Difference of elements from measurable set contains open interval 
Let $A\subset\mathbb{R}$ be a measurable set s.t $,m(A)>0$. Prove that the set $$B=\{x-y\mid x,y\in A\}$$contains nonempty open interval around 0. 

I thought to take an interval in $A$, $I=(x-\frac{\epsilon}{2},x+\frac{\epsilon}{2})\subset A$ and hence taking $y$ values from I we get an epsilon - neighborhood of $0$ but I'm quite not sure that I can assume the existence of such I.
How can I prove the existence of I?
 A: Another approach: WLOG, take $A$ to have finite measure. Let $\mathcal C$ be a covering of $A$ in disjoint open intervals such that 
$$
\left| \bigcup_{I \in \mathcal C} I\right| < \frac 43 |A|
$$
deduce that there exists an $I \in \mathcal C$ such that $|A \cap I| \geq \frac 34 |I|$.  Denote $A' = A \cap I$. Note that $A'$ is bounded.  Let $I = (a,b)$.
Now, suppose that $d \in \Bbb R$ is such that $(d+A') \cap A' = \emptyset$. Then
$$
2|A'| = |d+A'| + |A'| = |(d+A')\cup A'|
$$
If $d > 0$, then $(d+I)\cup I \subset (a,d+b)$.
If $d < 0$, then $(d+I) \cup I \subset (d+a,b)$.
In either case, we have $|(d + I) \cup I| \leq |d| + |I|$. That is, for any such $d \in \Bbb R$, we have
$$
2|A'| \leq |(d+I) \cup I| \leq |d| + |I|
$$
And if $(d+A') \cap A' = \emptyset$, then $|(d+A') \cup A'| = 2|A'|$.
So $(d+A') \cap A' = \emptyset$ implies that
$$
\frac 32|I| \leq 2|A'| \leq |d| + |I| \implies |d| \geq \frac 12|I|. 
$$
Contrapositively: if $|d| < \frac 12 |I|$, then $(d + A') \cap A' \neq \emptyset$.
That is, every $d \in (-|I|/2,|I|/2)$ can be written as $x-y$ for $x,y \in A' \subset A$.
A: HINT: Use the fact that for any $\epsilon>0$ there are a compact set $K\subseteq A$ and an open set $U\supseteq A$ such that $m(U)-m(K)<\epsilon$. Let 
$$d=\inf\{|x-y|:x\in K\text{ and }y\in\Bbb R\setminus U\}\;.$$


*

*Why is $d>0$?


Fix $\delta\in(0,d)$.


*

*Why is $K+(-\delta,\delta)\subseteq U$?


Suppose that $|r|<\delta$ and $K\cap(K+r)=\varnothing$.


*

*Calculate $m\big(K\cup(K+r)\big)$ in terms of $m(K)$.  

*Use the fact that $K\cup(K+r)\subseteq U$ to get an upper bound on $m\big(K\cup(K+r)\big)$ in terms of $m(K)$ and $\epsilon$.  

*Derive a contradiction by choosing $\epsilon$ sufficiently small.

A: Here is an interesting proof I came across sometime ago. It begins with a Lemma:

Let $F$ be a compact set and $U$ open such that $F \subset U$. Then $\exists$ an open neighbourhood $V$ of $0$ such that $V+F \subset U$

Proof:
For each $x\in F, \exists \epsilon > 0$ such that $(x-\epsilon,x+\epsilon) \subset U$. Let $V_x = (-\epsilon/2,\epsilon/2)$, then
$$
F \subset \bigcup_{x\in F} x+V_x
$$
Since $F$ is compact, $\exists x_1,x_2,\ldots, x_n\in F$ such that
$$
F \subset \bigcup_{i=1}^n x_i + V_{x_i}
$$
Take
$$
V := \bigcap_{i=1}^n V_{x_i}
$$
Then
$$
V+F \subset \cup_{i=1}^n V+(x_i + V_{x_i}) \subset \cup_{i=1}^n x_i+(V_{x_i}+V_{x_i}) \subset U
$$

Now we may prove the result:


*

*Since $A$ is measurable, and $m(A) > 0$, there is a compact set $F \subset A$ such that $m(F) > 0$. So, we may assume that $A$ itself is compact.

*Since $A$ is measurable and compact, $m(A) < \infty$ and $\exists U$ open such that $A\subset U$ and
$$
m(U) < 2m(A) \qquad\qquad \text{(1)}
$$

*By the Lemma, $\exists V$ an open neighbourhood of $0$ such that $V+A \subset U$. Now, we claim that $V \subset A-A$. If $v\in V$, then it suffices to prove that
$$
(v+A)\cap A \neq \emptyset
$$
(why?), so suppose $v+A\cap A = \emptyset$, then since $v+A \subset U$, we have
$$
m(U) \geq m(v+A) + m(A) \geq 2m(A)
$$
This contradicts (1) and we are done.
