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I'm reading about Hausdorff dimesion. Could you confirm if my understanding (through the proof of the proposition) is correct?

Let $(E, \rho)$ be a metric space. Let $d \in [0, +\infty)$ and $\delta>0$. $$ H_{\delta}^{d}(S) := \inf \left\{ \sum_{i=1}^{\infty}\left(\operatorname{diam} U_{i}\right)^{d} \,\middle\vert\, \bigcup_{i=1}^{\infty} U_{i} \supseteq S, \operatorname{diam} U_{i}<\delta\right\} \quad \forall S \subseteq E. $$

Then the $d$-dimensional Hausdorff measure is defined by $$ H^d(S) := \lim_{\delta \to 0^{+}} H_{\delta}^{d}(S) = \sup \{ H_{\delta}^{d}(S) \mid \delta > 0 \} \quad \forall S \subseteq E. $$

The Hausdorff dimension of $S \subseteq E$ is defined by $$ \dim_H (S) := \inf \{d \in [0, +\infty) \mid H^d (S) = 0\}. $$

In above formula, we adopt the usual convention that $\inf \emptyset = +\infty$.

I post my proof separately as below answer. This allows me to subsequently remove this question from unanswered list.

Analyst
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1 Answers1

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First, we have

Theorem 1: Let $S \subseteq E$ and $d \in [0, +\infty)$. If $H^d (S)<+\infty$, then ${H}^{t} (S)=0$ for all $t>d$.

Proof: For each $\delta>0$, there exists a cover $(U_i)$ of $S$ such that $\operatorname{diam} U_{i} < \delta$ for all $i$ and $$ \sum_{i=1}^{\infty} \left( \operatorname{diam} U_{i} \right)^{d} <H^d_\delta + 1. $$

We have \begin{align} H^t_\delta &\le \sum_{i=1}^{\infty} \left( \operatorname{diam} U_{i} \right)^{t} \\ &= \sum_{i=1}^{\infty} \left( \operatorname{diam} U_{i} \right)^{d} \left( \operatorname{diam} U_{i} \right)^{t-d}\\ &\le \delta^{t-d} \sum_{i=1}^{\infty} \left( \operatorname{diam} U_{i} \right)^{d} \quad \text{because} \,\, t-d > 0 \\ &< \delta^{t-d}( H^d_\delta + 1). \end{align}

The proof is complete by taking the limit $\delta \to 0^{+}$. By contrapositive, we also have

Theorem 2: Let $S \subseteq E$ and $d \in [0, +\infty)$. If $H^d (S) > 0$, then ${H}^{t} (S) = +\infty$ for all $t<d$.

It follows from Theorems 1 and 2 that the cardinality of the set $\{H^d (S) \mid d \in [0, +\infty)\}$ is at most $3$. Finally, we have

Proposition: $\dim_H (S) = \inf \{d \in [0, +\infty) \mid H^d (S) < +\infty\}$.

Proof: Let $d' := \inf \{d \in [0, +\infty) \mid H^d (S) < +\infty\}$. Clearly, $d' \le \dim_H (S)$. Assume the contrary that $d' < \dim_H (S)$. By definition of $d'$, there is $d' \le d_1 < \dim_H (S)$ such that $H^{d_1} (S) < +\infty$. Then by Theorem 1, we get $H^{d_2} (S) = 0$ for any $d_1 <d_2$, which contradicts the minimality of $\dim_H (S)$.

Analyst
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