My specific problem is with the Lebesgue measure, but out of curiosity I would also appreciate insights on general measure spaces.
Suppose I am given a non-measurable set $A$ with the property that there exists no zero measure set which contains all of $A$.
Is it then true that there exists a measurable subset of $A$ with non-null measure?
For my specific use case, the underlying space can even be finite-dimensional, in case that makes a difference. I am a graph theorist and woefully uneducated in measure theory.
Best Answer
In the case of Lebesgue measure there is a very easy counterexample. Let $N$ be any nonmeasurable subset of $\mathbb R$ and let $\mathcal B$ be a maximal collection of disjoint nonnull measurable subsets of $N$. Then $\mathcal B$ is countable, and the set $A=N\setminus\bigcup\mathcal B$ is nonmeasurable and contains no measurable subset of positive measure.
Edit. A commenter asked why $\mathcal B$ is countable. For $m,n\in\mathbb N$ let $\mathcal B_{m,n}=\{B\in\mathcal B:\mu(B\cap[-m,m])\ge\frac1n\}$. Since $\mathcal B=\bigcup_{m,n\in\mathbb N}\mathcal B_{m,n}$, it will suffice to show that $\mathcal B_{m,n}$ is countable. Assume for a contradiction that $\mathcal B_{m,n}$ is uncountable. Choose distinct sets $B_0,B_1,\dots,B_{2mn}\in\mathcal B_{m,n}$. Then $$\mu([-m,m])\ge\mu\left(\bigcup_{k=0}^{2mn}(B_k\cap[-m,m]\right)=\sum_{k=0}^{2mn}\mu(B_k\cap[-m,m])\ge\frac{2mn+1}n\gt2m$$ which is absurd.