Context: Towards the end of the proof that the n dimensional Hausdorff measure is equal to the Lebesgue measure on $\mathbb{R}^n$, there is an inequality I do not understand. $S$ is a ball in a covering of a set $A$.
Morgan claims that $$ \mathscr{L}(S)\geq \alpha_n(\frac{diam(S)}{2})^n $$ where diameter is defined as usual as the supremum of all distances between elements in the set and $\alpha_n$ is the Lebesgue n-volume of the unit ball.
I think this is probably just a basic fact from Lebesgue theory that I never learned. If it is not, I can provide more context.
The change-of-variables formula shows that for every Lebesgue measurable $M\subset \mathbb{R}^n$ and $c\in \mathbb{R}$ we have $\mathscr{L}(c\cdot M) = \lvert c\rvert^n\cdot \mathscr{L}(M)$. Using also the translation invariance, it follows that the Lebesgue measure of a ball of radius $r$ is $\alpha_n r^n$. So we have not only the inequality, in fact the equality
$$\mathscr{L}(S) = \alpha_n\biggl(\frac{\operatorname{diam}(S)}{2}\biggr)^n.$$