How to check that the surface measure is the weak limit of $\delta^{-1}\mathcal{L}^n|_{B(0,1+\delta)\setminus B(0,1)}$?

Question

We denote the unit sphere $\{x\in\mathbb{R}^n:|x|=1\}$ by $S^{n-1}.$ If $x\in\mathbb{R}^n\setminus\{0\}$, the polar coordinates of $x$ are \begin{align*} r=|x|\in(0,\infty),\quad \gamma=\dfrac x{|x|}\in S^{n-1}. \end{align*} The map $\Phi(x)=(r,\gamma)$ is a continuous bijection from $\mathbb{R}^n\setminus\{0\}$ to $(0,\infty)\times S^{n-1}$ whose (continuous) inverse is $\Phi^{-1}(r,\gamma)=r\gamma=x.$

We denote by $m_*$ be the image induced by $\Phi(x)$ from Lebesgue measure on $\mathbb{R}^n$, that is, \begin{align*} m_*(E)=m(\Phi^{-1}(E)). \end{align*} Then, $m_*$ is a Borel measure on $(0,\infty)\times S^{n-1}$. Moreover, we define the measure $\rho=\rho_n$ on $(0,\infty)$ by \begin{align*} \rho(E)=\int_Er^{n-1}dr. \end{align*}

There is a unique Borel measure $\sigma =\sigma_{n-1}$ on $S^{n-1} $ such that $ m_*= \rho \otimes \sigma.$ If $f$ is Borel measurable on $ \mathbb{R} ^n $ and $ f \geq 0 $ or $ f\in L^1(m),$ then
$$ \int_{\mathbb{R}^n}f(x)dx=\int_{S^{n-1}}\left(\int_0^\infty f(r\gamma)r^{n-1}dr\right)d\sigma(\gamma). $$

We call $\sigma$ the surface measure on $S^{n-1}$.

In page 35 of the book "Fourier analysis and Hausdorff dimenison" by Mattila, the author said: one check easily that $\sigma$ is the weak limit of the measures $\delta^{-1}\mathcal{L}^n|_{B(0,1+\delta)\setminus B(0,1)}$ as $\delta\to 0$.

How to check it? I tried it but I didn't succeed. Any comments would be appreciated!

Answer 1

$\newcommand\de\delta\newcommand\si\sigma\newcommand\L{\mathcal L}\newcommand\R{\mathbb R}$Take any continuous function $f\colon\R^n\to\R$. Then $f$ is uniformly continuous on $B(0,2)$. Let $\mu_\de:=\frac1\de\,\L^n|_{B_(0,1+\de)\setminus B(0,1)}$. Then for $\de\in(0,1)$ \begin{align} \int d\mu_\de\,f&=\frac1\de\int_{\R^n}dx\,f(x)\,1(1\le|x|<1+\de) \\ &=\frac1\de\int_{S^{n-1}}d\si(u)\,\int_0^\infty dr\,r^{n-1}f(ru) \,1(1\le r<1+\de) \\ &=\frac1\de\int_{S^{n-1}}d\si(u)\,\int_0^\infty dr\,r^{n-1}[f(u)+o(1)] \,1(1\le r<1+\de) \\ &\to\int_{S^{n-1}}d\si(u)f(u) \end{align} as $\de\downarrow0$, where $o(1)$ stands for $f(ru)-f(u)$, which converges to $0$ as $r\to1$ uniformly in $u\in S^{n-1}$. $\quad\Box$