$\newcommand\de\delta\newcommand\si\sigma\newcommand\L{\mathcal L}\newcommand\R{\mathbb R}$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\,f(ru) \,1(1\le r<1+\de) \\ &=\frac1\de\int_{S^{n-1}}d\si(u)\,\int_0^\infty dr\,[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 an expression in terms of $r$ and $u$ that converges to $0$ as $r\to1$ uniformly in $u\in S^{n-1}$. $\quad\Box$

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