What is the Borel $\sigma -$algebra of $\mathbb R^{[0,\infty)}$ ? I guess that $$\mathscr B(\mathbb R^{[0,\infty)})=\sigma \left\{\prod_{t\in [0,t)}O_t\mid \forall t\geq 0, O_t\text{ open }\text{ and } O_t\neq \mathbb R\text{ for a finite number of }t\right\}.$$
1) Is it correct ?
After, I think that $$\mathscr B(\mathbb R^{[0,\infty)})=\mathscr B(\mathbb R)^{[0,\infty)}$$ and that
$$\mathscr B(\mathbb R)^{[0,\infty)}=\sigma \left\{\prod_{t\in [0,t)}B_t\mid \forall t\geq 0, B_t\in \mathscr B(\mathbb R)\text{ and } B_t\neq \mathbb R\text{ for a finite number of }t\right\}.$$
2) Is this correct or not really ? I know if $\mathbb R^J$ and $J$ finite, then $\mathscr B(\mathbb R^J)=\mathscr B(\mathbb R)^J$, but is it also true if $J$ is infinite ?
1): NO! Your equality does not make sense. On the left you have a family of subsets of $\mathbb{R}^{[0,\infty)}$, and on the left you have the set of functions from $[0,\infty)$ to the Borel algebra of $\mathbb{R}$.
The Borel algebra of $\mathbb{R}^{[0,\infty)}$ is not the smallest $\sigma$-algebra that make all projections measurable; it is the smallest $\sigma$-algebra that contains all the open sets. In the case of uncountably many factors these two are different. The latter is much larger than the former. Every point in the product is closed and hence Borel but for each element, $B$, of the first algebra there is a countable subset $C$ of $[0,\infty)$ and a Borel subset $B'$ of the product $\mathbb{R}^C$ such that $B=B'\times\mathbb{R}^{[0,\infty)\setminus C}$.