# Integral Representation of Functions of Bounded Variation

Authors: Z. Lipcsey; I. M. Esuabana; J. A. Ugboh; I. O. Isaac
Journal: Journal of Mathematics
 (2019)

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## Abstract

Functions of bounded variations form important transition between absolute continuous and singular functions. With Bainov’s introduction of impulsive differential equations having solutions of bounded variation, this class of functions had eventually entered into the theory of differential equations. However, the determination of existence of solutions is still problematic because the solutions of differential equations is usually at least absolute continuous which is disrupted by the solutions of bounded variations. As it is known, iff:[a,bλ]→Rn is of bounded variation then f is the sum of an absolute continuous function fa and a singular function fs where the total variation of fs generates a singular measure τ and fs is absolute continuous with respect to τ. In this paper we prove that a function of bounded variation f has two representations: one is f which was described with an absolute continuous part with respect to the Lebesgue measure λ, while in the other an integral with respect to τ forms the absolute continuous part and t(τ) defines the singular measure. Both representations are obtained as parameter transformation images of an absolute continuous function on total variation domain [a,bν].


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## Body

## 1. Introduction

Bainov, Laksmikantam, and Semionov introduced a new class of differential equations [1] which, in addition to the usual dynamics, had impulse effects, acting on the system in zero time periods. Hence, the dynamics was composed of the usual interactions and singular interactions. This led to solutions of bounded variations. As recent publications show, finding solutions of integral equations in the spaces of bounded variations in various fields is coming to the focus of attention of international research in mathematics like solution of Volterra-Hammerstein integral equations. This research was supported among others by the Venezuelan Central Bank [2–4].The time dependence with the described composition of the dynamics led to difficulties in formulating efficient existence theorems, handling delay systems, and many other issues. The nature of bounded variation of the solutions was in contrast with such functions that constituted the solutions of differential equations (while not even being continuous).We want to establish a presentation of the functions of bounded variation which makes it possible to handle the solutions of bounded variations in terms of absolute continuous functions. By the origin of the problem outlined, we will focus onRn-valued functions of bounded variations. The combination of functions of bounded variation and singular functions in differential equations appears in the literature [5, 6].It is well known that a functionf of bounded variation on an interval [a,bλ] is the sum of an absolute continuous component fa and a component fs singular with respect to the Lebesgue measure λ [7, 8]. The total variation τfs of fs is then an ascending singular function.Letτ denote the measure defined by τfs which is singular with respect to λ. Now we form a measure ν≔λ+τ with respect to which both λ  &  τ are absolute continuous. The measure ν is generated by the strictly ascending function ν(t)≔t+τf(t). The strictly ascending function ν has strictly ascending left and a right continuous versions νλ,- and νλ,+, respectively [9], and maps the interval [a,bλ] into [ν(a),ν(bλ)]≕[a,bν] where bν≔a+⋁abλν.The transformation of[a,bν] into [a,bλ] is defined as follows. The function ν, being strictly ascending on [a,bλ], has a countable set Dλ⊂[a,bλ] of discontinuity points. Therefore, ∀t∈Dλ the inequality ν(t-0)<ν(t+0) holds, and ν is continuous in the rest of the points. Hence we define the transformation μ^λ as follows: μ^λ(t)≔s if ν(s-0)≤t≤ν(s+0) (see details in the paper).These three transformations enable us to formulate and prove presentations stated in the main result. The theorem states that a functionf:[a,bλ]→Rn is of bounded variation if and only if there exists an absolute continuous function φf(u)≔f(a)+∫auψfdν:[a,bν]→Rn with ψf∈L1[ν,[a,bν],Rn] such that(1)ft=φfμλ,+t=fa+∫aμλ,+tψfdν.The same main theorem states that a function of bounded variation has another representation where measure τ gives the absolute continuous component and t is an ascending function singular with respect to τ.


## 2. Integral Representation of Functions of Bounded Variation

### 2.1. Summary of Known Results

See the details of this section in [8, 10].Definition 1.
A functionf:[a,bλ]→R is a function of bounded variation if (2)⋁abf≔supΦ∈Γ∑tj∈Φtj<bftj+1-ftj<∞,where Γ≔Φ∣Φ≔tj∣tj∈[a,bλ],a=t0<tj<tj+1<tk=b,0≤j<k,1≤k<∞.The main target of this paper is the study of n-dimensional vector valued functions of bounded variation for somen≥1. We will introduce two types of variations for vector valued functions:Definition 2.
A functionf:[a,bλ]→Rn is a function of bounded variation if (3)⋁abf≔supΦ∈Γ⁡∑tj∈Φtj<bftj+1-ftj<∞.Theorem 3.
Iff:[a,bλ]→Rn is a function (f=(f1,f2,…fn),fi:[a,bλ]→R,∀1≤i≤n) then ∞>⋁abf≥⋁abfi, ∀1≤i≤n.We will focus our investigations on the properties of ascending functions; therefore we will now turn to the integral properties of ascending functions of bounded variation (equivalent with boundedness on the interval under consideration).


### 2.2. Total Variation of a Function of Bounded Variation

We will first establish that ascending functions play major role in our integral representation of functions of bounded variations.Theorem 4.
Letf:[a,bλ]→R be a function of bounded variation. Then the function φf(s)≔⋁asf, ∀s∈[a,bλ] fulfils φf(u)-φf(v)≥f(u)-f(v),u≥v,∀u,v∈[a,b]; hence φf+≔φf-f≥0 is an ascending function and f=φf-φf+.
Thus any function of bounded variation is differentiable almost everywhere and can be written asf′=df/dλ(4)fas≔fa+∫a,sf′dλ∀s∈a,bwhile fs≔f-fa is the singular part of f. Hence f=fa+fs.Proof.
See [8, 10] for proof.The singular functionfs is of bounded variation; hence assume the following.Definition 5.
τ f ( s ) ≔ ⋁ a s f s, ∀s∈[a,bλ].The singular functionfs is absolute continuous with respect to τf; hence fs(t)=∫atdfs/dτfdτf and therefore(5)ft=faa+fst=fa+∫atdfdλdλ+∫atdfsdτfdτf,∀t∈a,bλ.We introduce ν(t)≔t+τf(t),∀t∈[a,bλ] which is a strictly ascending function and ν:[a,bλ]→[a,bν] with bν≔b+τf(b). It follows from Definition 5 of τf that(6)νs=a+⋁ast+τf,∀s∈a,bλ.The functionν defines a measure ν≔λ+τ, such that both λ&τf are absolute continuous with respect to ν:(7)λA=∫Adλdνdν,∀A⊂a,bλBorelset,τA=∫Adτdνdν,∀A⊂a,bλBorelset.Using this in (5), we get an integral representation of the function f of bounded variation in terms of a single measure:(8)ft=fa+∫atdfdλdλdν+dfsdτdτdνdν.


#### 2.2.1. Properties of the Radon-Nikodym Derivatives

The integral in (8) is a reformulation of (5) with the sum of two measures: λ&τ. This is a measure generated from a right continuous ascending function ν. As stated in (7), both measures λ&τ are absolute continuous with respect to ν; therefore both can be written as an integral of the Radon-Nikodym derivatives [10] ρλ=dλ/dν&ρτ=dτ/dν. These have the following important properties.Lemma 6.
LetNτ≔dτ/dν>0&Nλ≔Nτ′, and then(1)
λ(Nτ)=0=τ(Nλ);(2)
ρλ+ρτ=d(λ+τ)/dν=dτ/dν+dλ/dν=1;(3)
ρλ(x)=1,∀x∈Nλ,ρτ(x)=1,∀x∈Nτ.​Proof.
We call our attention to the following relation: letA≔dτ/dνdλ/dν>0. Then the singularity of τ with respect to λ leads to the following relations:(9)∫Adτdνdλdνdν=∫Adλdνdτ=∫Adτdνdλ=∫Adτdλdλdνdλ≤∫Adτdλdλ=0,since τ is singular with respect to λ (if the fifth integral would be positive then A would be a set of positive ν-measure with positive λ&τ measures which contradicts the singularity of τ with respect to λ). Hence dτ/dνdλ/dν=0 almost everywhere with respect to ν. Therefore dτ/dλ=0 holds almost everywhere with respect to λ.


#### 2.2.2. The Relationship between the t-Scale and the Total Variations Scale

This section presents the main construction of this paper which connects the structure of the domain of the functionf:[a,bλ]→R of bounded variation and the range of the sum ν of the time parameter and the total variation τ of the singular component fs of f defined in Definition 5.The Structure from the Domain of a Function of Bounded Variation. The starting point here is the domain of the function f:[a,bλ]→R of bounded variation. The strictly ascending function ν(t)=t+τf(t)∈[a,bν] equivalently ν≔id[a,bλ]+τf:[a,bλ]→[a,bν] with bν≔bλ+τf(bλ) maps the domain [a,bλ] of f to the range [a,bν] of total variation scale. Both id[a,bλ]&τf are absolute continuous with respect to ν and the sets(10)Nλ≔dida,bdν=1&Nτ≔dτdν=1partition [a,bλ].The functionν as a strictly ascending on [a,bλ] has its left and right continuous versions.Definition 7.
​(11)μλ,-s≔νs-0,∀s∈a,bλ;μλ,-a≔a;μλ,-bλ≔bλ;μλ,+s≔νs+0,∀s∈a,bλ;μλ,+a≔a;μλ,+bλ≔bλ;​Definition 8.
We assign a closed interval to each point of[a,bλ]: ∅≠Θλ(s)≔[μλ,-(s),μλ,+(s)]⊂[a,bν] to each s∈[a,bλ].Definition 9.
LetDλν≔⋃t∈DfΘλ(t) and let Dλν≔Θλt∣t∈Df⊂P([a,bν]) (P([a,bν]) denote the power set of [a,bν]).Lemma 10.
Bothμλ,-,μλ,+:[a,bλ]→[a,bν] are strictly ascending and μλ,-(a)=μλ,+(a)=a  &  μλ,-(bλ)=μλ,+(bλ)=bν by a,bλ∉Df. Moreover both μλ,-,μλ,+ are continuous on [a,bλ]∖Df→[a,bν]∖Dλν.
The mapping defined in Definition8 assigns a closed bounded interval to each point in [a,bλ] such that ∀s,t∈[a,bλ],s≠t⇒Θλ(s)∩Θλ(t)=∅ and ⋃s∈[a,bλ]Θλ(s)=a,bν.Proof.
​(1)
Bothμλ,-,μλ,+:[a,bλ]→[a,bν] are strictly ascending and map a→a&bλ→bν by a,bλ∉Df. Let t,s∈[a,bλ] be such that t<s. τ is an ascending function; hence 0≤τ(s)-τ(t)⇒0<(s+τ(s))-(t+τ(t))⇒(t+τ(t))<(s+τ(s)) that proves the first statement.(2)
By the proved strictly ascending property ofν, ∀u,s,v∈[a,bλ],u<s<v⇒ν(u)<ν(s)<ν(v) holds. Hence supνu∣u<s=ν(s-0)≤ν(s)≤ν(s+0)=infνu∣u>s implies that the proposed interval is the smallest interval enclosing the ν-image of point s.Moreover, ift,s∈[a,bλ],t≠s⇒ either t<w=(t+s)/2<s or s<(t+s)/2=w<t holds. Assume the first option. Then (12)Θλt⊂a,inf⁡νu∣t<u⊂a,νw+t2νw+s2,b⊃sup⁡νu∣u<s,b⊃Θλs∅=a,w+t2∩w+s2,b⊃Θλt∩Θλs.(3)
Ifs∈[a,bλ]∖Df then idR is continuous and τ is continuous since s∉Df; hence ν is continuous at s.(4)
Letp∈[a,bν] then p∈[a,bν]∖Dλν or p∈Dλν. In the first case ∃sp∈[a,bλ]∖Dλ such that p=ν(sp). In the second case by Definition 9p∈Dλν≔⋃t∈DfΘλ(t); hence ∃!tp∈Dλ such that p∈Θλ(tp).Definition 11.
​(13)μ^λu≔s,∃s∈a,bλsuchthatu∈Θλs∀u∈a,bν.Lemma 12.
The mappingμ^λ is constant on Θλ(s),∀s∈[a,b], i. e., μ^λ(u)=μ^λ(v),∀u,v∈Θλ(s),∀s∈[a,bλ]. Moreover μ^λ:[a,bν]→[a,bλ] is continuous, ascending, and onto.Proof.
By (13) for u,v∈Θλ(s), it follows that ν(s-0)≤u,v≤ν(s+0)⇒μ^λ(u)=μ^λ(v)=s. The onto property follows from the proof of point (4) in Lemma 10.Lemma 13.
The relationships betweenμλ,-,μλ,+  &  μ^λ are as follows:(14)μ^λ∘μλ,-=ida,bλ;μ^λ∘μλ,+=ida,bλ.However, exchanging the two mappings will not give id[a,bν]. Indeed,(15)μλ,-∘μ^λ=ida,bν∖Dλν;μλ,+∘μ^λ=ida,bν∖Dλν.​Proof.
We prove it for the first case of (14) only. Let s∈[a,bλ]. Then μλ,-(s)=ν(s-0)=ν(s) if s∉Df. μ^λ(ν(s))=s since Θλ(s)={ν(s)} (when s is a continuity point, Θλ(s) is a one point set by Definition 11).
Ifs∈Df then μλ,-(s)=ν(s-0) and ν(s-0)∈Θλ(s) the left endpoint of which is μλ,-(s) (by the definition of Θλ in Lemma 10). Similarly, if s∈Df then μλ,+(s)=ν(s+0) and ν(s+0)∈Θλ(s) the right endpoint of which is μλ,+(s) (by the definition of Θλ in Lemma 10). Hence the first statement in (14) of the lemma is proved.
We now prove the first statement of (15). Let s∈[a,bν]∖Dλν. Then μ^λ(s)∈[a,bλ]∖Df since s∉Dλν; hence s∈Θλ(u) for a continuity point u∈[a,bλ]∖Df of ν. And μ^λ(s)=u⇒μλ,-(u)=s. This proves the lemma.The partitioning of[a,bλ] into the union Nλ∪Nτ can be transformed into the partitioning of [a,bν] as follows.Definition 14.
​(16)Nλν=μ^λ-1NλNτν=μ^λ-1NτThe Structure from the Total Variation. In this section we will construct functions of bounded variations from the scale [a,bν] and show that the time interval [a,bλ],τ,Nλ,Nτ, is obtainable from the structure defined below.Definition 15.
Let[a,bν] be a closed bounded interval and let {[a,bν],Bν,[a,bν],ν} be the Lebesgue measure defined by id[a,bν]:[a,bν]→[a,bν] on the Borel sets Bν,[a,bν] of [a,bν]. Let Nλν,Nτν⊂[a,bν] be Borel subsets of [a,bν] such that(17)νNλν×νNτν>0wherenonofthemisofzeromeasureNλν∩Nτν=∅;Nλν∪Nτν=a,bν.Since the conditions prescribed for time and τ in Definition 15 are symmetric, we will construct two representations with mappings below.Definition 16.
The definition of basic mappings for time representation is as follows:(1)
let the time scale interval be[a,bλ] with bλ≔a+∫abνχNλνdν;(2)
μ^λ(s)≔a+∫asχNλνdν∈[a,bλ], ∀s∈[a,bν];(3)
μλ,-(s)≔inf⁡μ^λ-1({s})∈[a,bν], ∀s∈[a,bλ];(4)
μλ(s)≔μλ,+(s)≔sup⁡μ^λ-1({s})∈[a,bν], ∀s∈[a,bλ].​Definition 17.
The definition of theτ-based mappings for τ-representation is as follows: (1)
let theτ-scale interval be [a,bτ] with bτ≔a+∫abνχNτνdν;(2)
μ^τ(s)≔a+∫asχNτνdν∈[a,bτ], ∀s∈[a,bν];(3)
μτ,-(s)≔inf⁡μ^τ-1({s})∈[a,bν]; ∀s∈[a,bτ];(4)
μτ(s)≔μτ,+(s)≔sup⁡μ^τ-1({s})∈[a,bν], ∀s∈[a,bτ].
We will treat the time case only since the other is the same word for word.(1) The Time, Singular Time τ and ν (1)
Ifμ^λ(s)=u,μ^λ(t)=v∈[a,bλ]&u<v⇒s<t∀s,t∈[a,bν] by Definition 16(2) (if the integral on an interval is positive then the length of the interval is positive.). (2)
By the axiom of choice,∃ν:[a,bλ]→[a,bν] is a strictly ascending function such that ν(t)∈μ^λ-1(t),∀t∈[a,bλ]. Then μλ,-(s)=ν(s-0)&μλ,+(s)=ν(s+0),∀s∈[a,bλ] which follows from the continuity of μ^λ. (3)
The ascending functionν:[a,bλ]→[a,bν] has a countable set of discontinuity points Dλ≔t∣νt-0<νt+0. Hence ν:[a,bλ]∖Dλ→[a,bν] is continuous. (4)
By ascending continuity ofμ^λ (absolute continuous for being an indefinite integral) Θλ(t)≔μ^λ-1({t}) is closed bounded interval by being preimage of a closed singleton ∀t∈[a,bλ]. Moreover Θλ(t)=[μλ,-(t),μλ,+(t)]∀t∈[a,bλ] and Θλ(t)∩Θλ(s)=∅,∀t,s∈[a,bλ],t≠s. (5)
The integral representation ofν(t),t∈[a,bλ]: using Definition 16(2) of μ^λ(s)≔a+∫asχNλνdν∈[a,bλ], ∀s∈[a,bν] and the definition of ν in point (2) gives the right continuous version of ν, i.e.,(18)νt=a+∫aνt+01dν=a+∫aνt+0χNλν+χNτνdν=∫aνt+0χNλνdν+∫aνt+0χNτνdνt≔∫aνt+0χNλνdν(19)τt≔∫aνt+0χNτνdννt=t+τt∀t∈a,bλ. Here we used the relations in (17) and defined τ:[a,bλ]→[0,bν-bλ] as ascending mapping. (6)
The ascending functionτ is singular with respect to λ and vice versa, and the ascending function λ is singular with repect to τ. Let u,v∈[a,bλ]∖Dλ. Then(20)∫νuνvdτdνdλdνdν=∫νuνvχNτνχNλνdν=0 (7)
Hence for anyφ,ψ∈L1([a,bλ],ν)(21)ft≔fa+∫atφχNλdν+∫atψχNτdν=fa+∫atφdλ+∫atψdτ is a function of bounded variation.With this we proved that the two representations mutually determine each other and what is more there are two representations of any function of bounded variation which is not absolute continuous.


