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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?

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Yes.

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 $a<b$ such that $m(E_{<a})=m(E_{>b})={1\over 3}$, 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.)

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