\[\[BP\], Proposition 6.2\]. To prove (1), first consider the case that $0<t:=\lim\sup_n\mathrm{Ind}(\mu_n)<\infty$. Then for sufficiently small $\epsilon>0$ we have $1<\lim\sup_n\mathrm{Ind}(\mu_n^{\uplus(t-\epsilon)})$ by Lemma \[3.4\], whence there exists a subsequence $\{\mu_{n_k}\}$ such that $\mu_{n_k}^{\uplus(t-\epsilon)}$ is $\boxplus$-infinitely divisible for all $k$. Since the set of $\boxplus$-infinitely divisible measures is weakly closed, we see that $\mathrm{Ind}(\mu^{\uplus(t-\epsilon)})\geq1$ by (3), which yields $t-\epsilon\leq\mathrm{Ind}(\mu)$. Letting $\epsilon\to0$ shows $t\leq\mathrm{Ind}(\mu)$. If $t=\infty$ then by similar arguments it is easy to see that $m\leq\mathrm{Ind}(\mu)$ for any $m>0$, and therefore $\mathrm{Ind}(\mu)=\infty$. The assertion (1) clearly holds if $t=0$, and hence the proof is complete.

It was shown in Proposition \[3.1\] that the subordination function for the $\boxplus$-convolution power appearing in (\[p-power\]) is in fact the reciprocal Cauchy transform of some $\boxplus$-infinitely divisible measure. The following theorem states that the converse is also true. For other related results about the $\boxplus$-infinite divisibility of the subordination functions, we refer the reader to \[\[G4\]\] and \[\[Nica\]\].

\[sub\] If $\mu\in\mathcal{M}$ then the following statements $(1)$ and $(2)$ are equivalent.

1.  [The measure $\mu$ is $\boxplus$-infinitely divisible.]{}

2.  [The function $F_\mu$ is the right inverse of some function $H\in\mathcal{H}$.]{}

If $(1)$ and $(2)$ hold then $F_\mu$ extends continuously to $\mathbb{C}^+\cup\mathbb{R}$, $H$ can be written as $$\label{H}
