FAMILIES OF VECTOR MEASURES OF UNIFORMLY
BOUNDED VARIATION
OLAV NYGAARD AND MÄRT PÕLDVERE
Abstract. We characterize families of vector measures of uniformly bounded variation and semivariation in terms of additivity properties. A classical theorem of Rickart and a simplified proof of Nikodým’s boundedness
theorem follow.
Throughout this note, X will be a Banach space with dual space X ∗ and unit
ball BX , Ω a non-empty set, F ⊂ P(Ω) an algebra, and {Fτ } = {Fτ : τ ∈ T } a
family of X-valued vector measures (i.e., finitely additive set functions) defined
on F. Recall [1, page 2] that the variation of a vector measure F : F → X is
the (finitely additive) set function |F | : F → X whose value on a set A ∈ F is
defined by


n

X
kF (Ai )k : n ∈ N, A1 , . . . , An ∈ F are pairwise disjoint .
|F | (A) = sup


j=1
If |F | (Ω) < ∞, then F is said to be of bounded variation. We shall say that
the family {Fτ } is of uniformly bounded variation, if supτ ∈T |Fτ | (Ω) < ∞.
Theorem 1. The following assertions are equivalent.
(i) The family {Fτ } is of uniformly bounded variation.
(ii) Whenever Aj ∈ F, j ∈ N, are pairwise disjoint sets, then
sup
∞
X
τ ∈T j=1
kFτ (Aj )k < ∞.
Proof. Only (ii)⇒(i) has to be proven. Denote Φ(A) := supτ ∈T |Fτ |(A), A ∈ F.
Let (ii) hold and suppose for contradiction that (i) fails. It suffices to find
indices τk ∈ T and pairwise disjoint sets Ek ∈ F such that |Fτk |(Ek ) ≥ k,
k ∈ N. To this end in turn, it suffices to show that, given k ∈ N and a set
A ∈ F with Φ(A) = ∞, there are disjoint measurable sets D, E ⊂ A and an
index τk ∈ T such that Φ(D) = ∞ and |Fτk |(E) ≥ k.
So, let k ∈ N and let A ∈ F satisfy Φ(A) = ∞. There are two alternatives:
1) There are an index τk ∈ T and a measurable set B ⊂ A such that
½
¾
kFτk (B)k ≥ max 2k, 2 sup kFτ (A)k = : δ.
τ ∈T
2000 Mathematics Subject Classification. Primary 46G10; Secondary 28B05.
Key words and phrases. Vector measures, uniformly bounded variation, Nikodým’s boundedness theorem.
The second named author was supported by Estonian Science Foundation Grant 5704.
1
2
O. NYGAARD AND M. PÕLDVERE
2) For all τ ∈ T and all measurable B ⊂ A, one has kFτ (B)k < δ.
1) In this case, denote C = A \ B. One has
kFτk (B)k
kFτk (B)k
=
≥ k.
2
2
Now A = B ∪ C, B ∩ C = ∅ and min{|Fτk |(B), |Fτk |(C)} ≥ k. Since Φ(A) = ∞,
at least one of the quantities Φ(B) and Φ(C) has to be ∞. If Φ(B) = ∞, put
D = B and E = C. If Φ(B) < ∞, then Φ(C) = ∞; in this case put D = C and
E = B.
2) Since Φ(A) = ∞, there are an index τk ∈PT and pairwise disjoint measurable sets B1 , . . . , Bn ⊂ A (n ∈ N) such that nj=1 kFτk (Bj )k ≥ 3δ. Clearly,
Pm
n ≥ 4. Let m be the least integer such that
kF (B )k ≥ δ. Since
Pm j=1 τk j
kFτk (Bj )k < δ for all 1 ≤ j ≤ n, we have j=1 kFτk (Bj )k < 2δ and hence
Pn
Sm
j=m+1 kFτk (Bj )k > δ. Denote B =
j=1 Bj and C = A\B. Then min{|Fτk |(B),
|Fτk |(C)} ≥ δ ≥ k. At least one of the quantities Φ(B) and Φ(C) has to be ∞.
So, if Φ(B) = ∞, put D = B and E = C. If Φ(B) < ∞, then Φ(C) = ∞; in
this case put D = C and E = B.
