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.