A GEOMETRIC AND STOCHASTIC PROOF OF THE TWIST POINT THEOREM

In this paper we will give a proof of the McMillan twist point theorem using geometry, potential theory and Ito's formula but not the Riemann mapping theorem.


Introduction
Let Ω be a simply connected domain and let z 0 be some fixed point in Ω.A point ξ ∈ ∂Ω is a cone point of Ω if there is an open triangle contained in Ω with a vertex at ξ. If, moreover, there is a unique φ ∈ (−π, π] such that for any > 0 there is δ > 0 for which It is said to be conformal at e iθ ∈ ∂D if, for z approaching e iθ inside any triangle in D with a vertex at e iθ , the limit lim z→e iθ f (z) exists and is not zero or infinite.An equivalent statement of the twist point theorem is that the set of points e iθ ∈ ∂D where f is neither conformal nor twisting has linear measure zero.In fact, this is the form in which the theorem was originally proved.McMillan's original reasoning in [11] relied on the properties of f which are inherited from the conformality of f .The purely geometric statement in terms of inner tangent points and twist points of ∂Ω is shown to be equivalent at the end of the same paper.
We will give another proof of the twist point theorem using geometric and stochastic arguments.The goal is to achieve a proof in the same spirit as the arguments in Chapter 5 of Bass's book, [3], and thus develop some geometric and probabilistic intuition about the phenomenon described by the theorem.In particular, we would like to understand McMillan's theorem as a statement about the interplay between the geometry of the boundary of Ω and the exit distribution of Brownian motion.In [1], the authors show that the almost everywhere characterization of cone points and twist points in terms of the derivative of a conformal mapping can be replaced by an equivalent characterization in terms of a certain harmonic function which can be defined directly on the domain using potential theory and not the Riemann mapping theorem.The proof of the equivalence in [1] uses the McMillan twist point theorem, but the results obtained there open an avenue to obtaining similar results in higher dimensions or for more general domains.We hope that the method of proof of the twist point theorem given here may do the same.
McMillan also showed in [11] that the set of cone points which are not interior tangent points has harmonic measure zero.We refer to [9, pp. 208-210] for a geometric and potential theoretic proof of this fact and will be satisfied with proving the following version of the twist point theorem.
where Cone denotes the set of cone points, Twist denotes the set of twist points and N is a set of harmonic measure zero.
The theorem is essentially potential theoretic as it is a statement about harmonic measure.In our proof, we will make use of various results which can be proved directly by potential theoretic means in Ω without recourse to a conformal mapping from Ω to D. In Sections 2 and the Appendix, we provide background and references for the results needed.In Section 3 we give the new proof of Theorem 1 above.

Preliminaries
Let Ω ⊂ C be a domain.The harmonic measure at z ∈ Ω of a Borel subset E ⊂ ∂Ω is denoted by ω(z, E, Ω) or by ω z (E) when there is no confusion about what domain is being considered.There are several equivalent definitions.
(1) ω(z, E, Ω) is the unique Perron-solution to the Dirichlet problem for the Laplacian in Ω with boundary data given by the characteristic function of E. (2) ω(z, E, Ω) is the probability that a Brownian motion started at z has its first exit from Ω in the set E. (3) ω(z, E, Ω) can be defined in terms of the trajectories of Green's function with pole at z as we will describe below.(4) If the domain Ω is simply connected then ω(z, E, Ω) is the normalized linear Lebesgue measure of the image of f (E) ⊂ ∂D by a Riemann mapping f of Ω to the unit disk which takes z to 0. This definition may be generalized to multiply connected domains by use of the universal covering map.
Only the last definition cannot be generalized to domains in higher dimensions.Notice that by Harnack's inequality, ω z1 ω z2 for any z 1 , z 2 ∈ Ω.We will therefore write ω(E) = 0 when ω z (E) = 0 for some z.
