Product rectangle kernels are dense in complex l two

Apply the representative theorem separately to the real and imaginary parts of $k$. The resulting base-measurable functions equal those components a.e.; their sets of infinite values are base-measurable and null. Replacing their infinite values by zero and combining them gives a finite complex product-measurable representative $k_0$ of $k$. Its norm is unchanged. AC supplies the countable choice required in this step.

The uncompleted product has finite mass $\mu (X)\nu (Y)$. On it choose a finite simple function $s={\sum }_{j=1}^m{c}_j{\mathbf{1}}_{E_j}$ with $\mathrm{\parallel }{k}_0-s{\mathrm{\parallel }}_2<\epsilon \mathrm{/}2$. Terms with zero coefficient may be removed. If no terms remain, the zero rectangle combination already approximates $k$ within $\epsilon$.

Otherwise put $B={\sum }_{j=1}^m\mathrm{\mid }{c}_j\mathrm{\mid }>0$ and $\delta =(\epsilon \mathrm{/}(2B){)}^2$. By the generating-algebra approximation choose, for each of these finitely many $j$, a finite disjoint rectangle union $R_j$ with $(\mu \otimes \nu )(E_j\mathrm{△}{R}_j)<\delta$. Since $\mathrm{\parallel }{\mathbf{1}}_{E_j}-{\mathbf{1}}_{R_j}{\mathrm{\parallel }}_2$ is the square root of that measure, the triangle inequality gives $\mathrm{\parallel }s-{\sum }_j{c}_j{\mathbf{1}}_{R_j}{\mathrm{\parallel }}_2<B\sqrt{\delta }=\epsilon \mathrm{/}2$. Each ${\mathbf{1}}_{R_j}$ is a finite sum of disjoint rectangle indicators. Combined with step 2.1 this proves density, on the completion as well because the norms of base-measurable functions agree with their completed norms.

If $⟨k,{\mathbf{1}}_{E\times F}⟩=0$ for every rectangle, conjugate-linearity gives $⟨k,R⟩=0$ for every finite complex rectangle combination $R$. For each $\epsilon >0$ choose such $R$ with $\mathrm{\parallel }k-R{\mathrm{\parallel }}_2<\epsilon$ by step 3.1. Then $\mathrm{\parallel }k{\mathrm{\parallel }}_2^2=\mathrm{\mid }⟨k,k-R⟩\mathrm{\mid }\le \mathrm{\parallel }k{\mathrm{\parallel }}_2\epsilon$. If $\mathrm{\parallel }k{\mathrm{\parallel }}_2>0$, division and arbitrarily small $\epsilon$ give a contradiction; thus $k=0$. When either factor has zero total measure, every class is zero and all assertions hold with $R=0$. No exchange of a universal test-function quantifier with an a.e. section quantifier is used.