prove the surjectivity of $\overline{\varphi}_\infty$ it is sufficient to consider $s\in S$. The argument is a variant of the proof of Lemma \[lem2.4quatro\]. Let $\ud{x}^\prime$ be a string such that for any $n\geq 1$ $$f_{x^\prime_{n-1}}\cirk \cdots \cirk
f_{x^\prime_0}(s)\in\overline{I}_{x^\prime_n}\,.$$ We use here the extension of $f_j$ to $\overline{I}_j$; we have a choice for $x^\prime_n$ whenever $f_{x^\prime_{n-1}}\cirk \cdots
\cirk f_{x^\prime_0}(s)\in S_0$. Suppose that $\overline{\varphi}_\infty(\ud{x}^\prime)<s$ and that $\overline{\varphi}_\infty(\ud{x}^\prime)<z<s$. Since $s,
\overline{\varphi}_\infty(\ud{x}^\prime)\in
\overline{I}_{x_0^\prime}$, we have $z\in I_{x_0^\prime}$ and therefore $\i(z)=x_0^\prime$. Moreover, $$\overline{\varphi}_\infty(\sigma\ud{x}^\prime)<Tz<f_{x^\prime_0}(s)\quad\text{if
$\delta(x_0^\prime)=1$}$$ or $$f_{x^\prime_0}(s)<Tz<\overline{\varphi}_\infty(\sigma\ud{x}^\prime)\quad\text{if
$\delta(x_0^\prime)=-1$}\,.$$ Iterating the argument we get $z\in X\backslash S$ and $\i(z)=\ud{x}^\prime$, contradicting the validity of the $\varphi$-expansion. Similarly we exclude the possibility that $\overline{\varphi}_\infty(\ud{x}^\prime)>s$, thus proving the surjectivity of the map $\overline{\varphi}_\infty$.

Suppose that $\overline{\varphi}_\infty\colon \tA^{\Z_+}\ra [0,1]$ is well-defined and surjective. Let $x\in X\backslash S$ and $\ud{x}=\i(x)$. Suppose that $x<
\overline{\varphi}_\infty(\ud{x})$. By Lemma \[lem2.4quatro\] any $z$, such that $x<z<\overline{\varphi}_\infty(\ud{x})$, does not have a $\varphi$-expansion. This contradicts the hypothesis that $\overline{\varphi}_\infty$ is surjective. Similarly we exclude the possibility that $x>
\overline{\varphi}_\infty(\ud{x})$.

\[thm2.1ter\] A $\varphi$-expansion is valid if and only if $\overline{\varphi}_\infty: \tA^{\Z_+}\ra [0,1]$ is well-defined, continuous and there exist $\ud{x}^+$ with $\overline{\varphi}_\infty(\ud{x}^+)=1$ and $\ud{x}^-$ with $\overline{\varphi}_\infty (\ud{x}^-)=0$.

Suppose that the $\varphi$-expansion is valid. By Theorem \[thm2.1bis\] $\overline{\varphi}_\infty$ is well-defined and 