of $tr^{-\alpha}$ we have $$n_{r,t}/(\kappa_dr^d)\ge t(v(\varepsilon)r^{-1})^{\alpha+d}(1+\varepsilon)^{(\alpha+d)/4}\ge (1+\varepsilon)^{(\alpha+d)/4},$$ the Poisson concentration property [@penrose Lemma 1.2] implies that $\sup_{r\le v(\varepsilon) t^{1/(\alpha+d)}} \P(N_r\ge n_{r,t})$ decays at least exponentially fast in $t^{d/(\alpha+d)}$ as $t\to\infty$. Similarly, taking into account $\E|U_1|^\alpha=d/(\alpha+d)$, we obtain that $$tr^{-\alpha}/(n_{r,t}\E |U_1|^\alpha)\ge(1+\varepsilon)^{(\alpha+d)/2},$$ using the classical theory of large deviations shows that also the expression $$\sup_{r\le v(\varepsilon) t^{1/(\alpha+d)}}\P\Big(\sum_{i=1}^{n_{r,t}}r^{\alpha}|U_i|^{\alpha}\ge t\Big)$$ decays at least exponentially fast in $t^{d/(\alpha+d)}$ as $t\to\infty$. , $$\begin{aligned}
\label{outEdgeEq1}
\lim_{t\to\infty} t^{s/(\alpha+d)}p^{(\alpha)}_{\out,t}=\lim_{t\to\infty}\int_{v(\varepsilon)t^{1/(\alpha+d)}}^\infty t^{s/(\alpha+d)}\P(D_r>t)\P_{R}(\d r),\end{aligned}$$ provided that the latter limit exists and is finite. In order to compute the right-hand side in , we split the integral into three parts, which are analyzed separately. To be more precise, put $$\begin{aligned}
I_1&=\int_{v(\varepsilon)t^{1/(\alpha+d)}}^{v(-\varepsilon)t^{1/(\alpha+d)}}t^{s/(\alpha+d)}\P(D_r>t)\P_R(\d r),\\
I_{2,1}&=-\int_{v(-\varepsilon)t^{1/(\alpha+d)}}^\infty t^{s/(\alpha+d)}\P(D_r\le t)\P_R(\d r),\end{aligned}$$ and $$I_{2,2}=\int_{v(-\varepsilon)t^{1/(\alpha+d)}}^\infty t^{s/(\alpha+d)}\P_R(\d r)=t^{s/(\alpha+d)}\P(R>v(-\varepsilon)t^{1/(\alpha+d)}).$$ First, note that $I_{2,2}$ tends to $v(-\varepsilon)^{-s}\beta$ as $t\to\infty$ and the latter expression tends to $(d\kappa_d/(\alpha+d))^{s/(\alpha+d)}$ as $\varepsilon\to0$. Hence, it suffices to show that the integrals $I_1$ and $I_{2,1}$ tend to $0$ as we let first $t\to\infty$ and then $\varepsilon\to0$. Indeed, $$I_1\le t^{s/(\alpha+d)}\big(\P(R>{v(\varepsilon)t^{1/(\alpha+d)}})-\P(R>{v(-\varepsilon)t^{1/(\alpha+d)}})\big),$$ and the right-hand side tends to $\beta(v(\varepsilon)^{-s}-v(-\varepsilon)^{-s})$ as $t\to\infty$, which vanishes as $\varepsilon\to0$. Finally, we observe that $$-I_{2,1}\le t^{s/(\alpha+d)}\P(D_{v(-\varepsilon)t^{1/(\alpha+d)}}\le t)\P(R>v(-\varepsilon)t^{1/(\alpha+d)}).$$ We conclude the proof of the theorem by noting that the expression $t^{s/(\alpha+d)}\P(R>v(-\varepsilon)t^{1/(\alpha+d)})$ remains bounded as $t\to\infty$, whereas $\P(D_{v(-\varepsilon)t^{1/(\alpha+d)}}\le t)$ 