$\{u_{n,t}\}$ as $t\to0^+$, we introduce some auxiliary lemmas which are the key steps to obtain $\{g_n(u_{n,t})\}$ uniformly integrable with respect to $t$. For $\lambda>0$, let us set $$\label{Slambda}
  S_\lambda=\{x\in \Omega:\mathbb{G}_{\alpha}[t^{-\alpha}|\nu_t|](x)>\lambda\}\quad{\rm and}\quad m(\lambda)=\int_{S_\lambda} \rho_{\partial\Omega}^\alpha(x)dx.$$

\[lm 0\] For $\nu\in\mathfrak{M}^b_{\partial\Omega}(\bar\Omega)$ and any $t\in(0, \sigma_0)$, there exists $c_6>0$ independent of $t$ such that $$\label{annex -0}
m(\lambda)\le c_6\lambda^{-\frac{N}{N-\alpha}}.$$

[**Proof.**]{} For $\Lambda>0$ and $y\in\partial\Omega_t$ with $t\in(0, \sigma_0/2)$, we denote $$A_\Lambda(y)=\{x\in\Omega\setminus\{y\}: G_\alpha(x,y)>\Lambda\}\ \ {\rm
{and}}\quad m_\Lambda(y)=\int_{A_\Lambda(y)}\rho_{\partial\Omega}^\alpha(x) dx.$$ For any $(x,y)\in\Omega\times\partial\Omega_t$, $x\neq y$, it infers by (\[annex 01\]) that $$\begin{aligned}
A_\Lambda(y)\subset \left\{x\in\Omega\setminus\{y\}: \frac{c_5
t^\alpha}{|x-y|^{N-\alpha}}>\Lambda\right\}\subset B_r(y),\end{aligned}$$ where $r=(\frac{c_5t^\alpha}{\Lambda})^{\frac1{N-\alpha}}$. Thus, $\rho_{\partial\Omega}(x)\le R_0$ for some $R_0>0$ such that $\Omega\subset B_{R_0}(0)$ and $$\label{annex 1xhw}
m_\Lambda(y)\le R_0^\alpha\int_{B_r(y)}dx\le  c_7t^{\frac{N\alpha}{N-\alpha}}\Lambda^{-\frac{N}{N-\alpha}},$$ where $c_7>0$ independent of $t$.

For $y\in\partial\Omega_t$, we have that $$\begin{aligned}
\int_{S_\lambda} G_\alpha(x,y)\rho_{\partial\Omega}^\alpha(x)dx\le
\int_{A_\Lambda(y)}G_\alpha(x,y)\rho_{\partial\Omega}^\alpha(x)dx+\Lambda\int_{S_\lambda}
\rho_{\partial\Omega}^\alpha(x)dx.\end{aligned}$$ By integration by parts, we obtain $$\begin{aligned}
\int_{A_\Lambda(y)}G_\alpha(x,y)\rho_{\partial\Omega}^\alpha(x)dx=\Lambda m_\Lambda(y)+ \int_\Lambda^\infty m_s(y)ds\le c_8t^{\frac{N\alpha}{N-\alpha}}\Lambda^{1-\frac{N}{N-\alpha}},\end{aligned}$$ where $c_8>0$ independent of $t$. Thus, $$\begin{aligned}
\int_{S_\lambda} G_\alpha(x,y)\rho_{\partial\Omega}^\alpha(x)dx\le
c_8t^{\frac{N\alpha}{N-\alpha}}\Lambda^{1-\frac{N}{N-\alpha}}+\Lambda \int_{S_\lambda}
\rho_{\partial\Omega}^\alpha(x)dx.\end{aligned}$$ Choose $\Lambda= t^\alpha(\int_{S_\lambda} \rho_{\partial\Omega}^\alpha(x)dx)^{-\frac{N-\alpha}{N}}$ and then $$\begin{aligned}
\int_{S_\lambda} G_\alpha(x,y)\rho_{\partial\Omega}^\alpha(x)dx\le c_9t^\alpha
(\int_{S_\lambda} \rho_{\partial\Omega}^\alpha(x)dx)^{\frac{\alpha}{N}},\end{aligned}$$ where $c_9=c_8+1$. Therefore, $$\begin{aligned}
 \int_{S_\lambda}
\mathbb{G}_\alpha [t^{-\alpha}|\nu_t|](x)\rho_{\partial\Omega}^\alpha(x)dx&=&\int_\Omega\int_{S_\lambda}
G_\alpha(x,y)\rho_{\partial\Omega}^\alpha(x)dxt^{-\alpha} d|\nu_t(y)|
\\&\le &c_9\int_\Omega d|\nu_t(y)|(\int_{S_\lambda} \rho_{\partial\Omega}^\alpha(x)dx)^{\frac{\alpha}{N}}
\\&\le& c_9\|\nu\|_{\mathfrak{M}^b(\bar\Omega)} (\int_{S_\lambda} \rho_{\partial\Omega}^\alpha(x)dx)^{\frac{\alpha}{N}}.\end{aligned}$$ As a consequence, $$\begin{aligned}
\lambda m(\lambda)\le c_9\|\nu\|_{\mathfrak{M}^b(\bar\Omega)}m(\lambda)^{\frac{\alpha}{N}},\end{aligned}$$ which implies (\[annex -0\]). This ends the proof. $\Box$

From Lemma \[lm 0\], it implies that $$\label{24-08-0}
{\|\mathbb{G}_\alpha[t^{-\alpha}|\nu_t|]\|}_{M^{\frac{N}{N-\alpha}}(\Omega,\rho_{\partial\Omega}^\alpha dx)}\le c_9\|\nu\|_{\mathfrak{M}^b(\bar\Omega)},$$ 