## Decomposability of Hausdorff measure

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?

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 Interesting question. For $s=0$ it is of course trivial. For $0 ## 1 Answer There are cardinality$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$measure, but I don't have a reference at hand and can't think of a proof. If this is the case, then the result is true. - This statement is true about Borel sets of infinite Hausdorff measure. See the article by J. Howroyd (Proc. London Math. Soc. (3) 70 (1995) 581-6O4) for the most general results known along this line. However, the result is false in general for$\mathcal{H}^s$-measurable sets; see section 439H of Fremlin's book. This is why I phrased the question in terms of Borel sets. What your argument shows is that a semifinite measure space whose$\sigma$-algebra has cardinality$\aleph_1\$ is decomposable. It answers the question under the assumption of the continuum hypothesis. – Bruce Blackadar Jan 26 2012 at 3:45
I carelessly misspoke slightly in the previous comment. The argument only works for semifinite measure spaces where each point is contained in a measurable set of finite measure. This of course holds for Hausdorff measure, but not in general; for a simple counterexample, see IX.1.9.3 and IX.1.9.6 in the Real Analysis manuscript on my website <wolfweb.unr.edu/homepage/bruceb/>; . – Bruce Blackadar Jan 27 2012 at 21:12