#### 2.2.3.ν-Integrable Functions on Time and ν-Scales

In this section we will construct everything for thet/λ scale. All steps applied here can be done for the τ scale also.Since the ascending functionsτ&ν are noncontinuous functions of t, the τ/ν measures have to be defined from the semiring of left closed, right open intervals with continuity endpoints.Definition 18.
Let the semiring of left closed, right open intervals with endpoints of continuity in[a,bλ] be (22)Pν,a,bλ,c≔u,v∣u,v∈a,bλ∖DλLet the smallest σ-algebra containing the semiring be (23)Bν,a,bλ,c≔σPν,a,bλ,c.​Definition 19.
Let the semiring of left closed right open intervals with endpoints of continuity in[a,bτ] be(24)Pν,a,bτ,c≔u,v∣u,v∈a,bτ∖DτLet the smallest σ-algebra containing the semiring be (25)Bν,a,bτ,c≔σPν,a,bτ,c.Definition 20.
Let the semiring of left closed right open intervals with endpoints from[a,bν]∖Dλν in [a,bν] for λ be(26)Pλ,a,bν,cν≔u,v∣u,v∈a,bν∖Dλνand the semiring of left closed right open intervals with endpoints from [a,bν]∖Dτν in [a,bν] for τ be (27)Pτ,a,bν,cν≔u,v∣u,v∈a,bν∖Dτν.The correspondingσ-algebras are defined as follows.Definition 21.
Let the smallestσ-algebra containing the semiring Pλ,[a,bν],cν be (28)Bλ,a,bν,cν≔σPλ,a,bν,cνand the smallest σ-algebra containing the semiring Pτ,[a,bν],cν be (29)Bτ,a,bν,cν≔σPτ,a,bν,cν.Now we can define the measures from the ascending functionν and from the ascending function id[a,bν].Definition 22.
​(30)νλs,t≔νt-νs∀s,t∈Pν,a,bλ,cντs,t≔νt-νs∀s,t∈Pν,a,bτ,cDefinition 23.
​(31)νλνs,t≔t-s∀s,t∈Pλ,a,bν,cνντνs,t≔t-s∀s,t∈Pτ,a,bν,cν.An important relationship is presented in the following lemma.Lemma 24.
FromPλ,[a,bν],cν⊂P([a,bν],ν) it follows that Bλ,[a,bν],cν⊂B([a,bν],ν).
FromPτ,[a,bν],cν⊂P([a,bν],ν) it follows that Bτ,[a,bν],cν⊂B([a,bν],ν).
Moreoverνλν(A)=ν(A),∀A∈Bλ,[a,bν],cν and ντν(A)=ν(A),∀A∈Bτ,[a,bν],cν.Proof.
LetA∈Bλ,[a,bν],cν. By the statement of this theorem Pλ,[a,bν],cν⊂P([a,bν],ν)⊂B([a,bν],ν). Then(32)νλνA=νλν∗A=inf⁡∑j=0∞νaj,bj∣A⊂⋃j=0∞aj,bj,aj,bjj=1∞⊂Pλ,a,bν,cν⊂Ba,bν,ν.Let us select a sequence {Sk}k=1∞ from the covers of A constructed in (32) as follows:(33)Sk∣Sk≔ak,j,bk,jj=1∞,A⊂⋃j=0∞ak,j,bk,j,12k>∑j=0∞νak,j,bk,j-νλν∗A).
Lemma  25.LetIt∣It∈Pλ,a,bν,cν,∀1≤t<∞and letH≔⋃t=1∞It.Then∃{Ds}s=1∞⊂R(Pλ,[a,bν],cν)such thatH=⋃s=1∞Ds,∀1≤n,m<∞,n≠m,Dn∩Dm=∅.
Proof. Let us work in the smallest ring R(Pλ,[a,bν],cν) containing the semiring Pλ,[a,bν],cν. Since It∈Pλ,[a,bν],cν⇒It∈R(Pλ,[a,bν],cν)∀1≤t<∞. Let Du≔Iu∖(⋃q=1u-1Iq)∈R(Pλ,[a,bν],cν), ∀1≤u<∞ such that H=⋃q=1∞Dq and Dn∩Dm=∅,∀1≤n,m<∞,n≠m, where we used the standard technique of changing an infinite union of ring elements into infinite union of pairwise disjoint ring elements [8, 11].□
Corollary  26.By the representation theorem ofR(Pλ,[a,bν],cν), ∃a pairwise disjoint system{Dj,i}i=1kj⊂Pλ,[a,bν],cνsuch thatDj=⋃i=1kjDj,i.HenceH=(⋃q=1∞Dq)=⋃q=1∞⋃i=1kqDq,i.ThereforeHis obtained as a countable pairwise disjoint union of left closed right open intervals with continuity endpoints. HenceH∈B([a,bν],ν)andνλν(H)=ν(H).
Let us apply these propositions to the sequence defined in (33).
LetAk≔⋃j=1∞[ak,j,bk,j)⊃A and let Bk≔⋂j=1kAj⊃A, ∀1≤k<∞. By the conditions of the theorem, Pλ,[a,bν],cν⊂P([a,bν],ν)⊂B([a,bν],ν)⇒∀0<k∈N, Ak, Bk∈Bλ,[a,bν],cν&Ak, Bk∈B([a,bν],ν).
Then letB≔⋂k=1∞Bk∈Bλ,[a,bν],cν and(34)νλνB∖A=νλνB-νλνA<νλνBn-νλνA<∑j=1∞νak,j,bk,j-νλν∗A)<12k,∀k∈N,where we used the definition of the covering intervals in (33). Hence B∖A∈Bλ,[a,bν],cν, &νλν(B∖A)=0⇒0=νλν(B∖A)=νλν∗(B∖A)≥ν∗(B∖A)=0. Recalling that any set of 0 outer measure is measurable we get B∖A∈B([a,bν],ν), hence with B∈B([a,bν],ν)⇒A∈B([a,bν],ν).(1) The Relationship between λ or τ Structures and ν -Ones. We will show that Pν,[a,bλ],c&Pλ,[a,bν],cν and Pν,[a,bτ],c&Pτ,[a,bν],cν are isomorphic semirings, respectively, and what is more, the isomorphisms preserve the respective measures.Lemma 27.
​(35)μλ,-u,v=νu-0,νv-0=νu,νv∈Pλ,a,bν,cν;νλνμλ,-u,v=νλννu,νv=νv-νu=νλu,v.by u,v∈[a,bλ]∖Dλ⇒ν(u),ν(v)∈[a,bν]∖Dλν.
Conversely, let[s,t)∈Pλ,[a,bν],cν and then(36)μ^λs,t=μ^λs,μ^λt∈Pν,a,bλ,c;νλμ^λs,t=νλμ^λs,μ^λt=νμ^λs-νμ^λt=t-s=νλνs,tby the identity ν(μ^λ)=id[a,bν]∖Dλν in (15) and Definition 23.Proof.
Essentially the lemma contains the references from which the statements of the lemma follow.


#### 2.2.4. The Relationships between the Semirings and Their Extensions

The lemma below describes the relationship between the continuity points ofν and the points in [a,bλ].Lemma 28.
∀ t ∈ [ a , b λ ]  ∃  { l t , j } j = 1 ∞, {ut,j}j=1∞⊂[a,bλ]∖Dλ such that (1)
ifa=t⇒lt,j=a,∀1≤j<∞; otherwise if a<t then lt,j<lt,j+1,∀0<j∈N and if t=b⇒ut,j=b,∀1≤j<∞; otherwise if t<b then ut,j>ut,j+1,∀0<j∈N;(2)
lt,j↗t for j→∞ and ut,j↘t for j→∞;(3)
ν(lt,j)↗μλ,-(t)=ν(t-0) for j→∞ and ν(ut,j)↘μλ,+(t)=ν(t+0) for j→∞.​Proof.
​(1)
Lett=a, then la,j≔a, ∀1≤j<∞. Similarly, if t=b then ub,j≔b, ∀1≤j<∞.(2)
Lett∈(a,b). Let U≔a<t<b≕V&j=1.(3)
SinceDλ is countable and the open intervals ∅≠(U,t)&(t,V)⊈Dλ⇒∃p∈(U,t)∖Dλ&q∈(t,V)∖Dλ (continuity points).(4)
Letlt,j≔p&ut,j≔q.(5)
LetU≔maxp+t/2,U+t/2,&V≔minq+t/2,V+t/2 and j=j+1.(6)
Go to(3).(7)
This process will stop after generating the sequences of continuity points which are strictly ascending/descending sequences converging tot.
By the strictly ascending property ofν on [a,bλ]∖Dλ the sequences {ν(lt,j)}j=1∞&{ν(ut,j)}j=1∞ converge to left and right limits of ν at t∈[a,bλ], respectively. This completes the proof.Lemma 29.
​(1)
∀s,t∈[a,bλ],s<t the following statements hold:
(a)
[s,t]∈Bλ,[a,bλ],c;(b)
[μλ,-(s),μλ,+(t)]=[ν(s-0),ν(t+0)]∈Bλ,[a,bν],cν.(2)
∀A∈Dλν⇒A∈Bλ,[a,bν],cν and A is an atom which means ∀B∈Bλ,[a,bν],cν either B⊂A⇒B=∅ or B⊇A. Equivalently a measurable set does not cut an atom: (37)A∩B=Aor∅.(3)
Iff is Bλ,[a,bν],cν-measurable then ∀A∈Dλν,f(s)=cA,∀s∈A, where cA∈R is a constant depending on A.(4)
An interval[s,t]∈Bλ,[a,bν],cν is in Dλν if and only if μ^λ([s,t])=[u,v]⊂Nτ&ρτ(s)=1&ρλ(s)=0∀s∈[u,v].​Proof.
​(1)
By Lemma28∀s,t∈[a,bλ], s<t, ∃{ls,j}j=1∞⊂[a,bλ]∖Dλ, &∃{ut,j}j=1∞⊂[a,bλ]∖Dλ such that the following applies.
(a)
Ifa<s then {ls,j}1≤j<∞ is strictly ascending and ls,j↗s for j→∞. For a=s the sequence is constant a.(b)
Ift<b then {ut,j}1≤j<∞ is strictly descending and ut,j↘t for j→∞. For b=t the sequence is constant b.(c)
By[ls,j,ut,j)∈Pν,[a,bλ],c⊂Bλ,[a,bλ],c∀1≤j<∞ it follows that [s,t]=⋂j=1∞[ls,j,ut,j)∈Bλ,[a,bλ],c (similarly for the intervals with closed b ending).(d)
limj→∞ν(ls,j)=ν(s-0) and limj→∞ν(ut,j)=ν(t+0).(e)
Hence[ν(ls,j),ν(ut,j))∈Pλ,[a,bν],cν⊂Bλ,[a,bν],cν and(38)⋂j=1∞νls,j,νut,j=νs-0,νt+0by (d) of this proof. Hence fromBλ,[a,bν],cν being σ-algebra it follows that(39)νs-0,νt+0∈Bλ,a,bν,cν.(2)
LetA∈Dλν⇒A=Θλ(t) for a t∈Dλ. Hence Θλ(t)=[ν(t-0),ν(t+0)]∈Bλ,[a,bν],cν by point (1) of this proof. We show now that if H⊂A then H is not measurable.Letp∈A∈Dλν⇒A=Θλ(t)=[ν(t-0),ν(t+0)] for a t∈Dλ. If s∈[a,bν]∖Dλν⊂[a,bν]∖A⇒p∈[u,v)∈Pν,[a,bλ],c⇒u,v∉A⇒u<ν(t-0)<ν(t+0)<v by u,v∈[a,bν]∖A. Hence ∅≠H⊂A⇒∃p∈H⇒p∈A. Then [u,v)∈Pν,[a,bλ],c⇒νλν(A)≤νλν∗(H) holds for the outer measure of H. Hence H is not measurable if H≠A by Caratheodory’s condition for outer measures ( νλν∗(Y)=νλν∗(Y∩H)+νλν∗(Y∖H),∀Y⊂[a,bν]) with Y=A:(40)νλν∗A<2×νλν∗A<νλν∗A∩H+νλν∗A∖Hproves the lemma.(3)
Letf:[a,bν]→R be measurable with respect to Bλ,[a,bν],cν. Let us assume that ∃A∈Dλν and ∃s,t∈A such that f(s)≠f(t). Then with α≔(fs+ft)/2 the sets U≔{f>α}∩A≠∅&V≔{f<α}∩A≠∅ are disjoint and hence proper nonempty subsets of the measurable set A; hence by statement 1 of this lemma, the sets U&V are not measurable; hence f is not measurable. If f:[a,bν]→Rn is measurable and then each component of f is measurable and hence constant on the elements of Dλν; hence the statement follows.(4)
(a)
Let[s,t]∈Bλ,[a,bν],cν∩Dλν. Then ∃u∈Dλ such that [s,t]=Θλ(u). Then by Definition 16 point (2)u=μ^λ(v)≔u+∫svχNλνdν, ∀v∈[s,t]. Hence 0=∫stχNλνdν, ⇒χNλν=0 almost everywhere. By statements (2) and (3) of this lemma, the measurable function χNλν= const on [s,t] follows χNλν=0 in [s,t]; hence χNτν=1 in [s,t].(b)
Conversely, let[s,t]∈Bλ,[a,bν],cν and let χNτν(v)=1,∀v∈[s,t]. Then χNλν(v)=0,∀v∈[s,t]. Hence by Definition 16 point (2)u=μ^λ(v)≔u+∫svχNλνdν=u+∫sv0dν, ∀v∈[s,t]. Hence Θλ(u)=[s,t] proves that [s,t]∈Dλν.​Lemma 30.
​
Case (1). Let s,t∈[a,bλ], s<t, and let χ[s,t]:[a,bλ]→{0,1} fulfil χ[s,t](u)=1 if u∈[s,t] and 0 otherwise. Then χ[s,t]∘μ^λ=χ[νλ,-(s),νλ,+(t)]:[a,bν]→{0,1}.
Case (2). Let s,t∈[a,bν]∖Dλν,s<t and let χ[s,t]:[a,bν]→{0,1} fulfil χ[s,t](u)=1 if u∈[s,t] and 0 otherwise. Then χ[s,t]∘μλ,-=χ[ν^λ(s),ν^λ(t)]=χ[νλ,--1(s),νλ,+-1(s)]:[a,bλ]→{0,1}.Proof.
Case (1). u∣χs,t∘μ^λu=1=u∣s≤μ^λu≤t. Applying Lemma 13 and the set of equations (14) to the left and right sides of the inequalities, respectively, leads to the following:(41)s≤μ^λu∣∘νλ,-⇔νλ,-s≤uμ^λu≤t∣∘νλ,+⇔u≤νλ,+t.Hence u∣s≤μ^λu≤t = u∣νλ,-s≤u≤νλ,-t has been proved and therefore case (1) holds.
Case (2). Let s,t∈[a,bν]∖Dλν,s<t. Then u∣χs,t∘μλ,-u=1=u∣s≤μλ,-u≤μλ,+u≤t. Applying Lemma 13 and the set of equations (15) leads to the following:(42)s≤μλ,-u∣∘μ^λ⇔μ^λs≤uμλ,+u≤t∣∘μ^λ⇔u≤μ^λt.Hence u∣s≤μλ,-u≤μλ,+u≤t=u∣μ^λs≤u≤μ^λt has been proved and therefore case (2) holds.Notation 31.
LetM[a,bλ],Bν,[a,bλ],c≔f∣f:a,bλ→Rn,Bν,a,bλ,c-measurable(43)Ma,bν,Bν,a,bλ,cνν=f∣f:a,bν→Rn,Bλ,a,bν,cν-measurable​


