Let $(X,d)$ be an arbitrary metric space and let $\Bbb B(x,r)$ denote the closed ball with center $x \in X$ and radius $r>0$. For $p\geq 0$, let $H^p$ denote the $p$ dimensional Hausdorff measure. Under which assumptions on $X$ and $p$ is $H^p(\Bbb B(x,r))< + \infty$? Is this always the case even if the Hausdorff dimension of $X$ is infinite?
1 Answer
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As a counterexample, let $X$ be an infinitedimensional normed space. For $\varepsilon<r/2$ it follows from BorsukUlam that you need more than $n$ closed sets of diameter $\varepsilon$ to cover the intersection of the boundary $B(0,r)$ with an $n$dimensional subspace (because these sets are free of antipodal points). Hence, the Hausdorff measure of $B(0,r)$ (actually already of its boundary) is infinite for every $p\ge0$.

$\begingroup$ Thanks for the answer! Follow up question: are you aware of a general method to define a finite(nontrivial) measure on the Borel sigma algebra of an arbitrary metric space? $\endgroup$– John DJun 20, 2021 at 11:05

1$\begingroup$ @JohnD: If your space is countable, then take counting measure. If it is uncountable and separable, then you can pull back Lebesgue measure via a Polishspace isomorphism to $\mathbb{R}$. If it is inseparable, then that sounds like a hard problem. (Do you consider a nontrivial measure on a separable strict subspace trivial?) $\endgroup$ Jun 21, 2021 at 0:37

1$\begingroup$ An infinitedimensional normed space is a famous example where no natural approach is known. In practice, all "usable" measures on such spaces have the property that most sets have infinite measure. If you want to have a ball with a finite measure in such a space, the measure becomes very unnatural (not translation invariant, extremely weighted, most sets have measure 0, etc.). $\endgroup$ Jun 21, 2021 at 17:56