Helms L.L. Potential Theory

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218 5 Dirichlet Problem for Unbounded Regions

Proof: Fix x ∈ Ω. By Lemma 5.4.6, there is a superharmonic function u

on Ω with u(x) < +∞ such that lim

y→z,y∈Ω

u(y)=+∞ for all z ∈ Z.

Since (1/n)u ∈ U

, 0 ≤ H

(x) ≤ (1/n)u(x) for all n ≥ 1 and therefore

(Z)=



dμ

= H

(x) = 0. The second assertion follows from the ﬁrst

and Equation (5.5).

Remark 5.4.11 Let Ω be an open subset of R

such that ∼ Ω is not polar.

The results of Theorem 5.4.2 then hold for such Ω.Toseethis,considera

closed ball B

−

= B

−

x,δ

⊂ Ω.Iff ∈ C

(∂

∞

Ω), let

g =



f on ∂

∞

on ∂B

x,δ

By considering upper and lower PWB solutions, it is easily seen that

≤ H

Ω∼B

−

≤ H

Ω∼B

−

≤ H

on Ω ∼ B

−

.Sincef is resolutive, H

= H

Ω∼B

−

on Ω ∼ B

−

.Thus,H

has

the same boundary properties as H

Ω∼B

−

at points of ∂

∞

Ω. Since the results

of Theorem 5.4.2 apply to Ω ∼ B

−

, they also apply to Ω.Notealsothatify ∈

∂Ω is a regular boundary point for Ω, then it is also a regular boundary point

for Ω ∼ B

−

and there is a barrier u at y on Ω ∼ B

−

having the additional

property that inf {u(z); z ∈ (Ω ∼ B

−

) ∼ Λ} > 0 for each neighborhood Λ of

y. It is easy to see that the barrier u has a superharmonic extension on Ω

satisfying inf {u(z); z ∈ Ω ∼ Λ} > 0 for each neighborhood Λ of y.

Theorem 5.4.12 Let Ω be a subset of R

such that ∼ Ω is nonpolar and let

x ∈ Ω.Ify

is a ﬁnite, regular boundary point, then lim

y→y

,y∈Ω

(x, y)= 0.

If n ≥ 3 and ∞∈∂

∞

Ω,thenlim

y→∞

(x, y)=0.Ifn =2and ∞ is a

regular boundary point for Ω,thenlim

y→∞,y∈Ω

(x, y)=0.

Proof: Suppose ﬁrst that n ≥ 3and∞∈∂

∞

Ω.SinceG

(x, ·) ≤ G(x, ·)

by Theorem 3.2.12, lim

y→∞

(x, y) = 0. Suppose now that y

is a ﬁnite,

regular boundary point for Ω. By Theorem 5.4.2 and the preceding remark,

there is a barrier u at y

on Ω such that inf {u(z); z ∈ Ω ∼ Λ} > 0 for all

neighborhoods Λ of y

.Chooseα>0 such that Λ

= {z ∈ Ω; G

(x, z) >α}

has compact closure in Ω.Choosingn ≥ 1sothatnu > α on ∂Λ

,nu ≥

(x, ·)on∂Λ

and therefore nu ≥ G

(x, ·)onΩ ∼ Λ

by Theorem 3.3.8.

It follows that lim

y→y

,y∈Ω

(x, y)=0.

It is possible to relate the swept measures δ

(x, ·) to the harmonic mea-

sures μ

by means of the following theorem.

Theorem 5.4.13 If u is a nonnegative superharmonic function on the

Greenian set Ω ⊂ R

, Λ is an open subset of Ω,and

5.4 Boundary Behavior 219

f =



u on Ω ∩ ∂

∞

0 on ∂

∞

Ω ∩ ∂

∞

Λ,

then f is resolutive and H

Ω∼Λ

on Λ.

