Since $\nu_r^{\boxplus\left(1+q_1/r\right)}=\mathbb{B}_{p,(1-q)r+q}(\mu)$ by (\[formula\]), Proposition \[prop2.1\](4) and (\[1\]) imply the assertion (4). Letting $r=1$ in (4) yields the last assertion.

Observe that if $\mu$ is $\boxplus$-infinitely divisible then $\mu\in\mathbb{B}(\mathcal{M})$, i.e., the measure $\mathbb{B}^{-1}(\mu)=(\mu^{\uplus2})^{\boxplus1/2}$ is defined. In order to investigate the measure of the form $\mu^{\boxplus p}$, $0<p<1$ (that is, $\mu=\nu^{\boxplus1/p}$ for some $\nu\in\mathcal{M}$ ), we need the following lemma. This lemma was also provided in \[\[Japan\]\]; however, the case $\mathrm{Ind}(\mu)=\infty$ was not considered there.

\[3.4\] For any measure $\mu\in\mathcal{M}$ and any number $q>0$, we have $$\mathrm{Ind}\left(\mu^{\uplus q}\right)=\frac{\mathrm{Ind}(\mu)}{q}.$$

First claim the inequality $\mathrm{Ind}(\mu^{\uplus q})\geq\mathrm{Ind}(\mu)/q$ holds. It clearly holds if $\mathrm{Ind}(\mu)=0$. Next, consider the case $\mathrm{Ind}(\mu)>0$. Then for any finite $r$ with $0<r<\mathrm{Ind}(\mu)$ pick a measure $\nu\in\mathcal{M}$ such that $\mu=\mathbb{B}_r(\nu)$, from which we obtain that $$\mathrm{Ind}\left(\mu^{\uplus q}\right)=\mathrm{Ind}\left(\left(\nu^{\boxplus(r+1)}\right)^{\uplus
\frac{q}{r+1}}\right)=\mathrm{Ind}\left(\mathbb{B}_{r/q}(\nu_1)\right)\geq
\frac{r}{q},$$ where $\nu_1\in\mathcal{M}$ and (\[t\]) is used in the second equality above. If $\mathrm{Ind}(\mu)=\infty$ then letting $r\uparrow\infty$ gives $\mathrm{Ind}(\mu^{\uplus
q})=\infty$ as well, which implies the desired inequality. Otherwise, letting $r\uparrow\mathrm{Ind}(\mu)<\infty$ yields the claim. By considering the identity $\mu=\left(\mu^{\uplus
q}\right)^{\uplus1/q}$ and using the claim, we obtain the opposite inequality, and then the proof is complete.

Now we are able to determine for what value of $p\in(0,1)$ the $p$-th $\boxplus$-convolution power of a measure is defined. 