Full text of DOI:10.1112/S0024609399006803

EVALUATION OF SUPERHARMONIC FUNCTIONS USING
LIMITS ALONG LINES
HIROAKI AIKAWA and STEPHEN J. GARDINER
1. Introduction
If u is a superharmonic function on R2, then
u(x, y) = lim inf u(x, t)
t→y
for all (x, y) ∈ R2. This follows from the fact that a line segment in R2 is non-thin
at each of its constituent points. (See Doob [1, 1.XI] or Helms [7, Chapter 10] for
an account of thin sets and the fine topology.) The situation is different in higher
dimensions. For example, if u is the Newtonian potential on R3 defined by
Z
u(x, y, z) = 1 {x2 + y2 + (z − t)2}−1/2|t| dt,
−1
then
ꢀ
+∞ (0 < |z| 6 1)
u(0, 0, z) =
2
(z = 0).
Corollary 2 below will show that, nevertheless, for nearly every vertical line L, the
value of a superharmonic function at any point X of L is determined by its lower
limit along L at X.
Throughout this paper, we let n > 3. A typical point of Rn will be denoted by
X or (X0, x), where X0 ∈ Rn−1 and x ∈ R. Given any function f : Rn → [−∞, +∞]
and any point X, we define the vertical cluster set of f at X by
CV (f; X) = {` ∈ [−∞, +∞] : there is a sequence (tm) in R \ {x}
such that tm → x and f(X0, tm) → `},
and the fine cluster set of f at X by
CF (f; X) = {` ∈ [−∞, +∞] : for each neighbourhood N of ` in [−∞, +∞],
the set f−1(N) is non-thin at X}.
Theorem. If f : Rn → [−∞, +∞], then there is a polar subset E0 of Rn−1 such
that CV (f; X) ∩ CF (f; X) = ? whenever X0 ∈ Rn−1 \ E0.
Received 23 February 1998; revised 31 March 1999.
1991 Mathematics Subject Classification 31B05.
This work was supported in part by Grant-in-Aid for Scientific Research (B) No. 09440062, Japanese
Ministry of Education, Science and Culture. The second author is grateful for the hospitality of the
Department of Mathematics at Shimane University.
Bull. London Math. Soc. 32 (2000) 209–213

210
hiroaki aikawa and stephen j. gardiner
In [2, Theorem 1], it was shown that there is a subset E0 of Rn−1 such that
E0 × {0} is a polar subset of Rn (whence E0 has Hausdorff dimension at most n − 2)
and CV (f; X) ⊆ CF (f; X) whenever X0 ∈ Rn−1 \ E0. In contrast, the exceptional set
E0 in our Theorem is polar in Rn−1, and hence has Hausdorff dimension at most
n − 3.
A function f which is continuous at a point with respect to the fine topology
will be called finely continuous at that point.
Corollary 1. If A ⊆ Rn and f : Rn → [−∞, +∞] is finely continuous at each
point of A, then there is a polar subset E0 of Rn−1 such that f(X) ∈ CV (f; X) whenever
X ∈ A \ (E0 × R).
Corollary 2. Let u be a superharmonic function on an open subset Ω of Rn.
Then there is a polar subset E0 of Rn−1 such that
u(X) = lim inf u(X0, t) (X ∈ Ω \ (E0 × R)).
t→x
Corollary 3. If Ω is a finely open subset of Rn, then there is a polar subset E0
of Rn−1 such that the set SXΩ0 = {t ∈ R : (X0, t) ∈ Ω} has no isolated points whenever
X0 ∈ Rn−1 \ E0.
The following example shows that the above results are sharp with regard to the
size of the exceptional set.
Example. Let E0 be any polar subset of Rn−1. Then there is a bounded
Newtonian potential u on Rn such that
u(X) < lim inf u(X0, t),
t→x
and hence
u(X) ∈ CV (u; X),
whenever X ∈ E0 ×Q. Also, there is a finely open set Ω such that SXΩ0 = {0} whenever
X0 ∈ E0.
The proof of the Theorem and Corollaries 1–3 will be given in Section 2, and
the details of the Example in Section 3.
2. Proof of the Theorem
We shall require the following.
Lemma 1. Let A0 ⊆ Rn−1 and (X0, x) ∈ Rn−1 × R, and let ε > 0. The following
are equivalent.
(a) A0 × R is thin at (X0, x) in Rn.
(b) A0 × (x, x + ε) is thin at (X0, x) in Rn.
(c) A0 × (x − ε, x) is thin at (X0, x) in Rn.
(d) A0 is thin at X0 in Rn−1
.
A proof of the equivalence of (a) and (d) above may be found in [5, Lemma 2]

