Does $\mathcal B(\mathbb R^I)=\mathcal B(\mathbb R)^{I}$? Let $I$ a set. I was wondering if $\mathcal B(\mathbb R^I)=\mathcal B(\mathbb R)^I$ always hold, i.e. if the borel $\sigma -$algebra of $\mathbb R^I$ coincides with the product $\mathcal B(\mathbb R)^I$. I know that if $I$ is finite, then it does. I'm not so sure in the case where $I$ is countable, or even if $I=[0,1]$ for example. I know that $$\mathcal B(\mathbb R)^I=\sigma \left\{\prod_{i\in I}A_i\mid A_i\in \mathcal B(\mathbb R), i\in I\right\}$$
whereas $$\mathcal B(\mathbb R^I)=\sigma \left\{\prod_{i\in I}A_i\mid A_i\text{ open}, A_i=\mathbb R \text{ but finitely many $i$}\right\}.$$
So, I would say that $\mathcal B(\mathbb R^I)\subset \mathcal B(\mathbb R)^I$, but I'm not sure if the inclusion is strict or if it's indeed equal.
 A: No, if $I$ is uncountable.  Indeed, members of
$\mathcal B(\mathbb R)^I$ depend on only countably many coordinates*, but a singleton point is closed in $\mathbb R^I$ and therefore a member of
$\mathbb B(\mathbb R^I)$.
$^*$ proof.  The collection of subsets of $\mathbb R^I$ that depend on countably many coordinates is a sigma-algebra, and contains the generating sets
for $\mathcal B(\mathbb R)^I$.

added
I used the usual definition for product sigma-algebra.  Given $(X_i, \mathcal F_i)$ for $i \in I$, define the product sigma-algebra $\bigotimes_i \cal F_i$ on the Cartesian product set $\prod_i X_i$ as:  The sigma-algebra generated by the sets of the form
$\prod_{i \in I} A_i$,
where $A_i = X_i$ for all but one $i \in I$ and $A_i \in \mathcal F_i$ for the final $i$.  This coincides with the notion of "product" from category theory.
Of course, if $I$ is countable, then by countable intersections this coincides with your definition
$$
\sigma \left\{\prod_{i\in I}A_i\mid A_i\in \mathcal F_i, i\in I\right\}
\tag2$$
But if $I$ is uncountable, it is (usually) different.  For example, in $\mathbb R^I$, a singleton is not in $\mathcal B(\mathbb R)^I$, even though it is of the form $(2)$.  Uncountable intersections are not postulated for a sigma-algebra.

Yes, if $I$ is countable.  Then $\mathbb R^I$ is metrizable, and a subbase for the topology is the collection of sets $\prod A_i$ with $A_i = \mathbb R$ for all but one $i$, and $A_i$ open for the final $i$.
Using that, taking finite intersections and countable unions, we see that
all open sets of $\mathbb R^I$ belong to $\mathcal B(\mathbb R)^I$.
