# If $E$ has Lebesgue measure $0$, must there exist a translate such that $E\cap E+x=\varnothing$?

If $y\in E\cap E+x$ then $y=h=h'+x \implies x=h-h'$ for $h,h'\in E$. Likewise, if $x=h-h'$, then $h\in E\cap E+x$. So that $E\cap E+x$ is non empty iff $x=h-h'$ where $h,h'\in E$. Thus, if there exists a real $x$ which cannot be expressed as above, the result is proven. The problem is I cannot prove that such an $x$ must exist. I have tried assuming that all real $x$ are of that form, but that gives me nothing except that $E$ must be uncountable and unbounded. Any hints?

## 2 Answers

Let $C$ be the usual Cantor set: a measure zero subset of $[0,1]$.

Let $C_\infty=\bigcup_{k=-\infty}^\infty(C+k)$, that is the union of the translates of $C$ by integers. Also $C_\infty$ has measure zero. Now $C$ has the property that $C+C=[0,2]$; every real in $[0,2]$ is the sum of two elements of $C$. Thus $C_\infty+C_\infty=\Bbb R$. So if $x\in R$ there are $a$, $b\in C_\infty$ with $x=a+b$. Then $-a\in C_\infty$ too, and $b=x-a\in C_\infty\cap(C_\infty +x)$.

The claim is false, i.e., in general there is no such $x$.

To see this, I like the following example which uses the Baire category theorem. There are probably other (easier?) examples that use something like the Cantor set.

Since $\Bbb{Q}$ is a null-set, there is by regularity of the Lebesgue measure for each $n \in \Bbb{N}$ an open set $U_n \supset \Bbb{Q}$ with $\mu(U_n) < 1/n$. It is clear that $E := \bigcap_n U_n$ is a null-set.

We want to show that there is no $x$ such that $E \cap (E + x) = \emptyset$. Since the map $\Bbb{R} \to \Bbb{R}, y \mapsto y + x$ is a homeomorphism, it suffices to prove the (much) stronger claim that there is no homeomorphism $\varphi : \Bbb{R} \to \Bbb{R}$ with $\varphi(E) \cap E = \emptyset$.

But $\varphi(E) \cap E = \bigcap_{n,m} (\varphi(U_n) \cap U_m )$ is a countable intersection of open, dense subsets of $\Bbb{R}$, so that Baire's category theorem implies that $\varphi(E) \cap E \neq \emptyset$.