The main ideas we will need in order to use definition (2) are the Strong Markov property for Brownian motion and Ito's formula.Let X denote a two dimensional Brownian motion started at z 0 and let f be a C 2 function defined in a domain Ω ⊂ C containing z 0 .Let τ = inf{t : In the form we will use it, Ito's formula says that for t < τ When f = u is harmonic, we obtain that is a continuous martingale.We will also need the fact that any continuous martingale is a time change of a Brownian motion, perhaps stopped at a stopping time.We refer to [3] for detailed background.We will require the following basic projection estimates of harmonic measure due to Beurling and Hall.The original proofs are in [6] and [10] respectively.Here we will give the statements as they appear in [13] where stochastic proofs and versions of the results for R 3 are given.
With the same notations, suppose that

Lemma 2.2 (Hall's Lemma
).There exists c > 0 such that for all compact K ⊂ {z : where and m 1 denotes one dimensional Lebesgue measure on R.
The next lemma is also standard, its proof being accomplished by a straightforward application of Hall's lemma and the strong Markov property.A complete probabilistic argument is given in [14] so we omit the proof here.Similar arguments have been used in many places.See for example the survey article [4] and the papers referred to there.Let D(z, R) denote a Euclidean disk with radius R and center z.In what follows, and throughout the rest of the paper, all domains are assumed to be simply connected unless stated otherwise.Lemma 2.3.Let Ω ⊂ C be a simply connected domain.Given > 0 there is K > 0 such that where dist denotes the Euclidean distance in the plane.
Two more main ingredients in our proof of Theorem 1 will be the use of definition (3) above and the use of Fatou's theorem on convergence at the boundary of harmonic functions.We refer to the paper [14] for a probabilistic and potential theoretic proof of Fatou's theorem in an arbitrary simply connected domain.We will need the Fatou theorem for convergence in the "Green cones" described below whereas it is proved in [14] only for convergence along Green lines.The lemmas necessary for the required extension are stated in this section and proved in the Appendix.
The definition (3) of harmonic measure in terms of the trajectories of the gradient of Green's function is from the paper [7] of Brelot and Choquet.Note that the other definitions become vacuous in case Ω has no Green function and assume from now on that all domains considered possess one.Let a ∈ Ω be fixed and let g a (x) be Green's function for Ω with pole at a.The Green lines starting at a are the maximal orthogonal trajectories of the level lines of g a which have a limit point at a.Each Green line has a well defined initial direction given by its unit tangent vector at a and each point on the unit sphere corresponds in this way to a Green line at a. Given a Borel subset E ⊂ ∂Ω, the Green's measure of E is defined to be the normalized Lebesgue measure of the set of unit tangent vectors on the sphere for which the corresponding Green lines terminate at points of E. Denote the Green's measure of E by g a (E) and take the normalization so that g a (∂Ω) = 1.With respect to Green's measure, almost every Green line terminates in a point of ∂Ω and such Green lines are called regular.Let g a and g a denote respectively the outer and inner Green measures and let ω a and ω a denote the outer and inner harmonic measures for general subsets of ∂Ω.Brelot and Choquet proved that for any subset A ⊂ ∂Ω so that harmonic measurability implies Green measurability with equality of the measures.Later, Arsove, in [2], proved the converse of this statement, thereby showing that the two measures are the same.
We will not go into the details of the arguments in [7] but we remark that the main idea behind them is to carefully apply Green's theorem.One defines a tube as the union of Green lines connecting a pair of neighborhoods on disjoint level surfaces of g a and uses Green's formula to show that the flux of the vector field of ∇g a is the same at each end of the tube.We let one end of a tube tend to ∂Ω and the other to the singularity of g a to get the equivalence of the normalized Lebesgue measure on a sphere centered at a with the harmonic measure of the ideal boundary points at the other end of the tube.Difficulties posed by critical points in multiply connected domains or in higher dimensions are circumvented by using the fact that the critical set of g a corresponds to a set of measure zero in the normalized spherical measure at a. Covering subsets of the boundary by ends of tubes and attending to the details leads to (1).Let (z 0 , z) denote the Green line connecting z 0 and z, formed by considering the Green function with pole at z 0 .In what follows we will make use of properties of (z 0 , z) which derive either from the general theory of existence and uniqueness for ordinary differential equations or from potential theoretic properties of g(z, z 0 ), such as the symmetry g(z, z 0 ) = g(z 0 , z).For example, we shall freely use the fact that (z 0 , z) = (z, z 0 ).We defer an outline of the argument for this to Lemma 4.1 of the Appendix.Definition 1.Let z 0 denote some fixed base point in a domain.The forward cone at z with aperture α is denoted by Λ α (z) and defined to be the union of all Green lines starting at z who's tangent vectors at z make an angle less than α 2 with the tangent vector of (z 0 , z) at z. Let dist(z, ∂Ω) denote the Euclidean distance from z to ∂Ω and let d = d(z, α) be the Euclidean distance from z to ∂Λ α (z) ∩ ∂Ω.In the following, the base point z 0 is fixed as in Definition 1.
Note that strict inequality may hold here since ∂Λ α (z) ∩ ∂Ω may contain other ideal boundary points besides the ones represented by the Green lines in Λ α (z).By Lemma 2.3 we may choose K > 0 such that In what follows, we will identify Green lines starting from some fixed base point z 0 ∈ Ω, with ideal boundary points of Ω.By the Moore triod theorem of plane topology, (see [12] or [16]), the number of points in the Euclidean boundary, ∂Ω, which are the endpoints of more than two regular Green lines is at most countable.In many situations, it is possible to reduce matters to the case in which there is exactly one Green line ending at each Euclidean boundary point by considering interior approximation by Jordan domains.
Let ζ ∈ ∂Ω and let ζ be a regular Green line ending at ζ, again with the fixed base point z 0 .For λ > 0, let ζ (λ) be the unique point on for all sufficiently small λ}.
We now collect a few facts about the behavior of Green lines, forward cones and Green cones.They are more or less obvious when considered as conformally transplanted from the unit disk.We provide proofs in the Appendix which depend only on the basic definitions to keep the paper free from reliance on conformal mapping.
Given a simply connected domain Ω ⊂ C and a fixed point z 0 ∈ Ω, choose an angle 0 < α < π and consider a regular Green line (z 0 , z 1 ) starting from z 0 and passing through some other point z 1 ∈ Ω. Denote the tangent vector to (z 0 , z 1 ) at z 1 in the direction of decreasing g by v.
Let Ω λ = {z : g(z 1 , z) > λ} and let I λ α (z 1 ) denote the set of endpoints on ∂Ω λ of the regular Green lines in Λ α (z 1 ).Let u λ (z) = ω(z, I λ α (z 1 ), Ω λ ) and let u(z) = lim λ→0 u λ (z).The existence of this limit is an exercise with the maximum principle.One can, for example, compare u λ (z) to the strictly increasing sequence v λ (z) where v λ (z) is the solution to the Dirichlet problem in ∩ Ω.We will require the following facts which are proved either in the Appendix or in the given references.
Let L(z 0 , z) denote a complete regular Green line starting from z 0 and passing through z to terminate at some ideal boundary point.Lemma 2.6.If z 2 ∈ L(z 0 , z 1 ) and g(z 0 , z 1 ) > g(z 0 , z 2 ) then From Lemma 2.6, it follows that if z ∈ Γ α ( ζ ) for some Green line ζ then the segment of L(z 0 , z) connecting z and z 0 is contained in Γ α ( ζ ).It also follows that if Let X t (η) denote Brownian motion started at the fixed base point Here the union is over all Green lines ending at points of K and we may assume that for each point of K there are at most two.
A version of the last lemma holds for subsets K of the ideal boundary, but the simpler statement here is sufficient for our purposes.Note that in the unit disk, the lemma is equivalent to the statement that a sawtooth sub-domain over a closed set K ⊂ ∂D has an interior tangent at almost every point of K.
Given a boundary point ζ, Lemma 2.9.Let 0 < α < π be fixed.There is a constant c > 0 only depending on α such that for any The following lemma is a slightly more general version of a lemma from [11].
