On the space of (s)-integrable functions for a monotonic measure

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Congxin Wu^∗, Xuekun Ren, Chong Wu

On the space of (s)-integrable functions for a monotonic measure

Dedicated to Professor Musielak on the occasion of his 85th birthday

Abstract. In this paper we introduce and study the space of real-valued (s)- integrable functions for a monotonic measure, and consider some relations between this space and the space of real-valued measurable functions.

2000 Mathematics Subject Classification: 28B05.

Key words and phrases: Monotonic measure, (s)-integral, space of real-valued (s)- integrable functions, space of real-valued measurable functions.

1. Introduction. In 2011 C. Wu, X.K. Ren and C.X. Wu [1] studied the space of all real-valued measurable functions for a monotonic measure and gave a necessary and sufficient condition to guarantee that this space is a topological linear space with a countable local base and the convergence in this space is equivalent to the convergence in measure. M. Sugeno [2] introduced fuzzy integrals (i.e. (s)- integrals) of a non-negative measurable function for a fuzzy measure in 1974, and in 2003 C.X. Wu and T. Mamadou [3] generalized the concept of (s)-integrals from non-negative measurable functions to extended real-valued measurable functions.

However, after [4], [5] more and more researchers have paid attention to monotonic measures, a generalization of fuzzy measures, and have obtained many correspon- ding conclusions in various directions, see for examples [6], [7], [8], [9], [10], [11], [12], [13].

Hence, the purpose of this paper is to investigate the space of all real-valued (s)- integrable functions for a monotonic measure and some relations between this space and the space of all real-valued measurable functions for a monotonic measure.

∗This paper is supported by the NNSF of P.R. China (No. 11126320), (No. 71271070).

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2. Preliminaries.

Definition 2.1 ([5], [12]) Let X be a nonempty set,P be a σ−algebra of subsets of X, µ : P

→ [0, ∞] is a monotonic measure iff (1) µ(φ) = 0 and (2) if A ⊂ B, then µ(A) ≤ µ(B). (X,P, µ)is said to be a monotonic measure space.

Definition 2.2 ([14], [11]) The monotonic measure µ is said to be double asymptotic null-additive if µ(An∪ Bⁿ)→ 0 (n → ∞) whenever {Aⁿ} ⊂P, {Bⁿ} ⊂P, µ(An)→ 0 and µ(Bⁿ)→ 0 (n → ∞).

Definition 2.3 ([5], [7]) The monotonic measure µ is said to be order continuous if E1 ⊃ E²⊃ · · · andT∞

n=1En= φ imply µ(En)→ 0 (n → ∞) whenever Eⁿ∈P (∀n ∈ N).

Definition 2.4 ([5], [12]) Let (X,P

, µ) be a monotonic measure space and f : X → [−∞, ∞] = R. If for any α ∈ R we have

{x ∈ X : f(x) ≥ α} ∈X , then we say that f is measurable on X.

Definition 2.5 ([5], [12]) Let f, fn (∀n ∈ N) be measurable on X. If for any ε > 0

µ{x ∈ X : |fⁿ(x)− f(x)| ≥ ε} → 0 (n → ∞),

then we say that {fⁿ} is µ−measure convergent to f on X, and is denoted by fn

−→ f.µ

Definition 2.6 ([3], [13], [15]) Let (X,P

, µ)be a monotonic measure space, f be measurable on X and

f⁺(x) =

(f (x) when f (x)≥ 0

0 when f (x) < 0, f⁻(x) =

(0 when f (x) > 0

−f(x) when f(x) ≤ 0 (x∈ X).

If f⁺ and f⁻ are (s)-integrable, namely (s)

Z

X

f⁺(x)dµ = sup

α>0

min{α, µ(N^α(f⁺))} < ∞,

(s) Z

X

f⁻(x)dµ = sup

α>0

min{α, µ(N^α(f⁻))} < ∞,

where Nα(f ) ={x ∈ X : f(x) ≥ α} (see [2]), then we say that f is (s)-integrable on X.

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Throughout this paper, (X,P, µ)is a monotonic measure space, µ is the monotonic measure, F(F) is the class of all real-valued (extended real-valued) measurable functions on X and F+(F+)is the class of all non-negative (extended non-negative) measurable functions on X.

