For a set of positive measure there is an interval in which its density is high, $\mu(E\cap I)> \rho \mu(I)$

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:

1. 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

2. 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.

This is more or less the same proof as in Paul Halmos' Measure Theory

For part (1):

First assume $\mu(E)<\infty$. Then, since Lebesgue measure is outer regular, we may, and do, choose an open set $U\supseteq E$ such that $$\rho \mu(U )<\mu(E).$$

Now write $U$ as the disjoint countable union of open intervals $U=\bigcup\limits_{n\in A} I_n$. Then $$ \rho\sum_{n\in A} \mu(I_n)= \rho\mu(U )<\mu(E)=\sum_{n\in A} \mu(I_n\cap E). $$ There must be an $n\in A$ with $\rho\mu(I_n)<\mu (E\cap I_n)$. Setting $I=I_n$ gives the desired open interval.

If $\mu(E)$ is infinite, choose an interval $J$ so that $0<\mu(E\cap J)<\infty$. As, above, there is an open interval $I$ with $\rho \mu(I)<\mu\bigl( (E\cap J) \cap I\bigr)\le \mu (E\cap I)$.

For part (2):

Take $\rho={3\over4}$ and choose an open interval $I$ so that

$$\rho\mu(I )<\mu (E\cap I ).$$ Take $$\tag{1}x\in( -{1\over 2}\mu(I), {1\over 2}\mu(I) ).$$

Consider the sets

$\ \ \ \ A=E\cap I$

and

$\ \ \ \ B= (E\cap I)+x$

We have $\mu(A)=\mu(B)$; so, if $A\cap B=\emptyset$, then $$ \mu(A\cup B)=2\mu(A) >{3\over 2} \mu(I). $$

But, by $(1)$, $A\cup B$ is contained in the interval $I\cup (I+x)$, which has measure at most $ {3\over2}\mu(I)$ (informally $I+x$ is the interval $I$ shifted to the right by at most half the length of $I$).

It follows that $A\cap B\ne\emptyset$. Let $y\in A\cap B$. Write $y=z+x$ for some $z\in E\cap I$. Then $x=y-z\in E-E$.

It follows that the interval $( -{1\over 2}\mu(I), {1\over 2}\mu(I) )\subset E$.

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.