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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 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?

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I think this is too elementary for this site. See mathoverflow.net/faq#whatnot –  Sergei Ivanov Apr 19 '11 at 11:46
    
I recall doing this as an exercise from Folland's Real Analysis book. –  BSteinhurst Apr 19 '11 at 13:58
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closed as too localized by Sergei Ivanov, Charles Matthews, Mark Meckes, Bill Johnson, Andres Caicedo Apr 19 '11 at 19:14

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1 Answer

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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.

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Care to explain the downvote? –  Martijn Apr 19 '11 at 16:50
    
@Martijn: it wasn't my downvote, but you did not answer the question. The OP asked to apply the density point theorem to the complement of the set, not to re-prove it ;) –  Sergei Ivanov Apr 19 '11 at 18:41
    
@Sergei Ivanov: you're right, I've misread the question. Thanks for pointing that out. –  Martijn Apr 20 '11 at 8:16
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