==== TABLE OF CONTENTS ====
An-explicit-exact-Hausdorff-gauge-for-SLE//abstract.tex
An-explicit-exact-Hausdorff-gauge-for-SLE//brownian-circuit.tex
An-explicit-exact-Hausdorff-gauge-for-SLE//captions.tex
An-explicit-exact-Hausdorff-gauge-for-SLE//dense-cover.tex
An-explicit-exact-Hausdorff-gauge-for-SLE//geometry.tex
An-explicit-exact-Hausdorff-gauge-for-SLE//introduction.tex
An-explicit-exact-Hausdorff-gauge-for-SLE//local-laws.tex
An-explicit-exact-Hausdorff-gauge-for-SLE//macros.tex
An-explicit-exact-Hausdorff-gauge-for-SLE//main.tex
An-explicit-exact-Hausdorff-gauge-for-SLE//mesh-tails.tex
An-explicit-exact-Hausdorff-gauge-for-SLE//retained-mass.tex
An-explicit-exact-Hausdorff-gauge-for-SLE//ribbon-front.tex
An-explicit-exact-Hausdorff-gauge-for-SLE//ribbon.tex
An-explicit-exact-Hausdorff-gauge-for-SLE//serpentine-ribbon.tex
An-explicit-exact-Hausdorff-gauge-for-SLE//sources.bib
===========================
==== BEGIN An-explicit-exact-Hausdorff-gauge-for-SLE//abstract.tex ====
For each fixed $0<\kappa<8$, let $d=1+\kappa/8$.
The gauge $h(r)=r^d(\log\log(1/r))^{(2-d)/2}$ at sufficiently small
radii almost surely gives positive finite Hausdorff measure to every
nontrivial positive-time compact segment of chordal $\mathrm{SLE}_\kappa$.
This gives an explicit solution to Schramm's Hausdorff-measure problem
in this parameter range. The entire trace has finite expected measure
in every bounded disk. The exponent-one iterated-logarithm gauge
suggested by Schramm is not sigma-finite on any such segment.
==== END An-explicit-exact-Hausdorff-gauge-for-SLE//abstract.tex ====
==== BEGIN An-explicit-exact-Hausdorff-gauge-for-SLE//brownian-circuit.tex ====
\begingroup
\definecolor{circuitblue}{RGB}{31,88,138}
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\node[anchor=west,font=\small\bfseries] at (0,9.85)
{(a) A forced circuit after scaling $x=0$, $r=1$};
\node[anchor=west,font=\small\bfseries] at (8.25,9.85)
{(b) The two crossings of $Q$};
\begin{scope}[shift={(3.35,6.2)},scale=.88]
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--(2.51,-1.5)--(2.6,-.8)--(2.55,0)--(2.62,0);
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(0,0) .. controls (1,-.5) and (2.75,-1.25) .. (3,-.6);
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at (-.25,-.68) {initial approach};
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\draw[black!65,dashed] (2.6,-.5) rectangle (3.4,.5);
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(3,-.65)--(2.94,-.3)--(3,0)--(3.08,.28)--(3,.66);
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(2.49,.02)--(2.75,-.03)--(3,0)--(3.28,.07)--(3.55,.02);
\node[anchor=west] at (3.45,.6) {$Q$};
\node[anchor=north,align=center,font=\footnotesize]
at (0,-3.5) {ordered corridor: one counterclockwise turn};
\end{scope}
\begin{scope}[shift={(11.2,6.6)}]
\fill[circuitblue!6] (-.4,-1.65) rectangle (.4,1.65);
\fill[circuitorange!8] (-1.65,-.4) rectangle (1.65,.4);
\draw[black!70] (-1.65,-1.65) rectangle (1.65,1.65);
\node[anchor=north east] at (1.62,1.64) {$Q$};
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(-.05,-1.85)--(.13,-1.42)--(-.1,-.95)--(.15,-.6)
--(-.1,-.27)--(0,0)--(.15,.37)--(-.16,.8)--(.08,1.25)--(0,1.9);
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(-1.88,.06)--(-1.37,-.09)--(-.91,.14)--(-.46,-.12)
--(0,0)--(.46,.12)--(.93,-.07)--(1.37,.09)--(1.89,.02);
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\node[anchor=south west,align=left,font=\footnotesize]
at (.36,.43) {a common point};
\draw[thin] (.38,.44)--(.05,.06);
\node[anchor=south,text=circuitblue] at (0,2.1) {first: bottom to top};
\node[anchor=north,text=circuitorange,align=center]
at (0,-1.98) {last: left to right};
\end{scope}
\node[anchor=west,font=\small\bfseries] at (0,2.28)
{(c) Covering all occupation of $E_r$ by time blocks};
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node[anchor=west] {$t$};
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\fill[circuitblue!10] (\a,.91) rectangle (\b,1.43);
\draw[circuitblue!55] (\a,.91) rectangle (\b,1.43);
}
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\draw[black!80,dashed] (10.95,.6)--(10.95,1.65);
\node[anchor=south] at (10.95,1.67) {$\zeta$};
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node[midway,fill=white,inner sep=1pt] {$Ar^2$};
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{outside $E_r$};
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{visits to $E_r$};
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\node[anchor=west,font=\footnotesize,text=black!65] at (0,-.03)
{Schematic; spatial corridors and time blocks are not to scale.};
\end{tikzpicture}
\endgroup
==== END An-explicit-exact-Hausdorff-gauge-for-SLE//brownian-circuit.tex ====
==== BEGIN An-explicit-exact-Hausdorff-gauge-for-SLE//captions.tex ====
\newcommand{\BrownianCircuitCaption}{%
Forcing exit near an arbitrary boundary. (a,b) The first path segment crosses the rectangle $Q$ from bottom to top. It then follows an ordered, sufficiently thin corridor for one counterclockwise turn around $\overline B(0,2)$ and returns across $Q$ from left to right. The two crossings intersect, creating a closed subpath with winding number one about the disk; its trace separates the disk from infinity. Self-intersections are allowed. Thus the path meets every connected unbounded complement that meets $\overline B(0,1)$. (c) At each entrance $\sigma_j$ after the preceding block, a block of length $Ar^2$ begins. The orange pieces represent occupation of $E_r=\{u\in D:\operatorname{dist}(u,\partial D)\le r\}$; the gaps have none. Conditional on each block start before exit $\zeta$, the probability of exit in that block is at least $\eta$. Hence the expected occupation is at most $Ar^2/\eta$.%
}
\newcommand{\RibbonFrontCaption}{%
One step in ribbon coordinates $P(s,x)$. The black past stays in the core $|x|<\epsilon$ and first reaches the front $s=k$ at $a$, with tip offset at most $\epsilon/4$. The blue access guide moves to the centerline and passes through the interior waypoint $P(k+1/2,0)$. The orange region is $P((k+1/3,k+2/3)\times(-\epsilon,\epsilon))$, where the next step's sausage area is tested before its first crossing of $s=k+1$. The guide depicts an available route, not the random continuation. The transverse scale and core width are exaggerated.%
}
\newcommand{\SerpentineRibbonCaption}{%
A dense visit through a narrow ribbon. (a) In normalized half-plane coordinates, the marked point $w$ is separated from a disk of radius $s_0/20$ centered at the horizontally displaced point $w'$. The connector starts vertically at $0$, reaches the first strand from below, and stays away from $w$; the final strand continues upward. (b) The disk contains $2m+1$ strands of length $2h_0$ joined by alternating exterior semicircles. Choosing ribbon half-width $\ell=h_0/(C_0m)$ gives order $m^2$ steps. The orange marks indicate selected disjoint middle test regions. Their sausage tests contribute order $m^2\ell^d\asymp m^{2-d}=m^p$. Both panels are schematic and not to scale; the array in (b) is enlarged, and the ribbon widths are exaggerated.%
}
==== END An-explicit-exact-Hausdorff-gauge-for-SLE//captions.tex ====
==== BEGIN An-explicit-exact-Hausdorff-gauge-for-SLE//dense-cover.tex ====
\section{Covers charged to a single sausage mesh}
\label{sec:dense-cover}
We prove finite expected $h$-measure of the entire trace in each bounded
region. Select disjoint disks carrying enough normalized sausage area,
all measured at one deterministic mesh. Their enlarged disks have a
uniform expected covering cost; the visited cells they miss have a cost
that tends to zero. Keeping that bound independent of the time and
interior cutoffs gives the whole-time spatial conclusion.
Use the parameters from \eqref{eq:parameters} and the gauge
from \eqref{eq:gauge}. Put
\(\Gamma=\gamma([0,\infty))\). For \(v>0\), define the random measure
\begin{equation}
\mu^v(A)=v^{-p}\int_A
{\bf1}_{\{\operatorname{dist}(w,\Gamma)<v\}}\,dA(w).
\label{eq:cover-sausage}
\end{equation}
The sausage is taken in the whole plane. We also use \(\mu_t^v\),
defined by replacing \(\Gamma\) with \(\gamma([0,t])\). In particular,
\(\mu_t^v\le\mu^v\).
\paragraph{The stopped estimates used by the cover.}
For an interior marked point \(z\), let \(R_t(z)\) be its conformal
radius in the surviving domain, and let
\[
M_t(z)=R_t(z)^{-p}\sin^{a_0}\theta_t(z),\qquad
M_0(z)=(2\operatorname{Im}z)^{-p}
\left(\frac{\operatorname{Im}z}{|z|}\right)^{a_0}.
\]
Let \(\mathbb P_z^*\) denote the finite-scale tilted laws. Lemmas~\ref{ll:finite-tilt}, \ref{ll:inverse-angle}, and~\ref{ll:approach}
give consistency at finite conformal-radius stops, conditional change
of law, and the following two bounds.
If \(\widehat S\) is one unit of the clock \(-\log R_t(z)\)
after a stop \(S\), then
\begin{equation}
\mathbb E_z^*\!\left[
\sin^{-a_0}\theta_{\widehat S}(z)\mid\mathcal F_S\right]\le C.
\label{eq:cover-angle}
\end{equation}
If \(S\) is a stop at which \(z\) is alive and
\(\rho\le e^{-1}R_S(z)\), then
\begin{equation}
\mathbb P\{\tau_\rho(z)<\infty\mid\mathcal F_S\}
\le C\rho^p M_S(z),
\label{eq:cover-one-point}
\end{equation}
where \(\tau_\rho\) here denotes the subsequent first stop with
conformal radius \(\rho\). Under the tilted law every finite clock
level is reached. The finite-stop identity takes the form
\begin{equation}
\mathbb E\!\left[
{\bf1}_{\{S<\infty\}}M_S(z)X\right]
=M_0(z)\mathbb E_z^*[X]
\label{eq:cover-finite-tilt}
\end{equation}
for nonnegative variables \(X\) measurable at a fixed conformal-radius
stop \(S\), with the killed density assigned value zero.
Proposition~\ref{rc:dense} supplies the following precise uniformity. Fix a
positive angle cutoff. There are constants \(b,c,C,C_2>0\), \(C_1>1\),
and an integer \(m_0\) such that, for every integer \(m\ge m_0\)
and every radius \(r>0\), there is a deterministic
\(\varepsilon(r,m)=r\delta_m>0\) with this property. At any state with
\(R(z)=r\) and angle sine above the cutoff, and for every prescribed
\(0<v\le\varepsilon(r,m)\), the tilted conditional probability is
at least \(c\exp(-Cm^2)\) that, before the radius falls to
\(r/C_1\), the future trace generates in \(B(z,C_2r)\) a
\(v\)-sausage of normalized area at least \(b r^d m^p\).
The constants and the cutoff \(\varepsilon(r,m)\) are uniform in
the entire stopped state. The event is observable by that clock
endpoint. This formulation permits a single deterministic physical
mesh to be used at all the tests below.
We also use the elementary visit geometry associated with the
surviving-domain convention. If \(\overline B(z,Lr)\subset\mathbb H\)
initially, with \(L\) a sufficiently large absolute constant, then a
visit to the square of side \(r\) centered at \(z\) requires a finite
live stop at conformal radius \(Lr\). Indeed, while \(z\) survives,
Koebe gives
\(R_t(z)\le4\operatorname{dist}(z,\partial D_t)\).
If \(z\) is swallowed without the trace first meeting
\(\overline B(z,2r)\), that ball is in the swallowed component and
the future trace cannot enter it. Otherwise, just before the first
such meeting the live radius has decreased to at most \(8r\);
continuity gives the earlier radius crossing, for example with
\(L=16\). The same argument shows that a trace visit within
distance \(v\) of \(z\) forces a live radius crossing at \(16v\),
provided \(\overline B(z,16v)\subset\mathbb H\) initially.
These statements concern the trace, rather
than the filled hull, and apply also when \(\kappa>4\).
\begin{lemma}[Uniform bounded-region charge]
\label{lem:cover-charge}
For every \(0<R<\infty\), there is \(C_R<\infty\), depending only on
\(R\) and \(\kappa\), such that
\[
\sup_{0<v\le1}\mathbb E\mu^v(B(0,R))\le C_R.
\]
\end{lemma}
\begin{proof}
Choose a fixed \(A>8e\). If \(y=\operatorname{Im}w>Av\),
the visit geometry and \eqref{eq:cover-one-point}, with a fixed
multiplicative slack in the target radius, give
\[
\mathbb P\{\operatorname{dist}(w,\Gamma)<v\}
\le C v^p M_0(w).
\]
For example, increasing \(A\) if necessary makes the target
\(16v\) at least one clock unit below the initial radius \(2y\).
Since \(\Gamma\subset\overline{\mathbb H}\), only the layer
\(-v<y\le Av\) contributes outside this range. Its area in
\(B(0,R)\) is at most \(C_Rv\). Tonelli's theorem therefore gives
\begin{align*}
\mathbb E\mu^v(B(0,R))
&\le C_Rv^{1-p}
+C\int_{B(0,R)\cap\mathbb H}M_0(w)\,dA(w)\\
&\le C_Rv^{1-p}
+C\int_{-R}^{R}\int_0^R(2y)^{-p}\,dy\,dx
\le C_R.
\end{align*}
Here \(p<1\) is exactly what makes the boundary integral finite.
The estimate concerns the whole trace at all times.
\end{proof}
\begin{proposition}[The upper Hausdorff bound]
\label{prop:dense-cover-upper}
For every \(0<R<\infty\),
\begin{equation}
\mathbb E\mathcal H^h
\bigl(\Gamma\cap\overline B(0,R)\bigr)\le C_R<\infty.
\label{eq:cover-all-time-expectation}
\end{equation}
Consequently, on one probability-one event,
\[
\mathcal H^h\bigl(\Gamma\cap([-m,m]+i[0,m])\bigr)<\infty
\qquad(m=1,2,\ldots).
\]
On that event every compact capacity-time segment has finite
\(\mathcal H^h\)-measure.
\end{proposition}
\begin{proof}
First fix a deterministic compact set
\(K\subset\mathbb H\cap\overline B(0,R)\).
Choose \(T_0>1+\log C_1\), and put
\[
r_j=e^{-T_0j},\qquad
J_0=\lfloor n/3\rfloor,\qquad J_1=\lfloor2n/3\rfloor,
\qquad m_n=\lfloor c_0\sqrt{\log n}\rfloor.
\]
The constant \(c_0>0\) will be chosen below. We omit finitely many
initial \(n\), so that \(m_n\ge m_0\) and every radius tested is
smaller than the initial conformal radius at the grid centers used.
\emph{One deterministic mesh.}
Choose a deterministic sequence \(v_n>0\) such that \(v_n\to0\)
and
\begin{equation}
v_n\le\min_{J_0\le j\le J_1}
\varepsilon(e^{-1}r_j,m_n).
\label{eq:cover-mesh-choice}
\end{equation}
For example, take half this finite minimum and also require
\(v_n\le2^{-n}\) and \(v_n\le v_{n-1}/2\). This choice is
deterministic because the dense-visit cutoff is uniform in the
state. Write \(\mu^{(n)}=\mu^{v_n}\).
Use the usual grid of closed squares of side \(r_n\), and keep
the finitely many squares meeting \(K\). A center \(z\) of such
a square has initial height bounded below once \(n\) is large.
For the argument at this one center, use \(\mathbb P_z^*\) only
through the finite clock levels required. Let \(S_j(z)\) be the
first live stop with \(R(z)=r_j\), and let \(S'_j(z)\) be the
stop one clock unit later, so that \(R(z)=e^{-1}r_j\).
All these stops are finite under \(\mathbb P_z^*\).
By \eqref{eq:cover-angle}, a fixed angle cutoff can be chosen
so that the conditional probability of a good angle at \(S'_j\),
given the complete history through \(S_j\), is at least \(1/2\).
The dense-visit input and \eqref{eq:cover-mesh-choice} then give
a conditional success probability at least
\[
c\exp(-Cm_n^2).
\]
On success, the prescribed mass has been generated before
\(S_{j+1}\), since \(T_0>1+\log C_1\). Choose a fixed
\(C_3\ge1\) so that all these mass tests lie in
\[
D_j(z)=\overline B(z,C_3r_j).
\]
There is a constant \(b_1>0\), independent of \(n,j,z\), such that
success implies
\begin{equation}
\mu_{S_{j+1}(z)}^{v_n}(D_j(z))
\ge b_1 r_j^d(\log n)^{p/2}.
\label{eq:cover-adapted-success}
\end{equation}
This uses \(m_n\ge(c_0/2)\sqrt{\log n}\) for large \(n\).
Let \(A_{n,j}(z)\) be the event that \(S_{j+1}(z)<\infty\) and
\eqref{eq:cover-adapted-success} holds. Under the tilted law all these
endpoint stops are finite. The event \(A_{n,j}(z)\) is measurable at
\(S_{j+1}(z)\), because its sausage uses only the trace through that stop.
Choose \(c_0\) so small that \(Cc_0^2<1/2\). For all sufficiently
large \(n\), the dense-visit lower bound therefore gives
\[
\mathbb P_z^*\!\left(A_{n,j}(z)\mid\mathcal F_{S_j(z)}\right)
\ge n^{-1/2}.
\]
Put
\[
F_n(z)=\bigcap_{j=J_0}^{J_1}A_{n,j}(z)^c.
\]
Every preceding test is observable when the next one starts.
Iterated conditional expectations give
\begin{equation}
\mathbb P_z^*(F_n(z))
\le(1-n^{-1/2})^{J_1-J_0+1}
\le C\exp(-c\sqrt n).
\label{eq:cover-failure}
\end{equation}
No independence is required between tests or between centers.
\emph{Returning to the original law.}
Call \(D_j(z)\) a candidate if
\begin{equation}
\mu^{(n)}(D_j(z))\ge b_1r_j^d(\log n)^{p/2}.
\label{eq:cover-candidate}
\end{equation}
This is a terminal-sausage test. A cell visited by the trace and
having no candidate must satisfy the adapted failure event \(F_n(z)\):
at each finite endpoint, \(\mu_{S_{j+1}}^{v_n}\le\mu^{(n)}\).
By the visit geometry, the visit also requires the
live stop \(\tau_{Lr_n}(z)<\infty\), for the fixed constant
\(L\) above. For large \(n\), that stop is after
\(S=S_{J_1+1}(z)\) with a gap of at least one clock unit.
Thus on the event on the left below all the test endpoints are finite,
and \(F_n(z)\) is measurable at \(S\).
Applying \eqref{eq:cover-one-point}, followed by
\eqref{eq:cover-finite-tilt}, gives
\begin{align}
&\mathbb P\{\text{the cell at }z\text{ is visited and has no candidate}\}
\notag\\
&\quad\le C r_n^p\,
\mathbb E\!\left[
{\bf1}_{\{S<\infty\}}M_S(z){\bf1}_{F_n(z)}\right]
\notag\\
&\quad= C r_n^p M_0(z)\mathbb P_z^*(F_n(z))
\le C r_n^p M_0(z)e^{-c\sqrt n}.
\label{eq:cover-missed-cell}
\end{align}
The full-sausage candidate information enters only through the
inclusion in the adapted failure event. All changes of law
occur at finite conformal-radius levels.
\emph{The finite cover and its charge.}
Order the finite collection of candidate disks by decreasing
radius, using a fixed deterministic order to break ties.
Select the first, discard all disks meeting it, and repeat.
The selected closed disks are pairwise disjoint. A discarded
disk has no greater radius than the selected disk meeting it,
so it is contained in the concentric triple of that selected
disk. For large \(n\), every cell is contained in each of its
candidate disks, since \(r_n/r_{J_1}\to0\). Consequently the
triples of the selected disks cover all cells having a candidate.
Uniformly for \(J_0\le j\le J_1\), formula
\eqref{eq:gauge} implies
\[
h(6C_3r_j)\le C r_j^d(\log n)^{p/2}.
\]
The qualification \eqref{eq:cover-candidate} and disjointness
therefore show that the total cost of these triples is at most
\begin{equation}
C\sum_{\text{selected }D}\mu^{(n)}(D)
\le C\mu^{(n)}(B(0,R+1)).
\label{eq:cover-selected-charge}
\end{equation}
All the candidate disks lie in this fixed larger disk once
\(n\) is large. No limiting measure or convergence of sausages
is needed in this estimate: every selected disk is charged to
the same measure \(\mu^{(n)}\).
Add every retained fine cell that is visited and has no candidate.
These cells, together with the selected triples, form a finite
cover of \(\Gamma\cap K\). There are at most \(C_Rr_n^{-2}\)
retained cells; their centers stay a positive distance from the
real line, so \(M_0(z)\le C_K\). Since
\(h(\sqrt2r_n)\le Cr_n^d(\log n)^{p/2}\),
\eqref{eq:cover-missed-cell} yields an expected added cost at most
\begin{equation}
C_Kr_n^{-2}r_n^d(\log n)^{p/2}r_n^p e^{-c\sqrt n}
=C_K(\log n)^{p/2}e^{-c\sqrt n}\longrightarrow0.
\label{eq:cover-cleanup}
\end{equation}
Here \(d+p=2\). The maximal diameter of the cover is at most
\[
\delta_n=\max\{6C_3r_{J_0},\sqrt2r_n\}\longrightarrow0.
\]
If \(Z_n(K)\) is its total cost, Lemma~\ref{lem:cover-charge}
and \eqref{eq:cover-selected-charge}--\eqref{eq:cover-cleanup}
show that
\begin{equation}
\limsup_{n\to\infty}\mathbb E Z_n(K)\le C_R.
\label{eq:cover-cost-expectation}
\end{equation}
The constant \(C_R\) does not depend on the distance of \(K\)
from the real line. Only the vanishing error and the discarded
initial values of \(n\) may depend on \(K\).
\emph{Measurability and the Hausdorff limit.}
For fixed \(v\), the distance to the whole trace is the infimum
over nonnegative rational times of \(|w-\gamma(t)|\). Thus all
the sausage integrals are measurable, by Tonelli's theorem.
There are finitely many candidate indicators and the greedy
order is deterministic. A fixed closed cell meets the trace
if and only if it meets \(\gamma([0,N])\) for some integer
\(N\); each event is measurable by continuity of \(\gamma\).
It follows that \(Z_n(K)\) is measurable.
For completeness, \(\mathcal H^h(\gamma([0,T])\cap K)\) is
also measurable for deterministic \(T<\infty\) and compact
\(K\). In defining the Hausdorff limit of this random compact
set, one may use finite covers whose members are finite unions
of rational open balls. There are only countably many such
covers. Their covering events are measurable: for an open
union \(U\), the event
\(\gamma([0,T])\cap K\subset U\) is the event that the compact
trace avoids the deterministic compact set \(K\setminus U\).
Arbitrarily small open enlargements, using continuity of \(h\),
show that these countable covering infima have the same limit
as the usual Hausdorff contents. This proves measurability.
For every fixed \(T\), the covers above cover
\(\gamma([0,T])\cap K\). Therefore, pathwise,
\[
\mathcal H^h(\gamma([0,T])\cap K)
\le\liminf_{n\to\infty}Z_n(K).
\]
Indeed every sufficiently late cover is admissible at any
fixed positive diameter cutoff. Fatou's lemma and
\eqref{eq:cover-cost-expectation} give
\begin{equation}
\mathbb E\mathcal H^h(\gamma([0,T])\cap K)\le C_R,
\label{eq:cover-compact-expectation}
\end{equation}
uniformly in both \(T\) and \(K\subset\mathbb H\cap\overline B(0,R)\).
\emph{The entire trace in a bounded region.}
To retain this uniformity when approaching both infinite time
and the real boundary, set
\[
K_q=\overline B(0,R)\cap\{\operatorname{Im}z\ge1/q\},
\qquad E_q=\gamma([0,q])\cap K_q
\quad(q=1,2,\ldots).
\]
The sets \(E_q\) are increasing compact sets, and
\[
\bigcup_{q\ge1}E_q=\Gamma\cap\overline B(0,R)\cap\mathbb H.
