In part of Kenneth Falconer's proof of the countable stability of Hausdorff dimension on p. 49, sect 3.2 of Fractal Geometry, I understand him to say that
$$\dim_H \bigcup_{i=1}^{\infty}F_i\leq \sup_{1\leq i\leq\infty}\{\dim_H F_i\}\;,$$
because when $s>\dim_H F_i$ for all $i$, ${\cal H}^s(F_i)=0$, and thus
$${\cal H}^s(\bigcup_{i=1}^{\infty} F_i) \leq \sum_{i=1}^{\infty}{\cal H}^s(F_i) = 0\;.$$
Here $\dim_H$ is Hausdoff dimension, and ${\cal H}^s$ is $s$-dimensional Hausdorff measure.
I understand everything after $s>\dim_H F_i$ above, but I'm confused about why ${\cal H}^s(\bigcup_{i=1}^{\infty} F_i) \leq \sum_{i=1}^{\infty}{\cal H}^s(F_i) = 0$ implies $\dim_H \bigcup_{i=1}^{\infty}F_i\leq \sup_{1\leq i\leq\infty}\{\dim_H F_i\}$.
I understand that Hausdorff dimension of a set $G$ is the Hausdorff measure for which ${\cal H}^s(G)$ is finite, such that for $s>\dim_H G$, the Hausdorff measure is $0$, and for $s<\dim_H G$, the Hausdorff measure is infinite. I don't see why the fact that ${\cal H}^s(\bigcup_{i=1}^{\infty} F_i) \leq 0$ for an $s$ that is larger than the dimension implies the first inequality above, which concerns a $\sup$ for Hausdorff dimensions that are possibly greater than $0$. I'm sure there must be something obvious that I'm not seeing. I've been thinking about it for a week and I'm still confused.
This answer gives a detailed proof of the part I already understand, but leaves my question unanswered.
Best Answer
Remember that $\dim_H(F)=\inf\{s:\mathcal{H}^s(F)=0\}$. So, $$ \begin{align*} \sup\limits_i \dim_H(F_i) = s &\Rightarrow \sum\limits_i \mathcal{H}^{s+\epsilon}(F_i) = 0, \forall\epsilon>0\\ &\Rightarrow \mathcal{H}^{s+\epsilon}(\bigcup_i F_i) = 0, \forall\epsilon>0\\ &\Rightarrow \dim_H(\bigcup_i F_i)\leq s \end{align*} $$
where the first implication is due to monotonicity of $\mathcal{H}^s$ in $s$ (i.e., if all the $F_i$ are of dimension $s$ or lower, then, by definition and monotonicity, $\mathcal{H}^{s+\epsilon}$ is $0$ for all of them) and the 2nd implication is due to your quoted inequality. Technically, to get the last implication, let $\epsilon\rightarrow 0$.
(Sort of) intuitively: when the $F_i$ have dimension $s$ or lower, then for all bigger $t>s$, your inequality tells you that $\mathcal{H}^t$ of their union must already be $0$, so the dimension of that union (by definition as the infimum) can't exceed $s$.