Uniform measures and countably additive measures
Uniform measures and countably additive measures
Jan Pachl
Toronto, Ontario, Canada
April 6, 2007
Abstract
Uniform measures are defined as the functionals on the space of bounded uniformly
continuous functions that are continuous on bounded uniformly equicontinuous sets. If
every cardinal has measure zero then every countably additive measure is a uniform mea-
sure. The functionals sequentially continuous on bounded uniformly equicontinuous sets
are exactly uniform measures on the separable modification of the underlying uniform
space.
1 Introduction
The functionals that we now call uniform measures were originally studied by Berezanskǐı [1],
Csiszár [2], Fedorova [3] and LeCam [17]. The theory was later developed in several directions
by a number of other authors; see the references in [19] and [21].
Uniform measures need not be countably additive, but they have a number of properties that
have traditionally been formulated and proved for countably additive measures, or countably
additive functionals on function spaces. The main result in this paper, in section 3, is that
countably additive measures are uniform measures on a large class of uniform spaces (on all
uniform spaces, if every cardinal has measure zero).
Section 4 deals with the functionals that behave like uniform measures on sequences of func-
tions; or, equivalently, like countably additive measures on bounded uniformly equicontinuous
sets. In the case of a topological group with its right uniformity, these functionals were defined
by Ferri and Neufang [6] and used in their study of topological centres in convolution algebras.
2 Notation
In the whole paper, linear spaces are assumed to be over the field R of reals. Uniform spaces are
assumed to be Hausdorff. Uniform spaces are described by uniformly continuous pseudometrics
([11], Chap. 15), abbreviated u.c.p.
When d is a pseudometric on a set X , define
Lip(d) = {f : X → R | |f(x)| ≤ 1 and |f(x)− f(x′)| ≤ d(x, x′) for all x, x′ ∈ X} .
http://arxiv.org/abs/0704.0885v1
Then Lip(d) is compact in the topology of pointwise convergence onX , as a topological subspace
of the product space RX .
When X is a uniform space, denote by Ub(X) the space of bounded uniformly continuous
functions f : X → R with the norm ‖f‖ = sup{ |f(x)| | x ∈ X}. Let Coz(X) be the set
of all cozero sets in X ; that is, sets of the form {x ∈ X | f(x) 6= 0} where f ∈ Ub(X). Let
σ(Coz(X)) be the sigma-algebra of subsets of X generated by Coz(X).
When d is a pseudometric on a set X , denote by O(d) the collection of open sets in the (not
necessarily Hausdorff) topology defined by d. Note that if d is a u.c.p. on a uniform space X
then O(d) ⊆ Coz(X).
Denote by M(X) the norm dual of Ub(X), and consider three subspaces of M(X):
1. Mu(X) is the space of those µ ∈ M(X) that are continuous on Lip(d) for every u.c.p. d
on X , where Lip(d) is considered with the topology of pointwise convergence on X . The
elements of Mu(X) are called uniform measures on X .
2. Mσ(X) is the space of µ ∈ M(X) for which there is a bounded (signed) countably additive
measure m on the sigma-algebra σ(Coz(X)) such that
µ(f) =
fdm for f ∈ Ub(X) .
3. Muσ(X) is the space of those µ ∈ M(X) that are sequentially continuous on Lip(d) for
each u.c.p. d. That is, limn µ(fn) = 0 whenever d is a u.c.p. on X , fn ∈ Lip(d) for
n = 1, 2, . . ., and limn fn(x) = 0 for each x ∈ X .
When X is a topological group G with its right uniformity, Muσ(X) is the space Leb
in the notation of [6].
Clearly Mu(X) ⊆ Muσ(X) for every uniform space X . By Lebesgue’s dominated conver-
gence theorem ([7], 123C), Mσ(X) ⊆ Muσ(X) for every X .
For any uniform space X , let cX be the set X with the weak uniformity induced by all
uniformly continuous functions fromX to R ([13], p. 129). Let eX be the cardinal reflection Xℵ1
([13], p. 52 and 129), also known as the separable modification of X . Thus eX is a uniform space
on the same set as X , and a pseudometric on X is a u.c.p. on eX if and only if it is a separable
u.c.p. on X . Note that Ub(X) = Ub(eX) = Ub(cX) and M(X) = M(eX) = M(cX).
Let ℵ be a cardinal number, and let A be a set of cardinality ℵ. As in [12], say that ℵ
has measure zero if m(A) = 0 for every non-negative countably additive measure m defined
on the sigma-algebra of all subsets of A and such that m({a}) = 0 for all a ∈ A. A related
notion, not used in this paper, is that of a nonmeasurable cardinal as defined by Isbell [13],
using two-valued measures m in the preceding definition.
It is not known whether every cardinal has measure zero. The statement that every cardinal
has measure zero is consistent with the usual axioms of set theory. A detailed discussion of this
and related properties of cardinal numbers can be found in [9] and [14].