\]
Gauge Hausdorff measure is a measure on Borel sets. Its
continuity from below and then monotone convergence, applied
to \eqref{eq:cover-compact-expectation}, give
\[
\mathbb E\mathcal H^h
(\Gamma\cap\overline B(0,R)\cap\mathbb H)
=\lim_{q\to\infty}\mathbb E\mathcal H^h(E_q)
\le C_R.
\]
Finally, a bounded real interval is covered by \(O_R(r^{-1})\)
intervals of length \(r\), whose total cost is
\[
O_R\!\left(r^{d-1}(\log\log(1/r))^{p/2}\right)\longrightarrow0,
\]
because \(d>1\). The real-axis part therefore has zero
\(\mathcal H^h\)-measure. This proves
\eqref{eq:cover-all-time-expectation}.
Apply the result at integer radii and intersect the resulting
countably many probability-one events. Every box in the
statement is contained in one of these disks. Every compact
time segment has bounded image by continuity of \(\gamma\),
so it too lies in such a disk. This proves both assertions
simultaneously.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:main}]
Section~\ref{retain:positivity} gives positivity simultaneously on all real
$0<s<t<\infty$ for the fixed parameter $\kappa$.
Proposition~\ref{prop:dense-cover-upper} gives the all-time bounded-box
estimate and finite measure on every compact time segment on another
probability-one event. Their intersection proves all assertions.
\end{proof}
==== END An-explicit-exact-Hausdorff-gauge-for-SLE//dense-cover.tex ====
==== BEGIN An-explicit-exact-Hausdorff-gauge-for-SLE//geometry.tex ====
\section{Boundary geometry and pointwise slit support}
\label{geom:section}
Boundary extension and uniqueness at a fresh tip identify the access
from a first disk hit. Pointwise smooth-slit support then supplies the
individual passages used later; uniformity over rough pasts will come
from the compactness argument in Section~\ref{rc:section}.
All closures at infinity in this section are spherical. A filled hull
generated by a continuous curve means the curve together with the components
that it disconnects from the target. The surviving domain is the component
towards the target. We use the continuous-trace description of chordal
Loewner evolution; the claims below concern its boundary geometry, not the
existence of that trace. For the fixed parameter $0<\kappa<8$, we use
the continuous generating trace given by Rohde--Schramm
\cite[Theorem~5.1]{rohdeschramm2005}. Its domain Markov property follows
from composition of the Loewner maps and Brownian strong Markov.
\begin{lemma}[Continuous extension for a past domain]
\label{geom:extension}
Let $\eta:[0,T]\to\overline{\mathbb H}$ be continuous, with $\eta(0)=0$,
and let $D$ be the unbounded component of
$\mathbb H\setminus\eta[0,T]$. Suppose $D$ is the surviving simply
connected domain of the trace. Its conformal uniformization
$F:\mathbb H\to D$ extends continuously from the closed half-plane in
the spherical topology. Boundary parameters of $F$ specify prime ends;
distinct parameters may have the same physical image.
\end{lemma}
\begin{proof}
Write
\[
A=\bigl(\widehat{\mathbb C}\setminus\mathbb H\bigr)
\cup\eta[0,T],
\qquad K=\widehat{\mathbb C}\setminus D.
\]
The image of a compact interval under a continuous map is a locally
connected continuum. This standard preservation statement is recalled in
\cite[Chapter~2, \S2.2, p.~19]{pommerenke1992}.
The other set in the union defining $A$ is a closed hemisphere, and the
two sets meet at $0$. Their union is locally connected: at a point of
their intersection take small connected relative neighborhoods in each
and unite them; at a point belonging to only one, first choose a
neighborhood disjoint from the other closed set. Thus $A$ is a locally
connected continuum.
The set $K$ is obtained from $A$ by adding complementary components.
Here is the elementary filling argument, stated locally in a Euclidean
chart. Given a prescribed small diameter, uniform local connectedness of
the compact set $A$ supplies a scale $\delta$ at which nearby points of
$A$ can be joined by a connected subset of that diameter. If $x,y$ are
points of the filled set with $|x-y|<\delta$ and
$\operatorname{dist}(x,A)>2\delta$, their joining segment avoids $A$
and remains in the same filled component as $x$. Otherwise choose
nearest points $a,b\in A$ to $x,y$. The segments $[x,a]$ and $[y,b]$
stay in their respective filled components until their endpoints, and
have lengths at most $2\delta$ and $3\delta$. Also
$|a-b|\le6\delta$. After reducing $\delta$, join $a$ to $b$ by a
small connected subset of $A$. The resulting connected set joins
$x$ to $y$ with the prescribed small diameter. A spherical version
follows by local charts, or by moving a point of $D$ to infinity.
This is also the filling statement in
\cite[\S2, Lemma~2.4, p.~7]{loridantluo2017}.
Consequently $K$ is locally connected. The Torhorst boundary theorem,
in the form recorded in
\cite[\S2, Lemma~2.6, p.~8]{loridantluo2017}, implies that
$\partial D$ is locally connected. The Continuity Theorem and Prime
End Theorem in \cite[Chapter~2, \S2.1, p.~18]{pommerenke1992} now give
the asserted continuous extension and its prime-end interpretation.
\end{proof}
\begin{lemma}[A fresh terminal point has one prime end]
\label{geom:fresh-tip}
In the preceding setting, suppose
$a=\eta(T)\in\mathbb H\cap\partial D$ and
$a\notin\eta[0,T)$. Then $F$ has exactly one boundary parameter with
image $a$. For the Loewner uniformization this parameter is the current
driving point. Every access in $D$ tending to $a$ approaches that same
prime end.
\end{lemma}
\begin{proof}
First, $A\setminus\{a\}$ is connected: it is the union of the closed
lower half-plane with the connected image $\eta[0,T)$, and these meet
at $0$. If $V$ is a filled complementary component of $A$, then
$\overline V\setminus\{a\}$ is connected. Indeed the connected set
$V$ is dense in this set in the punctured sphere. Moreover
$\partial V$ contains a point other than $a$. A nonempty proper open
component with boundary contained in a singleton would be both open
and closed in the connected punctured sphere, and hence would equal
that entire punctured sphere, which is impossible here. Thus
$\overline V\setminus\{a\}$ meets $A\setminus\{a\}$. It follows
that
\[
K\setminus\{a\}
=(A\setminus\{a\})\,
\cup\!\bigcup_{V\text{ filled}}
(\overline V\setminus\{a\})
\]
is connected. This argument permits arbitrary self-intersections of the
past away from $a$.
For completeness, the conformal consequence is a Jordan-curve
argument. Replace the half-plane by the disk. If two different circle
parameters $u,v$ had image $a$, join them by a simple crosscut in the
disk. Its image is a Jordan curve $J$ contained in $D\cup\{a\}$:
the interior of the crosscut is mapped injectively into $D$, and its
only coinciding endpoints have image $a$. Neither of the two open
circle arcs between $u$ and $v$ can map identically to $a$. If an open
arc did, Schwarz reflection of $F-a$ across that arc would give an
analytic extension vanishing on an interval, contrary to univalence.
Choose a boundary image different from $a$ on each arc. The two
crosscut components map to opposite sides of $J$, so those images lie
on opposite sides as well. Both belong to $K\setminus\{a\}$,
whereas $J\cap K=\{a\}$. This disconnects $K\setminus\{a\}$,
a contradiction. This is the separation argument of
\cite[proof of Theorem~1, pp.~426--427, equations~(2.2)--(2.3)]{pommerenke1985},
applied to the complement rather than only to its boundary.
The continuous extension is onto $\overline D$, so there is a boundary
preimage of $a$ and it is unique. Any sequence of domain points tending
to $a$ has inverse-image cluster points only in this singleton:
compactness of the closed disk and continuity of $F$ prove this claim.
This gives the assertion for every access. The trace formula identifies
the current driving point as a preimage of the tip, so it is the unique
one.
\end{proof}
\begin{lemma}[Access at a first disk hit]
\label{geom:first-hit-access}
Let $\overline{B(z,r)}$ lie in the initial surviving domain, and write
$\eta$ for the future trace. Let $\tau$ be its first time reaching
$\partial B(z,r)$.
Suppose $\tau<\infty$ and $z$ belongs to the surviving domain $D_\tau$.
Then, writing $a=\eta(\tau)$, the segment $(a,z]$ is contained in
$D_\tau$ and approaches the current tip prime end at $a$.
If $g:D_\tau\to\mathbb H$ sends tip to $0$, target to infinity,
and is scaled so that $|g(z)|=1$, the image of $[a,z]$ is a continuous
path from $0$ to modulus $1$ whose inverse image is contained in
$\overline{B(z,r)}$.
\end{lemma}
\begin{proof}
The open ball is disjoint from the trace through time $\tau$.
It is connected and contains the surviving point $z$, so all of it
belongs to the same surviving component. In particular $(a,z]\subset
D_\tau$. The point $a$ has not previously been visited: an earlier
visit would have been an earlier hit of the circle. It is also
disjoint from the initial hull by the hypothesis on the closed ball.
Lemma~\ref{geom:fresh-tip} therefore identifies the radial access
with the tip prime end, proving continuity of its image at $0$.
The other assertions follow from the normalization. One may stop this
path at its first modulus-one point if desired. No modulus of boundary
continuity, and no bound on its length, is needed.
\end{proof}
The same connected-component argument gives a useful convention about
killing: if a ball is disjoint from the trace when its center is
swallowed, its whole open interior is swallowed then. Subsequent
trace growth in the surviving domain cannot enter that interior.
\begin{lemma}[Smooth slits have continuous drivers]
\label{geom:smooth-driver}
Let $\beta$ be a compact regular smooth simple arc, starting at $0$,
initially vertical, and otherwise contained in $\mathbb H$. Its
successive prefixes, parametrized by half-plane capacity $2t$, are
generated by a continuous chordal Loewner driver on a compact interval.
\end{lemma}
\begin{proof}
Use initially any regular parameter $s$. The complements of the
prefixes converge in the kernel sense when $s$ varies, so their
hydrodynamically normalized maps and inverse maps converge locally
uniformly. Reflection near infinity, outside a ball containing the
whole slit, also gives convergence of their Laurent coefficients.
We record why the needed boundary convergence holds in this particular
family. The complements consisting of the closed lower half-plane
and a prefix of $\beta$ are uniformly locally connected, in fact
uniformly locally arcwise connected. On the simple compact arc,
uniform continuity of $\beta$ and of its inverse shows that nearby
points can be joined by a uniformly small subarc, which remains in
every prefix containing them. Points of the arc sufficiently close
to the real line are on a uniformly short initial part of the arc,
since every closed later part has positive height. The initial part
and the lower half-plane can thus be joined through $0$. This proves
the asserted uniform local connectivity, including at the attachment.
Here is the standard small-crosscut proof that it implies the boundary
convergence required here. Put the inverse maps on the disk using a
fixed half-plane-to-disk coordinate. Their spherical area integrals
are bounded by the area of the sphere. About any boundary parameter,
integrating over crosscuts formed by circles of radii between
$\delta$ and $\sqrt\delta$ gives a crosscut whose image has spherical
length at most $C/\sqrt{\log(1/\delta)}$, by Cauchy--Schwarz. Its
endpoint images can be joined in the complement by an arc with
uniformly small diameter. The resulting Jordan curve (or the image
crosscut alone when its endpoints coincide) cuts off a set of small
diameter on the side away from the image of an interior basepoint.
The latter images stay uniformly separated from the complements, by
kernel convergence and compactness of the parameter interval.
The disk cap at the boundary parameter maps into that small side.
This proves uniform equicontinuity at the boundary; interior
equicontinuity follows from local uniform convergence. Hence the
inverse maps converge uniformly on the closed disk in the spherical
metric. In particular their tip preimages converge: any cluster
point maps to the limiting tip, whose preimage is unique by
Lemma~\ref{geom:fresh-tip}. Returning to hydrodynamic half-plane
coordinates shows that the tip coordinate $U(s)$ is continuous.
At $s=0$ the same conclusion follows from the initially vertical slit.
For clarity, the usual infinitesimal mapping argument now gives the
equation and the parameter. If $g_s$ maps out a prefix and $s'>s$,
the inverse transition $h=g_s\circ g_{s'}^{-1}$ maps $\mathbb H$
into itself and has expansion $h(z)=z-m/z+O(|z|^{-2})$. The
maximum principle gives $\operatorname{Im}(h(z)-z)\ge0$; its
half-plane harmonic representation, with reflection outside the
bounded incremental hull, is
\[
h(z)=z-\int_{\mathbb R}\frac{d\nu(x)}{z-x},
\qquad \nu\ge0,\qquad \nu(\mathbb R)=m.
\]
Here $m$ is the increment of half-plane capacity. If $m=0$, the
representation gives $h(z)=z$, which is impossible for a nontrivial
added arc. Capacity is therefore strictly increasing, and its
continuity follows from the Laurent-coefficient convergence above.
As $s'\downarrow s$, the added arc mapped by $g_s$ shrinks to $U(s)$.
Indeed every approach to the fresh tip has that same inverse-image
limit by Lemma~\ref{geom:fresh-tip}. The support of $\nu$ consequently
shrinks to $U(s)$ as well. One elementary way to see the last statement
is to enclose the added hull in a half-disk of radius $\epsilon$
about $U(s)$. Its map-out function $Q=h^{-1}$ extends by reflection
to a normalized univalent function on $|z-U(s)|>\epsilon$.
Writing
\[
Q(U(s)+\epsilon\zeta)
=U(s)+\epsilon\left(\zeta+
\sum_{j\ge1}b_j\zeta^{-j}\right),
\]
Gronwall's exterior area theorem gives $\sum_{j\ge1}j|b_j|^2\le1$
\cite[Theorem~3.2.1 and Corollary~3.2.2, pp.~37--38]{beliaev2015}.
Cauchy--Schwarz at $\zeta=\pm2$ consequently gives
$|Q(U(s)\pm2\epsilon)-U(s)|<3\epsilon$.
The transition inverse is real analytic on the real rays beyond
these two images, so its representing measure is supported between
them.
Thus, uniformly on compact subsets of $\mathbb H$,
\[
h(z)=z-\frac{m}{z-U(s)}+o(m).
\]
Invert this transition and use capacity $2t$ as parameter. It follows
that
\[
\frac{\partial g_t(z)}{\partial t}
=\frac{2}{g_t(z)-U(t)}.
\]
The compact interval of capacity and continuity of $U$ prove the lemma.
\end{proof}
The following pointwise assertion is a special case of SLE support;
see Tran--Yuan \cite[Theorem~1.1 and Corollary~1.2]{tranyuan2020}.
We give a proof of the particular open-tube form needed here.
\begin{lemma}[Pointwise slit support]
\label{geom:slit-support}
Let $\beta$ be as in Lemma~\ref{geom:smooth-driver}, and let $q$ be its
terminal point. Let $N$ be a relative open neighborhood of the whole
slit in $\overline{\mathbb H}$, including a relative neighborhood of
$0$, and let $V$ be an open neighborhood of $q$. Chordal
$\mathrm{SLE}_{\kappa}$ satisfies
\[
\mathbb P\{\text{for some }t<\infty,
\ \gamma[0,t]\subset N\text{ and }\gamma(t)\in V\}>0.
\]
Suppose in addition that $G_1,\ldots,G_m$ are pairwise disjoint crosscuts
of $N$, each separating $0$ from all later crosscuts and from $V\cap N$.
On the displayed
event, the first hits of $G_1,\ldots,G_m$ occur in that order.
The probability asserted here is pointwise in the slit and tube; no
uniform positive lower bound over different slits or pasts is claimed.
\end{lemma}
\begin{proof}
Let $U$ be the deterministic slit driver on $[0,T]$. We show that a
sufficiently small uniform neighborhood of $U$ forces the required
geometric event whenever the perturbed driver generates a continuous
trace. Suppose $U_n\to U$ uniformly and let $K_n$ be their hulls at
time $T$. The height bound $\operatorname{Im}K_n\le2\sqrt T$ and
the driver-extrema horizontal bound place all hulls in a common
bounded rectangle.
For each point of
$\overline{\mathbb H}\setminus\beta[0,T]$, its limiting forward
Loewner solution stays a positive distance from the driver up to
$T$. This includes real points other than $0$: a simple slit
otherwise contained in $\mathbb H$ swallows no such point, and the
solution is reflected analytically across a surviving real interval.
Continuous dependence for the ODE supplies a whole surviving
neighborhood of each such point for all sufficiently large $n$.
Compactness of the fixed rectangle outside any prescribed neighborhood
of the slit now gives
\[
K_n\subset N\quad\hbox{for all large }n.
\]
The same argument applies to a smaller neighborhood when needed.
The reverse Loewner ODE gives local uniform convergence of the
inverse maps at time $T$: on compact subsets of $\mathbb H$ their
imaginary parts are bounded away from zero, so ordinary ODE
stability applies. The corresponding domains consequently converge
in the kernel sense to $\mathbb H\setminus\beta[0,T]$. For every
small $r>0$, this implies
\[
K_n\cap B(q,r)\ne\varnothing
\quad\hbox{for all large }n.
\]
Otherwise, along a subsequence the whole ball would survive. A
compact passage from a fixed interior basepoint through the limiting
domain to a point of that ball, together with the ball, would then
belong to the kernel, incorrectly making $q$ an interior kernel
point.
This hull intersection also forces an intersection by the trace.
Choose $r$ with $\overline{B(q,2r)}\subset V\cap\mathbb H$, and
fix $w\in B(q,2r)\setminus\beta[0,T]$. Upper hull containment
ensures $w\notin K_n$ eventually. If the perturbed trace avoided
$B(q,2r)$, that connected ball would lie in one component of its
complement. It meets $K_n$ by the preceding paragraph, so that
component would be swallowed. The whole ball, including $w$, would
then belong to $K_n$, a contradiction. Thus the trace visits $V$.
Since every earlier hull is contained in $K_n$, the trace stays in
$N$ up to that visit. This sequential argument supplies an actual
positive driver tolerance by contradiction.
Brownian motion has positive probability in each uniform
neighborhood of the continuous function $U/\sqrt\kappa$, which
starts at zero. This follows, for example, by polygonal approximation,
positive Gaussian increment probabilities, and positive Brownian-bridge
confinement probabilities on the finitely many resulting intervals.
Apply this to the driver $\sqrt\kappa B$ and use the almost-sure
continuous-trace property. Finally, the stated crosscut separation
forces a continuous path in $N$ from $0$ to $V$ to hit the crosscuts
in order; backtracking within $N$ causes no problem.
\end{proof}
\begin{corollary}[Restarted support in open sets]
\label{geom:restart-support}
At a finite stopping time $\tau$ for ordinary chordal SLE in
$\mathbb H$, fix a realized past for which the conditional continuation
law $\mathbb P_\tau$ is defined. Let $D$ be its surviving domain, and
let $F:\mathbb H\to D$ send $0$ to the current tip prime end and infinity
to the target. Use its continuous spherical extension on
$\overline{\mathbb H}$. Let
$\alpha:[0,1]\to\overline{\mathbb H}$ be a continuous simple path with
$\alpha(0)=0$ and $\alpha((0,1])\subset\mathbb H$.
If $O$ is relatively open in $\overline D$ and contains
$F(\alpha[0,1])$, and $V\subset D$ is an open neighborhood of
$F(\alpha(1))$, then the restarted half-plane trace $\widetilde\gamma$
satisfies
\[
\mathbb P_\tau\{\text{for some }t<\infty,
\ F(\widetilde\gamma[0,t])\subset O
\text{ and }F(\widetilde\gamma(t))\in V\}>0.
\]
The same statement holds for a Euclidean similarity of the evolution.
The positive probability may depend on the realized past, $\alpha$,
$O$, and $V$.
\end{corollary}
\begin{proof}
Lemma~\ref{geom:extension} makes $N=F^{-1}(O)$ a relative open
neighborhood of $\alpha[0,1]$, including a relative half-disk at $0$.
The path $\alpha((0,1])$ lies in one component of $N\cap\mathbb H$,
and a short vertical segment from $0$ enters that component. Polygonal
connectivity there joins this segment to a point of $F^{-1}(V)$.
Removing loops and smoothing corners inside $N$ gives a regular smooth
simple slit with an initially vertical segment and endpoint in
$F^{-1}(V)$.
Lemma~\ref{geom:slit-support}, applied under the restarted Brownian
Loewner law, gives the displayed event after mapping back by $F$.
For passages through successive physical fronts, one may choose $O$
as a thin tube with separating crosscuts and choose $V$ beyond the
last one; the final clause of that lemma then gives their order.
Any uniform lower bound over different pasts still requires a
separate argument.
\end{proof}
==== END An-explicit-exact-Hausdorff-gauge-for-SLE//geometry.tex ====
==== BEGIN An-explicit-exact-Hausdorff-gauge-for-SLE//introduction.tex ====
\section{Introduction}\label{sec:introduction}
A random curve whose dimension lies strictly between one and two
requires a finer notion of length than arc length. A dimension theorem
specifies the critical power of a small covering diameter, but that
power alone need not give a nonzero measure. For ordinary chordal
$\mathrm{SLE}_\kappa$, $0<\kappa<8$, Beffara identified the dimension
$d=1+\kappa/8$ \cite[main Theorem]{beffara2008}; Rezaei proved that its
$d$-dimensional Hausdorff measure is zero
\cite[Theorem~1.1]{rezaei2018}. The remaining question concerns the
correction to the critical power, not a further change of dimension.
A \emph{gauge} is a continuous nondecreasing function
$h:[0,\infty)\to[0,\infty)$ with $h(0)=0$ and $h(r)>0$ for $r>0$.
Its Hausdorff measure,
in the diameter convention used here, is
\begin{equation}\label{eq:hausdorff-definition}
\mathcal H^h(E)=\lim_{\delta\downarrow0}\inf\left\{
\sum_j h(\operatorname{diam}U_j):
E\subset\bigcup_jU_j,\quad \operatorname{diam}U_j\le\delta\right\},
\end{equation}
where covers are countable. We write $\mathcal H^h_\delta$ for the
infimum at a fixed cutoff. We seek a deterministic gauge that gives
positive finite measure to each nontrivial compact trace segment.
Hausdorff measure optimizes over spatial covers; ordered partitions
of a parametrized curve lead to a different notion of length.
Schramm asked for a sigma-finite Hausdorff measure of the SLE trace
and suggested the following candidate
\cite[Problem~7.1 and the following discussion]{schramm2007}:
\[
r^d\log\log(1/r).
\]
The article \emph{An exact Hausdorff gauge for SLE: A moment-integral
and finite-batch construction} already answered the existence question
for each fixed $0<\kappa<8$. Its deterministic moment integrals specify
coefficients, and finite irregular batches of radii specify a gauge
through a full infimum \cite[equations~(5.8) and~(6.4)]{openai2026exact}.
That gauge gives positive finite measure simultaneously on positive-time
compact segments and finite expected measure of the entire trace in
bounded regions \cite[Theorem~1.1 and Proposition~8.2]{openai2026exact}.
The present paper identifies a regular closed-form gauge and proves
its two Hausdorff inequalities from quantitative estimates at finite
scales. The stopped-law, mass, and covering arguments below do not
import a proof result from the companion. The relation between the
two exact gauges remains open.
The regular formula also settles the proposed exponent-one correction
negatively: its measure is not sigma-finite on any positive-time
compact segment.
\subsection{The result}
Fix $0<\kappa<8$ and set
\begin{equation}\label{eq:parameters}
d=1+\frac\kappa8,\qquad p=2-d=1-\frac\kappa8.
\end{equation}
Thus $1<d<2$ and $0<p<1$. Define, at all sufficiently small radii,
\begin{equation}\label{eq:gauge}
h(r)=r^d\bigl(\log\log(1/r)\bigr)^{p/2}.
\end{equation}
Set $h(0)=0$ and complete it continuously and nondecreasingly at
larger radii. The expression is increasing near zero, and its
completion does not affect \eqref{eq:hausdorff-definition}.
\begin{theorem}\label{thm:main}
Let $\gamma$ be ordinary chordal $\mathrm{SLE}_\kappa$ from $0$ to
$\infty$ in $\mathbb H$, parametrized by half-plane capacity $2t$,
and set $\Gamma=\gamma([0,\infty))$. For each fixed $0<\kappa<8$,
on one event of probability one,
\[
0<\mathcal H^h(\gamma([s,t]))<\infty
\qquad\text{for every real }0<s<t<\infty.
\]
For every deterministic $0<R<\infty$,
\[
\mathbb E\mathcal H^h(\Gamma\cap\overline B(0,R))
\le C_{\kappa,R}<\infty.
\]
In particular, on the same probability-one event,
\[
\mathcal H^h\bigl(\Gamma\cap([-m,m]+i[0,m])\bigr)<\infty
\qquad(m=1,2,\ldots).
