POTENTIAL THEORY and HARMONIC FUNCTIONS a Thesis Submitted to Kent State University in Partial Fulfillment of the Requirements F

POTENTIAL THEORY AND HARMONIC FUNCTIONS

A thesis submitted to Kent State University in partial fulﬁllment of the requirements for the degree of Master of Science

Hebah S. Alhwaitiy

December, 2015 Thesis written by

Hebah S. Alhwaitiy

B.S.,Tabouk University, 2009

M.S., KENT STATE UNIVERSITY, 2015

Approved by

Dr. Benjamin Jaye , Advisor

Dr. Andrew Tonge , Chair, Department of Mathematical

Sciences

Dr. James L. Blank , Dean, College of Arts and Sciences

ii TABLE OF CONTENTS

LIST OF FIGURES ...... iv

LIST OF TABLES ...... v

Acknowledgements ...... vi

1 Introduction ...... 1

2 Potential Theory and Harmonic Functions ...... 3

2.1 Harmonic Functions ...... 3

2.2 Subharmonic Function ...... 7

2.3 Proof of Theorem (2.1.5) ...... 14

2.4 Polar sets and Hausdorﬀ measure ...... 22

2.5 Harmonic Measure and Green,Functions ...... 32

2.6 Hausdorﬀ measure and harmonic measure ...... 39

BIBLIOGRAPHY ...... 46

iii Acknowledgements

I would like to thank Kent State University and Department of Mathematical

Seiences for letting me fulﬁll my dream of being a student here.

I will forever be thankful to my committe members- Dr. Dmitry Ryabogin , and

Dr. Artem Zvavitch for their encouragement, helpful career advice, and suggestions in general.

I would like to express my sincere gratitude to my advisor Dr. Benjamin Jaye for the continuous support my thesis study and related research for his patience, motivation, and immense knowledge. His guidance helped me in all the time of research and writting of this thesis. I could not have imagined having a better advisor and mentor for my master study.

I wish to express my sincere thanks to my parents, sisters, and brothers. They always supporting me and encourging me with their best wishes. I must express my very profound gratitude to my husband, Abdulrahman for all of his love and support without which I could not have managed to complete my thesis. I would like to express my gratitude and appreciation to my friends for their support, guidance, love and generosity. I am especially thankful to my the best friend Rawan. She has continually encouraged me to perform as strongly as I can. The product of this research paper would not be possible without all of them.

iv CHAPTER 1

Introduction

The purpose of this thesis is to see how measure theoretic ideas aid the study of harmonic functions in very general domains. We shall describe a general solution of the Dirichlet problem, and see how this gives rise to Harmonic Measure.

Many of the results in this expository thesis are taken from [1]. The main goal in the ﬁrst three sections is to introduce a wide class of a domain in which we can solve the Dirichlet problem for harmonic function. First, we consider the special case when the domain is a disc, where one can ﬁnd a solution directly using the Poisson integral. To solve the Dirichlet problem in a general domain, we appeal to the Perron method, which makes use of subharmonic functions.

A certain notion of exceptional set arises in solving the Dirichlet problem: the polar sets. In section 4 we introduce some Hausdorﬀ measure and describe how the polar sets have Hausdorﬀ dimension zero.

In the ﬁfth section, we use the solution of the Dirichlet problem to produce a special measure called Harmonic Measure. We study the properties of harmonic measure using apparatus from measure theory. The harmonic measure of a domain is closely connected to the Green function, and we will prove an integral formula for the Green function in terms of the harmonic measure. We use the same technique of

[1].

1 In the last section, we introduce two results that relate the harmonic meaure to

Hausdorﬀ measure. We follow the technique of [2] with a modiﬁcation that is resulted of changing the type of a domain. We shall only consider the special case of the Cantor set.

2 CHAPTER 2

Potential Theory and Harmonic Functions

2.1 Harmonic Functions

Deﬁnition 2.1.1. Let U ⊂ C be an open set. A function h : U → R is said to be harmonic if h ∈ C2(U) and ∆h = 0 on U, where the Laplace operator ∆ is given by

∂2h ∂2h ∆h := + . ∂x2 ∂y2

We say that we can solve the Dirichlet problem in a domain D ⊂ C, which is a connected open set, if for every φ ∈ C(∂D), we can ﬁnd a unique function u : D → R satisfying

(i) u is harmonic in D, and

(ii) limz∈D: z→ζ u(z) = φ(ζ) for each ζ ∈ ∂D.

We want to solve the Dirichlet problem in a wide class of domains. Let0s ﬁrst consider the special case when D is a disc

B(w, ρ) = {z ∈ C : |z − w| < ρ}.

We now introduce the notions of the Poisson Kernel P and Poisson integral PBφ.

Deﬁnition 2.1.2. (i) The Poisson kernel P : B(0, 1) × ∂B(0, 1) → R is deﬁned by

ζ+z 1−|z|2 P (z, ζ) := Re( ζ−z ) = |ζ−z|2 (|z| < 1, |ζ| = 1). 3 (ii) If B = B(w, ρ) and φ : ∂B → R is a Lebesgue-integrable function, then its

Poisson integral PBφ : B → R is deﬁned by

1 R 2π z−w iθ iθ PBφ(z) := 2π 0 P ( ρ , e )φ(w + ρe )dθ (z ∈ B)

We will now discuss the Dirichlet problem for the disc.

Theorem 2.1.3. (Dirichlet Problem for the disc) Let B = B(w, ρ). Then for φ ∈

C(∂B), the function PBφ is harmonic in B, and satisﬁes

limz→ζ PBφ(z) = φ(ζ) for ζ ∈ ∂B, z ∈ B.

Thus, PBφ solves the Dirichlet problem on B.

Proof. Suppose w = 0 and ρ = 1, so that B = B(0, 1). Then

Z 2π 1 iθ iθ PBφ(z) = P (z, e )φ(e )dθ 2π 0 Z 2π iθ 1 e + z iθ = Re( iθ )φ(e )dθ 2π 0 e + z Z 2π iθ 1 e + z iθ = Re iθ φ(e )dθ 2π 0 e − z then it is the real part of a holomorphic function of z. Hence it is harmonic on B.

Indeed, let h = Ref where f is holomorhic on B. Set f = h + i` then by cauchy-

Riemann equations we have that

∂h ∂` ∂h ∂` ∂x = ∂y and ∂y = − ∂x

Therefore, ∂2h ∂2h ∂2` ∂2` + = − = 0 ∂x2 ∂y2 ∂x∂y ∂y∂x

4 Thus h is harmonic function.

We now need to prove that limz→ζ PBφ(z) = φ(ζ)

Z 2π 1 iθ iθ |PBφ(z) − φ(ζ)| = P (z, e )(φ(e ) − φ(ζ))dθ 2π 0 1 Z 2π ≤ P (z, eiθ)|φ(eiθ) − φ(ζ)|dθ, 2π 0

where we have used that P (z, ξ) ≥ 0 if z ∈ B and that

1 Z 2π P (z, eiθ)dθ = 1. 2π 0

Fix ζ and let > 0. Since φ is a continuous function on the whole of ∂B, then for every ζ0 ∈ ∂B there exist δ > 0 such that

|φ(ζ0) − φ(ζ)| < whenever |ζ0 − ζ| < δ.

Hence,

= Re(2 − 1) = .

Next we shall show that there exists δ0 > 0 such that

0 |z − ζ| < δ ⇒ sup P (z, ζ0) < . ζ0∈∂B: |ζ0−ζ|≥δ

5 Indeed,

2 2 ζ0 + z ζ0 − z |ζ0| − |z| + (ζ0z − ζ0z) ( ) × ( ) = 2 ζ0 − z ζ0 − z |ζ0 − z| 2 2 |ζ0| − |z| + 2i=(ζ0z) = 2 |ζ0 − z|

So by Deﬁnition (2.1.2),

2 2 2 ζ0 + z |ζ0| − |z| 1 − |z| P (z, ζ0) = Re( ) = 2 = 2 . ζ0 − z |ζ0 − z| |ζ0 − z|

Where |z| < 1, |ζ0| = 1

δ δ Now we have ζ ﬁxed, and |ζ0 − ζ| > δ, if |z − ζ| < 2 then |z − ζ0| > 2 . Therfore,

1 − |z|2 4 P (z, ζ0) = 2 ≤ 2 (1 − |z|)(|z| + 1). |ζ0 − z| δ

0 8δ0 0 But since |z| + 1 ≤ 2 and |z| > 1 − |z − ζ| > 1 − δ , we have P (z, ζ0) ≤ δ2 < if δ is chosen small enough in terms of δ.

