Helms L.L. Potential Theory
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188 4 Negligible Sets
u
j
(x)=
J
δ
j
/3
(
ˆ
R
u
Ω
j
)(x)ifd(x, Ω
j
) ≤ δ
j
/3,x∈ Ω
ˆ
R
u
Ω
j
(x)ifd(x, Ω
j
) >δ
j
/3,x∈ Ω
where the operator J is defined by Equation (2.5.1). The sequence {u
j
} has
the required properties.
Lemma 4.6.12 If Ω is a Greenian subset of R
n
, Λ ⊂ Ω,andv is a non-
negative, finite-valued superharmonic function that is continuous on Λ,then
R
v
Λ
=inf{R
v
O
; O ∈O(Ω),O ⊃ Λ}.
Proof: ItcanbeassumedthatΩ is connected and that v>0onΩ.Letu be
any positive superharmonic function on Ω that majorizes v on Λ.Let>0,
fix x ∈ Ω,andchoosen ≥ 1 such that (1/n)R
v
Λ
(x) <.Then(1+
1
n
)u−v>0
on an open set O
n
⊃ Λ, and therefore
(1 +
1
n
)u ≥ R
v
O
n
≥ inf {R
v
O
; O ∈O(Ω),O ⊃ Λ}.
Thus,
R
v
Λ
(x)+ ≥ inf {R
v
O
(x); O ∈O,O ⊃ Λ}.
The result follows by letting → 0 and using the fact that the resulting
opposite inequality is always true.
Theorem 4.6.13 If {u
j
} is a sequence of nonnegative superharmonic func-
tions on the Greenian set Ω such that u =
∞
j=1
u
j
is superharmonic on Ω
and Λ ⊂ Ω,then
ˆ
R
u
Λ
=
∞
j=1
ˆ
R
u
j
Λ
.
Proof: Suppose first that the sequence is of length two. By Theorem 4.3.5,
ˆ
R
u
1
+u
2
Λ
≤
ˆ
R
u
1
Λ
+
ˆ
R
u
2
Λ
so that it suffices to prove the opposite inequality.
According to Theorem 4.6.9, there are nonnegative superharmonic functions
v
1
and v
2
such that
ˆ
R
u
1
+u
2
Λ
= v
1
+ v
2
with v
i
≤
ˆ
R
u
i
Λ
,i =1, 2. Assume that
Λ is open. Then
ˆ
R
u
1
Λ
= u
1
,
ˆ
R
u
2
Λ
= u
2
,and
ˆ
R
u
1
+u
2
Λ
= u
1
+ u
2
on Λ.Since
v
j
≤
ˆ
R
u
j
Λ
≤ u
j
,j =1, 2, and v
1
+ v
2
= u
1
+ u
2
on Λ, v
j
= u
j
,j =1, 2onΛ.
It follows that v
j
≥
ˆ
R
u
j
Λ
,j =1, 2, that
ˆ
R
u
1
+u
2
Λ
= v
1
+ v
2
≥
ˆ
R
u
1
Λ
+
ˆ
R
u
2
Λ
,and
therefore that
ˆ
R
u
1
+u
2
Λ
=
ˆ
R
u
1
Λ
+
ˆ
R
u
2
Λ
whenever Λ is open. Suppose now that
u
1
and u
2
are nonnegative, finite-valued, continuous superharmonic functions
on Ω and Λ is an arbitrary subset of Ω. For any O ∈O(Ω)withO ⊃
Λ, R
u
1
+u
2
O
= R
u
1
O
+ R
u
2
O
≥ R
u
1
Λ
+ R
u
2
Λ
≥
ˆ
R
u
1
Λ
+
ˆ
R
u
2
Λ
. By the preceding
lemma, R
u
1
+u
2
Λ
≥
ˆ
R
u
1
Λ
+
ˆ
R
u
2
Λ
. Taking the lower regularization of both sides,
ˆ
R
u
1
+u
2
Λ
≥
ˆ
R
u
1
Λ
+
ˆ
R
u
2
Λ
and the two are equal. Lastly, suppose that u
1
and u
2
are any nonnegative superharmonic functions on Ω and Λ is any subset of Ω.
