TITLE: The subset that $m(E \cap I) \geq \alpha m(I)$ has measure 1.
QUESTION [8 upvotes]: I'm stucked in this question:
If $E$ is a Lebesgue measurable set contained in the interval $[0,1]$ and there is a $\alpha>0$ such that $m(E \cap I) \geq \alpha m(I)$ for all open intervals $I \in [0, 1]$ then $m(E) = 1$.
How should I proceed to prove this?

REPLY [3 votes]: Following your comment about the Lebesgue Density Theorem, if you want to use it, you can argue as follows: 
It is clear that $0<\alpha <1.$ Consider $F:=[0,1]\setminus E.$ Since $I\cap (E\cup F)=(I\cap E)\cup (I\cap F)=I,$ we have $m(I\cap F)=m(I)-m(I\cap E)\le (1-\alpha )m(I).$ Then, $\frac{m(I\cap F)}{m(I)}\le 1-\alpha.$ Since this is true for all intervals $I\in [0,1],$ we conclude from the Lebesgue Density Theorem that $m(F)=0$ which now implies that $m(E)=1.$