\documentstyle{SibMatJ}
%\TestXML
\topmatter

  \Author           Tasoev
    \Initial        B.
    \Initial        B.
    \Gender         he
    \Email          tasoevbatradz\@yandex.ru
    \AffilRef       1
    \AffilRef       2
  \endAuthor

  \Affil 1
    \Organization   Vladikavkaz Scientific Center of the Russian Academy of Sciences
    \City           Village of Mikhailovskoye %\?Mikhailovskoye Village
    \Country        Russia
  \endAffil

  \Affil 2
    \Organization   Southern Mathematical Institute %\?of the Vladikavkaz Scientific Center of the Russian Academy of Sciences
    \City           Vladikavkaz
    \Country        Russia
  \endAffil

  \datesubmitted    November 4, 2025\enddatesubmitted
  \dateaccepted     January 16, 2026\enddateaccepted

  \UDclass
    517.982  %46B42, 46B04
  \endUDclass

  \thanks
    The research was supported by the Russian Science Foundation (project 24--71--10094, \Url*{https://rscf.ru/project/24-71-10094/}).
   \endthanks

  \title
    Order Structure of the Space of Partial Integral Operators
  \endtitle

  \abstract
    It is known that the space of kernel operators forms a band in the Dedekind complete vector lattice of order bounded operators,
    acting in ideal spaces of measurable functions. In this paper we investigate the order structure of the space of partial integral operators.
    We prove that the space of all absolute partial integral operators forms a band in the Dedekind complete vector lattice of order bounded operators acting in order dense ideal spaces of measurable functions.
  \endabstract

  \keywords
    kernel operator,
    partial integral operator,
    order bounded operator,
    vector lattice
  \endkeywords

\endtopmatter

\head
  1. Introduction
\endhead

 The theory of partial integral operators has numerous applications in many areas of mathematics (see [1--3]).
 Various properties of these operators are studied in  [4--7].
 It is well known that kernel operators forms %\?form
 a band in Dedekind complete vector lattice %\?a
 of order bounded operators, acting in ideal spaces of measurable functions
 (see, for example, [8,\,9]).
 The aim of this work is to investigate the order structure of the space of partial integral operators, acting in order dense ideal spaces of measurable functions.
 In \Par*{Theorem 3.7} we prove that the space of all absolute partial integral operators forms a band in the Dedekind complete vector lattice of order bounded operators acting
 in order dense ideal spaces of measurable functions.

 We use the standard notation and terminology of Aliprantis and Burkinshaw [10] for the theory of vector lattices
 (see also Meyer-Nieberg [8] and Zaanen [9]).
 Throughout the text we assume that all vector spaces are defined over the field of reals and all vector lattices are Archimedean.
 We let $:=$ denote the assignment by definition, while $\Bbb{N}$ and $\Bbb{R}$ symbolize the naturals and the reals.

 \head
   2. Preliminaries
 \endhead

 In this section we give some necessary preliminaries for further discussion.
 Let $(\Omega,\Sigma,\mu)$ and $(S,\Cal{F},m)$ be  measure spaces with  $\sigma$-finite complete measures $\mu$ and $m$ respectively,
 and let $(\Omega\times S,\Sigma\otimes\Cal{F},\mu\otimes m)$ be their product.
 We will denote by $\Cal{L}^0(\mu):=\Cal{L}^0(\Omega,\Sigma,\mu)$ the set of all real-valued 
 $\mu$-a.e.  finite $\Sigma$-measurable functions on $\Omega$,
 and $L^0(\mu):=L^0(\Omega,\Sigma,\mu)$ denotes the collection of all equivalence classes of functions from $\Cal{L}^0(\mu)$.
 As usual, functions differing on a set of measure zero are called equivalent. %\?\it
 Since $\mu$ is $\sigma$-finite measure, %\?
 in view of [8, Lemma 2.6] $L^0(\mu)$ is a super Dedekind complete vector lattice under the $\mu$-a.e.
 pointwise algebraic and lattice operations. In particular, for an arbitrary $f\in L^0(\mu)$ the symbol %\?и дальше
 $f\geqslant0$ means that
 $f(\omega)\geqslant 0$ for $\mu$-almost every $\omega\in\Omega$. We identify an arbitrary equivalence class  $\tilde{f}\in L^0(\mu)$ containing
 a function $f\in\Cal{L}^0(\mu)$ with  $f$.
 Recall that a linear subspace $E$ is called  an 
 {\it ideal space in $L^0(\mu)$\/} %\?{\it ideal space\/} in $L^0(\mu)$
 if for any $f\in L^0(\mu)$ and $g\in E$ the inequality $|f|\leqslant|g|$
 implies $f\in E$. A linear subspace $E$ is called and %\?
 {\it order dense\/} if for any $0<f\in L^0(\mu)$ there exists an element $g\in E$ such that $0<g\leqslant f$.

