# Lineability within Peano Curves, Martingales, and Integral Theory

Authors: Artur Bartoszewicz; Marek Bienias; Szymon Głąb
Journal: Journal of Function Spaces
 (2018)

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## Abstract

This paper is devoted to give several improvements of some known facts in lineability approach. In particular, we prove that (i) the set of continuous mappings from the unit interval onto the unit square contains a closed,c-semigroupable convex subset, (ii) the set of pointwise convergent martingales (Xn)n∈N with EXn→∞ is c-lineable, (iii) the set of martingales converging in measure but not almost surely is c-lineable, (iv) the set of sequences (Xn)n∈N of independent random variables, with EXn=0, ∑n=1∞varXn=∞, and the property that (X1+⋯+Xn)n∈N is almost surely convergent, is c-lineable, (v) the set of bounded functions f:[0,1]×[0,1]→R for which the assertion of Fubini’s Theorem does not hold is consistent with ZFC1-lineable (it is not 2-lineable), (vi) the set of unbounded functions f:[0,1]×[0,1]→R for which the assertion of Fubini’s Theorem does not hold (with infinite integral allowed) is c-lineable but not c+-lineable.


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## Body

## 1. Introduction

We start with a standard introduction to the idea of lineability (similar openings can be found, for example, in our paper with T. Natkaniec (see [1]) or in the papers of other authors mentioned below). For more than a decade now, many mathematicians have been looking at the largeness of some sets by constructing algebraic structures inside them. This approach is calledlineability. A comprehensive description of this concept as well as numerous examples and some general techniques can be found in the book [2] and references therein. Following R. Aron, A. Bartoszewicz, S. Głąb, V. Gurariy, D. Pérez-García, and J.B. Seoane-Sepúlveda, let us recall the following notion.Definition 1.
Letκ be a cardinal number. (1)
LetL be a vector space and A⊆L. We say that A is κ-lineable if A∪{0} contains a κ-dimensional subspace of L.(2)
LetL be a semigroup and A⊆L. We say that A is κ-semigroupable if A contains a κ-generated subsemigroup of L.This paper is devoted to the improvements of some known facts: in semigroupability (Section2) and lineability in probability theory (Section 3). In Sections 4 and 5 we present some new results in probability and integral theory. Along this paper we shall use standard set theoretical notation and notions. In particular, we identify an ordinal number α with the set of all ordinals β<α. Cardinal numbers are those ordinals α which are not equipotent with any β<α and by α+ we denote the cardinal successor of a cardinal number α. We use standard notation, i.e., N={1,2,3,…}.


## 2. Peano Curves and Semigroupability

In the19th century, inspired by the result that the unit interval [0,1] is of the same cardinality as the unit square [0,1]2, G. Peano constructed (see [3]) a continuous onto mapping F:[0,1]→[0,1]2. For topological spaces X,Y let us denote the set of all continuous surjections from X to Y by CS(X,Y). In [4], N. Albuquerque proved that for every pair n,m∈N the set CS(Rm,Rn) is c-lineable.LetI=[0,1]. In this section we are interested in the set (1)CSI,I2=f:I→I2:fis  cont.  andfI=I2.L. Bernal-González, M.C. Calderón-Moreno, and J. Prado-Bassas proved in [5] that the set CS(I,I2) is semigroupable; i.e., it contains ω-generated semigroup (with respect to coordinate-wise multiplication). Our goal is to prove the following.Theorem 2.
There exists a closed convex setX⊆CS(I,I2) such that it is c-semigroupable; i.e., X contains a c-generated semigroup.Proof.
Let us fix a functionF∈CS([0,1/2],I2) (its construction is a trivial modification of Peano’s example) and let A⊆[0,1]N be a set of cardinality c of sequences that converge to 0 and are free generators (see [6]). Let F={Fx:x∈A}, where, for x=(x(1),x(2),…)∈A, the function Fx:I→I2 is defined by the formula (2)Fxt=Ft,whent∈0,12t,xn,whent=1-1n+2forn∈N1,0,whent=1affine,elsewhere,(byaffine we mean line segments in the square I2 that continuously connect points F(1/2) with (2/3,x(1)) and (1-1/n+2,x(n)) with (1-1/n+3,x(n+1)) for n=1,2,…). Clearly Fx is continuous for every x∈A. Let Y be a semigroup generated by F; i.e., Y={Fx1k1·…·Fxnkn:x1,…,xn∈A,k1,…,kn∈N,n∈N}. We will show that X=conv(Y)¯ (the closure of the convex hull of Y) has the desired properties. It is easy to see that since A is a set of free generators, the set F is algebraically independent. Hence, Y is a c-generated semigroup.
Moreover,conv(Y)⊆CS(I,I2). Indeed, let f∈conv(Y). Then there are n∈N, f1,…,fn∈Y and α1,…,αn∈0,1 such that ∑i=1nαi=1, with f(t)=∑i=1nαifi(t) for t∈I. Clearly f is continuous. It is easy to see that, by the definition of the functions from F, there exist k1,…,kn∈N such that f∣0,1/2=∑i=1nαiFki. To show that f∈CS(I,I2), it is enough to prove that f0,1/2=I2. Let (a,b)∈I2 and let P:I→I be defined by P(t)=∑i=1nαixki. Since P(0)=0 and P(1)=1, the function P is onto and there exist x,y∈I such that P(x)=a and P(y)=b. Moreover, as F∈CS(0,1/2,I2), there is t∈[0,1/2] with F(t)=(x,y) and then f(t)=P(F(t))=(a,b) (for a function F=(F1,F2):I→I2,P:I→I and for t∈I by P(F(t)), we mean (P(F1(t)),P(F2(t)))∈I2). Hence, f∈CS(I,I2).
Finally, let us observe that the limit of any convergent sequence fromCS(I,I2) is still in CS(I,I2). Indeed, let (gn)n∈N be a sequence of elements of CS(I,I2) that converges uniformly to a function g:I→I2. Obviously g is continuous. Moreover, for any point w∈I2 and any n∈N there is xn∈I with gn(xn)=w. We may assume (by compactness of I) that (xn)n∈N is convergent to some point x0∈I. It is easy to see that limn→∞gn(xn)=g(x0) so g(x0)=w. Hence, g is also surjective. Thanks to this observation, we have proved that X=conv(Y)¯ is a closed, convex, and c-semigroupable subset of CS(I,I2).


