# The MultipleK-Riemann Integral

Authors: Oussama Kabbouch; Mustapha Najmeddine
Journal: Abstract and Applied Analysis
 (2021)

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## Abstract

The aim of this paper is to extend the notion ofK-Riemann integrability of functions defined over a,b to functions defined over a rectangular box of ℝn. As a generalization of step functions, we introduce a notion of K-step functions which allows us to give an equivalent definition of the K-Riemann integrable functions.


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## Body

## 1. Introduction

The motivation of this work results from the notion of theK-Riemann integral of a function defined over an interval I of ℝ recently introduced by Olbrys’ [1]. Our aim of this paper is to give an equivalent definition of the K-Riemann integrability by the use of K-step functions and extend this notion to functions defined over a rectangular box of ℝn.After introducing the notion of K-multiple integral, we give its properties and we show a K-version of Fubini’s theorem.Throughout this paper,K denotes a subfield of the field of real numbers ℝ. We denote the set of positive elements of K by K+. For two real numbers a<b, we denote
(1)a,bK=λa+1−λb:λ∈0,1∩K,a,bK=λa+1−λb:λ∈0,1∩K.If
(2)R=a1,b1×a2,b2×⋯×an,bnis a rectangular box of ℝn, we denote
(3)RK=a1,b1K×a2,b2K×⋯×an,bnK,R∘K=a1,b1K×a2,b2K×⋯×an,bnK.Following [1], we define the set of the K-partitions of the interval a,b by
(4)Pa,bK=⋃n≥1t0,t1,⋯,tn:a=t0<t1<⋯<tn=b,ti∈a,bK.As in [1], we give a generalization of the radially K-continuous and uniformly radially K-continuous functions and propositions easy to prove.We say that f is radially K-continuous if it is radially K-continuous at every point.Definition 1.
A functionf:R⟶ℝ is called radially K-continuous at a point x1,x2,⋯,xn∈R if for every y1,y2,⋯,yn∈R(5)fx1+α1y1−x1,⋯,xn+αnyn−xn⟶K+n∋α1,⋯,αn⟶0,⋯,0fx1,x2,⋯,xn.Definition 2.
We say that a functionf:R⟶ℝ is uniformly radially K-continuous if for any x1,x2,⋯,xn∈R and y1,y2,⋯,yn∈R, the mapping
(6)0,1∩Kn∋α1,⋯,αn↦fx1+α1y1−x1,⋯,xn+αnyn−xn,is uniformly continuous.Every continuous and any uniformly continuous function in the usual sense is radiallyK-continuous and uniformly radially K-continuous, respectively.Proposition 3.
(i)
A functionf:R⟶ℝ is radially K-continuous if and only if for every subrectangular box S of R, the function f∣SK is continuous(ii)
A functionf:R⟶ℝ is uniformly radially K-continuous if and only if for every subrectangular box S of R, the function f∣SK is uniformly continuous


