By definition, let's denote $n$-dimensional Hausdorff measure as $$H^{\alpha}{A} = \lim_{\epsilon \to 0} {H_{\epsilon}^{\alpha}{A}}$$ where $$H_{\epsilon}^{\alpha} = \inf\{\sum_{k=1}^{\infty}{diam(A_{k})^{\alpha}}| A \in \cup A_{k}, diam(A_{k}) < \epsilon\}$$

The outer Lebesgue measure is also defined in the following way: $$ \lambda_{n}^{*}(A) = \inf \{\sum_{j=1}^{\infty}{|P_{j}|}: A \subset \cup P_{j} \}.$$

Suppose that $H^{\alpha} = c_{n} \cdot \lambda_{n}^{*}$. How to find $c_{n}$?

Wikipedia says that for Lebesgue measurable sets the following equality holds $$\lambda_{d}(E) = 2^{-d} \cdot \beta_{d} H^{\alpha}(E)$$ where $$\beta_{d} = \frac{{\pi}^{\frac{d}{2}}}{Г(\frac{d}{2}+1)}$$ but the existence of such kind of equation does not help in finding the way how the $c_{n}$ was obtained.

Any sort of help would be much appreciated.

  • $\begingroup$ Does the $\alpha$ play the role of a parameter in the defintion of $H^{\alpha}$? Or is the superscript just standard notation? $\endgroup$ – Titus Sep 28 '15 at 21:46
  • $\begingroup$ And are the $P_j$ arbitrary sets in your definition? $\endgroup$ – Titus Sep 28 '15 at 21:47
  • $\begingroup$ @Titus $P_{j}$ are treated as elementary sets -- rectangles of the form $(a_{1}, b_{1}) \times (a_{2}, b_{2}) \times \ldots \times (a_{n}, b_{n})$ . $\alpha$ plays the role of a parameter, as you've mentioned, in order to avoid confusion, i've renamed it in $\lambda_{n}(E) = 2^{-d} \cdot \beta_{d} \cdot H^{\alpha}{(E)}$ $\endgroup$ – hyperkahler Sep 28 '15 at 21:51
  • $\begingroup$ So what does varying $\alpha$ do to the Hausdorff measure? I don't see it in the definition. Not sure I'll be much help really, I'm just interested in the problem and want to get everything straight; not trying to heckle you. $\endgroup$ – Titus Sep 29 '15 at 6:31
  • $\begingroup$ @Titus, thank you, there was a rude mistake in the definition of the Hausdorff measure. Now it looks much better. Nevertheless, seems that this problem can be treated as a general one, but i didn't found any relevant data about it apart from the piece of Wikipedia article. $\endgroup$ – hyperkahler Sep 29 '15 at 20:47

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