 Further $E$ and $F$ will denote arbitrary order dense ideal spaces in $L^0(\mu\otimes m):=L^0(\Omega\times S,\Sigma\otimes\Cal{F},\mu\otimes m)$,
 $E_+:=\{f\in E:\ f\geqslant 0\}$.
 A linear operator $T:E\rightarrow F$ is called {\it positive\/} (in symbol  %\?
 $T\geqslant0$) if  $Tf\in F_+$ for all  $f\in E_+$,
 and {\it regular\/} if it can be written as a difference of two positive operators.
 The set $L^r(E,F)$ of all regular operators from $E$ to $F$ is a Dedekind complete vector lattice.
 The symbol $L(E,F)_+$ denotes the set of all positive operators from $E$ to $F$.

 Let $(f_n)_{n=1}^\infty$ be a sequence in $E$ and $f\in E$. The symbol $0\leqslant f_n\uparrow f$  means that
 $0\leqslant f_n(\omega,t)\leqslant f_{n+1}(\omega,t)$ and $f(\omega,t)=\lim_n f(\omega,t)=\sup_n f_n(\omega,t)$ for $\mu\otimes m$-almost every
 $(\omega,t)\in\Omega\times S$ and for all $n\in\Bbb{N}$. Similarly, the symbol $f_n\downarrow f$  means that
 $f_n(\omega,t)\geqslant f_{n+1}(\omega,t)$ and $f(\omega,t)=\lim_n f(\omega,t)=\inf_n f_n(\omega,t)$ for $\mu\otimes m$-almost every
 $(\omega,t)\in\Omega\times S$ and for all $n\in\Bbb{N}$.

 \demo{Definition 2.1}
 A linear operator $T:E\rightarrow F$ is said to be {\it partial integral}, if there exists a~measurable function
 $k\in\Cal{L}^0(\Omega\times S\times S,\Sigma\otimes\Cal{F}\otimes\Cal{F},\mu\otimes m\otimes m)$
 such that
 $$
 Tf(\omega,t)=\int\limits_{S}k(\omega,t,s)f(\omega,s)dm(s)
 $$
 holds for all $f\in E$ and $\mu\otimes m$-almost every $(\omega,t)\in\Omega\times S$. The function $k$ is called
 the {\it kernel\/} of $T$.
 \enddemo

 \demo{Definition 2.2}
  A linear operator $T:E\rightarrow F$ is said to be {\it $L^\infty(\Omega,\Sigma,\mu)$-homogeneous}, if $T(hf)=hTf$ for all
 $f\in E$ and $h\in L^\infty(\Omega,\Sigma,\mu)$.
 Clearly, every partial integral operator is $L^\infty(\mu)$-homogeneous.
 \enddemo

 When is a positive $L^\infty(\Omega,\Sigma,\mu)$-homogeneous operator from $E$ to $F$ partial integral?
 The answer is given in the following result.

 \proclaim{Theorem 2.3}\Label{T2.3}
 Let $(\Omega,\Sigma,\mu)$ and $(S,\Cal{F},m)$ be measure spaces with $\sigma$-finite
 measures $\mu$ and $m$ respectively.
 Let $E$ and $F$ be order dense ideal spaces in $L^0(\Omega\times S,\Sigma\otimes\Cal{F},\mu\otimes m)$
 and let $T:E\rightarrow F$ be a~positive $L^\infty(\Omega,\Sigma,\mu)$-homogeneous operator. Then the following are equivalent:

 \Item (i) %(1)
 There exists a measurable function $k\in \Cal{L}^0(\Omega\times S\times S,\Sigma\otimes\Cal{F}\otimes\Cal{F})$
 such that $k\geqslant0$ $\mu\otimes m\otimes m$-a.e. and
 $$
 Tf(\omega,t)=\int\limits_S k(\omega,t,s)f(\omega,s)\ dm(s)
 \tag1
 $$
 holds for all $f\in E$ and $\mu\otimes m$-almost every $(\omega,t)\in\Omega\times S$.

