Take the tour ×

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?

share|improve this question
Interesting question. For $s=0$ it is of course trivial. For $0<s<n$ I would be ready to believe that such partition does not exist. For example with $n \ge 2$ and $s=1$ I would try decomposing any partition-candidate into rectifiable and unrectifiable parts and take an unrectifiable set inside the union of the rectifiable parts of the partition sets (or the other way around). Here there might be a problem with the regularity of the unions. –  Tapio Rajala Jan 25 '12 at 7:27
My first instinct is also that it is not possible if $0 < s < n$. But this might not be provable; it might be possible using some set-theoretic assumptions like maybe the continuum hypothesis. I have not found any discussion of this question in the literature. –  Bruce Blackadar Jan 25 '12 at 21:16
add comment

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.

share|improve this answer
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 '12 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 '12 at 21:12
add comment

Your Answer