evaluation of superharmonic functions using limits along lines
211
(see [6] for a general abstract result of this type). Clearly, (a) implies both (b) and (c).
Since superharmonicity is preserved by reflection, (b) is equivalent to (c). Hence it
remains to check that (b) and (c) together imply (a). Since the set
{Y ∈ A0 × {x} : 2−j 6 |Y − X| < 21−j
}
is contained in the image of
{Y ∈ A0 × (x, x + ε) : 2−j 6 |Y − X| < 21−j
}
under the canonical projection from Rn to Rn−1 × {x}, and Newtonian capacity
decreases under this projection, it follows from Wiener’s criterion (see [7, Theorem
10.21]) that A0 × {x} is thin at X = (X0, x). Hence A0 × (x − ε, x + ε) is thin at (X0, x),
being the union of three such sets, and (a) follows since thinness is a local property.
The following lemma is an easy consequence of the definition of the fine cluster
set CF (f; X).
Lemma 2. Let A ⊆ Rn and X ∈ Rn. If A is non-thin at X in Rn, then
CF (f; X) ∩ f(A) = ?,
where f(A) denotes the closure of f(A) in [−∞, +∞].
For the proof of the Theorem, we shall modify an argument of Hayman (see [8,
pp. 472–473]). Let L denote the collection of open intervals of R with endpoints
in Q, and let I denote the collection of finite unions of closed intervals of [−∞, +∞]
with endpoints in Q ∪ {−∞, +∞}. Also, we define
E0 = {X0 ∈ Rn−1 : CV (f; (X0, x)) ∩ CF (f; (X0, x)) = ? for some x ∈ R}.
Let Y 0 ∈ E0. Then there exists y in R such that CV (f; Y ) ∩ CF (f; Y ) = ?, where
Y = (Y 0, y). Since CV (f; Y ) and CF (f; Y ) are compact subsets of [−∞, +∞], we can
find I, J ∈ I and L ∈ L such that
CF (f; (Y 0, x)) ⊆ I for some value of x in L,
f(Y 0, x) ∈ J for all but at most one value of x in L,
I ∩ J = ?.
(1)
(2)
(3)
Given any I, J ∈ I and L ∈ L, we now define a subset E0(I, J, L) of Rn−1
by writing Y 0 ∈ E0(I, J, L) if Y 0 ∈ E0 and (1) and (2) both hold. The preceding
paragraph shows that
[
E0 =
E0(I, J, L),
(4)
where the union is over all choices of I, J and L which satisfy (3).
Now suppose (with the aim of obtaining a contradiction) that there is a con-
stituent member F0 = E0(I0, J0, L0) of the above union, and a point Z0 of F0, such
that F0 is non-thin at Z0. In view of (1), we can choose z in the open interval L0
such that CF (f; Z) ⊆ I0, where Z = (Z0, z). We choose a positive number ε such
that (z − ε, z + ε) ⊆ L0. Also, we define
F10 = {Y 0 ∈ F0 : f(Y 0, x) ∈ J0 whenever x ∈ (z, z + ε)}
and
F20 = {Y 0 ∈ F0 : f(Y 0, x) ∈ J0 whenever x ∈ (z − ε, z)}.
In view of (2), at least one of the sets F10 , F20 must be non-thin at Z0.

