Nonstandard Methods in Measure Theory

Abstract

Ideas and techniques from standard and nonstandard theories of measure spaces and Banach spaces are brought together to give a new approach to the study of the extension of vector measures. Applications of our results lead to simple new proofs for theorems of classical measure theory. The novelty lies in the use of the principle of extension by continuity (for which we give a nonstandard proof) to obtain in an unified way some notable theorems which have been obtained by Fox, Brooks, Ohba, Diestel, and others. The methods of proof are quite different from those used by previous authors, and most of them are realized by means of nonstandard analysis.

Dedicated to Professor Solomon Marcus in honour of the 90th anniversary of his birthday

1. Introduction

Let be a nonempty fixed set and a real-valued positive measure on a ring of subsets of ; the measure is assumed to be countably additive in the sense that if is a sequence of disjoint members of and if is also in , then . A fundamental problem in measure theory is that of finding conditions under which a countably additive measure on a ring can be extended to a countably additive measure on a wider class of sets containing . This problem is essentially solved by the Caratheodory process of generating an outer measure and taking the family of -measurable sets (see [1]); then the original measure can be extended to a which contains the generated by .

Suppose instead that is no longer real-valued, but it is a set function on taking values in a Banach space . If is countably additive in the above sense, in what circumstances is it still possible to extend to ? An obvious necessary condition is that should be bounded over ; that is, should be finite; for if the extension onto exists, then as a finite-valued measure on a sigma ring is well known to be bounded over its domain [2, ], so that is a fortiori bounded over . So, if is a bounded set function on a ring with values in a Banach space, what are the possibilities to obtain an extension?

One of the simplest methods is to consider the family of signed measure on obtaining from each element of the topological dual by the following real-valued mapping on , . Such set functions are bounded and countably additive over and as such can be subjected to the Jordan decomposition. Thus, the problem is reduced to the Caratheodory procedure, and the extension of is defined in terms of the elements of with the following properties: takes values in the algebraic dual of and is countably additive over in the weak topology. So, the extension problem reduces therefore to finding circumstances in which the range of is (identifying with its natural embedding into ). Whenever this is the case, for instance, in the case of reflexive spaces, the set function is countably additive in view of Pettis’ theorem [2]. This result was obtained for the first time by Fox [3].

Since vector-valued measures are important tools in integral representation as well as disintegration of measures, many authors have considered the extension problem when the range of is contained in a vector space . In the literature there are two main approaches to proving theorems concerning the extension of vector measures.

The first approach is due to the properties which have to be satisfied by the measure which follows to be extended. If is a Banach space, we can mention the solutions due to Gǎinǎ [4] (in the case when has finite variation), Dinculeanu [5, 6] (if is regular and of bounded variation), Arsene and Strǎtilǎ [7] (when is bounded above in norm by a positive measure), Dinculeanu and Kluvanek [8] (in the case when is absolutely continuous with respect to a positive measure), and Fox [9] ( satisfies a monotone-convergence condition).

The second approach relies on the conditions which have to be satisfied by the range of . In this category we have the results of Fox [3], Kluvanek [10], Ohba [11], or Gould [12]. Generalizing the notion of outer measure to vector-valued measures and imitating the -measurability procedure in order to obtain a Lebesgue extension of , Gould [12] showed that a necessary and sufficient condition for to have a Lebesgue extension is that the following property should hold for the image space of .

If is a sequence in whose norms have a positive lower bound, then there exists for arbitrary positive a finite subsequence such that .

This suggests a connection between weak completeness and property , and it is shown in [12] that all weakly complete spaces satisfy property . In Section 3 we present another proof of this result. It is easy to verify that does not hold property . Thus, is not weakly complete. More generally, Banach spaces which are infinite-dimensional function spaces with a supremum norm fail to satisfy property and therefore are not weakly complete. For Hilbert spaces property is satisfied. A direct proof of the fact that property is satisfied by is much harder.