Let d be a pseudometric on a set X . A collection W of nonempty subsets of X is uniformly
d-discrete if there exists ε > 0 such that d(x, x′) ≥ ε whenever x∈V, x′∈V ′, V, V ′∈W , V 6= V ′.
A set Y ⊆ X is uniformly d-discrete if the collection of singletons {{y} | y ∈ Y } is uniformly
d-discrete.
Let X be a uniform space. A set Y ⊆ X is uniformly discrete if there exists a u.c.p. d on X
such that Y is uniformly d-discrete. Say that X is a (uniform) D-space [18] if the cardinality
of every uniformly discrete subset of X has measure zero.
This generalizes the notion of a topological D-space as defined by Granirer [12] and further
discussed by Kirk [16] in the context of topological measure theory. A topological space T is a
D-space in the sense of [12] if and only if T with its fine uniformity ([13], I.20) is a uniform D-
space. If X is a uniform space and Y ⊆ X is uniformly discrete in X then Y is also uniformly
discrete in X with its fine uniformity. Therefore, if X is a topological D-space in the sense
of [12] then it is also a uniform D-space.
Since the countable infinite cardinal ℵ0 has measure zero, every uniform space X such that
X = eX is a D-space. Thus every uniform subspace of a product of separable metric spaces is
a D-space. Moreover, the statement that every uniform space is a D-space is consistent with
the usual axioms of set theory.
3 Measures on uniform D-spaces
The uniform spaces X for which Mu(X) ⊆ Mσ(X) were investigated by several authors [1] [3]
[4] [10] [17]. The opposite inclusion Mσ(X) ⊆ Mu(X) has not attracted as much attention.
Theorem 2 in this section characterizes the uniform spaces X for which Mσ(X) ⊆ Mu(X).
Lemma 1 Let d be a pseudometric on a set X, and ε > 0. Then there exist sets Wn of
nonempty subsets of X, n = 1, 2, . . ., such that
n=1 Wn is a cover of X;
2. for each n, Wn ⊆ O(d);
3. for each n, the d-diameter of each V ∈ Wn is at most ε;
4. each Wn is uniformly d-discrete.
The lemma is essentially the theorem of A.H. Stone about σ-discrete covers in metric spaces.
For the proof, see the proof of 4.21 in [15].
The next theorem is the main result of this paper. It generalizes a known result about
separable measures on completely regular topological spaces — Proposition 3.4 in [16].
Theorem 2 For any uniform space X, the following statements are equivalent:
(i) X is a uniform D-space.
(ii) Mσ(X) ⊆ Mu(X).
In view of Theorem 2 and the remarks in section 2, the statement that Mσ(X) ⊆ Mu(X)
for every uniform space X is consistent with the usual axioms of set theory.
Proof. This proof is adapted from the author’s unpublished manuscript [18].
To prove that (i) implies (ii), let X be a D-space. To show that Mσ(X) ⊆ Mu(X), it is
enough to show that µ ∈ Mu(X) for every non-negative µ ∈ Mσ(X), in view of the Jordan
decomposition of countably additive measures ([8], 231F). Take any µ ∈ Mσ(X), µ ≥ 0 and
any ε > 0. Let m be the non-negative countably additive measure on σ(Coz(X)) such that
µ(f) =
fdm for f ∈ Ub(X).
Let d be a u.c.p. on X , and {fα}α a net of functions fα ∈ Lip(d) such that limα fα(x) = 0
for every x ∈ X . Our goal is to prove that limα µ(fα) = 0.
For the given X , d and ε, let Wn be as in Lemma 1. If V ∈ Wn for some n then choose a
point xV ∈ V . Let Tn = {xV | V ∈ Wn} for n = 1, 2, . . ..
Fix n for a moment. For each subset W ′ ⊆ Wn we have
W ′ ∈ O(d) ⊆ Coz(X). Thus for
each S ⊆ Tn we may define m̃(S) = m(
{V ∈ Wn | xV ∈ S} ), and m̃ is a countably additive
measure defined on all subsets of Tn. Since the set Tn is uniformly discrete and X is a D-space,
it follows that the cardinality of Tn is of measure zero, and there exists a countable set Sn ⊆ Tn
such that
{V ∈ Wn | xV ∈ Tn \ Sn} ) = m̃(Tn \ Sn) = 0.
Denote P =
n=1 Sn and Y = {x ∈ X | d(x, P ) ≤ ε}. If V ∈ Wn for some n and xV ∈ P
then V ⊆ Y , by property 3 in Lemma 1. Therefore
X \ Y ⊆
{V ∈ Wn | xV ∈ Tn \ Sn}
and m(X \ Y ) = 0.