According to Theorem 4.6.11, there are increasing sequences {u
1j
} and {u
2j
}
of nonnegative, finite-valued, continuous superharmonic functions such that
u
i
= lim
j→∞
u
ij
,i =1, 2. Then,
ˆ
R
u
1j
+u
2j
Λ
=
ˆ
R
u
1j
Λ
+
ˆ
R
u
2j
Λ
by the preceding
4.7 Sweeping 189
step. By Theorem 4.6.10,
ˆ
R
u
1
+u
2
Λ
=
ˆ
R
u
1
Λ
+
ˆ
R
u
2
Λ
and the theorem is true for
sequences of length two. This result has a trivial extension to finite sequences.
Suppose now that {u
j
} is an infinite sequence. Then for each k ≥ 1,
ˆ
R
k
j=1
u
j
Λ
=
k
j=1
ˆ
R
u
j
Λ
.
By Theorem 4.6.10,
ˆ
R
u
Λ
=
∞
j=1
ˆ
R
u
j
Λ
.
4.7 Sweeping
If u = G
Ω
μ is the potential of a measure μ on the Greenian set Ω with
μ(Ω) < +∞ and Λ is an open subset of Ω,then
ˆ
R
u
Ω∼Λ
is again a potential
that is harmonic on Λ. By the Riesz decomposition theorem, Theorem 3.5.11,
ˆ
R
u
Ω∼Λ
= G
Ω
μ
Ω∼Λ
where μ
Ω∼Λ
(Λ)=0.Themeasureμ
Ω∼Λ
is obtained from
μ by “sweeping” the measure in Λ out of Λ. The operation taking μ into
μ
Ω∼Λ
is called the sweeping of μ,andμ
Ω∼Λ
is called the swept measure.
An iteration of the sweeping operation leads to a cumbersome notation.
For this reason, the notations
ˆ
R
u
Λ
and
ˆ
R
Λ
(u, ·) will be used interchangeably. If
x ∈ Ω, the unit measure concentrated on {x} will be denoted by δ
Ω
(x, ·).
Definition 4.7.1 If Λ is a subset of the Greenian set Ω, define G
Λ
Ω
(x, ·)and
a unique measure δ
Λ
Ω
(x, ·) by the equation
G
Λ
Ω
(x, y)=
ˆ
R
Λ
(G
Ω
(x, ·),y)=
Ω
G
Ω
(y,z) δ
Λ
Ω
(x, dz);
if μ is a measure on Ω,G
Λ
Ω
μ is defined by the equation
G
Λ
Ω
μ(x)=
G
Λ
Ω
(x, y) μ(dy).
The measure δ
Λ
Ω
(x, ·) is called the sweeping of δ
Ω
(x, ·)ontoΛ. That such a
measure exists is a consequence of Theorem 3.5.11.
Lemma 4.7.2 If Λ is any subset of the Greenian set Ω, then for any x ∈ Ω
(i) δ
Λ
Ω
(x, Ω) ≤ 1,
(ii) δ
Λ
Ω
(x, ·)=δ
Ω
(x, ·) and δ
Λ
Ω
(x, {x})=1if x is an interior point of Λ,
(iii) δ
Λ
Ω
(x, ·) has support in ∂Λ whenever x ∈ Ω ∼ int Λ.
Proof: (i)IfV is any open subset of Ω with compact closure V
−
⊂ Ω,
let u =
ˆ
R
1
V
= G
Ω
μ
V
.Since1≥ u = G
Ω
μ
V
=
G
Ω
(x, y) dμ
V
(y)and
G
Ω
(x, y) ≥
ˆ
R
Λ
(G
Ω
(x, ·),y), by Tonelli’s theorem
190 4 Negligible Sets
1 ≥
ˆ
R
Λ
(G
Ω
(x, ·),y) dμ
V
(y)
=
G
Ω
(y,z) δ
Λ
Ω
(x, dz) dμ
V
(y)
=
G
Ω
(z,y) dμ
V
(y) δ
Λ
Ω
(x, dz)
=
u(z) δ
Λ
Ω
(x, dz)
≥
V
u(z) δ
Λ
Ω
(x, dz)=δ
Λ
Ω
(x, V ).
