\subsection{Flux of the angular excluded set}
\label{sec:angular-failure-flux}

Changing from shear to angle removes one logarithmic factor
from the section measure. The remaining factor comes from
the smaller wall. We use the smaller of the radial loss bound
and the interval length. This bound is integrable against the
remaining factor, and the resulting flux is summable over all
small radial scales.

For $a>0$ introduce
\begin{equation}
 \Xi_a(\rho,\theta,t)
      =(\theta\rho,-\log\rho,-\log(2at)),
       \qquad \rho,t>0.
 \label{eq:angular-logarithmic-coordinates}
\end{equation}
The physical weak-wall label is $\zeta=2at$.
The derivative matrix and its determinant are
\begin{equation}
 D\Xi_a=
 \begin{pmatrix}
   \theta&\rho&0\\ -1/\rho&0&0\\ 0&0&-1/t
 \end{pmatrix},\qquad \det D\Xi_a=-1/t.
 \label{eq:angular-logarithmic-jacobian}
\end{equation}
The factor $\rho$ in the shear coordinate cancels the
radial logarithmic derivative $1/\rho$ exactly.
The map is injective: its second and third coordinates
determine $\rho,t$, and its first then determines $\theta$.
Thus, for any Borel set $A$ in the domain $\rho,t>0$,
its image is Borel and
\begin{equation}
 \Leb(\Xi_a(A))=\int_A\frac1t\dd\rho\dd\theta\dd t.
 \label{eq:angular-coordinate-volume}
\end{equation}
This follows from the injective Borel mapping theorem
and the change of variables formula.

Define the excluded set for the first wall order by
\begin{equation}
 E_a^+=\{(\theta\rho,-\log\rho,-\log\zeta):
       (\theta,\zeta)\in\Lambda_a,
       \ \rho\in\mathcal F_{\theta,\zeta}\}.
 \label{eq:angular-failure-section}
\end{equation}
When $a\leq1$ and $P>4$, every such label has $0<\zeta<2a$.
Consequently $E_a^+=\Xi_a(A_a)$, where
\[
 A_a=\{(\rho,\theta,t): |\theta|\leq L,\ 0<t<1,
       \ (\theta,2at)\in\Lambda_a,
       \ \rho\in\mathcal F_{\theta,2at}\}.
\]
Theorem~\ref{thm:angular-radial-statistics} makes $A_a$
Borel, and hence $E_a^+$ is Borel.

Put
\begin{equation}
 K_{\mathrm a}(L,\sigma)
       =640\sqrt3\,[102400(L+1)]^\sigma,
 \qquad
 C_1(K,\sigma)=\frac{1+\log K+2\sigma}{\sigma}.
 \label{eq:angular-flux-constants}
\end{equation}
For $L\geq0$ and $\sigma>0$, $K_{\mathrm a}\geq1$.
Since $M_L=8(L+1)$,
\begin{equation}
 B_{a,2at}=102400(L+1)a^{-2}t,\qquad
 640\sqrt3 B_{a,2at}^\sigma
                =K_{\mathrm a}a^{-2\sigma}t^\sigma.
 \label{eq:angular-normalized-budget}
\end{equation}

For any $K\geq1$, $\sigma>0$ and $0<a\leq1$, the
one-variable identity~\eqref{eq:clipped-logarithmic-one} gives
\begin{align}
 \int_0^1\frac{\min\{a,Ka^{-2\sigma}t^\sigma\}}t\dd t
   &=\frac a\sigma
        \bigl[1+\log K+(1+2\sigma)|\log a|\bigr]\notag\\
   &\leq C_1(K,\sigma)a(1+|\log a|).
 \label{eq:angular-clipped-logarithm}
\end{align}
Here $Ka^{-2\sigma}\geq a$. For the inequality, expand
$(1+\log K+2\sigma)(1+|\log a|)$; the difference
from the preceding bracket is $2\sigma+\log K\,|\log a|$.

