g(s)ds.$$ Notice that the above quantity on the right-hand side tends to $0$ when $\lambda\to\infty$. The conclusion follows: for any $\epsilon>0$ there exists $\lambda>0$ such that $$\frac{c_{24}}{\frac{N+\alpha}{N-\alpha}+1}\int_{\frac{\lambda}{k}}^\infty s^{-1-\frac{N+\alpha}{N-\alpha}}\tilde g(s)ds\leq \frac{\epsilon}{2}.$$ For $\lambda$ fixed, there exists $\delta>0$ such that $$\int_E\rho_{\partial\Omega}^\alpha(x) dx\leq \delta\Longrightarrow \tilde g\left(\frac{\lambda}{k}\right)\int_E\rho_{\partial\Omega}^\alpha(x) dx\leq\frac{\epsilon}{2},$$ which implies that $\{g_n\circ u_n\}$ is uniformly integrable in $L^1(\Omega,\rho_{\partial\Omega}^\alpha dx)$. Then $g_n\circ u_n\to g\circ u_\nu$ in $L^1(\Omega,\rho_{\partial\Omega}^\alpha dx)$ by Vitali convergence theorem.

Passing to the limit as $n\to +\infty$ in the identity (\[2.1.1000\]), it implies that $$\int_\Omega [u_\nu(-\Delta)^\alpha\xi+g(u_\nu)\xi]dx=\int_{\partial\Omega}\frac{\partial^\alpha\xi(x)}{\partial\vec{n}^\alpha_x}d\nu(x),\quad \forall\xi\in\mathbb{ X}_\alpha.$$ Then $u_\nu$ is a weak solution of (\[eq 1.1\]). Moreover, it follows by the fact $$-k\mathbb{G}_\alpha[\frac{\partial^\alpha \nu_-}{\partial \vec{n}^\alpha}]\le u_n\le  k\mathbb{G}_\alpha[\frac{\partial^\alpha \nu_+}{\partial \vec{n}^\alpha}]\quad {\rm in}\ \Omega.$$ which, together with $u_\nu=\lim_{n\to+\infty} u_n$, implies (\[1.33\]).

The arguments including $u_n=0$ in $\Omega^c\setminus{\rm supp}(\nu)$, uniqueness and regularity follow the proof of Proposition \[pr 2.01\]. $\Box$

The proof of the existence of weak solution is divided into two steps: the first step is to get weak solution $u_n$ to (\[eq 1.1\]) with truncated nonlinearity $g_n$ and then to prove the limit of $\{u_n\}$ as $n\to\infty$ is our desired weak solution. This is due to the estimate in Lemma \[lm 0\] where we only could get exponent $\frac{N}{N-\alpha}$ and in the second 