Question: Quite a while ago, I heard about a statement in measure theory, that goes as follows:

> 
> Let $A \subset \mathbb R^n$ be a Lebesgue-measurable set of positive measure. Then we follow that $A-A = \{ x-y \mid x,y\in A\}$ is a neighborhood of zero, i.e. contains an open ball around zero.
> 
> 
> 

I now got reminded of that statement as I have the homework problem (Kolmogorov, Introductory Real Analysis, p. 268, Problem 5):

> 
> Prove that every set of positive measure in the interval $[0,1]$ contains a pair of points whose distance apart is a rational number.
> 
> 
> 

The above statement would obviously prove the homework problem and I would like to prove the more general statement. I think that assuming the opposite and taking a sequence $\{x\_n\}$ converging to zero such that none of the elements are contained in $A$, we might be able to define an ascending/descending chain $A\_n$ such that the union/intersection is $A$ but the limit of its measures zero. I am in lack of ideas for the definition on those $A\_n$.

I am asking specifically not for an answer but a hint on the problem. Especially if my idea turns out to be fruitful for somebody, a notice would be great. Or if another well-known theorem is needed, I surely would want to know. Thank you for your help.

Answer: 
Here's an attempt at a hint for the first result you ask about: Assume without loss of generality that $A$ has finite measure. Let $f$ be the characteristic function of $A$ and let $\tilde{f}$ be the one of $-A$. The [convolution](http://en.wikipedia.org/wiki/Convolution) $g = f \ast \tilde{f}$ is continuous and $0$ is in the support of $g$.

**Added later:** One nice standard application is that every measurable homomorphism $\phi: \mathbb{R} \to \mathbb{R}$ is continuous. For more on that and related matters have a look at these two MO-threads: 

1. On [measurable homomorphisms $\mathbb{C} \to \mathbb{C}$](https://mathoverflow.net/questions/57616/).
2. On [measurable automorphisms of locally compact groups](https://mathoverflow.net/questions/64116/)

They might elucidate what is mentioned in another answer.

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**Update:**

What I wrote above is the way I prefer to prove this.

Another approach is to appeal to regularity of Lebesgue measure $\lambda$ (used in $1$ and $2$ below).

1. Since $A$ contains a compact set of positive measure, we can assume $A$ to be compact right away (as $B-B \subset A - A$ if $B \subset A$).
2. There is an open set $U \supset A$ such that $\lambda(U) \lt 2 \lambda(A)$.
3. Since $A$ is compact there is $I = (-\varepsilon, \varepsilon)^{n}$ such that $A + x \subset U$ for all $x \in I$.
4. Since $\lambda (U) \lt 2\lambda(A)$ we must have $\lambda((A + x) \cap A) \gt 0$.

This is of course very closely related to the argument given by Chandru1 below.
We need to use the following lemma:
$$Lemma \quad m(E)>0,\forall \ 0 \lt \alpha \lt 1,\exists \left(a,b\right)\subseteq \left[0,1\right] s.t. m\left(E\cap (a,b)\right)\ge \alpha \left(b-a \right)$$
For some $\alpha \in (\frac{3}{4},1), \exists I=(a,b)\ s.t.\ m((a,b)\cap E)\ge\alpha (b-a)$

Next, we prove that $O(0,\frac{b-a}{2}) \subset E-E$

It suffice to show that $\forall x\in O,(E\cap I)\cap (E\cap I+\{x\})\neq \varnothing$

Suppose $\exists x\_0\in O (E\cap I+\{x\_0\})= \varnothing$ then

$\frac{3}{2}(b-a)<2\alpha (b-a)\leq 2m(E\cap I)= m((E\cap I) \cup (E\cap I +\{x\}))\leq \frac{3}{2} (b-a)$

Which is a contradiction, and thus the first proposition is proved.