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Let $E$ be a set such that $m(E)>0$, $E \subset (0,1)$. Prove that there exists $c>0$ such that for some moving interval $I$, $$\lim_{m(I)\rightarrow 0}\frac{m(E\cap I)}{m(I)}=c$$

Proof: $m(E)=1$

My attempt:

I have proof E is dense in(0,1),proof by contradiction if E nowhere dense in (0,1), there exist $(a,b)$ such that $(a,b)\subset (0,1)-E$, then $m((a,b)\cap E)=0$,we get a conflict

I think Maybe need the theorem using in construct unmeasurable set: If m(E)>0, then for any $a \in (0,1)$,there exist an interval J,such that $m(E\cap J)>am(J)$

Lev Bahn
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jackson
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