Hausdorff measure (Folland exercise) I am trying to prove that in any metric space, zero dimensional Hausdorff measure is the counting measure. By looking at the definition  $$ H_{p,\delta}\left(E\right)=\inf\left\{ \sum_{j=1}^{\infty}\left(\text{diam}B_{j}\right)^{p}\thinspace:\thinspace E\subseteq\bigcup_{j=1}^{\infty}B_{j},\thinspace\thinspace\text{diam}\left(B_{j}\right)<\delta\right\}  $$
For $p=0$: $$ H_{p,\delta}\left(E\right)=\inf\left\{ \sum_{j=1}^{\infty}1\thinspace:\thinspace E\subseteq\bigcup_{j=1}^{\infty}B_{j},\thinspace\thinspace\text{diam}\left(B_{j}\right)<\delta\right\}  $$
Lettig $\delta \to 0 $, intuitivly obvious that we "count" the points in the set, because the diameter is going to $0$.
How can I prove it in a formal way?
Thanks in advance.
 A: If you are happy to use that $\mathcal H^p(A\cup B) =\mathcal H^p(A)+\mathcal H^p( B) $ for $A,B$ disjoint then it is enough to consider the case that $E$ is a singleton.
Before I write the proof, I want to mention a technical detail: we need to use the convention $0^0=0$, otherwise $\mathcal H^0_\delta (E) = +\infty$ for all $\delta >0$. The other (equivalent) way of thinking about it (and, in my opinion, the better way of writing it) would be to define the $p=0$ case as $$\mathcal H^0_\delta = \inf \bigg \{ N(\{U_k\}) \text{ s.t. } E \subset \bigcup_k U_k, \operatorname{diam}U_k <\delta \bigg \} $$ where $N(\{U_k\})$ denotes $k$ for which $U_k \neq \varnothing$.
Anyways, let $E=\{x_0\}$ and fix $\delta >0$. On one hand, let $\{U_k\}$ be a countable open cover of $E$ with $\operatorname{diam} U_k <\delta $. Then $\{x_0\}\subset \bigcup_kU_k$, so there exists some $N$ such that $U_N$ is non-empty. Hence, $ N(\{U_k\}) \geqslant 1 $. Since $\{U_k\}$ was arbitrary,we have that
$$\mathcal H^0_\delta(E) \geqslant 1. $$
On the other hand, a countable open cover is given by $U_1 = \{ x \in X \text{ s.t. } d(x,x_0)<\delta/4\}$, $U_k = \varnothing$, $k\geqslant 2$, so $$\mathcal H^0_\delta(E) \leqslant  N(\{U_k\})=1 . $$
Thus, $\mathcal H^0_\delta(E)=1$ for all $\delta >0$, so $\mathcal H^0(E) = \lim_{\delta \to 0^+}\mathcal H^0_\delta(E)=1$ as required.