Letting V ↑ Ω, δ
Λ
Ω
(x, Ω) ≤ 1. (ii)Ifx is an interior point of Λ,then
G
Ω
(y,z) δ
Λ
Ω
(x, dz)=
ˆ
R
Λ
(G
Ω
(x, ·),y)=G
Ω
(x, y)=
G
Ω
(y,z) δ
Ω
(x, dz)
so that δ
Λ
Ω
(x, ·)=δ
Ω
(x, ·) by the uniqueness of δ
Λ
Ω
(x, ·). (iii)Ifx is not an
interior point of Λ,then
ˆ
R
Λ
(G
Ω
(x, ·), ·) is harmonic on the interior of Λ and
on Ω ∼ Λ
−
so that δ
Λ
Ω
(x, ·) has its support in ∂Λ by Theorem 3.4.9.
By (iii)above,δ
Λ
Ω
(x, {x})=0ifx ∈ Ω ∼ Λ
−
so that δ
Λ
Ω
(x, {x})is0or
1 except possibly for x ∈ ∂Λ. It will be shown later in this section that the
same is true of points in ∂Λ.
Lemma 4.7.3 For each Borel set M,δ
Λ
Ω
(·,M) is a Borel measurable function
on Ω.
Proof: Let P be the set of bounded, continuous potentials u = G
Ω
μ on Ω
for which μ has compact support in Ω,andletF be the set of all differences
f = u − v, u, v ∈Phaving compact supports in Ω. Considering
ˆ
R
u
Λ
as an
operator, it can be extended to F by putting
ˆ
R
f
Λ
=
ˆ
R
u
Λ
−
ˆ
R
v
Λ
whenever f = u − v ∈F. If the function f ∈Fhas two representations
f = u − v and f = u
− v
,thenu + v
= u
+ v,
ˆ
R
u
Λ
+
ˆ
R
v
Λ
=
ˆ
R
u+v
Λ
=
ˆ
R
u
+v
Λ
=
ˆ
R
u
Λ
+
ˆ
R
v
Λ
by Theorem 4.6.13, and therefore
ˆ
R
u
Λ
−
ˆ
R
v
Λ
=
ˆ
R
u
Λ
−
ˆ
R
v
Λ
.Thus,
ˆ
R
f
Λ
is well-
defined. For each x ∈ Ω,themapf → L
x
(f)=
ˆ
R
f
Λ
(x) is easily seen to be
linear on F. Letting
F be the collection of uniform limits of sequences in
F,
F contains all continuous functions with compact support in Ω according
to Theorem 3.4.15. It will be shown next that L
x
has an extension to F
as a bounded linear functional. Let {f
j
} be a sequence in F that converges
uniformly to f. By Corollary 4.3.7,
4.7 Sweeping 191
sup
x∈Ω
|L
x
(f
i
) − L
x
(f
j
)|≤sup
x∈Ω
|
ˆ
R
f
i
Λ
(x) −
ˆ
R
f
j
Λ
(x)|≤|f
i
− f
j
|
from which it follows that the limit L
x
(f) = lim
j→∞
L
x
(f
j
) exists. It is easily
seen using a similar inequality that L
x
(f) does not depend upon the sequence
{f
j
}. Consider any nonnegative f ∈Fand a sequence {f
j
} in F with f
j
=
u
j
−v
j
which converges uniformly to f.If
j
= −inf
x∈Ω
(u
j
(x) −v
j
(x)), then
u
j
+
j
≥ v
j
,
ˆ
R
u
j
Λ
+
j
≥
ˆ
R
u
j
+
j
Λ
≥
ˆ
R
v
j
Λ
,
and therefore
L
x
(f) = lim
j→∞
L
x
(f
j
) = lim
j→∞
(
ˆ
R
u
j
Λ
−
ˆ
R
v
j
Λ
) ≥ lim
j→∞
(−
j
)=0.
This shows that L
x
is positive and linear on F.Since
|L
x
(f)| = lim
j→∞
|L
x
(f
j
)| = lim
j→∞
|
ˆ
R
u
j
Λ
−
ˆ
R
v
j
Λ
|≤ lim
j→∞
|u
j
−v
j
| = lim
j→∞
|f
j
| = |f |,
L
x
is a bounded, positive linear functional on F. By the Riesz representation
theorem, for each x ∈ Ω there is a measure ρ
Λ
Ω
(x, ·) on the Borel subsets of
Ω with ρ
Λ
Ω
(x, Ω) ≤ 1 such that
L
x
(f)=
f(y) ρ
Λ
Ω
(x, dy),f∈ F.
Since L
x
(f)=
ˆ
R
u
Λ
(x) −
ˆ
R
v
Λ
(x),L
x
(f) is a Borel measurable function for
all f ∈Fand all f ∈
F by approximation. Since the latter contains all
continuous functions with compact support in Ω, ρ
Λ
Ω
(x, M) is a Borel function
of x for each Borel set M ⊂ Ω. The conclusion of the theorem will follow if it
can be shown that δ
Λ
Ω
(x, ·)=ρ
Λ
Ω
(x, ·). Suppose u = G
Ω
μ is the potential of a
measure with compact support Γ ⊂ Ω.Let{Ω
j
} be an increasing sequence of
open subsets of Ω with compact closures such that Ω = ∪Ω
j
,Γ ⊂ Ω
1
,Ω
−
j
⊂
Ω
j+1
for all j ≥ 1. By Theorem 2.5.2, there is an increasing sequence {u
i
} of
continuous potentials for which u
i
= u on Ω ∼ Ω
1
and lim
i→∞
u
i
= u on Ω.
Moreover, for each i ≥ 1, there is an increasing sequence {u
ik
} of continuous
potentials such that lim
k→∞
u
ik
=
ˆ
R
u
Ω∼Ω
i
and u
ik
=
ˆ
R
u
Ω∼Ω
i
on Ω ∼ Ω
i+1
for all k ≥ 1. Then u
i
− u
ii
∈Fand
ˆ
R
u
i
Λ
(z) −
ˆ
R
u
ii
Λ
(z)=
(u
i
(y) − u
ii
(y)) ρ
Λ
Ω
(z,dy). (4.5)
Since {
ˆ
R
u
Ω∼Ω
i
} is a decreasing sequence of functions on Ω and each
ˆ
R
u
Ω∼Ω
i
is harmonic on Ω
i
, lim
i→∞
ˆ
R
u
Ω∼Ω
i
is a nonnegative harmonic function that is
dominated by the potential u. Since the greatest harmonic minorant of u is the
zero function,
lim
i→∞
ˆ
R
u
Ω∼Ω
i
(z)=0,z∈ Ω.
192 4 Negligible Sets
Therefore,
lim sup
i→∞
ˆ
R
u
ii
Λ
(z) ≤ lim sup
i→∞
u
ii
(z) ≤ lim
i→∞
ˆ
R
u
Ω∼Ω
i
(z)=0.
Since
ˆ
R
u
ii
Λ
(z)=
u
ii
(y) ρ
Λ
Ω
(z,dy), it follows from Equation (4.5) and
Theorem 4.6.10 that
ˆ
R
u
Λ
(z)=
u(y) ρ
Λ
Ω
(z,dy).
Taking u(x)=G
Ω
(x, ·),
ˆ
R
Λ
(G
Ω
(x, ·),z)=
G
Ω
(x, y) ρ
Λ
Ω
(z,dy).
Thus, ρ
Λ
Ω
(x, ·) satisfies the definition of δ
Λ
Ω
(x, ·) and the two are equal.
Theorem 4.7.4 If u is a nonnegative superharmonic function on the Gree-
nian set Ω and Λ ⊂ Ω,then
ˆ
R
u
Λ
(x)=
u(y) δ
Λ
Ω
(x, dy),x∈ Ω.
Proof: In the course of the preceding proof, it was shown that the result
is true if u is the potential of a measure having compact support in Ω.By
Corollary 4.3.6, there is a sequence of such potentials that increases to u.
The result follows from the Lebesgue monotone convergence theorem and
Theorem 4.6.10.
Lemma 4.7.5 If u is a nonnegative superharmonic function on the Greenian
set Ω and A ⊂ B ⊂ Ω,then
ˆ
R
B
(
ˆ
R
u
A
, ·)=
ˆ
R
A
(
ˆ
R
u
B
, ·)=
ˆ
R
u
A
.
Proof: Since A ⊂ B and
ˆ
R
u
B
≤ u,
ˆ
R
A
(
ˆ
R
u
A
, ·) ≤
ˆ
R
A
(
ˆ
R
u
B
, ·) ≤
ˆ
R
A
(u, ·)
and
ˆ
R
A
(
ˆ
R
u
A
, ·) ≤
ˆ
R
B
(
ˆ
R
u
A
, ·) ≤
ˆ
R
A
(u, ·).
