Let $(X,d)$ be a metric space and let $H^\alpha$ denote the $\alpha$-dimensional Hausdorff Borel measure on $X$, where $\alpha$ is the Hausdorff dimension of $X$. Are there any simple conditions on $X$ that allow me to conclude that $H^\alpha$ is locally finite, i.e. assigns finite measure to compact sets? Or perhaps it is true for all $X$?

If it helps feel free to assume $X$ is locally compact Polish and $d$-bounded sets are relatively compact.

For example, if $X=\mathbb{R}^d$ with the usual distance, then $\alpha=d$ and $H^d$ is $d$-dimensional Lebesgue measure, which is locally finite.

An example of the opposite, if $X$ is countable and has an accumulation point, then $\alpha=0$, $H^0$ is the counting measure on $X$, and the accumulation point has compact neighborhoods with infinitely many points.

ifthe question is appropriate for MO. $\endgroup$ – Michael Albanese Jan 23 '17 at 22:05