Let $(\Omega_1,\mathcal{A}_1,\mu_1)$ and $(\Omega_2,\mathcal{A}_2,\mu_2)$ be two measure spaces, where $\mu_1$ and $\mu_2$ are $\sigma$-finite. Consider the product space $(\Omega=\Omega_1\times\Omega_2,\mathcal{A}=\mathcal{A}_1\oplus\mathcal{A}_2,\mu=\mu_1\times\mu_2)$. I want to prove that the set $$\mathcal{C}=\{A\in\mathcal{A}:\,\mu(A)=\int_{\Omega_1}\int_{\Omega_2}1_A\,d\mu_2\,d\mu_1\}$$ is a monotone class (that is, closed under increasing and decreasing sequences of sets), using Lebesgue convergence theorems (but not Fubini).
My attempt: the fact that $\mathcal{C}$ is closed under unions is clear by the monotone convergence theorem and the fact that $\mu(\cup A_n)=\lim_n\mu(A_n)$. My problem arises when dealing with intersections. In this case, we may not have $\mu(\cap A_n)=\lim_n\mu(A_n)$, as we do not know whether $\mu(A_1)<\infty$. Here is where I need the $\sigma$-finiteness. Because of it, I can write $\Omega=\cup_{n=1}^{\infty}A_n$, where $Q(A_n)<\infty$ for all $n$.
Let $\{B_n\}_n\subseteq\mathcal{C}$ with $B=\cap_n B_n$. I want to show that $B\in\mathcal{C}$. We have $\cap_n (B_n\cap A_m)=B\cap A_m$, with $\mu(B_n\cap A_n)\leq \mu(A_m)<\infty$, so now we do have $\mu(B\cap A_m)=\lim_n\mu(B_n \cap A_m)$.
The idea now would be to have $B\cap A_m\in\mathcal{C}$ using a convergence theorem and then use the fact that $B=\cup_m(B\cap A_m)$ to obtain $B\in\mathcal{C}$.
For that purpose, I would like to prove that $B_n\cap A_m\in\mathcal{C}$ from the fact that $B_n\in\mathcal{C}$ and that $A_m$ can be chosen as a rectangle in $\mathcal{A}$. Any ideas?