+ \tilde{\xi}_t$, where $\sigma=\sqrt{\sigma^2}$ and $(B_t)_{t\geq 0}$ is a standard Brownian motion and independent of $\tilde{\xi}_t$. Hence $(\sigma B_t)_{t\geq 0}\overset{d}= (B_{\sigma^{2}t})_{t\geq 0}$ and thus $(\sigma B_t+ \tilde{\xi}_t)_{t\geq 0}\overset{d}= (B_{\sigma^{2}t} + \tilde{\tilde{\xi}}_{\sigma^2t})_{t\geq 0}$ where $\tilde{\tilde{\xi}}$ has characteristic triplet $(\gamma/\sigma^2, 0 , \nu/\sigma^2)$.\
This implies that for any subordinator $(\eta_t)_{t\geq 0}$, independent of $\xi$ and with ${\mathcal{L}}(\eta_1) \in D_\xi^+$ $$\int_{(0,\infty)} e^{-\xi_t} d \eta_t = \int_{(0,\infty)} e^{-(\sigma B_t + \tilde{\xi}_t)} d \eta_t \overset{d}= \int_{(0,\infty)} e^{-(B_{\sigma^{2}t}+ \tilde{\tilde{\xi}}_{\sigma^{2}t})} d \eta_t = \int_{(0,\infty)} e^{-(B_{t}+ \tilde{\tilde{\xi}}_{t})} d \eta_{t/\sigma^2}.$$ Thus ${\mathcal{L}}(\eta_{1/\sigma^2}) \in D_{B+\tilde{\tilde{\xi}}}^+$ and $\Phi^+_\xi({\mathcal{L}}(\eta_1))= \Phi^+_{B+\tilde{\tilde{\xi}}}({\mathcal{L}}(\eta_{1/\sigma^2}))$ from which we conclude the first assertion.

Now assume $\mu \in R^+(\gamma, \sigma^2, \nu)\cap {\mathrm{L}}$, then by [@BLMranges Thm. 3] $$\begin{aligned}
 f_{\gamma}(u) = & (\gamma - \frac{\sigma^2}{2}) u \psi_\mu'(u) + \frac{\sigma^2}{2} u^2\left( (\psi'_\mu(u))^2 - \psi''_\mu(u) \right) \\
 & + \int_{{\mathbb{R}}} \left( e^{\psi_\mu(u) - \psi_\mu(ue^{-y})} - 1 - u \psi'_\mu(u) y \mathds{1}_{|y|\leq 1}\right) \nu(dy), \quad u \geq  0,\end{aligned}$$ is the Laplace exponent of some subordinator, i.e. a BF. Observe that for $\gamma'\geq \gamma$ $$\begin{aligned}
f_{\gamma'}(u)= & f_{\gamma}(u) + (\gamma'-\gamma)  u \psi'_\mu(u).\end{aligned}$$ Since the set of BFs is a convex cone (cf. [@rene-book Cor. 3.8(i)]) and since by assumption $\mu\in{\mathrm{L}}^+$ such that $u \psi'_\mu(u)$ is a BF, $f_{\gamma'}(u)$ is again a BF. Hence $\mu\in 