In this post, we have shown 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 $\delta\to 0$.
For a radial function $f\in L^1(\Bbb R^n)$, i.e., $f(x) = \psi(|x|)$ for some $\psi:[0,\infty)\to \Bbb C$, Mattila proves the following expression for the Fourier transform of $f$: $$\widehat f(x) = c(n)|x|^{-(n-2)/2} \int_0^\infty \psi(s)J_{(n-2)/2}(2\pi|x|s) s^{n/2}\, ds \tag{3.33}$$ where $J_m$'s are Bessel functions for $m > -1/2$.
Applying the formula $(3.33)$ to the characteristic function of the annulus $B(0,1+\delta)\setminus B(0,1)$ and letting $\delta\to 0$, we get $$\widehat{\sigma^{n-1}}(x) = c(n)|x|^{(2-n)/2} J_{(n-2)/2}(2\pi|x|) \tag{3.41}$$
Applying the formula $(3.33)$ to the characteristic function of the annulus $B(0,1+\delta)\setminus B(0,1)$, and observing that $$\chi_{B(0,1+\delta) \setminus B(0,1)}(x) = \chi_{\{s>0: 1 < s\le 1+\delta\}}(|x|) \tag{1}$$ we get $$\widehat{\chi_{B(0,1+\delta) \setminus B(0,1)}}(x) = c(n)|x|^{-(n-2)/2} \int_1^{1+\delta} J_{(n-2)/2}(2\pi|x|s) s^{n/2}\, ds \tag{2}$$ Dividing both sides by $\delta$, we have $$\frac{\widehat{\chi_{B(0,1+\delta) \setminus B(0,1)}}(x)}{\delta} = \frac{c(n)|x|^{-(n-2)/2} \int_1^{1+\delta} J_{(n-2)/2}(2\pi|x|s) s^{n/2}\, ds}{\delta} \tag{3}$$ Certainly, $$\frac{c(n)|x|^{-(n-2)/2} \int_1^{1+\delta} J_{(n-2)/2}(2\pi|x|s) s^{n/2}\, ds}{\delta} \xrightarrow{\delta\to 0} c(n)|x|^{-(n-2)/2}J_{(n-2)/2}(2\pi|x|)$$ However, to get $(3.41)$, we may have to show $$\lim_{\delta\to 0} \frac{\widehat{\chi_{B(0,1+\delta) \setminus B(0,1)}}(x)}{\delta} = \widehat{\sigma^{n-1}}(x)$$ How can we do that? Thank you!
