On set-valued fuzzy integrals

Fuzzy Sets and Systems 56 (1993) 237-241 North-Holland

237

On set-valued fuzzy integrals Deli Zhang and Zixiao Wang Department of Mathematics, Jilin Prov. Inst. of Education, Changchun, Jilin, 130022, China Received May 1991 Revised November 1992

Abstract." In the paper, we extend fuzzy integrals of point-valued functions which were defined by Ralescu and Adams to set-valued functions. Some properties similar to Auman integrals are shown, and these include convexity, monotonicity, Fatou's lemma, Lebesgue convergence theorem, etc. Keywords: Fuzzy integral; set-valued function; set-valued fuzzy integral. 1. Introduction It is well known that set-valued functions have been used repeatedly in Economics [6]. Integrals of set-valued functions have been studied by A u m a n n [1], D e b r e u [4], and others. But they are all based on classical Lebesgue integrals. Since Sugeno [8] brought out the concepts of fuzzy measure and fuzzy integrals, it has been extended and made deeper by Ralescu and Adams [7], Wang [9], Wang [10, 11], Berres [2], etc. But the integrands are all point-valued functions. This paper's purpose is to define fuzzy integrals of set-valued functions (simply said to be set-valued fuzzy integrals). The method we will use here is just similar to Aumann's. Then it is a natural extension of fuzzy integrals of point-valued functions. In the paper, the following concepts and notations will be used. Symbol R ÷ denotes the interval [0, ~], P(R ÷) will denote the power set of R ÷. The triplet (X, ~1, m) is a classical complete and finite measure space (nonfuzzy). L e t / z :~1 ~ R ÷ be a fuzzy measure of Sugeno's sense, and in addition, we assume/z satisfies the following conditions, (i) /~ is null-additive, i.e.,/z(A) = 0 implies/z(A U B) = / z ( B ) ; (ii) /z << m, i.e., m ( A ) = 0 implies/z(A) = 0. The fuzzy integral of a measurable point-valued function f:X---~ R ÷ is defined as f4fd/z=

V

otER

+

(aAtz(AAF,,)),

where F~ = {x ~ X: f ( x ) >>-a}, A ~ M. Obviously, for a fuzzy m e a s u r e / z satisfying the above conditions, we have: If f~(x)= f2(x) for x ~ X m-a.e., then

fxfld~ = fx£d~. A set-valued function is a mapping F from X to P(R+)\{O}. By a measurable set-valued function F, it means that its graph is measurable, i.e., G r F = {(x, r) e X × R + : r e F ( x ) } e M

×B(R+),

where B ( R +) is the Borel field of R +.

Correspondence to: D. Zhang, Department of Mathematics, Jilin Prov. Inst. of Education, Changchun, Jilin, 130022, China. 0165-0114/93/$06.00 © 1993--Elsevier Science Publishers B.V. All rights reserved

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The rest of the paper consists of two parts. In Section 2, we will define set-valued fuzzy integrals at first. Then its properties will be shown. In Section 3, our main task is the discussion of convergence of sequence of set-valued fuzzy integrals. We will give Fatou's lemma and Lebesgue convergence theorem.

2. Definition and propositions In this section, by the similar way of Aumann's set-valued integrals, we first define set-valued fuzzy integrals, then we study it in detail. Let's start with the definition.

Definition. Let F be a set-valued function, A • ~/. Then the fuzzy integral of F on A is defined as

fAF d/X = {fAf d/X:f •S(F) } where S(F)= {f is measurable: selection of F.

f(x)EF(X), x ~ X m-a.e.}, i.e., the family of m-a.e, measurable

Instead of f x F d/x, we will write f F d/x. Obviously, f F d/x, may be empty. A set-valued function F is said to be integrable if f F d/x ¢ 0. Next, we will give a sufficient condition for which F is integrable.

Proposition 2.1. If F is a measurable set-valued function, then f F d/x ~ O. Proof. By the measurability of F, there exist a m-a.e, measurable selection of F. This implies

f F d /x ~ O. [] From this proposition, we can see that fuzzy integrals are special cases of the set-valued fuzzy integrals, since a point-valued function can be viewed as a special set-valued function.

Remark. With respect to an arbitrary fuzzy measure (not any restriction), we can also define the set-valued fuzzy integral by the same way. A set-valued function F is said to be convex-valued, if F(x) is convex for every x • X. The following proposition will show that the fuzzy integral of a convex-valued set-valued functions is convex.

Proposition 2.2. If a measurable set-valued function F is convex-valued, then f F d/x is convex. Proof. If f F d/x is a single point set, then it is convex. Otherwise, let Yl, Y2 • f F d/x, and Yl < Y2. Then there exist f~, f2 e S(F), such that

yl= f fld/x

and

y2= f f2d/x.

Further, let y • (Yl, Y2), we need to find

I y, f(x)=

fl, I.f2,

a f • S(F) with y = f f d/x. Define

x e (fl <~y <-f2), xe((flAf2)>y)U(fl>Y>f2), x • ((fl v f2) < y),

where (fl ~
Deli Zhang, Zixiao Wang / On set-valued fuzzy integrals

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It is easy to see that f • S(F), and furthermore, /x(f ~>y)/>/x(f2 ~>y) i>/x(j~ ~>y2) ~>y2~>y and /x(f > y ) = / x ( f l > y ) ~Yl) ~
[]

Further, let F be a set-valued function, co F(x) will denote the convex hull of F(x) for each x • X. Then the following holds:

Proposition 2.3. I f F is a measurable set-valued function, then co f F d~ = f co F d~. Proof. The relation ' ~ ' is obvious. To obtain the converse relation ' ~ ' , we assume y • f co F d/x. Then there exists f • S(co F) such that y = f f d/~. If f is in S(F), then y • co f F d/z. Otherwise, we assume y is not in S(F). Since by setting. G r E = G r F n{(x, r) • X × R+:f(x)<-r <- oo} is measurable, nonempty, then there exists a m-a.e, measurable selection fl of E. Further f~ • S(F), and

f f dtx <~f fl dtt. Similarly, we can find f2 •S(F) such that f f dlx >~f f2 dtz. Consequently, we have proved y • co f F d/x. [] The set-valued function F is said to be closed-valued if F(x) is closed for every x e X.

