Hausdorff measures of different dimensions are not Borel isomorphic
The author shows that Hausdorff measures of different dimensions are not Borel isomorphic. For every \(1 \leq n < \omega\), let \(\mathcal B _{n}\) denote the \(\sigma\)-algebra of Borel subsets of \(\mathbb R^{n}\), and for every \(0 \leq d \leq n\) let \(\mathcal H^{d}\) denote the \(d\)-dimensional Hausdorff measure on \(\mathbb R ^{n}\). The author show that for every \(1 \leq n < \omega\) and for every \(0 \leq d_{1} < d_{2} \leq n\) the measure spaces \((\mathbb R^{n}, \mathcal B _{n}, \mathcal H^{d_{1}})\) and \((\mathbb R^{n}, \mathcal B _{n}, \mathcal H^{d_{2}})\) are not Borel isomorphic; moreover, there is no Borel bijection \(f : \mathbb R^{n} \to \mathbb R^{n}\) such that for every Borel set \(B \subseteq \mathbb R^{n}\), \[ 0 < \mathcal H^{d_{1}}(B) < \infty \iff 0 < \mathcal H^{d_{2}}(f(B)) < \infty. \] This result follows from the following theorem. Below, \(\dim\) stands for the Hausdorff dimension. Theorem. Let \(1 \leq m,n < \omega\), let \(A \subseteq \mathbb R^{n}\) be a Borel set and let \(f : A \to \mathbb R^{m}\) be Borel measurable. Then for every \(0 \leq d \leq 1\) there is a Borel set \(B \subseteq A\) such that \[ \dim B = d \cdot \dim A \text{ and } \dim f(B) \leq d \cdot \dim f(A). \] The proof of this theorem is based on a random construction of perfect sets, which may be useful in other situations, as well.
scientific article; zbMATH DE number 5320442
scientific article