In the book Geometric Measure Theory, A Beginner's Guide by Frank Morgan there is a proof showing that $\mathscr{L}^n=\mathscr{H}^n$. I have some troubles in understanding the part $\mathscr{L}^n\le\mathscr{H}^n$.
By using Bieberbach inequality, Morgan states that $\mathscr{L}^n(S) \le \alpha_n(\frac{\mathrm{diam}S}{2})^n$ and it follow immediately that $\mathscr{L}^n\le\mathscr{H}^n$.
By looking at the definition of the Hausdorff measure, $$ \mathscr{H}^m(A)=\lim_{\delta\to0} \inf_{\substack{A\subset\bigcup S_j \\ \mathrm{diam}(S_j)\le \delta }} \sum \alpha_m ( \frac{\mathrm{diam}(S_j)}{2} )^m $$
I can clearly see that that $$ \mathscr{H}^n(S) \le \alpha_n(\frac{\mathrm{diam}S}{2})^n $$ But it tells nothing of relation between $\mathscr{L}^n$ and $\mathscr{H}^n$. Can you please help me prove the inequality ?
Originally in the book, there is a digresion on the Bieberbach inequality, where by using a symmetrization argument we can reshape $S$ and translate it to the origin, while preserving the Lebesgue measure. But from my understadning it tells us nothing about the relationship I'm looking for.