## 2.K-Step Functions and K-Riemann Integral

In the following definition, we introduce the notion of theK-step functions which are a generalization of step functions.Definition 4.
A functionf:a,b⟶ℝ is said to be a K-step function on a,b, if there exists a K-partition π=t0,t1,⋯,tm of a,b for which f is constant on each set ti,ti+1K. In this case, we say that π is a K-compatible partition of f on a,b.We denote the set of theK-step functions on a,b by EKa,b.Remark 5.
(i)
EveryK-step function on a,b is bounded on a,bK.(ii)
IfK=ℝ, we get the usual definition of a step function.Example 1.
(i)
Every constant function ona,b is a K-step function.(ii)
Consider the following functions:(7)f1x=1,x∈0,1ℚ,0,x∈0,1\0,1ℚ,andf2x=1,x∈0,3,0,x∈3,2.It is easy to show that f1 is a ℚ-step function but it is not a ℝ-step function on 0,1, and f2 is a ℝ-step function, but it is not a ℚ-step function on 0,1.Now, suppose thatf:a,b⟶ℝ is a K-step function on a,b and let π=t0,t1,⋯,tm be a K-compatible partition of f, with fx=λi whenever x∈ti−1,tiK for all i=1,2,⋯,m. Consider the following number:
(8)IKf,π=∑i=1mti−ti−1λi.We can easily show that the numberIKf,π does not depend on the choice of the K-compatible partition.Definition 6.
Letf:a,b⟶ℝ be a K-step function on a,b, and π a K-compatible partition of f. The real number IKf,π is called the K-integral of f on a,b and is denoted by
(9)∫abftdKt.As for the introduction of the usual Riemann integral, we extend the definition of theK-Riemann integral to a broader class of functions than K-step functions.Definition 7.
A functionf:a,b⟶ℝ is said to be K-Riemann integrable on a,b, if for every ε>0, there exist two K-step functions φ,ψ on a,b such that
(10)φx≤fx≤ψxforallx∈a,bK,∫abψt−φtdKt<ε.We denote the set of theK-Riemann integrable functions by RKa,b.Remark 8.
One can easily verify thatEKa,b is a part of RKa,b.Letf:a,b⟶ℝ be a bounded function on a,bK and set
(11)EK+f,a,b=∫abψtdKt:ψϵEKa,bandfx≤ψxforallxϵa,bK,EK−f,a,b=∫abφtdKt:ψ∈EKa,bandφx≤fxforallx∈a,bK.Note thatα≤βforallα∈EK−f,a,b and β∈EK+f,a,b.Proposition 9.
Letf:a,b⟶ℝ be a bounded function on a,bK. Then, f∈RKa,b if and only if infEK+f,a,b=supEK−f,a,b.Proof.
Suppose thatf∈RKa,b. Given ε>0, then there exist φ,ψ∈EKa,b such that
(12)φx≤fx≤ψxforallx∈a,bKand∫abψtdmathbbKt−∫abφtdKt<ε.
Since(13)∫abφtdKt≤supEK−f,a,b≤infEK+f,a,b≤∫abψtdKt,then,
(14)infEK+f,a,b−supEK−f,a,b<ε,
Therefore,(15)infEK+f,a,b=supEK−f,a,b.
Conversely, letε>0, then there are φ,ψ∈EKa,b such that
(16)φx≤fx≤ψxforallx∈a,bK,−∫abφtdKt≤−supEK−f,a,b+ε2and∫abψtdKt≤infEK+f,a,b+ε2.
Then,(17)∫abψtdKt−∫abφtdKt<ε.
Therefore,f∈RKa,b.Definition 10.
Letf∈RKa,b, the common value infEK+f,a,b=supEK−f,a,b is called the K-integral of f on a,b; it is denoted by
(18)∫abftdKt.In the case whenK=ℝ, we will use the standard symbol ∫abftdt instead of ∫abftdℝt.Now, we will show that our definition of theK-integral is equivalent to that recently introduced by Olbrys’ ([1], Definition 7)Proposition 11.
A functionf:a,b⟶ℝ is K-Riemann integrable on a,b in the sense of Definition 7 if and only if it is K-Riemann integrable on a,b in the sense of ([1], Definition 7) (meaning the two definitions are equivalents). Furthermore, when f∈RKa,b, we have
(19)infEK+f,a,b=supEK−f,a,b=∫ab¯ftdKt=∫ab¯ftdKt,where
(20)∫ab¯ftdKt=infπ=t0,t1,⋯,tm∈Pa,bK∑i=1mti−ti−1Mi:Mi=supx∈ti+ti+1fx,∫ab¯ftdKt=supπ=t0,t1,⋯,tm∈Pa,bK∑i=1m‍ti−ti−1mi:mi=supx∈ti+ti+1fx.Proof.
Suppose thatf∈RKa,b in the sense of ([1], Definition 7). Given ε>0, then there exists a partition π=t0,t1,⋯,tm of a,b such that
(21)∑i=1mti−ti−1Mi−∑i=1mti−ti−1mi<ε,where Mi=supx∈ti−1,tifx and mi=infx∈ti−1,tifx,i=1,2,⋯,m. We have thus built two functions φ,ψ∈EKa,b such that
(22)φx≤fx≤ψxforallx∈a,bK,∫abψtdKt−∫abφtdKt<ε,where φ and ψ are defined on a,bK by φb=ψb=0 and
(23)φx=mi,ψx=Miforallx∈ti−1,tiK,i=1,2,⋯,m.
Therefore,f is K-Riemann integrable on a,b in the sense of Definition 7.
Conversely, suppose thatf is K-Riemann integrable on a,b in the sense of Definition 7. Then, there exist φ and ψ∈EKa,b such that
(24)φx≤fx≤ψxforallx∈a,bK,∫abψtdKt−∫abφtdKt<ε.
Letπ′=a0,a1,⋯,an be a common K-compatible partition of φ and ψ, and let
(25)Bi=supx∈ai−1,aifxandAi=infx∈ai−1,aifx,i=1,2,⋯,m.
Then,(26)∫abφtdKt≤∑i=1mai−ai−1Ai≤∑i=1mai−ai−1Bi≤∫abψtdKt,which shows
(27)∑i=1mai−ai−1Bi−.∑i=1mai−ai−1Ai<∫abψtdKt−∫abφtdKt<ε.
This means thatf is K-Riemann integrable on a,b in the sense of ([1], Definition 7).
Finally, iff∈RKa,b, then the equalities (19) are ensured by Proposition 9 and the inequalities (26).