Consequently, if |z − ζ| < δ0, then

1 Z 1 Z 2π P (z, eiθ)|φ(eiθ) − φ(ζ)|dθ ≤ |φ(eiθ) − φ(ζ)|dθ 2π |eiθ−ζ|≥δ 2π 0 1 Z 2π ≤ |φ(eiθ)|dθ + |φ(ζ)| . 2π 0

Combining these facts, we deduce that if |z − ζ| < δ0 then

1 Z 2π 1 Z 2π P (z, eiθ)|φ(eiθ) − φ(ζ)|dθ ≤ (1 + |φ(eiθ)|dθ + |φ(ζ)|). 2π 0 2π 0

Consequently we deduce that

1 R 2π iθ |PBφ(z) − φ(ζ)| ≤ (1 + 2π 0 |φ(e )|dθ + |φ(ζ)|),

from which we conclude that PBφ(z) → φ(ζ) as z → ζ.

6 The class of domains that we shall consider the general Dirichlet problem need to

have “thick” boundaries. To quantify this we shall introduce the notions of Polar set.

Deﬁnition 2.1.4. A subset E ⊂ C is called polar if

RR log |z − w|dµ(z)dµ(w) = −∞ for every ﬁnite non-zero compactly supported Borel measure µ with supp(µ) ⊂ E.

We say that a property holds nearly everywhere (n.e.) if it holds everywhere

outside some polar set.

We shall show later (in Section 2.4) that any set of ﬁnite s-dimensional Hausdorﬀ

measure, s ∈ (0, 2], is non-polar.

Now we conclude this section by introducing result we shall prove concerning the

solution of generalized Dirichlet problem.

Theorem 2.1.5. (Solution of the Generalized Dirichlet Problem). Let D be a domain

in C such that ∂D is non-polar, and let φ : ∂D −→ R be a bounded function which is continuous n.e. on ∂D . Then there exists a unique bounded harmonic function h

on D such that

lim h(z) = φ(ζ) for n.e. ζ ∈ ∂D. z→ζ

2.2 Subharmonic Function

We shall use Perron’s method to prove Theorm (2.1.5). This method makes use

of subharmonic functions.

7 Deﬁnition 2.2.1. Let U ⊂ C be an open set. A function u : U → [−∞, ∞) is said to be upper semicontinuous if the set {z ∈ U : u(z) < α} is open in C for each α ∈ R. Also υ : U → (−∞, ∞] is lower semicontinuous if −υ is upper semicontinuous.

Equivalently, one may state the condition that {z ∈ C : u(z) < α} is open for all α as

lim sup u(z) ≤ u(ξ) for all ξ ∈ C. z→ξ This equivalent formulation will be used several times in what follows.

Deﬁnition 2.2.2. Let U ⊂ C be an open set. A function u : U → [−∞, ∞) is said to be subharmonic if it is upper semicontinuous and satisﬁes the local submean

inequality, that is, for every w ∈ U, there exists ρ > 0 such that

1 Z 2π u(w) ≤ u(w + reit)dt whenever 0 ≤ r < ρ. (2.2.1) 2π 0 We say that υ : U → (−∞, ∞] is superharmonic if −υ is subharmonic.

One main advantage that subharmonic functions have over harmonic functions is

that they can be cut and pasted together. We state a simple gluing lemma, whose

proof is immediate from the local submean value property.

Lemma 2.2.3. Suppose that u and v are subharmonic functions on U and V , where

V ⊂ U. If

lim sup v(z) ≤ u(ξ) for ξ ∈ U ∩ ∂V, z→ξ,z∈V then the function   max(u, v) in V  u in U\V is subharmonic in U.

8 The fact that the function satisﬁes the local submean value property at each point is immediate. Notice also that the condition of the lemma ensures that the function is upper semicontinuous.

We now move onto showing that the class of subharmonic functions contains the real valued harmonic functions.

Lemma 2.2.4. If u is harmonic function in D and B(w, ρ) ⊂ D, then

1 Z 2π u(w) = u(w + ρeiθ)dθ. 2π 0

1 R 2π iθ Proof. Fix w. Look at φ(ρ) = 2π 0 u(w + ρe )dθ. We want to show that the derivative of φ at ρ equals 0 (φ0(ρ) = 0) whenever B(w, ρ) ⊂ D. Since then we would have

φ(ρ) − u(w) = φ(ρ) − φ(0) Z ρ = φ0(s)ds = 0, 0 as required. To this end, write w = x + iy, and notice that

1 Z 2π d φ0(ρ) = u (x + ρ cos(θ) + i(y + ρ sin(θ))) dθ 2π 0 dρ 1 Z 2π ∂u ∂u = ( cos(θ) + sin(θ))dθ 2π 0 ∂x ∂y

By using the Divergence Theorem we see that the previous line equals

1 Z ∂2u ∂2u [ 2 (w + ρ) + 2 (w + ρ)]dm2(z) 2π B(0,1) ∂x ∂y

= 0

Therefore, φ0(ρ) = 0 as required.

9 In particular the lemma ensures that if u : D → R is harmonic, then it is subharmonic. Moreover, it shows that |u| is subharmonic.

Indeed, Since u harmonic function, then by lemma (2.2.4)

1 Z 2π 1 Z 2π iθ iθ |u(z)| = u(z + re )dθ ≤ |u(z + re )|dθ. 2π 0 2π 0

So |u| satisﬁes the local submean inequality. It is also continuous, and so is a subharmonic function.

Next we want recall the Strong Maximum Principle. We use the technique of [1].

Theorem 2.2.5. Let u be a subharmonic function on a domain D ∈ C.

(a) If u attains a global maximum on D, then u is constant.

(b) If D is bounded and lim supz→ζ u(z) ≤ 0 for all ζ ∈ ∂D, then u ≤ 0 on D.

Proof. (a) Suppose that u attains a maximum value N on D. Let us deﬁne K =

{z ∈ D : u(z) = N}. Since u is upper semicontinuous, the set K is relatively

closed in D. Also, if u(w) = N then the local submean inequality ensures that

the integral average of u is at least N on all small enough circles round w. But

since N is the maximal value of u, we ﬁnd that u is equal to N on all small

circles around w, and so K is open. We have that K is a non-empty relatively

open and closed subset of the connected set D, so K = D.

(b) The proof hinges on the following theorem, which is taken from [1] page (25).

Theorem 2.2.6. Let u be an upper semicontinuous function on C, and let K

be a compact subset of C. Then u is bounded above on K and attains its bound.

10 Proof. The sets {z ∈ C : u(z) < n, n ≥ 1} form an open cover of K, so have

a ﬁnite subcover. Hence u is bounded above on K. Let M = supK u. If the

1 value M was not attained, then the open sets {z ∈ : u(z) < M − n }, n ≥ 1 would form an open cover K, and hence have a ﬁnite subcover. But then we

1 would ﬁnd n such that u < M − n on C. This contradicts the deﬁnition of the

supremum as the least upper bound. Hence u(z) = M for some z ∈ C

We may not return to prove (b). Extend u to ∂D by deﬁning u(ζ) = lim supz→ζ u(z) (ζ ∈ ∂D). Then certainly u ≤ 0 on ∂D, and u is upper semicontinuous on the

compact set D. By theorem (2.2.6) u attains a maximum at some υ ∈ D. If

υ ∈ D, then by part (a) u is constant on D, and so u ≤ 0 on D. On other hand,

if υ ∈ ∂D, then by assumption u(υ) ≤ 0, so u ≤ 0 on D.

In the case when the domain is not bounded one also needs a condition at inﬁnity.

We state an extended maximum principle to cover this case.

Theorem 2.2.7. (Extended Maximum principle) Let D be a domain, and let u be a subharmonic function on D which is bounded above. If ∂D is non-polar and lim supz→ζ u(z) ≤ 0 for n.e ζ ∈ ∂D, then u ≤ 0 on D.