Let P (F) be the family of all subsets of F, and denote

S(f,1

n) ={g ∈ F : ρ(f, g) < 1

n} (∀f ∈ F, n ∈ N),

ρ(f, g) = (S) Z

X

|f(x) − g(x)|

1 +|f(x) − g(x)|dµ, Uf ={U ∈ P (F) : there exists n ∈ N such that f ∈ S(f,1

n)⊂ U}, T={U ⊂ F : ∀f ∈ U, U ∈ U^f}.

Theorem 2.7 [1] Let µ(X) < ∞. Then (F, T) is a topological linear space with respect to the operations: addition and scalar multiplication if and only if µ is double asymptotic null-additive, order continuous and satisfies condition (∗): for any natural number k > 1, any A ∈Pwith µ(A) < ¹_k, and any sequence {Bm} in P, if µ(Bm)→ 0, then there exists m⁰∈ N such that

µ(A∪ B^m) <1

k (∀m ≥ m⁰).

And in this case Uf is the neighborhood system of f relative to the topology T.

Theorem 2.8 [1] Under the assumptions in Theorem 2.7, the topological linear space (F, T) is pseudo-metrizable and in this space T-convergence is equivalent to the convergence in measure.

3. Main results.

Let

L(X,R, µ) = {f ∈ F : f is (s) − integrable on X}, L(X,R, µ) = {f ∈ F : f is (s) − integrable on X}.

Proposition 3.1 If f ∈ F, then f ∈ L(X, R, µ) if and only if two sets {α > 0 : µ(N^α(f⁺)) =∞}, {α > 0 : µ(N^α(f⁻)) =∞}

are bounded.

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Proof Sufficiency. If f 6∈ L(X, R, µ), then we can assume that f⁻ ∈ F⁺ is not (s)-integrable. Hence

(S) Z

X

f⁻(x)dµ = sup

α>0

min{α, µ(N^α(f⁻))} = ∞.

Thus there exist positive numbers αn increasing to ∞ such that

n→∞lim min{αⁿ, µ(Nαn(f⁻))} = ∞,

and it implies µ(Nαn(f⁻))→ ∞ (n → ∞). Clearly, by the monotonicity of µ we know that µ(Nαn(f⁻))is decreasing, so

µ(Nαn(f⁻)) =∞ (∀n ∈ N),

namely, {α > 0 : µ(Nα(f⁻)) =∞} is not a bounded set. This is a contradiction.

Necessity. Let f ∈ L(X, R, µ). Otherwise, no lose the generality, we suppose that the set {α > 0 : µ(N^α(f⁺)) = ∞} is unbounded. Therefore, for any n ∈ N there exists αn ≥ n such that

µ(Nαn(f⁺)) =∞ (∀n ∈ N).

Hence we have (s)

Z

X

f⁺(x)dµ≥ sup

n∈N

min{αⁿ, µ(Nαn(f⁺))}

= sup

n∈Nmin{αⁿ,∞} = sup

n∈Nαn =∞,

that is , f⁺ is not (s)-integrable. It is contradicted with f ∈ L(X, R, µ).

Proposition 3.2 Denote

F^∗={f ∈ F : max{µ({x ∈ X : f⁺(x) =∞}), µ({x ∈ X : f⁻(x) =∞})} < ∞}.

(Notice that F ⊂ F^∗. In fact, for any f ∈ F we have

{x ∈ X : f⁺(x) =∞} = {x ∈ X : f⁻(x) =∞} = φ, so

max{µ({x ∈ X : f⁺(x) =∞}), µ({x ∈ X : f⁻(x) =∞})} = 0 < ∞).

Then F^∗ = L(X,R, µ) if and only if there is no sequence {Aⁿ} ⊂Psuch that An% Aⁿ⁺¹ and µ(An) =∞ (∀n ∈ N).

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Proof Necessity. If the conclusion is false, then there exists a sequence {An} ⊂P such that An % Aⁿ⁺¹ and µ(An) = ∞ (∀n ∈ N). Take a positive sequence αⁿ increasing to ∞, and denote

f0(x) =

(αn when x∈ Aⁿ\ Aⁿ⁺¹ (∀n ∈ N), 0 otherwise.

