Helms L.L. Potential Theory

Подождите немного. Документ загружается.

178 4 Negligible Sets

Note that if X is a σ-compact metric space and F consists of compact

sets, then the B and B

of Sierpinski’s lemma are also compact.

Theorem 4.4.26 (Choquet [13]) If Ω is a Greenian set with paving K(Ω),

then the analytic sets are capacitable; in particular, the Borel subsets of Ω

are capacitable.

Proof: (Assuming Property B) Consider A ∈A(K(Ω)). If C

∗

(A)=0,

then C

∗

(A)=0andA is capacitable. It therefore can be assumed that

∗

(A) > 0. Let 0 ≤ α<C

∗

(A), and let A be the nucleus of the Suslin

scheme (n

,...,n

) → X

···n

∈K(Ω). Then

A =

n∈N

∞

∩ X

∩···).

Since

{n:n

≤b

}

∩ X

∩···) ↑ A as b

↑ +∞,

canbechosensothatC

∗

) >αby (ii) of Theorem 4.4.21. Suppose

,...,b

j−1

have been chosen so that C

∗

j−1

) >α.Since

{n:n

≤b

,i≤j}

∩ X

∩···) ↑ A

j−1

canbechosensothatC

∗

) >α. The sequence {b

} in Sierpinski’s Lemma

is thus deﬁned inductively. Since A

⊂ B

, C

∗

) ≥C

∗

) >α,j≥ 1.

Since B ∈K(Ω)andtheB

∈K(Ω), C(B

) >α,j≥ 1, and C(B) >αby

(iii) of Theorem 4.4.21. Since α is any number less than C

∗

(A), C

∗

(A)=

sup {C(B):B ∈K(Ω),B ⊂ A} = C

∗

(A)andA is capacitable.

4.5 Boundary Behavior of the Green Function

The proofs of the results of the preceding section, but not the results them-

selves, were conditional upon showing that the Greenian set Ω has Property

B. It is known that each ball B has Property B. The validity of the results of

the preceding section will be established here by showing that all Greenian

sets have Property B.

Lemma 4.5.1 If Ω is a bounded open set and x ∈ Ω,thenG

(x, ·)=u

−

on Ω.

Proof: Since u

∈ U

−H

≥ 0onΩ. From the deﬁnition of the Green

function, u

− H

≥ G

(x, ·)onΩ. Also from the deﬁnition, G

(x, ·)=

+ h

where h

is harmonic on Ω.SinceG

(x, ·) ≥ 0,u

≥−h

on Ω,and

4.5 Boundary Behavior of the Green Function 179

therefore −h

∈ L

;thatis,−h

≤ H

.Moreover,u

− H

≥ u

+ h

implies that −h

≥ H

.Thus,−h

= H

and G

(x, ·)=u

− H

Theorem 4.5.2 (Cartan [11]) AsubsetZ of a Greenian set Ω is polar if

and only if it has capacity zero.

Proof: (Assuming Property B) Let Z ⊂ Ω be a polar set. It follows from

(ii) of Theorem 4.4.21 that C

∗

(Z) = lim

j→∞

∗

(Z ∩Λ

) whenever {Λ

} is an

increasing sequence of open sets with Ω = ∪Λ

and with compact closures

−

⊂ Ω, j ≥ 1. It therefore can be assumed that Z ⊂ Λ where Λ is open

with compact closure Λ

−

⊂ Ω. By Theorems 4.2.9 and 4.2.10, there is a

superharmonic function u =+∞ on Z. By the Riesz decomposition theorem,

Theorem 3.5.11, there is a measure μ on Λ such that u = G

μ + h where h

is harmonic on Λ.Sinceu =+∞ on Z, G

μ =+∞ on Z.SinceG

≤ G

on Λ × Λ, G

μ =+∞ on Z.SinceG

μ is l.s.c., O

= {x ∈ Ω; G

μ(x) >

r},r > 0, is an open subset of Ω. Consider any compact set Γ ⊂ O

and

the capacitary distribution μ

corresponding to Γ and Ω. By the reciprocity

theorem, Theorem 3.5.1, and the fact that

= G

≤ 1onΩ,

C(Γ )=



dμ

≤



μdμ



dμ ≤

μ(Λ

−

It follows that C(O

) ≤ μ(Λ

−

)/r.SinceC

∗

(Z) ≤ inf

r>0

C(O

), C

∗

(Z)=0and

Z has capacity zero. As to the converse, assume that C

∗

(Z) = 0. Since a

countable union of polar sets is polar, it suﬃces to show that the intersection

of Z with any component of Ω is polar. Since the polar property is a local

property, it suﬃces to prove that the part of Z in a ball B with compact

closure B

−

⊂ Ω is polar. It can be assumed that Z ⊂ B.LetO be a nonempty

open subset of B with O

−

⊂ B.ThenbothZ ∩ O and Z ∩ (B ∼ O)have

outer capacity zero by (i) of Theorem 4.4.21. If it can be shown that each

of these two sets is polar, then it would follow that Z is polar. For each of

these two sets, there is a point not in the set and neighborhoods separating

the two. Assume that Z itself has this property; that is, there is a point

y ∈ B ∼ Z and open sets O

and O

such that (i)y ∈ O

⊂ B,(ii)Z ⊂

⊂ B,and(iii) O

∩ O

= ∅.SinceC

∗

(Z) = 0, for each j ≥ 1thereis

an open set Λ

with Z ⊂ Λ

⊂ O

and C(Λ

) < 1/j

.Alsoforeachj ≥ 1,

there is an increasing sequence of open sets {W

j,k

} with compact closures

−

j,k

⊂ W

j,k+1

⊂ Λ

and Λ

= ∪

k≥1

j,k

.Foreachj, k ≥ 1, let Γ

= W

−

j,k

and consider the corresponding capacitary potential

relative to Ω which

isequalto1onW

j,k

.SinceG

(y,·) is bounded outside any ball containing

y, sup

x∈O

(y,x) ≤ m for some m. Since the support of μ

is contained

in Γ

= W

−

j,k

⊂ Λ

⊂ O

(y)=



(y,x) dμ

(x) ≤ mμ

(Γ

) ≤ mC(Γ

) ≤ m/j

180 4 Negligible Sets

Since {

}

k≥1

is an increasing sequence, w

= lim

k→∞

is deﬁned

with w

(y) ≤ m/j

. Since the sequence {

}

k≥1

is uniformly bounded

by 1, w

is a nonnegative superharmonic function on Ω by Theorem 2.4.8.

Since

=1onW

j,k

, w

=1onΛ

⊃ Z. Letting w =



∞

j=1

,w(y)=



∞

j=1

(y) ≤



∞

j=1

m/j

< +∞. Since the partial sums of the series deﬁn-

ing w are nonnegative superharmonic functions, w is superharmonic on B by

Theorem 2.4.8. Since each w

=1onZ, w =+∞ on Z. This shows that Z

is polar.

Note that there is no mention of Property B in the statement of the

following theorem. All of the preceding results are applicable since

the proof is reduced to subsets of a ball that is known to have

Property B.

Theorem 4.5.3 Analytic inner polar sets are polar.

Proof: Let Z be an analytic inner polar set. Since Z = ∪

j≥1

(Z ∩ B

0,j

if it can be shown that each term is polar, then it would follow that Z is

polar, since a countable union of polar sets is polar by Theorem 4.2.12. It

is easily seen that each term is an analytic inner polar set. Thus, it can be

assumed that Z is an analytic inner polar subset of a ball B. Since each

compact subset Γ of Z is polar and C(Γ ) = 0 by the preceding theorem,

∗

(Z)=sup{C(Γ ); Γ ∈K(B),Γ ⊂ Z} = 0. Since analytic sets are capac-

itable by Theorem 4.4.26, C(Z) = 0. By the preceding theorem, Z is polar.

Completion of proof of Theorem 4.4.10 It has been shown that the set

of points in Ω where ˆu = u is an inner polar set. By the previous theorem,

the exceptional set is a polar set; that is, ˆu = u q.e.

The deﬁnition of a polar set was independent of any set Ω.Thisisalsotrue

of sets of capacity zero. If Z is a subset of a Greenian set Ω

of capacity zero

relative to Ω

, then it is a polar set; if it is also a subset of another Greenian

set Ω

, then it is a polar subset, and therefore of capacity zero relative to Ω

Theorem 4.5.4 If Ω is a Greenian set and x ∈ Ω,thenlim

y→z,y∈Ω

(x, y)

=0q.e. on ∂Ω.

Proof: By Corollary 2.6.31, there is an increasing sequence {Ω

} of regular

bounded open sets with compact closures in Ω such that Ω = ∪Ω

.Fix

x ∈ Ω. It can be assumed that x ∈ Ω

.Let

(x, ·)beequaltoG

(x, ·)

on Ω and equal to zero on ∼ Ω; similarly, deﬁne

to be G

(x, ·)on

and zero on ∼ Ω

.Then0≤

(x, ·) ≤

(x, ·). By Theorem 3.3.11,

(x, ·) ↑

(x, ·)onΩ. Consider a ball B = B

x,ρ

with B

−

x,ρ

⊂ Ω.By

Lemma 3.2.6,

(x, ·) is bounded on ∼ B and it follows that the sequence

{

(x, ·)} is uniformly bounded on ∼ B. Since each

(x, ·)isharmonic

4.5 Boundary Behavior of the Green Function 181

on Ω

∼ B and lim

y→z,y∈Ω

(x, y)=0forz ∈ ∂Ω

by Theorem 3.3.15,

(x, ·) is subharmonic on ∼ B.Letg = lim

j→∞

(x, ·)on∼ B,andlet

g(z) = lim sup

y→z

g(y),z ∈ R

, denote the upper regulariztion of g.Notethat

g =0on∂Ω. By Theorem 4.4.10, g =

g quasi everywhere. Since g = G

(x, ·)

on Ω ∼ B,forz ∈ ∂Ω

g(z) = lim sup

y→z

g(y) = lim sup

y→z,y∈Ω

g(y) = lim sup

y→z,y∈Ω

(x, y).

Since g =0on∂Ω,lim sup

y→z,y∈Ω

(x, y) = 0 quasi everywhere on ∂Ω.

Corollary 4.5.5 The set of irregular boundary points of a Greenian set Ω

is a polar set. Moreover, if h is a bounded harmonic function on Ω and

lim

y→x,y∈Ω

h(y)=0

for each regular boundary point x ∈ ∂Ω,thenh =0on Ω.

Proof: The ﬁrst assertion follows from Theorem 3.3.15. Let

f(x) = lim inf

y→x,y∈Ω

h(y),x∈ ∂Ω.

By the ﬁrst assertion, f = 0 except possibly for a polar subset of ∂Ω.By

Corollary 4.2.20, H

= H

=0onΩ.Sinceh ∈ U

,h ≥ 0onΩ. Similarly,

h ≤ 0onΩ.

According to the following theorem, all Greenian sets have Property B so

that all the results of this section and the preceding section apply

to all Greenian sets.

Theorem 4.5.6 Let Ω be a Greenian set. If μ is a measure with com-

pact support Γ ⊂ Ω,thenlim

y→z,y∈Ω

μ(y)=0q.e. on ∂Ω.IfΩ has

ﬁnite Lebesgue measure when n ≥ 3 or is bounded when n =2,then

lim

y→z,y∈Ω

Gf(y)=0q.e. whenever f is a bounded measurable function

on Ω.IfΩ is regular, then these statements hold at every point of ∂Ω.

Proof: Let Γ be any compact subset of Ω,andlet{Ω

} be the sequence of

components of Ω.ThenΓ ⊂∪

j=1

for some p ≥ 1. For 1 ≤ j ≤ p,letx

aﬁxedpointinΓ ∩Ω

. By the preceding theorem, for each such j there is a

polar set Z

⊂ ∂Ω such that lim

y→z,y∈Ω

,y) = 0 for all z ∈ ∂Ω ∼ Z

Letting Z = ∪Z

, lim

y→z,y∈Ω

,y)=0,z ∈ ∂Ω ∼ Z, 1 ≤ j ≤ p.For

1 ≤ j ≤ p,letO

be a neighborhood of Γ ∩Ω

with O

−

⊂ Ω

.ByHarnack’s

inequality, Theorem 2.2.2, for such j there is a constant k

such that

(x, y) ≤ k

,y),x∈ Γ ∩ Ω

,y ∈ Ω

∼ O

;

182 4 Negligible Sets

this inequality is trivially satisﬁed if y ∈ Ω

.Thus,

(x, y) ≤ k

,y),x∈ Γ ∩ Ω

,y ∈ Ω ∼∪

j=1

and therefore

(x, y) ≤ max

1≤j≤p

,y)),x∈ Γ, y ∈ Ω ∼∪

j=1

It follows that lim

y→z,y∈Ω

(·,y) = 0 uniformly on Γ for each z ∈ ∂Ω ∼ Z.

In particular, if the measure μ has support in Γ ,then

lim

y→z,y∈Ω

Gμ(y) = lim

y→z,y∈Ω



(x, y) dμ(x)=0,z∈ ∂Ω ∼ Z.

The proof of the second assertion is the same as that of Lemma 3.4.17 using, in

the notation of that lemma, the fact that lim

x→x

,x∈Ω

(x, z) = 0 uniformly

with respect to z ∈ Γ

provided x

∈ ∂Ω does not belong to a polar subset

of ∂Ω.

4.6 Applications

It is now possible to show that a nonempty open subset of R

is Greenian if

and only if it supports a nonconstant, positive superharmonic function.

Theorem 4.6.1 (Myrberg [47]) If Ω is a nonempty open subset of R

then the following three conditions are equivalent:

(i) ∼ Ω is not a polar set.

(ii) There is a nonconstant, positive superharmonic function on Ω.

(iii) Ω is Greenian.

Proof: The assertion (iii) ⇒ (i) is just Theorem 4.2.18. To show that (i) ⇒

(ii), assume that ∼ Ω is not polar. By Lemma 4.2.17, this assumption is

equivalent to assuming that ∂Ω is not polar. Let {B

} be a sequence of balls

covering R

. By Theorem 4.2.12, there is a j

≥ 1 such that Γ = ∂Ω ∩ B

−

is not polar. Letting B be any ball such that B

−

⊂ B, Γ is a compact

subset of B.Consider

,relativetoB, which is strictly positive on B by

Theorem 4.3.5 and harmonic on B ∼ Γ .SinceΓ is not polar, by Lemma 4.4.11

there is a point x ∈ Γ such that

(x)=R

(x) = 1. Thus, there are points

z ∈ Ω such that

(z) is arbitrarily close to 1. Since lim

y→z,y∈B

(y)=

lim

y→z,y∈B

(y) = 0 for all z ∈ ∂B by Theorem 3.4.16, the function

u(y)=



(y) y ∈ Ω ∩ (B ∼ Γ )

0 y ∈ Ω ∼ B

is subharmonic on Ω, takes on the value 0, and takes on values arbitrarily

close to 1, but not the value 1 by the maximum principle. The function

4.6 Applications 183

v =1− u on Ω is a nonconstant, positive superharmonic function on Ω.

It only remains to show that (ii) ⇒ (iii). Suppose there is a nonconstant,

positive superharmonic function u on Ω.Foreachj ≥ 1,Ω

= Ω ∩ B

0,j

is a

Greenian set by Theorem 3.2.10 and the sequence {Ω

} increases to Ω.By

the Riesz representation theorem, Theorem 3.5.11, for each j ≥ 1thereisa

measure μ

and a harmonic function h

≥ 0onΩ

such that u = G

+ h

on Ω

. As in the proof of Theorem 3.5.11, there is a measure μ on Ω such that

= μ|

. Extending G

(x, ·)toR

by putting G

(x, y)=0fory ∈ Ω

u(x)=G

μ(x)+h

(x) ≥



(x, y) dμ(y).

