On Decomposable Measures Induced by Metrics

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

  $$ ()
  with the convention (x, ∅) = 1.

• (ii)

  The “distance” from A to B is

  $$ ()
  with the convention d(∅, B) = 0.

• (iii)

  The Hausdorff distance, h(A, B), between A and B is

  $$ ()

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

  $$ ()
  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

  $$ ()

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

which implies that Consequently, we get that By the arbitrariness of y₀, we have that Then we have that which implies that Similarly, we can get that It follows that We conclude that (P(X), h) is a normalized pseudometric space.

Definition 3.2. Let (X, d) be a normalized metric space. Now, we define a set function μ on P(X) by

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

which shows the set function μ is monotonous. Thus, for any two sets A, B ∈ P(X), we have

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

which implies that

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 $$, 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 $$. 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 $$, there exists a positive integer i₀ such that $$. 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 $$, then lim _i→∞h(A_i, A) = 0.

Proof. Since A_i⊆A, it follows from Definition 2.1 that

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 it follows from Lemma 3.5 and lim _i→∞d(x_i, x) = 0 that 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 $$, then μ is continuous from below, that is, lim _i→∞μ(A_i) = μ(A).

Proof. By the definition of μ, we have that

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),

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

which implies that 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

Let $$ be a disjoint sequence set in 𝕊 with $$. By mathematical induction, we can get that for every positive integer n. Since μ is continuous from below and $$, we have which implies that μ|_𝕊 is a σ-max-decomposable measure.

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

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