Product rectangle kernels are dense in product L two

Each $Z_n$ is a measurable rectangle of finite product measure, $Z_n\subseteq Z_{n+1}$, and ${\bigcup }_n{Z}_n=X\times Y$; moreover $\rho (E\cap Z_n)\uparrow \rho (E)$ for every $E\in \mathcal{A}\otimes \mathcal{B}$ by [F3], so for $\rho (E)<+\mathrm{\infty }$ and real $\delta >0$ there is $n$ with $\rho (E\setminus Z_n)<\delta \mathrm{/}2$.

A local algebra on each exhausted rectangle. Fix $n$ and let ${\mathcal{G}}_n:=\{F\subseteq Z_n:F=E\cap Z_n\text{ for some }E\in \mathcal{A}\otimes \mathcal{B}\}$ be the trace sigma-algebra, and let ${\mathcal{C}}_n$ be the family of finite unions of rectangles $A\times B$ with $A\in \mathcal{A}$, $A\subseteq X_n$, $B\in \mathcal{B}$, $B\subseteq Y_n$. Then ${\mathcal{C}}_n$ is an algebra of subsets of $Z_n$: it contains $\mathrm{\varnothing }$, it is closed under finite unions by definition, and for $A\times B\subseteq Z_n$ the complement in $Z_n$ is $((X_n\setminus A)\times Y_n)\cup (A\times (Y_n\setminus B))$, a union of two rectangles inside $Z_n$, while complements of finite unions follow by De Morgan and the closure of products of intersections; every element of ${\mathcal{C}}_n$ is a finite disjoint union of rectangles by [F4]. Furthermore $\sigma ({\mathcal{C}}_n)={\mathcal{G}}_n$: the inclusion $\subseteq$ is clear since each generator of ${\mathcal{C}}_n$ lies in ${\mathcal{G}}_n$, and conversely $\{E\in \mathcal{A}\otimes \mathcal{B}:E\cap Z_n\in \sigma ({\mathcal{C}}_n)\}$ is a sigma-algebra containing every measurable rectangle, because $(A\times B)\cap Z_n=(A\cap X_n)\times (B\cap Y_n)\in {\mathcal{C}}_n$, hence it contains $\mathcal{A}\otimes \mathcal{B}$ and therefore ${\mathcal{G}}_n$. Finally the trace measure ${\rho }_n(F):=\rho (F)$ on ${\mathcal{G}}_n$ is a finite measure because ${\rho }_n(Z_n)=\mu (X_n)\nu (Y_n)<+\mathrm{\infty }$ by [F2].

Approximation of sets of finite product measure. Let $E\in \mathcal{A}\otimes \mathcal{B}$ with $\rho (E)<+\mathrm{\infty }$ and let $\delta >0$. Choose $n$ with $\rho (E\setminus Z_n)<\delta \mathrm{/}2$ by [step 1.1]; then $E\cap Z_n\in {\mathcal{G}}_n$, so [F5] applied to the finite measure space $(Z_n,{\mathcal{G}}_n,{\rho }_n)$ and its generating algebra ${\mathcal{C}}_n$ of [step 1.2] gives $C\in {\mathcal{C}}_n$ with ${\rho }_n((E\cap Z_n)\mathrm{△}C)<\delta \mathrm{/}2$. Since $E\mathrm{△}C\subseteq (E\setminus Z_n)\cup ((E\cap Z_n)\mathrm{△}C)$, [F3] gives $\rho (E\mathrm{△}C)<\delta$, and $C$ is a finite union of rectangles with $\mu (A)<+\mathrm{\infty }$, $\nu (B)<+\mathrm{\infty }$ as a subset of $Z_n$.

Indicator approximation. For $E$ and $C$ as in [step 2.1], ${\mathbf{1}}_C$ is a finite sum of rectangle kernels by [F4] and [step 2.1], and by [F8] the difference of the classes of ${\mathbf{1}}_E$ and ${\mathbf{1}}_C$ has squared $L^2(\rho )$-norm $\rho (E\mathrm{△}C)<\delta$.