If Ω were not Greenian, then G

↑ +∞ on Ω × Ω by Theorem 3.3.11, and

it would follow that u =+∞ on Ω, a contradiction. Thus, Ω is Greenian.

NowthatitisknownthatallGreeniansetshavePropertyB,itcanbe

shown that the set of points where

(x) = R

(x) is a negligible set.

Corollary 4.6.2 If u is a nonnegative superharmonic function on the Green-

ian set Ω and Λ ⊂ Ω,then

= R

q.e. on Ω.

Proof: Theorem 4.4.10.

Theorem 4.6.3 If u is a nonnegative superharmonic function on the Green-

ian set Ω and Λ ⊂ Ω,then

= R

on Ω ∼ Λ.

Proof: It is known from (ii) of Lemma 4.3.2 that R

= u on Λ,andfrom

the preceding corollary that

= R

quasi everywhere on Ω.LetZ be the

exceptional polar set, and let x

be any point of Ω ∼ Λ. By Theorem 4.3.11,

there is a positive superharmonic function w on Ω such that w =+∞ on

Λ ∩ Z and w(x

) < +∞.Then

+ w ≥ u on Λ for every >0, and it

follows from the deﬁnition of R

that

+ w ≥ R

.Sincew(x

) < +∞

and  is arbitrary,

) ≥ R

). Since the opposite inequality is always

true, there is equality at x

.Thus,

= R

on Ω ∼ Λ.

Corollary 4.6.4 If Λ ⊂ Ω, u and v are nonnegative superharmonic func-

tions on the Greenian set Ω with u = v q.e. on Λ,andZ is a polar subset of

Λ,then

Λ∼Z

and

Proof: By Theorem 4.3.5,

Λ∼Z

≤

. Consider any nonnegative superhar-

monic function w on Ω that majorizes u on Λ ∼ Z. By Theorem 4.3.11, if x

is any point of Λ ∼ Z, there is a positive superharmonic function v on Ω such

that v(x) < +∞ and v =+∞ on Z.Thenw + v majorizes u on Λ for every

>0. Therefore,

(x) ≤ w(x)+v(x), and letting  ↓ 0,

(x) ≤ w(x)

for all x ∈ Λ ∼ Z. Therefore,

≤ R

Λ∼Z

on Ω, and by taking the lower

regularization of both sides,

≤

Λ∼Z

and the two are equal. The second

equation follows from the ﬁrst.

184 4 Negligible Sets

According to Lemma 4.4.14, if {Γ

} is an increasing sequence of compact

subsets of the Greenian set Ω with compact limit Γ ,then

↑

for any

ﬁnite-value, nonnegative, continuous superharmonic function u on Ω.This

result can be improved.

Theorem 4.6.5 If u is a nonnegative superharmonic function on the Green-

ian set Ω, {Λ

} is an increasing sequence of subsets of Ω,andΛ = ∪Λ

,then

lim

j→∞

Proof: Since the sequence {

} is increasing and is majorized by the su-

perharmonic function

, the function v = lim

j→∞

is superharmonic

and v ≤

. By Corollary 4.6.2,

diﬀers from u on Λ

on at most a

polar subset of Λ

. Therefore, v diﬀers from u on Λ on at most a polar subset

Z of Λ.Consideranyx ∈ Λ ∼ Z. There is then a positive superharmonic

function w on Ω such that w(x) < +∞ and w =+∞ on Z. Then for every

>0,v+ w ≥ u on Λ, and therefore v + w ≥ R

≥

on Ω.IfB

−

x,δ

⊂ Ω,

then

A(v : x, δ)+A(w : x, δ) ≥ A(

: x, δ);

letting  ↓ 0, A(v : x, δ) ≥ A(

: x, δ) whenever B

−

x,δ

⊂ Ω. By Lemma 2.4.4,

v(x) ≥

(x) for all x ∈ Ω;thatis,v = lim

j→∞

≥

. It follows that

v = lim

j→∞

Lemma 4.6.6 If Λ is an open subset of the Greenian set Ω and f is a ∂Ω

resolutive boundary function, then the function

g =



on Ω ∩ ∂Λ

f on ∂Ω ∩ ∂Λ

is resolutive for Λ and H

= H

on Λ.