Since for any α > 0 there is an n0∈ N such that αⁿ0 ≥ α > αⁿ0−1(∀n ∈ N, α⁰= 0), we infer that

Nα(f0) = Nα_n0(f0) = [∞ n=n0

(An\Aⁿ⁺¹) = An0 ∈X ,

namely f0∈ F ⊂ F^∗.

On the other hand, we can get

µ(Nαn(f₀⁺)) = µ(Nαn(f0)) = µ(An) =∞ (∀n ∈ N).

Hence, by Proposition 3.1 we have f06∈ L(X, R, µ), a contradiction is obtained.

Sufficiency. If the conclusion is false, then there exists an f0 ∈ F^∗ and f0 6∈

L(X,R, µ). Thus, by Proposition 3.1 we can assume that {α > 0 : µ(N^α(f₀⁺)) =∞}

is unbounded. Hence there exists a positive sequence αn increasing to ∞ with µ(Nαn(f₀⁺)) =∞ (∀n ∈ N).

Denote

An= Nαn(f₀⁺) (∀n ∈ N), then

An∈X

, µ(An) =∞ and Aⁿ⊃ Aⁿ⁺¹ (∀n ∈ N).

If there exists a subsequence {Aⁿk} of {Aⁿ} with Aⁿk% Aⁿk+1 (k = 1, 2,· · · ), then a contradiction follows. Otherwise there is an n0∈ N such that Aⁿ = An0 whenever n≥ n⁰, and therefore

f₀⁺(x)≥ αⁿ (∀n ≥ n⁰, x∈ Aⁿ0), that is, f0⁺(x) =∞ (∀x ∈ Aⁿ⁰). Thus by the monotonicity of µ

µ({x ∈ X : f0⁺(x) =∞}) ≥ µ(Aⁿ0) =∞

namely f06∈ F^∗. This is a contradiction.

Corollary 3.3 If f ∈ F, then f ∈ L(X, R, µ) if and only if two sets {α > 0 : µ(N^α(f⁺)) =∞}, {α > 0 : µ(N^α(f⁻)) =∞}

are bounded.

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The proof is the same with Proposition 3.1.

The following corollary is a directed consequence from Proposition 3.2 and Co- rollary 3.3.

Corollary 3.4 F = L(X, R, µ) if and only if there is no sequence {An} ⊂Psuch that

An% Aⁿ⁺¹ and µ(An) =∞ (∀n ∈ N).

Remark 3.5 Evidently, the following example:

f0(x) =∞ (∀x ∈ X)

means that for the case of µ(X) = ∞ we have f⁰∈ F \ F^∗, namely F^∗$ F,

and it is easy to see that

f 6∈ L(X,R, µ) (∀f ∈ F\F^∗).

Theorem 3.6 If the monotonic measure µ satisfies the condition (): for any A, B ∈P

, µ(A) < ∞ and µ(B) < ∞ infer µ(A∪ B) < ∞, then L(X, R, µ) is a linear space.

Proof Clearly, we only need to prove that two operations: addition and scalar multiplication in L(X, R, µ) are closed.

(1) If f, g ∈ L(X, R, µ), then f + g ∈ L(X, R, µ).

In fact, by Corollary 3.3 four subsets in R:

{α > 0 : µ(N^α(f⁺)) =∞}, {α > 0 : µ(N^α(f⁻)) =∞}, {α > 0 : µ(N^α(g⁺)) =∞}, {α > 0 : µ(N^α(g⁻)) =∞}

are bounded. Let M < ∞ be a common upper bound of these four subsets. Since µ(Nα(f⁺)), µ(Nα(f⁻)), µ(Nα(g⁺))and µ(Nα(g⁻))are decreasing when α is increasing, we get

µ(Nα(f⁺)) <∞, µ(N^α(f⁻)) <∞, µ(Nα(g⁺)) <∞, µ(N^α(g⁻)) <∞ when α ≥ M + 1. Notice that

Nα((f + g)⁺) ={x ∈ X : f(x) + g(x) ≥ α, f(x) ≥ 0, g(x) ≥ 0}

[{x ∈ X : f(x) + g(x) ≥ α, f(x) > 0, g(x) < 0}

[{x ∈ X : f(x) + g(x) ≥ α, f(x) < 0, g(x) > 0}

⊂ N^α(f⁺+ g⁺)∪ N^α(f⁺)∪ N^α(g⁺)

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and

Nα(f⁺+ g⁺)⊂ N^α2(f⁺)∪ N^α2(g⁺).

Therefore, when α ≥ 2(M + 1) we obtain

µ(N^α₂(f⁺)) <∞, µ(N^α2(g⁺)) <∞.

Thus, by the condition () we have

µ(Nα(f⁺+ g⁺))≤ µ(N^α2(f⁺)∪ N^α2(g⁺)) <∞, µ(Nα(f⁺)∪ N^α(g⁺)) <∞.

By the condition () again, we infer that

µ(Nα((f + g)⁺))≤ µ(N^α(f⁺+ g⁺)∪ (N^α(f⁺)∪ N^α(g⁺))) <∞ for any α ≥ 2(M + 1), namely

{α > 0 : µ(N^α((f + g)⁺)) =∞}

is bounded. Similarly

{α > 0 : µ(N^α((f + g)⁻)) =∞}

is also bounded. Finally, by Corollary 3.3 we obtain f + g ∈ L(X, R, µ).

(2) If f ∈ L(X, R, µ), k ∈ R, then kf ∈ L(X, R, µ).

In fact, since for any x ∈ X

(−f)⁺(x) = f⁻(x), (−f)⁻(x) = f⁺(x),

then it is easy to see that −f ∈ L(X, R, µ) is true. Hence, we can assume k > 0.

By Corollary 3.3 we have M < ∞ such that

µ(Nα(f⁺)) <∞, µ(N^α(f⁻)) <∞ (∀α ≥ M + 1).

Obviously

Nα((kf )⁺) = Nα(kf⁺) = N^α_k(f⁺), Nα((kf )⁻) = Nα(kf⁻) = N^α_k(f⁻).

Thus

µ(Nα((kf )⁺)) <∞, µ(N^α(kf )⁻) <∞ (∀α ≥ k(M + 1)).

By Corollary 3.3 we get kf ∈ L(X, R, µ).

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Remark 3.7 The converse of Theorem 3.6 is not true.

Counterexample. Let X = N,P= P (X)and

µ(A) =









 P

k∈A 1

2^k when 16∈ A

1 when A ={1}

0 when A = φ

∞ otherwise

(∀A ∈X ).

Evidently, µ is a monotonic measure. Take A = {1} and B = {2}, then µ(A) = 1, µ(B) = ¹₄ and

µ(A∪ B) = ∞, so µ cannot satisfy the condition () in Theorem 3.6.

Now we prove that L(X, R, µ) is a linear space.

For any f, g ∈ L(X, R, µ) we can assume that there exist two sequences {fⁿ} and {gⁿ} in R such that

f (n) = fn, g(n) = gn (∀n ∈ N).

Clearly, for any α ≥ α⁰=|f¹| + |g¹| + 1 we have 1 6∈ N^α0((f + g)⁺)and Nα((f + g)⁺)⊂ N^α0((f + g)⁺). Thus

µ(Nα((f + g)⁺))≤ µ(N^α0((f + g)⁺))≤ X

k∈N\{1}

1

2^k ≤ 1 (∀α ≥ α⁰).

Therefore (s)

Z

X

(f + g)⁺(x)dµ

= sup

α>0min{α, µ(N^α((f + g)⁺))}

= sup

0<α<α0

min{α, µ(N^α((f + g)⁺))} ∨ sup

α≥α0

min{α, µ(N^α((f + g)⁺))}

≤ α⁰∨ sup

α≥α0

min{α, µ(N^α((f + g)⁺))}

≤ α⁰∨ sup

α≥α0

min{α, 1} = α⁰∨ 1 = α⁰<∞.

Similarly (s)R

X(f + g)⁻(x)dµ <∞. Hence f + g ∈ L(X, R, µ).

On the other hand, such as part (2) of the proof of Theorem 3.6, we obtain that for any f ∈ L(X, R, µ) and k ∈ R

kf∈ L(X, R, µ).

