Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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 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?

share|cite|improve this question
where can I find a reference about your "FACT"? I looked up the measure-theory book I own, but I did not find anything related. Please let me know. Thanks! – Qiang Li May 6 '11 at 1:34
This is exercise 25 on page 100 of Folland, Real Analysis, chapter "Differentiation on Euclidean Space" – Bruno Stonek May 14 '11 at 18:20
up vote 5 down vote accepted

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?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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