#### 2.2.5. Transpose of the Transformations between t and Total Variation Scales

Let us consider the transpose of a transformation of the domains of function spaces.Definition 32.
LetA,B&C be sets and let A→φB→hC with h∈{ψ:B→C}≕F(B,C) and let F(A,C)≔{ψ:A→C}. Then let(44)φ∗:h∈FB,C→h∘φ∈FA,C,∀h∈FB,C.Note that F(B,C)→φ∗F(A,C) against A→φB.Notation 33.
Notations for the lemma about the transpose mappings. LetA≔[a,bλ]&B≔[a,bν] and let C≔Rn. Let F[A:C]≔{f:[a,bλ]→Rn}≕F[[a,bλ]:Rn] be a set of functions.
LetF[B:C]≔f∣f:a,bν→RnsuchthatfisconstantonanyA∈Dλν≕F[[a,bν]:Rn],Lemma 34.
Letμ^λ:[a,bν]→[a,bλ] and μλ,-:[a,bλ]→[a,bν] be two mappings as defined above. Then the following relations hold:(45)μ^λ∗∘μλ,-∗=idFa,bν:Rnμλ,-∗∘μ^λ∗=idFa,bλ:RnMoreover μ^λ∗  &  μλ,-∗ are linear isomorphisms.Proof.
​(1)
Letf∈F[[a,bν]:Rn] and let μλ,-∗ be the transpose of μλ,-:[a,bλ]→[a,bν]. Then (46)μλ,-∗fs=fμλ,-s,∀s∈a,bλ.(2)
Letg∈F[[a,bλ]:Rn] and let μ^λ∗ be the transpose of μ^λ:[a,bν]→[a,bλ]. Then (47)μ^λ∗gv=gμ^λv,∀v∈a,bν.(3)
Let us consider now the compositionμ^λ∗∘μλ,-∗f:(48)μ^λ∗∘μλ,-∗f=μ^λ∗f∘μλ,-=f∘μλ,-∘μ^λ.By the first identity in (15) μλ,-∘μ^λ=id[a,bν]∖Dλν we discovered that ∀u∈[a,bν]∖Dλν⇒μ^λ∗∘μλ,-∗f(u)=f(u) holds.Ifu∈Dλν then u∈Θλ(t) for a t∈Dλ; hence f(μλ,-(t))=f(u),u,μλ,-(t)∈Θλ(t); hence f(u)=f(μλ,-(t))=const(t,f) (with f being constant on Θλ(t)).By (13) μ^λ(Θλ(t)) = {t}⇒∀u∈Θλ(t)⇒f(u) = f(μλ,-(μ^λ(u))) = f(μλ,-(t)) = const(t,f) completes the proof.(4)
Let us now consider the compositionμλ,-∗∘μ^λ∗g:(49)μλ,-∗∘μ^λ∗g=μλ,-∗g∘μ^λ=g∘μλ,-∘μ^λ=g∘ida,bλby (14). Hence μ^λ∗g(μλ,-(u))=g(μλ,-(μ^λ(s)))=g(u)∀u∈[a,bλ].(5)
IfRn is a vector space then F[[a,bλ]:Rn]&F[[a,bτ]:Rn] are vector spaces for the point-wise addition and multiplication by scalar. Equation (45) gives the linear isomorphism.​Corollary 35.
Let the notations of Lemma34 be changed as follows:F[[a,bλ]:Rn]→M[a,bλ],Bν,[a,bλ],cF[[a,bν]:Rn]→M[a,bν],Bν,[a,bλ],cνν.
Then Lemma34 withμ^λ∗:M[a,bλ],Bν,[a,bλ],c→M[a,bν],Bν,[a,bλ],cνν andμλ,-∗:M[a,bν],Bν,[a,bλ],cνν→M[a,bλ],Bν,[a,bλ],c holds. Precisely(50)μ^λ∗∘μλ,-∗=idMa,bλ,Bν,a,bλ,cμλ,-∗∘μ^λ∗=idMa,bν,Bν,a,bλ,cνν
Hence the mappingsμλ,-  &  μ^λ are linear isomorphisms between the two measurable function spaces.Proof.
The two continuous ascending mappingsμλ,-:[a,bλ] → [a,bν]&μ^λ:[a,bν] → [a,bλ] between the two domains [a,bλ]&[a,bν] are the same as in Lemma 34.
Letf∈M[a,bν],Bν,[a,bλ],cνν. f be measurable and μλ,- is continuous; hence(51)μλ,-∗f=f∘μλ,-∈Ma,bλ,Bν,a,bλ,cν.Let g∈M[a,bλ],Bν,[a,bλ],c be measurable and μ^λ is continuous; hence
μ ^ λ ∗ g = g ∘ μ ^ λ ∈ M [ a , b ν ] , B ν , [ a , b λ ] , c ν. By the properties proved in Lemma 34μ^λ∗g is constant on all A∈Dλν. Hence the statements of Lemma 34 are true (the two identities in (45)).LetL1[νλ,[a,bλ],c]≔f∣f:a,bλ→Rbeνλ-integrable and letL 1 [ ν λ ν , [ a , b ν ] , c ] ≔ f ∣ f : a , b ν → R   b e   ν λ ν -integrable. Then the following applies.Theorem 36.
By Corollary35 it follows that(52)μ^λ∗∘μλ,-∗=idL1νλν,a,bν,cμλ,-∗∘μ^λ∗=idL1νλ,a,bλ,cWhat is more,(53)∫abλfdνλ=∫abνμ^λ∗fdνλν,∀f∈L1νλ,a,bλ,c;∫abλμλ,-∗fdνλ=∫abνfdνλν∀f∈L1νλν,a,bν,cProof.
​(1)
By Definition22, let [s,t)∈Pν,[a,bλ],c be a measurable set and let its characteristic function be χ[s,t). Then(54)νλs,t≔νt-νs=∫abλχs,tdνλ=∫abνμ^λ∗χs,tdνλνBy Definition23 let [s,t)∈Pλ,[a,bν],cν be a measurable set and let its characteristic function be χ[s,t). Then(55)νλνs,t≔t-s=∫abλμλ,-∗χs,tdνλ=∫abνχs,tdνλν(2)
By Lemma34 for any subset A⊂[a,bλ]⇒μ^λ∗A⊂[a,bν] and(56)νλ∗A=νλν∗μ^λ∗Aholds for the outer measures(νλ)∗&(νλν)∗.Conversely∀A⊂[a,bν] such that ∀H∈Dλν either H∩A=∅ or H⊂A and then the relation holds: (νλ)∗(μλ,-∗A)=(νλν)∗(A).From this it follows that for measurable sets in bothσ-algebras the outer measures define the extended measures and therefore the measures on the σ-algebras are fulfilled. This is(57)νλνA=∫abλμλ,-∗χAdνλ=∫abνχAdνλν,∀A∈Bλ,a,bν,cννλA=∫abλχAdνλ=∫abνμ^λ∗χAdνλν,∀A∈Bν,a,bλ,cBy the linearity of the transpose mappings, the identities are valid for all nonnegative step-functions and from there for all integrable function. This completes the proof of the theorem.​Theorem 37.
Letf∈L1[νλ,[a,bλ],c]  &  t∈[a,bλ]. Then(58)∫atfdνλ=∫aμλ,+tμ^λ∗fdνλνLet f∈L1[νλν,[a,bν],c]  &  [a,s]∈Bλ,[a,bν],cν and s∈[a,bν]∖Dλν,→:(59)∫aμ^λsμλ,-∗fdνλ=∫asfdνλνProof.
​(1)
Letf∈L1[νλ,[a,bλ],c]&t∈[a,bλ]. Then(60)∫atfdνλ=∫abλχa,t×fdνλ=∫abνμ^λ∗χa,t×fdνλν=∫abν(μ^λ∗χa,t×μ^λ∗fdνλν,where the first statement in (53) of Theorem 36 was applied and the statement of case (1) in Lemma 30 gave the last equality from which finally we obtain the statement:(61)∫atfdνλ=∫abν(μ^λ∗χa,t×μ^λ∗fdνλν=∫aμλ,+tμ^λ∗fdνλν.This proves the first statement.(2)
Letf∈L1[νλν,[a,bν],c]&[a,s]∈Bν,[a,bλ],cν:(62)∫asfdνλν=∫abνχa,s×fdνλν=∫abνμλ,-∗χa,s×fdνλNow we will apply the statement of case(2) in Lemma 30 to prove the final result:(63)∫asfdνλν=∫abνμλ,-∗χa,s×fdνλ=∫abν(μλ,-∗χa,s×μλ,-∗fdνλ=∫abν(χa,μ^λs×μλ,-∗fdνλ=∫aμ^λsμλ,-∗fdνλ.​


#### 2.2.6. The Main Result

In Theorem37, we established the relationship between absolute continuous indefinite integrals of members of L1[νλ,[a,bλ],c] and the absolute continuous indefinite integrals of members of L1[νλν,[a,bν],c].Our next step is to establish the relationship between the indefinite integrals of the members ofL1[ν,[a,bν]] and the indefinite integrals of the members of L1[νλν,[a,bν]],c].We prove the following statements.(1)
Letf∈L1[ν,[a,bν]] and let νf,λ(A)≔∫Afdν, ∀A∈Bλ,[a,bν],cν.Letνf,τ(A)≔∫Afdν, ∀A∈Bτ,[a,bν],cν.(2)
FromPλ,[a,bν],cν⊂P([a,bν],ν) it follows that Bλ,[a,bν],cν⊂B([a,bν],ν).FromPτ,[a,bν],cν⊂P([a,bν],ν) it follows that Bτ,[a,bν],cν⊂B([a,bν],ν).(3)
The signed measureνf,λ:Bλ,[a,bν],cν→R is absolute continuous with respect to ν:Bλ,[a,bν],cν→R+ by A∈Bλ,[a,bν],cν&ν(A)=0⇒νf(A)=∫Afdν=0.The signed measureνf,τ:Bτ,[a,bν],cν→R is absolute continuous with respect to ν:Bτ,[a,bν],cν→R+ by A∈Bτ,[a,bν],cν&ν(A)=0⇒νf(A)=∫Afdν=0.(4)
By the absolute continuity stated in point(2) mentioned above, the Bλ,[a,bν],cν-measurable Radon Nikodym derivative exists:(64)f~ν,λ≔dνfdνλν⇔∫Afdν=∫Af~ν,λdνλν,∀A∈Bλ,a,bν,cν.By the absolute continuity stated in point(2) mentioned above, the Bτ,[a,bν],cν-measurable Radon Nikodym derivative exists:(65)f~ν,τ≔dνfdντν⇔∫Afdν=∫Af~ν,τdντν,∀A∈Bτ,a,bν,cν.​With the statements(1)-(4) mentioned above we have proved the following result.Theorem 38.
A functionf:[a,bλ]⊂R→Rn is of bounded variation on [a,bλ] if and only if the following conditions are fulfilled. (1)
There exists a closed interval[a,bν],a,bν∈R partitioned as union of two disjoint measurable sets [a,bν]=Nλν∪Nτν such that μ^λ(s)≔a+∫asχNλνdν∈[a,bλ],s∈[a,bν] is onto [a,bλ] (see the details in Definition 15, equation (17)).(2)
There exists an absolute continuous functionφf(u)≔f(a)+∫auψfdν:[a,bν]→Rn with ψf∈L1[ν,[a,bν],Rn] such that the following applies.
(a)
On the t-scale with the Radon - Nikodym derivativeψf^ν,λ defined in (64) is expressed by(66)ft=φfμλ,+t=fa+∫aμλ,+tψfdν=fa+∫aμλ,+tψf^dνλν=fa+∫atμλ,-∗ψf^ν,λdνλ,∀t∈a,bλ(b)
Theτ scale representation fτ of f with the Radon - Nikodym derivative ψf^ν,τ defined in (65) is expressed by(67)fτϑ=φfμτ,+ϑ=fa+∫aμτ,+ϑψfdν=fa+∫aμτ,+ϑψf^ν,τdντν=fa+∫aϑμτ,-∗ψf^ν,τdντ,∀ϑ∈0,bτwhere we used the notations and functions defined in Definition 15, equation (17) and the mappings defined in Definitions 16 and 17.


## 2.1. Summary of Known Results

See the details of this section in [8, 10].Definition 1.
A functionf:[a,bλ]→R is a function of bounded variation if (2)⋁abf≔supΦ∈Γ∑tj∈Φtj<bftj+1-ftj<∞,where Γ≔Φ∣Φ≔tj∣tj∈[a,bλ],a=t0<tj<tj+1<tk=b,0≤j<k,1≤k<∞.The main target of this paper is the study of n-dimensional vector valued functions of bounded variation for somen≥1. We will introduce two types of variations for vector valued functions:Definition 2.
A functionf:[a,bλ]→Rn is a function of bounded variation if (3)⋁abf≔supΦ∈Γ⁡∑tj∈Φtj<bftj+1-ftj<∞.Theorem 3.
Iff:[a,bλ]→Rn is a function (f=(f1,f2,…fn),fi:[a,bλ]→R,∀1≤i≤n) then ∞>⋁abf≥⋁abfi, ∀1≤i≤n.We will focus our investigations on the properties of ascending functions; therefore we will now turn to the integral properties of ascending functions of bounded variation (equivalent with boundedness on the interval under consideration).


