On Decomposable Measures Induced by Metrics

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, 11–13]. 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,𝑦∈𝑋}=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 𝑥∈𝑋, the “distance” from 𝑥 to 𝐵 is 𝑑(𝑥,𝐵)=𝑑(𝐵,𝑥)=inf_𝑦∈𝐵{𝑑(𝑥,𝑦)}(2.1) with the convention (𝑥,∅)=1.(ii)The “distance” from 𝐴 to 𝐵 is 𝑑(𝐴,𝐵)=sup_𝑥∈𝐴{𝑑(𝑥,𝐵)}(2.2) with the convention 𝑑(∅,𝐵)=0. (iii)The Hausdorff distance, ℎ(𝐴,𝐵), between 𝐴 and 𝐵 is ℎ(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,𝐹∈𝑅, 𝐸∪𝐹∈𝑅 and 𝐸^𝐶∈𝑅, where 𝐸^𝐶 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 𝜇∶𝑅→[0,1] with 𝜇(∅)=0 and 𝜇(𝑋)=1 is:(1)a max-decomposable measure, if and only if 𝜇(𝐴∪𝐵)=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 𝜇_𝑖∈ℕ𝐴_𝑖=sup𝜇𝐴^𝑖∶𝑖∈ℕ,(2.4) for each sequence (𝐴_𝑖)_𝑖∈ℕ of disjoint elements of 𝑅 (see [6]);(3)a max-superdecomposable measure if and only if 𝜇(𝐴∪𝐵)≥max{𝜇(𝐴),𝜇(𝐵)};(4)a 𝜎-max-superdecomposable measure if and only if 𝜇_𝑖∈ℕ𝐴_𝑖≥sup𝜇𝐴^𝑖∶𝑖∈ℕ.(2.5)

3. Main Results

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

Proof. It follows from Definition 2.1 that ℎ(∅,∅)=0 and ℎ(∅,𝐴)=1 for all nonempty subset 𝐴∈𝑃(𝑋). Then it is clear that ℎ(𝐴,𝐴)=0 and ℎ(𝐴,𝐵)=ℎ(𝐵,𝐴)≤1 for all 𝐴,𝐵∈𝑃(𝑋).
Let 𝐴,𝐵,𝐶∈𝑃(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 𝑥₀∈𝐴, 𝑦₀∈𝐵, and 𝑧₀∈𝐶, we have that 𝑑𝑥₀,𝑦₀+𝑑𝑦₀,𝑧₀≥𝑑𝑥₀,𝑧₀,(3.1) which implies that 𝑑𝐴,𝑦₀+𝑑𝑦₀,𝑧₀=inf_𝑥∈𝐴𝑑𝑥,𝑦₀+𝑑𝑦₀,𝑧₀≥inf_𝑥∈𝐴𝑑𝑥,𝑧₀=𝑑𝐴,𝑧₀.(3.2) Consequently, we get that sup_𝑦∈𝐵𝑑(𝐴,𝑦)+𝑑𝑦₀,𝑧₀≥𝑑𝐴,𝑦₀+𝑑𝑦₀,𝑧₀≥𝑑𝐴,𝑧₀.(3.3) By the arbitrariness of 𝑦₀, we have that sup_𝑦∈𝐵𝑑(𝐴,𝑦)+𝑑𝐵,𝑧₀=sup_𝑦∈𝐵𝑑(𝐴,𝑦)+inf_𝑦0∈𝐵𝑑𝑦₀,𝑧₀≥𝑑𝐴,𝑧₀.(3.4) Then we have that sup_𝑦∈𝐵𝑑(𝐴,𝑦)+sup_𝑧∈𝐶𝑑(𝐵,𝑧)≥sup_𝑦∈𝐵𝑑(𝐴,𝑦)+𝑑𝐵,𝑧₀≥𝑑𝐴,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 ℎ(𝐴,𝐵)+ℎ(𝐵,𝐶)=≥mmaaxx{{𝑑𝑑((𝐴𝐴,,𝐶𝐵)),,𝑑𝑑((𝐵𝐶,,𝐴𝐴))}}+=mℎa(𝐴x,{𝐶𝑑)(.𝐵,𝐶),𝑑(𝐶,𝐵)}(3.8) We conclude that (𝑃(𝑋),ℎ) is a normalized pseudometric space.