To prove this theorem, we shall state Theorem 3.6.1 from [1] without proof. This states that if U ⊂ C is an open set, and E is a closed polar subset of C. Then a function u that is subharmonic and bounded above on on U\E may be extended to a subharmonic function on U by the formula

u(ξ) = lim sup u(z), z→ξ: z∈U\E

11 for ξ ∈ U ∩ E.

Proof. Given > 0, deﬁne

E = {ζ ∈ ∂D\{∞} : lim supz→ζ u(z) ≥ }

Then by assumption, and the upper-semicontinuity of u, E is closed polar set. Now

Deﬁne υ on C\E by   max(u, ) on D υ =   on C\(D ∪ E)

The boundry condition on u ensures that υ is upper semicontinuous on C\E. As such, υ is a subharmonic function on C\E and is clearly bounded above on C\E.

Since E is a closed polar set, we extend u to a subharmonic function on C that is bounded above. Now we want show that υ is constant. To prove this we claim that every subharmonic function on C which is bounded above must be constant.

Fix z0 ∈ C, since υ is subharmonic function, we know that

lim supz→z0 υ(z) ≤ υ(z0).

Therefore, there exists δ > 0 and τ > 0 such that

sup υ(z) ≤ υ(z0) + τ whenever |z − z0| ≤ δ. (2.2.2) B(z0,δ)

Consider υe(z) = υ(z) − υ(z0) − [log |z − z0| − log δ] on C\B(z0, δ). We have that

υe(z) ≤ τ on ∂B(z0, δ) and υe(z) ≤ 0 for all large z. Then by maximum principle, applied with the bounded domain B(z0,R)\B(z0, δ) with R large enough so that

υe(z) ≤ 0 for z ∈ C\B(z0,R), yields that υe(z) ≤ τ on C\B(z0, δ). this hold for every 12 > 0. If we now let → 0, then we get υ(z) ≤ υ(z0) + τ on C\B(z0, δ) and by (2.2)

we get υ(z) ≤ υ(z0) + τ on C.

Let τ → 0 we get υ(z) ≤ υ(z0) on C so by strong maximum principle υ attains a global maximum on D so υ is constant.

Since υ = on ∂D\(E ∪ ∞), which is non-empty, it follows that υ ≡ . Hence u ≤ on D, and letting → 0 we have that u ≤ 0 on D

Next we introduce the following comparison lemma.

Lemma 2.2.8. Let u be a harmonic function on a bounded domain D, and let υ be

a subharmonic function on D. If u ≥ υ on ∂D, then u ≥ υ on D.

Proof. Since υ − u is subharmonic function and υ − u ≤ 0 on ∂D, so by apply

maximum principle we get that υ − u ≤ 0 on D.

Our candidate for the solution of the generalized Dirichlet Problem is the the

Perron function.

Deﬁnition 2.2.9. Let D be a proper subdomain of C∞, and let φ : ∂D → R be a bounded function. HDφ : D → R is the associated Perron function deﬁned by

HDφ = supuU u where U denotes the family of all subharmonic functions u on D such that

lim sup u(z) ≤ φ(ζ)(ζ ∈ ∂D) (2.2.3) z→ζ

13 2.3 Proof of Theorem (2.1.5)

In this section we write the proof of theorem (2.1.5). We follow the technique of [1] with some modiﬁcations. To prove the assertion (B) of the theorem, we ﬁrst introduce some notion.

Deﬁnition 2.3.1. Let D be a proper subdomain of C∞, and let ζ0 ∈ ∂D. A barrier at

ζ0 is a subharmonic function b deﬁned on D ∩ N, where N is an open neighbourhood

of ζ0, satisfying

b < 0 on D ∩ N and limz∈D∩N, z→ζ0 b(z) = 0.

A boundary point at which a barrier exists is called regular, otherwise it is irregular.

We shall use a well known theorem of Weiner, which states that the set of irregular

points is a polar set (see [1] Chapter 4).

We now begin the proof of Theorem (2.1.5).

Proof. Set h = HDφ

(A) We want to show that h is a bounded harmonic on D.

It suﬃces to prove harmonicity on each open disc B with B ⊂ D . Fix such a

disk B, and also a point w0 ∈ B. By deﬁnition of HDφ, we can ﬁnd (un)n≥1 ∈ U such

that

lim un(w0) = HDφ(w0) n→∞

Since φ is a bounded function we have that |φ(z)| ≤ N for every z. Then the

function u = −N is constant function and satisﬁes 2.2.1. So it is subharmonic func-

tion. By replacing un by the subharmonic function max(u1, ...., un, −N) if necessary

14 (cf. the Gluing lemma), we may suppose that u1 ≤ u2 ≤ .... on D, and so un converge to some function u. Now for each n, let uen denote the poisson modiﬁcation on un deﬁned by   PBun on B uen =  un on D\B

Claim 2.3.2. We want to show uen is harmonic on B.

Note that un is bounded (and hence Lebesgue integrable) on ∂B, so PBun makes sense. Indeed, since each un ≥ −N therefore uen ≥ −N. But then by Theorem (2.1.3) we have PBun(z) is harmonic on B, which proves the claim.

Claim 2.3.3. We want to show that uen is subharmoinc on D

We shall ultimately use the Gluing lemma to prove this claim. To this end we shall ﬁrst verify that that

lim sup PBun ≤ un(ζ)(ζ ∈ ∂B) (2.3.1) z∈B z→ζ

Fix n. As a maximum of upper semicontinuous functions, un is upper semicontinuous

function and so there exist continuous function ψk : ∂B → R such that ψk ≥ ψk+1

and limk→∞ ψk = un

Indeed, to show this result we set ψk : ∂B → R by

ψk(x) = supy∈∂B(un(y) − k|x − y|)

It is clear that ψ1 ≥ ψ2 ≥ · · · ≥ un. To show that ψk is continuous, ﬁx w, and notice

that by the deﬁnition of ψk, we have that for every > 0 there exists y such that

15 ψk(w) − ≤ un(y) − k|y − w| ≤ ψk(w)

Now we have,

ψk(x) ≥ un(y) − k|y − x|

= un(y) − k|y − w| + k|y − w| − k|y − x|

≥ ψk(w) − − k|x − w|

and consequently ψk(w) − ψk(x) ≤ + k|x − w|. Interchanging the roles of w and x in the preceeding argument yields |ψk(w) − ψk(x)| ≤ + k|x − w|, after which letting

→ 0 yields that ψk is (even Lipschitz) continuous.

It remains to show that ψk converge to un. Since un is upper semicontinuous, we have that lim supy→x un(y) ≤ un(x). Therefore for all > 0 there exists δ > 0 such that

supB(x,δ) un(y) ≤ un(x) + whenever |y − x| ≤ δ

Given > 0, if y∈ / B(x, δ), we have that if k is large enough, then

un(y) − k|y − x| ≤ supy∈∂B un(y) − kδ ≤ un(x) ≤ ψk(x)

Thus, for suﬃciently large k, we have

ψk(x) = supy∈B(x,δ)(un − k|x − y|) ≤ supy∈B(x,δ) un(y) ≤ un(x) +

Therfore limk→∞ ψk(x) = un(x).

To complete the veriﬁcation of (2.3.1), notice that (2.1.3) theorem tells us that

lim PBψk(z) = ψk(ζ) z→ζ

16 So for any k, lim supz→ζ PBun(z) ≤ limz→ζ PBψk(z) = ψk(ζ). Letting k → ∞ yields (2.3.1).

The second part of the proof of Claim (2.3.3) is to show that uen ≥ un on D. Once this is done, Claim (2.3.3) will follow from the Gluing lemma.

Since uen = un outside B, we just need prove the inequality on on B.

Now let us choose the same continuous function ψk ∈ C(∂B) such that ψk ↓ un.

Notice that the monotone convergence theorem ensures that

Z 2π 1 iθ iθ lim PBψk(z) = lim P (z, e )ψk(e )dθ k→∞ k→∞ 2π 0 Z 2π 1 iθ iθ = P (z, e )un(e )dθ 2π 0

= PBun(z)

We also have proved that limz→ζ PBun(z) ≤ un(ζ) and limz→ζ PBψk(z) = ψk(ζ).

