# Hausdorff measure

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??

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

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