\]
\end{theorem}
The spatial expectation concerns the whole trace at once, including
its real-axis points. It is not obtained by adding bounds over
successive finite time intervals. The event is for the fixed value
of $\kappa$.
\begin{corollary}\label{cor:larger-gauge}
Let $g(r)=r^d\log\log(1/r)$ at sufficiently small radii, with any
continuous nondecreasing completion. On the event in
Theorem~\ref{thm:main}, the restriction of $\mathcal H^g$ to every
$\gamma([s,t])$, $0<s<t<\infty$, is not sigma-finite; in particular
its total measure is infinite.
\end{corollary}
\begin{proof}
The ratio $h(r)/g(r)$ tends to zero. For each $\varepsilon>0$,
all sufficiently fine covers therefore give
$\mathcal H^h(A)\le\varepsilon\mathcal H^g(A)$.
Every set of finite $\mathcal H^g$-measure has zero
$\mathcal H^h$-measure. A countable cover of the segment by such sets
would contradict its positive $\mathcal H^h$-measure.
\end{proof}
Natural parametrization provides another notion of SLE length.
Lawler--Sheffield constructed it for
$0<\kappa<4(7-\sqrt{33})$ \cite[Theorem~3.1]{lawlersheffield2011};
Lawler--Zhou extended the construction to $0<\kappa<8$
\cite[Theorem~1]{lawlerzhou2013}; and Lawler--Rezaei identified it,
up to normalization, with $d$-dimensional Minkowski content
\cite[Theorem~1.1 and p.~5]{lawlerrezaei2015}.
For chordal SLE in the half-plane, Holden--Yuan identify positive
fine-mesh variation, up to a deterministic factor, with natural
parametrization, including segments that meet the real boundary
\cite[Theorems~1.7--1.8]{holdenyuan2026}.
Their gauge $r^d(\log\log(1/r))^{-(d-1)}$ is applied to ordered
partitions. The different logarithmic correction here belongs to
arbitrary spatial covers. Our argument requires no identification
of $\mathcal H^h$ with either natural parametrization or Minkowski
content.
\subsection{The two Hausdorff bounds}
For the lower Hausdorff bound we need a positive measure on a piece
of the trace whose mass in every small ball is at most a constant
times $h(r)$. For the upper bound we need spatial covers whose total
$h$-cost has a bounded expectation. Both tasks use finite-mesh
quantities, but they use them in different ways.
First consider a disk $A$ with a fixed free collar in the surviving
domain after a finite stopped past. This domain is the component
toward infinity after filling the components cut off by the trace.
At a live interior point $z$, the Green density is
$R^{-p}\sin^{8/\kappa-1}\theta$, where $R$ is its conformal radius
and $\theta$ is its argument in tip--target half-plane coordinates.
Stop this density at the point's own first live approach within
distance $e$, and assign zero if no such approach occurs. Denote
the resulting stopped weight by $W_e(z)$. The quantitative estimate is
\[
\mathbb P\left\{\int_AW_e(z)\,dA(z)>u\right\}
\le C\exp(-cu^{2/p})
\]
in unit-scale coordinates, for large $u$, uniformly in the permitted
past and final mesh. This controls a fixed spatial scale while the
approximation mesh decreases to zero. Rezaei and Zhan proved
fixed-$n$ multipoint approach bounds and finiteness of moments of natural
length \cite[Theorems~1.1--1.2]{rezaeizhan2017}. Their estimates already
allow arbitrary approach radii. The estimate here gives the quantitative
$e^{-cu^{2/p}}$ tail for the integrated stopped density, with constants
uniform over both the permitted pasts and the final meshes.
The exponent comes from testing a large mass first at an intermediate
distance $\delta$. There the deterministic density bound gives mass
of order $\delta^{-p}$. For each prescribed finer final mesh, the
later increments raise this bound by only a fixed factor outside an
event of probability at most $Ce^{-c\delta^{-2}}$, with constants
independent of that final mesh. Taking $\delta$ of order $u^{-1/p}$,
and using the pointwise cap for coarser meshes, produces the displayed
tail. Section~\ref{mesh:section} controls those increments
in two ways. For $\kappa\le4$, Green-kernel decrease bounds the bracket
of density increments that start at different times. For $\kappa>4$,
spatially separated windows and a swallowing estimate control the
surviving contributions.
For the lower bound, Section~\ref{retain:section} deletes dyadic
squares whose mass is too large. At scale $r=2^{-k}$ the threshold is
$Dr^d(\log k)^{p/2}$ for a sufficiently large fixed constant $D$.
Reaching the square costs $O(r^p)$, whereas
its mass scale is $r^d$ and there are $O(r^{-2})$ squares.
The powers cancel because $p+d=2$. The tail makes the remaining
loss summable. Positive first moments and bounded second moments
leave a positive retained mass with positive probability. A supported
weak subsequential limit satisfies the required bound on every ball;
restarting on deterministic capacity slots gives positivity on every
positive-time interval.
The upper bound asks for the opposite kind of local event: a disk
must occasionally contain \emph{more} sausage area than its usual
scale. For a curve segment $\eta$, its normalized sausage area at
mesh $v$ is
\[
v^{-p}\operatorname{Area}\{z:\operatorname{dist}(z,\eta)<v\}.
\]
Rezaei's covering proof already combines dense occupation, conditional
estimates after stopped histories, compactness, and prescribed spiral
passages \cite[Section~3, Lemmas~3.1--3.3]{rezaei2018}. Using the same
geometric idea, a serpentine ribbon with order $m$ strands provides order $m^2$
disjoint local tests at spatial scale $m^{-1}$. Successful tests
contribute total normalized area of order $m^{2-d}=m^p$. If each
step succeeds with a conditional probability bounded below uniformly
over every admissible rough past, the whole visit costs at most an
exponential in $m^2$: its probability is at least $ce^{-Cm^2}$.
Pointwise support for a fixed smooth path, as in Tran--Yuan
\cite[Theorem~1.1, Corollary~1.2, and Proposition~1.4]{tranyuan2020},
does not supply this common conditional probability over rough pasts.
Section~\ref{retain:section} first proves an ordinary-law lower-tail
statement for sausages on deterministic time intervals. In
Section~\ref{rc:section}, compactness of conformal maps along an interior
guide and a uniform localization estimate turn pointwise support into
one ribbon-step bound. The passage is described with fixed margins that
persist under convergence of the maps. A finite selection of rational
time intervals then transfers the sausage estimate, leaving a common
positive probability, mass threshold, and deterministic mesh cutoff.
For the final cover take $m$ proportional to $\sqrt{\log n}$ and
perform order $n$ tests at geometrically spaced radii $r_j$, with $j$ between
$n/3$ and $2n/3$. Conditional success bounds, not independence, make
total failure small. Since $\log n$ is comparable to
$\log\log(1/r_j)$, the generated mass matches the correction in
$h$. The deterministic mesh cutoff in the ribbon estimate lets
every test use one physical mesh. Disjoint selected disks can then
all be charged to the same whole-trace sausage measure.
Section~\ref{sec:dense-cover} carries out this cover. Its dense-occupation
and multiscale strategy follows Rezaei \cite[Section~3, especially
equations~(9)--(11)]{rezaei2018}, who selects maximal high-occupation
circles and charges them using bounded overlap. Here a greedy family
of disjoint disks is charged to one finite-mesh sausage measure, and
the quantitative ribbon probability specifies the logarithmic gauge.
To bound the cost of visited cells that the selected disks miss, the
proof uses test failures visible at finite stops, where the finite Green
changes of law apply. The disks themselves are selected afterward from
the whole-trace sausage. The final charge has a bounded expectation in
one fixed spatial disk, independently of time and height cutoffs.
Increasing both cutoffs then yields the all-time assertion.
The technical foundations appear in Sections~\ref{geom:section}
and~\ref{ll:section}. Starting from Rohde--Schramm's continuous trace
theorem \cite[Theorem~5.1]{rohdeschramm2005} and classical conformal
boundary theory, the first establishes access and slit support after a
rough past. The second develops the finite stopped Green laws from
the methods of Lawler--Zhou \cite[Section~2]{lawlerzhou2013} and
Lawler--Werness \cite[Sections~2.3--2.4]{lawlerwerness2013},
and proves the physical localization used by the ribbon argument.
The $\kappa\le4$ mesh proof uses the Green-decrease identity appearing in
Schramm--Sheffield's SLE$_4$/Gaussian-free-field covariance calculation
\cite[Lemmas~2.5--2.6]{schrammsheffield2013}, with the decomposition for
different starting times supplied here.
==== END An-explicit-exact-Hausdorff-gauge-for-SLE//introduction.tex ====
==== BEGIN An-explicit-exact-Hausdorff-gauge-for-SLE//local-laws.tex ====
\section{Finite one-point laws and physical localization}
\label{ll:section}
This section develops the one-point estimates used by both Hausdorff
bounds. Finite Green changes of law give approach probabilities and
show that the centered radius stops have a finite Loewner-time limit as
their radii tend to zero; the latter fact supplies finite-time mass in
Section~\ref{retain:section}. A conformal localization estimate then
controls the image of an initial trace uniformly over maps whose domains
may have narrow passages. It holds for every $0<\kappa<8$ and is an
input to the ribbon construction. When $4<\kappa<8$, combining it with
a swallowing estimate gives the physical-disk rule used in the mesh tail.
All constants below depend only on the fixed parameter $\kappa$, unless
another dependence is displayed.
Write
\[
p=1-\frac\kappa8,\qquad a_0=\frac8\kappa-1,
\qquad 0<\kappa<8.
\]
We use the original capacity normalization
$\partial_tg_t=2/(g_t-\sqrt\kappa B_t)$.
The conformal radius $R_D(z)$ is $|f'(0)|$ for a conformal map
$f:\mathbb D\to D$ with $f(0)=z$.
In particular, Koebe's Theorem gives
\begin{equation}
\operatorname{dist}(z,\partial D)\le R_D(z)
\le4\operatorname{dist}(z,\partial D).
\label{ll:koebe}
\end{equation}
These inequalities require no boundary regularity
\cite[Section~3.2]{beliaev2015}.
\subsection{Stopped Green tilts}
The following centered diffusion is the usual finite-radius Green
change of law; see \cite[Section~2]{lawlerzhou2013} and
\cite[Sections~2.3--2.4]{lawlerwerness2013}.
We include the exhaustion argument because our later stopping times
may follow arbitrarily irregular pasts.
Let $J:\mathbb H\to D$ be conformal. The boundary parameters $0$ and
infinity specify the starting and target prime ends of $D$; no continuous
boundary extension of $J$ is assumed here. Let $K_t$ and $g_t$ be the
filled hulls and map-out functions of ordinary half-plane SLE, and set
$D_t=J(\mathbb H\setminus K_t)$. Throughout this subsection, $t$ is
Loewner time in the initial half-plane, where the hull has capacity $2t$;
it need not be a capacity parametrization intrinsic to $D$.
For $z\in D$, put $\zeta=J^{-1}(z)$ and
$T_z=\inf\{t:z\notin D_t\}$, with $T_z=\infty$ if this set is empty.
We call the point live at times $t<T_z$. Until $T_z$, set
\[
Z_t=g_t(\zeta)-\sqrt\kappa B_t=X_t+iY_t,
\quad\theta_t=\arg Z_t,
\quad R_t=R_{D_t}(z)=\frac{2Y_t|J'(\zeta)|}{|g_t'(\zeta)|}.
\]
Define the radius clock by $s=\log(R_0/R_t)$.
Differentiation and It\^o's Formula give
\begin{align}
ds&=\frac{4Y_t^2}{|Z_t|^4}\,dt,
&\mathcal L&=\frac\kappa8\partial_\theta^2
+\frac{\kappa-4}{4}\cot\theta\,\partial_\theta,
\label{ll:clock}\\
M_t(z)&=R_t^{-p}\sin^{a_0}\theta_t,
&dM_t&=\sqrt\kappa a_0\frac{X_t}{|Z_t|^2}M_t\,dB_t.
\label{ll:density}
\end{align}
The sign of the last Brownian motion depends only on the driver
convention and has no later effect. For the displayed driver it is
the positive sign written here.
In particular,
\begin{equation}
|\text{Brownian coefficient of }M_t|
\le C R_t^{-p}\frac{\sin^{a_0}\theta_t}{|Z_t|}.
\label{ll:coefficient}
\end{equation}
The density is continued by zero after killing.
\begin{lemma}[Finite stopped centered law]\label{ll:finite-tilt}
For $0\le u<\infty$, let $\tau_u$ be the first time the live point has
$R_t=R_0e^{-u}$, with value infinity if this level is never reached.
Then
\[
\mathbb E\left[M_{\tau_u}(z)
\mathbf1_{\{\tau_u<\infty\}}\right]=M_0(z).
\]
The normalized weight defines a consistent law on the stopped path
through $\tau_u$. Under that law, in radius time,
\begin{equation}
d\theta_s=\frac{\sqrt\kappa}{2}\,dW_s
+\cot\theta_s\,ds,
\qquad
\mathcal L^*=\frac\kappa8\partial_\theta^2
+\cot\theta\,\partial_\theta.
\label{ll:tilt-generator}
\end{equation}
Both endpoints are unattainable at finite radius time. These claims
hold conditionally after any finite stopping time at which the point
is live, with the same constants and the restarted conformal map.
\end{lemma}
\begin{proof}
First stop at $\tau_u$ or when $\theta$ leaves $(1/N,\pi-1/N)$,
$Y$ falls below $1/N$, or Loewner time reaches $N$. The stochastic
logarithm coefficient is bounded there, since
\[
\frac{|X_t|}{|Z_t|^2}\le\frac1{2Y_t}\le\frac N2.
\]
Girsanov's Theorem therefore applies. Its added
angular drift is $(\kappa a_0/4)\cot\theta$; together with
\eqref{ll:clock} this gives \eqref{ll:tilt-generator}.
For $0<b<a_0$, let $V_b(\theta)=\sin^{-b}\theta$. Direct
differentiation gives
\begin{equation}
\mathcal L^*V_b
=V_b\left\{\frac{\kappa b}{8}
-\frac{\kappa b(a_0-b)}8\cot^2\theta\right\}
\le\frac{\kappa b}{8}V_b.
\label{ll:lyapunov}
\end{equation}
The stopped nonnegative supermartingale
$e^{-\kappa bs/8}V_b(\theta_s)$ bounds the probability of leaving
$(\delta,\pi-\delta)$ before $u$ by
$e^{\kappa bu/8}V_b(\theta_0)\sin^b\delta$.
This tends to zero. On every compact interval on which the angle
stays in $(0,\pi)$, the remaining variables satisfy
\begin{equation}
\frac{d\log Y_s}{ds}=-\frac1{2\sin^2\theta_s},
\qquad
\frac{dt}{ds}=\frac{Y_s^2}{4\sin^4\theta_s}.
\label{ll:physical-clock}
\end{equation}
They keep $Y$ positive and Loewner time finite there. Thus the
localized changed laws lose no probability before $u$ as the
cutoffs are removed. The localized density identities and monotone
convergence give the asserted expectation one. Repeating the same
argument between two finite levels gives consistency. Composition
of the Loewner maps and Brownian strong Markov give the conditional
statement. No change of law on a completed infinite-future sigma
field is involved.
\end{proof}
\begin{lemma}[Uniform inverse-angle moment]\label{ll:inverse-angle}
For every $u_0>0$ there is $C_{u_0}<\infty$ such that
\[
\sup_{\theta\in(0,\pi)}\sup_{s\ge u_0}
\mathbb E^*_{\theta}\sin^{-a_0}\theta_s\le C_{u_0}.
\]
In particular, after one unit of radius time the probability that
$\sin\theta$ exceeds a fixed positive constant is bounded below,
uniformly over the starting angle.
\end{lemma}
\begin{proof}
We give a short spectral proof which also avoids any ambiguity about
comparison inequalities for unbounded test functions. The invariant
probability measure for \eqref{ll:tilt-generator} is
\[
\pi(d\theta)=c_\kappa\sin^{8/\kappa}\theta\,d\theta.
\]
For $v=\cos\theta$, the generator and invariant measure become
\[
\mathcal A=\frac\kappa8(1-v^2)\partial_v^2
-\left(1+\frac\kappa8\right)v\partial_v,
\qquad
\varpi(dv)=\widetilde c_\kappa
(1-v^2)^{4/\kappa-1/2}\,dv.
\]
Integration by parts makes $\mathcal A$ symmetric on polynomials.
If $P_m$ are the orthonormal polynomials for $\varpi$, with $P_0=1$,
its action on the leading coefficient and symmetry imply
\[
\mathcal AP_m=-\lambda_mP_m,
\qquad \lambda_m=\frac\kappa8m^2+m.
\]
For completeness, choose $m+1$ separated intervals of length
$c/(m+1)$ in $[-1/2,1/2]$. Since the weight is bounded below there,
each contains a point where $|P_m|\le C\sqrt{m+1}$.
Lagrange interpolation at these points gives
\[
\|P_m\|_{\infty,[-1,1]}
\le \exp\{C(m+1)\log(m+2)\}.
\]
Consequently
\[
k_s(v,w)=1+\sum_{m\ge1}e^{-\lambda_ms}P_m(v)P_m(w)
\]
converges absolutely and uniformly for $s\ge u_0$, with a uniformly
bounded sum. It\^o's Formula for each bounded polynomial gives
$\mathbb E_vP_m(v_s)=e^{-\lambda_ms}P_m(v)$.
Polynomial density in $C([-1,1])$ identifies the actual transition
law with $k_s(v,w)\varpi(dw)$. Thus its density relative to $\pi$
is uniformly bounded for $s\ge u_0$. Finally,
\[
\int\sin^{-a_0}\theta\,\pi(d\theta)
=c_\kappa\int_0^\pi\sin\theta\,d\theta<\infty.
\]
Markov's inequality applied to $\sin^{-a_0}\theta$ proves the last
assertion. This is a direct proof in our normalization of the uniform
inverse-angle estimate in \cite[Lemma~2.9]{lawlerwerness2013}; see also
\cite[Lemma~2.2]{lawlerzhou2013}.
A Gegenbauer expansion for this diffusion is also used in
\cite[Proposition~4.2]{zhan2019optimal}. Here the interpolation estimate
suffices in place of explicit polynomial norm formulas.
\end{proof}
To pass from conformal radius to Euclidean clearance, write
$\delta_t=\operatorname{dist}(z,\partial D_t)$ while $t<T_z$.
This is an adapted continuous function of $t$ on the live interval.
Indeed, if $t_n\to t<T_z$, stability of the Loewner ODE gives locally
uniform convergence of the inverse maps. Composing them with $J$ and
normalizing at $z$ gives nondegenerate conformal maps onto $D_{t_n}$
converging to the corresponding map onto $D_t$. Thus these domains
converge in the kernel sense at $z$. Every closed centered disk of
radius less than $\delta_t$ eventually lies in $D_{t_n}$. Conversely,
if a fixed centered disk of radius greater than $\delta_t$ persisted
in a subsequence of the $D_{t_n}$, it would lie in the kernel $D_t$.
These two observations prove $\delta_{t_n}\to\delta_t$.
For $0<\varepsilon<\delta_0$, define the live clearance stop
\[
S_\varepsilon(z)=\inf\{t<T_z:\delta_t\le\varepsilon\},
\]
with value infinity if this set is empty. It is a stopping time, and
$\delta_{S_\varepsilon}=\varepsilon$ on its finite event. This definition
also applies when $J$ has no continuous boundary extension.
When $D_t$ is the surviving domain of a continuous physical trace
$\eta$ in an initial domain $D_0$, there is the more concrete identity
\begin{equation}\label{ll:trace-clearance}
\delta_t=\min\{\operatorname{dist}(z,\partial D_0),
\operatorname{dist}(z,\eta[0,t])\},\qquad t<T_z.
\end{equation}
In fact, $\partial D_t\subset\partial D_0\cup\eta[0,t]$, and both sets
on the right lie outside $D_t$; distance to the complement equals
distance to the boundary. Thus below the initial clearance,
$S_\varepsilon$ is exactly the first live approach of the trace to
distance $\varepsilon$. A first approach occurring only at killing
is not counted. The finite-past domains and their similarities used
later have this continuous-trace interpretation.
\begin{lemma}[Terminal zero and approach estimates]\label{ll:approach}
Let $0<r<R_0$ and freeze the density at the first live stop with $R=r$.
Almost surely, on the event that the live radius never reaches $r$, the
density has terminal value zero,
whether its original Loewner-time lifetime is finite or infinite.
Furthermore, for $u\ge1$,
\begin{equation}
\mathbb P\{\tau_u<\infty\}
\le C e^{-pu}\sin^{a_0}\theta_0.
\label{ll:radius-approach}
\end{equation}
For $0<\varepsilon\le\delta_0/(8\mathrm e)$,
\begin{equation}
\mathbb P\{S_\varepsilon(z)<\infty\}
\le C M_0(z)\varepsilon^p.
\label{ll:euclidean-approach}
\end{equation}
At this stop the density has the exact first moment
\begin{equation}\label{ll:clearance-first-moment}
\mathbb E\!\left[M_{S_\varepsilon}(z)
\mathbf1_{\{S_\varepsilon<\infty\}}\right]=M_0(z).
\end{equation}
Under the centered law, $S_\varepsilon$ occurs no later than the finite
radius stop with $R=\varepsilon$.
All statements have their conditional restarted versions.
\end{lemma}
\begin{proof}
Suppose the radius-time lifetime is finite. If the angle eventually
stayed in a compact subinterval of $(0,\pi)$, its SDE and
\eqref{ll:physical-clock} would extend it: the latter equations give
finite Loewner time, positive $Y$, and regular $Z$ there. Otherwise
the angle approaches an endpoint along a sequence. It must converge
to that endpoint. Indeed, repeated crossings of a fixed interior
interval in a finite remaining radius time would contradict uniform
continuity of the Brownian part and boundedness of the drift on that
interval. Thus the maximal angle solution ends at $0$ or $\pi$.
On a path that never reaches the radius $r$, its radius time is bounded
and its radius stays bounded below. Its density therefore tends to zero,
including when its original Loewner-time lifetime is infinite.
The density stopped at the radius cutoff is bounded by $r^{-p}$;
localization followed by bounded convergence is therefore legitimate.
Under the centered law, $M_{\tau_u}/M_0$
equals $e^{pu}\sin^{a_0}\theta_u/\sin^{a_0}\theta_0$.
Untilting at this finite stop and applying
Lemma~\ref{ll:inverse-angle} gives \eqref{ll:radius-approach}.
At $S_\varepsilon<\infty$, \eqref{ll:koebe} gives
$R_{S_\varepsilon}\le4\varepsilon$. The live radius must therefore
have crossed $8\varepsilon$ earlier. Since
$R_0\ge\delta_0\ge8\mathrm e\varepsilon$, the corresponding radius
time is at least one, and \eqref{ll:radius-approach} proves
\eqref{ll:euclidean-approach}.
Before $S_\varepsilon$, $R\ge\delta_t>\varepsilon$
by \eqref{ll:koebe}. The stopped density is consequently bounded by
$\varepsilon^{-p}$.
Under the centered law the clearance stop occurs no later than the
finite radius level $R=\varepsilon$, because
$\operatorname{dist}(z,\partial D)\le R$.
On a path with no clearance stop, the radius stays above
$\varepsilon$, so the preceding terminal-zero assertion applies.
Bounded optional stopping and bounded convergence now give
\eqref{ll:clearance-first-moment}. Strong Markov gives the conditional
versions with the restarted domain and its initial clearance.
\end{proof}
\begin{lemma}[Finite centered lifetime in Loewner time]\label{ll:finite-clock}
Under the consistent centered laws, $T^*(z)=\lim_{u\to\infty}\tau_u$
is finite almost surely.
\end{lemma}
\begin{proof}
The locally unique centered SDE in \eqref{ll:tilt-generator} has a global
solution $(\theta_s)_{s\ge0}$: neither endpoint is attained in finite
radius time, so the local solutions extend and agree on overlaps.
On this path reconstruct the other variables by
\[
Y_s=Y_0\exp\left\{-\frac12\int_0^s\sin^{-2}\theta_v\,dv\right\},
\qquad
t_s=\int_0^s\frac{Y_v^2}{4\sin^4\theta_v}\,dv.
\]
Put $X_s=Y_s\cot\theta_s$. Since $t_s$ is strictly increasing, a
continuous driver on every interval $[0,t_s]$ is recovered from
\[
U_{t_s}=\operatorname{Re}\zeta+
\int_0^s\frac{Y_v\cot\theta_v}{2\sin^2\theta_v}\,dv-X_s.
\]
This follows by integrating $\partial_t\operatorname{Re}g_t(\zeta)
=2X_t/|Z_t|^2$ and using $\operatorname{Re}g_t(\zeta)=U_t+X_t$.
Uniqueness of the centered SDE now identifies the restriction through
$s=u$ with the finite stopped law of Lemma~\ref{ll:finite-tilt}, with
$t_u=\tau_u$. This gives one probability space for the limit $T^*(z)$
and the argument below, without using a
density on an infinite-future sigma field or at the limiting time.
