The following implication that (2) implies (1) was proved in \[\[BN1\]\].

\[3.5\] Suppose that $\mu\in\mathcal{M}$ and fix a number $p\in(0,1)$. Then the following statements $(1)$-$(3)$ are equivalent:

1.  [$\mu^{\boxplus p}$ is defined, that is, $\mu=\nu^{\boxplus1/p}$ for some $\nu\in\mathcal{M}$;]{}

2.  [$1-p\leq\mathrm{Ind}(\mu)$;]{}

3.  [$\mu^{\uplus(1-p)}$ is $\boxplus$-infinitely divisible.]{}

If $(1)$-$(3)$ hold and $\Omega_p=F_{\mu^{\uplus(1-p)}}(\mathbb{C}^+)$ then the following statements are true.

1.  [The reciprocal Cauchy transform $F_\mu$ extends continuously to $\mathbb{C}^+\cup\mathbb{R}$.]{}

2.  [The identity $F_{\mu^{\boxplus
    p}}\left(F_{\mu^{\uplus(1-p)}}(z)\right)=F_\mu(z)$ holds for $z\in\mathbb{C}^+\cup\mathbb{R}$.]{}

3.  [The identity $F_{\mu^{\boxplus
    p}}(z)=F_\mu\left(F_{\mu^{\uplus(1-p)}}^{-1}(z)\right)$ holds for $z\in\overline{\Omega_p}$.]{}

4.  [The Voiculescu transform of $\mu^{\uplus(1-p)}$ can be expressed as $$\phi_{\mu^{\uplus(1-p)}}=E_{\left(\mu^{\boxplus
    p}\right)^{\uplus(1/p-1)}}=E_{\mathbb{B}^{-1}(\mu^{\uplus(1-p)})}.$$]{}

If $\nu^{\boxplus1/p}=\mu$ then by Lemma \[3.4\] we have $$\mathrm{Ind}(\mu)=p
\mathrm{Ind}\left(\mathbb{B}_{1/p-1}(\nu)\right)\geq
p\left(\frac{1}{p}-1\right)=1-p,$$ which shows that (1) implies (2). Applying Proposition \[3.1\] to $\nu$ and $1/p$ yields the assertions (3). If (3) holds, i.e., $\phi_{\mu^{\uplus(1-p)}}\in\mathcal{E}$ then the following function $$F(z)=\frac{-p\phi_{\mu^{\uplus(1-p)}}(z)+(1-p)z}{1-p},\;\;\;\;z\in\mathbb{C}^+,$$ belongs to $\mathcal{F}$, i.e., $F=F_\nu$ for some $\nu\in\mathcal{M}$, by which $F_{\mu^{\uplus(1-p)}}^{-1}$ can be written as $$\label{p<1o}
F_{\mu^{\uplus(1-p)}}^{-1}(z)=\frac{1}{p}z+\left(1-\frac{1}{p}\right)F_\nu(z),\;\;\;\;\;z\in\mathbb{C}^+.$$ Then by Proposition 2.1 and the definition of the $\uplus$-convolution power we have $$F_{\nu^{\boxplus1/p}}(z)=\frac{\frac{1}{p}F_{\mu^{\uplus(1-p)}}(z)-z}{\frac{1}{p}-1}
=F_\mu(z),\;\;\;\;\;z\in\mathbb{C}^+,$$ whence (1) holds. The assertion (a) holds since $\mu=\nu^{\boxplus1/p}$ and $1/p>1$, while (b)-(d) follow from the preceding discussions, (\[bijection\]), and Proposition \[prop2.1\](5).

The proof of the preceding proposition also gives the construction of the measure $\mu^{\boxplus 