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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> 
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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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> 
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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: 
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.
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.