 \Item (ii) %(2)
 For any sequence $(f_n)_{n=1}^\infty\subset E$ such that
 $0\leqslant f_n\leqslant f\in E$ for all $n\in\Bbb{N}$ and $f_n\rightarrow 0$ in measure $\mu\otimes m$ on every subset of finite measure,
 and for any set
 $C\in \Sigma\otimes\Cal{F}$ such that $\chi_CTf\in L^1(\Omega\times S,\Sigma\otimes\Cal{F},\mu\otimes m)$
 $$
 \int\limits_\Omega \chi_C(\omega,t) Tf_n(\omega,t)\ d\mu(\omega)\rightarrow 0
 $$
 holds for $m$-almost every $t\in S$.
 \endproclaim

 \demo{Proof}
 The proof given in [11, Theorem 1].
 \qed\enddemo

 \head
   3. Main Results
 \endhead

 In this section we investigate the order structure of the space of partial integral operators.
 As in \Sec*{Section 2},  $E$ and $F$ will denote arbitrary order dense ideal spaces in $L^0(\mu\otimes m):=L^0(\Omega\times S,\Sigma\otimes\Cal{F},\mu\otimes m)$.

 \proclaim{Lemma 3.1}\Label{L3.1}
 Let $T$ be a partial integral operator from $E$ into $F$ with kernel
 $k\in \Cal{L}^0(\Omega\times S\times S,\Sigma\otimes\Cal{F}\otimes\Cal{F})$.
 Then the following holds: %\?hold

 \Item (i)
 $T$ is positive if and only if $k\geqslant 0$  holds $\mu\otimes m\otimes m$-a.e. on $\Omega\times S\times S$.

 \Item (ii)
  $T=0$ if and only if $k=0$ holds $\mu\otimes m\otimes m$-a.e. on $\Omega\times S\times S$.
 \endproclaim

 \demo{Proof}
 \Par{L3.1}{(i)}:
 It is evident that if $k(\omega,t,s)\geqslant 0$  holds  
 $\mu\otimes m\otimes m$-almost everywhere  %\?a.e. еще есть, и есть полностью  дальше
 on $\Omega\times S\times S$ then $T$ is positive.
 For the converse, assume that $T$ is positive. By [9, Theorem 86.2] there exists a sequence
 $(C_n)_{n=1}^\infty\subset\Sigma\otimes\Cal{F}$ such that $C_1\subset C_2\subset\cdots$, 
 $\bigcup_{n=1}^\infty C_n=\Omega\times S$,
 $\mu\otimes m(C_n)<\infty$ and $\chi_{C_n}\in E$
 for all $n\in\Bbb{N}$.
 For an arbitrary $n\in\Bbb{N}$ denote by $T_n$ the operator from $E$ to $F$  defined by
 $$
 T_nf(\omega,t)=\int\limits_{S} \chi_{C_n}(\omega,t)k(\omega,t,s) f(\omega,s)\ dm(s)
 $$
 for all $f\in E$ and for $\mu\otimes m$-almost every  $(\omega,t)\in\Omega\times S$. 
 Then $T_n$ is positive partial integral operator %\?артикль
 from $E$ to $F$ with kernel $\chi_{C_n}k$ and $1_{\Omega\times S}\in E$,
 where $1_{\Omega\times S}$ is the equivalence class of the function identically equal to one everywhere on $\Omega\times S$.
 Since $\lim_n\chi_{C_n}(\omega,t)k(\omega,t,s)= k(\omega,t,s)$ for all $(\omega,t,s)\in\Omega\times S\times S$ it is enough to
 show that $\chi_{C_n}k\geqslant 0$ holds $\mu\otimes m\otimes m$-almost everywhere  
 for all $n\in\Bbb{N}$.

 Thus, without loss of generality we can assume that $\mu\otimes m(\Omega\times S)<\infty$ and
 $L^\infty(\Omega\times S,\Sigma\otimes\Cal{F},\mu\otimes m)\subset E$.
 From $1_{\Omega\times S}\in E$ it follows that the function $h:=\int_S |k(\omega,t,s)|\ dm(s)$ is finite 
 $\mu\otimes m$-almost everywhere  on $\Omega\times S$.
 So, $0\leqslant h\in L^0(\Omega\times S,\Sigma\otimes\Cal{F},\mu\otimes m)$. Denote by $U_n$ the following set:
 $$
 U_n:=\{(\omega,t)\in \Omega\times S: h(\omega,t)\leqslant n\}
 $$
 for all $n\in\Bbb{N}$. Then $\chi_{U_n}h\in L^1(\Omega\times S,\Sigma\otimes\Cal{F},\mu\otimes m)$ for all $n\in\Bbb{N}$ and $\chi_{U_n}\uparrow 1_{\Omega\times S}$ holds.
 For an arbitrary $n_0\in\Bbb{N}$ define the operator
 $T_{n_0}:L^\infty(\Omega\times S,\Sigma\otimes\Cal{F},\mu\otimes m)\rightarrow L^0(S,\Cal{F},m)$ by the formula
 $$
 (T_{n_0}f)(t):=\int\limits_{\Omega\times S} \chi_{U_{n_0}}(\omega,t)k(\omega,t,s)f(\omega,s)\ d\mu\otimes m(\omega,s)
 $$
 for all $f\in L^\infty(\Omega\times S,\Sigma\otimes\Cal{F},\mu\otimes m)\subset E$ and for $m$-almost every $t\in S$.
 Then $T_{n_0}$ is a positive kernel operator with kernel $\chi_{U_{n_0}}k$ and by [9, Theorem 91.1]   $\chi_{U_{n_0}}k\geqslant 0$
 $\mu\otimes m\otimes m$-almost everywhere on $\Omega\times S\times S$.
 Thus, $\chi_{U_{n }}k\geqslant 0$ $\mu\otimes m\otimes m$-almost everywhere on $\Omega\times S\times S$ for all $n\in\Bbb{N}$. Since $\chi_{U_n}\uparrow 1_{\Omega\times S}$ holds, we have $\chi_{U_n}k\rightarrow k$
 $\mu\otimes m\otimes m$-almost everywhere on $\Omega\times S\times S$.
 So, $k\geqslant 0$ $\mu\otimes m\otimes m$-almost everywhere on $\Omega\times S\times S$, as desired.

 \Par{L3.1}{(ii)}:
 Follows by applying the result in part \Par{L3.1}{(i)} to $T$ and $-T$.
 \qed\enddemo

 \proclaim{Lemma 3.2}
 Let $T$ and $S$ be leaner %\?linear
 operators from $E$ into $F$ such that $0\leqslant T\leqslant S$.
 If $S$ is an $L^\infty(\Omega,\Sigma,\mu)$-homogeneous operator then $T$ is also an $L^\infty(\Omega,\Sigma,\mu)$-homogeneous operator.
 \endproclaim

 \demo{Proof}
 Let $0\leqslant T\leqslant S$ with $L^\infty(\Omega,\Sigma,\mu)$-homogeneous $S$.
 First we show that $T\chi_A=\chi_A T$ holds for all $A\in\Sigma$, where $\chi_A$ is the characteristic function of $A\in\Sigma$.
 Fix an arbitrary $A\in\Sigma$. It follows from the inequalities $0\leqslant T\leqslant S$ and $L^\infty(\Omega,\Sigma,\mu)$-homogeneity of $S$
 that $0\leqslant\chi_{A^c} T\chi_A\leqslant\chi_{A^c} S \chi_A=S\chi_{A^c}\chi_A=0$,
 where $\chi_{A^c}=1-\chi_A=\chi_{\Omega\setminus A}$.
 Hence, $\chi_{A^c} T\chi_A=0$, and so we have $T\chi_A=\chi_A T\chi_A+\chi_{A^c} T\chi_A=\chi_A T\chi_A$.
 Thus,
 $$
 T\chi_A=\chi_A T\chi_A
 \tag2
 $$
 holds for all $A\in\Sigma$.
 It follows from \Tag(2) that $T\chi_{A^c} =\chi_{A^c}T\chi_{A^c}$ holds for all $A\in\Sigma$.
 That is,
 $T(1-\chi_A)=(1-\chi_A)T(1-\chi_A)=T(1-\chi_A)-\chi_A T+\chi_A T\chi_A$. Thus, $\chi_A T=\chi_A T\chi_A$
 and in view of  \Tag(2), we deduce that
 $$
 T\chi_A=\chi_A T
 \tag3
 $$
 holds for all $A\in\Sigma$. It follows from \Tag(3) that  $Tf_n=f_n T$ holds for every step function
 $f_n$ of the form $f_n=\sum_{i=1}^{l(n)}\alpha_i\chi_{A_i}$, where $A_i\in\Sigma$, $\alpha_i\in\Bbb{R}$, and $n,l(n)\in\Bbb{N}$.
 By [10, Theorem 2.8] or [12, Theorem~I.6.3] for every function $f\in L^\infty(\Omega,\Sigma,\mu)$ there exists a sequence
 of step functions $(f_n)_{n=1}^\infty$ such that $f_n$ converges uniformly to $f$. Since every positive operator is uniformly continuous, we have
 $Tf=\lim_{n} Tf_n=\lim_{n}f_nT=fT$ for all $f\in L^\infty(\Omega,\Sigma,\mu)$.
 \qed\enddemo

