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)?
[Math] Non Lebesgue measurable subsets with “large” outer measure
measure-theoryset-theory
Best Answer
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.