The formula \eqref{ll:lyapunov} at $b=a_0$ is
$\mathcal L^*V_{a_0}=pV_{a_0}$.
Thus $e^{-ps}V_{a_0}(\theta_s)$ is a nonnegative local martingale
and hence a supermartingale. Its stopped maximal inequality and
Lemma~\ref{ll:inverse-angle} imply, for integers $n\ge1$,
\[
\mathbb P^*\left\{\sup_{n\le s\le n+1}
V_{a_0}(\theta_s)>e^{a_0\eta n}\right\}
\le C e^{-a_0\eta n}.
\]
For any $0<\eta<1/4$, Borel--Cantelli therefore gives
$\sin^{-4}\theta_s\le C e^{4\eta s}$ eventually.
Also \eqref{ll:physical-clock} gives $Y_s\le Y_0e^{-s/2}$.
The integral of $Y_s^2/(4\sin^4\theta_s)$ over $s\in[0,\infty)$
is finite. Initial compact intervals have finite integrals by
endpoint nonattainment. This proves the assertion solely from
finite stopped laws.
\end{proof}
\subsection{Avoidance near the real line}
\begin{lemma}[Uniform small-target estimate]\label{ll:near-real}
There are $C,c>0$ such that ordinary chordal SLE in $\mathbb H$
satisfies
\[
\mathbb P\{\operatorname{dist}(v,\gamma)<\varepsilon\}
\le C\varepsilon^c,
\qquad \tfrac12\le|v|\le2,\quad
v\in\overline{\mathbb H},\quad 0<\varepsilon<\tfrac1{16}.
\]
The same estimate holds after a simultaneous spatial rescaling of
the annulus and target radius. The trace in this statement includes
all future times and its real boundary points.
\end{lemma}
\begin{proof}
A stronger, sharp boundary exponent was proved by Alberts and
Kozdron \cite[Theorem~3.2 and Corollary~5.1]{albertskozdron2008}.
Here is a direct weaker estimate sufficient for our use.
It suffices to treat sufficiently small $\varepsilon$, since increasing
$C$ covers the remaining values below $1/16$.
Fix real $x$ with $|x|$ bounded above and away from zero, and write
$T_x$ for its real Loewner collision time. Up to $T_x$, put
$X_t=g_t(x)-\sqrt\kappa B_t$ and
$h=\int_0^tX_u^{-2}\,du$. Changing the sign of a Brownian motion
if necessary gives
\[
\log|X|=\log|x|+(2-\kappa/2)h+\sqrt\kappa\,\widetilde B_h.
\]
For $w$ in the upper half-ball $|w-x|<4\varepsilon$, set
$D_t(w)=g_t(w)-g_t(x)$. While $|D_t(w)/X_t|\le\delta$, the ODE
gives
\[
\frac{d}{dh}\log|D_t(w)|
=-2\operatorname{Re}\frac{X_t}{X_t+D_t(w)}
\le-2+C\delta.
\]
Choose $\delta>0$ so small that
$\mu=4-\kappa/2-C\delta>0$. Until the first failure of the ratio
bound, simultaneously for all these $w$,
\[
\log\frac{|D_t(w)|}{|X_t|}
\le\log\frac{4\varepsilon}{|x|}
-\mu h-\sqrt\kappa\,\widetilde B_h.
\]
For sufficiently small $\varepsilon$,
$H=\log(\delta|x|/(4\varepsilon))$ is positive. The Brownian motion
with negative drift reaches that level with probability
$\exp(-2\mu H/\kappa)$. Thus with failure
probability at most $C\varepsilon^{2\mu/\kappa}$ the ratio never
reaches $\delta$. One may first apply the argument to compact
sub-half-balls and countably many points. Analyticity and continuous
dependence extend it simultaneously to the whole open half-ball.
Put $U=B(x,4\varepsilon)\cap\mathbb H$. On this event every point of
$U$ stays live before $T_x$, so the trace avoids $U$ there. A
capacity-parametrized generating trace cannot stay entirely on
$\mathbb R$ during a nondegenerate time interval: adding only real
points leaves its filled half-plane hull unchanged, contrary to its
strict capacity increase. If the trace visited
$B(x,\varepsilon)\cap\overline{\mathbb H}$ at a time $0<t\le T_x$,
continuity would place a short preceding time interval inside
$B(x,4\varepsilon)\cap\overline{\mathbb H}$. That interval contains a
time when the trace is in $\mathbb H$, contradicting avoidance of $U$.
The initial point $0$ is outside the smaller ball.
If $T_x<\infty$, then $|X_t|\to0$, and the ratio bound shows that every
point of $U$ collides at $T_x$. Thus $U$ is disjoint from the surviving
domain at that time. The smaller relative half-ball is disjoint even
from its closure: each of its points has an upper-half-plane neighborhood
contained in $U$. Future trace growth stays in that closure and cannot
visit the smaller half-ball. If $T_x=\infty$, the preceding argument
already excludes every finite visit. This proves the real-center estimate.
For $0\le\operatorname{Im}v\le\sqrt\varepsilon$, a visit within
$\varepsilon$ of $v$ gives a visit within $2\sqrt\varepsilon$ of
$\operatorname{Re}v$, to which the real estimate applies.
For $\operatorname{Im}v>\sqrt\varepsilon$, the radius $2\varepsilon$
is below $\delta_0/(8\mathrm e)$ once $\varepsilon$ is sufficiently
small, so \eqref{ll:euclidean-approach} applies. Here
\[
M_0(v)=(2\operatorname{Im}v)^{-p}
\left(\frac{\operatorname{Im}v}{|v|}\right)^{a_0}
\le C,
\]
because $a_0-p>0$.
An actual visit to $B(v,\varepsilon)$ requires a live first hit of
$\partial B(v,2\varepsilon)$. If the center were killed at or before
that first hit, the still untouched open ball would be entirely swallowed
and could not be entered later. By \eqref{ll:trace-clearance}, the live
hit gives $S_{2\varepsilon}(v)<\infty$. The approach estimate therefore
proves the result for all $v$.
\end{proof}
\subsection{A conformal localization estimate}
The next argument converts conformal area distortion into a
probability estimate. Its constants require no modulus of boundary
continuity for the conformal map. This is the uniformity needed
when a previous trace has formed narrow passages.
\begin{lemma}[Rectangle energy estimate]\label{ll:rectangle}
Let $H>0$, let $Q$ be a rectangle of fixed bounded aspect ratio, and let
$u\in W^{1,2}(Q)\cap C(\overline Q)$ satisfy $0\le u\le H$
and $u_Q\le H/2$. Write $E=\int_Q|\nabla u|^2$.
For any random set $\mathcal X$ for which every dyadic level-$k$
subrectangle of $Q$ is hit with probability at most $C_02^{-\alpha k}$,
\[
\mathbb P\{\mathcal X\cap\{u=H\}\ne\varnothing\}
\le C\frac E{H^2}\exp\left(-c\frac{H^2}{E}\right)
\]
whenever $E\le c_0H^2$. The constants depend only on the aspect
ratio, $C_0$, and $\alpha$. If $E=0$, the probability is zero.
\end{lemma}
\begin{proof}
We first record a two-dimensional average estimate. If $Q_k$ is
any nested chain of dyadic subrectangles with $Q_0=Q$, then
\begin{equation}
|u_{Q_k}-u_Q|\le C\sqrt{(k+1)E}.
\label{ll:nested-means}
\end{equation}
One proof is to reflect $u$ across the sides of $Q$ and use
concentric annuli of radii comparable to $2^{-j}$ about the center
of $Q_k$. The difference between means on consecutive annuli is
at most $C$ times the square root of the energy on their union,
by the scale-invariant Poincar\'e inequality. These unions have
bounded overlap. Cauchy--Schwarz over the $k+O(1)$ annuli gives
\eqref{ll:nested-means}. Poincar\'e on a ball of comparable radius
compares the innermost annular mean with $u_{Q_k}$ and the outermost
mean with $u_Q$. The reflected region uses a bounded number of
copies of $Q$, so its energy is at most $CE$.
Choose $N=\lfloor c_1H^2/E\rfloor$ with $c_1>0$ sufficiently small.
Decreasing $c_0$ if needed ensures $N\ge1$ and, by
\eqref{ll:nested-means}, every mean through level $N$ is at most
$3H/4$. If $u(x)=H$, choose a nested chain containing $x$.
Continuity implies that its means tend to $H$. Hence, for some
$j\ge1$, the mean changes between levels $N+j-1$ and $N+j$ by at
least $c_2H/j^2$, where $c_2$ is a sufficiently small numerical
constant. Otherwise the total change is less than $H/4$.
Poincar\'e on the parent rectangle bounds such a mean difference
by $C(\int_{Q_{N+j-1}}|\nabla u|^2)^{1/2}$.
Since level-$N+j-1$ parents have disjoint interiors, at most
$CEj^4/H^2$ of them can have this property.
They cover $\{u=H\}$ after taking the union over $j$.
The assumed hitting bound and a union bound give
\[
C\frac E{H^2}\sum_{j\ge1}j^4
2^{-\alpha(N+j-1)}
\le C\frac E{H^2}e^{-cH^2/E}.
\]
Dyadic boundary points can be assigned either adjoining chain; closed
rectangles and the stated hitting bound cover them as well.
\end{proof}
\begin{proposition}[Uniform conformal localization]\label{ll:localization}
Let $F:\mathbb H\to\mathbb C$ be univalent and admit a continuous
extension to $\overline{\mathbb H}$ with values in the sphere.
Suppose a continuous path $\eta$ in $\overline{\mathbb H}$ starts
at $0$, reaches modulus $1$, and satisfies
$F(\eta)\subset\overline B(q,r)$, where $r>0$.
There exist $b,c,C>0$ and $L_0>1$, depending only on $\kappa$, such
that for every $L\ge L_0$,
ordinary chordal SLE satisfies
\[
\mathbb P\left\{F(\gamma[0,\sigma_{L^b}])
\not\subset B(q,Lr)\right\}
\le CL^{-c},
\qquad
\sigma_A=\inf\{t:|\gamma(t)|=A\}.
\]
If the modulus stop is not reached, use the entire future trace.
All four constants are independent of $F,q,r$, and of the path $\eta$.
Boundary excursions are included.
\end{proposition}
\begin{proof}
Put $H=\log L$ and use strip coordinates $z=e^{v+i\phi}$,
$0<\phi<\pi$. Set
\[
U(v,\phi)=\min\{H,\log^+(|F(e^{v+i\phi})-q|/r)\}.
\]
This function is continuous on each closed finite strip rectangle,
including at a boundary pole of $F$, where it equals $H$.
Conformal invariance of Dirichlet energy and univalence give
\begin{equation}
\int_{\mathbb R\times(0,\pi)}|\nabla U|^2
\le\int_{r<|w-q|<Lr}\frac{dA(w)}{|w-q|^2}
=2\pi H.
\label{ll:total-energy}
\end{equation}
For every $v<0$, the path assumption supplies a point on that
vertical section where $U=0$. One-dimensional Poincar\'e therefore
bounds the mean of $U$ on
$Q_j=[j,j+1]\times[0,\pi]$, $j<0$, by $C\sqrt{E_j}$, where
$E_j=\int_{Q_j}|\nabla U|^2$.
Let $A(v)=\pi^{-1}\int_0^\pi U(v,\phi)\,d\phi$.
For some $v_0\in[-1,0]$, $A(v_0)\le C\sqrt H$.
Cauchy--Schwarz applied to $A'$ gives, for $v\ge0$,
\[
A(v)\le C\sqrt H+C\sqrt{(v+1)H}.
\]
Choose a fixed $b>0$ small enough. There is then $H_0$, depending only
on the constants above, such that for $H\ge H_0$,
the mean on every slab with $j\le bH$ is at most $H/2$.
Under the exponential map, a dyadic level-$k$ rectangle in $Q_j$
lies in a Euclidean disk of radius $C2^{-k}e^j$ whose center has
modulus comparable to $e^j$. Lemma~\ref{ll:near-real} and Brownian
scaling therefore bound its probability of being hit by the full
SLE trace by $C2^{-\alpha k}$, uniformly in $j$.
After increasing $H_0$ if necessary, \eqref{ll:total-energy} also gives
$E_j\le2\pi H\le c_0H^2$.
Lemma~\ref{ll:rectangle} consequently bounds the probability of a
hit on $\{U=H\}$ in $Q_j$ by
$C(E_j/H^2)e^{-cH^2/E_j}$.
The infinitely many slabs to the left cause no loss: since
$E_j\le2\pi H$ and $\sum_jE_j\le2\pi H$,
\[
\sum_{j\le bH}C\frac{E_j}{H^2}e^{-cH^2/E_j}
\le\frac C H e^{-c'H}\le CL^{-c'}.
\]
Set $L_0=\exp(H_0)$. Before $\sigma_{L^b}$ the trace has $v\le bH$.
If its image leaves
$B(q,Lr)$, continuity gives a point with $U=H$ in one of these
slabs. The start itself maps into $\overline B(q,r)$.
This proves the result, including points on either strip edge.
\end{proof}
\subsection{Swallowing and a rule in physical disks}
\begin{lemma}[Uniform swallowing before a large excursion]
\label{ll:swallowing}
Suppose $4<\kappa<8$. For ordinary chordal SLE and every
$z\in\mathbb H$ with $|z|=1$, let $T_z$ be its swallowing time.
There are $C,c>0$ such that
\[
\mathbb P\{\sigma_A\le T_z\}\le CA^{-c},\qquad A\ge2,
\]
uniformly over the angle of $z$.
\end{lemma}
\begin{proof}
Use the clock $h=-\log(Y/Y_0)$ and put $P=X/Y$.
Its generator is
\[
\mathcal G=\frac\kappa4(1+P^2)\partial_P^2+2P\partial_P,
\qquad
\frac{dt}{dh}=\frac{Y_0^2}{2}e^{-2h}(1+P_h^2).
\]
Choose $0<\beta<1-4/\kappa$ and set
$V(P)=(1+P^2)^{\beta/2}$. Direct differentiation yields
\[
\frac{\mathcal GV}{V}
=\frac{2\beta P^2+(\kappa\beta/4)
[1+(\beta-1)P^2]}{1+P^2}.
\]
Its limit at infinity is
$\rho=2\beta+\kappa\beta(\beta-1)/4<\beta$.
Choose $\rho<\lambda<\beta$. Then
$\mathcal GV\le\lambda V+C$, and localized It\^o followed by
Gronwall gives
\begin{equation}
\mathbb EV(P_h)\le Ce^{\lambda h}V(P_0).
\label{ll:swallow-moment}
\end{equation}
The coefficients of the SDE for $P$ have linear growth, so its
solution exists for every finite $h$. Every such $h$ is reached by
the original time change before its lifetime: a finite terminal
$h$ would leave $P$ finite and $Y=Y_0e^{-h}>0$, while the displayed
clock gives finite Loewner time. The original Loewner equation
would then extend beyond its supposed lifetime, a contradiction.
Both drift and diffusion coefficients of $\log V(P_h)$ are bounded.
The exponential maximal inequality for its martingale part on a
unit interval therefore gives
\begin{equation}
\mathbb E\left[\sup_{n\le h\le n+1}V(P_h)\right]
\le Ce^{\lambda n}V(P_0).
\label{ll:swallow-sup}
\end{equation}
One can see the conditional version directly: the martingale part
has bracket at most a constant on that interval, hence its maximum
has uniformly bounded exponential moment; the bounded drift only
changes the multiplicative constant.
Because $0<\beta/2<1$, subadditivity of $x^{\beta/2}$ and
\eqref{ll:swallow-sup} show that the terminal Loewner clock satisfies
\begin{align*}
\mathbb E T_\infty^{\beta/2}
&\le C Y_0^\beta\sum_{n\ge0}e^{-\beta n}
\mathbb E\sup_{n\le h\le n+1}V(P_h)\\
&\le C Y_0^\beta V(P_0)
\sum_{n\ge0}e^{-(\beta-\lambda)n}
\le C|z|^\beta=C.
\end{align*}
This finite clock ends at collision. Indeed $Y_h=Y_0e^{-h}\to0$,
and the same unit-interval estimate gives
\[
\mathbb E\sup_{n\le h\le n+1}|X_h|^\beta
\le C|z|^\beta e^{-(\beta-\lambda)n}.
\]
Markov and Borel--Cantelli imply $X_h\to0$.
Thus $T_\infty=T_z$, rather than a spurious endpoint of the time
change, and $\mathbb P\{T_z>A\}\le CA^{-\beta/2}$.
A hull driven through time $t$ lies below height $2\sqrt t$ and
between the minimum and maximum of its driver. The height bound
follows from $d(Y^2)/dt\ge-4$ for a point not yet swallowed;
the horizontal bound follows because a real part beyond the driver
range moves away from that range under the ODE. Consequently, for
large $A$, a trace excursion to modulus $A$ by time $A$ forces
$\sup_{t\le A}|\sqrt\kappa B_t|\ge cA$.
The Brownian reflection bound makes the probability of this event
at most $Ce^{-c'A}$. Splitting according to $T_z>A$ proves the
claim, after changing constants for bounded $A$.
\end{proof}
\begin{proposition}[The conditional physical-disk rule]
\label{ll:disk-rule}
Suppose $4<\kappa<8$. There exist $C,c>0$ and $L_0\ge2$, depending only
on $\kappa$, with the following property. Consider ordinary chordal SLE
in $\mathbb H$,
possibly after a finite stopping time with a past already drawn,
or a fixed similarity of such an evolution. Suppose
$\overline B(z,r)$ is strictly inside the current surviving domain.
Let $\tau$ be the subsequent first hit of $\partial B(z,r)$.
For $L\ge L_0$, on $\{\tau<\infty,\ z\in D_\tau\}$, set
\[
\sigma=\inf\{t\ge\tau:|\gamma(t)-z|\ge Lr\},
\]
with value infinity if there is no such time. Then
\[
\mathbb P\{\sigma<\infty,\ z\in D_\sigma
\mid\mathcal F_\tau\}\le CL^{-c}.
\]
The event in the display is an exit while the point is still live.
The constants are independent of the past, the initial disk, and the exit
time.
\end{proposition}
\begin{proof}
The first hit is a fresh interior tip. By
Lemma~\ref{geom:first-hit-access}, the radial physical segment from
this tip to $z$ has its interior in $D_\tau$ and approaches the
tip's prime end. Map $D_\tau$ to $\mathbb H$, sending tip to zero,
target to infinity, and normalize the image of $z$ to have modulus
one. Lemma~\ref{geom:extension} gives the continuous spherical
boundary extension of its inverse $F$. The radial segment supplies
exactly the path assumption of Proposition~\ref{ll:localization},
with the physical ball $\overline B(z,r)$.
Increase $L_0$ so that Proposition~\ref{ll:localization} applies and
$L^b\ge2$ for $L\ge L_0$. Except on an event of conditional probability
$CL^{-c_1}$, the
mapped future trace stays in $B(z,Lr)$ up to the parameter-plane
modulus $L^b$. Except on an event of conditional probability
$C(L^b)^{-c_2}$, Lemma~\ref{ll:swallowing} swallows the normalized
marked point before that modulus is reached. Outside their union,
the physical point is swallowed before an exit from $B(z,Lr)$.
The domain Markov property supplies the same ordinary half-plane
law for every past. This proves the stated uniform bound.
\end{proof}
==== END An-explicit-exact-Hausdorff-gauge-for-SLE//local-laws.tex ====
==== BEGIN An-explicit-exact-Hausdorff-gauge-for-SLE//macros.tex ====
\newcommand{\HH}{\mathbb H}
\newcommand{\DD}{\mathbb D}
\newcommand{\PP}{\mathbb P}
\newcommand{\EE}{\mathbb E}
\newcommand{\dd}{\mathop{}\!\mathrm{d}}
\newcommand{\ind}{\mathbf 1}
\newcommand{\SLE}{\mathrm{SLE}}
\newcommand{\Haus}{\mathcal H}
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\newcommand{\diam}{\operatorname{diam}}
\newcommand{\dist}{\operatorname{dist}}
\newcommand{\Var}{\operatorname{Var}}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}
\numberwithin{equation}{section}
\setlist[enumerate]{itemsep=3pt,topsep=5pt}
\setlist[itemize]{itemsep=3pt,topsep=5pt}
\allowdisplaybreaks[2]
==== END An-explicit-exact-Hausdorff-gauge-for-SLE//macros.tex ====
==== BEGIN An-explicit-exact-Hausdorff-gauge-for-SLE//main.tex ====
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\input{macros}
\input{figures/captions}
\title{An explicit exact Hausdorff gauge for SLE\\[4pt]
\large A regular formula from quantitative tails and dense visits}
\author{OpenAI}
\date{September 26, 2026}
\begin{document}
\maketitle
\begin{abstract}
\input{sections/abstract}
\end{abstract}
\input{sections/introduction}
\input{sections/geometry}
\input{sections/local-laws}
\input{sections/mesh-tails}
\input{sections/retained-mass}
\input{sections/ribbon}
\input{sections/dense-cover}
\phantomsection
\addcontentsline{toc}{section}{References}
\bibliographystyle{plain}
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\RaggedRight
\bibliography{refs/sources}
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\end{document}
==== END An-explicit-exact-Hausdorff-gauge-for-SLE//main.tex ====
==== BEGIN An-explicit-exact-Hausdorff-gauge-for-SLE//mesh-tails.tex ====
\section{A uniform tail for spatially integrated stopped densities}
\label{mesh:section}
We prove the estimate that determines the logarithmic correction in the
gauge. The essential feature is uniformity in the final mesh: a large
threshold is tested first at an intermediate mesh, and the remaining
increments must be controlled down to any prescribed finer mesh.
Throughout this section, $0<\kappa<8$ is fixed,
\[
p=1-\kappa/8,\qquad d=2-p,\qquad a_0=8/\kappa-1.
\]
We use the capacity normalization
$\partial_tg_t=2/(g_t-\sqrt\kappa B_t)$. The initial domains considered
here are surviving domains of an ordinary half-plane trace at finite
stopping times, and their Euclidean similarities. Conditional statements below refer
to the ordinary chordal law in that domain, with its marked tip and
target. We write $D_t$ for its subsequent surviving domain.
\paragraph{Inputs used here.}
Lemmas~\ref{ll:finite-tilt} and~\ref{ll:approach} supply the nonnegative density
\[
M_t(z)=R_{D_t}(z)^{-p}\sin^{a_0}\theta_t(z),
\]
continued by zero on killing. Here $R$ is the full conformal radius and
$\theta$ is the argument of the image point in tip--target half-plane
coordinates. We use the following precise parts of that law.
First, on a region where the distance from $z$ to the domain boundary is
bounded below, $M_t(z)$ is a bounded stopped martingale. If a smaller
distance is never reached, its terminal contribution is zero. Second,
at a live stopping time, the probability of a subsequent approach to
distance $u$, for $u$ smaller than a fixed multiple of the current
distance to the boundary, is at most $C u^pM_t(z)$. The expected density
at that approach is $M_t(z)$. All these assertions use only finite
stopped changes of law and monotone exhaustion. The coefficient of
$dB_t$ in $M_t(z)$ has absolute value at most
\begin{equation}
C R_{D_t}(z)^{-p}\sin^{a_0}\theta_t(z)/|Z_t(z)|,
\qquad Z_t(z)=G_t(z)-\sqrt\kappa B_t.
\label{mesh:coefficient-input}
\end{equation}
Here $G_t$ includes the initial uniformization: if the initial domain
is $J(\mathbb H)$, then $G_t=g_t\circ J^{-1}$, with $g_t$ in the
half-plane Loewner coordinates of Section~\ref{ll:section}.
When $\kappa>4$, Proposition~\ref{ll:disk-rule} supplies one more
estimate. Suppose $\overline B(z,r)$ is contained in the initial
domain. At its first circle hit, if $z$ is still live, the conditional
probability that the next exit of $B(z,Lr)$ occurs with $z$ still live
is at most $CL^{-c}$ for all sufficiently large $L$. The outer disk need
not be free. The domain class is exactly the one stated above.
After a translation, fix $A=B(0,1)$ with a fixed enlargement contained in
the initial domain. All the points in $A$ consequently have a uniform
initial boundary clearance, which we denote by $\rho_{\rm col}>0$.
For $0<r<\rho_{\rm col}$, let
$\tau_r(z)$ be the first live approach of the trace to distance $r$
from $z$, with value infinity if there is none, and set
\[
W_r(z)=M_{\tau_r(z)}(z)\mathbf1_{\{\tau_r(z)<\infty\}},
\qquad I_r(z)=\mathbf1_{\{\tau_r(z)<\infty\}}.
\]
The maps $(z,\omega)\mapsto\tau_r(z),W_r(z),I_r(z)$ have jointly
measurable versions: before killing this follows from continuous trace
distance and the continuous Loewner maps. The weights $W_r$ and
indicators $I_r$ are zero when no live approach occurs; the stopping
time then has value $\infty$. One may first localize the maps on
compact subsets of the surviving domain and then exhaust.
