If $\mu (A) > 0$, is there an open interval $I$, such that $\mu (A\cap I) > (1-\epsilon)\mu (I) $?

Question

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.

Answer 1

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

Answer 2

$A$ has a measurable superset $B$ whose Lebesgue measure equals the outer measure of $A$. In other words, $B-A$ has inner measure $0$. Then, for any interval $I$, $(B\cap I)-(A\cap I)$ also has inner measure $0$, and so $\mu(A\cap I)=\mu(B\cap I)$. Apply the Lebesgue density theorem to $B$ and get the result you want for $A$.