if $\lambda(A)>1$, then $\exists x, y \in A , (x - y) \in \mathbb{N}$

Question

Let $A$ be a set in $\mathbb{R}$ of Lebesgue measure greater than 1. Prove that there exist $x, y \in A$ such that $x - y$ is a positive integer.

I started , (similar of the proof of vitali set non-measurabiliy), by defining equivalent classes $A_x = \{ y\in \mathbb{R} | (x - y) \in \mathbb{N} ,x \in A\}$.

let $B$ the set that contains exactly one element from each class. I should be claiming that this set $B$ has positive measure , using that $\lambda(A)>1$ . Is this logic working?

Answer 1

Define $A_m=A \cap [m,m+1)$ and $K_m=A_m-m $ for $m\in \Bbb{Z}$

Note that $K_m \subseteq [0,1],\forall m \in \Bbb{Z}$

If all sets $K_m$ are disjoint then $$1=\lambda([0,1])\geq \lambda(\bigcup_{m\in \Bbb{Z}}K_m)=\sum_{m \in \Bbb{Z}}\lambda(K_m)$$ $$=\sum_{m \in \Bbb{Z}}\lambda(A_m)=\lambda(A)>1$$ which is a contradiction.

So exist $m \neq n$ such that $(A_m-m) \cap( A_n-n) \neq \emptyset$

Thus exists $x \in (A_m-m) \cap( A_n-n)$

Thus $x=a_1-m$ and $x=a_2-n$ so $a_1-a_2=n-m$ and $a_2-a_1=m-n$

So one of $m-n,n-m$ is a positive integers