On a finite approach, \eqref{ll:trace-clearance} gives
\[
\operatorname{dist}(z,\partial D_{\tau_r(z)})=r.
\]
Thus at $\tau_r(z)$ the restarted estimates of Lemma~\ref{ll:approach}
apply to the finer distance $qr$ whenever $0<q\le1/(8\mathrm e)$.
\begin{proposition}[Integrated mesh tail]\label{mesh:tail}
There are constants $C,c>0$ and $0<\delta_0<\rho_{\rm col}$, depending only on $\kappa$ and
the fixed collar, such that, for $0<r_f\le\delta\le\delta_0$,
\begin{equation}
\mathbb P\left\{\int_A W_{r_f}(z)\,dA(z)>C\delta^{-p}\right\}
\le C\exp(-c\delta^{-2}).
\label{mesh:tail-equation}
\end{equation}
The constants are uniform over the stated initial marked domains, including
conditioning on any past for which the collar is free. They do not
depend on the final distance $r_f$ or on an upper bound for the
distance to the tip. The same estimate holds if only approaches by a
fixed or stopping time are counted.
\end{proposition}
The last assertion follows pointwise by deleting contributions. We
will also use the following form, valid for every prescribed final
distance $0<e<\rho_{\rm col}$ with constants independent of $e$:
\begin{equation}
\mathbb P\left\{\int_AW_e\,dA>u\right\}
\le C\exp(-cu^{2/p}),\qquad u\ge u_0.
\label{mesh:normalized-tail}
\end{equation}
Here is the exact passage from \eqref{mesh:tail-equation} to this
form. Write $C_{\rm t}$ for the threshold constant in that equation.
The pointwise bound $W_e\le e^{-p}$ at a first live approach gives
$\int_AW_e\,dA\le C_{\rm d}e^{-p}$. Choose
$C_0\ge\max(C_{\rm t},C_{\rm d})$, and for
$u\ge u_0:=C_0\delta_0^{-p}$ set
$\delta=(C_0/u)^{1/p}$. If $e\le\delta$, then
\eqref{mesh:tail-equation} applies and its threshold is at most $u$.
If $e>\delta$, the deterministic bound is strictly less than
$C_{\rm d}\delta^{-p}\le u$, so the event is empty. These two cases
prove \eqref{mesh:normalized-tail}. Scaling space by $R$ multiplies
the density by $R^{-p}$ and area by $R^2$. Thus on a disk of radius
$R$, the same bound has threshold $uR^d$ and permits final distances
$0<e<R\rho_{\rm col}$.
\subsection{Brownian occupation of a boundary layer}
The bracket estimate will use a geometric fact about the Dirichlet
Green function. The Green occupation identity is classical; we use
the killed Brownian version described in
\cite[\S3.3, Definitions~3.28 and~3.31 and Theorem~3.32, pp.~78--80]{mortersperes2010}. The uniform boundary-layer
estimate below requires an additional argument and no smoothness of
the boundary.
\begin{lemma}[Boundary-layer occupation]\label{mesh:layer}
Let $D$ be a simply connected plane domain with connected unbounded
complement, and let $K_D$ be its Dirichlet Green function in the
normalization
\[
K_{\mathbb H}(z,w)=\log\left|\frac{z-\overline w}{z-w}\right|.
\]
For every $r>0$ and every $z\in D$,
\begin{equation}
\int_{\{w\in D:\operatorname{dist}(w,\partial D)\le r\}}
K_D(z,w)\,dA(w)\le Cr^2.
\label{mesh:layer-bound}
\end{equation}
\end{lemma}
\begin{proof}
Figure~\ref{fig:brownian-circuit} illustrates the forced circuit and
the occupation blocks used below. We first record a uniform exit probability. If $x\in D$ is within
distance $r$ of $D^c$, planar Brownian motion starting at $x$ has
probability at least $\eta>0$ of exiting $D$ before time $Ar^2$, for
absolute constants $A,\eta$. To see this, scale and translate so that
$x=0,r=1$. Brownian motion has a fixed positive probability of making,
within a fixed time, a circuit in a fixed annulus that separates
$\overline B(0,2)$ from infinity. This is a finite open crossing event:
use overlapping rectangles along a circular route and two transverse
crossings of the final rectangle to force intersection with the
initial portion, and hence a separating closed subcurve. Its
probability is positive by Brownian scaling, the Markov property and
positive transition probabilities through the prescribed rectangles.
If the path has not exited earlier, that circuit must meet $D^c$:
$D^c$ is connected, meets $\overline B(0,1)$, and is unbounded. This
proves the asserted exit probability.
Let $\zeta$ be Brownian exit time from $D$, and let $E_r$ be the layer
in the integral. Start a time block of length $Ar^2$ at the first
entrance into $E_r$. After a block ends, start the next block at the
next entrance into $E_r$, unless exit has already occurred. Entrance
times are stopping times; one may equally first use the open layer of
width $2r$ to avoid any convention at its outer boundary. Conditional
on every block start before $\zeta$, the probability of exit during
the block is at least $\eta$. Thus the expected number of blocks is
at most $1/\eta$. All time spent in $E_r$ before exit is included in
these blocks, since the gaps contain no visit to $E_r$. Consequently
\[
\mathbb E_z\int_0^\zeta\mathbf1_{E_r}(B_t)\,dt
\le Ar^2/\eta.
\]
The Green occupation formula gives \eqref{mesh:layer-bound}, up to the
absolute constant determined by the Brownian generator. This formula
can first be applied in bounded exhaustions of $D$ and then passed to
the limit by monotone convergence. The occupation estimate itself
shows that the layer integral remains finite.
\end{proof}
\begin{figure}[tbp]
\centering
\input{figures/brownian-circuit}
\caption{\BrownianCircuitCaption}
\label{fig:brownian-circuit}
\end{figure}
\clearpage
\subsection{Asynchronously started brackets for
\texorpdfstring{$\kappa\le4$}{kappa at most four}}
Schramm--Sheffield use Dirichlet Green decrease in their SLE$_4$
and Gaussian-free-field covariance calculation
\cite[Lemmas~2.5--2.6]{schrammsheffield2013}.
Here each point starts and stops at its own first approach times, so
the spatial bracket estimate needs the additional later-start
decomposition below.
Fix $0<q\le1/(8\mathrm e)$. For the moment consider one
slab from first distance $r$ to first distance $qr$. A point is
\emph{active} after $\tau_r(z)$ until $\tau_{qr}(z)$ or killing. The
one-point martingale is bounded on this interval by $C r^{-p}$, where
$C$ may depend on the fixed ratio $q$. Because $a_0\ge1$, its Brownian
coefficient $b_t(z)$, set to zero outside this interval, satisfies
\begin{equation}
|b_t(z)|\le C r^{-p}\frac{Y_t(z)}{|Z_t(z)|^2}.
\label{mesh:active-coefficient}
\end{equation}
Indeed the initial collar and the fact that the smaller trace
distance has not yet been reached give $R_{D_t}(z)\ge qr$, whereas
$\sin^{a_0}\theta\le\sin\theta$.
For two live distinct points put $K_t(z,w)=K_{D_t}(z,w)$. The common
Brownian translation cancels in the differences appearing in its
half-plane formula. Direct differentiation of the Loewner equation
therefore gives
\begin{equation}
-\partial_tK_t(z,w)
=4\frac{Y_t(z)}{|Z_t(z)|^2}\frac{Y_t(w)}{|Z_t(w)|^2}.
\label{mesh:green-decrease}
\end{equation}
Consider pairs whose active intervals overlap. Assign such a pair to
its later starting point, say $z$, so that
$\tau_r(w)\le\tau_r(z)$. The contribution of the overlapping
interval to the integral of $|b_t(z)b_t(w)|$ is at most
\begin{equation}
Cr^{-2p}K_{D_{\tau_r(z)}}(z,w).
\label{mesh:pair-bound}
\end{equation}
Here we used positivity of the Green function at the end of the
overlap. Infinite terminal intervals follow by monotone convergence.
At time $\tau_r(z)$ the earlier point $w$ is still in the domain and
lies within distance $r$ of its boundary. This is because it was
within that distance at its own start and the domains decrease.
For each fixed $z$, its starting time does not depend on the dummy
integration variable $w$. Lemma~\ref{mesh:layer} thus applies in the
single domain $D_{\tau_r(z)}$ and gives
\[
\int_{\{w:\text{$w$ is active at $\tau_r(z)$}\}}
K_{D_{\tau_r(z)}}(z,w)\,dA(w)\le Cr^2.
\]
Integrating in $z\in A$ and allowing both orders of the pair yields
the pathwise bound
\begin{equation}
\int_0^\infty\left|\int_A b_t(z)\,dA(z)\right|^2dt
\le\int_{A^2}\int_0^\infty |b_t(z)b_t(w)|\,dt\,dA(z)dA(w)
\le Cr^{2-2p}.
\label{mesh:bracket-bound}
\end{equation}
The spatial diagonal has area measure zero. Ties of the starting
times may be assigned either order or counted twice; the upper bound
is unchanged. Points with no start contribute zero.
We spell out the stochastic integration step. For $s\in\{r,qr\}$,
let
\[
M_t^{(s)}(z)=M_{t\wedge\tau_s(z)}(z),
\qquad N_t(z)=M_t^{(qr)}(z)-M_t^{(r)}(z),
\]
using the zero continuation of $M$ at killing. The collar and the
distance cutoff give $0\le M_t^{(s)}\le Cs^{-p}$.
If the distance approach never occurs, the live conformal radius
stays above $s$. The radius terminal-zero assertion of
Lemma~\ref{ll:approach} then makes the zero continuation continuous
at killing and gives $M_\infty^{(s)}=0$. At a finite approach the
process is frozen at $W_s$, so in all cases $M_\infty^{(s)}=W_s$.
Localization and bounded convergence therefore make each
stopped process a bounded continuous martingale. Since
$\tau_{qr}\ge\tau_r$, the difference $N_t(z)$ is zero before its
start and has coefficient $b_t(z)$ precisely while the point is
active. This formula also covers a missing start and an infinite
terminal interval.
Choose a version of $b_t(z)$ jointly measurable for the product of
the Borel sigma field on $A$ and the predictable sigma field.
Compact localization of the continuous adapted Loewner coefficients
and the indicators of the stochastic intervals supplies this version. The
uniform bound $|N_t(z)|\le Cr^{-p}$ bounds its expected quadratic
variation by $Cr^{-2p}$. On the finite area set $A$, this gives
\[
\int_A\left(\mathbb E\int_0^\infty |b_t(z)|^2dt\right)^{1/2}
dA(z)<\infty.
\]
The $L^2$ stochastic Fubini theorem, or simple spatial approximation
and the It\^o isometry, now identifies
$N_t=\int_A N_t(z)dA(z)$ with the Brownian integral of
$\int_A b_t(z)dA(z)$. It converges in $L^2$ at infinity. The same
uniform bound permits the pointwise terminal identity to pass through
the spatial integral, giving
\[
N_\infty=\int_A(W_{qr}-W_r)\,dA.
\]
All uses of Fubini in \eqref{mesh:bracket-bound} concern nonnegative
functions and are valid pathwise. Thus no interchange of
uncontrolled signed infinite-time integrals is needed.
The exponential martingale inequality and
\eqref{mesh:bracket-bound} imply
\begin{equation}
\mathbb P\left\{\int_A(W_{qr}-W_r)\,dA>x\right\}
\le\exp\left(-\frac{c x^2}{r^{2-2p}}\right).
\label{mesh:slab-tail}
\end{equation}
This estimate, notably, permits different starting times for all
points in the spatial integral.
\subsection{Summing the slabs for
\texorpdfstring{$\kappa\le4$}{kappa at most four}}
Given $r_f\le\delta$, choose $N\ge0$ and
$r_0\in(q\delta,\delta]$ so that $r_j=r_0q^j$ and $r_N=r_f$.
This harmless adjustment ensures that even the final slab has ratio
exactly $q$. The initial bound
$S_0:=\int_AW_{r_0}\,dA\le C\delta^{-p}$ is deterministic.
Apply \eqref{mesh:slab-tail} with
$x_j=\delta^{-p}/(j+1)^2$. The exceptional probability is at most
\[
\exp\left(-c\delta^{-2}
\frac{q^{-2(1-p)j}}{(j+1)^4}\right).
\]
The sum of these probabilities over all $j\ge0$ is bounded by
$C\exp(-c'\delta^{-2})$: an exponential sequence divided by a fixed
power has a positive minimum and then grows geometrically. Outside
this union of exceptions,
\[
S_N\le S_0+\sum_{j=0}^{N-1}x_j\le C'\delta^{-p}.
\]
This proves Proposition~\ref{mesh:tail} for $\kappa\le4$.
\subsection{An integrated concentration estimate for disjoint windows}
The following elementary concentration argument allows the order of the
tests to depend on the curve. In particular, it does not condition on a
permutation of the test locations.
\Needspace{26\baselineskip}
\begin{lemma}[Disjoint-window concentration]\label{win:concentration}
Let $D\subset B(0,1)$ be Borel, let $0<r\le R$ and $K>0$ with $3R\le1$, and let
$\gamma$ be a continuous adapted curve. For each $z\in D$, suppose that
a test starts at a stopping time $\sigma_z$, with $\sigma_z=\infty$ if it
never starts. At a finite start,
\[
|\gamma(\sigma_z)-z|=r.
\]
Suppose its ending time $\tau_z\ge\sigma_z$ is a stopping time and
$\gamma(t)\in B(z,R)$ for $\sigma_z\le t<\tau_z$.
Let $Y_z$ be an $\mathcal F_{\tau_z}$-measurable output satisfying
$0\le Y_z\le K$, and let $\mu_z\ge0$ be $\mathcal F_{\sigma_z}$-measurable,
with
\[
\mathbb E[Y_z\mid\mathcal F_{\sigma_z}]\le\mu_z
\quad\hbox{on }\{\sigma_z<\infty\}.
\]
Assume the outputs and upper means are jointly measurable in $(z,\omega)$,
are zero when there is no start, and are uniformly bounded. If a started
window never ends, assume its output is zero. Then
\begin{equation}
\mathbb E\exp\left(\lambda\int_D(Y_z-\mu_z)\,dA(z)\right)
\le \exp(CK^2\lambda^2 R^2),\qquad \lambda\ge0,
\label{win:mgf}
\end{equation}
and hence, for $b>0$,
\begin{equation}
\mathbb P\left\{\int_D(Y_z-\mu_z)\,dA(z)>b\right\}
\le \exp\left(-\frac{c b^2}{K^2R^2}\right).
\label{win:tail}
\end{equation}
Here $C,c>0$ are absolute constants. The same conclusion holds
conditionally on an initial sigma field if all the hypotheses and
conditional-mean estimates hold after that conditioning.
\end{lemma}
\begin{proof}
Set $a=3R$. Fix a shift $\xi\in Q=[0,a)^2$, and consider the finitely
many centers of $(\xi+a\mathbb Z^2)\cap D$. Their number is at most
$N\le C a^{-2}$. Use the convention
$\mathcal F_\infty=\sigma(\bigcup_{t<\infty}\mathcal F_t)$.
While a window at $z$ is running, a window at any
other lattice center $w$ cannot start: otherwise
\[
|z-w|\le |z-\gamma(\sigma_w)|+|\gamma(\sigma_w)-w|
<R+r\le2R<3R.
\]
Order the finite starts by always waiting for the next start among the
remaining centers. The minimum of the remaining stopping times is a
stopping time, and its index is measurable at that time. Distinct finite
starts cannot tie, by the same geometric argument. The preceding
window has ended by the next start, so its output is already known.
This is an adaptive enumeration, not conditioning on an eventual order.
For its $k$th actual test write $Z_k=Y_k-\mu_k$. Conditional Hoeffding's
lemma gives, for $\theta\ge0$,
\[
\mathbb E[e^{\theta Z_k}\mid\mathcal F_{\sigma_k}]
\le \exp(K^2\theta^2/8).
\]
Indeed $Y_k$ has conditional range of length at most $K$, and
$\mathbb E[Y_k\mid\mathcal F_{\sigma_k}]-\mu_k\le0$.
For the selected test this inequality follows by partitioning over
the finitely many possible center indices, each known at the start.
The filtration at the end of test $k-1$ is contained in that at the
start of test $k$. Iterating conditional expectations therefore yields
\[
\mathbb E\exp\left(\theta\sum_{k=1}^{N}Z_k\right)
\le \exp(NK^2\theta^2/8).
\]
Here missing tests are padded with zero increments. This also covers
infinite waiting. If a started window lasts forever, its confinement
prevents every later lattice start, its own increment is $-\mu_k\le0$,
and all later increments are zero. Equivalently, one can use terminal
sigma fields at the infinite stopping times in the same argument.
Extend $f(z)=Y_z-\mu_z$ by zero outside $D$, and set
\[
T_\xi=a^2\sum_{z\in\xi+a\mathbb Z^2}f(z).
\]
Taking $\theta=\lambda a^2$ gives
$\mathbb E e^{\lambda T_\xi}\le\exp(CK^2\lambda^2a^2)$.
The exact averaging identity is
\[
\int_D f(z)\,dA(z)=\frac1{a^2}\int_Q T_\xi\,d\xi.
\]
Jensen's inequality, followed by Tonelli's theorem, proves
\eqref{win:mgf}. There is no probabilistic independence assumption
on the shift. Optimizing the exponential Markov bound in $\lambda$
proves \eqref{win:tail}.
\end{proof}
\subsection{The mesh tail for
\texorpdfstring{$4<\kappa<8$}{kappa between four and eight}}
\begin{proof}[Proof of Proposition~\ref{mesh:tail} for $4<\kappa<8$]
Use the first approach times $\tau_r(z)$, the densities $W_r(z)$,
and the indicators $I_r(z)$ already defined. The pointwise bound is
$W_r\le Cr^{-p}I_r$. Lemma~\ref{ll:approach} gives, at a finite
$\tau_r(z)$ and for one fixed $0<q\le1/(8\mathrm e)$,
\begin{equation}
\mathbb P\{I_{qr}(z)=1\mid\mathcal F_{\tau_r(z)}\}
\le C(qr)^pW_r(z).
\label{win:onepoint-input}
\end{equation}
The conditional expected density at that finer approach is $W_r(z)$;
requiring it to occur before an additional stop can only decrease this
expectation. Proposition~\ref{ll:disk-rule} gives constants
$C_*,c_*,L_0>0$ for which a live exit from $B(z,Lr)$ has conditional
probability at most $C_*L^{-c_*}$ for $L\ge L_0$.
These are exactly the mean bounds needed for the three window outputs
below. Their constants are uniform over the permitted initial pasts.
Choose $r_0\in[q\delta,\delta]$ and an integer $n\ge0$ such that
$r_f=r_0q^n$, and set $r_j=r_0q^j$. Thus every consecutive ratio is
exactly $q$, including the last; the starting mesh changes only by a
fixed factor. Fix $0<\varepsilon<1-p$. For the moment let
$L_*\ge L_0$ be arbitrary and set
\[
L_j=L_*q^{-\varepsilon j}.
\]
During the estimates below suppose that $\delta<\rho_{\rm col}$ and
$3L_*\delta\le1$. After deriving the recurrences we
will choose $L_*$ and then $\delta_0$ to enforce these requirements.
At every level,
\[
L_jr_j=L_*r_0q^{(1-\varepsilon)j}\le L_*\delta\le\tfrac13,
\]
as required in Lemma~\ref{win:concentration}. Also each
$\overline B(z,r_j)$ lies in the original collared initial domain, and a finite
$\tau_{r_j}(z)$ is its first circle hit with $z$ live. These are the
hypotheses on the inner disk in Proposition~\ref{ll:disk-rule} at every level.
At a finite start $\tau_{r_j}(z)$, stop the window at the first of
the next live $r_{j+1}$-approach, exit from $B(z,L_jr_j)$, and death
of $z$. Let $H_j(z)$ indicate a next approach achieved during this
window, let
\[
V_j(z)=r_j^pW_{r_{j+1}}(z)H_j(z),
\]
and let $B_j(z)$ indicate an outer exit before that next approach,
with $z$ still live. Set all three outputs to zero if no window
starts. A window which never ends has all three outputs zero.
The outputs are bounded by a constant depending only on the fixed
parameters, since $V_j\le Cq^{-p}$.
First exits of the continuous process $t\mapsto|\gamma(t)-z|$ and death times
from the live Loewner maps are jointly measurable after compact
localization. Thus these outputs and the upper means below have the
jointly measurable versions required by Lemma~\ref{win:concentration}.
The three conditional upper means at the start, also set to zero for
no start, are respectively
\begin{equation}
\begin{split}
\mu^H_j(z)&=C(qr_j)^pW_{r_j}(z),\\
\mu^V_j(z)&=r_j^pW_{r_j}(z),\\
\mu^B_j(z)&=C_*L_j^{-c_*}I_{r_j}(z).
\end{split}
\label{win:means}
\end{equation}
The pointwise bound on $W_{r_j}$ and $L_j\ge L_0$ make these upper
means uniformly bounded as well.
The first two follow from Lemma~\ref{ll:approach}. The bad event is a
subset of the local-rule event, so the last bound follows from Proposition~\ref{ll:disk-rule}. Lattice windows have disjoint time intervals by the
tip-containment argument in Lemma~\ref{win:concentration}, including
when a window ends early by killing.
For each of the three outputs apply that lemma at level
\[
b_j=\frac{(r_j/\delta)^p}{(j+1)^2}\le1.
\]
Except on an event of probability at most
\begin{equation}
C\exp\left(-\frac{c b_j^2}{(L_jr_j)^2}\right),
\label{win:level-failure}
\end{equation}
the corresponding integral is at most the integral of its upper mean
plus $b_j$. The exponent satisfies
\begin{equation}
\frac{b_j^2}{(L_jr_j)^2}
=\frac{\delta^{-2}}{L_*^2}
\left(\frac{r_0}{\delta}\right)^{-2+2p}
\frac{q^{-2(1-p-\varepsilon)j}}{(j+1)^4}
\ge \frac{\delta^{-2}}{L_*^2}
\frac{q^{-2(1-p-\varepsilon)j}}{(j+1)^4}.
\label{win:failure-exponent}
\end{equation}
Because $q^{-2(1-p-\varepsilon)}>1$, the sum of
\eqref{win:level-failure} over all $j\ge0$ and the three outputs is
at most $C\exp(-c\delta^{-2}/L_*^2)$. For example, the last quotient in
\eqref{win:failure-exponent} is bounded below by a positive constant
times $j+1$, which reduces the sum to a geometric series. This bound
is independent of $n$ and hence of $r_f$.
It remains to propagate the simultaneous integral bounds. A next
approach missed by its window must occur after an outer exit at which
the center was still live. Death or an infinite window cannot be
followed by a missed live approach. Consequently, pointwise,
\begin{equation}
\begin{split}
I_{r_{j+1}}&\le H_j+B_j,\\
W_{r_{j+1}}&\le r_j^{-p}V_j+C(qr_j)^{-p}B_j.
\end{split}
\label{win:missed-output}
\end{equation}
Define
\[
S_j=\int_A W_{r_j}\,dA,\qquad
A_j=r_j^{-p}\int_A I_{r_j}\,dA.
\]
On the simultaneous good event, integrating
\eqref{win:missed-output} and using \eqref{win:means} gives
\begin{equation}
\begin{split}
S_{j+1}&\le S_j+\alpha_j A_j+e_j,\\
A_{j+1}&\le C_1 S_j+\alpha_j A_j+e_j,
\end{split}
\label{win:recurrences}
\end{equation}
where constants can be enlarged so that
\[
\alpha_j=C_2L_*^{-c_*}q^{\varepsilon c_*j},\qquad
e_j=\frac{C_3\delta^{-p}}{(j+1)^2}.
\]
Here $r_j^{-p}b_j=\delta^{-p}/(j+1)^2$ exactly; factors of $q^{-p}$
are fixed constants. The constants $C_1,C_2,C_3$ depend only on
$q$ and the constants in the local laws, not on $L_*$ or the final mesh.
The pointwise initial bounds and
$r_0\ge q\delta$ give $S_0,A_0\le C\delta^{-p}$.
For completeness, the two recurrences have a bound uniform in their
length. Choose $\Lambda\ge\max(1,2C_1)$ and then $L_*$ so large that
$\alpha_0=C_2L_*^{-c_*}\le1/2$. Finally choose
$0<\delta_0<\rho_{\rm col}$ with $3L_*\delta_0\le1$. The preceding
estimates now apply whenever $r_f\le\delta\le\delta_0$.
Put $U_j=\max(S_j,A_j/\Lambda)$. The second recurrence
implies
\[
A_{j+1}/\Lambda
\le(C_1/\Lambda+\alpha_j)U_j+e_j/\Lambda\le U_j+e_j,
\]
while the first implies
$S_{j+1}\le(1+\Lambda\alpha_j)U_j+e_j$. Thus
\[
U_{j+1}\le(1+\Lambda\alpha_j)U_j+e_j,
\]
and iteration gives
\[
U_j\le\left(U_0+\sum_{i\ge0}e_i\right)
\exp\left(\Lambda\sum_{i\ge0}\alpha_i\right)
\le C\delta^{-p}.
