that $$\begin{aligned}
            \mathcal{E}_1\left(A(\epsilon)^p\mathds{1}_{\left\{A(\epsilon)>\epsilon^{-n}\right\}}\right)
            &\leq\mathcal{E}_1\left(A(\epsilon)^{pq}\right)^{1/q}\times\mathcal{P}(A(\epsilon)>\epsilon^{-n})^{1/q'}.
        \end{aligned}$$ Since $pq<\omega_+/\omega_-$ we know that $\mathcal{E}_1(A(\epsilon)^{pq})<\infty$ by Lemma 2.3 in [@BBCK16]. Markov’s Inequality shows that the probability on the right-hand side is bounded from above by $$\begin{aligned}
            \mathcal{E}_1(A(\epsilon))^{1/q'}\times\epsilon^{n/q'}.
        \end{aligned}$$ Since $n$ was chosen arbitrary large, this concludes the proof.

The proof above would work for any $p(\epsilon)$ converging to 1 from above as $\epsilon\to 0$, but will shall use it with this specific choice of $p(\epsilon)$ in order to have a technical result, that we record now for later use.

\[Lemma equiv p(epsilon)\] For $p=p(\epsilon)$ as in Lemma \[Lemma speed A\^p\], it holds that $$\begin{aligned}
            \epsilon^{1+\frac{p\omega_-}{|\alpha|}}\overline{\Lambda}(\epsilon^{1/|\alpha|})=o\left(\epsilon^{p(1+\frac{\omega_-}{|\alpha|})}\overline{\Lambda}(\epsilon^{1/|\alpha|})^{p}\right),\quad\text{as }\epsilon\to 0+.
        \end{aligned}$$

By $\eqref{Equation assumption rho}$ and Theorem 1.4.1(iii) in [@BGT87], we know that there exists a function $\ell:{\left(0\,,\infty\right)}\to{\left(0\,,\infty\right)}$ with slow variation at $0$ such that $\overline{\Lambda}(\epsilon)\sim\epsilon^{-\rho}\ell(\epsilon)$. We write $$\begin{aligned}
            &\log\left(\frac{\epsilon^{1+\frac{p\omega_-}{|\alpha|}}\overline{\Lambda}(\epsilon^{1/|\alpha|})}{\epsilon^{p(1+\frac{\omega_-}{|\alpha|})}\overline{\Lambda}(\epsilon^{1/|\alpha|})^{p}}\right)
            =(1-p)\log(\epsilon)+(1-p)\log\left(\overline{\Lambda}(\epsilon^{1/|\alpha|})\right)\\
            &\hspace{3.8cm}\underset{\epsilon\to 0+}{\sim}(1-p)\log(\epsilon)+\frac{(1-p)\rho}{\alpha}\log(\epsilon)+(1-p)\log(\ell(\epsilon^{1/|\alpha|})).
        \end{aligned}$$ Since $\ell$ has slow variation, the terms with $\log(\epsilon)$ dominates. By definition of $p$ and since $|\alpha|<\rho$ by , we see that $(1-p)(1+\rho/\alpha)\log(\epsilon)=(1+\rho/\alpha)\sqrt{|\log(\epsilon)|}\to -\infty$ as $\epsilon\to 0+$, which entails the claim.

The almost sure behaviour
-------------------------

Throughout this subsection, we shall only work under $\mathcal{P}_1$, since the other cases follow from the self-similarity. Indeed, under $\mathcal{P}_x$, we have $(A(t))_{t\geq 0}\stackrel{d}{=}(x^{\omega_-}A(t x^\alpha))_{t\geq 0}$ under 