Lebesgue measure of a sphere 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.
 A: Let $\mu$ be the $n$-dimensional Lebesgue measure and $V_n$ the measure of the unit ball.
By isotropy, $\mu(B(c,r)) = \mu(B(0,r))$, so let $B_r=\{\overline{x}:\|\overline{x}\|=r\}$. 
By the smoothness and regularity of $B_r$:
$$ \mu(\partial B_r) = \lim_{\varepsilon\to 0}\left[\mu(B_{r+\varepsilon})-\mu(B_{r-\varepsilon})\right]=V_n\cdot\lim_{\varepsilon\to 0}\left[(r+\varepsilon)^n-(r-\varepsilon)^n\right]=\color{red}{0}.$$
A: A key point of Lebesgue measure is that it develops from our intuitive idea of volume of a box. Using this and Riemann integrals one can find the exact formula for the volume of $B(0,r)$, but it is enough to deduce that $\mu(B(0,r))=k\,r^n$ for a certain constant $k$ (explicitly, $k=\pi^{n/2}/\Gamma(\frac n2+1)$, but that's not important here). 
Then, for any $\varepsilon>0$ small enough, $S^{n-1}\subset B(0,r+\varepsilon)\setminus B(0,r-\varepsilon)$, so 
\begin{align}
\mu(S^{n-1})&\leq\mu(B(0,r+\varepsilon)\setminus B(0,r-\varepsilon))
=\mu(B(0,r+\varepsilon))-\mu(B(0,r-\varepsilon))\\ \ \\
&=k\,[(r+\varepsilon)^n-(r-\varepsilon)^n]
=2k\,\varepsilon\,\sum_{j=0}^{n-1}(r+\varepsilon)^j(r-\varepsilon)^{n-1-j}\\ \ \\
&\leq4k\,n\,r^{n-1}\,\varepsilon\,.
\end{align}
As $\varepsilon$ was arbitrary, $\mu(S^{n-1})=0$. 
A: 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.