## 2.2. Total Variation of a Function of Bounded Variation

We will first establish that ascending functions play major role in our integral representation of functions of bounded variations.Theorem 4.
Letf:[a,bλ]→R be a function of bounded variation. Then the function φf(s)≔⋁asf, ∀s∈[a,bλ] fulfils φf(u)-φf(v)≥f(u)-f(v),u≥v,∀u,v∈[a,b]; hence φf+≔φf-f≥0 is an ascending function and f=φf-φf+.
Thus any function of bounded variation is differentiable almost everywhere and can be written asf′=df/dλ(4)fas≔fa+∫a,sf′dλ∀s∈a,bwhile fs≔f-fa is the singular part of f. Hence f=fa+fs.Proof.
See [8, 10] for proof.The singular functionfs is of bounded variation; hence assume the following.Definition 5.
τ f ( s ) ≔ ⋁ a s f s, ∀s∈[a,bλ].The singular functionfs is absolute continuous with respect to τf; hence fs(t)=∫atdfs/dτfdτf and therefore(5)ft=faa+fst=fa+∫atdfdλdλ+∫atdfsdτfdτf,∀t∈a,bλ.We introduce ν(t)≔t+τf(t),∀t∈[a,bλ] which is a strictly ascending function and ν:[a,bλ]→[a,bν] with bν≔b+τf(b). It follows from Definition 5 of τf that(6)νs=a+⋁ast+τf,∀s∈a,bλ.The functionν defines a measure ν≔λ+τ, such that both λ&τf are absolute continuous with respect to ν:(7)λA=∫Adλdνdν,∀A⊂a,bλBorelset,τA=∫Adτdνdν,∀A⊂a,bλBorelset.Using this in (5), we get an integral representation of the function f of bounded variation in terms of a single measure:(8)ft=fa+∫atdfdλdλdν+dfsdτdτdνdν.


### 2.2.1. Properties of the Radon-Nikodym Derivatives

The integral in (8) is a reformulation of (5) with the sum of two measures: λ&τ. This is a measure generated from a right continuous ascending function ν. As stated in (7), both measures λ&τ are absolute continuous with respect to ν; therefore both can be written as an integral of the Radon-Nikodym derivatives [10] ρλ=dλ/dν&ρτ=dτ/dν. These have the following important properties.Lemma 6.
LetNτ≔dτ/dν>0&Nλ≔Nτ′, and then(1)
λ(Nτ)=0=τ(Nλ);(2)
ρλ+ρτ=d(λ+τ)/dν=dτ/dν+dλ/dν=1;(3)
ρλ(x)=1,∀x∈Nλ,ρτ(x)=1,∀x∈Nτ.​Proof.
We call our attention to the following relation: letA≔dτ/dνdλ/dν>0. Then the singularity of τ with respect to λ leads to the following relations:(9)∫Adτdνdλdνdν=∫Adλdνdτ=∫Adτdνdλ=∫Adτdλdλdνdλ≤∫Adτdλdλ=0,since τ is singular with respect to λ (if the fifth integral would be positive then A would be a set of positive ν-measure with positive λ&τ measures which contradicts the singularity of τ with respect to λ). Hence dτ/dνdλ/dν=0 almost everywhere with respect to ν. Therefore dτ/dλ=0 holds almost everywhere with respect to λ.


### 2.2.2. The Relationship between the t-Scale and the Total Variations Scale

This section presents the main construction of this paper which connects the structure of the domain of the functionf:[a,bλ]→R of bounded variation and the range of the sum ν of the time parameter and the total variation τ of the singular component fs of f defined in Definition 5.The Structure from the Domain of a Function of Bounded Variation. The starting point here is the domain of the function f:[a,bλ]→R of bounded variation. The strictly ascending function ν(t)=t+τf(t)∈[a,bν] equivalently ν≔id[a,bλ]+τf:[a,bλ]→[a,bν] with bν≔bλ+τf(bλ) maps the domain [a,bλ] of f to the range [a,bν] of total variation scale. Both id[a,bλ]&τf are absolute continuous with respect to ν and the sets(10)Nλ≔dida,bdν=1&Nτ≔dτdν=1partition [a,bλ].The functionν as a strictly ascending on [a,bλ] has its left and right continuous versions.Definition 7.
​(11)μλ,-s≔νs-0,∀s∈a,bλ;μλ,-a≔a;μλ,-bλ≔bλ;μλ,+s≔νs+0,∀s∈a,bλ;μλ,+a≔a;μλ,+bλ≔bλ;​Definition 8.
We assign a closed interval to each point of[a,bλ]: ∅≠Θλ(s)≔[μλ,-(s),μλ,+(s)]⊂[a,bν] to each s∈[a,bλ].Definition 9.
LetDλν≔⋃t∈DfΘλ(t) and let Dλν≔Θλt∣t∈Df⊂P([a,bν]) (P([a,bν]) denote the power set of [a,bν]).Lemma 10.
Bothμλ,-,μλ,+:[a,bλ]→[a,bν] are strictly ascending and μλ,-(a)=μλ,+(a)=a  &  μλ,-(bλ)=μλ,+(bλ)=bν by a,bλ∉Df. Moreover both μλ,-,μλ,+ are continuous on [a,bλ]∖Df→[a,bν]∖Dλν.
The mapping defined in Definition8 assigns a closed bounded interval to each point in [a,bλ] such that ∀s,t∈[a,bλ],s≠t⇒Θλ(s)∩Θλ(t)=∅ and ⋃s∈[a,bλ]Θλ(s)=a,bν.Proof.
​(1)
Bothμλ,-,μλ,+:[a,bλ]→[a,bν] are strictly ascending and map a→a&bλ→bν by a,bλ∉Df. Let t,s∈[a,bλ] be such that t<s. τ is an ascending function; hence 0≤τ(s)-τ(t)⇒0<(s+τ(s))-(t+τ(t))⇒(t+τ(t))<(s+τ(s)) that proves the first statement.(2)
By the proved strictly ascending property ofν, ∀u,s,v∈[a,bλ],u<s<v⇒ν(u)<ν(s)<ν(v) holds. Hence supνu∣u<s=ν(s-0)≤ν(s)≤ν(s+0)=infνu∣u>s implies that the proposed interval is the smallest interval enclosing the ν-image of point s.Moreover, ift,s∈[a,bλ],t≠s⇒ either t<w=(t+s)/2<s or s<(t+s)/2=w<t holds. Assume the first option. Then (12)Θλt⊂a,inf⁡νu∣t<u⊂a,νw+t2νw+s2,b⊃sup⁡νu∣u<s,b⊃Θλs∅=a,w+t2∩w+s2,b⊃Θλt∩Θλs.(3)
Ifs∈[a,bλ]∖Df then idR is continuous and τ is continuous since s∉Df; hence ν is continuous at s.(4)
Letp∈[a,bν] then p∈[a,bν]∖Dλν or p∈Dλν. In the first case ∃sp∈[a,bλ]∖Dλ such that p=ν(sp). In the second case by Definition 9p∈Dλν≔⋃t∈DfΘλ(t); hence ∃!tp∈Dλ such that p∈Θλ(tp).Definition 11.
​(13)μ^λu≔s,∃s∈a,bλsuchthatu∈Θλs∀u∈a,bν.Lemma 12.
The mappingμ^λ is constant on Θλ(s),∀s∈[a,b], i. e., μ^λ(u)=μ^λ(v),∀u,v∈Θλ(s),∀s∈[a,bλ]. Moreover μ^λ:[a,bν]→[a,bλ] is continuous, ascending, and onto.Proof.
By (13) for u,v∈Θλ(s), it follows that ν(s-0)≤u,v≤ν(s+0)⇒μ^λ(u)=μ^λ(v)=s. The onto property follows from the proof of point (4) in Lemma 10.Lemma 13.
The relationships betweenμλ,-,μλ,+  &  μ^λ are as follows:(14)μ^λ∘μλ,-=ida,bλ;μ^λ∘μλ,+=ida,bλ.However, exchanging the two mappings will not give id[a,bν]. Indeed,(15)μλ,-∘μ^λ=ida,bν∖Dλν;μλ,+∘μ^λ=ida,bν∖Dλν.​Proof.
We prove it for the first case of (14) only. Let s∈[a,bλ]. Then μλ,-(s)=ν(s-0)=ν(s) if s∉Df. μ^λ(ν(s))=s since Θλ(s)={ν(s)} (when s is a continuity point, Θλ(s) is a one point set by Definition 11).
Ifs∈Df then μλ,-(s)=ν(s-0) and ν(s-0)∈Θλ(s) the left endpoint of which is μλ,-(s) (by the definition of Θλ in Lemma 10). Similarly, if s∈Df then μλ,+(s)=ν(s+0) and ν(s+0)∈Θλ(s) the right endpoint of which is μλ,+(s) (by the definition of Θλ in Lemma 10). Hence the first statement in (14) of the lemma is proved.
We now prove the first statement of (15). Let s∈[a,bν]∖Dλν. Then μ^λ(s)∈[a,bλ]∖Df since s∉Dλν; hence s∈Θλ(u) for a continuity point u∈[a,bλ]∖Df of ν. And μ^λ(s)=u⇒μλ,-(u)=s. This proves the lemma.The partitioning of[a,bλ] into the union Nλ∪Nτ can be transformed into the partitioning of [a,bν] as follows.Definition 14.
​(16)Nλν=μ^λ-1NλNτν=μ^λ-1NτThe Structure from the Total Variation. In this section we will construct functions of bounded variations from the scale [a,bν] and show that the time interval [a,bλ],τ,Nλ,Nτ, is obtainable from the structure defined below.Definition 15.
Let[a,bν] be a closed bounded interval and let {[a,bν],Bν,[a,bν],ν} be the Lebesgue measure defined by id[a,bν]:[a,bν]→[a,bν] on the Borel sets Bν,[a,bν] of [a,bν]. Let Nλν,Nτν⊂[a,bν] be Borel subsets of [a,bν] such that(17)νNλν×νNτν>0wherenonofthemisofzeromeasureNλν∩Nτν=∅;Nλν∪Nτν=a,bν.Since the conditions prescribed for time and τ in Definition 15 are symmetric, we will construct two representations with mappings below.Definition 16.
The definition of basic mappings for time representation is as follows:(1)
let the time scale interval be[a,bλ] with bλ≔a+∫abνχNλνdν;(2)
μ^λ(s)≔a+∫asχNλνdν∈[a,bλ], ∀s∈[a,bν];(3)
μλ,-(s)≔inf⁡μ^λ-1({s})∈[a,bν], ∀s∈[a,bλ];(4)
μλ(s)≔μλ,+(s)≔sup⁡μ^λ-1({s})∈[a,bν], ∀s∈[a,bλ].​Definition 17.
The definition of theτ-based mappings for τ-representation is as follows: (1)
let theτ-scale interval be [a,bτ] with bτ≔a+∫abνχNτνdν;(2)
μ^τ(s)≔a+∫asχNτνdν∈[a,bτ], ∀s∈[a,bν];(3)
μτ,-(s)≔inf⁡μ^τ-1({s})∈[a,bν]; ∀s∈[a,bτ];(4)
μτ(s)≔μτ,+(s)≔sup⁡μ^τ-1({s})∈[a,bν], ∀s∈[a,bτ].
We will treat the time case only since the other is the same word for word.(1) The Time, Singular Time τ and ν (1)
Ifμ^λ(s)=u,μ^λ(t)=v∈[a,bλ]&u<v⇒s<t∀s,t∈[a,bν] by Definition 16(2) (if the integral on an interval is positive then the length of the interval is positive.). (2)
By the axiom of choice,∃ν:[a,bλ]→[a,bν] is a strictly ascending function such that ν(t)∈μ^λ-1(t),∀t∈[a,bλ]. Then μλ,-(s)=ν(s-0)&μλ,+(s)=ν(s+0),∀s∈[a,bλ] which follows from the continuity of μ^λ. (3)
The ascending functionν:[a,bλ]→[a,bν] has a countable set of discontinuity points Dλ≔t∣νt-0<νt+0. Hence ν:[a,bλ]∖Dλ→[a,bν] is continuous. (4)
By ascending continuity ofμ^λ (absolute continuous for being an indefinite integral) Θλ(t)≔μ^λ-1({t}) is closed bounded interval by being preimage of a closed singleton ∀t∈[a,bλ]. Moreover Θλ(t)=[μλ,-(t),μλ,+(t)]∀t∈[a,bλ] and Θλ(t)∩Θλ(s)=∅,∀t,s∈[a,bλ],t≠s. (5)
The integral representation ofν(t),t∈[a,bλ]: using Definition 16(2) of μ^λ(s)≔a+∫asχNλνdν∈[a,bλ], ∀s∈[a,bν] and the definition of ν in point (2) gives the right continuous version of ν, i.e.,(18)νt=a+∫aνt+01dν=a+∫aνt+0χNλν+χNτνdν=∫aνt+0χNλνdν+∫aνt+0χNτνdνt≔∫aνt+0χNλνdν(19)τt≔∫aνt+0χNτνdννt=t+τt∀t∈a,bλ. Here we used the relations in (17) and defined τ:[a,bλ]→[0,bν-bλ] as ascending mapping. (6)
The ascending functionτ is singular with respect to λ and vice versa, and the ascending function λ is singular with repect to τ. Let u,v∈[a,bλ]∖Dλ. Then(20)∫νuνvdτdνdλdνdν=∫νuνvχNτνχNλνdν=0 (7)
Hence for anyφ,ψ∈L1([a,bλ],ν)(21)ft≔fa+∫atφχNλdν+∫atψχNτdν=fa+∫atφdλ+∫atψdτ is a function of bounded variation.With this we proved that the two representations mutually determine each other and what is more there are two representations of any function of bounded variation which is not absolute continuous.


