\subsection{Iteration and statistics on entrance graphs} \label{sec:era-tree-statistics} The rapid decrease of the transverse amplitude makes the errors of all later returns summable. The graphs constructed in Section~\ref{sec:era-successors} let us apply the same return estimate after every finite history. Their inverse branches then fit the measure argument of Section~\ref{sec:coding-measure}. Fix an admissible graph $\Gamma$ at amplitude $B$, where $B$ is smaller than the thresholds of Theorem~\ref{thm:era-curve-step} and Section~\ref{sec:era-selected-return}. Use the amplitude sequence \eqref{eq:era-amplitude-sequence}, and put \begin{equation} \beta_n=B_n^\sigma,\qquad N_n=\lfloor1/\beta_n\rfloor. \label{eq:era-tree-margins} \end{equation} Define $\Gamma_\varnothing=\Gamma$. If $|\mathbf a|=n$ and $1\leq a\leq N_n$, let $\Gamma_{\mathbf a a}$ be the successor graph for the branch with $a$ cores, as in Theorem~\ref{thm:era-curve-step}. For the other digits take \[ \Gamma_{\mathbf a a}(r) =(r,B_{n+1}/2,B_{n+1}/2),\qquad r>2. \] The latter graphs are admissible at amplitude $B_{n+1}$; the corresponding branches will have empty domains. Induction proves that every $\Gamma_{\mathbf a}$ is admissible at amplitude $B_{|\mathbf a|}$. Indeed $0N_n. \end{cases} \label{eq:era-tree-domains} \end{equation} On a nonempty domain, let $h_{\mathbf a,a}$ be the inverse phase branch of $\Gamma_{\mathbf a}$ with $a$ cores. Empty domains contribute no image or constraint. Theorem~\ref{thm:era-curve-step} gives, for $y\in D_{\mathbf a,a}$, \begin{align} |h_{\mathbf a,a}(y)-g_a(y)|&\leq CB_n/a^2\leq CB_n,\notag\\ \left|\log\frac{|h_{\mathbf a,a}'(y)|}{|g_a'(y)|}\right| &\leq5C\sqrt{B_n}. \label{eq:era-tree-errors} \end{align} Each branch is $C^1$ with negative derivative, its image lies in $(1/(a+1),1/a)$, and images with different digits are disjoint. Thus these are branches of the kind considered in Section~\ref{sec:coding-measure}, with root interval $(0,1)$. Use $K_{\mathbf a},H_{\mathbf a},C_{\mathbf a}$ and $S$ for their terminal domains, compositions, cells and surviving set, as defined there. The excluded terminal set $A_{\mathbf a}$ in \eqref{eq:stopped-domain} is a subset of $(0,1)\setminus\mathcal V_{\Gamma_{\mathbf a}}(B_n)$. Indeed the union of the images of the nonempty domains in \eqref{eq:era-tree-domains} is precisely the retained set \eqref{eq:era-retained-phases} for $\Gamma_{\mathbf a}$. Theorem~\ref{thm:era-retained-phases} therefore gives \begin{equation} \lambda(A_{\mathbf a})\leq10B_{|\mathbf a|}^{\sigma}. \label{eq:era-tree-stopping} \end{equation} The intersections in \eqref{eq:actual-compositions} remain part of this construction: a successor graph contributes only those points whose entire preceding history lies in the actual branch domains. For $x\in(0,1)$, set \begin{equation} Z_{\mathbf a}(x) =\bigl(B_{|\mathbf a|},\Gamma_{\mathbf a}(2/x)\bigr). \label{eq:era-tree-state} \end{equation} The successor matching identities give the following two relations: \begin{align} \mathcal R_a\bigl(Z_{\mathbf a}(h_{\mathbf a,a}(y))\bigr) &=Z_{\mathbf a a}(y) &&(y\in D_{\mathbf a,a}),\notag\\ Z_{\mathbf a}(x)\in\mathcal V_a &\ \Longleftrightarrow\ x\in h_{\mathbf a,a}(D_{\mathbf a,a}) &&(x\in(0,1)). \label{eq:era-tree-one-step} \end{align} The first is Theorem~\ref{thm:era-curve-step}, including the identity $B_{n+1}=B_n^{4/3}$ in the amplitude coordinate. For a retained digit, the second is Lemma~\ref{lem:era-selection-image}. For an unretained digit both sides are false, by the digit condition in \eqref{eq:era-selected-entrance} and the empty branch domain. \begin{proposition} \label{prop:era-tree-realization} For every finite word $\mathbf a$ and $y\in K_{\mathbf a}$, \begin{equation} F_{\mathbf a}\bigl(Z_\varnothing(H_{\mathbf a}(y))\bigr) =Z_{\mathbf