Proposition 5.5.2.label Let $X$ be a set, $\cali{E}\suf 2^{X}$ and $\rho:\cali{E}\to[0,\infty]$ such that $\emp\in\cali{E}$ and $\rho(\emp)=0$. Define
\begin{align*}\mu^{*}:2^{X}\to[0,\infty]\\ E&\mto\inf \curl{\sums{}{n\in\N^+}\rho(E_n):\curl{E_n}_{n\in\N^+}\suf\cali{E},\cups{}{n\in\N^+}E_n\supf E}\end{align*}
then $\mu^{*}$ is an outer measure.[1]
Proof.
- (OM1)
$\emp\in\cali{E}$ and $\rho(\emp)=0$
- (OM2)
Let $E,F\suf X$ with $E\suf F$ then any cover of $F$ is a cover of $E$ hence $\mu^{*}(E)\leq \mu^{*}(F)$ by definition of infimum.
- (OM3)
Let $\curl{E_n}_{n\in\N^+}\suf X$ then WLOG $\sums{}{n\in\N^+}\mu^{*}(E_{n})<\infty$. Let $\e>0$. For each $n\in\N^{+}$, let $\curl{F_{n,k}}_{k\in\N^+}\suf\cali{E}$ such that $\cups{}{k\in\N^+}F_{n,k}\supf E_{n}$ and $\sums{}{n\in\N^+}\rho(F_{n,k})<\mu^{*}(E_{n})+2^{-n}\e$ then
\begin{align*}\mu^{*}\parens{\cups{}{n\in\N^+}E_n}&\sums{}{(n,k)\in\N^+\ti\N^+}\rho(F_{n,k})=\sums{}{n\in\N^+}\sums{}{k\in\N^+}\rho(F_{n,k})\\&\leq \sums{}{n\in\N^+}\parens{\mu^*(E_n)+2^{-n\e}}\leq \e+\sums{}{n\in\N^+}\mu^{*}(E_{n})\end{align*}
$\square$
- under the convention $\inf(\emp)=\infty$keyboard_return
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