Given a set $A$ with $\mu (A) > 0$, is there an open interval $I$ such that $\mu (A\cap I) > (1-\epsilon)\mu (I)$ or more precisely does the Lebesgue density theorem holds for arbitrary sets?
Here, $\mu $ is Lebesgue outer measure and $\epsilon\in(0,1)$ is small. Of course, this is true if $A$ is measurable according to Lebesgue density theorem.
Best Answer
Cover $ A $ with a countable union of intervals $\bigcup_iI_i \supseteq A$ such that $\mu (A) + \epsilon \sum_i \mu (I_i)> \sum_i \mu (I_i)$
$\sum_i \mu (A \bigcap I_i) \ge \mu (A) > (1-\epsilon) \sum_i \mu (I_i)$
This implies there exists $ I_i $ such that $\mu (A \bigcap I_i) >(1-\epsilon) \mu (I_i)$
However this does not imply Lebesgue density theorem