# Non Lebesgue measurable subsets with “large” outer measure

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)?

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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.

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Thanks. This is what I was looking for. –  Haim Jan 25 '10 at 18:02
Yes, I believe so - since subsets of a null set $A$ (i.e., $m(A)=0$) are not necessarily measurable, but will obviously still have outer measure 0, given any measurable set $E$ you "should" (i.e., I think so, but not sure) be able to find a non-measurable subset $S$ of a null set $A$ inside $E$, remove $S$, and since $m(E)=m(E-S)+m(S)$ and $m(S)$ isn't anything, we must have that $m(E-S)$ isn't anything, while we still have $m^{\ast}(E)=m^{\ast}(E-S)+m^{\ast}(S)=m^{\ast}(E-S)$.