Given a complete measure space $(X, \mathcal{X}, \mu)$, and a subset $A \subseteq X$ of positive outer measure, does there necessarily exist a subset $E$ of a $A$ which is measurable for which $\mu(E) > 0$?
This feels intuitively like it should be true, but I don't see how to show it, since it involves inner approximations and outer measure is generally defined by reference to outer approximations. One thought I had was to exploit the idea of inner regularity for Borel measures, but I think that definition is typically only stated as the ability to approximate measurable sets from within. However, I want to find a positive-measure and measurable subset of a potentially non-measurable set.
Thanks in advance for your help!
EDIT: Since somebody asked, by the outer measure I mean
$$\mu^*(A) : = \inf \left\{ \sum_{n = 1}^\infty \mu(E_n) : E_n \in \mathcal{X}, A \subseteq \bigcup_{n = 1}^\infty E_n \right\} . $$
Best Answer
Not necessarily. In fact, we can prove that for all non-measurable $A$ such that $\mu^*(A)<\infty$ there is some non-measurable $B\subseteq A$ such that all measurable subsets of $B$ are null sets and (necessarily) $\mu^*(B)>0$.
Let $A$ be as in the hypothesis and let $s=\sup_{C\subseteq A\\ C\in \mathcal X}\mu^*(C)$, consider $s_n\nearrow s$ and $C_n\subseteq A$ measurable such that $\mu^*(C_n)=s_n$. Call $C=\bigcup_{n\in\Bbb N} C_n$. $C\in \mathcal X$ and it's clear by monotonicity and by the definition of $s$ that $\mu^*(C)=\mu(C)=s$.
Now, let's call $B=A\setminus C$. $B\notin \mathcal X$, otherwise $A=C\cup B$ would be too. In particular, $\mu^*(B)>0$. By Carathéodory's condition, $\mu^*(B)=\mu^*(A)-\mu^*(C)=\mu(A)-s$
If $B$ had a measurable subset $H$ of positive measure, then $C\cup H$ would be a measurable subset of $A$ of measure $$\mu(C\cup H)=\mu(C)+\mu(H)-\mu(C\cap H)=\mu(C)+\mu(H)-\mu(\emptyset)=\mu(C)+\mu(H)>s$$ Which is against the definition of $s$.