Lemma 2.10.Let E ⊂ ∂Ω be a Borel set, E c its complement.Suppose that for each ζ ∈ E there is an α < π 2 , a c > 0, and a sequence at ω z0 a.e.ζ ∈ E. This claim is a simple consequence of Fatou's theorem in the disk and a conformal invariance argument.To keep our argument free of the use of conformal invariance we refer to [14] where a stochastic and potential theoretic proof of the claim is given.Assuming the claim, note that by Lemma 2.9 and Harnack's inequality, there is c > 0 and infinitely many In our proof of the twist point theorem we will need the following technical lemma on harmonic measure in order to make an approximation argument.Proof: We have where g is Green's function for Ω, µ is the equilibrium mass supported on L ρ for the Green potential and we see by inspection that for ξ ∈ L ρ (ξ 0 ), ∇ z g is a vector which points into the half plane H from z. Since µ is a positive probability measure supported on L ρ , the same is true for ∇ z ω at z. Consequently, ω increases along [w 0 , ζ 0 ] in the direction toward L ρ .
Finally, we record for reference the following result of Dahlberg from [8].
Theorem 2. Let D ⊂ R n be a Lipschitz domain.Then a Borel measurable set E ⊂ ∂D is of harmonic measure zero with respect to D if and only if E is of vanishing (n − 1)-dimensional Hausdorff measure.
In fact, it is shown in [8] that there are constants α > 1  2 and β > 0 depending on the Lipschitz constant of the domain D, such that ( 2) for subsets F of ∂D.Dahlberg's proof is potential theoretic, holds for any dimension n ≥ 3 and can be modified to work for n = 2.For a stochastic proof of Dahlberg's results see [3,Chapter 3].

The twist point theorem
As before, Ω is a simply connected domain and z 0 ∈ Ω is a fixed base point.Combining the above results shows that almost every ζ ∈ ∂Ω is either a cone point or a twist point, and it suffices to prove Theorems 3 and 4.
We come to the proof of Theorem 3.
Proof: Let σ denote an open line segment contained in Ω with one end point at z 0 and the other at some point of ∂Ω.Let g denote the Green's function for Ω with pole at z 0 .A calculation shows that U = arg(g x +ig y ) is harmonic in a neighborhood of any point in Ω\{z 0 }.Since Ω is simply connected, U can be defined as a continuous harmonic function in Ω \ σ.Let w 0 be some fixed point in Ω \ σ.U (z) < +∞ for almost every Green line (with respect to the Green's measure at z 0 in Ω).If not, we could find a compact set K of positive ω z0 measure and a corresponding set of Green lines, ζ , with positive Green measure such that lim inf Then with positive probability X τ ∈ K and τ = τ 0 where τ 0 is the first exit time from Ω starting at w 0 and τ is as before.By Lemma 2.8, we would then have lim t→τ U (X t ) = +∞ with positive probability.
With the claim established, we consider subsets of Green lines (and ideal boundary points) for which lim inf for some fixed M > 0.
Let z ∈ Γ α (ζ) be a point such that U (z) ≤ M and assume, as we may, that d(z, ∂Ω) < |z − z 0 |/1000.We will first describe a general construction of a boundary set F with large ω z measure such that the twisting of (z 0 , z) around z is nearly the same as the twisting around points of F .We will then further specialize the choice of z so that the twisting of a Green line can be related to the change in U along the Green line.Finally, we will apply Lemma 2.10 to finish the proof.
To proceed as indicated, let ζ * denote a nearest boundary point to z and let: The Beurling projection theorem shows that ω(z, F, Ω) > c > 0 for some absolute constant c.We construct a domain Ω ρ as in Lemma 2.11 with z and ζ * playing the roles of w 0 and ζ 0 respectively.Let g ρ denote the Green's function for Ω ρ with pole at z 0 .Then g ρ and its first derivatives converge uniformly on compact subsets Ω\{z 0 } to g(•, z 0 ) as ρ → 0. With sufficiently small ρ > 0, the Green lines (z 0 , z) for both Ω ρ and Ω are uniformly close and we now work with Ω ρ .The set F is as before but may be considered to include the segment L ρ (ζ * ).
