Uniform s-boundedness and convergence results for measures with values in complete l-groups

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1 Uniform s-boundedness and convergence results for measures with values in complete l-groups A. Boccuto D. Candeloro ABSTRACT. Absolute continuity, s-boundedness and extensions are studied, in the context of the so-called RD-convergence, for set functions taking values in Dedekind complete l-groups. Subsequently, we obtain results of uniform s-boundedness for RD-convergent sequences of measures (Vitali-Hahn-Saks-Nikodým theorem) and deduce a Schur-type theorem for measures defined on P(IN ). A.M.S. SUBJECT CLASSIFICATION (1995): 28B15, 28B05, 28B10, 46G10. KEY WORDS: l-groups, uniform s-boundedness, Vitali-Hahn-Saks-Nikodým theorems, Schur theorems. 1 Introduction The concept of s-boundedness is crucial in many problems of measure theory, both for finitely additive and for σ-additive measures. We mention here only some papers dealing with this topic, just to stress the wide range of applicability of this concept: [1, 4, 6, 8, 9, 10, 11, 12, 13, 14, 17, 22, 23, 27, 36]. One of the most interesting results in this area is the well-known Vitali-Hahn- Saks-Nikodým theorem, together with its manifold consequences, such as the well- Authors Address: Dipartimento di Matematica e Informatica, via Vanvitelli,1 I PERUGIA (ITALY) boccuto@dipmat.unipg.it, candelor@dipmat.unipg.it tel , fax

2 2 A. Boccuto--D. Candeloro known Schur theorem, concerning convergence in l 1, or the Dunford-Pettis theorem in L 1 (just to avoid confusion, we point here that Schur s theorem shows equivalence between weak and strong convergence for sequences in l 1, and the Dunford-Pettis result we are dealing with states that a sequence in L 1 converges strongly if and only if it is bounded in norm, and convergent weakly and in measure (to the same limit)). Despite the various generalizations and extensions of the Vitali-Hahn-Saks-Nikodým theorem (VHSN), to the case of measures taking values in topological groups and semigroups, it seems that just some partial results have been obtained for latticevalued measures (see [1, 4]): this is so, because of the lacking of a topology (in general), and hence new techniques are needed. Here we deal with l-group-valued measures, and make use of the so-called D- convergence, rather than (O)-convergence, that is the order convergence as introduced in [2], pp. 244 and 314: in fact, although the two kinds of convergence coincide for many spaces of interest, the former is easier to handle. However, in the crucial definitions (s-boundedness, σ-additivity, convergence, and so on), we need a special form of D-convergence, which we call RD-convergence. The results we have found, concerning some consequences of s-boundedness and uniform s-boundedness, and a version of the VHSN theorem with some of its consequences, correspond, in the topological setting, to the same concepts given for metrizable groups. For instance, we shall give some consequences of s-boundedness, and its relations with absolute continuity. One of these results is that every s-bounded finitely additive measure is bounded: this is not contrary to the well-known example of [33], because of our different definition of s-boundedness (actually, this is one of the reasons we decided to choose it). Other relevant results, concerning uniform s-boundedness, are: a sort of uniform extension, for measures σ-additive on an algebra; a property of uniform absolute continuity, holding for a sequence of uniformly bounded, uniformly s-bounded finitely additive measures, each absolutely continuous with respect to some real-valued non-negative finitely additive measure. As to the Vitali-Hahn-Saks-Nikodým theorem, we shall give a version for σ-

3 Uniform s-boundedness and convergence results 3 additive measures, defined on a σ-algebra A of subsets of an abstract set Ω, and taking values in a complete l-group R. Consequences are then found, concerning a form of the Schur s lemma, and also a finitely additive version of the VHSN theorem. 2 Definitions We shall introduce now the main definitions we need, together with some preliminary results. We first recall the meaning of D-convergence, in an l-group R, and the concept of RD-convergence. Definition 2.1 An abelian group (R, +) is called l-group if it is endowed with a compatible ordering, and is a lattice with respect to it. An l-group R is said to be complete if every nonempty subset of R, bounded from above, has supremum in R. For a reference about the basic facts on l-groups, see [2], Chapter XIII, pp Definition 2.2 Given a sequence (r n ) in R, we say that (r n ) D-converges to an element r R if there exists a bounded double sequence (a i,j ) in R, such that a i,j 0 for each i IN, that is a i,j a i,j+1 i, j IN and a i,j = 0 i IN j IN (such a sequence will be called a regulator or D-sequence from now on), satisfying the following condition: mapping φ : IN IN, there exists an integer n 0 such that r n r a i,φ(i), for all n n 0. In this case, we write D lim n r n = r. From now on, we shall denote by Φ the set of all mappings φ : IN IN, involved in the previous definition.

4 4 A. Boccuto--D. Candeloro Definition 2.3 We say that R is weakly σ-distributive if for every bounded double sequence (b i,j ) i,j with b i,j b i,j+1 i, j one has: b i,j = ( ) b i,φ(i) j=1 (see [38]). φ Φ It is easy to check that an l-group is weakly σ-distributive if and only if for every D-sequence (a i,j ) one has: (1) ( ) a i,φ(i) = 0. (2) φ Φ For the theory of weakly σ-distributive l-groups, see [30], pp It s easy to prove that the common (O)-convergence implies D-convergence, while the converse is true in weakly σ-distributive spaces. For example, concerning the first implication, if a sequence (r n ) order converges to r in R, then there exists a monotone decreasing sequence (a n ) in R, with n a n = 0 and r n r a n n. (Just take a n = k=n r k r.) Set a i,j = a j for all i, j. It is easy to check that (a i,j ) is the required regulator, with respect to which we have D-convergence of the sequence (r n ) n to r (see [18]). Throughout the paper, we shall always assume that R is weakly σ-distributive, however we prefer to use D-convergence for our definitions and results, because it is more flexible (from a technical point of view) than (O)-convergence. Definition 2.4 Let S denote any nonempty abstract set, and assume that {(r n (s)) n IN : s S} is a family of sequences in R, each depending on an element s S. We say that the sequences (r n (s)) are RD-convergent to some element r(s), if there exists a regulator (a i,j ) such that, for every φ Φ and every s S there exists an integer n 0 (depending both on φ and on s), such that for all n n 0. In this case, we write r n (s) r(s) a i,φ(i), RD lim n r n (s) = r(s) for all s S.