## 3. Lineability within Martingales Theory

This section is devoted to some lineability problems in the martingales setting. Let us start with a short introduction to the probability and martingales theory. Till the end of the section let(Ω,S,P) be a probability space (i.e., Ω is a nonempty set, S is a σ-algebra of subsets of Ω, and P:S→[0,1] is a probability measure). A function X:Ω→R is calleda random variable, whenever it is S-measurable (i.e., X-1[B]∈S for every Borel B⊆R); itsexpected value is computed as follows: (3)EX=∫ΩXdPand itsvariance is as follows: (4)var⁡X=EX2-EX2.There are different notions of convergence of a sequence of random variables. We recall those that will be needed later.Definition 3.
LetX and Xn (n∈N) be random variables. We say that the sequence (Xn)n∈N(1)
is pointwise convergent toX, whenever Xn(t)→X(t) for every t∈Ω,(2)
is convergent almost everywhere toX, whenever Xn(t)→X(t) almost surely, i.e., (5)Pt∈Ω:Xnt→Xt=1,(3)
is convergent in measure toX, whenever (6)∀ε>0Pt∈Ω:Xnt-Xt≥ε→0.It is clear that pointwise convergence implies convergence almost surely and it implies convergence in measure. Moreover, none of these implications can be reversed in general.Definition 4.
LetX:Ω→R be a random variable with finite expected value and let H⊆S be a σ-algebra. Theconditional expectation of X with respect to H, denoted by EX∣H, is any H-measurable function Ω→R that satisfies (7)∫HEX∣HdP=∫HXdPfor every H∈H.It is known that the conditional expectation ofX with respect to H is defined uniquely up to a set of measure zero. In the sequel the following properties of conditional expectation will be useful (see [7]).Remark 5.
(1)
LetX,Y:Ω→R be random variables, let G⊆S be a σ-algebra, and assume that X is G-measurable. If X and XY have finite expected values, then (8)EXY∣G=XEY∣G,almostsurely.(2)
LetX:Ω→R be a random variable and G1⊆G2⊆S are σ-algebras. Then (9)EEX∣G2∣G1=EX∣G1.(3)
LetX:Ω→R be a random variable and let G⊆S be a σ-algebra such that σ(X) and G are independent (i.e., for all A∈σ(X) and B∈G we have P(A∩B)=P(A)·P(B), where σ(X) is the smallest σ-algebra with respect to which X is measurable). Then (10)EX∣G=EX;i.e., it is a constant random variable.Definition 6.
We say that a sequence(Fn)n∈N of σ-algebras forms a filtration whenever (11)F1⊆F2⊆⋯⊆S.Let us state the definition of a martingale.Definition 7.
Let(Fn)n∈N be a filtration and (Xn)n∈N be a sequence of random variables. We say that (Xn)n∈N is a martingale adapted to the filtration (Fn)n∈N if for every n∈N(1)
Xn is Fn-measurable,(2)
EXn<∞,(3)
EXn+1∣Fn=Xn almost everywhere.Remark 8.
The set of all martingales adapted to a fixed filtration forms a real vector space (with respect to natural addition and scalar multiplication).In [8], the authors proved the following theorem:Theorem 9 (Thm. 1, [8]).
