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Work in the prime end closure of the marked domain, so that its two conformal boundary points are distinct even if their physical boundary representatives coincide. A chordal filling is a closed connected set joining those points, meeting the conformal boundary nowhere else, and full relative to the two boundary arcs between them: complementary pockets that do not reach either boundary arc have been added. In the normalized complex upper half-plane, its two complementary sides are simply connected and abut the negative and positive real axes. For a proper boundary-to-boundary curve, chordal filling means adjoining all such trapped pockets to its range. Independent curves are combined by taking their union and then applying this same chordal filling operation.

An admissible marked subdomain $V\subset U$ contains neighborhoods of the two marked boundary points and has a simply connected component joining them. The restriction property says that, conditional on the random chordal filling $K$ being contained in $V$, its law is the chordal filling law for $(V,z_0,z_1)$. Equivalently, mapping the conditional chordal filling conformally back to $(U,z_0,z_1)$ recovers the original law. The contained-set events are understood away from the shared marked endpoints.

By transporting the family under conformal maps, it suffices to work in $(\mathbb{H},0,\infty )$. Let $A$ be an admissible compact H-hull whose closure avoids zero, and normalizeIn terms of the single normalized measure $\mu$, restriction is preciselyThe Schwarz reflection principle extends ${\Phi }_A$ analytically near zero, with ${\Phi }_A^{\prime }(0)>0$, so the stated conditioning events have positive probability for the two laws considered below. This is the chordal restriction property for filled sets. An unbounded marked subdomain can be approximated by such compact-hull removals; conformal maps transport the formulation to general domains.

We need a determination principle for filled sets, since a probability of one avoidance event alone would not establish equality of laws. Choose countably many thin polygonal compact H-hulls with rational data, attached to real boundary intervals away from zero. For each chordal filling, every point of its complement has a path within one of the two complementary sides to the appropriate real boundary interval. A sufficiently thin tube along such a path avoids the chordal filling and contains a neighborhood of that point. Hence the union of interiors of all avoided test tubes is exactly the complement.

Their joint avoidance probabilities are also determined by single-hull probabilities. For a finite union of test compact H-hulls, either the union separates zero from infinity, in which case a connected chordal filling cannot avoid it, or its unbounded complementary component contains both marked endpoints. In the latter case fill the union's bounded pockets to get an admissible compact H-hull $C$. A connected set joining zero to infinity and avoiding the union cannot enter one of these separated pockets, so it avoids $C$ as well. Thus joint avoidance is either impossible or is the avoidance event for $C$. The inclusion-exclusion principle gives every finite joint probability distribution of the tube-avoidance indicators, and these indicators determine the random complement and chordal filling. This proves avoidance probabilities determine a chordal filling law.

In the given avoidance formulas, the argument of the conformal map must be the retained open domain, rather than the random closed filling. We express that domain as $\mathbb{H}\setminus A$ to keep the two objects distinct. The formulas then imply for the two filled processesThe chordal filling operation does not change avoidance of an admissible compact H-hull: every piece of that compact H-hull is attached to the real boundary away from the endpoints, and so cannot be inside a trapped pocket while its generating curve avoids the compact H-hull. These laws are conformally invariant as unparameterized chordal filling laws. For SLE this follows from Question 1; for the Brownian half-plane excursion, Brownian scaling of its independent one- and three-dimensional coordinates shows that its unparameterized law is unchanged by positive dilations, and mapping the Brownian half-plane excursion to another marked domain gives the conformally natural law. Conformal invariance of planar Brownian motion describes the corresponding change of time, which does not alter the chordal filling.

Here is the conditional-law calculation for either exponent $\alpha \in \{5/8,1\}$. For another admissible compact H-hull $B$ in the image complex upper half-plane, setThe complement of $C$ is mapped to $\mathbb{H}$ by this composition, so $C$ is again admissible; both maps fix zero and have derivative one at infinity. The chain rule gives ${\Phi }_C^{\prime }(0)={\Phi }_B^{\prime }(0){\Phi }_A^{\prime }(0)$. ThereforeThese are exactly the original chordal filling's avoidance probabilities for every $B$. The determination principle proves equality of the conditional image law with the original law. Hence both generated chordal fillings have restriction, with exponents $5/8$ and $1$ respectively. This is a complete implication from the two allowed avoidance formulas; the restriction conclusion was not assumed.

