Last step in proof of countable stability of Hausdorff dimension

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.

• Looking quickly at this (I'm in the middle of something else), I think one approach is to assume, for a later contradiction, that $\sup_{1\leq i\leq\infty}\{\dim_H F_i\} < \dim_H \bigcup_{i=1}^{\infty}F_i.$ As a consequence of this strict inequality, it follows that you can sandwich a real number $s$ strictly between these two quantities, namely $\sup_{1\leq i\leq\infty}\{\dim_H F_i\} < s < \dim_H \bigcup_{i=1}^{\infty}F_i,$ and I believe it's not difficult to get a contradiction from this. For instance, note that $\dim_H F_i < s$ for each $i,$ so $\dots$ Sep 15, 2020 at 16:24
• Thanks @DaveL.Renfro. Thinking about that now.
– Mars
Sep 16, 2020 at 12:55