I need to prove that if $A\subseteq\mathbb{R}^n$ is measurable with positive measure, then it contains a non-measurable set. I produced a proof but I think it is wrong since I didn't use the fact that $A$ is measurable, only that its outer measure is positive. Here it is.
Let $Q_j$ be the collection of closed unit cubes with integer base points, i.e. of the form $[a_1,a_1+1]\times\dots\times[a_n,a_n+1]$ for $a_i$ integers. Then $A = \bigcup (A\cap Q_j)$. Now suppose that $m_*(A\cap Q_j)=0$ for all $j$. Then, by subadditivity of the outer measure, $m_*(A) = 0$. In our case, since $m_*(A) > 0$, there must exists a unit cube $K$ such that $m_*(A\cap K) > 0$. By translating the set $A$, we can assume wlog that $K$ is the unit cube at the origin $[0,1]^n$.
Now it's just an adaptation of the Vitali set argument. Consider the classes of $\mathbb{R}^n/\mathbb{Q}^n$ that intersect $A\cap K$. Take exactly one representative from each of these classes and use them to form the set $V$. Clearly it's possible to choose the representatives so that $V$ is contained in $A\cap K$. Now let $\{r_k\}$ be an enumeration of the rationals in $[-1,1]^n$ and let $V_k = V + r_k$. For each $k$, we have $V_k \subseteq K+r_k$, so $V_k\subseteq [-1,2]^n$. We also have that $A\cap K \subseteq \bigcup V_k$ because if $x\in A\cap K$, then there exists a $x_\alpha\in V$ such that $x\sim x_\alpha$ and therefore $x\in V_k$ for some $k$. Putting these inclusions together and using monotonicity of the outer measure, $0 < m_*(\bigcup V_k)\leq 3^n$ where the strict inequality is because $m_*(A\cap K)>0$. If we assume that $V$ is measurable, we have that $0 < \sum m_*(V) \leq 3^n$, which is impossible. Since $V$ is contained in $A\cap K$, we have produced a non-measurable set contained in $A$.
Is this proof valid? Did I use the measurability of $A$ without knowing it?
Best Answer
(adapted and extended from comments)
I didn't look at the details of what you've done, but note that if $A$ is non-measurable, then $A$ itself is a non-measurable subset of $A$ (with positive outer measure, as I assume we're dealing with Lebesgue outer measure and Lebesgue measure). And if you want a proper subset of $A$ that is non-measurable, then use a set formed by removing some points from $A,$ say at most countably many points from $A.$ Therefore, it is only in the case when $A$ is measurable (with positive measure) that the result is not trivial.
However, a natural question is whether there are any (not obvious) restrictions on the inner and outer measures of such subsets. In light of the intermediate value property of measures (sometimes called Lyapunov's theorem) -- see the mathoverflow questions A result of Sierpiński on non-atomic measures & Reference for a strong intermediate value theorem for measures, and see this 25 November 2005 sci.math post -- and in light of results such as found in the MSE question A non-measurable set of given outer measure (especially bof's answer), the following conjecture -- probably known, but might be worth investigating -- seems reasonable.
Conjecture: Let $A$ be a subset of $\mathbb R,$ and let $m_*$ and $m^*$ be Lebesgue inner measure and Lebesgue outer measure. Let $r,\,s$ be extended real numbers such that $0 \leq r \leq m_*(A)$ and $r \leq s \leq m^*(A).$ Then there exists a set $B$ such that $B \subseteq A$ and $m_*(B) = r$ and $m^*(B) = s.$
In the above, note that $A$ is measurable if and only if $m_*(A) = m^*(A),$ and $B$ is measurable if and only if $m_*(B) = m^*(B)$ (in each of these $=$ can be replaced with $\geq.)$ For those interested, some basic facts about inner/outer Lebesgue measure are given in this 5 February 2006 sci.math post (minor correction here).