# Borel $\sigma$-algebra on $\mathbb{R}^n$

I'm reading Shiryaev's Probability Theory, and I am unable to understand the proof of a result concerning the Borel $$\sigma$$-algebra on $$\mathbb{R}^n$$.

He defines the Borel $$\sigma$$-algebra $$\mathcal B(\mathbb{R}^n)$$ on $$\mathbb{R}^n$$ as the smallest $$\sigma$$-algebra containing the collection $$\mathcal{S}$$ of all rectangles $$S = S_1 \times \ldots \times S_n, \quad S_k = (a_k, b_k].$$

He then claims and proceeds to prove that $$\mathcal{B}(\mathbb{R}^n) = \mathcal{B}(\mathbb{R}) \otimes \ldots \otimes \mathcal{B}(\mathbb{R}),$$ where $$\mathcal{B}(\mathbb{R}) \otimes \ldots \otimes \mathcal{B}(\mathbb{R})$$ denotes the smallest $$\sigma$$-algebra generated by the Borel rectangles $$B = B_1 \times \ldots B_n, \quad B_k \in \mathcal{B}(\mathbb{R}),$$ i.e.\ $$\mathcal{B}(\mathbb{R}) \otimes \ldots \otimes \mathcal{B}(\mathbb{R}) = \sigma(\mathcal{B}(\mathbb{R}) \times \ldots \times \mathcal{B}(\mathbb{R}))$$

I don't understand the line of reasoning in the proof.

Before continuing, I would like to point out that I found some posts about this on MSE, namely [1] and [2], both of which provide quite nice arguments for this proof. However, I would like to understand how the proof in the book works.

Shiryaev only proves it for $$n=2$$, which clearly suffices by induction. First off, it is trivial that $$\mathcal B(\mathbb{R}^2) \subset \mathcal B(\mathbb{R}) \otimes B(\mathbb{R}),$$ since every rectangle is also a Borel rectangle. The converse is what confuses me. He denotes $$\tilde{\mathcal{B}}_1 = \mathcal{B}_1 \times \mathbb{R}, \quad \tilde{\mathcal{B}}_2 = \mathbb{R} \times \mathcal{B}_2,$$ and $$\tilde{\mathcal{S}}_1 = \mathcal{S}_1 \times \mathbb{R}, \quad \tilde{\mathcal{S}}_2 = \mathbb{R} \times \mathcal{S}_2,$$ where $$\mathcal{S}_{1}$$ and $$\mathcal{S}_{2}$$ are the systems of half-open intervals that form the sides of the rectangles from before. He then claims that for all $$B_1 \times B_2 \in \mathcal{B}(\mathbb{R}) \times \mathcal{B}(\mathbb{R})$$, we have \begin{align} B_1 \times B_2 = \tilde B_1 \cap \tilde B_2 \in \tilde{\mathcal{B}}_1 \cap \tilde{\mathcal{B}}_2, \end{align} where $$\tilde B_1 = B_1 \times \mathbb{R}$$ and $$\tilde B_2 = \mathbb{R} \times B_2$$, which I understand. However, then he proceeds by saying $$\tilde{\mathcal{B}}_1 \cap \tilde{\mathcal{B}}_2 \color{red}{=} \sigma(\tilde{\mathcal{S}}_1) \cap \tilde B_2 = \sigma(\tilde{\mathcal{S}}_1 \cap \tilde B_2) \color{red}{\subset} \sigma(\tilde{\mathcal{S}}_1 \cap \tilde{\mathcal{S}}_2) = \sigma(\mathcal{S}_1 \times \mathcal{S}_2) = \mathcal{B}(\mathbb{R}^2).$$ The steps marked in red are the ones that I cannot wrap my head around. Could someone clarify this for me?

• On the first mark: do you agree with/understand $\tilde{\mathcal B}_1=\sigma(\tilde{\mathcal S}_1)$? – drhab Jan 4 at 14:39
• @drhab Yes, I can see that being true. – MisterRiemann Jan 4 at 14:44
• It's a typo. The $B$ should be $\mathcal B$. This proof seems overly laborious especially for the $n=2$ case. – Matematleta Jan 4 at 16:25
• Thank you. The crux of this is to show that $\left \{ A\times \mathbb R :A\in \mathscr B(\mathbb R)\right \}\subseteq \mathscr B(\mathbb R^{2})$. You can do this by using projections, or just by checking directly, which is what Sirayev is doing I guess. – Matematleta Jan 4 at 17:14
• @MisterRiemann As you say: every topological space goes along with a Borel $\sigma$-algebra. I think we are on the same line here. – drhab Jan 6 at 9:51