Proof. By [Reference Aliprantis and Burkinshaw4, Theorem 2.21 (2)] we get that $\Psi$ is order continuous. By [Reference Aliprantis and Burkinshaw4, Theorem 2.15] we have that $\Psi$ and $\Psi^{-1}$ are positive. Hence, it suffices to show that for an order interval we have $\Psi([0,x])=[0,\Psi(x)]$. The inclusion $\Psi([0,x])\subseteq[0,\Psi(x)]$ is clear, since $\Psi$ is positive while the reverse inclusion follows similarly since $\Psi^{-1}$ is positive.

Proof. Let $(f_n)_{n=1}^\infty$ be a dense sequence in $M$. For each $c \in \mathbb{Q}$ and $n \in \mathbb{N}$ define $A_{n,c} = f^{^{-1}}_n((c, \infty))$. Let $\tilde{\Sigma} = \sigma(\{A_{n,c}: n \in \mathbb{N}, c \in \mathbb{Q} \})$, which is clearly countably generated. It is clear that $L_1(X, \tilde{\Sigma}, \mu)$ is a separable sublattice of $L_1(X, \Sigma, \mu)$ which contains $M$.

In the strictly localizable case, let $\Omega = \bigcup_{n \in \mathbb{N}} \bigcup_{c \in \mathbb{Q}} A_{n,c}$ and define $\hat{\Sigma} = \{\Omega \cap B: B \in \tilde{\Sigma}\}$ so that $\Omega$ is $\sigma$-finite and $\hat{\Sigma}$ is a countably generated $\sigma$-algebra. Again, it is clear that $L_1(\Omega, \hat{\Sigma}, \mu)$ is a separable sublattice of $L_1(X, \Sigma, \mu)$ which contains $M$.

The complementation result follows, for example, from [Reference Lacey29, Chapter 6, Theorem 6].

Proof. Let $(\Theta_n)_{n \in \mathbb{N}}$ be a partition of $\Theta$ satisfying $0 \lt \nu(\Theta_n) \lt \infty$ and define $g = \sum_{n = 1}^\infty 2^{-n} \nu(\Theta_n)^{-1} \mathbb{1}_{\Theta_n}$. By construction $\int_\Theta g(y) d\nu(y) = 1$. It is now straightforward to verify that if we define $\nu_1$ by $\nu_1(B) = \nu(B \cap (Y \backslash \Theta))$ for $B \in \Gamma$ and $d\nu_2 = g d\nu$ then the measure $\tilde{\nu} = \nu_1 + \nu_2$ satisfies $\tilde{\nu}(\Theta) = 1$ and the map $\Psi: L_1(Y, \Gamma, \nu) \to L_1(Y, \Gamma, \tilde{\nu})$ given by $f \mapsto \mathbb{1}_{Y \backslash \Theta}f + \mathbb{1}_{\Theta} f/g$ is an isometric lattice isomorphism. Obviously $(Y, \Gamma, \tilde{\nu})$ is strictly localizable, since $(Y, \Gamma, \nu)$ was. That $\Psi$ commutes with the multiplication operator can be easily seen from the construction above.

Proof. Since $(X, \Sigma, \mu)$ is finite, we have that $\mathbb{1}_X \in L_1(X, \Sigma, \mu)$. It follows by using Remark 3.12 and Tonelli’s Theorem that

\begin{equation*} \int_{X \times Y}|k(x,y)| d(\mu(x) \otimes d\nu(y)) = \int_{Y} \int_X |k(x,y)| d\mu(x) d\nu(y) = \lVert|K| \mathbb{1}_X \rVert \lt \infty, \end{equation*}

that is $k \in L_1(X \times Y, \Sigma \otimes \Gamma, \mu \otimes \nu)$.

1. (i) We start by proving $\tilde{K} = K \Phi$ is an integral operator and show that the truncated approximations are transformed in the natural way. Without loss of generality, decomposing $K = K^+ - K^-$ and considering $K^+$ and $K^-$ separately, we may assume that $K$ is positive. Let $k: X \times Y \to \mathbb{R}$ be the kernel of $K$. By Proposition 3.14 we have $k \in L_1(X \times Y, \Sigma \otimes \Gamma, \mu \otimes \nu)$. Therefore, we can choose $(k^{(m)})_{m \in \mathbb{N}}$ a sequence of simple functions $k^{(m)}: X \times Y \to \mathbb{R}$ converging to $k$ in norm,

   \begin{equation*} k^{(m)} = \sum_{l=1}^{N_m} a_{m,l}\cdot\mathbb{1}_{A_{m,l}\times B_{m,l}}. \end{equation*}

   Define the simple functions $\tilde{k}^{(m)}: \tilde{X} \times Y \to \mathbb{R}$ by

   \begin{equation*} \tilde{k}^{(m)} = \sum_{l=1}^{N_m} a_{m,l}\cdot\mathbb{1}_{\phi^{-1}(A_{m,l})\times B_{m,l}}. \end{equation*}

   Since the sequence $(k^{(m)})_{m \in \mathbb{N}}$ is Cauchy, so is $(\tilde{k}^{(m)})_{m \in \mathbb{N}} \subseteq L_1(\tilde{X} \times Y, \tilde{\Sigma} \otimes \Gamma, \tilde{\mu} \otimes \nu)$, thus it is convergent. Denote by $\tilde{k}$ the limit of this sequence and define the integral operator $\tilde{H} \tilde{f}(y) = \int_{\tilde{X}} \tilde{k}(\tilde{x}, y) \tilde{f}(\tilde{x}) d\tilde{\mu}(\tilde{x})$, we will show that $\tilde{H} = \tilde{K}$, so that $\tilde{K}$ is in fact an integral operator.

   Passing to a subsequence, we can assume that $k^{(m)}$ and $\tilde{k}^{(m)}$ converge pointwise a.e. Define $k_y = k(\cdot, y)$, $\tilde{k}_y = \tilde{k}(\cdot, y)$, $k^{(m)}_y = k^{(m)}(\cdot, y)$, $\tilde{k}^{(m)}_y = \tilde{k}^{(m)}(\cdot, y)$. It follows that for $\nu$-almost all $y$ we have that $k^{(m)}_y \to k_{y}$ pointwise $\mu$-a.e and $\tilde{k}^{(m)}_y \to \tilde{k}_{y}$ pointwise $\tilde{\mu}$-a.e.

   It is easy to see that for any $c \in \mathbb{R}$ we have

   \begin{equation*} \{\tilde{k}_{y}^{(m)} \gt c\}=\phi^{-1}(\{k_{y}^{(m)} \gt c\}), \end{equation*}
   where the equality is taken in the equivalence classes modulo null sets. Therefore, for $c \in \mathbb{R}$ we get
   \begin{align*} \{\tilde{k}_{y} \gt c\}&=\bigcup_{n_{0}=1}^{\infty}\bigcap_{m=n_0}^{\infty}\{\tilde{k}^{(m)}_{y} \gt c\} =\bigcup_{n_{0}=1}^{\infty}\bigcap_{m=n_0}^{\infty}\phi^{-1}(\{k^{(m)}_{y} \gt c\})\\ &=\phi^{-1}(\bigcup_{n_{0}=1}^{\infty}\bigcap_{m=n_0}^{\infty}\{k^{(m)}_{y} \gt c\}) =\phi^{-1}(\{k_{y} \gt c\}). \end{align*}

   Thus, for $f = \Phi \tilde{f}$ we have

   \begin{align*} (\tilde{H}_{n}\tilde{f})(y)&=\int_{\{\tilde{k}_y \gt n\}}\tilde{k}(\tilde{x},y)\tilde{f}(\tilde{x})d\tilde{\mu}(\tilde{x}) =\int_{\phi^{-1}(\{k_y \gt n\})}\tilde{k}(\tilde{x},y)\tilde{f}(\tilde{x})d\tilde{\mu}(\tilde{x})\\ &=\int_{\{k_y \gt n\}}k(x,y)f(x)d\mu(x)=(K_n f)(y), \end{align*}
   that is $\tilde{H}_n \tilde{f} = K_n \Phi \tilde{f}$. In particular, for $n = 0$ we get that $\tilde{H} = K\Phi = \tilde{K}$, as desired. Moreover, this proves that the truncated kernels transform in the natural way. The preceding shows, in particular, that right multiplication by $\Phi$ preserves integral operators.

   To show that $\Phi$ also preserves singular operators, it suffices to verify that right multiplication by $\Phi$ defines a lattice isomorphism on the space of operators. Once this is established, it follows that right multiplication by $\Phi$ preserves projection bands. Moreover, as we showed before, multiplication by $\Phi$ is a bijection from $\mathcal{L}_\kappa(L_1(X,\Sigma,\mu),L_1(Y,\Gamma,\nu))$ onto $\mathcal{L}_\kappa(L_1(\tilde{X},\tilde{\Sigma},\tilde{\mu}),L_1(Y,\Gamma,\nu))$, and the result now follows immediately from Theorem 3.15.

   Thus, we are only left to show that right multiplication by $\Phi$,

   \begin{equation*} \mathscr{B}(L_1(X, \Sigma, \mu), L_1(Y, \Gamma, \nu)) \to \mathscr{B}(L_1(\tilde{X}, \tilde{\Sigma}, \tilde{\mu}), L_1(Y, \Gamma, \nu)), T \mapsto T\Phi \end{equation*}
   is a lattice isomorphism. Observe that since $\Phi$ is a lattice isomorphism, it is interval preserving by Proposition 3.2. By [Reference Aliprantis and Burkinshaw4, Theorem 2.16 (2)], it follows that right multiplication by $\Phi$ is a lattice homomorphism, while it is clearly bijective and thus a lattice isomorphism.

2. (ii) We proceed by first defining the simple functions $\hat{k}^{(m)} : X \times \tilde{Y} \to \mathbb{R}$ by

   \begin{equation*} \hat{k}^{(m)} = \sum_{l=1}^{N_m} a_{m,l} \,\mathbb{1}_{A_{m,l} \times \psi(B_{m,l})}. \end{equation*}

   As above, the sequence $(\hat{k}^{(m)})_{m \in \mathbb{N}} \subseteq L_1(X \times \tilde{Y}, \Sigma \otimes \tilde{\Gamma}, \mu \otimes \tilde{\nu})$ is Cauchy, thus it is convergent; denote by $\hat{k}$ the limit of this sequence. Let $\hat{H}$ be the integral operator with kernel $\hat{k}$, and we will show that $\hat{H} = \hat{K}$, so that $\hat{K}$ is an integral operator.

   The argument then proceeds analogously to that for (i), but now applied to the adjoint operator. We can define

   \begin{equation*} k_x = k(x, \cdot), \quad \hat{k}_x = \hat{k}(x, \cdot), \quad k_x^{(m)} = k^{(m)}(x, \cdot), \quad \hat{k}_x^{(m)} = \hat{k}^{(m)}(x, \cdot). \end{equation*}

   Arguing as above, it is straightforward to see that $\{\hat{k}_{x} \gt c\} = \psi(\{k_{x} \gt c\})$. Consider the adjoint $(\hat{H}_n)^*: L_\infty (\tilde{Y}, \tilde{\Gamma}, \tilde{\nu}) \to L_\infty(X, \Sigma, \mu)$. Similarly as before, $\Psi$ is a positive and order continuous operator, and thus it is interval preserving by Proposition 3.2. By [Reference Aliprantis and Burkinshaw4, Theorem 2.19], $\Psi^*$ is a lattice homomorphism, while it is clearly an isometry and thus it is a lattice isometric isomorphism. For any $\tilde{g} \in L_\infty(\tilde{Y}, \tilde{\Gamma}, \tilde{\nu})$, define $g = \Psi^* \tilde{g}$, we have

   \begin{align*} ((\hat{H}_{n})^{*}\tilde{g})(x)&=\int_{\{\hat{k}_x \gt n\}}\hat{k}(x,\tilde{y})\tilde{g}(\tilde{y})d\tilde{\nu}(\tilde{y}) =\int_{\psi(\{k_x \gt n\})}\hat{k}(x,\tilde{y})\tilde{g}(\tilde{y})d\tilde{\nu}(\tilde{y})\\ &=\int_{\{k_y \gt n\}}k(x,y)g(y)d\nu(y)=((K_{n})^{*}g)(y) = (((K_n)^*\Psi^{*})\tilde{g})(x), \end{align*}
   and therefore $(\hat{H}_{n})^{*}=(K_n)^*\Psi^{*}=(\Psi K_n)^{*}$, so that $\hat{H}_{n}=\Psi K_n$. Taking $n = 0$ gives that $\hat{H} = \Psi K = \hat{K}$, so that $\hat{K}$ is an integral operator and the truncated approximations transform as claimed. The preceding shows that left multiplication by $\Psi$ preserves integral operators.

   It remains to show that $\Psi$ also preserves singular operators. As before, since left multiplication by $\Psi$ is a bijection from $\mathcal{L}_\kappa(L_1(X, \Sigma, \mu), L_1(Y, \Gamma, \nu))$ onto $\mathcal{L}_\kappa(L_1(X, \Sigma, \mu), L_1(\tilde{Y}, \tilde{\Gamma}, \tilde{\nu}))$, it suffices to verify that left multiplication by $\Psi$ defines a lattice isomorphism on the space of operators.

   Thus, we need to show that left multiplication by $\Psi$,

   \begin{equation*} \mathscr{B}(L_1(X, \Sigma, \mu), L_1(Y, \Gamma, \nu)) \to \mathscr{B}(L_1(X, \Sigma, \mu), L_1(\tilde{Y}, \tilde{\Gamma}, \tilde{\nu})), T \mapsto \Psi T \end{equation*}
   is a lattice isomorphism. $\Psi$ is order continuous and thus by [Reference Aliprantis and Burkinshaw4, Theorem 2.17] left multiplication by $\Psi$ is a lattice homomorphism, while it is clearly bijective and thus a lattice isomorphism. This completes the proof.

Proof. Applying Theorem 3.7, we can find standard measure spaces $(\tilde{X}, \tilde{\Sigma}, \tilde{\mu})$ and $(\tilde{Y}, \tilde{\Gamma}, \tilde{\nu})$, as well as isometric lattice isomorphisms

\begin{equation*} \Phi: L_1(\tilde{X}, \tilde{\Sigma}, \tilde{\mu}) \to L_1(X, \Sigma, \mu) \quad~\text{and} \quad \Psi: L_1(Y, \Gamma, \nu) \to L_1(\tilde{Y}, \tilde{\Gamma}, \tilde{\nu}), \end{equation*}

mapping characteristic functions to characteristic functions. Decompose $T = S + K$ into its singular and integral parts and consider the operator

\begin{equation*} \tilde{T} = \Psi T \Phi = \Psi S \Phi + \Psi K \Phi = \tilde{S} + \tilde{K} \end{equation*}

where $\tilde{S} = \Psi S \Phi$ is a singular operator and $\tilde{K} = \Psi K \Phi$ is an integral operator by Lemma 4.3.

Let $(\tilde{\nu}_{\tilde{x}})_{\tilde{x}\in \tilde{X}}$ be the representation of the operator $\tilde{T}^{*}$, with $(\tilde{\nu}_{\tilde{x}}^{s})_{\tilde{x}\in \tilde{X}}$ its singular part and $\tilde{k}(\cdot,\cdot)$ the kernel induced by the absolutely continuous part. The operator $\tilde{S}^{*}$ is also singular (see [Reference Weis41, Theorem 5.3]) and its measure representation is equal to $(\tilde{\nu}_{\tilde{x}}^{s})_{\tilde{x}\in \tilde{X}}$ since the decomposition into singular and absolutely continuous part of the measure representation is unique and the space of integral operators is a projection band (see Theorem 3.15 and [Reference Weis41, Remark 2.3]).

By Lemma 4.3, the truncated integral approximations $\tilde{K}_n$ of $\tilde{K}$ can be expressed as $\tilde{K}_n = \Psi K_n \Phi$ where $K_n$ are the truncated integral approximations of $K$.

