If $A\subset \mathbb R^n, m(A)>1$, then $\exists x,y\in A$ with $x-y\in \mathbb Z^n$ Let $A\subset \mathbb R^n$ with $m(A)>1$, where $m$ is the $n$-dimensional Lebesgue measure. Prove that $\exists$ distinct $x,y \in A$ with $x-y\in \mathbb Z^n$.
This is supposed to be a follow-up to this problem: If $A,B\subset (0,1)$ with $m(A)>\frac{1}{2}, m(B)>\frac{1}{2}$, then there exist $x\in A$ and $y\in B$ with $x+y=1$. Still, even though I've solved the simpler one, I fail to see how exactly this would help...
 A: Here is a somewhat heuristic solution, I'll leave it to you to formalize. 
Let $A \subset \mathbf{R}^n$ such that $m(A)>1$. Consider each "cell" of the lattice $\mathbf{Z}^n$ in $\mathbf{R}^n$. That is, each unit box $(b_1,b_1+1]\times (b_2,b_2+2] \times \dots \times (b_n,b_n+1]$, $b_i \in \mathbf{Z}$. Call such a box $B_{\nu}$ where $\nu$ is the appropriate multiindex $(b_1,\dots,b_n)\in \mathbf{Z}^n$. 
Now, take each $A \cap B_{\nu}$ and shift it to the fundamental unit box, that is, $(0,1]^n$. Since $m$ is not affected by such linear translations, we know that the measure of the union of all shifted $A\cap B_{\nu}$ is equal to the measure of $A$ itself, which is greater than 1. Hence, there is $\nu_1, \nu_2$ distinct such that $(A \cap B_{\nu_1} - \nu_1)\cap(A \cap B_{\nu_2}-\nu_2) \neq \emptyset$, since $m((0,1]^n)=1$. (This is the hardest part of the proof, what happens if all the pieces $A\cap B_{\nu}$ are disjoint?)
That is, there is overlap among all the shifted pieces of $A$. Now, $\nu_1-\nu_2$ is a point of the lattice $\mathbf{Z}^n$. Let $z \in (A \cap B_{\nu_1} - \nu_1)\cap(A \cap B_{\nu_2}-\nu_2)$. Then, $z+\nu_1 \in A$ and $z +\nu_2 \in A$. Let $x=z+\nu_1, y=z+\nu_2$. Then, $x-y=z+\nu_1-z-\nu_2=\nu_1-\nu_2 \in \mathbf{Z}^n$. And, $x \neq y$, so the proof is done.