There is a probability space in which the set of pointwise convergent sequences of random variables{Xn:n∈N} with EXn→∞ is ω-lineable.Our goal is to state a more general result than the above in the setting of martingales.Theorem 10.
Let(Ω,S,P) be a probability space such that there is infinitely many pairwise disjoint sets of positive measures. Then there exist a filtration (Fn)n∈N and c-dimensional vector space V of martingales adapted to the filtration (Fn)n∈N such that for every nonzero X=(Xn)n∈N∈V we have that (Xn)n∈N is pointwise convergent but EXn→∞.Proof.
LetA1,A2,…,B1,B2,…∈S be a partition of Ω (i.e., Ω is a disjoint union of all Ai’s with all Bi’s) such that P(Ai),P(Bi)>0 for all i∈N. Let Fn=σ(A1,…,An,B1,…,Bn), for n∈N. Clearly (Fn)n∈N forms a filtration. For α∈(0,1) let us define X(α)=(Xn(α))n∈N in the following way: for t∈Ω(12)
X
n
α
t
=
1
k
α
P
A
k
,
w
h
e
n
t
∈
A
k
,
for
k
≤
n
-
1
k
α
P
B
k
,
w
h
e
n
t
∈
B
k
,
for
k
≤
n
0
e
l
s
e
w
h
e
r
e
. Observe that X(α) is a martingale adapted to the filtration (Fn)n∈N. Indeed, for any n∈N, (1)
Xn(α) is an Fn-measurable;(2)
EXnα=2∑k=1n1/kα<∞;(3)
as forM=⋃k=1n(Ak∪Bk) we have that ∫Ω∖MXn+1(α)=0 and Xn+1αM=XnαM, so for every C∈Fn we have that ∫CXn+1(α)=∫CXn(α). Hence, EXn+1α∣Fn=Xn(α).
LetV=span{X(α):α∈(0,1)}. Then V is a vector space of martingales adapted to the filtration (Fn)n∈N. Clearly its dimension equals c as {X(α):α∈(0,1)} is linearly independent. We will show that V has the desired properties. Let X=(Xn)n∈N∈V be a nonzero martingale of the form (Xn)n∈N=(∑i=1mciXn(αi))n∈N for some m∈N, c1,…,cm∈R∖{0} and 0<α1<α2<⋯<αm<1. It is easy to see that (Xn)n∈N converges pointwise to the variable Y:Ω→R defined as follows: for t∈Ω(13)Yt=∑i=1mcikαiPAk,whent∈Ak,fork∈N∑i=1m-cikαiPBk,whent∈Bk,fork∈N. On the other hand, we have the following estimation for every n∈N: (14)EXn≥Ec1Xnα1-c2Xnα2-⋯-cmXnαm=∑k=1n2c1kα1-2c2kα2-⋯-2cmkαm=∑k=1n2c1kα11-c2c1kα2-α1-⋯-cmc1kαm-α1.Since α1<α2<⋯<αm, there is K∈N such that for every k≥K(15)1-c2c1kα2-α1-⋯-cmc1kαm-α1≥12.In particular, for every n≥K we have (16)∑k=Kn2c1kα11-c2c1kα2-α1-⋯-cmc1kαm-α1≥∑k=Knc1kα1→n→∞∞.Hence, (17)EXn→∞,as required.In the martingales setting one may find another somehow surprising example, namely a martingale such that it converges to0 in measure but not almost surely (see [9]). The next theorem is based on this construction but provides a large structure of such martingales.Theorem 11.
Let(Ω,S,P)=([0,1]N,Bor([0,1]N),P) be a probability space, where Bor([0,1]N) is the σ-algebra of Borel subsets of  [0,1]N and P:Bor([0,1]N)→[0,1] is the product measure (countable product of the Lebesgue measure on [0,1]). There exist a filtration (Fn)n∈N and a c-dimensional vector space V of martingales adapted to the filtration (Fn)n∈N such that every nonzero (Xn)n∈N∈V converges to 0 in measure but no almost surely.Proof. For