$\mathcal{P}_1$. It means that if Theorem \[Theorem area GF\] is true under $\mathcal{P}_1$, we easily deduce that under $\mathcal{P}_x$, $$\begin{aligned}
        \frac{\epsilon^{-(1+\frac{\omega_-}{|\alpha|})}}{\overline{\Lambda}\left(\epsilon^{1/|\alpha|}\right)}A(\epsilon)
        &\stackrel{d}{=}x^{\alpha}\frac{\overline{\Lambda}\left((\epsilon x^\alpha)^{1/|\alpha|}\right)}{\overline{\Lambda}\left(\epsilon^{1/|\alpha|}\right)}\cdot\frac{(\epsilon x^{\alpha})^{-(1+\frac{\omega_-}{|\alpha|})}}{\overline{\Lambda}\left((\epsilon x^\alpha)^{1/|\alpha|}\right)}A(\epsilon x^{\alpha}),\quad\text{under }\mathcal{P}_1\\
        &\!\!\!\!\underset{\epsilon\to 0+}{\sim}x^{\rho+\alpha}\lim_{\epsilon\to 0+}\frac{(\epsilon x^{\alpha})^{-(1+\frac{\omega_-}{|\alpha|})}}{\overline{\Lambda}\left((\epsilon x^\alpha)^{1/|\alpha|}\right)}A(\epsilon x^{\alpha}),\hspace{1.3cm}\text{under }\mathcal{P}_1,
    \end{aligned}$$ where we used the assumption of regular variation of $\overline{\Lambda}$ at $0+$, as written in .

Recall that $\xi$ is the Lévy process associated with $X$ by Lamperti’s transformation, and that under $\mathcal{P}_1$, $\chi_\emptyset$ is distributed as $X$. Let $\sigma:s\mapsto\int_0^s e^{-\alpha\xi(u)}\mathrm{d}u$ be the inverse change time of $(\tau_t)_{t\geq 0}$. Define the compensated process $(M_t)_{t\geq 0}$ as $$\begin{aligned}
        M_t
        :=&\sum_{s\leq t}(\Delta_- e^{\xi(s)})^{\omega_-}A_s\left((\sigma_t-\sigma_s)(\Delta_-e^{\xi(s)})^\alpha\right)-S_t,\\
        \shortintertext{where the $A_s$'s are i.i.d. copies of $A$ under $\mathcal{P}_{1}$, independent from $\xi$, with}
        S_t:=&\int_0^t\mathrm{d}se^{\omega_-\xi(s)}\int_{{\left(-\infty\,,0\right)}}\Lambda(\mathrm{d}y)(1-e^y)^{\omega_-}\mathbb{P}_1^-\left(I\leq (1-e^y)^{\alpha}e^{\alpha\xi(s)}(\sigma_t-\sigma_s)|\mathcal{F}_s\right),
    \end{aligned}$$ where $I$ is independent from $\xi$ and $(\mathcal{F}_s)_{s\geq 0}$ is the natural filtration of $\xi$.

\[Lemma M martingale and M(tau)=A-S\] The process $(M_t)_{t\geq 0}$ is a martingale. Moreover, identifying $\chi_{\emptyset}$ with $X=\exp(\xi_\tau)$, we have that $M_{\tau_t}=A(t)-S_{\tau_t}$ for all $t\geq 0$.

Thanks to Lemma \[Lemma expect A(epsilon) = distr function of I\] and independence of the $A_s$’s with $\xi$, one sees that $(S_t)_{t\geq 0}$ is the predictable compensator of the series in $(M_t)_{t\geq 0}$, so that the latter is a martingale. We then write $$\begin{aligned}
            M_{\tau_{t}}
            &=\sum_{s\leq \tau_t}(\Delta_- 