$$\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R}) = \mathcal{B}(\mathbb{R}^2)$$ To proof this, $$\mathcal{B}(\mathbb{R}^2)=\sigma\big[(a,b)\times(c,d): a,b,c,d \in \mathbb{R}\big]$$
and $(a,b)\times(c,d)$ is a set of $\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$. Hence, $$\mathcal{B}(\mathbb{R}^2)\subseteq \mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$$
My question is proof of other direction. There is a proof on the Billingsley's Probability and Measure and here. But in the proof of Billingsley I can not understand how to he show that $[B : A\times B \in \mathcal{B}(\mathbb{R}^2)]$ is a Borel set.
In the proof of other site, I can not understand the notion of Lebesgue-null set.
Is there any simple proof of my problem?
If $V$ is open then $\{A: A\times V \in \mathbb B (R^{2})\}$ is a sigma algebra containing all open sets. (Because $A \times V$ is open when $A$ is open). Hence it contains all Borel sets. This proves that $A \times V \in \mathbb B (R^{2})$ whenever $V$ is open. Next fix a Borel set $A$ in $\mathbb R$ and consider $\{B: A\times B \in \mathbb B (R^{2})\}$. This is a sigma algebra which contains all open sets, so it contains all Borel sets. This proves that $A\times V \in \mathbb B (R^{2})$ whenever $A$ and $B$ are Borel sets in $\mathbb R$.
[If you have difficulty in verifying that above two families are sigma algebras I will be glad to help].