Suppose $E_{0}$ is a topological space, with topology $\tau_{0}$. Let $\mathcal{F}_{0}$ be its Borel $\sigma$-algebra, i.e. the $\sigma$-algebra generated by its open sets. Now, take $I$ a countable set and set $E := \prod_{i\in I}E_{0}$. We can equip $E$ with both its product topology $\tau$ and its product $\sigma$-algebra $\mathcal{F}$. It is a well-known result that if $E$ is metrizable (or metric) and separable, then the product $\sigma$-algebra $\mathcal{F}$ coincides with the Borel $\sigma$-algebra on $E$. In the theory of Gibbs measures, however, we usually take $E_{0}$ to be metric (or metrizable) and compact, not necessarily separable. However, I'm not sure that the product topology and the product $\sigma$-algebra, in this case, are compatible, i.e. if the product $\sigma$-algebra is the $\sigma$-algebra generated by the open sets of $E$ when it's equipped with its product topology. I think we need separability here, but some references do not explicitly assume $E_{0}$ separable.
Question: If $E_{0}$ is a compact metric space, is the product $\sigma$-algebra the $\sigma$-algebra generated by open sets of $E$ when it's equipped with the product topology?
Best Answer
In fact, I don't see a reason to worry, as all compact (or Lindelöf) metric (or metrisable) spaces are separable (just take a finite (or countable) subcover of the cover of all $\frac{1}{n}$-balls, for every $n$, and collect the centres used in all these subcovers to get a countable dense subset of the space).
So compact is more than enough to apply your result on separable metric spaces.