I'm writing a project on fractals and I got as far with Hausdorff dimension to say that, for some surface $S \subset \mathbb{R}^n$ and $r,s\in \mathbb{R}_{\ge 0}$ such that $r>s$ that $$0 \leq H^r(S) \leq \lim_{\delta \to 0^+} \delta^{r-s}H_\delta^s(S).$$ Naturally, this would imply that $H^r(S) = 0$ whenever $H_\delta^s(S) < \infty$. My layman's question is, how can we show that $H_\delta^s(S) = \infty$ for some $s$? I know that a disk in $\mathbb{R}^2$ would have $H_\delta^s(S) = \infty$ for $s \in (0, 2)$, for instance, but how can I prove that only knowing that $H_\delta^s(S) = \inf \{ \sum_{i} \text{diam}(U_i)^s \}$?
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@Xander Henderson wrote a good answer about this once but I can't find it right now – FShrike Dec 06 '23 at 23:26
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1I found it. https://math.stackexchange.com/questions/2586056/why-does-hausdorff-measure-go-to-zero-as-diameter-power-increases. This helps a ton, thank you ! – accountmadefor1question Dec 07 '23 at 00:05