and for a.s. positive random variables in the recent article [@maischenkscherer] as well as in Corollary \[cor:densitySD\] below.

\[prop:SD\] Let $\mu\in {\mathcal{P}}^+$ be a probability measure with Laplace exponent $\psi_\mu(u)$, $u\geq 0$. Then the following statements are equivalent.

1.  $\mu\in {\mathrm{L}}^+$.

2.  $\psi_{\mu_c}(u):=\psi_\mu(u)-\psi_\mu(cu)$ is a Bernstein function for all $c\in(0,1)$.

3.  $-\psi_{\mu_c}(u)= \psi_\mu(cu) - \psi_\mu(u)$ is a BF for all $c>1$.

4.  $u\cdot \psi_\mu'(u)$ is a BF.

5.  $\mu={\mathcal{L}}(\int_{(0,\infty)} e^{-t}dX_t)$ for some subordinator $(X_t)_{t\geq 0}$ with ${\mathbb{E}}[\log^+(X_1)]<\infty$.

Equivalence of (i) and (ii) is well known and follows immediately from the definition of selfdecomposability and the fact that $\mu_c$ as in is infinitely divisible (see e.g. [@sato Prop. 15.5]). Further by [@rene-book Cor. 3.8(iii)] (ii) implies that also $\psi_{\mu_c}(c^{-1}u)=\psi_\mu(c^{-1}u)-\psi_\mu(u)$, $c\in(0,1)$ is a BF, i.e. (iii). The converse can be seen similarly.\
We continue proving that (ii) implies (iv). Assume (ii), then for all $c\in(0,1)$ $$\frac{\psi_\mu(u)-\psi_\mu(u-(1-c)u)}{(1-c)}$$ is a BF in $u$. Thus $$u\psi_\mu'(u)=\lim_{c\to 1} \frac{\psi_\mu(u)-\psi_\mu(u-(1-c)u)}{(1-c)}$$ is a BF, too ([@rene-book Cor. 3.8(ii)]), which shows (iv).\
Now assume (iv) and set $$\label{eq-SD-bernsteinableitung}
 \psi_X(u):=u\psi'_\mu(u), \quad u\geq 0,$$ then $\psi_X$ is a BF with $\psi_X(0)=0$ and hence there exists a subordinator $(X_t)_{t\geq 0}$ with Laplace exponent $\psi_X$. Now by [@BLMranges Thm. 5.1 (ii)] (setting $\sigma=0$) 