For independent chordal fillings, avoidance of $A$ by their filled union is equivalent to avoidance by every component. Independence multiplies the probabilities, and thus adds their restriction exponents of a chordal filling. ConsequentlyThe determination principle applies once more, givingThe equality concerns the filled, unparameterized closed sets. The numerical reason is $8\cdot (5/8)=5\cdot 1=5$.

Start the planar Brownian motion $X_t+i{Y}_t$ at $-x$ and stop at $\tau =\inf \{t:X_t=0\}$. The complex exponential is holomorphic with nonzero derivative. By conformal invariance of planar Brownian motion, the process $e^{X_t+i{Y}_t}$, run on its conformal Brownian clockis planar Brownian motion started at $\epsilon =e^{-x}$, until its first exit from the unit disk. The clock is strictly increasing before $\tau$, and $A_{\tau }\le \tau <\infty$ almost surely. It is legitimate to use the exponential although it is not globally injective: the Brownian coordinate brackets depend locally on the nonzero derivative, and the clock construction gives the stopped planar Brownian law. Before $\tau$, $|e^{X_t+i{Y}_t}|<1$; at $\tau$ it equals one. Hence the disk exit time in the new clock is exactly $A_{\tau }$.

The continuous lifted argument of this path, starting from angle zero, is $Y_t$. Its total argument at exit is therefore $Y_{\tau }$, which has distribution $xC$ by part (b). If the winding count records only completed integer turns, it differs from $Y_{\tau }/(2\pi )$ by a remainder $R_{\epsilon }$ with $|R_{\epsilon }|\le 1$. If fractional turns are retained, that remainder is zero. In either convention, since $\log \epsilon =-x$,The first term has distribution $-C/(2\pi )$ for every $x$, and the second tends to zero deterministically in absolute value as $x\to \infty$. Symmetry of the Cauchy distribution removes the minus sign, so Slutsky theorem gives the small-radius Brownian winding law:

We use the following standard planar Brownian motion facts: its exit probability distribution is harmonic measure; conformal maps preserve that probability distribution after an increasing time change; and the Poisson kernel for the upper half-plane gives its real-boundary exit density. For rough hulls the boundary is interpreted through prime ends, with boundary correspondence at harmonic-measure-almost every exit point. This avoids imposing a smoothness or local-connectedness assumption on $K$.

Under $g_K$, the exit event through $K$ corresponds to a measurable set $E_K\subset \mathbb{R}$ of boundary images. It is bounded: the mapping-out function extends by Schwarz reflection along the real line outside a sufficiently large interval enclosing $K$, and its boundary near infinity corresponds to the original real boundary near infinity. All prime ends representing $K$ therefore have images in a bounded interval. The boundary correspondence assertion can equivalently be formulated using the almost-everywhere limits of $g_K^{-1}$; changing null sets does not change $E_K$'s measure.

Write $g_K(iy)=u_y+i{v}_y$. The hydrodynamic normalization at infinity gives $u_y=O(y^{-2})$ and $v_y=y-a_K/y+O(y^{-2})$. By conformal invariance of planar Brownian motion and the Poisson kernel for the upper half-plane,On the bounded set $E_K$, the integrand multiplied by $y$ tends uniformly to one. Therefore the limit exists and is finite:This is the harmonic capacity from infinity in the upper half-plane with the normalization $c(K)=\mathrm{cap}(K)/\pi$; it is different from half-plane capacity. If real boundary points are retained in the definition of $K$, include their boundary images in $E_K$ as well; the argument is unchanged.

