Product of $\sigma -$algebra. Let $\mathbb F$ and $\mathbb G$ two $\sigma -$algebra. How is defined $\mathcal F\times \mathcal G$ ? Is it $$\{A\times B\mid (A,B)\in \mathcal F\times \mathcal G\}\ \ ?$$
I have to prove that $$\mathcal B(\mathbb R^2)=\mathcal B(\mathbb R)\times \mathcal B(\mathbb R)$$ where $\mathcal B$ is for borelian. I want to do as follow:
Let $A\in \mathcal B(\mathbb R^2)$. Then,
$$A=\bigcup_{i\in \mathbb N}O_i\times U_i=\left(\bigcup_{i}O_i\right)\times \left(\bigcup_{i}V_i\right)\in \mathcal B(\mathbb R)\times \mathbb B(\mathbb R),$$
where $O_i$ and $V_i$ are open set. Reciprocally, if $B\in \mathcal B(\mathbb R)\times \mathbb B(\mathbb R)$, then, $$B=\left(\bigcup_{i}O_i\right)\times\left(\bigcup_{i}V_i\right)=\bigcup_{i}O_i\times V_i\in \mathcal B(\mathbb R^2).$$
Therefore $$\mathcal B(\mathbb R^2)=\mathcal B(\mathbb R)\times \mathcal B(\mathbb R).$$
Is it correct ? 
 A: First of all see the comments of Sassatelli.
If $\langle X,\mathcal F\rangle$ and $\langle Y,\mathcal G\rangle$ are measurable spaces then $\langle X\times Y,\mathcal F\times\mathcal G\rangle$ denotes the measurable space that is characterized by the fact that $\mathcal F\times\mathcal G$ is the smallest $\sigma$-algebra on $X\times Y$ such that the projections $p_1:X\times Y\to X$ and $p_2:X\times Y\to Y$ are measurable.
Now let $\mathbb R^2$ be equipped with $\sigma$-algebra $\mathcal B(\mathbb R^2)$. Then the projections are both continuous hence measurable. Proved is now that: $$\mathcal B(\mathbb R)\times\mathcal B(\mathbb R)\subseteq\mathcal B(\mathbb R^2)$$
If $U$ is an open subset of $\mathbb R^2$ then we can write: $$U=\bigcup\{(a,b)\times(c,d)\in\wp(U)\mid a,b,c,d\in\mathbb Q\}$$ This is a countable union and if the projections are indeed measurable then $(a,b)\times(c,d)=p_1^{-1}((a,b))\cap p_2^{-1}((c,d))$ is measurable. 
This together proves that $U\in\mathcal B(\mathbb R)\times\mathcal B(\mathbb R)$, and since $\mathcal B(\mathbb R^2)$ is generated by the collection of open sets we are allowed to conclude that: $$\mathcal B(\mathbb R^2)\subseteq\mathcal B(\mathbb R)\times\mathcal B(\mathbb R)$$
