0

I would like to prove that $\mathcal B(\mathbb R)^\infty \subsetneq\mathcal B(\mathbb R^\infty)$, where $\mathbb R^\infty =\prod_{i=1}^\infty \mathbb R$, $\mathcal B(A)$ is the Borel $\sigma -$algebra of $A$ and $\mathcal B(\mathbb R)^\infty $ is the product $\sigma -$algebra on $\mathbb R^\infty $.

I don't really know how to proceed. I know that $\mathcal B(\mathbb R)^\infty =\sigma (\mathcal C)$ where $\mathcal C$ is the set of the cylinder, i.e. set of the form $$C_{i_1,...,i_n}(B_1,...,B_n)=\{x\in \mathbb R^\infty\mid x_{i_1}\in B_1,...,x_{i_n}\in B_n\}$$ where $B_i$ are Borel sets in $\mathbb R$.

Now, I know that $\mathcal B(\mathbb R^\infty )$ is the $\sigma -$algebra generated by open sets of $\mathbb R^\infty $. A subbasis is given by $$\left\{\prod_{i=1}^\infty U_i\mid U_i\text{ open in $\mathbb R$ and }U_i\neq \mathbb R\text{ for a finite number of $i$}\right\}.$$

For the inclusion, it's clear since if $$\left\{\prod_{i=1}^nB_{i}\mid C_{j_1,...,j_n}(B_1,...,B_n)\in \mathcal B(\mathbb R^\infty )\right\},$$ is a $\sigma -$algebra that contain open set of $\mathbb R^n$, and thus, cylinders are in $\mathcal B(\mathbb R^\infty )$.

Now, how can I prove that the inclusion is strict ?

John
  • 858
  • 2
    If you are providing $\mathbb R^{\infty}$ with the product topology the I think the inclusion is not strict. Where did you find this statement? – Kavi Rama Murthy Jun 10 '19 at 12:16
  • @KaviRamaMurthy: In the edit of the answer in this post (but they use $\mathbb R^{[0,\infty )}$ instead of $\mathbb R^\infty $, but maybe it's not the same. – John Jun 10 '19 at 12:26
  • $\mathbb R^{\infty}$ is very different from $\mathbb R^{[0,\infty)}$. Please edit the question. – Kavi Rama Murthy Jun 10 '19 at 12:29

1 Answers1

1

If $\{(X_n, \tau_n)\}_{n=1}^{\infty}$ is a sequence of second countable topological space then if $\prod\limits_{n=1}^{\infty}X_n$ is equipped with the product topology, $\mathcal{B}(\prod\limits_{n=1}^{\infty}X_n)$ does equal $\prod\limits_{n=1}^{\infty}\mathcal{B}(X_n)$.

Jonathan Hole
  • 3,564