Proposition 2.4. If a measurable set-valued function F is closed-valued, then f F d~ is closed. Proof. It is easy to see that S(F) is a closed subset of the space (R+) × (with the usual product topology). Then S(F) is compact. Since fuzzy integral ' f . d/z' is continuous with respect to function, then f F d/x is compact. Therefore, it is closed. [] Corollary. Let F be a measurable interval-valued function, i.e., F(x) = [f-(x), f+(x)]. Then (i) f - , f + • S(F); (ii) f F dtx = [f f - dt~, f f+ d/x]. Proof. Since F is closed-valued and measurable, by Castaing representation, then there exist {fn} ~ S(F) such that

F(x) = cl{fn(x)}

(cl is closure).

Then

f - ( x ) = inf{fn(x)} and f+(x) = sup{f~(x)}. Consequently, f - , f + are measurable, (i) is proved. (ii) is a direct result of Proposition 2.2 and 2.4.

[]

The above properties are all related to 'convexity'. We will deal with 'monotonicity' in the following. Let A, B • P(I). A <~B means that: (i) For each x 0 • A , there exists yo•B such that Xo<~Y0; (ii) For each Y0• B, there exists Xo • A such that Xo ~
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Proposition 2.5. Let F be a measurable set-valued function, A, B e ~1. Then A ~ B implies

fAFdtz~fBFdtz. Proposition 2.6. Let F1 and F2 be two measurable set-valued functions. Then F1<~F2 implies

f Fl dt~ <~f F2dtz. The proof of Proposition 2.5 is direct, and the proof of Proposition 2.6 is similar to the proof of Proposition 2.3, here we omit it.

3. Convergence theorem In this section, it is assumed that the fuzzy measure ~ is finite, i . e . , / z ( X ) < oo. Then Fatou's lemma and the Lebesgue convergence theorem will be shown. We begin with the concept of convergence of a sequence of elements in P(R+). Let {An} c p(R ÷) be a sequence. We define

Limsup Zn = {x: x = Lim x~, xn e Z~ (n ~ l ) and Liminf A . = {x: x = Lim x,, Xn e An (n/> 1)}. If LimsupAn=LiminfAn Lim An = A or An --> A.

=A, we say that {An} is convergent to A, and it is simply written as

Theorem 3.1 (Fatou's lemma). If {Fn} is a sequence of measurable set-valued functions, then

Limsup f & d~ c f Limsup & d~. Proof. Let y ~ Limsup f Fn d~. Then from the definition of limit superior we have that y = Lim~_~ooyn, and Yn E f Fn d/~ (n I> 1). So Yn, = ffn~ dl~ -~ y (k --~ oo) where fn, e S(F~,). Since {fn,} is included in the compact space (R+) × (with the usual product topology), we can find a suhsequence {fro} of { f j , with {fm} convergent. So ffmdl~-~y (m-~o~). Since the limits are unique, therefore we have that y = Lira f fm d/~, and this implies that y e f Limsup F~ d/~. []

Theorem 3.2. If {F~} is a sequence of measurable set-valued function, then

f Liminf f, d~ c Liminf f Fn d~. Proof. Let y e f L i m i n f F , d~. Then y = ffdtz, where f e S(LiminfF~). Write X~ R + = R + × R + × • • -, then ×~ R + is a complete metric space (with the metric induced by the usual product topology). For each x e X , define a subset G(x) of ×~ R + by

G(x) = {(Yl, Y2,-..): Yn eF,,(X) (n >i 1), and Limy~ = f(x)}. Then the statement f E S(LiminfF~) is equivalent to the statement that G is a set-valued mapping. Furthermore, we can easily prove that G is measurable (for a metric space the measurability is defined

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the same). Then there exists a m-a.e, measurable selection g of G, that is, a sequence of measurable function {fn}, such that fn • S( Fn) (n >>-1) and Limf,(x) = f (x). This implies f fn dlz --* f f dtz = y. Consequently y e Liminf f F, d/z. [] From the above two theorems, we can easily obtain the following Lebesgue convergence theorem. Theorem 3.3. If Fn(x) --. F(x) for x ~ X m-a.e., and all the F~ are measurable, then f Fn dtx ~ f F dl.t. Proof. Since F(x) = Liminf F~(x) = Limsup F,(x), then by Theorems 3.1 and 3.2:

f F d~ = f Liminf Fn dtz c Liminf f Fn d~ cLimsupfF,,dtxcfLimsupF,,dtx=fFdtx. Hence, Lim f F~ d/x exists and equals to f F dtx.

[]

Remark. We can also prove Proposition 2.4 in Section 2 by using Theorem 3.1. That is, if we set F~ = F, then Limsup Fn = F. So

LimsupfFnd~cfLimsupF~d~z=fFd~. This implies f F d/z is closed.

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