TITLE: Why $\mathcal B(\mathbb R)^{[0,\infty )}\subsetneq \mathcal B(\mathbb R^{[0,\infty )})$?
QUESTION [1 upvotes]: After reading the answer of Jason Swanson in this post, I saw that $\mathcal B(\mathbb R))^{[0,\infty )} \subsetneq \mathcal B(\mathbb R^{[0,\infty )})$ where $\mathbb R^{[0,\infty )}$ is endow with the product topology. To understand why, I asked this question yesterday, but made a confusion between $\mathbb R^{[0,\infty )}$ and $\mathbb R^\infty $. It looks like $\mathcal B(\mathbb R^\infty )=\mathcal B(\mathbb R)^\infty $. So it was finally not a good question. In this more general case, I can show that indeed $\mathcal B(\mathbb R)^{[0,\infty )}\subset \mathcal B(\mathbb R^{[0,\infty )})$, but I can't prove that the inequality is strict. Does someone knows an example of set $E$ that is in $\mathcal B(\mathbb R^{[0,\infty )})$ but not in $\mathcal B(\mathbb R)^{[0,\infty )}$ ? Moreover, in which situation will we prefer use $\mathcal B(\mathbb R)^{[0,\infty )}$ ? And in which situation will we prefer use $\mathcal B(\mathbb R^{[0,\infty )})$ ?

REPLY [1 votes]: Broad hints: let $p_t:\mathbb R^{[0,\infty)} \to \mathbb R$, $t \geq 0$, be the projection maps. Say that a set $E$ in $\mathbb R^{[0,\infty)} $ depends only on countable number of coordinates if there exists a countable set $\{t_1,t_2,...\}$ and a set $A$ in $\mathbb  R^{\infty}$ such that $E=(p_{t_1},p_{t_2},...)^{-1}(A)$. Show that sets depending only on coutable number of coordinates form a sigma algebra and conclude that $\mathbb B  (\mathbb R)^{[0,\infty)}$ is contained in this sigma algebra. But there exist sets in $\mathbb B  (\mathbb R^{[0,\infty)})$ which 'depend on all coordinates',say $\{f: [0,\infty) \to \mathbb R: f(t)>0 \, \forall t\}$.
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