α∈1/3,1/2 and n∈N define a random variable Zn(α):[0,1]N→R by the formula: (18)Znαωkk∈N=1,ifωn<12nα-1,ifωn>1-12nα0,elsewhere,for (ωk)k∈N∈[0,1]N. Clearly, for any α∈1/3,1/2 and any natural numbers i≠j the random variables Zi(α) and Zj(α) are independent.Now, let us define a sequence(Xn(α))n∈N in the following way: (19)X1α=Z1αXnα=Znα1Xn-1α=0+nαXn-1αZnα1Xn-1α≠0,forn≥2.We will show that (Xn(α))n∈N forms a martingale adapted to the filtration (Fn)n∈N, where (20)Fn=A×0,1N∖1,…,n:A∈Bor0,1n,forn∈N.For every n∈N, we have that (1)
Xn(α) is Fn measurable by the definition,(2)
EXnα≤(n!)α<∞ as EX1α=1 and Xnα≤nαXn-1α, for every n>1,(3)
by Remark5, we have the following equalities (almost sure): (21)EXn+1α∣Fn=EZn+1α1Xnα=0∣Fn+En+1αXnαZn+1α1Xnα≠0∣Fn=1Xnα=0EZn+1α∣Fn+n+1αXnα1Xnα≠0EZn+1α∣Fn=1Xnα=0EZn+1α+n+1αXnα1Xnα≠0EZn+1α=0+n+1αXnα1Xnα≠01n+1α=Xnα.Let V=span{(Xn(α))n∈N:α∈1/3,1/2}, which is a vector space of martingales adapted to the filtration (Fn)n∈N. We will show that V has the desired properties. Let X=(Xn)n∈N∈V be a nonzero martingale; i.e., there are m∈N, c1,…,cm∈R∖{0} and 1/3<αm<⋯<α2<α1≤1/2 such that X=∑i=1mciX(αi); that is, Xn=∑i=1mciXn(αi) for every n∈N. Observe that, by the definition of Xn(α), we have that for every t∈Ω(22)Xnαt=0⇔Znαt=0,hence (23)Znαmt=0⇒∀i≤mZnαit=0⇒∀i≤mXnαit=0⇒Xnt=0.Therefore (24)PXn=0≥PZnαm=0=1-1nαmand in particular P(Xn=0)→1 with n→∞; i.e., (Xn)n∈N converges to 0 in measure. To prove the second assertion, let us firstly observe the following.Claim 1.
For everyα∈1/3,1/2 and any n∈N with n≥3, we have(25)PXnα≤nα=1-1nn-1n-2αand the event Xnα≤nα depends, at most, on the events Ziα=j for i∈{n,n-1,n-2} and j∈{0,1}.Proof.
By an easy inductive argument we may prove that(26)XnαΩ⊆0,±1,±nα,±nn-1α,…,±n!α,so for t∈Ω, (27)Xnαt≤nα⇔Xnαt=0∨Xnαt=1∨Xnαt=nα.By the definition, we have the following: for t∈Ω, (28)Xnαt=0⇔Znαt=0,Xnαt=1⇔Znαt=1∧Xn-1αt=0⇔Znαt=1∧Zn-1αt=0,Xnαt=nα⇔Znαt=1∧Xn-1αt=1⇔Znαt=1∧Zn-1αt=1∧Zn-2αt=0.Clearly the above events are independent so (29)PXnα≤nα=PZnα=0+PZnα=1∧Zn-1α=0+PZnα=1∧Zn-1α=1∧Zn-2α=0=1-1nα+1nα1-1n-1α+1nα1n-1α1-1n-2α=1-1nn-1n-2α.The second assertion is a direct consequence of the proof.LetAn={t∈Ω:Xnα1t≥nα1∧∀i≤mXnαit≤nαi} for n∈N. By the Claim, we have the following estimation: (30)PAn≥PXnα1≥nα1-∑i≤m1nn-1n-2αi=PZnα1=1∧Xn-1α1≠0-∑i≤m1nn-1n-2αi=PZnα1=1∧Zn-1α1=1-∑i≤m1nn-1n-2αi=1nα11n-1α1-∑i≤m1nn-1n-2αiand ∑n=1∞P(A3n)=∞ (as αi∈1/3,1/2 for i≤m). Moreover, by the Claim, the events A3,A6,A9,… are independent so by the Borel–Cantelli Lemma⋆Plimsupn→∞⁡A3n=1.Since αm<⋯<α2<α1, there exists M∈N such that for all n≥M we have (31)c1nα1-∑i=2mcinαi>c1nα12>1.Let n≥M and t∈An. Then (32)Xnt≥c1Xnα1t-∑i=2mciXnαit≥t∈Anc1nα1-∑i=2mcinαi>1. In particular, by ⋆, (33)PXn≠0for infinitely manyn=1; hence (Xn)n∈N does not tend to 0 almost surely. In particular, V is c-dimensional vector space.(End of Proof)


## 4. Kolmogorov’s Theorem

We start this section with recalling a well-known theorem due to Kolmogorov (see [7]).Theorem 12 (Kolmogorov’s theorem, [7]).
Let(Ω,S,P) be a probability space and (Xn)n∈N be a sequence of independent random variables such that EXn=0 for all n∈N and ∑n=1∞varXn<∞. Then, for Sn=X1+⋯+Xn the sequence (Sn)n∈N is convergent almost surely.Examples of sequences of independent random variables with expected values equal to0 and divergent series of their variations such that (X1+⋯+Xn)n∈N does converge almost surely are known (see [9]). Here we show that there is a c-dimensional vector space of such sequences.Theorem 13.
There exist a probability space(Ω,S,P) and a c-dimensional vector subspace V⊆(L2(Ω,S,P))N of sequences of independent random variables, such that for every nonzero X=(Xn)n∈N∈V we have EXn=0 for every n∈N, ∑n=1∞varXn=∞, and (X1+⋯+Xn)n∈N converges almost surely.Proof.
Let(Ω,S,P) be a probability space such that there exists a sequence of independent random variables (Zn)n∈N such that (34)PZn=n=PZn=-n=12n2andPZn=0=1-1n2.Clearly, EZn=0 and varZn=1 for any n∈N, so ∑n=1∞varXn=∞. Let {Iα:α<c} be an almost disjoint (i.e., an intersection of any two different members is finite) family of infinite subsets of N. For α<c and n∈N, define a random variable Xn(α) by the formula: (35)Xnα=Zn,ifn∈Iα0,ifn∉Iαand observe that EXn(α)=0 and varXn(α)≤1. Hence, Xn(α)∈L2(Ω,S,P) for any α<c and n∈N. Let V be the subspace of (L2(Ω,S,P))N spanned by {(Xn(α))n∈N:α<c}. We will show that V has the desired properties. Let X=(Xn)n∈N∈V be of the form (Xn)n∈N=(∑i=1mciXn(αi))n∈N, for some m∈N, c1,…,cm∈R∖{0}, and α1<⋯<αm<c. Clearly, (Xn)n∈N is a sequence of independent random variables (by the independence of (Zn)n∈N) and EXn=0 for all n∈N. Moreover, for infinitely many naturals, namely, from the set Iα1∖⋃i=2m(Iα1∩Iαi), we have varXn=c1varZn=c1 so ∑n=1∞varXn=∞. In particular, V is c-dimensional vector space. As {Iα:α<c} is almost disjoint, there is M∈N such that the sets Iαi∖{1,…,M}(i∈{1,…,m}) are pairwise disjoint. Let n>M and observe that (36)PXn≠0≤PZn≠0=1n2;therefore ∑n=1∞P(Xn≠0)<∞. By the Borel-Cantelli Lemma (37)PXn≠0for infinitely manyn=0so (38)PXn=0for almost alln=1.In particular, (X1+⋯+Xn)n∈N converges almost surely.


## 5. Failure of Fubini’s Theorem

The famous Theorem of Fubini states that for any integrable functionf:[0,1]×[0,1]→R the following equality holds: (39)∫0,1∫0,1fx,ydxdy=∫0,1∫0,1fx,ydydx(from now onintegrable, measurable, almost every (a.e.) meansLebesgue integrable, Lebesgue measurable, from a co-null set). The idea of looking for examples of functions f:[0,1]×[0,1]→R (measurable unbounded or nonmeasurable) attracted an attention of many mathematicians (see, for example, [10–13]). In this section, we will present some lineability results (both positive and negative) in this area.


### 5.1. The Case of Bounded Functions

Letf:[0,1]×[0,1]→R be a bounded function. We say that for f theIterated Integral Property Failure holds when(IPPF)(i)
[0,1]∋x↦f(x,y) is integrable for almost every y∈[0,1],[0,1]∋y↦f(x,y) is integrable for almost every x∈[0,1],(ii)
[0,1]∋x↦φf∫0,1f(x,y)dy, defined a.e., is integrable,[0,1]∋y↦ψf∫0,1f(x,y)dx, defined a.e., is integrable,(iii)
∫0,1φf(x)dx≠∫0,1ψf(y)dy.It is known that existence of a bounded functionf:[0,1]×[0,1]→R with (IIPF) is independent from ZFC (we refer the reader to [14] for the details). In this situation, we may obtain some results consistently with ZFC.Theorem 14.
It is consistent withZFC that the set of bounded functions f:[0,1]×[0,1]→R with(IIPF) is 1-lineable (or star-like) but not 2-lineable.Proof.
LetS⊆[0,1]×[0,1] be the so-called Steinhaus set, i.e., a set with all horizontal sections of measure zero and all vertical sections of full measure (existence of such a set is consistent with ZFC; for example, it can be constructed under the Continuum Hypothesis; see [14]). It is easy to see that, for the characteristic function fS:[0,1]×[0,1]→R of S and any a∈R∖{0}, the function afS has (IIPF).
To prove the second assertion, letf,g:[0,1]×[0,1]→R be bounded and satisfy (IIPF). Denote (the real numbers) Af=∫0,1φf(x)dx, Bf=∫0,1ψf(y)dy and Ag, Bg in analogous way. Then Af≠BfAg≠Bg. It is easy to see that for nonzero numbers a=Ag-Bg/Af-Bf, b=-1, and the function h=af+bg, we have (40)∫0,1φhxdx-∫0,1ψhydy=Ag-BgAf-BfAf-Ag-Ag-BgAf-BfBf+Bg=0,so h does not have (IIPF) and the proof is finished.Remark 15.
A modification of a function with (IIPF) on the set of measure zero still has (IIPF). Hence, there are2c distinct bounded functions with (IIPF).


