Helms L.L. Potential Theory

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158 4 Negligible Sets

Proof: It will be shown ﬁrst that the exceptional set is a polar set. Letting

Z denote the exceptional set,

Z = {x ∈ ∂Ω; lim inf

y→x,y∈Ω

u(y) <α}

∞

n=1



x ∈ ∂Ω; |x|≤n, lim inf

y→x,y∈Ω

u(y) ≤ α −

Since each of the sets of the countable union is a compact subset of Z,each

is polar and so Z is polar. It therefore can be assumed that the exceptional

set is a polar set. If Ω is bounded, then u ≥

= α by Theorem 4.2.19.

Suppose now that Ω is unbounded. The rest of the proof requires separate

consideration of the n =2andn ≥ 3 cases. Consider ﬁrst the n =2case.

Since Ω has a Green function, ∂Ω is not a polar set by Theorem 4.2.18. It

follows that there is a point x

∈ ∂Ω such that lim inf

y→x

,y∈Ω

u(y) ≥ α.Ifη

is any positive number, then there is a ball B = B

,δ

such that u(y) >α−η

for y ∈ Ω ∩ B

−

. By suitable choice of a constant c, the function v(y)=

log |y − x

| + c can be made harmonic and nonnegative on Λ = Ω ∼ B

−

and will satisfy the condition lim

|y|→+∞,y∈Λ

v(y)=+∞.Foreach>0, the

function u + v has the properties that lim inf

y→x,y∈Λ

(u + v)(y) ≥ α −η for

x ∈ ∂Λ except possibly for a polar set and lim

|y|→+∞,y∈Λ

(u + v)(y)=+∞.

Thus, for all suﬃciently large integers j,(u + v) ≥ α − η on ∂B

∩ Λ,

and therefore lim inf

y→x,y∈Λ∩B

(u + v) ≥ α − η for all x ∈ ∂(Λ ∩ B

)

except possibly for a polar set. It follows from the bounded case that u+v ≥

α − η on Λ ∩ B

for all large j. Therefore, u + v ≥ α − η on Λ. Letting

 → 0,u ≥ α − η on Λ. But since u>α− η on Ω ∩ B

−

,u ≥ α − η on Ω.

Since η is an arbitrary positive number, u ≥ α on Ω. Consider the n ≥ 3

case with lim inf

|y|→+∞,y∈Ω

u(y) ≥ α.Foreachη>0, there is a positive

integer j

such that u(y) ≥ α − η for all y ∈ Ω ∼ B

0,j

,j ≥ j

.Then

lim inf

y→x,y∈Ω∩B

0,j

u(y) ≥ α − η for x ∈ ∂(Ω ∩ B

0,j

) except possibly for

a polar set. By the ﬁrst part of the proof for bounded sets, u ≥ α − η on

Ω ∩ B

0,j

for every j ≥ j

. It follows that u ≥ α − η on Ω and that u ≥ α on

Ω since η is an arbitrary positive number.

4.3 Reduction of Superharmonic Functions

In this section, the notion of “balayage,” due to Poincar´e, will be developed.

As conceived by Poincar´e, “balayage” or “sweeping out” is just the smoothing

process used to deﬁne the reduction u

∞,Λ

of the superharmonic function u.

Let Ω be a Greenian subset of R

,letu be a nonnegative superharmonic

function on Ω,andifΛ is any subset of Ω let

= {v ∈S(Ω); v ≥ 0onΩ,v ≥ u on Λ}.

4.3 Reduction of Superharmonic Functions 159

Deﬁnition 4.3.1 R

=inf{v; v ∈ Φ

} is called the reduction of u over Λ

relative to Ω.

In using the notation R

, it will be understood that u is a superharmonic

function.

Clearly, if u and v are nonnegative superharmonic functions on Ω which

agree on Λ,thenR

= R

. R

, as the inﬁmum of a family of superharmonic

functions, need not be superharmonic. Consider, for example, the fundamen-

tal harmonic function u

for R

with pole 0. Let Ω = R

and Λ = {0}.Then

=+∞ on Λ, zero elsewhere, and therefore not l.s.c.

