$\exists x,y \in A, x \neq y$ such that $x-y \in \mathbb{Z}$ I am trying to prove a few assertions from measure theory and, one of them, is to prove the existence of a pair of elementos in $A$ such that $x-y \in \mathbb{Z}$, with $x \neq y$. In order to find a contradiction, I assumed the opposite. A could find a few conclusions, but, the last part, which is the one I write below, I could not show. Could anyone give me a hand in it? [$\mu$ means the Lebesgue measure in $\mathbb{R}$]

Let $A \subset \mathbb{R}$ is a set where it is not possible to find a pair $x,y \in A, x\neq y$. such that $x-y \in \mathbb{Z}$, with $A \in \mathcal{B}(\mathbb{R})$ and $\mu(A)>1$. . Let $f: A \rightarrow \mathbb{R}$ is such that $x \longmapsto x-\lfloor x \rfloor$. Let $A_n:= A\cap [n,n+1)$. Then, $$\mu(f(A_n)) = \mu(A_n)\ \forall n \in \mathbb{Z}$$



Conclusions I've made: $f$ is injective and  $f(A_n)\cap f(A_j) = \emptyset \ \forall n \neq j$.


 A: You are on the right track, and you need very little to finish off your solution. But I will rewrite a bit, since I don't feel you need to define and (explicitly) use $f$.
So, as you did, let $A_n:= A\cap [n,n+1)$. Then $A$ is the disjoint union of all the $A_n$, hence $\mu(A)=\sum_{n\in\mathbb Z}\mu(A_n)$.
Next (instead of using $f$), define $B_n=A_n-n:=\{a-n:a\in A_n\}$, Then $B_n$ is a translation of $A_n$ and $B_n\subset[0,1)$. (Since the Lebesgue measure $\mu$ is translation-invariant, or just obvious:) We have that $\mu(B_n)=\mu(A_n)$ for each $n\in\mathbb Z.$
Assume (toward a contradiction) that it was not possible to find different $x,y\in A$ with $x-y\in\mathbb Z.$ Then $B_n\cap B_m=\varnothing$ whenever $n\not=m$
(indeed, if $p\in B_n\cap B_m$ then $x:=p+n\in A_n\subseteq A$ and $y:=p+m\in A_m\subseteq A$, and $x-y=n-m\in\mathbb Z\setminus\{0\}$, a contradiction).
Let $B$ be the union of all $B_n$. Clearly $B\subseteq[0,1)$ hence $\mu(B)\le1.$ But, since $B_n\cap B_m=\varnothing$ whenever $n\not=m,$ we have that $\mu(B)=\sum_{n\in\mathbb Z}\mu(B_n)=\sum_{n\in\mathbb Z}\mu(A_n)=\mu(A)>1,$ a contradiction which completes the proof. (You must put the assumption that $\mu(A)>1$ in the statement of your problem, not just in your attempted solution or in the comments.)
