If $(X, \mathcal{A}, \mu)$ is a measure space, define $\mu^*(A) = \inf\{\mu(B) : A \subset B, B \in \mathcal{A} \}$ for all subsets $A$ of $X$. Show that $\mu^*$ is an outer measure. Show that each set in $\mathcal{A}$ is $\mu^*$-measurable and $\mu^*$ agrees with the measure $\mu$ on $\mathcal{A}$.
To show $\mu^*$ is an outer measure.
Show $\mu^*(\emptyset) = 0$
Let $B = \emptyset$, then $A \subset B, B \in \mathcal{A}$ and $\mu(\emptyset) = 0$, therefore $\mu^*(\emptyset) = 0$.
if $C \subset D$, then $\mu^*(C) \leq \mu^*(D).$
Since $C \subset D$, then every $B \in \mathcal{A}$ that covers $D$ covers $C$ as well, $C \subset D \subset B$. Since the infimum that is defined on $\mu^*(C)$ has more members than $\mu^*(D)$, then $\mu^*(C) \leq \mu^*(D).$
I'm having trouble with showing this condition of an outer measure.
$\mu^*(\cup_{i=1}^\infty{A_i}) \leq \sum_{i=1}^\infty{\mu^*(A_i)}$ whenever $A_1, A_2, ...$ are subsets of $X$.
As well as showing the each set is $\mu^*$-measurable.
Fix $\epsilon>0$. For all $i\ge 1$, $\exists B_i\in \mathcal{A}$ s.t. $\mu(B_i)\le \mu^*(A_i)+\epsilon2^{-i}$. Then $$ \mu^*\left(\bigcup_{i\ge 1}A_i\right)\le \mu^*\left(\bigcup_{i\ge 1}B_i\right)\le \sum_{i\ge 1}\mu^*(A_i)+\epsilon. $$
It's clear that $\mu^*\mid_{\mathcal{A}}=\mu$ (by monotonicity of $\mu$). If $A\in\mathcal{A}$, then for any $E\subset X$ and $\epsilon>0$, $\exists B\in\mathcal{A}$ s.t. $\mu^*(E)+\epsilon\ge \mu(B)$ and $$ \mu^*(E)+\epsilon\ge \mu(B\cap A)+\mu(B\cap A^c)\ge \mu^*(E\cap A)+\mu^*(E\cap A^c). $$