\]
Both sums converge and their constants are independent of the final
mesh. With $L_*$ fixed, the summed failure probability is
$C\exp(-c\delta^{-2})$. Thus $S_n\le C\delta^{-p}$ outside that event,
proving \eqref{mesh:tail-equation} for $4<\kappa<8$ and completing the proposition.
\end{proof}
==== END An-explicit-exact-Hausdorff-gauge-for-SLE//mesh-tails.tex ====
==== BEGIN An-explicit-exact-Hausdorff-gauge-for-SLE//retained-mass.tex ====
\section{Retained mass and a lower bound for fine sausages}
\label{retain:section}
We now turn the uniform mesh tail into a positive measure with the
required small-ball bound. A subsequential weak limit of the retained densities will give the
lower Hausdorff bound. A separate prelimit estimate at the end of the
section supplies the sausage mass used by the ribbon construction.
Use the parameters and density $W_e$ of Section~\ref{mesh:section}.
The gauge $h$ is the one in \eqref{eq:gauge}. For each fixed $L\ge1$,
$h(Lr)/h(r)\to L^d$ as $r\downarrow0$, so these values differ by
bounded factors at small scales. This permits conformal changes of
coordinates on compact interior sets.
\subsection{A positive first moment before a finite time}
Fix a closed disk $K\Subset\mathbb H$ with a fixed free collar. Define
\[
\nu_e^T(dz)=\mathbf1_K(z)W_e(z)
\mathbf1_{\{\tau_e(z)\le T\}}\,dA(z).
\]
Let $\delta_K=\operatorname{dist}(K,\partial\mathbb H)>0$.
Use the enlargement of $K$ by $\delta_K/2$ in the scaled tail
\eqref{mesh:normalized-tail}, so its permitted final distances include
$0<e<\delta_K/2$. Choose $0<e_0\le\delta_K/(8\mathrm e)$, which
lies in this range and in the range of Lemma~\ref{ll:approach} for
every $z\in K$.
We first show that a deterministic finite $T$ can be chosen so that
\begin{equation}
\inf_{0<e<e_0}\mathbb E\nu_e^T(K)\ge m_0>0,
\qquad
\sup_{0<e<e_0}\mathbb E\bigl[\nu_e^T(K)^2\bigr]\le m_2<\infty.
\label{retain:first-second}
\end{equation}
The second bound follows by integrating the uniform upper tail
\eqref{mesh:normalized-tail}, after scaling to $K$. We justify the first bound,
including the finite physical time needed for a tilted approach.
For an integer $m\ge1$, let $\rho_m(z)$ be the first time the live
point has conformal radius $R_0(z)e^{-m}$, and write $\mathbb P_z^{*,m}$ for
the finite centered law through $\rho_m(z)$ from
Lemma~\ref{ll:finite-tilt}. If $R_0(z)e^{-m}\le e$, the distance stop
$\tau_e(z)$ occurs by $\rho_m(z)$ under this law: Koebe's inequality
gives distance to the boundary at most the conformal radius, and the
initial boundary is farther than $e$. The finite stopped density
identity therefore gives
\[
\mathbb E\!\left[W_e(z)\mathbf1_{\{\tau_e(z)\le T\}}\right]
=M_0(z)\mathbb P_z^{*,m}\{\tau_e(z)\le T\}
\ge M_0(z)\mathbb P_z^{*,m}\{\rho_m(z)\le T\}.
\]
By consistency, the probabilities on the right decrease with $m$.
Define their limit by
\[
q_T(z)=\lim_{m\to\infty}
\mathbb P_z^{*,m}\{\rho_m(z)\le T\}.
\]
By continuity from above on the consistent centered path, this is
the probability that $T^*(z)=\lim_m\rho_m(z)$ is at most $T$.
Lemma~\ref{ll:finite-clock} therefore gives $q_T(z)\uparrow1$ as
$T\to\infty$. Here the initial uniformization is the identity, so
the lemma's Loewner time is the original capacity time.
Each probability under a finite stopped law is a
measurable kernel in $z$ by the localized construction, so the same
is true of $q_T$. Taking the decreasing
limit above and integrating over $K$ yields
\[
\mathbb E\nu_e^T(K)\ge\int_K M_0(z)q_T(z)\,dA(z).
\]
The right side increases to $\int_KM_0(z)dA(z)>0$ as $T\to\infty$.
This proves \eqref{retain:first-second} for a sufficiently large
deterministic $T$, using density identities only at finite stops.
Brownian scaling places the same experiment in every prescribed
initial capacity interval of length $\Delta>0$. Scaling space by
$c=\sqrt{\Delta/T}$ scales capacity time by $c^2$, the permitted mesh
cutoff by $c$, and the integrated density by $c^d$. Thus the first
moment lower bound becomes $c^dm_0$ and the second moment upper bound
becomes $c^{2d}m_2$, leaving the ratio $m_0^2/m_2$ unchanged.
\subsection{Removing squares with excessive mass}
Take the deterministic sequence $e_N=2^{-N}$ and write
$\nu_N=\nu_{e_N}^T$. Dyadic squares are taken half open and form a
partition at each level. Let $k_0$ be a starting level to be fixed below. For a
square $Q$ of side $l=2^{-k}$, $k_0\le k\le N$, declare it excessive
when
\[
\nu_N(Q)>H_k,
\qquad H_k=D l^d(\log k)^{p/2},
\]
where $D$ will be chosen large. We claim that
\begin{equation}
\mathbb E\left[\nu_N(Q)\mathbf1_{\{\nu_N(Q)>H_k\}}\right]
\le C l^{p+d}k^{-\beta},
\qquad \beta=cD^{2/p},
\label{retain:bad-square}
\end{equation}
after reducing the positive constant $c$ if necessary. Constants
are uniform in $N\ge k$ and in the squares meeting $K$.
Let $z_Q$ be the center of $Q$, let $\sigma_Q$ be the first hit of
$\partial B(z_Q,10l)$, and set
\[
E_Q=\{\sigma_Q<\infty,\ z_Q\in D_{\sigma_Q}\}.
\]
Require $k_0$ to be large enough that $2^{-k_0}<e_0$ and
$10l\le\operatorname{dist}(z_Q,\partial\mathbb H)/(8\mathrm e)$
for every square under consideration. In particular every
$\overline B(z_Q,10l)$ under consideration is initially interior.
We first check that only $E_Q$ can contribute to $\nu_N(Q)$.
Since $Q\subset B(z_Q,l)$ and $e_N\le l$, such a contribution requires
the trace to enter $B(z_Q,2l)$ after first hitting the larger circle.
If $z_Q$ is swallowed before or at that first hit, the open larger
disk is still disjoint from the trace at the swallowing time. The
connected-component observation after Lemma~\ref{geom:first-hit-access}
then shows that its whole interior is swallowed, so no later trace
can enter it.
Thus $\nu_N(Q)>0$ implies $E_Q$.
On $E_Q$, the hit is a first live approach to $z_Q$ at distance $10l$.
Lemma~\ref{ll:approach} therefore gives
$\mathbb P(E_Q)\le Cl^p$, uniformly because $M_0(z_Q)$ is bounded on
a fixed compact neighborhood of $K$. Also the open disk
$B(z_Q,10l)$ is contained in $D_{\sigma_Q}$: it is trace-free,
connected, and contains the live center. In particular
$A_Q=B(z_Q,2l)$ contains $Q$ and its closed enlargement
$\overline B(z_Q,6l)$ remains free.
Conditionally on $E_Q$ and the stopped past, the future density in
physical coordinates is the same $M_t$ as before the restart: its
conformal radius and tip--target angle depend only on the same marked
physical domain. There were no earlier $e_N$-approaches in
$Q$. The normalized tail \eqref{mesh:normalized-tail}, applied in
$A_Q$ at scale $2l$, with the time cutoff deleting contributions,
therefore gives
\[
\mathbb P\{\nu_N(Q)>u l^d\mid\mathcal F_{\sigma_Q}\}
\le C\exp(-cu^{2/p})
\]
on $E_Q$, for $u\ge u_0$, uniformly in the final mesh $e_N\le l$;
here $u_0$ is enlarged by a fixed factor to account for the radius $2l$.
The case settled by the deterministic bound in the derivation of
\eqref{mesh:normalized-tail} includes meshes too coarse for its
intermediate test. Provided $D(\log k_0)^{p/2}\ge u_0$, integrate
this tail above
$u=D(\log k)^{p/2}$, including the usual threshold-times-tail term.
Any polynomial prefactor is absorbed by reducing $c$. Multiplication
by the entrance probability $Cl^p$ proves
\eqref{retain:bad-square}.
Let $E_N$ be the union of all excessive squares at levels
$k_0,\ldots,N$, and retain the measure
\[
\widehat\nu_N=\mathbf1_{E_N^c}\nu_N.
\]
There are at most $Cl^{-2}$ squares meeting $K$ at level $k$.
Since $p+d=2$, summation of \eqref{retain:bad-square} gives
\[
\mathbb E\bigl[\nu_N(K)-\widehat\nu_N(K)\bigr]
\le C\sum_{k\ge k_0}k^{-\beta}.
\]
Choose $D$ so that $\beta>1$, then choose $k_0$ large enough to meet
the geometric and threshold requirements above and make this sum at most $m_0/2$.
It follows from \eqref{retain:first-second} that
\[
\mathbb E\widehat\nu_N(K)\ge m_0/2,
\qquad\mathbb E\widehat\nu_N(K)^2\le m_2.
\]
Paley--Zygmund therefore supplies deterministic $a_*,p_*>0$ such
that
\begin{equation}
\mathbb P\{\widehat\nu_N(K)\ge a_*\}\ge p_*
\quad\text{for all sufficiently large }N.
\label{retain:positive-event}
\end{equation}
At every tested level each square has retained mass at most $H_k$:
an excessive square has retained mass zero, and otherwise restriction
can only reduce its mass. In particular the finitely many squares
at level $k_0$ meeting $K$ give a deterministic bound on the total
retained mass, uniform in $N$.
\subsection{A supported weak limit and its ball bound}
The event in \eqref{retain:positive-event} occurs infinitely often
with probability at least $p_*$. Indeed each union over $N\ge n$
has probability at least $p_*$, and these unions decrease to the
limsup event. On that event choose a subsequence on which the mass
is at least $a_*$. Measures on the fixed compact disk $K$ with the
deterministic total-mass bound are weakly sequentially compact.
Thus a further subsequence converges weakly to a measure $\nu$ with
$\nu(K)\ge a_*$. No measurable selection of this subsequence is
needed: the pathwise existence of such a measure will imply a
measurable positive-Hausdorff-measure event.
The support of $\nu$ lies in $K\cap\gamma([0,T])$. Every
$\widehat\nu_N$ is carried by the closed $e_N$-neighborhood of that
compact trace, and a compact set disjoint from the trace has an open
neighborhood disjoint from all these carriers for large $N$.
For completeness, weak convergence causes no difficulty at dyadic
boundaries. Given a sufficiently small $r$, choose a fixed dyadic
side length $l$ with $r\le l<2r$. For all sufficiently large $N$
this level was tested, and every open ball $B(x,r)$ meets only a
bounded number of its half-open squares. Therefore
\[
\widehat\nu_N(B(x,r))\le C D l^d(\log\log(1/l))^{p/2}
\le C' h(r).
\]
Portmanteau for this open ball gives
\begin{equation}
\nu(B(x,r))\le C'h(r).
\label{retain:ball}
\end{equation}
This argument works for every $x,r$ on each path with a weak limit;
it does not assume zero mass on grid boundaries. It also shows that
$\nu$ has no atoms.
Any set of positive diameter $r$ is contained in an open ball of
radius at most $2r$, centered at one of its points. The bounded
scaling property of $h$ and \eqref{retain:ball} consequently imply
$\nu^*(U)\le C h(\operatorname{diam}U)$ for all sufficiently small
sets, where $\nu^*$ is the outer measure associated with $\nu$;
singleton sets have zero mass. Countable subadditivity over an arbitrary
countable cover of the trace proves
\[
\mathcal H^h(K\cap\gamma([0,T]))\ge \nu(K)/C>0.
\]
We have proved a fixed positive probability of positive Hausdorff
measure on a compact interior piece, in every prescribed nonzero
initial time interval after scaling.
\subsection{Restarting in every positive-time interval}\label{retain:positivity}
Fix $0\le s<t<\infty$ and divide $[s,t]$ into $n$ deterministic
capacity intervals. At a left endpoint $t_i$, the hydrodynamic map
$g_{t_i}-\sqrt\kappa B_{t_i}$ turns the future into an ordinary
chordal half-plane evolution driven by
$\sqrt\kappa(B_{t_i+u}-B_{t_i})$. Its capacity time is exactly $u$:
hydrodynamic composition adds the coefficient $2u$. Thus no random
time change is hidden in this subdivision.
Apply the preceding scaled interior experiment in each fresh
half-plane interval. Its success probability is at least $p_*$,
conditionally on the preceding past, with $p_*$ independent of the
interval length. Success provides a compact subset strictly inside
the half-plane. The inverse map is conformal in a neighborhood of
this compact subset. A finite cover by interior disks gives upper
and lower metric distortion bounds locally, and hence preserves
positivity of $\mathcal H^h$ by the bounded scaling property of $h$.
One may restrict to one disk carrying positive measure if needed.
The mapped compact subset belongs to the original trace during that
same capacity interval.
Consequently failure of positivity on the whole $[s,t]$ implies
failure of all $n$ experiments, and has probability at most
$(1-p_*)^n$. Letting $n\to\infty$ shows positivity almost surely for
this fixed interval. The event is measurable because the trace is
a random compact set and Hausdorff measure for a continuous gauge
can be computed using the countable family of finite covers by
finite unions of rational open balls, with arbitrarily small
enlargements. Intersecting over rational $0<s<t$ gives a single
probability-one event on which every real $0<s<t<\infty$ has
positive measure: each such interval contains a nontrivial rational
subinterval. The event is for the one fixed value of $\kappa$.
\subsection{A prelimit lower-tail statement for ordinary sausages}
The first and second moments also give an input for the later ribbon
argument without passing to a limiting measure. For a compact time
interval $J$ write
\[
X_e(J)=e^{-p}\operatorname{Area}
\{z\in\mathbb C:\operatorname{dist}(z,\gamma(J))<e\}.
\]
\begin{proposition}[Small sausage values are unlikely]
\label{retain:sausage}
For every deterministic $J=[s,t]$ with $0\le s<t<\infty$,
\begin{equation}
\lim_{a\downarrow0}\limsup_{e\downarrow0}
\mathbb P\{X_e(J)<a\}=0.
\label{retain:sausage-tail}
\end{equation}
Equivalently, for every $\varepsilon>0$ there are $a,e_0>0$ such
that $\mathbb P\{X_e(J)<a\}<\varepsilon$ for all $0<e<e_0$.
The same conclusion can be required simultaneously for any fixed
finite list of such intervals, with total exceptional probability
as small as desired.
\end{proposition}
\begin{proof}
In the initial half-plane experiment, Paley--Zygmund applied directly
to \eqref{retain:first-second} gives constants $a_{\rm s},p_{\rm s}>0$
such that $\mathbb P\{\nu_{e/2}^T(K)>a_{\rm s}\}\ge p_{\rm s}$
for all sufficiently small $e$. Since
$W_{e/2}(z)\le Ce^{-p}$ and the density is supported within distance
$e/2$ of the trace up to $T$, this implies a positive lower bound on
$e^{-p}$ times its sausage area inside $K$. Use a slightly enlarged
compact disk $K^+\Subset\mathbb H$ to ensure that all witnessing
trace points lie in $K^+$ for small $e$. Scaling gives this
experiment on any prescribed positive-length half-plane slot,
with the same lower bound $p_{\rm s}$ on its success probability.
Divide $J$ into $n$ deterministic slots and let $F_i$ be the inverse
hydrodynamic map at the start of slot $i$. The corresponding test
disks $K_i\Subset K_i^+\Subset\mathbb H$ are deterministic after
scaling by that slot's length. There are almost surely finite upper
and strictly positive lower bounds on $|F_i'|$ throughout $K_i^+$,
for each $i$. Because the list is finite, for every $\eta>0$ we can
choose deterministic $0<b<L<\infty$ such that, except on an event
of probability at most $\eta$, all these derivatives lie between
$b$ and $L$. Enlarge $L$ to include the local Lipschitz constants on
the relevant compact disks. No independence between this event and
the slot experiments is asserted or needed.
On that event take each half-plane mesh to be $ce$, where the
deterministic $c>0$ is small enough that image distances less than
$ce$ become physical distances less than $e$. If its slot
experiment succeeds, conformal area change, with Jacobian at least
$b^2$, gives $X_e(J)\ge a_{n,\eta}$, where $a_{n,\eta}>0$ is
deterministic. This conclusion holds for all sufficiently small
$e$, with a deterministic cutoff depending on the finite list.
The half-plane experiments have conditional success probability at
least $p_{\rm s}$, so their simultaneous failure has probability at most
$(1-p_{\rm s})^n$. Hence
\[
\limsup_{e\downarrow0}\mathbb P\{X_e(J)<a_{n,\eta}\}
\le\eta+(1-p_{\rm s})^n.
\]
First choose $n$ large and then $\eta$ small. Monotonicity in the
threshold $a$ proves \eqref{retain:sausage-tail}. For a fixed finite
list, apply this result to each interval and take the minimum
threshold and cutoff, followed by a union bound.
\end{proof}
Only deterministic intervals are asserted in
Proposition~\ref{retain:sausage}. Its use for a random passage first
selects, by countable exhaustion, a finite list of rational closed
intervals and compact interior neighborhoods with positive margins.
The proposition then removes an arbitrarily small unconditional
exceptional event. On a passage whose selected segment lies inside
one of those neighborhoods, sufficiently fine sausages of the segment
are counted there. No conditional sausage inequality at an arbitrary
random interval is required.
==== END An-explicit-exact-Hausdorff-gauge-for-SLE//retained-mass.tex ====
==== BEGIN An-explicit-exact-Hausdorff-gauge-for-SLE//ribbon-front.tex ====
\begingroup
\definecolor{ribbonblue}{RGB}{31,88,138}
\definecolor{ribbonorange}{RGB}{181,85,22}
\begin{tikzpicture}[x=1cm,y=1cm,font=\small,
line cap=round,line join=round,>=Latex]
\path[use as bounding box] (-.9,-1.05) rectangle (12.25,3.85);
\fill[ribbonblue!3] (0,0) rectangle (12,3);
\draw[black!40] (0,0)--(12,0) (0,3)--(12,3);
\fill[black!8] (0,1.15) rectangle (12,1.85);
\draw[black!35,dotted] (0,1.15)--(12,1.15) (0,1.85)--(12,1.85);
\fill[ribbonorange!20] (5.5,1.15) rectangle (7.5,1.85);
\draw[ribbonorange] (5.5,1.15) rectangle (7.5,1.85);
\draw[dashed,black!50] (3.5,-.2)--(3.5,3.3) (9.5,-.2)--(9.5,3.3);
\node[above] at (3.5,3.3) {front $s=k$};
\node[above] at (9.5,3.3) {$s=k+1$};
\draw[very thick] (0,1.4)--(.5,1.65)--(1,1.36)--(1.6,1.7)
--(2.2,1.42)--(2.8,1.64)--(3.5,1.57);
\fill (3.5,1.57) circle (2pt);
\node[above left] at (3.5,1.7) {$a$};
\draw[thick,ribbonblue,->] (3.5,1.57)--(5,1.5)--(11.7,1.5);
\fill[white] (6.5,1.5) circle (2pt);
\draw[ribbonblue] (6.5,1.5) circle (2pt);
\node[below,align=center] at (6.5,.95) {waypoint\\$P(k+1/2,0)$};
\draw[thin,black!60] (6.5,.94)--(6.5,1.4);
\node[above,text=ribbonorange] at (6.5,2.25) {middle test region};
\draw[ribbonorange] (6.5,2.2)--(6.5,1.88);
\node[above,align=center] at (1.55,2.05) {past in the core\\$|x|<\epsilon$};
\node[above,text=ribbonblue] at (10.65,1.85) {access guide $\alpha$};
\node[left] at (0,3) {$x=1$};
\node[left] at (0,0) {$x=-1$};
\draw[<->] (3.5,-.65)--(9.5,-.65)
node[midway,fill=white,inner sep=2pt] {one unit in $s$};
\end{tikzpicture}
\endgroup
==== END An-explicit-exact-Hausdorff-gauge-for-SLE//ribbon-front.tex ====
==== BEGIN An-explicit-exact-Hausdorff-gauge-for-SLE//ribbon.tex ====
\section{Uniform passages and dense visits}
\label{rc:section}
The covering argument needs a visit that creates unusually much sausage
area. We construct such a visit by traversing a narrow serpentine ribbon.
The essential estimate is uniform over the history already drawn inside
the ribbon. A positive probability for each fixed smooth template would
not give the required bound as the number of traversals increases.
Throughout this section $0<\kappa<8$ is fixed,
$d=1+\kappa/8$, and $p=2-d$. For a curve segment $\eta$, write
$N_v(\eta)=\{z:\operatorname{dist}(z,\eta)<v\}$.
We use Corollary~\ref{geom:restart-support}, Proposition~\ref{ll:localization}, and Proposition~\ref{retain:sausage} in the following forms. First, a smooth simple
slit can be followed in an open tube through ordered separating
crosscuts, with positive probability; the same assertion holds at a
stopped restart. Second, suppose a conformal map $F$ from $\mathbb H$ has a
continuous spherical extension to the closed half-plane and an access
curve from $0$ to modulus $b_0$ whose image is in $B(a,r/L)$, then
for all $L\ge L_0$,
\begin{equation}\label{rc:localization-input}
\mathbb P\{F(\gamma[0,\sigma_{b_0L^b}])
\not\subset B(a,r)\}\le C L^{-c}.
\end{equation}
Here $\sigma_u$ is the first time the parameter-plane trace reaches modulus
$u$; the containment is interpreted up to infinite time if that stop is
not reached. The constants $b,c,C>0$ and $L_0>1$ are uniform over the
maps and access curves, and the estimate includes parameter-boundary
visits. Third, for
each deterministic compact capacity-time interval $J$ of positive length,
\begin{equation}\label{rc:sausage-input}
\lim_{a\downarrow0}\limsup_{v\downarrow0}
\mathbb P\{v^{-p}\operatorname{Area}(N_v(\gamma(J)))<a\}=0.
\end{equation}
Only finite lists of deterministic intervals will be used. In
particular, we do not condition this lower-tail estimate on a prescribed
passage or apply it directly to a random interval.
Rezaei's cover argument already uses conditional bounds after stopped
histories, compactness, and passages through prescribed spirals and tubes
\cite[Section~3, Lemmas~3.1--3.3]{rezaei2018}. Here we need one
probability bound for every past admitted by a ribbon step, together
with one deterministic small-mesh cutoff for those pasts. Iterating
this quantitative step will give the explicit lower bound
$c\exp(-Cm^2)$ for the dense visit.
\subsection{Geometry at the front of a ribbon}
Let $\sigma:[0,\infty)\to\overline{\mathbb H}$ be a unit-speed simple
curve, with Lipschitz tangent, curvature at most $1/10$, and normal $n$.
Assume that
\[
P(s,x)=\sigma(s)+xn(s),\qquad s\ge0,\quad |x|\le1,
\]
is an embedded closed ribbon, lies in $\mathbb H$ when $s>0$, escapes
every bounded set as $s\to\infty$, and is straight vertical for
$0\le s\le4$, with $P(0,0)=0$. The coordinate maps on patches of
length four have one fixed bi-Lipschitz constant. These geometric
bounds are part of the admissible class. Fix $\epsilon=1/50$.
Consider any stopped chordal past that first reaches $s=k$, $k\ge1$,
has remained in $|x|<\epsilon$ up to that stop, meets the real boundary
only at the ribbon's base $P(\{0\}\times(-\epsilon,\epsilon))$, and ends at
$a=P(k,x_a)$ with $|x_a|\le\epsilon/4$. The future portion of the
ribbon is in the surviving domain $D$: its interior is disjoint from the
past and has a route to the target along the ribbon. The same holds for
the strips outside the core on each side of the preceding local patch.
\begin{lemma}[One ribbon step]\label{rc:step}
There are $q,c,v_0>0$, depending only on $\kappa$ and the fixed ribbon
constants, with the following property. For every stopped past
satisfying the preceding ribbon conditions and every deterministic
$0<v<v_0$, the conditional
probability is at least $q$ that the next first crossing of $s=k+1$
occurs in finite time, with tip offset $|x|<\epsilon/4$, while the
trace segment $\gamma_{\rm step}$ from the current
front to that next crossing stays in
\[
k-1/4<s<k+1,\qquad |x|<\epsilon,
\]
before that crossing, and that
\begin{equation}\label{rc:step-mass}
v^{-p}\operatorname{Area}
\left(N_v(\gamma_{\rm step})\cap
P((k+1/3,k+2/3)\times(-\epsilon,\epsilon))\right)\ge c.
