Last step in proof of countable stability of Hausdorff dimension

Question

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.

Answer 1

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 $s$.

Answer 2

I think this 2nd part of the proof would fall into place if we pick $\xi = \sup_{i=1,\infty} \dim_H(F_i)$.

Then, $\xi \ge \dim_H(F_i) = \inf \{s | \mathcal{H}^s (F_i) = 0\}, i=1,2,\dots$

Hence, $\mathcal{H}^{\xi}(F_i) = 0, i=1,2,\dots$. Now, assuming $\{F_i\}$ is (or can be partitioned into) a countable collection of disjoint Borel sets, then by Eq. (2.3) in Falconer's book,

$\mathcal{H}^{\xi}(F_i) = 0, i=1,2,\dots \implies \mathcal{H}^{\xi}(\bigcup\limits_{i=1}^{\infty} F_i) = 0$

Therefore, $\inf \{s | \mathcal{H}^s (\bigcup\limits_{i=1}^{\infty}F_i) = 0\} \le \xi$;

i.e., $\dim_H(\bigcup\limits_{i=1}^{\infty} F_i) \le \xi = \sup_{i=1,\infty} \dim_H(F_i)$