## 3.K-Multiple Integral

This section is devoted toK-multiple Riemann integral. For the theory of the classical multiple Riemann integral, see for instance [2].LetR=a1,b1×a2,b2×⋯×an,bn be a closed rectangular box in ℝn, where ai,bi are real numbers such that ai<bi for all i=1,2,⋯,n.We denoted by(28)volR=∏i=1nbi−ai,RK≔a1,b1K×a2,b2K×⋯×an,bnK.Definition 12.
A rectangular boxS=a1′,b1′×a2′,b2′×⋯×an′,bn′ is said to be a K-subrectangular box of R if ai′,bi′∈ai,biK for all i=1,2,⋯,n.Example 2.
0,3/4×0,1/2 is a ℚ-subrectangular box of 0,1×0,1.
0,1×0,2 is not a ℚ-subrectangular box of 0,3×0,2.Definition 13.
AK-partition of R is a finite collection π=Rii=1m of K-subrectangular box of R such that
(29)R=∪mi=1RiandR∘i∩R∘j=∅,i≠j,where R∘i is the interior of Ri.Definition 14.
(i)
Letf:R⟶ℝ be a bounded function over RK. f is said to be a K-step function on R if there exists a K-partition π=Rii=1m such that f is constant on R∘iK for all i=1,2,⋯,m. In this case, π is said to be a K-compatible partition of f on R
We denoted the set of theK-step functions on R by EKR.
(ii)
AK-refinement of a K-partition Rii=1m of R is another K-partition Sjj=1r of R such that each Sj is a K-subrectangular box of some RiLetf:R⟶ℝ be a K-step function and π=Rii=1m be a K-compatible partition of f such that fx=λi for all x∈Ri∘K,i=1,2,⋯,m. We define the number
(30)IKf,π≔∑i=1mλivolRi.We will first show thatIKf,π does not depend on the choice of the K-compatible partition π.Lemma 15.
Letf:R⟶ℝ be a K-step function and π be a K-compatible partition of f. Then, for every K-refinement π′ of π, we have
(31)IKf,π=IKf,π′.Proof.
Letπ=Rii=1m be a K-compatible partition of f such that fx=λi for all x∈Ri∘K,i=1,2,⋯,m and π′=Sjj=1r be a K-refinement of π. For all i∈1,2,⋯,m, there is i1,i2,⋯,iki∈1,2,⋯,r such that Ri=∪kij=1Sij. Then,
(32)IKf,π=∑i=1mλivol∪kij=1Sij=∑i=1mλi∑j=1kivolSij=∑i=1m∑j=1kiλivolSij=IKf,π′.Corollary 16.
Letf:R⟶ℝ be a K-step function and π be a K-compatible partition of f. The number IKf,π does not depend on the choice of the K-compatible partition π.Proof.
Letπ=Rii=1m and π′=Sjj=1r be two K-compatible partitions of f, then π′′=Tij1≤i≤m1≤j≤r, where Tij=Ri∩Sj, is a K-refinement of π and π′. Therefore, by using Lemma 15, we get
(33)IKf,π=IKf,π′=IKf,π′′.Definition 17.
Letf:R⟶ℝ be a K-step function and π be a K-compatible partition of f. We called IKf,π the K-Riemann integral of f on R and we denote it by
(34)∫RfxdKx.In the following proposition, we present some basic properties ofK-Riemann multiple integral of K-step functions.Proposition 18.
Letf,g:R⟶ℝ be two K-step functions on R and α,β∈ℝ. Then,
(1)
The functionαf+βg is a K-step function on R; moreover,(35)∫Rαfx+βgxdKx=α∫RfxdKx+β∫RgxdKx(2)
Iffx≥0,x∈RK, then ∫RfxdKx≥0(3)
The absolute valuef is a K-step function on R and ∫RfxdKx≤∫RfxdKx(4)
Ifπ=Sii=1r is a K-partition on R, then ∫RfxdKx=∑i=1r∫SifxdKxProof.
(1) Letπ=Rii=1mrespπ′=Sjj=1r be a K-compatible partition of frespg, then π′′=Tij1≤i≤m1≤j≤r, where Tij=Ri∩Sj, is a K-compatible partition of αf+β; moreover, we have
(36)∫Rαfx+βgxdKx=IKαf+βg,π″=αIKf,π″+βIKg,π″=αIKf,π+βIKg,π′=α∫RfxdKx+β∫RgxdKx.
(2) and (3) are evident.
(4) Letπ=Rii=1m be a K-compatible partition of f such that fx=λi for all x∈Ri∘K,i=1,2,⋯,m. Then,
(37)∫RfxdKx=∑i=1mλivolRi=∑i=1mλi∑j=1rvolRi∩Sj=∑j=1r∑i=1mλivolRi∩Sj.
SinceRi∩Sji=1m is a K-compatible partition of f on Sj. Then,
(38)∫RfxdKx=∑j=1r∫SjfxdKx