For a masterful study of measures with values in a topological group we refer the reader to Sion [13] or Drewnowski [14–16]. When is a commutative complete topological group and is of bounded variation, then there is a very nice extension theorem by Takahashi [17].

The starting point in nonstandard theory of measure spaces is a paper [18] by Loeb. He gave a way to construct new rich standard measure spaces from internal measure spaces. This construction has been used in recent years to establish new standard results in a variety of different areas. Some of these results can be found in [19] or [20].

Also, nonstandard analysis has proved to be a natural framework for studying vector measures and Banach spaces. The central construction in this approach is the notion of nonstandard hull introduced by Luxemburg [21]. This notion is not only a useful tool in studying vector measures and Banach spaces, but also a construction arising naturally throughout nonstandard analysis. For a deeper discussion of nonstandard hulls and their applications we refer the reader to the survey paper [22] of Henson and Moore.

Živaljević [23] has pursued the extension problem using the nonstandard hull of . Osswald and Sun [24] treated the same problem from a different point of view; the extension of additive vector measures has been made using the internal control measures. Furthermore, the authors present a different approach of Loeb’s [18] in order to construct a countably additive vector measure from internal, locally convex space-valued measure.

In this work we study the extension of vector valued set functions in the framework of nonstandard analysis.

The plan of this paper is as follows.

Section 2 is devoted to some preliminary results on standard vector measures. Using concurrent relations, we also obtain a result concerning the concentration of -bounded vector measures on a specific set from the nonstandard extension of . Moreover, the principle of extension by continuity [2] will be proved using nonstandard techniques.

In Section 3 we give a nonstandard characterization of the absolute continuity and an extension theorem for vector measures. The proof still uses nonstandard arguments. Since reflexive spaces are weakly complete and the weakly complete spaces satisfy property , we can rederive the Fox’s result [3]. Some results of Gould [12] will be reproved in a different manner. For this, we use a result of Diestel et al. (see [25] or [26]) on -bounded measures. To obtain these results, Gould has used Pettis’ theorem. Our approach does not use this result.

In Section 4 we address the issues of the existing control measures. We tackle this subject by using the extension of set functions and linking these extensions to control measures.

Section 5 deals with the extension of set functions with finite semivariation. The results in this section (for which we present a nonstandard proof) were originally obtained by Lewis [27].

The last section shows that the extension of set functions with finite variation is a particular case of the extension of set functions with finite semivariation.

We adopt the nonstandard framework of [28]. The nonstandard model used in this paper is assumed to be sufficiently saturated for our needs. In what follows, denotes the set , where is the extension of in our model.

2. Preliminaries

In this section we collect some basic definitions, notations, and elementary results about standard vector measures. Also, the principle of extension by continuity will be proved using nonstandard techniques.

The terminology concerning families of sets, set functions, and so forth, will be, in general, that of [2] or [1]. Let denote the nonnegative reals, and let denote the set of positive integers. Sets are denoted as ; means the empty set. Notation for set operations is that commonly used, in particular means . Everywhere in the sequel denotes a ring of subsets of a nonempty fixed set ; the cases should be a -ring or -ring will be explicitly specified. The complement (in ) of a set is denoted by . Symbols and for sequences of sets or of reals have their usual meaning. A set function will be called a submeasure (subadditive measure in the terminology of Orlicz [29]) if , whenever and , if and . The ring is an abelian group with respect to the symmetric difference operation and each submeasure generates a semimetric on the group by the Frechet-Nikodym ecart . This semimetric is invariant in the sense that for sets . Therefore, any submeasure generates a topology on the group , and a base of neighborhoods is given by the family of sets . In the semimetric space the set operations are continuous [5] (we also present a nonstandard proof of this result). Requiring from a topology in a ring to possess this property one obtains the topological ring of sets, a natural generalization of the so-called spaces of measurable sets, introduced by M. Fréchet and O. Nikodym, in which a distance between two sets is defined as the measure of their symmetric difference.

Let be a Banach space, and let be a set function from to . We say that is a finitely additive vector measure, or simply a vector measure, if whenever and are disjoint members of then . If, in addition, in the norm topology of for all disjoint sequences of members of such that , then is termed a countably additive vector measure or simply, is countably additive. We use the terminology from [30] and call a set function from to strongly bounded (often abbreviated -bounded) if , whenever is a disjoint sequence. We say that is order continuous () if, for each sequence such that , we have . We will denote by the class of all subsets of which have the property that for every . Then, forms an ideal in , and is an algebra. We will denote by the power set of , and for any we put .

If is a vector measure from to , we call a vector measure space. The inner quasi-variation of an arbitrary subset of is defined by ; a set is -bounded if is finite [12]. If is finite, then will be called a bounded vector measure. Clearly, the inner quasi-variation is a submeasure on .

Abstraction of the condition of strong boundedness on a ring is the concept of the Rickart submeasure. Thus, a sequence is called dominated if there exists a set such that , for . The submeasure is said to be Rickart on the ring if for each dominated, disjoint sequence , we have . Note that every finite additive submeasure on the ring is Rickart.

An extensive research of topological rings of sets generated by Rickart families of submeasures, with applications to vector measures, was initiated by Oberle [31] and developed by Bogdanowicz and Oberle [32]. The topological point of view was realized by Drewnowski [14–16].

On the other hand, essential properties of finite or countably additive vector measures are reflected on the properties of corresponding submeasures. This enables us to use submeasures as a convenient tool in various questions concerning vector measures. For instance, Walker [33] has used the corresponding submeasures to study uniform sigma additivity or equicontinuity.

Throughout this section we assume that is countably additive and -bounded. Then, we know that is a bounded vector measure, and is -bounded. Furthermore, for any sequences of members of we have , so the countable additivity of implies order continuity of the submeasure [14–16, 26, 34]. For any set , is a directed set, where if and only if for . Then, the generalized sequence is a Cauchy net. By the completeness of we put for any subset of the outer quasi-variation of given by . Consequently, there exists a unique set function such that for every set we have . We may refer to and as the inner and the outer measures generated by .

Since is bounded we can define on the ecart function , where is the symmetric difference of and . Then is a semimetric space. We denote by the closure of in the space ; if is a topological space and , the closure of in is denoted by . For a subset of and a positive integer, let . If is a neighbourhood of in we put .

The below proposition is straightforward, so we omit its proof.

Definition 1. A ring is called -dense if for any subset of and for any positive real number , there is , such that .

Definition 2. The sets of will be called equivalent if , where is the ecart function .

Proposition 3. Let be a vector measure. If is -bounded, then is -dense.

Lemma 4. (i) Let subsets of . Then and .
(ii) For subset of we have and on .
(iii) Let and subset of . Then .
(iv) If and then .
(v) If and then .
(vi) If then .

Lemma 5. (i) If then .
(ii) If , then .

The above properties of the inner and the outer measures generated by are well known. Now we are going to prove a result which is needed in the sequel.

Theorem 6. (i)    is a sigma algebra and .
(ii) If and then .
(iii) If and then .

Proof. (i) Let . There is a set such that . Then . Since is -dense, there is such that and . But , because of .
It is easy to see that . Hence , so .
Let now . There are such that , . We have , , so . Therefore. is algebra.
Suppose is a sequence of elements of and set . It remains to prove that . Since is algebra we can take . Choose and neighbourhoods of zero in such that .
Let be a sequence of neighbourhoods of zero in so that for all . For every we can choose such that . If , a trivial verification shows that . By Lemma 5(i) we have whence . Setting , we have , and . Lemma 5(i) implies so . Moreover, , and by Lemma 4(ii) according to . It follows that for large enough. We observe that for each and, on account of Lemma 5(i), we conclude that Thus, for large enough, , so .
Let and fix . Proposition 3 now gives so that and . According to Lemma 4(ii), since we get Thus, .
(ii) If and , we verify immediately that Lemma 5(i) implies From this we deduce, for large enough, that , so as .
(iii) From we see that . According to (i) and (ii), is a sequence of elements of and as . We remark at once that , and the proof is complete.

Because the main tool in our approach is the principle of extension by continuity, we give a nonstandard proof of it. For a standard proof we refer the reader to [2].

Theorem 7 (principle of extension by continuity). Let and be metric spaces, and let be complete. If is a dense subset of and is uniformly continuous, then has a unique uniformly continuous extension .

Proof. Let , then there is a with . Moreover, there is a sequence of points of with . For each we have . But is uniformly continuous on , so implies . For a positive real number we put . Then is internal by the definition principle, and each belongs to . In particular, for some we have . Hence is a prenearstandard point, and so is nearstandard. Let , where is the standard part map. Then is obviously well defined and extends . To verify that is uniformly continuous, let be an arbitrary positive real number. The assumption implies that there is a positive real number such that By the transfer principle, Let with . There are such that , . Then , so . Hence . By the definition of we have , . Since we also have we get that , whence is uniformly continuous.
Let be another function with the same properties, and . Then there is with . Since , are uniformly continuous we have and . By the transfer principle , so that . As and are standard, it follows that , and the proof is complete.

The section closes with an outcome about the concentration of -bounded measures on a set of . By we denote the set of all -bounded additive measures from to .

Proposition 8. There exists a set of so that for all of , disjoint of and for every belonging to , we have .

Proof. Let be a relation on defined by if and only if for all , disjoint of we have . We see at once that is , which is clear from Proposition 3. We verify that is a concurrent relation. Indeed, if , there exists such that for all , disjoint of we have , . Setting , we have and for all , which is our assertion. By the concurrence theorem [35], there is a set such that for all . Fix . Then, for all , disjoint of we have . Let us regard as ran and the proof is complete.

3. Extension of a Vector Valued Measure

We adopt here the main framework of nonstandard analysis from [28]. We also give the definition, nonstandard formulation, and some of the basic properties of the absolutely continuous concept. In this section denotes a submeasure from to and stands for the Frechet-Nikodym ecart associated with . We have seen in Section 2 that is a semimetric space. We recall that the sets of are equivalent if .

Definition 9. A vector measure is called absolutely continuous with respect to , or simply -continuous, if for any positive real number there is a positive real number , such that for any , if . In this case we say that is a control submeasure of , and we denote that by .

A nonstandard formulation of absolutely continuous may be stated as follows.

Lemma 10. A vector measure on is absolutely continuous with respect to , if and only if for all with we have in the monad of zero in .

Proof. Assume is a control submeasure of and let be an arbitrary positive real number. If is some element in the internal algebra with , we have by the transfer principle that . Since is arbitrary, we obtain .
To prove the converse, fix with , and for a given positive real number , let be an infinitesimal such that . Then, for any with we get . Thus, we have shown Now apply the transfer principle to obtain the desired condition for absolute continuity.

Remark 11. As we have mentioned the set operations are continuous [5] in the semimetric space . Here a nonstandard proof is given.

Lemma 12. The maps defined by the equalities are uniformly continuous.

Proof. We denote by one of the operations . For every set of we have . Hence, If , , then , . Thus, we conclude , and the set operations are uniformly continuous.

Lemma 13. A vector measure on is absolutely continuous with respect to , if and only if is uniformly continuous on .

Proof. Suppose and . By Lemma 10   and . But is a vector measure, so and . Therefore, , which means that is uniformly continuous.
Conversely, let with . In this case , and by uniform continuity of we have in the monad of zero in . Then, the result follows by Lemma 10.

Theorem 14. Let be a ring of subsets of such that and let be a vector measure on . Suppose is dense in and is absolutely continuous with respect to . Then has a unique vector measure extension . Furthermore, this extension is uniformly continuous on .