In the general case, R

fails to be superharmonic on Ω only because of

the l.s.c. requirement. Since u ∈ Φ

and elements of Φ

are nonnegative,

0 ≤ R

≤ u.Thus,R

> −∞ and R

cannot be identically +∞ on any

component of Ω since this is true of u. R

can be seen to be super-mean-

valued on Ω as follows. If B

x,δ

⊂ Ω and v ∈ Φ

,thenv(x) ≥ L(v : x, δ) ≥

L(R

: x, δ). Thus, R

(x) ≥ L(R

: x, δ) (assuming that R

is measurable).

This shows that only the l.s.c. requirement stands in the way of R

being

superharmonic on Ω.

Lemma 4.3.2 The lower regularization

of R

is superharmonic on Ω.

Moreover,

(i) 0 ≤

≤ R

≤ u on Ω,

(ii) u = R

on Λ,

(iii) u = R

on the interior of Λ,and

(iv) R

on Ω ∼ Λ

−

and both are harmonic on Ω ∼ Λ

−

Proof: The ﬁrst three assertions are obvious. To prove (iv), it is only

necessary to show that R

is harmonic on Ω ∼ Λ

−

. As in the proof

of Theorem 2.6.2, Φ

is a saturated family of superharmonic functions on

Ω ∼ Λ

−

and R

is therefore harmonic on Ω ∼ Λ

−

by the same theorem.

According to the preceding lemma,

diﬀers from R

only on the bound-

ary of Λ. At such points, it may be that

< R

.Ifu

is the funda-

mental harmonic function on R

with pole 0,Ω = R

,andΛ = {0},then

(0) < R

(0) = +∞. It will be shown later that, in general,

diﬀers

from R

only on a polar set.

Remark 4.3.3 If Ω has components {Ω

} and Λ ⊂ Ω, it is easy to see that

Λ∩O

on O

,j ≥ 1.

Deﬁnition 4.3.4

is called the regularized reduction of u over Λ rel-

ative to Ω.

Theorem 4.3.5 If u and v are nonnegative superharmonic functions on a

Greenian set Ω,then

160 4 Negligible Sets

(i) Λ ⊂ Σ ⊂ Ω ⇒

≤

(ii) u ≤ v ⇒

≤

(iii) λ>0 ⇒

λu

= λ

(iv)

u+v

≤

(v)

is positive or identically zero on each component of Ω,

(vi) Λ open subset of Ω ⇒

= R

on Ω,and

(vii) Γ a compact subset of Ω ⇒

is a potential.

Proof: If Λ ⊂ Σ,thenΦ

⊂ Φ

and R

≤ R

.Assertion(i) follows by

taking the lower regularization of each side of this inequality. Assertions (ii)

and (iii) are proved in the same way. If f ∈ Φ

and g ∈ Φ

,thenf +g ∈ Φ

u+v

and

u+v

≤ R

u+v

≤ f + g,where

u+v

,f, and g are superharmonic on Ω.

For the time being, ﬁx g.IfB

−

x,δ

⊂ Ω,then

u+v

: x, δ) ≤ A(f : x, δ)+A(g : x, δ)

≤ f(x)+A(g : x, δ).

Taking the inﬁmum over f ∈ Φ

u+v

: x, δ) ≤ R

(x)+A(g : x, δ).

Since superharmonic functions are integrable on compact subsets of Ω by

Theorem 2.4.2, the two averages are continuous on Ω

= {y ∈ Ω; d(y, ∼

Ω) >δ}.Thus,

u+v

: x, δ) ≤

(x)+A(g : x, δ), on Ω

Letting δ ↓ 0,

u+v

(x) ≤

(x)+g(x), on Ω.

Repeating this argument,

u+v

≤

, on Ω.

Assertion (v) follows from the minimum principle. Consider assertion (vi). By

the preceding lemma,

= R

= u on Λ.Thus,

∈ Φ

and R

≤

Ω. Since the opposite inequality is always true,

= R

on Ω. To show that

is a potential, it suﬃces to show that the greatest harmonic minorant of

is the zero function by Corollary 3.5.12. Assume ﬁrst that u is bounded

on Γ .LetΩ

be any component of Ω and let x

∈ Ω

. By Theorem 3.3.7 and

Corollary 3.3.10, the greatest harmonic minorant of G

, ·)|

is the zero

function. Since G

, ·) is superharmonic and strictly positive on Ω

,ithas

a strictly positive inﬁmum on Ω

∩ Γ .Forsomeλ>0,λG

, ·) majorizes

u on Ω

∩ Γ and the function

4.3 Reduction of Superharmonic Functions 161

u =



λG

, ·)onΩ

u on Ω ∼ Ω

belongs to Φ

; therefore, 0 ≤

≤ R

≤ λG

, ·)onΩ

and it follows

that the greatest harmonic minorant of

on Ω

is the zero function. Since

is any component of Ω, the greatest harmonic minorant of

on Ω is

the zero function and

is a potential by Riesz’s representation theorem,

Theorem 3.5.11. Suppose now that u is unbounded on Γ . By the Riesz rep-

resentation theorem, u = v + h where v is a potential and h is harmonic on

Ω.Since0≤

≤ R

≤ v and v is a potential,

is a potential. Since h

is bounded on Γ,

is a potential on Ω by the ﬁrst part of the proof. By

(iv),

≤

. Since the sum of two potentials is again a potential,

is a potential.

Corollary 4.3.6 If u is a nonnegative superharmonic function on the Green-

ian set Ω, then there is an increasing sequence of potentials {u

} having

compact supports such that lim

j→∞

= u.

Proof: Let Ω

} be an increasing sequence of open sets with compact closures

−

⊂ Ω such that Ω = ∪Ω

.Foreachj ≥ 1,u

−

is a potential that

agrees with u on Ω

and u

↑ u.

Corollary 4.3.7 If u and v are nonnegative, bounded, superharmonic func-

tions on Ω and Λ ⊂ Ω,then

sup

x∈Ω

(x) −

(x)|≤sup

x∈Ω

|u(x) − v(x)|.

Proof: Let c =sup

x∈Ω

|u(x) − v(x)|.Thenv − u ≤ c,

≤

u+c

≤

+c. Interchange u and v and combine the two inequalities to obtain

the result.

Theorem 4.3.8 If Γ is a compact subset of a connected Greenian set Ω,

then

is identically zero on Ω if and only if Γ is a polar set.

Proof: Suppose

is identically zero on Ω.Since

= R

on Ω ∼ Γ, R

is identically zero on Ω ∼ Γ .Letx

be any point of Ω ∼ Γ .Foreach

n ≥ 1, choose u

∈ Φ

such that u

) < 1/2

,andletu =



∞

n=1

.By

Theorem 2.4.8, u is either superharmonic or identically +∞ on Ω.Since

u(x



∞

n=1

) ≤ 1, u is superharmonic on Ω.Sinceu

≥ 1on

Γ, u =+∞ on Γ and Γ is therefore polar. Suppose now that Γ is a compact

polar subset of Ω.Letu =

on Ω ∼ Γ .Sinceu is harmonic on Ω ∼ Γ ,it

has a unique superharmonic extension to Ω by Theorem 4.2.15, namely

;

by Corollary 4.2.16, the extension is harmonic on Ω. Letting

= G

μ, μ

must be the zero measure by Corollary 3.4.6 and

is identically zero on Ω.

162 4 Negligible Sets

Theorem 4.3.9 If Γ is a compact nonpolar subset of a connected Greenian

set Ω having a Green function, then there is a measure μ with support in Γ

such that G

μ is positive, continuous, and bounded on Ω.

Proof: Since Γ is nonpolar, the potential

is positive at some point of Ω by

the preceding lemma and therefore positive on Ω by the minimum principle.

Since

= G

ν for some measure ν with support in Γ , by Theorem 3.6.2

there is a measure μ with support in Γ such that G

μ is positive and con-

tinuous on Ω.SinceG

μ ≤ G

ν =

≤ 1, G

μ is bounded on Ω.

Lemma 4.3.10 If u is superharmonic on B

y,ρ

and 0 <r<ρ, then there

is a superharmonic function ˜u that agrees with u on B

y,r

and is ﬁnite on

∼ B

y,r

Proof: By reducing ρ slightly it can be assumed that u is bounded below on

−

y,ρ

and superharmonic on a neighborhood of B

y,ρ

. It also can be assumed

that u ≥ 0onB

−

y,ρ

.LetΓ = B

−

y,r

and consider

, the regularized reduction

of u over Γ relative to B

y,ρ

.Now

= u on B

y,r

and is harmonic on

y,ρ

∼ B

−

y,r

. By the preceding theorem,

is the potential of a measure

having support in B

−

y,r

⊂ B

y,ρ

. By Theorem 3.4.16, lim

z→z

(z)=0for

all z

∈ ∂B

y,ρ

. Deﬁne

w(z)=



(z)ifz ∈ B

y,ρ

0ifz ∈ ∂B

y,ρ

By Theorem 1.9.4, w can be continued harmonically across ∂B

y,ρ

;thatis,

there is a number δ<ρ− r and a function ˜w deﬁned on B

−

y,ρ+δ

such that

˜w agrees with w on ∂B

y,ρ

and is harmonic on B

y,ρ+δ

∼ B

−

y,ρ− δ

.Nowchoose

α and β such that αu

+ β agrees with ˜w on ∂B

y,ρ

and αu

+ β ≤ ˜w on

y,ρ+δ/2

∼ B

y,ρ

.Then

˜u =



˜w on B

−

y,ρ

αu

+ β on ∼ B

−

y,ρ

has the desired properties.

Theorem 4.3.11 If Z ⊂ R

is a polar set and x

∈ Z, then there is a su-

perharmonic function u such that u =+∞ on Z and u(x

) < +∞; moreover,

if Z is a polar subset of the Greenian set Ω and x

∈ Ω ∼ Z, then there is a

potential u on Ω with these properties.

Proof: Let v be a superharmonic function taking on the value +∞ on Z.

For each y ∈ Z,letB

y,ρ

be a ball such that x

∈ B

y,ρ

. A countable

number of balls corresponding to a sequence {y

} in Z suﬃces to cover Z.

Let B

be the ball corresponding to y

. By the preceding lemma, there is

a superharmonic function v

such that v

=+∞ on B

∩ Z and is ﬁnite

4.4 Capacity 163

outside B

−

.Sincev

is bounded below on B

−

, there is a constant α

such

that v

+ α

≥ 0onB

−

.Nowlet{β

} be a sequence of positive numbers

such that



)+α

) converges. By Theorem 2.4.8, the function u =



+ α

) has the desired properties. As to the second assertion, let μ

be the measure associated with the function u by the Riesz representation

theorem, Theorem 3.5.11. Consider the sequence of balls {B

} where B

,j ≥ 1. Since u(x

) is ﬁnite and u(x

)=G

μ|

)+h

)bythe

same theorem, where h

is harmonic on B

, G

μ|

) is ﬁnite for each

j ≥ 1. Since G

μ|

)=G

μ|

where

is harmonic on B

)

Theorem 3.4.8, G

μ|

) is also ﬁnite for each j ≥ 1. Thus, for each j ≥ 1

a c

> 0 can be chosen so that c

μ|

(Ω) < 1/2

and c

μ|

) < 1/2

Since



∞

j≥1

μ|

is a ﬁnite measure on Ω, G





∞

j=1

μ|



is a potential

on Ω by Theorem 3.4.4. This fact along with the inequality



j=1

μ|

) ≤ G

(

∞



j=1

μ|

)

implies that the sequence of superharmonic functions on the left increases

to a potential function v on Ω.Notethatv(x



∞

j=1

μ|

) ≤ 1.