### 5.2. The Case of Unbounded Functions

It is natural to ask a question about the (IIPF) condition for an unbounded functionf:[0,1]×[0,1]→R (an example of such function exists in ZFC; see [14]). Let us start with an easy observation, based on an analogous argument as in Theorem 14.Theorem 16.
The set of unbounded functionsf:[0,1]×[0,1]→R with(IIPF) is 1-lineable (or star-like) but not 2-lineable.Proof.
Letf:[0,1]×[0,1]→R be an unbounded function with (IIPF) (see [14]). It is easy to check that for any a∈R∖{0} we have that af has (IIPF), what proves 1-lineability. The second assertion is a consequence of a property (ii) in (IIPF), where we actually assume that the integrals ∫0,1φf(x)dx, ∫0,1ψf(y)dy (even for an unbounded f) are finite and the same argument as in the proof of Theorem 14 works.Having in mind the above result, in the context of an unbounded functionf:[0,1]×[0,1]→R, it is natural to reformulate the (IIPF) condition as follows:(IPPF∞)(i)
[0,1]∋x↦f(x,y) is integrable for almost every y∈[0,1],[0,1]∋y↦f(x,y) is integrable for almost every x∈[0,1],(ii)
[0,1]∋x↦φf∫0,1f(x,y)dy, defined a.e., is measurable and the integral ∫0,1φf(x)dx exists (is finite or infinite),[0,1]∋y↦ψf∫0,1f(x,y)dx, defined a.e., is measurable and the integral ∫0,1ψf(y)dy exists (is finite or infinite),(iii)
∫0,1φf(x)dx≠∫0,1ψf(y)dy.Theorem 17.
The set of unbounded functionsf:[0,1]×[0,1]→R with(IIPF∞) is c-lineable.Proof.
Letα∈(2,∞) and let an=1/2n-1 for n∈N. Define a function f(α):[0,1]×[0,1]→R in the following way: for any n∈N and any x∈an+1,an(41)fαx,y=-α2n,wheny∈0,an+1α2n,wheny∈an+1,an0,wheny∈an,1and f(α)(0,y)=0 for y∈[0,1]. Let α>2. Observe that φf(α)(x)=0, for any x∈[0,1]. On the other hand, for any n∈N and any y∈an+1,an, we have (42)ψfαy=α2n2n-∑k=1n-1α2k2k=Aαn-Aα1-Aαn-11-Aα=Aαn+1-2Aαn+AαAα-1>0,where Aα=α2/2. For the simplicity, let us denote Bα(n)=Aαn+1-2Aαn+Aα/Aα-1.
LetV=span{f(α):α∈(2,∞)}; we will show that every nonzero element of V has (IIPF∞). Let f∈V be of the form f=∑i=1mcif(αi) for some m∈N,c1,…,cm∈R∖{0} and α1>⋯>αm>2. Clearly, for all x∈[0,1], we have that φf(x)=0 as φf(αi)(x)=0 for every i∈{1,…,m}. Therefore, ∫0,1φf(x)dx=0. On the other hand, we have (43)∫0,1ψfydy=∑n=1∞12n∑i=1mciBαin=∑n=1∞∑i=1mciBαin2n.Observe that, as α1>α2>⋯>αm, (44)BαinBα1n→n→∞0,for any i∈{2,…,m}. Hence, there is M∈N such that for any n≥M(45)-12<2c2Bα2nc1Bα1n+⋯+2cmBαmnc1Bα1n<12.Let n≥M. We have two possibilities: (1)
Ifc1>0, then ∑i=1mciBαin/2n>c1Bα1(n)/2n+1. Indeed, (46)∑i=1mciBαin2n-c1Bα1n2n+1=c1Bα1n2n+1+c2Bα2n2n+⋯+cmBαmn2n=c1Bα1n2n+11+2c2Bα2nc1Bα1n+⋯+2cmBαmnc1Bα1n>0.(2)
Ifc1<0, then ∑i=1mciBαin/2n<c1Bα1(n)/2n+1. Indeed, (47)∑i=1mciBαin2n-c1Bα1n2n+1=c1Bα1n2n+1+c2Bα2n2n+⋯+cmBαmn2n=c1Bα1n2n+11+2c2Bα2nc1Bα1n+⋯+2cmBαmnc1Bα1n<0.Moreover, Bα1(n)/2n+1→n→∞∞, because α1>2. In particular, (48)∫0,1ψfydy=∞,whenc1>0-∞,whenc1<0.The above implies that f satisfies (IIPF∞) and that V is c-dimensional.Finally, we show that the last theorem is optimal in the sense of cardinalities.Theorem 18.
The set of unbounded functionsf:[0,1]×[0,1]→R with(IIPF∞) is not c+-lineable.Proof.
Suppose on the contrary that the set isc+-lineable and let fξ::[0,1]×[0,1]→R, for ξ<c+, have (IIPF∞) and be such that for any m∈N,c1,…,cm∈R∖{0} and any ξ1<⋯<ξm<c+ the function f=∑i=1mcifξi has (IIPF∞). By the assumption, for any ξ<c+, there exist (defined a.e.) measurable functions φfξ,ψfξ. It is known that any measurable function is equal to a Borel function almost everywhere; i.e., there exist Borel functions φ¯fξ,ψ¯fξ such that φ¯fξ=φfξ almost everywhere and ψ¯fξ=ψfξ almost everywhere. As there are c many distinct Borel functions, there exist ξ<ζ<c+ such that φ¯fξ=φ¯fζ,ψ¯fξ=ψ¯fζ. Clearly, for the function f=fξ-fζ we have φf=φfξ-φfζ=φ¯fξ-φ¯fζ=0 almost everywhere and ∫0,1φf(x)dx=0. Similarly, ∫0,1ψf(y)dy=0 and we reach a contradiction.