Proof: Let v be a nonnegative superharmonic function on Ω with v ≥ u

on Ω ∼ Λ. Then lim inf

y→x,y∈Λ

v(y) ≥ f(x)forx ∈ ∂

∞

Λ ∩ Ω and

lim inf

y→x,y∈Λ

v(y) ≥ 0=f(x)forx ∈ ∂

∞

Ω ∩ ∂

∞

Λ.Thus,v ∈ U

.It

follows that R

Ω∼Λ

≥ H

≥ 0onΛ.SinceR

Ω∼Λ

≤ u and the latter

is ﬁnite a.e. on Λ, f is resolutive relative to Λ, H

≤ R

Ω∼Λ

, and therefore

≤

Ω∼Λ

. Now consider any w ∈ U

.Sinceu|

∈ U

, min (w, u|

) ∈ U

Deﬁne ˜w on Ω by

˜w =



min (w, u|

)onΛ

u on Ω ∼ Λ.

Then ˜w is superharmonic on Ω, majorizes u on Ω ∼ Λ, and therefore ˜w ≥

Ω∼Λ

≥

Ω∼Λ

on Ω.Sincew ≥ ˜w on Λ,

= H

≥

Ω∼Λ

on Λ

and it follows that H

Ω∼Λ

on Λ.

Remark 5.4.14 If Λ

−

⊂ Ω, then by Theorem 4.7.4



udμ



u(y) δ

Ω∼Λ

(x, dy),x∈ Λ.

Applying Lemma 5.3.14 to Ω ∼ Λ

−

, it follows that μ

= δ

Ω∼Λ

(x, ·) whenever

−

⊂ Ω.

Let Ω be a Greenian subset of R

. Consider the following extension g

(x, ·)

of G

(x, ·)toR

∞

for x ∈ Ω.

(x, y)=

⎧

⎨

⎩

(x, y)ify ∈ Ω

0ify ∈ R

∼ Ω

−

lim sup

z→y,z∈Ω

(x, z)ify ∈ ∂

∞

Ω.

In the case of unbounded Ω, the deﬁnition of g

(x, ∞) is consistent with the

behavior of G

(x, ·)at∞. According to Theorem 4.5.4, g

(x, ·)=0q.e.on

∂Ω.IfΩ is unbounded, g

(x, ∞)=0inthen ≥ 3case;inthen =2case,

∞ is a polar subset of ∂

∞

Ω according to Theorem 5.3.8 so that q

(x, ·)=0

q.e. on ∂

∞

Ω. The function g

(x, ·) has the following properties:

(i) g

(x, ·)=0q.e.on∂

∞

Ω ,

(ii) g

(x, ·) is subharmonic on R

∼{x},and

(iii) if n ≥ 3orn =2and∞ is a regular boundary point for Ω,then

(x, ∞)=0.

220 5 Dirichlet Problem for Unbounded Regions

Theorem 5.4.15 If Λ is an open subset of the Greenian set Ω ⊂ R

,then

(i) for each x ∈ Ω, the function g

(x, ·)|

∂

∞

is a resolutive boundary func-

tion, and

(ii) g

(x, y)=G

(x, y) − H

(y,·)

(x),x∈ Λ, y ∈ Ω .

Proof: For x ∈ Ω,let

(y)=



(x, y)ify ∈ Ω ∩ ∂

∞

0ify ∈ ∂

∞

Ω ∩ ∂

∞

Λ.

By the preceding theorem, Theorem 5.4.13, the function f

is resolutive on

∂

∞

Λ.Sinceg

=0q.e.on∂

∞

Ω,f

= g

(x, ·)q.e.on∂

∞

Λ so that g

(x, ·)

is resolutive by Theorem 5.4.10 and (i) is true. The proof of (ii) will be split

into two cases.

Case (i):(y ∈ Ω ∼ ∂

∞

Λ). Since G

(y,·)|

∈ U

(y,·)|

≥ H

on Λ.

Since H

is a harmonic minorant of G

(y,·)onΛ, ghm

(y,·) ≥ H

Λ.Consideranyv ∈L

.Thenforz ∈ ∂

∞

lim sup

x→z,x∈Λ

ghm

(y,x) ≥ lim sup

x→z,x∈Λ

v(x) ≥ f

(x),

ghm

(y,·) ∈L

and therefore ghm

(y,·) ≤ H

on Λ.Thus,

ghm

(y,·)=H

on Λ.

