question regarding the existence of measurable subset with positive measure It is well-known that any measurable set with a positive measure has a non-measurable subset. But how about the related question: whether there always exists a measurable subset with positive measure in a non-measurable set.
Intuitively I would say yes. But not sure how to provide a proof. Please help prove or refute it.
Thanks.
 A: A key definition here is "inner measure".  The Lebesgue inner measure of a set $A$ is the supremum of the measures of the Lebesgue measurable subsets of $A$ (equivalently, of the closed subsets of $A$).  You are asking whether every nonmeasurable set has positive inner measure, and as user8268 has shown, the answer is no.  
A set of real numbers $A$ of finite outer measure is measurable if and only if its Lebesgue inner measure $m_*(A)$ is equal to its Lebesgue outer measure $m^*(A)$.  If $m^*(A)=m_*(A)$, then there is a $G_\delta$ set $G$ and an $F_\sigma$ set $F$ such that $F\subseteq A\subseteq G$ and $m(F)=m_*(A)=m^*(A)=m(G)$.  Then $A\setminus F\subseteq G\setminus F$ is a null set, and therefore $A=F\cup(A\setminus F)$ is measurable.
Let $A$ be a nonmeasurable set with finite inner measure.  Let $F\subset A$ be an $F_\sigma$ such that $m(F)=m_*(A)$.  Then $A\setminus F$ is a nonmeasurable set with inner measure $0$.  Thus, every nonmeasurable set contains nonmeasurable subsets with no measurable subsets of positive measure.
A: Vitali's non-measurable subset $V\subset S^1=\mathbb{R}/\mathbb{Z}$ (obtained by choosing one representative from each class modulo $\mathbb{Q}/\mathbb{Z}$) doesn't have such a subset. If $A\subset V$ is measurable then $A+q\,$'s are measurable and disjoint for different $q\in\mathbb{Q}/\mathbb{Z}$, hence $\sum_q\mu(A+q)\leq \mu(S^1)=1$, and $\mu(A+q)=\mu(A)$ for every $q$, hence $\mu(A)=0$.
A: Edit Originally answered this as yes, but obviously the answer is no (poor proofreading skills, sorry).
The key here is that Lebesgue measure is regular. Any of the equivalent formulations of regularity gives you that any set contains a measurable subset with the same inner measure. Of course, if the outer and inner measures of a set coincide, then the set is measurable. So any non-measurable set $A$ must have positive outer measure. Its inner measure is the sup of the measures of its compact subsets, so $A$ contains an $F_\sigma$ set with the same inner measure, so their difference has inner measure zero and is non-measurable. This gives the result.
