Stone Weierstrass Overkill in the Measurable Setting? If $\mu$ is Lebesgue measure on the Borel sigma algebra $\mathcal{B}$ of $[0,1]$. Establishing that the linear span in $L^{2}([0,1]\times[0,1],\mathcal{B}\otimes\mathcal{B},d(\mu \times \mu))$ of the set of measurable functions of the form $(f \otimes g)(x,y)=f(x)g(y)$ where $f\in L^{\infty}([0,1],d\mu)$ and $g\in L^{\infty}([0,1],d\mu)$ is dense is a standard application of the Stone-Weierstrass Theorem. 
I've always felt that the presence of so many characteristic functions in the dense algebra mentioned above ought to make the use of Stone-Weierstrass overkill to obtain the above result.
I wonder if it is easy to use the outer regularity of the Lebesgue measure on $[0,1]$ along with the fact that any open subset of $[0,1]$ is a countable union of pairwise disjoint open intervals to directly prove the above result without the use of the Stone-Weierstrass theorem, perhaps by proving that functions can be approximated adequately using simple functions supported on elementary sets (finite unions of pairwise disjoint measurable rectangles).
Question: What is an easy elementary proof of the above density result that does not require or reduce essentially to the use of the Stone-Weierstrass theorem?
Filling in the details of the proposed strategy of Davide below would be great!
 A: $1.$ The functional monotone class theorem mentioned by Martin works indeed perfectly well to prove the result. You can also prove it directly with the "standard" monotone class theorem, as follows: consider the family of all Borel sets $A\subset [0,1]\times [0,1]$ whose characteristic function can be approximated by linear combinations of functions of the form $f\otimes g$, and check that this is a monotone class containing all Borel rectangles. Although I didnt fill in all the details, I'm pretty sure that the "approximation by simple functions" result mentioned by Davide can be proved in the same way (and I don't see any way of avoiding a monotone class argument).
$2.$ You can also prove the result by duality. Let $\Phi\in L^2([0,1]\times [0,1])$ be orthogonal to all $f\otimes g$: we want to show that $\Phi=0$.
Fix once and for all a representative for $\Phi$ (i.e. a true Borel function). Note that for almost every $x\in [0,1]$, the function $\Phi_x(y)=\Phi(x,y)$ is in $L^2([0,1])$. By Fubini's theorem, it is enough to show that $\Phi_x(y)$ is a.e. $0$, for almost every $x\in [0,1]$.
If you take any $f\in L^\infty([0,1])$, then 
$u_f(y)=\int_{[0,1]} \Phi(x,y)f(x)dx$ is well-defined for all $y\in [0,1]$, the function $u_f$ is measurable (which is not trival) and in $L^2([0,1])$ (which is easy); and by Fubini, $u_f$ is orthogonal to all $g\in L^\infty([0,1])$. So $u_f(y)$ is $0$ almost everywhere (because $L^\infty$ is dense in $L^2$ or, if you prefer, because $\int_B u_f=0$ for any Borel set $B\subset [0,1]$). This means the following: for any fixed $f\in L^\infty$, we have $\langle \Phi_x,f\rangle_{L^2}=0$ for almost every $x\in [0,1]$. Now take a sequence $(f_n)_{n\in\mathbb N}$ which is dense in $L^2([0,1])$ (which is possible since your $L^2$ is separable and $L^\infty$ is dense in $L^2$): this leads you to the conclusion that for almost every $x\in [0,1]$, the function $\Phi_x$ is orthogonal to all $f_n$, i.e. $\Phi_x=0$ as an element of $L^2([0,1])$.
A: First we prove that characteristic functions of measurable sets of $[0,1]^2$ can be approximated in $L^2$ by the characteristic function of a disjoint finite union $\bigsqcup_{i\in F}A_i\times B_i$, with $A_i,B_i$ measurable subsets of $[0,1]$. 
Then we prove it for simple functions, and for functions of $L^2$.
