VOL. 80 1999 NO. 1
COMPLETENESS OF L 1 SPACES OVER FINITELY ADDITIVE PROBABILITIES
BY
S. G A N G O P A D H Y A Y AND B. V. R A O (CALCUTTA)
1. Introduction. Finitely additive measures—though not as technically amenable as their countably additive counterparts—seem to lead to inter- esting, and at times peculiar, mathematical problems. For an exposition of the theory of finitely additive measures, see Dunford and Schwartz [5]
or Bhaskara Rao and Bhaskara Rao [1]. The bold initiative of Dubins and Savage [4] to develop gambling in a finitely additive setup and the beautiful existence theorem of Purves and Sudderth [18] in infinite product spaces paved the way to develop a substantial part of classical probability theory in a finitely additive setup—called the strategic setup.
The strong law of large numbers was treated in Purves and Sudderth [18] and Chen [3]; the law of iterated logarithm in Chen [2]; the central limit theorem in Ramakrishnan [22] and Karandikar [15]; Markov chains and po- tential theory in Ramakrishnan [20, 21, 23]; random walks in Karandikar [16]
and S. Gangopadhyay and Rao [7, 8]; martingales in Dubins and Savage [4], Purves and Sudderth [18]; Komlos type theorems in Halevy and Bhaskara Rao [11]. The zero-one laws of L´evy and Kolmogorov hold as shown in Purves and Sudderth [18, 19] whereas the Hewitt–Savage zero-one law needs mod- ification as shown in Purves and Sudderth [19], Gangopadhyay and Rao [9].
It is interesting to note that an appropriate finitely additive version of this zero-one law already appears in the fundamental paper of Hewitt and Savage [14], but of course, not in the strategic setup.
More recently (as pointed out to us by J. K. Ghosh), Heath and Sudderth [12, 13] and Lane and Sudderth [17] have advocated that it is beneficial to use finitely additive priors in some problems of statistical inference. In fact the prescription of [13] is simple: A Bayesian who seeks to avoid incoherent inferences might be advised to abandon improper countably additive priors and use only finitely additive priors.
1991 Mathematics Subject Classification: 28A25, 28A35, 60A10.
Key words and phrases: L 1 space; Hewitt–Yosida decomposition; Sobczyk–Hammer decomposition; strategic products.
[83]
This paper is concerned with the completeness of the space L 1 (γ) of integrable functions over a finitely additive nonnegative bounded measure γ defined on a σ-field of subsets of a set. In the mathematical literature [1, 5, 14, 26] the domain for such a γ is a field of sets. Our interest is not in finitely additive measures per se, but in the development of prob- ability. The natural domain for γ then is a σ-field and that is what we take.
By the Hewitt–Yosida theorem γ = γ 0 + γ 1 where γ 0 is countably ad- ditive and γ 1 is purely finitely additive. We show (Theorem 3) that L 1 (γ) is complete iff L 1 (γ 1 ) is complete and γ 0 , γ 1 are supported on disjoint sets.
By the Sobczyk–Hammer theorem γ 1 = γ 2 + γ 3 where γ 2 is discrete and γ 3 is strongly continuous. We show (Theorem 6) that L 1 (γ 1 ) is complete iff L 1 (γ 2 ), L 1 (γ 3 ) are complete and γ 2 , γ 3 are supported on disjoint sets. We have γ 2 = P
i a i δ i where each δ i is a 0-1 valued measure. We show (Theorem 4) that L 1 (γ 2 ) is complete iff the δ i are uniformly singular.
Next we consider finite strategic products, γ = N k
i=1 γ i . We show (The- orem 9) that if L 1 (γ i ) is complete for each i then L 1 (γ) is complete. If L 1 (γ) is complete then L 1 (γ 1 ) is complete, but for i > 1, L 1 (γ i ) need not be complete. We have partial results for infinite strategic products.