Define $V_n = S + K_n$ and $\tilde{V}_n = \tilde{S} + \tilde{K}_n = \Psi V_n \Phi$. Since $\lVert|V_n|^*\rVert = \lVert|\tilde{V}_n|^*\rVert$ we get

\begin{align*} \operatorname*{ess\,sup}_{x\in X}\{|S|^{*}\mathbb{1}_{Y}(x)+ \int_{Y}|k_{n}(x,y)|d\nu(y)\}&=\lVert|S|^{*}+|K_n|^{*}\rVert\\ &=\lVert|V_{n}|^{*}\rVert = \lVert|\tilde{V_n}|^{*}\rVert = \lVert|\tilde{S}|^{*}+|\tilde{K}_{n}|^{*}\rVert\\ &=\operatorname*{ess\,sup}_{\tilde{x}\in \tilde{X}}\{\lVert\tilde{\nu}_{\tilde{x}}^{s}\rVert + \int_{Y}|\tilde{k}_{n}(\tilde{x},\tilde{y})|d\tilde{\nu}(\tilde{y})\}. \end{align*}

From the definition of $\Delta$, it is easy to see that $\Delta(T) = \Delta(\tilde{T})$. Applying Theorem 4.1 to $\tilde{T}$ gives

\begin{align*} \Delta(T) &= \Delta(\tilde{T}) \overset{\text{Thm.~4.1}}{=}\inf_{n \in \mathbb{N}}\operatorname*{ess\,sup}_{\tilde{x}\in \tilde{X}} \{\lVert\tilde{\nu}_{\tilde{x}}^{s}\rVert + \int_{Y}|\tilde{k}_{n}(\tilde{x},\tilde{y})|d\tilde{\nu}(\tilde{y})\} \\ &=\inf_{n \in \mathbb{N}}\operatorname*{ess\,sup}_{x\in X}\{|S|^{*}\mathbb{1}_{Y}(x)+ \int_{Y}|k_{n}(x,y)|d\nu(y)\}. \end{align*}

This completes the first part of the proof. If $S$ is a singular operator, the previous result immediately gives $\Delta(S) = \lVert S^*\rVert = \lVert S\rVert$.

1. (i) We will first show that condition (a) of Definition 4.5 holds. For that, let $M\subseteq L_1(X,\Sigma, \mu)$ be a separable set. By Proposition 3.9, we can find a countably generated $\sigma$-algebra $\Sigma_1\subseteq \Sigma$, such that the sublattice $E_1 := L_1(X, \Sigma_1, \mu) \subseteq L_1(X, \Sigma, \mu)$ contains $M$. Let $(f_n)_{n \in \mathbb{N}}$ be a norm dense subset in $E_1^+$. Using the Riesz-Kantorovich formula (5), we have

   \begin{equation*} |T|f_n = \sup \{T g: -f_n \leq g \leq f_n, g \in L_1(X, \Sigma, \mu) \}. \end{equation*}

   By Proposition 3.1, $L_1(X,\Sigma,\mu)$ has the countable sup property and therefore, for each $n \in \mathbb{N}$, we can find a sequence $(g_{n}^{(r)})_{r \in \mathbb{N}} \subseteq L_1(X, \Sigma, \mu)$ such that

   \begin{equation*} |T|f_n = \sup_{r \in \mathbb{N}} \{T g_{n}^{(r)}: -f_n \leq g_{n}^{(r)} \leq f_n \}. \end{equation*}

   Again, by Proposition 3.9, we can find a countably generated $\sigma$-algebra $\Sigma_2$ with $\Sigma_1\subseteq\Sigma_2$, so that $E_2 = L_1(X, \Sigma_2, \mu)$ satisfies

   \begin{equation*} \{g_{n}^{(r)}: n,r \in \mathbb{N} \} \cup \{f_n: n \in \mathbb{N} \}\subseteq E_2, \end{equation*}
   and thus
   \begin{equation*} |T|f_n = \sup_{r \in \mathbb{N}} \{Tg_{n}^{(r)}: -f_n \leq g_{n}^{(r)} \leq f_n \} \leq \sup\{T h: -f_n \leq h \leq f_n, h \in L_1(X, \Sigma_2, \mu) \}. \end{equation*}

   It follows that for every $n \in \mathbb{N}$ we have

   \begin{equation*} |T| f_n = |T J_{E_2}|f_n, \end{equation*}
   and since the sequence $(f_n)_{n \in \mathbb{N}}$ is dense in $E_1^+$ we get
   \begin{equation*} |T| f = |T J_{E_2}|f \end{equation*}
   for all $f \in E_1$.

   Arguing inductively, we can construct an increasing sequence $(\Sigma_m)_{m \in \mathbb{N}}$ of countably generated $\sigma$-algebras such that the sublattices $(E_m)_{m \in \mathbb{N}} = (L_1(X, \Sigma_m, \mu))_{m \in \mathbb{N}}$ satisfy

   \begin{equation*} M \subseteq E_1 \subseteq E_2 \subseteq E_3 \subseteq \ldots \subseteq E_{m-1} \subseteq E_m \subseteq \ldots \end{equation*}
   and
   \begin{equation*} |T| f = |T J_{E_m}|f \end{equation*}
   for all $f \in E_{m-1}$.

   Observe that if we let $\tilde{\Sigma}$ be the smallest $\sigma$-algebra on $X$ containing $\bigcup_{m \in \mathbb{N}} \Sigma_m$ then $\tilde{\Sigma}$ is countably generated. Let $E = L_1(X, \tilde{\Sigma}, \mu)$ and note that $\bigcup_{m \in \mathbb{N}} E_m$ is dense in $E$. For each $m \in \mathbb{N}$ and $f \in E_m^+$, it follows that

   \begin{equation*} |TJ_E| f \geq |TJ_{E_{m+1}}| f = |T| f. \end{equation*}

   Since this holds for any $f \in E^+_{m}$, it follows that $|TJ_E|f = |T|f$ for all $f \in E_m$. As the union of the $E_m$ is dense in $E$, we get that $|TJ_E|f = |T|f$ for all $f \in E$; in other words, $|TJ_E| = |T|J_E$. By construction $M \subseteq E = L_1(X, \tilde{\Sigma}, \mu)$, so that $\mathcal{R}_T$ satisfies condition (a) of Definition 4.5.

   We now prove condition (b) of Definition 4.5. Let $(G_n)_{n \in \mathbb{N}} := (L_1(X, \tilde{\Sigma}_n, \mu))_{n \in \mathbb{N}} \subseteq \mathcal{R}_T$ be an increasing sequence, and let

   \begin{equation*} G = \overline{\bigcup_{n\in\mathbb{N}} G_n}. \end{equation*}

   Since $(G_n)_{n \in \mathbb{N}}$ is increasing, it is easy to see that $G = L_1(X, \hat{\Sigma}, \mu)$, where $\hat{\Sigma}$ is the smallest $\sigma$-algebra containing $\bigcup_{n \in \mathbb{N}} \tilde{\Sigma}_n$; in particular, $\hat{\Sigma}$ is countably generated. Therefore, to verify that $G \in \mathcal{R}_T$, we only need to check that

   \begin{equation*} |TJ_G| = |T| J_G. \end{equation*}

   This argument is identical to the one previously done for $E$, this time using that $\bigcup_{n \in \mathbb{N}} G_n$ is dense in $G$, together with the fact that

   \begin{equation*} |T|f = |TJ_{G_n}|f \end{equation*}
   for all $f \in G_{n}$ and $G_{n} \in \mathcal{R}_T$. Details are omitted.

2. (ii) Follows immediately from Proposition 4.7.

Proof. We start by proving part (a) of Definition 4.5. Let $A$ be a separable subset of $L_1(X,\Sigma,\mu)\times L_1(Y,\Gamma,\nu)$. Without loss of generality, we may assume that $A = M \times N$ for some separable sets $M \subseteq L_1(X,\Sigma,\mu)$ and $N \subseteq L_1(Y,\Gamma,\nu)$. Indeed, let $(f_n, g_n)_{n\in\mathbb{N}}$ be a dense sequence in $A$, and let $M$ and $N$ be the sublattices generated by the $f_n$ and $g_n$, respectively. Then $A \subseteq M \times N$. Therefore, to verify part (a), it suffices to show that for every pair of separable subsets $M \subseteq L_1(X,\Sigma,\mu)$ and $N \subseteq L_1(Y,\Gamma,\nu)$, there exists $E \times F \in \mathcal{R}^S$ such that $M \times N \subseteq E \times F$.

