# Does every set of positive measure contain an uncountable null set?

If $$E$$ is Lebesgue measurable and $$m(E)>0$$, does there exist an uncountable $$C\subset E$$ with $$m(C)=0$$?
This seems intuitively clear but I cannot prove it. Since $$E$$ has positive measure it contains a nonmeasurable set $$V$$, and every measurable subset of $$V$$ is null, but I could not show $$V$$ contains uncountable measurable subsets. I also tried using the fact that for $$\alpha\in(0,1)$$ there is an interval $$I$$ s.t $$m(E\cap I)\ge \alpha m(I)$$, attempting to construct a Cantor set inside $$I$$ with an uncountable intersection with $$E$$ but I was unsuccessful. Any hints? Is this even true?

By regularity, if $$m(E)>0$$ then there is some closed $$C\subseteq E$$ with $$m(C)>0$$. So WLOG assume $$E$$ is closed (and has measure $$1$$).
A variant of the Cantor set construction now lets us "thin" $$E$$ out to a closed uncountable $$D\subseteq E$$ with measure zero. E.g. at our first step we fix $$a such that $$m(E_{b})={1\over 3}$$, and cut down to $$E\setminus (a,b)$$.
(How does the closedness of $$E$$ matter? Well, we need to argue that the $$D$$ we construct is in fact as desired. Trivially this $$D$$ is null, since we've shrunk its measure appropriately at each stage; knowing that $$D$$ is closed tells us that everything that should be in $$D$$, is.)