If $C\in \sigma \{A\times B\mid A\in \mathcal A, B\in \mathcal B\}$, are there $A_n\times B_n$ s.t. $1_{A_n\times B_n}\nearrow 1_C$? Let $(X,\mathcal A)$ and $(Y,\mathcal B)$ measure spaces. Consider the product space $(X\times Y,\mathcal A\otimes \mathcal B)$, where $$\mathcal A\otimes \mathcal B:=\sigma \{A\times B\mid A\in \mathcal A, B\in \mathcal B\}.$$
If $C\in \mathcal A\otimes \mathcal B$ are there $A_n\in \mathcal A$ and $B_n\in \mathcal B$ s.t. $$\boldsymbol 1_{A_n\times B_n}\nearrow \boldsymbol 1_C \ \ ?$$ 
An equivalent question could be : if $\mathcal F=\sigma (\mathcal A)$, and if $F\in \mathcal F$, can I find $A_n\in A$ s.t. $\boldsymbol 1_{A_n}\nearrow \boldsymbol 1_F$, and I'm not sure if this is really true... I'm asking this question because I'm trying to prove $$\mathbb E[\Phi(U,V)\mid \mathcal A]=\mathbb E[\Phi(u,V)]|_{u=U},$$ where $U$ is a r.v. on $X$ and $V$ a r.v. on $Y$. I was able to prove this for $\Phi(x,y)=f(x)g(y)$, and I would like to generalize this.
 A: No, it's in general not true, consider for instance $X=Y$ and the diagonal $C=\{(x,x); x \in X\}$. (Just as a side remark: There are "bad" measure spaces where the diagonal does not belong to the $\mathcal{A} \otimes \mathcal{A}$ but for many measure spaces it does - so just pick one of these nice measure spaces, e.g. $X=Y=\mathbb{R}$ with the Borel sets, to get a counterexample.)
There is a somewhat weaker statement which is true: For any $C \in \mathcal{A} \otimes \mathcal{B}$ there exist sequences $(A_n)_{n \in \mathbb{N}} \subseteq \mathcal{A}$ and $(B_n)_{n \in \mathbb{N}} \subseteq \mathcal{B}$ such that $$C \in \sigma(A_n \times B_n; n \in \mathbb{N}).$$ This result can be quite helpful when dealing with sets in the product $\sigma$-algebra.
To come back to your original question: Define $$\mathcal{D} := \{C \in \mathcal{A} \otimes \mathcal{B}; \mathbb{E}(1_C(U,V) \mid \mathcal{A}) = \mathbb{E}(1_C(u,V)) |_{u=U}\}.$$
You have already shown that $A \times B \in \mathcal{D}$ for any $A \in \mathcal{A}$, $B \in \mathcal{B}$. Check that $\mathcal{D}$ is a Dynkin system. Since $\mathcal{A} \times \mathcal{B}$ is a $\cap$-stable generator of $\mathcal{A} \otimes \mathcal{B}$, this implies $\mathcal{A} \otimes \mathcal{B} = \sigma(\mathcal{A} \times \mathcal{B}) \subset \mathcal{D}$.
A: No, imagine $X=Y=\mathbb R$ (with the Borel-$\sigma$-algebra) and $C$ is the union of two rectangles (that are not accidentally lined up or something like this), then it can never be approximated by sets of the form $A\times B$.
I think intuition is more important than a proof at this point, so I drew a picture which shows how sets of the form $A\times B$ look like compared to $C$.
(Your assertion with the expected values might still be true, I didn't look into it.)

A: This will only be true if one of the measure spaces is trivial. Indeed, suppose that $\mathcal{A}\neq \{\emptyset, X\}$ and $\mathcal{B}\neq \{\emptyset, Y\}$ and pick sets $A\in \mathcal{A}\backslash \{\emptyset, X\}$ and $B\in \mathcal{B}\backslash \{\emptyset, Y\}$. Then $A \times B^C\cup A^C \times B\in \mathcal{A} \otimes\mathcal{B}$. If there exists a sequences $\{A_n\}_n\subset \mathcal{A}$ and $\{B_n\}_n\subset \mathcal{B}$ such that $1_{A_n \times B_n}\uparrow 1_{A \times B^C\cup A^C \times B}$, then if $p_1$ denotes the projection $X \times Y \rightarrow X$ onto the first factor and $p_2$ denotes the projection $X \times Y \rightarrow Y$ onto the s second factor, we will have $1_{A_n}=1_{p_1(A_n \times B_n)} \uparrow 1_{p_1(A \times B^C\cup A^C \times B)}=1_X$ and similarly $1_{B_n}\uparrow 1_Y$. This implies that $A_n \times B_n$ will eventually contain a point in $A\times B$ which contradicts the assumption that $1_{A_n \times B_n}\leq 1_{A \times B^C\cup A^C \times B}$ for all $n$. Hence $A \times B^C\cup A^C \times B$ cannot be approximated from below by cartesian products.
