Coneability of Anti-Fubini Functions and Other Lineability Properties

Abstract

It is proved that the family of measurable real functions on the product measure space satisfying (or not) the conclusion of Fubini’s Theorem (along with a number of related families) is algebraically large, in the sense that it contains large convex cones or even large vector subspaces (except for zero) under rather general assumptions. Several earlier related results are improved, mainly regarding the cardinality of the generator set of the corresponding cones or subspaces.

1 Introduction, Motivation and Notation

In [6] the first and second authors study the topological and algebraic genericity of measurable real-valued functions on the product measure space that satisfy (or not) the conclusions of Fubini’s Theorem. Throughout the paper, given a set X, we will consider functions f on X that take values on the real line. For a measure space $(X,{\mathcal {A}},\mu )$, we denote the space of ($\mu$-classes of) measurable functions on X by $L^0(X)$, and the space of integrable functions (i.e., $\int _X |f|d\mu <\infty$) by $L^1(X)$. Note that we are identifying two functions  f, g  in  $L^0(X)$  when they are equal outside a $\mu$-null set. When this identification is not considered, then the set of measurable functions will be denoted by  ${\mathcal {L}}^0(X)$.

We will endow the space $L^0(X)$ with the topology of local convergence in measure, defined as follows (see [17, Chapter 24]): a net $(f_\alpha )_{\alpha \in I} \subset L^0(X)$  is said to converge to a function $f \in L^0(X)$ if, for every  $\varepsilon > 0$ and every $A \in {\mathcal {A}}$ with $\mu (A) < \infty$, one has  $\displaystyle {\lim \nolimits _{\alpha \in I}} \, \mu (\{ x \in A: \, |f_\alpha (x) - f(x)| > \varepsilon \}) = 0$.

Let $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$ be two measure spaces. The product $\sigma$-algebra ${\mathcal {A}} \otimes {\mathcal {B}}$ is the smallest $\sigma$-algebra on $X\times Y$ that contains all $A\times B$, where $A\in {\mathcal {A}}$ and $B\in {\mathcal {B}}$. A product measure on the measurable space $(X\times Y,{\mathcal {A}}\otimes {\mathcal {B}})$ is a measure $\tau$ that satisfies $\tau (A\times B)=\mu (A)\nu (B)$ for all $A\in {\mathcal {A}}$ and $B\in {\mathcal {B}}$. The standard product measure on $(X\times Y,{\mathcal {A}}\otimes {\mathcal {B}})$ can be constructed in the following way:

Consider the outer measure $(\mu \otimes \nu )^* :{\mathcal {P}}(X\times Y) \rightarrow [0,\infty ]$ defined as:

$$\begin{aligned} (\mu \otimes \nu )^* (E) = \inf \left\{ \sum _{i=1}^\infty \mu (A_i)\nu (B_i) :E\subseteq \bigcup _{i=1}^\infty (A_i\times B_i),\ A_i\in {\mathcal {A}},\ B_i \in {\mathcal {B}}, \,\, \forall i\in {\mathbb {N}}\right\} . \end{aligned}$$

The restriction of  $(\mu \otimes \nu )^*$  to the $\sigma$-algebra  ${\mathcal {A}} \otimes {\mathcal {B}}$  will be denoted by $\mu \otimes \nu$  and defines the standard product measure on $(X\times Y,{\mathcal {A}}\otimes {\mathcal {B}})$. Recall that if $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$ are $\sigma$-finite, then a product measure $\tau$ on $(X\times Y,{\mathcal {A}}\otimes {\mathcal {B}})$ is uniquely determined by Caratheodory’s Theorem (i.e., $\tau =\mu \otimes \nu$). However, if $(X,{\mathcal {A}},\mu )$ or $(Y,{\mathcal {B}},\nu )$ are not $\sigma$-finite, then there are distinct product measures $\tau _1$ and $\tau _2$ on $(X\times Y,{\mathcal {A}}\otimes {\mathcal {B}})$. For instance, this is the case for the standard product measure and the complete locally determined product $\pi$ defined as follows: for all $C\in {\mathcal {A}} \otimes {\mathcal {B}}$,

$$\begin{aligned} \pi (C)=\sup \{(\mu \otimes \nu )(C\cap (A\times B)) :\mu (A),\nu (B)<\infty \}. \end{aligned}$$

If $f :X\times Y \rightarrow {\mathbb {R}}$ is ${\mathcal {A}}\otimes {\mathcal {B}}$-measurable, then, for each $x\in X$ and $y\in Y$, the section functions $f_y:x\in X \mapsto f(x,y)\in {\mathbb {R}}$ and $f^x :y\in Y \mapsto f(x,y)\in {\mathbb {R}}$ are ${\mathcal {A}}$-measurable and ${\mathcal {B}}$-measurable, respectively. Define $\widehat{f_y}:=f_y$ ($:=0$, resp.) for those $y\in Y$ for which $f_y$ is (is not, resp.) $\mu$-integrable (analogously for $\widehat{f^x}$). Given two complete measure spaces $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$, we will denote the completion of a product measure $\tau$ on $(X\times Y,\widehat{{\mathcal {A}}\otimes {\mathcal {B}}})$, where $\widehat{{\mathcal {A}}\otimes {\mathcal {B}}}$ is the completion of the $\sigma$-algebra ${\mathcal {A}}\otimes {\mathcal {B}}$, by ${\widehat{\tau }}$.

Fubini’s Theorem, as stated by the first and second author in [6, Theorem 1], is given below.

Theorem 1.1

(Fubini’s Theorem, version 1). If $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$ are $\sigma$-finite complete measure spaces and $f\in L^1 (X\times Y)$, where in $X\times Y$ the completion of the standard product measure space $(X\times Y,\widehat{{\mathcal {A}}\otimes {\mathcal {B}}},\widehat{\mu \otimes \nu })$ is considered, then

• (a) $f^x\in L^1(Y)$ for $\mu$-a.e. $x\in X$, and $f_y\in L^1(X)$ for $\nu$-a.e. $y\in Y$.

• (b) The functions $x\mapsto \int _Y \widehat{f^x} d\nu$, $y\mapsto \int _X \widehat{f_y} d\mu$ belong respectively to $L^1(X)$, $L^1(Y)$.

• (c) We have the equality

  $$\begin{aligned} \int _X \left( \int _Y \widehat{f^x} d\nu \right) d\mu =\int _Y \left( \int _X \widehat{f_y} d\mu \right) d\nu \end{aligned}$$

  (1.1)

Moreover, as $f\in L^1(X\times Y)$, we also have that the double integral

$$\begin{aligned} \int _{X\times Y} f \, d\mu \otimes \nu \end{aligned}$$

(1.2)

equals the iterated integrals in (1.1).