Proof: Consider any u ∈ U

.Thenu ≥ H

on Ω.Forx ∈ Ω ∩ ∂Λ,

lim inf

z→x,z∈Λ

u(z) ≥ lim inf

z→x,z∈Λ

(z)=H

(x);

while for x ∈ ∂Ω ∩ ∂Λ, lim inf

z→x,z∈Λ

u(z) ≥ lim inf

z→x,z∈Ω

u(z) ≥ f(x).

Therefore, u ∈ U

and H

≥ H

on Λ. Similarly, Hg

≥ H

and it follows

that 0 ≤

− H

≤ H

− H

.Sincef is ∂Ω resolutive, the latter diﬀer-

ence is equal to zero and so g is ∂Λ resolutive. Using the fact that f is ∂Ω

resolutive, H

≤ H

on Λ, and therefore H

= H

The operation on the positive superharmonic function u described below

is an extension of the lowering operation of Deﬁnition 2.4.10.

Lemma 4.6.7 If Λ is an open subset of the Greenian set Ω having compact

closure Λ

−

⊂ Ω and u is a positive superharmonic function on Ω, then the

lower regularization ˆv of the function

4.6 Applications 185

v =



u on Ω ∼ Λ

on Λ

is superharmonic on Ω.

Proof: Since u ≥

≥ H

≥ 0onΛ, H

is deﬁned by Theorem 2.9.1. Let

B be a ball with B

−

⊂ Ω.Itwillbeshownﬁrstthatv ≥ H

on B.Since

≤ u on Λ, v ≤ u on Ω and H

≤ H

≤ u = v on B ∼ Λ.Nowlet

w =



v on Ω ∼ B

on B.

Note that w = H

≤ H

≤ u = v on ∂Λ ∩B;thatis,w ≤ v on ∂(B ∩Λ). It

follows that H

B∩Λ

≤ H

B∩Λ

on B ∩ Λ.SinceH

= H

B∩Λ

and H

B∩Λ

= H

on B ∩Λ by the preceding lemma, it follows that H

≤ H

on B ∩Λ.Since

v = H

on Λ, H

≤ v on B.Notingthatˆv ≤ v, H

ˆv

≤ H

≤ v on B.Since

ˆv

is continuous on B, H

ˆv

≤ ˆv on B for any ball B with B

−

⊂ Ω and ˆv is

superharmonic on Ω.

Theorem 4.6.8 If u is a positive superharmonic function on the Greenian

set Ω and Λ is an open subset of Ω with compact closure Λ

−

⊂ Ω,then

Ω∼Λ

= H

on Λ;moreover,if{Λ

} is an increasing sequence of such sets

with Ω = ∪Λ

,thenu is a potential if and only if lim

j→∞

=0on Ω.

Proof: If

v =



u on Ω ∼ Λ

on Λ,

then ˆv is superharmonic on Ω.Sinceu is l.s.c. on ∂Λ, there is a sequence

} in C

(∂Λ) such that f

↑ u on ∂Λ.Since

lim inf

z→y,z∈Λ

v(z) = lim inf

z→y,z∈Λ

(z) ≥ lim inf

z→y,z∈Λ

(z)=f

(y)

for all y ∈ ∂Λ that are regular boundary points for Λ, lim inf

z→y,z∈Λ

v(z) ≥

u(y) for all y ∈ ∂Λ except possibly for a polar set according to Corollary 4.5.5.