a}(y). \label{eq:era-tree-prefix-state} \end{equation} For $x\in(0,1)$, \begin{align} Z_\varnothing(x)\in\mathcal D_{\mathbf a} &\ \Longleftrightarrow\ x\in C_{\mathbf a},\notag\\ Z_\varnothing(x)\in\mathcal S &\ \Longleftrightarrow\ x\in S. \label{eq:era-tree-survival-identity} \end{align} On these corresponding surviving sets, the digits, continued fraction codes and phases agree. \end{proposition} \begin{proof} The prefix identity is immediate for the empty word. Suppose it holds for $\mathbf a$. If $y\in K_{\mathbf a a}$, then $y\in D_{\mathbf a,a}$ and $h_{\mathbf a,a}(y)\in K_{\mathbf a}$. Since $H_{\mathbf a a}=H_{\mathbf a}\circ h_{\mathbf a,a}$, the induction hypothesis identifies the state after $\mathbf a$ with $Z_{\mathbf a}(h_{\mathbf a,a}(y))$. The second identity in \eqref{eq:era-tree-one-step} says that this state belongs to $\mathcal V_a$, so the next composition is defined. The first identity then gives \begin{align*} F_{\mathbf a a} \bigl(Z_\varnothing(H_{\mathbf a a}(y))\bigr) &=\mathcal R_a\bigl(Z_{\mathbf a}(h_{\mathbf a,a}(y))\bigr)\\ &=Z_{\mathbf a a}(y). \end{align*} We prove the first equivalence in \eqref{eq:era-tree-survival-identity} by the same induction. It holds for the empty word. If $Z_\varnothing(x)\in\mathcal D_{\mathbf a}$, the induction hypothesis gives a unique $z\in K_{\mathbf a}$ with $H_{\mathbf a}(z)=x$. The prefix identity then says $F_{\mathbf a}(Z_\varnothing(x))=Z_{\mathbf a}(z)$. The next digit $a$ is selected exactly when $z\in h_{\mathbf a,a}(D_{\mathbf a,a})$. Writing $z=h_{\mathbf a,a}(y)$, this is exactly the condition $y\in K_{\mathbf a a}$ and $x=H_{\mathbf a a}(y)$. This proves the equivalence for $\mathbf a a$ in both directions. Taking the union at each depth and then the intersection over depths gives the second equivalence. Let $x\in S$ and put $p=Z_\varnothing(x)$. The unique allowed word of length $n$ for $p$ is the word of the unique cell of depth $n$ containing $x$. Their digits and codes are therefore equal. For that word $\mathbf a$, let $y_n(x)\in K_{\mathbf a}$ satisfy $H_{\mathbf a}(y_n(x))=x$. Equation~\eqref{eq:era-tree-prefix-state} gives \[ p_n=\bigl(B_n,\Gamma_{\mathbf a}(2/y_n(x))\bigr). \] Its radial coordinate is $2/y_n(x)>2$, so $x_n(p)=2/(2/y_n(x))=y_n(x)$, as required. \end{proof} For fixed $B,\Gamma$, define \begin{equation} G_{\Gamma,B} =\{x\in(0,1):(B,\Gamma(2/x))\in\mathcal S_{\mathrm{stat}}\}. \label{eq:era-curve-statistical-set} \end{equation} The equality of codes and surviving sets in Proposition~\ref{prop:era-tree-realization} gives the exact identity \begin{equation} G_{\Gamma,B}=\{x\in S:\psi(x)\in\mathcal B\}. \label{eq:era-typical-set-identity} \end{equation} Here $\psi$ is the code associated with the inverse branches, as in Section~\ref{sec:coding-measure}. Thus the set on which the branch measure argument gives statistics is the restriction of the same Borel set of entrance states for every graph. \begin{theorem}[Statistics on entrance graphs] \label{thm:era-graph-statistics} For the fixed parameters $\eta,J,\sigma$, there are constants $0\frac12, \qquad \lambda\bigl((I\cap(0,1))\setminus G_{\Gamma,B}\bigr) \leq\min\{\lambda(I),80B^\sigma\} \label{eq:era-graph-statistical-loss} \end{equation} for every Borel set $I\subset\R$. For every $x\in G_{\Gamma,B}$, the selected returns starting at $p=(B,\Gamma(2/x))$ continue indefinitely, and all six limits of Theorem~\ref{thm:main} hold for their digits $a_n(p)$ and phases $x_n(p)$. \end{theorem} \begin{proof} First choose $B_{\mathrm{cap}}$ for Proposition~\ref{prop:era-selected-borel} and Lemma~\ref{lem:era-selection-image}. This fixes the Borel set $\mathcal S_{\mathrm{stat}}$. Choose $B_*>0$ no greater than $B_{\mathrm{cap}}$ and the graph iteration threshold, and small enough that \begin{equation*} B^\sigma<\frac1{160},\qquad 10C\sqrt B+4CB^{4/3}\leq\log2 \quad(0