For the scaling statement, Brownian scaling says that $r^{-1}{B}_{r^{2}t}$ started from $iy/r$ is again standard planar Brownian motion. The exit event scales with the hull, soSet $v=y/r$, multiply by $y=rv$ and take the limit. It follows thatThe same conclusion follows geometrically from $E_{rK}=r{E}_K$ and the formula above.

The conformal invariance of planar Brownian motion is a time-change statement. If $\varphi :D\to D^{\prime }$ is a conformal bijection of planar domains and $B$ is planar Brownian motion started in $D$, let ${\tau }_D$ be its exit time and defineBefore exit this clock is strictly increasing. Its inverse $C_u$ makes $\varphi (B_{C_u})$ a planar Brownian motion started at $\varphi (B_0)$, up to its transformed lifetime $A_{{\tau }_D}$. The image of the stopped trajectory is therefore invariant in law after this change of time. A holomorphic map with nonzero derivative gives the corresponding local statement even without global injectivity.

Here is the proof. Write $\varphi =u+iv$ and $B=(B^1,B^2)$. The real and imaginary parts are harmonic. The Itô formula therefore gives the two continuous local martingalesstopped on compact subsets of $D$. The Cauchy-Riemann equations implyThus both coordinate quadratic variations equal $A_t$ and their quadratic covariation is zero. After inverse time change, the covariance matrix of the two continuous local martingales is $u{I}_2$. The multidimensional Lévy characterization of multidimensional Brownian motion states that a continuous vector local martingale starting at zero with this covariance matrix is standard vector Brownian motion. Apply it after subtracting $\varphi (B_0)$, and exhaust $D$ by compact subsets. This proves the theorem, with killing at the transformed exit lifetime. The clock is the conformal Brownian clock; omitting it would generally give an incorrect speed.

The exponential map is a conformal bijection from the stripto the wedge containing the starting point $1$. The restriction $\alpha +\beta <2\pi$ makes it injective. A planar Brownian motion $U+iV$ started at zero in this strip maps, after the conformal Brownian clock, to the required Brownian motion in the wedge.

Let $\sigma$ be the first time $V$ exits $(-\alpha ,\beta )$. It is finite almost surely, and $V_{t\wedge \sigma }$ is a bounded martingale. The optional stopping theorem givesThe strip exit maps to the corresponding bounding ray. The exponential clock is finite at $\sigma$, because its integrand is continuous on the compact time interval $[0,\sigma ]$. Thus it does not remove or reverse the exit event. We obtainThis is the Brownian wedge-exit probability obtained directly from the strip, using conformal invariance of planar Brownian motion.

Here the map direction is opposite to Question 2: write $\varphi :U\to D$, with $\varphi (0)=0$ and $\varphi (1)=1$. The separation condition makes $U$ agree with $D$ near $1$. Schwarz reflection across that analytic boundary arc therefore defines the boundary derivative, whose modulus is positive.

For ordinary Brownian motion $W$ started at $0$, let ${\sigma }_U$ be its first exit from $U$. Take sufficiently short arcs $I_{ϵ}$ around $1$. They are also arcs of $\partial U$. The event that $W$ stays in $U$ until its disc exit and exits in $I_{ϵ}$ is exactly the event $W_{{\sigma }_U}\in I_{ϵ}$. Therefore the conditional avoidance probability isConformal invariance of planar Brownian motion gives ${\omega }_U(0,I_{ϵ})={\omega }_D(0,\varphi (I_{ϵ}))$. At the disc center harmonic measure is normalized arc length, so the ratio tends to $|{\varphi }^{\prime }(1)|$.

