p}$ whenever it is defined for $p\in(0,1)$. Indeed, by (\[p&lt;1o\]) the right inverse $\omega_p$ of the function $H_p(z)=pz+(1-p)F_\mu(z)$ ($H_p=F_{\mu^{\uplus(1-p)}}$) satisfies the relation $$\label{p<1} F_{\mu^{\boxplus
p}}(z)=\frac{p\omega_p(z)-z}{p-1},\;\;\;\;z\in\mathbb{C}^+,$$ and we have $F_{\mu^{\boxplus p}}(z)=F_\mu(\omega_p(z))$ for $z\in H_p(\mathbb{C}^+)$.

The following proposition can be proved by \[\[BN1\], Proposition 3.1\]. It can be also obtained by using (\[formula\]), (\[general\]), and (\[p&lt;1\]), and we leave the proof for the reader.

\[3.6\] Let $\mu\in\mathcal{M}$ and for $p,q>0$ let $q'=1+pq-p$ and $p'=pq/q'$.

1.  [If $\mu^{\boxplus p}$ is defined and $q'>0$ then we have the following identity $$\label{formula2}\left(\mu^{\boxplus p}\right)^{\uplus q}=\left(\mu^{\uplus q'}\right)^{\boxplus
    p'}.$$]{}

2.  The formula $(\ref{formula})$ holds for either

    1.  [$p\geq1$ or]{}

    2.  [$1-\mathrm{Ind}(\mu^{\uplus q})\leq p<1$ and $1+pq-q>0$.]{}

It was proved in \[\[BN1\]\] that $\mu\in\mathbb{B}_t(\mathcal{M})$ for any finite $t$ with $0\leq t\leq\mathrm{Ind}(\mu)$. In the following proposition we give an explicit expression for the measure $\mu_t$ so that $\mu=\mathbb{B}_t(\mu_t)$. The reader should be aware of that this conclusion holds under the essential condition that $t$ has to be finite and this condition may not be noticed without caution.

If $\mu\in\mathcal{M}$ then for any finite number $t$ with $-1<t\leq\mathrm{Ind}(\mu)$ there exists a unique measure $\mu_t\in\mathcal{M}$ such that $$\label{N}
\mu=\left(\mu_t^{\boxplus(1+t)}\right)^{\uplus\frac{1}{1+t}},$$ in which case $\mu_t$ can be expressed as $$\label{N1}
\mu_t=\left(\mu^{\uplus(t+1)}\right)^{\boxplus\frac{1}{t+1}}.$$ Particularly, we have $\mu=\mathbb{B}_t(\mu_t)$ 