Since f

= g

(y,·)q.e.on∂

∞

Λ,ghm

(y,·)=H

(y,·)

on Λ.By

Theorem 3.4.8, for x ∈ Λ, G

(y,x)=G

(y,x)+h where h is a nonneg-

ative harmonic function on Λ.SinceghmG

(y,·)=0onΛ, ghm G

(y,·)=h

on Λ by Lemma 3.3.6. Therefore, for x ∈ Λ,

(x, y)=G

(y,x)

= G

(y,x) − ghm

(y,x)

= G

(x, y) − H

(y,·)

(x)

and (ii)holdsinthey ∈ Ω ∼ ∂

∞

Λ case.

Before discussing the next case, it will be shown that

Ω∼Λ

(y,·),x)=

Ω∼Λ

(x, ·),y),x∈ Λ, y ∈ Ω. (5.6)

For x, y ∈ Λ, it follows from Case (i) that G

(x, y)=G

(x, y) −H

(y,·)

(x).

Since G

and G

are symmetric functions, H

(y,·)

(x) is a symmetric func-

tion on Λ × Λ.Sinceg

(y,·)=G

(y,·)onΩ ∩ ∂

∞

Λ and g

(y,·)=0quasi

everywhere on ∂

∞

Ω ∩ ∂

∞

Λ,forx, y ∈ Λ

5.4 Boundary Behavior 221

Ω∼Λ

(y,·),x)=H

(y,·)

(x)=H

(x,·)

(y)=

Ω∼Λ

(x, ·),y) (5.7)

by Theorem 5.4.13. Suppose now that x ∈ Λ and y ∈ Ω ∼ Λ. Letting Λ

Λ ∪ B

y,1/n

and using the preceding equation,

Ω∼Λ

(y,·),x)=

Ω∼Λ

(x, ·),y).

By Theorem 4.6.5,

Ω∼(Λ∪{y})

(y,·),x)=

Ω∼(Λ∪{y})

(x, ·),y).

Since Ω ∼ (Λ∪{y})diﬀersfromΩ ∼ Λ only by a polar set, by Corollary 4.6.4

the preceding equation holds with Ω ∼ (Λ ∪{y}) replaced by Ω ∼ Λ.Thus,

Equation (5.6) has been established.

Case (ii):(y ∈ Ω ∩ ∂

∞

Λ). Suppose ﬁrst that y is a regular boundary point

for Λ.Since

Ω∼Λ

(y,·), ·)=R

Ω∼Λ

(y,·), ·)q.e.onΩ and polar sets

have Lebesgue measure zero, there is a set Z ⊂ Ω of Lebesgue measure

zero such that

Ω∼Λ

(y,·),x)=R

Ω∼Λ

(y,·),x)forx ∈ Ω ∼ Z.By

Theorem 2.4.7,

Ω∼Λ

(x, ·),y) = lim inf

z→y,z∈Z∪{y}

Ω∼Λ

(x, ·),z).

The lim inf on the right can be calculated by taking the minimum of the

lim inf on Λ ∼ (Z ∪{y}) and the lim inf on (Ω ∼ Λ) ∼ (Z ∪{y}). Since y is

a regular boundary point for Λ,

lim inf

z→y,z∈Λ∼(Z∪{y})

Ω∼Λ

(x, ·),z) = lim

z→y,z∈Λ∼(Z∪{y}

(x,·)

(z)=G

(x, y)

by continuity of G

(

x, ·)aty.Moreover,sinceR

Ω∼Λ

(x, ·), ·)=G

(x, ·)

on Ω ∼ Λ,

lim inf

z→y,z∈(Ω∼Λ)∼(Z∪{y})

Ω∼Λ

(x, ·),z) = lim inf

z→y,z∈(Ω∼Λ)∼(Z∪{y})

(x, z)

= G

(x, y).

Therefore,

Ω∼Λ

(x, ·),y)=G

(x, y),x ∈ Λ, whenever y is a regular

boundary point for Λ. By Theorem 5.4.13, Equation (5.6), Equation (5.7),

and the above equation,

(x, y) − H

(y,·)

(x)=G

(x, y) −

Ω∼Λ

(y,·),x)

= G

(x, y) −

Ω∼Λ

(x, ·),y)

= G

(x, y) − G

(x, y)=0

222 5 Dirichlet Problem for Unbounded Regions

for x ∈ Λ and y ∈ Ω ∩ ∂

∞

Λ a regular boundary point for Λ;sinceG

is a

symmetric function,

(y,·)

(x)=H

(x,·)

(y)

for such x and y.Forx, z ∈ Λ,byCase(i),

(x, z)=G

(x, z) − H

(z,·)

(x)

= G

(x, z) − H

(x,·)

(z).

Letting z → y and using the fact that y is a regular boundary point for Λ,

(x, y)=G

(x, y) − H

(x,·)

(y)

= G

(x, y) − H

(y,·)

(x).

Thus for x ∈ Λ,

(x, y)=G

(x, y) − H

(y,·)

(x),

except possibly for a polar subset of Ω ∩∂

∞

Λ. Combining the two cases, the

above equation holds for all y ∈ Ω except possibly for a polar subset. Since

(x, ·) is subharmonic on R

∼{x} and G

(x, ·)isharmoniconΩ ∼{x},

both sides are subharmonic on Ω ∼{x} and the equation holds for all y ∈ Ω

by Theorem 2.4.7.

Theorem 5.4.16 If Ω is a Greenian subset of R

and Λ ⊂ Ω,then

(x, y)=

(x, ·),y)=

(y,·),x)=G

(y,x),x,y∈ Ω.

(5.8)

Proof: Suppose ﬁrst that Λ is closed. Then Ω ∼ Λ is open and

(x, ·),y)=

(y,·),x),x,y∈ Ω ∼ Λ

by Equation (5.6). If either x ∈ Λ or y ∈ Λ, replace Λ by Λ

= Λ ∼

x,1/n

∪B

y,1/n

)sothat

(x, ·),y)=

(y,·),x); letting n →∞,

Λ∼{x,y}

(x, ·),y)=

Λ∼{x,y}

(y,·),x) by Theorem 4.6.5. Since Λ dif-

fers from Λ ∼{x, y} by a polar set, Equation (5.8) holds whenever Λ is a

closed set. It then follows from Theorem 4.6.5 that the equation holds for F

sets and, in particular, for open sets. Lastly, consider any Λ ⊂ Ω and any

open set O ⊂ Ω, O ∈O(Ω). By Equation (5.8), (vi) of Theorem 4.3.5, and

Lemma 4.6.12,

(x, ·),y)=R

(y,·),x),x,y∈ Ω.

Since R

(x, ·),y)=

(x, ·),y)andR

(y,·),x)=

(y,·),x)forx, y ∈ Ω ∼ Λ by Theorem 4.6.3,

(x, ·),y)=

(y,·),x),x,y∈ Ω ∼ Λ.

5.5 Intrinsic Topology 223

If x ∈ Λ or y ∈ Λ, apply this result to Λ ∼{x, y} and use the fact that this

set diﬀers from Λ by a polar set to arrive at Equation (5.8).

Theorem 5.4.17 If u is the potential of a measure μ on the Greenian set Ω

and Λ ⊂ Ω,then

= G

(μ, ·)=G

μ = δ

(·,u).

Proof: The ﬁrst and fourth terms are equal by Theorem 4.7.4. By Tonelli’s

theorem,

μ(x)=



(x, y) μ(dy)=



(y,z) δ

(x, dz) μ(dy)



u(z) δ

(x, dz)=

(x)

so that the ﬁrst and third terms are equal. By the symmetry of G

(x, y)

established in the preceding theorem,

(μ, ·)(x)=



(x, y) δ

(μ, dy)=



(x, y)



(z,dy) μ(dz)







(x, y) δ

(z,dy)



μ(dz)=



(z,x) μ(dz)



(x, z) μ(dz)=G

μ(x).

