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

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!

• I think you can prove it by estimating the integral of a continuous function over $B(0, 1 + \delta) \setminus B(0, 1)$ by its values on $\partial B(0, 1)$, using the formula for the volume of a ball to bound the error. Commented Jun 2 at 4:11
• @Nate River Would you like to give the details? Many Thanks! Commented Jun 2 at 13:38
• Hm, I am thinking of how to use my method to compute it without resorting to spherical coordinates (and thus using circular reasoning), and coming up with nothing, except maybe an ugly partition of unity thing. Commented Jun 2 at 13:40

$$\newcommand\de\delta\newcommand\si\sigma\newcommand\L{\mathcal L}\newcommand\R{\mathbb R}$$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 $$u\in S^{n-1}$$. $$\quad\Box$$