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We have

dim$F := \inf \left\{s > 0 : \mathcal{H}^s (F) = 0\right\}$.

My question is, with dim$F$ defined as the value where the Hausdorff measure equals zero, then how can

$\mathcal{H}^{\text{dim}F}(F) \neq 0$ ?

I know it is true that $0 \leq \mathcal{H}^{\text{dim}F}(F) \leq \infty$, but I don't understand how that makes sense with the given definition.

Thanks in advance

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  • $\begingroup$ For some sets $X\subset\mathbb{R}$, it is the case that $\inf(X)\notin X$... $\endgroup$ Apr 25 '15 at 17:43
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$\dim F$ is not defined as a value for which the Hausdorff dimension equals 0. It's defined as the infimum of a set of such values, which does not mean it has to be a member of the set itself.

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An example. $F = [0,7]$ with its usual metric. Then:

$\mathcal H^s(F) = 0 $ for $s>1$,

$\mathcal H^1(F) = 7$,

$\mathcal H^s(F) = \infty$ for $0<s<1$.

Thus, according to your definition, $\dim F = \inf\;(1,\infty) = 1$.

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