Since u = G

μ|

+ h

where h

is harmonic on B

, G

μ|

=+∞ on

∩Z for each j ≥ 1. Since G

μ|

= G

μ|

where

is harmonic on

, G

μ|

=+∞ on B

∩ Z for each j ≥ 1. Thus, v =



∞

j=1

μ|

+∞ on Z.

Corollary 4.3.12 There are discontinuous ﬁnite-valued superharmonic

functions.

Proof: Let Z be a nonclosed polar set and let x be a limit point of Z that

is not in Z. By the preceding theorem, there is a superharmonic function v

such that v(x) < +∞ and v =+∞ on Z. Letting u =min(v, v(x)+1),uis a

ﬁnite-valued superharmonic function that takes on the value v(x)+1 on Z.

4.4 Capacity

If an electrical charge is placed on a conductor in the interior of a grounded

sphere, the charge will distribute itself on the surface of the conductor in

such a way that the potential within the conductor is constant. The ratio

of the total charge to the value of this constant is known as the capacity of

the conductor. As the capacity depends only upon the ratio, the capacity is

equal to the total charge if the potential is equal to 1 within the conductor.

164 4 Negligible Sets

In the terminology of the preceding section, if the conductor is represented

by a compact subset Γ of a ball B,thepotentialis

= G

μ,whereμ is

some measure on ∂Γ, and the capacity of Γ is just μ(∂Γ). The mathematical

concept of capacity was deﬁned ﬁrst by Wiener [63] for compact sets in the

n ≥ 3 case in the following way. Let Γ be a compact set and let u be a

harmonic function on ∼ Γ corresponding to the boundary function 1 on ∂Γ

and satisfying lim

|x|→∞

u(x) = 0. Then Wiener deﬁned the capacity C(Γ )of

Γ by the equation

C(Γ )=−

4π



udσ

where Σ is a smooth surface encompassing Γ .

Throughout this section, Ω will be a Greenian subset of R

and Γ ,with

or without subscripts, will denote a compact subset of Ω. The collection of

compact subsets of Ω will be denoted by K(Ω) and the collection of open

subsets by O(Ω). When u =1onΩ, the superscript 1 will be omitted from

and

. As a function of Γ , the maps Γ → R

and Γ →

will be

denoted by R



and



, respectively.

Deﬁnition 4.4.1 For each Γ ∈K(Ω),

is called the capacitary po-

tential of Γ . The unique measure μ

for which

= G

is called the

capacitary distribution for Γ .Thecapacity of Γ relative to Ω, denoted

by C(Γ ), is deﬁned by the equation C(Γ )=μ

(Γ )withC(∅)=0.

The deﬁnition of capacity of a compact set Γ agrees with Wiener’s deﬁni-

tion for the case Ω = R

,n≥ 3.

Example 4.4.2 If Γ is a compact subset of Ω = R

,n ≥ 3, and B is any

ball containing Γ ,then

C(Γ )=−

(n − 2)



∂B

n(x)

(x) dσ(x)

where

is relative to Ω. This formula for calculating C(Γ )canbederived

as follows. In this case,

(x)=



|x − y|

n−2

(dy).

Using Tonelli’s theorem,

−

(n − 2)



∂B

n(x)

(x) dσ(x)

= −

(n − 2)



∂B

n(x)



|x − y|

n−2

(dy) dσ(x)

= −

(n − 2)



∂B

n(x)

|x − y|

n−2

dσ(x)μ

(dy).

4.4 Capacity 165

Letting B

y,δ

⊂ B

−

y,δ

⊂ B and applying Green’s identity, Theorem 1.2.2, to

the region B ∼ B

−

y,δ

, the last expression is equal to

−

(n − 2)



∂B

y,δ

n(x)

|x − y|

n−2

dσ(x)μ

(dy)=μ

(Γ )=C(Γ ).

