Measurable Steinhaus sets do not exist for finite sets or the integers   in the plane
  A Steinhaus set $S \subseteq \RR^d$ for a set $A \subseteq \RR^d$ is a set such that $S$ has exactly one point in common with $\tau A$, for every rigid motion $\tau$ of $\RR^d$. We show here that if $A$ is a finite set of at least two points then there is no such set $S$ which is Lebesgue measurable.   An old result of Komj\'ath says that there exists a Steinhaus set for $A = \ZZ\times\Set{0}$ in $\RR^2$. We also show here that such a set cannot be Lebesgue measurable. 
Mihail N. Kolountzakis and Michael Papadimitrakis
math.MG math.CA