By Proposition 3.9, we can find a countably generated $\sigma$-algebra $\Sigma_1\subseteq\Sigma$ such that the sublattice $E_1 = L_1(X, \Sigma_1, \mu) \subseteq L_1(X, \Sigma, \mu)$ contains $M$. Similarly, we can find a countably generated $\sigma$-algebra $\Gamma_1 \subseteq \Gamma$ such that the sublattice $F_1 = L_1(Y, \Gamma_1, \nu) \subseteq L_1(Y, \Gamma, \nu)$ satisfies $S[E_1]+ N \subseteq F_1$.

Observe that if $K: E_1 \to F_1$ is an integral operator with kernel $k: X \times Y \to \mathbb{R}$, then Proposition 3.14 guarantees that $k \in G_1 = L_1(X \times Y, \Sigma_1 \otimes \Gamma_1, \mu \otimes \nu)$. Find a dense sequence $(k^{(m)})_{m \in \mathbb{N}}$ of simple functions in $G_1$. We note that, since simple functions are essentially bounded, $k^{(m)}$ is the kernel of an integral operator $K^{(m)}: L_1(X, \Sigma, \mu) \to L_1(Y, \Gamma, \nu)$.

Define a sequence of operators $V_s: L_1(X, \Sigma, \mu) \to L_1(Y, \Gamma, \nu)$ by

\begin{equation*} V_{2m - 1} = S - K^{(m)}~\text{for} s = 2m-1~\text{and} V_{2m} = S + K^{(m)}~\text{for} s = 2m. \end{equation*}

By Lemma 4.8 (ii) we can find a countably generated $\sigma$-algebra $\tilde{\Sigma}_1 \subseteq \Sigma$ such that the sublattice $\tilde{E}_1 = L_1(X, \tilde{\Sigma}_1, \mu)$ contains $E_1$ and

\begin{equation*} |V_s J_{\tilde{E}_1}| = |V_s|J_{\tilde{E}_1} \end{equation*}

for all $s \in \mathbb{N}$. In other words

(7)\begin{equation} |SJ_{\tilde{E}_{1}} \pm K^{(m)}J_{\tilde{E}_{1}}|=|S \pm K^{(m)}|J_{\tilde{E}_{1}} \end{equation}

for all $m \in \mathbb{N}$.

Since $(k^{(m)})_{m \in \mathbb{N}}$ is dense in $G_1$, for any integral operator $K$ with kernel $k \in G_1$ we can find a subsequence of $(k^{(m)})_{m \in \mathbb{N}}$, which for simplicity we still denote by $k^{(m)}$, such that $\lVert k^{(m)} - k\rVert \to 0$. Fix $f \in \tilde{E}_1^+$ an essentially bounded function. We have

\begin{align*} \big||SJ_{\tilde{E}_1} + K^{(m)} J_{\tilde{E}_1}|f - |SJ_{\tilde{E}_1} + K J_{\tilde{E}_1}|f \big| &\leq |SJ_{\tilde{E}_1} + K^{(m)} J_{\tilde{E}_1} - ( SJ_{\tilde{E}_1} + K J_{\tilde{E}_1})|f \\ &= |K^{(m)} J_{\tilde{E}_1} - K J_{\tilde{E}_1}|f, \end{align*}

and thus

(8)\begin{align} \big\lVert |SJ_{\tilde{E}_1} + K^{(m)} J_{\tilde{E}_1}|f - &|SJ_{\tilde{E}_1} + K J_{\tilde{E}_1}|f \big\rVert \leq \big\lVert |K^{(m)}J_{\tilde{E}_1}- K J_{\tilde{E}_1}|f \big\rVert \nonumber \\ &\leq \int_Y \int_{X} |k^{(m)}(x,y) - k(x,y)| f(x) d\mu(x) d\nu(y) \to 0,\end{align}

since $\lVert k^{(m)} - k\rVert \to 0$ and $f$ is essentially bounded.

Combining (7) and (8), we obtain

(9)\begin{align}|S + K| f \geq |SJ_{\tilde{E}_{1}} + KJ_{\tilde{E}_{1}}|f &\overset{(8)}{=} \lim_{m \to \infty} |SJ_{\tilde{E}_{1}} + K^{(m)}J_{\tilde{E}_{1}}|f \nonumber\\ &\overset{(7)}{=} \lim_{m \to \infty}|S + K^{(m)}|f = |S + K|f,\end{align}

where the last limit exists using the same argument as before. Since this is true for any essentially bounded positive function, we obtain

\begin{equation*} |SJ_{\tilde{E}_{1}} + KJ_{\tilde{E}_{1}}| = |S + K|J_{\tilde{E}_{1}}; \end{equation*}

an identical argument shows $|SJ_{\tilde{E}_{1}} - KJ_{\tilde{E}_{1}}| = |S - K|J_{\tilde{E}_{1}}$.

Therefore, (7) holds with $K^{(m)}$ replaced by any integral operator $K$ with kernel in $G_1$. We can now find a countably generated $\sigma$-algebra $\tilde{\Gamma}_1 \subseteq \Gamma$ such that the sublattice $\tilde{F_1} = L_1(Y, \tilde{\Gamma}_1, \nu)$ satisfies $S[\tilde{E}_1] + F_1 \subseteq \tilde{F}_1$.

Arguing in the same manner and using Proposition 3.9, we can find a countably generated $\sigma$-algebra $\Sigma_2 \subseteq \Sigma$ such that $\tilde{E}_1 \subseteq L_1(X, \Sigma_2, \mu) = E_2$ and

\begin{equation*} |SJ_{E_2} \pm K J_{E_2}| = |S \pm K|J_{E_2} \end{equation*}

for all integral operators with kernel in $\tilde{G}_1 = L_1(X \times Y, \tilde{\Sigma}_1 \otimes \tilde{\Gamma}_1, \mu \otimes \nu)$. As before, we can choose a countably generated $\sigma$-algebra $\Gamma_2 \subseteq \Gamma$ such that the sublattice $F_2 = L_1(Y, \Gamma_2, \nu)$ satisfies $S[E_2] + \tilde{F}_1 \subseteq F_2$.

Inductively, we can construct a sequence of increasing countably generated $\sigma$-algebras $(\Sigma_m)_{m \in \mathbb{N}}$, $(\tilde{\Sigma}_m)_{m \in \mathbb{N}}$, $( \Gamma_m)_{m \in \mathbb{N}}$ and $(\tilde{\Gamma}_m)_{m \in \mathbb{N}}$, each contained in $\Sigma$ and $\Gamma$, respectively, such that

1. (a) $M \subseteq E_1 \subseteq \tilde{E}_1 \subseteq E_2 \subseteq \tilde{E}_2 \subseteq \ldots$,

2. (b) $N\subseteq F_1 \subseteq \tilde{F}_1 \subseteq F_2 \subseteq \tilde{F}_2 \subseteq\ldots$,

3. (c) $G_1 \subseteq \tilde{G}_1 \subseteq G_2 \subseteq \tilde{G}_2 \subseteq \ldots$,

4. (d) $S[E_{2m}] + \tilde{F}_{2m-1} \subseteq F_{2m}$ for all $m \geq 1$,

5. (e) $|SJ_{E_{2m}} \pm KJ_{E_{2m}}| = |S \pm K|J_{E_{2m}}$ for all integral operators $K$ with kernel $k \in \tilde{G}_{2m-1}$ and all $m \geq 1$,

where

\begin{equation*} \begin{aligned} E_m &= L_1(X, \Sigma_m, \mu), \quad & \tilde{E}_m &= L_1(X, \tilde{\Sigma}_m, \mu), \\ F_m &= L_1(Y, \Gamma_m, \nu), \quad & \tilde{F}_m &= L_1(Y, \tilde{\Gamma}_m, \nu), \\ G_m &= L_1(X \times Y, \Sigma_m \otimes \Gamma_m, \mu \otimes \nu), \quad & \tilde{G}_m &= L_1(X \times Y, \tilde{\Sigma}_m \otimes \tilde{\Gamma}_m, \mu \otimes \nu). \end{aligned} \end{equation*}

Let $\tilde{\Sigma} \subseteq \Sigma$ be the $\sigma$-algebra generated by the $\Sigma_m$ and $\tilde{\Gamma} \subseteq \Gamma$ the $\sigma$-algebra generated by the $\Gamma_m$, and observe that both $\tilde{\Sigma}$ and $\tilde{\Gamma}$ are countably generated $\sigma$-algebras. Define $E = L_1(X, \tilde{\Sigma}, \mu)$ and $F = L_1(Y, \tilde{\Gamma}, \nu)$.

