# 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

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 $$s$$.

• Just wanted to thank you for this, feltshire. I haven't yet had time to fully get my head back into Falconer and work through implications of your answer. I'll vote/accept after I do that.
– Mars
Aug 27, 2022 at 18:00

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

• Thanks Andrew! This works for me. I think there's a typo, though. You wrote "Eq. (2.3)"; I think you meant (3.2) [p. 45], i.e. that ${\cal H}^s(F)=\lim_{\delta\rightarrow 0}{\cal H}^s_\delta(F).$
– Mars
Dec 25, 2022 at 22:17
• The two answers so far are equally helpful; I'm accepting @feltshire's because it was first.
– Mars
Dec 25, 2022 at 22:19