While reading proofs (for ex. this) about measure theory I am inclined to think that it is implicitly intended that the $n$-dimensional Lebesgue measure of a hypersphere $\mathbf{S}^{n-1}$, i.e. of the boundary of a ball of the form $$B(c,r)=\{x\in\mathbb{R}^n:\|x-c\|\le r\}$$where $c\in\mathbb{R}^n$ is the centre of the ball and $r>0$ its radius, is null.
I know the concepts of inner and outer regularity of the Lebesgue measure, but I cannot use them to prove that $\mu(\mathbf{S}^{n-1})=0$. How can it be done? I heartily thank any answerer.
Best Answer
Suppose $\mu(S) > 0.$ Because $\mu(rE) = r^n\mu(E)$ for any measurable $E\subset \mathbb R^n$ and $r>0,$ we have $\mu(rS)\ge \mu(S)$ for $r\ge 1.$ Now $\{1\le |x|\le 2\},$ a compact subset of finite measure, contains the pairwise disjoint compact sets $S_k= (1+1/k)S, k = 1, 2, \dots $ This implies $\mu(\{1\le |x|\le 2\}) \ge \sum_k \mu(S_k) = \infty,$ contradiction.