Measurable Steinhaus sets do not exist for finite sets or the integers in the plane

TL;DR

This paper proves that for finite sets with at least two points and for the integer lattice in the plane, no Lebesgue measurable Steinhaus set exists that intersects each rigid motion of the set exactly once.

Contribution

It establishes the non-existence of Lebesgue measurable Steinhaus sets for finite point sets and for the integer lattice in the plane, extending previous results.

Findings

No Lebesgue measurable Steinhaus set for finite sets with at least two points.

No Lebesgue measurable Steinhaus set for the integer lattice in the plane.

Extends classical results on Steinhaus sets to measurable category.

Abstract

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.