Since v = u on Ω ∼ Λ, lim inf

z→y,z∈Ω∼Λ

v(z) = lim inf

z→y,z∈Ω∼Λ

u(z) ≥

u(y), and therefore ˆv(y) = lim inf

z→y,z∈Ω

v(z) ≥ u(y) for all y ∈ ∂Λ except

possibly for a polar set. Since ˆv = v = u on Ω ∼ Λ

−

, ˆv ≥ u on Ω ∼ Λ except

possibly for a polar set. It follows from Corollary 4.6.4 that ˆv ≥

Ω∼Λ

If w is any nonnegative superharmonic function on Ω that majorizes u on

Ω ∼ Λ,thenw|

∈ U

and w ≥ v on Λ; therefore, w ≥ v on Ω.This

shows that R

Ω∼Λ

≥ v on Ω, and consequently that

Ω∼Λ

≥ ˆv on Ω.Asthe

opposite inequality was just proved,

Ω∼Λ

=ˆv = v = H

on Λ.Nowlet

{Λ

} be an increasing sequence of open subsets of Ω with compact closures

−

⊂ Ω such that Λ

↑ Ω. It was just shown that H

Ω∼Λ

on Λ

.Since

Ω∼Λ

j+1

≤

Ω∼Λ

by Theorem 4.3.5, H

j+1

≤ H

on Λ

for each j ≥ 1.

186 4 Negligible Sets

The sequence {H

} is then a decreasing sequence of nonnegative harmonic

functions (with due regard given to domains), and h = lim

j→∞

deﬁnes a

nonnegative harmonic function on Ω with u ≥ h ≥ 0. If

h is any nonnegative

harmonic minorant of u on Ω,thenH

≥ H

h on Λ

,j ≥ 1, and h ≥

that is, h is the greatest harmonic minorant of u on Ω, and the result follows

from Corollary 3.5.12.

Theorem 4.6.9 (G. Mokobodzki and D. Sibony [46]) If u, u

,andu

are nonnegative superharmonic functions on the Greenian set Ω with u ≤

+ u

, then there are unique nonnegative superharmonic functions ˜u

, ˜u

on Ω for which ˜u

≤ u

, ˜u

≤ u

,andu =˜u

+˜u

Proof: Letting ˜u

be the lower regularization of

inf {v ∈S(Ω); v ≥ 0,u≤ v + u

u ≤ ˜u

+ u

q.e. on Ω by Theorem 4.4.10. Since polar sets have Lebesgue

measure zero by Theorem 4.2.6, u ≤ ˜u

+ u

a.e. on Ω and u ≤ ˜u

+ u

Ω by Lemma 2.4.4; moreover, ˜u

is superharmonic on Ω by Theorem 2.4.9.

Letting ˜u

be the lower regularization of

inf {v ∈S(Ω); v ≥ 0,u≤ ˜u

+ v},

u ≤ ˜u

+˜u

a.e. on Ω and therefore u ≤ ˜u

+˜u

on Ω as above; moreover,

˜u

is superharmonic on Ω as above. Since u isamemberofeachofthesets

deﬁning ˜u

and ˜u

,u ≥ ˜u

a.e. on Ω and u ≥ ˜u

a.e. on Ω so that u ≥ ˜u

on Ω and u ≥ ˜u

on Ω. It will be shown now that u ≥ ˜u

+˜u

on Ω.Since

u(x) − ˜u

(x) may be indeterminant, redeﬁne u − ˜u

at x so that

(u − ˜u

)(x)=



u(x) − ˜u

(x)if˜u

(x) < +∞

lim inf

y→x,˜u

(y)<+∞

(u(y) − ˜u

(y)) if ˜u

(x)=+∞.

Then u − ˜u

is l.s.c. at points x for which ˜u

(x)=+∞. It will be shown

now that u − ˜u

is l.s.c. at points x for which ˜u

(x) < +∞.Considerany

ball B = B

x,δ

with B

−

⊂ Ω. Letting ˜u

=(˜u

)

,i =1, 2 and using the

inequality u ≤ ˜u

+˜u

u = u

+(u −u

) ≤ ˜u

+(˜u

+ u − u

) ≤ ˜u

+(˜u

+ u − u

The function v =˜u

+ u − u

is superharmonic on B with v ≥ ˜u

on B.

