$x\in\partial\Omega$, there exists a unique point $x_t\in\partial\Omega_t$ such that $|x-x_t|=\rho_{\partial\Omega_t}(x)$, where $t\in(0,\sigma_0)$ and $\rho_{\partial\Omega_t}(x)={\rm dist}(x,\partial\Omega_t)$. Then for any Borel set $E\subset \partial\Omega$, there exists unique $E_t\subset \partial\Omega_t$ such that $E_t=\{x_t: x\in E\}$.

In what follows, we always assume that $t\in[0,\sigma_0]$.

Denote by $\nu_t$ a Radon measure generated by $\nu$ as $$\nu_t(E_t)=\nu(E),$$ and then $\nu_t$ is a bounded Radon measure with support in $\partial\Omega_t$ and $$\nu_t(E)=\nu_t(E\cap\partial\Omega_t),\qquad \forall E\subset\bar\Omega\  {\rm Borel\ set}.$$ In the distribution sense, we have that $$\label{2.1.2}
\langle\nu_t,f\rangle=\int_{\partial\Omega_t}f(x)d\nu_t(x)=\int_{\partial\Omega}f(x+t\vec{n}_x)d\nu(x),\qquad \forall f\in C_0(\Omega).$$ Then we observe that $$\label{2.1.0}
\{x_t: x\in{\rm supp}(\nu)\}={\rm supp}(\nu_t) \quad{\rm and}\quad{\|\nu_t\|}_{\mathfrak{M}^b(\bar\Omega)}={\|\nu\|}_{\mathfrak{M}^b(\bar\Omega)}.$$

Now we are able to show an approximation of $\frac{\partial^\alpha\nu}{\partial \vec{n}^\alpha}$.

\[pr 2.1\] The sequence of Radon measures $\{t^{-\alpha}\nu_t\}_t$ converges to $\frac{\partial^\alpha\nu}{\partial \vec{n}^\alpha}$ as $t\to0^+$ in the following distribution sense: $$\lim_{t\to0^+}\int_{\partial\Omega_t}\xi(x)t^{-\alpha}d\nu_t(x)=\int_{\partial\Omega}\frac{\partial^\alpha\xi(x)}{\partial \vec{n}^\alpha_x}d\nu(x),\qquad \forall\xi\in \mathbb{X}_\alpha.$$

[**Proof.**]{} It follows from [@RS Proposition 1.1], that $\xi\in C^\alpha({\mathbb{R}}^N)$ if $\xi\in \mathbb{X}_\alpha$. This together with the fact that supp$(\xi)\subset\bar\Omega$, $\frac{\partial^\alpha\xi(x)}{\partial \vec{n}_x^\alpha}$ is well-defined for any $x\in\partial\Omega$ and for $x_t\in\partial\Omega_t$, implies that there exists a unique $x\in\partial\Omega$ such that $$x_t=x+t\vec{n}_x\quad{\rm and}\quad |x-x_t|=\rho_{\partial\Omega}(x_t),$$ then $$\xi(x+t\vec{n}_x)t^{-\alpha}=\frac{\xi(x+t\vec{n}_x)-\xi(x)}{t^\alpha},$$ which implies that $$\xi(\cdot+t\vec{n})t^{-\alpha}\to \frac{\partial^\alpha\xi(\cdot)}{\partial \vec{n}^\alpha}\quad {\rm as}\quad t\to0^+\quad {\rm in}\quad C(\bar\Omega).$$ Along with (\[2.1.2\]), we have that $$\arraycolsep=1pt
\begin{array}{lll}
|\int_{\partial\Omega_t}\xi(x)t^{-\alpha}d\nu_t(x)-\int_{\partial\Omega}\frac{\partial^\alpha\xi(x)}{\partial \vec{n}_x^\alpha}d\nu(x)|
\\[3mm]\phantom{-----}
=|\int_{\partial\Omega}\xi(x+t\vec{n}_x)t^{-\alpha}d\nu(x)-\int_{\partial\Omega}\frac{\partial^\alpha\xi(x)}{\partial \vec{n}_x^\alpha}d\nu(x)|
\\[3mm]\phantom{-----}
\le\int_{\partial\Omega}|\xi(x+t\vec{n}_x)t^{-\alpha}-\frac{\partial^\alpha\xi(x)}{\partial \vec{n}_x^\alpha}|d|\nu(x)|
\\[3mm]\phantom{-----}\to0\quad{\rm 