Finite Hausdorff Measure Implies Zero Higher Dimensional Measure

Theorem
Let $n \in \N_{>0}$.

Let $F \subseteq \R^n$ be a subset of the real Euclidean space.

Let $s \in \R_{\ge 0}$.

Let $\map {\HH^s} \cdot$ denote the $s$-dimensional Hausdorff measure.

Then:
 * $\map {\HH^s} F < +\infty \implies \forall t \in \R_{>s} : \map {\HH^t} F = 0$

Proof
For each $\delta$-cover $\sequence {U_i}$ of $F$:

Thus:
 * $\map {\HH^t_\delta} F \le \delta^{t - s} \map {\HH^s_\delta} F$

Therefore: