Helms L.L. Potential Theory

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Chapter 4

Negligible Sets

4.1 Introduction

In the early development of potential theory, the singularities of superhar-

monic functions gave rise to the concept of a polar set. At the same time,

the classic notion of capacity of a conductor in a grounded sphere became an

object of interest to mathematicians. Both of these concepts are fully devel-

oped in this chapter and will culminate in the characterization of polar sets

as sets of capacity zero. An essential part of this development is Choquet’s

theory of capacities which has important applications to stochastic processes

as well as potential theory. These concepts are used to settle questions per-

taining to equilibrium distribution of charges, the existence of Green function

for regions in R

, and the boundary behavior of Green functions. It will be

shown that an open subset of R

has a Green function if and only if the

region supports a positive superharmonic function.

4.2 Superharmonic Extensions

In the proof of Zaremba’s theorem, Theorem 2.6.29, it was possible to con-

clude that a harmonic function u was nonnegative on Ω knowing that

lim inf

y→x,y∈Ω

u(y) ≥ 0,x∈ ∂Ω,

except possibly for a single point in ∂Ω. This suggests that a single point

is in some sense negligible. By examining the proof more carefully, it can

be seen that the same proof would work if there were a nonnegative super-

harmonic function taking on the value +∞ at those points x ∈ ∂Ω where

lim inf

y→x,y∈Ω

u(y) ≥ 0 fails to hold. This suggests that the set of points

L.L. Helms, Potential Theory, Universitext, 149

 Springer-Verlag London Limited 2009

150 4 Negligible Sets

where a superharmonic function takes on the value +∞ constitutes a negli-

gible set as far as boundary behavior of potentials is concerned.

Deﬁnition 4.2.1 AsetZ ∈ R

is a polar set ifthereisanopensetΛ ⊃ Z

and a superharmonic function u on Λ such that u =+∞ on Z (and perhaps

elsewhere); Z is an inner polar set if each compact subset of Z is a polar

set.

Deﬁnition 4.2.2 A statement concerning points of R

is said to hold quasi

everywhere or q.e. if the set of points where the statement fails to hold is a

polar set; the statement is said to hold inner quasi everywhere or inner

q.e. if the set of points where it fails to hold is an inner polar set.

Example 4.2.3 Let x

be a ﬁxed point of R

.ThenZ = {x

} is a polar set

since u

is superharmonic on R

and u

)=+∞.Also,u

=0q.e.

Example 4.2.4 Let {x

} be a sequence of distinct points in R

.Then

Z = {x

,...} is a polar set. To see this, let y be a ﬁxed point in ∼ Z.

Choose constants c

> 0 such that



|y −x

< +∞.

Deﬁning u(x)=



∞

j=1

|x − x

−1

, u is superharmonic on R

by Theorem 2.4.8 since it is the limit of an increasing sequence of super-

harmonic functions and u(y) < +∞. Clearly, u =+∞ on Z.

Theorem 4.2.5 If Z is a polar set, then it is a subset of a G

polar set.

Moreover, any subset of a polar set is a polar set.

Proof: Let u be a superharmonic function on a neighborhood Λ of Z which

is +∞ on Z.Since{x; u(x)=+∞} = ∩

({x; u(x) >j} and each of the

latter sets is open by the l.s.c. of u, {x; u(x)=+∞} is a G

set. The second

statement is immediate from the deﬁnition of polar set.

Theorem 4.2.6 If Z ⊂ R

is a polar set, then Z has n-dimensional Lebesgue

measure zero.

Proof: Let u be superharmonic on the open set Λ ⊃ Z and +∞ on Z.Since

u is ﬁnite a.e. by Theorem 2.4.2, Z has Lebesgue measure zero.

Example 4.2.7 A line segment I ⊂ R

of positive length is not a polar set.

To see this, assume that there is a superharmonic function u on a neigh-

borhood Λ of I with u =+∞ on I. It follows from the averaging principle

that superharmonicity is invariant under rotations about a point. If x is an

interior point of I,chooseρ>0 such that B

x,ρ

⊂ Λ and choose a subinterval

4.2 Superharmonic Extensions 151

of I with an end point at x of length less that ρ/

√

2. Rotating I

about

x,anintervalI

can be constructed that is perpendicular to I

and has an

end point at x; a superharmonic function u

can be constructed on the ball

containing I

and I

which is +∞ on I

by means of the rotation about x.

Continuing in this way, a square and four superharmonic functions can be

constructed on a neighborhood of the square with at least one of the four

functions taking on the value +∞ on each side of the square. The sum v of

the four functions is then superharmonic on a neighborhood of the square and

is equal to +∞ on the boundary of the square. By the minimum principle,

v must be +∞ on the square, contradicting the fact that a superharmonic

function is ﬁnite a.e. on its domain. Thus, I is not polar.

Example 4.2.8 A line segment in R

is a polar set. Consider, for example,

the line segment I joining (0, 0, 0) to (1, 0, 0), one-dimensional Lebesgue mea-

sure μ on I, and the Newtonian potential

(x)=



|x − z|

dμ(z),x∈ R



((x

− z)

+ x

)

1/2

dz.

By Theorem 3.4.4, U

is superharmonic. If x =(x

, 0, 0) ∈ I,then

(x)=



− z|

dz =+∞

so that U

=+∞ on I. This shows that I is a polar subset of R

The deﬁnition of a polar set Z requires that a superharmonic function u be

deﬁned only on a neighborhood of Z and that it be inﬁnite on Z.Ontheone

hand, this deﬁnition makes it easy to conﬁrm that some sets are polar, but

on the other hand, makes it diﬃcult to show that the union of two disjoint

polar sets is again polar. A global alternative is essential.

Theorem 4.2.9 If Z ⊂ R

is a polar set, then there is a superharmonic

function u such that u =+∞ on Z.

Proof: Let Λ be a neighborhood of Z on which there is deﬁned a superhar-

monic function u such that u =+∞ on Z. It can be assumed that u>0on

Λ by replacing Λ with the open set Λ ∩{y; u(y) > 0}, if necessary. Letting

= Λ ∩ B

0,j

,j ≥ 1,Z = ∪

(Λ

∩ Z). By a change of notation, if nec-

essary, it can be assumed that each Λ

is nonempty. Since Λ

is bounded,

it has a Green function G

by Theorem 3.2.10. By the Riesz decompo-

sition theorem,Theorem 3.5.11, there is a μ

associated with u such that

u = G

+ h

,whereh

is harmonic on Λ

.Sinceu =+∞ on Λ

∩ Z and

≤ G

0,j

, the latter is +∞ on Λ

∩ Z. Deﬁning

152 4 Negligible Sets

(x)=



0,j

log

|x − y|

dμ

(y),x∈ R

diﬀers from the logarithmic potential of μ

only by a constant and is

superharmonic by Theorem 3.4.10. Referring to Deﬁnition 1.5.2,

log

|x − y|

≥ G

0,j

(x, y),x,y∈ B

0,j

so that w

=+∞ on Λ

∩Z,sinceG

0,j

=+∞ on Λ

∩Z.Insummary,(i)

is superharmonic, (ii) w

≥ 0onB

0,j

, and (iii) w

=+∞ on Λ

∩Z.Since

each w

is ﬁnite a.e. on B

0,1

,thereisapointx

∈ B

0,1

such that 0 <w

) <

+∞ for all j ≥ 1. A sequence of positive numbers {b

} therefore can be

chosen so that the series



) converges. Deﬁning v =



,v =

+∞ on Z. It remains only to show that v is a superharmonic function. By

Theorem 2.4.8, v is either superharmonic or identically +∞ on R

.The

former is true since v(x

) < +∞.

Theorem 4.2.10 If Z ⊂ R

,n ≥ 3, is a polar set, then there is a Borel

measure ν on R

such that the potential Gν =+∞ on Z.

Proof: By hypothesis, there is an open set Λ ⊃ Z and a superharmonic

function u on Λ with u =+∞ on Z. As in the preceding proof, it can be

assumed that u>0onΛ.Letμ be the measure associated with u and Λ by

the Riesz decomposition theorem. Extend μ to R

by putting μ(∼ Λ)=0.

Then u = G

μ + h,whereh is harmonic on Λ.Let{Λ

} be a sequence

of nonempty open sets having compact closures such that R

= ∪

.If

= μ|

,thenμ

(Λ

)=μ(Λ

) < +∞ for each j ≥ 1. Also extend μ

to R

by putting μ

(∼ Λ

) = 0. Letting u

= Gμ

, u

≥ 0 and is superharmonic

since μ

)=μ

(Λ

) < +∞.Since

u = G

μ + h = G

+ G

μ|

Λ∼Λ

+ h on Λ

and μ|

Λ∼Λ

(Λ

)=0,

u = G

+ h

on Λ ∩ Λ

where h

is harmonic on Λ

by Theorem 3.4.7. Since u =+∞ on Z, G

+∞ on Z ∩Λ

. By Theorem 3.4.8,

= G

Λ∩Λ

on Λ ∩ Λ

where

is harmonic on Λ∩Λ

and it follows that G

Λ∩Λ

=+∞ on Z ∩Λ

Applying Theorem 3.4.8 again,

= G

Λ∩Λ

+ h

∗

on Λ ∩ Λ

4.2 Superharmonic Extensions 153

where h

∗

is harmonic on Λ ∩Λ

and it follows that G

=+∞ on Z ∩Λ

Since u

= Gμ

≥ G

=+∞ on Z ∩ Λ

, u

=+∞ on Z ∩ Λ

.Lety

be a ﬁxed point of Λ

such that u

(y) < +∞ for all j ≥ 1, and let {c

}

be a sequence of positive numbers such that



(y)converges.Putv =



.Sincetheu

’s are nonnegative superharmonic functions and v(y) <

+∞, v is a nonnegative superharmonic function on R

by Theorem 2.4.8 and

v =+∞ on Z. By the Riesz decomposition theorem, there is a Borel measure

ν on R

and a harmonic function H such that v = Gν + H.Sincev =+∞

on Z, Gν =+∞ on Z.

Since superharmonic functions are ﬁnite a.e. according to Theorem 2.4.2,

polar sets have Lebesgue measure zero. The following theorem describes polar

sets as small sets in another way.

Theorem 4.2.11 If Z is a polar set, its intersection with any sphere has

surface area zero.

Proof: By the two preceding theorems, there is a superharmonic function u

such that u =+∞ on Z.IfB = B

x,ρ

is any ball, then L(u : x, ρ) is ﬁnite by

Theorem 2.4.3. It follows that Z ∩ ∂B has surface area zero.

Theorem 4.2.12 If {Z

} is a sequence of polar sets in R

,then∪

is a

polar set.

Proof: For each j ≥ 1, let u

be superharmonic and u

=+∞ on Z

.Since

the u

’s are ﬁnite a.e.,thereisapointy ∈ B

0,1

such that u

(y) < +∞ for all

j ≥ 1. Since u

is l.s.c., inf

|x|≤j

(x) is ﬁnite. Thus, a sequence of positive

numbers {b

} canbechosensuchthat



(y) − inf

|x|≤j

(x))

converges. Let

u =



− inf

|x|≤j

(x)),

and consider a ﬁxed ball B

0,k

. All terms except the ﬁrst k − 1oftheseries

deﬁning u are nonnegative on B

0,k

and the series

∞



j=k

− inf

|x|≤j

(x))

is superharmonic on B

0,k

by Theorem 2.4.8. Since the sum of the ﬁrst k − 1

termsissuperharmoniconB

0,k

,uis superharmonic on B

0,k

for every positive

integer k and therefore superharmonic on R

.Sinceu

=+∞ on Z

, u =+∞

on Z.

In the following discussion, {B

; i ≥ 1} will be a countable base for the

Euclidean topology on R

consisting of balls.

154 4 Negligible Sets

Remark 4.2.13 Polarity is a local property;thatis,ifforeachx ∈ Z

there is a ball B

containing x such that Z ∩ B

is polar, then Z is polar.

This can be shown as follows. For each x ∈ Z,thereisaballB

such that

x ∈ B

⊂ B

with Z ∩ B

polar since it is a subset of the polar set

Z ∩ B

. It therefore can be assumed that each B

belongs to the countable

base {B

; i ≥ 1}.SinceZ ⊂∪(Z ∩B

) and the latter is a countable union

of polar sets, Z is a subset of a polar set by the preceding theorem and is

therefore polar. Similarly, inner polarity is a local property; that is, if for each

x ∈ Z there is a ball B

containing x such that Z ∩B

is inner polar, then Z

is inner polar. To see this, let Γ be a compact subset of Z.Asabove,itcan

be assumed that each B

is a member of the countable base {B

; i ≥ 1}.

Suppose B

= B

.SinceΓ ∩B

is inner polar and each Γ ∩B

−

−1/m

,m≥

1, is a compact subset of Γ ∩ B

,Γ ∩ B

= ∪

∞

m=1

Γ ∩ B

−

x−j,r

−1/m

is a polar

set. Since Γ ⊂ (Γ ∩ B

),Γ is polar and therefore Z is inner polar.

The preceding results provide a quantitative measure of the smallness of

polar sets. The results will be applied to the problem of extending superhar-

monic functions across polar sets and to the existence of Green functions in

the two-dimensional case.

Lemma 4.2.14 Let Ω be an open set, and let Z be a relatively closed polar

subset of Ω.Ifx ∈ Z and {x

} is a sequence of distinct points in Ω ∼ Z such

that x = lim

j→∞

, then there is a superharmonic function v on Ω such that

v =+∞ on Z and v(x

) < +∞ for all j ≥ 1.

Proof: For each j ≥ 1, a number δ

> 0 can be chosen so that B

−

,δ

⊂ Ω ∼ Z

and B

−

,δ

∩ B

−

,δ

= ∅ for i = j. It is easy to see that lim

j→∞

=0.Since

Z is polar, there is a superharmonic function v such that v =+∞ on Z.By

Lemma 2.4.11, the function



PI(v : B

)onB

v on Ω ∼ B

where B

= B

,δ

, is superharmonic on Ω.Thenv = v

=+∞ on Z and

) < +∞. Inductively deﬁne a sequence of superharmonic functions {v

}

by putting



PI(v

j−1

: B

)onB

j−1

on Ω ∼ B

where B

= B

,δ

. The sequence {v

} is then a decreasing sequence of su-

perharmonic functions on Ω, v

=+∞ on Z,andv

) < +∞ for all j ≥ 1.

Consider the function ˜v = lim

j→∞

. Given any neighborhood of x,thev

’s

agree outside the neighborhood except possibly for a ﬁnite number of terms

of the sequence. This means that ˜v is superharmonic on Ω ∼{x}.Toshow

that ˜v is superharmonic on Ω, it suﬃces to show that ˜v is l.s.c. at x, ˜v being

locally super-mean-valued at x since ˜v(x)=+∞. By the l.s.c. of v at x and

4.2 Superharmonic Extensions 155

the fact that v(x)=+∞, for each integer k ≥ 1thereisaneighborhoodΛ

of x such that v>kon Λ

. Moreover, it can be assumed that there is an

integer j

such that B

−

⊂ Λ

for all j ≥ j

. It follows that v

≥ k on Λ

for

all j ≥ j

, and therefore that ˜v ≥ k on Λ

.Thus,+∞ =˜v(x) = lim

y→x

˜v(y);

that is, ˜v is continuous (in the extended sense) at x.

The following theorem was ﬁrst proved by Schwarz [56] for a bounded

harmonic function on a deleted neighborhood of a point, then by Bouligand [5]

for bounded harmonic functions and Z compact polar, and then by Vasilesco

[61] for Z a compact polar set.

Theorem 4.2.15 (Brelot [7]) Let Ω be an open set and let Z be a relatively

closed polar subset of Ω.Ifu is superharmonic on Ω ∼ Z and locally bounded

below on Ω, then it has a unique superharmonic extension to Ω.

