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Let $ \mathscr{H}^m $ be the m-dimensional Hausdorff measure in $ R^n $. Let $ m<k $ and $ A \subset R^n $. If $ \mathscr{H}^m(A) < \infty $ then $ \mathscr{H}^k (A) =0 $. How can i prove this fact?

Thank you

Andrew Uzzell
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    Isn't that a fairly straightforward argument directly based on the definition of outer Hausdorff measure? Basically, $\varepsilon^k=\varepsilon^{k-m}\cdot\varepsilon^m$, and take sums … Even more accurately, if $d<\varepsilon$ then $d^k<\varepsilon^{m-k}\cdot d^m$. – Harald Hanche-Olsen Apr 11 '13 at 09:47
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    Yes, after a little reflection i've seen that the proof follows immediately from the definition, as you have suggested. Thanks –  Apr 11 '13 at 14:22

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