I was doing a theorem and I got stuck in this part.
Let us $(\Omega_1,S_1,\mu_1)$ and $(\Omega_2,S_2,\mu_2)$ be sigma-finite measure space i.e $\exists A_n\in F_1(S_1\times S_2)$ such that $A_n\uparrow$ and $\Omega_1=\cup_{n\geq1}A_n$ and $\mu_1(A_n)<\infty$ where $F_1(S_1\times S_2)$ is algebra generated by $S_1\times S_2$ and similarly $\exists B_n\in F_2(S_1\times S_2)$ such that $B_n \uparrow$ and $\Omega_2=\cup_{n\geq1}B_n$ and $\mu_1(B_n)<\infty$ where $F_2(S_1\times S_2)$ is algebra generated by $S_{1} \times S_{2}$
Let assume $E_n=A_n\times B_n\in S_1\times S_2$ then $\cup_{n\geq 1}E_n=\Omega_1\times \Omega_2$
I understood how to prove this but my question is what if I don't take $A_n \uparrow$ and $B_n \uparrow$? will the statement still be true? If not what will be the problem? since in sigma-finite measure, I can take $A_n's$ such that they are disjoint and $\cup_{n\geq1}A_n=\Omega_1$ and $\mu_1(A_n)<\infty$.