Let 𝜇 be a normalized measure on an algebra 𝑅⊆𝑃(𝑋) and 𝜇^∗ be the outer measure induced by 𝜇. Let 𝜌∶𝑃(𝑋)×𝑃(𝑋)→ℝ⁺ be defined by the equation 𝜌(𝐴,𝐵)=𝜇^∗(𝐴Δ𝐵), where the symmetric difference of 𝐴 and 𝐵 is defined by 𝐴Δ𝐵=(𝐴∩𝐵^𝐶)∪(𝐴^𝐶∩𝐵). Then (𝑃(𝑋),𝜌) is a normalized pseudometric space and 𝜇^∗(𝐴)=𝜌(𝐴,∅) for all 𝐴∈𝑃(𝑋) [15]. Now, we consider the converse of this process for the normalized pseudometric space (𝑃(𝑋),ℎ). Since ℎ(𝐴,∅)=1 for all nonempty subset 𝐴∈𝑃(𝑋), it would not get any nontrivial results if the set function 𝜇 is defined by 𝜇(𝐴)=ℎ(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−ℎ(𝑋,𝐴),(3.9) for all 𝐴∈𝑃(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,𝐵∈𝑃(𝑋) with 𝐴⊆𝐵. By the definition of 𝜇, we have that 𝜇(𝐴)==11−−msuap_𝑥∈𝑋xsup_𝑥∈𝑋𝑑(𝑥,𝐴),sup_𝑦∈𝐴𝑑(𝑋,𝑦)inf_𝑦∈𝐴≤1−sup_𝑥∈𝑋𝑑(𝑥,𝑦)inf_𝑦∈𝐵=1−maxsu𝑑p_𝑥∈𝑋(𝑥,𝑦)=𝜇(𝐵),𝑑(𝑥,𝐵),sup_𝑦∈𝐵𝑑(𝑋,𝑦)(3.10) which shows the set function 𝜇 is monotonous. Thus, for any two sets 𝐴,𝐵∈𝑃(𝑋), we have 𝜇(𝐴∪𝐵)≥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 (𝐴_𝑖)_𝑖∈ℕ of elements of 𝑃(𝑋) and every positive integer 𝑛, by mathematical induction we have that 𝜇_𝑖∈ℕ𝐴_𝑖≥max𝜇𝐴¹,𝜇𝐴₂,…,𝜇𝐴_𝑛,(3.12) which implies that 𝜇_𝑖∈ℕ𝐴_𝑖≥sup𝜇𝐴^𝑖∶𝑖∈ℕ.(3.13)

Lemma 3.5. Let (𝑋,𝑑) be a normalized metric space. If (𝐴_𝑖)_𝑖∈ℕ is an increasing sequence in 𝑃(𝑋) such that ⋃^∞𝑖=1𝐴_𝑖=𝐴, then lim_𝑖→∞𝑑(𝑥,𝐴_𝑖)=𝑑(𝑥,𝐴) for any point 𝑥∈𝑋.

Proof. Since 𝐴_𝑖⊆𝐴, it follows from Definition 2.1 that 𝑑(𝑥,𝐴_𝑖)=inf_{𝑦∈𝐴𝑖}𝑑(𝑥,𝑦)≥𝑑(𝑥,𝐴). If lim_𝑖→∞𝑑(𝑥,𝐴_𝑖)=𝑎>𝑏=𝑑(𝑥,𝐴), then for the decreasing sequence (𝑑(𝑥,𝐴_𝑖))_𝑖∈ℕ, we have 𝑑(𝑥,𝑦)≥𝑎 for all 𝑦∈𝐴_𝑖, 𝑖∈ℕ.
On the other hand, from 𝑑(𝑥,𝐴)=inf_𝑦∈𝐴𝑑(𝑥,𝑦)=𝑏, it follows that there exists a point 𝑦₀∈𝐴 such that 𝑑(𝑥,𝑦₀)≤(𝑎+𝑏)/2. Since ⋃^∞𝑖=1𝐴_𝑖=𝐴, there exists a positive integer 𝑖₀ such that 𝑦₀∈𝐴_𝑖0. Thus we get that 𝑑(𝑥,𝑦₀)≥𝑎 which contradicts 𝑑(𝑥,𝑦₀)≤(𝑎+𝑏)/2. We conclude that lim_𝑖→∞𝑑(𝑥,𝐴_𝑖)=𝑑(𝑥,𝐴) for any point 𝑥∈𝑋.

Lemma 3.6. Let (𝑋,𝑑) be a normalized compact metric space. If (𝐴_𝑖)_𝑖∈ℕ is an increasing sequence in 𝑃(𝑋) such that ⋃^∞𝑖=1𝐴_𝑖=𝐴, then lim_𝑖→∞ℎ(𝐴_𝑖,𝐴)=0.

