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This is a home work problem that I am stuck on even though it feels like it should be easy:

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

1 Answer 1

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

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