## 5.1. The Case of Bounded Functions

Letf:[0,1]×[0,1]→R be a bounded function. We say that for f theIterated Integral Property Failure holds when(IPPF)(i)
[0,1]∋x↦f(x,y) is integrable for almost every y∈[0,1],[0,1]∋y↦f(x,y) is integrable for almost every x∈[0,1],(ii)
[0,1]∋x↦φf∫0,1f(x,y)dy, defined a.e., is integrable,[0,1]∋y↦ψf∫0,1f(x,y)dx, defined a.e., is integrable,(iii)
∫0,1φf(x)dx≠∫0,1ψf(y)dy.It is known that existence of a bounded functionf:[0,1]×[0,1]→R with (IIPF) is independent from ZFC (we refer the reader to [14] for the details). In this situation, we may obtain some results consistently with ZFC.Theorem 14.
It is consistent withZFC that the set of bounded functions f:[0,1]×[0,1]→R with(IIPF) is 1-lineable (or star-like) but not 2-lineable.Proof.
LetS⊆[0,1]×[0,1] be the so-called Steinhaus set, i.e., a set with all horizontal sections of measure zero and all vertical sections of full measure (existence of such a set is consistent with ZFC; for example, it can be constructed under the Continuum Hypothesis; see [14]). It is easy to see that, for the characteristic function fS:[0,1]×[0,1]→R of S and any a∈R∖{0}, the function afS has (IIPF).
To prove the second assertion, letf,g:[0,1]×[0,1]→R be bounded and satisfy (IIPF). Denote (the real numbers) Af=∫0,1φf(x)dx, Bf=∫0,1ψf(y)dy and Ag, Bg in analogous way. Then Af≠BfAg≠Bg. It is easy to see that for nonzero numbers a=Ag-Bg/Af-Bf, b=-1, and the function h=af+bg, we have (40)∫0,1φhxdx-∫0,1ψhydy=Ag-BgAf-BfAf-Ag-Ag-BgAf-BfBf+Bg=0,so h does not have (IIPF) and the proof is finished.Remark 15.
A modification of a function with (IIPF) on the set of measure zero still has (IIPF). Hence, there are2c distinct bounded functions with (IIPF).