This shows that the second and third terms are equal. Lastly,

μ(x)=



(x, y) μ(dy)=



(y,z) δ

(x, dz) μ(dy)



u(z) δ

(x, dz)=

(x)

so that the ﬁrst and third terms are equal.

5.5 Intrinsic Topology

In view of the central role of superharmonic functions in potential theory, a

topology deﬁned by such functions would be better suited to potential theory

problems than the metric topology.

Deﬁnition 5.5.1 The ﬁne topology is the smallest topology on R

for

which all superharmonic functions are continuous in the extended sense.

Since each fundamental harmonic function u

is superharmonic on R

and {y ∈ R

; u

(y) >α} is a ball, balls with arbitrary centers and radii are

224 5 Dirichlet Problem for Unbounded Regions

ﬁnely open. It follows that the metric topology is coarser than the ﬁne topol-

ogy. In particular, open sets are ﬁnely open, continuous functions are ﬁnely

continuous, etc. Since it was shown in Corollary 4.3.12 that there are ﬁnite-

valued discontinuous superharmonic functions, the ﬁne topology is strictly

ﬁner than the metric topology. If u is any superharmonic function on R

and

α, β are any real numbers, sets of the type {x; u(x) >α} and {x; u(x) <β}

must be ﬁnely open; sets of the ﬁrst type are open but sets of the second type

must be adjoined to the metric topology. The collection of such sets can be

taken as a subbase for the ﬁne topology. If u is superharmonic on the open

set Ω ⊂ R

,thenu is ﬁnely continuous on Ω. To see this, let x ∈ Ω and let B

be any ball containing x with B

−

⊂ Ω. By the Riesz representation theorem,

u diﬀers from a Green potential G

ν by a (continuous) harmonic function

on B;sinceG

ν diﬀers from the superharmonic Newtonian potential U

a harmonic function on B, u diﬀers from a superharmonic function on R

a harmonic function on B.Thus,u is ﬁnely continuous at x.

The ﬁne closure of a set A ⊂ R

will be denoted by cl

A,theﬁne

interior by int

A,theﬁne limit by f-lim, the ﬁne limsup by f-limsup, etc.

If O is ﬁnely open, the notation O

will be used as a reminder that is ﬁnely

open; if Γ is ﬁnely closed, Γ

will be used as a reminder of this fact. A set Λ

will be called a ﬁne neighborhood of a point x ∈ Λ if it is a superset of a

ﬁnely open set containing x.

Theorem 5.5.2 If Z is a polar subset of R

,thenZ has no ﬁne limit points;

in particular, the points of Z are ﬁnely isolated points.

Proof: Assume that x ∈ R

is a ﬁne limit point of Z.Thenx is a limit point

of Z. Since this is true if and only if x is a limit point of Z ∼{x} and the

latter is polar, it can be assumed that x ∈ Z. By Theorem 4.3.11, there is a

superharmonic function u such that u(x) < lim inf

y→x,y∈Z

u(y)=r<+∞

for any r>0. Thus, there is a neighborhood O of x such that u(y) >

(r + u(x)) for y ∈ O ∩Z; but since u is ﬁnely continuous at x, there is a ﬁne

neighborhood O

of x such that u(y) <

(r + u(x)) for y ∈ O

. Therefore,

(O ∩ Z) ∩ O

=(O ∩O

) ∩ Z = ∅.SinceO ∩ O

is a ﬁne neighborhood of x

that does not intersect Z, x is not a ﬁne limit point of Z, a contradiction.

Example 5.5.3 Consider the sequence of points {x

} in R

with x

(1/j, 0),j ≥ 1. There is a ﬁne neighborhood of (0, 0) that contains no points

of the sequence.

5.6 Thin Sets

Consider an open set Ω and a point x ∈ Ω.Thenx is a regular boundary

point for the Dirichlet problem if there is a cone in ∼ Ω having x as its vertex.

