\subsection{Finite flux of the excluded set}
\label{sec:small-section-finite-failure}

There are two bounds for the excluded part of a radial interval:
its total length and the statistical bound proportional to
$(n_2n_3)^\sigma$. Their minimum is integrable against the
logarithmic wall measure. We first compute this integral,
then apply it to the common statistical family and sum over
all sufficiently small shear scales.

\begin{lemma}
\label{lem:clipped-logarithmic-integral}
For $\ell,b,\sigma>0$, put $z=\max\{\log(b/\ell),0\}$.
Then
\begin{align}
 \int_0^1\frac{\min\{\ell,bx^\sigma\}}x\dd x
    &=\frac{\min\{\ell,b\}}\sigma(1+z),
 \label{eq:clipped-logarithmic-one}\\
 \int_0^1\!\int_0^1
        \frac{\min\{\ell,b(xy)^\sigma\}}{xy}\dd y\dd x
    &=\frac{\min\{\ell,b\}}{\sigma^2}
                       \left(1+z+\frac{z^2}2\right).
 \label{eq:clipped-logarithmic-two}
\end{align}
\end{lemma}

\begin{proof}
The integrands are nonnegative. In one variable they are
bounded above by $bx^{\sigma-1}$, which is integrable at
zero. If $b\leq\ell$, clipping never occurs, and direct
power integration proves both formulas.

Suppose $b>\ell$, and set $c=(\ell/b)^{1/\sigma}\in(0,1)$.
For the first formula split at $c$. The integrand equals
$bx^{\sigma-1}$ below $c$ and $\ell/x$ above it. Hence
the integral is $\ell/\sigma+\ell\log(1/c)$, giving
\eqref{eq:clipped-logarithmic-one}.

For the second formula, integrate first in $y$ and divide
by $x$. The resulting function is
\[
 \begin{cases}
  (b/\sigma)x^{\sigma-1},&0<x<c,\\
  (\ell/\sigma-\ell\log c+\ell\log x)/x,&c<x<1.
 \end{cases}
\]
The first part has integral $\ell/\sigma^2$; the second
has integral $-(\ell/\sigma)\log c
+(\ell/2)(\log c)^2$. Each is finite. Substituting
$\log c=-z/\sigma$ proves~\eqref{eq:clipped-logarithmic-two}.
\end{proof}

For $K\geq1$ and $\sigma>0$ define
\begin{equation}
 \begin{split}
 \mathcal L(a;K,\sigma)
   &=\int_0^1\!\int_0^1
       \frac{\min\{a,Ka^{-2\sigma}(xy)^\sigma\}}{xy}
          \dd y\dd x,\\
 C_{\log}(K,\sigma)
   &=\frac{1+k+k^2/2}{\sigma^2},\qquad
       k=\log K+1+2\sigma.
 \end{split}
 \label{eq:logarithmic-layer-cost}
\end{equation}
For $0<a\leq1$, the preceding lemma gives
\begin{equation}
 \mathcal L(a;K,\sigma)
       \leq C_{\log}(K,\sigma)\,
                         a(1+|\log a|)^2.
 \label{eq:logarithmic-layer-bound}
\end{equation}
To see the uniform constant, write $L=1+|\log a|\geq1$.
The logarithm in~\eqref{eq:clipped-logarithmic-two} is
\[
 z=\log K+(1+2\sigma)|\log a|\leq kL.
\]
It follows that $1+z+z^2/2\leq(1+k+k^2/2)L^2$,
which proves~\eqref{eq:logarithmic-layer-bound}.

Use the common label set $\Lambda_c$, amplitude $B_\nu$
and excluded set $\mathcal F_\nu$ from
Section~\ref{sec:small-section-common-statistics}. Make
their dependence on the scale explicit by setting
\begin{equation}
 \begin{split}
 E_a^+&=\{(\rho,-\log n_2,-\log n_3):
               \nu\in\Lambda_c,\ \rho\in\mathcal F_\nu\},\\
 E_a&=E_a^+\cup\mathsf s E_a^+,\qquad
 K_\sigma=640\sqrt3\,800^\sigma.
 \end{split}
 \label{eq:statistical-failure-section}
\end{equation}
Here the return defining $\mathcal F_\nu$ uses the
fixed time $t_0=(q/2)\log(1/a)$. In particular both signs
use the same statistical cap and the same set
$\mathcal S_{\mathrm{stat}}$.

\begin{theorem}[Finite flux on a shear layer]
\label{thm:finite-failure-layer}
Fix admissible statistical parameters $\eta,J,\sigma$
and their cap as in Theorem~\ref{thm:common-shear-statistical-loss}.
There is $h_*>0$ such that, for $0<h\leq h_*$,
$q>3$, $D>4$ and $P>6D+2$, all sufficiently small
$a>0$ have the following properties, uniformly for
$0\leq\varepsilon\leq1/128$. The sets $E_a^+$ and
$E_a$ are Borel and
\begin{align}
 \mu_h(E_a^+)&\leq
       3C_{\log}(K_\sigma,\sigma)a(1+|\log a|)^2,
 \label{eq:positive-failure-flux}\\
 \mu_h(E_a)&\leq
       6C_{\log}(K_\sigma,\sigma)a(1+|\log a|)^2<\infty.
 \label{eq:signed-failure-flux}
\end{align}
For $\nu\in\Lambda_c$, $\rho\in[a,2a]$ and
$\varsigma\in\{1,-1\}$, if
\begin{equation}
 (\varsigma\rho,-\log n_2,-\log n_3)\notin E_a,
 \label{eq:signed-failure-exclusion}
\end{equation}
then the original orbit through $x_\varsigma(\rho,n_2,n_3)$
has successive complete eras satisfying all six statistical
limits of Theorem~\ref{thm:main}, with entrance times
tending to infinity.
\end{theorem}