 \proclaim{Lemma 3.3}
 Let $0\leqslant T \leqslant U$ be positive operators from $E$ into $F$.
 If $U$ is a partial integral operator then $T$ is also a partial integral operator.
 \endproclaim

 \demo{Proof}
 Let $0\leqslant T \leqslant U$ be positive operators from $E$ into $F$ with $U$ partial integral. %\?артикль
 Since $U$ is $L^\infty(\Omega,\Sigma,\mu)$-homogeneous, it follows from \Par*{Lemma 3.2} that $T$ is $L^\infty(\Omega,\Sigma,\mu)$-homogeneous.
 Therefore, it suffices to show that assertion \Par{T2.3}{(ii)}  %(2)
 of \Par*{Theorem 2.3} holds. In order to do this, let $(f_n)_{n=1}^\infty$ be a sequence in $E$ such that
 $0\leqslant f_n\leqslant f\in E$ for all $n\in\Bbb{N}$ and $f_n\rightarrow 0$ in measure $\mu\otimes m$ on every subset of finite measure,
 and let  $C\in\Sigma\otimes\Cal{F}$ satisfy $\chi_CTf\in L^1(\Omega\times S,\Sigma\otimes\Cal{F},\mu\otimes m)$.
 We must show that
 $$
 \int\limits_\Omega \chi_C(\omega,t) Tf_n(\omega,t)\ d\mu(\omega)\rightarrow 0
 $$
 holds for $m$-almost every $t\in S$. Since $Uf\geqslant 0$, by [12, Corollary IV.3.1] there exists an increasing sequence
 $(C_i)_{i=1}^\infty\subset\Sigma\otimes\Cal{F}$ such that,
 $\chi_{C_i}Uf\in L^1(\Omega\times S,\Sigma\otimes\Cal{F},\mu\otimes m)$ for all $i\in\Bbb{N}$
 and $0\leqslant\chi_{C_i} Uf \uparrow Uf$. For an arbitrary $n\in \Bbb{N}$ we have
 $0\leqslant\chi_C Tf_n-\chi_{C_i}\chi_C Tf_n=(1-\chi_{C_i})\chi_C Tf_n\leqslant (1-\chi_{C_i})Uf\downarrow 0$.
 Thus,  $0\leqslant\chi_{C_i}\chi_C Tf_n \uparrow_i \chi_C Tf_n$ holds uniformly over all $n\in\Bbb{N}$, and so
 by the dominated convergence theorem for an arbitrary $\varepsilon>0$
 there exists $i(\varepsilon)\in\Bbb{N}$ such that
 $$
 0\leqslant\int\limits_\Omega \chi_C(\omega,t) Tf_n(\omega,t)\ d\mu(\omega)-\int\limits_\Omega \chi_{C_i}(\omega,t)\chi_C(\omega,t)Tf_n(\omega,t)\ d\mu(\omega)\leqslant\varepsilon
 $$
 for all $i\geqslant i(\varepsilon)$ and for all $n\in\Bbb{N}$. %\?for all один раз
 Consequently, applying \Par*{Theorem 2.3} to the partial integral operator $U$ and
 taking into account that $\chi_{C_{i(\varepsilon)}}Uf\in L^1(\Omega\times S,\Sigma\otimes\Cal{F},\mu\otimes m)$, we get
 $$
 \aligned
 0
&\leqslant\int\limits_\Omega \chi_C(\omega,t) Tf_n(\omega,t)\ d\mu(\omega)
 \\
&=
 \int\limits_\Omega \chi_C(\omega,t) Tf_n(\omega,t)\ d\mu(\omega)-\int\limits_\Omega \chi_{C_{i(\varepsilon)}}(\omega,t)\chi_C(\omega,t) Tf_n(\omega,t)\ d\mu(\omega)
 \\
&\qquad +
 \int\limits_\Omega \chi_{C_{i(\varepsilon)}}(\omega,t)\chi_C(\omega,t) Tf_n(\omega,t)\ d\mu(\omega)\leqslant\varepsilon+
 \int\limits_\Omega \chi_{C_{i(\varepsilon)}}(\omega,t) Uf_n(\omega,t)\ d\mu(\omega)\rightarrow\varepsilon
 \endaligned
 $$
 as $n\rightarrow\infty$ for $m$-almost every $t\in S$.
 Thus, $\int_\Omega \chi_C(\omega,t) Tf_n(\omega,t)\ d\mu(\omega)\rightarrow 0$ holds for $m$-almost every $t\in S$.
 \qed\enddemo