Therefore

PBψk(ψk) ≥ un

on ∂B. Consequently, by the strong maximum principle we have un − PBψk ≤ 0 on

B, in other words un ≤ PBψk on B. Letting k → ∞, we see that un ≤ PBun as k → ∞ on B as required. Claim (2.3.3) is therefore proved.

We are now ready to complete the proof of property (A). We have ue1 ≤ ue2 ≤ ue3 ≤

··· on D, and we claim that ue = limn→∞ uen satisﬁes:

(i) ue ≤ HDφ on D;

(ii) ue(w0) = HDφ(w0);

17 (iii) ue is harmonic on B.

Indeed, since each uen is subharmonic on D, and it is clear that

lim supz→ζ uen(z) = lim supz→ζ un(z) ≤ φ(ζ)(ζ ∈ ∂D ), we have that uen ∈ U, hence uen ≤ HDφ for all n, therefore ue ≤ HDφ which proves (i).

To prove (ii), recall that we showed that uen ≥ un, so

ue(w0) = limn→∞ uen(w0) ≥ limn→∞ un(w0) = HDφ(w0).

Since the reverse inequality follows from (i), we therefore get (ii).

For (iii), we merely note that uen = PBun, and passing to the limit (using the monotone convergence theorem), we get ue = PBu, so ue is harmonic in B.

Claim 2.3.4. Now we want to show that ue = HDφ on B.

To this end, take an arbitrary point w ∈ B, and choose (υn)n≥1 ∈ U such that

υn(w) → HDφ(w). Replacing υn by max(u1, ··· , un, υ1, ··· , υn), we can suppose that

υ1 ≤ υ2 ≤ υ3 ≤ · · · and υn ≥ un on D. Let υen denote the poisson modiﬁcation of υn.

Then as before υen ↑ υe satisﬁes:

(i) υe ≤ HDφ on D;

(ii) υe(w) = HDφ(w);

(iii) υe is harmonic on B.

In particular,(i) implies that

18 υe(w0) ≤ HDφ(w0) = ue(w0)

On other hand, υen ≥ uen for each n, so υe ≥ ue. Thus the function ue − υe,which is

harmonic on B , attains a maximum value of 0 at w0. By maximum principle, this

implies that ue − υe ≡ 0 on B. Then it follows that

ue(w) = υe(w) = HDφ(w).

As w was chosen arbitrarily, we have ue = HDφ on B so HDφ is harmonic and bounded. This completes the proof of the claim, and with it, of property (A).

(B) We need to prove that limz→ζ HDφ(z) = φ(ζ) for n.e. ζ ∈ ∂D. Here we follow

the argument from [1] pages 89–90 closely.

Put E1 to be the set of irregular boundary point of D and E2 is the set of points

of discontinuity of φ. Let us ﬁx ζ ∈ ∂D\(E1 ∪E2). So ζ is regular and φ is continuous

at ζ. To prove property (B), it shall suﬃce to show that limz→ξ HDφ(z) = φ(ξ).

Let > 0. Since φ is continuous at ζ, there exists an open neighbourhood N of ζ

such that

ζ0 ∈ ∂D ∩ N ⇒ |φ(ζ0) − φ(ζ)| < .

We now construct a barrier b. We claom that there exists a function b that is

subharmonic in D and satisﬁes

b < 0 on D, b ≤ −1 on D\N, and lim infb→ζ b(z) ≥ −

Indeed, since ζ is regular, by deﬁnition (2.3.1) there exists a neighbourhood Ne of ζ and a subharmonic functon b deﬁned on D ∩ Ne, satisfying

19 b < 0 on D ∩ Ne and limz→ζ b(z) = 0

Let us choose ρ such that B = B(ζ, ρ) satisﬁes B(ζ, ρ) ⊂ Ne ∩N. Then the normalized Lebesgue measure on ∂B is a regular measure (see the Borel regularity theorem proved later in this thesis), and so we can ﬁnd a compact set S ⊂ D ∩ ∂B such that

F = D ∩ ∂B\S has measure < . Since F is a (relatively) open subset of ∂B, the characteristic function χF continuous on F , and thus limz→ξ PB1F (z) = 1 for ξ ∈ F

Now set m = − supS(b) (so that m > 0) and put

b eb = − PB(1). m

Then for ξ ∈ S, b(ξ) ≤ −m, so eb(ξ) ≤ −1. On the other hand, if ξ ∈ F , then limz→ξ PB1F (z) = 1, so whenever ξ ∈ D ∩ ∂B,

lim supeb(z) ≤ −1. (2.3.2) z→ξ z∈D∩B

Let us deﬁne b on D by   max(−1,eb) on D ∩ B b =  −1 on D\B.

Notice that b is subharmonic on D by the Gluing lemma because the boundary condition on eb guarantees that b is upper semicontinuous. But also notice that

b < 0 on D and b ≤ −1 on D\N

Finally, we observe that

b(z) lim infz→ζ b(z) ≥ limz→ζ ( m − PB1F ) = 0 − PB1F (ζ) > −.

20 With this barrier constructed, we can readily complete the proof. With ξ a ﬁxed regular point, set u = φ(ζ) − + (M + φ(ζ)) b, where M = sup∂D |φ|. Then u is

subharmonic on D, and if ζ0 ∈ ∂D then    φ(ζ) − + 0 if ζ0 ∈ ∂D ∩ N0  lim sup u(z) ≤ ≤ φ(ζ0) (2.3.3) z→ζ0  φ(ζ) − − (M + φ(ζ)), if ζ0 ∈ ∂D\N0 

Hence u ≤ HDφ on D. Consequently,

lim infz→ζ HDφ(z) ≥ lim infz→ζ u(z) ≥ φ(ζ) − (1 + M + φ(ζ)),

but since was chosen arbitrarily, we get that

lim infz→ζ HDφ(z) ≥ φ(ζ).

Now repeating the argument with replaced φ by −φ, we have limz→ζ HD(−φ)(z) ≥

−φ(ζ).

However, the maximum principle ensures that HD(−φ) = −HD(φ). Indeed, let

ν be the corresponding family for −φ. Then, given u ∈ U and v ∈ ν, their sum is

subharmonic on D and

lim supz→ζ (u + v) ≤ φ(ζ) − φ(ζ) = 0

Hence by the maximum principle u + v ≤ 0 on D. Taking suprema over all such u

and v, we get HDφ + HD(−φ) ≤ 0 on D. So HDφ ≤ −HD(−φ), then we have that

lim supz→ζ HDφ ≤ − lim supz→ζ HD(−φ) ≤ φ(ζ)

and therefore

21 φ(ζ) ≤ lim infz→ζ HDφ ≤ lim supz→ζ HDφ ≤ φ(ζ)

Therefore, limz→ζ HDφ = φ(ζ)

(C) To prove h is a uniqueness, suppose that h1 and h2 are two solutions. Then h1 − h2 is a bounded harmonic function on D satisfying

limz→ζ (h1 − h2)(z) = limz→ζ h1(z) − limz→ζ h2(z) = φ(ζ) − φ(ζ) = 0

Now by applying the extended maximum principle to ±(h1 −h2), we get h1 = h2.

2.4 Polar sets and Hausdorﬀ measure

Theorem 2.4.1. Let µ be a ﬁnite Borel measure on C with compact support, and suppose that ZZ log |z − w|dµ(z)dµ(w) > −∞

Then µ(E) = 0 for every Borel polar set E.

Proof. Fix a Borel set E and suppose that µ(E) > 0. We need show that E is

not polar. Since µ is a ﬁnite Borel measure, it is regular (see the Borel regularity

Theorem), and so we can choose a compact subset K of E with µ(K) > 0. Now

set µe = µ|K and d = diam(supp(µ)). Then clearly, log |z − w| ≤ log d for every z, w ∈ supp(µ), and so

ZZ ZZ [log |z − w| − log d]dµ(z)dµ(w) ≥ [log |z − w| − log d]dµ(z)dµ(w). K×K C×C

22 But µe is a ﬁnite non-zero measure whose support is a compact subset of E, so ZZ Z Z log |z − w|dµe(z)dµe(w) = [log |z − w| − log d]dµ(z)dµ(w) K K + (log d)µ(K)2 Z Z ≥ [log |z − w| − log d] dµ(z)dµ(w) C C + µ(K)2 log d Z Z 2 = log |z − w|dµ(z)dµ(w) − µ(C) log d C C + µ(K)2 log d > −∞, and so by Deﬁniton (2.1.4) E is not polar.