It suffices to show that
ˆ
R
A
(u, ·)=
ˆ
R
A
(
ˆ
R
A
(u, ·). Since
ˆ
R
u
A
= u q.e. on A by
Corollary 4.6.2,
ˆ
R
A
(u, ·)=
ˆ
R
A
(
ˆ
R
u
A
, ·) according to Corollary 4.6.4.
Theorem 4.7.6 If u is a nonnegative superharmonic function on the Gree-
nian set Ω,A, B ⊂ Ω,andu
=min(
ˆ
R
u
A
,
ˆ
R
u
B
),then
R
u
A∪B
+ R
u
A∪B
= R
u
A
+ R
u
B
4.7 Sweeping 193
and
ˆ
R
u
A∪B
+
ˆ
R
u
A∪B
=
ˆ
R
u
A
+
ˆ
R
u
B
.
Proof: On A ∼ B,R
u
A
= u ≥ R
u
B
so that u + u
= u + R
u
B
= R
u
A
+ R
u
B
.
Interchanging A and B,u+u
= R
u
A
+R
u
B
on B ∼ A.OnA∩B,R
u
A
= u = R
u
B
so that u + u
= u + u = R
u
A
+ R
u
B
. Therefore, u + u
=
ˆ
R
u
A
+
ˆ
R
u
B
q.e. on
A ∪B. By the preceding lemma and Corollary 4.6.4,
ˆ
R
u
A∪B
+
ˆ
R
u
A∪B
=
ˆ
R
A∪B
(u + u
, ·)
=
ˆ
R
A∪B
(
ˆ
R
u
A
+
ˆ
R
u
B
, ·)
=
ˆ
R
A∪B
(
ˆ
R
u
A
, ·)+
ˆ
R
A∪B
(
ˆ
R
u
B
, ·)
=
ˆ
R
u
A
+
ˆ
R
u
B
on A ∪ B. By Theorem 4.6.3, on Ω ∼ (A ∪ B),
R
u
A∪B
+ R
u
A∪B
= R
u
A
+ R
u
B
.
Since this equation holds on A ∪ B, it holds on Ω.
Definition 4.7.7 If u is a Borel measurable function on the Greenian set Ω,
define the function δ
Λ
Ω
(·,u) by putting
δ
Λ
Ω
(x, u)=
u(y) δ
Λ
Ω
(x, dy)
provided the integral is defined for all x ∈ Ω.Ifμ is a Borel measure on Ω,
define the measure δ
Λ
Ω
(μ, ·) on the Borel subsets M of Ω by putting
δ
Λ
Ω
(μ, M )=
δ
Λ
Ω
(x, M) μ(dx)
provided the integral is defined for all M.
Lemma 4.7.8 For each x and each Borel subset M of the Greenian set
Ω,δ
·
Ω
(x, M) is subadditive; that is,
δ
A∪B
Ω
(x, M) ≤ δ
A
Ω
(x, M)+δ
B
Ω
(x, M).
Proof: Using the notation in the proof of the Lemma 4.7.3, let f = u
1
−
u
2
≥ 0whereu
1
and u
2
are bounded, continuous potentials of measures with
compact support and let u
i
=min(R
u
i
A
, R
u
i
B
),i=1, 2. By Theorem 4.7.6 and
Theorem 4.7.4,
δ
A∪B
Ω
(x, u
i
)+δ
A∪B
Ω
(x, u
i
)=δ
A
Ω
(x, u
i
)+δ
B
Ω
(x, u
i
),i=1, 2.
194 4 Negligible Sets
Since u
1
≥ u
2
,u
1
≥ u
2
and
δ
A∪B
Ω
(x, u
1
− u
2
) ≤ δ
A
Ω
(x, u
1
− u
2
)+δ
B
Ω
(x, u
1
− u
2
).
Thus,
δ
A∪B
Ω
(x, f) ≤ δ
A
Ω
(x, f)+δ
B
Ω
(x, f)
for all nonnegative f ∈Fhaving compact support. Since any continuous
function on Ω with compact support can be approximated uniformly by a
sequence of such functions according to Theorem 3.4.15, the last inequality
holds for all continuous functions with compact support. Using the usual
methods of measure theory to pass from continuous functions to Borel sets,
the result holds for all Borel sets.
