surjective so that there exist $\ud{x}^+$ and $\ud{x}^-$ with $\overline{\varphi}_\infty(\ud{x}^+)=1$ and $\overline{\varphi}_\infty(\ud{x}^-)=0$. Suppose that $\ud{x}^n\downarrow\ud{x}$ and set $y:=\overline{\varphi}_\infty(\ud{x})$, $x_n:=\overline{\varphi}_\infty(\ud{x}^n)$. By Lemma \[lem2.3\] the sequence $\{x_n\}$ is monotone decreasing; let $x:=\lim_nx_n$. Suppose that $y<x$ and $y<z<x$. Since $y<z<x_n$ for any $n\geq
1$ and $\lim_n\ud{x}^n=\ud{x}$, we prove, as in the beginning of the proof of Lemma \[lem2.4quatro\], that $z\in X\backslash S$. The validity of the $\varphi$-expansion implies that $z=\overline{\varphi}_\infty(\i(z))$. By Lemma \[lem2.4quinto\] $$\ud{x}\preceq\i(z)\preceq\ud{x}^n\,.$$ Since these inequalities are valid for any $z$, with $y<z<x$, the validity of $\varphi$-expansion implies that we have strict inequalities, $\ud{x}\prec\i(z)\prec \ud{x}^n$. This contradicts the hypothesis that $\lim_{n\ra\infty}\ud{x}^n=\ud{x}$. A similar argument holds in the case $\ud{x}^n\uparrow\ud{x}$. Hence $$\lim_{n\ra\infty}\ud{x}^n=\ud{x}\implies \lim_{n\ra\infty}
\overline{\varphi}_\infty(\ud{x}^n)=\overline{\varphi}_\infty(\ud{x})\,.$$

Conversely, suppose that $\overline{\varphi}_\infty \colon
\tA^{\Z_+}\ra [0,1]$ is well-defined and continuous. Then, given $\delta>0$ and $\ud{x}\in\tA^{\Z_+}$, $\exists n$ so that $$0\leq\sup\{\overline{\varphi}_\infty(\ud{x}^\prime)\colon
x^\prime_j=x_j\;j=0,\ldots,n-1\}-
\inf\{\overline{\varphi}_\infty(\ud{x}^\prime)\colon
x^\prime_j=x_j\;j=0,\ldots,n-1\}\leq \delta\,.$$ We set $$\ud{x}^{n,-}:=x_0\cdots x_{n-1}\ud{x}^-\quad\text{and}\quad
\ud{x}^{n,+}:=x_0\cdots x_{n-1}\ud{x}^+\,.$$ For any $x\in X\backslash S$ we have the identity , $$x=\varphi\big(\i_0(x)+\varphi(\i_1(x)+\ldots
+\varphi(\i_{n-1}+T^{n}x))\big)=
\overline{\varphi}_{n}(\i_0(x),\ldots,\i_{n-1}(x)+T^{n}x)\,.$$ If $\delta(\i_0(x)\cdots\i_{n-1}(x))=1$, then $$\begin{aligned}
\overline{\varphi}_\infty(\ud{x}^{n,-}):&= \overline{\varphi}_n(
\i_0(x),\ldots,\i_{n-1}(x)+\overline{\varphi}_\infty(\ud{x}^-))\\
&=\overline{\varphi}_n(
\i_0(x),\ldots,\i_{n-1}(x))\\
&\leq
\overline{\varphi}_{n}(\i_0(x),\ldots,\i_{n-1}(x)+T^{n}x)\\
&\leq \overline{\varphi}_{n}(
\i_0(x),\ldots,\i_{n-1}(x)+1)\\
&=\overline{\varphi}_{n}(
\i_0(x),\ldots,\i_{n-1}(x)+\overline{\varphi}_\infty(\ud{x}^+))=:
\overline{\varphi}_\infty(\ud{x}^{n,+})\,.\end{aligned}$$ If $\delta(\i_0(x)\cdots\i_{n-1}(x))=-1$, then the inequalities are reversed. Letting $n$ going to infinity, we get $\overline{\varphi}_\infty(\i(x))=x$.

\[rem2.4\] When the maps $f_0$ and $f_{k-1}$ are increasing, then we can take 