Example 4.4.3 If D = B

−

x,δ

⊂ Ω = R

,n≥ 3, then

(y)=



1ify ∈ D

n−2

|x−y|

n−2

if y ∈ D

so that

C(D)=−

(n − 2)



∂B

x,2δ

n(y)

(y) dσ(y)=δ

n−2

Thus, the capacity of a closed ball in R

is its radius.

Example 4.4.4 If A = B

−

x,δ

∼ B

x,ρ

, 0 <ρ<δ, is an annulus in Ω =

,n≥ 3, then

is the same as the

of the preceding example so that

C(A)=δ

n−2

. This is in keeping with the intuitive notion that an electrical

charge placed on the annulus will gravitate to the outer sphere.

As Lebesgue measure theory commences with the area of simple regions,

it is possible to extend the concept of capacity commencing with compact

sets in an analogous way.

Deﬁnition 4.4.5 If E is a subset of Ω,theinner capacity of E, denoted

by C

∗

(E), is deﬁned by

∗

(E)=sup{C(K); K ∈K(Ω),K ⊂ E}.

Clearly, C

∗

(K)=C(K)ifK ∈K(Ω), and C

∗

is a nonnegative extended real-

valued set function on the class of all subsets of Ω with C(∅)=C

∗

(∅)=0.

Deﬁnition 4.4.6 If E is a subset of the Greenian set Ω,theouter capacity

of E, denoted by C

∗

(E), is deﬁned by

∗

(E)=inf{C

∗

(O):O ∈O(Ω),O ⊃ E}.

The outer capacity C

∗

is a nonnegative extended real-valued set function on

the class of all subsets of Ω.

166 4 Negligible Sets

Deﬁnition 4.4.7 A subset E of Ω is capacitable if C

∗

(E)=C

∗

(E), in

which case the capacity C(E) is deﬁned to be the common value.

As is the case for Lebesgue measure, every Borel subset of Ω is capacitable.

The proof of this fact requires several steps and will be completed only in the

next section. The initial steps are limited to Greenian sets Ω ⊂ R

having a

certain property, known at this stage to be valid only for balls. The theorems

will be stated, however, for any Greenian set as it will be shown in the next

section that every Greenian set has this property.

Property B: The open set Ω has a Green function and lim

y→x,y∈Ω

μ(y)=

0 quasi everywhere on ∂Ω for all measures μ having compact support in Ω.

If Ω is an unbounded open subset of R

, then lim

|y|→+∞,y∈Ω

μ(y)=0

since G

is dominated by the Green function for R

,namely x − y 

−n+2

According to Theorem 3.4.16, balls are known to have Property B. It will be

shown in the next section that every open set having a Green function has

Property B. Some of the proofs of the following results do not require that

this property hold. If this property is required, it will be indicated at the

beginning of the proof. The following theorem is known as the domination

principle.

Theorem 4.4.8 (Maria [42], Frostman [22]) Let Ω be a Greenian set, μ

ameasureonΩ,andv a positive superharmonic function on Ω.IfG

μ<

+∞a.e.(μ) on the support of μ and G

μ ≤ v inner q.e. on the support of μ,

then G

μ ≤ v on Ω.

Proof: (Assuming Property B) According to Theorem 3.6.3, there is a se-

quence of measures {μ

} such that each μ

in the restriction of μ to a compact

subset of the support of μ, G

is ﬁnite-valued and continuous on Ω,and

↑ G

μ.SinceG

≤ G

μ<+∞a.e.(μ), G

< +∞a.e.(μ

)

and G

μ ≤ v inner quasi everywhere on the support of μ

.Ifitcanbe

shown that G

≤ v on Ω, it would then follow that G

μ ≤ v on Ω.It

therefore suﬃces to prove the result assuming that μ has compact support

Γ ⊂ Ω and G

μ is ﬁnite-valued and continuous on Ω.Ifx is a point of Γ

for which G

μ(x) ≤ v(x), then

lim inf

y→x,y∈Ω

v(y) ≥ v(x) ≥ G

μ(x) = lim

y→x,y∈Ω

μ(y).

