Let (Phi: mathbb{R}^{d} ightarrow M subset mathbb{R}^{n}, d leqslant n), be a (C^{1})-diffeomorphism. (i) Show that (lambda_{M}:=Phileft(|operatorname{det}

Question:

Let \(\Phi: \mathbb{R}^{d} ightarrow M \subset \mathbb{R}^{n}, d \leqslant n\), be a \(C^{1}\)-diffeomorphism.

(i) Show that \(\lambda_{M}:=\Phi\left(|\operatorname{det} D \Phi| \lambda^{d}ight)\) is a measure on M. Find a formula for \(\int_{M} u d \lambda_{M}\).

(ii) Show that for a dilation \(\theta_{r}: \mathbb{R}^{n} ightarrow \mathbb{R}^{n}, x \mapsto r x, r>0\), we have

\[\int_{M} u(r \xi) r^{n} d \lambda^{n}(\xi)=\int_{\theta_{r}(M)} u(\xi) d \lambda_{M}(\xi)\]

(iii) Let \(M=\{\|x\|=r\}=\mathbb{S}_{r}^{n-1}\) be a sphere in \(\mathbb{R}^{n}\), so that \(d=n-1\). Show that for every integrable \(u \in \mathcal{L}^{1}\left(\mathbb{R}^{n}ight)\) and \(\sigma:=\lambda_{M}\)

\[
\begin{aligned}
\int u(x) \lambda^{n}(d x) & =\int_{(0, \infty)} \int_{\{\|x\|=r\}} u(x) \sigma(d x) \lambda^{1}(d r) \\
& =\int_{(0, \infty)} \int_{\{\|x\|=1\}} r^{n-1} u(r x) \sigma(d x) \lambda^{1}(d r)
\end{aligned}
\]

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