MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

closed as too localized by Sergei Ivanov, Charles Matthews, Mark Meckes, Bill Johnson, Andrés E. Caicedo Apr 19 '11 at 19:14

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

I think this is too elementary for this site. See – 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
up vote 0 down vote accepted

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.

share|cite|improve this answer
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

Not the answer you're looking for? Browse other questions tagged or ask your own question.