Distinct points with integer difference in Lebesgue measurable set

Question

Suppose $A \subset \textbf{R}$ has Lebesgue measure greater than $1$. Show that $A$ contains a pair of points separated by an integer.

I have seen the result with Lebesgue measurable sets in $[0,1]$ with points separated by a rational number. I have been trying to generalize this argument but cannot make it work. Can anyone provide a hint for this problem?

Answer 1

Suppose $A\subset\mathbb{R}$ is such that

• $m(A) > 1$.$\\[4pt]$
• No two elements of $A$ differ by an integer.

Our goal is to derive a contradiction.

Let $f:A\to [0,1)$ be given by $f(x)=x-\lfloor{x}\rfloor$.

Since no two elements of $A$ differ by an integer, it follows that $f$ is injective.

For each $n\in\mathbb{Z}$, let $A_n=A\cap [n,n+1)$.

Then $f(A_n)$ is a translation of $A_n$ by $-n$, hence $$m(f(A_n))=m(A_n)$$ for all $n\in\mathbb{Z}$.

Since $f$ is injective, it follows that $\{f(A_n)\mid n\in\mathbb{Z}\}$ is a collection of pairwise disjoint sets.

But then we get $$ m(A)= m\left(\bigcup_{n\in\mathbb{Z}}A_n\right) =\sum_{n\in\mathbb{Z}}m(A_n) =\sum_{n\in\mathbb{Z}}m(f(A_n)) =m\left(\bigcup_{n\in\mathbb{Z}}f(A_n)\right) \le m([0,1)) = 1 $$ contradiction.