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