it is also not proper to pose $$\arraycolsep=1pt
\begin{array}{lll}
(-\Delta)^\alpha  u+g(u)=0\quad &{\rm in}\quad \Omega,
\\[2mm]\phantom{------\ }
u=\mu\quad&{\rm in}\quad \Omega^c
\end{array}$$ with $\mu$ being a Radon measure in $\Omega^c$ concentrated on $\partial\Omega$. In fact, letting functions $\{\mu_n\}\subset C^1_0(\Omega^c)$ converging to $\mu$, the solution $u_n$ of $$\arraycolsep=1pt
\begin{array}{lll}
(-\Delta)^\alpha  u+g(u)=0\quad &{\rm in}\quad \Omega,
\\[2mm]\phantom{------\ }
u=\mu_n\quad&{\rm in}\quad \Omega^c,
\end{array}$$ is equivalent to the solution of $$\arraycolsep=1pt
\begin{array}{lll}
(-\Delta)^\alpha  u+g(u)=G_{\mu_n}\quad &{\rm in}\quad \Omega,
\\[2mm]\phantom{------\ }
u=0\quad&{\rm in}\quad \Omega^c,
\end{array}$$ where $$G_{\mu_n}(x)=\int_{\Omega^c}\frac{\mu_n(y) }{|x-y|^{N+2\alpha}}dy,\qquad  x\in\Omega,$$ see [@CFQ]. It could be seen that $$\int_\Omega [u_n(-\Delta)^\alpha\xi+ g(u_n)\xi]dx=\int_{\Omega}G_{\mu_n}\xi dx,\quad \forall\xi\in C^2_0(\Omega),$$ Then the limit of $\{u_n\}$ as $n\to\infty$ wouldn’t be a weak solution as we desired, similar to (\[1.1.1.0\]).

Therefore, a totally different point of view has to be found to propose the fractional elliptic problem involving measure concentrated on boundary. Our idea is inspired by the study of elliptic equations with fractional Laplacian and Radon measure inside of $\Omega$ in [@CV1], where the authors considered the equations $$\label{1.22}
 \arraycolsep=1pt
\begin{array}{lll}
 (-\Delta)^\alpha  u+h(u)=\nu\quad & \rm{in}\quad\Omega,\\[2mm]
 \phantom{   (-\Delta)^\alpha  +h(u)}
u=0\quad & \rm{in}\quad \Omega^c
\end{array}$$ for $\nu\in\mathfrak{M}(\Omega,\rho_{\partial\Omega}^\beta)$ with $ \beta\in[0,\alpha]$ the space of Radon measure $\nu$ in $\Omega$ satisfying $$\int_\Omega\rho_{\partial\Omega}^\beta(x) d|\nu(x)|<+\infty.$$ A function $u$ is said to be a weak 