R^+(\gamma', \sigma^2, \nu)$ by [@BLMranges Thm. 3].

Finally, assume $\mu \in R^+(\gamma, \sigma^2, \nu)$ where $\nu((0,\infty))=0$ and $\int_{[-1,0)}|x|\nu(dx)<\infty$ and set for $\lambda\in(0,1]$ $$\begin{aligned}
 g_{\lambda}(u) = & (\gamma_\lambda - \frac{\sigma^2}{2}) u \psi_\mu'(u) + \frac{\sigma^2}{2} u^2\left( (\psi'_\mu(u))^2 - \psi''_\mu(u) \right) \\
 & + \int_{{\mathbb{R}}_-} \left( e^{\psi_\mu(u) - \psi_\mu(ue^{-y})} - 1 \right) \lambda \nu(dy), \quad u\geq  0,\end{aligned}$$ where $\gamma_\lambda:=\gamma -  \lambda\int_{[-1,0)} x \nu(dx)$, then $g_1(u)$ is a BF by assumption. For any $\lambda<1$ we observe that for $u >0$ $$\begin{aligned}
 g_{\lambda}(u) =& g_1(u)+(1-\lambda)\int_{[-1,0)} x \nu(dx)u \psi_\mu'(u) +  (1-\lambda) \int_{{\mathbb{R}}_-} \left( 1- e^{\psi_\mu(u) - \psi_\mu(ue^{-y})} \right) \nu(dy).\end{aligned}$$ Since $\xi$ is spectrally negative, $\mu$ is selfdecomposable and thus $\psi_\mu(ue^{-y})-\psi_\mu(u)$ is a BF for any negative $y$ by Proposition \[prop:SD\] (it is the Laplace exponent of the $e^{-y}$-factor of $\mu$). Hence $e^{\psi_\mu(u) - \psi_\mu(ue^{-y})}$ is CM and we can write $$e^{\psi_\mu(u) - \psi_\mu(ue^{-y})} = \int_{(0,\infty)} e^{-ut} \mu_{e^{-y}}(dt).$$ Thus for $u>0$ $$\begin{aligned}
 g_{\lambda}(u) =& g_1(u)+(1-\lambda)\int_{[-1,0)} x \nu(dx)u \psi_\mu'(u)+ (1-\lambda) \int_{{\mathbb{R}}_-} \int_{(0,\infty)} \mu_{e^{-y}}(dt)\nu(dy) \left( 1- e^{-ut} \right).\end{aligned}$$ Since $u \psi_\mu'(u)$ is a BF by Proposition \[prop:SD\] and since all appearing integrals exist, we conclude that $g_\lambda(u)+(\gamma' - \gamma)u \psi_\mu'(u)$ is again a BF. Hence $\mu\in R^+(\gamma', \sigma^2, \lambda\nu)$ which proves .

\[prop:notnested\] Let $(\xi_t)_{t\geq 0}$ be a subordinator 