  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.