To justify that this limit is the avoidance probability for the point-conditioned diffusion, one may use the stopped transform directly. Its probability of reaching a smooth boundary arc at $1$ before any other boundary of $U$ is the mass at that pole in the transformed harmonic measure. All other exit points have weights $h(w)$ and represent failure. Equivalently integrate the Poisson-kernel density at $1$ for $U$ and divide by that for $D$. This gives the boundary-density ratio above, without assuming that a full-path avoidance event is a continuity set for arbitrary weak convergence. The kernels transform by the boundary Jacobian:ConsequentlyThere is no extra factor of $|{\Phi }_U^{\prime }(0)|$. The starting point is fixed and the conditioning concerns boundary harmonic-measure density. In this normalization the boundary derivative is a positive real, so it may also be written ${\Phi }_U^{\prime }(1)$. It is at most one by the probability interpretation. Given avoidance, the mapped path is the same conditioned Brownian motion up to conformal time change; this is radial restriction for a conditioned Brownian path.

Use the intrinsic boundary of a simply connected domain, so different approaches to the two sides of a slit remain distinct. The mapping-out function $g_K$ extends as a homeomorphism from this intrinsic compactification to that of the complex upper half-plane. Set $T=T(H)$. The imaginary coordinate of planar Brownian motion hits zero in finite time almost surely, and $T$ is no larger than that time. Thus $T<\infty$ and continuity gives a finite Euclidean exit point.

By conformal invariance of planar Brownian motion, $g_K(B_t)$ is a planar Brownian motion in the complex upper half-plane, run with clockThe terminal clock cannot be infinite: that would make the transformed Brownian motion stay in the upper half-plane forever. It cannot stop while the transformed path is in the interior either, since continuity of $g_K^{-1}$ would then put the original exit point inside $H$. Hence the terminal clock is precisely the transformed Brownian exit time. The transformed path converges to a real boundary point, and applying the extended inverse proves almost sure convergence to a point of the intrinsic boundary.

Write $g_K(x+iy)=u+iv$, and let $E=g_K(S)\cap \mathbb{R}$. The point at infinity has zero harmonic measure. Conformal invariance of planar Brownian motion and the Poisson kernel for the upper half-plane giveThe hydrodynamic normalization at infinity gives $g_K(z)=z+O(1/z)$, so along the specified approach $v/y\to 1$ and $u/y\to 0$. For each fixed real $t$,If $\mathrm{Leb}(E)<\infty$, dominated convergence applies because $y/v$ is eventually bounded. If $\mathrm{Leb}(E)=\infty$, Fatou's lemma makes the limit infinite. This proves the harmonic-measure asymptotic at infinitywith the equality understood in the extended nonnegative reals.

For a simple Loewner trace, the domain mapped by $g_t$ is $D_t=\mathbb{H}\setminus \gamma [0,t]$. For a non-simple Loewner trace, use instead the unbounded component $D_t=\mathbb{H}\setminus K_t$; the map $g_t$ is not defined on swallowed bounded components. This is the necessary domain interpretation when $\kappa >4$.

The function $\mathrm{Im}\log (g_t(z)-{\xi }_t)$ is harmonic in $D_t$ because the logarithm is holomorphic on the complex upper half-plane. Let $S_t^{-}$ and $S_t^{+}$ be the intrinsic boundary sets mapped to $(-\infty ,{\xi }_t)$ and $({\xi }_t,\infty )$ respectively. The Dirichlet problem has boundary dataFor a simple Loewner trace these are the left bank of the Loewner trace together with the negative real boundary, and the right bank together with the positive real boundary. Left and right refer to the orientation from the starting point towards the tip. The tip and infinity correspond to discontinuities of the data; no unique limit is imposed there. They have zero harmonic measure.

More precisely, the bounded solution isby the Poisson kernel for the upper half-plane and conformal invariance of planar Brownian motion. This establishes the boundary values in the intrinsic sense and uniqueness among bounded solutions of the Dirichlet problem, without treating the two banks as one Euclidean boundary point.

Set $\beta =\pi /\alpha$ and use the branch $\varphi (z)=z^{\beta }$ with argument in $(-\alpha /2,\alpha /2)$. This is a conformal map from the wedge to the right half-plane, fixes the starting point $1$, and sends the outer circle of radius $r$ to that of radiusThe two wedge sides map to the imaginary axis.

