An explicit exact Hausdorff gauge for SLE

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. Write

The image of a compact interval under a continuous map is a locally connected continuum. This standard preservation statement is recalled in [13, Chapter 2, §2.2, p. 19]. 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 [9, §2, Lemma 2.4, p. 7].

Consequently K is locally connected. The Torhorst boundary theorem, in the form recorded in [9, §2, Lemma 2.6, p. 8], implies that \partial D is locally connected. The Continuity Theorem and Prime End Theorem in [13, Chapter 2, §2.1, p. 18] now give the asserted continuous extension and its prime-end interpretation.□

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

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 [12, proof of Theorem 1, pp. 426–427, equations (2.2)–(2.3)], 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.□

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 2.2 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.□

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 2.2. 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

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 2.2. 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

Gronwall’s exterior area theorem gives \sum_{j\ge1}j|b_j|^2\le1 [3, Theorem 3.2.1 and Corollary 3.2.2, pp. 37–38]. 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,

Invert this transition and use capacity 2t as parameter. It follows that

The compact interval of capacity and continuity of U prove the lemma.□

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

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

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.□

Proof. Lemma 2.1 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 2.5, 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.□

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

Girsanov’s Theorem therefore applies. Its added angular drift is (\kappa a_0/4)\cot\theta; together with (3.2) this gives (3.5).

For 0<b<a_0, let V_b(\theta)=\sin^{-b}\theta. Direct differentiation gives

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

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 (3.5) is

For v=\cos\theta, the generator and invariant measure become

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

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

Consequently

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,

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 [7, Lemma 2.9]; see also [8, Lemma 2.2]. A Gegenbauer expansion for this diffusion is also used in [20, 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 (3.7) 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 3.2 gives (3.9). At S_\varepsilon<\infty, (3.1) 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 (3.9) proves (3.10).

Before S_\varepsilon, R\ge\delta_t>\varepsilon by (3.1). 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 (3.11). Strong Markov gives the conditional versions with the restarted domain and its initial clearance.□

Proof. The locally unique centered SDE in (3.5) 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

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

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 3.1, 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 (3.6) 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 3.2 imply, for integers n\ge1,

For any 0<\eta<1/4, Borel–Cantelli therefore gives \sin^{-4}\theta_s\le C e^{4\eta s} eventually. Also (3.7) 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 [1, 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

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

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,

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 (3.10) applies. Here

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 (3.8), 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

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 (3.12). 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 (3.12), 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

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

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

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,

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 3.5 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, (3.13) also gives E_j\le2\pi H\le c_0H^2. Lemma 3.6 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,

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

Choose 0<\beta<1-4/\kappa and set V(P)=(1+P^2)^{\beta/2}. Direct differentiation yields

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

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

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 (3.15) show that the terminal Loewner clock satisfies

This finite clock ends at collision. Indeed Y_h=Y_0e^{-h}\to0, and the same unit-interval estimate gives

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 2.3, 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 2.1 gives the continuous spherical boundary extension of its inverse F. The radial segment supplies exactly the path assumption of Proposition 3.7, with the physical ball \overline B(z,r).

Increase L_0 so that Proposition 3.7 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 3.8 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

The Green occupation formula gives (4.4), 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

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 kth actual test write Z_k=Y_k-\mu_k. Conditional Hoeffding’s lemma gives, for \theta\ge0,

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

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

Taking \theta=\lambda a^2 gives \mathbb E e^{\lambda T_\xi}\le\exp(CK^2\lambda^2a^2). The exact averaging identity is

Jensen’s inequality, followed by Tonelli’s theorem, proves (4.10). There is no probabilistic independence assumption on the shift. Optimizing the exponential Markov bound in \lambda proves (4.11).□

Proof of Proposition 4.1 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 3.3 gives, at a finite \tau_r(z) and for one fixed 0<q\le1/(8\mathrm e),

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 3.9 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

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,

as required in Lemma 4.3. 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 3.9 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

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 4.3.

The three conditional upper means at the start, also set to zero for no start, are respectively

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 3.3. The bad event is a subset of the local-rule event, so the last bound follows from Proposition 3.9. Lattice windows have disjoint time intervals by the tip-containment argument in Lemma 4.3, including when a window ends early by killing.