It should be remarked that the completeness of L 1 spaces is not only interesting in its own right (see [6, 10] and reference therein) but is also related to the existence of Radon–Nikodym derivatives. As noted in [6, 10]
the completeness of L 1 (γ) is equivalent to the completeness of L p (γ) for any p with 1 ≤ p < ∞. It is interesting to note [6] that L ∞ (γ) is always complete.
2. Preliminaries. Throughout we consider a nonnegative finitely addi- tive bounded set function γ defined on a σ-field F of subsets of a space Ω.
Even though our motivation and interest is only in probabilities, it is conve- nient to deal with bounded measures. Recall that [1, 5] L 1 (γ) is the collection of F-measurable real-valued functions f on Ω such that
T
|f | dγ < ∞, with the usual pseudometric.
This paper can be regarded as a continuation of [6]. Our starting point is the following theorem:
Theorem 1 (see [6]). The following are equivalent:
1. L 1 (γ) is complete.
2. (F, d) is complete where d is the usual pseudometric on F given by d(A, B) = γ(A △ B).
3. Given any sequence {A n } of sets in F, there exists a set A ∈ F such that γ(A n \ A) = 0 for each n, and γ(A) ≤ P
γ(A n ).
4. Given any sequence {A n } of sets in F and ε > 0, there exists a set A ∈ F such that
γ(A n − A) = 0 for each n, and γ(A) ≤ X
γ(A n ) + ε.
5. Given any sequence {A n } of sets in F there is a sequence {B n } of sets in F such that
B n ⊂ A n , γ(A n − B n ) = 0 for each n, and γ [
B n
≤ X
γ(B n ) = X γ(A n ).
6. Given a sequence {A n } of pairwise disjoint sets in F, there exists a sequence {B n } in F such that
B n ⊂ A n , γ(A n − B n ) = 0 for each n, and γ [
B n
= X
γ(B n ) = X γ(A n ).
7. Given an increasing sequence {A n } of sets in F, there exists a set A ∈ F such that
γ(A n − A) = 0 for each n, and γ(A) = lim
n γ(A n ).
This theorem is not stated in this form in [6], but a proof can be based on the proofs of Lemma 2.2, Theorem 2.4 and Remark 2.5 of [6]. See also [10] where a notion of self-separability of γ was introduced and was shown to be equivalent to the completeness of Banach space valued L 1 spaces.
Outer measure was used in [6], but the above pleasing form of the result is a consequence of the fact that the domain of γ is now a σ-field.
As a useful consequence of the criteria given above, we have the following Theorem 2. (a) Suppose γ = γ 1 + γ 2 .
(i) If L 1 (γ) is complete then so are L 1 (γ 1 ) and L 1 (γ 2 ).
(ii) If L 1 (γ 1 ) and L 1 (γ 2 ) are complete then L 1 (γ) is not necessarily complete.
(iii) If L 1 (γ 1 ) and L 1 (γ 2 ) are complete and γ 1 , γ 2 are supported on disjoint sets then L 1 (γ) is complete.
(b) Suppose Ω 0 ∈ F, 0 < γ(Ω 0 ) < γ(Ω). Then L 1 (γ) is complete iff L 1 (γ 1 ) and L 1 (γ 2 ) are complete where γ 1 , γ 2 are the restrictions of γ to Ω 0
and Ω − Ω 0 respectively.
(c) Suppose L 1 (γ) is complete. Let F 0 be a sub-σ-field of F which includes all γ null sets that are in F. Let γ 0 be γ restricted to F 0 . Then L 1 (γ 0 ) is complete.
P r o o f. (c) and (aiii) follow from criterion 6 of Theorem 1. (b) follows
from (ai) and (aiii).
To show (aii) take Ω = {1, 2, . . .}, F = power set of Ω; γ 1 is the countably additive measure γ 1 (n) = 1/2 n , n = 1, 2, . . .; γ 2 is a diffuse 0-1 valued measure on F giving 0 to singletons; γ = γ 1 + γ 2 . Then both L 1 (γ 1 ) and L 1 (γ 2 ) are complete. However, criterion 6 of Theorem 1 fails for A n = {n}, showing that L 1 (γ) is not complete.
