TITLE: The set of differences for a set of positive Lebesgue measure
QUESTION [44 upvotes]: 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.

REPLY [21 votes]: 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 $g = f \ast \tilde{f}$ 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: 

On measurable homomorphisms $\mathbb{C} \to \mathbb{C}$.
On measurable automorphisms of locally compact groups

They might elucidate what is mentioned in another answer.

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).

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 $B \subset A$).
There is an open set $U \supset A$ such that $\lambda(U) \lt 2 \lambda(A)$.
Since $A$ is compact there is $I = (-\varepsilon, \varepsilon)^{n}$ such that $A + x \subset U$ for all $x \in I$.
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.