For $A\subset \mathbb R^n$ let $\mathcal H^s(A)$ be the $s$-dimensional Hausdorff measure with respect to the Euclidean metric. The Hausdorff dimension of $A$ is given by $$\dim_H(A) = \inf\{s>0\mid\mathcal H^s(A)=0\}.$$ One of the go-to curiosities in measure theory courses is the fact the middle-third Cantor $C$ set has Hausdorff dimension $\log2/\log3$. This value is easily determined using a self-similarity argument assuming that $0<\mathcal H^s(C)<\infty$ for some $s$. However, verifying this assumption for $s=\log2/\log3$ is not quite as easy. Thinking about this and other self-similar objects got me wondering about a more general question:
For non-empty, compact $K\subset\mathbb R^n$ of Hausdorff dimension $d$, is it automatic that $0<\mathcal H^d(K)<\infty$? In other words, do non-empty compacta in $\mathbb R^n$ have finite, non-zero Hausdorff measure in their Hausdorff dimension?
At first, I thought that the answer must be yes for the extreme cases $d=0$ and $d=n$ where the relevant Hausdorff measures are the counting measure and the Lebesgue measure (up to scaling). However, at some point I realized that I had implicitly assumed that Hausdorff dimension 0 or n would imply discreteness or non-empty interior (which may or may not be true, probably not).
I'm now at the point where I've realized that I have already spent much more time thinking about this than I should have. Hence, I resort to the infinite wisdom of the internet. I believe that the answer to the above question is no in the stated generality, but I'm basically at a loss. Maybe someone on here has something up their sleeve.
