It is well known that for any set A in R^d there exists a measurable set E such that E contains A and m\*(A)=m\*(E). Is it possible to go the other direction?
In other words, is it true that for any measurable set E (such that m(E)>0) there is a non-measurable subset A such that m\*(A)=m\*(E)?



Non Lebesgue measurable subsets with "large" outer measure
A set $E$ with positive Lebesgue measure can be decomposed as a union $E = A \cup B$ where each of $A$ and $B$ have zero inner measure, and therefore each of $A$ and $B$ are nonmeasurable with $m^\*(A) = m^\*(B) = m(E)$. 


An example for this construction is a [Bernstein set](http://en.wikipedia.org/wiki/Bernstein_set).