### 3.1.K-Riemann Integrable Functions Defined over R

Definition 19.
Letf:R⟶ℝ be a function, we said that f is a K-Riemann integrable function, if for every ε>0, there exist two K-step functions φ and ψ on R such that
(39)φx≤fx≤ψxforallx∈RK,∫R∫Kψx−φx.We denote the set of theK-Riemann integrable functions by RKR.Remark 20.
(i)
EveryK-Riemann integrable function on R is bounded on RK(ii)
EKR is a part of RKRNow, suppose thatf:R⟶ℝ is a bounded function on RK. Set
(40)EK+f,R=∫RψxdKx:ψ∈εKRandfx≤ψxforallx∈RK,EK−f,R=∫RφxdKx:φ∈εKRandφx≤fxforallx∈RK.Note thatα≤β for all α∈EK−f,R and β∈EK+f,R.Theorem 21.
Letf:R⟶ℝ be a bounded function on RK. Then, f∈RKR if and only if infEK+f,R=supEK−f,R.Proof.
Suppose thatf∈RKR. Given ε>0, there exists φ and ψ in EKR such that
(41)φx≤fx≤ψxforallx∈RK,∫RψxdKx−∫RφxdKx<ε.
Then,(42)∫RφxdKx≤supEK−f,R≤infEK+f,R≤∫RψxdKx,which shows that
(43)infEK+f,R−supEK−f,R<ε,for all ε>0. Therefore,
(44)infEK+f,R=supEK−f,R.
Conversely, suppose thatinfEK+f,R=supEK−f,R. Let ε>0, by the definition of the greatest lower bound and the least upper bound, there are φ,ψ∈EKR such that
(45)φx≤fx≤ψxforallx∈RK,−∫RφxdKx≤−supEK−f,R+ε2,∫RψxdKx≤infEK+f,R+ε2.
Then,(46)∫RψxdKx−∫RφxdKx<ε.
Therefore,f∈RKR.Definition 22.
Letf∈RKR, the common value infEK+f,R=supEK−f,R is called the K-Riemann integral of f on R, and we denote it by
(47)∫RfxdKx.In the case whenK=ℝ, we will use the standard symbol ∫Rfxdx instead of ∫Rfxdℝx.Proposition 23.
Letf:R⟶ℝ be a function. Then, f is K-Riemann integrable on R if and only if there exist two sequences of K-step functions φmm and ψmm such that
(48)φmx≤fx≤ψmxm∈ℕ,x∈RK,limm↦∞∫Rψmx−φmxdKt=0.Moreover, we have(49)∫RfxdKx=limm→∞∫RφmxdKx=limm→∞∫RψmxdKx.Before presenting a version of the Fubini theorem for K-Riemann integrable functions. We start by giving a lemma that is essential for the proof of this result.Proof.
It runs in a similar way as for the case of one variable. Suppose thatf is K-integrable. Then, infEK+f,R=supEK−f,R. For all m∈ℕ∗, there exist two K-step functions φm and ψm on R such that
(50)φm≤f≤ψmonRKand∫Rψm−φmxdKx<1m.
Hence,limm↦+∞∫Rψm−φmxdKx=0. Now, let us show that
(51)∫RfxdKx=limm→∞∫RφmxdKx.
Letm∈ℕ. We have
(52)∫RφmxdKx≤∫RfxdKx≤∫RψmxdKx.
Then,(53)0≤∫RfxdKx−∫RφmxdKx≤∫Rψmx−φmxdKx.
Hence,(54)0≤limm→∞∫RfxdKx−∫RφmxdKx≤limm→∞∫Rψmm−φmxdKx=0.
The converse is trivial.Lemma 24.
LetK1⊆K2 be two subfields of ℝ and let f:R⟶ℝ be a K2-step function. Then, f is a K1-Riemann integrable function. Moreover, we have
(55)∫RfxdK1x=∫RfxdK2x.Proof.
LetRii=1m be a K2-compatible partition of f such that fx=λi for all x∈RiK∘. For all i∈1,2,⋯,m, there is a sequence of K1-subrectangular Sipp of R such that
(56)Sip⊂Sip+1,p∈ℕ,limp⟶∞Sip=∪∞p=0Sip=Ri,i∈1,2,⋯,m.
Set(57)Tp=clR\∪mi=1Sip,where clR\∪mi=1Sip is the closure of R\∪mi=1Sip. Then, there is a sequence Tjpj=1kp of K1-subrectangular of R such that
(58)Tp=∪kpj=1TjpandTjp∘∩Tj′p∘=∅,j≠j′.
Now, consider the functions(59)ψpx=fx,x∈∪mj=1SjpK1,M=supx∈RK2fx,else,φpx=fx,x∈∪mj=1SjpK1,m=infx∈RK2fx,else.We have π=Tjpj=1kp∪Sjpj=1m is a K1-compatible partition of φp and ψp on R; moreover,
(60)φpx≤fx≤ψpxforallx∈RK1,∫RψPx−φpxdK1x=∑j=1kpM−mvolTjp=M−mvolTp.Since Tpp is a decreasing sequence, then limp⟶∞Tp=∩∞p=0Tp=∅ then limp⟶∞volTp=0. Hence, f is K1-Riemann integrable. Moreover,
(61)limp→∞∫RψpxdK1x=limp→∞∑j=1kpMvolTjp+∑j=1mλjvolSjp=∑j=1mλjlimp→∞volSjp=∑j=1mλjvolRj=∫RfxdK2x.Theorem 25.
LetK1⊆K2 be two subfields of ℝ. Then, RK2R⊂RK1R and
(62)∫RfxdK1x=∫RfxdK2xforallf∈RK2R.Proof.
Letf∈RK2R, given ε>0, there exist φ,ψ∈EK2R such that
(63)φx≤fx≤ψxforallx∈ℝK2,∫ℝψx−φxdK2x<ε.By Lemma 24, there exist two sequences φ~nn and ψ~nn of K1-step functions such that
(64)φ~nx≤φx≤ψx≤ψ~nx,x∈RK1,with
(65)limn⟶∞∫ℝφ~nxdK1x=∫ℝφxdK2xandlimn⟶∞∫ℝψ~nxdK1x=∫abφxdK2x.Then, there is n0∈ℕ such that
(66)−∫ℝφ~n0xdK1x+ε≤−∫ℝφxdK2x,∫Rψ~n0xdK1x−ε≤∫RψxdK2x.Hence,
(67)∫Rψ~n0xdK1x−∫Rφ~n0xdK1x<ε.Therefore, f∈RK1R and
(68)∫RfxdK1x=∫RfxdK2x.Corollary 26.
Letf:R⟶ℝ be a uniformly radially K-continuous function, then it is K-Riemann integrable on any subrectangular S of R.Proof.
LetS be a subrectangular of R. By Proposition 3, f is uniformly continuous on SK. Since clSK=S, where clSK is the closure of SK, then there exists a unique continuous function g:S⟶ℝ such that gx=fx,x∈SK. Since g is Riemann integrable on S then, using Theorem 25, g is K-Riemann integrable on S; moreover,
(69)∫SfxdKx=∫Sgxdx.Lemma 27.
Letf:R=R1×R2⟶ℝ be a K-step function on R. Then, there is a K-partition R1ii=1r of R1 and a K-partition R2jj=1s of R2 such that the function
(70)fx:R2⟶ℝ,y↦fx,yis a K-step function and R2jj=1s is a K-compatible partition of fx whenever x∈∪ri=1Ri∘1K. Moreover, the function
(71)Fx=∫R2fx,ydKy,x∈∪ri=1Ri∘1K,0,elseis a K-step function on R1 and
(72)∫R1FxdKx=∫Rfx,ydKxdKy.Proof.
Letπ=Tii=1m be a K-partition of R, then there exists a K-refinement T1i×T2j1≤i≤r1≤j≤s of π such that T1ii=1r and T2jj=1s are two K-partitions of R1 and R2, respectively. Since f is a K-step function on R, then there are two K-partitions R1ii=1r and R2jj=1s of R1 and R2, respectively, such that R1i×R2j1≤i≤r1≤j≤s is a K-partition of R and f is constant on each R1i∘K×R2j∘K; hence, fx is a K-step function on R2.Now, we can say that F is a K-step function on R1.Let us show the last equality in the theorem. Consider the two K-partitions R1ii=1r and R2jj=1s given in the first step, then for all i=1,2,⋯,r and j=1,2,⋯,s, there exist λij such that fx,y=λij whenever x,y∈R1i∘K×R2j∘K; therefore, we have
(73)Fx=∑j=1sλijvolR2j,x∈R∘1Ki,i=1,2,⋯,r,0,else,∫R1FxdKx=∑i=1r∑j=1sλijvolR2jvolR1i=∑i=1r∑j=1sλijvolR2j×R1i=∫Rfx,ydKxdKy.Theorem 28 (Fubini theorem).
Letf:R=R1×R2⟶ℝ be a K-Riemann integrable function on R. If the function
(74)fx:R2⟶ℝ,y↦fx,yis K-Riemann integrable on R2 for all x∈R1K, then the function
(75)Fx=∫R2fx,ydKy,x∈R1K,0,elseis a K-Riemann integrable function on R1 and
(76)∫Rfx,ydKxdKy=∫R1FxdKx.Proof.
Letε>0, there exist two K-step functions φ and ψ on R such that
(77)φx,y≤fx,y≤ψx,yforallx,y∈R1K×R2K,∫Rψx,y−φx,ydKxdKy<ε.
Now, we can find twoK-step functions Φ and Ψ such that
(78)∫R1ΦxdKx=∫Rφx,ydKxdKy,∫R1ΨxdKx=∫Rψx,ydKxdKy,Φx≤Fx≤Ψx,x∈R1K.
Since(79)∫R1Ψ−ΦxdKx=∫Rψ−φx,ydKxdKy<ε,then F is K-Riemann integrable on R; moreover,
(80)∫R1FxdKx=∫Rfx,ydKxdKy.


### 3.2. Biconvex Function

Now, we give the notion ofK-biconvex function and we will prove that any K-biconvex function is K-Riemann integrable.Since a function is Jensen-convex if and only if it is ℚ-convex, for more details, see [3, 4]. Then, we can easily show that a function is Jensen-biconvex if and only if it is ℚ-biconvex.Definition 29.
A mappingf:ℝ×ℝ⟶ℝ is said to be K-bilinear, if it satisfies
(i)
fλx+y,z=λfx,z+fy,z(ii)
fx,λy+z=λfx,y+fx,zfor every λ∈K and for every x,y,z∈ℝ.Definition 30.
A functionf:a,b×c,d⟶ℝ is said to be
(1)
Jensen-biconvex if(i)
fx+y/2,z≤fx,z+fy,z/2 for every x,y∈a,b and z∈c,d(ii)
fx,y+z/2≤fx,y+fx,z/2 for every x∈a,b and y,z∈c,d(2)
K-biconvex if(i)
fλx+1−λy,z≤λfx,z+1−λfy,z, for every x,y∈a,b and z∈c,d and λ∈0,1K(ii)
fx,λy+1−λz≤λfx,y+1−λfx,z for every x∈a,b and y,z∈c,d and λ∈0,1KNow, we claim that everyK-biconvex function is K-Riemann integrable. For the proof, we need the following lemma (see [5] p.143)Lemma 31.
LetI⊂ℝ be an open interval and g:I⟶ℝ be a convex function. Then,
(81)gx−gx−ndn≤gx−gx−mdm≤gx+md−gxm≤gx+nd−gxn,for every x∈I,d∈ℝ and m,n∈ℕ such that 0<m<n and x±nd∈I.Theorem 32.
Letf:R⟶ℝ be a bounded and K-biconvex function, and let S be a closed rectangular box such that S⊂R∘. Then, f is uniformly continuous on SK.Proof.
PutS=a,b×c,d⊂R∘, there exists a positive number ρ∈K+ such that
(82)a1,b1×c1,d1⊂R∘,with a1=a−ρb−a,b1=b+ρb−a,c1=c−ρd−c and d1=d+ρd−c. We show easily that
(83)a,bK×c,dK⊂a1,b1K×c1,d1K.It is sufficient to show that a,bK⊂a1,b1K. If u∈a,bK, then there exists λ∈0,1∩K such that u=λa+1−λb. Put ν=ρ+λ/1+2ρ, then ν∈0,1∩K. Moreover,
(84)νa1+1−νb1=ρ+λ1+2ρa−ρb−a+1+ρ−λ1+2ρb+ρb−a=u.Then,
(85)a,bK×c,dK⊂a1,b1K×c1,d1K.On the other hand, since f is bounded, there exist two real numbers m≤M such that
(86)m≤fx,y≤Mforallx,y∈R.Now, given an arbitrary ε>0, put n=2M−m/ε+1theintegralpartof2M−m/ε+1. We have n∈ℕ and
(87)2M−mn<ε.Further, put η=ρ/nminb−a,d−c. Take arbitrary x,y,x′,y′∈a,bK×c,dK such that
(88)x,y−x′,y′∞=maxx−x′,y−y′<η.Then, there exist αi,βi∈0,1∩K, i=1,2 such that
(89)x=α1a+1−α1b,x′=α2a+1−α2b,y=β1c+1−β1d,y′=β2c+1−β2d.Whence
(90)x−x′=α2−α1b−a,y−y′=β2−β1d−c,and by (88), we get
(91)nα1−α2<ρ,(92)nβ1−β2<ρ.Consider the numbers
(93)μ12=ρ+α1±nα2−α11+2ρ,λ12=ρ+β1±nβ2−β11+2ρ.By (92), we have
(94)0≤α11+2ρ<μ12<α1+2ρ1+2ρ≤1,0≤β11+2ρ<λ12<β1+2ρ1+2ρ≤1.Then, μ12,λ12∈0,1∩K. Now,
(95)μ12a1+1−μ12b1=x±nx−x′∈a1,b1K,λ12c1+1−μ12d1=y±ny−y′∈c1,d1K.Since the functions
(96)fy:I⟶ℝ,x↦fx,y,andfx′:J⟶ℝ,y↦fx′,yare K-convex, taking in Lemma 31x"=x,d=x−x′, and m=1, we get
(97)fx,y−fx−nd,yn≤fx,y−fx′,y≤fx′+nd,y−fx′,yn,and taking y"=y,d′=y−y′, and m=1, we find that
(98)fx′,y−fx′,y−nd′n≤fx′,y−fx′,y′≤fx′,y+nd′−fx′,yn.Since
(99)m≤fx,y≤Mforallx,y∈R,then
(100)−2M−mn≤fx,y−fx′,y′≤2M−mn,which mean, by using (87), that
(101)−ε<fx,y−fx′,y′<ε,Hence, f is uniformly continuous on SK.Corollary 33.
Letf:R⟶ℝ be a bounded and K-biconvex function. Then, f is K-Riemann integrable on any subrectangular S such that S⊂R∘.Proposition 34.
Letf:ℝ×ℝ⟶ℝ be a K-bilinear function. Then, it is K-Riemann integrable on every rectangular box R=a,b×c,d. Moreover,
(102)∫Rfx,ydKxdKy=fa+b2,c+d2b−ad−c.Proof.
f is a K-bilinear function; then, it is a K-biconvex function; hence, f is a K-Riemann integrable function. Since
(103)fy:a,b⟶ℝ,y∈c,d,y↦fx,y,fx:c,d⟶ℝ,x∈a,b,y↦fx,yare K-convex function, then by ([1], Theorem 7), they are K-Riemann integrable functions; moreover,
(104)∫cdfx,ydKy=d−cfx,c+d2,x∈a,b,∫abfx,ydKx=b−afa+b2,y,y∈c,d.By Fubini Theorem 3.18, we get
(105)∫Rfx,ydKxdKy=∫ab∫cdfx,ydKydKx=∫abd−cfx,c+d2dKx=fa+b2,c+d2b−ad−c.Remark 35.
In this last part, we only took the case of two variables. By the same way, we can extend these results to the case ofn variables.