  \proclaim{Lemma 3.4}
  Let $(\Omega,\Sigma,\mu)$ and $(S,\Cal{F},m)$ be measure spaces with $\sigma$-finite
 measures $\mu$ and $m$ respectively.
 Let $E$ and $F$ be order dense ideals in $L^0(\Omega\times S,\Sigma\otimes\Cal{F},\mu\otimes m)$
 and let $k$ be a~positive $\Sigma\otimes\Cal{F}\otimes\Cal{F}$-measurable function with values in
 $\Bbb{R}\cup\{+\infty\}$ such that
 $$
 \int\limits_S k(\omega,t,s)f(\omega,s)\ dm(s)<\infty
 $$
 holds for $\mu\otimes m$-almost every $(\omega,t)\in\Omega\times S$ and for all $f\in E$. Then $k<\infty$
 $\mu\otimes m\otimes m$-a.e., that is, $k\in \Cal{L}^0(\Omega\times S\times S,\Sigma\otimes\Cal{F}\otimes\Cal{F},\mu\otimes m\otimes m)$.
 \endproclaim

 \demo{Proof}
 Put $P:=\{(\omega,t,s)\in\Omega\times S\times S: k(\omega,t,s)=\infty\}$.
 Assume first that $1\in E$, where
 $1(\omega,s)=1$ for all  $(\omega,s)\in\Omega\times S$.  Then  $\int_S  k(\omega,t,s)\ dm(s)<\infty$ holds
 for $\mu\otimes m$-almost every $(\omega,t)\in\Omega\times S$.
 Consequently, $k(\omega,t,s)<\infty$ holds for $m$-almost every $s\in S$ and for $\mu\otimes m$-almost every $(\omega,t)\in\Omega\times S$.
 In other words, the section $P_{\omega,t}:=\{s\in S:(\omega,t,s)\in P\}$ has measure zero for
 $\mu\otimes m$-almost every $(\omega,t)\in\Omega\times S$. Thus, by Fubini's theorem we have
 $$
 \mu\otimes m\otimes m(P)=\int\limits_{\Omega\times S}m(P_{\omega,t})\ d\mu\otimes m(\omega,t)=0,
 $$
 and so $k<\infty$ $\mu\otimes m\otimes m$-a.e.

 Now consider the general case.  By [9, Theorem 86.2] there exists a sequence
 $(C_n)_{n=1}^\infty\subset\Sigma\otimes\Cal{F}$ such that $C_1\subset C_2\subset\cdots$, $\bigcup_{n=1}^\infty C_n=\Omega\times S$,
 $\mu\otimes m(C_n)<\infty$ and $\chi_{C_n}\in E$
 for all $n\in\Bbb{N}$. Fix an arbitrary $n\in\Bbb{N}$. Then we have
 $$
 \int\limits_S \chi_{C_n}(\omega,t)k(\omega,t,s)f(\omega,s)\ dm(s)<\infty
 $$
 for $\mu\otimes m$-almost every $(\omega,t)\in\Omega\times S$ and all $f\in L^\infty(\Omega\times S,\Sigma\otimes\Cal{F},\mu\otimes m)$.
 Consequently,  since $L^\infty(\Omega\times S,\Sigma\otimes\Cal{F},\mu\otimes m)$ is an order dense ideal,
 $\chi_{C_n}k<\infty$ $\mu\otimes m\otimes m$-a.e. for all $n\in\Bbb{N}$. In other words,
 $k<\infty$ $\mu\otimes m\otimes m$-a.e. on $C_n\times S$ for all $n\in\Bbb{N}$.
 Thus, $k<\infty $ $\mu\otimes m\otimes m$-a.e.
 \qed\enddemo