\end{equation}
All three constants are uniform over the ribbon and the entire stopped
past. For a ribbon of half-width $\ell$, the same assertion has mesh cutoff
$v_0\ell$ and lower bound $c\ell^d$ in \eqref{rc:step-mass}.
\end{lemma}
The goal is thus a single positive probability and a single mesh cutoff
for this entire class of pasts. We first establish uniform conformal
control of an access guide; the proof of Lemma~\ref{rc:step} then
transfers an interior sausage test through those conformal maps.
Use the guide
\[
\alpha(t)=P(k+t,x_a(1-4t)_+),\qquad 0\le t\le2,
\]
and continue along the centerline to infinity. It is a simple access
curve from the fresh tip to the target. Freshness follows from the first
crossing of the front; the fresh-tip prime-end statement makes its
initial access agree with the driving prime end. On this guide,
\begin{equation}\label{rc:guide-clearance}
\operatorname{dist}(\alpha(t),\partial D)\ge c\min(t,1),
\qquad |\alpha(t)-a|\le Ct\quad(0<t\le2).
\end{equation}
Indeed a local coordinate ball ahead of the front is free. Its radius
can be chosen proportional to $t$ until the guide has advanced a fixed
distance, and fixed thereafter. Nonlocal portions of the past cannot
enter these balls: they would meet the interior of this ribbon patch,
contrary to embedding. Every ball is in the surviving component because
it connects to the future ribbon.
Set $w=P(k+1/2,0)$. If $g:D\to\mathbb H$ sends the tip to $0$ and the
target to infinity, then
\begin{equation}\label{rc:waypoint-angle}
\sin\arg g(w)\ge s_*>0.
\end{equation}
Here is a uniform harmonic-measure proof. The complete access guide
separates $D$ into two sides corresponding to the prime-end boundary
arcs mapped to the positive and negative real axes. An exit approached
from one side without crossing the guide uses that arc, even when its
physical exit point represents several prime ends. From $w$, on either
chosen side of the guide, take a route in the local ribbon to offset
$x=1/2$ or $x=-1/2$,
and backwards to a transverse rectangle centered at $s=k-1/2$.
These routes have uniformly bounded length and fixed positive clearance
from the core and the opposite side of the guide. Across that rectangle
take a transverse route from the chosen offset to beyond the opposite
edge of the core. The past has a longitudinal crossing of the rectangle:
take its portion between the last crossing of its rear edge preceding
the first crossing of its forward edge. This connected crossing lies
in $|x|<\epsilon$ and separates the two transverse ends. Thus a
continuous transverse crossing must meet the past before it reaches the
other end.
Starting at $w$, first run Brownian motion in a fixed small free ball
about $w$ until it reaches a smaller disk strictly on the chosen side of
the guide. There is a uniformly positive probability to do so before
leaving the free ball. Crossings of the guide during this initial move
are allowed. From that smaller disk, Brownian motion has a uniformly
positive chance to follow the first route and then cross the transverse
rectangle. One can specify this event in finitely many overlapping
coordinate disks and rectangular crossings with uniformly bounded aspect
ratios. During this latter part, until the first domain exit, its route
avoids the access guide. Its exit therefore belongs to the chosen
prime-end arc. The strong Markov property gives a lower bound for the harmonic
measure of each arc from $w$. In half-plane coordinates these two
harmonic measures are $\arg g(w)/\pi$ and $1-\arg g(w)/\pi$, proving
\eqref{rc:waypoint-angle}. The argument uses no regularity of the past
inside the core.
After imposing $|g(w)|=1$, the conformal radius of $w$ is bounded above
and below, by \eqref{rc:guide-clearance} and the boundary point $a$.
Consequently the inverse maps have nondegenerate interior compact
limits. The next two lemmas turn this interior normalization into a uniform
passage. First, the parameter guide approaches the driving point
uniformly. A small-image semicircle then joins an early localized trace
to the limiting guide. Finally, a finite choice of rational time
intervals transfers the sausage estimate. This will produce one
probability, mass threshold, and mesh cutoff valid for every past.
The estimates above verify the first lemma's hypotheses. Its limiting
map, guide and omitted boundary point then supply the hypotheses of
the second lemma.
\begin{figure}[tbp]
\centering
\input{figures/ribbon-front}
\caption{\RibbonFrontCaption}
\label{fig:ribbon-front}
\end{figure}
Figure~\ref{fig:ribbon-front} shows the guide and the middle region
whose sausage mass appears in Lemma~\ref{rc:step}.
The compactness argument at the initial tip requires control of a guide,
but no common modulus of continuity for the inverse maps on the boundary.
We give that control explicitly. Hyperbolic length in
$\mathbb H$ is normalized as $|dz|/\operatorname{Im}z$.
\Needspace{24\baselineskip}
\begin{lemma}[Compactness along a guide]\label{es:guide-compactness}
Let $F_n:\mathbb H\to D_n$ be conformal, with inverse $g_n$, and let
$\alpha_n:[0,T]\to\overline{D_n}$ be $C$-Lipschitz guides satisfying
\[
\begin{gathered}
\alpha_n(0)=a_n\in\partial D_n,\qquad
\alpha_n(t)\in D_n\quad(t>0),\\
\operatorname{dist}(\alpha_n(t),\partial D_n)
\ge c\min(t,1).
\end{gathered}
\]
Assume that $a_n$ stays in a bounded set and that
$\beta_n=g_n\circ\alpha_n$, initially defined for $t>0$, extends
continuously to $\beta_n(0)=0$. Fix $t_w\in(0,T]$, put
$w_n=\alpha_n(t_w)$, and suppose that
\[
|g_n(w_n)|=1,\qquad \operatorname{Im}g_n(w_n)\ge s>0.
\]
There is a subsequence on which $\alpha_n$ converges uniformly to a
guide $\alpha$, $F_n$ converges locally uniformly on $\mathbb H$ to a
nonconstant univalent map $F$, and $\beta_n$ converges locally uniformly
on $(0,T]$ to $\beta$, where $F\circ\beta=\alpha$. In addition,
\begin{equation}\label{es:uniform-tip-collapse}
\lim_{h\downarrow0}\sup_n\sup_{0\le t\le h}|\beta_n(t)|=0.
\end{equation}
Consequently $\beta(0+)=0$. If $a_n\to a$, then $a\notin F(\mathbb H)$
and $F(\beta(t))\to a$ as $t\downarrow0$.
\end{lemma}
\begin{proof}
Write $z_n=g_n(w_n)$. These points lie in a compact subset of
$\mathbb H$. The conformal-radius identity and Koebe's theorem give
\[
2\operatorname{Im}z_n\,|F_n'(z_n)|
=R_{D_n}(w_n),\qquad
c\min(t_w,1)\le R_{D_n}(w_n)\le4Ct_w.
\]
Here the upper bound uses $a_n\in\partial D_n$ and
$|w_n-a_n|\le Ct_w$. Thus the inverse derivatives are bounded above
and away from zero at an interior compact set. Since $w_n$ is bounded,
the growth and distortion theorems for normalized univalent functions
give a subsequence converging locally uniformly to a nonconstant
univalent $F$. Arzel\`a--Ascoli also gives uniform convergence of the
physical guides.
For each fixed $\delta>0$, the part of $\alpha_n$ joining $w_n$ to any
$\alpha_n(t)$, $\delta\le t\le T$, has uniformly bounded hyperbolic
length. Indeed the hyperbolic density in $D_n$ is at most twice the
reciprocal of the distance to its boundary, and the guides have speed
at most $C$. Conformal invariance therefore confines
$\beta_n([\delta,T])$ to a fixed compact subset of $\mathbb H$.
The same estimates, or the conformal-radius identity for $g_n'$,
give equicontinuity there. Subsequence limits satisfy
$F(\beta(t))=\alpha(t)$; univalence identifies them consistently for
all $\delta>0$.
It remains to prove \eqref{es:uniform-tip-collapse}. Suppose otherwise.
There are $h_n\downarrow0$ and $\varepsilon>0$ such that
$\beta_n([0,h_n])$ reaches modulus $\varepsilon$. After taking the
subsequence just described, there are a fixed disk $K\Subset\mathbb H$
and $r_0>0$ for which
\begin{equation}\label{es:waypoint-separation}
|F_n(z)-a_n|\ge r_0\qquad(z\in K)
\end{equation}
for all sufficiently large $n$. To see this, take $K$ sufficiently
small about the limit of $z_n$: the points $w_n$ have distance at least
$c\min(t_w,1)$ from $a_n$, and local convergence and distortion preserve
a smaller positive separation on $K$.
Put $T_n=\log(r_0/(Ch_n))$, which tends to infinity. In the strip
$S=\mathbb R\times(0,\pi)$ set
\[
U_n(v,\phi)=\min\left\{T_n,
\log^+\frac{|F_n(e^{v+i\phi})-a_n|}{Ch_n}\right\}.
\]
Conformal invariance of Dirichlet energy and area change imply
\begin{equation}\label{es:collapse-energy}
E_n:=\int_S|\nabla U_n|^2\,dv\,d\phi
\le\int_{Ch_n<|w-a_n|<r_0}\frac{dA(w)}{|w-a_n|^2}
=2\pi T_n.
\end{equation}
Every section $v<\log\varepsilon$ contains a zero of $U_n$:
the continuous curve $\beta_n([0,h_n])$ crosses that modulus, and
$|\alpha_n(t)-a_n|\le Ch_n$ throughout the corresponding physical
guide segment. Angular Cauchy--Schwarz consequently gives, for almost
every such $v$,
\[
\int_0^\pi U_n(v,\phi)^2\,d\phi
\le\pi^2\int_0^\pi|\partial_\phi U_n(v,\phi)|^2\,d\phi.
\]
In particular this bounds the squared $L^2$ norm on the fixed unit
slab $(\log\varepsilon-1,\log\varepsilon)\times(0,\pi)$ by
$\pi^2E_n$.
For completeness, this bound propagates to every fixed bounded
rectangle $J\times(0,\pi)$ containing that slab. Let
$m_n(v)=\pi^{-1}\int_0^\pi U_n(v,\phi)\,d\phi$.
There is a section $v_n$ in the initial slab with
$|m_n(v_n)|^2\le\pi E_n$, while
\[
\int_J|m_n'(v)|^2\,dv\le E_n/\pi.
\]
Horizontal Cauchy--Schwarz bounds $|m_n(v)|^2$ on $J$ by a constant
depending only on $J$ and $\varepsilon$, times $E_n$.
The one-dimensional Poincar\'e inequality for
$U_n(v,\cdot)-m_n(v)$ then gives
\begin{equation}\label{es:anchored-ltwo}
\int_{J\times(0,\pi)}U_n^2\le C_{J,\varepsilon}E_n.
\end{equation}
Choose $J$ so that $J\times(0,\pi)$ also contains $\log K$, using the
logarithm with argument in $(0,\pi)$. By
\eqref{es:waypoint-separation}, $U_n=T_n$ on the
fixed positive-area set $\log K$. Equations
\eqref{es:collapse-energy}--\eqref{es:anchored-ltwo} imply
\[
|\log K|T_n^2\le C_{J,\varepsilon}2\pi T_n,
\]
a contradiction. This proves uniform collapse.
Finally $F_n-a_n$ has no zeros in $\mathbb H$. Hurwitz's theorem and
nonconstancy show that $F-a$ has no zeros there. Since
$F(\beta(t))=\alpha(t)\to a$, the omitted point $a$ belongs to
$\partial F(\mathbb H)$.
\end{proof}
\begin{lemma}[A small image semicircle]\label{es:good-semicircle}
Let $F$ be univalent on $\mathbb H$, and let $a\notin F(\mathbb H)$.
Suppose $\beta:[0,T]\to\overline{\mathbb H}$ is continuous,
$\beta(0)=0$, $\beta(t)\in\mathbb H$ for $t>0$, and
$F(\beta(t))\to a$ as $t\downarrow0$.
Fix $r>0$ and $L>1$, and write $A=\log L$. Choose $t_0>0$ so that
\[
|F(\beta(t))-a|\le r/L\quad(0<t\le t_0),
\qquad b_0=|\beta(t_0)|>0.
\]
In every unit logarithmic slab $(q,q+1)$ lying below $\log b_0$,
there is $B\in(e^q,e^{q+1})$ such that, provided $A>2\pi^2$,
\begin{equation}\label{es:semicircle-bound}
\sup_{0<\theta<\pi}|F(Be^{i\theta})-a|
\le r\exp\{-A+\pi\sqrt{2A}\}.
\end{equation}
Moreover
\begin{equation}\label{es:sector-bound}
\sup_{\substack{B/8\le|z|\le B\\
\operatorname{Im}z\ge |z|/2}}
|F(z)-a|
\le r\exp\{-A+\pi\sqrt{2A}+6\}.
\end{equation}
For $A\ge64$ both upper bounds are smaller than $r/4$.
\end{lemma}
\begin{proof}
On $S=\mathbb R\times(0,\pi)$ use
\[
U(v,\phi)=\min\left\{A,
\log^+\frac{|F(e^{v+i\phi})-a|}{r/L}\right\}.
\]
As in \eqref{es:collapse-energy},
$\int_S|\nabla U|^2\le2\pi A$. The initial guide supplies a zero on
every section $v<\log b_0$. In the specified unit slab there is thus
a section $v=\log B$ whose angular energy is at most $2\pi A$.
Starting at a zero on that section and applying Cauchy--Schwarz gives
\[
U(\log B,\phi)\le\sqrt{\pi\cdot2\pi A}
=\pi\sqrt{2A}\qquad(0<\phi<\pi).
\]
The section may be chosen where the Sobolev restriction is absolutely
continuous; continuity in the open semicircle gives the bound at every
interior point. Since $\pi\sqrt{2A}<A$, the upper truncation is
inactive and \eqref{es:semicircle-bound} follows.
Koebe's quarter theorem, applied after mapping the disk to
$\mathbb H$ at $z$, gives
\[
\frac{\operatorname{Im}z}{2}|F'(z)|
\le\operatorname{dist}(F(z),\partial F(\mathbb H))
\le |F(z)-a|.
\]
Therefore $\log|F-a|$ is $2$-Lipschitz for hyperbolic distance.
If $z=\rho Be^{i\theta}$ is in the sector in
\eqref{es:sector-bound}, then $1/8\le\rho\le1$ and
$\sin\theta\ge1/2$, so
\[
\cosh d_{\mathbb H}(z,iB)
=\frac{1+\rho^2}{2\rho\sin\theta}
\le\rho+\rho^{-1}\le65/8<\cosh3.
\]
Comparing with $iB$ in \eqref{es:semicircle-bound} proves
\eqref{es:sector-bound}. For $A\ge64$,
$-A+\pi\sqrt{2A}+6<-\log4$.
\end{proof}
Here is the geometric consequence used to splice an initial passage
to the limiting guide. Suppose $\beta$ in
Lemma~\ref{es:good-semicircle} is simple, and fix $\delta\in(0,T)$.
Choose the logarithmic slab so far to the left that its upper radius
is smaller than
\[
\min\left\{b_0/2,\ \min_{\delta\le t\le T}|\beta(t)|\right\}.
\]
The last time $t_B\le T$ with $|\beta(t_B)|=B$ then satisfies
$t_B<\delta$, and the remaining guide lies strictly outside that
semicircle. Put $b=B/4$. If an initial hull stays in
\[
\{z:|\operatorname{Re}z|<b/4,\quad
0\le\operatorname{Im}z\le b\}
\]
and ends at $\zeta=x+ib$, its modulus is less than $B/2$.
From $\zeta$ move vertically upwards to the semicircle $|z|=B$,
follow the semicircle to $\beta(t_B)$, and then follow
$\beta([t_B,T])$. The vertical piece belongs to the sector in
\eqref{es:sector-bound}; the semicircular piece obeys
\eqref{es:semicircle-bound}. Both therefore map into $B(a,r/4)$
when $\log L\ge64$. The vertical piece is above the initial hull,
and the other pieces are outside its containing disk. These pieces
form a simple continuation route; corners can be smoothed in its
open neighborhoods. Each fixed such route after $\zeta$ lies in a
compact subset of $\mathbb H$. Consequently local uniform
convergence of the maps is sufficient to transfer all strict
continuation restrictions along that route.
\subsection{The uniform passage estimate}
We now prove Lemma~\ref{rc:step}. If uniform constants failed,
a sequence of admissible states and vanishing meshes would have a
nondegenerate conformal limit. A passage in that limit has positive
probability. Selecting finitely many compact sets, margins and
rational time intervals transfers it to the sequence; subtracting
the unconditional sausage exception preserves a fixed positive
passage-and-mass probability.
\begin{proof}[Proof of Lemma~\ref{rc:step}]
Suppose the uniform assertion were false. Then for each integer $n$
there would be admissible data and $0<v_n<1/n$ such that the probability
of passage together with normalized mass at least $1/n$ is less than
$1/n$. Denote their ribbon maps and front positions by $P_n$ and
$k_n$. Translate the physical starting tips to a bounded region, keeping
the same notation, and put
\[
V=[-1/2,3/2]\times[-1,1],\qquad Q_n(u,x)=P_n(k_n+u,x).
\]
On a subsequence these charts converge uniformly, together with the
centerline tangents, to a chart $Q$ with the same bi-Lipschitz bound.
The limiting guide through relative coordinate $u=5/4$ is therefore
simple. Lemma~\ref{es:guide-compactness}, applied to this part of the
guide, gives locally uniform convergence of the normalized inverse maps
$F_n$ to a nonconstant univalent $F$. The parameter guides converge away
from time zero to a simple guide $\beta$ with $\beta(0+)=0$. Write
$a_n\to a$ for the physical tips.
Use one ordinary parameter-plane SLE $\gamma$ for all $n$. For each
fixed $n$, the image $F_n\circ\gamma$, after the corresponding capacity
rescaling, has the law of the conditional continuation in that state.
This realizes the events below on one probability space; no joint
relation between the original stopped states is needed.
Choose $r>0$ so small that, for all sufficiently large $n$ and also for
the limiting chart,
\[
\begin{aligned}
\overline B(a_n,2r)&\subset
Q_n\bigl((-1/8,1/4)\times(-\epsilon/2,\epsilon/2)\bigr),\\
\overline B(a,r)&\subset
Q\bigl((-1/8,1/4)\times(-\epsilon/2,\epsilon/2)\bigr).
\end{aligned}
\]
The centered tips and the common bi-Lipschitz bound permit this choice.
In particular, a trace in the first ball has not yet reached the middle
region or the next front.
We fix the probability of the initial vertical passage before choosing
the localization scale. Let $\tau_b$ be the first time ordinary
parameter-plane SLE attains height $b$. Slit support gives a constant
$A<\infty$ for which
\[
E_b=\left\{\tau_b\le Ab^2,\quad
|\operatorname{Re}\gamma(t)|<b/4\ (0\le t\le\tau_b)\right\}
\]
has probability $q_0>0$. Brownian scaling makes $q_0$ independent of
$b>0$, and $E_b\in\mathcal F_{\tau_b}$. Fix $L\ge L_0$ large enough that
$\log L\ge64$ and $CL^{-c}<q_0/2$, using the constants in
\eqref{rc:localization-input}.
A sufficiently short fixed physical guide segment lies in
$B(a_n,r/L)$ for every large $n$. Its endpoint has parameter modulus
bounded below by some $b_0>0$: this follows from interior convergence
at that fixed positive guide time. Decrease $b_0$ so that the same
statement holds for the limiting guide, and truncate the guides at
their first modulus-$b_0$ points when using them as localization
access curves. Keep the full limiting guide for the continuation.
Lemma~\ref{es:good-semicircle}
supplies a radius $B<b_0/2$ whose semicircle and central sector map
under $F$ into $B(a,r/4)$. Choose its logarithmic slab sufficiently
far to the left that the last crossing of $|z|=B$ by $\beta$ occurs
while its physical image is still in $B(a,r/4)$. The rest of the guide,
through $u=5/4$, then lies outside that semicircle.
Put $b=B/4$, $\tau=\tau_b$, and $E=E_b$. On $E$ the trace through
$\tau$ stays in
\[
|\operatorname{Re}z|<b/4,\qquad 0\le\operatorname{Im}z\le b.
\]
Its filled hull stays there too, since the part of $\mathbb H$ outside
the closed rectangle is connected to infinity. The terminal tip is fresh
and has an upward vertical access ray. Let $G_n$ be the event that the
physical image under $F_n$ of this initial trace stays in $B(a_n,r)$.
It is $\mathcal F_\tau$-measurable. Lemma~\ref{geom:extension} gives
continuous boundary extensions of the inverse maps, so the short access
curve and \eqref{rc:localization-input} give
\[
\mathbb P(E\setminus G_n)\le CL^{-c}.
\]
Indeed, on $E$ the trace through $\tau$ has modulus less than
$B/2<b_0$, so the localization estimate controls this entire prefix.
Our choice of $L$ yields $\mathbb P(E\cap G_n)\ge q_0/2$.
Given a realization in $E$, use the geometric splice after
Lemma~\ref{es:good-semicircle}: move upwards from its tip to the
semicircle of radius $B$, follow the semicircle to the last crossing
of $\beta$, and then follow $\beta$ through $u=5/4$. This simple route
avoids the previous hull. Its initial pieces map into the early ball
$B(a,r/4)$; thereafter it follows the guide, whose longitudinal
coordinate increases and whose offset is zero by $u=1/4$.
We describe a passage event with margins that persist under uniform
changes of the maps. For $\delta=1/\nu<\min(\epsilon/8,1/12)$, where
$\nu$ is a positive integer, define the compact coordinate sets
\[
V_\delta=\left\{(u,x)\in V:
\begin{gathered}
-1/4+\delta\le u\le3/2-\delta,\quad |x|\le\epsilon-\delta,\\
u\le1-\delta\ \text{or}\ |x|\le\epsilon/4-\delta
\end{gathered}\right\},
\]
\[
\begin{aligned}
P_\delta&=V_\delta\cap\{u\le1-2\delta\},&
O_\delta&=V_\delta\cap\{u\ge1+\delta\},\\
M_\delta&=[1/3+\delta,2/3-\delta]
\times[-\epsilon+\delta,\epsilon-\delta].
\end{aligned}
\]
The set $V_\delta$ keeps the whole suffix in the core and centers it
near the new front. The sets $P_\delta$, $M_\delta$, and $O_\delta$
specify a prefix before that front, a middle interval, and a point
beyond the front. The closed alternative in $V_\delta$ is useful:
these restrictions are closed conditions on a continuous path.
Let
\[
K_j=\{z:|\operatorname{Re}z|\le j,\quad
1/j\le\operatorname{Im}z\le j\},\qquad j=1,2,\ldots.
\]
Let $C$ be the union, over $j,\nu$ as above and rational
$0<t_-<t_+<t_{\rm out}$, of the following events:
\[
\begin{gathered}
\tau<t_-,\qquad \gamma([\tau,t_{\rm out}])\subset K_j,\\
F(\gamma([\tau,t_{\rm out}]))\subset Q(V_\delta),\qquad
F(\gamma([\tau,t_+]))\subset Q(P_\delta),\\
F(\gamma([t_-,t_+]))\subset Q(M_\delta),\qquad
F(\gamma(t_{\rm out}))\in Q(O_\delta).
\end{gathered}
\]
A witness is false unless its $K_j$ containment holds, so $F$ is
evaluated only in $\mathbb H$. All target sets in the display are
compact. By continuity, the interval containments can be checked on
dense rational times, together with $\gamma(\tau)$. Each witness is
therefore $\mathcal F_{t_{\rm out}}$-measurable. Their countable union
$C$ depends on the fixed limit map and chart, independently of $n$,
$G_n$, and the route used to establish its probability.
For each realization in $E$, choose a sufficiently thin tube around
the spliced route. Its image under $F$ lies in the interior of
$Q(V_\delta)$ for some allowed $\delta$. Choose two ordered middle
gates so that the $F$-image of the tube portion from its initial tip
through the second gate lies in the interior of $Q(P_\delta)$, and
the $F$-image of the portion between the gates lies in the interior
of $Q(M_\delta)$. Choose the terminal parameter open set with
$F$-image in the interior of $Q(O_\delta)$. The parameter-plane closure
of the post-$\tau$ tube, including its initial tip, lies in some $K_j$.
