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??

  • 2
    $\begingroup$ Hilbert Cube $[0,1]^{\mathbb N}$ has infinite topological dimension, so no matter how you metrize it, the result has infinite Hausdorff dimension. $\endgroup$ Oct 13, 2021 at 15:11

1 Answer 1


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 $H^{1/2}(X) = \infty$.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.