For a boundary point to be irregular, ∼ Ω must somehow be thinner than a

cone at x. The concept of “thinness” of a set at a point expresses this more

precisely.

5.6 Thin Sets 225

Deﬁnition 5.6.1 AsetΛ is thin at x if x is not a ﬁne limit point of Λ.

Theorem 5.6.2 A polar set Z is thin at every point of R

Proof: By Theorem 5.5.2, no point of R

is a ﬁne limit point of Z.

In particular, a line segment in R

is thin at each of its points since it is

polar by Example 4.2.8. Also note that Λ is thin at x whenever x is not a

limit point of Λ.

Theorem 5.6.3 AsetΛ is thin at a limit point x of Λ if and only if there

is a superharmonic function u on a neighborhood U of x such that

u(x) < lim inf

y→x,y∈Λ∩U∼{x}

u(y).

Proof: Itcanbeassumedthatx ∈ Λ. Suppose that Λ is thin at x.Thereis

then a ﬁne neighborhood O

of x such that O

∩ Λ = ∅.Sincex is a limit

point of Λ, O

is not a metric neighborhood of x. Since the class of sets of

the form {y; v(y) <α} and {y; v(y) >β} for arbitrary α, β and arbitrary

superharmonic functions v constitute a subbase for the ﬁne topology, it can

be assumed that O

has the form O

= ∩

i=1

{y; u

(y) <α

}∩U where U is

a neighborhood of x and the u

are superharmonic. Since x ∈ O

(x) <α

for i =1,...,j and >0 can be chosen so that 0 <<min

1≤i≤j

(α

−u

(x)).

With this choice of , it can be assumed that O

= ∩

i=1

{y; u

(y) <u

(x)+

}∩U.Nowletu =



i=1

which is superharmonic. It remains only to show

that u(x) < lim inf

y→x,y∈Λ∼{x}

u(y). Since the u

are l.s.c. at x,thereisa

neighborhood V ⊂ U of x such that u

(y) >u

(x) − (/j)for1≤ i ≤ j

and y ∈ V .Consideranyy ∈ Λ ∩ V .SinceO

∩ Λ = ∅,y ∈ O

and so

(y) ≥ u

(x)+ for some k.Thus,

u(y)=



i=1

(y) >



i=1

(x) −

j − 1

 +  = u(x)+



;

that is, u(y) >u(x)+/j for all y ∈ Λ∩V . Therefore, lim inf

y→x,y∈Λ∼{x}

u(y)

>u(x), as was to be proved. Conversely, suppose there is a superharmonic

function u on a neighborhood U of x such that

u(x) < lim inf

y→x,y∈Λ∩U∼{x}

u(y)=r.

It can be assumed that r<+∞ since u can be replaced by min (u, u(x)+1).

Consider O

= {y; u(y) <

(u(x)+r). By choice of r, there is a neighborhood

V ⊂ U of x such that u(y) >

(u(x)+r)fory ∈ Λ ∩V .ThenO

∩(Λ ∩V )=

∩ V ) ∩ Λ = ∅.SinceO

∩ V is a ﬁne neighborhood of x, x is not a ﬁne

limit point of Λ and Λ is thin at x.

Note that the function u constructed in the proof of the necessity is su-

perharmonic on R

226 5 Dirichlet Problem for Unbounded Regions

Theorem 5.6.4 AsetΛ is thin at a limit point x of Λ if and only if there

is a superharmonic function u such that

u(x) < lim

y→x,y∈Λ∼{x}

u(y)=+∞. (5.9)

If x belongs to the Greenian set Ω, then there is a potential u = G

μ satis-

fying (5.9) for which μ has compact support in Ω.

Proof: The suﬃciency was proved in the preceding theorem. To prove the

necessity, suppose Λ is thin at x. As usual, it can be assumed that x ∈ Λ.

