# question regarding the existence of measurable subset with positive measure

It is well-known that any measurable set with a positive measure has a non-measurable subset. But how about the related question: whether there always exists a measurable subset with positive measure in a non-measurable set.

Intuitively I would say yes. But not sure how to provide a proof. Please help prove or refute it.

Thanks.

-
Are you familiar with the concept of completeness? – cardinal Apr 17 '11 at 5:20
@cardinal: right, I know the concept. so...? – Qiang Li Apr 17 '11 at 5:24
I think what cardinal's getting at is that a generalized version of this question would have the answer no for any incomplete measure, as seen by taking any nonmeasurable subset of a set of measure $0$. – Jonas Meyer Apr 17 '11 at 5:28
I'm probably too tired to be thinking or reading straight. However, let $B \subset A$ such that $\mu(A) = 0$ and $B$ is non measurable. Then any subset $C \subset B$ that is measurable must have measure zero by monotonicity. You seem to be asking something quite strong, i.e., that for every non measurable set, there exists some measurable subset and furthermore that it have positive measure. – cardinal Apr 17 '11 at 5:31
@cardinal: Lebesgue measure is complete. All subsets of a set of measure 0 are measurable with measure 0. By the way, this shows there are measurable sets that are non-Borel, by a cardinality argument. – Andrés Caicedo Apr 17 '11 at 5:36

Vitali's non-measurable subset $V\subset S^1=\mathbb{R}/\mathbb{Z}$ (obtained by choosing one representative from each class modulo $\mathbb{Q}/\mathbb{Z}$) doesn't have such a subset. If $A\subset V$ is measurable then $A+q\,$'s are measurable and disjoint for different $q\in\mathbb{Q}/\mathbb{Z}$, hence $\sum_q\mu(A+q)\leq \mu(S^1)=1$, and $\mu(A+q)=\mu(A)$ for every $q$, hence $\mu(A)=0$.

-

A key definition here is "inner measure". The Lebesgue inner measure of a set $A$ is the supremum of the measures of the Lebesgue measurable subsets of $A$ (equivalently, of the closed subsets of $A$). You are asking whether every nonmeasurable set has positive inner measure, and as user8268 has shown, the answer is no.

A set of real numbers $A$ of finite outer measure is measurable if and only if its Lebesgue inner measure $m_*(A)$ is equal to its Lebesgue outer measure $m^*(A)$. If $m^*(A)=m_*(A)$, then there is a $G_\delta$ set $G$ and an $F_\sigma$ set $F$ such that $F\subseteq A\subseteq G$ and $m(F)=m_*(A)=m^*(A)=m(G)$. Then $A\setminus F\subseteq G\setminus F$ is a null set, and therefore $A=F\cup(A\setminus F)$ is measurable.

Let $A$ be a nonmeasurable set with finite inner measure. Let $F\subset A$ be an $F_\sigma$ such that $m(F)=m_*(A)$. Then $A\setminus F$ is a nonmeasurable set with inner measure $0$. Thus, every nonmeasurable set contains nonmeasurable subsets with no measurable subsets of positive measure.

-

The key here is that Lebesgue measure is regular. Any of the equivalent formulations of regularity gives you that any set contains a measurable subset with the same inner measure. Of course, if the outer and inner measures of a set coincide, then the set is measurable. So any non-measurable set $A$ must have positive outer measure. Its inner measure is the sup of the measures of its compact subsets, so $A$ contains an $F_\sigma$ set with the same inner measure, so their difference has inner measure zero and is non-measurable. This gives the result.
The inner measure is the sup of the measures of compact subsets, right? So there is an $F_\sigma$ subset whose measure is the outer measure of your set. – Andrés Caicedo Apr 17 '11 at 5:34
There is an $F_\sigma$ in the set whose measure is the inner measure of the set, but if the outer measure is finite this will be less than the outer measure for a nonmeasurable set. – Jonas Meyer Apr 17 '11 at 5:50