By construction, $M \subseteq E$ and $N \subseteq F$ so that $M \times N \subseteq E \times F$. Therefore, to finish, we need to show that $E \times F \in \mathcal{R}^S$.

Note that that since $\bigcup_{m \in \mathbb{N}} E_m$ is dense in $E$ and $S[E_{2m}] \subseteq F_{2m}$, it is easy to check that $S[E] \subseteq F$. Therefore, to show that $E \times F \in \mathcal{R}^S$, we only need to verify that $S\big|_E: E \to F$ is still a singular operator.

Define $G = L_1(X \times Y, \tilde{\Sigma} \otimes \tilde{\Gamma}, \mu \otimes \nu)$ and let $H: E \to F$ be an integral operator with kernel $h \in G$. Observe that $\tilde{G}_{m} \subseteq \tilde{G}_{m+1}$ and $\bigcup_{m \in \mathbb{N}} \tilde{G}_m$ is norm dense in $G$, and thus we can find a sequence of simple functions $h^{(m)} \in \tilde{G}_{2m-1}$ such that $\lVert h^{(m)} - h\rVert \to 0$. Denote by $H^{(m)}: E \to F$ the integral operators with kernel $h^{(m)}$.

Fix $n \in \mathbb{N}$ and an essentially bounded positive function $f \in E^+_n$. Arguing as in the proof of (8) we have

\begin{align*} \big\lVert |SJ_E - H^{(m)}J_E|f - |SJ_E - HJ_E|f \big\rVert \to 0, ~\text{and} \lVert|S - H^{(m)}|f - |S - H|f\rVert \to 0 \end{align*}

since $\lVert h^{(m)} - h\rVert \to 0$ and $f$ is essentially bounded.

Thus, for all $m$ with $2m \geq n$ we get

\begin{align*} |S - H|f &\geq |SJ_E - HJ_E|f = \lim_{m \to \infty} |SJ_{E} - H^{(m)} J_E|f \\ &\geq \lim_{m \to \infty} |SJ_{E_{2m}} - H^{(m)} J_{E_{2m}}|f \overset{(e)}{=} \lim_{m \to \infty}|S - H^{(m)}|f = |S - H|f, \end{align*}

that is $|S-H|f = |SJ_E - HJ_E|f = |SJ_E - H|f$. A similar argument also gives $|SJ_E + H|f = |S + H|f$.

Since $S$ is singular and $H$ is an integral operator, we get

\begin{equation*} |SJ_E - H|f = |S - H|f = |S + H|f = |SJ_E + H|f, \end{equation*}

for all essentially bounded functions $f \in E_n$. Since $n \in \mathbb{N}$ was arbitrary and $\bigcup_{n \in \mathbb{N}} E_n$ is dense in $E$, the above equation holds for all $f \in E$. In other words, we have $|SJ_E| \wedge |H| = 0$ for all integral operators $H$ with kernel in $G$, which, by (6), is equivalent to $SJ_E$ being a singular operator. This shows that $SJ_E$ is singular, so $E \times F \in \mathcal{R}^S$, and hence condition (a) of Definition 4.5 is satisfied.

To prove part (b) of Definition 4.5, consider an increasing sequence

\begin{equation*} (E_n \times F_n)_{n \in \mathbb{N}} := \big(L_1(X, \Sigma_n, \nu) \times L_1(Y, \Gamma_n, \nu)\big)_{n \in \mathbb{N}} \subseteq \mathcal{R}^S, \end{equation*}

where the reader should be careful not to confuse $\Sigma_n$, $\Gamma_n$, $E_n$, and $F_n$ here with those used in the construction above. We aim to show that

\begin{equation*} \overline{\bigcup_{n \in \mathbb{N}} E_n \times F_n } = \overline{\bigcup_{n \in \mathbb{N}} E_n } \times \overline{\bigcup_{n \in \mathbb{N}} F_n } \in \mathcal{R}^S. \end{equation*}

Define

\begin{equation*} E = \overline{\bigcup_{n \in \mathbb{N}} E_n } \quad~\text{and} \quad F = \overline{\bigcup_{n \in \mathbb{N}} F_n }, \end{equation*}

so that the previous condition can simply be expressed as $E \times F \in \mathcal{R}^S$.

Observe that since $(E_n \times F_n)_{n \in \mathbb{N}}$ is an increasing family, the same holds for $(E_n)_{n \in \mathbb{N}}$ and $(F_n)_{n \in \mathbb{N}}$. In particular, it is easy to see that this implies

\begin{equation*} E = L_1(X, \tilde{\Sigma}, \mu) \quad~\text{and} \quad F = L_1(Y, \tilde{\Gamma}, \nu), \end{equation*}

where $\tilde{\Sigma} \subseteq \Sigma$ and $\tilde{\Gamma} \subseteq \Gamma$ are the smallest $\sigma$-algebras containing $\bigcup_{n \in \mathbb{N}} \Sigma_n$ and $\bigcup_{n \in \mathbb{N}} \Gamma_n$, respectively.

Since $\bigcup_{n \in \mathbb{N}} E_n$ is dense in $E$ and $S(E_n) \subseteq F_n$, it is clear that $S(E) \subseteq F$. To see that $E \times F \in \mathcal{R}^S$, we only need to check that $ S\big|_{E}: E \to F~\text{is singular}$. This follows similarly as before, using the density of $\bigcup_{n \in \mathbb{N}} E_n$ in $E$ together with the fact that $E_n \times F_n \in \mathcal{R}^S$. Since the arguments are similar, and condition (b) will not be needed in future results, we omit the details.

Proof. It is enough to show $\Delta(S) \geq \lVert S\rVert$, since the other inequality is always true. Let $(f_n)_{n \in \mathbb{N}}$ be a sequence of norm one functions such that $\lVert Sf_n\rVert \to \lVert S\rVert$ and define $M = \operatorname{span} \{f_n: n \in \mathbb{N}\}$. By Lemma 4.9, the family $\mathcal{R}^S$ is a rich family of closed separable subspaces of $L_1(X, \Sigma, \mu) \times L_1(Y, \Gamma, \nu)$ and thus by Definition 4.5 (a) applied to the separable set $M\times\{0\}$, we can find countably generated $\sigma$-algebras $\tilde{\Sigma}$ and $\tilde{\Gamma}$ such that $M \subseteq L_1(X, \tilde{\Sigma}, \mu) =: E$, $S[E] \subseteq L_1(Y, \tilde{\Gamma}, \nu) =: F$ and $S\big|_E: E \to F$ is singular. We note that $\lVert SJ_E\rVert = \lVert S\rVert$ since $M \subseteq E$. Theorem 4.4 applied to $SJ_E: E \to F$ gives $\Delta(S J_E) = \lVert SJ_E\rVert$ and therefore

\begin{equation*} \Delta(S) \geq \Delta(SJ_E) = \lVert SJ_E\rVert = \lVert S\rVert, \end{equation*}

finishing the proof.

Proof. Without loss of generality, we can assume that the operator $K$ is positive. We divide the proof into three cases.

1. Case 1: (Standard measure spaces) Suppose first that $(X, \Sigma, \mu)$ and $(Y, \Gamma, \nu)$ are standard measure spaces. In this case, the result was proven in the first proposition of [Reference Weis42].

2. Case 2: (Measure spaces of finite measure with countably generated $\sigma$-algebra) Suppose now that $(X, \Sigma, \mu)$ and $(Y, \Gamma, \nu)$ are measure spaces of finite measure with $\Sigma$ and $\Gamma$ countably generated. By Theorem 3.7 we can find standard measure spaces $(\tilde{X}, \tilde{\Sigma}, \tilde{\mu})$ and $(\tilde{Y}, \tilde{\Gamma}, \tilde{\nu})$ and measure algebra isomorphisms $\phi: \tilde{\Sigma}_{\tilde{\mu}} \to \Sigma_\mu$ and $\psi: \Gamma_\nu \to \tilde{\Gamma}_{\tilde{\nu}}$; denote by $\Phi: L_1(\tilde{X}, \tilde{\Sigma}, \tilde{\mu}) \to L_1(X, \Sigma, \mu)$ and $\Psi: L_1(Y, \Gamma, \nu) \to L_1(\tilde{Y}, \tilde{\Gamma}, \tilde{\nu})$ the induced isometric lattice isomorphisms.

