obtain that for any $n\in{\mathbb{N}}$, $$\label{3.5}
v_n\le \mathbb{G}_{\Omega}[\nu]\le \mathbb{G}_{{\mathbb{R}}^N}[\nu]\quad {\rm a.e.\ in}\ \ {\mathbb{R}}^N.$$

[*Step 3. Existence of weak solution.*]{} By (\[3.3\]) and (\[3.5\]), we see that the limit of $\{v_n\}$ exists, denoted it by $u_\nu$. Hence, $$\label{2.1.1}
0\le u_\nu\le \mathbb{G}_{\Omega}[\nu]\le \mathbb{G}_{{\mathbb{R}}^N}[\nu]\quad {\rm a.e.\ in}\ \ {\mathbb{R}}^N.$$ It follows by Lemma \[lm 2.2\] that $\mathbb{G}_{{\mathbb{R}}^N}[\nu]\in L^1({\mathbb{R}}^N, d\omega)$ and then $u_\nu\in L^1({\mathbb{R}}^N, d\omega)$. Thus, $v_n\to u_\nu$ in $L^1({\mathbb{R}}^N, d\omega)$ as $n\to\infty$. Moreover,

-   1.  $\{h(v_n)\}_{n\in{\mathbb{N}}}$ is an increasing sequence of functions and $h(v_n)\to h(u_\nu)$ a.e. in ${\mathbb{R}}^N$;

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    1.  it implies by $\Omega^c\subset\mathcal{O}_n^c$ and $v_n=0$ in $\mathcal{O}_n^c$ that $u_\nu=0$ in $\Omega^c$.

For $\xi\in \mathbb{X}_{\Omega}$, there exists $N_0>0$ such that for any $n\ge N_0$, $${\rm supp}(\xi)\subset \bar\mathcal{O}_n,$$ which implies that $\xi\in \mathbb{X}_{\mathcal{O}_n}$ and then $$\label{3.6}
\int_{\Omega} [v_n(-\Delta)^\alpha\xi+h(v_n)\xi]dx=\int_{\Omega} \xi d\nu_{n}.$$ By Lemma \[lm 2.1\], $$|(-\Delta)^\alpha\xi(x)|\le \frac{c_{9}{\|\xi\|}_{L^\infty(\Omega)}}{1+|x|^{N+2\alpha}},\quad x\in \Omega.$$ Thus, $$\label{3.7}
 \lim_{n\to\infty}\int_{\Omega} v_n(x)(-\Delta)^\alpha\xi(x) dx=\int_{\Omega} u_\nu(x)(-\Delta)^\alpha\xi(x) dx.$$

By (\[2.1.1\]) and increasing monotonicity of $h$, it follows $h(u_\nu)\le h(\mathbb{G}_{{\mathbb{R}}^N}[\nu])$ a.e. in ${\mathbb{R}}^N$. By Lemma \[lm 2.2\], we find that $h(u_\nu)\in L^1({\mathbb{R}}^N,d\omega)$. As a result, for any $n\ge N_0$, $$h(v_n)\to h(u_\nu)\quad{\rm in}\quad L^1({\mathbb{R}}^N,d\omega).$$ Consequently, $$\label{3.8}
 \lim_{n\to\infty}\int_{\Omega} h(v_n)\xi(x) dx=\int_{\Omega} h(u_\nu)\xi(x) dx.$$ It is obvious that $$\label{3.9}
 \lim_{n\to\infty}\int_{\Omega} \xi(x)d\nu_{n}=\int_{\Omega} \xi(x) d\nu.$$ Combining (\[3.7\]), (\[3.8\]) with (\[3.9\]) and taking 