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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?

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  • $\begingroup$ It's better to write $\Omega_i = \bigsqcup_{n\ge 1} A^i_n$ with $\mu_i(A_n^i)<\infty$, $i=1,2$, so that $\Omega = \bigsqcup_{n,m\ge 1} A^1_n\times A^2_m$. Can you finish? $\endgroup$ – zhoraster Oct 27 '16 at 19:01
  • $\begingroup$ @zhoraster Do you mean that the $A_m$ from my question can be chosen as a rectangle? I don't see how I can use that to prove $B_n\cap A_m\in\mathcal{C}$. Could you give me another hint? $\endgroup$ – user39756 Oct 27 '16 at 19:14
  • $\begingroup$ Well, this proof can be found in most textbooks when proving Fubini. $\endgroup$ – Cave Johnson Oct 28 '16 at 3:24
  • $\begingroup$ @CaveJohnson Could you tell me a reference where that $\mathcal{C}$ is proved to be a monotone class? $\endgroup$ – user39756 Oct 28 '16 at 6:23
  • $\begingroup$ Chapter 8 of Rudin's Real and Complex Analysis has a proof, where he established Fubini's theorem :) $\endgroup$ – Cave Johnson Oct 28 '16 at 8:49

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