are continuous in $\bar O$ and satisfy $$(-\Delta)^\alpha  u_1+|u_1|^{p-1}u_1\ge0 \ {\rm in}\  O\quad {\rm and} \quad (-\Delta)^\alpha  u_2+|u_2|^{p-1}u_2\le0 \ {\rm in}\  O.$$ If $u_1\ge u_2$ a.e. in $O^c$, then $u_1\ge u_2$ in $O$.

By the Comparison Principle, we have the following result:

\[lm 2.3\] Assume that $f\in C^1({\mathbb{R}}^N)$ is a nonnegative function, $h$ is a continuous and nondecreasing function and $\mathcal{O}_1,\mathcal{O}_2$ are bounded $C^2$ domain such that $\mathcal{O}_1\subset \mathcal{O}_2$. Let $w_1$ and $w_2$ be the solutions of (\[2.1\]) with $\mu=f$ in $\mathcal{O}=\mathcal{O}_1$ and $\mu=f$ in $\mathcal{O}=\mathcal{O}_2$, respectively. Then $$w_1\le w_2\quad {\rm in}\quad {\mathbb{R}}^N.$$

[**Proof.**]{} Since $\mathcal{O}_1\subset \mathcal{O}_2$ and $f\ge0$, it follows by Lemma \[teo CP\] that $w_2\ge 0$ in $\mathcal{O}_2$. Suppose on the contrary that $$\min_{x\in {\mathbb{R}}^N}(w_2-w_1)(x)<0,$$ there would exist $x_0\in \mathcal{O}_1$ such that $$(w_2-w_1)(x_0)=\min_{x\in {\mathbb{R}}^N}(w_2-w_1)(x).$$ Then $$\arraycolsep=1pt
\begin{array}{lll}
(-\Delta)^\alpha(w_2-w_1)(x_0)=-\lim_{\epsilon\to0^+}\int_{{\mathbb{R}}^N\setminus B_\epsilon(x_0)}\frac{(w_2-w_1)(z)-(w_2-w_1)(x_0)}{|z-x_0|^{N+2\alpha}}dz<0
\end{array}$$ and $h(w_2(x_0))\leq h(w_1(x_0))$, which implies a contradiction since $w_1$ and $w_2$ satisfy $(-\Delta)^\alpha u(x_0)+h(u(x_0))=f(x_0).$ The proof is completed.$\Box$

\[lm 2—1\] Suppose that $p>0$, ${\mathcal{O}}$ is a bounded $C^2$ domain in ${\mathbb{R}}^N$, $g\in L^1({\mathcal{O}}^c,d\omega)$ is $C^2$ in $\{z\in{\mathcal{O}}^c, {\rm dist}(z,\partial{\mathcal{O}})\leq \delta \}$ with $\delta>0$. Then there exists a unique classical solution $u$ of $$\label{3.2.4}
\left\{ \arraycolsep=1pt
\begin{array}{lll}
 (-\Delta)^{\alpha} u(x)+|u|^{p-1}u(x)=0,\ \ \ \ &
x\in{\mathcal{O}},\\[2mm]
u(x)=g(x),\ &x\in{\mathcal{O}}^c.
\end{array}
\right.$$

[**Proof.**]{} 