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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?

RitterSport
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    I think it would be more appropriate to interpret it as $\infty$ – Hagen von Eitzen Jul 02 '17 at 17:14
  • @HagenvonEitzen Why do you say your interpretation is more appropriate? – RitterSport Jul 02 '17 at 17:16
  • @HagenvonEitzen I guess yours is the standard interpretation of the infimum of the empty. set.https://math.stackexchange.com/questions/432295/infimum-and-supremum-of-the-empty-set – RitterSport Jul 02 '17 at 17:20
  • Well, if you cannot find a way to cover a metric space with countable union then it is a "huge" space. So, it should e assigned infinite measure. Isn't it that simple?! – Behnam Esmayli Apr 25 '21 at 04:02

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

Emanuele Paolini
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