Answer to a question of Kolmogorov

Abstract

More than 80 years ago Kolmogorov asked the following question. Let $E\subseteq {\mathbb{R}}^2$ be a measurable set with ${\lambda }^2(E)<\infty$, where ${\lambda }^2$ denotes the two-dimensional Lebesgue measure. Does there exist for every $\epsilon >0$ a contraction $f:E\to {\mathbb{R}}^2$ such that ${\lambda }^2(f(E))\ge {\lambda }^2(E)-\epsilon$ and $f(E)$ is a polygon? We answer this question in the negative by constructing a bounded, simply connected open counterexample. Our construction can easily be modified to yield the analogous result in higher dimensions.

arXiv comment5 pages