Before its first passage through the second gate, a path following the
embedded tube remains in its prefix portion. It visits an open region
between the gates, so continuity gives a nondegenerate rational interval
$[t_-,t_+]$ there before that second passage. A later visit to the
terminal open set supplies a rational $t_{\rm out}>t_+$. These times
satisfy the displayed conditions. Map out the initial hull. Prime-end
access and continuity at the driving point permit a smooth simple slit in the
restarted half-plane following the tube. Slit support and the domain
Markov property give
\[
q(\omega):=\mathbb P(C\mid\mathcal F_\tau)(\omega)>0
\quad\text{on }E
\]
outside the usual null set. This argument needs no measurable choice
of a route, since $C$ itself is the fixed countable union just defined.
The value of $q$ may depend on the past.
Choose $q_1>0$ so that
$\mathbb P(E\cap\{q<q_1\})<q_0/4$. Since $G_n$ is
$\mathcal F_\tau$-measurable, for every large $n$,
\begin{equation}\label{rc:varying-good-events}
\mathbb P(E\cap G_n\cap C)
=\mathbb E[\mathbf1_{E\cap G_n}q]
\ge q_1q_0/4=:q_2>0.
\end{equation}
Choose a finite subfamily of witnesses whose omission from $C$ has
probability less than $q_2/4$ on $E\cap C$. The same absolute loss
bound holds after intersecting with any of the varying events $G_n$.
We next transfer a retained witness to the $n$th chart. If $z$ is on
its suffix, write $F(z)=Q(v)$ with $v\in V_\delta$. Such $v$ has
distance at least $\delta$ from $\partial V$. If $K$ is the common
bi-Lipschitz constant, then $Q_n(v)$ has distance at least $\delta/K$
from the Jordan boundary $Q_n(\partial V)$. On the finitely many
retained sets,
\[
\sup_{z\in K_j}|F_n(z)-F(z)|
+\sup_{v\in V}|Q_n(v)-Q(v)|\longrightarrow0.
\]
For all sufficiently large $n$, the displayed sum is less than
$\delta/(4K)$ for every retained witness. Thus $F_n(z)$ lies in the
ball about $Q_n(v)$ of radius $\delta/K$. That ball stays inside
the Jordan domain $Q_n(\operatorname{int}V)$, so the inverse coordinates
are defined. The inverse Lipschitz bound then gives
\[
\bigl|Q_n^{-1}(F_n(z))-v\bigr|<\delta/4
\]
uniformly along every retained suffix.
The prefix on $G_n$ has relative coordinate $u_n<1/4$, whereas the
point at $t_{\rm out}$ has $u_n>1$. Continuity forces a first crossing
of $u_n=1$ before $t_{\rm out}$. The full $V_\delta$ containment gives
the rear and core bounds until this crossing. At the crossing, the
limit coordinate satisfies $u\ge1-\delta/4>1-\delta$, so the closed
alternative in $V_\delta$ forces $|x_n|<\epsilon/4$. The
$P_\delta$ containment gives $u_n\le1-7\delta/4<1$ through $t_+$.
Thus $J=[t_-,t_+]$ precedes the first crossing. Its image lies inside
the $n$th middle region with a positive physical margin, by the
$M_\delta$ bounds and the common bi-Lipschitz control.
For each retained $K_j$, use the larger rectangle $K_{2j}$.
Derivative convergence gives deterministic constants
$0<c_*<C_*<\infty$ such that
\[
c_*\le |F_n'|\le C_*
\]
there for all sufficiently large $n$. Fix $\lambda>0$ sufficiently
small. On a retained witness, since $v_n\to0$, the
$\lambda v_n$-sausage of its interval lies in $K_{2j}$; the segment
from a sausage point to its trace point also lies there. The upper
derivative bound maps this
sausage into the physical $v_n$-sausage, and the middle-region margin
keeps its image inside that region. Change of area gives
\begin{equation}\label{rc:compact-sausage-transfer}
v_n^{-p}\operatorname{Area}(\hbox{physical sausage in the middle})
\ge c_*^2\lambda^p
(\lambda v_n)^{-p}\operatorname{Area}
(N_{\lambda v_n}(\gamma(J))).
\end{equation}
Apply \eqref{rc:sausage-input} to the finite list of deterministic
intervals. It supplies a fixed positive lower mass threshold $a_*$ with
total failure probability less than $q_2/4$ at all sufficiently small
$v_n$. This is an unconditional estimate, so it can be subtracted
from \eqref{rc:varying-good-events}. After the finite extraction and
this sausage exception, the remaining probability is at least
$q_2-q_2/4-q_2/4=q_2/2$. On that event,
\eqref{rc:compact-sausage-transfer} gives the fixed mass threshold
$c_*^2\lambda^p a_*$. This event is contained in the step event stated
in the lemma: its sausage comes from $J$, which is before the first
crossing, and the transferred path has all the required bounds.
That step event is measurable from the trace stopped at the first
crossing. Although the auxiliary witness $C$ may inspect a later suffix
through $t_{\rm out}$, it is used only to prove this inclusion and
lower-bound the probability of the stopped event. The later suffix is
discarded before any iteration. For all sufficiently large $n$, both
the fixed mass threshold and $q_2/2$ exceed $1/n$, contradicting the
choice of the data.
This proves the existence of deterministic $q,c,v_0>0$ uniformly over
admissible states, at each prescribed deterministic mesh below $v_0$.
Scaling spatial coordinates by $\ell$ multiplies normalized sausage area by
$\ell^{2-p}=\ell^d$, proving the final assertion.
\end{proof}
\subsection{A dense visit in bounded conformal-radius time}
\begin{proposition}[Dense visit]\label{rc:dense}
Fix $s_0>0$. There are constants $c,C,C_2>0$, $C_1>1$, and an integer
$m_0$ with the following property. Let $S$ be a stopping time of ordinary
chordal SLE in $\mathbb H$, and let $T_z$ be the lifetime of an interior
marked point $z$. The assertion below holds conditionally almost surely
on $\{S<\infty,\ S<T_z\}$, in the surviving state at $S$ or its image
under a translation and positive dilation, with the transformed marked
point still denoted by $z$.
Suppose the marked point in that state has conformal radius $r$ and angle
sine at least $s_0$. Let $\rho$ be the first subsequent live stop at which
its conformal radius is $r/C_1$, with $\rho=\infty$ if that level is not
reached, and use the finite centered law of Lemma~\ref{ll:finite-tilt}
through $\rho$. For each integer $m\ge m_0$ there is a deterministic
$\delta_m>0$, independent of the stopped state, such that, for every
prescribed deterministic physical mesh $0<v\le r\delta_m$, there is an
$\mathcal F_\rho$-measurable success event with conditional centered
probability at least
\begin{equation}\label{rc:dense-probability}
c\exp(-Cm^2)
\end{equation}
on which the experiment ends at a finite stopping time $T<\rho$ and
generates sausage mass at least
\begin{equation}\label{rc:dense-mass}
v^{-p}\operatorname{Area}
\bigl(N_v(\gamma_{\rm experiment})\cap B(z,C_2r)\bigr)
\ge c r^d m^p.
\end{equation}
\end{proposition}
\begin{proof}
Normalize the stopped half-plane map so that $w=g(z)$ lies on the upper
unit semicircle. Write $y=\operatorname{Im}w\ge s_0$. Initially work
under the ordinary chordal law in these coordinates. Let $w'$ be
$w$ displaced horizontally by $s_0/4$ on the side away from the imaginary
axis, choosing either side if $\operatorname{Re}w=0$. The closed disk
of radius $s_0/20$ about $w'$ is inside $B(w,y/2)$ and stays a fixed
positive distance from $w$. Denote this disk by $\mathcal D$.
Set $h_0=s_0/200$. In this disk place $2m+1$ vertical segments from
height $y-h_0$ to $y+h_0$, with equally spaced horizontal coordinates
in an interval of length $2h_0$ centered at $\operatorname{Re}w'$.
Join consecutive ends by exterior semicircles, alternately above and
below the array, and traverse the segments in serpentine order. The
semicircles have radii comparable to $h_0/m$. Approach the first segment
from below along a smooth simple connector from $0$, initially vertical,
whose horizontal displacement is made between heights $s_0/8$ and
$s_0/2$. Continue the last segment vertically upwards to infinity.
The connector may be chosen from a family with uniform smooth bounds,
with $w$ a fixed positive distance from it; the sign choice gives two
such compact families as $w$ varies.
Choose
\[
\ell=\frac{h_0}{C_0m}
\]
with $C_0$ a sufficiently large fixed constant. The normal ribbon of
width $\ell$ about this entire guide is embedded and, in units of
$\ell$, has the geometry required above. Indeed the nonadjacent
straight passes have separation comparable to $h_0/m\gg\ell$;
the turn radii are at least $10\ell$; and the smooth connector has a
fixed-width embedded ribbon on its original scale and straight
attachments. The last segment and approach lie in different columns.
The ribbon starts with a vertical segment of length at least $4\ell$.
Let $s_{\rm out}$ be arclength in the normalized half-plane from the
base to the point where the final vertical continuation exits $\mathcal D$,
and set
\[
N=\left\lceil s_{\rm out}/\ell\right\rceil+1.
\]
We end the experiment at the first crossing of the front at arclength
$N\ell$. This front lies between $\ell$ and $2\ell$ beyond the disk
exit. Since $s_{\rm out}=O(m)$ and $\ell\asymp m^{-1}$, the number of
unit steps satisfies $N\le Cm^2$. At least $cm^2$ middle strips before
this front lie wholly inside $\mathcal D$.
Figure~\ref{fig:serpentine-ribbon} shows the displaced disk and the
serpentine array within it.
\begin{figure}[!htbp]
\centering
\input{figures/serpentine-ribbon}
\caption{\SerpentineRibbonCaption}
\label{fig:serpentine-ribbon}
\end{figure}
Let $F$ be the inverse of the original stopped map. Choose a fixed
$\chi>0$, depending only on $s_0$, so that the closed
$\chi$-neighborhood of $\mathcal D$ is inside $B(w,y/2)$. Such a
choice is uniform because the disk center is at distance $s_0/4$
from $w$ and its radius is $s_0/20$, while $y\ge s_0$.
On this neighborhood, Koebe distortion gives
\[
c r\le |F'|\le C r,
\qquad F(\hbox{that neighborhood})\subset B(z,C_2r).
\]
Here $|F'(w)|=r/(2y)$ and $s_0\le y\le1$; all constants are uniform
in the stopped domain. For the prescribed physical mesh $v$, put
\[
\eta=\lambda v/r,
\]
where $\lambda>0$ is small enough for the upper Lipschitz comparison.
Choose the deterministic $\delta_m>0$ so that
\[
\lambda\delta_m<\min\{v_0\ell,\chi/2\}.
\]
Then $v\le r\delta_m$ makes the same parameter mesh $\eta$ admissible
at every ribbon step. It also keeps the segment from any sausage point
counted in $\mathcal D$ to a witnessing trace point inside the
distortion neighborhood.
Use vertical-slit support for the initial step from the real boundary,
requiring confinement and centered arrival only. It has a fixed
positive probability by scaling. At each subsequent first front
crossing apply Lemma~\ref{rc:step} at this same prescribed mesh
$\eta$. Impose its passage-and-mass event determined at that crossing,
although only the mass contributions inside $\mathcal D$ will be
counted. The auxiliary later suffix used to prove the lemma is not
retained. Given previous successes, the stopped history is therefore
admissible for the next step. Denote the intersection of these events
by $E_\eta$.
Iterated conditional probabilities give
\[
\mathbb P(E_\eta)\ge q_{\mathrm{init}}q^N
\ge c\exp(-Cm^2).
\]
On $E_\eta$, the disjoint middle strips inside the disk yield
normalized sausage area at the chosen mesh $\eta$ of at least
\begin{equation}\label{rc:layout-sum}
c m^2\ell^d\ge c m^{2-d}=c m^p.
\end{equation}
Disjointness concerns the regions in which area is counted, so it
continues to hold if a sausage also meets another piece of the curve.
On this $\eta$-dependent success event, the sausage comparison in
\eqref{rc:compact-sausage-transfer} maps the area counted in
\eqref{rc:layout-sum} into the physical $v$-sausage. Conformal area
change multiplies normalized area by at least $c r^d$, so
\eqref{rc:dense-mass} follows. The choice of $\delta_m$ above depends
only on $m$ and the fixed constants, uniformly in the stopped state.
It remains to check the radius cutoff and the change of law. Through
completion, the prescribed trace and its filling remain in a narrow
neighborhood of the connector, array disk and short terminal extension.
More precisely, choose a closed obstacle consisting of $\mathcal D$,
a thin neighborhood of the simple connector, and a thin neighborhood of
the terminal stub from the disk exit through the completion front.
The stub has length at most $2\ell$. The connector and stub attach to
disjoint small arcs of the disk, and their neighborhoods are otherwise
disjoint. This obstacle has
connected complement in $\mathbb H$, so every filled prefix hull of a
trace contained in it is contained in it as well.
The point $w$ is separated from this set and the real boundary by a
fixed positive distance. The connector is below $w$ until its final
vertical portion, whose horizontal distance from $w$ is bounded below
by a fixed multiple of $s_0$; the array and terminal extension have the
same separation. Thus $w$ can move vertically to height $2$, then
horizontally to either $x=3$ or $x=-3$, and then down to a small real
interval about that coordinate, with uniformly positive clearance from
the prescribed trace except at the real endpoint. Small adjustments
of these polygonal routes give fixed-width neighborhoods. They also
connect $w$ and a fixed ball around it to infinity, so these sets are
in the surviving domain throughout the experiment. The two chosen
real intervals remain unvisited and unswallowed; their Loewner
coordinates stay on their respective sides of the driving point by
continuity.
Brownian tracking along the two routes therefore gives
a uniform lower bound for the two boundary-arc harmonic measures in
every successful intermediate domain. Hence the
marked-point angle sine stays at least $s_1>0$. Its conformal radius
stays above $c_1>0$ in parameter coordinates. Conformal covariance
therefore keeps its physical radius above $c_2r$. Choose $C_1>1/c_2$
with strict slack.
Let $T$ be the first crossing of the completion front. It is a stopping
time and is finite on $E_\eta$. On $E_\eta$, the one-point density at
this stop,
$M=R^{-p}\sin^{a_0}\theta$, $a_0=8/\kappa-1$, satisfies
\[
\frac{M_T}{M_0}
=\left(\frac{R_T}{r}\right)^{-p}
\left(\frac{\sin\theta_T}{\sin\theta_0}\right)^{a_0}
\ge s_1^{a_0}.
\]
Stop the density at the radius cutoff and use the finite stopped
change of law, first also stopping at a deterministic capacity horizon
and then increasing that horizon. Since success occurs in finite
time and before the radius cutoff, it follows that its tilted
probability is at least $s_1^{a_0}$ times its ordinary probability.
This proves \eqref{rc:dense-probability} and
\eqref{rc:dense-mass}. All tests concern the already generated
sausage at finite stops; no limiting marked-point visit is used.
\end{proof}
==== END An-explicit-exact-Hausdorff-gauge-for-SLE//ribbon.tex ====
==== BEGIN An-explicit-exact-Hausdorff-gauge-for-SLE//serpentine-ribbon.tex ====
\begingroup
\definecolor{ribbonblue}{RGB}{31,88,138}
\definecolor{ribbonorange}{RGB}{181,85,22}
\begin{tikzpicture}[x=1cm,y=1cm,font=\small,
line cap=round,line join=round,>=Latex]
\path[use as bounding box] (-.2,-.55) rectangle (14.7,6.05);
\begin{scope}[shift={(0,-6)}]
\node[anchor=west,font=\small\bfseries] at (0,11.85)
{(a) Marked point and displaced disk};
\node[anchor=west,font=\small\bfseries] at (7.25,11.85)
{(b) The array inside the disk (enlarged)};
\draw[black!70] (.1,6.55)--(6.6,6.55);
\node[anchor=north east] at (6.55,6.49) {$\mathbb R$};
\node[anchor=west] at (.16,10.95) {$\mathbb H$};
\fill (1.65,9.65) circle (2pt);
\node[anchor=south east,align=right] at (1.56,9.7)
{$w=g(z)$\\[-2pt]$|w|=1$};
\draw[black!45,dashed] (3.6,9.65) circle (1.03);
\fill[ribbonblue!4] (3.6,9.65) circle (1.03);
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\node[fill=white,inner sep=1.5pt] at (5.45,9.46) {$w'$};
\draw[thin,black!60] (5.22,9.49)--(3.65,9.63);
\draw[<->,thin] (1.68,10.92)--(3.6,10.92)
node[midway,fill=white,inner sep=1.4pt] {$s_0/4$};
\draw[thin,black!45] (1.65,10.16)--(1.65,11.08);
\draw[thin,black!45] (3.6,10.68)--(3.6,11.08);
\node[anchor=west,align=left,font=\footnotesize] at (4.58,10.27)
{radius\\$s_0/20$};
\draw[thin,black!50] (4.57,10.29)--(4.3,10.41);
\draw[ribbonblue,line width=1.1pt]
(.65,6.55)--(.65,7.15)
.. controls (.65,7.95) and (2.89,7.63) .. (2.89,8.5)
--(2.89,9.08)
--(2.89,10.05) arc[start angle=180,end angle=0,radius=.12]
--(3.13,9.08) arc[start angle=180,end angle=360,radius=.12]
--(3.37,10.05) arc[start angle=180,end angle=0,radius=.12]
--(3.61,9.08) arc[start angle=180,end angle=360,radius=.12]
--(3.85,10.05) arc[start angle=180,end angle=0,radius=.12]
--(4.09,9.08) arc[start angle=180,end angle=360,radius=.12]
--(4.33,10.05)--(4.33,11.27);
\draw[ribbonblue,-{Latex[length=2mm]}] (4.33,11.02)--(4.33,11.33);
\node[anchor=west,font=\footnotesize] at (4.43,11.18) {future guide};
\fill (.65,6.55) circle (1.8pt);
\node[anchor=north] at (.65,6.43) {$0$};
\node[anchor=north,align=center,font=\footnotesize] at (2.65,7.47)
{smooth connector\\initially vertical};
\begin{scope}[shift={(7.45,6.92)}]
\fill[ribbonblue!3] (-.15,-.08) rectangle (4.95,3.81);
\node[anchor=west,font=\footnotesize,text=black!65] at (.03,3.7)
{$2m+1$ vertical strands};
\draw[ribbonblue!15,line width=4pt]
(.2,-.32)--(.2,2.65) arc[start angle=180,end angle=0,radius=.26]
--(.72,.5) arc[start angle=180,end angle=360,radius=.26]
--(1.24,2.65) arc[start angle=180,end angle=0,radius=.26]
--(1.76,.5) arc[start angle=180,end angle=360,radius=.26]
--(2.28,2.65) arc[start angle=180,end angle=0,radius=.26]
--(2.8,.5) arc[start angle=180,end angle=360,radius=.26]
--(3.32,2.65) arc[start angle=180,end angle=0,radius=.26]
--(3.84,.5) arc[start angle=180,end angle=360,radius=.26]
--(4.36,3.37);
\draw[ribbonblue,line width=.85pt]
(.2,-.32)--(.2,2.65) arc[start angle=180,end angle=0,radius=.26]
--(.72,.5) arc[start angle=180,end angle=360,radius=.26]
--(1.24,2.65) arc[start angle=180,end angle=0,radius=.26]
--(1.76,.5) arc[start angle=180,end angle=360,radius=.26]
--(2.28,2.65) arc[start angle=180,end angle=0,radius=.26]
--(2.8,.5) arc[start angle=180,end angle=360,radius=.26]
--(3.32,2.65) arc[start angle=180,end angle=0,radius=.26]
--(3.84,.5) arc[start angle=180,end angle=360,radius=.26]
--(4.36,3.37);
\foreach \x in {.2,.72,1.24,1.76,2.28,2.8,3.32,3.84,4.36}
\foreach \y in {.87,1.23,1.59,1.95,2.31}
\fill[ribbonorange] (\x-.037,\y-.043) rectangle (\x+.037,\y+.043);
\draw[ribbonblue,-{Latex[length=1.7mm]}] (.2,.06)--(.2,.35);
\draw[ribbonblue,-{Latex[length=1.7mm]}] (4.36,3.06)--(4.36,3.45);
\draw[<->,thin] (.2,-.53)--(4.36,-.53)
node[midway,fill=white,inner sep=1.4pt] {$2h_0$};
\draw[<->,thin] (5.07,.5)--(5.07,2.65)
node[midway,fill=white,inner sep=1.5pt,rotate=90] {$2h_0$};
\node[anchor=west,text=ribbonorange,align=left,font=\footnotesize]
at (5.38,2.85) {disjoint\\middle\\regions};
\draw[ribbonorange,thin] (5.25,2.85)--(4.4,2.31);
\end{scope}
\node[anchor=north,font=\small,align=center] at (10.25,5.97)
{$\ell=h_0/(C_0m)\asymp m^{-1},\qquad N\asymp m^2$};
\end{scope}
\end{tikzpicture}
\endgroup
==== END An-explicit-exact-Hausdorff-gauge-for-SLE//serpentine-ribbon.tex ====
==== BEGIN An-explicit-exact-Hausdorff-gauge-for-SLE//sources.bib ====
@misc{openai2026exact,
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year = {2026}
}
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archivePrefix = {arXiv},
primaryClass = {math.GN},
doi = {10.48550/arXiv.1703.05914},
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}
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address = {Cambridge},
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isbn = {978-0-521-76018-8},
doi = {10.1017/CBO9780511750489},
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With an appendix by Oded Schramm and Wendelin Werner}
}
@article{beffara2008,
author = {Beffara, Vincent},
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volume = {36},
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pages = {1421--1452},
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arXiv version: \href{https://arxiv.org/abs/math/0211322v3}{arXiv:math/0211322v3}, 27 August 2008}
}
@incollection{schramm2007,
author = {Schramm, Oded},
title = {Conformally invariant scaling limits: an overview and a collection of problems},
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editor = {Sanz-Sol{\'e}, Marta and Soria, Javier and Varona, Juan L. and Verdera, Joan},
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arXiv version: \href{https://arxiv.org/abs/math/0602151v2}{arXiv:math/0602151v2}, 10 February 2006}
}
@article{rezaei2018,
author = {Rezaei, Mohammad A.},
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}
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}
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author = {Lawler, Gregory F. and Rezaei, Mohammad A.},
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@article{lawlersheffield2011,
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22 June 2009, titled ``The natural parametrization for the Schramm--Loewner evolution''}
}
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author = {Lawler, Gregory F. and Zhou, Wang},
title = {{SLE} curves and natural parametrization},
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number = {3A},
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Electronic reprint: \href{https://arxiv.org/abs/1006.4936v2}{arXiv:1006.4936v2},
27 May 2013; reprint pagination differs from the journal}
}
@article{rohdeschramm2005,
author = {Rohde, Steffen and Schramm, Oded},
title = {Basic properties of {SLE}},
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series = {2},
volume = {161},
number = {2},
year = {2005},
pages = {883--924},
doi = {10.4007/annals.2005.161.883},
url = {https://annals.math.princeton.edu/wp-content/uploads/annals-v161-n2-p07.pdf},
note = {\href{https://doi.org/10.4007/annals.2005.161.883}{doi:10.4007/annals.2005.161.883}. Published Theorem 5.1, p. 899; normalization in Section 2.1, pp. 886--887}
}
@misc{beliaev2015,
author = {Beliaev, Dmitry},
title = {Conformal Maps and Geometry},
year = {2015},
howpublished = {Lecture notes, University of Oxford},
url = {https://people.maths.ox.ac.uk/belyaev/TCC/lecture_notes.pdf},
note = {\href{https://people.maths.ox.ac.uk/belyaev/TCC/lecture_notes.pdf}{Author lecture notes}. Version dated June 8, 2015; Section 3.2, pp. 37--43}
}
@article{lawlerwerness2013,
author={Lawler, Gregory F. and Werness, Brent M.},
title={Multi-point {Green}'s functions for {SLE} and an estimate of {Beffara}},
journal={The Annals of Probability}, volume={41}, number={3A},
year={2013}, pages={1513--1555}, doi={10.1214/11-AOP695},
url={https://arxiv.org/abs/1011.3551v4},
note={\href{https://doi.org/10.1214/11-AOP695}{doi:10.1214/11-AOP695}. Numbered references use \href{https://arxiv.org/abs/1011.3551v4}{arXiv:1011.3551v4}}
}
@article{rezaeizhan2017,
author={Rezaei, Mohammad A. and Zhan, Dapeng},
title={Higher moments of the natural parameterization for {SLE} curves},
journal={Annales de l'Institut Henri Poincar{\'e}, Probabilit{\'e}s et Statistiques},
volume={53}, number={1}, year={2017}, pages={182--199},
doi={10.1214/15-AIHP712}, url={https://arxiv.org/abs/1412.3700v2},
note={\href{https://doi.org/10.1214/15-AIHP712}{doi:10.1214/15-AIHP712}. Numbered references use \href{https://arxiv.org/abs/1412.3700v2}{arXiv:1412.3700v2}}
}
==== END An-explicit-exact-Hausdorff-gauge-for-SLE//sources.bib ====