## 5.2. The Case of Unbounded Functions

It is natural to ask a question about the (IIPF) condition for an unbounded functionf:[0,1]×[0,1]→R (an example of such function exists in ZFC; see [14]). Let us start with an easy observation, based on an analogous argument as in Theorem 14.Theorem 16.
The set of unbounded functionsf:[0,1]×[0,1]→R with(IIPF) is 1-lineable (or star-like) but not 2-lineable.Proof.
Letf:[0,1]×[0,1]→R be an unbounded function with (IIPF) (see [14]). It is easy to check that for any a∈R∖{0} we have that af has (IIPF), what proves 1-lineability. The second assertion is a consequence of a property (ii) in (IIPF), where we actually assume that the integrals ∫0,1φf(x)dx, ∫0,1ψf(y)dy (even for an unbounded f) are finite and the same argument as in the proof of Theorem 14 works.Having in mind the above result, in the context of an unbounded functionf:[0,1]×[0,1]→R, it is natural to reformulate the (IIPF) condition as follows:(IPPF∞)(i)
[0,1]∋x↦f(x,y) is integrable for almost every y∈[0,1],[0,1]∋y↦f(x,y) is integrable for almost every x∈[0,1],(ii)
[0,1]∋x↦φf∫0,1f(x,y)dy, defined a.e., is measurable and the integral ∫0,1φf(x)dx exists (is finite or infinite),[0,1]∋y↦ψf∫0,1f(x,y)dx, defined a.e., is measurable and the integral ∫0,1ψf(y)dy exists (is finite or infinite),(iii)
∫0,1φf(x)dx≠∫0,1ψf(y)dy.Theorem 17.
The set of unbounded functionsf:[0,1]×[0,1]→R with(IIPF∞) is c-lineable.Proof.
Letα∈(2,∞) and let an=1/2n-1 for n∈N. Define a function f(α):[0,1]×[0,1]→R in the following way: for any n∈N and any x∈an+1,an(41)fαx,y=-α2n,wheny∈0,an+1α2n,wheny∈an+1,an0,wheny∈an,1and f(α)(0,y)=0 for y∈[0,1]. Let α>2. Observe that φf(α)(x)=0, for any x∈[0,1]. On the other hand, for any n∈N and any y∈an+1,an, we have (42)ψfαy=α2n2n-∑k=1n-1α2k2k=Aαn-Aα1-Aαn-11-Aα=Aαn+1-2Aαn+AαAα-1>0,where Aα=α2/2. For the simplicity, let us denote Bα(n)=Aαn+1-2Aαn+Aα/Aα-1.
LetV=span{f(α):α∈(2,∞)}; we will show that every nonzero element of V has (IIPF∞). Let f∈V be of the form f=∑i=1mcif(αi) for some m∈N,c1,…,cm∈R∖{0} and α1>⋯>αm>2. Clearly, for all x∈[0,1], we have that φf(x)=0 as φf(αi)(x)=0 for every i∈{1,…,m}. Therefore, ∫0,1φf(x)dx=0. On the other hand, we have (43)∫0,1ψfydy=∑n=1∞12n∑i=1mciBαin=∑n=1∞∑i=1mciBαin2n.Observe that, as α1>α2>⋯>αm, (44)BαinBα1n→n→∞0,for any i∈{2,…,m}. Hence, there is M∈N such that for any n≥M(45)-12<2c2Bα2nc1Bα1n+⋯+2cmBαmnc1Bα1n<12.Let n≥M. We have two possibilities: (1)
Ifc1>0, then ∑i=1mciBαin/2n>c1Bα1(n)/2n+1. Indeed, (46)∑i=1mciBαin2n-c1Bα1n2n+1=c1Bα1n2n+1+c2Bα2n2n+⋯+cmBαmn2n=c1Bα1n2n+11+2c2Bα2nc1Bα1n+⋯+2cmBαmnc1Bα1n>0.(2)
Ifc1<0, then ∑i=1mciBαin/2n<c1Bα1(n)/2n+1. Indeed, (47)∑i=1mciBαin2n-c1Bα1n2n+1=c1Bα1n2n+1+c2Bα2n2n+⋯+cmBαmn2n=c1Bα1n2n+11+2c2Bα2nc1Bα1n+⋯+2cmBαmnc1Bα1n<0.Moreover, Bα1(n)/2n+1→n→∞∞, because α1>2. In particular, (48)∫0,1ψfydy=∞,whenc1>0-∞,whenc1<0.The above implies that f satisfies (IIPF∞) and that V is c-dimensional.Finally, we show that the last theorem is optimal in the sense of cardinalities.Theorem 18.
The set of unbounded functionsf:[0,1]×[0,1]→R with(IIPF∞) is not c+-lineable.Proof.
Suppose on the contrary that the set isc+-lineable and let fξ::[0,1]×[0,1]→R, for ξ<c+, have (IIPF∞) and be such that for any m∈N,c1,…,cm∈R∖{0} and any ξ1<⋯<ξm<c+ the function f=∑i=1mcifξi has (IIPF∞). By the assumption, for any ξ<c+, there exist (defined a.e.) measurable functions φfξ,ψfξ. It is known that any measurable function is equal to a Borel function almost everywhere; i.e., there exist Borel functions φ¯fξ,ψ¯fξ such that φ¯fξ=φfξ almost everywhere and ψ¯fξ=ψfξ almost everywhere. As there are c many distinct Borel functions, there exist ξ<ζ<c+ such that φ¯fξ=φ¯fζ,ψ¯fξ=ψ¯fζ. Clearly, for the function f=fξ-fζ we have φf=φfξ-φfζ=φ¯fξ-φ¯fζ=0 almost everywhere and ∫0,1φf(x)dx=0. Similarly, ∫0,1ψf(y)dy=0 and we reach a contradiction.



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