How can I prove that if $\mathcal F$ is a $\sigma$-algebra on a set $X$ and if $J\subset \mathbb R$ is s.t. $|J|=\infty$, then the product $\sigma$-algebra is $$\mathcal F^J=\sigma \left\{\prod_{j\in J}B_j\mid \forall j\in J, B_j\in \mathcal F, B_j\neq X\text{ for a finite number of }j\right\}.$$
To me, $\mathcal F^J$ is the smallest $\sigma$-algebra s.t. projection are continuous ie $$\mathcal F^J:=\sigma \{\pi_j^{-1}(F)\mid F\in \mathcal F, j\in J\}=\sigma \{X\times ...\times F\times ...\times X\times ...\mid F\in \mathcal F\}$$ where $F$ is at the $j$-th place and $\pi_j$ is the projection on the $j$-th variable. So only one $B_i$ in $\prod_{j\in J}B_j$ is different of $X$. So for me $\prod_{j\in J}B_j$ with $B_j\neq X$ for a countable number of $j$ is in $\mathcal F^J$ (not only finite number of $j$). Could someone give me some explanations ?
I know that $X\times ...\times X\times ...$ is not really adapted for $\prod_{j\in J}X$ when $J$ is not countable, but I think it's more visual here... So I abused a little bit of the notation to be more clear :)