In the paragraph after Fubini’s Theorem as stated in [6, Theorem 1], the authors claim that the property of $\sigma$-finiteness cannot be dropped as proven in [20, Example 17.15]. However, this is partially true since the example provided in [20] also satisfies that $f\notin L^1(X\times Y)$. If $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$ are ($\sigma$-finite or not) complete measure spaces and $f\in L^1 (X\times Y)$, where in $X\times Y$ the (standard or not) complete measure space $(X\times Y,\widehat{{\mathcal {A}}\otimes {\mathcal {B}}},{\widehat{\tau }})$ is considered, the conclusions of Fubini’s Theorem are still satisfied (see [14, Theorem 14.1]). For completeness, we state this alternative version of Fubini’s Theorem in Theorem 1.2 below.

Theorem 1.2

(Fubini’s Theorem, version 2). If $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$ are complete measure spaces and $f\in L^1 (X\times Y)$, where in $X\times Y$ the complete product measure space $(X\times Y,\widehat{{\mathcal {A}}\otimes {\mathcal {B}}},{\widehat{\tau }})$ is considered, then

• (a) $f^x\in L^1(Y)$ for $\mu$-a.e. $x\in X$, and $f_y\in L^1(X)$ for $\nu$-a.e. $y\in Y$.

• (b) The functions $x\mapsto \int _Y \widehat{f^x} d\nu$, $y\mapsto \int _X \widehat{f_y} d\mu$ belong respectively to $L^1(X)$, $L^1(Y)$.

• (c*) We have the equalities

  $$\begin{aligned} \int _{X\times Y} f d\mu \otimes \nu = \int _X \left( \int _Y \widehat{f^x} d\nu \right) d\mu =\int _Y \left( \int _X \widehat{f_y} d\mu \right) d\nu . \end{aligned}$$

Therefore, we have that the conclusions of Fubini’s Theorem (a), (b) and (c) are satisfied in the more general setting when $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$ are complete measure spaces and $f\in L^1(X\times Y)$. Moreover, observe that if at least one of the measure spaces is not $\sigma$-finite, then we can consider a product measure different from the standard product measure.

Throughout [6] the first and second authors consider measure spaces $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$ that are $\sigma$-finite and complete. Under these assumptions they study the algebraic and topological genericity of the following families of functions and some of their set-theoretical relations:

• $L_a:=\{f\in L^0(X\times Y) :f \text { satisfies (a) in Theorem}$ 1.1$\}$.

• $L_{ab}:=\{f\in L^0(X\times Y) :f \text { satisfies (a)-(b) in Theorem}$ 1.1$\}$.

• $L_{abc}:=\{f\in L^0(X\times Y) :f \text { satisfies (a)-(b)-(c) in Theorem}$ 1.1$\}$.

• $a{\mathcal {F}}:=L_{ab}{\setminus } L_{abc}$, known as the family of anti-Fubini functions.

• $p{\mathcal {F}}:=L_{abc}{\setminus } L^1(X\times Y)$, known as the family of pseudo-Fubini functions.

Observe that we have the following relations of containment:

$$\begin{aligned} L^1(X\times Y)\subseteq L_{abc} \subseteq L_{ab}\subseteq L_{a}\subseteq L^0(X\times Y). \end{aligned}$$

A corresponding chain of inclusions takes place if we consider  $f = g$  only if  $f(z) = g(z)$  for all  $z \in X \times Y$:

$$\begin{aligned} {\mathcal {L}}^1(X\times Y) \subseteq {\mathcal {L}}_{abc} \subseteq {\mathcal {L}}_{ab} \subseteq {\mathcal {L}}_{a} \subseteq {\mathcal {L}}^0(X\times Y). \end{aligned}$$

We shall also consider the sets  $a {{\mathcal {L}}}{{\mathcal {F}}}:= {\mathcal {L}}_{ab} {{\setminus }} {\mathcal {L}}_{abc}$  and  $p \mathcal{L}\mathcal{F}:= {\mathcal {L}}_{abc} {{\setminus }} {\mathcal {L}}^1(X\times Y)$.

In the special case  $X = {\mathbb {R}}= Y, \, \mu = \nu =$ the unidimensional Lebesgue measure, the set  $p {\mathcal {F}}$  is thoroughly studied in [7, 8, 10].

In this paper, we extend the results dealing with the algebraic and topological genericity obtained in [6] to the case when $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$ are (not necessarily $\sigma$-finite) complete measure spaces as well as give a partial answer to an open problem posed in part 4 of [6, Section 4]. Observe that there are measures spaces that are not contemplated in [6] that are considered in this paper. For instance, as a consequence of the main results, we analyze the lineability of the above families of functions when we consider the counting measure on ${\mathbb {R}}$, which is a complete measure space that is not $\sigma$-finite.

The following theorem from Set Theory will be used throughout this paper. To state it, let us define what is known as an independent family of sets.

Definition 1.3

Let X be a nonempty set. We say that a family ${\mathcal {Y}}\subset {\mathcal {P}}(X)$ is independent if for any distinct sets $A_1,\ldots ,A_n\in {\mathcal {Y}}$ we have that

$$\begin{aligned} A_1^{\varepsilon _1} \cap \cdots \cap A_n^{\varepsilon _n} \ne \varnothing , \end{aligned}$$

where $Y^1:=Y$ and $Y^0=X{\setminus } Y$ for all $Y\in {\mathcal {Y}}$.

Theorem 1.4

(Fichtenholz–Kantorovich–Hausdorff Theorem, [16, 18]). If X is a set of infinite cardinality $\kappa$, then there exists an independent family ${\mathcal {Y}}$ of subsets of X of cardinality $2^\kappa$.

For more information on Fichtenholz-Kantorovich-Hausdorff Theorem, we refer the interested reader to [19].

Next, we present two concepts related to the existence of large algebraic structures within nonlinear sets. The first one can be found in [3] (see, also, [1, 2, 4, 13]). The second one, as far as we know, has not been defined formally up to now, being a yuxtaposition of properties of coneability (see [11]) and convex lineability (see [9]). Recall that the cone generated by a subset  A  of a real vector space is the set  $\textrm{Cone} (A):= \{\alpha x: \, \alpha \in {\mathbb {R}}, \, x \in A \}$, while the convex cone generated by  A  is the set  $\textrm{Cone} (\textrm{Conv} (A))$, where  $\textrm{Conv} (A)$  denotes the convex hull of  A.

Definition 1.5

Assume that X is a topological vector space, $\alpha$ is a cardinal number and $A \subset X$. Then A is said to be $\alpha$-dense-lineable if there is an $\alpha$-dimensional dense vector subspace V of X such that $V \subset A \cup \{0\}$.