In Ω ρ , the segment L ρ (ζ * ) corresponds, under connection by Green lines, to a forward cone at z with some opening angle depending on ρ and Ω.Each point of the segment will be the end point of two Green lines under this correspondence; one for each ideal boundary point.By Lemma 2.5, ω(•, L ρ , Ω ρ ) increases along the Green line which is at the center of this forward cone in the direction toward L ρ and therefore (using Lemma 2.6) decreases along the Green line in the opposite direction (at angle π) which we denote by 1 .By Lemma 2.11, 1 does not intersect the segment [z, ζ * ] except at the initial point z.Notice also that, by uniqueness of Green lines, either 1 = (z 0 , z) or 1 does not intersect (z 0 , z) except at z. Perturbing z slightly if necessary, we may assume the latter case holds.
Let arg 1 (• − z) denote a continuous branch of the argument defined in Ω \ 1 such that −π < arg 1 (z 0 − z) ≤ π and for a given point ζ ∈ F , let arg(• − ζ) denote a continuous branch of the argument defined in Ω such that We claim there is an absolute constant k > 0 such that for all w sufficiently close to z on (z 0 , z) and for all ζ ∈ F .To see this, divide (z 0 , z) into three disjoint pieces (z 0 , z) = I ∪ II ∪ III where Considering some regular parametrization of (z 0 , z) from z 0 to z, let ∆ I , ∆ II and ∆ III denote the respective contributions from each piece to the difference (4).
It is evident that ∆ I is bounded.The boundedness of ∆ II is a consequence of the fact that Ω is simply connected.The Green line (z 0 , z) To see the boundedness of ∆ III we use the following topological argument which is made routine in [11].
Let γ denote a maximal subarc of (z 0 , z) which is contained in D d/8 (ζ * ).As before, (z 0 , z) is considered to be given by a regular parametrization from z 0 to z.Let s 0 and s 1 denote the endpoints of γ on S with (s 0 , s 1 ) respecting the order given by the parametrization of (z 0 , z).The points s 0 and s 1 may lie on different components of S ∩Ω.With ( 4) established, we wish to show that the set nested disks with rational radii and centers.Writing E as a countable union of subsets we may assume that E satisfies the conditions below.The requirement that D 2 ⊂ D 1 allows for control of the twisting of the Green lines and this explains why conditions v), vi), vii) repeat the conditions i), ii), iii).
Proof of Theorem 2: Given ζ ∈ E M and z ∈ Γ α (ζ) such that arg(z−ζ) < M , we will construct a set of boundary points F ⊂ ∂Ω with ω z (F ) > c > 0 such that for any ζ ∈ F either condition (2) or (3) in the definition of E M fails.By Lemma 2.10, this will complete the proof.We make a general construction which will be referred to as P, (for "picture").The construction of P will be moved around the domain Ω by translations, rotations and dilations as needed.
Let 1 > 0 be small and fixed and let Ω be simply connected.Suppose that (0, 1) ∈ ∂Ω and that (x, y) : The domain Ω is a simply connected Lipschitz subdomain of Ω with Lipschitz constant independent of Ω and it is the union of triangles in T 1 whose interiors are not contained in the interior of any larger triangle in T 1 .We will call these maximal triangles in Ω .Note that right angle vertices lying in Ω of maximal triangles in Ω form a set of isolated points in Ω.Let V denote the set of right angle vertices of maximal triangles which are in ∂Ω.We say that case 0 occurs if ω (− 1 2 , 1 2 ) (V, Ω ) ≥ and otherwise that we are in case I.The set ∂Ω ∩ Ω is a union of countably many segments of edges of maximal triangles.In case I, the Dahlberg estimate (1) implies that ω (− 1 2 , 1 2 ) (ρ, Ω) > c > 0, where ρ denotes the set of right hand edge segments, looking from (0, 0) to (0, 1), of ∂Ω ∩ Ω.
