originally been proven in [@MaejimaBarndorff Thm. C(iii)]. Nevertheless, we can now give a completely different and shorter proof as we shall do.

\[prop:bondessontoGGC\] Let $\xi_t=t$. Then $$\Phi_\xi({\mathrm{BO}}\cap {\mathrm{ID}}_{\log}) = {\mathrm{T}}$$ In particular, the distributions in ${\mathrm{BO}}\cap {\mathrm{ID}}_{\log}$ with finite Stieltjes measure are mapped surjectively on the generalized Gamma convolutions with $k(0+)<\infty$.

Assume $\mu\in {\mathrm{T}}\subset {\mathrm{L}}^+$, then there exists a Lévy process $X$ with ${\mathcal{L}}(X_1)\in {\mathrm{ID}}^+_{\log}$ such that $\Phi_\xi(X_1)=\mu$, i.e. $X$ and $\mu$ are related via or . Hence from and $$\begin{aligned}
 \psi_X(u)&= a u + u \int_{(0,\infty)} e^{-u t} k(t) dt 
= a u + u \int_{(0,\infty)} e^{-u t} \int_{[0,\infty)} e^{-t x} d \rho (x) dt\end{aligned}$$ for some unique measure $\rho$ with $\rho(\{0\})= \lim_{t\to \infty} k(t)=0$. Using Tonelli we can proceed $$\begin{aligned}
 \psi_X(u) &= a u +  \int_{(0,\infty)} u \int_{(0,\infty)} e^{-u t} e^{-t x} dt\, d \rho (x) = a u +  \int_{(0,\infty)} \frac{u}{u + x} d \rho(x).\end{aligned}$$ Hence $\psi_X(u)$ is a CBF (see e.g. [@rene-book Remark 6.4]) such that ${\mathcal{L}}(X_1)\in {\mathrm{BO}}$ by [@rene-book Def. 9.1].\
Conversely, assume that $X$ is a Lévy process such that ${\mathcal{L}}(X_1)\in {\mathrm{BO}}\cap {\mathrm{ID}}_{\log}$. Then $\Phi_\xi({\mathcal{L}}(X_1))$ exists and the same computation backwards proves that $\Phi_\xi({\mathcal{L}}(X_1)) \in {\mathrm{T}}$.\
The remaining assertion follows directly from an inspection 