Corollary 2.4.2. Every Borel polar set has Lebesgue measure zero.

RR Proof. Let ﬁx ρ > 0 , dµ := dA|B(0,ρ). We need show that log |z−w|dµ(z)dµ(w) >

−∞. For z ∈ B(0, ρ),

Z Z Z 2ρ −1 log |z − w|dA(w) = [ ( )dt + log 2ρ]dA(w) B(0,ρ) B(0,ρ) |z−w| t Z Z 2ρ −1 Z = χ{|z−w|

However,

Z 2ρ Z 1 Z 2ρ 1 dA(w)( )dt = area(B(z, t) ∩ B(0, ρ))( )dt ≤ 2πρ2. B(0,ρ) 0 t 0 t |z−w|

Hence we have Z Z log |z − w|dµ(z)dµ(w) ≥ (−2πρ2 + πρ2 log 2ρ)πρ2 > −∞ B(0,ρ) B(0,ρ)

So by Theorem (2.4.1) we have that E ∩ B(0, ρ) has Lebesgue measure zero for any polar set E, and by letting ρ → ∞ we get the result.

23 We next introduce the definition of Hausdorff measure. For s > 0, and δ > 0 we define ∞ s nX s [ o Hδ (E) = inf diam(Uj) : E ⊂ Uj, diam(Uj) ≤ δ . j=1 j Notice that Hs (E) ≥ Hs (E) if δ ≤ δ . We set δ1 δ2 1 2

s s H (E) = sup Hδ (E). δ>0

Instead of take the inﬁmum over all sets, one can instead consider the covering by

balls. We deﬁne

s s Hballs = lim Hδ,balls(E), δ→0 where

s X s Hδ,balls = inf{ rj : E ⊂ ∪jB(xj, rj), rj ≤ δ}. j Since any set U is contained in an open ball B(x, r) where r is any number greater

than diam(U), we get that there is a constant C > 0 such that

s s Hballs(E) ≤ CH (E) for any set E.

Let recall two lemmas. First, the Mass Distintution principle.

Lemma 2.4.3. Suppose that µ is a Borel measure on X, and E ⊂ X satisﬁes µ(E) >

s 0, if there is a constant C0 > 0 so that µ(B(x, r)) ≤ C0r for every x ∈ X, r > 0,

then

Hs(E) ≥ µ(E) . C0

s Proof. Suppose that there exsits C0 > 0 such that µ(B(x, r)) ≤ C0r for every ball

s s s B(x, r). Since H (E) ≥ CHballs(E) it suﬃces to prove the estimate with H replaced

s by Hballs.

24 P∞ Take an arbitrarily cover of E by balls B(xj, rj) then µ(E) ≤ j=1 µ(B(xj, rj)).

s But by assumption we have µ(B(xj, rj)) ≤ C0rj , and so

P∞ s P∞ µ(B(xj ,rj )) 1 P∞ µ(E) r ≥ = µ(B(xj, rj)) ≥ j=1 j j=1 C0 C0 j=1 C0

So for any δ > 0 we have µ(E) ≤ Hs (E) ≤ Hs(E). C0 δ,balls

We shall also need a result on weak convergent of measures.

Lemma 2.4.4. Suppose that E ⊆ C is a compact set. If µn is a sequense of Borel

measures supported on E such that supn µn(E) < +∞. Then there is a subsequence

of the measures µn that converges weakly to a Borel measure µ in the sense that

R R lim fdµn = lim fdµ for every f ∈ C(C).

This lemma is a consequence of the separability of C(E), along with the Riesz

Representation theorem (see A.4 of [1]).

Now we will show that how this lemma allows one to estimate the dimension of

1 the Cantor square. Let λ < 2

k k Then k generation squares Qk,j are a collection of 4 squares of sidelength λ . Let

4k k k Eλ = ∩k ∪j=1 Qk,j. For every k, we may cover Eλ by the 4 squares of sidelength λ , so

4k s X s Hδ (Eλ) ≤ (diamQk,j) j=1 √ s = ( 2λk) 4k

√ provided that 2λk ≤ δ.

25 Consequently, 4k s X s H (Eλ) ≤ lim sup (diamQk,j) k→∞ j=1 √ = ( 2)s lim sup λks4k. k→∞ √ k sk 1 s log 4 s s Now if 4 λ = 1 then s = log 1 4 ⇔ ( λ ) = 4 ⇔ s = 1 . Therefore H (Eλ) ≤ ( 2) . λ log λ

It follows that dimEλ ≤ s.

s To show that H (Eλ) > 0, it suﬃces to construct a Borel measure µ such that

µ(Eλ) = 1. Let µ0 = χQ0 m2 where m2 is two dimensian lebesgue measure, and

k(s−2) P4k µk = λ χ k m2 j=1 Qj

So, for every k, 4k 2 k(s−2) X k µk([0, 1] ) = λ m2(Qj ) j=1

= λk(s−2)4k(λk)2

= 4kλks = 1.

ks Also, if l ≥ k then µl(Qk,j) = λ . Indeed,

4l l(s−2) X µl(Qk,j) = λ m2(Ql,n ∩ Qk,j). n=1 (l−k) 2l There are 4 squares each of them has m2 measure λ . Thus,

l(s−2) (l−k) 2l µl(Qk,j) = λ 4 λ

= λls4(l−k) = λksλ(l−k)s4(l−k) = λks.

2 2 So, each measure µk is a probability measure on [0, 1] (i.e. µk([0, 1] ) = 1). Since

2 [0, 1] is a compact metric space, there is a subsequence µl that converges weakly to

a measure µ. Now we claim that

26 (i) µ(Eλ) = 1

s 2 (ii) If r > 0 there exists c0 > 0 such that µ(B(x, r)) ≤ c0r for every x ∈ [0, 1] .

To show (i), we want show that µl(Qk,j) → µ(Qk,j) as l → ∞ where l > k. We know that for every f ∈ C([0, 1]2)

R R fdµl → fdµ as l → ∞

Take some function f ∈ C([0, 1]2) such that   1 on Qk,j f =  0 on Qk,l for l 6= j

Notice that f ≥ χQk,j , and therefore, Z Z µ(Qk,j) ≤ fdµ = lim fdµl l→∞ Z

≤ lim sup χQk,j dµl l→∞

ks = µl(Qk,j) = λ .

ks So we have µ(Qk,j) ≤ λ .

For the converse direction, note that since µ is a ﬁnite Borel measure, for each

> 0, there is an open set U ⊃ Qk,j such that

µ(Qk,j) ≥ µ(U) −

2 Now pick f ∈ C([0, 1] ) such that supp(f) ⊂ U but f ≡ 1 on Qk,j Then, Z µ(Qk,j) ≥ fdµ − Z = lim fdµl − ≥ lim sup µl(Qk,j) − l→∞ l→∞

27 ks Since > 0 was arbitrary, we get converse direction. Therefore, µ(Qk,j) = λ . Hence,

4k k ks µ(E) = µ(∪j=1Qk,j) = 4 λ = 1.

To show (ii), ﬁrst note that if r > 1 then µ(B(x, r)) ≤ µ(R2) = 1 ≤ rs. If r < 1,

n+1 n consider a ball B(x, r) intersecting Eλ, and set λ ≤ r < λ for some n ∈ Z. But then B(x, r) can intersct at most four Qn,j, so

µ(B(x, r)) ≤ 4λks 4 ≤ λs(k+1) λs 4rs ≤ λs = 16rs

s 1 Thus by Lemma (2.4.3) we get H (Eλ) ≥ 16 ≥ 0 and so

dim(Eλ) ≥ s. (2.4.1)

so we get dimEλ = s.

The second lemma is F rostman,lemma.

Lemma 2.4.5. Let E ⊆ Rn be a compact set then Hs(E) > 0 if and only if there is a measure µ with µ(E) = 1 such that µ(B(x, r)) ≤ crs for every ball B(x,r).

