Furthermore, the weak solution of (\[1.1.1\]) is approached by the classical solutions of (\[1.1.1\]) replacing $\mu$ by a sequence of regular functions $\{\mu_n\}$, which converge to $\mu$ in the distribution sense. Then this subject has been vastly expanded in recent works, see the papers of Marcus and Véron [@MV1; @MV2; @MV3; @MV4], Bidaut-Véron and Vivier [@BV] and reference therein.

A very challenging question consists in studying the analogue elliptic problem involving fractional Laplacian defined by $$(-\Delta)^\alpha  u(x)=\lim_{\varepsilon\to0^+} (-\Delta)_\varepsilon^\alpha u(x),$$ where $$(-\Delta)_\varepsilon^\alpha  u(x)=-\int_{{\mathbb{R}}^N\setminus B_\varepsilon(x)}\frac{ u(z)-
u(x)}{|z-x|^{N+2\alpha}} dz$$ for $\varepsilon>0$. The main difficulty comes from how to define the boundary type data. Given a Radon measure $\mu$ defined in $\partial\Omega$, it is ill-posed that $$\arraycolsep=1pt
\begin{array}{lll}
(-\Delta)^\alpha  u+g(u)=0\quad &{\rm in}\quad \Omega,
\\[2mm]\phantom{------\ }
u=\mu\quad&{\rm on}\quad \partial\Omega,
\\[2mm]\phantom{------\ }
u=0\quad&{\rm in}\quad \bar\Omega^c.
\end{array}$$ Indeed, let $\{\mu_n\}$ be a sequence of regular functions defined in $\partial\Omega$ converging to the measure $\mu$ and a surprising result is that there is just zero solution for $$\arraycolsep=1pt
\begin{array}{lll}
(-\Delta)^\alpha  u+g(u)=0\quad &{\rm in}\quad \Omega,
\\[2mm]\phantom{------\ }
u=\mu_n\quad&{\rm on}\quad \partial\Omega,
\\[2mm]\phantom{------\ }
u=0\quad&{\rm in}\quad \bar\Omega^c,
\end{array}$$ which is in sharp contrast with Laplacian case, where (\[1.1.1\]) replacing $\mu$ by $\mu_n$ admits a unique nontrivial solution. On the other hand, 