Inner measure

An inner measure is a set function $${\displaystyle \varphi :2^{X}\to [0,\infty ],}$$ defined on all subsets of a set ${\displaystyle X,}$ that satisfies the following conditions:

• Null empty set: The empty set has zero inner measure (see also: measure zero); that is, $${\displaystyle \varphi (\varnothing )=0}$$
• Superadditive: For any disjoint sets ${\displaystyle A}$ and ${\displaystyle B,}$ $${\displaystyle \varphi (A\cup B)\geq \varphi (A)+\varphi (B).}$$
• Limits of decreasing towers: For any sequence ${\displaystyle A_{1},A_{2},\ldots }$ of sets such that ${\displaystyle A_{j}\supseteq A_{j+1}}$ for each ${\displaystyle j}$ and ${\displaystyle \varphi (A_{1})<\infty }$ $${\displaystyle \varphi \left(\bigcap _{j=1}^{\infty }A_{j}\right)=\lim _{j\to \infty }\varphi (A_{j})}$$
• If the measure is not finite, that is, if there exist sets ${\displaystyle A}$ with ${\displaystyle \varphi (A)=\infty }$, then this infinity must be approached. More precisely, if ${\displaystyle \varphi (A)=\infty }$ for a set ${\displaystyle A}$ then for every positive real number ${\displaystyle r,}$ there exists some ${\displaystyle B\subseteq A}$ such that $${\displaystyle r\leq \varphi (B)<\infty .}$$

Let ${\displaystyle \Sigma }$ be a σ-algebra over a set ${\displaystyle X}$ and ${\displaystyle \mu }$ be a measure on ${\displaystyle \Sigma .}$ Then the inner measure ${\displaystyle \mu _{*}}$ induced by ${\displaystyle \mu }$ is defined by $${\displaystyle \mu _{*}(T)=\sup\{\mu (S):S\in \Sigma {\text{ and }}S\subseteq T\}.}$$

Essentially ${\displaystyle \mu _{*}}$ gives a lower bound of the size of any set by ensuring it is at least as big as the ${\displaystyle \mu }$-measure of any of its ${\displaystyle \Sigma }$-measurable subsets. Even though the set function ${\displaystyle \mu _{*}}$ is usually not a measure, ${\displaystyle \mu _{*}}$ shares the following properties with measures:

1. ${\displaystyle \mu _{*}(\varnothing )=0,}$
2. ${\displaystyle \mu _{*}}$ is non-negative,
3. If ${\displaystyle E\subseteq F}$ then ${\displaystyle \mu _{*}(E)\leq \mu _{*}(F).}$