# Note on the Regularity of Nonadditive Measures

Authors: Toshikazu Watanabe; Syuji Yamada; Tamaki Tanaka
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We consider the regularity for nonadditive measures.
We prove that the non-additive measures which satisfy Egoroff's theorem and have pseudometric generating property possess Radon property (strong regularity) on a complete or a locally compact, separable metric space.


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## Body

## 1. Introduction

The relations of continuity and regularity of nonadditive measures are considered in several papers [1–4]. In [5], Li et al. investigated the regularity in nonadditive measures. They proved that the null-additive fuzzy measures possess a Radon property (strong regularity) on a complete metric space. In [6], Kawabe also investigated the regularity in fuzzy measures taking value in Riesz spaces. He proved that every weakly null-additive Riesz space valued fuzzy measure on a complete or a locally compact, separable metric space is Radon, provided that the Riesz space has the multiple Egoroff property.On the other hand Li and Mesiar [7] proved the regularity of nonadditive monotone measures. They proved that the equivalence condition of Egoroff’s theorem implies regularity for the nonadditive measures by using pseudometric generating property of a set function. For information on real valued nonadditive measures, see [8–10].In this paper, as notes, we prove that Egoroff’s theorem implies Radon property (strong regularity) for nonadditive measures which have pseudometric generating property on a complete or a locally compact, separable metric space.


## 2. Preliminaries

LetR be the set of real numbers and N the set of natural numbers. In what follows, let (X,ℱ) be a measurable space.Definition 1.
A set functionμ:ℱ→R is called a nonadditive measure if it satisfies the following two conditions:(1)
μ(∅)=0,(2)
ifA, B∈ℱ and A⊂B, then μ(A)≤μ(B).Definition 2.
Letμ:ℱ→R be a nonadditive measure. (1)
μ is said to be continuous from above if for any {An}⊂ℱ and A∈ℱ satisfying An↘A and there exists n0 with μ(An0)<∞ it holds that limn→∞μ(An)=μ(A).(2)
μ is said to be continuous from below if for any {An}⊂ℱ and A∈ℱ satisfying An↗A it holds that limn→∞μ(An)=μ(A).(3)
μ is said to be fuzzy measure if it is continuous from above and below.(4)
μ is said to be strongly order continuous if it is continuous from above at measurable sets of measure 0; that is, for any {An}⊂ℱ and A∈ℱ satisfying An↘A and μ(A)=0 it holds that limn→∞μ(An)=0.(5)
μ is said to be weakly null-additive if μ(A∪B)=0 whenever A,B∈ℱ and μ(A)=μ(B)=0.(6)
μ has property (S) if for any sequence {An}⊂ℱ with limn→∞μ(An)=0 there exists a subsequence {Ank} such that μ(∩i=1∞∪k=i∞Ank)=0; see [11].(7)
μ is said to be autocontinuous from above if limn→∞μ(A∪Bn)=μ(A) for each A∈ℱ and {Bn}⊂ℱ with limn→∞μ(Bn)=0.(8)
μ is said to be autocontinuous from below if limn→∞μ(A∖Bn)=μ(A) for each A∈ℱ and {Bn}⊂ℱ with limn→∞μ(Bn)=0.(9)
μ is said to be autocontinuous if it is autocontinuous from above and below.Definition 3.
Letμ:ℱ→R be a nonadditive measure. (1)
A double sequence{Am,n}⊂ℱ is said to be a μ-regulator if it satisfies the following two conditions:
(D1)
Am,n⊃Am,n′ whenever n≤n′,(D2)
μ(∪m=1∞∩n=1∞Am,n)=0.(2)
μ satisfies the Egoroff condition if for any μ-regulator {Am,n} and for every ɛ>0 there exists a sequence {nm} of natural numbers such that μ(∪m=1∞Am,nm)<ɛ.Remark 4.
A nonadditive measureμ satisfies the Egoroff condition if (and only if) for any double sequence {Am,n}⊂ℱ satisfying (D2) and the following (D1′) it holds that for every ɛ>0 there exists a sequence {nm} of natural numbers such that μ(∪m=1∞Am,nm)<ɛ:(D1′)
Am,n⊃Am′,n′ whenever m≥m′ and n≤n′.


## 3. Compact Measure and Regularity of Measure

In this section, we pick up several results for compact nonadditive measures and regularity of measures.Definition 5.
Letμ:ℱ→R be a nonadditive measure.(1)
A nonempty family𝒦 of subsets of X is called a compact system if for any sequence {Kn}⊂𝒦 with ∩n=1∞Kn=∅ there is n0∈N such that ∩n=1n0Kn=∅; see [12].(2)
We say thatμ is compact if there exists a compact system 𝒦 such that for each A∈ℱ there are sequences {Kn}⊂𝒦 and {Bn}⊂ℱ such that Bn⊂Kn⊂A for all n∈N and limn→∞μ(A∖Bn)=0.Remark 6.
(
1
) The family of all compact subsets of a Hausdorff space is a compact system.
(
2
) The family of all finite unions of sets in a compact system is also compact [13, Lemma 1.4]. Therefore, in (2) of the above definition, the compact system 𝒦 and the sequences {Kn}⊂𝒦 and {Bn}⊂ℱ may be chosen so that 𝒦 is closed for finite unions and both {Kn} and {Bn} are increasing.By [6, Theorem 1], the following result follows.Theorem 7.
Letμ:ℱ→R be a nonadditive measure. If μ is compact and autocontinuous, then it is continuous from above and below.Proof.
Sinceμ is compact and autocontinuous, by [6, Theorem 1], the assertion follows.In what follows, let(X,d) be a metric space. Denote by ℬ(X) the σ-field of all Borel subsets of X, that is, the σ-field generated by the open subsets of X. A nonadditive measure defined on ℬ(X) is called a nonadditive Borel measure on X.Definition 8.
μ is said to have pseudometric generating property if for each ɛ>0 there exists δ>0 such that for any A, B∈ℬ(X), μ(A)∨μ(B)<δ implies μ(A∪B)<ɛ.Proposition 9.
Ifμ satisfies pseudometric generating property, then it is weakly null-additive.Proof.
It is easy to see from the definition.Definition 10.
Letμ:ℬ(X)→R be a nonadditive Borel measure on X.
μ is called regular if for any ɛ>0 and A∈ℬ(X), there exist a closed set Fɛ and an open set Gɛ such that Fɛ⊂A⊂Gɛ and μ(Gɛ∖Fɛ)<ɛ.Li and Mesiar [7] also investigated the regularity on monotone measures. The following follows.Lemma 11.
LetX be a metric space and μ:ℬ(X)→R a nonadditive Borel measure on X. If μ has the Egoroff condition and pseudometric generating property, then μ is regular.Corollary 12.
LetX be a metric space and μ:ℬ(X)→R a nonadditive Borel measure on X. If μ has property (S), is strong order continuous, and is weakly null-additive, then μ is regular.By Li and Yasuda [14, Theorem 1], we also have the following.Corollary 13.
LetX be a metric space. If μ:ℬ(X)→R is weakly null-additive fuzzy Borel measure on X, then it is regular. Moreover if μ is null-additive, we have
(1)μ(A)=sup{μ(F)∣F⊂A,Fisclosedset}=inf{μ(G)∣G⊃A,Gisopenset}.Corollary13 above is a special case of [6, Theorem 5] and [15, Theorem 3].For more information on regularity of nonadditive measures, see [5, 6].


## 4.  Radon Measure

In this section, as main results, if we assume that a nonadditive Borel measure satisfies the equivalence condition of Egoroff’s theorem and pseudometric generating property on a complete or a locally compact, separable metric space, then it is Radon.Definition 14.
Letμ be a nonadditive Borel measure on X. (1)
μ is said to be Radon (strongly regular) if for each A∈ℬ(X) there are sequences {Kn}n∈N of compact sets and {Gn}n∈N of open sets such that Kn⊂A⊂Gn for all n∈N and limn→∞μ(Gn∖Kn)=0.(2)
μ is said to be tight if there is a sequence {Kn}n∈N of compact sets such that limn→∞μ(X∖Kn)=0.Remark 15.
Sequences of sets in the above definition may be chosen so that{Gn}n∈N is decreasing, while {Fn}n∈N and {Kn}n∈N are increasing.Proposition 16.
LetX be a Hausdorff space. Let μ be a nonadditive Borel measure on X which is weakly null-additive and strongly order continuous. Then, the following two conditions are equivalent:(i)
μis Radon (strongly regular),(ii)
μis regular and tight.Proof.
See [6, Proposition 2].It is known that every finite Borel measure on a complete or a locally compact, separable metric space is Radon; see [16, Theorem 3.2] and [17, Theorems 6 and 9, Chapter II, Part I]. Its counterpart in nonadditive measure theory can be found in [5, 9], which states that every Borel fuzzy measure on a complete separable metric space is tight, so that it is Radon if it is null-additive; see also [3, Theorem 2.3]. The following two theorems contain those previous results; see also [18, Theorem 12].Theorem 17.
LetX be a complete separable metric space and μ:ℬ(X)→R a nonadditive Borel measure on X. If μ is weakly null-additive and satisfies the Egoroff condition, then it is tight. Moreover, if μ has pseudometric generating property and satisfies the Egoroff condition, then it is Radon.To prove the theorem, we need the following; see [7, Proposition 3.7].Proposition 18.
Letμ:ℱ→R be a nonadditive measure. Then (i) implies (ii). (i)
μis weakly null-additive and satisfies the Egoroff condition.(ii)
For eachɛ>0and double sequence{Am,n}⊂ℱsatisfyingAm,n↘∅asn→∞for eachm∈N,there exists a sequence{nm}of natural numbers such thatμ(∪m=1∞Am,nm)<ɛ.Proof of Theorem17.
Sinceμ satisfies the Egoroff condition, by [19, Proposition 3], it is strongly order continuous. By Proposition 16 and Lemma 11, we have only to prove that μ is tight. Let {si}i∈N be a countable dense subset of X. For each m,i∈N, denote by Bm(si)¯ the closed ball with center si and radius 1/m. For each m,n∈N, put Am,n:=X∖∪i=1nBm(si)¯. Then, for any ɛ>0 and m∈N, we have Am,n↘∅, so that by Proposition 18, there exists a sequence {nm} of natural numbers such that
(2)μ(∪m=1∞Am,nm)<ɛ.
Put Pɛ:=∩m=1∞∪i=1nmBm(si)¯. Then, each Pɛ is closed and totally bounded, so that it is compact. Since X∖Pɛ=∪m=1∞Am,nm, it follows from (2) that μ(X∖Pɛ)<ɛ. Thus μ is tight.Corollary 19.
LetX be a complete separable metric space and μ:ℬ(X)→R a nonadditive Borel measure on X. If μ is weakly null-additive, strongly order continuous, and has property (S), then it is Radon.Proof.
It follows from Theorem17 since μ has pseudometric generating property [7, Proposition 5.1] and satisfies the Egoroff condition [19, Proposition 2].Corollary 20.
LetX be a complete separable metric space and μ:ℬ(X)→R a fuzzy measure on X. If μ is weakly null-additive, then it is Radon.Proof.
It follows from Theorem17 since μ satisfies the Egoroff condition [7, Proposition 3.1] and it is regular [14, Theorem 1].Remark 21.
Corollary20 above is a special case of [6, Theorem 5] and [15, Theorem 3].Theorem 22.
LetX be a locally compact, separable metric space and μ:ℬ(X)→R a nonadditive Borel measure on X. If μ is weakly null-additive and satisfies the Egoroff condition, then it is tight. Moreover, if μ has pseudometric generating property and satisfies Egoroff condition, then it is Radon.Proof.
By Lemma11 and Proposition 16, we have only to prove the tightness of μ. Denote by ℋ the family of all open and relatively compact subsets of X. The local compactness of X implies that ℋ is an open cover of X. Since X is strongly Lindelöf, that is, every open cover of any open subset of X has a countable subcover [17, Proposition 3 and Theorem 6, Chapter II, Part I], there is a sequence {Hm}m∈N⊂ℋ such that X=∪m=1∞Hm. Put Kn:=∪m=1nHm¯ for all n∈N, where A¯ denotes the closure of a set A. Then Kn is compact and X∖Kn↘∅. Since μ is strongly order continuous [19, Proposition 3], limn→∞μ(X∖Kn)=0. Thus μ is tight.Corollary 23.
LetX be a locally compact, separable metric space and μ:ℬ(X)→R a nonadditive Borel measure on X. If μ is weakly-null-additive, strongly order continuous, and has property (S), then μ is Radon.Corollary 24.
LetX be a locally compact, separable metric space and μ:ℬ(X)→R a fuzzy Borel measure on X. If μ is weaklly null-additive, then μ is Radon.Remark 25.
Corollary24 above is a special case of [6, Theorem 6] and [15, Theorem 4].



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