I am trying to prove that the Hausdorff measure is, in fact, a measure.

Definition. $\mu$ is a measure on $\mathbb{R}^{m}$ if $\mu$ assigns a non-negative number (possibly $\infty$), to each subset of $\mathbb{R}^{m}$ such that

  • $\mu(\emptyset)=0$;
  • $\mu(S)\leq\mu(T)$ if $S\subseteq T$;
  • if $S_{1},S_{2},\ldots$ is a countable or finite sequence of sets, then $\mu\left(\bigcup_{i=1}^{\infty}S_{i}\right)\leq\sum_{i=1}^{\infty}\mu(S_{i})$ with equality if $S_{i}$ are disjoint Borel sets.

We call $\mu(S)$ the measure of $S$.

Definition. Suppose $F\subset\mathbb{R}^{n}$ and $s\in\mathbb{R}_{\geq0}$. For $\delta>0$, consider all $\delta$-covers of $F$ and minimise the sum of the $s$th powers of the diameters. As $\delta\to0$, the class of permissible covers of $F$ is reduced. So $$\mathcal{H}_{\delta}^{s}(F)=\inf\left\{\sum_{i=1}^{\infty}|U_{i}|^{s}:\{U_{i}\} \text{ is a } \delta \text{-cover of } F\right\}$$ increases and so approaches a limit. The $s$-dimensional Hausdorff measure $$\mathcal{H}^{s}(F)=\lim_{\delta\to0}\mathcal{H}_{\delta}^{s}(F)$$ exists for any $F\subset\mathbb{R}^{n}$ although it can be, and often is, either $0$ of $\infty$.

What I've done: Obviously $\mathcal{H}^{s}(\emptyset)=0$, and it is intuitive that if $E\subseteq F$, then $\mathcal{H}^{s}(E)\leqslant\mathcal{H}^{s}(F)$. What I am not sure how to do it show that $$\mathcal{H}^{s}\left(\bigcup_{i=1}^{\infty}F_{i}\right)\leqslant\sum_{i=1}^{\infty}\mathcal{H}^{s}(F_{i})$$ with equality if $\{F{i}\}$ are disjoint Borel sets.

  • $\begingroup$ In your definition of a measure, do you really want to define $\mu$ on each subset of $\mathbb{R}^m$? Or perhaps only on Borel subsets? $\endgroup$ – Lee Mosher Jul 9 '17 at 14:04
  • $\begingroup$ Theorem 2.19 here (books.google.de/…) might be helpful. $\endgroup$ – PhoemueX Jul 9 '17 at 14:09
  • $\begingroup$ @LeeMosher only Borel subsets are necessary, but it is my understanding that they are well-defined on all subsets. $\endgroup$ – JSharpee Jul 9 '17 at 14:15
  • $\begingroup$ @PhoemueX I will have a read through. Thank you. $\endgroup$ – JSharpee Jul 9 '17 at 14:15
  • $\begingroup$ Generally speaking, it is not the case that measures are well-defined on all subsets. That's not even true for the standard Lebesgue measure on the real line (assuming the axiom of choice). $\endgroup$ – Lee Mosher Jul 9 '17 at 14:16

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