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?

  • 1
    $\begingroup$ On the first mark: do you agree with/understand $\tilde{\mathcal B}_1=\sigma(\tilde{\mathcal S}_1)$? $\endgroup$
    – drhab
    Commented Jan 4, 2019 at 14:39
  • $\begingroup$ @drhab Yes, I can see that being true. $\endgroup$
    – MSDG
    Commented Jan 4, 2019 at 14:44
  • 1
    $\begingroup$ It's a typo. The $B$ should be $\mathcal B$. This proof seems overly laborious especially for the $n=2$ case. $\endgroup$ Commented Jan 4, 2019 at 16:25
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Jan 4, 2019 at 17:14
  • 1
    $\begingroup$ @MisterRiemann As you say: every topological space goes along with a Borel $\sigma$-algebra. I think we are on the same line here. $\endgroup$
    – drhab
    Commented Jan 6, 2019 at 9:51


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