Finally, we prove (ai) as follows: Towards verifying criterion 6 of The- orem 1, suppose {A n } is a sequence of disjoint sets in F. Since L 1 (γ) is complete, get a sequence {B n } as stated there. In particular, γ 1 (A n −B n ) = 0 = γ 2 (A n − B n ) for each n. Moreover
γ [ B n
= γ 1
[ B n + γ 2
[ B n , X γ(B n ) = X
γ 1 (B n ) + X
γ 2 (B n ).
By choice of {B n }, the left sides of the equations above are the same. So must be the right sides. But γ 1 ( S
B n ) ≥ P
γ 1 (B n ) and γ 2 ( S
B n ) ≥ P
γ 2 (B n ) so that equality must hold at both places. In other words, the same sequence {B n } witnesses that criterion 6 of Theorem 1 holds for both γ 1 and γ 2 .
Remark 1 . Theorem 2(ai) can equivalently be stated as follows: If L 1 (γ) is complete and γ 1 ≤ γ (inequality being understood setwise) then L 1 (γ 1 ) is also complete.
3. Yosida–Hewitt decomposition. Recall that a finitely additive pos- itive measure µ on (Ω, F) is said to be purely finitely additive if λ being a positive countably additive measure with λ(A) ≤ µ(A) for all A ∈ F implies that λ ≡ 0.
The celebrated decomposition theorem due to Yosida and Hewitt [26]
(see also [24]) says that any finitely additive positive measure γ on (Ω, F) can be decomposed as γ = γ 1 + γ 2 where γ 1 is countably additive and γ 2 is purely finitely additive. Moreover such a decomposition is unique.
Theorem 3. Let γ = γ 1 + γ 2 be the Yosida–Hewitt decomposition of γ.
Then L 1 (γ) is complete iff γ 1 , γ 2 are supported on disjoint sets and L 1 (γ 2 ) is complete.
P r o o f. If γ 1 , γ 2 satisfy the conditions of the theorem, then L 1 (γ) is complete in view of Theorem 2(aiii).
Conversely, assume that L 1 (γ) is complete. The idea is the following:
We shall express Ω = A ∪ B ∪ C where A, B, C ∈ F are pairwise disjoint, γ 1 (A) = 0, γ 2 (B) = 0, and if S ∈ F, S ⊂ C then γ 1 (S) > 0 iff γ 2 (S) > 0.
Assume the decomposition for a moment. We claim that γ 2 (and hence
γ 1 ) is null on C. If not, γ 2 being purely finitely additive we can get C n ⊂ C
with C n ↑ C and lim n γ 2 (C n ) < γ 2 (C). By criterion 7 of Theorem 1, get e C
such that γ(C n − e C) = 0 for each n and γ( e C) = lim n γ(C n ). In particular,
γ 1 (C n − e C) = 0 for each n so that γ( e C) ≥ lim n γ 1 (C n ). In case γ 2 (C − e C) = 0 we have γ 2 ( e C) ≥ γ 2 (C) > lim n γ 2 (C n ), implying that γ( e C) > lim n γ(C n ), a contradiction. Thus γ 2 (C − e C) > 0. But then γ 1 (C − e C) > 0. Since C n − e C ↑ C − e C and γ 1 is countably additive we conclude that γ 1 (C − e C) = lim n γ 1 (C n − e C) = 0, again a contradiction. Thus γ 2 must be null on C. So must be γ 1 . Thus γ 1 and γ 2 are supported on B and A respectively. The proof is completed by using Theorem 2(b).
We now proceed to exhibit the stated decomposition. Consider C = {S ∈ F : γ 1 (S) > 0 and γ 2 (S) = 0},
β = sup{γ 1 (S) : S ∈ C}.
Completeness of L 1 (γ) implies that this supremum is indeed attained. To see this, pick S n ∈ C with γ 1 (S n ) ↑ β. Since C is closed under finite unions we can assume that S n increases with n. By criterion 7 of Theorem 1, get B such that γ(B) = β and γ(S n − B) = 0 for each n. There is no loss to assume that B ⊂ S
n S n . First observe that γ 1 (B) + γ 2 (B) = γ(B) = β.
