Existence of x,y so that x-y is in Z-{0}.

[WWGD]

Hi, let E be a measurable subset of the Real line with m(E)>1 . I want to show
there are x,y in E so that x-y is in $\mathbb Z-{0}$. My idea is to restrict the
quotient $\mathbb R / \mathbb Z |_E$. This quotient cannot be contained in
[0,1], since m([0,1])=1 and m(E)>1. From this I want to show that there must be
more than one representative of each clase of the quotient in E , but I am having
trouble tightening up the argument. Any ideas?

Thanks.

 

[jbunniii]

Suppose there are no such $x,y \in E$.

For each integer $k$, let $E_k = E \cap [k, k+1)$. Then the $E_k$'s are measurable and pairwise disjoint, with $\sum m(E_k) = m(E)$.

Now consider the translated sets $F_k = E_k - k$, in other words, $F_k$ is $E_k$ translated by $k$, so $F_k \subset [0,1)$.

If $F_k \cap F_m \neq \emptyset$ for some $k \neq m$, then there are $x_k \in E_k$ and $x_m \in E_m$ such that $x_k - k = x_m - m$, so $x_k - x_m \in \mathbb{Z} \setminus \{0\}$, contrary to our assumption above. So the $F_k$'s are pairwise disjoint.

Now Lebesgue measure is translation-invariant, so each $F_k$ is measurable with $m(F_k) = m(E_k)$. By countable additivity,
$$m\left(\bigcup_{k \in \mathbb{Z}} F_k\right) = \sum_{k \in \mathbb{Z}} m(F_k) =\sum_{k \in \mathbb{Z}} m(E_k) = m(E) > 1$$
But on the other hand, each $F_k \subset [0,1)$ for each $k$, and therefore also $\bigcup_{k \in \mathbb{Z}} F_k \subset [0,1)$, which means that
$$m\left(\bigcup_{k \in \mathbb{Z}} F_k\right) \leq 1$$
So we have a contradiction.