212
hiroaki aikawa and stephen j. gardiner
We consider first the case where F10 is non-thin at Z0. By Lemma 1, the set
F10 × (z, z + ε) is non-thin at Z in Rn. Hence CF (f; Z) ∩ J0 = ?, by Lemma 2. This
yields a contradiction, since CF (f; Z) ⊆ I0 and I0 ∩ J0 = ? by (3).
If, instead, F20 is non-thin at Z0, then we note from Lemma 1 that F20 × (z − ε, z)
is non-thin at Z, and argue similarly.
From the above contradiction we conclude that F0 must be thin at each of its
points and hence is polar (see [1, 1.XI.6]). Since this is true for each constituent set
of the union in (4), and since this is a countable union, the set E0 is polar.
This concludes the proof of the Theorem.
If f is finely continuous at a point X, then CF (f; X) = f(X). Thus Corollary 1
is an immediate consequence of the Theorem.
Corollary 2 is deduced as follows. We extend u to be defined on all of Rn by
assigning arbitrary values on Rn \ Ω. Since u is finely continuous at all points of Ω,
it follows from Corollary 1 that there is a polar subset E0 of Rn−1 such that
u(X) ∈ CV (u; X) whenever X ∈ Ω \ (E0 × R).
However, by the lower semicontinuity of u,
u(X) 6 ` for all ` ∈ CV (u; X) (X ∈ Ω).
Thus u(X) ∈ CV (u, X) if and only if
u(X) = lim inf u(X0, t),
t→x
and the result follows.
Corollary 3 is obtained from Corollary 1 by defining f to be the characteristic
function valued 1 on Ω and 0 elsewhere.
3. Details of the Example
The Example was suggested by a construction in [3, Example 4] (or see [4,
Example 1.13]). Let E0 be any non-empty polar subset of Rn−1. Then there is a
superharmonic function v on Rn−1 such that v = +∞ on E0. Let A be the open
subset of Rn defined by
2−n
A = {(X0, x) : v(X0) > |x| },
and let µ be the Riesz measure associated with the superharmonic function defined
on Rn by
(X0, x) −→ v(X0).
In view of the invariance of the above function under vertical translation, µ cannot
charge any point of Rn. Since
v(X0)
v(X0)
lim inf
> lim inf
> 1 > 0 = µ({(Y 0, 0)}) (Y 0 ∈ Rn−1),
X → (Y , 0) |X − (Y 0, 0)|
X → (Y , 0) |x|
2−n
2−n
0
0
X ∈ A
X ∈ A
the set A is thin at each point of Rn−1 × {0} (see [1, 1.XI.4 (c)]).
Now let w be a bounded continuous potential on Rn which determines thinness
(see [1, 1.XI.10]); that is, w has the property that for any set F ⊂ Rn,
Rw (X) < w(X) if and only if F is thin at X,
F
b

evaluation of superharmonic functions using limits along lines
213
F
b
where Rw is the regularized reduction of w on F relative to superharmonic functions
on Rn. Let u0 = Rw . Then u0 is a bounded potential on Rn, and u0 < w on
A
b
Rn−1 × {0}. However, since v = +∞ on E0, it is clear that E0 × (R \ {0}) ⊆ A, so
u0 = w on E0 × (R \ {0}). It follows that
u0(X0, x) = w(X0, x) → w(X0, 0) > u0(X0, 0) (x → 0, X0 ∈ E0).
(5)
Finally, let (qm) be an enumeration of Q, and let
X
u(X0, x) =
2
−mu0(X0, x − qm) ((X0, x) ∈ Rn).
m
It follows from the dominated convergence theorem that for any k ∈ N and X0 ∈ E0,
X
X
2
−mu0(X0, x − qm) →
2
−mu0(X0, qk − qm)
m=k
m=k
as x → qk. Hence
X
lim inf u(X0, x) = 2−k lim inf u0(X0, x) +
2
−mu0(X0, qk − qm)
x→qk
x→0
m=k
X
> 2−ku0(X0, 0) +
2
−mu0(X0, qk − qm) = u(X0, qk),
(6)
m=k
by (5). Since (6) holds for every k and every X0 ∈ E0, the first assertion of the
Example is established. Also, if we define Ω to be the fine interior of Rn \ A, then it
is clear that SXΩ0 = {0} whenever X0 ∈ E0.
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´
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4. S. J. Gardiner, Harmonic approximation, London Math. Soc. Lecture Note Ser. 221 (Cambridge
University Press, 1995).
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¨
¨
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Department of Mathematics
Shimane University
Matsue 690-8504
Japan
Department of Mathematics
University College Dublin
Dublin 4
Ireland