROOF

TODO