Answer to a question of Kolmogorov

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 $\varepsilon>0$ a contraction $f\colon E\to \mathbb{R}^{2}$ such that $\lambda^{2}(f(E))\geq \lambda^{2}(E)-\varepsilon$ 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.