Let $\Omega\subset\mathbb R$ be open and bounded. For continuous functions $C(\Omega\times\Omega)$, the Stone-Weierstrass theorem shows that the products $a(x)b(y)$ of univariate continuous functions in $C(\Omega)$ span a dense subspace. However, for bounded functions in $L^\infty(\Omega\times\Omega)$, the only subspace I can think of is spanned by the indicator functions $\chi_A(x,y)$, where $A\subset\Omega\times\Omega$ is measurable.
My question: is there a more convenient dense subspace of $L^\infty(\Omega\times\Omega)$? In particular, do the products $a(x)b(y)$ of $L^\infty(\Omega)$ functions span a dense subspace?
Motivation: I am considering the $L^2$ continuity of pseudo-differential operators with $L^\infty$ symbols $a$: $$ Au(x)=\int e^{i(x-y)\cdot\xi}a(x,y,\xi)u(y)dyd\xi. $$ If $a(x,y,\xi)\sim \sum a(x)b(y)c(\xi)$, such as if $a(x,y,\xi)$ is bounded and continuous, then $L^2$ continuity follows immediately from the properties of $L^\infty$ multiplier operators. But without this product structure, I am not sure where to go.
$L^\infty$ functions can be approximated by simple functions, i.e., finite linear combinations of characteristic functions.
see Uniform convergence of simple functions to a bounded function $f$
The function $$ f(x,y) =\begin{cases} 1 & x\le y\\ 0 & x>y \\ \end{cases} $$ cannot be approximated by functions of the type $a(x)b(y)$. Let me consider $f$ as measurable function on $[0,1]^2$. In order to have $\sup_{(x,y)}|f(x,y)-a(x)b(y)|<\epsilon$ it is necessary to have this for the corners of the square, i.e., $$ |1-a(0)b(0)|,|1-a(0)b(1)|,|1-a(1)b(1)|, |a(1)b(0)|< \epsilon. $$ This implies that all values $a(0),b(0),a(1),b(1)$ are not zero. And $a(0)$ is close to $b(0)^{-1}$, $b(1)^{-1}$, $a(1)$. Hence $a(1)b(0)$ is close to $1$, contradiction.