We claim that $v_{n}\le v_{n+1}$ a.e. in ${\mathbb{R}}^N$.*]{} In fact, let $\tilde v_{n+1}$ be the solution of (\[2.1\]) with $\mathcal{O}=\mathcal{O}_{n+1}$ and $\mu=\nu_{n}$. By Proposition \[pr 2.1\], $$\label{3.1}
\tilde v_{n+1}\le v_{n+1}\quad{\rm a.e.\ in}\ {\mathbb{R}}^N.$$

Choosing a sequence nonnegative functions $\{f_{n,m}\}_{m\in{\mathbb{N}}}\subset C^1_0(\mathcal{O}_{n})$ such that $f_{n,m}\rightharpoonup \nu_{n}$ as $m\to\infty$ in the distribution sense, we make zero extension of $f_{n,m}$ into $C_0^1(\mathcal{O}_{n+1})$ and denote the extension by $\tilde f_{n,m}$. Let $v_{n,m}$ and $\tilde v_{n+1,m}$ be solutions of (\[2.1\]) with $\mu=f_{n,m}$, $\mathcal{O}=\mathcal{O}_n$ and $\mu=\tilde f_{n,m}$, $\mathcal{O}=\mathcal{O}_{n+1}$, respectively. Lemma \[lm 2.3\] implies that $$v_{n,m}\le \tilde v_{n+1,m}\quad{\rm in}\ {\mathbb{R}}^N.$$ Together with the facts that $v_{n,m}\to v_n$ a.e. in ${\mathbb{R}}^N$ and $\tilde v_{n+1,m}\to \tilde v_{n+1}$ a.e. in ${\mathbb{R}}^N$ as $m\to\infty$, we obtain that $$\label{3.2}
v_{n}\le \tilde v_{n+1}\quad{\rm a.e.\ in}\ \ {\mathbb{R}}^N.$$ It follows by (\[3.1\]) and (\[3.2\]) that for any $n\in{\mathbb{N}}$, $$\label{3.3}
 v_{n}\le v_{n+1}\quad {\rm a.e.\ in}\ \ {\mathbb{R}}^N.$$

[*Step 2. Uniform bounds of $\{v_{n}\}$.*]{} We deduce by (\[2.2\]) that $$\label{3.4}
0\le v_{n}\le \mathbb{G}_{\mathcal{O}_{n}}[\nu_{n-1}]\quad {\rm a.e.\ in}\ \ {\mathbb{R}}^N.$$ Observing that for any $n\in{\mathbb{N}}$, $$G_{\mathcal{O}_{n}}(x,y)\le G_{\Omega}(x,y)\le  G_{{\mathbb{R}}^N}(x,y),\quad {\rm for\ any}\ x,y\in{\mathbb{R}}^N,x\not=y$$ and $\nu_{n-1}\le \nu$, we have that $$\begin{aligned}
 \mathbb{G}_{\mathcal{O}_{n}}[\nu_{n-1}](x)
   \le \int_{\Omega}G_{\Omega}(x,y)d\nu(y) =\mathbb{G}_{\Omega}[\nu](x)\le \mathbb{G}_{{\mathbb{R}}^N}[\nu](x),\quad x\in{\mathbb{R}}^N,\end{aligned}$$ where we make zero extension of $\nu$ such that $\nu\in\mathfrak{M}^b({\mathbb{R}}^N)$. Therefore, by (\[3.4\]) we 