Definition 1.6

Assume that X is a vector space over ${\mathbb {R}}$, $\alpha$ is a cardinal number and $A \subset X$. Then A is said to be $\alpha$-convex-coneable if there is a linearly independent set $S \subset X$ with $\textrm{card}(S) = \alpha$ such that $\textrm{Cone} (\textrm{Conv} (S)) \subset A \cup \{0\}$.

2 Main Results

Given a measure space $(X,{\mathcal {A}},\mu )$, let us consider the following properties (Q) and (S) from [6], together with $({\overline{Q}})$, that is stronger than (Q):

(Q):

There are countably many mutually disjoint sets $(A_n)_{n\in {\mathbb {N}}}\subset {\mathcal {A}}$ such that $0<\mu (A_n)<\infty$ for every $n\in {\mathbb {N}}$.

$({\overline{Q}})$:

There are countably many mutually disjoint sets $(A_n)_{n\in {\mathbb {N}}}\subset {\mathcal {A}}$ such that $0<\mu (A_n)<\infty$ for every $n\in {\mathbb {N}}$ and $\sum _{n=1}^\infty \frac{1}{\mu (A_n)}<\infty$.

(S):

There is a countable family ${\mathcal {A}}_1\subset {\mathcal {A}}$ such that, for any $A\in {\mathcal {A}}$ with $\mu (A)<\infty$ and $\varepsilon >0$, there exists $A_0\in {\mathcal {A}}_1$ with $\mu (A\triangle A_0)<\varepsilon$, where $A\triangle B:=(A{\setminus } B)\cup (B{\setminus } A)$ denotes the symmetric difference.

Let us make a small digression at this point. Plainly, condition $({\overline{Q}})$ implies (Q) but the reverse is not true: take, for instance, the Lebesgue measure on  [0, 1]. On the contrary, both counting measure on  ${\mathbb {N}}$  and Lebesgue measure on an unbounded interval of  ${\mathbb {R}}$  satisfy $({\overline{Q}})$. By adapting the proof of [20, Theorem 14.23], it can be shown that $({\overline{Q}})$ is equivalent to the property that  $\sup \{\mu (A): \, A \in {\mathcal {A}}$ and $\mu (A) < \infty \} = \infty$, which in turn is equivalent to the fact that  $L^q(X) \not \subset L^p(X)$  for some (or any) pair of extended real numbers satisfying  $1 \le p < q \le \infty$; see also [21, 22].

In [6, Proposition 2] the first and second authors prove that, in a complete $\sigma$-finite measure space $(X,{\mathcal {A}},\mu )$, we have that $L^0(X){\setminus } L^1(X)\ne \varnothing$ if, and only if, $L^0(X){\setminus } L^1(X)$ is dense in $L^0(X)$ if, and only if, $L^0(X){\setminus } L^1(X)$ is residual in $L^0(X)$. The latter part is obtained by Baire Category Theorem since $L^0(X)$ is an F-space, i.e., a completely metrizable topological vector space (a Baire space). However, if the complete measure space is not $\sigma$-finite, then $L^0(X)$ is not metrizable, which does not guarantee that $L^0(X){\setminus } L^1(X)$ is residual in $L^0(X)$. (See [17].) Nonetheless, the following Proposition 2.1 can still be deduced from the proof of [6, Proposition 2] for arbitrary complete measure spaces.

Proposition 2.1

If $(X,{\mathcal {A}},\mu )$ is a complete measure space, then the following assertions are equivalent:

• (i) $L^0(X){\setminus } L^1(X)\ne \varnothing$.

• (ii) $L^0(X){\setminus } L^1(X)$ is a dense $G_\delta$-subset of $L^0(X)$.

The same proof of [6, Theorem 2] guarantees the following Theorem 2.2.

Theorem 2.2

Let $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$ be complete measure spaces. Assume that in $X\times Y$ a complete product measure space $(X\times Y,\widehat{{\mathcal {A}}\otimes {\mathcal {B}}},{\widehat{\tau }})$ is considered. Then the set  $a{\mathcal {F}}$ is not 2-lineable.

The special case  $X = [0,1] = Y$  with the Lebesgue measure was thoroughly analyzed in [5, § 5].

The following Theorem 2.3 is a generalization of [6, Theorem 3]. The idea of the proof is similar to the one in [6, Theorem 3].

Theorem 2.3

Let $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$ be complete measure spaces, and $(X\times Y,\widehat{{\mathcal {A}}\otimes {\mathcal {B}}},{\widehat{\tau }})$ a complete product measure space. If $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$ satisfy property (Q), then the set $a{\mathcal {F}}$ contains a set ${\mathcal {S}}$ with the following properties:

• (1) $\textrm{card}({\mathcal {S}}) = {\mathfrak {c}}$,

• (2) ${\mathcal {S}}$ is linearly independent,

• (3) $\textrm{span}({\mathcal {S}}){\setminus } \{0\}\subset L^0(X\times Y){\setminus } L^1(X\times Y)$, and

• (4) the convex cone generated by  ${\mathcal {S}}$ is contained in $a {\mathcal {F}} \cup \{0\}$.

Consequently, the set $a{\mathcal {F}}$ is ${\mathfrak {c}}$-convex-coneable.

Proof

By property (Q), there are countably many mutually disjoint $(A_n)_{n\in {\mathbb {N}}}\subset {\mathcal {A}}$ and $(B_n)_{n\in {\mathbb {N}}}\subset {\mathcal {B}}$ such that $0<\mu (A_n),\nu (B_n)<\infty$ for all $n\in {\mathbb {N}}$.

By Theorem 1.4, there exists a family of independent subsets ${\mathcal {N}}$ of ${\mathbb {N}}$ of cardinality ${\mathfrak {c}}$.

For every $N\in {\mathcal {N}}$, denote $N=\{n_{1,N}<n_{2,N}<\cdots< n_{k,N} < \cdots \}$ and let us define $f_N:X\times Y \rightarrow {\mathbb {R}}$ as:

$$\begin{aligned} f_N:=-\frac{\chi _{A_{n_{1,N}}\times B_{n_{2,N}}}}{\mu (A_{n_{1,N}})\nu (B_{n_{2,N}})}+\sum _{k=2}^\infty \left( \frac{\chi _{A_{n_{k,N}}\times B_{n_{k-1,N}}}}{\mu (A_{n_{k,N}})\nu (B_{n_{k-1,N}})} - \frac{\chi _{A_{n_{k,N}}\times B_{n_{k+1,N}}}}{\mu (A_{n_{k,N}})\nu (B_{n_{k+1,N}})} \right) . \end{aligned}$$

As the sets $A_n$’s and $B_n$’s are mutually disjoint, observe that each function $f_N$ is well defined.