 \demo{Definition 3.5}
 A partial integral operator $T:E\rightarrow F$ with kernel
 $k\in\Cal{L}^0(\Omega\times S\times S,\Sigma\otimes\Cal{F}\otimes\Cal{F},\mu\otimes m\otimes m)$  is called
 an {\it absolute partial integral operator\/}
 if the operator $[T]$ defined by
 $$
 [T]f(\omega,t)=\int\limits_{S}|k(\omega,t,s)|f(\omega,s)dm(s)
 $$
 for $\mu\otimes m$-almost every $(\omega,t)\in\Omega\times S$, %\?парную , поставить, наверное, перед defined
 maps $E$ into $F$.
\enddemo

 Denote by $L_{pk}(E,F)$ the space of all absolute partial  integral operators from $E$ into $F$.

 \proclaim{Lemma 3.6}
 If $T, U\in L_{pk}(E,F)$ are absolute partial integral  operators from $E$
 into $F$ with kernels $k_T$ and $k_U$ respectively, then $T\vee U$ is   an absolute partial  integral operator with kernel equal to the
 pointwise supremum of $k_T$ and $k_U$ $\mu\otimes m\otimes m$-a.e. In particular, $L_{pk}(E,F)$ is a vector sublattice of $L^r(E,F)$.
 \endproclaim

 \demo{Proof}
 Let $T,U\in L_{pk}(E,F)$ with kernels $k_T$ and $k_U$ respectively. Then $T+U$ is a partial integral operator with kernel $k_T+k_U$.
 It follows from $|k_T+k_U|\leqslant|k_T|+|k_U|$ that $T+U$ is absolute, and so
 $L_{pk}(E,F)$ is a vector subspace of $L^r(E,F)$. In view of $T\vee U=\frac{1}{2}(T+U-|T-U|)$,
 it is sufficient to show that $|T|$ is a partial  integral operator from $E$ to $F$ and $|T|=[T]$, where
 $$
 [T]f(\omega,t)=\int\limits_{S}|k_T(\omega,t,s)|f(\omega,s)dm(s)
 $$
 for all $f\in E$ and for $\mu\otimes m$-almost every $(\omega,t)\in\Omega\times S$. Since $[T]\geqslant T$ and $[T]\geqslant-T$, we have
 $[T]\geqslant |T|$. Consequently, it follows from \Par*{Lemma 3.3} that $|T|$ is a partial  integral operator from $E$ to $F$.
 It remains to show that  $[T]\leqslant |T|$. Let $k_1$ be the kernel of $|T|$. Since $|T|\geqslant T$ and $|T|\geqslant -T$,
 it follows from \Par*{Lemma~3.1} that $k_1\geqslant k_T$ and $k_1\geqslant -k_T$, and so $k_1\geqslant |k_T|$.
 Therefore, by \Par*{Lemma 3.1} again $|T|\geqslant[T]$, which implies $|T|=[T]$ holds, %\?
 as desired.
 \qed\enddemo

 Recall that a vector lattice $X$ is called {\it super Dedekind complete},
 if whenever an arbitrary subset $D\subset X$ has a supremum, then there exists at most a countable subset $C$ of $D$ with $\sup C =\sup D$.
 The following theorem is the main result of this paper.

 \proclaim{Theorem 3.7}
 Let $(\Omega,\Sigma,\mu)$ and $(S,\Cal{F},m)$ be measure spaces with $\sigma$-finite
 measures $\mu$ and $m$ respectively. Let $E$ and $F$ be order dense ideal spaces in $L^0(\Omega\times S,\Sigma\otimes\Cal{F},\mu\otimes m)$.
 Then the space $L_{pk}(E,F)$ of all absolute partial  integral operators from $E$ into $F$ is a band in $L^r(E,F)$ and is super Dedekind complete.
 Moreover, if $T,U\in L_{pk}(E,F)$ with kernels $k_T$ and $k_U$ respectively,  then $k_T\vee k_U$ is the kernel of $T\vee U$.
 \endproclaim

 \demo{Proof}
 It follows from \Par{Lemma 3.3}{Lemmas 3.3} and \Par{Lemma 3.6}{3.6} that $L_{pk}(E,F)$ is an ideal in $L^r(E,F)$.
 If $T,U\in L_{pk}(E,F)$ with kernels $k_T$ and $k_U$ respectively, then by \Par*{Lemma 3.6} $k_T\vee k_U$ is the kernel of $T\vee U$.

