Lobachevskii Journal of Mathematics Vol.6, 2000, 47–50 \begin{matrix}c \\ ◯ \end{matrix}O. E. Tikhonov

O. E. Tikhonov PARTIAL MEASURES

(submitted by D.Kh.Mustari)

Abstract. We study σ-additive set functions defined on a hereditary subclass of a σ-algebra and taken values in the extended real line. Analogs of the Jordan decomposition theorem and the Radon-Nikodym theorem are obtained.

Throughout this note, A stands for a σ-algebra of subsets of a set Ω, \overline{ℝ} denotes the extended real line ℝ ∪ {−∞} ∪ {+∞}. By a measure on A we mean a σ-additive set function μ : A → \overline{ℝ} with μ(∅) = 0. Note that in fact a measure can take at most one of the values +∞, −∞.

Definition. A set function μ : D(μ) → \overline{ℝ} defined on a nonvoid subclass D(μ) of a σ-algebra A is called a partial measure on A if for every B ∈ D(μ) the restriction μ|_{A∩B} is a measure on the σ-algebra A∩B = {A∩B : A ∈ A}. A partial measure is called maximal if it has no proper extension to a partial measure on A.

By making use of the Zorn lemma, it is easy to see that every partial measure can be extended to a maximal one.

Example 1. Let μ_{1}, μ_{2} be positive measures on a A. Then the formula (μ_{1}−μ_{2})(A) = μ_{1}(A)−μ_{2}(A) defines a partial measure μ_{1}−μ_{2}. (D(μ_{1}−μ_{2}) = {A ∈ A : μ_{1}(A) − μ_{2}(A) is well-posed}.)

Example 2. Let (Ω, A, ℙ) be a probability space, ξ : Ω → \overline{ℝ} be a random variable. Then the formula μ_{ξ}(A) = ∫_{A} ξ dℙ defines a maximal partial measure μ_{ξ} on A. (D(μ_{ξ}) = {A ∈ A : ξ is quasiintegrable over A}.)

For a partial measure μ on A, define two classes: F^{+}(μ) = {F ∈ D(μ) : A ∋ A ⊂ F ⟹ μ(A) ≥ 0}, F^{−}(μ) = {F ∈ D(μ) : A ∋ A ⊂ F ⟹ μ(A) ≤ 0}.

Key words and phrases.Set function, σ-algebra, Jordan decomposition theorem, Radon-Nikodym theorem.

1991 Mathematical Subject Classification.28A10.

Lemma 1. Let {F_{i} : i ∈ I} and {F′_{j} : j ∈ J} be two finite or countable families of mutually disjoint sets of F^{+}(μ) such that ⋃_{i∈I} F_{i} = ⋃_{j∈J} F′_{j}. Then ∑_{i∈I} μ(F_{i}) = ∑_{j∈J} μ(F′_{j}).

Proof. Since F_{i} = ⋃_{j∈J}(F_{i} ∩ F′_{j}) for every i ∈ I, F′_{j} = ⋃_{i∈I}(F_{i} ∩ F′_{j}) for every j ∈ J, and F_{i}, F′_{j} ∈ D(μ) we have

∑_{i∈I} μ(F_{i}) = ∑_{i∈I} ∑_{j∈J} μ(F_{i} ∩ F′_{j}) = ∑_{j∈J} ∑_{i∈I} μ(F_{i} ∩ F′_{j}) = ∑_{j∈J} μ(F′_{j}).

Lemma 2. Let a partial measure μ on A be maximal. If {F_{i} : i ∈ I} is a finite or countable family of sets of F^{+}(μ) then ⋃_{i∈I} F_{i} ∈ F^{+}(μ).

Proof. Clearly, it suffices to prove that all the unions ⋃_{i∈I} F_{i} of finite or countable families of mutually disjoint sets of F^{+}(μ) belong to F^{+}(μ). Extend μ to such unions by means of the formula μ̃(⋃_{i∈I} F_{i}) = ∑_{i∈I} μ(F_{i}). (By Lemma 1, μ̃ is well-defined.) The extension obtained is a partial measure on A. Really, if A = ⋃_{i∈I} F_{i} (I is finite or countable, F_{i} ∈ F^{+}(μ) for any i ∈ I, F_{i} ∩ F_{j} = ∅ for i ≠ j) and A = ⋃_{k∈K} A_{k} (K is finite or countable, A_{k} ∈ A for any k ∈ K, A_{k} ∩ A_{l} = ∅ for k ≠ l) then

μ̃(A) = ∑_{i∈I} μ( ⋃_{k∈K}(F_{i} ∩ A_{k})) = ∑_{i∈I} ∑_{k∈K} μ(F_{i} ∩ A_{k}) = ∑_{k∈K} ∑_{i∈I} μ(F_{i} ∩ A_{k})

= ∑_{k∈K} μ̃(A_{k}).