¤
kFτk (C)k ≥ kFτk (B)k − kFτk (A)k ≥ kFτk (B)k −
Corollary 2. A vector measure F : F → X is of bounded variation if and only
if whenever Aj ∈ F, j ∈ N, are pairwise disjoint sets, then
∞
X
kF (Aj )k < ∞.
j=1
We now turn to uniformly bounded families of vector measures. Recall [1,
page 2] that the semivariation of a vector measure F : F → X is the (finitely
subadditive) set function kF k : F → X whose value on a set A ∈ F is defined
by kF k (A) = supx∗ ∈BX ∗ |x∗ F | where |x∗ F | is the variation of the real-valued
vector measure x∗ F . By [1, page 4, Proposition 11], one has, for all A ∈ F,
sup kF (B)k ≤ kF k (A) ≤ 4 sup kF (B)k ,
A⊃B∈F
A⊃B∈F
thus F is bounded (i.e., it has bounded range) if and only if kF k (Ω) < ∞, and
the family {Fτ } is uniformly bounded if and only if supτ ∈T kFτ k (Ω) < ∞.
Theorem 3. The following assertions are equivalent.
(i) The family {Fτ } is uniformly bounded.
(ii) Whenever Aj ∈ F, j ∈ N, are pairwise disjoint sets and x∗ ∈ X ∗ , then
sup
∞
X
τ ∈T j=1
|x∗ Fτ (Aj )| < ∞.
Proof. Only (ii)⇒(i) has to be proven. Let (ii) hold. Then, whenever x∗ ∈ X ∗ ,
Theorem 1 says that the family of scalar-valued vector measures {x∗ Fτ }τ ∈T is of
∗
∗
∗
uniformly bounded variation.
S∞ Denote Cj = {x ∈ X : supτ ∈T |x Fτ | (Ω) ≤ j},
∗
j ∈ N. Notice that X = j=1 Cj with each Cj being norm (and even weak∗ )
closed. Baire’s theorem says that there is some Cm containing a ball, and since
Cm is absolutely convex, a ball centered at the origin, i.e., δBX ∗ ⊂ Cm for some
FAMILIES OF VECTOR MEASURES OF UNIFORMLY BOUNDED VARIATION
δ > 0. This means that, for all x∗ ∈ BX ∗ , one has supτ ∈T |x∗ Fτ | (Ω) ≤
supτ ∈T kFτ k (Ω) ≤ m
δ .
m
δ ,
3
thus
¤
Corollary 4. A vector measure F : F → X is bounded if and only if whenever
Aj ∈ F, j ∈ N, are pairwise disjoint sets and x∗ ∈ X ∗ , then
∞
X
|x∗ F (Aj )| < ∞.
j=1
Recall [1, page 7] that a vector measure F : F → X is said to be strongly
additive,
if whenever Aj ∈ F, j ∈ N, are pairwise disjoint sets, then the series
P∞
j=1 F (Aj ) converges in norm. The following classical result of Rickart [1,
page 9, Corollary 19] is immediate from Corollary 4.
Corollary 5. If a vector measure F : F → X is strongly additive, then it is
bounded.
The converse to Corollary 5 fails in general: By the Diestel-Faires theorem
[1, page 20, Theorem 2], X does not contain any isomorphic copies of c0 if and
only if every bounded X-valued vector measure is strongly additive. It is well
known that the classical Bessaga-PeÃlczyński test (see e.g. [1, page 22]) for “c0 freeness” — X does not contain any isomorphic copies of c0 if and only if every
weakly unconditionally Cauchy series in X is (unconditionally) convergent —
can be derived from the Diestel-Faires theorem. Notice that the Diestel-Faires
result above, in turn, follows immediately from the Bessaga-PeÃlczyński test via
Corollary 4.
Theorem 3 enables one to simplify the standard Darst proof of the Nikodým
boundedness theorem [1, page 14, Theorem 1].
Theorem 6 (Nikodým boundedness theorem). Suppose that F is a σ-algebra.
Let each Fτ be bounded and let the family {Fτ } be setwise bounded (i.e.,
supτ ∈T kFτ (A)k < ∞ for all A ∈ F). Then the family {Fτ } is uniformly
bounded.