By the preceding comment, there is a superharmonic function v such that

v(x) < lim

y→x,y∈Λ∼{x}

v(y). If the right side is +∞, there is nothing to

prove. Assume that the right side is ﬁnite. Let B = B

x,r

be any ball. By

the Riesz decomposition theorem, Theorem 3.5.11, there is a measure ν on

B and a harmonic function h on B such that v = G

ν + h on B.Sinceh

is continuous at x, it can be assumed that v = G

ν on B.Let{

} be a

decreasing sequence of positive numbers. Since

v(x)=



x,

(x, y) dν(y)+



B∼B

x,

(x, y) dν(y) < +∞

and the second integral on the right increases to v(x),

lim

j→∞



x,

(x, y) dν(y) = lim

j→∞



(x, y) dν

(y)=0

where ν

(M)=ν(M ∩ B

x,

) for each Borel set M ⊂ B.Bypassingto

a subsequence, if necessary, it can be assumed that



(x) < +∞.

The function u =



is then superharmonic on B and ﬁnite at x.Let

δ = lim inf

y→x,y∈Λ

(v(y) − v(x)) > 0andletv

= G

.Thenv = v

+ h

where h

is harmonic on B

x,

, and it follows that

lim inf

y→x,y∈Λ

(y) − v

(x))

≥ lim inf

y→x,y∈Λ

(y)+h

(y)) + lim inf

y→x,y∈Λ

(−h

(y) − v

(x))

= lim inf

y→x,y∈Λ

(v(y) −v(x)) ≥ δ

for all j ≥ 1andthat

lim inf

y→x,y∈Λ

⎛

⎝



j=1

(y) −



j=1

(x)

⎞

⎠

≥ pδ

5.6 Thin Sets 227

for all p ≥ 1. Therefore,

lim inf

y→x,y∈Λ



j=1

(y) ≥



j=1

(x)+pδ ≥ pδ

for all p ≥ 1. Since u =



≥



j=1

, lim inf

y→x,y∈Λ

u(y) ≥ pδ for all

p ≥ 1. Therefore, lim inf

y→x,y∈Λ

u(y)=+∞.Nowextendu to R

using

Lemma 4.3.10. Suppose now that x belongs to the Greenian set Ω,andlet

B = B

x,r

be a ball with B

−

⊂ Ω. Choose a decreasing sequence {

} of

positive numbers as above for which 

<r. By Tonelli’s theorem, u =



= G

(



)onB. Letting μ =



on B, G

μ has the same

properties as u = G

μ since the two diﬀer by a harmonic function on B.

Corollary 5.6.5 If Λ is thin at a limit point x of Λ,then

lim

r→0

σ(∂B

x,r

∩Λ)

σ(B

x,r

)

=0.

Proof: By the preceding theorem, there is a superharmonic function u such

that u(x) < lim

y→x,y∈Λ∼{x}

u(y)=+∞. It can be assumed that u is non-

negative on a neighborhood of x. For suﬃciently small r>0,

u(x) ≥ L(u : x, r)=

n−1



∂B

x,r

u(y) dσ(y)

≥

n−1



∂B

x,r

∩Λ

u(y) dσ(y)

≥

σ(∂B

x,r

∩ Λ)

σ(∂B

x,r

)



inf

y∈B

x,r

∩Λ

u(y)



Since u(x) < +∞ and the last factor tends to +∞ as r → 0,

lim

r→0

σ(∂B

x,r

∩ Λ)

σ(∂B

x,r

)

=0.

This section will be concluded by showing that thinness is invariant under

certain transformations. A map τ : R

→ R

is measurable if τ

−1

(M)isa

Borel subset of R

for each Borel set M ⊂ R

; if, in addition, |τx − τy|≤

|x − y| for all x, y ∈ R

,thenτ is called a contraction map.

Theorem 5.6.6 If Λ is thin at x and τ : R

→ R

is a contraction mapping

such that |τx− τy| = |x − y| for all y ∈ R

,thenτ(Λ) is thin at τx.

Proof: If x is not a limit point of Λ, it is easy to see that τx is not a limit

point of τ(Λ) and thinness is preserved. It therefore can be assumed that x is

a limit point of Λ and that x ∈ Λ. By Theorem 5.6.4, there is a superharmonic