Consider the function

w =



min (˜u

,v)onB

˜u

on Ω ∼ B.

Since v ≥ ˜u

and ˜u

≥ ˜u

on B, w ≥ ˜u

on B so that w is a nonnegative

superharmonic function on Ω by Lemma 2.4.11. Since u ≤ ˜u

+˜u

and

4.6 Applications 187

u ≤ ˜u

+ v, u ≤ ˜u

+min(˜u

,v)=˜u

+ w on B and u ≤ ˜u

+˜u

=˜u

+ w on

Ω ∼ B,u ≤ ˜u

+ w on Ω.Sincew is a nonnegative superharmonic function,

w ≥ ˜u

and v ≥ w ≥ ˜u

on B, it follows that

− ˜u

≤ u − ˜u

(4.4)

on B. Note that the function on the left is continuous on B. Therefore,

(x) − ˜u

(x) ≤ lim inf

y→x,y∈Ω

(u − ˜u

)(y).

Since superharmonic functions are integrable on spheres,

(x) − ˜u

(x)=(u − ˜u

)

(x)=L(u : x, δ) − L(˜u

: x, δ) → u(x) − ˜u

(x)

as δ → 0. It follows from Inequality (4.4) that

u(x) − ˜u

(x) ≤ lim inf

y→x,y∈Ω

(u − ˜u

)(y)

and u−˜u

is l.s.c. at x whenever ˜u

(x) < +∞ and therefore l.s.c. on Ω.From

Inequality (4.4),

u(x) − ˜u

(x) ≥ (u − ˜u

)

(x)=PI(u − ˜u

,B)(x)=L(u − ˜u

: x, δ),

so that u − ˜u

satisﬁes the averaging principle and is therefore a nonnegative

superharmonic function on Ω.Sinceu =(u −˜u

)+˜u

, u −˜u

≥ ˜u

a.e. on Ω.

Since the opposite inequality was established above, u =˜u

+˜u

on Ω.To

prove uniqueness, assume that there are nonnegative superharmonic functions

˜v

and ˜v

such that u =˜v

+˜v

, ˜v

≤ u

and ˜v

≤ u

.Thenu ≤ ˜v

+ u

and

therefore ˜u

≤ ˜v

a.e. on Ω by deﬁnition of ˜u

and consequently on all of Ω.

Similarly, ˜u

≤ ˜v

on Ω.Since˜u

+˜u

=˜v

+˜v

, ˜u

=˜v

,i=1, 2onΩ.

Theorem 4.6.10 If {u

} is an increasing sequence of nonnegative super-

harmonic functions on the Greenian set Ω with superharmonic limit u and

Λ ⊂ Ω,then

= lim

j→∞

Proof: Since

≤

,j ≥ 1,w = lim

j→∞

≤

and is superharmonic

on Ω by Theorem 2.4.8. Since

= R

= u

q.e. on Λ, w = u q.e. on Λ;

that is, there is a polar set Z ⊂ Λ such that w = u on Λ ∼ Z.Since

Λ∼Z

by Corollary 4.6.4 and

Λ∼Z

≤ R

Λ∼Z

by Lemma 4.3.2,

Λ∼Z

≤ R

Λ∼Z

= R

Λ∼Z

≤ w = lim

j→∞

Lemma 4.6.11 If u is a nonnegative superharmonic function on the open

set Ω, then there is an increasing sequence {u

} of nonnegative, ﬁnite-valued,

continuous superharmonic functions on Ω such that u = lim

j→∞

on Ω.

Proof: Let {Ω

} be an increasing sequence of open subsets of Ω with compact

closures in Ω such that Ω = ∪Ω

.Then

= u on Ω

and is harmonic on

Ω ∼ Ω

−

.Notethat

≤

j+1

on Ω for all j ≥ 1. For each j ≥ 1, let

=min(d(Ω

, ∼ Ω), 1/j)) and let