Let $A\subseteq\Bbb R$ with $\lambda^*(A)>0$. Show that there exists a nonmeasurable $B\subseteq\Bbb R$ s.t. $B\subseteq A$

Let $A$ be a subset of $\Bbb R$ with $\lambda^*(A)>0$. Show that there exists a nonmeasurable subset $B$ of $\Bbb R$ s.t. $B$ is a subset of $A$

I'm a little confused where to start with this one. If $A$ is nonmeasurable there is nothing to prove so we have to assume that $A$ is measurable, more specifically a nonzero measure. Define an equivalence class next?? not sure....

• What exactly is $\lambda^*(A)$ here? – aschepler Dec 9 '13 at 1:30
• @aschepler: Presumably Lebesgue outer measure. – Brian M. Scott Dec 9 '13 at 1:31
• yes, @aschepler, it means the outer measure generated by the Lebesgue measure on R – cele Dec 9 '13 at 1:56
• Have you seen the construction of a Vitali set? – Brian M. Scott Dec 9 '13 at 1:58
• @BrianM.Scott have seen it in text but have not done too much work using it. my thinking here is use the Axiom of choice to show there exists a subset B of A containing precisely one member from each equivalence class. let {r1, r2, ....} be an enumeration of the rationals [-1,1] and let Bn=rn+B. Im still working on how to show this and finish from here – cele Dec 9 '13 at 2:03