Let $E \subset [0,1]$ be measurable set. Suppose for each interval $I \subset [0,1]$, $m(E \bigcap I)>1/4 m(I) $. Show that $m(E)=1$.
Any hints would be appreciated.
Let $E \subset [0,1]$ be measurable set. Suppose for each interval $I \subset [0,1]$, $m(E \bigcap I)>1/4 m(I) $. Show that $m(E)=1$.
Any hints would be appreciated.
The hypothesis implies that $m(I\cap E^c)<\frac 34m(I)$ for all interval $I$. So we have to show that $m(E^c)=0$. Fix $\delta_0$. By outer regularity of Lebesgue measure, we can find $O$ open such that $m(O)-m(E^c)<\delta$ and $O\supset E^c$. We can write $O$ as a countable disjoint union of intervals $\{I_j\}$. This gives $m(E^c)=m\left(E^c\cap\bigsqcup_{j\in\Bbb N}I_j\right)<\frac 34\sum_{j\in\Bbb N}m(I_j)<\frac 34(\delta +m(E^c)),$ hence $m(E^c)\leq 3\delta.$ As $\delta$ was arbitrary, we are done.