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