e^{\xi(s)})^{\omega_-}A_s\left((t-\sigma_s)(\Delta_-e^{\xi(s)})^\alpha\right)-S_{\tau_t}\\
            &=\sum_{s\leq t}(\Delta_- e^{\xi(\tau_s)})^{\omega_-}A_{\tau_s}\left((t-s)(\Delta_-e^{\xi(\tau_s)})^\alpha\right)-S_{\tau_t}\\
            &=A(t)-S_{\tau_t},
        \end{aligned}$$ by Lemma \[Lemma Markov-branching property of A\] (where we reindexed the $A_s$’s), as claimed.

The main idea of the current section is to use Lemma \[Lemma M martingale and M(tau)=A-S\], namely the martingale property of $M$ and its relation to $A$, to show that $A(\epsilon)\sim S_{\tau_\epsilon}$. In this direction, we first establish the following:

\[Lemma predictable compensator as E(A(epsilon))\] $\mathcal{P}_1$-almost surely, it holds that $$\begin{aligned}
            S_{\tau_{\epsilon}}&\underset{\epsilon\to 0}{\sim}\frac{|\alpha|\rho}{(\omega_--\rho)(\omega_-+|\alpha|-\rho)}\mathbb{E}_1^-(I^{-\frac{\omega_--\rho}{|\alpha|}})\overline{\Lambda}\left(\epsilon^{1/|\alpha|}\right)\epsilon^{1+\frac{\omega_-}{|\alpha|}}.
        \end{aligned}$$

Recall . After a change of variables and replacing $\exp(\xi(\tau_s))$ by $X(s)$, we have that $$\begin{aligned}
            S_{\tau_\epsilon}
            &=\int_0^\epsilon\mathrm{d}sX(s)^{\omega_-+\alpha}\int_{{\left(-\infty\,,0\right)}}\Lambda(\mathrm{d}y)(1-e^y)^{\omega_-}\mathbb{P}_1^-\left(I\leq (1-e^y)^{\alpha}X(s)^{\alpha}(\epsilon-s)|\mathcal{F}_{\tau_s}\right)
        \end{aligned}$$ It is well known that the law of $I$ is absolutely continuous, see e.g. [@BLM08] Theorem 3.9. Hence, since $X(s)\sim 1$ as $s\to 0+$, one sees that $$\begin{aligned}
            S_{\tau_\epsilon}
            &\underset{\epsilon\to 0+}{\sim}\int_0^\epsilon\mathrm{d}s\int_{{\left(-\infty\,,0\right)}}\Lambda(\mathrm{d}y)(1-e^y)^{\omega_-}\mathbb{P}_1^-\left(I\leq(\epsilon-s)(1-e^y)^\alpha\right),\quad \mathcal{P}_1\text{-almost surely.}
        \end{aligned}$$ We express the right-hand side in the form $$\begin{aligned}
    &\int_{{\left(-\infty\,,0\right)}}\Lambda(\mathrm{d}y)(1-e^y)^{\omega_-}\int_0^\epsilon\mathrm{d}s\mathbb{P}_1^-(I\leq s(1-e^y)^{\alpha})\\
    &\hspace{2cm}=\int_{{\left(-\infty\,,0\right)}}\Lambda(\mathrm{d}y)(1-e^y)^{\omega_-+|\alpha|}\int_0^{\epsilon(1-e^y)^\alpha}\mathrm{d}s\mathbb{P}_1^-(I\leq s)\\
    &\hspace{2cm}=C_1(\epsilon)+C_2(\epsilon),\end{aligned}$$ where $C_1(\epsilon)$ is the part of the first integral with domain restricted to $E_1:={\left(-\infty\,,\log\left(1-\epsilon^{\frac{1-\delta}{|\alpha|}}\right)\right)}$ for $\delta>0$ arbitrary small, and $C_2(\epsilon)$ is that restricted to $E_2:={\left[\log\left(1-\epsilon^{\frac{1-\delta}{|\alpha|}}\right)\,,0\right)}$. We first address $C_1(\epsilon)$. Note that for $y\in E_1$, we have that $s\leq\epsilon (1-e^y)^\alpha\leq \epsilon^\delta$. Appealing to lemmas \[Lemma expect A(epsilon) = distr function of I\] and \[Lemma asymp expect A(espilon)\], there exists 