Hausdorff measure

Remarks

The limit exists in $[0, \infty]$ because $\mathcal{H}^s_\delta(A)$ grows as $\delta$ decreases. The map $\mathcal{H}^s$ is a metric outer measure, so all Borel sets are measurable and its restriction to them is a measure. Some authors multiply by a constant so that $\mathcal{H}^n$ equals Lebesgue measure on $\R^n$; without it, $\mathcal{H}^n$ is a constant multiple of Lebesgue measure.

Example: $\mathcal{H}^0$ is counting measure. On $\R$, $\mathcal{H}^1$ is Lebesgue measure: covers by intervals realize the infimum, and the length of an interval is its diameter.

Non-example: the quantity without the limit $\delta \to 0$, the Hausdorff content $\mathcal{H}^s_\infty$ with no bound on diameters, is not a measure on Borel sets. In $\R^2$ with $s = 1$, the segments $A = [0, 1] \times \set{0}$ and $B = [0, 1] \times \set{1}$ each have content $1$. But $A \cup B$ lies in the unit square, of diameter $\sqrt 2$, so its content is at most $\sqrt 2 < 2$.