This may come off as a naive question but it's something I've always wondered. We define the Hausdorff measure as $\mathcal H^s(A) = \lim_{\delta\rightarrow0}\mathcal H^s_\delta(A)$ for some $A\subseteq\mathbb R^n$ where $\mathcal H^s_\delta(A) = \inf\left\lbrace \sum_\alpha \left(\frac{\mathrm{diam}(E_{\alpha})}{2}\right)^s \right\rbrace$ where $(E_\alpha)$ is a finite or countable cover of $A$ such that $\mathrm{diam}(E_\alpha) < \delta$.
According to intuition, decreasing the diameter of the covering sets should make the measurement more precise so I understand why the measure is defined as it is. However it's easy to prove that $\mathcal H^s_\delta$ is decreasing in $\delta$ so I've been wondering if there's any sense in defining $h^s(A) = \lim_{\delta\rightarrow\infty}\mathcal H^s_\delta(A)$.
If I'm not mistaken $h^s(A)$ should always be finite, is there any way I can interpret its value?