Please help me with the following proof on Borel σ-algebra 



Please help me with the following proof on Borel $\sigma$-algebra
$$\mathcal{B}(\mathbb{R^2})=\sigma(\{(a,b)\times(c,d); a,b,c,d\in \mathbb{R}\})$$
show $\mathcal{B}(\mathbb{R^2})=\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$, where $\mathcal{B}(\mathbb{R})$ is the Borel $\sigma$-algebra.


 A: Hint:
$$
\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})=\sigma(\{A\times B\,:\,A\in\mathcal{B}(\mathbb{R}),\,B\in \mathcal{B}(\mathbb{R})\})
$$
Now show both inclusions.

To show that $\mathcal{B}(\mathbb{R})\times\mathcal{B}(\mathbb{R})\subseteq \mathcal{B}(\mathbb{R}^2)$, we could the use the following result:
Let $(X,\mathcal{E})$ and $(Y,\mathcal{F})$ be measurable spaces. Then the product sigma-algebra is by definition given by
$$
\mathcal{E}\otimes\mathcal{F}=\sigma(\{A\times B: A\in\mathcal{E},\,B\in\mathcal{F}\}).
$$

Suppose $\mathcal{C}$ and $\mathcal{D}$ are generators for the two sigma-algebras $\mathcal{E}$ and $\mathcal{F}$, i.e. $\sigma(\mathcal{C})=\mathcal{E}$ and $\sigma(\mathcal{D})=\mathcal{F}$ and assume that there exists sequencec $(C_n)$ and $(D_n)$ of sets from $\mathcal{C}$ and $\mathcal{D}$ respectively such that
  $$
X=\bigcup_n C_n\quad\text{and}\quad Y=\bigcup_nD_n.
$$
  Then
  $$
\mathcal{E}\otimes \mathcal{F}=\sigma(\left\{A\times B:A\in\mathcal{C},\,B\in\mathcal{D}\right\}).
$$

Now choose $\mathcal{C}$ and $\mathcal{D}$ according to your setup.