Thus,

lim inf

y→x,y∈Ω

(v −G

μ)(y) ≥ 0

for x ∈ Γ except possibly for an inner polar set. This inequality also holds

for x ∈ ∂Ω except possibly for a polar set and, in case Ω is an unbounded

subset of R

,n≥ 3, x = ∞. By Theorem 4.2.21, v −G

μ ≥ 0onΩ.

4.4 Capacity 167

Remark 4.4.9 Let F = {u

; i ∈ I} be a left-directed family of locally uni-

formly bounded below superharmonic functions on Ω,andletˆu be the lower

regularization of u =inf

i∈I

. Then there is a decreasing sequence {w

} in

F such that ˆu is the lower regularization of inf

j≥1

. This can be seen as fol-

lows. By Theorem 2.2.8, there is a countable set I

⊂ I such that if g is l.s.c.

on Ω and g ≤ inf

j≥1

,theng ≤ inf

i∈I

. Letting w

= u

and inductively

replacing u

by a w

∈Fsatisfying w

≤ min (u

j−1

), the decreasing

sequence {w

} in F has the property that inf

i∈I

≤ inf

j≥1

≤ inf

j≥1

Letting w =inf

j≥1

, ˆu ≤ ˆw; but since ˆw is l.s.c. and ˆw ≤ inf

j≥1

≤

inf

j≥1

, ˆw ≤ inf

i∈I

. Therefore, ˆw ≤ ˆu and the two are equal; that is, ˆu

is the lower regularization of a decreasing sequence in F.

It is necessary to prove the following theorem in two steps, the ﬁrst estab-

lishing the conclusion inner quasi everywhere. The proof will be revisited.

Theorem 4.4.10 (Cartan [11]) Let F = {u

; i ∈ I} be a family of locally

uniformly bounded below superharmonic functions on an open set Ω,andlet

u =inf

i∈I

.Thenˆu = u q.e. on Ω.

Partial proof:(ˆu = u inner quasi everywhere) Since adjoining the min-

ima of all ﬁnite subsets of F to F has no eﬀect on u or ˆu, it can be assumed

that F is left-directed. Since the conclusion is a local property, it can be as-

sumed that Ω is a ball and that all u

are nonnegative on Ω. By the preceding

remark, it can be assumed that F is a decreasing sequence {u

} of nonneg-

ative superharmonic functions with u = lim

j→∞

. Furthermore, it suﬃces

to prove that u =ˆu inner quasi everywhere on a slightly smaller concentric

ball B ⊂ Ω.Also,eachu

can be replaced by

, the regularized reduction

of u

on B relative to Ω,sinceu

on B.Since

is a potential on

Ω,

= G

for some measure μ

with support in B

−

. It therefore can be

assumed that u

= G

,j ≥ 1. By taking a concentric ball slightly larger

than B and the reduction of 1 over it, a measure λ can be found so that

λ =1onB

−

. By the reciprocity theorem, for j ≥ 1,

−



1 dμ



λdμ



dλ ≤



Gμ

dλ;

by reversing the steps, the latter integral is equal to μ

−

), and therefore,

−

) ≤ μ

−

) < +∞ for all j ≥ 1. Thus, the set {μ

−

):j ≥ 1} is

bounded, and there is a subsequence of the sequence {μ

} that converges in

the w

∗

- topology to some measure μ with support in B

−

. It can be assumed

that the sequence itself has this property. If m is any positive number, then

u = lim

j→∞

≥ lim

j→∞



min (G

(·,y),m) dμ

(y)=



min (G

(·,y),m) dμ(y).

Letting m ↑∞,u≥ G

μ on Ω. Consider the set M = {u>ˆu},letΓ be any

compact subset of M, and assume there is a measure ν with support in Γ