Let ζ j denote the right hand endpoint of I j as seen from inside Ω and let J j be the segment of I j with |J j | = 1 10 |I j | and right hand endpoint ζ j .Let p j be the point in Ω with |p j −ζ j | = |J j | and with the segment [p j , ζ j ] perpendicular to J j .With the endpoint ζ j fixed, rotate J j and p j clockwise about ζ j , as far as possible, until the rotated segment hits another boundary point.Call the resulting segment J * j and the resulting point p * j .The amount of rotation can be zero.Note that 2 ) (I j , Ω ) for all j for some fixed constant c 1 > 0. The description of P is complete.We return to the setting of the theorem and let z and ζ be as in the first paragraph of the proof.Let ζ 0 be a nearest boundary point to z and choose a point p 0 on the segment joining z to ζ 0 .We move P so that (0, 0) and (0, 1) correspond respectively to p 0 and ζ 0 and we require of p 0 that the circle {w : |w−p 0 | = |ζ −p 0 |} intersects the (moved) rectangle R only on its short edges and that conditions necessary for P to lie inside Ω are satisfied.We can do this with ( 6) for some constant c 2 > 0 depending on .In this initial step, we write V 0 = V and if case 0 occurs, we put F = V 0 .If case I occurs, we obtain the segments J * j and corresponding points p * j and ζ * j .We repeat the construction of P for each j with (0, 0) corresponding to p * j and (0, 1) corresponding to ζ * j .For each j, if case 0 occurs we find a set of cone points V = V j and if case I occurs we find segments J * (j1,j2) and the corresponding points p * (j1,j2) and ζ * (j1,j2) .Let α denote a multi index and |α| its length.Choose a large integer K and stop the construction when multiindices reach length KM .For such indices α we find a set F α with (7) ω pα (F α ) > c > 0 by using the Beurling projection theorem as in the proof of Theorem 1.By Harnack's inequality and induction and the construction shows that with α = (j 1 , . . ., j KM ), a simple curve connecting p 0 , p (j1) , p (j1,j2) , p (j1,j2,j3) , . . ., p α , winds clockwise around the points of F α at least 10M times if K is sufficiently large.Another simple topological argument shows that arg(p α − ζ ) < −M so that ∆c is comparable to the length of the initial segment and the level curve of g 1 passing through the endpoint of the first increment is asymptotically circular as ∆c → 0. Let d denote the Euclidean length of (z 2 , z 1 ) and let N be the greatest integer not exceeding d ∆c .The above discussion shows that for j = 1, . . ., N the distance between the j th increments of the two methods is O(j(∆c) 2 ) and so the two methods converge to the same curve (z 2 , z 1 ).But the second construction shows that a tangent vector to (z 2 , z 1 ) at any point is perpendicular to both a level curve of g 2 and a level curve of g 1 .By symmetry, we then have (z 2 , z 1 ) = (z 1 , z 2 ).The argument can clearly be continued to construct a slightly longer Green line (z 0 , z 3 ) = (z 3 , z 0 ) which contains (z 2 , z 1 ) as a proper subset.For a general pair of points z 1 = z 2 , we can cover (z 2 , z 1 ) by small disks in which our initial assumptions hold and obtain (z 2 , z 1 ) = (z 1 , z 2 ) by piecing together smaller subarcs.Lemma 4.2.∇u |∇u| = v.Proof: By a limiting argument, it suffices to work in the domain Ω = {g z1 > λ} where λ > 0 is small and fixed.The Green function for this domain is g z1 − λ.Basic properties of the Green function imply that for large λ 0 , the level curves {g z1 = λ 0 } are asymptotically circular.It follows that, given > 0, we may choose λ 0 sufficiently large so that within the domain {g z1 > λ 0 }, the harmonic measure of Λ α (z 1 )∩{g z1 = λ 0 } has a gradient at z 1 pointing within angle of the vector v. Now, given δ > 0, there is η > 0 such that if |z − z 1 | < η, then the level curve {g z = λ 0 } lies in a δ-neighborhood of {g z1 = λ 0 }.By continuous dependence of trajectories on the initial point (in the domain {g z1 > λ}), a sufficiently small choice of η > 0 guarantees that the subset of {g z = λ 0 } corresponding to I α (z 1 ) via the Green lines at z lies in a δ-neighborhood of the subset of {g z1 = λ 0 } which corresponds to I α (z 1 ) via Green lines at z 1 .For any z such that |z − z 1 | < η, the harmonic measure u(z) is equal to the harmonic measure within {g z > λ 0 } of the subset of {g z = λ 0 } corresponding to I α (z 1 ) via Green lines at z.If η > 0 and therefore δ > 0 is sufficiently small, it is then clear that the direction of greatest increase of u must be within angle 2 of v. Lemma 4.3.If z 2 ∈ (z 0 , z 1 ) and g(z 0 , z 1 ) > g(z 0 , z 2 ) then Λ α (z 2 ) ⊂ Λ α (z 1 ).