5 Uniform s-boundedness and convergence results 5 The next lemma gives us the possibility to deduce RD-convergence from simple D-convergence, in case S is countable ([21], [30]). This result essentially motivates our choice to use D-convergence, rather than (O)-convergence: in other words, it says that RD-convergence, though involving some kind of uniformity, is weaker than the corresponding concept for (O)-convergence, and consequently D-convergence is more flexible than (O)-convergence. Lemma 2.5 Let (a k i,j) be any countable family of regulators. Then for each fixed element u R there exists a regulator (a i,j ) such that, for every φ Φ one has An easy consequence is the following Proposition 2.6 Assume that (r s n) n IN ( ) u a k i,φ(i+k) a i,φ(i). k=1 is a sequence D-converging to 0, for each s IN. If r s n u for some u R, the family (r s n) n IN is RD-converging to 0. Definition 2.7 Let F be any algebra of subsets of a set Ω. Assume that µ : F R is any finitely additive bounded measure. We put: µ + (A) = sup{µ(b) : B F, B A}, µ (A) = inf{µ(b) : B F, B A}, v(µ)(a) = sup{ µ(b) : B F, B A} for all A F. µ + is called the positive variation of µ, µ is called the negative variation of µ, while v(µ) is called the semivariation of µ. It is easy to see that: µ + and µ are positive finitely additive measures, µ + µ = µ, and v(µ) µ + + µ 2v(µ). Definition 2.8 Let F be any algebra of subsets of a set Ω. Assume that µ : F R is any finitely additive measure. We say that µ is σ-additive if there exists a regulator

6 6 A. Boccuto--D. Candeloro (a i,j ) such that, for every sequence C k in F, C k, and for every φ Φ, there exists k 0 IN such that µ(f ) a i,φ(i), for all F F, F C k0. In case µ : F R is σ-additive and F is a σ-algebra, µ will be said to be a measure (Usually, a σ-algebra will be denoted by the symbol A). Remark 2.9 Clearly, if µ is bounded and σ-additive, then µ + and µ are. However, we shall see later that σ-additivity on a σ-algebra always implies boundedness. We now introduce our concept of s-boundedness. Definition 2.10 Let µ : F R be any finitely additive measure. We say that µ is s-bounded if there exists a regulator (a i,j ) such that, for every disjoint sequence (H k ) in F, and every φ Φ, there exists an integer k such that µ(h k ) a i,φ(i) (3) holds, for all k k. Remark 2.11 In the above definition, formula (3) can be equivalently replaced by sup{ µ(f ) : F F, F H k } a i,φ(i) (4) for all k k. Indeed, if (4) were not true for some disjoint sequence (H k ) and some mapping φ Φ, there would exist an increasing sequence (k h ) in IN and a corresponding sequence (F h ) of subsets of H kh such that µ(f h ) a i,φ(i) for all h, and this contradicts s-boundedness of µ. Moreover, we note that the concept of s-boundedness formulated in Definition 2.10 is different than the one of exhaustivity given in [33], p. 391: indeed the two concepts differ exactly for the kind of convergence involved, because in [33] simple D-convergence is concerned, while here (RD)-convergence is used. More precisely, it turns out that µ is s-bounded according to this last definition if and only if for

7 Uniform s-boundedness and convergence results 7 every disjoint sequence (H k ) in F there exists a regulator (a i,j ) such that, for every φ Φ, there exists an integer k such that µ(h k ) a i,φ(i) holds, for all k k. Proposition 2.12 Let µ : F R be any finitely additive bounded measure. Then µ is s-bounded if and only if µ + + µ is. Proof. Of course, the only non-trivial implication is the direct one: assuming that µ is s-bounded, then µ + and µ are. From the remark above, it follows immediately that s-boundedness (and boundedness) of µ imply that both µ + and µ are s- bounded, thus µ + + µ is too. We shall see later that s-boundedness actually implies boundedness, hence in the previous Proposition the boundedness condition is superfluous, at least for the direct implication. An easy result is the following. Proposition 2.13 Assume that µ : A R is a measure. Then µ is s-bounded. Proof. Given any disjoint sequence (H n ) in A, set: E n = k=n H k ; then (E n ) is a decreasing sequence in A, E n. Thus, the assertion follows from σ-additivity, and the inclusion: H n E n. The next result will be useful later. Proposition 2.14 Let µ : F R be any s-bounded finitely additive measure. Then, for every monotone sequence (A n ) in F, there exists in R the limit D lim µ(a n ). (5) n Proof. Without loss of generality, we can suppose that (A n ) is decreasing. Let (a i,j ) be the regulator related to s-boundedness of µ, and let s show that it works for (5). If this is not true, there exists a mapping φ Φ such that µ(a nk ) µ(a nk +p k ) a i,φ(i)