\begin{proposition}
\label{prop:angular-failure-flux}
Choose the statistical parameters and sufficiently small
scales as in Theorem~\ref{thm:angular-radial-statistics}.
If also
\[
 0<h\leq1/8,\qquad a\leq h/16,\qquad 2La\leq1/8,
\]
then
\begin{equation}
 \mu_h(E_a^+)\leq
       6L C_1(K_{\mathrm a},\sigma)a(1+|\log a|).
 \label{eq:angular-positive-failure-flux}
\end{equation}
The Borel symmetric set $E_a=E_a^+\cup\mathsf s E_a^+$
satisfies
\begin{equation}
 \mu_h(E_a)\leq
       12L C_1(K_{\mathrm a},\sigma)a(1+|\log a|).
 \label{eq:angular-two-order-failure-flux}
\end{equation}
Every point of either selected wall order outside $E_a$
has the statistical entrance of
Theorem~\ref{thm:angular-radial-statistics}.
\end{proposition}

\begin{proof}
Use~\eqref{eq:angular-coordinate-volume} and Tonelli's
theorem, integrating first in $\rho$. For a valid label,
\eqref{eq:angular-failed-length} and
\eqref{eq:angular-normalized-budget} bound that integral
by $\min\{a,K_{\mathrm a}a^{-2\sigma}t^\sigma\}/t$.
For an invalid label the integral is zero. Enlarging the
label domain to $[-L,L]\times(0,1)$ therefore gives
\[
 \Leb(E_a^+)\leq
   2L\int_0^1\frac{\min\{a,K_{\mathrm a}a^{-2\sigma}t^\sigma\}}t\dd t.
\]
All integrals are nonnegative, so no finiteness assumption
is needed for this calculation.

On $E_a^+$, $|s|\leq2La\leq1/8$ and both weak walls
are at most $2a\leq h/8$. The density calculation in
Lemma~\ref{lem:small-section-flux-bounds} uses precisely
these smallness bounds and gives $w_h\leq3$ on the chart.
More generally, any Borel set satisfying these bounds
has $\mu_h$-measure at most three times its Lebesgue
measure, since $\mu_h$ is restricted to $\mathcal D_h$.
Multiplying the preceding integral by three and applying
\eqref{eq:angular-clipped-logarithm} proves
\eqref{eq:angular-positive-failure-flux}.

Reflection preserves $\mu_h$ by
\eqref{eq:section-reflection-measure}, giving
\eqref{eq:angular-two-order-failure-flux}.
If $y=(\theta\rho,-\log\rho,-\log\zeta)$ is selected,
membership in $E_a^+$ is equivalent to
$\rho\in\mathcal F_{\theta,\zeta}$, by injectivity of
the coordinate map. Since $\mathsf sE_a=E_a$, a point outside
$E_a$ has its radial parameter outside the corresponding
$\mathcal F_{\theta,\zeta}$ in either wall order. The two orbit identities of
Section~\ref{sec:angular-actual-return} finish the proof.
\end{proof}

For $a_j=d2^{-(j+1)}$ with $0<d\leq1$, the elementary
geometric sums used in~\eqref{eq:dyadic-logarithmic-sum} give
\begin{equation}
 \sum_{j\geq0}a_j(1+|\log a_j|)
       =d(1+|\log d|+2\log2).
 \label{eq:angular-dyadic-logarithm}
\end{equation}
Indeed, $|\log a_j|=|\log d|+(j+1)\log2$, and
$\sum2^{-(j+1)}=1$, $\sum(j+1)2^{-(j+1)}=2$.
Thus the Borel set $E_d^{\mathrm a}=\bigcup_{j\geq0}E_{a_j}$
has finite flux
\begin{equation}
 \mu_h(E_d^{\mathrm a})\leq
   12L C_1(K_{\mathrm a},\sigma)
                   d(1+|\log d|+2\log2).
 \label{eq:angular-dyadic-failure-flux}
\end{equation}
Every scale uses its own fixed time $T_j=(q/2)\log(1/a_j)$
and the same statistical parameters and cap. A single
sufficiently small upper bound on $d$ makes all the
preceding estimates valid at every $a_j$.