## 3.1.K-Riemann Integrable Functions Defined over R

Definition 19.
Letf:R⟶ℝ be a function, we said that f is a K-Riemann integrable function, if for every ε>0, there exist two K-step functions φ and ψ on R such that
(39)φx≤fx≤ψxforallx∈RK,∫R∫Kψx−φx.We denote the set of theK-Riemann integrable functions by RKR.Remark 20.
(i)
EveryK-Riemann integrable function on R is bounded on RK(ii)
EKR is a part of RKRNow, suppose thatf:R⟶ℝ is a bounded function on RK. Set
(40)EK+f,R=∫RψxdKx:ψ∈εKRandfx≤ψxforallx∈RK,EK−f,R=∫RφxdKx:φ∈εKRandφx≤fxforallx∈RK.Note thatα≤β for all α∈EK−f,R and β∈EK+f,R.Theorem 21.
Letf:R⟶ℝ be a bounded function on RK. Then, f∈RKR if and only if infEK+f,R=supEK−f,R.Proof.
Suppose thatf∈RKR. Given ε>0, there exists φ and ψ in EKR such that
(41)φx≤fx≤ψxforallx∈RK,∫RψxdKx−∫RφxdKx<ε.
Then,(42)∫RφxdKx≤supEK−f,R≤infEK+f,R≤∫RψxdKx,which shows that
(43)infEK+f,R−supEK−f,R<ε,for all ε>0. Therefore,
(44)infEK+f,R=supEK−f,R.
Conversely, suppose thatinfEK+f,R=supEK−f,R. Let ε>0, by the definition of the greatest lower bound and the least upper bound, there are φ,ψ∈EKR such that
(45)φx≤fx≤ψxforallx∈RK,−∫RφxdKx≤−supEK−f,R+ε2,∫RψxdKx≤infEK+f,R+ε2.
Then,(46)∫RψxdKx−∫RφxdKx<ε.
Therefore,f∈RKR.Definition 22.
Letf∈RKR, the common value infEK+f,R=supEK−f,R is called the K-Riemann integral of f on R, and we denote it by
(47)∫RfxdKx.In the case whenK=ℝ, we will use the standard symbol ∫Rfxdx instead of ∫Rfxdℝx.Proposition 23.
Letf:R⟶ℝ be a function. Then, f is K-Riemann integrable on R if and only if there exist two sequences of K-step functions φmm and ψmm such that
(48)φmx≤fx≤ψmxm∈ℕ,x∈RK,limm↦∞∫Rψmx−φmxdKt=0.Moreover, we have(49)∫RfxdKx=limm→∞∫RφmxdKx=limm→∞∫RψmxdKx.Before presenting a version of the Fubini theorem for K-Riemann integrable functions. We start by giving a lemma that is essential for the proof of this result.Proof.
It runs in a similar way as for the case of one variable. Suppose thatf is K-integrable. Then, infEK+f,R=supEK−f,R. For all m∈ℕ∗, there exist two K-step functions φm and ψm on R such that
(50)φm≤f≤ψmonRKand∫Rψm−φmxdKx<1m.
Hence,limm↦+∞∫Rψm−φmxdKx=0. Now, let us show that
(51)∫RfxdKx=limm→∞∫RφmxdKx.
Letm∈ℕ. We have
(52)∫RφmxdKx≤∫RfxdKx≤∫RψmxdKx.
Then,(53)0≤∫RfxdKx−∫RφmxdKx≤∫Rψmx−φmxdKx.
Hence,(54)0≤limm→∞∫RfxdKx−∫RφmxdKx≤limm→∞∫Rψmm−φmxdKx=0.
The converse is trivial.Lemma 24.
LetK1⊆K2 be two subfields of ℝ and let f:R⟶ℝ be a K2-step function. Then, f is a K1-Riemann integrable function. Moreover, we have
(55)∫RfxdK1x=∫RfxdK2x.Proof.
LetRii=1m be a K2-compatible partition of f such that fx=λi for all x∈RiK∘. For all i∈1,2,⋯,m, there is a sequence of K1-subrectangular Sipp of R such that
(56)Sip⊂Sip+1,p∈ℕ,limp⟶∞Sip=∪∞p=0Sip=Ri,i∈1,2,⋯,m.
Set(57)Tp=clR\∪mi=1Sip,where clR\∪mi=1Sip is the closure of R\∪mi=1Sip. Then, there is a sequence Tjpj=1kp of K1-subrectangular of R such that
(58)Tp=∪kpj=1TjpandTjp∘∩Tj′p∘=∅,j≠j′.
Now, consider the functions(59)ψpx=fx,x∈∪mj=1SjpK1,M=supx∈RK2fx,else,φpx=fx,x∈∪mj=1SjpK1,m=infx∈RK2fx,else.We have π=Tjpj=1kp∪Sjpj=1m is a K1-compatible partition of φp and ψp on R; moreover,