\begin{proof}
Decrease $h_*$ to at most $1/8$, and the scale threshold
so that $a\leq\min\{h/8,1\}$. All labels satisfy
$0<n_i<a$, and Lemma~\ref{lem:small-section-flux-bounds}
places the entire family inside $\mathcal D_h$.
The joint Borel property from
Theorem~\ref{thm:common-shear-statistical-loss} and
\eqref{eq:logarithmic-radial-family} show that $E_a^+$
is Borel. Its reflection and their union are Borel too.

On each radial interval use
\eqref{eq:common-statistical-weighted-loss} with the
actual density $w_h\leq3$. Together with the total
mass bound $3a$, this gives
\begin{equation*}
 \mu_{h,n_2,n_3}(\mathcal F_\nu)
       \leq3\min\{a,640\sqrt3\,B_\nu^\sigma\}.
\end{equation*}
Under $n_2=ax$, $n_3=ay$, the amplitude bound becomes
\[
 B_{(ax,ay)}=800a^{-2}xy,\qquad
 640\sqrt3\,B_{(ax,ay)}^\sigma
            =K_\sigma a^{-2\sigma}(xy)^\sigma.
\]
Insert this in~\eqref{eq:logarithmic-radial-flux}, and
enlarge the nonnegative integral from the selected labels
to the whole unit square. The result is
$\mu_h(E_a^+)\leq3\mathcal L(a;K_\sigma,\sigma)$.
Since $K_\sigma\geq1$, inequality
\eqref{eq:logarithmic-layer-bound} proves
\eqref{eq:positive-failure-flux}. Reflection invariance
\eqref{eq:section-reflection-measure} proves
\eqref{eq:signed-failure-flux}.

For positive shear, exclusion from $E_a$ implies
$\rho\notin\mathcal F_\nu$. For negative shear it
implies $\rho\notin\mathcal F_{(n_3,n_2)}$, since
\[
 \mathsf s(\rho,-\log n_3,-\log n_2)
           =(-\rho,-\log n_2,-\log n_3).
\]
These are exactly the two excluded sets in
\eqref{eq:signed-shear-statistical-data}. The actual
entrance and complete-era conclusions of
Section~\ref{sec:small-section-signed-return} apply.
\end{proof}

For $0<d\leq1$ let $a_j=d\,2^{-(j+1)}$, $j\geq0$.
The elementary geometric moments
\[
 \sum_{j\geq0}2^{-(j+1)}=1,\qquad
 \sum_{j\geq0}(j+1)2^{-(j+1)}=2,\qquad
 \sum_{j\geq0}(j+1)^2 2^{-(j+1)}=6
\]
give
\begin{equation}
 \begin{split}
 \sum_{j\geq0}a_j(1+|\log a_j|)^2
  &=d\big[(1-\log d)^2+4(1-\log d)\log2
                                      +6(\log2)^2\big]\\
  &\leq11d(1+|\log d|)^2.
 \end{split}
 \label{eq:dyadic-logarithmic-sum}
\end{equation}
For the inequality use $0<\log2\leq1\leq1-\log d$.

\begin{corollary}
\label{cor:dyadic-statistical-failure}
Under the choices of Theorem~\ref{thm:finite-failure-layer},
take $d>0$ below its scale threshold and $d\leq1$.
For fixed $\varepsilon\in[0,1/128]$, form each $E_{a_j}$
with its own cutoff $(q/2)\log(1/a_j)$ and set
$E_d=\bigcup_{j\geq0}E_{a_j}$. Then $E_d$ is Borel and
\begin{equation}
 \mu_h(E_d)\leq\sum_{j\geq0}\mu_h(E_{a_j})
   \leq66C_{\log}(K_\sigma,\sigma)
                    d(1+|\log d|)^2<\infty.
 \label{eq:dyadic-statistical-failure-flux}
\end{equation}
Every selected initial point from any of these layers
outside $E_d$ has the complete-era statistics stated in
Theorem~\ref{thm:finite-failure-layer}.
\end{corollary}

\begin{proof}
Each $a_j$ lies below the common scale threshold.
Countable subadditivity,~\eqref{eq:signed-failure-flux}
and~\eqref{eq:dyadic-logarithmic-sum} prove the flux bound.
A point outside the union is outside its own layer's
excluded set, so the last assertion follows from the
one-layer theorem.
\end{proof}

The two logarithmic wall directions thus leave a finite
total excluded flux after summation over the shear scales.
The next task is dynamical: a positive-measure family that
continues to cross the small section cannot remain confined
to a finite-flux excluded set under strictly expanding returns.