The function $d$ defined in (0.3) may not be a metric unless\begin{equation}\phi_p(a_1,\dots, a_p)\le \phi_{p+1} (a_1,\dots, a_{p+1}),\quad\text{for all } (a_i)^{p+1}_1 \in \Gamma^{p+1},\ p=1,\dots.\tag{A}\end{equation}Let$$\tilde\phi_p(a_1,\dots, a_p)=\max_{1\le i\le p}\phi_1 (a_1,\dots, a_i),\quad (a_i)^p_1\in\Gamma^p,\ p=1,\dots.$$Then $\tilde\phi$ satisfies (A) and (0.1). Note that\begin{equation}\phi_p(a_1,\dots, a_p) \le\tilde\phi_p (a_1,\dots, a_p)\le p \max_{1\le i\le n} \phi_1(a_i).\tag{B}\end{equation}In (0.3) replace the equality $d(a,b)=e^{-\phi_p(a_1,\dots,a_p)}$ by $d(a,b)=e^{-\tilde\phi_p(a_1,\dots,a_p)}$. Then $d(a,b)\le \max(d(a,c), d(c,b))$. Let\begin{gather*}\tilde\psi_p(x):=\max_{1\le i\le p} \psi_i(x),\ x\in\Gamma^\infty,\quad \tilde\alpha_p(\mu) :=\int \tilde\psi_p\,d\mu,\ p=1, \dots\\\tilde\alpha(\mu) :=\lim_{p\rightarrow\infty} \frac{\tilde\alpha_p(\mu)}{p}.\end{gather*}Then (B) yields$$\lim_{p\rightarrow\infty} \frac{\psi_p(x)}{p} = s\Rightarrow \lim_{p\rightarrow\infty} \frac{\tilde\psi_p(x)}{p}=s.$$
