Higher Dimensional Hausdorff Measure than Euclidean Space is Zero

Theorem
Let $\R^n$ be the $n$-dimensional Euclidean space.

Let $s \in \R_{>n}$.

Then:
 * $\map {\HH ^s} {\R^n} = 0$

where $\map {\HH^s} \cdot$ denotes the $s$-dimensional Hausdorff measure on $\R^n$

Proof
Let $s \in \R_{>n}$.

Consider the $n$-cube:
 * $Q := {\closedint 0 1}^n$

Then:
 * $\ds \R^n = \bigcup _{x \mathop \in \Z^n} Q + x$

As $\map {\HH ^s} \cdot$ is countably subadditive and translation invariant, it suffices to show:
 * $\map {\HH ^s} Q = 0$

Let $N \in \N_{>0}$.

Let $\CC_N$ be the set of $n$-cubes:
 * $\ds \closedint {\dfrac {i_1 - 1} {2^N} } {\dfrac {i_1} {2^N} } \times \closedint {\dfrac {i_2 - 1} {2^N} } {\dfrac {i_2} {2^N} } \times \cdots \times \closedint {\dfrac {i_n - 1} {2^N} } {\dfrac {i_n} {2^N} }$

where $i_1,\ldots ,i_n \in \set {1, 2, 3, \ldots, 2^N}$.

By Diameter of N-Cube:
 * $\forall Z \in \CC_N : \size Z = \dfrac {\sqrt n} {2^N}$

where $\size Z$ denotes the diameter of $Z$.

Thus: