$\lim_{\delta\to 0} \delta^{-1}\int \phi\, dL^n\vert_{B(0, 1+\delta)\setminus B(0,1)} = \int \phi \, d\sigma^{n-1}$ for every $\phi\in C_0(\Bbb R^n)$ Pertti Mattila's Fourier Analysis and Hausdorff Dimension makes the following claim:

One checks easily that $\sigma^{n-1}$ is the weak limit of the measures $\delta^{-1} \mathcal L^n\vert_{B(0, 1+\delta)\setminus B(0,1)}$ as $\delta\to 0$.

In other words, we want to show that for every $\phi\in C_0(\Bbb R^n)$,
$$\lim_{\delta\to 0} \delta^{-1}\int \phi\, d\mathcal L^n\vert_{B(0, 1+\delta)\setminus B(0,1)} = \int \phi \, d\sigma^{n-1}$$
How does one do this? Thanks for any help!

Note $1$: The surface measure $\sigma^{n-1}$ on the sphere $S^{n-1}$ is defined in Folland text, in the following way:

Consider the homeomorphism $\Phi:\Bbb R^n\setminus\{0\} \to (0,\infty)\times S^{n-1}$ given by $\Phi(x) = (r,x')$ where $r = |x|$ and $x' = \frac{x}{|x|}$. Let $m_\ast$ be the Borel measure induced on $(0,\infty)\times S^{n-1}$ by the Lebesgue measure on $\Bbb R^n$, via $\Phi$, i.e., $$m_\ast(E) := m(\Phi^{-1}(E))$$
We define the measure $\rho = \rho_n$ on $(0,\infty)$ by $$\rho(E) := \int_E r^{n-1}\, dr$$
There exists a unique Borel measure $\sigma = \sigma_{n-1}$ on $S^{n-1}$ such that $$m_\ast = \rho \times\sigma$$
If $f$ is Borel measurable on $\Bbb R^n$ and $f\ge 0$, or $f\in L^1(m)$, then $$\int_{\Bbb R^n} f(x)\, dx = \int_0^\infty \int_{S^{n-1}} f(rx') r^{n-1}\, d\sigma(x') \, dr$$


Note $2$: $\mathcal L^n$ and $m$ both denote the $n$-dimensional Lebesgue measure. I did not bother using same notation everywhere, since the first half of the post is borrowed from Folland's book, while the second half is from Mattila's book.

 A: Suppose $\phi\in\mathcal{C}_0(\mathbb{R}^n)$. Then
$$\begin{align}
\frac{1}{\delta}\int_{B(0;1+\delta)\setminus B(0;1)} \phi(x)\,dx&=\int_{\mathbb{S}_{n-1}}\frac{1}{\delta}\Big(\int^{1+\delta}_1 r^{n-1}\phi(r\mathbf{u})\,dr\Big)\sigma_{n-1}(d\mathbf{u})\\
&=\int_{\mathbb{S}_{n-1}}\frac{1}{\delta}\Big(\int^{1+\delta}_1 r^{n-1}\big(\phi(r\mathbf{u})-\phi(\mathbf{u})\big)\,dr\Big)\sigma_{n-1}(d\mathbf{u})+\\
&\qquad \qquad +\frac{(1+\delta)^n-1}{n\delta}\int_{\mathbb{S}_{n-1}}\phi(\mathbf{u})\,\sigma_{n-1}(d\mathbf{u})
\end{align}$$
Since $\phi\in\mathcal{C}_0$, $\phi$ is uniformly continuous. Hence, the conclusion follows by passsing to the limit $\delta\rightarrow0$ by standard $\varepsilon-\delta$ arguments. For example, given $\varepsilon>0$ there is $\delta_\varepsilon>0$ such that $|x-y|<\delta_\varepsilon$ implies $|\phi(x)-\phi(y)|<\varepsilon$. Thus for $\delta<\delta_\varepsilon$,
$|\phi(r\mathbf{u})-\phi(\mathbf{u})|<\varepsilon$ for all $1\leq r<1+\delta$ and $u\in\mathbb{S}_{n-1}$; hence
$$
\left|\int_{\mathbb{S}_{n-1}}\frac{1}{\delta}\Big(\int^{1+\delta}_1 r^{n-1}\big(\phi(r\mathbf{u})-\phi(\mathbf{u})\big)\,dr\Big)\sigma_{n-1}(d\mathbf{u})\right|\leq
\frac{(1+\delta)^n-1}{n\delta}\varepsilon\sigma_{n-1}(\mathbb{S}_{n-1})$$
All this shows that
$$\lim_{\delta\rightarrow0}\int_{\mathbb{S}_{n-1}}\frac{1}{\delta}\Big(\int^{1+\delta}_1 r^{n-1}\big(\phi(r\mathbf{u})-\phi(\mathbf{u})\big)\,dr\Big)\sigma_{n-1}(d\mathbf{u})=0$$