Lemma 4.7.9 (Zero-one Law) If Ω is a Greenian set, x ∈ Ω, Λ ⊂ Ω,and
u is a nonnegative superharmonic function on Ω,then
(i) δ
Λ
Ω
(x, Z)=0for all polar sets Z not containing x,
(ii) δ
Λ
Ω
(x, {x})=0or 1,
(iii) if δ
Λ
Ω
(x, {x})=0,thenδ
Λ∩U
Ω
(x, {x})=0for all neighborhoods U of x,
and
lim
U↓{x}
ˆ
R
u
Λ∩U
=0 on Ω
lim
U↓{x}
ˆ
R
Λ
(
ˆ
R
u
U
,x)= 0
whenever u(x) < +∞ ,and
(iv) if δ
Λ
Ω
(x, {x})=1,thenδ
Λ∩U
Ω
(x, {x})=1and
ˆ
R
u
Λ∩U
(x)=u(x) for all
neighborhoods U of x.
Proof: (i)LetZ be a polar set not containing x. By Theorem 4.3.11, there is
a nonnegative superharmonic function v with v(x) < +∞ such that v =+∞
on Z.Since
+∞ >v(x) ≥
ˆ
R
v
Λ
(x)=
v(y) δ
Λ
Ω
(x, dy) ≥
Z
v(y) δ
Λ
Ω
(x, dy),
δ
Λ
Ω
(x, Z)=0.(ii)Considerδ
Λ∩U
Ω
(x, {x}) as a function of the neighborhood
U of x.IfU
1
and U
2
are any neighborhoods of x with U
1
⊂ U
2
, then by the
preceding lemma
δ
Λ∩U
2
Ω
(x, {x}) ≤ δ
Λ∩U
1
Ω
(x, {x})+δ
Λ∩(U
2
∼U
1
)
Ω
(x, {x}).
Since the support of δ
Λ∩(U
2
∼U
1
)
Ω
(x, ·) is in the closure of U
2
∼ U
1
, the second
term on the right is zero. Thus, as a function of U, δ
Λ∩U
Ω
(x, {x}) increases as
U decreases. By Lemma 4.7.5,
ˆ
R
Λ∩U
(G
Ω
(x, ·)) =
ˆ
R
Λ∩U
(
ˆ
R
Λ
(G
Ω
(x, ·))
4.7 Sweeping 195
so that
G
Ω
(y,z) δ
Λ∩U
Ω
(x, dz)=
G
Ω
(y,z) δ
Λ
Ω
(w, dz) δ
Λ∩U
Ω
(x, dw)
=
G
Ω
(y,z)
δ
Λ
Ω
(w, dz) δ
Λ∩U
Ω
(x, dw).
Since the function on the left side of this equation is a potential, by the
uniqueness assertion of the Riesz decomposition theorem, Theorem 3.5.11,
δ
Λ∩U
Ω
(x, ·)=
δ
Λ
Ω
(w, ·) δ
Λ∩U
Ω
(x, dw).
Since δ
Λ∩U
Ω
(x, ·) vanishes on polar sets not containing x by (i),
δ
Λ∩U
Ω
(x, {x})=δ
Λ
Ω
(x, {x})δ
Λ∩U
Ω
(x, {x}). (4.6)
Taking U = Ω, it follows that δ
Λ
Ω
(x, {x})=0or1,proving(ii). If
δ
Λ
Ω
(x, {x}) = 0, then Equation (4.6) implies that δ
Λ∩U
Ω
(x, {x}) = 0 for all
neighborhoods U of x;ifδ
Λ
Ω
(x, {x})=1,thenδ
Λ∩U
Ω
(x, {x}) = 1 for all neigh-
borhoods of x since δ
Λ∩U
Ω
(x, {x}) decreases as U increases. It remains only
to prove the assertions concerning the reduced function. If δ
Λ
Ω
(x, {x})=1,
then
ˆ
R
u
Λ∩U
(x)=
u(y) δ
Λ∩U
Ω
(x, dy)=u(x).
