measure of boundary of unit sphere in $\mathbb{R}^d$ is zero

Question

I've seen some solutions to this problem but I'm wondering what is incorrect about an argument like this:

$S = \{x \in \mathbb{R}^d: |x| = 1\}$, then $\delta S = \{x \in \mathbb{R}^d: |\delta x| = 1\}$, and so

\begin{align*} \{ x \in \mathbb{R}^d: |\delta ||x| = 1\} & = \{ x \in \mathbb{R}^d: |x| = 1 / |\delta | \} \end{align*}

As we take $\delta $ large, then $\delta S$ becomes the set in which $|x| = 0$, i.e. the origin in $\mathbb{R}^d$ , which has zero measure, and since: $m(\delta S) = \delta^d m(S)$ where $m$ is the Lebesgue measure, it follows that $m(S) = 0$

Answer 1

$\delta S$ is for any $\delta>0$ a sphere with radius $\delta>0$. No matter how big $\delta >0$ is choosen, it does not become the set $\{0\}$.

Answer 2

The argument you give is wrong, as already noted. One can show that $m(S)=0$ by considering the sets $\delta S$.

For example, say $(\delta_j)\subset(1,2)$ and $\delta_j\ne\delta_k$ for $j\ne k$. Let $$A=\bigcup_j\delta_j S.$$ Then $$m(\delta_j S)=\delta_j^n m(S)\ge m(S);$$since the $\delta_j S$ are disjoint, if $m(S)>0$ this would show that $m(A)=\infty$, which is false since $A$ is bounded.

(Here of course I'm taking $\delta S=\{\delta x:x\in S\}$ as usual...)