The set ${\mathcal {S}}:= \{f_N:N\in {\mathcal {N}} \}$ is as needed. Indeed, first it is obvious that $f_N$ is measurable since it is the pointwise limit of simple functions. Second, take $x_0\in X$ arbitrary. Then,

$$\begin{aligned} (f_N)^{x_0}={\left\{ \begin{array}{ll} 0 &{} \quad \text {if } x_0\notin A_{n_{k,N}} \text { for all } k\in {\mathbb {N}},\\ -\frac{\chi _{B_{n_{2,N}}}}{\mu (A_{n_{1,N}})\nu (B_{n_{2,N}})} &{} \quad \text {if } x_0\in A_{n_{1,N}}\\ \frac{\chi _{A_{n_{k,N}}\times B_{n_{k-1,N}}}}{\mu (A_{n_{k,N}})\nu (B_{n_{k-1,N}})} - \frac{\chi _{A_{n_{k,N}}\times B_{n_{k+1,N}}}}{\mu (A_{n_{k,N}})\nu (B_{n_{k+1,N}})} &{} \quad \text {if } x_0\in A_{n_{k,N}} \text { for some } k\ge 2. \end{array}\right. } \end{aligned}$$

Hence, $(f_N)^{x_0}\in L^1(Y)$ in any case. Moreover, observe that

$$\begin{aligned} \int _X\left( \int _Y (f_N)^{x} d\nu \right) d\mu =-1. \end{aligned}$$

Analogously, we have that  $(f_N)_{y_0} \in L^1(X)$  for every  $y_0\in Y$  but

$$\begin{aligned} \int _Y \left( \int _X (f_N)_{y} d\mu \right) = 1. \end{aligned}$$

Hence, we have ${\mathcal {S}}\subset a{\mathcal {F}}$. Now take $N_1,\ldots , N_m \in {\mathcal {N}}$ distinct, $a_1,\ldots ,a_m\in {\mathbb {R}}{\setminus } \{0\}$ and consider $F:=\sum _{i=1}^m a_i f_{N_i}$. Let us prove thirdly that ${\mathcal {S}}$ satisfies (1) and (2). Assume that $F\equiv 0$. By construction, there exists $n_1 \in N_1 {\setminus } \bigcup _{i=2}^m N_i$ with $n_1>\min \left\{ N_1 {\setminus } \bigcup _{i=2}^m N_i \right\}$. Thus, there exists $k_1\in {\mathbb {N}}{\setminus } \{1\}$ such that $n_{k_1,N_1}=n_1$. Hence, by taking $\langle x,y\rangle \in A_{n_{k_1,N_1}}\times B_{n_{k_1-1,N_1}}$ arbitrary, we have that

$$\begin{aligned} f_{N_1}(x,y)=a_1 \frac{1}{\mu (A_{n_{k_1,N_1}})\nu (B_{n_{k_1-1,N_1}})}, \end{aligned}$$

by the pairwise disjointness of the sets $(A_n)_{n\in {\mathbb {N}}}$ and $(B_n)_{n\in {\mathbb {N}}}$. But we have reached a contradiction since

$$\begin{aligned} {{\widehat{\tau }}}(A_{n_{k_1,N_1}}\times B_{n_{k_1-1,N_1}})=\mu (A_{n_{k_1,N_1}})\nu (B_{n_{k_1-1,N_1}})>0, \end{aligned}$$

that is, we have proven (1) and (2).

Now let us prove that $F\notin L^1(X\times Y)$. To do so, since

$$\begin{aligned} \text {card}\left( N_1{\setminus } \bigcup _{i=2}^m N_i \right) =\aleph _0, \end{aligned}$$

(see, for instance, the observation right after Definition 1.3 of [15]), we can consider $N_1{\setminus } \bigcup _{i=2}^m N_i=\{n_{1,1}<n_{2,1}<\cdots<n_{k,1}<\cdots \}$ and define

$$\begin{aligned} M:=\bigcup _{k=2}^\infty A_{n_{k,1}} \times B_{n_{k-1,1}}. \end{aligned}$$

Observe that $M\in \widehat{{\mathcal {A}}\otimes {\mathcal {B}}}$, which implies that

$$\begin{aligned} F\cdot \chi _M=\sum _{k=2}^\infty \frac{\chi _{A_{n_{k,1}} \times B_{n_{k-1,1}}}}{\mu (A_{n_{k,1}}) \nu (B_{n_{k-1,1}})} \in L^0(X\times Y). \end{aligned}$$

Moreover, if $F\in L^1(X\times Y)$, then we would have that $F\cdot \chi _M\in L^1(X\times Y)$ but

$$\begin{aligned} \int _{X\times Y} |F\cdot \chi _M| d{\widehat{\tau }}&=|a_1|\sum _{k=2}^\infty \int _{A_{n_{k,1}} \times B_{n_{k-1,1}}} \frac{1}{\mu (A_{n_{k,1}}) \nu (B_{n_{k-1,1}})} d{\widehat{\tau }}\\&=|a_1|\sum _{k=2}^\infty 1=\infty . \end{aligned}$$

Hence, we have proved (3).

Finally, let  $F = \sum _{i=1}^m a_i f_{N_i}$  be a nonzero element of the convex cone generated by  ${\mathcal {S}}$, so that we can assume that  $a_i > 0$  (resp. $a_i < 0$)  for every $i\in \{1,\ldots ,m \}$. It is clear that $F^{x_0}\in L^1(Y)$ for every $x_0\in X$ and $F_{y_0}\in L^1(X)$ for every $y_0\in Y$, but also

$$\begin{aligned} \int _X \left( \int _Y F^x \, d\nu \right) d\mu = - \sum _{i=1}^m a_i\ne \sum _{i=1}^m a_i=\int _Y \left( \int _X F_y \, d\mu \right) d\mu . \end{aligned}$$

Hence $F\in a{\mathcal {F}}$, which yields (4). $\square$

Corollary 2.4

Let $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$ be complete measure spaces, and $(X\times Y,\widehat{{\mathcal {A}}\otimes {\mathcal {B}}},{\widehat{\tau }})$ a complete product measure space. If $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$ satisfy property (Q), then $a{\mathcal {F}}$ is ${\mathfrak {c}}$-convex-coneable.