Proof: Note ﬁrst that Ω ∼ Z is dense in Ω, for if this were not the case, then

Z would have a nonempty interior and therefore positive Lebesgue measure,

a contradiction. Since Ω ∼ Z is dense in Ω, a function ˜u canbedeﬁnedonΩ

by putting ˜u = lim inf

y→x,y∈Ω∼Z

u(y)forx ∈ Ω.Sinceu is locally bounded

below on Ω,˜u cannot take on the value −∞. Clearly, ˜u is l.s.c. on Ω.Since

˜u = u on the open set Ω ∼ Z,˜u is superharmonic on Ω ∼ Z and therefore

not identically +∞ on any component of Ω. To show that ˜u is superharmonic

on Ω, it remains only to show that ˜u is locally super-mean-valued on Ω.This

is true at points of the open set Ω ∼ Z since ˜u = u on this set. Consider

any x ∈ Z and let {x

} be a sequence of distinct points in Ω ∼ Z such that

lim

j→∞

= x and lim

j→∞

˜u(x

)=˜u(x). Since Z is a polar set, there is a

superharmonic function v such that v =+∞ on Z. By the preceding lemma,

itcanbeassumedthatv(x

) < +∞ for each j ≥ 1. Since ˜u cannot take on

the value −∞ and is l.s.c. on Ω,˜u + v is deﬁned and l.s.c. on Ω for every

>0. For y ∈ Z and B

−

y,ρ

⊂ Ω, +∞ =(˜u + v)(y) ≥ L(˜u + v : y, ρ). This

shows that ˜u + v is superharmonic on Ω.Sincev(x

) < +∞ and

˜u(x

)+v(x

) ≥ A(˜u : x

,δ)+A(v : x

,δ),

letting  → 0

˜u(x

) ≥ A(˜u : x

,δ)=



,δ

(y)˜u(y) dy.

Since ˜u is bounded below on the compact closure of a neighborhood of x,a

δ can be chosen so that all but a ﬁnite number of the integrands on the right

are bounded below. By Fatou’s Lemma,

˜u(x) = lim

j→∞

˜u(x

) ≥ A(˜u : x, δ)

for all suﬃciently small δ. This concludes the proof that ˜u is superharmonic

on Ω. To prove uniqueness, let

u be a second such extension of u.SinceZ

156 4 Negligible Sets

has Lebesgue measure zero, ˜u = ua.e.on Ω and A(˜u : x, δ)=A(u : x, δ)

whenever B

x,δ

⊂ Ω. Letting δ ↓ 0, ˜u(x)=u(x) for all x ∈ Ω by Lemma 2.4.4.

Corollary 4.2.16 Let Ω be an open set, Z a relatively closed polar subset of

Ω,andu a function that is harmonic on Ω ∼ Z and locally bounded on Ω.

Then u has a unique harmonic extension.

Proof: By the preceding theorem, u has a superharmonic extension u



Ω. Applying the same result to −u, it has a subharmonic extension u



on Ω

with u



= u



on Ω ∼ Z.SinceZ has Lebesgue measure zero, A(u



: x, δ)=

A(u



: x, δ) whenever B

x,δ

⊂ Ω. Letting δ ↓ 0,u



(x)=u



(x) for all x ∈ Ω

and u



is therefore harmonic on Ω.

Lemma 4.2.17 If Ω is an open set, then ∂Ω is polar if and only if ∼ Ω is

polar.

Proof: Since ∂Ω ⊂∼ Ω, ∼ Ω polar implies that ∂Ω is polar and the suﬃ-

ciency is proved. Assume that ∂Ω is polar. By Theorems 4.2.9 and 4.2.10,

there is a superharmonic function u such that u =+∞ on ∂Ω.IfΛ is any

component of int(∼ Ω), then u =+∞ on ∂Λ ⊂ ∂(int(∼ Ω)) = ∂Ω and

therefore u =+∞ on Λ by the minimum principle, Corollary 2.3.6. Since Λ

is any component of int(∼ Ω),u=+∞ on int(∼ Ω)andthesameistrueon

its boundary. Thus, u =+∞ on ∼ Ω,provingthat∼ Ω is polar.

The concept of polar set can shed some light on the existence of a Green

function for a subset of R

Theorem 4.2.18 If Ω is an open subset of R

having a Green function, then

∼ Ω is not a polar set (equivalently, ∂Ω is not a polar set).

