Example of a set in $\mathcal{B}(\mathbb R^{I})$ but not in $\mathcal{B}(\mathbb R)^{I}$ for $I$ uncountable

Question

Let $I$ be an uncountable set and consider the Borel sigma algebra on $\mathbb R^{I}$, i.e. $\mathcal{B}(\mathbb R^{I})=\sigma(\tau_{I})$ where $\tau_{I}$ is the product topology on $\mathbb R^{I}$. By definition, the projections $p_{i}$ are continuous for all $i\in I$ and thus the projections are $\sigma(\tau_{I})$-measurable. Since the product sigma algebra $\mathcal{B}(\mathbb R)^{I}$ is the smallest sigma algebra such that the projections are measurable with respect to $\mathcal{B}(\mathbb R)^{I}$. Therefore we have $\mathcal{B}(\mathbb R)^{I} \subseteq \mathcal{B}(\mathbb R^{I})$.

Now I am interested in finding a set $A$ such that is in $\mathcal{B}(\mathbb R^{I})$ but not in $\mathcal{B}(\mathbb R)^{I}$ to show that the inclusion is strict. Any ideas?

Answer 1

Any set in $\mathbb B(\mathbb R)^{I}$ depends only on a countable number of coordinates. This means if $E \in \mathbb B(\mathbb R)^{I}$ then there exists a countable subset $(i_n)$ of $I$ such that $E=\{f: (f(i_1),f(i_2),...) \in F$ for some $F$ in $\mathbb R^{\infty}$. [To prove this show that the collection of all sets with this property is a sigma algebra]. Hence any set which does not depend only on a countable number of coordinates will be an example. You can take $\{f: |f(i)| \leq 1 \forall i \in I\}$.