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Journal of Applied Mathematics
Volume 2012 (2012), Article ID 701206, 8 pages
http://dx.doi.org/10.1155/2012/701206
Research Article

On Decomposable Measures Induced by Metrics

1College of Mathematics and Physics, Chongqing University of Posts and Telecommunications, Nanan, Chongqing 400065, China
2School of Information Engineering, Guangdong Medical College, Guangdong, Dongguan 523808, China

Received 27 April 2012; Accepted 9 June 2012

Academic Editor: C. Conca

Copyright © 2012 Dong Qiu and Weiquan Zhang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We prove that for a given normalized compact metric space it can induce a 𝜎-max-superdecomposable measure, by constructing a Hausdorff pseudometric on its power set. We also prove that the restriction of this set function to the algebra of all measurable sets is a 𝜎-max-decomposable measure. Finally we conclude this paper with two open problems.

1. Introduction

The classical measure theory is one of the most important theories in mathematics, and it was extended, generalized, and deeply examined in many directions [1]. Nonadditive measure [2, 3] is an extension of the measure in the sense that the additivity of the measure is replaced with a weaker condition, the monotonicity. There are many kinds of nonadditive measures [1, 4]: the Choquet capacity, the decomposable measure [5, 6], the 𝜆-additive measure, the belief measure, the plausibility measure, and so fourth. Many important types of nonadditive measures occur in various branches of mathematics, such as potential theory [7], harmonic analysis, fractal geometry [8], functional analysis [9], the theory of nonlinear differential equations, and in optimization [1, 4, 10]. The Hausdorff distance introduced by Felix Hausdorff in the early 20th century as a way to measure the distance has many applications [8, 1113]. In this paper, we will give a method for inducing a 𝜎-max-superdecomposable measure from a given normalized compact metric space, by defining a Hausdorff pseudometric on the power set. Furthermore, we will prove that the restriction of the 𝜎-max-superdecomposable measure to the algebra of all measurable sets is a 𝜎-max-decomposable measure.

2. Preliminaries

Most notations and results on metric space and measure theory which are used in this paper can be found in [4, 14]. For simplicity, we consider only the normalized metric spaces (𝑋,𝑑), that is, diam𝑋=sup{𝑑(𝑥,𝑦)𝑥,𝑦𝑋}=1. But it is not difficult to generalize the results obtained in this paper to the bounded metric spaces. Let 𝑃(𝑋) be the space of all subsets of X. A distance function, called the Hausdorff distance, on 𝑃(𝑋) is defined as follows.

Definition 2.1 (see [14]). Let (𝑋,𝑑) be a normalized metric space, and let 𝐴 and 𝐵 be elements in 𝑃(𝑋).(i)If 𝑥𝑋, the “distance” from 𝑥 to 𝐵 is 𝑑(𝑥,𝐵)=𝑑(𝐵,𝑥)=inf𝑦𝐵{𝑑(𝑥,𝑦)}(2.1) with the convention (𝑥,)=1.(ii)The “distance” from 𝐴 to 𝐵 is 𝑑(𝐴,𝐵)=sup𝑥𝐴{𝑑(𝑥,𝐵)}(2.2) with the convention 𝑑(,𝐵)=0. (iii)The Hausdorff distance, (𝐴,𝐵), between 𝐴 and 𝐵 is (𝐴,𝐵)=max{𝑑(𝐴,𝐵),𝑑(𝐵,𝐴)}.(2.3)

A nonempty subset 𝑅 of 𝑃(𝑋) is called an algebra if for every 𝐸,𝐹𝑅, 𝐸𝐹𝑅 and 𝐸𝐶𝑅, where 𝐸𝐶 is the complement of 𝐸. A 𝜎-algebra is an algebra which is closed under the formation of countable unions [4].

Definition 2.2. Let 𝑅 be an algebra. A set function 𝜇𝑅[0,1] with 𝜇()=0 and 𝜇(𝑋)=1 is:(1)a max-decomposable measure, if and only if 𝜇(𝐴𝐵)=max{𝜇(𝐴),𝜇(𝐵)}, for each pair (𝐴,𝐵) of disjoint elements of 𝑅 (see [6]);(2)a 𝜎-max-decomposable measure, if and only if 𝜇𝑖𝐴𝑖𝜇𝐴=sup𝑖,𝑖(2.4) for each sequence (𝐴𝑖)𝑖 of disjoint elements of 𝑅 (see [6]);(3)a max-superdecomposable measure if and only if 𝜇(𝐴𝐵)max{𝜇(𝐴),𝜇(𝐵)};(4)a 𝜎-max-superdecomposable measure if and only if 𝜇𝑖𝐴𝑖𝜇𝐴sup𝑖.𝑖(2.5)

3. Main Results

Theorem 3.1. Let (𝑋,𝑑) be a normalized metric space. Then (𝑃(𝑋),) is a normalized pseudometric space.

Proof. It follows from Definition 2.1 that (,)=0 and (,𝐴)=1 for all nonempty subset 𝐴𝑃(𝑋). Then it is clear that (𝐴,𝐴)=0 and (𝐴,𝐵)=(𝐵,𝐴)1 for all 𝐴,𝐵𝑃(𝑋).
Let 𝐴,𝐵,𝐶𝑃(𝑋). If at least one of the three sets is empty, then one can easily prove the triangle inequality. Thus, without loss of generality, suppose that the three sets are not empty. For any three points 𝑥0𝐴, 𝑦0𝐵, and 𝑧0𝐶, we have that 𝑑𝑥0,𝑦0𝑦+𝑑0,𝑧0𝑥𝑑0,𝑧0,(3.1) which implies that 𝑑𝐴,𝑦0𝑦+𝑑0,𝑧0=inf𝑥𝐴𝑑𝑥,𝑦0𝑦+𝑑0,𝑧0inf𝑥𝐴𝑑𝑥,𝑧0=𝑑𝐴,𝑧0.(3.2) Consequently, we get that sup𝑦𝐵𝑑𝑦(𝐴,𝑦)+𝑑0,𝑧0𝑑𝐴,𝑦0𝑦+𝑑0,𝑧0𝑑𝐴,𝑧0.(3.3) By the arbitrariness of 𝑦0, we have that sup𝑦𝐵𝑑(𝐴,𝑦)+𝑑𝐵,𝑧0=sup𝑦𝐵𝑑(𝐴,𝑦)+inf𝑦0𝐵𝑑𝑦0,𝑧0𝑑𝐴,𝑧0.(3.4) Then we have that sup𝑦𝐵𝑑(𝐴,𝑦)+sup𝑧𝐶𝑑(𝐵,𝑧)sup𝑦𝐵𝑑(𝐴,𝑦)+𝑑𝐵,𝑧0𝑑𝐴,𝑧0,(3.5) which implies that 𝑑(𝐵,𝐴)+𝑑(𝐶,𝐵)𝑑(𝐶,𝐴).(3.6) Similarly, we can get that 𝑑(𝐵,𝐶)+𝑑(𝐴,𝐵)𝑑(𝐴,𝐶).(3.7) It follows that (𝐴,𝐵)+(𝐵,𝐶)=max{𝑑(𝐴,𝐵),𝑑(𝐵,𝐴)}+max{𝑑(𝐵,𝐶),𝑑(𝐶,𝐵)}max{𝑑(𝐴,𝐶),𝑑(𝐶,𝐴)}=(𝐴,𝐶).(3.8) We conclude that (𝑃(𝑋),) is a normalized pseudometric space.

Let 𝜇 be a normalized measure on an algebra 𝑅𝑃(𝑋) and 𝜇 be the outer measure induced by 𝜇. Let 𝜌𝑃(𝑋)×𝑃(𝑋)+ be defined by the equation 𝜌(𝐴,𝐵)=𝜇(𝐴Δ𝐵), where the symmetric difference of 𝐴 and 𝐵 is defined by 𝐴Δ𝐵=(𝐴𝐵𝐶)(𝐴𝐶𝐵). Then (𝑃(𝑋),𝜌) is a normalized pseudometric space and 𝜇(𝐴)=𝜌(𝐴,) for all 𝐴𝑃(𝑋) [15]. Now, we consider the converse of this process for the normalized pseudometric space (𝑃(𝑋),). Since (𝐴,)=1 for all nonempty subset 𝐴𝑃(𝑋), it would not get any nontrivial results if the set function 𝜇 is defined by 𝜇(𝐴)=(𝐴,). Thus, we give the following definition.

Definition 3.2. Let (𝑋,𝑑) be a normalized metric space. Now, we define a set function 𝜇 on 𝑃(𝑋) by 𝜇(𝐴)=1(𝑋,𝐴),(3.9) for all 𝐴𝑃(𝑋).

Theorem 3.3. Let (𝑋,𝑑) be a normalized metric space. Then the set function 𝜇 is a max-superdecomposable measure on 𝑃(𝑋).

Proof. It is easy to see 𝜇()=0 and 𝜇(𝑋)=1. Let 𝐴,𝐵𝑃(𝑋) with 𝐴𝐵. By the definition of 𝜇, we have that 𝜇(𝐴)=1maxsup𝑥𝑋𝑑(𝑥,𝐴),sup𝑦𝐴𝑑(𝑋,𝑦)=1sup𝑥𝑋inf𝑦𝐴𝑑(𝑥,𝑦)1sup𝑥𝑋inf𝑦𝐵𝑑(𝑥,𝑦)=1maxsup𝑥𝑋𝑑(𝑥,𝐵),sup𝑦𝐵𝑑(𝑋,𝑦)=𝜇(𝐵),(3.10) which shows the set function 𝜇 is monotonous. Thus, for any two sets 𝐴,𝐵𝑃(𝑋), we have 𝜇(𝐴𝐵)max{𝜇(𝐴),𝜇(𝐵)}.(3.11)

Theorem 3.4. Let (𝑋,𝑑) be a normalized metric space. Then the set function 𝜇 is a 𝜎-max-superdecomposable measure on 𝑃(𝑋).

Proof. Due to the monotonicity of 𝜇, for each sequence (𝐴𝑖)𝑖 of elements of 𝑃(𝑋) and every positive integer 𝑛, by mathematical induction we have that 𝜇𝑖𝐴𝑖𝜇𝐴max1𝐴,𝜇2𝐴,,𝜇𝑛,(3.12) which implies that 𝜇𝑖𝐴𝑖𝜇𝐴sup𝑖.𝑖(3.13)

Lemma 3.5. Let (𝑋,𝑑) be a normalized metric space. If (𝐴𝑖)𝑖 is an increasing sequence in 𝑃(𝑋) such that 𝑖=1𝐴𝑖=𝐴, then lim𝑖𝑑(𝑥,𝐴𝑖)=𝑑(𝑥,𝐴) for any point 𝑥𝑋.

Proof. Since 𝐴𝑖𝐴, it follows from Definition 2.1 that 𝑑(𝑥,𝐴𝑖)=inf𝑦𝐴𝑖𝑑(𝑥,𝑦)𝑑(𝑥,𝐴). If lim𝑖𝑑(𝑥,𝐴𝑖)=𝑎>𝑏=𝑑(𝑥,𝐴), then for the decreasing sequence (𝑑(𝑥,𝐴𝑖))𝑖, we have 𝑑(𝑥,𝑦)𝑎 for all 𝑦𝐴𝑖, 𝑖.
On the other hand, from 𝑑(𝑥,𝐴)=inf𝑦𝐴𝑑(𝑥,𝑦)=𝑏, it follows that there exists a point 𝑦0𝐴 such that 𝑑(𝑥,𝑦0)(𝑎+𝑏)/2. Since 𝑖=1𝐴𝑖=𝐴, there exists a positive integer 𝑖0 such that 𝑦0𝐴𝑖0. Thus we get that 𝑑(𝑥,𝑦0)𝑎 which contradicts 𝑑(𝑥,𝑦0)(𝑎+𝑏)/2. We conclude that lim𝑖𝑑(𝑥,𝐴𝑖)=𝑑(𝑥,𝐴) for any point 𝑥𝑋.

Lemma 3.6. Let (𝑋,𝑑) be a normalized compact metric space. If (𝐴𝑖)𝑖 is an increasing sequence in 𝑃(𝑋) such that 𝑖=1𝐴𝑖=𝐴, then lim𝑖(𝐴𝑖,𝐴)=0.

Proof. Since 𝐴𝑖𝐴, it follows from Definition 2.1 that 𝐴𝑖,𝐴=maxsup𝑥𝐴𝑖𝑑(𝑥,𝐴),sup𝑥𝐴𝑑𝑥,𝐴𝑖=sup𝑥𝐴𝑑𝑥,𝐴𝑖.(3.14) If lim𝑖(𝐴𝑖,𝐴)=𝑎>0, then for the decreasing sequence ((𝐴𝑖,𝐴))𝑖, we have (𝐴𝑖,𝐴)𝑎 for all 𝑖. Consequently there exists a point 𝑥𝑖𝐴 for each 𝐴𝑖 such that 𝑑(𝑥𝑖,𝐴𝑖)>𝑎/2. Since 𝑋 is a compact metric space, passing to subsequence if necessary, we may assume that the sequence (𝑥𝑖)𝑖 converges to a point 𝑥 in the closure of 𝐴 and lim𝑖𝑑(𝑥𝑖,𝐴𝑖)=𝑏𝑎/2. However since ||𝑑𝑥𝑖,𝐴𝑖||||𝑑𝑥𝑑(𝑥,𝐴)𝑖,𝐴𝑖𝑑𝑥,𝐴𝑖||+||𝑑𝑥,𝐴𝑖||𝑥𝑑(𝑥,𝐴)𝑑𝑖+||𝑑,𝑥𝑥,𝐴𝑖||,𝑑(𝑥,𝐴)(3.15) it follows from Lemma 3.5 and lim𝑖𝑑(𝑥𝑖,𝑥)=0 that lim𝑖𝑑𝑥𝑖,𝐴𝑖=𝑑(𝑥,𝐴)=0.(3.16) This is a contradiction. Thus we have lim𝑖(𝐴𝑖,𝐴)=0.

Lemma 3.7. Let (𝑋,𝑑) be a normalized compact metric space. If (𝐴𝑖)𝑖 is an increasing sequence in 𝑃(𝑋) such that 𝑖=1𝐴𝑖=𝐴, then 𝜇 is continuous from below, that is, lim𝑖𝜇(𝐴𝑖)=𝜇(𝐴).

Proof. By the definition of 𝜇, we have that ||𝜇𝐴𝑖||=||𝐴𝜇(𝐴)𝑖||𝐴,𝑋(𝐴,𝑋)𝑖,𝐴,(3.17) for all 𝑖. By Lemma 3.6, we have that lim𝑖𝜇(𝐴𝑖)=𝜇(𝐴).

Definition 3.8. A set 𝐸 in 𝑃(𝑋) is 𝜇-measurable if, for every set 𝐴 in 𝑃(𝑋), 𝜇𝜇(𝐴)=max(𝐴𝐸),𝜇𝐴𝐸𝐶.(3.18)

Theorem 3.9. If 𝕊 is the class of all 𝜇-measurable sets, then 𝕊 is an algebra.

Proof. It is easy to see that ,𝑋𝕊, and that if 𝐸𝕊 then 𝐸𝐶𝕊. Let 𝐸,𝐹𝕊 and 𝐴𝑃(𝑋). It follows that 𝜇𝜇(𝐴(𝐸𝐹))=max(𝐴(𝐸𝐹)𝐹),𝜇𝐴(𝐸𝐹)𝐹𝐶=max𝜇(𝐴𝐹),𝜇𝐴𝐸𝐹𝐶,(3.19) which implies that max𝜇(𝐴(𝐸𝐹)),𝜇𝐴(𝐸𝐹)𝐶=max𝜇(𝐴𝐹),𝜇𝐴𝐸𝐹𝐶,𝜇𝐴(𝐸𝐹)𝐶=max𝜇(𝐴𝐹),𝜇𝐴𝐹𝐶𝐸,𝜇𝐴𝐹𝐶𝐸𝐶=max𝜇(𝐴𝐹),𝜇𝐴𝐹𝐶=𝜇(𝐴).(3.20) Thus, 𝕊 is closed under the formation of union.

Theorem 3.10. The restriction of set function 𝜇 to 𝕊, 𝜇|𝕊, is a 𝜎-max-decomposable measure.

Proof. Let 𝐸1,𝐸2 be two disjoint sets in 𝕊. It follows that 𝜇𝐸1𝐸2𝜇𝐸=max1𝐸2𝐸1𝐸,𝜇1𝐸2𝐸𝐶1𝜇𝐸=max1𝐸,𝜇2.(3.21) Let {𝐸𝑖}𝑖=1 be a disjoint sequence set in 𝕊 with 𝑖=1𝐸𝑖=𝐸𝕊. By mathematical induction, we can get that 𝜇𝑛𝑖=1𝐸𝑖𝜇𝐴=max1𝐴,𝜇2𝐴,,𝜇𝑛(3.22) for every positive integer 𝑛. Since 𝜇 is continuous from below and lim𝑛𝑛𝑖=1𝐸𝑖=𝐸, we have 𝜇(𝐸)=lim𝑛𝜇𝑛𝑖=1𝐸𝑖=lim𝑛𝜇𝐸max1𝐸,𝜇2𝐸,,𝜇𝑛𝜇𝐸=sup𝑖,𝑖(3.23) which implies that 𝜇|𝕊 is a 𝜎-max-decomposable measure.

4. Concluding Remarks

For any given normalized compact metric space, we have proved that it can induce a 𝜎-max-superdecomposable measure, by constructing a Hausdorff pseudometric on its power set. We have also proved that the restriction of the set function to the algebra of all measurable sets is a 𝜎-max-decomposable measure. However, the following problems remain open.

Problem 1. Is 𝜇 a 𝜎-subadditive measure on 𝑃(𝑋)?

Problem 2. Is the class of all 𝜇-measurable sets a 𝜎-algebra?

Acknowledgments

The authors thank the anonymous reviewers for their valuable comments. This work was supported by The Mathematical Tianyuan Foundation of China (Grant no. 11126087) and The Science and Technology Research Program of Chongqing Municipal Educational Committee (Grant no. KJ100518).

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