solution GV\] We say that $u$ is a weak solution of (\[eq 1.1\]), if $u\in
L^1(\Omega)$, $g(u)\in L^1(\Omega,\rho_{\partial\Omega}^\alpha dx)$ and $$\int_\Omega [u(-\Delta)^\alpha\xi+\epsilon g(u)\xi]dx=k\int_{\partial\Omega}\frac{\partial^\alpha\xi(x)}{\partial\vec{n}_x^\alpha}d\nu(x),\qquad \forall\xi\in \mathbb{X}_\alpha.$$

We notice that $\mathbb{X}_\alpha\supset C_0^2(\Omega)$ is the test functions space when we study semilinear fractional elliptic equations involving measures, which plays the same role as $C^{1,1}_0(\Omega)$ for dealing with second order elliptic equations with measures, see [@CV1; @CV2; @CV3; @CY]. Moreover, it follows from [@RS Proposition 1.1] that $\xi $ is $C^\alpha$ ($\alpha$-Hölder continuous) in ${\mathbb{R}}^N$ if $\xi\in \mathbb{X}_\alpha$.

Denote by $G_\alpha$ the Green kernel of $(-\Delta)^\alpha$ in $\Omega\times\Omega$ and by $\mathbb{G}_\alpha[\cdot]$ the Green operator defined as $$\mathbb{G}_\alpha[\frac{\partial^\alpha\nu}{\partial\vec{n}^\alpha}](x)=\lim_{t\to0^+}\int_{\partial\Omega} G_\alpha(x,y+t\vec{n}_y)t^{-\alpha}d\nu(y).$$

Now we are ready to state our first result for problem (\[eq 1.1\]).

\[teo 1\] Assume that $\epsilon=1$, $k>0$, $\nu\in\mathfrak{M}^b_{\partial\Omega}(\bar\Omega)$ and $g$ is a continuous nondecreasing function satisfying $g(0)\ge0$ and $$\label{g1}
 \int_1^\infty [g(s)-g(-s)]s^{-1-\frac{N+\alpha}{N-\alpha}}ds<+\infty.$$ Then

$(i)$ problem (\[eq 1.1\]) admits a unique weak solution $u_\nu$;

$(ii)$ the mapping $\nu\to u_\nu$ is increasing and $$\label{1.33}
-k\mathbb{G}_\alpha[\frac{\partial^\alpha\nu_-}{\partial\vec{n}^\alpha}](x)\le u_\nu(x)\le k\mathbb{G}_\alpha[\frac{\partial^\alpha\nu_+}{\partial\vec{n}^\alpha}](x),\qquad x\in\Omega,$$ where $\nu_+,\nu_-$ are the positive and negative decomposition of $\nu$ such that $\nu=\nu_+-\nu_-$;

$(iii)$ if we assume additionally that $g$ is $C^{\beta}$ locally in ${\mathbb{R}}$ with $\beta>0$, then $u_\nu$ is a classical solution of $$\label{1.1}
 \arraycolsep=1pt
\begin{array}{lll}
 (-\Delta)^\alpha   u+g(u)=0\quad 