For each of the three outputs apply that lemma at level

Except on an event of probability at most

the corresponding integral is at most the integral of its upper mean plus b_j. The exponent satisfies

Because q^{-2(1-p-\varepsilon)}>1, the sum of (4.14) over all j\ge0 and the three outputs is at most C\exp(-c\delta^{-2}/L_*^2). For example, the last quotient in (4.15) 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,

Define

On the simultaneous good event, integrating (4.16) and using (4.13) gives

where constants can be enlarged so that

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

while the first implies S_{j+1}\le(1+\Lambda\alpha_j)U_j+e_j. Thus

and iteration gives

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 (4.2) for 4<\kappa<8 and completing the proposition.□

Proof. In the initial half-plane experiment, Paley–Zygmund applied directly to (5.1) 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

First choose n large and then \eta small. Monotonicity in the threshold a proves (5.5). 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

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 (6.6). 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

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

Conformal invariance of Dirichlet energy and area change imply

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,

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

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

Choose J so that J\times(0,\pi) also contains \log K, using the logarithm with argument in (0,\pi). By (6.7), U_n=T_n on the fixed positive-area set \log K. Equations (6.8)–(6.9) imply

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

As in (6.8), \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

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 (6.10) follows.

Koebe’s quarter theorem, applied after mapping the disk to \mathbb H at z, gives

Therefore \log|F-a| is 2-Lipschitz for hyperbolic distance. If z=\rho Be^{i\theta} is in the sector in (6.11), then 1/8\le\rho\le1 and \sin\theta\ge1/2, so

Comparing with iB in (6.10) proves (6.11). For A\ge64, -A+\pi\sqrt{2A}+6<-\log4.□

Proof of Lemma 6.1. 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

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 6.2, 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,

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

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 (6.1).

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 6.3 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

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 2.1 gives continuous boundary extensions of the inverse maps, so the short access curve and (6.1) give

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 6.3: 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

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

Let C be the union, over j,\nu as above and rational 0<t_-<t_+<t_{\rm out}, of the following events:

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

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,

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 nth 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,

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

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 nth 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

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

Apply (6.2) 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 (6.12). 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, (6.13) 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

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

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

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

where \lambda>0 is small enough for the upper Lipschitz comparison. Choose the deterministic \delta_m>0 so that

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 6.1 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

On E_\eta, the disjoint middle strips inside the disk yield normalized sausage area at the chosen mesh \eta of at least

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 (6.13) maps the area counted in (6.16) into the physical v-sausage. Conformal area change multiplies normalized area by at least c r^d, so (6.15) 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

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 (6.14) and (6.15). 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 (7.3), with a fixed multiplicative slack in the target radius, give

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

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

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

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 (7.2), 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 (7.6) then give a conditional success probability at least

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

There is a constant b_1>0, independent of n,j,z, such that success implies

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 (7.7) 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

Put

Every preceding test is observable when the next one starts. Iterated conditional expectations give

No independence is required between tests or between centers.

Returning to the original law. Call D_j(z) a candidate if

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 (7.3), followed by (7.4), gives

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 (1.3) implies

The qualification (7.9) and disjointness therefore show that the total cost of these triples is at most

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}, (7.10) yields an expected added cost at most

Here d+p=2. The maximal diameter of the cover is at most

If Z_n(K) is its total cost, Lemma 7.1 and (7.11)–(7.12) show that

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,

Indeed every sufficiently late cover is admissible at any fixed positive diameter cutoff. Fatou’s lemma and (7.13) give

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

The sets E_q are increasing compact sets, and

Gauge Hausdorff measure is a measure on Borel sets. Its continuity from below and then monotone convergence, applied to (7.14), give

Finally, a bounded real interval is covered by O_R(r^{-1}) intervals of length r, whose total cost is

because d>1. The real-axis part therefore has zero \mathcal H^h-measure. This proves (7.5).

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 1.1. Section 5.4 gives positivity simultaneously on all real 0<s<t<\infty for the fixed parameter \kappa. Proposition 7.2 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.□