By conformal invariance of planar Brownian motion, the image of the stopped path is planar Brownian motion after the increasing conformal Brownian clock $A(t)={\int }_0^t|{\varphi }^{\prime }(B_s){|}^{2}ds$. This clock does not change which boundary portion is reached first. Localization away from the vertex justifies the map even when its derivative is unbounded there; the vertex is a polar point for planar Brownian motion and has zero hitting probability from $1$.

Consequentlywhere the probability on the right is for the right half-plane. The power-map reduction for Brownian exit from a wedge also works at $\alpha =2\pi$, when the wedge is the plane slit along the negative real axis.

Scale the disk to the unit disc, so that the starting point is $a=1/r\in (0,1)$. Use the Möbius transformationwhich maps the disk onto itself, sends $a$ to $0$, and sends $1$ to $1$. By conformal invariance of planar Brownian motion, the exit image is the circular exit of a Brownian motion starting at $0$. Rotational invariance of planar Brownian motion makes that exit uniform in angle.

The endpoints of the right semicircle satisfyIts image is the arc through $1$ between these points, of angular length $2\phi$. Thus the Möbius calculation of circular Brownian exit givesSince $E_{+}$ and $E_{-}$ partition the exit almost surely, part (b) yieldsThe last equality uses $2\arctan (1/r)\in (0,\pi /2)$, so the double-angle tangent identity uses the stated principal branch. As a check, the boundary angle derivative of ${\psi }_a$ is $(1-a^2)/|e^{i\theta }-a{|}^2$; integrating this circular exit density over the right semicircle gives exactly the integral supplied in the question.

Use $U_t=2W_t$ and write $Z_t=X_t+i{Y}_t$ up to its interior-point swallowing time for a Loewner chain $T_z$. The Chordal Loewner equation givesThe branch of the complex logarithm with argument in $(0,\pi )$ is well-defined while $t<T_z$. The Itô formula yields the crucial cancellationThe $2/Z_t^2$ drift is cancelled by the quadratic variation term, because $\kappa =4$. Thus both $L_t=\log |Z_t|$ and ${\theta }_t=\arg Z_t$ are continuous local martingales, withThis is the logarithmic martingale for SLE4.

We justify absence of a finite swallowing time rather than presuming that the logarithm survives forever. On a finite horizon $H$, $0<Y_t\le \mathrm{Im}z$. Also $|X_t|$ is bounded pathwise before $T_z\wedge H$. Indeed, while $|X_t|\ge 1$ its drift $2X_t/|Z_t{|}^2$ has absolute value at most $2$; on each excursion outside $[-1,1]$, integrate from its starting point and bound the Brownian oscillation on $[0,H]$. For example,Therefore $L_t$ is bounded above pathwise on this interval.

Suppose $T_z\le H$. By the Dambis-Dubins-Schwarz theorem, $L$ is a Brownian motion run at its own quadratic variation. If that clock diverged as $t\uparrow T_z$, Brownian oscillation would make $L$ unbounded above, contradicting the preceding bound. The one-sided bound criterion for a martingale clock therefore gives a finite clock limit and a finite real limit for $L_t$. Hence $|Z_t|$ is bounded away from zero near $T_z$.

Now $Y_t=Y_0\exp (-2{\int }_0^t|Z_s{|}^{-2}ds)$ has a strictly positive limit at $T_z$. The drift in $X$ is integrable there, so continuity of $W$ gives a finite limit for $X$ as well. The limiting point is in $\mathbb{H}$ and away from the Loewner driver singularity, and the differential equation extends past $T_z$, a contradiction. Thus $T_z=\infty$ almost surely. A point of the Loewner trace at a finite time belongs to that time's hull, soThis proves the fixed-interior-point avoidance of SLE4 without using simplicity as an input.

The angle remains in $(0,\pi )$ at all finite times. The bounded local martingale criterion upgrades its local martingale equation to a genuine martingale:In particular this is the SLE4 angle martingale, and it converges almost surely and in $L^1$ by bounded martingale convergence theorem.

