r_m, u_{m-1})\le\frac{F_M(0, r_m,
  0)}{1-e^{-M}}.$$ Thus, we can replace $F_M(0, r_m, 0)$ with $F_M(r_{m-1}, r_m, u_{m-1})$ in (\[red\]) at the expense of an extra factor in the right-hand side and obtain (\[co2\]).

From Corollary \[corbase\] we see that for each ${\varrho}\in(0,\infty)$ measures $\Pi^{\varrho}_n$ on ${\mathbb{N}}^n$, $n\in{\mathbb{N}}$, defined by $$\label{pi}
  \Pi^{\varrho}_n(\{r_1,r_2,\dots,r_n\}):=\prod_{i=1}^n (1-q)^{r_i}{\varrho}F_M(r_{i-1}, r_i, u_{i-1}),\ (r_1,r_2,\dots,r_n)\in{\mathbb{N}}^n,$$ form an “almost” consistent family of “almost” probability measures.

We can sharpen (\[rho\]) as follows.

\[propcalc\] Let $q$ be defined as in Corollary \[corbase\] and ${\varrho}=p/(1-p)$. Then $$\lim_{{\varrho}\to
    0}{\varrho}\log(1/q)=2(e^M-1)\ \text{ and }\
  \lim_{M\to\infty}e^{-M}\log(1/q)=2/{\varrho}.$$

In view of (\[rho\]) we only need to show that for every ${\varepsilon}>0$ there is ${\varrho}_0>0$ such that ${\varrho}\log(1/q)\le
  2(1+{\varepsilon})(e^M-1)$ for all ${\varrho}\in(0,{\varrho}_0)$. Assume the contrary, i.e. that there is ${\varepsilon}\in(0,1/2)$ such that for every ${\varrho}_0>0$ there is ${\varrho}\in(0,{\varrho}_0)$, for which ${\varrho}\log(1/q)> 2(1+{\varepsilon})(e^M-1)$. Then $$\sum_{r=1}^\infty
  \frac{{\varrho}(1-q)^r}{2(e^M-1)r}>1+{\varepsilon}.$$ We shall show that this contradicts to the fact that the total mass of $\Pi^{\varrho}_n$ is bounded above by $1$ uniformly in $n\in{\mathbb{N}}$ and ${\varrho}\in(0,\infty)$. Recall that for all $\ell\ge 0,\ r\ge 1$ and $u\in[0,1]$ $$F_M(\ell,r,u)\ge
  F_M(0,r,0)=\frac{1}{2re^M}=\frac{1-e^{-M}}{2r(e^M-1)}$$ and that, by (\[Fb\]), for all $\ell,r> R$, where $R$ is sufficiently large, $$F_M(\ell,r,u)\ge \frac{(1-{\varepsilon}/4)}{2r(e^M-1)}.$$ Given $r_1,r_2,\dots,r_n$, we split the set of indices into $A_R:=\{i\in\{1,2,\dots,n\}:\,\min\{r_{i-1},r_i\}\le R\}$ and $\bar{A}_R:=\{1,2,\dots,n\}\setminus A_R$. Then $$\begin{gathered}
    