Moreover, it is easy to check that the unions under consideration are sets of the class F^{+}(μ̃). To complete the proof, it suffices to observe that μ̃ = μ by the maximality of μ.

Theorem 1. Let μ be a maximal partial measure on a σ-algebra A. Then the formulas

μ^{+}(A) = sup_{F ∈F^{+}(μ)} μ(A ∩ F), μ^{−}(A) = sup_{F ∈F^{−}(μ)}(−μ(A ∩ F)) (A ∈ A)

define two positive measures on A such that μ = μ^{+} − μ^{−}. Moreover, if μ_{1} is a positive measure on A with μ(A) ≤ μ_{1}(A) for all A ∈ D(μ) then μ^{+} ≤ μ_{1}. Similarly, if μ_{2} is a positive measure on A with −μ(A) ≤ μ_{2}(A) for all A ∈ D(μ) then μ^{−} ≤ μ_{2}.

Proof. Observe, first, that for every F ∈ F^{+}(μ) the formula μ_{F} (A) = μ(A ∩ F) defines a positive measure μ_{F} on A. Also, from Lemma 2 it follows that the class F^{+}(μ) is upwards directed by inclusion. Moreover, the condition F_{1} ⊂ F_{2} (F_{1}, F_{2} ∈ F^{+}(μ)) implies μ_{F_{1}}(A) ≤ μ_{F_{2}}(A) for all A ∈ A. It follows that the set function μ^{+} on A is a pointwise limit of an increasing net of positive measures, therefore, μ^{+} is a positive measure in turn. Similarly, μ^{−} is a positive measure on A.

Next, let A ∈ D(μ). By the Hahn decomposition theorem, A can be represented as a union, A = A^{+} ∪ A^{−}, of two disjoint sets A^{+} ∈ F^{+}(μ) and A^{−} ∈ F^{−}(μ). Clearly, μ^{+}(A) = μ(A^{+}) and μ^{−}(A) = −μ(A^{−}). Thus, μ(A) = μ^{+}(A)−μ^{−}(A) for any A ∈ D(μ). Since μ^{+} −μ^{−} is a partial measure on A (see Example 1) the maximality of μ yields μ = μ^{+} − μ^{−}.

Now, let μ_{1} is a positive measure on A with μ(A) ≤ μ_{1}(A) for all A ∈ D(μ). Take A ∈ A. For every F ∈ F^{+}(μ) we have μ(A ∩ F) ≤ μ_{1}(A ∩ F) ≤ μ_{1}(A). Therefore, μ^{+}(A) = sup_{F ∈F^{+}(μ)} μ(A ∩ F) ≤ μ_{1}(A). Thus, we have proved the

extremal property for μ^{+}. The similar extremal property for μ^{−} can be proved in the same manner.

Corollary 1. Let μ be a maximal partial measure on a σ-algebra A and A be a set of A \ D(μ). Then there exist two subsets A′, A′′ of A such that A′ ∈ F^{+}(μ) and μ(A′) = +∞, A′′ ∈ F^{−}(μ) and μ(A′′) = −∞.

Proof. Since A ∉ D(μ) = D(μ^{+} − μ^{−}) we have μ^{+}(A) = +∞ and μ^{−}(A) = +∞. By the definition of μ^{+}, we can find a sequence {A_{n}} ⊂ F^{+}(μ) of subsets of A, such that μ(A_{n}) ≥ n. By Lemma 2, the union A′ = ⋃_{n=1}^{∞} A_{n} is a set of F^{+}(μ), and, obviously, μ(A′) = +∞. The set A′′ can be constructed in the same manner.

Remark. The preceding theorem is an analog of the Jordan decomposition theorem. Example 3 below shows that the direct analog of the Hahn decomposition does not hold for general maximal partial measures. Nevertheless, the latter analog does hold provided that a maximal partial measure under consideration is absolutely continuous with respect to a certain “good” measure (see the proof of Theorem 2 below).

