Relationship between Borel Sigma algebras

Question

In general, is it true that $\mathcal{B}(\mathbb{R^n})=(\mathcal{B}(\mathbb{R}))^n:= \mathcal{B}(\mathbb{R}) \times \mathcal{B}(\mathbb{R})\times \dots \times\mathcal{B}(\mathbb{R})$?

I'd have to say yes, since we have $U_1,U_2\dots U_n$ open sets in $\mathbb{R}$ then their product is open in $\mathbb{R^n}$, so I'd suppose that an argument involving the sigma algebra generated by product of open sets and and the sigma algebra of all open sets in $\mathbb{R^n} $ should be the same. I'd appreciate some help in formalizing the argument though

Thanks

Answer 1

If $\mathcal V\subseteq\wp(X)$ for some set $X$ then in this answer $\sigma(\mathcal V)$ denotes the smallest $\sigma$-algebra and $\tau(\mathcal V)$ denotes the smallest topology that contains $\mathcal V$.

Important is that:$$\mathcal V\text{ countable}\implies\tau(\mathcal V)\subseteq\sigma(\mathcal V)\tag1$$

This because every open set can be written as a countable union of finite intersections of $\mathcal V$.

Let $\mathcal O(\mathbb R^n)$ denote the topology on $\mathbb R^n$ and let $$\mathcal V:=\left\{ U_{1}\times\cdots\times U_{n}\mid U_{1},\dots,U_{n}\in\mathcal W\right\} \subseteq\mathcal{O}\left(\mathbb{R}\right)^{n}\subseteq\mathcal{B}\left(\mathbb{R}\right)^{n}$$ where $\mathcal W$ denotes a countable base of $\mathcal O(\mathbb R)$.

Then $\mathcal V$ is a countable base of $\mathcal O(\mathbb R^n)$ and we have: $$\mathcal{B}\left(\mathbb{R}^{n}\right)=\sigma\left(\mathcal{O}\left(\mathbb{R}^{n}\right)\right)=\sigma\left(\tau\left(\mathcal{V}\right)\right)\subseteq\sigma\left(\sigma\left(\mathcal{V}\right)\right)=\sigma\left(\mathcal{V}\right)\subseteq\mathcal{B}\left(\mathbb{R}\right)^{n}$$ The inclusion in it is an application of $(1)$.

The other side $\mathcal B(\mathbb R)^n\subseteq\mathcal B(\mathbb R^n)$ is not difficult.

Answer 2

This answer shows that every open set $U \subseteq \mathbb{R}^n$ is a countable union of open rectangles $\langle a_1, b_1\rangle \times\cdots\times \langle a_n, b_n\rangle$. Each interval $\langle a_k, b_k\rangle \in \mathcal{B}(\mathbb{R})$ so the open rectangles are in $\mathcal{B}(\mathbb{R}) \times \cdots \times \mathcal{B}(\mathbb{R})$.

$$U = \bigcup_{i\in\mathbb{N}} \underbrace{\langle a_1^i, b_1^i\rangle \times\cdots\times \langle a_n^i, b_n^i\rangle}_{\mathcal{B}(\mathbb{R}) \times \cdots \times \mathcal{B}(\mathbb{R})} \in \mathcal{B}(\mathbb{R}) \times \cdots \times \mathcal{B}(\mathbb{R})$$

Hence $$\mathcal{B}(\mathbb{R}^n) = \sigma(\{U \subseteq \mathbb{R}^n : U \text{ open}\}) \subseteq \mathcal{B}(\mathbb{R}) \times \cdots \times \mathcal{B}(\mathbb{R})$$

For the converse inclusion, define $$\mathcal{A}_i = \{E \subseteq \mathbb{R} : \underbrace{\mathbb{R}^{i-1} \times E \times \mathbb{R}^{n-i}}_{\subseteq\mathbb{R}^n} \in \mathcal{B}(\mathbb{R}^n)\}$$

Check that $\mathcal{A}_i$ is a $\sigma$-algebra and notice that $\mathcal{A}_i$ clearly contains all open sets in $\mathbb{R}$. It follows that $\mathcal{B}(\mathbb{R}) \subseteq \mathcal{A}_i$.

Now, for $B_1, \ldots, B_n \in \mathcal{B}(\mathbb{R})$ we have $B_i \in \mathcal{A}_i$ so

$$B_1 \times \cdots \times B_n = \underbrace{(B_1 \times \mathbb{R}^{n-1})}_{\in \mathcal{B}(\mathbb{R}^n)} \cap \underbrace{(\mathbb{R} \times B_2 \cap \mathbb{R}^{n-2})}_{\in \mathcal{B}(\mathbb{R}^n)} \cap \cdots \cap \underbrace{(\mathbb{R}^{n-1} \times B_n)}_{\in \mathcal{B}(\mathbb{R}^n)} \in \mathcal{B}(\mathbb{R}^n)$$

We conclude $\mathcal{B}(\mathbb{R}) \times \cdots \times \mathcal{B}(\mathbb{R}) \subseteq \mathcal{B}(\mathbb{R}^n)$.

Answer 3

It suffices to prove that $\sigma(\mathcal{B}({\bf{R}})\times\mathcal{B}({\bf{R}}))=\sigma(\mathcal{I}^{2})=\mathcal{B}({\bf{R}}^{2})$, here $\mathcal{I}^{2}$ is the set of all open boxes.

Denoting $\mathcal{I}^{1}$ the set of all open intervals, then $\mathcal{I}^{2}=\mathcal{I}^{1}\times\mathcal{I}^{1}$ in the canonical sense.

It suffices to prove that $\sigma(\sigma(\mathcal{I}^{1})\times\sigma(\mathcal{I}^{1}))=\sigma(\mathcal{I}^{1}\times\mathcal{I}^{1})$.

The crucial point is to prove that $\sigma(\mathcal{I}^{1})\times\sigma(\mathcal{I}^{1})\subseteq\sigma(\mathcal{I}^{1}\times\mathcal{I}^{1})$.

Let $E\in\mathcal{I}^{1}$, and $\pi_{1}$ the first projection, as we know that $\pi_{1}^{-1}(\sigma(\mathcal{I}^{1}))=\sigma(\pi_{1}^{-1}(\mathcal{I}^{1}))$, then $\sigma(\mathcal{I}^{1})\times E=\sigma(\mathcal{I}^{1}\times E)\subseteq\sigma(\mathcal{I}^{1}\times\mathcal{I}^{1})$, so $\sigma(I^{1})\times\mathcal{I}^{1}\subseteq\sigma(\mathcal{I}^{1}\times\mathcal{I}^{1})$.

Similarly, we let $\pi_{2}$ to be the second projection, then $\pi_{2}^{-1}(\sigma(\mathcal{I}^{1}))=\sigma(\pi_{2}^{-1}(\mathcal{I}^{1}))$. For $F\in\sigma(\mathcal{I}^{1})$, we have $F\times\sigma(\mathcal{I}^{1})=\sigma(F\times\mathcal{I}^{1})\subseteq\sigma(\mathcal{I}^{1}\times\mathcal{I}^{1})$ by the result just proved.