@
Applied General Topology

c© Universidad Politécnica de Valencia

Volume 4, No. 2, 2003

pp. 243–253

Quasi-pseudometric properties of the
Nikodym-Saks space

Jesús Ferrer
∗

Dedicated to Professor S. Naimpally on the occasion of his 70th birthday.

Abstract. For a non-negative finite countably additive measure
µ defined on the σ-field Σ of subsets of Ω, it is well known that a
certain quotient of Σ can be turned into a complete metric space Σ(Ω),
known as the Nikodym-Saks space, which yields such important results
in Measure Theory and Functional Analysis as Vitali-Hahn-Saks and
Nikodym’s theorems. Here we study some topological properties of
Σ(Ω) regarded as a quasi-pseudometric space.

2000 AMS Classification: 54E15, 54E55.
Keywords: quasi-pseudometric space, Nikodym-Saks space.

1. Introduction.

All throughout this paper we shall assume that the measure space (Ω, Σ,µ)
corresponds to a non-negative finite countably additive measure µ defined in
the σ-algebra Σ of subsets of Ω, 0 < µ(Ω) < ∞.

Following the terminology of [5, p.156], by Σ(Ω) we denote the quotient space
obtained after identifying the measurable sets A,B such that their symmetric
difference A 5 B has zero measure. For the sake of convenience, we shall
not use any special symbol to distinguish between the elements of Σ and the
equivalence classes in Σ(Ω).

It is shown in [5, p.156], and also in [3, p.86] and [7, p.208], that the function
µ(A 5 B) defines a metric in Σ(Ω) such that it becomes a complete metric
space. This property allows one to apply Baire category arguments to obtain
important results in convergence of measures such as the theorems of Vitali-
Hahn-Saks and Nikodym.

∗Supported in part by MCyT and FEDER Project BFM2002-01423



244 J. Ferrer

Besides, the set operations in Σ(Ω) are well defined and are continuous
respect to the metric considered, so that (Σ(Ω),5,∩) may also be regarded as
a topological ring. Generalizations of this can be found in [4].

The purpose of this paper is to notice that the metric space Σ(Ω) admits
a quasi-pseudometric structure which determines in the standard way both
the topology and the order given by set-inclusion. We shall as well revisit
some topological properties of Σ(Ω), such as completeness, compactness and
connectedness from a quasi-pseudometric perspective.

2. Nikodym-Saks’ complete quasi-pseudometric ordered space.

Given A,B ∈ Σ(Ω), we define

q(A,B) := µ(B \A).

It is immediate to verify that q is a quasi-pseudometric in Σ(Ω). By q−1 and
q∗ we denote the conjugate quasi-pseudometric and the metric associated to q,
respectively, that is

q−1(A,B) = q(B,A) = µ(A\B),

q∗(A,B) = q(A,B) + q−1(A,B) = µ(B \A) + µ(A\B) = µ(A5B).
It is also quite simple to see that the set operations ∪ and ∩ are continuous
in the quasi-pseudometric space (Σ(Ω),q). Also, the mapping A −→ Ω \ A
is a quasi-uniform isomorphism from (Σ(Ω),q) onto (Σ(Ω),q−1). In the same
manner, one may easily see that µ : (Σ(Ω),q) −→ R is quasi-uniformly upper
semicontinuous, i.e., given ε > 0, there is δ > 0 such that, whenever q(A,B) <
δ, we have µ(B) −µ(A) < ε.

Noticing that set-inclusion ⊆ is an ordering compatible with the equivalence
relation defined in Σ, we may regard (Σ(Ω),⊆) as an ordered space. Again,
following the terminology of [6], we have the following result.

Proposition 2.1. (Σ(Ω),q∗,⊆) is a metric ordered space determined by the
quasi-pseudometric q.

Proof. It all reduces to see that the graph of the order relation ⊆ coincides
with ∩ε>0V −1ε , where

V −1ε = {(A,B) ∈ Σ(Ω) × Σ(Ω) : q
−1(A,B) < ε}.

This is simple, since (A,B) ∈ ∩V −1ε>0 if and only if q
−1(A,B) = 0, which is

equivalent to q(B,A) = µ(A\B) = 0. That is, A\B = ∅ and so A ⊆ B. �

Notice that the order defined by set-inclusion coincides with the so called
”specialization order” defined by the quasi-pseudometric q, i.e.,

A ⊆ B ⇔ q(B,A) = 0 ⇔ B ∈{A},

that is, ”it takes no effort to move from B to A, so A must be lower”.
We introduce a couple of definitions by means of which we shall show that

Nikodym-Saks’ space is complete from a quasi-pseudometric perspective.



Nikodym-Saks space 245

Definition 2.2. In a quasi-pseudometric space (X,q), we say that a subset A
is quasi-bounded provided there is an element x0 ∈ A such that the set of
reals {q(x,x0) : x ∈ A} is bounded.

It is plain that the associated pseudometric space (X,q∗) is bounded if
and only if the quasi-pseudometric spaces (X,q) and (X,q−1) are both quasi-
bounded.

Definition 2.3. By a quasi-pseudometric ordered space we mean a triple
(X,q,≤) such that the quasi-pseudometric q determines the topological or-
dered space (X,q∗,≤) in the sense given in [6]. Thus, we say that the quasi-
pseudometric ordered space (X,q,≤) is orderly quasi-complete whenever
every quasi-bounded sequence (xn)∞n=1 satisfies the following two conditions:

1) (xn)∞n=1 admits a supremum (least upper bound) and an infimum (great-
est lower bound) in (X,≤).

2) For each n, if yn := inf{xj : j ≥ n}, then

q(yn,xn) ≤
∞∑
j=n

q(xj+1,xj).

Following the terminology introduced in [10], if (X,q) is a quasi-pseudometric
space, a sequence (xn)∞n=1 in X is said to be right-k-Cauchy whenever, given
ε > 0, there is k ∈ N such that, for n ≥ m ≥ k, we have q(xn,xm) < ε. We
say that (X,q) is right-k-sequentially complete provided every right-k-Cauchy
sequence converges.

Proposition 2.4. If (X,q,≤) is orderly quasi-complete, then (X,q) is right-
k-sequentially complete.

Proof. Let (xn)∞n=1 be a right-k-Cauchy sequence in (X,q). We define induc-
tively an increasing sequence (kj)∞j=1 of positive integers such that, for each j,
if n ≥ m ≥ kj, then q(xn,xm) < 2−j. Now, since (xkj )∞j=1 is quasi-bounded,
if, for each j, yj := inf{xki : i ≥ j}, and y := sup{yj : j ≥ 1} = limjxkj ,
then, for each j, using condition 2 of the former definition,

q(y,xkj ) ≤ q(y,yj) + q(yj,xkj ) = q(yj,xkj )

≤
∞∑
i=j

q(xki+1,xki) <
∞∑
i=j

2−i = 2−j+1.

Finally, for ε > 0, let j0 be such that 2−j0+1 < ε/2. Then, for n ≥ kj0 , we take
j1 ≥ j0 with kj1 ≥ n, and so

q(y,xn) ≤ q(y,xkj1 ) + q(xkj1 ,xn) ≤ 2
−j1+1 + 2−j0 < ε.

�

Corollary 2.5. Nikodym-Saks’ quasi-pseudometric space (Σ(Ω),q) is right-k-
sequentially complete.



246 J. Ferrer

Proof. After Proposition 2.2, it all reduces to see that (Σ(Ω),q,⊆) is orderly
quasi-complete. For any sequence (An)∞n=1 in Σ(Ω), it is clear that ∪∞n=1An
and ∩∞n=1An are in Σ(Ω) and they correspond to supn An, infn An, respectively.
Now, for each n, let Bn := ∩∞j=nAj, then

q(B,An) = µ(An\B) = µ(∪∞j=n(An\Aj)) ≤ µ(∪
∞
j=n(Aj\Aj+1))

=
∞∑
j=n

µ(Aj\Aj+1) =
∞∑
j=n

q(Aj+1,Aj).

�

Again following [6], a quasi-pseudometric space (X,q) is bicomplete when
its associated pseudometric space (X,q∗) is complete. The completeness of
Nikodym-Saks’ space can now be reobtained by means of quasi-pseudometrics.

Corollary 2.6. Nikodym-Saks’ quasi-pseudometric space (Σ(Ω),q) is bicom-
plete.

Proof. Let (An)∞n=1 be a Cauchy sequence in (Σ(Ω),q
∗). For each j ∈ N, there

is kj ∈ N such that, if n,m ≥ kj, then q∗(An,Am) < 2−j. So, if n,m ≥ kj, we
have

q(An,Am) < 2
−j, q−1(An,Am) < 2

−j,

thus obtaining, after what we did previously, that, if B := limjAkj and C :=
limjAkj , then (An)

∞
n=1 q-converges to B. Now, since (An)

∞
n=1 is q

−1-right-
k-Cauchy, it follows that (Ω\An)∞n=1 is q-right-k-Cauchy and, given that, for
each j,

q(Ω\An, Ω\Am) = q−1(An,Am) < 2−j, n,m ≥ kj,
we have that (Ω\An)∞n=1 q-converges to limj(Ω\Akj ) = Ω\C. Hence (An)∞n=1
q−1-converges to C. Hence, since B ⊆ C, and taking limits in
µ(C \B) = q(B,C) ≤ q(B,An) + q(An,C) = q(B,An) + q−1(C,An),

it follows that µ(B \ C) = µ(C \ B) = 0. That is, B = C in Σ(Ω), and so
(An)∞n=1 converges in (Σ(Ω),q

∗). �

Again after [6, p.84], we recall that a quasi-uniformity U in a space X is
convex with respect to the order ≤ whenever, given U ∈ U, there is V ∈ U
such that V ⊆ U, and, for each x ∈ X, V (x) = {y ∈ X : (x,y) ∈ V} is convex
respect to ≤, i.e., a ≤ c ≤ b, a,b ∈ V (x), imply c ∈ V (x).

After Proposition 4.19 of [6, p.84], in light of our previous result, we know
that (Σ(Ω),q∗) is a convex metric space in the sense before defined. Never-
theless, we cannot conclude, as it happens in many metric convex spaces, that
every ball V ∗ε (A) = {X ∈ Σ(Ω) : q∗(A,X) < ε} has to be a convex set, as our
next result proves.

Proposition 2.7. Let Ω = [0, 1] and let λ represent the Lebesgue measure.
Then, for each 0 < ε < 1/2, there is a measurable set A such that the ball
V ∗ε (A) is not convex with respect to set-inclusion.



Nikodym-Saks space 247

Proof. Let ε/2 < a < 1 − 3ε
2

. We consider the following measurable sets

A = [a,a +
11ε
8

], X = [a,a +
ε

8
] ∪ [a +

�

4
,a +

ε

2
] ∪ [a + ε,a +

11ε
8

],

Y = [a−
ε

2
,a +

3ε
2

], Z = [a−
ε

2
,a +

ε

2
] ∪ [a + ε,a +

11ε
8

]

q∗(A,X) = λ(X \A) + λ(A\X) = λ(A\X) =
ε

8
+
ε

2
=

5ε
8
< ε,

q∗(A,Y ) = λ(Y \A) + λ(A\Y ) = λ(Y \A) =
ε

2
+
ε

8
=

5ε
8
< ε,

q∗(A,Z) = λ(Z \A) + λ(A\Z) =
ε

2
+
ε

2
= ε.

Hence, we have X ⊆ Z ⊆ Y , X,Y ∈ V ∗ε (A), but Z /∈ V ∗ε (A). �

The following result will be needed afterwards. Let us recall first that a
subset F of an ordered set (X,≤) is said to be inductive whenever every totally
ordered subset of F has an upper bound in X.

Proposition 2.8. Every non-empty closed subset of (Σ(Ω),q∗) is inductive
with respect to set-inclusion.

Proof. Let F be a non-empty closed set in (Σ(Ω),q∗). Let (Fi)i∈I be a totally
ordered subset of F. Since µ is finite, there exists ρ = supi∈Iµ(Fi). We now
start an inductive process by taking i1 ∈ I such that µ(Fi1 ) > ρ−

1
2
. Assuming

already found Fi1 ⊆ Fi2 ⊆ ... ⊆ Fin in F such that, for j = 1, 2, ...,n, µ(Fij ) >
ρ− 1

2j
, we proceed to find Fin+1 with the same properties. If µ(Fin) > ρ−

1
2n+1

,
then we set Fin+1 := Fin; if µ(Fin) ≤ ρ −

1
2n+1

, we find Fin+1 in F such that
µ(Fin+1 ) > ρ −

1
2n+1

, then Fin ⊆ Fin+1 , otherwise, since we are dealing with
totally ordered elements, we would have Fin+1 ⊆ Fin, and, µ(Fin+1 ) ≤ µ(Fin) ≤
ρ − 1

2n+1
, which is a contradiction. We have thus constructed an increasing

sequence (Fin)
∞
n=1 in F, with µ(Fin) > ρ−

1
2n

, n ∈ N. We set F := ∪∞n=1Fin.
Then, F ∈ Σ(Ω), and, for each n,

q∗(F,Fin) = µ(F\Fin) = µ(∪
∞
j=1(Fij\Fin)) ≤ µ(∪

n
j=1(Fij\Fin))+µ(∪

∞
j=n+1(Fij\Fin))

= µ(∪∞j=n+1(Fij \Fin)) ≤ µ(∪
∞
j=n+1(Fij \Fij−1 )) =

∞∑
j=n+1

µ(Fij \Fij−1 )

=
∞∑

j=n+1

(µ(Fij ) −µ(Fij−1 )) ≤
∞∑

j=n+1

(ρ− (ρ−
1

2j−1
)) =

∞∑
j=n+1

1
2j−1

=
1

2n−1
.

Hence, (Fin)
∞
n=1 converges to F in (Σ(Ω),q

∗). Since F is closed, it follows that
F ∈ F. We show finally that F is an upper bound for the chain (Fi)i∈I. Give
i ∈ I, we consider two possibilities:

If there is n0 ∈ N such that Fi ⊆ Fin0 , then it is clear that Fi ⊆ Fin0 ⊆ F.



248 J. Ferrer

On the contrary, if Fi is not contained in Fin, n ∈ N, then again the total
ordering guarantees that Fin ⊆ Fi, n ∈ N, and so F ⊆ Fi. Then, since
ρ ≥ µ(Fi) ≥ µ(F) = limnµ(Fin) ≥ ρ, we have

µ(Fi 5F) = µ(Fi \F) = µ(Fi) −µ(F) = ρ−ρ = 0.
That is, Fi = F. �

3. Connectedness and compactness in Nikodym-Saks’ space.

In this section we study the topological properties of connectedness and
compactness in the space (Σ(Ω),q∗), observing that such properties are directly
related with the degree of atomicity of the measure µ. We shall introduce again
some notation. For A ∈ Σ(Ω), by Σ(A) we denote the collection of elements of
Σ(Ω) contained in A, we shall also refer to this collection as a lower interval;
similarly Σ+(A) will stand for all the measurable supersets of A and we will
refer to this as an upper interval. Let us recall that E ∈ Σ(Ω) is called an atom
when it has positive measure and the lower interval Σ(E) is reduced to {∅,E}.
When a measure µ does not have any atom then it is said to be non-atomic. It is
convenient to recall that a measure can only admit at most a countable amount
of disjoint atoms, when Ω admits a countable partition formed by atoms, then
µ is said to be purely atomic.

Before characterizing the connectedness of (Σ(Ω),q∗) in terms of atoms,
let us notice that the quasi-pseudometric spaces (Σ(Ω),q) and (Σ(Ω),q−1) are
always connected: Let us suppose that (Σ(Ω),q) admits two disjoint open sets
A, B covering Σ(Ω). Then, one of them, say A, contains Ω, so there is δ > 0
such that Vδ(Ω) ⊆ A. But, Vδ(Ω) = Σ(Ω). Hence, B = ∅.

Proposition 3.1. The following are equivalent.
(i) No upper interval Σ+(A), A 6= ∅, is q−1-open.
(ii) µ is non-atomic.
(iii) (Σ(Ω),q∗) is connected.
(iv) For each A ∈ Σ(Ω) and α ∈ [0,µ(A)], there is B ∈ Σ(A)

such that
µ(B) = α.

Proof. (i) ⇒ (ii). Assume that E is an atom. We show that the upper inteval
Σ+(E) is q−1-open. Given A ∈ Σ+(E), let δ = µ(E) > 0. If X ∈ V −1δ (A),
then q−1(A,X) < δ implies µ(A\X) < δ. But,
µ(E \X) = µ(E \X \A) + µ(E ∩A\X) ≤ µ(E \A) + µ(A\X) < δ.

Hence, µ(E\X) = 0, otherwise, since E is an atom, we would have µ(E\X) =
µ(E) = δ. Thus, X ∈ Σ+(E).
(ii) ⇒ (iii). Let us assume that (Σ(Ω),q∗) is disconnected. So, let A, B be
two non-empty disjoint closed sets covering Σ(Ω). We may suppose without
restriction that Ω ∈ A. Applying Proposition 4 to the closed set B and after
Zorn’s lemma, there is a maximal element M in (B,⊆). Since B is also open,
there is δ > 0 such that V ∗δ (M) ⊆ B. Now, since Ω \M has non-zero measure



Nikodym-Saks space 249

(otherwise, Ω = M would also be in B), the fact that µ is atom-free guarantees,
see [1, p.24], that

inf{µ(E) : ∅ 6= E ∈ Σ(Ω \M)} = 0.
Thus, we may find E ∈ Σ(Ω \M) such that 0 < µ(E) < δ. Let A := M ∪E.
Then, q∗(M,A) = µ(E) < δ, and so A ∈ B with M ⊆ A, M 6= A, contradicting
the maximality of M.
(iii) ⇒ (iv). The continuous mapping X −→ X ∩A maps Σ(Ω) onto Σ(A).
Thus, if Σ(Ω) is connected, so is Σ(A). Now, since µ is continuous, it follows
that µ(Σ(A)) = [0,µ(A)].
(iv) ⇒ (i). Let us assume there is E ∈ Σ(Ω), E 6= ∅, such that the upper
interval Σ+(E) is q−1-open. Clearly, since µ(E) > 0, there is 0 < δ < µ(E)
such that V −1δ (E) ⊆ Σ+(E). By hypothesis, there is A ∈ Σ(E) for which
µ(A) = δ/2. Let X := E \A. Then, µ(E \X) = µ(A) = δ/2 < δ implies that
X ∈ V −1δ (E) and consequently E ⊆ X, a contradiction, since µ(E\X) 6= 0. �

By recalling that a finite countably additive measure λ in (Ω, Σ) is µ-
continuous whenever limµ(X)→0 λ(X) = 0, we have that in this case the identity
mapping is well defined and continuous from (Σ(Ω),q∗µ) into (Σ(Ω),q

∗
λ). There-

fore the following result is straightforward.

Corollary 3.2. If λ is a finite countably additive µ-continuous measure and µ
is non-atomic, then so is λ.

We study in the following the compactness of Nikodym-Saks’ space. As we
did in the connectedness part, it is curious to notice that the quasi-pseudometric
spaces (Σ(Ω),q) and (Σ(Ω),q−1) are always compact; just recall that, for in-
stance, any q-open cover of Σ(Ω) must have a member containing Ω, conse-
quently, this open set has to be Σ(Ω).

We show next that, in some sense, the compactness of (Σ(Ω),q∗) does not get
along with the connectedness. As a matter of fact, we show that compactness
is equivalent to µ being purely atomic. We need, in order to do so, to introduce
some more notation.

Given a sequence (An)∞n=1 in Σ(Ω), for each n, if (i1, i2, ..., in) ∈{0, 1}n, we
define the following sets

A(i1,i2,...,in) := (∩{Aj : 1 ≤ j ≤ n,ij = 1}) ∩ (∩{Ω \Aj : 1 ≤ j ≤ n,ij = 0}).
It is plain that, for each k = 1, 2, ...,n, the collections

{A(i1,i2,...in) : ik = 1}, {A(i1,i2,...in) : ik = 0},
are partitions of Ak and Ω \Ak, respectively.
Lemma 3.3. If µ is non-atomic, then there is a sequence (An)∞n=1 in Σ(Ω)
such that, for each n,

µ(A(1,1,...,1)) = µ(A(0,0,...,0)) = (
1
4

+
1

2n+1
)µ(Ω),

µ(A(i1,i2,...,in)) =
1

2n+1
µ(Ω), (i1, i2, ...in) /∈{(0, 0, ..., 0), (1, 1, ..., 1)}.



250 J. Ferrer

Proof. We give an inductive sketch of proof. The first set A1 appears courtesy of
the non-atomicity of the measure µ and Proposition 3.1. Once already obtained
A1,A2, ...,An, again Proposition 3.1 guarantees the existence of measurable sets

B(i1,i2,...,in) ⊆ A(i1,i2,...,in), (i1, i2, ...in) ∈{0, 1}
n,

such that

µ(B(i1,i2,...,in)) =
1

2n+2
µ(Ω), (i1, i2, ..., in) 6= (1, 1, ..., 1),

µ(B(1,1,...,1)) = (
1
4

+
1

2n+2
)µ(Ω).

Defining An+1 := ∪{B(i1,i2,...,in) : (i1, i2, ...in) ∈ {0, 1}
n}, the induction pro-

cess is done. �

Proposition 3.4. Nikodym-Saks’ space (Σ(Ω),q∗) is compact if and only if µ
is purely atomic.

Proof. Assuming (Σ(Ω),q∗) is compact, it all reduces to show that every set
A ∈ Σ(Ω), with µ(A) > 0, contains atoms. If this were not so, then the
restricted measure µ|A would be non-atomic in the restricted Nikodym-Saks’
space Σ(A). Applying the former lemma, we would find a sequence (An)∞n=1 ⊆
Σ(A) satisfying the conditions there stated. It may be easily seen that the
distance q∗(An,Am) = 14µ(A), n 6= m. Hence, such a sequence cannot admit
any Cauchy subsequence, which contradicts the fact that Σ(Ω) is compact.

Conversely, let (En)∞n=1 be an atomic partition of Σ(Ω). We show that
(Σ(Ω),q∗) is homeomorphic to Cantor’s space 2N. Consider the one-to-one and
onto mapping T : 2N −→ Σ(Ω) such that

T(x) := ∪{En : xn = 1}, x ∈ 2N.
Given x ∈ 2N, ε > 0, let A = T(x). Since

µ(A) =
∑
{µ(En) : xn = 1}, µ(Ω \A) =

∑
{µ(En) : xn = 0},

there are two finite subsets F,F ′ of N such that x(F) = 1, x(F ′) = 0, and∑
{µ(En) : n ∈ F} > µ(A) −

ε

2
,
∑
{µ(En) : n ∈ F ′} > µ(Ω \A) −

ε

2
.

It follows that the set V = {y ∈ 2N : y(F) = 1,y(F ′) = 0} is a neighborhood of
x such that µ(A5T(y)) < ε, y ∈ V . Finally, to see that T−1 is also continuous,
it suffices to notice that, for each p ∈ N, the intervals Σ+(Ep) and Σ(Ω \Ep)
are closed subsets of Σ(Ω). �

4. Nikodym-Saks’ space as a topological group.

We know that Σ(Ω) is an abelian nilpotent group and that the symmetric dif-
ference 5 is continuous, thus Σ(Ω) may be regarded also as a topological group.
Besides, for each A ∈ Σ(Ω), Σ(Ω) can be expressed as the topological direct
sum of the closed subgroups Σ(A) and Σ(Ω\A), i.e., Σ(Ω) = Σ(A)⊕Σ(Ω\A).
Our aim in this section is to show that Nikodym-Saks’ space can be decom-
posed, in a unique way, as the topological direct sum of a connected subgroup



Nikodym-Saks space 251

(the component of the zero element ∅) plus a compact totally disconnected
subgroup.

Proposition 4.1. There exists a unique measurable set M such that Σ(Ω) =
Σ(M) ⊕ Σ(Ω \M), with Σ(M) connected and Σ(Ω \M) compact.

Proof. Let A denote the countable collection, possibly empty, of atoms of Σ(Ω).
Since the set M := Ω \ (∪{E : E ∈ A}) contains no atoms, it follows after
Proposition 3.1 that the topological group Σ(M) is connected. Now, assuming
M 6= Ω (otherwise, Σ(Ω \ M) = {∅}, clearly compact), we have that the
restricted measure µ|Ω\M is purely atomic, Proposition 3.4 then shows that
Σ(Ω \M) is compact. Notice also that in this case, since Σ(Ω \M) is a copy
of a Cantor space, it is totally disconnected. �

Proposition 4.2. The connected subgroup Σ(M) before obtained coincides with
the connected component of ∅.

Proof. Let us denote by C the connected component of Σ(Ω) containing ∅. It
is well known that C is a closed topological subgroup of Σ(Ω). Hence, since
∅ ∈ Σ(M) and Σ(M) is connected, it follows that Σ(M) ⊆ C. We show the
reverse inclusion.

