# The Borel sigma-algebra on $\mathbb{R}^2$.

Let $Bo(\mathbb{R}^2)$, be the borel-sigma algebra on $\mathbb{R}^2$, it is the sigma algebra generated by the open sets on $\mathbb{R}^2$.

Let $\mathcal{B}$ be the borel sigma algebra on $\mathbb{R}$. Let $\mathcal{B}\otimes\mathcal{B}$ be the sigma-algebra on $\mathbb{R}^2$, generated by the set $\{B_1\times B_2|,B_1,B_2 \in \mathcal{B} \}.$

I am trying to prove that $Bo(\mathbb{R}^2)=\mathcal{B}\otimes\mathcal{B}$. Now the implication $Bo(\mathbb{R}^2)\subset \mathcal{B}\otimes \mathcal{B}$, is ok:

$\mathcal{B}\otimes \mathcal{B}$ is a sigma algebra, and for any two intervals in $I_1,I_2 \subset\mathbb{R}$, $I_1,I_2 \in \mathcal{B}$, so $I_1\times I_2 \in \mathcal{B}\otimes\mathcal{B}$.Since any open set in $\mathbb{R}^2$ is a countable union of open rectangles, $\mathcal{B}\otimes \mathcal{B}$ must contain the open sets of $\mathbb{R}^2$, since it is a $\sigma$-algebra, it must contain the smallest sigma algebra on $\mathbb{R}^2$ containing the open sets, which is $Bo(\mathbb{R}^2)$, hence $Bo(\mathbb{R}^2)\subset \mathcal{B}\otimes \mathcal{B}$.

But how do I prove the reverse implication, that is: $\mathcal{B}\otimes\mathcal{B}\subset Bo(\mathbb{R}^2)$? If I can prove that for any two sets $B_1,B_2 \in \mathcal{B}$, $B_1 \times B_2\in Bo(\mathbb{R}^2)$, I will be done. But how do I prove this? All we know about the sets $B_1,B_2$ is that they are in $\mathcal{B}$, which is the smallest sigma-algebra on the real line, containing the open sets. but this doesn't really tell us much about how the sets $B_1,B_2$ are.

Do you guys have any tips?

• Do you know that $\mathcal{B}\otimes\mathcal{B}$ could be generated by $\{U_1\times U_2: U_1, U_2\ are \ open\}$ ? – mac Oct 6 '15 at 16:29
• Any open set in the plane is a union of open rectangles. Hence it is a countable union of open rectangles (consider, say, rectangles such that both coordinates of each vertex are rational). – David C. Ullrich Oct 6 '15 at 16:52