$$\begin{aligned}
\int_E G_{{\mathbb{R}}^N}(x,y) d\omega(x)\le
\int_{A_\lambda(y)} G_{{\mathbb{R}}^N}(x,y) d\omega(x)+\lambda\int_Ed\omega.\end{aligned}$$ By (\[gra 0003wx2\]), we have that $$\begin{aligned}
\int_{A_\lambda(y)}  G_{{\mathbb{R}}^N}(x,y)d\omega(x)=\lambda m_\lambda(y)+ \int_{\lambda}^\infty m_s(y)ds\le c_7\lambda^{1-p_\alpha^*},\end{aligned}$$ for some $c_7>0$, then it results that $$\begin{aligned}
\int_E  G_{{\mathbb{R}}^N}(x,y)d\omega(x)\le c_7 \lambda^{1-p_\alpha^*}+\lambda \int_Ed\omega.\end{aligned}$$ Choosing $\lambda=(\int_Ed\omega)^{-\frac1{p_\alpha^*}}$, we obtain $$\begin{aligned}
\int_E  G_{{\mathbb{R}}^N}(x,y)d\omega(x)\le (c_7+1)\left(\int_E d\omega\right)^{\frac{p_\alpha^*-1}{p_\alpha^*}},\quad \forall y\in {\mathbb{R}}^N.\end{aligned}$$ Therefore, $$\label{none}\begin{array}{lll}\displaystyle
\displaystyle \int_E \mathbb{G}_{{\mathbb{R}}^N}[|\nu|](x)d\omega(x)=\int_{{\mathbb{R}}^N}\int_E
 G_{{\mathbb{R}}^N}(x,y)  d\omega(x) d|\nu(y)|
\\[3mm]\phantom{---------}
\displaystyle\le (c_7+1)\int_{{\mathbb{R}}^N} d|\nu(y)|\left(\int_E d\omega\right)^{\frac{p^*_\alpha-1}{p^*_\alpha}}
\\[3mm]\phantom{---------}
\displaystyle\le (c_7+1)\|\nu\|_{\mathfrak{M}^b({\mathbb{R}}^N)} \left(\int_E
d\omega\right)^{\frac{p^*_\alpha-1}{p^*_\alpha}}.
\end{array}$$ As a consequence, $$\begin{aligned}
\| \mathbb{G}_{{\mathbb{R}}^N}[|\nu|]\|_{M^{p^*_\alpha}({\mathbb{R}}^N,d\omega)}\le
c_5\|\nu\|_{\mathfrak{M}^b({\mathbb{R}}^N)},\end{aligned}$$ which ends the proof.$\Box$

Now we use Marcinkiewicz type estimate to prove the following lemma, which is the key-stone in the proof of Theorem \[teo 1\].

\[lm 2.2\] Assume that $\nu\in \mathfrak{M}_+^b({\mathbb{R}}^N)$ and $h:{\mathbb{R}}_+\to{\mathbb{R}}_+$ is a continuous nondecreasing function satisfying (\[1.4\]). Then $$\mathbb{G}_{{\mathbb{R}}^N}[\nu],\ h(\mathbb{G}_{{\mathbb{R}}^N}[\nu]) \in L^1({\mathbb{R}}^N, d\omega).$$

[**Proof.**]{} On the one hand, using Fubini’s lemma, we have that $$\arraycolsep=1pt
\begin{array}{lll}
  {\|\mathbb{G}_{{\mathbb{R}}^N}[\nu]\|}_{L^1({\mathbb{R}}^N,d\omega)}=c_{N,\alpha}\int_{{\mathbb{R}}^N}\int_{{\mathbb{R}}^N}\frac1{1+|x|^{N+2\alpha}}\frac{1}{|x-y|^{N-2\alpha}}d\nu(y)dx \\[3mm]\phantom{--------\ }
= c_{N,\alpha}\int_{{\mathbb{R}}^N}\int_{{\mathbb{R}}^N}\frac1{1+|x|^{N+2\alpha}}\frac{1}{|x-y|^{N-2\alpha}}dx d\nu(y)
 \\[3mm]\phantom{--------\ }
\le c_{N,\alpha}\int_{{\mathbb{R}}^N}[\int_{B_1(y)}\frac{1}{|x-y|^{N-2\alpha}}dx+\int_{B_1^c(y)}\frac1{1+|x|^{N+2\alpha}}dx ]d\nu(y)
 \\[3mm]\phantom{--------\ }
 <+\infty,
\end{array}$$ that is, $\mathbb{G}_{{\mathbb{R}}^N}[\nu]\in L^1({\mathbb{R}}^N, d\omega).$

On the other hand, let $S_\lambda=\{x\in B_R(0):\mathbb{G}_{{\mathbb{R}}^N}[\nu](x)>\lambda\}$ and $g(\lambda)=\int_{S_\lambda} d\omega$, where $\lambda\ge1$. We observe that $$\label{2.4}
\displaystyle\begin{array}{lll}
\displaystyle
\int_{{\mathbb{R}}^N}h(\mathbb{G}_{{\mathbb{R}}^N}[\nu])d\omega=\int_{S^c_{\lambda}}h(\mathbb{G}_{{\mathbb{R}}^N}[\nu])d\omega+\int_{S_{\lambda}}
h(\mathbb{G}_{{\mathbb{R}}^N}[\nu])d\omega
\\[4mm]\phantom{-\int_{E}|g(u_{n_k})|\rho^{\beta}dx}
\leq h(\lambda)\int_{{\mathbb{R}}^N}d\omega+\int_{S_{\lambda}}h(\mathbb{G}_{{\mathbb{R}}^N}[\nu])d\omega
\\[4mm]\phantom{-\int_{E}|g(u_{n_k})|\rho^{\beta}dx}
= h(\lambda)\int_{{\mathbb{R}}^N} d\omega+h(\lambda)g(\lambda)+\int_\lambda^{\infty}g(s)dh(s).
\end{array}$$ Since $$\int_{\lambda}^\infty g(s) dh(s)=\lim_{T\to\infty}\int_{\lambda}^T g(s)dh(s)$$ and $\mathbb{G}_{{\mathbb{R}}^N}[\nu]\in M^{p^*_{\alpha}}({\mathbb{R}}^N,d\omega)$, it derives from Proposition \[general\] and Proposition \[pr 1\] with $q=1$, $\kappa=p^*_\alpha$, $E=S_\lambda$ 