# Show that The Hausdorff Measure is a Measure

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.

• 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? – Lee Mosher Jul 9 '17 at 14:04
• Theorem 2.19 here (books.google.de/…) might be helpful. – PhoemueX Jul 9 '17 at 14:09
• @LeeMosher only Borel subsets are necessary, but it is my understanding that they are well-defined on all subsets. – JSharpee Jul 9 '17 at 14:15
• @PhoemueX I will have a read through. Thank you. – JSharpee Jul 9 '17 at 14:15
• 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). – Lee Mosher Jul 9 '17 at 14:16