### 2.2.3.ν-Integrable Functions on Time and ν-Scales

In this section we will construct everything for thet/λ scale. All steps applied here can be done for the τ scale also.Since the ascending functionsτ&ν are noncontinuous functions of t, the τ/ν measures have to be defined from the semiring of left closed, right open intervals with continuity endpoints.Definition 18.
Let the semiring of left closed, right open intervals with endpoints of continuity in[a,bλ] be (22)Pν,a,bλ,c≔u,v∣u,v∈a,bλ∖DλLet the smallest σ-algebra containing the semiring be (23)Bν,a,bλ,c≔σPν,a,bλ,c.​Definition 19.
Let the semiring of left closed right open intervals with endpoints of continuity in[a,bτ] be(24)Pν,a,bτ,c≔u,v∣u,v∈a,bτ∖DτLet the smallest σ-algebra containing the semiring be (25)Bν,a,bτ,c≔σPν,a,bτ,c.Definition 20.
Let the semiring of left closed right open intervals with endpoints from[a,bν]∖Dλν in [a,bν] for λ be(26)Pλ,a,bν,cν≔u,v∣u,v∈a,bν∖Dλνand the semiring of left closed right open intervals with endpoints from [a,bν]∖Dτν in [a,bν] for τ be (27)Pτ,a,bν,cν≔u,v∣u,v∈a,bν∖Dτν.The correspondingσ-algebras are defined as follows.Definition 21.
Let the smallestσ-algebra containing the semiring Pλ,[a,bν],cν be (28)Bλ,a,bν,cν≔σPλ,a,bν,cνand the smallest σ-algebra containing the semiring Pτ,[a,bν],cν be (29)Bτ,a,bν,cν≔σPτ,a,bν,cν.Now we can define the measures from the ascending functionν and from the ascending function id[a,bν].Definition 22.
​(30)νλs,t≔νt-νs∀s,t∈Pν,a,bλ,cντs,t≔νt-νs∀s,t∈Pν,a,bτ,cDefinition 23.
​(31)νλνs,t≔t-s∀s,t∈Pλ,a,bν,cνντνs,t≔t-s∀s,t∈Pτ,a,bν,cν.An important relationship is presented in the following lemma.Lemma 24.
FromPλ,[a,bν],cν⊂P([a,bν],ν) it follows that Bλ,[a,bν],cν⊂B([a,bν],ν).
FromPτ,[a,bν],cν⊂P([a,bν],ν) it follows that Bτ,[a,bν],cν⊂B([a,bν],ν).
Moreoverνλν(A)=ν(A),∀A∈Bλ,[a,bν],cν and ντν(A)=ν(A),∀A∈Bτ,[a,bν],cν.Proof.
LetA∈Bλ,[a,bν],cν. By the statement of this theorem Pλ,[a,bν],cν⊂P([a,bν],ν)⊂B([a,bν],ν). Then(32)νλνA=νλν∗A=inf⁡∑j=0∞νaj,bj∣A⊂⋃j=0∞aj,bj,aj,bjj=1∞⊂Pλ,a,bν,cν⊂Ba,bν,ν.Let us select a sequence {Sk}k=1∞ from the covers of A constructed in (32) as follows:(33)Sk∣Sk≔ak,j,bk,jj=1∞,A⊂⋃j=0∞ak,j,bk,j,12k>∑j=0∞νak,j,bk,j-νλν∗A).
Lemma  25.LetIt∣It∈Pλ,a,bν,cν,∀1≤t<∞and letH≔⋃t=1∞It.Then∃{Ds}s=1∞⊂R(Pλ,[a,bν],cν)such thatH=⋃s=1∞Ds,∀1≤n,m<∞,n≠m,Dn∩Dm=∅.
Proof. Let us work in the smallest ring R(Pλ,[a,bν],cν) containing the semiring Pλ,[a,bν],cν. Since It∈Pλ,[a,bν],cν⇒It∈R(Pλ,[a,bν],cν)∀1≤t<∞. Let Du≔Iu∖(⋃q=1u-1Iq)∈R(Pλ,[a,bν],cν), ∀1≤u<∞ such that H=⋃q=1∞Dq and Dn∩Dm=∅,∀1≤n,m<∞,n≠m, where we used the standard technique of changing an infinite union of ring elements into infinite union of pairwise disjoint ring elements [8, 11].□
Corollary  26.By the representation theorem ofR(Pλ,[a,bν],cν), ∃a pairwise disjoint system{Dj,i}i=1kj⊂Pλ,[a,bν],cνsuch thatDj=⋃i=1kjDj,i.HenceH=(⋃q=1∞Dq)=⋃q=1∞⋃i=1kqDq,i.ThereforeHis obtained as a countable pairwise disjoint union of left closed right open intervals with continuity endpoints. HenceH∈B([a,bν],ν)andνλν(H)=ν(H).
Let us apply these propositions to the sequence defined in (33).
LetAk≔⋃j=1∞[ak,j,bk,j)⊃A and let Bk≔⋂j=1kAj⊃A, ∀1≤k<∞. By the conditions of the theorem, Pλ,[a,bν],cν⊂P([a,bν],ν)⊂B([a,bν],ν)⇒∀0<k∈N, Ak, Bk∈Bλ,[a,bν],cν&Ak, Bk∈B([a,bν],ν).
Then letB≔⋂k=1∞Bk∈Bλ,[a,bν],cν and(34)νλνB∖A=νλνB-νλνA<νλνBn-νλνA<∑j=1∞νak,j,bk,j-νλν∗A)<12k,∀k∈N,where we used the definition of the covering intervals in (33). Hence B∖A∈Bλ,[a,bν],cν, &νλν(B∖A)=0⇒0=νλν(B∖A)=νλν∗(B∖A)≥ν∗(B∖A)=0. Recalling that any set of 0 outer measure is measurable we get B∖A∈B([a,bν],ν), hence with B∈B([a,bν],ν)⇒A∈B([a,bν],ν).(1) The Relationship between λ or τ Structures and ν -Ones. We will show that Pν,[a,bλ],c&Pλ,[a,bν],cν and Pν,[a,bτ],c&Pτ,[a,bν],cν are isomorphic semirings, respectively, and what is more, the isomorphisms preserve the respective measures.Lemma 27.
​(35)μλ,-u,v=νu-0,νv-0=νu,νv∈Pλ,a,bν,cν;νλνμλ,-u,v=νλννu,νv=νv-νu=νλu,v.by u,v∈[a,bλ]∖Dλ⇒ν(u),ν(v)∈[a,bν]∖Dλν.
Conversely, let[s,t)∈Pλ,[a,bν],cν and then(36)μ^λs,t=μ^λs,μ^λt∈Pν,a,bλ,c;νλμ^λs,t=νλμ^λs,μ^λt=νμ^λs-νμ^λt=t-s=νλνs,tby the identity ν(μ^λ)=id[a,bν]∖Dλν in (15) and Definition 23.Proof.
Essentially the lemma contains the references from which the statements of the lemma follow.


### 2.2.4. The Relationships between the Semirings and Their Extensions

The lemma below describes the relationship between the continuity points ofν and the points in [a,bλ].Lemma 28.
∀ t ∈ [ a , b λ ]  ∃  { l t , j } j = 1 ∞, {ut,j}j=1∞⊂[a,bλ]∖Dλ such that (1)
ifa=t⇒lt,j=a,∀1≤j<∞; otherwise if a<t then lt,j<lt,j+1,∀0<j∈N and if t=b⇒ut,j=b,∀1≤j<∞; otherwise if t<b then ut,j>ut,j+1,∀0<j∈N;(2)
lt,j↗t for j→∞ and ut,j↘t for j→∞;(3)
ν(lt,j)↗μλ,-(t)=ν(t-0) for j→∞ and ν(ut,j)↘μλ,+(t)=ν(t+0) for j→∞.​Proof.
​(1)
Lett=a, then la,j≔a, ∀1≤j<∞. Similarly, if t=b then ub,j≔b, ∀1≤j<∞.(2)
Lett∈(a,b). Let U≔a<t<b≕V&j=1.(3)
SinceDλ is countable and the open intervals ∅≠(U,t)&(t,V)⊈Dλ⇒∃p∈(U,t)∖Dλ&q∈(t,V)∖Dλ (continuity points).(4)
Letlt,j≔p&ut,j≔q.(5)
LetU≔maxp+t/2,U+t/2,&V≔minq+t/2,V+t/2 and j=j+1.(6)
Go to(3).(7)
This process will stop after generating the sequences of continuity points which are strictly ascending/descending sequences converging tot.
By the strictly ascending property ofν on [a,bλ]∖Dλ the sequences {ν(lt,j)}j=1∞&{ν(ut,j)}j=1∞ converge to left and right limits of ν at t∈[a,bλ], respectively. This completes the proof.Lemma 29.
​(1)
∀s,t∈[a,bλ],s<t the following statements hold:
(a)
[s,t]∈Bλ,[a,bλ],c;(b)
[μλ,-(s),μλ,+(t)]=[ν(s-0),ν(t+0)]∈Bλ,[a,bν],cν.(2)
∀A∈Dλν⇒A∈Bλ,[a,bν],cν and A is an atom which means ∀B∈Bλ,[a,bν],cν either B⊂A⇒B=∅ or B⊇A. Equivalently a measurable set does not cut an atom: (37)A∩B=Aor∅.(3)
Iff is Bλ,[a,bν],cν-measurable then ∀A∈Dλν,f(s)=cA,∀s∈A, where cA∈R is a constant depending on A.(4)
An interval[s,t]∈Bλ,[a,bν],cν is in Dλν if and only if μ^λ([s,t])=[u,v]⊂Nτ&ρτ(s)=1&ρλ(s)=0∀s∈[u,v].​Proof.
​(1)
By Lemma28∀s,t∈[a,bλ], s<t, ∃{ls,j}j=1∞⊂[a,bλ]∖Dλ, &∃{ut,j}j=1∞⊂[a,bλ]∖Dλ such that the following applies.
(a)
Ifa<s then {ls,j}1≤j<∞ is strictly ascending and ls,j↗s for j→∞. For a=s the sequence is constant a.(b)
Ift<b then {ut,j}1≤j<∞ is strictly descending and ut,j↘t for j→∞. For b=t the sequence is constant b.(c)
By[ls,j,ut,j)∈Pν,[a,bλ],c⊂Bλ,[a,bλ],c∀1≤j<∞ it follows that [s,t]=⋂j=1∞[ls,j,ut,j)∈Bλ,[a,bλ],c (similarly for the intervals with closed b ending).(d)
limj→∞ν(ls,j)=ν(s-0) and limj→∞ν(ut,j)=ν(t+0).(e)
Hence[ν(ls,j),ν(ut,j))∈Pλ,[a,bν],cν⊂Bλ,[a,bν],cν and(38)⋂j=1∞νls,j,νut,j=νs-0,νt+0by (d) of this proof. Hence fromBλ,[a,bν],cν being σ-algebra it follows that(39)νs-0,νt+0∈Bλ,a,bν,cν.(2)
LetA∈Dλν⇒A=Θλ(t) for a t∈Dλ. Hence Θλ(t)=[ν(t-0),ν(t+0)]∈Bλ,[a,bν],cν by point (1) of this proof. We show now that if H⊂A then H is not measurable.Letp∈A∈Dλν⇒A=Θλ(t)=[ν(t-0),ν(t+0)] for a t∈Dλ. If s∈[a,bν]∖Dλν⊂[a,bν]∖A⇒p∈[u,v)∈Pν,[a,bλ],c⇒u,v∉A⇒u<ν(t-0)<ν(t+0)<v by u,v∈[a,bν]∖A. Hence ∅≠H⊂A⇒∃p∈H⇒p∈A. Then [u,v)∈Pν,[a,bλ],c⇒νλν(A)≤νλν∗(H) holds for the outer measure of H. Hence H is not measurable if H≠A by Caratheodory’s condition for outer measures ( νλν∗(Y)=νλν∗(Y∩H)+νλν∗(Y∖H),∀Y⊂[a,bν]) with Y=A:(40)νλν∗A<2×νλν∗A<νλν∗A∩H+νλν∗A∖Hproves the lemma.(3)
Letf:[a,bν]→R be measurable with respect to Bλ,[a,bν],cν. Let us assume that ∃A∈Dλν and ∃s,t∈A such that f(s)≠f(t). Then with α≔(fs+ft)/2 the sets U≔{f>α}∩A≠∅&V≔{f<α}∩A≠∅ are disjoint and hence proper nonempty subsets of the measurable set A; hence by statement 1 of this lemma, the sets U&V are not measurable; hence f is not measurable. If f:[a,bν]→Rn is measurable and then each component of f is measurable and hence constant on the elements of Dλν; hence the statement follows.(4)
(a)
Let[s,t]∈Bλ,[a,bν],cν∩Dλν. Then ∃u∈Dλ such that [s,t]=Θλ(u). Then by Definition 16 point (2)u=μ^λ(v)≔u+∫svχNλνdν, ∀v∈[s,t]. Hence 0=∫stχNλνdν, ⇒χNλν=0 almost everywhere. By statements (2) and (3) of this lemma, the measurable function χNλν= const on [s,t] follows χNλν=0 in [s,t]; hence χNτν=1 in [s,t].(b)
Conversely, let[s,t]∈Bλ,[a,bν],cν and let χNτν(v)=1,∀v∈[s,t]. Then χNλν(v)=0,∀v∈[s,t]. Hence by Definition 16 point (2)u=μ^λ(v)≔u+∫svχNλνdν=u+∫sv0dν, ∀v∈[s,t]. Hence Θλ(u)=[s,t] proves that [s,t]∈Dλν.​Lemma 30.
​
Case (1). Let s,t∈[a,bλ], s<t, and let χ[s,t]:[a,bλ]→{0,1} fulfil χ[s,t](u)=1 if u∈[s,t] and 0 otherwise. Then χ[s,t]∘μ^λ=χ[νλ,-(s),νλ,+(t)]:[a,bν]→{0,1}.
Case (2). Let s,t∈[a,bν]∖Dλν,s<t and let χ[s,t]:[a,bν]→{0,1} fulfil χ[s,t](u)=1 if u∈[s,t] and 0 otherwise. Then χ[s,t]∘μλ,-=χ[ν^λ(s),ν^λ(t)]=χ[νλ,--1(s),νλ,+-1(s)]:[a,bλ]→{0,1}.Proof.
Case (1). u∣χs,t∘μ^λu=1=u∣s≤μ^λu≤t. Applying Lemma 13 and the set of equations (14) to the left and right sides of the inequalities, respectively, leads to the following:(41)s≤μ^λu∣∘νλ,-⇔νλ,-s≤uμ^λu≤t∣∘νλ,+⇔u≤νλ,+t.Hence u∣s≤μ^λu≤t = u∣νλ,-s≤u≤νλ,-t has been proved and therefore case (1) holds.
Case (2). Let s,t∈[a,bν]∖Dλν,s<t. Then u∣χs,t∘μλ,-u=1=u∣s≤μλ,-u≤μλ,+u≤t. Applying Lemma 13 and the set of equations (15) leads to the following:(42)s≤μλ,-u∣∘μ^λ⇔μ^λs≤uμλ,+u≤t∣∘μ^λ⇔u≤μ^λt.Hence u∣s≤μλ,-u≤μλ,+u≤t=u∣μ^λs≤u≤μ^λt has been proved and therefore case (2) holds.Notation 31.
LetM[a,bλ],Bν,[a,bλ],c≔f∣f:a,bλ→Rn,Bν,a,bλ,c-measurable(43)Ma,bν,Bν,a,bλ,cνν=f∣f:a,bν→Rn,Bλ,a,bν,cν-measurable​


