$\mathcal{M}$ whose reciprocal Cauchy transform satisfies $$F_{\mu_0}(z)=-\int_\mathbb{R}\frac{s}{s^2+1}\;d\nu(s)+z+\int_\mathbb{R}\frac{1+sz}{s-z}\;d\rho(s).$$ Consequently, $p(\lambda,\nu)$ is a $\boxplus$-infinitely divisible measure with an atom at $0$ of mass $1-\lambda$ for $\lambda<1$ and no atom for $\lambda\geq1$, $$\phi_{p(\lambda,\nu)}(z)=\lambda E_{\mu_0}(z)=\lambda z\int_\mathbb{R}\frac{s}{z-s}\;d\nu(s),$$ and $$\mathrm{Ind}(p(\lambda,\nu))=\frac{\mathrm{Ind}(\mu_0)+\lambda}{\lambda}.$$

Since $s^2/(s^2+1)\in L^1(d\nu)$, the measure $\rho$ is finite and positive, and the limit $$\lim_{n\to\infty}\int_\mathbb{R}\frac{ns}{s^2+1}\;d\mu_n(s)=\int_\mathbb{R}\frac{\lambda s}{s^2+1}\;d\nu(s)$$ exists. Moreover, it is easy to see that $$\frac{ns^2}{s^2+1}\;d\mu_n(s)\to \lambda d\rho(s)$$ weakly. By Theorem \[thm2.2\], the measure $\mu_n^{\boxplus n}$ converges weakly to $p(\lambda,\nu)$, which satisfies $$\phi_{p(\lambda,\nu)}(z)=\int_\mathbb{R}\frac{\lambda s}{s^2+1}\;d\nu(s)+
\int_\mathbb{R}\frac{\lambda(1+sz)}{z-s}\;d\rho(s)=\lambda
z\int_\mathbb{R}\frac{s}{z-s}\;d\nu(s).$$ On the other hand, the definition of $\mu_0$ and Proposition 3.1 show that $$\phi_{\mathbb{B}_{1+\lambda,1/(1+\lambda)^*}
(\mu_0)}=\lambda E_{\mu_0}=\phi_{p(\lambda,\nu)}.$$ Then by Lemma \[3.4\] and (\[Nindicator\]) we have $$\mathrm{Ind}(p(\lambda,\nu))=\mathrm{Ind}(\mathbb{B}_\lambda(\mu_0)^{\uplus \lambda})
=\frac{\mathrm{Ind}(\mu_0)+\lambda}{\lambda}.$$ Next, we apply Theorem \[3.10\](3) to locate the atom of $p(\lambda,\nu)$. Since $\phi_{p(\lambda,\nu)}(0)=0$, $0\in\partial
F_{p(\lambda,\nu)}(\mathbb{C}^+)$. Moreover, by the dominated convergence theorem we obtain $$\lim_{\epsilon\downarrow0}\frac{\phi_{p(\lambda,\nu)}(i\epsilon)-\phi_{p(\lambda,\nu)}(0)}{i\epsilon}
=\lambda\lim_{\epsilon\downarrow0}\int_\mathbb{R}\frac{s}{i\epsilon-s}\;d\nu(s)=-\lambda,$$ which gives the desired result. This completes the proof.

Since the $\boxplus$-divisibility indicator is zero for any measure with finite support, we have the following result.

We have $p(\lambda,\delta_a)=\mathbb{B}_{1+\lambda,1/(1+\lambda)^*}(\mu_0)$, where $\mu_0=(\delta_0+\delta_{2a})/2$, $\mathrm{Ind}(p(\lambda,\delta_a))=1$, and $\phi_{p(\lambda,\delta_a)}(z)=a\lambda z/(z-a)$.

We finish this section with an interesting observation. If $\mathrm{Ind}(\mu)>1$ then $\mathrm{Ind}(\mathbb{B}^{-1}(\mu))=\mathrm{Ind}(\mu)-1>0$ by (\[Nindicator\]) and (\[bijection\]). This implies that $\phi_\mu=E_{\mathbb{B}^{-1}(\mu)}$ has a continuous extension to $\mathbb{C}^+\cup\mathbb{R}$ by Proposition \[3.5\], whence we have the following proposition.