Proof. On account of Theorem 7, there is a unique uniformly continuous map which extends . It remains to prove that whenever and are disjoint members of then . By assumption, we can choose such that , . According to Lemma 12, we have , . Uniform continuity of implies , , and . The transfer principle leads to that is finitely additive on , and
Now
which is due to the fact that and are disjoint.
We conclude from (14) and (15) that which clearly forces This completes the proof.

Theorem 15. Let be a -bounded and countably additive vector measure. Then there exists a unique countably additive vector measure extending .

Proof. We claim that is absolutely continuous with respect to . Indeed, if and the transfer principle leads to . Since we have in the monad of zero in , which is our assertion. Theorem 14 shows that there exists a vector measure absolutely continuous with respect to which extends . What is left to prove is that is countably additive. Consider a sequence of elements of such that . Therefore, by Theorem 6(iii). Then for , . Therefore, applying Theorem 14 and Lemma 10, we have in the monad of zero in . Then as , which completes the proof.

The following corollary can be found in [11, Theorem 1] or [36].

Corollary 16. Let be a ring and the -ring generated by . A countably additive vector measure can be extended uniquely to a countably additive vector measure if and only if is -bounded.

The next two results were proved by Gould [12] and they are important consequences of Corollary 16. He has shown that in every weakly complete Banach space property holds [12, Theorem 3.1]. Even though our construction is adapted from [12], Proposition 17 yields an elegant proof of this result.

Proposition 17. Let be a weakly complete Banach space and a bounded set function from to . If is a countably additive vector measure, then has a unique countably additive extension .

Proof. We begin by proving that is -bounded. If were not -bounded, we would have that contains a subspace isomorphic to (see, for instance, [26, ]). Since every closed linear subspace of weakly complete Banach space is weakly complete and is not weakly complete [2, ], we obtain a contradiction. Now it suffices to apply Corollary 16, and the proof is complete.

Remark 18. The reader should observe the similarity between this proposition and some of the criteria, established by Gould [12, Theorem 2.5, ], for the extension of set functions with values in weakly complete Banach spaces. It is worth noting that in the work of Gould [12] it suffices to require that be locally bounded over and the extension is made onto a family which is a -hereditary ring containing (if is a given family of sets, a ring is said to be -hereditary if every member of which is a subset of some member of is also a member of ). Our requirements are stronger, but the final conclusion is somewhat more general. Moreover, we do not use Pettis’ theorem for the proof.

We are now ready to prove [12, Theorem 3.1, ].

Proposition 19. If is a weakly complete Banach space then property holds.

Proof. Suppose that the proposition was false. Then we could find positive , , and a sequence in so that for all and for every finite subsequence of . Let be the ring of finite sets of positive integers, and let denote the set function taking each finite set into the vector . Clearly is a bounded vector measure from to . Furthermore, is countably additive, since there are no infinite disjoint nonempty sequences in whose union is in . Proposition 17 yields that there is a countably additive extension onto , which extends . In particular, the set of all the positive integers belongs to . Hence, , and the convergence of this series contradicts the hypothesis that for all .

For Banach spaces reflexivity and semireflexivity are equivalent, and either implies weak completeness. Thus, the rederivation of Fox’s theorem [3] is a consequence of Proposition 17 (see also [11, Corollary 2, ]).

Corollary 20. A bounded countably additive vector measure on a field, taking its values in a reflexive Banach space, extends uniquely to a countably additive vector measure on the generated sigma-field.

Theorem 21. Let be a Banach space for which property holds. Assume is a countably additive vector measure. Then has a countably additive extension if and only if is bounded.

Proof.
Necessity. Assume that has a countably additive extension . Then as a finite-valued measure on a -ring is bounded over its domain [2, , 4.5], so is bounded over .
To prove the sufficiency for such we verify that is -bounded. Indeed, if were not -bounded, we would have a sequence of disjoint members of and such that for all . By property for all positive there is a finite subsequence such that . This contradicts that is bounded, and the theorem follows.

4. Existence of Control Measure

In [8], the authors show that the Bartle-Dunford-Schwartz theorem [37] does not work if we replace the -ring by a ring and ask whether the result remains valid for -rings. The theorem states that for every countably additive measure defined on sigma algebra there exists a positive control measure such that if and only if , where is the semivariation of . By a counterexample, it is shown in [38] that this result could not work even if the measure is defined on -rings. So, we have to impose additional conditions for obtaining this goal. In [38], one also shows that the theorem works if the space is separable. Now, if we pass to the conditions imposed on the measure and no the space in which it has values, we will prove the following two Brooks’ results [36].

Theorem 22. Let be a countably additive vector measure. Then is -bounded if and only if there exists a positive countably additive bounded set function defined on such that

To prove this theorem Brooks uses Orlicz-Pettis theorem and two results of Porcelli [39] about some embedding theorems and their implications in weak convergence, respectively, compactness in the space of finitely additive measure. He also uses a result of Leader [40] from the theory of spaces for finitely additive measures. Brooks and Dinculeanu in [41], by extending a result of Dieudonné [42], prove the assertion for finitely additive and locally strongly additive measures. Traynor [43] gave an elementary proof of this result for strongly additive measures. Using this result, he shows that for strongly additive and countably additive measures on algebra, the existence of a finite control measure is equivalent to the relatively weak compactness of range of measures, which is equivalent to the existence of a countably additive extension on the sigma algebra generated by [11, 26].

There is some interest in the extension measure theoretic approach given here. The proof that we will give uses the extension of to -ring generated by , and then we apply the Bartle-Dunford-Schwartz theorem. Thus, we avoid some deep results in vector measures and unconditionally convergent series.

Proof of Theorem 22. First assume that is countably additive and -bounded. Theorem 15 gives a countably additive vector measure , which extends . We know that is algebra and (see Theorem 6(i)). By the Bartle-Dunford-Schwartz theorem [37], there exists a positive countably additive bounded set function on such that Define if , and as claimed.
For the converse, let be a sequence of pairwise disjoint members of . Since is bounded and countably additive there exists such that for all so is convergent. Then as , and implies as , which completes the proof.

Corollary 23. Let be countably additive. Then there is a countably additive bounded set function defined on which is a control measure for if and only if is -bounded.

The next result can be found in [11, Theorem 2] or in [26, Theorem 2, ].

Corollary 24. Let be a ring and the -ring generated by . Every countably additive vector measure can be extended uniquely to a vector measure if and only if one of the following conditions is satisfied:(i)there exists a positive bounded measure on such that (ii) is -bounded;(iii)the range of the vector measure is relatively weakly compact.

Proof. The equivalence of (i) and (ii) is given by Theorem 22.
To prove (ii) ⇒ (iii), we note by Corollary 16 that there is a countably additive vector measure , which extends . Applying the Bartle-Dunford-Schwartz theorem [37, Theorem 2.9], the set is relatively weakly compact, and so is the set .
The implication (iii) ⇒ (ii) is proved by Kluvánek in [44, Theorem 5.3].

5. Extension of Set Function with Finite Semivariation

Lewis in [27] used Caratheodory’s method about the extension of set functions to perform the extension of set functions with finite semivariation. While the circumstances are somewhat similar to the extension of set functions with finite variation, the techniques employed from there have carried over to that situation studied by Dinculeanu [5]. In this section we give a nonstandard proof of the central result of Lewis. This in turn is applied in Section 6 to achieve the extension of set functions with finite variation, so we can unify the extension of set function with finite semivariation with the extension of set function with finite variation.

Let be a ring of subsets of a universal space . For , Banach spaces we denote by the Banach space of all bounded linear operators , and let be a set function with finite semivariation. That is, if we assume that is finite, where we take the supremum over all finite subdivisions of which consist of elements of and all elements of unit norm in . Let be the hereditary -ring generated by . We use the semivariation of to define an outer measure on in the obvious way. Thus, for is where the infimum is taken over all countable -coverings of . Clearly is an outer measure on . Let be the set of all elements in so that if , we have By virtue of a well-known result, is a sigma ring [5]. will be the largest class of subsets of so that forms an ideal in ; that is, if for each we have . For define where the supremum is taken over all such . will be the -ring of all elements in with finite measure. The main result which will be proved by nonstandard means is the extension theorem of to a uniquely set function defined on with values in . A standard proof can be found in [27].

Definition 25. We say that a finitely additive set function is variationally semiregular provided that if is a decreasing sequence of sets in whose intersection is empty and , then . Halmos [1] says that is continuous from above at .

For , in , define to be . Then defines a semimetric on . In [27] the following two results are proved which we mention without proof.

Lemma 26. Suppose that is variationally semiregular with finite semivariation on . Let be the outer measure on . Then on .

Lemma 27. If , then is -dense in .

We now move on to the nonstandard proof of our problem.

Theorem 28. Let be variationally semiregular. If is with finite semivariation on and , then there exists a unique extension of such that satisfies the following conditions:(a) is countably additive on ;(b) on ;(c) has finite semivariation ;(d) extends .

Proof. (a) Let . Obviously, is finite and countably additive on . From this we obtain that if is a sequence of members of such that , then for all . By Lemma 27 the assumptions of Theorem 14 are satisfied, so there exists a unique finite additive extension of which is absolutely continuous with respect to . Then is uniformly continuous on (Lemma 13). Lemma 10 now leads to for all , which is our claim.
(b) Let . According to Lemma 27 there exists so that . It is well known that if is absolutely continuous with respect to , then is absolutely continuous with respect to . We conclude from Lemma 13 that is uniformly continuous, hence that . Since and agree on , and agree on . By using the transfer principle they agree on , so that . We also have that and agree on (Lemma 26), so a repeated application of the transfer principle enables us to write on . We get that and agree on , because they agree on and (apply again the transfer principle). It follows that and finally that . Now implies , so ; with both parts being standard this clearly forces on .
(c) On account of (b) we have that has finite semivariation on .
(d) Since is countably additive on , (b) we obtain that is countably additive on . As is countably additive on and we have by Lemma 26 that on . Thus, is countably additive on . Therefore, extends , both being countably additive on , respectively, and on .

6. Extension of Set Function with Finite Variation

Finally, we deal with the extension of set functions with finite variation. As mentioned in Section 1, we are concerned in this section with the study of how the extension of set functions with finite semivariation implies the extension of set functions with finite variation. We use the same notations as in the previous section, and we will reduce the problem to the previous case. It is known that for any finite additive set function with , we can choose the spaces and so that the semivariation of relative to these spaces is equal to the variation of . If is a normed space, we have a well-known connection between semivariation and variation [5].

Proposition 29. The semivariation of the set function , relatively to the spaces and , is equal to the variation of .

This result will be used in the next theorem. But first we make some additional observations related to the precedence theorems and lemmas. Note that if we choose so that is embedded in , not only the variation and the semivariation are equal but also .

Lemma 27 of Section 5 shows a density property of the ring in the for the topology induced by the semimetric . We now expand sigma additive vector measures with finite variation from the ring to which is wider than the generated by . Moreover, if the function is with finite variation and countably additive it is automatically variationally semiregular. Then, Theorem 28 leads to the following result [5].

Theorem 30. Let be a Banach space, a ring of sets, and a countably additive measure with finite variation. Then can be extended uniquely to a vector measure such that(a) is countably additive on ;(b) on ;(c) has finite variation;(d) extends .

Remark 31. In the construction presented by Lewis [27] the author uses a technique similar to that of the extension of set functions with finite variation. Basically, this technique carries over to that of Caratheodory process. Here, noting that variation and semivariation are equal in some particular cases, we could get the extension of set functions with finite variation as a particular case of the extension of set functions with finite semivariation. In addition to these results, nonstandard proofs of these classical measure theory results are found to be more intuitive and easier than the standard proofs.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The author wishes to express his gratitude to the anonymous referee for a number of helpful comments.