### 2.2.5. Transpose of the Transformations between t and Total Variation Scales

Let us consider the transpose of a transformation of the domains of function spaces.Definition 32.
LetA,B&C be sets and let A→φB→hC with h∈{ψ:B→C}≕F(B,C) and let F(A,C)≔{ψ:A→C}. Then let(44)φ∗:h∈FB,C→h∘φ∈FA,C,∀h∈FB,C.Note that F(B,C)→φ∗F(A,C) against A→φB.Notation 33.
Notations for the lemma about the transpose mappings. LetA≔[a,bλ]&B≔[a,bν] and let C≔Rn. Let F[A:C]≔{f:[a,bλ]→Rn}≕F[[a,bλ]:Rn] be a set of functions.
LetF[B:C]≔f∣f:a,bν→RnsuchthatfisconstantonanyA∈Dλν≕F[[a,bν]:Rn],Lemma 34.
Letμ^λ:[a,bν]→[a,bλ] and μλ,-:[a,bλ]→[a,bν] be two mappings as defined above. Then the following relations hold:(45)μ^λ∗∘μλ,-∗=idFa,bν:Rnμλ,-∗∘μ^λ∗=idFa,bλ:RnMoreover μ^λ∗  &  μλ,-∗ are linear isomorphisms.Proof.
​(1)
Letf∈F[[a,bν]:Rn] and let μλ,-∗ be the transpose of μλ,-:[a,bλ]→[a,bν]. Then (46)μλ,-∗fs=fμλ,-s,∀s∈a,bλ.(2)
Letg∈F[[a,bλ]:Rn] and let μ^λ∗ be the transpose of μ^λ:[a,bν]→[a,bλ]. Then (47)μ^λ∗gv=gμ^λv,∀v∈a,bν.(3)
Let us consider now the compositionμ^λ∗∘μλ,-∗f:(48)μ^λ∗∘μλ,-∗f=μ^λ∗f∘μλ,-=f∘μλ,-∘μ^λ.By the first identity in (15) μλ,-∘μ^λ=id[a,bν]∖Dλν we discovered that ∀u∈[a,bν]∖Dλν⇒μ^λ∗∘μλ,-∗f(u)=f(u) holds.Ifu∈Dλν then u∈Θλ(t) for a t∈Dλ; hence f(μλ,-(t))=f(u),u,μλ,-(t)∈Θλ(t); hence f(u)=f(μλ,-(t))=const(t,f) (with f being constant on Θλ(t)).By (13) μ^λ(Θλ(t)) = {t}⇒∀u∈Θλ(t)⇒f(u) = f(μλ,-(μ^λ(u))) = f(μλ,-(t)) = const(t,f) completes the proof.(4)
Let us now consider the compositionμλ,-∗∘μ^λ∗g:(49)μλ,-∗∘μ^λ∗g=μλ,-∗g∘μ^λ=g∘μλ,-∘μ^λ=g∘ida,bλby (14). Hence μ^λ∗g(μλ,-(u))=g(μλ,-(μ^λ(s)))=g(u)∀u∈[a,bλ].(5)
IfRn is a vector space then F[[a,bλ]:Rn]&F[[a,bτ]:Rn] are vector spaces for the point-wise addition and multiplication by scalar. Equation (45) gives the linear isomorphism.​Corollary 35.
Let the notations of Lemma34 be changed as follows:F[[a,bλ]:Rn]→M[a,bλ],Bν,[a,bλ],cF[[a,bν]:Rn]→M[a,bν],Bν,[a,bλ],cνν.
Then Lemma34 withμ^λ∗:M[a,bλ],Bν,[a,bλ],c→M[a,bν],Bν,[a,bλ],cνν andμλ,-∗:M[a,bν],Bν,[a,bλ],cνν→M[a,bλ],Bν,[a,bλ],c holds. Precisely(50)μ^λ∗∘μλ,-∗=idMa,bλ,Bν,a,bλ,cμλ,-∗∘μ^λ∗=idMa,bν,Bν,a,bλ,cνν
Hence the mappingsμλ,-  &  μ^λ are linear isomorphisms between the two measurable function spaces.Proof.
The two continuous ascending mappingsμλ,-:[a,bλ] → [a,bν]&μ^λ:[a,bν] → [a,bλ] between the two domains [a,bλ]&[a,bν] are the same as in Lemma 34.
Letf∈M[a,bν],Bν,[a,bλ],cνν. f be measurable and μλ,- is continuous; hence(51)μλ,-∗f=f∘μλ,-∈Ma,bλ,Bν,a,bλ,cν.Let g∈M[a,bλ],Bν,[a,bλ],c be measurable and μ^λ is continuous; hence
μ ^ λ ∗ g = g ∘ μ ^ λ ∈ M [ a , b ν ] , B ν , [ a , b λ ] , c ν. By the properties proved in Lemma 34μ^λ∗g is constant on all A∈Dλν. Hence the statements of Lemma 34 are true (the two identities in (45)).LetL1[νλ,[a,bλ],c]≔f∣f:a,bλ→Rbeνλ-integrable and letL 1 [ ν λ ν , [ a , b ν ] , c ] ≔ f ∣ f : a , b ν → R   b e   ν λ ν -integrable. Then the following applies.Theorem 36.
By Corollary35 it follows that(52)μ^λ∗∘μλ,-∗=idL1νλν,a,bν,cμλ,-∗∘μ^λ∗=idL1νλ,a,bλ,cWhat is more,(53)∫abλfdνλ=∫abνμ^λ∗fdνλν,∀f∈L1νλ,a,bλ,c;∫abλμλ,-∗fdνλ=∫abνfdνλν∀f∈L1νλν,a,bν,cProof.
​(1)
By Definition22, let [s,t)∈Pν,[a,bλ],c be a measurable set and let its characteristic function be χ[s,t). Then(54)νλs,t≔νt-νs=∫abλχs,tdνλ=∫abνμ^λ∗χs,tdνλνBy Definition23 let [s,t)∈Pλ,[a,bν],cν be a measurable set and let its characteristic function be χ[s,t). Then(55)νλνs,t≔t-s=∫abλμλ,-∗χs,tdνλ=∫abνχs,tdνλν(2)
By Lemma34 for any subset A⊂[a,bλ]⇒μ^λ∗A⊂[a,bν] and(56)νλ∗A=νλν∗μ^λ∗Aholds for the outer measures(νλ)∗&(νλν)∗.Conversely∀A⊂[a,bν] such that ∀H∈Dλν either H∩A=∅ or H⊂A and then the relation holds: (νλ)∗(μλ,-∗A)=(νλν)∗(A).From this it follows that for measurable sets in bothσ-algebras the outer measures define the extended measures and therefore the measures on the σ-algebras are fulfilled. This is(57)νλνA=∫abλμλ,-∗χAdνλ=∫abνχAdνλν,∀A∈Bλ,a,bν,cννλA=∫abλχAdνλ=∫abνμ^λ∗χAdνλν,∀A∈Bν,a,bλ,cBy the linearity of the transpose mappings, the identities are valid for all nonnegative step-functions and from there for all integrable function. This completes the proof of the theorem.​Theorem 37.
Letf∈L1[νλ,[a,bλ],c]  &  t∈[a,bλ]. Then(58)∫atfdνλ=∫aμλ,+tμ^λ∗fdνλνLet f∈L1[νλν,[a,bν],c]  &  [a,s]∈Bλ,[a,bν],cν and s∈[a,bν]∖Dλν,→:(59)∫aμ^λsμλ,-∗fdνλ=∫asfdνλνProof.
​(1)
Letf∈L1[νλ,[a,bλ],c]&t∈[a,bλ]. Then(60)∫atfdνλ=∫abλχa,t×fdνλ=∫abνμ^λ∗χa,t×fdνλν=∫abν(μ^λ∗χa,t×μ^λ∗fdνλν,where the first statement in (53) of Theorem 36 was applied and the statement of case (1) in Lemma 30 gave the last equality from which finally we obtain the statement:(61)∫atfdνλ=∫abν(μ^λ∗χa,t×μ^λ∗fdνλν=∫aμλ,+tμ^λ∗fdνλν.This proves the first statement.(2)
Letf∈L1[νλν,[a,bν],c]&[a,s]∈Bν,[a,bλ],cν:(62)∫asfdνλν=∫abνχa,s×fdνλν=∫abνμλ,-∗χa,s×fdνλNow we will apply the statement of case(2) in Lemma 30 to prove the final result:(63)∫asfdνλν=∫abνμλ,-∗χa,s×fdνλ=∫abν(μλ,-∗χa,s×μλ,-∗fdνλ=∫abν(χa,μ^λs×μλ,-∗fdνλ=∫aμ^λsμλ,-∗fdνλ.​


### 2.2.6. The Main Result

In Theorem37, we established the relationship between absolute continuous indefinite integrals of members of L1[νλ,[a,bλ],c] and the absolute continuous indefinite integrals of members of L1[νλν,[a,bν],c].Our next step is to establish the relationship between the indefinite integrals of the members ofL1[ν,[a,bν]] and the indefinite integrals of the members of L1[νλν,[a,bν]],c].We prove the following statements.(1)
Letf∈L1[ν,[a,bν]] and let νf,λ(A)≔∫Afdν, ∀A∈Bλ,[a,bν],cν.Letνf,τ(A)≔∫Afdν, ∀A∈Bτ,[a,bν],cν.(2)
FromPλ,[a,bν],cν⊂P([a,bν],ν) it follows that Bλ,[a,bν],cν⊂B([a,bν],ν).FromPτ,[a,bν],cν⊂P([a,bν],ν) it follows that Bτ,[a,bν],cν⊂B([a,bν],ν).(3)
The signed measureνf,λ:Bλ,[a,bν],cν→R is absolute continuous with respect to ν:Bλ,[a,bν],cν→R+ by A∈Bλ,[a,bν],cν&ν(A)=0⇒νf(A)=∫Afdν=0.The signed measureνf,τ:Bτ,[a,bν],cν→R is absolute continuous with respect to ν:Bτ,[a,bν],cν→R+ by A∈Bτ,[a,bν],cν&ν(A)=0⇒νf(A)=∫Afdν=0.(4)
By the absolute continuity stated in point(2) mentioned above, the Bλ,[a,bν],cν-measurable Radon Nikodym derivative exists:(64)f~ν,λ≔dνfdνλν⇔∫Afdν=∫Af~ν,λdνλν,∀A∈Bλ,a,bν,cν.By the absolute continuity stated in point(2) mentioned above, the Bτ,[a,bν],cν-measurable Radon Nikodym derivative exists:(65)f~ν,τ≔dνfdντν⇔∫Afdν=∫Af~ν,τdντν,∀A∈Bτ,a,bν,cν.​With the statements(1)-(4) mentioned above we have proved the following result.Theorem 38.
A functionf:[a,bλ]⊂R→Rn is of bounded variation on [a,bλ] if and only if the following conditions are fulfilled. (1)
There exists a closed interval[a,bν],a,bν∈R partitioned as union of two disjoint measurable sets [a,bν]=Nλν∪Nτν such that μ^λ(s)≔a+∫asχNλνdν∈[a,bλ],s∈[a,bν] is onto [a,bλ] (see the details in Definition 15, equation (17)).(2)
There exists an absolute continuous functionφf(u)≔f(a)+∫auψfdν:[a,bν]→Rn with ψf∈L1[ν,[a,bν],Rn] such that the following applies.
(a)
On the t-scale with the Radon - Nikodym derivativeψf^ν,λ defined in (64) is expressed by(66)ft=φfμλ,+t=fa+∫aμλ,+tψfdν=fa+∫aμλ,+tψf^dνλν=fa+∫atμλ,-∗ψf^ν,λdνλ,∀t∈a,bλ(b)
Theτ scale representation fτ of f with the Radon - Nikodym derivative ψf^ν,τ defined in (65) is expressed by(67)fτϑ=φfμτ,+ϑ=fa+∫aμτ,+ϑψfdν=fa+∫aμτ,+ϑψf^ν,τdντν=fa+∫aϑμτ,-∗ψf^ν,τdντ,∀ϑ∈0,bτwhere we used the notations and functions defined in Definition 15, equation (17) and the mappings defined in Definitions 16 and 17.


## 2.2.1. Properties of the Radon-Nikodym Derivatives

The integral in (8) is a reformulation of (5) with the sum of two measures: λ&τ. This is a measure generated from a right continuous ascending function ν. As stated in (7), both measures λ&τ are absolute continuous with respect to ν; therefore both can be written as an integral of the Radon-Nikodym derivatives [10] ρλ=dλ/dν&ρτ=dτ/dν. These have the following important properties.Lemma 6.
LetNτ≔dτ/dν>0&Nλ≔Nτ′, and then(1)
λ(Nτ)=0=τ(Nλ);(2)
ρλ+ρτ=d(λ+τ)/dν=dτ/dν+dλ/dν=1;(3)
ρλ(x)=1,∀x∈Nλ,ρτ(x)=1,∀x∈Nτ.​Proof.
We call our attention to the following relation: letA≔dτ/dνdλ/dν>0. Then the singularity of τ with respect to λ leads to the following relations:(9)∫Adτdνdλdνdν=∫Adλdνdτ=∫Adτdνdλ=∫Adτdλdλdνdλ≤∫Adτdλdλ=0,since τ is singular with respect to λ (if the fifth integral would be positive then A would be a set of positive ν-measure with positive λ&τ measures which contradicts the singularity of τ with respect to λ). Hence dτ/dνdλ/dν=0 almost everywhere with respect to ν. Therefore dτ/dλ=0 holds almost everywhere with respect to λ.


## 2.2.2. The Relationship between the t-Scale and the Total Variations Scale

This section presents the main construction of this paper which connects the structure of the domain of the functionf:[a,bλ]→R of bounded variation and the range of the sum ν of the time parameter and the total variation τ of the singular component fs of f defined in Definition 5.The Structure from the Domain of a Function of Bounded Variation. The starting point here is the domain of the function f:[a,bλ]→R of bounded variation. The strictly ascending function ν(t)=t+τf(t)∈[a,bν] equivalently ν≔id[a,bλ]+τf:[a,bλ]→[a,bν] with bν≔bλ+τf(bλ) maps the domain [a,bλ] of f to the range [a,bν] of total variation scale. Both id[a,bλ]&τf are absolute continuous with respect to ν and the sets(10)Nλ≔dida,bdν=1&Nτ≔dτdν=1partition [a,bλ].The functionν as a strictly ascending on [a,bλ] has its left and right continuous versions.Definition 7.
​(11)μλ,-s≔νs-0,∀s∈a,bλ;μλ,-a≔a;μλ,-bλ≔bλ;μλ,+s≔νs+0,∀s∈a,bλ;μλ,+a≔a;μλ,+bλ≔bλ;​Definition 8.
We assign a closed interval to each point of[a,bλ]: ∅≠Θλ(s)≔[μλ,-(s),μλ,+(s)]⊂[a,bν] to each s∈[a,bλ].Definition 9.
LetDλν≔⋃t∈DfΘλ(t) and let Dλν≔Θλt∣t∈Df⊂P([a,bν]) (P([a,bν]) denote the power set of [a,bν]).Lemma 10.
Bothμλ,-,μλ,+:[a,bλ]→[a,bν] are strictly ascending and μλ,-(a)=μλ,+(a)=a  &  μλ,-(bλ)=μλ,+(bλ)=bν by a,bλ∉Df. Moreover both μλ,-,μλ,+ are continuous on [a,bλ]∖Df→[a,bν]∖Dλν.
The mapping defined in Definition8 assigns a closed bounded interval to each point in [a,bλ] such that ∀s,t∈[a,bλ],s≠t⇒Θλ(s)∩Θλ(t)=∅ and ⋃s∈[a,bλ]Θλ(s)=a,bν.Proof.
​(1)
Bothμλ,-,μλ,+:[a,bλ]→[a,bν] are strictly ascending and map a→a&bλ→bν by a,bλ∉Df. Let t,s∈[a,bλ] be such that t<s. τ is an ascending function; hence 0≤τ(s)-τ(t)⇒0<(s+τ(s))-(t+τ(t))⇒(t+τ(t))<(s+τ(s)) that proves the first statement.(2)
By the proved strictly ascending property ofν, ∀u,s,v∈[a,bλ],u<s<v⇒ν(u)<ν(s)<ν(v) holds. Hence supνu∣u<s=ν(s-0)≤ν(s)≤ν(s+0)=infνu∣u>s implies that the proposed interval is the smallest interval enclosing the ν-image of point s.Moreover, ift,s∈[a,bλ],t≠s⇒ either t<w=(t+s)/2<s or s<(t+s)/2=w<t holds. Assume the first option. Then (12)Θλt⊂a,inf⁡νu∣t<u⊂a,νw+t2νw+s2,b⊃sup⁡νu∣u<s,b⊃Θλs∅=a,w+t2∩w+s2,b⊃Θλt∩Θλs.(3)
Ifs∈[a,bλ]∖Df then idR is continuous and τ is continuous since s∉Df; hence ν is continuous at s.(4)
Letp∈[a,bν] then p∈[a,bν]∖Dλν or p∈Dλν. In the first case ∃sp∈[a,bλ]∖Dλ such that p=ν(sp). In the second case by Definition 9p∈Dλν≔⋃t∈DfΘλ(t); hence ∃!tp∈Dλ such that p∈Θλ(tp).Definition 11.
​(13)μ^λu≔s,∃s∈a,bλsuchthatu∈Θλs∀u∈a,bν.Lemma 12.
The mappingμ^λ is constant on Θλ(s),∀s∈[a,b], i. e., μ^λ(u)=μ^λ(v),∀u,v∈Θλ(s),∀s∈[a,bλ]. Moreover μ^λ:[a,bν]→[a,bλ] is continuous, ascending, and onto.Proof.
By (13) for u,v∈Θλ(s), it follows that ν(s-0)≤u,v≤ν(s+0)⇒μ^λ(u)=μ^λ(v)=s. The onto property follows from the proof of point (4) in Lemma 10.Lemma 13.
The relationships betweenμλ,-,μλ,+  &  μ^λ are as follows:(14)μ^λ∘μλ,-=ida,bλ;μ^λ∘μλ,+=ida,bλ.However, exchanging the two mappings will not give id[a,bν]. Indeed,(15)μλ,-∘μ^λ=ida,bν∖Dλν;μλ,+∘μ^λ=ida,bν∖Dλν.​Proof.
We prove it for the first case of (14) only. Let s∈[a,bλ]. Then μλ,-(s)=ν(s-0)=ν(s) if s∉Df. μ^λ(ν(s))=s since Θλ(s)={ν(s)} (when s is a continuity point, Θλ(s) is a one point set by Definition 11).
Ifs∈Df then μλ,-(s)=ν(s-0) and ν(s-0)∈Θλ(s) the left endpoint of which is μλ,-(s) (by the definition of Θλ in Lemma 10). Similarly, if s∈Df then μλ,+(s)=ν(s+0) and ν(s+0)∈Θλ(s) the right endpoint of which is μλ,+(s) (by the definition of Θλ in Lemma 10). Hence the first statement in (14) of the lemma is proved.
We now prove the first statement of (15). Let s∈[a,bν]∖Dλν. Then μ^λ(s)∈[a,bλ]∖Df since s∉Dλν; hence s∈Θλ(u) for a continuity point u∈[a,bλ]∖Df of ν. And μ^λ(s)=u⇒μλ,-(u)=s. This proves the lemma.The partitioning of[a,bλ] into the union Nλ∪Nτ can be transformed into the partitioning of [a,bν] as follows.Definition 14.
​(16)Nλν=μ^λ-1NλNτν=μ^λ-1NτThe Structure from the Total Variation. In this section we will construct functions of bounded variations from the scale [a,bν] and show that the time interval [a,bλ],τ,Nλ,Nτ, is obtainable from the structure defined below.Definition 15.
Let[a,bν] be a closed bounded interval and let {[a,bν],Bν,[a,bν],ν} be the Lebesgue measure defined by id[a,bν]:[a,bν]→[a,bν] on the Borel sets Bν,[a,bν] of [a,bν]. Let Nλν,Nτν⊂[a,bν] be Borel subsets of [a,bν] such that(17)νNλν×νNτν>0wherenonofthemisofzeromeasureNλν∩Nτν=∅;Nλν∪Nτν=a,bν.Since the conditions prescribed for time and τ in Definition 15 are symmetric, we will construct two representations with mappings below.Definition 16.
The definition of basic mappings for time representation is as follows:(1)
let the time scale interval be[a,bλ] with bλ≔a+∫abνχNλνdν;(2)
μ^λ(s)≔a+∫asχNλνdν∈[a,bλ], ∀s∈[a,bν];(3)
μλ,-(s)≔inf⁡μ^λ-1({s})∈[a,bν], ∀s∈[a,bλ];(4)
μλ(s)≔μλ,+(s)≔sup⁡μ^λ-1({s})∈[a,bν], ∀s∈[a,bλ].​Definition 17.
The definition of theτ-based mappings for τ-representation is as follows: (1)
let theτ-scale interval be [a,bτ] with bτ≔a+∫abνχNτνdν;(2)
μ^τ(s)≔a+∫asχNτνdν∈[a,bτ], ∀s∈[a,bν];(3)
μτ,-(s)≔inf⁡μ^τ-1({s})∈[a,bν]; ∀s∈[a,bτ];(4)
μτ(s)≔μτ,+(s)≔sup⁡μ^τ-1({s})∈[a,bν], ∀s∈[a,bτ].
We will treat the time case only since the other is the same word for word.(1) The Time, Singular Time τ and ν (1)
Ifμ^λ(s)=u,μ^λ(t)=v∈[a,bλ]&u<v⇒s<t∀s,t∈[a,bν] by Definition 16(2) (if the integral on an interval is positive then the length of the interval is positive.). (2)
By the axiom of choice,∃ν:[a,bλ]→[a,bν] is a strictly ascending function such that ν(t)∈μ^λ-1(t),∀t∈[a,bλ]. Then μλ,-(s)=ν(s-0)&μλ,+(s)=ν(s+0),∀s∈[a,bλ] which follows from the continuity of μ^λ. (3)
The ascending functionν:[a,bλ]→[a,bν] has a countable set of discontinuity points Dλ≔t∣νt-0<νt+0. Hence ν:[a,bλ]∖Dλ→[a,bν] is continuous. (4)
By ascending continuity ofμ^λ (absolute continuous for being an indefinite integral) Θλ(t)≔μ^λ-1({t}) is closed bounded interval by being preimage of a closed singleton ∀t∈[a,bλ]. Moreover Θλ(t)=[μλ,-(t),μλ,+(t)]∀t∈[a,bλ] and Θλ(t)∩Θλ(s)=∅,∀t,s∈[a,bλ],t≠s. (5)
The integral representation ofν(t),t∈[a,bλ]: using Definition 16(2) of μ^λ(s)≔a+∫asχNλνdν∈[a,bλ], ∀s∈[a,bν] and the definition of ν in point (2) gives the right continuous version of ν, i.e.,(18)νt=a+∫aνt+01dν=a+∫aνt+0χNλν+χNτνdν=∫aνt+0χNλνdν+∫aνt+0χNτνdνt≔∫aνt+0χNλνdν(19)τt≔∫aνt+0χNτνdννt=t+τt∀t∈a,bλ. Here we used the relations in (17) and defined τ:[a,bλ]→[0,bν-bλ] as ascending mapping. (6)
The ascending functionτ is singular with respect to λ and vice versa, and the ascending function λ is singular with repect to τ. Let u,v∈[a,bλ]∖Dλ. Then(20)∫νuνvdτdνdλdνdν=∫νuνvχNτνχNλνdν=0 (7)
Hence for anyφ,ψ∈L1([a,bλ],ν)(21)ft≔fa+∫atφχNλdν+∫atψχNτdν=fa+∫atφdλ+∫atψdτ is a function of bounded variation.With this we proved that the two representations mutually determine each other and what is more there are two representations of any function of bounded variation which is not absolute continuous.


