Measure-preserving map of sphere

Question

I'm reading Stein and Shakarchi's Real Analysis, trying to do some exercises and puzzled about ex4 of Chapter 6. Let $r$ be a rotation of $\mathbb{R}^d$. As I know it's proved that the mapping $\pi : x \mapsto r(x)$ preserves Lebesgue measure. The problem is to show that it induces a measure-preserving map of the sphere $S^{d-1}$. I tried to prove it like this: It can be shown that $S^{d-1} \subseteq ker(r)$ and there is a natural projection $\varphi: \mathbb{R}^d-\{0\} \rightarrow S^{d-1}\, ,\, x \mapsto \frac{x}{|x|}$. How can I show that the mapping $\varphi \circ\pi$ preserves measure? From this problem, $d\mu(r(x))=d\mu(x)$. Is this true because $d\mu (r(\frac{x}{|x|})) = d\mu(\frac{1}{|x|} x)$? Is is sufficient to prove this problem?

Answer 1

What a coincidence! I just have finished this exercise today!

You need to refer to section 5 in Chapter 6 to find the definition of "measure-preserving transformation". In this exercise, our measure space is $(X,\mathcal M,\mu)=(S^{d-1},\mathcal M,\sigma)$. Given any set $E\subset S^{d-1}$ we let $\widetilde{E}=\{x\in\mathbb R^d: x/|x|\in E,0<|x|<1\}$. We shall say $E\in\mathcal M$ exactly when $\widetilde{E}$ is a Lebesgue measurable subset of $\mathbb R^d$, and define $\sigma(E)=d\cdot m(\widetilde{E})$.

Let $E\in\mathcal M$, then by definition, $\sigma(E)=d\cdot m(\widetilde{E})$ and $\sigma(rE)=d\cdot m(\widetilde{rE})$. But $r\widetilde{E}=\widetilde{rE}$ since \begin{align*} x\in \widetilde{rE}&\iff x=\rho\theta\ \ \text{for some }\rho\leq1,\theta\in rE\\&\iff x=\rho r(\alpha) \ \text{for some }\alpha\in E\\&\iff x=r(\rho\alpha)\\&\iff x\in r\widetilde{E}. \end{align*} Thus $\sigma(rE)=d\cdot m(r\widetilde{E})=d\cdot m(\widetilde{E})=\sigma(E)$, so $r$ preserves measures on $S^{d-1}$.