   Define $\tilde{K} = \Psi K \Phi$, which is an integral operator with truncated integral approximations $\tilde{K}_n = \Psi K_n \Phi$ by Lemma 4.3. Fix any $\varepsilon \gt 0$.

   By the previous case, for each $n \in \mathbb{N}$, we can find $\tilde{G}_n \subseteq \tilde{Y}$ with $\tilde{\nu}(\tilde{G}_n) \lt \frac{2\lVert\tilde{K}\rVert}{n}$ and $\tilde{M}_n \in \tilde{\Sigma}$ of positive measure such that

   \begin{equation*} \tilde{K}^* \mathbb{1}_{\tilde{G}_n}(\tilde{x}) = \int_{\tilde{G}_n}\tilde{k}(\tilde{x},\tilde{y})d\tilde{\nu}(\tilde{y})\geq \int_{\tilde{Y}}\tilde{k}_{n}(\tilde{x},\tilde{y})d\tilde{\nu}(\tilde{y})-\varepsilon = \tilde{K}^* \mathbb{1}_{\tilde{Y}}(\tilde{x})- \varepsilon, \end{equation*}
   for all $\tilde{x} \in \tilde{M}_n$. In other words,
   \begin{equation*} \chi_{\tilde{M}_n} \tilde{K}^* \mathbb{1}_{\tilde{G}_n} \geq \chi_{\tilde{M}_n}\tilde{K}_n^* \mathbb{1}_{\tilde{Y}} - \varepsilon\mathbb{1}_{\tilde{M}_n}, \end{equation*}
   equivalently
   \begin{equation*} \chi_{\tilde{M}_n} \Phi^* K^* \Psi^* \mathbb{1}_{\tilde{G}_n} \geq \chi_{\tilde{M}_n} \Phi^* K^*_n \Psi^* \mathbb{1}_{\tilde{Y}} - \varepsilon\mathbb{1}_{\tilde{M}_n}. \end{equation*}

   Using the definition of $\Psi$, we get

   \begin{equation*} \chi_{\tilde{M}_n} \Phi^* K^* \mathbb{1}_{\psi^{-1}(\tilde{G}_n)} \geq \chi_{\tilde{M}_n} \Phi^* K^*_n \mathbb{1}_{Y} - \varepsilon\mathbb{1}_{\tilde{M}_n}, \end{equation*}
   and since $\Phi$, and thus $\Phi^*$, is a lattice isomorphism, this is equivalent to
   \begin{equation*} (\Phi^*)^{-1}\chi_{\tilde{M}_n} \Phi^* K^* \mathbb{1}_{\psi^{-1}(\tilde{G}_n)} \geq (\Phi^*)^{-1}\chi_{\tilde{M}_n} \Phi^* K^*_n \mathbb{1}_{Y} - \varepsilon(\Phi^*)^{-1}\mathbb{1}_{\tilde{M}_n}. \end{equation*}

   Since $\Phi$ satisfies $\chi_{\tilde{M}_n}\Phi^* = \Phi^* \chi_{\phi(\tilde{M}_n)}$ and $(\Phi^*)^{-1}\mathbb{1}_{\tilde{M}_n} = \mathbb{1}_{\phi(\tilde{M}_n)}$, we get

   \begin{equation*} \chi_{\phi(\tilde{M}_n)} K^* \mathbb{1}_{\psi^{-1}(\tilde{G}_n)} \geq \chi_{\phi(\tilde{M}_n)} K^*_n \mathbb{1}_{Y} - \varepsilon\mathbb{1}_{\phi(\tilde{M}_n)}. \end{equation*}

   Setting $M_n = \phi(\tilde{M}_n)$ and $G_n = \psi^{-1}(\tilde{G}_n)$ this gives

   \begin{equation*} \chi_{M_n} K^* \mathbb{1}_{G_n} \geq \chi_{M_n} K^*_n \mathbb{1}_{Y} - \varepsilon \mathbb{1}_{M_n}. \end{equation*}

   In other words, we have

   \begin{equation*} \int_{G_n} k(x,y) d\nu(y) = K^* \mathbb{1}_{G_n}(x) \geq K_n^* \mathbb{1}_Y - \varepsilon = \int_{Y} k_n(x,y) d\nu(y) - \varepsilon \end{equation*}
   for all $x \in M_n$.

   Since $\tilde{M}_n$ has positive measure, then so does $M_n$, while

   \begin{equation*} \nu(G_n) = \nu(\psi^{-1}(\tilde{G}_n)) = \tilde{\nu}(\tilde{G}_n) \lt \frac{2 \lVert\tilde{K}\rVert}{n} = \frac{2 \lVert K\rVert}{n}, \end{equation*}
   which finishes this case.

3. Case 3: (Finite measure) Finally, consider the case when $(X, \Sigma, \mu)$ and $(Y, \Gamma, \nu)$ are measure spaces of finite measure. By Proposition 3.14, we have that $k \in L_1(X \times Y, \Sigma \otimes \Gamma, \mu \otimes \nu)$. Let $k^{(m)}$ be a sequence of simple functions defined on rectangles,

   \begin{equation*} k^{(m)} = \sum_{l=1}^{N_m} a_{m, l} \mathbb{1}_{A_{m,l} \times B_{m,l}}, \end{equation*}
   such that $\lVert k^{(m)} - k\rVert \to 0$. Let $\tilde{\Sigma}$ be the $\sigma$-algebra generated by $\{A_{m,l}: m, l \in \mathbb{N} \}$ and $\tilde{\Gamma}$ be the $\sigma$-algebra generated by $\{B_{m,l}: m, l \in \mathbb{N} \}$. Observe that $\tilde{\Sigma}$ and $\tilde{\Gamma}$ are countably generated and $k \in L_1(X \times Y, \tilde{\Sigma} \otimes \tilde{\Gamma}, \mu \otimes \nu)$. Now, $K$ can be seen as an integral operator $K: L_1(X, \tilde{\Sigma}, \mu) \to L_1(Y, \tilde{\Gamma}, \nu)$. By the previous case, for any $\varepsilon \gt 0$ and any $n \in \mathbb{N}$ we can find measurable sets $G_n \subseteq Y$ with $\nu(G_n) \lt \frac{2\lVert K\rVert}{n}$ and $M_n \subseteq X$ of positive measure such that
   \begin{equation*} \int_{G_n} k(x,y) d\nu(y) \geq \int_{Y} k_n(x,y) d\nu(y) - \varepsilon \end{equation*}
   for all $x \in M_n$. This finishes the proof.

Proof. We have the decomposition

\begin{equation*} L_1(Y, \Gamma, \nu) = \Bigl(\bigoplus_{j \in J} L_1(Y_j, \Gamma_j, \nu)\Bigr)_{\ell_1}. \end{equation*}

Since $T$ is weakly continuous, the image of any weakly compact set under $T$ is again weakly compact. By [Reference Kačena, Kalenda and Spurný24, Lemma 7.2 (ii)], this implies that the image of a weakly compact set is contained in a countable sum of the subspaces $L_1(Y_j, \Gamma_j, \nu)$. Finally, since $\mu$ is finite, $L_1(X, \Sigma, \mu)$ is weakly compactly generated, and the result follows.

Proof. It is clear that for any $T: L_1(X, \Sigma, \mu) \to L_1(Y, \Gamma, \nu)$ we have

\begin{equation*} \alpha_\Theta(T) \leq \lVert T\rVert. \end{equation*}

Hence for any weakly compact operator $W: L_1(\Sigma, X, \mu) \to L_1(\Theta, \gamma, \nu)$ it follows that

\begin{equation*} \alpha_\Theta(T) = \alpha_\Theta(T + W) \leq \lVert T + W\rVert. \end{equation*}

Since this is true for any weakly compact operator $W: L_1(\Sigma, X, \mu) \to L_1(\Theta, \gamma, \nu)$, taking the infimum yields $\alpha_\Theta(T) \leq \lVert T\rVert_{w, \Theta}$.