Define gα(x) = supβ≥α |fβ(x)| for x ∈ X . Then gα ∈ Lip(d), gα ≥ gβ for α ≤ β, and
limα gα(x) = 0 for every x ∈ X .
Since the set P is countable, there is an increasing sequence of indices α(n), n = 1, 2, . . .,
such that limn gα(n)(x) = 0 for every x ∈ P , hence limn gα(n)(x) ≤ ε for every x ∈ Y . Thus
|µ(fα)| ≤ lim
µ(gα) ≤ lim
µ(gα(n)) = lim
gα(n)dm+
gα(n)dm
≤ εm(X)
which proves that limα µ(fα) = 0.
To prove that (ii) implies (i), assume that X is not a D-space. Thus there is a u.c.p. d on
X , a subset P ⊆ X and a non-negative countably additive measure m defined on all subsets of
P such that
• d(x, y) ≥ 1 for x, y ∈ P , x 6= y;
• m(x) = 0 for each x ∈ P ;
• m(P ) = 1.
Define µ(f) =
fdm for f ∈ Ub(X). Clearly µ ∈ Mσ(X).
For any set S ⊆ P , define the function fS ∈ Lip(d) by fS(x) = min(1, d(x, S)) for x ∈ X .
Then fS(x) = 0 for x∈S and fS(x) = 1 for x∈P \ S. Let F be the directed set of all finite
subsets of P ordered by inclusion. We have limS∈F fS(x) = fP (x) for each x ∈ X , µ(fS) = 1
for every S ∈ F , and µ(fP ) = 0. Thus µ 6∈ Mu(X). �
The inclusion Mσ(X) ⊆ Mu(cX) in the following corollary is Theorem 2.1 in [5].
Corollary 3 If X is any uniform space then Mσ(X) ⊆ Mu(eX) ⊆ Mu(cX).
Proof. As is noted above, eX is a D-space for any X . Thus Mσ(eX) ⊆ Mu(eX) by
Theorem 2. From the definitions of Mσ(X), eX and cX we get Mσ(X) = Mσ(eX) and
Mu(eX) ⊆ Mu(cX). �
Corollary 3 follows also from Theorem 4 in the next section: Mσ(X) ⊆ Muσ(X) = Mu(eX).
4 Countably uniform measures
In this section we compare the spaces Muσ(X) and Mu(X).
Theorem 4 If X is any uniform space then Muσ(X) = Mu(eX).
Proof. To prove that Muσ(X) ⊆ Mu(eX), note that if a pseudometric d is separable then
Lip(d) with the topology of pointwise convergence is metrizable, and therefore sequential con-
tinuity on Lip(d) implies continuity.
To prove that Mu(eX) ⊆ Muσ(X), take any µ ∈ Mu(eX). Let d be a u.c.p. on X ,
fn ∈ Lip(d) for n = 1, 2, . . ., and limn fn(x) = 0 for each x ∈ X . Define a pseudometric d̃ on X
d̃(x, y) = sup
|fn(x) − fn(y)| for x, y ∈ X .
Then d̃ is a separable u.c.p. on X , hence a u.c.p. on eX , and fn ∈ Lip(d̃ ) for n = 1, 2, . . ..
Therefore limn µ(fn) = 0. �
In view of Theorem 4, spaces Muσ(X) have all the properties of general Mu(X) spaces.
For example, every Mu(X) is weak∗ sequentially complete [19], and the positive part µ
every µ ∈ Mu(X) is in Mu(X) [1] [3] [17]. Therefore the same is true for Muσ(X).
By Theorem 4, if X = eX then Mu(X) = Muσ(X) (cf. [6], 2.5(iii)). To see that the
equality Mu(X) = Muσ(X) does not hold in general, first consider a uniform space X that is
not a uniform D-space. Since Mσ(X) ⊆ Muσ(X), from Theorem 2 we get Mu(X) 6= Muσ(X).
However, that furnishes an actual counterexample only if there exists a cardinal that is not of
measure zero. Next we shall see that, even without assuming the existence of such a cardinal,
there is a space X such that Mu(X) 6= Muσ(X).
Let X̂ denote the completion of a uniform space X . Pelant [20] constructed a complete
uniform space X for which eX is not complete. For such X , there exists an element x ∈ êX \X .
Every f ∈Ub(X) = Ub(eX) uniquely extends to f̂ ∈Ub(êX). Let δx ∈M(X) be the Dirac
measure at x; that is, δx(f) = f̂(x) for f ∈Ub(X). Then δx∈Mu(eX), therefore δx∈Muσ(X)
by Theorem 4. On the other hand, δx 6∈ Mu(X), since δx is a multiplicative functional on
Ub(X) and x 6∈ X̂ ([19], section 6). Thus Mu(X) 6= Muσ(X).
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	Introduction
	Notation
	Measures on uniform D-spaces
	Countably uniform measures