Proof. Suppose that Hs(E) > 0. Then Hs(E ∩ Q) > 0 for some cube Q of sidelength

s n 1. By translation invariance of H we may assume that E∩Q0,(Q0 = [0, 1] ,) satisﬁes

s n H (Q0 ∩ K) > 0. By replacing E by E ∩ Q0, we shall assume that E ⊆ [0, 1] . Then

P∞ s there is a constant α > 0 such that j (diam(Qj)) ≥ α whenever Qj is a collection

∞ ∞ P s √α of cubes with E ⊂ ∪j=1Qj. Thus j=1 `(Qj) ≥ ( n)s .

28 −m For each m, let Dm be the collection of dyadic cubs of sidelength `(Q) = 2 .

(m) Deﬁne a measure µm by

(m) X χQ s µm = `(Q) mn mn(Q) Q∈Dm Q∩E6=∅ s X mn(Q) n = χQmn mn(Q) Q∈Dm Q∩E6=∅

On other words, for each Q ∈ Dm

(m) s 1 µm | = `(Q) χ m if Q ∩ E 6= ∅, Q mn(Q) Q n and also,

(m) µm |Q = 0 if Q ∩ E = ∅.

So if Q ∩ E 6= ∅ then

(m) s mn(Q) s µm (Q) = `(Q) = `(Q) . mn(Q)

(m) (m) Next we modify µm on the cubes in Dm−1 to get a new measure µm−1. Deﬁne for

−(m−1) Q ∈ Dm−1, `(Q) = 2 by

(m) (m) (m) −(m−1)s s µm−1|Q = µm |Q if µm (Q) ≤ 2 = `(Q) and also,

(m) s 1 (m) (m) s µm−1|Q = `(Q) (m) µm |Q if µm (Q) > `(Q) µm (Q)

(m) s So if µm (Q) > `(Q) we have

(m) (m) s µm (Q) s µm−1(Q) = `(Q) (m) = `(Q) for Q ∈ Dm−1 µm (Q)

29 (m) s (m) Still we have µm−1(Q) ≤ `(Q) for every Q ∈ Dm. But now also have µm−1(Q) ≤

s `(Q) for every Q ∈ Dm−1. Thus

(m) s µm−1(Q) ≤ `(Q) for every Q ∈ Dm ∪ Dm−1

(m) (m) Now given µk , for Q ∈ Dk−1 we get µk−1 by setting

(m) (m) (m) s µk−1|Q = µk |Q if µk (Q) ≤ `(Q)

And,

(m) `(Q)s (m) (m) s µk−1|(Q) = (m) µk |Q if µk (Q) > `(Q) µk (Q)

(m) (m) Set µ = µ0 , the measure obtained after going all the way up to the zeroth dyadic generation, then

(m) s m µ (Q) ≤ `(Q) for every Q ∈ ∪k=0Dk.

m m s For x ∈ E, there is a cube Q ∈ ∪k=1Dk such that µ (Q) = `(Q) . Recall that, if Q, Q0 are any two dyadic cubes whose interiors interesect, then either Q ⊆ Q0 or

0 m Q ⊆ Q. Consequently, for each x ∈ E, we may consider the largest cube Qx ∈ ∪0 Dk

s 0 satisfying µ(Qx) = `(Qx) . Then if x, x ∈ E and Qx ∩Qx0 6= φ we have by maximality that Qx = Qx0 . Consider the collection of largest such dyadic cubes Q1, ··· ,QN , this collection is disjoint and E ⊂ ∪Qj, and so we have

(m) X (m) µ (∪jQj) = µ (Qj) j

X s = `(Qj) j α ≥ √ ( n)s

30 Furthermore, E ⊂ [0, 1]n and

µ(m)(E) ≤ µ(m)([0, 1]n) ≤ 1

So there is a subsequence of the measure µ that converges weakly to a measure µ

satisfying

√α 1 ≥ µ(E) ≥ ( n)s

Let B(x, r) be a ball with r < 1,then B(x, r) is contained in a union of 2n-dyadic cubes of sidelength between r and 2r but if k is large enough, the µ(k)-measure of each these cubes Q0 satisﬁes

µ(k)(Q0) ≤ `(Q0)s

So we have

µ(k)(B(x, r)) ≤ 2n(2r)s = 2n+srs

But then µ(B(x, r)) ≤ 2n+srs for any ball B(x, r).

Converesely, suppose there is µ with µ(E) = 1 such that µ(B(x, r)) ≤ crs for

every ball B(x, r). Then by Lemma (2.4.3) we get the result.

Now lets prove that Borel polar sets have s-dimensional Hausdorﬀ measure zero

for each s > 0, and thus are of Hausdorﬀ dimension zero.

Indeed, suppose that E is a compact set with Hs(E) > 0 so by the Frost-

man lemma there exists a measure µ with µ(E) = 1 and supp(µ) ⊂ E such that

31 µ(B(x, r)) ≤ crs for any B(x, r) ⊂ C. We shall show E is non-polar by verifying the condition of Theorem 2.4.1. Fix ρ large enough so that E ⊂ B(0, ρ). Then

Z Z Z 2ρ −1 log |z − w|dµ(w) = ( ( )dt + log 2ρ)dµ(w) B(0,ρ) B(0,ρ) |z−w| t Z Z 2ρ −1 = | ( χ|z−w|

So we have that

Z Z (2ρ)α log |z − w|dµ(z)dµ(w) ≥ −c + ρα log(2ρ) > −∞. B(0,ρ) B(0,ρ) α

Therefore E is not polar.

2.5 Harmonic Measure and Green,Functions

Deﬁnition 2.5.1. Let D be a proper subdomain of C, and denote by B(∂D) the

σ-algebra of Borel subsets of ∂D. A harmonic measure for D is a function ωD :

D × B(∂D) → [0, 1] such that:

(a) for each z ∈ D, the map B 7→ ωD(z, B) is a Borel probability measure on ∂D;

32 (b) if φ : ∂D → R is a continuous function, then HDφ = PDφ on D, where PDφ is the generalized Poisson integral of φ on D, given by

R PDφ(z) = ∂D φ(ζ)dωD(z, ζ)(z ∈ D).

Now we prove theorem about the existence and the uniqueness of harmonic measure.

Theorem 2.5.2. Let D be a domain in C such that ∂D is bounded and non-polar.Then

there exists a unique harmonic measure ωD for D.

Proof. First we shall prove the existence of ωD. Denote the space of continuous function ψ : ∂D → R by C(∂D). We know that the function HDψ : C(∂D) 7→ R is a

solution to the generalized Dirichlet problem. Now let ψ1, ψ2 ∈ C(∂D) and β1, β2 ∈ R,

then β1HDψ1 + β2HDψ2 is also a solution to the generalized Dirichlet problem. Since

the solution is unique we have that

HD(β1ψ1 + β2ψ2) = β1HDψ1 + β2HDψ2 on D.

If ψ ≥ 0 on ∂D, then by the maximum principle HDψ ≥ 0 on D. So for each z ∈ D,

the map ψ 7→ HDψ(z) is a positive linear functional on C(∂D), and hence by the

Riesz Representation therom there exists a Borel measure µz on ∂D such that

R HDψ(z) = ∂D ψdµz (ψ ∈ ∂D)

If ψ ≡ 1 on ∂D then also HDψ ≡ 1 on D. Thus µz is a probability measure. Put

ωD(z, B) = µz(B) for z ∈ D, and B ∈ B(∂D). Then properties (a) and (b) in

Deﬁnition (2.5.1) hold.

33 We now move onto prove the uniqueness. Let µ1 and µ2 be two ﬁnite Borel probability measure such that

Z Z ψdµ1 = ψdµ2 (2.5.1)

for any ψ ∈ C(C). Let be u be an open set. We want to show that µ1(u) = µ2(u).

For k ∈ N, deﬁne  1 if dist(x, uc) ≥ 1 c  k ψK (x) = min(k dist(x, u ), 1) =  0 outside u.