Corollary 2.5

Let $({\mathbb {R}},{\mathcal {P}}({\mathbb {R}}),\mu )$ denote the counting measure space on ${\mathbb {R}}$ and $\tau$ a product measure on $({\mathbb {R}}^2,\widehat{{\mathcal {P}}({\mathbb {R}})\otimes {\mathcal {P}}({\mathbb {R}})})$. Assume that $({\mathbb {R}}^2,\widehat{{\mathcal {P}}({\mathbb {R}})\otimes {\mathcal {P}}({\mathbb {R}})},{\widehat{\tau }})$ is the completion of the product measure space. Then $L^0({\mathbb {R}}^2){\setminus } L^1({\mathbb {R}}^2)$ is ${\mathfrak {c}}$-lineable and  $a{\mathcal {F}}$ is ${\mathfrak {c}}$-convex-coneable.

Although it is impossible to find a vector space of dimension at least 2 inside the family $a{\mathcal {F}}\cup \{0\}$ as Theorem 2.2 states, Corollary 2.4 shows that we are able to find cones of dimension ${\mathfrak {c}}$ inside $a{\mathcal {F}}\cup \{0\}$. Now it is natural to ask if the dimension of the cone inside $a{\mathcal {F}}\cup \{0\}$ can be improved. For instance, we wonder whether $a{\mathcal {F}}\cup \{0\}$ is $2^{{\mathfrak {c}}}$-convex-coneable and under which conditions.

We study the latter in Theorem 2.6 below as well as the convex coneability of the family $p \mathcal{L}\mathcal{F}$.

Theorem 2.6

Let $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$ be complete measure spaces, $\kappa$ and $\lambda$ infinite cardinal numbers, and $(X\times Y,\widehat{{\mathcal {A}}\otimes {\mathcal {B}}},{\widehat{\tau }})$ a complete product measure space.

• (i) Assume that there is a family ${\mathcal {N}}$ of independent subsets of X (resp. Y) of cardinality $2^\kappa$ such that each $N\in {\mathcal {N}}$ is $\mu$-null (resp. $\nu$-null), and there exists an $M\in {\mathcal {B}}{\setminus } \{\varnothing \}$ (resp. $M\in {\mathcal {A}}{\setminus } \{\varnothing \}$) with $\nu (M)=0$ (resp. $\mu (M)=0$).

  • (1) If $a\mathcal{L}\mathcal{F} \ne \varnothing$, then $a\mathcal{L}\mathcal{F}$ is $2^\kappa$-convex-coneable.

  • (2) If $p\mathcal{L}\mathcal{F} \ne \varnothing$, then $p\mathcal{L}\mathcal{F}$ is $2^\kappa$-convex-coneable.

• (ii) Assume that there are, respectively, families ${\mathcal {N}}$ and ${\mathcal {M}}$ of independent subsets of X and Y of cardinality $2^\kappa$ and $2^\lambda$ such that each $N\in {\mathcal {N}}$ is $\mu$-null and every $M\in {\mathcal {M}}$ is $\nu$-null.

  • (1) If $a\mathcal{L}\mathcal{F} \ne \varnothing$, then $a\mathcal{L}\mathcal{F}$ is $\max \{2^\kappa ,2^\lambda \}$-convex-coneable.

  • (2) If $p\mathcal{L}\mathcal{F} \ne \varnothing$, then $p\mathcal{L}\mathcal{F}$ is $\max \{2^\kappa ,2^\lambda \}$-convex-coneable.

Proof

We will prove only part (ii) since part (i) can be done in a similar way.

Take $f \in a\mathcal{L}\mathcal{F}$ for part (1) and $f \in p \mathcal{L}\mathcal{F}$ for part (2). Let $\{N_\xi :\xi <2^{\kappa } \}$ and $\{M_\zeta :\zeta <2^\lambda \}$ be enumerations, without repetition, of ${\mathcal {N}}$ and ${\mathcal {M}}$, respectively. For every $\langle \xi ,\zeta \rangle \in 2^\kappa \times 2^\lambda$, let us define $f_{\langle \xi ,\zeta \rangle } :X\times Y \rightarrow {\mathbb {R}}$ as:

$$\begin{aligned} f_{\langle \xi ,\zeta \rangle }:=f\restriction {[(X\times Y) {\setminus } (N_\xi \times M_\zeta )]}+\chi _{(N_\xi \times M_\zeta )}. \end{aligned}$$

Observe that each $f_{\langle \xi ,\zeta \rangle }$ is $\widehat{{\mathcal {A}}\otimes {\mathcal {B}}}$-measurable since f is $\widehat{{\mathcal {A}}\otimes {\mathcal {B}}}$-measurable and $N_\xi \times M_\zeta$ is ${{\widehat{\tau }}}$-null.

In the case of part (i), define $f_\xi :=f\restriction {[(X\times Y) {\setminus } (N_\xi \times M)]}+\chi _{(N_\xi \times M)}$ provided that the family of independent subsets is contained in X and M is nonempty and $\nu$-null. The case when the family of independent subsets is contained in Y and M is nonempty and $\mu$-null is done in a similar way.

Let us prove first that the functions in  ${\mathcal {S}}:= \{f_{\langle \xi ,\zeta \rangle } :\langle \xi , \zeta \rangle \in 2^{\kappa }\times 2^\lambda \}$ are linearly independent. With this aim, let  $F \in \textrm{span} \, ({\mathcal {S}})$  so that there are  $n \in {\mathbb {N}}$, $a_1, \dots ,a_n \in {\mathbb {R}} {\setminus } \{0\}$  and  $\langle \xi _i, \zeta _i \rangle \in 2^{\kappa } \times 2^{\kappa }$ ($i = 1, \dots ,n$) (without loss of generality, we can assume  $\xi _1 \le \cdots \le \xi _n<2^{\kappa }$, $\zeta _1\le \cdots \le \zeta _n < 2^\lambda$)  such that  $F = \sum _{i=1}^n a_i f_{\langle \xi _i,\zeta _i \rangle }$. Also, if $\xi _i=\xi _j$ for some $i\ne j$, then $\zeta _i\ne \zeta _j$ (analogously, if $\zeta _i=\zeta _j$ for $i\ne j$, then $\xi _i\ne \xi _j$). Assume that  F  is identically zero. Let  A  be the set of pairwise distinct ordinals $<2^\kappa$ such that $A=\{\xi _1,\ldots ,\xi _n \}$. Analogously, let  B  be the set of pairwise distinct ordinals $<2^\lambda$ such that $B=\{\zeta _1,\ldots ,\zeta _n \}$. Clearly, $1\le \text {card}(A),\text {card}(B)\le n$.

• Case 1: If $\text {card}(A)=n$, then take $\langle x,y\rangle \in \left( N_{\xi _1}{\setminus } \bigcup _{i=2}^n N_{\xi _i} \right) \times M_{\zeta _1}$. (Observe that $N_{\xi _1}{\setminus } \bigcup _{i=2}^n N_{\xi _i}$ and $M_{\zeta _1}$ are nonempty by construction.) Then $0=F(x,y)=a_1$, a contradiction.