8 8 A. Boccuto--D. Candeloro for suitable sequences (n k ) and (p k ) in IN, such that n k+1 > n k + p k for all k. Setting now B k = A nk \ A nk +p k, we see that the sets B k are pairwise disjoint, and µ(a nk ) µ(a nk +p k ) = µ(b k ) a i,φ(i) for all k, which contradicts s-boundedness relative to the regulator (a i,j ). We now show that s-boundedness, for a finitely additive measure, always implies boundedness. Lemma 2.15 Let µ : F R be any s-bounded finitely additive measure, and let (a i,j ) be the regulator corresponding to s-boundedness. Denote by u any majorant for (a i,j ), and assume that an element B F exists, satisfying µ(b) u. Then there exists an element A F, A B, such that µ(a) u, but such that the set {µ(e) : E A} is bounded in R. Proof. If the set {µ(e) : E B} is bounded in R, we take A = B, and we have finished. Otherwise, there exists a set B 1 F, B 1 B, such that µ(b 1 ) u + µ(b). Therefore, we get µ(b \ B 1 ) u. If the set {µ(e) : E B 1 } is bounded in R, we can take A = B 1, and we have finished. Otherwise, there exists a set B 2 F, B 2 B 1, such that µ(b 2 ) u + µ(b 1 ). Therefore, we get µ(b 1 \ B 2 ) u. Proceeding in this fashion, either we find a subset B n B, fulfilling the requirement of the lemma, or we construct a decreasing sequence (B n ) such that µ(b n \ B n+1 ) u, for all n. As the sets (B n \ B n+1 ) are pairwise disjoint, and u is a majorant for the regulator (a i,j ), we have a contradiction. Thus, only the first case can occur, and the proof is finished. Theorem 2.16 Let µ : F µ is bounded. R be any s-bounded finitely additive measure. Then Proof. By contradiction, we assume that the set {µ(a) : A F} is unbounded. Define u as in the lemma 2.15, and choose any set B 1 F, such that µ(b 1 ) u.

9 Uniform s-boundedness and convergence results 9 Thanks to lemma 2.15, there exists in F a set A 1 B 1, such that µ(a 1 ) u, but such that {µ(e) : E A 1 } is bounded in R. Now, the set {µ(e) : E Ω \ A 1 } is not bounded in R, hence there exists an element B 2 F, B 2 Ω \ A 1, such that µ(b 2 ) u. Then, by 2.15, there exists in F an element A 2 B 2, with µ(a 2 ) u, but such that {µ(e) : E A 2 } is bounded in R. Now, the set {µ(e) : E Ω \ (A 1 A 2 )} is not bounded in R, hence there exists an element B 3 F, B 3 Ω \ (A 1 A 2 ), such that µ(b 3 ) u. So we can find an element A 3 F, disjoint from A 1 A 2, such that µ(a 3 ) u,... Proceeding in this way, we obtain a disjoint sequence (A n ) in F, such that µ(a n ) u, for all n, which contradicts the meaning of u in the condition of s-boundedness. Example 3 in Swartz ([33]) displays a measure that is s-bounded according to Swartz s use of the term, but not bounded. However this does not contradict our Theorem 2.16 because the Swartz measure is not s-bounded in the sense of Definition Example 2.17 We shall see now, by means of an example, that boundedness usually does not imply s-boundedness (as defined in 2.10). Take Ω = [0, 1], A = the Borel σ-field, and R = L 0 ([0, 1], λ), where λ denotes the usual Lebesgue measure. Set µ(a) = χ A, for every A A. Then, it s clear that µ is a bounded finitely additive measure. If µ were s-bounded, a bounded regulator (a i,j ) should exist in L 0, such that, for every φ Φ and for every disjoint sequence (H n ) in A, one should have a i,φ(i) (x) 1 for almost all x n n 0 H n, for some integer n 0. Now, for every x [0, 1] it s easy to construct a disjoint sequence (H n ) in A, such that for all n 0 the set {x} ( n n 0 H n ) contains a neighbourhood of x. This means that, for each fixed φ Φ, for each x [0, 1] there exists δ > 0 such that a i,φ(i) (t) 1 for almost all t ]x δ, x + δ[. By compactness, we deduce that a i,φ(i) 1. Thus, ( ) a i,φ(i) 1. (6) φ Φ But in L 0 the topology corresponding to convergence in measure is locally convex,

10 10 A. Boccuto--D. Candeloro Hausdorff and σ-compatible, that is such that any monotone increasing sequence (b n ) with least upper bound b converges to b with respect to this topology. Thus, thanks to Corollary J of [38], we get that L 0 is weakly σ-distributive. This contradicts (6). Remark 2.18 As to the uniform boundedness principle, a counterexample is given in [33], Example 5, showing that even a sequence of pointwise bounded measures may fail to be uniformly bounded (we stress here the fact that those measures are also σ-additive according with our definition 2.8). 3 Uniform s-boundedness We now introduce other important concepts. Definition 3.1 Let (µ n ) be any sequence of finitely additive s-bounded measures on F. We say that they are uniformly s-bounded if there exists a regulator (a i,j ) such that, whenever (H k ) is any sequence of pairwise disjoint members of F, and whenever φ is arbitrarily chosen from Φ, it s possible to find an element k 0 IN, such that for all k k 0. sup v(µ n )(H k ) n IN a i,φ(i) (7) Remark 3.2 Contrary to intuition, uniform s-boundedness for a sequence of measures does not imply uniform boundedness, even in the case R = IR; for instance, take Ω = {1}, F = P(Ω), and define µ n in this way: µ n ({1}) = n. Clearly, the µ n s are uniformly s-bounded (as Ω is finite), and also uniformly σ-additive, but they are not uniformly bounded. A general result of uniform boundedness is the following. Theorem 3.3 Let (µ n ) be a uniformly s-bounded sequence of finitely additive measures in an algebra F. Then they are uniformly bounded if and only if, for each set F F, the set {µ n (F ) : n IN } is bounded in R.

11 Uniform s-boundedness and convergence results 11 Proof. Clearly, only one implication must be proved. So, set u(f ) = sup{ µ n (F ) : n IN }, and assume by contradiction that the set {u(f ) : F F } is unbounded. Let now (a i,j ) be the regulator related to uniform s-boundedness, and denote by u any majorant for it. Then, there exists a set H 1 F, such that u(h 1 ) u. Let s show that H 1 can be chosen so that the set {v(µ n )(H 1 ) : n IN } is bounded. For, otherwise there exists a set B 1 H 1, B 1 F, such that u(b 1 ) u+ u(h 1 ), and subsequently an element B 2 B 1, B 2 F, such that u(b 2 ) u+u(b 1 ), and so on: this yields a decreasing sequence (B k ) in F, and a subsequence (µ nk ) such that µ nk (B k+1 ) u + µ nk (B k ) and so µ nk (B k \ B k+1 ) u for all k, which contradicts uniform s-boundedness. So, let s take H 1 F, with u(h 1 ) u, but such that the set {v(µ n )(H 1 ) : n IN } is bounded; we then must deduce that in Ω \ H 1 there exists a set H 2, H 2 F, such that u(h 2 ) u. Again, we can get that the set {v(µ n )(H 2 ) : n IN } is bounded. Proceeding in this fashion, we construct a disjoint sequence (H k ), such that u(h k ) u for all k, which contradicts uniform s-boundedness relative to the regulator (a i,j ). Easier results are the following. Proposition 3.4 Assume that (µ n ) is a uniformly s-bounded sequence of finitely additive measures on F, and let (a i,j ) be a regulator related to this property. For every monotone sequence (S h ) h in F, the limit r n := RD lim h µ n (S h ) exists uniformly in n, and the regulator (a i,j ) works for this property. Proof. The existence of the limit is due to 2.14.

12 12 A. Boccuto--D. Candeloro If the last assertion were not true, then, by virtue of the Cauchy condition (see also [5], Theorem 2.16), there would exist some (say) decreasing sequence (S h ), some φ Φ and increasing sequences (n k ), (p k ), (h k ) in IN, such that h k + p k < h k+1 and µ nk (S hk \ S hk +p k ) a i,φ(i) for all k. This contradicts uniform s-boundedness relative to the regulator (a i,j ). A consequence involves σ-additivity. Corollary 3.5 Given a sequence (µ n ) of uniformly s-bounded σ-additive measures on an algebra F, they are uniformly σ-additive, i.e. there exists a regulator (a i,j ) such that, for every decreasing sequence (F k ) in F, F k, and every φ Φ, it s possible to find an integer k 0 such that v(µ n )(F k0 ) a i,φ(i), for all n. 4 Absolute continuity and s-boundedness In this section, we shall outline some connections between uniform s-boundedness and absolute continuity, for a sequence of measures, which are similar, up to a certain extent, to the classic ones (see also [10]). From now on, let IR 0 + be the set of all non-negative real numbers. First, we introduce the concept of absolute continuity in our setting. Definition 4.1 Let µ be any finitely additive measure on F. Given any other finitely additive measure ν : F IR 0 +, we say that µ is absolutely continuous with respect to ν (and write µ ν) if there exists a D-sequence (a i,j ) such that, whenever (C k ) is a sequence from F, satisfying lim ν(c k ) = 0, for every φ Φ an integer k can be found, such that µ(c k ) a i,φ(i), for all k k. In case ν is fixed, and (µ n ) is a sequence of finitely additive measures on F, uniform absolute continuity of (µ n ) with respect to ν can be defined in a similar way, but clearly the integer k must be independent of n. If a finitely additive measure µ is absolutely continuous with respect to some realvalued positive measure ν, it s clear that µ is s-bounded in any case, and σ-additive

13 Uniform s-boundedness and convergence results 13 if ν is. Moreover, if (µ n ) is a sequence of finitely additive measures on F, uniformly absolutely continuous with respect to ν, then the measures (µ n ) are uniformly s- bounded. We shall show later a converse to this statement. In the next proposition an equivalent formulation of absolute continuity is introduced: only one implication is stated, because the reverse one is trivial. Proposition 4.2 Let µ be any finitely additive measure on F and assume µ is absolutely continuous with respect to some finitely additive measure ν : F IR 0 +. Then, there exists a regulator (a i,j ) such that, for every φ Φ a positive number δ can be found, satisfying the following implication: A F, ν(a) δ = µ(a) a i,φ(i). (8) Proof. Let (a i,j ) be the regulator related to absolute continuity, and let s see that it works for the assertion. By contradiction, if this is not the case, there exists a mapping φ Φ such that for every integer k a set C k can be found in F, with ν(c k ) k 1, and satisfying µ(c k ) a i,φ(i). Of course, this contradicts absolute continuity, hence the proof is finished. In the σ-additive case, a further definition is available. Proposition 4.3 Assume that µ : A R and ν : A IR + 0 µ ν if and only if are σ-additive. Then ν(a) = 0 = µ(a) = 0 (9) holds, for A A. Proof. Using weak σ-distributivity of R, we see easily that absolute continuity of µ with respect to ν entails the implication (9). Conversely, assume that (9) holds, and let (a i,j ) be the regulator involved in Definition 2.8. We shall prove that (a i,j ) works for absolute continuity. Indeed, let (C k ) be any sequence in A, with lim k ν(c k ) = 0, and assume by contradiction that there exists φ Φ such that v(µ)(c k ) a i,φ(i) for infinitely many integers k. By taking a subsequence if necessary, we may assume that v(µ)(c k ) a i,φ(i) for all k, and k=1 ν(c k ) <. Now, set B k = j=k C j,

14 14 A. Boccuto--D. Candeloro and B = k IN B k. It s clear that lim k ν(b k ) = 0, hence ν(b) = 0. Therefore µ(b) = 0, by (9). As the sequence (B k \ B) decreases to, there exists k such that v(µ)(b k \ B) = v(µ)(b k ) a i,φ(i) for all k k. As C k B k, this is impossible. One of the main connections between s-boundedness and absolute continuity states that uniform s-boundedness of a family of measures (µ n ), each absolutely continuous with respect to a fixed measure ν, implies uniform absolute continuity. We shall prove this result, first in the σ-additive case. Later, a more general result will be proved, by using an extension theorem. Theorem 4.4 Let (µ n ) be any sequence of uniformly s-bounded measures, defined on a σ-algebra A, each absolutely continuous with respect to a measure ν : A IR + 0. Then the measures µ n are uniformly absolutely continuous with respect to ν. Proof. Without loss of generality, we can assume that the measures µ n are positive. Now, let (a i,j ) be the D-sequence related with uniform s-boundedness, and let s show that (a i,j ) works for uniform absolute continuity. If we deny the assertion, there exist a sequence (C k ) in A, a mapping φ Φ, and an increasing sequence of integers (n k ), such that ν(c k ) < 2 k for each k, while µ nk (C k ) a i,φ(i) (10) for each k. Setting A k = C j, we see that ν(a k ) 2 1 k, and therefore j=k D lim k µ n (A k ) = 0 for all n. Thanks to Proposition 3.4 and taking into account of weak σ-distributivity of R, there exists an integer k 0 such that sup µ n (C k ) n IN for all k k 0, thus contradicting (10). a i,φ(i)

15 Uniform s-boundedness and convergence results 15 An important implication of absolute continuity can be found in the Carathéodory extension of a σ-additive measure. Indeed, if ν : F IR 0 + is any σ- additive measure on the algebra F, it s well-known that ν can be uniquely extended to a measure ν on the σ-algebra A(F) generated by F. In case µ : F R is absolutely continuous with respect to ν, we already observed that µ is countably additive too. Thus, we can obtain, and describe, the measure extension of µ to A(F)(See also [3] and [38]). This will be clarified in the sequel, where we shall assume (without loss of generality) that µ is positive. Definition 4.5 In the situation above, denote by F σ the family of all countable unions of (pairwise disjoint) members from F. Then, for all B F σ, B = n=1 F n, and assuming µ to be positive, we set n µ(b) = sup µ(f j ). n IN j=1 Moreover, µ is well-defined, because of the countable additivity of µ. Now, for any element A A(F), we set µ(a) = inf{ µ(b) : B F σ, A B}. Theorem 4.6 The function µ : A(F) R is a measure, whose restriction to Fcoincides with µ, and is absolutely continuous with respect to ν. Moreover, a regulator (a i,j ) exists, such that, for all A A(F), and each φ Φ an element F F exists, such that µ(a F ) a i,φ(i). Proof. As already observed, µ is well-defined: this also implies that µ(f ) = µ(f ) whenever F F. Now, as µ ν, there exists a regulator (a i,j ), such that, for every φ Φ there exists δ > 0 such that, whenever F F satisfies ν(f ) < δ, then µ(f ) a i,φ(i). Now, fix A A(F), with ν(a) < δ/2, and choose B F σ such that A B and ν(b) < δ. Writing B = n IN F n, where (F n ) is some increasing sequence in F, we see that ν(f n ) < δ for each n, and therefore µ(f n ) a i,φ(i)

16 16 A. Boccuto--D. Candeloro for all n. This implies that µ(b) a i,φ(i) and so µ(a) a i,φ(i). In this way we have proved that µ ν. Now, to prove the last assertion, let (a i,j ) be as above, and fix arbitrarily A in A(F). Fix also φ Φ, and choose δ corresponding to φ, in the condition of absolute continuity of µ with respect to ν: then, an element F F exists, such that ν(f A) < δ. By absolute continuity, we get µ(f A) a i,φ(i). We still have to prove that µ is σ-additive: this will be an immediate consequence of absolute continuity, as soon as we show that µ is finitely additive. To this end, let A 1 and A 2 be two disjoint members of A(F ), and let (a i,j ) be the regulator of absolute continuity of µ with respect to ν. Fix any mapping φ Φ, and let δ be the corresponding number related to absolute continuity. It s easy to check that we can find two disjoint members of F, say F 1 and F 2, such that ν(f l A l ) < δ/2 for l = 1, 2. Hence, if we set F = F 1 F 2, A = A 1 A 2, we get ν(f A) < δ, and so µ(f A) a i,φ(i). From this, it s easy to deduce that µ(a) ( µ(a 1 ) + µ(a 2 )) 3 a i,φ(i). As φ is arbitrary, thanks to weak σ-distributivity, we get additivity of µ, and the proof is finished. Theorem 4.6 becomes particularly meaningful when it can be applied to a sequence of uniformly s-bounded finitely additive measures. We recall that, as above, the symbol A(F) denotes the σ-algebra generated by the algebra F, while the symbol F σ denotes the family of all countable unions of sets from F. We note that J. D. M. Wright in [38] proved that a Riesz space R is weakly σ-distributive if and only if for every set G, for each algebra F and for every σ-additive (with respect to the order convergence) map m : F R there exists a R-valued σ-additive extension ν of m, defined on the σ-algebra A(F). However, in our context, σ-additivity is intended differently, that is, in terms of D-convergence, with respect to the same regulator. Theorem 4.7 Let (µ n ) be a uniformly bounded, uniformly s-bounded sequence of σ- additive R-valued measures on an algebra F, and assume that they are all absolutely

