$$\begin{aligned}
            \mathcal{E}_1(A(\epsilon))\underset{\epsilon\to 0}{\sim}\frac{|\alpha|\rho}{(\omega_--\rho)(\omega_-+|\alpha|-\rho)}\mathbb{E}_1^-(I^{\frac{\omega_--\rho}{\alpha}})\epsilon^{1+\frac{\omega_-}{|\alpha|}}\overline{\Lambda}(\epsilon^{1/|\alpha|}),
        \end{aligned}$$ where the expectation on the right-hand side is finite.

Let $\Pi_\alpha^-$ be the Lévy measure of $|\alpha|\eta^-$. The behaviour of $\mathbb{P}_1^-(I\leq t)$ as $t\to 0$ is given in Theorem 7 of [@AR15]. More precisely, provided that $\Pi_\alpha^-({\left(-\infty\,,-(x+y)\right)})/\Pi_\alpha^-({\left(-\infty\,,-x\right)})\sim e^{-\gamma y}$ as $x\to\infty$ with $\mathbb{E}_1^-(\exp(|\alpha|\gamma\eta^-(1)))<1$ (corresponding to $|\alpha|\gamma\in{\left(0\,,\omega_+-\omega_-\right)}$ by ), it holds that $$\begin{aligned}
            \mathbb{P}_1^-(I\leq t)
            \underset{t\to 0}{\sim}\frac{\mathbb{E}_1^-(I^{-\gamma})}{1+\gamma}t\Pi_\alpha^-({\left(-\infty\,,\log(1/t)\right)}).
        \end{aligned}$$ By , we have that $$\begin{aligned}
            \Pi_\alpha^-({\left(-\infty\,,-x\right)})
            &\underset{x\to \infty}{\sim}\overline{\Lambda}\left(e^{x/\alpha}\right)e^{\omega_-x/\alpha}\frac{\rho}{\omega_--\rho}.
        \end{aligned}$$ We thus see that $\Pi_\alpha({\left(-\infty\,,-x-y\right)})/\Pi_\alpha({\left(-\infty\,,-x\right)})\to \exp(-\frac{\omega_--\rho}{|\alpha|} y)$ as $x\to\infty$. It remains to check that $\omega_--\rho<\omega_+-\omega_-$, which is true since $2\omega_--\omega_+<\rho<\omega_-$ by . We therefore get by [@AR15] Theorem 7 that $$\begin{aligned}
            \mathbb{P}_1^-(I\leq \epsilon)
            &\underset{\epsilon\to 0+}{\sim}\left(1+\frac{\omega_--\rho}{|\alpha|}\right)^{-1}\mathbb{E}_1^-\left(I^{-\frac{\omega_--\rho}{|\alpha|}}\right)\epsilon\Pi_\alpha^-({\left(-\infty\,,\log(\epsilon)\right)})\\
            &\underset{\epsilon\to 0+}{\sim}\frac{|\alpha|\rho}{(\omega_--\rho)(\omega_-+|\alpha|-\rho)}\mathbb{E}_1^-\left(I^{-\frac{\omega_--\rho}{|\alpha|}}\right)\epsilon^{1+\frac{\omega_-}{|\alpha|}}\overline{\Lambda}(\epsilon^{1/|\alpha|}),
        \end{aligned}$$ as claimed, where the expectation is finite.

We shall need a bit more than the speed of convergence to 0 of the first moment.

\[Lemma speed A\^p\] For $p=p(\epsilon):=1+1/\sqrt{|\log(\epsilon)|}$, for all $0<q<1+\frac{\omega_--\rho}{|\alpha|}$ it holds that $$\begin{aligned}
            \mathcal{E}_1\left(A(\epsilon)^p\right)=o\left(\epsilon^q\right),\quad\text{as}\quad\epsilon\to 0.
        \end{aligned}$$

Let $n\geq 1$ be arbitrary large. We write $$\begin{aligned}
            \mathcal{E}_1\left(A(\epsilon)^p\right)
            &\leq\mathcal{E}_1\left(A(\epsilon)\mathds{1}_{\left\{A(\epsilon)\leq \epsilon^{-n}\right\}}\right)\times \epsilon^{-n(p-1)}+\mathcal{E}_1\left(A(\epsilon)^p\mathds{1}_{\left\{A(\epsilon)>\epsilon^{-n}\right\}}\right).
        \end{aligned}$$ By Lemma \[Lemma asymp expect A(espilon)\], the first term is of order $\epsilon^{1+\frac{\omega_-}{-\alpha}-n(p-1)}\overline{\Lambda}(\epsilon^{1/|\alpha|})$, with $p-1=1/\sqrt{-\log(\epsilon)}$ going to 0 with $\epsilon$. It remains to bound the second term. Take $q,q'>1$ such that $1/q+1/q'=1$ and $pq<\omega_+/\omega_-$, Hölder’s Inequality then yields 