• Case 2: If $1\le \text {card}(A)<n$, then we can assume, without loss of generality, that $\xi _1=\xi _i$ for some $i\ne 1$. Take the set of all indexes $I\subseteq \{1,\ldots ,n \}$ such that $\xi _1=\xi _i$ for every $i\in I$. (Notice that $1\in I$.) By construction, we have that $\zeta _i\ne \zeta _j$ for every distinct $i,j\in I$. Hence, there exists $\langle x,y\rangle \in \left( N_{\xi _1}{\setminus } \bigcup _{i\in \{1,\ldots ,n \}{\setminus } I } N_{\xi _i} \right) \times \left( M_{\zeta _1}{\setminus } \bigcup _{i\in I} M_{\zeta _i} \right)$. Thus, $0=F(x,y)=a_1$, a contradiction.

Furthermore, we have also proven that the functions in ${\mathcal {S}}$ are pairwise distinct, which implies that $\text {card}({\mathcal {S}}) = 2^\kappa \cdot 2^\lambda =\max \{2^\kappa ,2^\lambda \}$.

Assume now that our f belongs to $a \mathcal{L}\mathcal{F}$ in order to prove part (1). It is enough to prove that $F:= \sum _{i=1}^n a_i f_{\langle \xi _i,\zeta _i \rangle } \in \textrm{Cone}(\textrm{Conv} ({\mathcal {S}}))$, where $a_i > 0$ (or $a_i < 0$) for every $i\in \{1,\ldots ,n \}$, belongs to $a \mathcal{L}\mathcal{F}$. It is clear that $F \in {\mathcal {L}}^0(X \times Y)$ as F is a linear combination of $\widehat{{\mathcal {A}}\otimes {\mathcal {B}}}$-measurable functions. Now, since $f \in a \mathcal{L}\mathcal{F}$, there exist a $\mu$-null set $N_f$ and a $\nu$-null set $M_f$ such that $f^x\in {\mathcal {L}}^1(Y)$ for every $x\in X{\setminus } N_f$ and $f_y \in {\mathcal {L}}^1(X)$ for every $y \in Y {\setminus } M_f$. Hence, for every $x\in X{\setminus } N_f$, we have

$$\begin{aligned} \int _Y |F^x| d\nu = \int _Y \left| \sum _{i=1}^n a_i f^x \right| d\nu = \left| \sum _{i=1}^n a_i\right| \int _Y |f^x| d\nu <\infty , \end{aligned}$$

where we have used the fact that $\bigcup _{i=1}^n M_{\zeta _i}$ is $\nu$-null and $f\in a \mathcal{L}\mathcal{F}$. Analogously, for every $y \in Y{\setminus } M_f$,

$$\begin{aligned} \int _X |F_y| d\mu = \int _X \left| \sum _{i=1}^n a_i f_y \right| d\mu = \left| \sum _{i=1}^n a_i\right| \int _X |f_y| d\mu <\infty , \end{aligned}$$

since $\bigcup _{i=1}^n N_{\xi _i}$ is $\mu$-null and $f\in a\mathcal{L}\mathcal{F}$. Therefore $F^x\in {\mathcal {L}}^1(Y)$ for any $x \in X {\setminus } N_f$ and $F_y\in {\mathcal {L}}^1(X)$ for any $y \in Y{\setminus } M_f$. Moreover, as $f \in a \mathcal{L}\mathcal{F}$,

$$\begin{aligned} \int _X \left| \int _Y {\widehat{F}}^x d\nu \right| d\mu = \int _X \left| \int _Y \sum _{i=1}^n a_i {\widehat{f}}^x d\nu \right| d\mu = \left| \sum _{i=1}^n a_i \right| \int _X \left| \int _Y {\widehat{f}}^x d\nu \right| d\mu < \infty , \end{aligned}$$

and

$$\begin{aligned} \int _Y \left| \int _X {\widehat{F}}_y d\mu \right| d\nu = \int _Y \left| \int _X \sum _{i=1}^n a_i {\widehat{f}}_y d\nu \right| d\mu = \left| \sum _{i=1}^n a_i \right| \int _Y \left| \int _X {\widehat{f}}_y d\nu \right| d\mu < \infty . \end{aligned}$$

But

$$\begin{aligned} \int _X \left( \int _Y {\widehat{F}}^x d\nu \right) d\mu&= \sum _{i=1}^n a_i \int _X \left( \int _Y {\widehat{f}}^x d\nu \right) d\mu \\&\ne \sum _{i=1}^n a_i \int _Y \left( \int _X {\widehat{f}}_y d\nu \right) d\mu = \int _Y \left( \int _X {\widehat{F}}_y d\mu \right) d\nu , \end{aligned}$$

since  $\sum _{i=1}^n a_i\ne 0$  and  $f \in a \mathcal{L}\mathcal{F}$.

Assume for the rest of the proof that  $f \in p \mathcal{L}\mathcal{F}$  in order to prove part (2). As in part (1) we have that $F \in {\mathcal {L}}_{ab}$, but now we have that  F  satisfies part (c) of Fubini’s Theorem since  $f \in p \mathcal{L}\mathcal{F}$. Finally, as  $f \in p \mathcal{L}\mathcal{F}$, $\sum _{i=1}^n a_i\ne 0$ and $\bigcup _{i=1}^n N_{\xi _i} \times M_{\zeta _i}$ is ${{\widehat{\tau }}}$-null, we have

$$\begin{aligned} \int _{X\times Y} |F| \,d{{\widehat{\tau }}} = \int _{X\times Y} \left| \sum _{i=1}^n a_i f \right| \,d{{\widehat{\tau }}} = \left| \sum _{i=1}^n a_i \right| \int _{X\times Y} |f| \,d {{\widehat{\tau }}} = \infty . \end{aligned}$$

Thus, F is not ${{\widehat{\tau }}}$-integrable, which shows that  $F \in p \mathcal{L}\mathcal{F}$. $\square$

Corollary 2.7

On the Lebesgue measure space over  ${\mathbb {R}}^2$ we have that  $a \mathcal{L}\mathcal{F}$ and $p\mathcal{L}\mathcal{F}$ are $2^{{\mathfrak {c}}}$-convex-coneable.