Proof: Assume that ∼ Ω is polar and ﬁx x ∈ Ω.SinceG

(x, ·) ≥ 0on

Ω, by Theorem 4.2.15 it has a nonnegative superharmonic extension on R

denoted by the same symbol. By Theorem 2.5.3, L(G

(x, ·):x, δ)isaconcave

function φ of −log δ.If(ξ)=aξ + b is any supporting line for the convex

set {(ξ,η); η ≤ φ(ξ), −∞ <ξ<+∞},then

L(G

(x, ·):x, δ)=φ(−log δ) ≤−a log δ + b.

Since L(G

(x, ·):x, δ) ↑ +∞ as δ ↓ 0,a must be nonnegative; on the other

hand if a>0, then −a log δ + b is negative for large δ and therefore a =0;

that is, every supporting line for the above convex set is horizontal. This

implies that φ is a constant function. Thus, L(G

(x, ·):x, δ)isaconstant

function on (0, +∞), a contradiction since L(G

(x, δ):x, δ) ↑ +∞ as δ ↓ 0.

Therefore, ∼ Ω cannot be polar.

In deﬁning the upper class U

for a function f on the boundary ∂Ω,itwas

required of a function u ∈ U

that it satisfy the condition lim inf

y→x,y∈Ω

u(y)

≥ f(x) for all x ∈ ∂Ω. This requirement can be relaxed so that the inequality

4.2 Superharmonic Extensions 157

holds only quasi everywhere on ∂Ω.LetU

and L

be deﬁned in the same way

as U

and L

, respectively, except that the boundary condition is required

to hold only quasi everywhere.

Theorem 4.2.19 If f is a boundary function for the bounded open set Ω,

then

=inf{u; u ∈ U

} and H

=sup{u; u ∈ L

Proof: Consider any u ∈ U

.Thenu is superharmonic or identically +∞ on

each component of Ω, u is bounded below on Ω, and lim inf

y→x,y∈Ω

u(y) ≥

f(x) for all x ∈ ∂Ω ∼ Z,whereZ is a polar subset of ∂Ω.Letv be a

superharmonic function such that v =+∞ on Z.SinceΩ is bounded, it can

be assumed that v ≥ 0onΩ. Then for any >0, lim inf

y→x,y∈Ω

(u+ v)(y) ≥

f(x) for all x ∈ ∂Ω. Therefore, u +v ≥

on Ω.IfH

is identically −∞ on

a component of Ω,thenu ≥

on that component; if H

is identically +∞

on a component of Ω,thenu ≥

a.e. on that component and, consequently,

u ≥

on that component; if H

is harmonic on a component of Ω and

y,δ

is a ball in that component, then

A(u : y, δ)+A(v : y, δ) ≥

(y)

and it follows that u ≥

on that component by letting  → 0ﬁrstand

then letting δ ↓ 0. Therefore, inf {u; u ∈ U

}≥H

. Since inf {u; u ∈ U

}≤

inf {u; u ∈ U

} = H

, H

=inf{u; u ∈ U

Generally speaking, the values of a boundary function can be changed on

a polar subset of the boundary without aﬀecting the Dirichlet solution.

Corollary 4.2.20 Let Ω be a bounded open set. If f and g are functions on

∂Ω such that f = g on ∂Ω except possibly on a polar subset of ∂Ω,then

= H

and H

= H

; if, in addition, f is resolutive, then g is also and

= H

on Ω.

Proof: If u ∈ U

, then lim inf

y→x,y∈Ω

u(y) ≥ f (x)=g(x) quasi everywhere

on ∂Ω and u ∈ U

;thatis,U

⊂ U

so that H

≤ H

. Interchanging f and

= H

The following theorem exhibits an essential diﬀerence between the n =2

and n ≥ 3cases.Inthen = 2 case, the point at inﬁnity is a negligible point;

but in the n ≥ 3 case, the point at inﬁnity cannot be ignored.

Theorem 4.2.21 Let Ω be a Greenian set, let u be superharmonic and

bounded below on Ω, and suppose there is a constant α such that

lim inf

y→x,y∈Ω

u(y) ≥ α

for all x ∈ ∂Ω except possibly for an inner polar set. If (i)n =2,or(ii)n ≥ 3

and Ω is bounded, or (iii)n ≥ 3,Ω is unbounded, and lim inf

|y|→+∞,y∈Ω

u(y)

≥ α,thenu ≥ α on Ω.