TITLE: For a set of positive measure there is an interval in which its density is high, $\mu(E\cap I)> \rho \mu(I)$
QUESTION [15 upvotes]: In my self-study, I came across the following two interesting, related results:
Let $E$ be Lebesgue measurable, with $\mu(E)>0$ (here $\mu$ denotes the Lebesgue measure). Then:

for any $0<\rho<1$, there exists an open interval $I$ such that $\mu(E \cap I)> \rho \cdot \mu(I)$. In turn we should be able to use (1) with $\rho > 3/4$ to prove that 
the set $E-E = \{x-y : x, y \in E\}$ contains an (open) interval centered at $0$ (in particular, if $\rho > 3/4$, the text I am using suggests that $(-\frac{1}{2} \mu(I), \frac{1}{2}\mu(I)) \subseteq E-E$).

I would like to see if anyone visiting today would be up for proving (1) or (2) (inclusive-or). I have tried to work out (1) a few times, but none of my attempts have been satisfactory.

REPLY [3 votes]: Here is another proof of (1) which I found slightly more intuitive. It is very similar to the one provided by David. 
Let $\rho \in (0, 1)$ be fixed. Suppose towards a contradiction that every open interval $I$ satisfies $\mu(E \cap I) \leq \rho \mu(I)$. I'll do the case when $\mu(E) < \infty$ (the case when $\mu(E)=\infty$ by applying the finite case on a subset of $E$). 
Begin with the outer regularity of Lebesgue measure which allows us to pick, for any $\epsilon > 0$, an open set $U \supset E$ such that $\mu(E) \geq \mu(U) - \epsilon$. Since $U$ is an open set, $U=\bigcup_{j=1}^{\infty} I_j$ where the $I_j$'s are disjoint open intervals. Observe that $E = \bigcup_{j=1}^{\infty} E \cap I_j$, and each $E\cap I_j$ is disjoint. Hence,
$$
  \mu(E) = \mu( \bigcup_{j=1}^{\infty} E \cap I_j ) = \sum_{j=1}^{\infty} \mu( E \cap I_j ) \leq \sum_{j=1}^{\infty} \rho \mu( I_j ) = \rho \sum_{j=1}^{\infty} \mu( I_j ) = \rho \mu( U ) \:.
$$
We have learned that
$$
  \mu(U) - \epsilon \leq \mu(E) \leq \rho \mu(U) \:.
$$
Taking $\epsilon \rightarrow 0$ we have $\mu(U) \leq \rho \mu(U) < \mu(U)$, a contradiction.