Example 3. Let a σ-algebra A of subsets of Ω contains all the one-point subsets of Ω and does not coincide with the class P(Ω) of all the subsets of Ω. (For instance, we may take the Borel σ-algebra of [0, 1] as A.) Let B ∈ P(Ω) \ A. Define the set function μ as follows:

μ(A) = \{ \begin{matrix}0 & if A = ∅ , \\ +∞ & if A ∋ A ⊂ B and A ≠ ∅ , \\ −∞ & if A ∋ A ⊂ Ω \ B and A ≠ ∅ . \end{matrix}

It is easy to see that μ is a maximal partial measure on A and that there exists no C ∈ F^{+}(μ) with Ω \ C ∈ F^{−}(μ).

In the rest of the note, the triple (Ω, A, ℙ) stands for a probability space, the abbreviation a. s. means “almost surely with respect to ℙ”. The following theorem is an analog of the Radon-Nikodym one.

Theorem 2. Let a maximal partial measure μ on A be absolutely continuous with respect to a probability measure ℙ (i. e., ℙ(A) = 0 implies A ∈ D(μ) and μ(A) = 0). Then there exists an a. s. unique random variable ξ : Ω → \overline{ℝ} such that μ = μ_{ξ}, where μ_{ξ}(A) = ∫_{A} ξ dℙ (see Example 2).

Proof. Set Ω^{+} = ess sup F^{+}(μ). Recall that ess sup F^{+}(μ) is a set of A such that for any A ∈ A

∀F ∈ F^{+}(μ) (F ⊂_{a. s.} A) ⟺ ess sup F^{+}(μ) ⊂_{a. s.} A

(see, e. g., [1, Proposition II.4.1]). From the proof of the proposition just referred, it follows that Ω^{+} can be constructed as a union of a countable family of sets of F^{+}(μ). Therefore, Ω^{+} ∈ F^{+}(μ) by Lemma 2.

Our next goal is to show that Ω^{−} = Ω \ Ω^{+} is a set of F^{−}(μ). Suppose, on the contrary, that Ω^{−} ∉ F^{−}(μ). Then there exists A ⊂ Ω^{−} (A ∈ A) such that either μ(A) > 0 or A ∉ D(μ). In each of the cases, we can find a subset A′ ⊂ A with A′ ∈ F^{+}(μ) and μ(A′) > 0 (see Corollary 1 for the second case). Hence ℙ(A′) > 0. This contradicts the definition of Ω^{+} = ess sup F^{+}(μ).

By making use of the Radon-Nikodym theorem (see, e. g., [1, Proposition IV.1.4]), we can find two positive random variables ξ^{+} and ξ^{−} such that ξ^{+}(ω) = 0 for ω ∈ Ω^{−}, ξ^{−}(ω) = 0 for ω ∈ Ω^{+}, μ(A) = ∫_{A} ξ^{+} dℙ if Ω^{+} ⊃ A ∈ A, μ(A) = − ∫_{A} ξ^{−} dℙ if Ω^{−} ⊃ A ∈ A. Writing ξ = ξ^{+} − ξ^{−} we have for A ∈ D(μ)

μ(A) = μ(A ∩ Ω^{+}) + μ(A ∩ Ω^{−}) = ∫_{A∩Ω^{+}} ξ^{+} dℙ − ∫_{A∩Ω^{−}} ξ^{−} dℙ = ∫_{A} ξ dℙ.

Hence, μ = μ_{ξ} by the maximality of μ.

It remains to prove the claim about the uniqueness. Let η be a random value such that μ_{η} = μ. Then, by the uniqueness assertion in the Radon-Nikodym theorem, we have η|_{Ω^{+}} =_{a. s.} ξ|_{Ω^{+}} and η|_{Ω^{−}} =_{a. s.} ξ|_{Ω^{−}}, hence η =_{a. s.} ξ.

References

1. J. Neveu, Bases mathématiques du calcul des probabilités, Masson et Cie, Paris, 1964; Russian transl., Mir, Moscow, 1969.

Research Institute of Mathematics and Mechanics, Kazan State University, Universitetskaya Str. 17, Kazan, Tatarstan, 420008 Russia

E-mail address: Oleg.Tikhonov@ksu.ru

Received May 1, 2000