7
$\begingroup$

If $A$ is a subset of $\mathbb{R}$ with Lebesgue measure strictly greater than $0$, does it follow then that are there $a$ and $b$ such that the measure of $[a,b]\cap A$ is $b-a$?

Thank you.

  • 5
    @Asaf Karagila: Post as an answer?2010-12-05

2 Answers 2

4

For the sake of completeness, the answer is no; a counterexample is given by the Smith-Volterra-Cantor set, or fat Cantor set.

4

What is actually true is this: for every set $C$ of positive measure and every $\epsilon < 1$ there is some open interval $(a,b)$ such that $\mu(C \cap (a,b)) \geq \epsilon |b-a|$.

I have always viewed this as an instance of one of Littlewood's three principles for analysis: a measurable set is almost an open set.

  • 0
    It's called the Lebesgue density theorem and is often proved in textbooks. http://en.wikipedia.org/wiki/Lebesgue's_density_theorem is a little cryptic because it covers the theorem for $\mathbb{R}^n$.2011-09-27