Let $E\subset\mathbb{R}^n$ be such that for the boundary of $E$ holds $\partial E=\{(1+u(x))x \mid x\in \partial B_1(0)\}$, where $u:\partial B_1(0)\to (-1,\infty)$ is a function of class $C^1$, and $\partial B_1(0)$ denotes the boundary of the closed unit ball in $\mathbb{R}^n$. Denoting $|E|$ as the Lebesgue measure of $E$ and letting $B:=B_1(0)$ for short, why is $$|E|=\frac{1}{n}\int\limits_{\partial B}(1+u(x))^nd\mathcal{H}^{n-1}\; ?$$ Here $\mathcal{H}^{n-1}$ denotes the $(n-1)$-dimensional Hausdorff measure.
My try: Let $v:\mathbb{R}^n\setminus \{0\}\to (-1,\infty)$ given by $v(x)=u(\frac{x}{\|x\|}) $. Let $T:\mathbb{R}^n\to \mathbb{R}^n$ given by $T(x)=(1+v(x))x$. Then, I am not sure if $T$ bijective (I think not necessarily, perhabs one has to restrict the map to $B$), but applying the change-of-variation theorem $|E|=\int_{\mathbb{R}^n} (\chi_E\circ T)(x)|det DT(x)|dx=\int_{\mathbb{R}^n} \chi_{T^{-1}(E)}|det DT(x)|dx=\int_B|det DT(x)|dx$, where $\chi$ stands for the characteristic function and $T^{-1}(E)=B$. Now, $DT(x)=I(1+v(x))+x\otimes \nabla v(x)$, where $I$ denotes the unit matrix and $x\otimes \nabla v(x)$ the outer product of the two vectors. Using that the determinant of any outer product is zero, it is $|E|=\int_B(1+v(x))^ndx$, and I think theat $v$ is the function $u$ in the formula (which I want to obtain). Then, one has to use the Coarea formula https://en.wikipedia.org/wiki/Coarea_formula to go further. But here, I don't know how to apply the formula concretely. My questions are:
1) Is $T(B)=E?$ Or how to fix it, if not? 2) How to finish the proof using the Coarea formula?
I appreciate any help.
Using the first formula here with $f=\mathbf{1}_E$ and Fubini-Tonelli you get $$ |E| = \int_{\partial B_1} \left( \int_0^{1+u(x)} r^{n-1}\,dr \right) d\mathcal H^{n-1}(x) = \frac1n \int_{\partial B_1} \bigl(1+u(x)\bigr)^n \,d\mathcal H^{n-1}(x) . $$
ADDENDUM
I change the integration on $\partial B_r$ to an integration on $\partial B_1$: the substitution is $y=rx\in\partial B_r$ with $x\in\partial B_1$; this introduces a scaling factor of $r^{n-1}$.
Notice that $\mathbf{1}_E(rx)=[\![r\leq1+u(x)]\!]$, where $[\![\,\cdot\,]\!]$ denotes the Iverson bracket.
$$ \begin{split} |E| &= \int_0^\infty \int_{\partial B_r} \mathbf{1}_E(y) \,d\mathcal H^{n-1}(y) \,dr \\ &= \int_0^\infty \int_{\partial B_1} \mathbf{1}_E(rx) r^{n-1}\,d\mathcal H^{n-1}(x) \,dr \\ &= \int_0^\infty \int_{\partial B_1} [\![r\leq1+u(x)]\!] r^{n-1}\,d\mathcal H^{n-1}(x) \,dr \\ &= \int_{\partial B_1} \int_0^{1+u(x)}r^{n-1} \,dr \,d\mathcal H^{n-1}(x) \\ &= \frac1n \int_{\partial B_1} \bigl(1+u(x)\bigr)^n \,d\mathcal H^{n-1}(x) . \end{split} $$