Secondly, γ(S n − B) = 0 and hence γ 1 (S n − B) = 0 for each n, so that γ 1 (B) ≥ γ 1 (S n ) for all n, implying that γ 1 (B) ≥ β. These two observations show that γ 1 (B) = β and γ 2 (B) = 0.
In an analogous manner, consider
D = {S ∈ F : γ 2 (S) > 0 and γ 1 (S) = 0}.
Since γ 1 is countably additive, D is closed under countable unions and hence there is a set A such that γ 1 (A) = 0 and γ 2 (A) = sup{γ 2 (S) : S ∈ D}. By construction it is clear that γ 1 (A∩B) = 0 = γ 2 (A∩B). Thus we can assume A ∩ B = ∅. Set C = Ω − {A ∪ B} to complete the proof.
Remark 2 . Since γ 1 in the theorem above is countably additive, clearly L 1 (γ 1 ) is complete.
4. Discrete measures. Recall that a finitely additive nonnegative mea- sure γ on (Ω, F) is said to be discrete if γ = P
a i δ i where the δ i are distinct 0-1 valued measures and a i > 0. Since we are considering only bounded measures, clearly P
a i < ∞.
Theorem 4. Let γ = P
a i δ i be discrete. Then L 1 (γ) is complete iff the δ i are uniformly singular , that is, there are pairwise disjoint sets A i ∈ F with δ i (A i ) = 1 for each i.
P r o o f. If {δ i } are uniformly singular then criterion 6 of Theorem 1 applies to show that L 1 (γ) is complete.
To prove the converse, assume that {δ i } are not uniformly singular. Let
us say that δ i can be separated if there is a set A i in F such that δ i (A i ) = 1
and δ j (A i ) = 0 for j 6= i. If each δ i can be separated, witnessed by say A i ,
then setting B n = A n − S
i<n A i , we observe that δ i (B i ) = 1 for each i, showing that δ i are uniformly singular. Thus there is a δ i , say δ 1 , which cannot be separated. We shall construct a sequence of pairwise disjoint sets (A n ) n≥1 such that (i) if j > 1 then δ j (A i ) = 1 for some i and (ii) δ 1 (A i ) = 0 for each i. If this is done, then we claim that criterion 6 of Theorem 1 fails for this sequence. Indeed, suppose we have sets B i ⊂ A i with γ(A i − B i ) = 0 for each i. By properties (i) and (ii) we have P
i≥1 γ(B i ) = P
i>1 a i . If δ 1 ( S
B i ) = 0 then property (i) implies that δ 1 can be separated, which is not the case. Thus δ 1 ( S
B i ) = 1, implying that γ( S
B i ) = P
i≥1 a i > P γ(B i ).
We now proceed to exhibit sets A i as stated. Pick B 0 such that δ 1 (B 0 )
= 1. Set A 1 = B 0 c and S 1 = {j : δ j (B 0 ) = 1}. Then S 1 is infinite because δ 1
cannot be separated. Pick the first integer j 1 ∈ S 1 and write B 0 = B 1 ∪ A 2 , a disjoint union with δ 1 (B 1 ) = 1 and δ j 1 (A 2 ) = 1. To do this, just note that the δ i , being 0-1 valued, are pairwise singular. Then S 2 = {j : δ j (B 1 ) = 1}
is again infinite. Proceed inductively by picking the first j in S n at the nth stage. This completes the proof of the theorem.
The following theorem, a slight extension of Theorem 4, will be needed later. Theorem 4 corresponds to the case when γ 0 is absent.
Theorem 5. Suppose γ = P
i≥0 a i γ i with a i > 0 and P
a i < ∞. Assume that γ i , i ≥ 1, are 0-1 valued. If L 1 (γ) is complete then γ i , i ≥ 1, are uniformly singular.
P r o o f. Apply Theorems 2 and 4.
