Consider a growth-fragmentation $\mathbf{X}$ such that and are satisfied. Then, for every $x>0$, it holds that $$\begin{aligned}
            \frac{\epsilon^{-(1+\frac{\omega_-}{|\alpha|})}}{\overline{\Lambda}(\epsilon^{1/|\alpha|})}A(\epsilon)
            \underset{\epsilon\to 0}{\longrightarrow}\frac{|\alpha|\rho}{(\omega_--\rho)(\omega_-+|\alpha|-\rho)}\mathbb{E}_1^-(I^{\frac{\omega_--\rho}{\alpha}})x^{\alpha+\rho},
        \end{aligned}$$ $\mathcal{P}_x$-almost surely and in $\mathbb{L}^1$, where the expectation is finite.

Since $\alpha<0$, we know from [@BBCK16] Proposition 4.1 that for any $x>0$, $\mathcal{P}_x^+$ is absolutely continuous with respect to the restriction of $\mathcal{P}_x$ to the natural filtration of the growth-fragmentation and the spine up to time $t>0$. Therefore the above convergence also holds $\mathcal{P}_x^+$-a.s.

In Section 6, we shall see that the following interesting result easily follows from self-similarity.

\[Theorem stationary area from 0\] Suppose that holds. Then under $\mathcal{P}_0^+$, the process $$\begin{aligned}
            \left(e^{\frac{\omega_-}{\alpha}u}A\left(e^u\right)\right)_{u\in\mathbb{R}}
        \end{aligned}$$ is stationary. In particular, the law of $t^{\frac{\omega_-}{\alpha}}A(t)$ does not depend on $t>0$.

The next two propositions provide some bound about the almost sure behaviour of this area and shows that almost surely, $A(t)$ does not deviate from $t^{\omega_-/|\alpha|}$ by more than a power of $|\log(t)|$:

\[Theorem upper envelope A(t)\] Suppose that and $\kappa(\omega_++\omega_-+\alpha)<\infty$ hold. For all $\delta>0$, we have that $$\begin{aligned}
            \limsup_{t\to 0+\text{ or }\infty}|\log(t)|^{-1-\delta}t^{\frac{\omega_-}{\alpha}}A(t)=0,\quad\mathcal{P}_0^+\text{-a.s.}
        \end{aligned}$$

\[Proposition lower envelope A(t)\] Suppose that and hold. Then, there exists $q_0>0$ such that for all $q>q_0$, we have that $$\begin{aligned}
            \liminf_{t\to 0+\text{ or }\infty}|\log(t)|^{q}t^{\frac{\omega_-}{\alpha}}A(t)=\infty,\quad\mathcal{P}_0^+\text{-a.s.}
        \end{aligned}$$

The 