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??

## 1 Answer

Not true for all $X$.

Taking the gauge function $\phi(t) = t^{1/2}/\log|t|$, construct a Cantor set $X$ in $\mathbb R$ using Hausdorff's original method so that $H^\phi(X) = 1$. Then the Hausdorff dimension of $X$ is $1/2$, but the "natural measure" on it is $H^\phi$, not $H^{1/2}$. Now every open set $U \subseteq X$ has $0 < H^\phi(U)<\infty$ and thus $H^{1/2}(U) = \infty$. The whole space $X$ is compact, yet $H^{1/2}(X) = \infty$.

Reference

*Hausdorff, F.*, **Dimension und äußeres Maß.**, Math. Ann. 79, 157-179 (1918). ZBL46.0292.01.

English translation in

*Edgar, Gerald A. (ed.)*, Classics on fractals, Studies in Nonlinearity. Boulder, CO: Westview Press (ISBN 0-8133-4153-1/pbk). xii, 366 p. (2004). ZBL1062.28007.