Inequality for Hausdorff Measure

Question

I am reading Falconer's book "Fractal Geometry" and he mentions that we can assume that "$0<\mathcal{H}^{s}(F)<\infty$" where $\mathcal{H}^{s}$ is the $s$-dimensional Hausdorff measure. I get the feeling the he means that we can only do this in certain situations.

He uses it (along with the scaling property of the Hausdorff measure) in his heuristic calculation of the Hausdorff dimension of the Middle Third Cantor set as well as other self-similar sets.

I was just wondering if someone could tell me if and when we are allowed to assume that $s=\dim_{H}(F)$ and so $0<\mathcal{H}^{s}(F)<\infty$.

Answer 1

I think that the issue here is not really one of computation or mathematics, but of terminology and/or pedagogy. The full quote, which reads

Assuming that at the critical value $s = \dim_H F$ we have $0 < \mathcal{H}^s(F) < \infty$ (a big assumption, but one that can be justified) we may divide by $\mathcal{H}^s(F)$ to get $2(\frac{1}{3})^2$ or $s = \log(2)/\log(3)$. [original emphasis]

is found in a subsection titled "Heuristic calculation." Heuristic arguments are informal and non-rigorous, but are meant to give some insight into how one should think about a problem. Heuristics are meant to give intuition.

Thus I think that the way to understand this passage is to read Falconer as saying "If this division is justified, then we may conclude that $s=\log(2)/\log(3)$." At the time that Falconer introduces this assumption, he has not yet justified it (though he knows that it can be justified and, indeed, justifies it in the next subsection).

As such, I think that the correct answer to your question is that we cannot generally assume that the Hausdorff measure in the critical dimension is finite. If you know ahead of time that it is, or if you are trying to make a heuristic argument that needn't be fully justified, you might temporarily make such an assumption, but it would required rigorous justification post facto.