## 2.2.3.ν-Integrable Functions on Time and ν-Scales

In this section we will construct everything for thet/λ scale. All steps applied here can be done for the τ scale also.Since the ascending functionsτ&ν are noncontinuous functions of t, the τ/ν measures have to be defined from the semiring of left closed, right open intervals with continuity endpoints.Definition 18.
Let the semiring of left closed, right open intervals with endpoints of continuity in[a,bλ] be (22)Pν,a,bλ,c≔u,v∣u,v∈a,bλ∖DλLet the smallest σ-algebra containing the semiring be (23)Bν,a,bλ,c≔σPν,a,bλ,c.​Definition 19.
Let the semiring of left closed right open intervals with endpoints of continuity in[a,bτ] be(24)Pν,a,bτ,c≔u,v∣u,v∈a,bτ∖DτLet the smallest σ-algebra containing the semiring be (25)Bν,a,bτ,c≔σPν,a,bτ,c.Definition 20.
Let the semiring of left closed right open intervals with endpoints from[a,bν]∖Dλν in [a,bν] for λ be(26)Pλ,a,bν,cν≔u,v∣u,v∈a,bν∖Dλνand the semiring of left closed right open intervals with endpoints from [a,bν]∖Dτν in [a,bν] for τ be (27)Pτ,a,bν,cν≔u,v∣u,v∈a,bν∖Dτν.The correspondingσ-algebras are defined as follows.Definition 21.
Let the smallestσ-algebra containing the semiring Pλ,[a,bν],cν be (28)Bλ,a,bν,cν≔σPλ,a,bν,cνand the smallest σ-algebra containing the semiring Pτ,[a,bν],cν be (29)Bτ,a,bν,cν≔σPτ,a,bν,cν.Now we can define the measures from the ascending functionν and from the ascending function id[a,bν].Definition 22.
​(30)νλs,t≔νt-νs∀s,t∈Pν,a,bλ,cντs,t≔νt-νs∀s,t∈Pν,a,bτ,cDefinition 23.
​(31)νλνs,t≔t-s∀s,t∈Pλ,a,bν,cνντνs,t≔t-s∀s,t∈Pτ,a,bν,cν.An important relationship is presented in the following lemma.Lemma 24.
FromPλ,[a,bν],cν⊂P([a,bν],ν) it follows that Bλ,[a,bν],cν⊂B([a,bν],ν).
FromPτ,[a,bν],cν⊂P([a,bν],ν) it follows that Bτ,[a,bν],cν⊂B([a,bν],ν).
Moreoverνλν(A)=ν(A),∀A∈Bλ,[a,bν],cν and ντν(A)=ν(A),∀A∈Bτ,[a,bν],cν.Proof.
LetA∈Bλ,[a,bν],cν. By the statement of this theorem Pλ,[a,bν],cν⊂P([a,bν],ν)⊂B([a,bν],ν). Then(32)νλνA=νλν∗A=inf⁡∑j=0∞νaj,bj∣A⊂⋃j=0∞aj,bj,aj,bjj=1∞⊂Pλ,a,bν,cν⊂Ba,bν,ν.Let us select a sequence {Sk}k=1∞ from the covers of A constructed in (32) as follows:(33)Sk∣Sk≔ak,j,bk,jj=1∞,A⊂⋃j=0∞ak,j,bk,j,12k>∑j=0∞νak,j,bk,j-νλν∗A).
Lemma  25.LetIt∣It∈Pλ,a,bν,cν,∀1≤t<∞and letH≔⋃t=1∞It.Then∃{Ds}s=1∞⊂R(Pλ,[a,bν],cν)such thatH=⋃s=1∞Ds,∀1≤n,m<∞,n≠m,Dn∩Dm=∅.
Proof. Let us work in the smallest ring R(Pλ,[a,bν],cν) containing the semiring Pλ,[a,bν],cν. Since It∈Pλ,[a,bν],cν⇒It∈R(Pλ,[a,bν],cν)∀1≤t<∞. Let Du≔Iu∖(⋃q=1u-1Iq)∈R(Pλ,[a,bν],cν), ∀1≤u<∞ such that H=⋃q=1∞Dq and Dn∩Dm=∅,∀1≤n,m<∞,n≠m, where we used the standard technique of changing an infinite union of ring elements into infinite union of pairwise disjoint ring elements [8, 11].□
Corollary  26.By the representation theorem ofR(Pλ,[a,bν],cν), ∃a pairwise disjoint system{Dj,i}i=1kj⊂Pλ,[a,bν],cνsuch thatDj=⋃i=1kjDj,i.HenceH=(⋃q=1∞Dq)=⋃q=1∞⋃i=1kqDq,i.ThereforeHis obtained as a countable pairwise disjoint union of left closed right open intervals with continuity endpoints. HenceH∈B([a,bν],ν)andνλν(H)=ν(H).
Let us apply these propositions to the sequence defined in (33).
LetAk≔⋃j=1∞[ak,j,bk,j)⊃A and let Bk≔⋂j=1kAj⊃A, ∀1≤k<∞. By the conditions of the theorem, Pλ,[a,bν],cν⊂P([a,bν],ν)⊂B([a,bν],ν)⇒∀0<k∈N, Ak, Bk∈Bλ,[a,bν],cν&Ak, Bk∈B([a,bν],ν).
Then letB≔⋂k=1∞Bk∈Bλ,[a,bν],cν and(34)νλνB∖A=νλνB-νλνA<νλνBn-νλνA<∑j=1∞νak,j,bk,j-νλν∗A)<12k,∀k∈N,where we used the definition of the covering intervals in (33). Hence B∖A∈Bλ,[a,bν],cν, &νλν(B∖A)=0⇒0=νλν(B∖A)=νλν∗(B∖A)≥ν∗(B∖A)=0. Recalling that any set of 0 outer measure is measurable we get B∖A∈B([a,bν],ν), hence with B∈B([a,bν],ν)⇒A∈B([a,bν],ν).(1) The Relationship between λ or τ Structures and ν -Ones. We will show that Pν,[a,bλ],c&Pλ,[a,bν],cν and Pν,[a,bτ],c&Pτ,[a,bν],cν are isomorphic semirings, respectively, and what is more, the isomorphisms preserve the respective measures.Lemma 27.
​(35)μλ,-u,v=νu-0,νv-0=νu,νv∈Pλ,a,bν,cν;νλνμλ,-u,v=νλννu,νv=νv-νu=νλu,v.by u,v∈[a,bλ]∖Dλ⇒ν(u),ν(v)∈[a,bν]∖Dλν.
Conversely, let[s,t)∈Pλ,[a,bν],cν and then(36)μ^λs,t=μ^λs,μ^λt∈Pν,a,bλ,c;νλμ^λs,t=νλμ^λs,μ^λt=νμ^λs-νμ^λt=t-s=νλνs,tby the identity ν(μ^λ)=id[a,bν]∖Dλν in (15) and Definition 23.Proof.
Essentially the lemma contains the references from which the statements of the lemma follow.


## 2.2.4. The Relationships between the Semirings and Their Extensions

The lemma below describes the relationship between the continuity points ofν and the points in [a,bλ].Lemma 28.
∀ t ∈ [ a , b λ ]  ∃  { l t , j } j = 1 ∞, {ut,j}j=1∞⊂[a,bλ]∖Dλ such that (1)
ifa=t⇒lt,j=a,∀1≤j<∞; otherwise if a<t then lt,j<lt,j+1,∀0<j∈N and if t=b⇒ut,j=b,∀1≤j<∞; otherwise if t<b then ut,j>ut,j+1,∀0<j∈N;(2)
lt,j↗t for j→∞ and ut,j↘t for j→∞;(3)
ν(lt,j)↗μλ,-(t)=ν(t-0) for j→∞ and ν(ut,j)↘μλ,+(t)=ν(t+0) for j→∞.​Proof.
​(1)
Lett=a, then la,j≔a, ∀1≤j<∞. Similarly, if t=b then ub,j≔b, ∀1≤j<∞.(2)
Lett∈(a,b). Let U≔a<t<b≕V&j=1.(3)
SinceDλ is countable and the open intervals ∅≠(U,t)&(t,V)⊈Dλ⇒∃p∈(U,t)∖Dλ&q∈(t,V)∖Dλ (continuity points).(4)
Letlt,j≔p&ut,j≔q.(5)
LetU≔maxp+t/2,U+t/2,&V≔minq+t/2,V+t/2 and j=j+1.(6)
Go to(3).(7)
This process will stop after generating the sequences of continuity points which are strictly ascending/descending sequences converging tot.
By the strictly ascending property ofν on [a,bλ]∖Dλ the sequences {ν(lt,j)}j=1∞&{ν(ut,j)}j=1∞ converge to left and right limits of ν at t∈[a,bλ], respectively. This completes the proof.Lemma 29.
​(1)
∀s,t∈[a,bλ],s<t the following statements hold:
(a)
[s,t]∈Bλ,[a,bλ],c;(b)
[μλ,-(s),μλ,+(t)]=[ν(s-0),ν(t+0)]∈Bλ,[a,bν],cν.(2)
∀A∈Dλν⇒A∈Bλ,[a,bν],cν and A is an atom which means ∀B∈Bλ,[a,bν],cν either B⊂A⇒B=∅ or B⊇A. Equivalently a measurable set does not cut an atom: (37)A∩B=Aor∅.(3)
Iff is Bλ,[a,bν],cν-measurable then ∀A∈Dλν,f(s)=cA,∀s∈A, where cA∈R is a constant depending on A.(4)
An interval[s,t]∈Bλ,[a,bν],cν is in Dλν if and only if μ^λ([s,t])=[u,v]⊂Nτ&ρτ(s)=1&ρλ(s)=0∀s∈[u,v].​Proof.
​(1)
By Lemma28∀s,t∈[a,bλ], s<t, ∃{ls,j}j=1∞⊂[a,bλ]∖Dλ, &∃{ut,j}j=1∞⊂[a,bλ]∖Dλ such that the following applies.
(a)
Ifa<s then {ls,j}1≤j<∞ is strictly ascending and ls,j↗s for j→∞. For a=s the sequence is constant a.(b)
Ift<b then {ut,j}1≤j<∞ is strictly descending and ut,j↘t for j→∞. For b=t the sequence is constant b.(c)
By[ls,j,ut,j)∈Pν,[a,bλ],c⊂Bλ,[a,bλ],c∀1≤j<∞ it follows that [s,t]=⋂j=1∞[ls,j,ut,j)∈Bλ,[a,bλ],c (similarly for the intervals with closed b ending).(d)
limj→∞ν(ls,j)=ν(s-0) and limj→∞ν(ut,j)=ν(t+0).(e)
Hence[ν(ls,j),ν(ut,j))∈Pλ,[a,bν],cν⊂Bλ,[a,bν],cν and(38)⋂j=1∞νls,j,νut,j=νs-0,νt+0by (d) of this proof. Hence fromBλ,[a,bν],cν being σ-algebra it follows that(39)νs-0,νt+0∈Bλ,a,bν,cν.(2)
LetA∈Dλν⇒A=Θλ(t) for a t∈Dλ. Hence Θλ(t)=[ν(t-0),ν(t+0)]∈Bλ,[a,bν],cν by point (1) of this proof. We show now that if H⊂A then H is not measurable.Letp∈A∈Dλν⇒A=Θλ(t)=[ν(t-0),ν(t+0)] for a t∈Dλ. If s∈[a,bν]∖Dλν⊂[a,bν]∖A⇒p∈[u,v)∈Pν,[a,bλ],c⇒u,v∉A⇒u<ν(t-0)<ν(t+0)<v by u,v∈[a,bν]∖A. Hence ∅≠H⊂A⇒∃p∈H⇒p∈A. Then [u,v)∈Pν,[a,bλ],c⇒νλν(A)≤νλν∗(H) holds for the outer measure of H. Hence H is not measurable if H≠A by Caratheodory’s condition for outer measures ( νλν∗(Y)=νλν∗(Y∩H)+νλν∗(Y∖H),∀Y⊂[a,bν]) with Y=A:(40)νλν∗A<2×νλν∗A<νλν∗A∩H+νλν∗A∖Hproves the lemma.(3)
Letf:[a,bν]→R be measurable with respect to Bλ,[a,bν],cν. Let us assume that ∃A∈Dλν and ∃s,t∈A such that f(s)≠f(t). Then with α≔(fs+ft)/2 the sets U≔{f>α}∩A≠∅&V≔{f<α}∩A≠∅ are disjoint and hence proper nonempty subsets of the measurable set A; hence by statement 1 of this lemma, the sets U&V are not measurable; hence f is not measurable. If f:[a,bν]→Rn is measurable and then each component of f is measurable and hence constant on the elements of Dλν; hence the statement follows.(4)
(a)
Let[s,t]∈Bλ,[a,bν],cν∩Dλν. Then ∃u∈Dλ such that [s,t]=Θλ(u). Then by Definition 16 point (2)u=μ^λ(v)≔u+∫svχNλνdν, ∀v∈[s,t]. Hence 0=∫stχNλνdν, ⇒χNλν=0 almost everywhere. By statements (2) and (3) of this lemma, the measurable function χNλν= const on [s,t] follows χNλν=0 in [s,t]; hence χNτν=1 in [s,t].(b)
Conversely, let[s,t]∈Bλ,[a,bν],cν and let χNτν(v)=1,∀v∈[s,t]. Then χNλν(v)=0,∀v∈[s,t]. Hence by Definition 16 point (2)u=μ^λ(v)≔u+∫svχNλνdν=u+∫sv0dν, ∀v∈[s,t]. Hence Θλ(u)=[s,t] proves that [s,t]∈Dλν.​Lemma 30.
​
Case (1). Let s,t∈[a,bλ], s<t, and let χ[s,t]:[a,bλ]→{0,1} fulfil χ[s,t](u)=1 if u∈[s,t] and 0 otherwise. Then χ[s,t]∘μ^λ=χ[νλ,-(s),νλ,+(t)]:[a,bν]→{0,1}.
Case (2). Let s,t∈[a,bν]∖Dλν,s<t and let χ[s,t]:[a,bν]→{0,1} fulfil χ[s,t](u)=1 if u∈[s,t] and 0 otherwise. Then χ[s,t]∘μλ,-=χ[ν^λ(s),ν^λ(t)]=χ[νλ,--1(s),νλ,+-1(s)]:[a,bλ]→{0,1}.Proof.
Case (1). u∣χs,t∘μ^λu=1=u∣s≤μ^λu≤t. Applying Lemma 13 and the set of equations (14) to the left and right sides of the inequalities, respectively, leads to the following:(41)s≤μ^λu∣∘νλ,-⇔νλ,-s≤uμ^λu≤t∣∘νλ,+⇔u≤νλ,+t.Hence u∣s≤μ^λu≤t = u∣νλ,-s≤u≤νλ,-t has been proved and therefore case (1) holds.
Case (2). Let s,t∈[a,bν]∖Dλν,s<t. Then u∣χs,t∘μλ,-u=1=u∣s≤μλ,-u≤μλ,+u≤t. Applying Lemma 13 and the set of equations (15) leads to the following:(42)s≤μλ,-u∣∘μ^λ⇔μ^λs≤uμλ,+u≤t∣∘μ^λ⇔u≤μ^λt.Hence u∣s≤μλ,-u≤μλ,+u≤t=u∣μ^λs≤u≤μ^λt has been proved and therefore case (2) holds.Notation 31.
LetM[a,bλ],Bν,[a,bλ],c≔f∣f:a,bλ→Rn,Bν,a,bλ,c-measurable(43)Ma,bν,Bν,a,bλ,cνν=f∣f:a,bν→Rn,Bλ,a,bν,cν-measurable​