Proof: By a limiting argument, we may again assume that we are in a Jordan domain.Let 0 = (z 0 , z 1 ) and let 1 and 2 denote the Green lines starting at z 1 which make an angle of α/2 with 0 .The trajectories 1 and 2 form the sides of Λ α (z 1 ).Consider the set of trajectories starting at z 1 lying in Λ α (z 1 ) between 1 and 0 and denote the part of ∂Ω corresponding to them by H.We claim that the harmonic measure of H increases along 0 in the direction of the forward cone Λ α (z 1 ).A similar argument will show that the harmonic measure of the other half of Λ α (z 1 )∩∂Ω corresponding to the trajectories between 0 and 2 also increases along 0 in the same way.This claim implies the lemma since if Λ α (z 2 ) ⊂ Λ α (z 1 ) for some z 2 on (z 0 , z 1 ) ∩ Λ α (z 1 ) with g(z 0 , z 2 ) < g(z 0 , z 1 ) then the harmonic measure of one of the halves of Λ α (z 1 ) would have decreased from z 1 to z 2 .
To prove the claim, let z ∈ 0 ∩ Λ α (z 1 ) and let 3 be a Green line starting at z and ending in the same ideal boundary point as 1 .( 3 exists by a simple limiting argument.)Let θ < π be the angle between the forward tangent vectors to 0 and 3 at z and consider the trajectory 4 starting at z, lying between 0 and 3 in Λ α (z 1 ), and at angle θ 2 from 0 .We know (by Lemma 2.5) that ω = ω(•, H, Ω) has gradient tangent to 4 at z. Consequently the level curve of ω passing through z is orthogonal to 4 at z.It follows that ω increases on 0 in the direction of decreasing g(•, z 0 ).Proof: We may again assume that Ω is a Jordan domain.Note that Lemma 2.6 implies that for any w ∈ C λ we have Λ π (w) ∩ C λ = {w}.In fact, the angle between (z 0 , w) and the complete Green line L which forms the edge of Λ π (w) is π 2 .Lemma 2.6 implies that there can be no other intersection (say at a point w ) at angle π 2 between L and another Green line starting at z 0 .If there were, there would be a forward cone centered on L at w and another forward cone centered on L at w with the same opening angle but whose edges intersected at z 0 , contradicting Lemma 2.6.Considering a Green line starting at z 0 which intersects C λ at a point very close to w, we see (in the same way by Lemma 2.6) that the forward angle of intersection with L must be less than √ K so that can not be arbitrarily small.

ξ 2 −
+ re iθ : |θ − φ| < π , r < δ ⊂ Ω then ξ is called an interior tangent point of Ω.A boundary point ξ ∈ ∂Ω is a twist point if lim inf z∈Ω→ξ arg(z − ξ) = −∞ and lim sup z∈Ω→ξ arg(z − ξ) = +∞ where arg(z − ξ) denotes a continuous branch of the argument defined in Ω with −π < arg(z 0 − ξ) ≤ π.The McMillan twist point theorem states that the set of boundary points of Ω which are neither inner tangent points nor twist points has harmonic measure zero.(See Section 2 for a definition of harmonic measure.)To consider the same theorem from a function theoretic point of view, let f : D → Ω with f (0) = z 0 be a Riemann mapping and let Arg denote a continuously defined argument.The conformal mapping f is said to be twisting at the point e iθ ∈ ∂D if lim inf z∈D→e iθ Arg(f (z) − f (e iθ )) = −∞ and lim sup z∈D→e iθ Arg(f (z) − f (e iθ )) = +∞.The function f is, by definition, conformal at every interior point of D.

