Let $(X,d)$ be a metric space and let $H^\alpha$ denote the $\alpha$-dimensional Hausdorff measure on $X$, where $\alpha$ is the Hausdorff dimension of $X$. Is there any simple condition 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 $X$?

I am trying to generalize that if $X=\mathbb{R}^d$ with the usual distance, then $\alpha=d$ and $H^d$ is $d$-dimensional Lebesgue measure, which is locally finite.

In general, this should be false since if I pick the Hilbert cube, which is compact, it has infinite hausdorff measure for any hausdorff dimension, correct?

Let $(X,d)$ be a metric space and let $H^\alpha$ denote the $\alpha$-dimensional Hausdorff measure on $X$, where $\alpha$ is the Hausdorff dimension of $X$. Is there any simple condition on $X$ that allow me to conclude that $H^\alpha$ is locally finite??