Let (Ω1,A1,μ1)(\Omega_1,\mathcal{A}_1,\mu_1) and (Ω2,A2,μ2)(\Omega_2,\mathcal{A}_2,\mu_2) be two measure spaces, where μ1\mu_1 and μ2\mu_2 are σ\sigma-finite. Consider the product space (Ω=Ω1×Ω2,A=A1⊕A2,μ=μ1×μ2)(\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 C={A∈A:μ(A)=∫Ω1∫Ω21Adμ2dμ1}\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 C\mathcal{C} is closed under unions is clear by the monotone convergence theorem and the fact that μ(∪An)=limnμ(An)\mu(\cup A_n)=\lim_n\mu(A_n). My problem arises when dealing with intersections. In this case, we may not have μ(∩An)=limnμ(An)\mu(\cap A_n)=\lim_n\mu(A_n), as we do not know whether μ(A1)<∞\mu(A_1)<\infty. Here is where I need the σ\sigma-finiteness. Because of it, I can write Ω=∪∞n=1An\Omega=\cup_{n=1}^{\infty}A_n, where Q(An)<∞Q(A_n)<\infty for all nn. Let {Bn}n⊆C\{B_n\}_n\subseteq\mathcal{C} with B=∩nBnB=\cap_n B_n. I want to show that B∈CB\in\mathcal{C}. We have ∩n(Bn∩Am)=B∩Am\cap_n (B_n\cap A_m)=B\cap A_m, with μ(Bn∩An)≤μ(Am)<∞\mu(B_n\cap A_n)\leq \mu(A_m)<\infty, so now we do have μ(B∩Am)=limnμ(Bn∩Am)\mu(B\cap A_m)=\lim_n\mu(B_n \cap A_m). The idea now would be to have B∩Am∈CB\cap A_m\in\mathcal{C} using a convergence theorem and then use the fact that B=∪m(B∩Am)B=\cup_m(B\cap A_m) to obtain B∈CB\in\mathcal{C}. For that purpose, I would like to prove that Bn∩Am∈CB_n\cap A_m\in\mathcal{C} from the fact that Bn∈CB_n\in\mathcal{C} and that AmA_m can be chosen as a rectangle in A\mathcal{A}. Any ideas? ================= ================= =================