(
See Chapter 3 of [9] for example.)Given z ∈ [w 0 , ζ 0 ] ∩ Ω ρ let H denote the half plane with normal vector [w 0 , ζ 0 ] whose interior contains ζ 0 and whose boundary contains z.Since g(z, ξ) For each ζ ∈ ∂Ω which is the endpoint of a regular Green line starting at z 0 , let arg(• − ζ) denote the continuous branch of the argument in Ω which satisfies −π < arg(z 0 − ζ) ≤ π.We will prove Theorem 1 in two steps.First we show we let E M ⊂ ∂Ω denote the set of boundary points ζ ∈ ∂Ω satisfying (1) arg(z − ζ) < M for infinitely many z ∈ Γ α (ζ), (2) ζ is not a cone point of ∂Ω, (3) arg(z − ζ) ≥ −M for each z ∈ Ω and we prove Theorem 4. ω(E M ) = 0. Taken together, Theorems 3 and 4 show that the set of non-cone points ξ for which lim inf z→ζ z∈Ω arg(z − ξ) = −∞ has harmonic measure zero, because it is contained in the union of E M as M ranges through the positive integers.The same arguments show that ω that with E M defined as the set of points with (1) arg(z − ζ) > −M for infinitely many z ∈ Γ α (ζ); (2) ζ is not a cone point of ∂Ω; (3) arg(z − ζ) ≤ M for each z ∈ Ω; we have ω(E M ) = 0.

8 C
) S = w : |ζ * − w| = d ⊂ S = the unique crosscut of Ω which intersects D(z, d) F ⊂ ∂Ω = the set of ideal boundary points separated from z by C.
Consider the bounded component, denoted by V , of (γ ∪ S) c which does not contain ζ.We can define arg 2 (w − ζ) as a continuous function of w on the closure of V so that it agrees with arg(w −ζ) on γ and we see that (arg(s 1 − ζ) − arg(s 0 − ζ)) is equal to the change in arg 2 (w − ζ) on the circular arc ∂V ∩ S. Summing the contributions of the various maximal arcs of (z 0 , z) ∩ D d/8 (ζ * ) to the total change in arg(• − ζ) and respecting the signs of the changes shows that the total change in arg(• − ζ) on III is bounded by 2π while the total change in arg 1 (• − z) on III is evidently bounded by a smaller constant.
2 and let T 0 denote the collection of isosceles right triangles with right angle vertex in (x, y) : −1 ≤ x ≤ 0, y ≥ 1 − 2 and base on the x-axis.Let T 1 be the collection of triangles in T 0 whose interiors are contained in Ω and let Ω = T ∈T1 T .

π 2
and therefore, locally, L must lie in the unbounded component of the complement of C λ except for the point {w}.Combining this with the previous observation, we see that L must lie completely outside of C λ in {z : g(z, z 0 ) < λ} except for the point {w}.Consequently, the only local extrema of g(•, z 0 ) restricted to Green lines (z 1 , z 2 ) are local maxima.By the local geometry of the Green lines and C λ , the set C λ ∩ Γ α ( ζ ) includes an open arc of C λ containing the point ζ ∩ C λ .Let J λ ⊂ C λ be a maximal arc of C λ ∩ Γ α ( ζ ) containing ζ ∩ C λ and denote its endpoints by J λ = (a λ , b λ ) with the standard orientation the endpoints of t 1 and t 2 must also pass through D. By the Beurling projection theorem and the maximum principle, we would then have π − α 4π < C X t is Brownian motion started at w 0 then U (X t ) is a continuous martingale by Ito's formula.With τ denoting the first exit time from Ω \ σ, the set of paths for which