(60)φpx≤fx≤ψpxforallx∈RK1,∫RψPx−φpxdK1x=∑j=1kpM−mvolTjp=M−mvolTp.Since Tpp is a decreasing sequence, then limp⟶∞Tp=∩∞p=0Tp=∅ then limp⟶∞volTp=0. Hence, f is K1-Riemann integrable. Moreover,
(61)limp→∞∫RψpxdK1x=limp→∞∑j=1kpMvolTjp+∑j=1mλjvolSjp=∑j=1mλjlimp→∞volSjp=∑j=1mλjvolRj=∫RfxdK2x.Theorem 25.
LetK1⊆K2 be two subfields of ℝ. Then, RK2R⊂RK1R and
(62)∫RfxdK1x=∫RfxdK2xforallf∈RK2R.Proof.
Letf∈RK2R, given ε>0, there exist φ,ψ∈EK2R such that
(63)φx≤fx≤ψxforallx∈ℝK2,∫ℝψx−φxdK2x<ε.By Lemma 24, there exist two sequences φ~nn and ψ~nn of K1-step functions such that
(64)φ~nx≤φx≤ψx≤ψ~nx,x∈RK1,with
(65)limn⟶∞∫ℝφ~nxdK1x=∫ℝφxdK2xandlimn⟶∞∫ℝψ~nxdK1x=∫abφxdK2x.Then, there is n0∈ℕ such that
(66)−∫ℝφ~n0xdK1x+ε≤−∫ℝφxdK2x,∫Rψ~n0xdK1x−ε≤∫RψxdK2x.Hence,
(67)∫Rψ~n0xdK1x−∫Rφ~n0xdK1x<ε.Therefore, f∈RK1R and
(68)∫RfxdK1x=∫RfxdK2x.Corollary 26.
Letf:R⟶ℝ be a uniformly radially K-continuous function, then it is K-Riemann integrable on any subrectangular S of R.Proof.
LetS be a subrectangular of R. By Proposition 3, f is uniformly continuous on SK. Since clSK=S, where clSK is the closure of SK, then there exists a unique continuous function g:S⟶ℝ such that gx=fx,x∈SK. Since g is Riemann integrable on S then, using Theorem 25, g is K-Riemann integrable on S; moreover,
(69)∫SfxdKx=∫Sgxdx.Lemma 27.
Letf:R=R1×R2⟶ℝ be a K-step function on R. Then, there is a K-partition R1ii=1r of R1 and a K-partition R2jj=1s of R2 such that the function
(70)fx:R2⟶ℝ,y↦fx,yis a K-step function and R2jj=1s is a K-compatible partition of fx whenever x∈∪ri=1Ri∘1K. Moreover, the function
(71)Fx=∫R2fx,ydKy,x∈∪ri=1Ri∘1K,0,elseis a K-step function on R1 and
(72)∫R1FxdKx=∫Rfx,ydKxdKy.Proof.
Letπ=Tii=1m be a K-partition of R, then there exists a K-refinement T1i×T2j1≤i≤r1≤j≤s of π such that T1ii=1r and T2jj=1s are two K-partitions of R1 and R2, respectively. Since f is a K-step function on R, then there are two K-partitions R1ii=1r and R2jj=1s of R1 and R2, respectively, such that R1i×R2j1≤i≤r1≤j≤s is a K-partition of R and f is constant on each R1i∘K×R2j∘K; hence, fx is a K-step function on R2.Now, we can say that F is a K-step function on R1.Let us show the last equality in the theorem. Consider the two K-partitions R1ii=1r and R2jj=1s given in the first step, then for all i=1,2,⋯,r and j=1,2,⋯,s, there exist λij such that fx,y=λij whenever x,y∈R1i∘K×R2j∘K; therefore, we have
(73)Fx=∑j=1sλijvolR2j,x∈R∘1Ki,i=1,2,⋯,r,0,else,∫R1FxdKx=∑i=1r∑j=1sλijvolR2jvolR1i=∑i=1r∑j=1sλijvolR2j×R1i=∫Rfx,ydKxdKy.Theorem 28 (Fubini theorem).
Letf:R=R1×R2⟶ℝ be a K-Riemann integrable function on R. If the function
(74)fx:R2⟶ℝ,y↦fx,yis K-Riemann integrable on R2 for all x∈R1K, then the function
(75)Fx=∫R2fx,ydKy,x∈R1K,0,elseis a K-Riemann integrable function on R1 and
(76)∫Rfx,ydKxdKy=∫R1FxdKx.Proof.
Letε>0, there exist two K-step functions φ and ψ on R such that
(77)φx,y≤fx,y≤ψx,yforallx,y∈R1K×R2K,∫Rψx,y−φx,ydKxdKy<ε.
Now, we can find twoK-step functions Φ and Ψ such that
(78)∫R1ΦxdKx=∫Rφx,ydKxdKy,∫R1ΨxdKx=∫Rψx,ydKxdKy,Φx≤Fx≤Ψx,x∈R1K.
Since(79)∫R1Ψ−ΦxdKx=∫Rψ−φx,ydKxdKy<ε,then F is K-Riemann integrable on R; moreover,
(80)∫R1FxdKx=∫Rfx,ydKxdKy.