5. Sobczyk–Hammer decomposition. Recall that a finitely additive nonnegative measure γ on (Ω, F) is said to be strongly continuous if given ε > 0, there is a finite decomposition Ω = S
A i with γ(A i ) < ε for each i. The well known decomposition theorem due to Sobczyk and Hammer [25] says that any finitely additive positive measure γ on (Ω, F) can be decomposed as γ = γ 1 +γ 2 where γ 1 is discrete and γ 2 is strongly continuous.
Further such a decomposition is unique.
Theorem 6. Let γ = γ 1 + γ 2 be the Sobczyk–Hammer decomposition of γ. Then L 1 (γ) is complete iff L 1 (γ 1 ), L 1 (γ 2 ) are complete and γ 1 , γ 2 are supported on disjoint sets.
P r o o f. The “if” part is a consequence of Theorem 2. To prove the converse, assume that L 1 (γ) is complete. By Theorem 2 again, L 1 (γ 1 ) and L 1 (γ 2 ) are also complete. We only need to show now that γ 1 , γ 2 are sup- ported on disjoint sets.
First assume that γ 1 is 0-1 valued. By using the strong continuity of γ 2 , one can obtain, for each n, a set A n such that γ 1 (A c n ) = 1 and γ 2 (A c n ) <
1/2 n . We can also assume that A n increases with n. We have lim n γ(A n ) =
lim n γ 2 (A n ) = γ 2 (Ω). By Theorem 1(7) get B ⊂ S
A n with γ(B) = γ 2 (Ω) and γ(A n − B) = 0 for each n. In particular γ 2 (A n − B) = 0 for each n so that γ 2 (B) ≥ γ 2 (A n ∩ B) = γ 2 (A n ), which increases to γ 2 (Ω). Thus γ 2 (B) = γ 2 (Ω). But since γ(B) = γ 2 (Ω) we conclude that γ 1 (B) = 0. In other words γ 1 , γ 2 are supported on B c and B respectively.
To treat the general case, assume that the discrete part is γ 1 = P
i≥1 a i δ i , where a i > 0, P
a i < ∞ and the δ i are distinct 0-1 valued measures.
By Theorem 5, the δ i are uniformly singular so that Ω can be written as a disjoint union S
i≥1 A i with δ i (A i ) = 1 for each i. By Theorem 2, L 1 (a i δ i + γ 2 ) is complete for each i and hence by earlier para we can get B i ⊂ A i such that δ i (B i ) = 1 and γ 2 (B i ) = 0. Since L 1 (γ) is complete, by Theorem 1(6) we can get C i ⊂ B i with γ(B i − C i ) = 0 for each i and γ( S
C i ) = P
γ(C i ). In particular for each i, δ i (B i − C i ) = 0 so that δ i (C i ) = δ i (B i ) = 1. Since γ 2 (C i ) = 0 and γ 1 (C i ) = a i for each i, we have
X γ(C i ) = X
γ 1 (C i ) + X
γ 2 (C i ) = X a i , γ [
C i
= γ 1 [ C i
+ γ 2 [ C i
≥ X
a i + γ 2 [ C i
. Since the left sides are equal we conclude that γ 2 ( S
C i ) = 0. In other words γ 1 is supported on S
C i and γ 2 is supported on its complement, as claimed.
Combining Theorems 3–6 we obtain the following two versions of the main characterization theorem.
Theorem 7. Let γ be a finitely additive probability on (Ω, F). Then L 1 (γ) is complete iff Ω has a decomposition Ω = Ω 0 ∪ Ω 1 ∪ Ω 2 such that (i) γ restricted to Ω 0 is countably additive, (ii) γ restricted to Ω 1 is dis- crete and is a combination of uniformly singular 0-1 probabilities and (iii) γ restricted to Ω 2 is strongly continuous and its L 1 space is complete.
