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)?
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
0
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.
answered Jan 25, 2010 at 14:39
$\begingroup$
$\endgroup$
3
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 $m^{\ast}(E)=m^{\ast}(E-S)+m^{\ast}(S)=m^{\ast}(E-S)$.
answered Jan 25, 2010 at 14:26
-
$\begingroup$ Perhaps I'm missing something here, but isn't the Lebesgue measure complete? i.e. for every measurable set A with m(A)=0, each subset of A is Lebesgue-measurable with measure zero. $\endgroup$– HaimCommented Jan 25, 2010 at 14:31
-
$\begingroup$ Subsets of null sets are always Lebesgue measurable, but not necessarily Borel measurable. Perhaps you should clarify what you mean by "measurable" in your answer. $\endgroup$ Commented Jan 25, 2010 at 14:31
-
$\begingroup$ Ah, right - I meant Lebesgue measure in my answer, but got mixed up. Clarifying which measure you mean is a good idea, as it will help avoid confusions such as mine :) $\endgroup$ Commented Jan 25, 2010 at 14:41