Let $(X,\mathcal{A},\mu)$ be a measure space, and define $$\mu^*(A):=\inf\{\mu(B) : A\subset B, B\in\mathcal{A}\}$$ for all $A\subset X$.
Problem: Show that each set in $\mathcal{A}$ is $\mu^*$-measurable.
I have shown that $\mu^*$ is an outer measure. $A\in\mathcal{A}$ is $\mu^*$-measurable iff for any $E\subset X$, $\mu^*(E)=\mu^*(A\cap E)+\mu^*(A^c\cap E)$. I know that $\mu^*(E)\leq\mu^*(A\cap E)+\mu^*(A^c\cap E)$ (by definition of an outer measure). But how can I show the other inequality?
By definition of infimum, for any $\epsilon>0$, there exists a set $E_{\epsilon}\in \mathcal{A}$ s.t. $E\subset E_{\epsilon}$ and $\mu^*(E_{\epsilon})\le\mu^*(E)+\epsilon$ (assuming that $\mu^*(E)< \infty$). Then $$ \mu^*(E)+\epsilon\ge \mu(E_{\epsilon}\cap A)+\mu(E_{\epsilon}\cap A^c)\ge \mu^*(E\cap A)+\mu^*(E\cap A^c) $$ because $\mu$ is additive on $\mathcal{A}$.