This proves (iv). Lastly, suppose δ
Λ∩U
Ω
(x, {x})=0andu(x) < +∞.Toshow
that lim
U↓{x}
ˆ
R
u
Λ∩U
=0onΩ it suffices to consider a decreasing sequence of
balls {B
j
} with centers at x, with closures in Ω,and∩B
j
= {x}.Then
ˆ
R
u
Λ∩B
j
is a decreasing sequence of potentials on Ω with limit w = lim
j→∞
ˆ
R
u
Λ∩B
j
which is harmonic on Ω ∼{x}. Restricting w to Ω ∼{x},byBˆocher’s
theorem, Theorem 3.2.2, it has a nonnegative superharmonic extension ˜w to
Ω of the form cG
Ω
(x, ·)+h,whereh is harmonic on Ω.SincecG
Ω
(x, ·)+
h ≤
ˆ
R
u
Λ∩B
j
on Ω ∼{x} and both functions are superharmonic, the same
inequality holds on Ω. By the Riesz decomposition theorem, Theorem 3.5.11,
h is the greatest harmonic minorant of ˜w and is therefore nonnegative. Since
h ≤
ˆ
R
u
Λ∩B
j
and the latter is a potential, h =0onΩ.Thus, ˜w = cG
Ω
(x, ·) ≤
ˆ
R
u
Λ∩B
j
≤ u on Ω.Sinceu(x) < +∞,c =0and ˜w = w =0onΩ ∼{x}.
Since
u(y) δ
Λ
Ω
(x, dy)=
ˆ
R
u
Λ
(x) ≤ u(x) < +∞, the sequence {
ˆ
R
u
Λ∩B
j
} is
dominated by the δ
Λ
Ω
(x, ·) integrable function u so that by Lemma 4.7.5 and
the result just proved
196 4 Negligible Sets
lim
j→∞
ˆ
R
u
Λ∩B
j
(x) = lim
j→∞
ˆ
R
Λ
(
ˆ
R
u
Λ∩B
j
)(x)
= lim
j→∞
ˆ
R
u
Λ∩B
j
(z) δ
Λ
Ω
(x, dz)
= lim
j→∞
Ω∼{x}
ˆ
R
u
Λ∩B
j
(z) δ
Λ
Ω
(x, dz)
=0.
The same argument can be applied to the sequence {
ˆ
R
u
B
j
} to show that
lim
j→∞
ˆ
R
u
B
j
=0onΩ ∼{x}
and that the sequence is dominated by the δ
Λ
Ω
(x, ·) integrable function u
so that
lim
j→∞
ˆ
R
Λ
(
ˆ
R
u
B
j
,x) = lim
j→∞
ˆ
R
u
B
j
(y) δ
Λ
Ω
(x, dy)=0.
Chapter 5
Dirichlet Problem for Unbounded Regions
5.1 Introduction
The principle problem associated with unbounded regions is the lack of
uniqueness of the solution to the Dirichlet problem. To achieve uniqueness,
the point at infinity ∞ will be adjoined to R
n
with the enlarged space de-
noted by R
n
∞
. This will require redefinition of harmonic and superharmonic
functions. The Dirichlet problem for the exterior of a ball will be solved by
a Poisson type integral. Using this result, it will be shown that the Perron-
Wiener-Brelot method can be used to solve the Dirichlet problem for un-
bounded regions.
Poincar´e’s exterior ball condition and Zaremba’s exterior cone condition
are sufficient conditions for a finite boundary point to be a regular boundary
point for the Dirichlet problem. Both conditions preclude the boundary point
from being “too surrounded” by the region. On the other hand, the Lebesgue
spine is an example of a region that does “surround” a boundary point too
much; in some sense, the complement of the region is “thin” at the boundary
point. A concept of thinness will be explored and related to a topology on
R
n
∞
finer than the metric topology which is more natural from the potential
theoretic point of view. The words “open,” “neighborhood,” “continuous,”
etc., will be prefixed by “fine” or “finely” when used in this context.
5.2 Exterior Dirichlet Problem
Consider the one-point compactification R
n
∞
= R
n
∪{∞} of R
n
. The topology
for R
n
∞
is obtained by adjoining to a base for the Euclidean topology of R
n
the class of sets of the form R
n
∼ Γ ,whereΓ is a compact subset of R
n
in
the usual topology. In this chapter, Ω will denote an open subset of R
n
∞
that
might include the point at infinity. As usual, if Ω is an open subset of R
n
,
L.L. Helms, Potential Theory, Universitext, 197
c
Springer-Verlag London Limited 2009