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. ◻

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 (5) this gives (8).

For \(0<b<a_0\), let \(V_b(\theta)=\sin^{-b}\theta\). Direct differentiation gives \[ \mathcal L^*V_b =V_b\left\{\frac{\kappa b}{8} -\frac{\kappa b(a_0-b)}8\cot^2\theta\right\} \le\frac{\kappa b}{8}V_b. \tag{9}\] 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 \[ \frac{d\log Y_s}{ds}=-\frac1{2\sin^2\theta_s}, \qquad \frac{dt}{ds}=\frac{Y_s^2}{4\sin^4\theta_s}. \tag{10}\] 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. ◻

Proof. We give a short spectral proof which also avoids any ambiguity about comparison inequalities for unbounded test functions. The invariant probability measure for (8) 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. Itô’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 (Lawler and Werness 2013, Lemma 2.9); see also (Lawler and Zhou 2013, Lemma 2.2). A Gegenbauer expansion for this diffusion is also used in (Zhan 2019, Proposition 4.2). Here the interpolation estimate suffices in place of explicit polynomial norm formulas. ◻

Proof. Suppose the radius-time lifetime is finite. If the angle eventually stayed in a compact subinterval of \((0,\pi)\), its SDE and (10) 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 10 gives (12). At \(S_\varepsilon<\infty\), (4) 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 (12) proves (13).

Before \(S_\varepsilon\), \(R\ge\delta_t>\varepsilon\) by (4). 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 (14). Strong Markov gives the conditional versions with the restarted domain and its initial clearance. ◻

Proof. The locally unique centered SDE in (8) 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 9, 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 (9) 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 10 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 (10) 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. ◻

Proof. A stronger, sharp boundary exponent was proved by Alberts and Kozdron (Alberts and Kozdron 2008, Theorem 3.2 and Corollary 5.1). 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 (13) 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 (11), the live hit gives \(S_{2\varepsilon}(v)<\infty\). The approach estimate therefore proves the result for all \(v\). ◻

Proof. We first record a two-dimensional average estimate. If \(Q_k\) is any nested chain of dyadic subrectangles with \(Q_0=Q\), then \[ |u_{Q_k}-u_Q|\le C\sqrt{(k+1)E}. \tag{15}\] 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 Poincaré inequality. These unions have bounded overlap. Cauchy–Schwarz over the \(k+O(1)\) annuli gives (15). Poincaré 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 (15), 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 \(H/4\). Poincaré 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. ◻

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 \[ \int_{\mathbb R\times(0,\pi)}|\nabla U|^2 \le\int_{r<|w-q|<Lr}\frac{dA(w)}{|w-q|^2} =2\pi H. \tag{16}\] For every \(v<0\), the path assumption supplies a point on that vertical section where \(U=0\). One-dimensional Poincaré 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 13 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, (16) also gives \(E_j\le2\pi H\le c_0H^2\). Lemma 14 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. ◻

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 Itô followed by Gronwall gives \[ \mathbb EV(P_h)\le Ce^{\lambda h}V(P_0). \tag{17}\] 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 \[ \mathbb E\left[\sup_{n\le h\le n+1}V(P_h)\right] \le Ce^{\lambda n}V(P_0). \tag{18}\] 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 (18) 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 \(A\). ◻

Proof. The first hit is a fresh interior tip. By Lemma 5, 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 3 gives the continuous spherical boundary extension of its inverse \(F\). The radial segment supplies exactly the path assumption of Proposition 15, with the physical ball \(\overline B(z,r)\).

Increase \(L_0\) so that Proposition 15 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 16 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. ◻

Proof. Figure 1 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 (22), 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. ◻

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 (28). There is no probabilistic independence assumption on the shift. Optimizing the exponential Markov bound in \(\lambda\) proves (29). ◻

Proof of Proposition 18 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 11 gives, at a finite \(\tau_r(z)\) and for one fixed \(0<q\le1/(8\mathrm e)\), \[ \mathbb P\{I_{qr}(z)=1\mid\mathcal F_{\tau_r(z)}\} \le C(qr)^pW_r(z). \tag{30}\] 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 17 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 20. 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 17 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 20.

The three conditional upper means at the start, also set to zero for no start, are respectively \[ \begin{split} \mu^H_j(z)&=C(qr_j)^pW_{r_j}(z),\\ \mu^V_j(z)&=r_j^pW_{r_j}(z),\\ \mu^B_j(z)&=C_*L_j^{-c_*}I_{r_j}(z). \end{split} \tag{31}\] 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 11. The bad event is a subset of the local-rule event, so the last bound follows from Proposition 17. Lattice windows have disjoint time intervals by the tip-containment argument in Lemma 20, 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 \[ C\exp\left(-\frac{c b_j^2}{(L_jr_j)^2}\right), \tag{32}\] the corresponding integral is at most the integral of its upper mean plus \(b_j\). The exponent satisfies \[ \frac{b_j^2}{(L_jr_j)^2} =\frac{\delta^{-2}}{L_*^2} \left(\frac{r_0}{\delta}\right)^{-2+2p} \frac{q^{-2(1-p-\varepsilon)j}}{(j+1)^4} \ge \frac{\delta^{-2}}{L_*^2} \frac{q^{-2(1-p-\varepsilon)j}}{(j+1)^4}. \tag{33}\] Because \(q^{-2(1-p-\varepsilon)}>1\), the sum of (32) over all \(j\ge0\) and the three outputs is at most \(C\exp(-c\delta^{-2}/L_*^2)\). For example, the last quotient in (33) 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, \[ \begin{split} I_{r_{j+1}}&\le H_j+B_j,\\ W_{r_{j+1}}&\le r_j^{-p}V_j+C(qr_j)^{-p}B_j. \end{split} \tag{34}\] 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 (34) and using (31) gives \[ \begin{split} S_{j+1}&\le S_j+\alpha_j A_j+e_j,\\ A_{j+1}&\le C_1 S_j+\alpha_j A_j+e_j, \end{split} \tag{35}\] 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 (20) for \(4<\kappa<8\) and completing the proposition. ◻

