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 $F$:

\(\ds \sum \size {U_i}^t\)

\(=\)

\(\ds \sum \size {U_i}^s \size {U_i}^{t - s}\)

\(\ds \)

\(\le\)

\(\ds \delta^{t - s} \sum \size {U_i}^s\)

Thus:

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

Therefore:

\(\ds \map {\HH^t} F\)

\(=\)

\(\ds \lim_{\delta \to 0^+} \map {\HH^t_\delta} F\)

\(\ds \)

\(\le\)

\(\ds \lim_{\delta \to 0^+} \delta^{t - s} \map {\HH^s_\delta} F\)

\(\ds \)

\(=\)

\(\ds \lim_{\delta \to 0^+} \delta^{t - s} \cdot \lim_{\delta \to 0^+} \map {\HH^s_\delta} F\)

\(\ds \)

\(=\)

\(\ds 0 \cdot \map {\HH^s} F\)

\(\ds \)

\(=\)

\(\ds 0\)

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