It remains to identify the limiting angle using the assumed simple path tending to infinity. Orient that path from $0$ to infinity. Its left component is the one adjacent to the negative real half-axis. Under $g_t$, the left boundary of the slit domain maps to $(-\infty ,U_t)$ and the right boundary to $(U_t,\infty )$. The harmonic measure of the former as seen from $z$ isThis follows by conformal invariance of planar Brownian motion: in the upper half-plane, $\arg (w-U_t)/\pi$ is the bounded harmonic function with values $1$ on the left half-axis and $0$ on the right.

To justify the limiting boundary classification, condition on a simple proper realization of the path and use an independent planar Brownian motion from $z$. It exits the upper half-plane in finite time almost surely, so its path up to that time is compact. The curve tends to infinity, so its intersection with this compact set is contained in a finite initial curve segment. Once that segment has been drawn, the Brownian path exits the slit domain through its left boundary exactly when $z$ is in the final left component: a path from that component cannot reach the right boundary without crossing the curve, and the reverse assertion holds on the right. Endpoints have zero harmonic measure. Bounded convergence of these exit indicators provesTaking expectations in the bounded angle martingale gives the SLE4 left-passage probabilityFor the imaginary axis it is $1/2$; near the negative real axis it tends to $1$, fixing the orientation of “left”.

Interpret $f$ as a conformal bijection onto $D^{\prime }$. If it is merely an injective map into a larger target, the correct exit domain is $f(D)$ instead. Let $\zeta$ be the exit time from $D$, write $f=u+iv$, and identify planar Brownian motion with $B^1+i{B}^2$. The Cauchy-Riemann equations give $u_x=v_y$, $u_y=-v_x$, and both $u,v$ are harmonic functions. The Itô formula, localized on compact subsets of $D$, yieldsThus both coordinates are continuous local martingales, withSince $f^{\prime }$ never vanishes, $A$ is strictly increasing before $\zeta$. Let ${\eta }_s=\inf \{t:A_t>s\}$, defined for $s<A_{\zeta -}$. The optional time-change theorem gives continuous local martingale coordinates for ${\beta }_s=f(B_{{\eta }_s})$, with bracket matrix $sI$. The Lévy characterization of multidimensional Brownian motion makes $\beta$ a standard planar Brownian motion started at $f(z)$, up to its lifetime. Therefore conformal invariance of planar Brownian motion takes the formTo identify the lifetime, exhaust $D$ by relatively compact open sets $D_n$ with nested closures. The images $f(D_n)$ exhaust $D^{\prime }$ because $f$ is a homeomorphism onto $D^{\prime }$. Each stopped transformed path exits $f(D_n)$ at precisely the clock value of its exit from $D_n$. Increasing these exit times gives the exit lifetime of Brownian motion from $D^{\prime }$, including a possibly infinite lifetime. This identifies $A_{\zeta -}$ with that lifetime and needs no global extension of $f$ to the boundary.

Use the open strip $S=\{x+iy:0<y<\pi \}$; the printed closed strip denotes the region together with its exit boundary. Starting at $z=x+iy$, the conformal bijection $e^z$ sends $S$ onto the complex upper half-plane and the bottom edge $a\in \mathbb{R}$ onto $v=e^a>0$. The upper edge maps to the negative half-axis. The transformed initial point is $e^x\cos y+i{e}^x\sin y$.

By conformal invariance of planar Brownian motion, transform the given Poisson kernel for the upper half-plane and include the Jacobian determinant $dv/da=e^a$. The bottom-boundary harmonic measure of a strip isThis is the unconditional exit density on the bottom edge, not the density conditional on using that edge. Its total mass isIndeed substitution $v=e^{a-x}$ gives the positive-half-axis mass of the Poisson kernel, equal to $\frac{1}{2}+{\pi }^{-1}\arctan (\cot y)=1-y/\pi$ for $0<y<\pi$. To obtain the conditional bottom-edge density, divide by $1-y/\pi$. The top-edge density is $\sin y/[2\pi (\cosh (a-x)+\cos y)]$ and has mass $y/\pi$, a useful normalization check.

