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

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

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> Prove that every set of positive measure in the interval $[0,1]$ contains a pair of points whose distance apart is a rational number.
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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: 
If you need to find information on the subject, the first proof of the fact that the set of differences contains a neighbourhood of the origin is (for Lebesgue measure on the line) due to Steinhaus. There is a substantial collection of generalizations of the result.
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.