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?



Finiteness of Hausdorff measure of balls
As a counterexample, let $X$ be an infinite-dimensional normed space. For $\varepsilon<r/2$ 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 $p\ge0$.