Proof

By [6] we have that $a \mathcal{L}\mathcal{F} \ne \varnothing$ (analogously for $p\mathcal{L}\mathcal{F}$). Recall now that the standard ternary Cantor set ${\mathfrak {C}}$ is a null set and a Polish space (i.e., a separable completely metrizable topological space). It is known that in every uncountable Polish space there is a family of ${\mathfrak {c}}$-many disjoint Bernstein sets (see [12, pp. 105, exercise 1]).¹ Hence, let ${\mathcal {B}}\subset {\mathfrak {C}}$ be such family and take $\{B_\xi :\xi <2^{{\mathfrak {c}}} \}$ a family of independent subsets of ${\mathcal {B}}$ of cardinality $2^{{\mathfrak {c}}}$. For every $\xi <2^{{\mathfrak {c}}}$, take $N_\xi :=\bigcup B_\xi$. Since the Lebesgue measure is complete, we have that each $N_\xi$ has measure 0. Therefore by taking $M=\{0\}$ and applying Theorem 2.6 part (i), we have the desired result. $\square$

It is important to mention that the same proof as in the first part of (1) in [6, Theorem 4] provides us with the following theorem.

Theorem 2.8

Let $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$ be complete measure spaces, and $(X\times Y,\widehat{{\mathcal {A}}\otimes {\mathcal {B}}},{\widehat{\tau }})$ a complete product measure space. If $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$ satisfy property (Q), then $p{\mathcal {F}}$ is ${\mathfrak {c}}$-lineable.

To finish this paper, we will study the algebraic and topological genericity of the families  $L_a{\setminus } L_{ab}$ and ${\mathcal {L}}_a {\setminus } {\mathcal {L}}_{ab}$; giving a partial answer to a problem posted in part 4 of [6, Section 4].

Theorem 2.9

Let $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$ be complete measure spaces, $(X\times Y,\widehat{{\mathcal {A}}\otimes {\mathcal {B}}},{\widehat{\tau }})$ a complete product measure space, and $\kappa$ and $\lambda$ infinite cardinal numbers.

• (i) Assume that there is a family ${\mathcal {N}}$ of independent subsets of X (resp. Y) of cardinality $2^\kappa$ such that each $N\in {\mathcal {N}}$ is $\mu$-null (resp. $\nu$-null), and there exists an $M\in {\mathcal {B}}{\setminus } \{\varnothing \}$ (resp. $M\in {\mathcal {A}}{\setminus } \{\varnothing \}$) with $\nu (M)=0$ (resp. $\mu (M)=0$). If ${\mathcal {L}}_a {\setminus } {\mathcal {L}}_{ab} \ne \varnothing$, then ${\mathcal {L}}_a {\setminus } {\mathcal {L}}_{ab}$ is $2^\kappa$-convex-coneable.

• (ii) Assume that there are, respectively, families ${\mathcal {N}}$ and ${\mathcal {M}}$ of independent subsets of X and Y of cardinality $2^\kappa$ and $2^\lambda$ such that each $N\in {\mathcal {N}}$ is $\mu$-null and every $M\in {\mathcal {M}}$ is $\nu$-null. If ${\mathcal {L}}_a {\setminus } {\mathcal {L}}_{ab} \ne \varnothing$, then ${\mathcal {L}}_a{\setminus } {\mathcal {L}}_{ab}$ is $\max \{2^\kappa ,2^\lambda \}$-convex-coneable.

• (iii) If $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$ satisfy property $({\overline{Q}})$, then $L_a{\setminus } L_{ab}$ is ${\mathfrak {c}}$-lineable. In addition, if $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$ are $\sigma$-finite and satisfy condition (S), then $L_a{\setminus } L_{ab}$ is ${\mathfrak {c}}$-dense-lineable in $L^0(X \times Y)$, where $L^0(X\times Y)$ is endowed with the topology of local convergence in measure.

Proof

We will prove parts (ii) and (iii) since part (i) can be done using similar arguments from the proof of part (ii) (compare the proof of Theorem 2.6).

For part (ii), let ${\mathcal {N}}$ and ${\mathcal {M}}$ be as in the assumptions and take  $f \in {\mathcal {L}}_a {\setminus } {\mathcal {L}}_{ab}$. Let $\{N_\xi :\xi <2^\kappa \}$ and $\{M_\zeta :\zeta <2^\lambda \}$ be enumerations, without repetition, of ${\mathcal {N}}$ and ${\mathcal {M}}$, respectively. For every $\langle \xi ,\zeta \rangle \in 2^\kappa \times 2^\lambda$, define $f_{\langle \xi ,\zeta \rangle } :X\times Y \rightarrow {\mathbb {R}}$ as

$$\begin{aligned} f_{\langle \xi ,\zeta \rangle }:= f \restriction [(X\times Y) {\setminus } (N_\xi \times M_\zeta )] + \chi _{N_\xi \times M_\zeta }. \end{aligned}$$

As in the proof of Theorem 2.6, the functions $\{f_{\langle \xi ,\zeta \rangle } :\langle \xi ,\zeta \rangle \in 2^\kappa \times 2^\lambda \}$ are linearly independent and $\text {span} \{f_{\langle \xi ,\zeta \rangle } :\langle \xi ,\zeta \rangle \in 2^\kappa \times 2^\lambda \} \subseteq {\mathcal {L}}_a$. Let us show that $F \notin {\mathcal {L}}_{ab}$, where  $F:= \sum _{i=1}^n a_i f_{\langle \xi _i,\zeta _i \rangle }$ with $n\in {\mathbb {N}}$, $a_1,\ldots ,a_n>0$ (resp. $a_1,\ldots ,a_n<0$), $\xi _i\le \cdots \le \xi _n < 2^\kappa$, $\zeta _1\le \cdots \le \zeta _n < 2^\lambda$ and, if $\xi _i=\xi _j$ for $i\ne j$, then $\zeta _i\ne \zeta _j$ (analogously, if $\zeta _i=\zeta _j$ for $i\ne j$, then $\xi _i \ne \xi _j$). Assume, without loss of generality, that  $X \ni x \mapsto \int _Y \widehat{f^x} d\nu$  is not in  ${\mathcal {L}}^1(X)$. Let $N_f\in {\mathcal {A}}$ be $\mu$-null such that $f^x \in {\mathcal {L}}^1(Y)$  for every  $x \in X {\setminus } N_f$. Then, for every $x\in X{\setminus } N_f$, since the $N_{\xi _i}$’s and $M_{\zeta _i}$’s are, respectively, $\mu$-null and $\nu$-null, and $\sum _{i=1}^n a_i\ne 0$, we have that

$$\begin{aligned} \int _X \left| \int _Y \widehat{F^x} d\nu \right| d\mu = \int _X \left| \int _Y \sum _{i=1}^n a_i \widehat{f^x} d\nu \right| d\mu = \left| \sum _{i=1}^n a_i \right| \cdot \int _X \left| \int _Y \widehat{f^x} d\nu \right| d\mu = \infty . \end{aligned}$$

