# Can Hausdorff measures really be defined in any metric space?

I've read in several places (including wikipedia) that in a metric space $X$ the Hausdorff $s$-measure of a set $A \subseteq X$ is defined to be $$H^s(A) = \lim_{\delta \to 0} \inf\left\{ \sum_{i=1}^{\infty} (\text{diam}(E_i))^s : \bigcup_{i=1}^{\infty} E_{i} \supseteq A, \text{diam}(E_i) \leq \delta \right\}$$

This seems to assume that the metric space has the property that \begin{gather} \text{$\forall \delta > 0$ $\exists$ a countable collection of sets $E_1,E_2,\ldots \subseteq X$ such that} \\ \bigcup_{i=1}^{\infty} E_{i} \supseteq X \quad \text{ and } \quad \text{diam}(E_i) \leq \delta. \end{gather}

Separable (hence also compact) metric spaces have this property.

But what about other metric spaces?

Is it standard to just interpret the infimum as $0$ or $+\infty$ if there are no valid covers?

• I think it would be more appropriate to interpret it as $\infty$ – Hagen von Eitzen Jul 2 '17 at 17:14
• @HagenvonEitzen Why do you say your interpretation is more appropriate? – RitterSport Jul 2 '17 at 17:16
• @HagenvonEitzen I guess yours is the standard interpretation of the infimum of the empty. set.math.stackexchange.com/questions/432295/… – RitterSport Jul 2 '17 at 17:20

Yes, of course you can set $H^s(A)=+\infty$ if no such convering can be found.