## 3.2. Biconvex Function

Now, we give the notion ofK-biconvex function and we will prove that any K-biconvex function is K-Riemann integrable.Since a function is Jensen-convex if and only if it is ℚ-convex, for more details, see [3, 4]. Then, we can easily show that a function is Jensen-biconvex if and only if it is ℚ-biconvex.Definition 29.
A mappingf:ℝ×ℝ⟶ℝ is said to be K-bilinear, if it satisfies
(i)
fλx+y,z=λfx,z+fy,z(ii)
fx,λy+z=λfx,y+fx,zfor every λ∈K and for every x,y,z∈ℝ.Definition 30.
A functionf:a,b×c,d⟶ℝ is said to be
(1)
Jensen-biconvex if(i)
fx+y/2,z≤fx,z+fy,z/2 for every x,y∈a,b and z∈c,d(ii)
fx,y+z/2≤fx,y+fx,z/2 for every x∈a,b and y,z∈c,d(2)
K-biconvex if(i)
fλx+1−λy,z≤λfx,z+1−λfy,z, for every x,y∈a,b and z∈c,d and λ∈0,1K(ii)
fx,λy+1−λz≤λfx,y+1−λfx,z for every x∈a,b and y,z∈c,d and λ∈0,1KNow, we claim that everyK-biconvex function is K-Riemann integrable. For the proof, we need the following lemma (see [5] p.143)Lemma 31.
LetI⊂ℝ be an open interval and g:I⟶ℝ be a convex function. Then,
(81)gx−gx−ndn≤gx−gx−mdm≤gx+md−gxm≤gx+nd−gxn,for every x∈I,d∈ℝ and m,n∈ℕ such that 0<m<n and x±nd∈I.Theorem 32.
Letf:R⟶ℝ be a bounded and K-biconvex function, and let S be a closed rectangular box such that S⊂R∘. Then, f is uniformly continuous on SK.Proof.
PutS=a,b×c,d⊂R∘, there exists a positive number ρ∈K+ such that
(82)a1,b1×c1,d1⊂R∘,with a1=a−ρb−a,b1=b+ρb−a,c1=c−ρd−c and d1=d+ρd−c. We show easily that
(83)a,bK×c,dK⊂a1,b1K×c1,d1K.It is sufficient to show that a,bK⊂a1,b1K. If u∈a,bK, then there exists λ∈0,1∩K such that u=λa+1−λb. Put ν=ρ+λ/1+2ρ, then ν∈0,1∩K. Moreover,
(84)νa1+1−νb1=ρ+λ1+2ρa−ρb−a+1+ρ−λ1+2ρb+ρb−a=u.Then,
(85)a,bK×c,dK⊂a1,b1K×c1,d1K.On the other hand, since f is bounded, there exist two real numbers m≤M such that
(86)m≤fx,y≤Mforallx,y∈R.Now, given an arbitrary ε>0, put n=2M−m/ε+1theintegralpartof2M−m/ε+1. We have n∈ℕ and
(87)2M−mn<ε.Further, put η=ρ/nminb−a,d−c. Take arbitrary x,y,x′,y′∈a,bK×c,dK such that
(88)x,y−x′,y′∞=maxx−x′,y−y′<η.Then, there exist αi,βi∈0,1∩K, i=1,2 such that
(89)x=α1a+1−α1b,x′=α2a+1−α2b,y=β1c+1−β1d,y′=β2c+1−β2d.Whence
(90)x−x′=α2−α1b−a,y−y′=β2−β1d−c,and by (88), we get
(91)nα1−α2<ρ,(92)nβ1−β2<ρ.Consider the numbers
(93)μ12=ρ+α1±nα2−α11+2ρ,λ12=ρ+β1±nβ2−β11+2ρ.By (92), we have
(94)0≤α11+2ρ<μ12<α1+2ρ1+2ρ≤1,0≤β11+2ρ<λ12<β1+2ρ1+2ρ≤1.Then, μ12,λ12∈0,1∩K. Now,
(95)μ12a1+1−μ12b1=x±nx−x′∈a1,b1K,λ12c1+1−μ12d1=y±ny−y′∈c1,d1K.Since the functions
(96)fy:I⟶ℝ,x↦fx,y,andfx′:J⟶ℝ,y↦fx′,yare K-convex, taking in Lemma 31x"=x,d=x−x′, and m=1, we get
(97)fx,y−fx−nd,yn≤fx,y−fx′,y≤fx′+nd,y−fx′,yn,and taking y"=y,d′=y−y′, and m=1, we find that
(98)fx′,y−fx′,y−nd′n≤fx′,y−fx′,y′≤fx′,y+nd′−fx′,yn.Since
(99)m≤fx,y≤Mforallx,y∈R,then
(100)−2M−mn≤fx,y−fx′,y′≤2M−mn,which mean, by using (87), that
(101)−ε<fx,y−fx′,y′<ε,Hence, f is uniformly continuous on SK.Corollary 33.
Letf:R⟶ℝ be a bounded and K-biconvex function. Then, f is K-Riemann integrable on any subrectangular S such that S⊂R∘.Proposition 34.
Letf:ℝ×ℝ⟶ℝ be a K-bilinear function. Then, it is K-Riemann integrable on every rectangular box R=a,b×c,d. Moreover,
(102)∫Rfx,ydKxdKy=fa+b2,c+d2b−ad−c.Proof.
f is a K-bilinear function; then, it is a K-biconvex function; hence, f is a K-Riemann integrable function. Since
(103)fy:a,b⟶ℝ,y∈c,d,y↦fx,y,fx:c,d⟶ℝ,x∈a,b,y↦fx,yare K-convex function, then by ([1], Theorem 7), they are K-Riemann integrable functions; moreover,
(104)∫cdfx,ydKy=d−cfx,c+d2,x∈a,b,∫abfx,ydKx=b−afa+b2,y,y∈c,d.By Fubini Theorem 3.18, we get
(105)∫Rfx,ydKxdKy=∫ab∫cdfx,ydKydKx=∫abd−cfx,c+d2dKx=fa+b2,c+d2b−ad−c.Remark 35.
In this last part, we only took the case of two variables. By the same way, we can extend these results to the case ofn variables.


## 4. Conclusion

We plan, thereafter, to extend the definition ofk-integrability to functions defined on bounded subsets and to generalize Fubini’s theorem.



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