Hausdorff-Besicovitch Dimension is Well-Defined

Theorem

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

Let $\map {\HH^s} F$ be the $s$-dimensional Hausdorff measure on $\R^n$ of $F$ for each $s \in \R_{\ge 0}$.

Then:

\(\ds \)

\(\)

\(\ds \inf \set {s \in \R_{\ge 0} : \map {\HH^s} F = 0}\)

\(\ds \)

\(=\)

\(\ds \sup \set {s \in \R_{\ge 0} : \map {\HH^s} F = +\infty}\)

\(\ds \)

\(\in\)

\(\ds \closedint 0 n\)

In particular, the Hausdorff-Besicovitch dimension is well-defined.

Proof

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