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arXiv · 1604.06454

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

Mihail N. Kolountzakis·Michael Papadimitrakis
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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.

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Mihail N. Kolountzakis, Michael Papadimitrakis. 2016-04-21. Measurable Steinhaus sets do not exist for finite sets or the integers in the plane. https://doi.org/10.1112/blms.12069

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