# When is Hausdorff measure locally finite?

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.

• – Michael Albanese Jan 23 '17 at 4:51
• @MichaelAlbanese, Is crossposting not allowed or in bad taste? I will delete the question on MSE if that is the case. – nullUser Jan 23 '17 at 21:17
• You shouldn't crosspost at the same time. Someone could take a lot of time to write an answer on one site when an answer already exists on the other. It is better to choose one site and post there. Sometimes people post on MSE and if they don't receive any answers after a week or so, they crosspost to MO; that's considered fine if the question is appropriate for MO. – Michael Albanese Jan 23 '17 at 22:05
• Question is no longer cross-posted, and I received no answers there during the time that it was cross-posted. I apologize that I wasn't familiar with cross-posting etiquette. – nullUser Jan 28 '17 at 19:12
• No worries. Now that they are no longer relevant, I think we should delete our comments. Once you have deleted yours, I'll delete mine. – Michael Albanese Jan 28 '17 at 19:23