\subsection{Statistics in radial families}
\label{sec:era-radial-families}

To use the statistical set on a transverse section, we integrate
along curves arriving at core entrances. A lower bound for the
derivative of the entrance phase transfers the loss estimate to the
original curve coordinate. The common set
$\mathcal S_{\mathrm{stat}}$ then makes these estimates compatible
across a measurable family of curves.

Throughout this subsection, fix $\eta,J,\sigma$ and the constants
$B_*,B_{\mathrm{cap}}$ of
Theorem~\ref{thm:era-graph-statistics}. We write $\lambda$ for
Lebesgue measure on $\R$. All weights are nonnegative, and their
integrals may be infinite.

The following elementary estimate records the dependence on the
phase derivative and on a density.

\begin{lemma}
\label{lem:radial-weighted-loss}
Let $I=[a,b]$, let $c>0$, and suppose that $f:\R\to\R$ is
differentiable at every point of $I$, with
\[
 f'(\rho)\geq c,\qquad f(\rho)\in(0,1)
       \quad(\rho\in I).
\]
Let $G\subset(0,1)$ be Borel and suppose
$\lambda((0,1)\setminus G)\leq L$. For every Borel set
$A\subset I$ and every nonnegative Borel weight $\omega$ satisfying
$\omega\leq M$ on $A$, where $M\geq0$, the set
$A\setminus f^{-1}(G)$ is Borel and
\begin{equation}
 \int_{A\setminus f^{-1}(G)}\omega(\rho)\dd\rho
   \leq
   \min\left\{\int_A\omega(\rho)\dd\rho,\frac{M}{c}L\right\}.
 \label{eq:radial-weighted-loss}
\end{equation}
\end{lemma}

\begin{proof}
Continuity and the mean value theorem imply that $f$ is strictly
increasing on $I$. In particular it is injective there.
For a Borel set $E\subset I$, the one-dimensional change of
variables formula gives
\[
 \lambda(f(E))
     =\int_E|f'(\rho)|\dd\rho
     \geq c\,\lambda(E).
\]
Consequently, whenever $f(E)\subset H$,
\begin{equation}
 \lambda(E)\leq
   \min\left\{(b-a)_+,\frac1c\lambda(H)\right\},
 \qquad (b-a)_+=\max\{b-a,0\}.
 \label{eq:radial-deleted-length}
\end{equation}
The continuity of $f|_I$ shows that
$E=A\setminus f^{-1}(G)$ is Borel. Its image lies in
$(0,1)\setminus G$, so $\lambda(E)\leq L/c$. Thus
\[
 \int_E\omega\dd\rho\leq M\lambda(E)\leq\frac{M}{c}L.
\]
The other bound in \eqref{eq:radial-weighted-loss} follows from
$E\subset A$.
\end{proof}

Here differentiability at an endpoint of $I$ means ordinary
differentiability, as in the return estimates used below.

Suppose now that $\Gamma$ is admissible at amplitude
$0<B\leq B_*$ and that an entrance state $Y(\rho)$ has the
representation
\begin{equation}
 Y(\rho)=\Gamma(2/f(\rho)),\qquad \rho\in I,
 \label{eq:radial-entrance-representation}
\end{equation}
where $f$ satisfies the hypotheses of
Lemma~\ref{lem:radial-weighted-loss}. Define the excluded set
\begin{equation}
 F_I=\{\rho\in I:(B,Y(\rho))\notin
                       \mathcal S_{\mathrm{stat}}\}.
 \label{eq:radial-statistical-failure}
\end{equation}
This set records the points left outside the chosen statistical
entrance construction. Such a point may enter the statistical set
at a later return; the capture argument will use precisely this
possibility.
By \eqref{eq:era-curve-statistical-set} and
\eqref{eq:radial-entrance-representation},
\begin{equation}
 F_I=I\setminus f^{-1}(G_{\Gamma,B}).
 \label{eq:radial-failure-identity}
\end{equation}
This identity proves measurability even if $Y$ is only specified
on $I$.

\begin{corollary}
\label{cor:radial-statistics}
The set $F_I$ is Borel. For every Borel $A\subset I$ and every
nonnegative Borel weight $\omega\leq M$ on $A$,
\begin{equation}
 \int_{A\cap F_I}\omega(\rho)\dd\rho
   \leq
   \min\left\{\int_A\omega(\rho)\dd\rho,
                       \frac{80M}{c}B^\sigma\right\}.
 \label{eq:radial-statistical-loss}
\end{equation}
For every $\rho\in I\setminus F_I$, the selected digits and phases
starting at $(B,Y(\rho))$ satisfy all six limits of
Theorem~\ref{thm:main}.
\end{corollary}

\begin{proof}
Theorem~\ref{thm:era-graph-statistics} gives
\[
 \lambda((0,1)\setminus G_{\Gamma,B})\leq80B^\sigma.
\]
Apply Lemma~\ref{lem:radial-weighted-loss} with
$G=G_{\Gamma,B}$ and use
\eqref{eq:radial-failure-identity}. For $\rho\notin F_I$,
$f(\rho)\in G_{\Gamma,B}$, and the same theorem gives the
statistical conclusion at
$(B,\Gamma(2/f(\rho)))=(B,Y(\rho))$.
\end{proof}