17 Uniform s-boundedness and convergence results 17 continuous with respect to a σ-additive non-negative measure ν : F IR. Then there exists a sequence (π n ) of measures on A(F ), each (π n ) extending µ n, uniformly absolutely continuous with respect to the extension ν, and a regulator (β i,j ) such that, for every A A(F ) and every φ Φ there exists an element F F satisfying for all n. v(π n )(F A) β i,ϕ(i) Proof. As in the previous case, we shall assume that the measures µ n are positive. For each n IN, let µ n be as in 4.6, and set π n = µ n. According to the hypotheses, applying 4.6 and 2.5, we see that there exists a regulator (a i,j ) such that: i) for each element φ Φ and each n IN there exists a real number δ > 0 such that for A A(F), and: ν(a) < δ = π n (A) a i,φ(i) ii) whenever (F k ) is a disjoint sequence in F, for any mapping φ Φ there exists k 0 such that for all n, and all k k 0. µ n (F k ) a i,φ(i) Again applying 2.5, we can deduce the existence of a further D-sequence (b i,j ) such that, for every φ Φ we get: where u = sup{µ n (Ω) : n IN }. ( ) u a i,φ(i+k) b i,φ(i), (11) k=1 We shall show that the D-sequence (2b i,j ) works for uniform s-boundedness of the measures π n. Proceeding by contradiction, let s assume that there exists a disjoint sequence (A k ) in A(F), and an element φ Φ such that π nk (A k ) 2 b i,φ(i) (12)

18 18 A. Boccuto--D. Candeloro for a suitable subsequence (n k ). Thanks to 4.6, for each k there exists an element F k F, satisfying the following condition: for all r = 1, 2,..., k. Now, set, for each k: π nr (F k A k ) a i,φ(i+k) F 1 = F 1, F 2 = F 2 \ F 1,..., F k = F k \ ( k 1 ) F r r=1 Clearly, the sets F k are pairwise disjoint, and belong to F. Moreover, by induction it s possible to prove that for each k. Thanks to (13) we see that,... k A k Fk (A r F r ) (13) r=1 ( k ) π nk (A k Fk ) = u π nk (A k Fk ) u a i,φ(i+r) r=1 and therefore, using also (11), we get π nk (A k ) µ nk (Fk ) + π nk (A k Fk ) µ nk (Fk ) + b i,φ(i) k. (14) Now, as the sets (F k ) are in F, and the measures µ n are uniformly s-bounded, there exists an integer k 0 such that µ n (Fk ) a i,φ(i) b i,φ(i) for all n and all k k 0. From (14) we deduce that π nk (A k ) 2 b i,φ(i) for all k k 0, which is impossible, by (12). So far, we have shown that the extensions π n are uniformly s-bounded in A(F). Now, we can apply Theorem 4.4 to the σ-additive measures π n, absolutely continuous with respect to ν: hence we deduce that the measures π n are uniformly absolutely continuous; finally, the last assertion follows from this result, by using the regulator (β i,j ) = (2b i,j ). We can state now an analogous result as 4.4 for finitely additive measures. k

19 Uniform s-boundedness and convergence results 19 Theorem 4.8 Let (µ n ) be a sequence of uniformly bounded, uniformly s-bounded R-valued finitely additive measures on an algebra F. If the measures µ n are absolutely continuous with respect to the same finitely additive measure ν : F IR 0 +, then they are uniformly absolutely continuous. Proof. Let S be the Stone space associated with F, i.e. a totally disconnected, Hausdorff compact space, such that the algebra G of its clopen subsets is algebraically isomorphic to F (see [32]). If we denote by u : G F such isomorphism, then it s possible to transfer the measures ν and µ n to G by setting: uν(g) = ν(u(g)), uµ n (G) = µ n (u(g)) whenever G G. By the mere definition, it s clear that all uµ n s are absolutely continuous with respect to uν, and that they are uniformly s-bounded. Moreover, by the particular structure of G, it turns out that uν, and therefore all uµ n s, are σ-additive. Thus, by applying Theorem 4.7, we deduce that the measures uµ n s are uniformly absolutely continuous with respect to uν. Coming back to F yields uniform absolute continuity of the original measures µ n with respect to ν. 5 The Vitali-Hahn-Saks-Nikodým theorem In this section we establish our main theorem, i.e. a version of the Vitali-Hahn- Saks-Nikodým theorem, (5.5), in the σ-additive case, and assuming RD-convergence (which will be introduced in a moment). Some consequences will be deduced from it, e.g. a Schur-type theorem, and also a finitely additive version of 5.5. Though the main concepts have already been introduced, we need some slight generalizations, in order to obtain more applicable results. Definition 5.1 Let µ : F R be any finitely additive bounded measure, on an algebra F. Assume now that a lattice G is given, contained in F. We define the G -semivariation of µ as the set function v G (µ) defined by: v G (µ)(a) = sup{ µ(g) : G G, G A}, A F.

