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.
Best Answer
Assume $A$ is the set contained in [0,1] with positive measure, say $m(A) > 0$ with $m$ the Lebesgue measure on $[0,1]$. Let Q be the set of all rational numbers in $[0,1]$ Since $\mathbb{Q}$ is countable, it can be presented as
$$\mathbb{Q}=\{p_1, p_2, ..., p_n, ...\}$$
Let
$$A_n = A+ p_n = \{x+p_n\mid x\in A\}.$$
If there exists a pair of integers $n$ and $m$ such that $A_n$ and $A_m$ intersect, then the claim of this proposition is proved. If no such pair exists, then the set of $\{A_n\}$ are all disjoint. Since the union of this family of sets is contained in $[0,2]$ and since $m(A) > 0$, we have
$$2 = m([0,2]) \geq m( \bigcup A_n ) = \sum\limits_{n \in \mathbb{N}} m(A_n) =\infty\cdot m(A) = \infty .$$