So 0 ≤ ψk < 1 and ψk ≤ ψk+1, and ψk → χu as k → ∞. Indeed, we always have that

ψk(x) ≤ χu(x). On other hand, if x ∈ u there exists rx > 0 such that B(x, rx) ⊂ u that mean dist(x, u) ≥ r , so ψ (x) = 1 if k ≥ 1 . So for each x, ψ (x) = 1 for x k rx k suﬃciently large k.

By the monotone convergence therom we have that

R R limk→∞ ψkdµ1 = χudµ1 = µ1(u), and similarly

R R limk→∞ ψkdµ2 = χudµ2 = µ2(u).

By (2.5.1) we get that

µ1(u) = µ2(u)

To complete the proof we shall need the following theorem.

34 Theorem 2.5.3. (Borel regularity theorem) Suppose that µ is a ﬁnite Borel measure, then for each Borel set E, and every > 0 , there is an open set U ⊃ E and a closed set K ⊂ E such that

µ(U\K) < .

Proof. Deﬁne F to be the class of Borel sets such that if E ∈ F for > 0 there exist

an open set U and a closed set K such that

K ⊂ E ⊂ U and µ(U\K) < .

We need to show that F is σ-algebra containing the open sets. We ﬁrst show that F

c 1 contains the open sets. Let U be open, and set Kj = {x ∈ U : dist(x, U ) ≥ j }. This is closed and we have

S j Kj = U, Kj ⊆ Kj+1.

so, as µ is ﬁnite,

Sn µ(U\ j=1 Kj) → 0 as n → ∞.

SN Thus, there exists N such that µ(U\ j=1 Kj) < , and so U ∈ F. We now check that F is a σ-algebra. It is clear from the deﬁnition that F is closed under taking

complements. Consequently, it suﬃes F is closed under countable unions. Suppose

that Ej ∈ F for every j. Let > 0. Then for every j > 1 there exist an open set Uj

and a closed set Kj such that

Kj ⊂ Ej ⊂ Uj and µ(Uj\Kj) < 2j+1

35 S S Let U = j Ujand E = j Ej. Choose N large enough so that

SN µ(U\ j Uj) < 2

SN Consider the set K = j Kj, this is closed. But now we have K ⊂ E ⊂ U, and

N N [ [ µ(U\K) ≤ µ(U\ Uj) + µ( Uj\K) j=1 j=1 N [ ≤ + µ( (U \K )) 2 j j j N X ≤ + µ(U \K ) 2 j j j=1 N X 1 ≤ + 2 2 2j j=1 = + = 2 2

S∞ So E = j=1 E ∈ F. Hence F contains all the Borel sets.

Now back to the proof of theorem (2.5.2). By Borel regularity theorem, let E be

Borel, > 0, and choose u be open set such that E ⊂ u and µ1(u\E) < . But then

µ1(E) ≤ µ1(u) = µ2(u) < µ2(E) +

So µ1(E) ≤ µ2(E). But interchanging roles gives µ1(E) = µ2(E)

Deﬁnition 2.5.4. Let D be a proper subdomain of C.A Green,sfunction for D is

a map gD : D × D → (−∞, ∞], such that for each w ∈ D:

(a) gD(., w) is harmonic on D\w, and bounded out side each neighbourhood of w.

(b) gD(w, w) = ∞, and as z → w,

36   log |z| + O(1), w = ∞, gD(z, w) =  − log |z − w| + O(1), w 6= ∞;

Lemma 2.5.5. Let D1 and D2 be domains in C with non-polar boundaries. If D1 ⊂

D2 then

gD1 (z, w) ≤ gD2 (z, w) (z, w ∈ D1)

Proof. Deﬁne

u(z) = gD1 (z, w) − gD2 (z, w)(z, w ∈ D1\w).

Notice that u is subharmonic on D1\w. Also u(z) is bounded above outside each neighbourhood of w, and as z → w,

z−w u(z) = log | z−w | + O(1) = O(1)

so u is bounded above on D1\w. Since gD2 ≥ 0,

lim supz→ζ u(z) ≤ limz→ζ gD1 (z, w) = 0 for n.e. ζ ∈ ∂D1

Hence by the extended maximum principle, u ≤ 0 on D1\w, and this yields that

gD1 (z, w) ≤ gD2 (z, w)(z, w ∈ D1)

We now give an important result which provides a link between the Green’s function and harmonic measure. The proof follows that of Theorem 4.4.7 of [1] closely.

37 Theorem 2.5.6. Let D be a bounded domain in C. Then

R gD(z, w) = ∂D log |ζ − w|dwD(z, ζ) − log |z − w| (z, w ∈ D).

Proof. Given w ∈ D, deﬁne φw : ∂D → R by

φw(ζ) = log |ζ − w| (ζ ∈ ∂D).

Since PDφw is harmonic and bounded on D, and limz→ζ PDφw(z) = φw(ζ) for n.e.ζ ∈

∂D, we have that

(z, w) → gD(z, w) = PDφw(z) − log |z − w| satisﬁes the properties (a),(b), and (c) of deﬁnition (2.5.4)

To show that this function is the Green function, so we claim that the Green function is unique. Suppose that g1 and g2 are two Green function for D. Given w ∈ D, deﬁne

h(z) = g1(z, w) − g2(z, w)(z ∈ D\w)

Then h is harmonic and bounded on D\w, and also from Lemma 2.5.5 we have that limz→ζ h(z) = 0 for n.e ζ ∈ ∂D. Thus the extended maximum principle yields that h ≡ 0 on D\w. Since this hold for each w ∈ D so g1 = g2 on D × D.

Theorem 2.5.7. (Symmetry Theorem) Let D be a bounded subdomain of C such that ∂D is non-polar. Then

gD(z, w) = gD(w, z)(z, w ∈ D)

Proof. Let w ∈ D. Let deﬁne υ on D\w by

38 υ(z) = gD(z, w) − gD(w, z)(z ∈ D\w)

With the aid of the previous theorem, we have

R υ(z) = gD(z, w) + log |z − w| − ∂D log |ζ − z|dωD(w, ζ)(z ∈ D\w).

The function υ is subharmonic on D\w and bounded above. On the other hand, from

the deﬁnition of υ, we have that

lim supz→ζ υ(z) ≤ limz→ζ gD(z, w) = 0 for n.e ζ ∈ ∂D,

and so by the extended maximum principle υ ≤ 0 on D\w. We therefore arrive at

gD(z, w) ≤ gD(w, z)(z ∈ D).

Since w is arbitrary, we get the result.

2.6 Hausdorﬀ measure and harmonic measure

We conclude the thesis with two results that relate the Harmonic measure to

Hausdorﬀ measure.

The ﬁrst lemma is a simpliﬁed version of Bourgain’s lemma [2] for the Cantor set

E = E 1 (the choice of λ for which Eλ is one dimensional). 4

Lemma 2.6.1. Let E = E 1 be the Cantor set of C. If z0 ∈ E, r > 0, then there 4 exists δ > 0 such that

ωC\E(z, B(z0, r)) ≥ c whenever z ∈ B(z0, δr).

39 Proof. Consider the ﬁnite Borel measure µ that we constructed with supp(µ) ⊂ E

n n n satisfying µ(Qj ) = `(Qj ), where Qj are the squares that comprise the j-th generation of E.

Then there is a constant C0 > 0 such that µ(B(z, r)) ≤ C0r for every z ∈ C and r > 0, and also µ(B(z, r)) ≥ 1 r whenever z ∈ supp(µ) and r ∈ (0, 10). C0

Let δ > 0, z0 ∈ E, r ∈ (0, 10). Consider the function

Z 2r u(z) = log dµ(ζ)(ζ ∈ B(z0, δr)), B(z0,δr) |z − ζ| which is harmonic out of supp µ and satisﬁes that

(i) For z ∈ B(z0, δr) then |z − ζ| ≤ |z − z0| + |z0 − ζ| = 2δr for every ζ ∈ B(z0, δr).

