Prove a seemingly simple equality

Question

If $$C_1 = \{A \times \mathbb{R} \times \mathbb{R} \times \mathbb{R} \times \cdots\ \vert A \text{ is an open set in }\mathbb{R}\}$$ and $$C_2 = \{B \times \mathbb{R} \times \mathbb{R}\times \mathbb{R} \times \cdots \vert B \in \mathbb{B}(\mathbb{R})\}$$ where $\mathbb{B}(\mathbb{R})$ is the Borel $\sigma$-algebra on $\mathbb{R}$. It is very clear that $\sigma(C_1) = C_2$, but I am unable to prove it explicitly.

Thanks, rjp

Does $\sigma(C_1)$ denote the Borel $\sigma$-algebra on $C_1$? — Alex Becker, May 17 '12 at 07:06
@Alex $\sigma(C_1)$ denotes the smallest $\sigma$-algebra containing $C_1$. — rjp, May 17 '12 at 07:09
I think that "bottom-up" description of the smallest $\sigma$-algebra might be useful here. This is given e.g. in Halmos, p.26 for $\sigma$-rings, it should be easy to modify this to $\sigma$-algebras. See also this answer for some general comments about top-down and bottom-up approach. — Martin Sleziak, May 17 '12 at 07:14
Top-down also works: Obviously $\sigma(C_1) \subseteq C_2$. Now let $D := {B \in \mathbb B(\mathbb R) \mid B \times \prod_{\mathbb N}\mathbb R \in \sigma(C_1)}$ and show that $D$ is a $\sigma$-algebra containig all open sets. — martini, May 17 '12 at 07:19
@martini: Thanks for the very nice answer. I think this is a very neat trick! — rjp, May 17 '12 at 07:40
@martini That is indeed a simpler approach than what I suggested. Perhaps you could post it as an answer. What should one do when one's question has been answered in the comments? — Martin Sleziak, May 17 '12 at 07:57

score 1 · Answer 1 · answered May 17 '12 at 08:03

here, as often in such cases, we can use the so called "good sets principle", i. e. we take the collection of all sets which are "good" \[ D := \{ B \in \mathbb B(\mathbb R) \mid B \times \mathbb R \times \mathbb R\times \cdots \in \sigma(C_1) \} \] and prove, that all sets are good in proving that $D$ is a $\sigma$-algebra (more or less obvious here) which contains the open sets (as $C_1 \subseteq \sigma(C_1)$). Now we can conclude $D = \mathbb B(\mathbb R)$ and hence $\sigma(C_1) = C_2$ as wanted.

Prove a seemingly simple equality

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