The proof is complete.

Furthermore, for the above counterexample the sequence of subsets in X:

An={1, n + 1, n + 2, · · · } (∀n ∈ N)

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is in P and

An% Aⁿ⁺¹and µ(An) =∞ (∀n ∈ N).

By Corollary 3.4 we know that in this case L(X, R, µ) is a proper subset of F.

Corollary 3.8 If µ(X) < ∞, then L(X, R, µ) is a linear space.

Proof By the assumption µ(X) < ∞ we know that the condition () in Theorem 3.6 can be satisfied naturally.

By Corollary 3.8, clearly, Theorem 2.8 and the sufficiency of Theorem 2.7 are also true for the subs pace L(X, R, µ) of F. When we check the proof of the necessity of Theorem 2.7 in [1] carefully, it is not hard to find that we only need to handle the proof of the necessity of Lemma 2 in [1], namely, we must to prove the function

f (x) =

(n + 1 when x∈ Eⁿ\Eⁿ⁺¹ (∀n ∈ N) 0 when x∈ X\E¹

is (s)-integrable on X, where

{Eⁿ} ⊂X

, E1⊃ E²⊃ · · · ,

\∞ n=1

En = φ

and there exists c > 0 such that

lim_n→∞µ(En) = c.

In fact, obviously, there exists n0∈ N such that n⁰≥ c + 1 and µ(En)≤ c + 1 (∀n ≥ n⁰).

Therefore, for any α ≥ n⁰+ 1

Nα(f )≤ Nⁿ0+1(f ) = [∞ k=n0

Ek\E^k+1= En0,

that is,

µ(Nα(f ))≤ µ(Eⁿ⁰)≤ c + 1 (∀α ≥ n⁰+ 1).

Thus (S)

Z

X

f (x)dµ = sup

0<α<n0+1min{α, µ(N^α(f ))} ∨ sup

α≥n0+1

min{α, µ(N^α(f ))}

≤ min{n⁰+ 1, µ(E1)} ∨ sup

α≥n0+1

min{α, µ(N^α(f ))}

≤ min{n⁰+ 1, µ(E1)} ∨ sup

α≥n0+1

min{α, c + 1}

≤ min{n⁰+ 1, µ(E1)} ∨ (c + 1) < ∞.

Hence f ∈ L(X, R, µ).

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Finally, we get the following two theorems for the space L(X, R, µ).

Theorem 3.9 Let (X,P

, µ) be a monotonic measure space and µ(X) < ∞. Then (L(X,R, µ), T) is a topological linear space with respect to operations: addition and scalar multiplication if and only if µ is double asymptotic null-additive, order continuous and satisfies condition (∗): for any natural number k > 1, any A ∈Pwith µ(A) < ¹_k, and any sequence {B^m} in P, if µ(Bm)→ 0, then there exists m⁰∈ N such that

µ(A∪ B^m) <1

k (∀m ≥ m⁰).

And in this case Uf is the neighborhood system of f relative to the topology T. Here T={U ∈ L(X, R, µ) : ∀f ∈ U, U ∈ U^f},

Uf ={U ∈ P (L(X, R, µ)) : there exists n ∈ N such that f ∈ S(f,1

n)⊂ U}, S(f, 1

n) ={g ∈ L(X, R, µ) : ρ(f, g) < 1

n} (∀n ∈ N) and

ρ(f, g) = (S) Z

X

|f(x) − g(x)|

1 +|f(x) − g(x)|dµ (∀f, g ∈ L(X, R, µ)).

Theorem 3.10 Under the assumptions in Theorem 3.9, the topological linear space (L(X,R, µ), T) is pseudo-metrizable and in this space T−convergence is equivalent to the convergence in measure.

References

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Congxin Wu

Department of Mathematics, Harbin Institute of Technology Harbin, 150001, P.R.China

E-mail: wucongxin@hit.edu.cn Xuekun Ren

Department of Mathematics, Harbin Institute of Technology Harbin, 150001, P.R.China

E-mail: renxuekun@126.com Chong Wu

School of Management, Harbin Institute of Technology Harbin, 150001, P.R.China

E-mail: wuchong@hit.edu.cn

(Received: 8.10.2013)