Therefore we have

Z 2r u(z) = log dµ(ζ) B(z0,δr) |z − ζ| Z 2r ≥ log dµ(ζ) B(z0,δr) 2rδ 1 = log(1/δ)µ(B(z0, δr)) ≥ (δr) log(1/δ). C0

(ii) For z ∈ B(z0, δr)

Z 2r Z Z 2r 1 u(z) = log dµ(ζ) = dtdµ(ζ) B(z0,δr) |z − ζ| B(z0,δr) |z−ζ| t Z Z 2r 1 = χ{|z−ζ|

40 1 (iii) For z ∈ ∂B(z0, r), ξ ∈ B(z0, δr) the provided that δ ≤ 2 we have that,

r r |z − ζ| ≥ |z − z | − |ζ − z | ≥ r − δr ≥ r − = , 0 0 2 2

Thus

Z 2r Z u(z) = log dµ(ζ) ≤ log(4)dµ(ζ) ≤ C3δr. B(z0,δr) |z − ζ| B(z0,δr)

While also, |z − ξ| ≤ 2r, which ensures that u(z) ≥ 0.

Since u is harmonic outside of B(z0, δr) by maximum principle we have that

max u(z) ≤ max( max u(z), max u(z)) ≤ C4rδ log(1/δ). B(z0,r)\B(z0,δr) ∂B(z0,r) ∂B(z0,δr)

So max u ≤ C rδ log(1/δ) B(z0,r) 4

Now set υ(z) = u(z)−max∂B u(z) then υ satisﬁes that maxB u(z)

(i) υ is harmonic outside of B(z0, δr)

(ii) υ(z) ≤ 0 on ∂B(z0, r)

(iii) υ(z) = u(z)−max∂B u(z) ≤ 1 maxB u(z)

(iv) For z ∈ B(z0, δr) we have

δr 1 υ(z) ≥ ( log(1/δ) − C3δr)( ) ≥ c5, C0 C4δr log(1/δ)

for some constant c5 > 0 that is chosen independently of δ, provided that δ has

been chosen small enough.

Let ψ be any continuous function such that

41 ψ ≡ 1 on B(z0, r) and ψ ≥ 0 on C.

Deﬁne

R w(z) = ψ(ζ)dωC\E(z, ζ) for z ∈ B(z0, r).

Since w is harmonic measure therefore w(z) = 1 n.e on E ∩ B(z0, r). Also, w is harmoinc outside E. So by maximum principle w ≥ v on B(z0, r). Thus w ≥ c5 on

B(z0, δr). By choosing infumum over such functions, we get that

ωC\E(z, B(z0, r)) ≥ c5 for z ∈ B(z0, δr).

We ﬁnish this thesis with a lemma concerning how the harmonic measure of a ball changes with the choice of pole. This lemma has recently proved useful in deep studies of the harmonic measure, see [3].

Lemma 2.6.2. Let Ω ∈ C be an open set with Ωc bounded and non-polar. Let ζ ∈ ∂Ω.

Let ζ ∈ ∂Ω, r > 0, and B = B(ζ, r). Fix M > 1, suppose there is xB ∈ Ω so that

r B0 = B(xB, M ) satisﬁes 4B0 ⊂ Ω ∩ B. Then

1 ωΩ(x, B) ≥ C(M)ωΩ(xB,B)|gΩ(x, xB) − gΩ(x, z)| where x ∈ Ω\B0, z ∈ 2 B0.

1 Proof. Without loss of generality suppose ωΩ(xB,B) > 0. For x ∈ ∂B0, z ∈ 2 B0, we

42 have by Theorem 2.5.6 that Z |gΩ(x, xB) − gΩ(x, z)| = | log |ζ − xB|dωΩ(x, ζ) − log |x − xB| ∂Ω Z + log |x − z| − log |ζ − z|dωΩ(x, ζ)| ∂Ω Z |x − z| |ζ − z| = log( ) − log( )dωΩ(x, ζ) |x − xB| ∂Ω |ζ − xB|

1 Since x ∈ ∂B , we have |x−x | = r , and since z ∈ B it follows that |x −z| < 0 B M 2 0 B r 2M . Thus r r r |x − z| ≥ |x − x | − |x − z| ≥ − = . B B M 2M 2M Also we have

r r 3r |x − z| ≤ |x − x | + |x − z| = + = . B B M 2M 2M

Therefore we have 1 |x − z| 3 ≤ ≤ 2 |x − xB| 2

r Since ζ ∈ ∂Ω and 4B0 ⊂ Ω ∩ B. Then 4B0 ∩ ∂Ω = φ, |ζ − xB| ≥ 4 M , and

r 1 |xB − z| < 2M ≤ 8 |ζ − xB|

Therefore

1 9 |ζ − z| ≤ |ζ − x | + |x − z| ≤ |ζ − x | + |ζ − x | = |ζ − x |, B B B 8 B 8 B and also

1 7 |ζ − z| ≥ |ζ − x | − |x − z| ≥ |ζ − x | − |ζ − x | = |ζ − x |. B B B 8 B 8 B

Consequently, we have that 7 |ζ − z| 9 ≤ ≤ 8 |ζ − xB| 8 43 Combining these bounds, we get that

Z |x − z| |ζ − z| |gΩ(x, xB) − gΩ(x, z)| = log − log dωΩ(x, ζ) |x − xB| ∂Ω |ζ − xB| 3 8 ≤ log + log 2 7 12 = log ≤ log 3 < 2 7

2r 2r From the Harnack inequality for 2B0 = B(xB, M ) we have for R < M that

2r 2r ( M ) − R ( M ) + R 2r ωΩ(xB,B) ≤ ωΩ(x, B) ≤ 2r ωΩ(xB,B) ( M ) + R ( M ) − R

2r ( M )+R Fix c = 2r so we have ( M )−R ω (x, B) c−1 ≤ Ω ≤ c (2.6.1) ωΩ(xB,B)

1 So for every ball B0 with 2B0 ⊂ Ω and if z ∈ 2 B0, x ∈ ∂B0, we have

1 ωΩ(x, B) |gΩ(x, xB) − gΩ(x, z)| ≤ 2 = (2c)( ) ≤ (2c) (2.6.2) c ωΩ(xB,B)

Now let us deﬁne

ωΩ(x,B) u(x) = |gΩ(x, xB) − gΩ(x, z)| − (2c) ωΩ(xB ,B)

which is subharmonic in Ω\B0. We want prove that u(x) ≤ 0 in Ω\B0. We will do

so by using the extended maximum principle. First note that we have that u(x) ≤ 0

on ∂B0. Also by deﬁniton of Green function we have that

|gΩ(x, xB) − gΩ(x, z)| = 0

on ∂Ω for every x outside a polar set, so u(x) ≤ 0 n.e on ∂Ω.

44 To apply the maximum principle, it remains to show that lim sup|x|→∞ u(x) ≤ 0.

Since the Green function is symmetric,

|gΩ(x, xB) − gΩ(x, z)| = |gΩ(xB, x) − gΩ(z, x)| Z = | log |ζ − x|dωΩ(xB, ζ) ∂Ω Z − log |x − xB| − log |ζ − x|dωΩ(z, ζ) + log |x − z|| ∂Ω |x − z| Z = | log + log |ζ − x|dωΩ(xB, ζ) |x − xB| ∂Ω Z − log |ζ − x|dωΩ(z, ζ)|. ∂Ω

Notice that ω(xB, ζ) and ω(z, ζ) are probability measures supported on the compact set ∂Ω. But for any two probability measures supported on ∂Ω, we have that

Z Z | log |x − ζ1|dµ(ζ1) − log |x − ζ2|dν(ζ2)| → 0 as |x| → ∞ (2.6.3)

To see this, note that for each ζ1, ζ2 ∈ ∂Ω, we have that

|x − ζ | 1 → 1 uniformaly as |x| → ∞ (2.6.4) |x − ζ2|

But then

RR |x−ζ1| | log dµ(ζ1)dν(ζ2)| → 0 as |x| → ∞, |x−ζ2| as required. It follows that

|gΩ(x, xB) − gΩ(x, z)| → 0 as |x| → ∞.

So lim sup|x|→∞ u(x) ≤ 0 as required.

45 BIBLIOGRAPHY

[1] T. Ramsford, Potential Theory in the Complex Plane, 1995. [2] J. Bourgain, On the Hausdorﬀ measure of harmonic measure in higher dimension, Invent. Math. 87, 477–483 (1987) [3] J. Azzam, M. Mourgoglou, X. Tolsa, Rectiﬁability of harmonic measure in domains with porous boundaries, arXiV : 1505. 0608 (2015).