Theorem 8. Let γ be a finitely additive probability on (Ω, F). Then L 1 (γ) is complete iff Ω has a decomposition Ω = S
i≥0 Ω i such that (i) γ restricted to Ω 0 is countably additive, (ii) γ restricted to each Ω i , i ≥ 2, is at most two-valued , (iii) γ restricted to Ω 1 is strongly continuous and its L 1
space is complete and (iv) for each A ∈ F, γ(A) = P ∞
i=0 γ(Ω i ∩ A).
In Theorem 7 we had a finite decomposition so that the last condition of Theorem 8 was not imposed. We conclude this section with a few remarks.
Remark 3 . Here is an example of a sequence of 0-1 measures which are not uniformly singular. Set Ω = {0, 1, 2, · · ·}. For i ≥ 2, let γ i be a diffuse 0-1 measure concentrated on the powers of the ith prime. Let
C = {A : γ i (A) = 1 for all but finitely many i ≥ 2}.
Extend C to an ultrafilter and denote by γ 1 the corresponding 0-1 measure.
Then {γ i : i ≥ 1} are not uniformly singular. Note that all these γ i are
purely finitely additive. If we did not want this, we could have taken point masses and any diffuse 0-1 measure. Also note that all these γ i are defined on the power set of Ω. The same construction can be carried out on any (Ω, F) provided F is infinite.
Remark 4 . Given any sequence {γ i } of 0-1 measures on (Ω, F) there exists an infinite subsequence which is uniformly singular. We inductively construct a sequence of disjoint sets A i ≥ 1 and indices n i , i ≥ 1, such that γ n i (A i ) = 1 for all i. Just make sure that at the kth stage infinitely many γ i are concentrated on the complement of S
i≤k A i . Remark 5 . If γ = P
2 −i γ i where γ i are as in Remark 3, then L 1 (γ) is not complete—though γ is defined on the power set of Ω.
Remark 6 . Given (Ω, F) where F is infinite it is always possible to obtain strongly continuous γ on F such that L 1 (γ) is not complete. F being infinite, the general case can be reduced to Ω = N and F the power set.
This is what we treat. Let µ be any extension of the density charge defined on arithmetic progressions. Then µ is clearly strongly continuous. Fix a decomposition of N into disjoint sets A n , n ≥ 1, with µ(A n ) > 0 for each n.
Let
F = {B : µ(B ∩ A n ) = µ(A n ) for all but finitely many n}.
For each k, 1 ≤ k ≤ n, and each sequence (ε 1 , . . . , ε k ) of 0’s and 1’s fix a subset A n ε 1 ...ε k of A n with positive µ measure such that A n ε 1 ...ε k is the disjoint union of A n ε
1 ...ε k 0 and A n ε
1 ...ε k 1 , and A n is the disjoint union of A n 0 and A n 1 . This can be done by the strong continuity of µ. For k ≥ 1 define B ε 1 ...ε k = S
n≥k A n ε 1 ...ε k . Extend F restricted to B ε 1 ···ε k to an ultrafilter on S
n≥k A n . Let η ε 1 ...ε k be the associated 0-1 measure supported on S
n≥k A n , defined on the power set of N. For k ≥ 1, let η k be the average of the 2 k measures η ε 1 ...ε k . Fix any Banach limit ℓ and set η(A) = ℓ{η k (A) : k ≥ 1}
for A ⊂ N. Let γ = 1 2 η + 1 2 µ. It is not difficult to see that γ is strongly continuous. Since η k (A n ) = 0 for k > n, it follows that η(A n ) = 0 for any n. Using this, one can argue that condition 6 of Theorem 1 fails for the sequence {A n } so that L 1 (γ) is not complete.
Remark 7 . Later we shall see examples of strongly continuous γ for which L 1 (γ) is complete. However, we are unable to decide if there are extensions of density charges for which the L 1 space is complete. (The referee has kindly informed us that he has examples of such extensions.)
6. Finite strategic products. When dealing with finitely additive mea-
sures, products are not in general well defined on product σ-fields. However,
there is one situation where the product measures are well defined by suc-
cessive integration as was done by Dubins and Savage [4]. For this we need
to consider measures defined on power sets. For 1 ≤ i ≤ k, let γ i be a finitely additive probability defined on the power set of Ω i . Let Ω = N k
i=1 Ω i . On the power set of Ω define
γ(A) =
\
. . .
