Question

Consider $s$-dimensional Hausdorff measure $\mathcal{H}^s$ on the Borel sets in $\mathbb{R}^n$.
$\mathcal{H}^s$ is not $\sigma$-finite if $s < n$, but it is semifinite (on Borel sets!)
Is it known whether $\mathcal{H}^s$ can be decomposable, i.e. can there be a partition of $\mathbb{R}^n$ into disjoint Borel sets ${X_i:i\in I}$ ($I$ necessarily uncountable) such that $\mathcal{H}^s(X_i)<\infty$ for all $i$ and, for every Borel set $E$, $\mathcal{H}^s(E)=\sum\limits_{i\in I}\mathcal{H}^s(E\cap X_i)$? Does the answer depend on any set-theoretic assumptions?

Answer 1

There are cardinality $c$ Borel sets of finite ${\cal H}^s$ measure in ${\mathbb R}^n$. Assuming the Continuum Hypothesis, well-order these by the first uncountable ordinal as $E_\alpha$ for $\alpha \in A$, and define $X_\alpha = E_\alpha \backslash \bigcup_{\beta < \alpha} E_\beta$. Then $X_\alpha$ are disjoint Borel sets of finite ${\cal H}^s$ measure. For any Borel set $E$ of finite ${\cal H}^s$ measure, $E = E_\alpha = \bigcup_{\beta \le \alpha} (E \cap X_\beta)$ for some $\alpha$, and ${\cal H}^s(E) = \sum_{\beta \le \alpha} {\cal H}^s(E \cap X_\beta)$ with $E \cap X_\beta = \emptyset$ for all $\beta > \alpha$.

Hmm, I'm not sure about the case ${\cal H}^s(E) = \infty$: I think it's true that $E$ has Borel subsets of arbitrarily large finite ${\cal H}^s$ measure, but I don't have a reference at hand and can't think of a proof. If this is the case, then the result is true.