Proof. Suppose for contradiction that the family {Fτ } is not uniformly bounded.
Then, by Theorem 3, there are x∗ ∈ BX ∗ , pairwise disjoint sets Aj ∈ F, j ∈ N,
∗
and a sequence (τk )∞
k=1 ⊂ T such that, with µk = x Fτk , one has
lim
k→∞
∞
X
|µk (Aj )| = ∞.
j=1
Step 1. We may assume that limk→∞ |µk (Ak )| = ∞.
∞
Indeed, P
we can choose increasing sequences of indices (nk )∞
k=1 and (νk )k=1
νk
such that j=νk−1 +1 |µnk (Aj )| ≥ 4k for all k ∈ N (here ν0 = 0). Given k ∈ N
and defining Ik = {νk−1 + 1, . . . , νk }, at least one of the sets
{j ∈ Ik : Re µnk (Aj ) ≥ 0} , {j ∈ Ik : Re µnk (Aj ) < 0} ,
{j ∈ Ik : Im µnk (Aj ) ≥ 0} , {j ∈ Ik : Im µnk (Aj ) < 0} ,
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O. NYGAARD AND M. PÕLDVERE
¯
¯
S
¯
¯
which we denote by Jk , must satisfy ¯µnk ( j∈Jk Aj )¯ ≥ k. It remains to relabel
S
µnk by µk and j∈Jk Aj by Ak .
Step 2. By passing to subsequences, we may clearly assume that, for all
k ∈ N, k ≥ 2, one has
|µk (Ak )| −
k−1
X
|µk (Aj )| ≥ k + 1.
j=1
The rest of the proof follows the last part of the proof presented in [1, pages
15–16]. We include it for completeness.
³S
´
∞
Step 3. We may assume that, for all k ∈ N, one has |µk |
A
< 1.
j
j=k+1
Indeed, put n1 = 1 and partition N \ {n1 } into pairwise disjoint infinite sets
Ni , i ∈ N. Since
∞
X
∞ [
³[
´
³[
´
|µn1 |
Aj ≤ |µn1 |
Aj ≤ |µn1 |(Ω) < ∞,
i=1
i=1 j∈Ni
j∈Ni
S
it follows that some Ni , which we denote by Ñ1 , satisfies |µn1 |( j∈Ñ1 Aj ) < 1.
Put n2 = min Ñ1 and repeat the preceding argument with N \ {n1 } replaced by
Ñ1 \ {n2 } and µn1 by µn2 to obtain an infinite set Ñ2 ⊂ Ñ1 \ {n2 } satisfying
S
|µn2 |( j∈Ñ2 Aj ) < 1. Put n3 = min Ñ2 . Continuing in an obvious manner, we
obtain indices n1 < n2 < . . .. It remains to relabel Anj by Aj and µnj by µj ,
j ∈ N.
S
Step 4. To arrive at a contradiction, define A = ∞
j=1 Aj and observe that,
for all k ∈ N, k ≥ 2,
¯
¯
¯ ³ [
k−1
∞
´¯¯
X
¯
|µk (A)| ≥ |µk (Ak )| −
|µk (Aj )| − ¯¯µk
Aj ¯¯
¯
¯
j=1
j=k+1
≥ k + 1 − |µk |
∞
³ [
´
Aj ≥ k + 1 − 1 = k,
j=k+1
and thus
sup kFτ (A)k ≥ sup |x∗ Fτk (A)| = sup |µk (A)| = ∞.
τ ∈T
k∈N
k∈N
¤
Acknowledgement. The authors are indebted to the referee for comments
and suggestions that improved the exposition.
References
[1] J. Diestel and J. J. Uhl, jr., Vector Measures, Math. Surveys, vol. 15, Amer. Math.
Soc., Providence, 1977.
FAMILIES OF VECTOR MEASURES OF UNIFORMLY BOUNDED VARIATION
5
Department of Mathematics, Agder University College, Servicebox 422,, 4604
Kristiansand, Norway.
E-mail address: [email protected]
URL: http://home.hia.no/~olavn/
Institute of Pure Mathematics, University of Tartu, J. Liivi 2, 50409 Tartu,
Estonia
E-mail address: [email protected]