Let A ∈ C. Assume A \ M 6= ∅. Hence, after what we have seen before,
Σ(A \ M) is a totally disconnected subgroup. But, the mapping τ : C −→
Σ(A \ M) such that τ(X) = X ∩ (A \ M) is continuous, and so τ(C) is
connected in Σ(A \ M). Thus, τ(C) must be a singleton, but this is not so
since ∅,A\M ∈ τ(C). �

We finish by studying the properties of compactness and connectedness re-
lated with their local properties.

Corollary 4.3. Σ(Ω) is compact if and only if it is locally compact.

Proof. Since Σ(Ω) is Hausdorff, we need only show the sufficiency part. So, if
Σ(Ω) is locally compact, we may find δ > 0 such that the closed ball V

∗
δ(∅) =

{X ∈ Σ(Ω) : q∗(∅,X) ≤ δ} is compact. Let, as before, M ∈ Σ(Ω) be such
that Σ(M) is the connected component of ∅. It all reduces to see that M = ∅.
If not, since Σ(M) is connected, we can find, after Proposition 3.1, A ∈ Σ(M)
such that 0 < µ(A) ≤ δ, and so Σ(A) ⊆ V

∗
δ(∅). Hence Σ(A) is compact

and, after Proposition 3.4, A must contain atoms. This would imply that, M
containing A would also contain atoms, thus contradicting the definition of
M. �

Corollary 4.4. If Σ(Ω) is connected, then it is locally connected.

Proof. Let us assume that Σ(Ω) is connected. We first show that every ball
centered at ∅ is connected. This is a simple consequence of the facts

V ∗δ (∅) = ∪{Σ(X) : X ∈ V
∗
δ (∅)}, ∅ ∈∩{Σ(X) : X ∈ V

∗
δ (∅)},



252 J. Ferrer

and that, for any X, Σ(X) is connected. Now, being in a topological group,
since every ball centered at A is a translation of the form

V ∗δ (A) = A 5V
∗
δ (∅),

it follows that V ∗δ (A) is also connected. �

As it was pointed out by Professor Paolo De Lucia one can easily find exam-
ples of locally connected disconnected (but not totally disconnected) Nikodym-
Saks spaces.

Example 4.5. Consider the measure space (Ω, Σ,µ) given by (λ represents the
Lebesgue measure in [0, 1] and we denote by L the class of Lebesgue-measurable
subsets of [0, 1]):

Ω = [0, 1] ∪{2}, Σ = L ∪{A∪{2} : A ∈ L},
µ|L = λ, µ(A∪{2}) = λ(A) + 1, A ∈ L.

Clearly, since {2} is a µ-atom, the corresponding Nikodym-Saks’ space Σ(Ω)
is not connected. Notice that it is neither totally disconnected, since Σ([0, 1])
is the connected component of ∅. And, after recalling that Σ(Ω) = Σ([0, 1]) ⊕
Σ({2}), it can be easily shown that Σ(Ω) is locally connected.

Acknowledgements. The author wishes to thank the referee for the inter-
esting suggestions and comments on the paper. Also, it means a pleasure for
this author to express his gratitude to Virtu Bertó for her patience and help in
the preparation of the talk

References

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(1979).

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(1977).

[3] Diestel, J.: Sequences and Series in Banach Spaces, Springer-Verlag, New York (1984).
[4] Drewnowski, L.: Topological Rings, Continuous Set Functions, Integration, Bull. Acad.

Polon. Scien. 20 , 269-276 (1972). 1981.

[5] Dunford, N.; Schwartz, J.T.: Linear Operators (I), Wiley, New York (1976).
[6] Fletcher, P.; Lindgren, W.: Quasi-Uniform Spaces, Marcel Dekker, (1982).

[7] Halmos, P.R.: Measure Theory, Springer-Verlag, New York (1974).
[8] Munroe, M.E.: Measure and Integration, Addison-Wesley, London (1968).

[9] Oxtoby, J.C.: Measure and Category, Springer-Verlag, New York (1980).

[10] Reilly, I.L.; Subrahmanyam, P.V.; Vamanamurthy, M.K.: Cauchy sequences in quasi-
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[11] Williamson, J.H.: Integración Lebesgue, Tecnos , Madrid (1973).

Received November 2001

Revised November 2002



Nikodym-Saks space 253

Jesús Ferrer

Departamento de Análisis Matemático, Universidad de Valencia, Dr. Moliner
50, 46100 Burjasot (Valencia), Spain

E-mail address : Jesus.Ferrer@uv.es