Here is the fact in measure theory:


FACT : Let $E$ be a Lebesgue measurable subset of $\mathbb{R}^n$. Almost every $x\in E$ satisfies $\lim\limits\_{m(B)\to 0,~x\in B}\frac{m(B\cap E)}{m(B)}=1$ i.e. limit is taken over the ball $B$ containing $x$ with shrinking it.


Using this fact, I want to prove that


If a Lebesgue masureale subset $E$ of $[0,1]$ satisfies $m(E\cap I)\geq \alpha m(I)$ for some $\alpha>0$, for all interval $I$ in $[0,1]$, then $E$ has measure 1.


How can I use the fact to prove the last assertion? 



Question on measure theory
Clearly, you can only have this for $alpha\le 1$. Let $F:=[0, 1]\setminus E$, and take a covering $\{I\_k:k=1,..., n\}$ of $F$ of subintervals of $[0, 1]$ such that $\sum\_{k=1}^n m(I\_k)\le m(F)-\epsilon$ for some $\epsilon>0$. By assumption $m(F\cap I)\le (1-\alpha)m(I)$. Then


$(1-\alpha)(m(F)-\epsilon)\ge(1-\alpha)\sum\_{k=1}^n m(I)\ge\sum\_{k=1}^n m(F\cap I\_k)$


Since the union of $F\cap I\_k$ still contains $F$, the right hand side is at least $m(F)$. Then, since $\epsilon>0$ was arbitrary, $(1-\alpha)m(F)\ge m(F)$, which means that $m(F)$ must be zero and $m(E)$ must be one.