\
1 A (x 1 , . . . , x k ) dγ k (x k ) . . . dγ 1 (x 1 ).
This γ is called the strategic product of the γ i , 1 ≤ i ≤ k. Carefully note the order of integration. For more on this, see [4]. The reader should note that even though the probabilities are defined on power sets, their L 1 spaces may still be incomplete (see Remarks 5 and 6 of the previous section).
Theorem 9. Let γ be the strategic product of γ i , 1 ≤ i ≤ k.
(a) If L 1 (γ i ) is complete for each i, then so is L 1 (γ).
(b) If L 1 (γ) is complete, then so is L 1 (γ 1 ).
(c) L 1 (η k ) is complete iff L 1 (η) is complete. Here η k is the k-fold strate- gic product of η.
P r o o f. (c) is immediate from (a) and (b). For simplicity we assume that k = 2 in what follows.
To prove (b) we verify condition 7 of Theorem 1 for the measure γ 1 . So let {A n } be an increasing sequence of subsets of Ω 1 . Set B n = A n × Ω 2 . Using the fact that L 1 (γ) is complete and criterion 7 of Theorem 1 get B ⊂ Ω 1 ×Ω 2
such that γ(B n \ B) = 0 for each n and lim n→∞ γ(B n ) = γ(B). Set A = {x ∈ Ω 1 : γ 2 (B x ) > 1/2}.
Firstly, for each n, γ 1 (A n \ A) ≤ 2
\
A n \A
γ 2 (B n \ B) x dγ 1 (x) ≤ 2γ(B n \ B) = 0.
Secondly, a similar computation gives γ 1 (A \ A n ) ≤ γ(B \ B n ), so that lim n→∞ γ 1 (A n ) ≥ γ 1 (A), which together with the first observation implies that lim n→∞ γ 1 (A n ) = γ 1 (A).
We now prove (a). Using ideas from [6] we verify condition 7 of Theorem 1 for γ. Let {A n } be an increasing sequence of subsets of Ω 1 × Ω 2 . Passing to a subsequence if necessary we can and do assume that γ(A n+1 \A n ) < 1/2 2n for each n ≥ 1. This of course implies that for each n ≥ 1,
γ 1 {x 1 : γ 2 ((A n+1 ) x 1 ) > γ 2 ((A n ) x 1 ) + 1/2 n } < 1/2 n .
Let the set in braces be denoted by B n . Fix k ≥ 1. By completeness of L 1 (γ 1 ) and condition 3 of Theorem 1 applied to the sequence {B n : n ≥ k}
we obtain C k ⊂ Ω 1 such that
γ 1 (C k ) ≤ 1/2 k−1 and γ 1 (B n \ C k ) = 0 ∀n ≥ k.
By taking successive intersections we can assume that C k decreases with k.
Set C ∞ = T
k C k . Clearly γ 1 (C ∞ ) = 0. If x 1 6∈ C ∞ then let k(x 1 ) be the
first integer k such that x 1 6∈ C k . Let D 1 = n
x 1 : x 1 6∈ C ∞ , x 1 ∈ [
n≥k(x 1 )
B n
o ,
D 2 = n
x 1 : x 1 6∈ C ∞ , x 1 6∈ [
n≥k(x 1 )
B n
o .
For x 1 ∈ D 1 let n(x 1 ) be the first integer n ≥ k(x 1 ) such that x 1 ∈ B n . For each x 1 ∈ Ω 1 by completeness of L 1 (γ 2 ) get a set A(x 1 ) ⊂ Ω 2 such that
γ 2 ((A n ) x 1 \ A(x 1 )) = 0 for each n and lim
n γ 2 ((A n ) x 1 ) = γ 2 (A(x 1 )).
Define
A = [
x 1 ∈D 1
({x 1 } × (A n(x 1 ) ) x 1 ) ∪ [
x 1 ∈D 2
({x 1 } × A(x 1 )).
