# Finiteness of Hausdorff measure of balls

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?

As a counterexample, let $$X$$ be an infinite-dimensional normed space. For $$\varepsilon it follows from Borsuk-Ulam that you need more than $$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$$.
• @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 Polish-space 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?) Jun 21 at 0:37