Sigma algebra on a Cartesian product $\Omega_1$ and $\Omega_2$ are countable sets. With $\mathcal P(\cdot)$ we denote a power set of a set.
We need to proof that: 
$$\mathcal P(\Omega_1)\otimes \mathcal P(\Omega_2)=\mathcal P(\Omega_1 \times \Omega_2)$$ 
where $\mathcal P(\Omega_1)\otimes \mathcal P(\Omega_2)=\sigma (\{B_1 \times B_2 | B_1 \in \Omega_1,B_2 \in \Omega_2 \})$.
 A: These sigma algebras are generated by singleton sets since the ground set is countable. The singleton sets $\{b_1\}\times \{b_2\}=\{b_1\times b_2\}$ are also contained in the product sigma algebra. A set $A\in\mathcal P(\Omega_1\times \Omega_2)$ is a countable union of singleton sets, hence is in the product sigma algebra. The opposite containment is obvious, so we are done.
A: 1) Let show that $\mathcal P(\Omega_1)\otimes \mathcal P(\Omega_2) \subset \mathcal P(\Omega_1 \times \Omega_2)$. 
Clear enough $\{B_1 \times B_2 | B_1 \in \Omega_1,B_2 \in \Omega_2 \} \subset  \mathcal P(\Omega_1 \times \Omega_2)$, thus $\sigma(\{B_1 \times B_2 | B_1 \in \Omega_1,B_2 \in \Omega_2 \}) = \mathcal P(\Omega_1)\otimes \mathcal P(\Omega_2) \subset \sigma(\mathcal P(\Omega_1 \times \Omega_2))=\mathcal P(\Omega_1 \times \Omega_2)$. 
2) Let show that $\mathcal P(\Omega_1)\otimes \mathcal P(\Omega_2) \supset \mathcal P(\Omega_1 \times \Omega_2)$.
$A \in \mathcal P(\Omega_1 \times \Omega_2)$, $A$ is a countable set of tuples $(\omega_1,\omega_2)$ such that $\omega_1 \in \Omega_1$ and $\omega_2 \in \Omega_2$. Thus $A = \bigcup_{n=1}^{\infty}\{(\omega_{1,n},\omega_{2,n})\}=\bigcup_{n=1}^{\infty} A_n$ where $A_i\bigcap A_j=\emptyset$. $\forall A_i=\{(\omega_{1,i},\omega_{2,i})\}=\{\omega_{1,i}\} \times \{\omega_{2,i}\} \in \{B_1 \times B_2 | B_1 \in \Omega_1,B_2 \in \Omega_2 \} \subset \mathcal P(\Omega_1)\otimes \mathcal P(\Omega_2)$. Because $\mathcal P(\Omega_1)\otimes \mathcal P(\Omega_2)$ is a $\sigma$-algebra, from $A_i \in \mathcal P(\Omega_1)\otimes \mathcal P(\Omega_2)$ follows that $\bigcup_{n=1}^{\infty} A_n \in \mathcal P(\Omega_1)\otimes \mathcal P(\Omega_2)$. So we have shown $\mathcal P(\Omega_1)\otimes \mathcal P(\Omega_2) \supset \mathcal P(\Omega_1 \times \Omega_2)$.
3) From 1) and 2) follows the equality.    