Concerning part (iii), assume first that $(X,{\mathcal {A}},\mu )$ and $(Y,{\mathcal {B}},\nu )$ satisfy property $({\overline{Q}})$. Then there are sequences $(A_n)_{n=1}^\infty \subset {\mathcal {A}}$, $(B_n)_{n=1}^\infty \subset {\mathcal {B}}$ with $0<\mu (A_n), \nu (B_n)<\infty$ for each $n\in {\mathbb {N}}$, $A_n\cap A_k = \varnothing = B_n \cap B_k$ provided that $n,k\in {\mathbb {N}}$ are distinct, and $\sum _{n=1}^\infty \frac{1}{\mu (A_n)}, \sum _{n=1}^\infty \frac{1}{\nu (B_n)}<\infty$. By Theorem 1.4, there exists a family ${\mathcal {N}}$ of independent subsets of ${\mathbb {N}}$ such that $\text {card}({\mathcal {N}})={\mathfrak {c}}$. For every $N\in {\mathcal {N}}$, let us define $f_N :X\times Y \rightarrow {\mathbb {R}}$ as

$$\begin{aligned} f_N:= \sum _{n\in N} \frac{\chi _{A_n \times B_n}}{\mu (A_n)\nu (B_n)}. \end{aligned}$$

It is obvious that $f_N$ is $\widehat{{\mathcal {A}}\otimes {\mathcal {B}}}$-measurable.

Let us prove that the functions $\{f_N :N\in {\mathcal {N}} \}$ are linearly independent. Take $F:=\!\sum _{i=1}^m a_i f_{N_i}$, where $m\in {\mathbb {N}}$, $a_1,\ldots ,a_m\in {\mathbb {R}}{\setminus } \{0\}$ and $N_1,\ldots ,N_m\in {\mathcal {N}}$ are pairwise distinct. Assume that $F\equiv 0$. By construction, there is an $n_1\in N_1{\setminus } \bigcup _{i=2}^m N_i$. Hence, for every $\langle x,y\rangle \in A_{n_1}\times B_{n_1}$, we have that $0=F(x,y)=a_1$, a contradiction.

Let us prove that $F \in {\mathcal {L}}_a$. Since ${\mathcal {L}}_a$ is a vector space, it is enough to show that each $f_N \in {\mathcal {L}}_a$ for every $N\in {\mathcal {N}}$. Fix $N\in {\mathcal {N}}$ and $x \in X$:

• If $x\in A_n$ for some $n\in N$ (in which case n is unique), then

  $$\begin{aligned} \int _Y \left| (f_N)^x \right| d \nu = \int _Y \left| \sum _{n\in N}\frac{\chi _{B_n}}{\mu (A_n)\nu (B_n)} \right| d\nu = \sum _{n\in N} \frac{1}{\mu (A_n)} \le \sum _{n=1}^\infty \frac{1}{\mu (A_n)} < \infty . \end{aligned}$$

• If $x\notin \bigcup _{n\in N} A_n$, then $(f_N)^x \equiv 0$.

Thus, $(f_N)^x\in {\mathcal {L}}^1 (Y)$ for any $x\in X$ and $N \in {\mathcal {N}}$. Similarly, $(f_N)_y \in {\mathcal {L}}^1 (X)$ for any $y\in Y$ and $N \in {\mathcal {N}}$.

Now let us show that $F \notin {\mathcal {L}}_{ab}$. Consider the infinite set $M_1:=N_1 {\setminus } \bigcup _{i=2}^m N_i$ and define $X_1:=\bigcup _{n\in M_1} A_n$. Hence,

$$\begin{aligned} \int _X \left| \int _Y F^x d \nu \right| d\mu&= \int _{X{\setminus } X_1} \left| \int _Y F^x d\nu \right| d\mu + \int _{X_1} \left| \int _Y F^x d\nu \right| d\mu \ge \int _{X_1} \left| \int _Y F^x d\nu \right| d\mu \\&= \int _{X_1} \left| \int _Y a_1 (f_{N_1})^x \right| d\mu = |a_1| \sum _{n\in M_1} 1 = \infty . \end{aligned}$$

Finally, assume for the rest of the proof that  $(X,{\mathcal {A}},\mu )$  and  $(Y,{\mathcal {B}},\nu )$  are $\sigma$-finite and satisfy condition (S). We have that $B:=L^1(X\times Y)$ is dense by [6, Proposition 2], and $L^0(X\times Y)$ is separable by the proof of [6, Theorem 4(2)]. Moreover, observe that $A+B\subseteq A$, where $A:=L_a{{\setminus }} L_{ab}$. Therefore $L_a{\setminus } L_{ab}$ is ${\mathfrak {c}}$-dense-lineable in $L^0(X\times Y)$ (see, for instance, [3, § 7.3]) and (iii) is proved.

$\square$

Corollary 2.10

On the Lebesgue measure space over ${\mathbb {R}}^2$ we have that ${\mathcal {L}}_a {\setminus } {\mathcal {L}}_{ab}$ is $2^{{\mathfrak {c}}}$-convex-coneable, and $L_a {\setminus } L_{ab}$ is ${\mathfrak {c}}$-dense-lineable in $L^0({\mathbb {R}}^2)$.

Proof

Recall that the Lebesgue measure on ${\mathbb {R}}$ satisfies property $({\overline{Q}})$ by the second paragraph of Sect. 2, but also (S) (see [6], the paragraph before Theorem 4). The result follows from Theorem 2.9. $\square$

Corollary 2.11

Let $({\mathbb {N}},{\mathcal {P}}({\mathbb {N}}),\mu )$ denote the counting measure space on ${\mathbb {N}}$ and $\tau$ a product measure on $({\mathbb {N}}^2,\widehat{{\mathcal {P}}({\mathbb {N}})\otimes {\mathcal {P}}({\mathbb {N}})})$. Assume that $({\mathbb {N}}^2,\widehat{{\mathcal {P}}({\mathbb {N}})\otimes {\mathcal {P}}({\mathbb {N}})},{\widehat{\tau }})$ is the completion of the product measure space. Then ${\mathcal {L}}_a {\setminus } {\mathcal {L}}_{ab}$ is $2^{{\mathfrak {c}}}$-convex-coneable, and  $L_a {\setminus } L_{ab}$  is ${\mathfrak {c}}$-dense-lineable in $L^0({\mathbb {N}}^2)$.

Proof

Once again, the counting measure on ${\mathbb {N}}$ satisfies property $({\overline{Q}})$ by the second paragraph of Sect. 2 as well as (S) (see [6], the paragraph before Theorem 4). Hence, the corollary is proved by Theorem 2.9. $\square$