Proof. For convenience, let

\begin{equation*} \eta = \inf_{n\in\mathbb{N}}\operatorname*{ess\,sup}_{x\in X}\{|S^*|\mathbb{1}_{Y}(x)+\int_{Y\backslash \Theta}|k(x,y)|d\nu(y)+\int_{\Theta}|k_{n}(x,y)|d\nu(y)\}. \end{equation*}

We first show that $\alpha_\Theta(T) \leq \eta$. For this, choose a sequence $B_m \subseteq \Theta$ of measurable sets with $\nu(B_m) \to 0$ and $\lVert\chi_{Y \backslash \Theta}T + \chi_{B_m}T\rVert \to \alpha_{\Theta}(T)$. Since $\chi_{Y \backslash \Theta}$ and $\chi_{B_m}$ are band projections, we have

\begin{equation*} |\chi_{Y \backslash \Theta}T + \chi_{B_m} T| = (\chi_{Y \backslash \Theta} + \chi_{B_m})|T|. \end{equation*}

Therefore, by decomposing $T$ into its positive and negative parts and arguing over these, we can assume without loss of generality that $T$ is a positive operator. In particular, $K$ and $S$ are also positive operators.

We have

\begin{align*} \lVert\chi_{Y \backslash \Theta} T + \chi_{B_m}T\rVert &= \lVert T^*\chi^*_{Y \backslash \Theta} + T^* \chi^*_{B_m}\rVert \\ &= \lVert S^*\chi^*_{Y \backslash \Theta} + S^* \chi^*_{B_m} + K^*\chi^*_{Y \backslash \Theta} + K^* \chi^*_{B_m}\rVert\\ &= \operatorname*{ess\,sup}_{x \in X} \{S^*\mathbb{1}_{Y \backslash \Theta}(x) + S^* \mathbb{1}_{B_m}(x) + K^*\mathbb{1}_{Y \backslash \Theta}(x) + K^* \mathbb{1}_{B_m}(x)\}. \end{align*}

Since $S^*$ is a positive operator, we have

\begin{equation*} S^* \mathbb{1}_{Y \backslash \Theta} + S^* \mathbb{1}_{B_m} \leq S^* \mathbb{1}_Y \end{equation*}

and thus

\begin{equation*} \lVert\chi_{Y \backslash \Theta} T + \chi_{B_m}T\rVert \leq \operatorname*{ess\,sup}_{x \in X} \{S^*\mathbb{1}_{Y}(x) + K^*\mathbb{1}_{Y \backslash \Theta}(x) + K^* \mathbb{1}_{B_m}(x)\}. \end{equation*}

For each fixed $n \in \mathbb{N}$ we have

\begin{align*} K^* \mathbb{1}_{B_m}(x) &= \int_{B_m} k(x,y) d \nu(y) \leq \int_{B_m} k_n(x,y) d \nu(y) + n \cdot \nu(B_m) \\ &\leq \int_{\Theta} k_n(x,y) d \nu(y) + n \cdot \nu(B_m), \end{align*}

and thus

\begin{equation*} \lVert\chi_{Y \backslash \Theta} T + \chi_{B_m}T\rVert \leq \operatorname*{ess\,sup}_{x \in X} \{S^*\mathbb{1}_{Y}(x) \,+ \int_{Y \backslash \Theta} k(x,y) d\nu(y) \,+ \int_{\Theta} k_n(x,y) d \nu(y) + n \cdot \nu(B_m)\}. \end{equation*}

Letting $m \to \infty$ gives

\begin{equation*} \alpha_\Theta(T) \leq \operatorname*{ess\,sup}_{x \in X} \{S^*\mathbb{1}_{Y}(x) + \int_{Y \backslash \Theta} k(x,y) d\nu(y) + \int_{\Theta} k_n(x,y) d \nu(y) \}, \end{equation*}

for each $n \in \mathbb{N}$. Taking the infimum over $n \in \mathbb{N}$ gives $\alpha_\Theta(T) \leq \eta$.

To see the converse inequality, define the operators

\begin{equation*} V_n = S^* \chi^*_{Y \backslash \Theta} + K^* \chi^*_{Y \backslash \Theta} + S^* \chi^*_\Theta + K^*_n \chi^*_\Theta, \end{equation*}

where $K_n$ are the truncated integral approximations of $K$. Observe that since $K$ is a positive operator, so are each of its truncated integral approximations $K_n$, and thus $V_n$ is positive for each $n \in \mathbb{N}$. Note that we have

\begin{equation*} \lVert V_n\rVert = \operatorname*{ess\,sup}_{x \in X} V_n\mathbb{1}_Y(x), \end{equation*}

and

\begin{equation*} \lim_{n \to \infty} \lVert V_n\rVert = \eta. \end{equation*}

Since $(X, \Sigma, \mu)$ is strictly localizable, we can find measurable sets $M_n \subseteq X$ of finite positive measure such that

\begin{equation*} V_n \mathbb{1}_{Y}(x) \geq \operatorname*{ess\,sup}_{x \in X} V_n\mathbb{1}_Y(x) - \frac{1}{n} = \lVert V_n\rVert - \frac{1}{n}, \end{equation*}

for all $x \in M_n$. For each $n \in \mathbb{N}$, let $\Sigma_n = \{A \cap M_n: A \in \Sigma\}$ and recall that $J_{L_1(M_n, \Sigma_n, \mu)}$ is the inclusion operator of the projection band $L_1(M_n, \Sigma_n, \mu)$ into $L(X, \Sigma, \mu)$. Consider the operators

\begin{equation*} R_n: L_1(M_n, \Sigma_n, \mu) \to L_1(\Theta, \gamma, \nu) \end{equation*}

given by $R_n = \chi_{\Theta}SJ_{L_1(M_n, \Sigma_n, \mu)}$. We note that since $L_1(M_n, \Sigma_n, \mu)$ and $L_1(\Theta, \gamma, \nu)$ are projection bands, then $R_n$ is still a singular operator. By Proposition 4.10 we have $\Delta(R_n) = \lVert R_n\rVert = \lVert R_n^*\rVert$. For each $n \in \mathbb{N}$, choose a measurable set $B_n \subseteq \Theta$ such that $\nu(B_n) \lt 1/n$ and

\begin{equation*} \operatorname*{ess\,sup}_{x \in X} R_n^* \mathbb{1}_{B_n}(x) = \lVert R_n^* \chi^*_{B_n}\rVert = \lVert\chi_{B_n}R_n\rVert \geq \Delta(R_n) - \frac{1}{n} = \lVert R_n^*\rVert - \frac{1}{n}. \end{equation*}

We can find $\tilde{M}_n\subseteq M_n$ of positive measure such that for all $x \in \tilde{M}_n$ we have

\begin{equation*} R_n^* \mathbb{1}_{B_n}(x) \geq \operatorname*{ess\,sup}_{x \in X} R_n^* \mathbb{1}_{B_n}(x) - \frac{1}{n}. \end{equation*}

Since $\lVert R_n^*\rVert = \operatorname*{ess\,sup}_{x \in X} R_n^* \mathbb{1}_{\Theta}(x)$, the previous inequalities give

\begin{equation*} R_n^* \mathbb{1}_{B_n}(x) \geq \lVert R_n^*\rVert - \frac{2}{n} = R_n^* \mathbb{1}_{\Theta}(x) - \frac{2}{n}, \end{equation*}

for all $x \in \tilde{M}_n$. This implies that

\begin{equation*} S^* \mathbb{1}_{B_n}(x) \geq S^* \mathbb{1}_{\Theta}(x) - \frac{2}{n} \end{equation*}

for all $x \in \tilde{M}_n$.

