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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 measurable 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?

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    This is exercise 25 on page 100 of Folland, Real Analysis, chapter "Differentiation on Euclidean Space"2011-05-14

1 Answers 1

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Hint: look at the set $F = [0,1]\setminus E$. It is measurable. What do you know about $m(I\cap F)$ for any interval $I$? Apply now the fact to $F$. What can you conclude?