Fix any integer n ≥ 1.
Claim . γ(A n \ A) = 0.
Firstly, N = C ∞ ∪ ( S
1≤k≤m<n (B m \C k )) is γ 1 null. Secondly, if x 1 ∈ D 1
and n(x 1 ) ≤ n then k(x 1 ) ≤ n(x 1 ) < n and so x 1 ∈ N ; whereas if n(x 1 ) ≥ n then clearly (A n \ A) x 1 = ∅. Thirdly, if x 1 ∈ D 2 then by choice of A(x 1 ) we have γ 2 (A n \ A) x 1 = 0. These three observations prove the claim.
Claim . lim n γ(A n ) = γ(A) (or equivalently, in view of the earlier claim, lim n γ(A \ A n ) = 0.)
To see this, fix any k ≥ 1. Set E k = S k i=1
S k
n=i (B n \C i )∪C k . Then firstly, γ 1 (E k ) = γ 1 (C k ) ≤ 1/2 k−1 . Secondly, suppose x 1 6∈ E k . Then k(x 1 ) ≤ k and n(x 1 ) > k.
If x 1 ∈ D 1 then using the fact that x 1 6∈ B i for k ≤ i < n(x 1 ) we get γ 2 (A \ A k ) x 1 = γ 2 ((A n(x 1 ) ) x 1 \ (A k ) x 1 )
≤ X
k≤i<n(x 1 )
γ 2 ((A i+1 ) x 1 \ (A i ) x 1 ) < 1 2 k−1 .
If x 1 ∈ D 2 then pick l > k (depending on x 1 ) so that γ 2 (A(x 1 )\(A l ) x 1 ) <
1/2 k−1 . Using the fact that x 1 6∈ B i for all i ≥ k(x 1 ) we conclude that γ 2 (A \ A k ) x 1 ≤ X
k≤i<l
γ 2 ((A i+1 ) x 1 \ (A i ) x 1 ) + γ 2 (A(x 1 ) \ (A l ) x 1 ) < 1 2 k−2 . These two observations show that γ(A\A k ) ≤ 1/2 k−3 , proving the claim.
This completes the proof of the theorem.
Remark 8 . If L 1 (γ 1 ×γ 2 ) is complete then L 1 (γ 2 ) need not be complete.
In fact if γ 1 is 0-1 valued and not countably additive then L 1 (γ 1 × γ 2 )
is complete for any γ 2 . To see this, we verify condition 4 of Theorem 1.
Let {A n } be a sequence of subsets of Ω 1 × Ω 2 and ε > 0. Fix a partition N 1 , N 2 , · · · of Ω 1 such that γ 1 (N i ) = 0 for each i. Fix n ≥ 1. Observe that γ 1 {x : γ 2 (A n ) x < γ(A n ) + ε/2 n+1 } = 1 where γ = γ 1 × γ 2 . If the set in braces is denoted by C n then take
B n = n
(x, y) : x ∈ C n , x 6∈ [
k≤n
N k o
∩ A n .
We show that these sets satisfy condition 4 of Theorem 1. Since A n \ B n ⊂ C n c ∪( S
k≤n N k )×Ω 2 , we have γ(A n \B n ) = 0. To see γ( S
B n ) ≤ P
γ(B n ) we first observe that for any fixed x there is exactly one k such that x ∈ N k . So if n ≥ k, then (B n ) x is empty. Therefore, for all x there exists a k, depending on x, such that γ 2 ( S ∞
n=1 B n ) x = γ 2 ( S k
n=1 B n ) x ≤ P k
n=1 γ 2 (B n ) x . Hence, γ(∪ ∞ n=1 B n ) =
\
γ 2 [ ∞
n=1
B n
x dγ 1 (x)
≤
\
X ∞ n=1
γ 2 (B n ) x dγ 1 (x) ≤
\
X ∞ n=1
γ 2 (A n ) x 1 C n (x) dγ 1 (x)
<
\
