# Proof about Borel sigma-algebra on $\mathbb{R}^2$

I'm trying to understand the proof that the Borel $\sigma-$algebra generated by the rectangles in $\mathbb{R^2}$, $\mathcal{B}(\mathbb{R}^2)$ is equal to the $\sigma-$algebra generated by product of borel sets $\mathcal{B}(\mathbb{R}) \times \mathcal{B}(\mathbb{R}):= \sigma ( {B_1 \times B_2}, B_i \in \mathcal{B}(\mathbb{R}))$.

I get that since every rectange is a Borel set $\mathcal{B}(\mathbb{R}^2) \subset \mathcal{B}(\mathbb{R}) \times \mathcal{B}(\mathbb{R})$.

It's the other inclusion that I don't understand:

In my notes from the lecture our teacher defined these collection of sets,

$\mathcal{S}_1=\{ (a_1,b_1] \times (a_2,b_2]\}$

$\mathcal{S}_2=\{B_1 \times B_2; B_i \in\mathcal{B}(\mathbb{R}) \}$

$\widetilde{B}_1=B_1 \times \mathbb{R}, B_1 \in \mathcal{B}(\mathbb{R})$

$\widetilde{B}_2= \mathbb{R} \times B_2, B_2 \in \mathcal{B}(\mathbb{R})$

Then $\mathcal{B}(\mathbb{R}^2 )=\sigma(\mathcal{S}_1)$ and $\mathcal{B}(\mathbb{R}) \times \mathcal{B}(\mathbb{R})=\sigma(\mathcal{S}_2)$

Here is the first thing that I don't get: In my notes I've written $\widetilde{B}_1=B_1\times\mathbb{R} \in \sigma(\mathcal{S}_1) \times \mathbb{R} \stackrel{\text{this is the equality I don't understand}}{=} \sigma(\mathcal{S}_1 \times \mathbb{R})=\sigma(\widetilde{\mathcal{S}}_1)$

Then $\widetilde{\mathcal{S}}_1 \cap \widetilde{\mathcal{S}}_2=\mathcal{S}_1 \times \mathcal{S}_1=\mathcal{S}$ (I think the last equallity is by definition).

And then the last thing I don't understand is the inclusion: $\sigma(\widetilde{\mathcal{S}}_1\cap \widetilde{\mathcal{B}}_2) \subseteq \sigma(\widetilde{\mathcal{S}}_1\cap \widetilde{\mathcal{S}}_2)$.

If somebody could help me clarify these things I'd be very greatful!

The inclusion $\mathscr B(\mathbb R^2)\subseteq \mathscr B(\mathbb R)\times \mathscr B(\mathbb R)$ is clear. For the other direction, note that if $A,B\in \mathscr B(\mathbb R)$, then $A\times B=(A\times \mathbb R)\cap (\mathbb R\times B)=\pi_1^{-1}(A)\cap \pi_2^{-1}(B)\in \mathscr B(\mathbb R^2)$ because the projections are continuous, hence measurable. Now since the $\sigma-$algebra generated by the rectangles $A\times B$ is precisely $\mathscr B(\mathbb R)\times \mathscr B(\mathbb R)$, we are done.
• I just have one question; doesn't the measurability of the projections depend on the topology on $\mathbb{R}$ and $\mathbb{R}^2$? – user202542 Dec 24 '16 at 9:13
• The Borel $\sigma-$algebra on a space $X$ is the smallest one that contains the open sets in $X$ so yes it does. A function is Borel-measurable if the preimage of every open set is Borel. So, for example, if we take $\mathbb R^2$ with the indiscrete topology, then $\pi_1^{-1}((-\infty, ])$ is not Borel. – Matematleta Dec 24 '16 at 16:16