Showing that the preimage of the projection mappings $E_1\times\cdots\times E_n\mapsto E_i, E_1,\dots,E_n\in B(R)$ is contained in $B(R^n)$

Question

Preamble: The main proof that $\mathcal{B}(\mathbb{R}^n) = \sigma(C), C = \{E_1\times\cdots\times E_n\mid E_1,\dots,E_N \in \mathcal{B}(\mathbb{R})\}$ is given here: Is $\mathcal B(\mathbb R^n)$ generated by $\{E_1\times\cdots\times E_n:E_1,\ldots,E_n\in\mathcal B(\mathbb R)\}$?. What I am interested in is how $p^{-1}_i(E)\in \mathcal{B}(\mathbb{R}^n)$ is proven in the aforementioned proof. Namely, the prover takes the set $\{A\subset \mathbb{R}\mid p^{-1}_i(A) \in \mathcal{B}(\mathbb{R}^n)\}$, and then argues why this will yield that $p^{-1}_i(E)\in \mathcal{B}(\mathbb{R}^n)$: i.e. we take the collection most suitable for us (preimages of the projections which are included in the desired set), and then argue why this collection holds the Borel sigma-algebra $\mathcal{B}(\mathbb{R})$.

Question: What I am wondering is how you could show the same inclusion by viewing $\sigma(C)$ as the smallest sigma-algebra such that the projection mappings onto the $i$th factor, $p_i(E_1\times\dots\times E_n) = E_i$ are measurable, so that by definition for any $E \in \mathcal{B}(\mathbb{R}): p_i^{-1}(E) = \{A_1\times\cdots\times E\times\cdots\times A_n\mid A_1,\dots,A_n \in \mathcal{B}(\mathbb{R})\} \in \sigma(C)$. It is not entirely obvious to me how you could show that each of these pre-images are contained in $\mathcal{B}(\mathbb{R}^n)$.

Thoughts: Perhaps arguing that other than the $i$th factor $E$, the other factors can be generated by e.g. closed intervals of $\mathbb{R}$, so that considering the componentwise nature of intersections, unions and differences in $\mathcal{B}(\mathbb{R}^n)$ we have that $p_i^{-1}(E) \subset \mathcal{B}(\mathbb{R}^n)$? But to be honest this is a pretty hand-wavy proof at best, and stating that each of the factors can be generated in $\mathcal{B}(\mathbb{R}^n)$ as they are Borel sets sounds like a circular reasoning.

So is there a way to proof this same inclusion with this different edge, or will it rail to be essentially the same proof in the end?

Answer 1

The original proof is unclear. Here is a simple proof.

Let $\mathcal{A}=\{E_1\times\cdots\times E_n:E_1,\ldots,E_n\in\mathcal B(\mathbb R)\}$. Then since $\pi_i(\sigma(\mathcal{C}))$ is a sigma field containing the open sets of $\mathbb R$, $E_i\subset \pi_i(\sigma(\mathcal{C}))$ and $\mathcal{A}\subset \sigma(\mathcal{C})$. Thus $\sigma(\mathcal{A})\subset \sigma\sigma(\mathcal{C})=\sigma(\mathcal{C})$.

$\pi_i$ is the standard notation for the $i^{th}-$ projection.