the $\varphi$-code (for Lemma \[lem2.4\] see e.g. [@CoE]).

\[lem2.4\] The $\varphi$-code $\i$ is $\preceq$-order-preserving on $X\backslash S$: $x\leq y$ implies $\i(x)\preceq\i(y)$.

Let $x<y$. Either $\i_0(x)<\i_0(y)$, or $\i_0(x)=\i_0(y)$; in the latter case, the strict monotonicity of $f_{\i_0(x)}$ implies $$\begin{aligned}
\varphi^{-1}(x)&=\i_0(x) +T(x)< \varphi^{-1}(y)=\i_0(x)
+T(y)\quad\text{if
$\delta(\i_0(x))=+1$}\\
\varphi^{-1}(x)&=\i_0(x) +T(x)> \varphi^{-1}(y)=\i_0(x)
+T(y)\quad\text{if $\delta(\i_0(x))=-1$.}\end{aligned}$$ Repeating this argument we get $\i(x)\preceq\i(y)$.

\[lem2.4bis\] The $\varphi$-code $\i$ is continuous[^3] on $X\backslash S$.

Let $x\in X\backslash S$ and $\{x^n\}\subset X\backslash S$, $\lim_n x^n=x$. Let $x\in I_{j_0}$. For $n$ large enough $x^n\in
I_{j_0}$ and $\i_0(x^n)=\i_0(x)=j_0$. Let $j_1:=\i_1(x)$; we can choose $n_1$ so large that for $n\geq n_1$ $Tx_{n}\in I_{j_1}$. Hence $\i_0(x^n)=j_0$ and $\i_1(x^n)=j_1$ for all $n\geq n_1$. By induction we can find an increasing sequence $\{n_m\}$ such that $n\geq n_m$ implies $\i_j(x)=\i_j(x^{n})$ for all $j=0,\ldots,m$.

The next lemmas give the essential properties of the map $\overline{\varphi}_\infty$.

\[lem2.2\] Let $\ud{x}\in\tA^{\Z_+}$. Then there exist $y_\uparrow(\ud{x})$ and $y_\downarrow(\ud{x})$ in $[0,1]$, such that $y_\uparrow(\ud{x})\leq y_\downarrow(\ud{x})$; $y_\uparrow(\ud{x})$ and $y_\downarrow(\ud{x})$ are the only possible cluster points of the sequence $\{\overline{\varphi}_n(\ud{x})\}_n$.\
Let $x\in X\backslash S$ and set $\ud{x}:=\i(x)$. Then $$a_j\leq y_\uparrow(\ud{x})\leq x\leq y_\downarrow(\ud{x})\leq
a_{j+1}\quad \text{if $x_0=j$}\,.$$ If the $\varphi$-expansion is valid, then each $y\in X\backslash
S$ has a unique $\varphi$-expansion[^4], $$y=\overline{\varphi}_\infty(\ud{x})\in X\backslash S\iff
\ud{x}=\i(y)\,.$$

Consider the map $$t\mapsto 