# Show that the n-dimensional Hausdorff measure of an $n$-dimensional cube is positive and finite.

Show that the n-dimensional Hausdorff measure of an $n$-dimensional cube is positive and finite. I can easily show that if it is finite then the $n+1$ dimensional measure is $0$ and the $n-1$ dimensional measure is $\infty$, but I'm not sure how to show that it is at exactly $n$ that the positive finite case occurs. Can anyone provide any tips?

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I would start by showing that for some constant $c$ depending only on $n$, you may obtain the inequalities $c \mathcal{H}_n(E) \le m(E) \le 2^n c \mathcal{H}_n(E)$, where $\mathcal{H}_n$ is the n-dimensional Hausdorff measure. By taking E to be a cube, this will prove your result. – anonymous Nov 28 '12 at 15:36
The easy one is "finite". For large $k$, decompose the $n$-cube into $k^n$ cubes with side divided by $k$. Use this to estimate the Haudorff measure. In the limit, get an inequality showing the measure is finite. – GEdgar Nov 28 '12 at 15:37
The harder direction is "positive". For that you can use the existence of Lebesgue measure to get a lower bound. – GEdgar Nov 28 '12 at 15:39

More detail for my comments from 2 years ago. Since the OP merely asked for "tips" I didn't provide a complete proof then.

finite
Take an $n$-dimensional cube $C$ of side $a>0$. Fix $\epsilon>0$. Let $k$ be a positive integer. Then $C$ is the union of $k^n$ cubes of side $a/k$. Each of those small cubes has diameter $a\sqrt{n}/k$. Choose $k$ large enough that this is $< \epsilon$. So, using this cover of $C$, we get $$\mathcal{H}^n_\epsilon(C) \le k^n \left(\frac{a\sqrt{n}}{k}\right)^n = a^n n^{n/2}.$$ This is true for all $\epsilon>0$, so $\mathcal{H}^n(C) \le a^n n^{n/2}$. Finite.

positive
Write $m$ for the $n$-dimensional Lebesgue outer measure. Note: If $A$ is a set of diameter $t$, then $A$ is contained in a closed ball of radius $t$. (Namely any such ball whose center is a point of $A$.) Let $k$ be the measure of a ball of radius $1$. So if $A$ has diameter $t$, then $m(A) \le kt^n = k({\rm diam }\;A)^n$. Now let $C$ be a cube with side $a$. Consider a countable cover $C \subseteq \bigcup_{i=1}^\infty A_i$. Then $$a^n = m(C) \le \sum_{i=1}^\infty m(A_i) \le k\sum_{i=1}^\infty ({\rm diam}\;A_i)^n$$ so $$\sum_{i=1}^\infty ({\rm diam}\;A_i)^n \ge \frac{a^n}{k} .$$ This is true for all countable covers of $C$, so $\mathcal{H}^n(C) \ge a^n/k$. Positive.

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Tao suggests a proof in his answer. However, it may be that the fact in question is part of the proof of the relationship between Hausdorff and Lebesgue measures. So using it may be circular reasoning? – GEdgar Nov 20 '14 at 2:16
This was very enlightening. For some reason the relation between the Lebesgue measure and $\operatorname{diam}$ escaped me. Thank you. (I have to wait to award bounty). – user157227 Nov 20 '14 at 2:19

you can use the relationship between Hausdorff n-dimensional measure and Lebesgue Measure . http://en.wikipedia.org/wiki/Hausdorff_measure

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