Problem about not intersection stable generators 
Disprove: Let $(\Omega_1,\cal F_1)$ and $(\Omega_2,\cal F_2)$ measure spaces and let $\cal D_1$ ,$\cal D_2$ generators of $\cal F_1$, $\cal F_2$ respectively, then $\cal D:=\{A \times B \mid A \in D_1, B \in D_2\}$ generates $\cal F_1 \otimes \cal F_2$. 

I think if $\cal D_1$ and $\cal D_2$ are intersection stable , then this statement is true. But whithout this condition?
A counter example is appreciated!
 A: Let me start with a general rule:
If $f:X\to Y$ and $\mathcal B\subseteq\wp(Y)$  then $\sigma(f^{-1}(\mathcal B))=f^{-1}(\sigma(\mathcal B))$.
Here $\sigma(\mathcal B)$ is a notation for the smallest $\sigma$-algebra that contains $\mathcal B$ as a subcollection.
See here for a proof of that rule.

$\cal F_1 \otimes \cal F_2$ is the smallest $\sigma$-algebra such that the projections $\pi_i:\Omega_1\times\Omega_2\to\Omega_i$ are both measurable.
Note that $\pi_i^{-1}(\mathcal D_i)\subseteq\mathcal D$ for $i=1,2$.
Then with the rule we find for $i=1,2$: $$\pi_i^{-1}(\mathcal F_i)=\pi_i^{-1}(\sigma(\mathcal D_i))=\sigma(\pi_i^{-1}(\mathcal D_i))\subseteq\sigma(\mathcal D)$$ 
So the functions $\pi_i$ are both measurable wrt $\mathcal\sigma(D)$ which means that $\cal F_1 \otimes \cal F_2\subseteq\sigma(\mathcal D)$.
Further it is evident that $\mathcal D\subseteq\cal F_1 \otimes \cal F_2$ so that also $\sigma(\mathcal D)\subseteq\cal F_1 \otimes \cal F_2$.
Proved is now:$$\sigma(\mathcal D)=\cal F_1 \otimes \cal F_2$$