Proof. In the initial half-plane experiment, Paley–Zygmund applied directly to (36) 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 (40). For a fixed finite list, apply this result to each interval and take the minimum threshold and cutoff, followed by a union bound. ◻

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 \(F\). Arzelà–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 (46). 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 \[ |F_n(z)-a_n|\ge r_0\qquad(z\in K) \tag{47}\] 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 \[ E_n:=\int_S|\nabla U_n|^2\,dv\,d\phi \le\int_{Ch_n<|w-a_n|<r_0}\frac{dA(w)}{|w-a_n|^2} =2\pi T_n. \tag{48}\] 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 Poincaré inequality for \(U_n(v,\cdot)-m_n(v)\) then gives \[ \int_{J\times(0,\pi)}U_n^2\le C_{J,\varepsilon}E_n. \tag{49}\] Choose \(J\) so that \(J\times(0,\pi)\) also contains \(\log K\), using the logarithm with argument in \((0,\pi)\). By (47), \(U_n=T_n\) on the fixed positive-area set \(\log K\). Equations (48)–(49) 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 \(\partial F(\mathbb H)\). ◻

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 (48), \(\int_S|\nabla U|^2\le2\pi A\). 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 (50) 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 (51), 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 (50) proves (51). For \(A\ge64\), \(-A+\pi\sqrt{2A}+6<-\log4\). ◻

Proof of Lemma 22. 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 23, 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 (41).

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 24 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 3 gives continuous boundary extensions of the inverse maps, so the short access curve and (41) 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 24: 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 \(n\), \[ \mathbb P(E\cap G_n\cap C) =\mathbb E[\mathbf1_{E\cap G_n}q] \ge q_1q_0/4=:q_2>0. \tag{52}\] 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 \[ v_n^{-p}\operatorname{Area}(\hbox{physical sausage in the middle}) \ge c_*^2\lambda^p (\lambda v_n)^{-p}\operatorname{Area} (N_{\lambda v_n}(\gamma(J))). \tag{53}\]

Apply (42) 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 (52). 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, (53) 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. ◻

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 3 shows the displaced disk and the serpentine array within it.

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 22 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 \[ c m^2\ell^d\ge c m^{2-d}=c m^p. \tag{56}\] 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 (53) maps the area counted in (56) into the physical \(v\)-sausage. Conformal area change multiplies normalized area by at least \(c r^d\), so (55) 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 (54) and (55). All tests concern the already generated sausage at finite stops; no limiting marked-point visit is used. ◻

Proof. Choose a fixed \(A>8e\). If \(y=\operatorname{Im}w>Av\), the visit geometry and (59), 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. ◻

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.

One deterministic mesh. Choose a deterministic sequence \(v_n>0\) such that \(v_n\to0\) and \[ v_n\le\min_{J_0\le j\le J_1} \varepsilon(e^{-1}r_j,m_n). \tag{62}\] 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 (58), 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 (62) 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 \[ \mu_{S_{j+1}(z)}^{v_n}(D_j(z)) \ge b_1 r_j^d(\log n)^{p/2}. \tag{63}\] 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 (63) 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 \[ \mathbb P_z^*(F_n(z)) \le(1-n^{-1/2})^{J_1-J_0+1} \le C\exp(-c\sqrt n). \tag{64}\] No independence is required between tests or between centers.

Returning to the original law. Call \(D_j(z)\) a candidate if \[ \mu^{(n)}(D_j(z))\ge b_1r_j^d(\log n)^{p/2}. \tag{65}\] 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 (59), followed by (60), gives \[\begin{align*} &\mathbb P\{\text{the cell at }z\text{ is visited and has no candidate}\} \\ &\quad\le C r_n^p\, \mathbb E\!\left[ {\bf1}_{\{S<\infty\}}M_S(z){\bf1}_{F_n(z)}\right] \\ &\quad= C r_n^p M_0(z)\mathbb P_z^*(F_n(z)) \le C r_n^p M_0(z)e^{-c\sqrt n}. \tag{66}\end{align*}\] 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.

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 (3) implies \[h(6C_3r_j)\le C r_j^d(\log n)^{p/2}.\] The qualification (65) and disjointness therefore show that the total cost of these triples is at most \[ C\sum_{\text{selected }D}\mu^{(n)}(D) \le C\mu^{(n)}(B(0,R+1)). \tag{67}\] 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 \(h(\sqrt2r_n)\le Cr_n^d(\log n)^{p/2}\), (66) yields an expected added cost at most \[ C_Kr_n^{-2}r_n^d(\log n)^{p/2}r_n^p e^{-c\sqrt n} =C_K(\log n)^{p/2}e^{-c\sqrt n}\longrightarrow0. \tag{68}\] 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 26 and (67)–(68) show that \[ \limsup_{n\to\infty}\mathbb E Z_n(K)\le C_R. \tag{69}\] 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 \(K\).

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 (69) give \[ \mathbb E\mathcal H^h(\gamma([0,T])\cap K)\le C_R, \tag{70}\] uniformly in both \(T\) and \(K\subset\mathbb H\cap\overline B(0,R)\).

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 (70), 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 (61).

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. ◻

Proof of Theorem 1. Section 5.4 gives positivity simultaneously on all real \(0<s<t<\infty\) for the fixed parameter \(\kappa\). Proposition 27 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. ◻