20 20 A. Boccuto--D. Candeloro Moreover, we say that µ is G -s-bounded if there exists a regulator (a i,j ) in R such that, whenever (H k ) is any disjoint sequence from G, for all φ Φ there exists an integer k 0 such that µ(h k ) a i,φ(i) for all k k 0. Also a concept of uniform G -s-boundedness can be given, for a sequence of measures, in a similar way as in Definition 3.1, provided that v(µ n ) now is replaced by v G (µ n ). We note that, in some applications, F is often the family of all Borelian subsets of a topological space, and G the lattice of all open sets. Some of the theorems proved in this section will be useful in order to prove Dieudonné-type theorems, which will be done in a forthcoming paper. We now state the RD-convergence condition we shall be using below, for sequences of measures. Definition 5.2 Let (µ n ) be any sequence of finitely additive measures on an algebra F, and let G be any lattice in F. We say that the measures µ n are RD-convergent in G to a set function µ : G R, if there exists a regulator (a i,j ) such that, whenever A is a fixed element of G, and φ is any fixed element from Φ, an integer n 0 can be found, in such a way that µ n (A) µ(a) a i,φ(i), for all n n 0. In case G = F, RD-convergence in F will be simply called RD-convergence. One can see that RD-convergence is a pointwise convergence, but requiring that the regulator does not depend on the set. The concept of uniform RD-convergence is stated in the same fashion, but with n 0 independent of A. In case G = F, uniform D-convergence in F will be simply called uniform D-convergence. RD-convergence can be shown to be equivalent to a Cauchy condition: Proposition 5.3 The measures (µ n ) RD-converge in G to some µ if and only if there exists a regulator (a i,j ) such that, for every A G and every φ Φ there exists

21 Uniform s-boundedness and convergence results 21 an integer n 0 such that µ n (A) µ n+p (A) a i,φ(i), for all n n 0, and all p IN. Theorem 5.4 Let (µ n ) be a sequence of uniformly bounded σ-additive measures, defined on a σ-algebra A, and taking values in R. Let now G be any lattice in A, closed under countable disjoint unions. If the measures (µ n ) are RD-convergent in G, then they are uniformly G -s-bounded. Proof. Thanks to Lemma 2.5, we can formulate the property of σ-additivity of the measures µ n in the following way: there exists a regulator (a i,j ) such that, for every decreasing sequence (B h ) in A, B h, for every n IN, and every φ Φ it s possible to find an integer h 0 such that v(µ n )(B h0 ) a i,φ(i) (15) (Here and in the sequel, v = v A, and we will not say it explicitly). As to the RD-convergence, it can be formulated as a Cauchy property, i.e.: there exists a regulator (b i,j ) such that, for every A G, and every φ Φ, an integer n 0 can be found, such that µ n (A) µ n+p (A) b i,φ(i) (16) for all n n 0 and every p IN. We shall prove uniform s-boundedness in the following way: setting c i,j = a i,j b i,j, then (6c i,j ) is a regulator, and we will prove that for every disjoint sequence (H k ) in G, and for every φ Φ there exists k 0 IN such that sup v n IN G (µ n )(H k ) 6 c i,φ(i) (17) for all k k 0. Assume by contradiction that this is not the case. Then there exists a disjoint

22 22 A. Boccuto--D. Candeloro sequence (H r ) in G, and some φ Φ, such that sup v n IN G (µ n )(H r ) 6 c i,φ(i) (18) for all r IN. Before going on, let s denote by c the element c i,φ(i). As (18) holds for all r, we deduce that, corresponding to H 1 an integer n 1 can be found, such that v G (µ n1 )(H 1 ) 6c, and hence there exists A 1 G, with A 1 H 1, such that µ n1 (A 1 ) 6c. Now, applying RD-convergence, we find an integer n 1 > n 1 such that µ n (A 1 ) µ n+p (A 1 ) b i,φ(i) for all n n 1 and for all p IN. By the σ-additivity of µ 1, µ 2,..., µ n 1, we find an integer r 1 > 1 such that [v(µ 1 ) v(µ 2 )... v(µ n 1 )] H r r r 1 a i,φ(i). Now, as H r1 satisfies (18), there exists n 2 > n 1 such that v G (µ n2 )(H r1 ) 6c (we must have n 2 > n 1 because of the last formula). Then there exists A 2 G, with A 2 H r1, such that µ n2 (A 2 ) 6c. Denoting by G 2 the family of all unions of sets of the type A 1, A 2, (i.e. G 2 = {A 1, A 2, A 1 A 2 }), and applying RD-convergence, we find an integer n 2 > n 2 such that µ n (E) µ n+p (E) b i,φ(i) for all n n 2, for all p IN, and for all E G 2. Now, by applying the σ-additivity condition to µ 1, µ 2,..., µ n 2, we find an integer r 2 > r 1 such that [v(µ 1 ) v(µ 2 )... v(µ n 2 )] a i,φ(i). r r 2 H r Again, as H r2 satisfies (18), there exists n 3 > n 2 such that v G (µ n3 )(H r2 ) 6c, and hence there exists A 3 G, with A 3 H r2, such that µ n3 (A 3 ) 6c.

