On tensor products of (essentially) bounded measurable functions

Question

Let $F:\mathbb R^2\rightarrow\mathbb R$ be an essentially bounded measurable function ($\mathbb R^d$ is equipped with its standard Lebesgue measure) and assume that $F(x,y)=F(y,x)$. I would like to write $$ F(x,y)=\sum_{\nu, \mu} a_{\mu,\nu}\mathbf 1_{A_\mu}(x)\mathbf 1_{A_\nu}(y), \text{ with $a_{\mu,\nu}=a_{\nu,\mu}$ and $\bigcup_\mu A_\mu=\mathbb R$, measure$(A_\mu\cap A_\nu)=\delta_{\mu, \nu}$.} $$ Is it possible?

Answer 1

Let $\mu$ be a probability measure on $\mathbb R$ equivalent to the Lebesgue measure (for instance, the Gaussian measure). Then $A_\mu$ is a partition of $(\mathbb R,\mu)$ modulo null sets, and is therefore countable; so $\mu,\nu$ range over a countable set. Furthermore, throwing away intersections of zero measure, one can assume that the sets $A_\mu$ are disjoint. But then $F$ is equal in $L^\infty$ to a function which has the following property: for almost all $y\in\mathbb R$, the function $$ x\mapsto F(x,y) $$ is a step function (there will be only one $\nu_0$ which contributes to the sum). This is, of course, quite a severe restriction, so by far not all functions will have this property.