Proof. Since 𝐴_𝑖⊆𝐴, it follows from Definition 2.1 that ℎ𝐴_𝑖,𝐴=maxsup_{𝑥∈𝐴𝑖}𝑑(𝑥,𝐴),sup_𝑥∈𝐴𝑑𝑥,𝐴_𝑖=sup_𝑥∈𝐴𝑑𝑥,𝐴_𝑖.(3.14) If lim_𝑖→∞ℎ(𝐴_𝑖,𝐴)=𝑎>0, then for the decreasing sequence (ℎ(𝐴_𝑖,𝐴))_𝑖∈ℕ, we have ℎ(𝐴_𝑖,𝐴)≥𝑎 for all 𝑖∈ℕ. Consequently there exists a point 𝑥_𝑖∈𝐴 for each 𝐴_𝑖 such that 𝑑(𝑥_𝑖,𝐴_𝑖)>𝑎/2. Since 𝑋 is a compact metric space, passing to subsequence if necessary, we may assume that the sequence (𝑥_𝑖)_𝑖∈ℕ converges to a point 𝑥 in the closure of 𝐴 and lim_𝑖→∞𝑑(𝑥_𝑖,𝐴_𝑖)=𝑏≥𝑎/2. However since ||𝑑𝑥_𝑖,𝐴_𝑖−𝑑(𝑥,𝐴)||≤≤𝑑||𝑑𝑥𝑥_𝑖𝑖,𝐴_𝑖−𝑑𝑥,𝐴_𝑖||+||𝑑𝑥,𝐴_𝑖−𝑑(𝑥,𝐴)||,𝑥+||𝑑𝑥,𝐴_𝑖−𝑑(𝑥,𝐴)||,(3.15) it follows from Lemma 3.5 and lim_𝑖→∞𝑑(𝑥_𝑖,𝑥)=0 that lim_𝑖→∞𝑑𝑥_𝑖,𝐴_𝑖=𝑑(𝑥,𝐴)=0.(3.16) This is a contradiction. Thus we have lim_𝑖→∞ℎ(𝐴_𝑖,𝐴)=0.

Lemma 3.7. Let (𝑋,𝑑) be a normalized compact metric space. If (𝐴_𝑖)_𝑖∈ℕ is an increasing sequence in 𝑃(𝑋) such that ⋃^∞𝑖=1𝐴_𝑖=𝐴, then 𝜇 is continuous from below, that is, lim_𝑖→∞𝜇(𝐴_𝑖)=𝜇(𝐴).

Proof. By the definition of 𝜇, we have that ||𝜇𝐴_𝑖−𝜇(𝐴)||=||ℎ𝐴_𝑖,𝑋−ℎ(𝐴,𝑋)||≤ℎ𝐴_𝑖,𝐴,(3.17) for all 𝑖∈ℕ. By Lemma 3.6, we have that lim_𝑖→∞𝜇(𝐴_𝑖)=𝜇(𝐴).

Definition 3.8. A set 𝐸 in 𝑃(𝑋) is 𝜇-measurable if, for every set 𝐴 in 𝑃(𝑋), 𝜇(𝐴)=max𝜇(𝐴∩𝐸),𝜇𝐴∩𝐸^𝐶.(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 ∅,𝑋∈𝕊, and that if 𝐸∈𝕊 then 𝐸^𝐶∈𝕊. Let 𝐸,𝐹∈𝕊 and 𝐴∈𝑃(𝑋). It follows that 𝜇(𝐴∩(𝐸∪𝐹))==mmaaxx𝜇𝜇((𝐴𝐴∩∩𝐹(𝐸),∪𝜇𝐹𝐴)∩∩𝐹𝐸)∩,𝜇𝐹_𝐶𝐴∩(𝐸∪𝐹)∩𝐹^𝐶,(3.19) which implies that max𝜇(𝐴∩(𝐸∪𝐹)),𝜇𝐴∩(𝐸∪𝐹)^𝐶==mmaaxx𝜇𝜇((𝐴𝐴∩∩𝐹𝐹)),,𝜇𝜇𝐴𝐴∩∩𝐹𝐸_𝐶∩𝐹^𝐶,𝜇𝐴∩(𝐸∪𝐹)^𝐶=max𝜇(𝐴∩𝐹),𝜇𝐴∩𝐹_𝐶∩𝐸,𝜇𝐴∩𝐹^𝐶∩𝐸^𝐶=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 𝜇𝐸₁∪𝐸₂=max𝜇𝐸₁∪𝐸₂∩𝐸₁,𝜇𝐸₁∪𝐸₂∩𝐸^𝐶1=max𝜇𝐸₁,𝜇𝐸₂.(3.21) Let {𝐸_𝑖}^∞𝑖=1 be a disjoint sequence set in 𝕊 with ⋃^∞𝑖=1𝐸_𝑖=𝐸∈𝕊. By mathematical induction, we can get that 𝜇^𝑛_𝑖=1𝐸_𝑖=max𝜇𝐴¹,𝜇𝐴₂,…,𝜇𝐴_𝑛(3.22) for every positive integer 𝑛. Since 𝜇 is continuous from below and lim_𝑛→∞⋃^𝑛𝑖=1𝐸_𝑖=𝐸, we have 𝜇(𝐸)=lim_𝑛→∞𝜇^𝑛_𝑖=1𝐸_𝑖=lim^𝑛→∞max𝜇𝐸₁,𝜇𝐸₂,…,𝜇𝐸_𝑛=sup𝜇𝐸_𝑖∶𝑖∈ℕ,(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).