Let $\Lambda$ be a measurable space of labels, and equip $\R$
with its Borel $\sigma$-algebra. For each
$\ell\in\Lambda$, choose an interval
$I_\ell=[a_\ell,b_\ell]$, an amplitude
$B_\ell\in(0,B_*]$, and a graph $\Gamma_\ell$ admissible at
amplitude $B_\ell$. Assume that $\ell\mapsto B_\ell$ is
measurable. Let $D\subset\Lambda\times\R$ be measurable, with
$(\ell,\rho)\in D$ only if $\rho\in I_\ell$. An entrance family
$Y(\ell,\rho)$ is required to satisfy
\begin{equation}
 \begin{gathered}
   (\ell,\rho)\longmapsto(B_\ell,Y(\ell,\rho))
        \quad\hbox{is measurable on }D,\\
   Y(\ell,\rho)=\Gamma_\ell(2/f_\ell(\rho))
        \quad(\rho\in I_\ell),
 \end{gathered}
 \label{eq:radial-family-entrances}
\end{equation}
where each $f_\ell:\R\to\R$ is differentiable at every point
of $I_\ell$ and, for one constant $c>0$,
\begin{equation}
 f_\ell'(\rho)\geq c,\qquad
 f_\ell(\rho)\in(0,1)
       \quad(\ell\in\Lambda,\ \rho\in I_\ell).
 \label{eq:radial-family-slope}
\end{equation}
The graphs and phase functions are used separately on each fiber;
joint measurability is imposed on the entrance state in
\eqref{eq:radial-family-entrances}.

Define one excluded set in the full family by
\begin{equation}
 F=\{(\ell,\rho)\in D:
                (B_\ell,Y(\ell,\rho))\notin
                \mathcal S_{\mathrm{stat}}\}.
 \label{eq:radial-family-failure}
\end{equation}
For any $A\subset\Lambda\times\R$, write
$A_\ell=\{\rho:(\ell,\rho)\in A\}$.

\begin{theorem}[Integration over entrance families]
\label{thm:radial-family-statistics}
Under \eqref{eq:radial-family-entrances} and
\eqref{eq:radial-family-slope}, the set $F$ is measurable.
For every $(\ell,\rho)\in D\setminus F$, the selected returns
starting at $(B_\ell,Y(\ell,\rho))$ satisfy all six statistical
limits of Theorem~\ref{thm:main}.

Let $\mu$ be a $\sigma$-finite measure on $\Lambda$, let $A\subset D$ be
measurable, and let
$\omega:\Lambda\times\R\to[0,\infty]$ be measurable.
Suppose that $M:\Lambda\to[0,\infty)$ is measurable and
\[
 \omega(\ell,\rho)\leq M_\ell
       \quad((\ell,\rho)\in A).
\]
Put
\begin{equation}
 m_A(\ell)=\int_{A_\ell}\omega(\ell,\rho)\dd\rho,
 \qquad
 L_\ell=\frac{80M_\ell}{c}B_\ell^\sigma.
 \label{eq:radial-family-mass}
\end{equation}
Then
\begin{align}
 \int_{A\cap F}\omega\dd(\mu\times\lambda)
   &\leq\int_\Lambda\min\{m_A(\ell),L_\ell\}\dd\mu(\ell)
       \notag\\
   &\leq
     \min\left\{\int_A\omega\dd(\mu\times\lambda),
                         \int_\Lambda L_\ell\dd\mu(\ell)\right\}.
 \label{eq:radial-family-integral}
\end{align}
\end{theorem}

\begin{proof}
Since $\mathcal S_{\mathrm{stat}}$ is Borel, its complement has
a measurable inverse image under the restricted state map in
\eqref{eq:radial-family-entrances}. The domain $D$ is measurable,
so this inverse image is also measurable in
$\Lambda\times\R$. This is precisely $F$.

Fix $\ell$. The representation of $Y$ gives
\[
 F_\ell
   =D_\ell\setminus f_\ell^{-1}(G_{\Gamma_\ell,B_\ell}).
\]
If $(\ell,\rho)\in D\setminus F$, then
$f_\ell(\rho)\in G_{\Gamma_\ell,B_\ell}$.
Theorem~\ref{thm:era-graph-statistics} therefore supplies
the statistical conclusion for the stated entrance.

For the integral estimate, the section $A_\ell$ is Borel and lies
in $I_\ell$. Since $A\subset D$,
\begin{equation*}
 (A\cap F)_\ell
    =A_\ell\setminus f_\ell^{-1}
                          (G_{\Gamma_\ell,B_\ell}).
\end{equation*}
Corollary~\ref{cor:radial-statistics}, with weight
$\omega(\ell,\cdot)$ and bound $M_\ell$, gives
\begin{equation}
 \int_{(A\cap F)_\ell}\omega(\ell,\rho)\dd\rho
       \leq\min\{m_A(\ell),L_\ell\}.
 \label{eq:radial-family-fiber-bound}
\end{equation}

For a measurable set $E\subset\Lambda\times\R$,
the function $\ind_E\omega$ is measurable and nonnegative.
Both $\mu$ and Lebesgue measure are $\sigma$-finite. Tonelli's
theorem shows that the fiber integral is measurable and that
\begin{equation}
 \int_E\omega\dd(\mu\times\lambda)
     =\int_\Lambda
          \left(\int_{E_\ell}\omega(\ell,\rho)\dd\rho\right)
          \dd\mu(\ell).
 \label{eq:radial-family-tonelli}
\end{equation}
Apply this identity to $E=A\cap F$ and integrate
\eqref{eq:radial-family-fiber-bound}. This proves the first
inequality in \eqref{eq:radial-family-integral}. For the second,
the integrand is bounded both by $m_A(\ell)$ and by $L_\ell$.
Another application of \eqref{eq:radial-family-tonelli}, now
with $E=A$, completes the proof.
\end{proof}

The first bound in \eqref{eq:radial-family-integral} retains the
mass of every restricted fiber. In particular, integrability of
$M_\ell B_\ell^\sigma$ gives finite weighted mass of the excluded set.
This is the form needed for the section flux, whose density grows
near the boundary. We next construct
the entrance family from initial data on a fixed small section
and prove the amplitude and phase derivative estimates required
above.