Density in the product space. Let $h$ be a class in $L^2(\rho ;\mathbb{C})$ and let $\epsilon >0$. By [F6] with $p=2$ there is a complex finite simple function $s$ with $\mathrm{\parallel }h-s{\mathrm{\parallel }}_2<\epsilon \mathrm{/}2$. If $s=0$, take the zero rectangle combination. Otherwise write $s={\sum }_{j<m}{c}_j{\mathbf{1}}_{E_j}$ using only its nonzero values, so every $c_j\mathrm{\ne }0$ and every $E_j$ has finite measure; put $B:={\sum }_{j<m}\mathrm{\mid }{c}_j\mathrm{\mid }>0$ and $\delta :=(\epsilon \mathrm{/}(2B){)}^2$. For each $j$, [step 3.1] gives a set $C_j$ that is a finite union of finite-measure rectangles and satisfies $\mathrm{\parallel }{\mathbf{1}}_{E_j}-{\mathbf{1}}_{C_j}{\mathrm{\parallel }}_2<\epsilon \mathrm{/}(2B)$. Then $R:={\sum }_j{c}_j{\mathbf{1}}_{C_j}$ is a finite complex linear combination of rectangle kernels and the triangle inequality gives $\mathrm{\parallel }h-R{\mathrm{\parallel }}_2\le \mathrm{\parallel }h-s{\mathrm{\parallel }}_2+B{\max }_j\mathrm{\parallel }{\mathbf{1}}_{E_j}-{\mathbf{1}}_{C_j}{\mathrm{\parallel }}_2<\epsilon$.

Density in the completed space. Let $h$ be a class in $L^2(\overline{\mu \times \nu };\mathbb{C})$ and let $\epsilon >0$. By [F6] applied in the completed measure space there is a complex finite simple function $s$ with $\mathrm{\parallel }h-s{\mathrm{\parallel }}_{\overline{\mu \times \nu }}<\epsilon \mathrm{/}2$. If $s=0$, take the zero rectangle combination. Otherwise, using the same nonzero-value representation and coefficient bookkeeping as in [step 4.1], write $s={\sum }_{j<m}{c}_j{\mathbf{1}}_{E_j}$ with every $c_j\mathrm{\ne }0$ and every $E_j$ of finite completed measure, and put $B:={\sum }_{j<m}\mathrm{\mid }{c}_j\mathrm{\mid }>0$ and ${\delta }^{\mathrm{\prime }}:=(\epsilon \mathrm{/}(2B){)}^2$. For each $j$, [F7] applied to the indicator of $E_j$ provides $A_j\in \mathcal{A}\otimes \mathcal{B}$ with $\overline{\mu \times \nu }(E_j\mathrm{△}{A}_j)=0$, hence $\rho (A_j)=\overline{\mu \times \nu }(A_j)=\overline{\mu \times \nu }(E_j)<+\mathrm{\infty }$ by [F2]. By [step 2.1] there is a set $C_j$ that is a finite union of finite-measure rectangles with $\rho (A_j\mathrm{△}{C}_j)<{\delta }^{\mathrm{\prime }}$, so the classes satisfy $\mathrm{\parallel }{\mathbf{1}}_{E_j}-{\mathbf{1}}_{C_j}{\mathrm{\parallel }}_{\overline{\mu \times \nu }}^2=\overline{\mu \times \nu }(E_j\mathrm{△}{C}_j)\le \overline{\mu \times \nu }(E_j\mathrm{△}{A}_j)+\rho (A_j\mathrm{△}{C}_j)<{\delta }^{\mathrm{\prime }}$ by [F3] and [F8]. The triangle inequality in the completed space therefore gives $\mathrm{\parallel }h-{\sum }_j{c}_j{\mathbf{1}}_{C_j}{\mathrm{\parallel }}_{\overline{\mu \times \nu }}<\epsilon$, and the approximant is a finite complex linear combination of rectangle kernels.

Steps 4.1 and 5.1 give the two density assertions of the statement, for an arbitrary class and arbitrary positive tolerance in each of the two spaces; all approximations are finite complex linear combinations of rectangle kernels with $\mu (A)<+\mathrm{\infty }$ and $\nu (B)<+\mathrm{\infty }$.