Use the Green function of killed planar Brownian motion, normalized as the density of expected occupation with respect to area:for nonnegative measurable $f$. Equivalently, $G_D(z,w)={\int }_0^{\infty }{p}_D(t,z,w)dt$, where $p_D$ is the killed Brownian transition density. With generator $\frac{1}{2}\Delta$, the distributional normalization is $-\frac{1}{2}{\Delta }_w{G}_D(z,w)={\delta }_z(w)$, and the singularity is $-{\pi }^{-1}\log |w-z|$ plus a locally harmonic function. This fixes the normalization of the Dirichlet Green function explicitly.

By conformal invariance of planar Brownian motion and its conformal Brownian clock,The Jacobian determinant of a conformal map is $|{\varphi }^{\prime }(w){|}^2$. Changing the area variable to $w^{\prime }=\varphi (w)$ givesUniqueness of the occupation density proves the desired equality almost everywhere. Both functions are continuous and harmonic away from their pole, so it holds at every $w\ne z$. The Green function is conformally invariant:If the Dirichlet Green function is instead normalized for $-\Delta$, both kernels are divided by two and the invariance statement is unchanged.

Set $g_K(iy)=u_y+i{v}_y$. The hydrodynamic normalization at infinity givesThe real interval under consideration lies outside the hull, so the reflected boundary map sends it monotonically to $(g_K(x),g_K(b))$. Conformal invariance of planar Brownian motion transports its harmonic measure to the half-plane. By the Poisson kernel for the upper half-plane,The endpoints are fixed and finite. On this interval,uniformly, using $v_y/y\to 1$ and $u_y\to 0$. Integrating gives the interval-length limit:

The implication is false. Take the thin rectanglesEach is a compact H-hull, and $\mathrm{diam}(A_{\epsilon })=\sqrt{4+{\epsilon }^2}\to 2$. We verify that its half-plane capacity nevertheless tends to zero.

Let $K=\{z\in \mathbb{H}:|z|\le 2\}$, which contains all these rectangles. At the Brownian exit time $\tau$ from $\mathbb{H}\setminus A_{\epsilon }$, the exit height is at most $\epsilon$, and a positive exit height requires hitting $K$ before $\mathbb{R}$. ThereforeFor $y>2$, conformal invariance of planar Brownian motion and $g_K(z)=z+4/z$ identify the probability on the right with the harmonic measure of $[-4,4]$ from $i(y-4/y)$ in $\mathbb{H}$. Integrating the Poisson kernel for the upper half-plane givesThe Brownian representation of half-plane capacity now proves the explicit estimateChoose $\epsilon =1/n$ to obtain the required sequence. This is an instance of half-plane capacity of a low rectangle: shrinking height can make half-plane capacity small while horizontal extent stays fixed.

Let $B$ be planar Brownian motion started at $z\in D$, let ${\tau }_D$ be its exit time, and let $\varphi :D\to D^{\prime }$ be conformal. DefineThe conformal invariance of planar Brownian motion states thatis Brownian motion started at $\varphi (z)$ and stopped when it exits $D^{\prime }$. Thus conformal maps preserve Brownian paths after this random time-change.

Put $F(z)=z+z^{-1}$. On the unit semicircle, $F(e^{i\theta })=2\cos \theta$, andThe conformal invariance of planar Brownian motion and the Poisson kernel for the upper half-plane therefore give the exit density with respect to $d\theta$:As $z\to \infty$ in $\mathbb{H}$,uniformly in $\theta \in [0,\pi ]$. Hence

The map $z\mapsto z^{\pi /\alpha }$ takes a wedge of opening $\alpha$ to a half-plane and the circular boundary of radius $r$ to radius $r^{\pi /\alpha }$. Conformal invariance of planar Brownian motion preserves which boundary part is reached first, although it changes the clock. Use the branch defined by the wedge argument; the origin is a polar point for planar Brownian motion, so localization handles the vertex.