 Consider an increasing net $(T_\alpha)_{\alpha\in A}\subset L_{pk}(E,F)$ such that $0\leqslant T_\alpha\uparrow T$
 for some $T\in L^r(E,F)$. It is sufficient to show that $T\in L_{pk}(E,F)$. Let $k_\alpha$ be the kernel of $T_\alpha$ for all $\alpha\in A$.
 Then $(k_\alpha)_{\alpha\in A}$
 is increasing net %\?an
 in $L^0(\Omega\times S\times S,\Sigma\otimes\Cal{F}\otimes\Cal{F},\mu\otimes m\otimes m)_+$.
 Since $\mu\otimes m\otimes m$ is $\sigma$-finite measure, %\?
 by [8, Lemma 2.6.1]
 $L^0(\Omega\times S\times S,\Sigma\otimes\Cal{F}\otimes\Cal{F},\mu\otimes m\otimes m)$ is super Dedekind complete.

 Let $P(\Omega\times S\times S):=P(\Omega\times S\times S, \Sigma\otimes\Cal{F}\otimes\Cal{F},\mu\otimes m\otimes m)$ be the set of all
 $\Sigma\otimes\Cal{F}\otimes\Cal{F}$-measurable functions $k$ on $\Omega\times S\times S$ assuming non-negative  %\?
 values in $\Bbb{R}\cup\{+\infty\}$.
 Functions differing only on a~$\mu\otimes m\otimes m$-null set are identified. The set $P(\Omega,\times S\times S)$ is partial ordered as usual, i.e.,
 $k_1\leqslant k_2$ means that $k_1(\omega,t,s)\leqslant k_2(\omega,t,s)$ holds $\mu\otimes m\otimes m$-almost everywhere on
 $\Omega\times S\times S$. Then $P(\Omega\times S\times S)$ is a~lattice.

 It follows from [9, Lemma 94.4] that there exist an increasing sequence
 $(\alpha_n)_{n=1}^\infty\subset A$ and a function $0\leqslant k\in P(\Omega\times S\times S)$ such that
 $k=\sup_n k_{\alpha_n}=\sup_{\alpha\in A} k_\alpha$ holds in $P(\Omega\times S\times S)$.
 So, we have $0\leqslant k_{\alpha_n}\uparrow k$  and
 $$
 (T_{\alpha_n}f)(\omega,t)=\int\limits_S k_{\alpha_n}(\omega,t,s)f(\omega,s)\ dm(s)\leqslant Tf(\omega,t)
 $$
 holds  for all $f\in E$, $n\in\Bbb{N}$, and for $\mu\otimes m$-almost every $(\omega,t)\in\Omega\times S$.
 Consequently, by Levi's theorem there exists
 $$
  \lim_{n}\int\limits_S k_n(\omega,t,s)f(\omega,s)\ dm(s)=\int\limits_S k(\omega,t,s)f(\omega,s)\ dm(s)
  $$
 for all $0\leqslant f\in E$ and for $\mu\otimes m$-almost every $(\omega,t)\in\Omega\times S$. If we define
 $T_k:E\rightarrow F$ by
 $$
 T_kf(\omega,t)=\int\limits_S k(\omega,t,s)f(\omega,s)\,dm(s),
 $$
 then by \Par*{Lemma 3.4} $k<\infty$ $\mu\otimes m\otimes m$-a.e., and so $0\leqslant T_k\in L_{pk}(E,F)$.
 Since $T_kf=\sup_{n}T_{\alpha_n}f$ holds for all $f\in E$ and $T_{\alpha_n}\leqslant T$ for all $n\in\Bbb{N}$,  we have $T_k\leqslant T$ in $L^r(E,F)$.
 On the other hand, since $k\geqslant k_\alpha$ $\mu\otimes m\otimes m$-a.e. for all $\alpha\in A$ it follows from \Par*{Lemma 3.1} that
 $T_k\geqslant T_\alpha$  for all $\alpha\in A$. Consequently, $T_k\geqslant T$, which implies $T_k=T$.
 Therefore, $T\in L_{pk}(E,F)$. Super Dedekind completeness of $L_{pk}(E,F)$ follows from equalities $T=\sup_{\alpha\in A}=\sup_{n\in\Bbb{N}} T_{\alpha_n}$.
 \qed\enddemo

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