# How can the Hausdorff measure be nonzero?

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.

• For some sets $X\subset\mathbb{R}$, it is the case that $\inf(X)\notin X$... Apr 25 '15 at 17:43

$\dim F$ 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.
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, $\dim F = \inf\;(1,\infty) = 1$.