Again, for each $n \in \mathbb{N}$, let $\tilde{\Sigma}_n = \{A \cap \tilde{M}_n: A \in \Sigma\}$ and $J_{L_1(\tilde{M}_n, \tilde{\Sigma}_n, \mu)}$ be the inclusion operator of the projection band $L_1(\tilde{M}_n, \tilde{\Sigma}_n, \mu)$. Define the integral operator $L_n: L_1(\tilde{M}_n, \tilde{\Sigma}_n, \mu) \to L_1(\Theta, \gamma, \nu)$ by $L_n = \chi_{\Theta} K J_{L_1(\tilde{M}_n, \tilde{\Sigma}_n, \mu)}$. By Lemma 4.11, we can find a measurable set $G_n \subseteq \Theta$ with $\nu(G_n) \lt \frac{2\lVert L_n\rVert}{n} \leq \frac{2\lVert K\rVert}{n}$ and a measurable set $\hat{M}_n \subseteq \tilde{M}_n$ of positive measure such that

\begin{equation*} \int_{G_n} k(x,y) d\nu(y) \geq \int_{\Theta} k_n(x,y) d\nu(y) - \frac{1}{n} \end{equation*}

for all $x \in \hat{M}_n$. Let $H_n = G_n \cup B_n$ and note that $\lim_{n \to \infty} \nu(H_n) = 0$.

For all $x \in \hat{M}_n$ we have

\begin{align*} (T^* \chi^*_{Y \backslash \Theta} + T^* \chi^*_{H_n}) \mathbb{1}_Y(x) &= S^* \mathbb{1}_{Y \backslash \Theta}(x) + K^* \mathbb{1}_{Y \backslash \Theta}(x) + S^* \mathbb{1}_{B_n \cup G_n}(x) + K^* \mathbb{1}_{B_n \cup G_n}(x) \\ &= S^* \mathbb{1}_{Y \backslash \Theta}(x) + K^* \mathbb{1}_{Y \backslash \Theta}(x) + S^* \mathbb{1}_{B_n \cup G_n}(x) \\ &\quad + \int_{B_n \cup G_n} k(x,y) d\nu(y) \\ &\geq S^* \mathbb{1}_{Y \backslash \Theta}(x) + K^* \mathbb{1}_{Y \backslash \Theta}(x) + S^* \mathbb{1}_{B_n}(x) \\ &\quad + \int_{\Theta} k_n(x,y) d\nu(y) - \frac{1}{n} \\ &\geq S^* \mathbb{1}_{Y \backslash \Theta}(x) + K^* \mathbb{1}_{Y \backslash \Theta}(x) + S^* \mathbb{1}_{\Theta}(x) \\ &\quad + \int_{\Theta} k_n(x,y) d\nu(y) - \frac{3}{n} \\ &\geq (S^* \mathbb{1}_{Y \backslash \Theta}(x) + K^* \mathbb{1}_{Y \backslash \Theta}(x) + S^* \mathbb{1}_\Theta(x) + K^*_n \mathbb{1}_\Theta(x)) - \frac{3}{n} \\ &= V_n \mathbb{1}_Y(x) - \frac{3}{n} \geq \lVert V_n\rVert - \frac{1}{n} - \frac{3}{n} = \lVert V_n\rVert - \frac{4}{n}. \end{align*}

Since

\begin{align*} \lVert\chi_{Y \backslash \Theta} T + \chi_{H_n}T\rVert = \lVert T^* \chi^*_{Y \backslash \Theta} + T^*\chi^*_{H_n}\rVert = \operatorname*{ess\,sup}_{x \in X} \{(T^* \chi^*_{Y \backslash \Theta} + T^* \chi^*_{H_n}) \mathbb{1}_Y(x)\} \end{align*}

and $\hat{M}_n$ has positive measure, we get

\begin{equation*} \lVert\chi_{Y \backslash \Theta} T + \chi_{H_n}T\rVert \geq \lVert V_n\rVert - \frac{4}{n}, \end{equation*}

and since $\nu(H_n) \to 0$, it follows that

\begin{equation*} \alpha_\Theta(T) \geq \limsup_{n \to \infty }\lVert\chi_{Y \backslash \Theta} T + \chi_{H_n}T\rVert \geq \lim_{n \to \infty} (\lVert V_n\rVert - \frac{4}{n}) = \eta, \end{equation*}

which proves the other inequality and finishes the proof.

Proof. Express $T = S + K$ where $S$ is a singular operator and $K$ is an integral operator with kernel $k$. Consider the measurable function $g: X \to \mathbb{R}$ given by

\begin{equation*} g(x) = |S^*|\mathbb{1}_Y(x) + \int_{Y \backslash \Theta} |k(x,y)|d \nu(y) \end{equation*}

and the function $f: X \times [0, \infty] \to \mathbb{R}$ given by

\begin{equation*} f(x,s)= g(x) + \int_{\Theta}(|k(x,y)| - s)^+ d\nu(y). \end{equation*}

Observe that $f$ is a measurable function on $X \times [0,\infty]$, while for fixed $x \in X$, the dominated convergence theorem gives that $f$ is continuous in $s$. Define $A = \{x \in X: |T^*|\mathbb{1}_Y(x) \gt \alpha_{\Theta}(T)\}$, so that by (10) we have

\begin{equation*} f(x, 0) = |T^*| \mathbb{1}_Y(x) \gt \alpha_{\Theta}(T) \end{equation*}

for all $x \in A$.

Observe that if $k_s$ is the truncated kernel of $K$, then $(|k(x,y)| - s)^+ \leq |k_s(x,y)|$. It follows that

\begin{equation*} f(x,s) \leq g(x) + \int_{\Theta} |k_s(x,y)|d\nu(y) \leq \operatorname*{ess\,sup}_{x \in X} \{g(x) + \int_{\Theta} |k_s(x,y)|d\nu(y)\}. \end{equation*}

Recall that by Proposition 4.17 we have

\begin{align*} \alpha_{\Theta}(T)&=\inf_{n\in\mathbb{N}}\operatorname*{ess\,sup}_{x\in X}\{|S^*|\mathbb{1}_{Y}(x)+\int_{Y\backslash \Theta}|k(x,y)|d\nu(y)+\int_{\Theta}|k_{n}(x,y)|d\nu(y)\}\\ &= \inf_{n\in\mathbb{N}}\operatorname*{ess\,sup}_{x\in X}\{g(x)+\int_{\Theta}|k_{n}(x,y)|d\nu(y)\}, \end{align*}

and thus $\lim_{s \to \infty} f(x,s) \leq \alpha_{\Theta}(T)$ for all $x \in A$.

Recall that, for fixed $x \in X$, the function $f(s,x)$ is continuous and decreasing in $s$. Therefore, for every $x \in A$ there exists some $s \in [0,\infty]$ such that $f(x,s) = \alpha_\Theta(T)$. Define the function $s: X \to [0, \infty]$ by

\begin{equation*} s(x) = \begin{cases} \displaystyle \inf\{s \in [0,\infty]: f(x,s) = \alpha_\Theta(T)\} &~\text{if} x \in A, \\ 0 &~\text{if} x \in X \backslash A. \end{cases} \end{equation*}

Since $A$ is a measurable set and the function $f$ is measurable, it follows that $s$ is a measurable function. We can also find a measurable unimodular function $\pi: X \times Y \to \mathbb{R}$ such that $k(x,y) = \pi(x,y) |k(x,y)|$ for all $(x,y) \in X \times Y$. Finally define the function $h: X \times Y \to \mathbb{R}$ by

\begin{equation*} h(x,y) = \begin{cases} \displaystyle \pi(x,y)[|k(x,y)| \wedge s(x)] &~\text{if} (x,y) \in X \times \Theta, \\ 0 &~\text{if} (x,y) \in X \times (Y \backslash \Theta). \end{cases} \end{equation*}

In particular, note that $h(x,y) = 0$ whenever $x \in X \backslash A$. Furthermore, observe that $h$ is a measurable function, corresponding to the kernel of an integral operator $H: L_1(X, \Sigma, \mu) \to L_1(\Theta, \gamma, \nu) \subseteq L_1(Y, \Gamma, \nu)$, in other words

\begin{equation*} (Hf)(y) = \int_{X} h(x,y) f(x) d\mu(x). \end{equation*}

Indeed, that $h$ is measurable is clear from the definition, while it corresponds to a kernel of an integral operator since $k$ was the kernel of an integral operator.

We claim that $H$ is weakly compact. Assume the contrary, so that there exists $C \gt 0$ such that for all $n \in \mathbb{N}$ we have