Journal of Applied Mathematics

Volume 2012, Article ID 701206, 8 pages

http://dx.doi.org/10.1155/2012/701206

## On Decomposable Measures Induced by Metrics

^{1}College of Mathematics and Physics, Chongqing University of Posts and Telecommunications, Nanan, Chongqing 400065, China^{2}School 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 -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 -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 -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, 11–13]. In this paper, we will give a method for inducing a -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 -max-superdecomposable measure to the algebra of all measurable sets is a -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 , that is, . But it is not difficult to generalize the results obtained in this paper to the bounded metric spaces. Let be the space of all subsets of X. A distance function, called the Hausdorff distance, on is defined as follows.

*Definition 2.1 (see [14]). * Let be a normalized metric space, and let and be elements in .(i)If , the “distance” from to is
with the convention .(ii)The “distance” from to is
with the convention . (iii)The Hausdorff distance, , between and is

A nonempty subset of is called an algebra if for every , and , where is the complement of . A -algebra is an algebra which is closed under the formation of countable unions [4].

*Definition 2.2. *Let be an algebra. A set function with and is:(1)a max-decomposable measure, if and only if , for each pair of disjoint elements of (see [6]);(2)a -max-decomposable measure, if and only if
for each sequence of disjoint elements of (see [6]);(3)a max-superdecomposable measure if and only if ;(4)a -max-superdecomposable measure if and only if

#### 3. Main Results

Theorem 3.1. *Let be a normalized metric space. Then is a normalized pseudometric space. *

* Proof. * It follows from Definition 2.1 that and for all nonempty subset . Then it is clear that and for all .

Let . 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 , , and , we have that
which implies that
Consequently, we get that
By the arbitrariness of , we have that
Then we have that
which implies that
Similarly, we can get that
It follows that
We conclude that is a normalized pseudometric space.

Let be a normalized measure on an algebra and be the outer measure induced by . Let be defined by the equation , where the symmetric difference of and is defined by . Then is a normalized pseudometric space and for all [15]. Now, we consider the converse of this process for the normalized pseudometric space . Since for all nonempty subset , it would not get any nontrivial results if the set function is defined by . Thus, we give the following definition.

*Definition 3.2. *Let be a normalized metric space. Now, we define a set function on by
for all .

Theorem 3.3. *Let be a normalized metric space. Then the set function is a max-superdecomposable measure on .*

* Proof. * It is easy to see and . Let with . By the definition of , we have that
which shows the set function is monotonous. Thus, for any two sets , we have

Theorem 3.4. *Let be a normalized metric space. Then the set function is a -max-superdecomposable measure on .*

* Proof. * Due to the monotonicity of , for each sequence of elements of and every positive integer , by mathematical induction we have that
which implies that

Lemma 3.5. *Let be a normalized metric space. If is an increasing sequence in such that , then for any point . *

* Proof. * Since , it follows from Definition 2.1 that . If , then for the decreasing sequence , we have for all , .

On the other hand, from , it follows that there exists a point such that . Since , there exists a positive integer such that . Thus we get that which contradicts . We conclude that for any point .

Lemma 3.6. *Let be a normalized compact metric space. If is an increasing sequence in such that , then .*

* Proof. * Since , it follows from Definition 2.1 that
If , then for the decreasing sequence , we have for all . Consequently there exists a point for each such that . Since is a compact metric space, passing to subsequence if necessary, we may assume that the sequence converges to a point in the closure of and . However since
it follows from Lemma 3.5 and that
This is a contradiction. Thus we have .

Lemma 3.7. *Let be a normalized compact metric space. If is an increasing sequence in such that , then is continuous from below, that is, .*

* Proof. * By the definition of , we have that
for all . By Lemma 3.6, we have that .

*Definition 3.8. *A set in is -measurable if, for every set in ,

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

* Proof. * It is easy to see that , and that if then . Let and . It follows that
which implies that
Thus, is closed under the formation of union.

Theorem 3.10. *The restriction of set function to , , is a -max-decomposable measure. *

* Proof. * Let be two disjoint sets in . It follows that
Let be a disjoint sequence set in with . By mathematical induction, we can get that
for every positive integer . Since is continuous from below and , we have
which implies that is a -max-decomposable measure.

#### 4. Concluding Remarks

For any given normalized compact metric space, we have proved that it can induce a -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 -max-decomposable measure. However, the following problems remain open.

*Problem 1. *Is a -subadditive measure on ?

*Problem 2. *Is the class of all -measurable sets a -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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