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:

\(\ds \map {\HH ^s} Q\)

\(\le\)

\(\ds \sum_{Z \mathop \in \CC_N} {\size Z}^s\)

\(\ds \)

\(=\)

\(\ds 2^{nN} \paren {\dfrac {\sqrt n} {2^N} }^s\)

\(\ds \)

\(=\)

\(\ds {\sqrt n}^s 2^{N \paren {n - s} }\)

\(\ds \)

\(\to\)

\(\ds 0\)

as $N \to +\infty$

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