## 2.2.5. Transpose of the Transformations between t and Total Variation Scales

Let us consider the transpose of a transformation of the domains of function spaces.Definition 32.
LetA,B&C be sets and let A→φB→hC with h∈{ψ:B→C}≕F(B,C) and let F(A,C)≔{ψ:A→C}. Then let(44)φ∗:h∈FB,C→h∘φ∈FA,C,∀h∈FB,C.Note that F(B,C)→φ∗F(A,C) against A→φB.Notation 33.
Notations for the lemma about the transpose mappings. LetA≔[a,bλ]&B≔[a,bν] and let C≔Rn. Let F[A:C]≔{f:[a,bλ]→Rn}≕F[[a,bλ]:Rn] be a set of functions.
LetF[B:C]≔f∣f:a,bν→RnsuchthatfisconstantonanyA∈Dλν≕F[[a,bν]:Rn],Lemma 34.
Letμ^λ:[a,bν]→[a,bλ] and μλ,-:[a,bλ]→[a,bν] be two mappings as defined above. Then the following relations hold:(45)μ^λ∗∘μλ,-∗=idFa,bν:Rnμλ,-∗∘μ^λ∗=idFa,bλ:RnMoreover μ^λ∗  &  μλ,-∗ are linear isomorphisms.Proof.
​(1)
Letf∈F[[a,bν]:Rn] and let μλ,-∗ be the transpose of μλ,-:[a,bλ]→[a,bν]. Then (46)μλ,-∗fs=fμλ,-s,∀s∈a,bλ.(2)
Letg∈F[[a,bλ]:Rn] and let μ^λ∗ be the transpose of μ^λ:[a,bν]→[a,bλ]. Then (47)μ^λ∗gv=gμ^λv,∀v∈a,bν.(3)
Let us consider now the compositionμ^λ∗∘μλ,-∗f:(48)μ^λ∗∘μλ,-∗f=μ^λ∗f∘μλ,-=f∘μλ,-∘μ^λ.By the first identity in (15) μλ,-∘μ^λ=id[a,bν]∖Dλν we discovered that ∀u∈[a,bν]∖Dλν⇒μ^λ∗∘μλ,-∗f(u)=f(u) holds.Ifu∈Dλν then u∈Θλ(t) for a t∈Dλ; hence f(μλ,-(t))=f(u),u,μλ,-(t)∈Θλ(t); hence f(u)=f(μλ,-(t))=const(t,f) (with f being constant on Θλ(t)).By (13) μ^λ(Θλ(t)) = {t}⇒∀u∈Θλ(t)⇒f(u) = f(μλ,-(μ^λ(u))) = f(μλ,-(t)) = const(t,f) completes the proof.(4)
Let us now consider the compositionμλ,-∗∘μ^λ∗g:(49)μλ,-∗∘μ^λ∗g=μλ,-∗g∘μ^λ=g∘μλ,-∘μ^λ=g∘ida,bλby (14). Hence μ^λ∗g(μλ,-(u))=g(μλ,-(μ^λ(s)))=g(u)∀u∈[a,bλ].(5)
IfRn is a vector space then F[[a,bλ]:Rn]&F[[a,bτ]:Rn] are vector spaces for the point-wise addition and multiplication by scalar. Equation (45) gives the linear isomorphism.​Corollary 35.
Let the notations of Lemma34 be changed as follows:F[[a,bλ]:Rn]→M[a,bλ],Bν,[a,bλ],cF[[a,bν]:Rn]→M[a,bν],Bν,[a,bλ],cνν.
Then Lemma34 withμ^λ∗:M[a,bλ],Bν,[a,bλ],c→M[a,bν],Bν,[a,bλ],cνν andμλ,-∗:M[a,bν],Bν,[a,bλ],cνν→M[a,bλ],Bν,[a,bλ],c holds. Precisely(50)μ^λ∗∘μλ,-∗=idMa,bλ,Bν,a,bλ,cμλ,-∗∘μ^λ∗=idMa,bν,Bν,a,bλ,cνν
Hence the mappingsμλ,-  &  μ^λ are linear isomorphisms between the two measurable function spaces.Proof.
The two continuous ascending mappingsμλ,-:[a,bλ] → [a,bν]&μ^λ:[a,bν] → [a,bλ] between the two domains [a,bλ]&[a,bν] are the same as in Lemma 34.
Letf∈M[a,bν],Bν,[a,bλ],cνν. f be measurable and μλ,- is continuous; hence(51)μλ,-∗f=f∘μλ,-∈Ma,bλ,Bν,a,bλ,cν.Let g∈M[a,bλ],Bν,[a,bλ],c be measurable and μ^λ is continuous; hence
μ ^ λ ∗ g = g ∘ μ ^ λ ∈ M [ a , b ν ] , B ν , [ a , b λ ] , c ν. By the properties proved in Lemma 34μ^λ∗g is constant on all A∈Dλν. Hence the statements of Lemma 34 are true (the two identities in (45)).LetL1[νλ,[a,bλ],c]≔f∣f:a,bλ→Rbeνλ-integrable and letL 1 [ ν λ ν , [ a , b ν ] , c ] ≔ f ∣ f : a , b ν → R   b e   ν λ ν -integrable. Then the following applies.Theorem 36.
By Corollary35 it follows that(52)μ^λ∗∘μλ,-∗=idL1νλν,a,bν,cμλ,-∗∘μ^λ∗=idL1νλ,a,bλ,cWhat is more,(53)∫abλfdνλ=∫abνμ^λ∗fdνλν,∀f∈L1νλ,a,bλ,c;∫abλμλ,-∗fdνλ=∫abνfdνλν∀f∈L1νλν,a,bν,cProof.
​(1)
By Definition22, let [s,t)∈Pν,[a,bλ],c be a measurable set and let its characteristic function be χ[s,t). Then(54)νλs,t≔νt-νs=∫abλχs,tdνλ=∫abνμ^λ∗χs,tdνλνBy Definition23 let [s,t)∈Pλ,[a,bν],cν be a measurable set and let its characteristic function be χ[s,t). Then(55)νλνs,t≔t-s=∫abλμλ,-∗χs,tdνλ=∫abνχs,tdνλν(2)
By Lemma34 for any subset A⊂[a,bλ]⇒μ^λ∗A⊂[a,bν] and(56)νλ∗A=νλν∗μ^λ∗Aholds for the outer measures(νλ)∗&(νλν)∗.Conversely∀A⊂[a,bν] such that ∀H∈Dλν either H∩A=∅ or H⊂A and then the relation holds: (νλ)∗(μλ,-∗A)=(νλν)∗(A).From this it follows that for measurable sets in bothσ-algebras the outer measures define the extended measures and therefore the measures on the σ-algebras are fulfilled. This is(57)νλνA=∫abλμλ,-∗χAdνλ=∫abνχAdνλν,∀A∈Bλ,a,bν,cννλA=∫abλχAdνλ=∫abνμ^λ∗χAdνλν,∀A∈Bν,a,bλ,cBy the linearity of the transpose mappings, the identities are valid for all nonnegative step-functions and from there for all integrable function. This completes the proof of the theorem.​Theorem 37.
Letf∈L1[νλ,[a,bλ],c]  &  t∈[a,bλ]. Then(58)∫atfdνλ=∫aμλ,+tμ^λ∗fdνλνLet f∈L1[νλν,[a,bν],c]  &  [a,s]∈Bλ,[a,bν],cν and s∈[a,bν]∖Dλν,→:(59)∫aμ^λsμλ,-∗fdνλ=∫asfdνλνProof.
​(1)
Letf∈L1[νλ,[a,bλ],c]&t∈[a,bλ]. Then(60)∫atfdνλ=∫abλχa,t×fdνλ=∫abνμ^λ∗χa,t×fdνλν=∫abν(μ^λ∗χa,t×μ^λ∗fdνλν,where the first statement in (53) of Theorem 36 was applied and the statement of case (1) in Lemma 30 gave the last equality from which finally we obtain the statement:(61)∫atfdνλ=∫abν(μ^λ∗χa,t×μ^λ∗fdνλν=∫aμλ,+tμ^λ∗fdνλν.This proves the first statement.(2)
Letf∈L1[νλν,[a,bν],c]&[a,s]∈Bν,[a,bλ],cν:(62)∫asfdνλν=∫abνχa,s×fdνλν=∫abνμλ,-∗χa,s×fdνλNow we will apply the statement of case(2) in Lemma 30 to prove the final result:(63)∫asfdνλν=∫abνμλ,-∗χa,s×fdνλ=∫abν(μλ,-∗χa,s×μλ,-∗fdνλ=∫abν(χa,μ^λs×μλ,-∗fdνλ=∫aμ^λsμλ,-∗fdνλ.​


## 2.2.6. The Main Result

In Theorem37, we established the relationship between absolute continuous indefinite integrals of members of L1[νλ,[a,bλ],c] and the absolute continuous indefinite integrals of members of L1[νλν,[a,bν],c].Our next step is to establish the relationship between the indefinite integrals of the members ofL1[ν,[a,bν]] and the indefinite integrals of the members of L1[νλν,[a,bν]],c].We prove the following statements.(1)
Letf∈L1[ν,[a,bν]] and let νf,λ(A)≔∫Afdν, ∀A∈Bλ,[a,bν],cν.Letνf,τ(A)≔∫Afdν, ∀A∈Bτ,[a,bν],cν.(2)
FromPλ,[a,bν],cν⊂P([a,bν],ν) it follows that Bλ,[a,bν],cν⊂B([a,bν],ν).FromPτ,[a,bν],cν⊂P([a,bν],ν) it follows that Bτ,[a,bν],cν⊂B([a,bν],ν).(3)
The signed measureνf,λ:Bλ,[a,bν],cν→R is absolute continuous with respect to ν:Bλ,[a,bν],cν→R+ by A∈Bλ,[a,bν],cν&ν(A)=0⇒νf(A)=∫Afdν=0.The signed measureνf,τ:Bτ,[a,bν],cν→R is absolute continuous with respect to ν:Bτ,[a,bν],cν→R+ by A∈Bτ,[a,bν],cν&ν(A)=0⇒νf(A)=∫Afdν=0.(4)
By the absolute continuity stated in point(2) mentioned above, the Bλ,[a,bν],cν-measurable Radon Nikodym derivative exists:(64)f~ν,λ≔dνfdνλν⇔∫Afdν=∫Af~ν,λdνλν,∀A∈Bλ,a,bν,cν.By the absolute continuity stated in point(2) mentioned above, the Bτ,[a,bν],cν-measurable Radon Nikodym derivative exists:(65)f~ν,τ≔dνfdντν⇔∫Afdν=∫Af~ν,τdντν,∀A∈Bτ,a,bν,cν.​With the statements(1)-(4) mentioned above we have proved the following result.Theorem 38.
A functionf:[a,bλ]⊂R→Rn is of bounded variation on [a,bλ] if and only if the following conditions are fulfilled. (1)
There exists a closed interval[a,bν],a,bν∈R partitioned as union of two disjoint measurable sets [a,bν]=Nλν∪Nτν such that μ^λ(s)≔a+∫asχNλνdν∈[a,bλ],s∈[a,bν] is onto [a,bλ] (see the details in Definition 15, equation (17)).(2)
There exists an absolute continuous functionφf(u)≔f(a)+∫auψfdν:[a,bν]→Rn with ψf∈L1[ν,[a,bν],Rn] such that the following applies.
(a)
On the t-scale with the Radon - Nikodym derivativeψf^ν,λ defined in (64) is expressed by(66)ft=φfμλ,+t=fa+∫aμλ,+tψfdν=fa+∫aμλ,+tψf^dνλν=fa+∫atμλ,-∗ψf^ν,λdνλ,∀t∈a,bλ(b)
Theτ scale representation fτ of f with the Radon - Nikodym derivative ψf^ν,τ defined in (65) is expressed by(67)fτϑ=φfμτ,+ϑ=fa+∫aμτ,+ϑψfdν=fa+∫aμτ,+ϑψf^ν,τdντν=fa+∫aϑμτ,-∗ψf^ν,τdντ,∀ϑ∈0,bτwhere we used the notations and functions defined in Definition 15, equation (17) and the mappings defined in Definitions 16 and 17.



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*Source: 1065946-2019-07-08.xml*