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The following contents are from my class lecture:

Let $(\Omega_1,\mathcal{A}_1)$ and $(\Omega_2,\mathcal{A}_2)$ be two measurable space, and let $\pi_i:\Omega_1\times\Omega_2\rightarrow\Omega_i$ be the $i$th projection. Set $$\mathcal{C}:=\pi_1^{-1}(\mathcal{A}_1)\cup \pi_2^{-1}(\mathcal{A}_2)$$ $$\mathcal{R}:=\{A_1\times A_2:A_1\in\mathcal{A}_1,A_2\in\mathcal{A}_2\}$$ $$\mathcal{U}=\left\{\sum_{j\in J}{R_j}:J\ \text{finite}, {R}_j\in\mathcal{R}\ \text{for each }j\right\}$$ I was ask to show that $\mathcal{U}$ is a field generated by $\mathcal{C}$, but I'm quite confused about what $\mathcal{C}$ looks like. In my opinion, $\mathcal{C}$ is like $$\mathcal{C}=\left\{A_1\times B_2:A_1\in\mathcal{A}_1,B_2\subset\Omega_2\right\}\cup \left\{B_1\times A_2:A_2\in\mathcal{A}_2,B_1\subset \Omega_1\right\}$$ but then, I think that I don't even have $\mathcal{C}\subset \mathcal{U}$.

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First, I wanted to check that there's a typo: \begin{equation} \mathcal{U} = \left\{ \bigcup_j R_j : \dots \right \} \end{equation}

Second, $\mathcal{C}$ contains all sets of the form $(A_1 \times \Omega_2)$ and $(\Omega_1 \times A_2)$ where $A_1 \in \mathcal{A}_1$ and $A_2 \in \mathcal{A}_2$. (Take one element $A_1 \in \mathcal{A}_1$. Then note that $\pi_1^{-1}(A_1) = (A_1 \times \Omega_2)$.)

Intuitively, one intersection would give you any element in $\mathcal{R}$ and finite unions would get you $\mathcal{U}$. You can argue the necessity of any set in $\mathcal{U}$, and you can easily show it is a field, so you're done.

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