23 Uniform s-boundedness and convergence results 23 Proceeding in this way, we find sequences (r l ), (n l ), (n l), (A l ) such that for all l, n l+1 > n l > n l, r l+1 > r l, A l+1 H rl µ nl (A l ) 6c (19) for all l, [v(µ 1 ) v(µ 2 )... v(µ n l )] H r r r l a i,φ(i) (20) for all l, µ n (E) µ n+p (E) b i,φ(i) (21) n n l, p IN, E G l, where G l is the family of all unions of sets from {A 1,..., A l }. Now, set A = A l l=1 and let s show that the Cauchy condition (16) at this set leads to a contradiction. For every fixed l IN, we have µ n l (A) µ nl+1 (A) = = [µ n l (A 1 A 2... A l ) µ nl+1 (A 1 A 2... A l )]+ +[µ n l (A l+1 ) µ nl+1 (A l+1 )]+ + µ n l A s µ nl+1 A s. (22) s=l+2 s=l+2 Now, we have µ n l (A 1 A 2... A l ) µ nl+1 (A 1 A 2... A l ) b i,φ(i)

24 24 A. Boccuto--D. Candeloro because of (21), and µ n l (A l+1 ) v(µ n l ) H r r r l a i,φ(i), µ n A l s a i,φ(i), s=l+2 µ n l+1 A s a i,φ(i), s=l+2 each holding in view of (20). Now, the Cauchy condition at A gives an integer l 0 such that µ n l (A) µ nl+1 (A) c holds, for all l l 0. Isolating µ nl+1 (A l+1 ) from the equation (22), we get µ nl+1 (A l+1 ) 5c, for all l l 0, which contradicts (19). Corollary 5.5 Let (µ n ) be a sequence of uniformly bounded σ-additive measures, defined on a σ-algebra A, and taking values in R. If the measures (µ n ) are RDconvergent in A, then they are uniformly σ-additive, and the limit of the sequence is a σ-additive measure. Proof. Uniform s-boundedness follows from 5.4, taking A = G. Now, uniform σ-additivity is a consequence of 3.5. Finally, from this and weak σ-distributivity of R, σ-additivity of the limit follows easily. We now prove a Schur-type theorem. Corollary 5.6 Let Ω = IN, and A = P (IN ). Let (µ n ) be any sequence of uniformly bounded σ-additive measures, defined on A and taking values in R. Then RD-convergence of the measures µ n to some limit µ 0 implies uniform D-convergence to µ 0.

25 Uniform s-boundedness and convergence results 25 Proof. Thanks to 5.5, the measures µ n are uniformly σ-additive, and µ 0 is σ- additive. Hence, there exists a regulator (a i,j ) such that, as soon as (H k ) is a disjoint sequence in A, to each mapping φ Φ an integer k 0 corresponds, such that v(µ n ) H k a i,φ(i) k k 0 for all n IN {0}. Moreover, thanks to Lemma 2.5 there exists a regulator (b i,j ) such that, for every integer h and every φ Φ an integer n 0 corresponds, such that µ n ({q}) µ 0 ({q}) b i,φ(i) q h as soon as n n 0. Setting A i,j = a i,j b i,j, we shall prove that for each φ Φ there exists n such that µ n (F ) µ 0 (F ) 3 A i,φ(i) for all F A, and all n n. Indeed, fix φ Φ, and take H k = {k} for each k IN. Hence, by uniform σ-additivity, an integer k 0 corresponds, such that v(µ n )({k 0, k 0 + 1,...}) a i,φ(i) for any n IN {0}. Now, an integer n exists, such that µ n ({q}) µ 0 ({q}) b i,φ(i) q k 0 holds, as soon as n n. Thus, fixed arbitrarily F A, we have, for each n n : µ n (F ) µ 0 (F ) µ n (F {1,..., k 0 }) µ 0 (F {1,..., k 0 }) + +v(µ n )(F {k 0 + 1, k 0 + 2,...}) + v(µ 0 )(F {k 0 + 1, k 0 + 2,...}) b i,φ(i) + 2 a i,φ(i) 3 A i,φ(i). This concludes the proof. A finitely additive version of 5.4 will now be given: it s well-known that in general the result is false, when the domain of the measures is not a σ-algebra: so we shall

26 26 A. Boccuto--D. Candeloro assume that our measures, though finitely additive, are defined on a σ-algebra. However, we also need a further assumption, i.e. that our measures are absolutely continuous with respect to some finitely additive real-valued non-negative measure. Corollary 5.7 Let (µ n ) be a sequence of finitely additive measures, uniformly bounded, defined on a σ-algebra A, and RD-convergent to some limit µ 0. If the measures µ n are absolutely continuous with respect to a finitely additive measure ν : A IR 0 +, then they are uniformly s-bounded, and uniformly absolutely continuous. Proof. Thanks to 4.8, the last assertion follows from uniform s-boundedness, so we shall prove this only. Thanks to 2.5, we can see that there exists a regulator (a i,j ) such that, for each disjoint sequence (H k ) in A, every φ Φ and every n IN an integer k 0 exists, such that v(µ n )(H k ) a i,φ(i) for all k k 0. Moreover, thanks to RD-convergence, there exists a regulator (b i,j ) such that, for every A A, and every φ Φ, an integer n 0 can be found, such that µ n (A) µ 0 (A) b i,φ(i) for all n n 0. We shall prove uniform s-boundedness in the following way: setting c i,j = a i,j b i,j, then (6c i,j ) is a regulator, and we claim that it works for uniform s-boundedness. If this is not the case, then there exist: a disjoint sequence (H k ) in A, a mapping φ Φ, and a subsequence (µ nk ), such that µ nk (H k ) 6 c i,φ(i) (23) for all k. Now, there exists a subsequence (H kr ) such that ν is σ-additive in the σ-field B generated by the sets H kr (see [15]). This implies also that the measures µ n are σ-additive in B. Now, by using 5.5 we see that (6c i,j ) is a regulator for the uniform s-boundedness of the measures µ n in B, and this contradicts (23). This concludes the proof.

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