Let $E$ be a set, and $(X_e,\mathcal{B}_e)_{e \in E}$ be a family of measurable spaces (that is, for each $e$, $X_e$ is a set and $\mathcal{B}_e$ is a $\sigma$-algebra of subsets of $X_e$).
Let $X := \Pi_e X_e$, endowed with the $\sigma$-algebra $\mathcal{B}$ generated by the cylinders. If $C$ is a cylinder, and $F$ is a subset of $E$, let's say that "$C$ is supported in $F$" if $C = \{\omega \in X \ \vert \ \forall e \in F\quad \omega_e \in B_e\}$ for some $B_e \in \mathcal{B}(e)$.
If $B \in \mathcal{B}$, and $F$ is a subset of $E$, let's say that $B$ does not depend on what's happening in $F$ if, for every $\omega,\omega' \in X$, $\left(\omega_{|_{F^c}} = \omega'_{|_{F^c}} \ and \ \ \omega \in B\right) \Rightarrow \omega' \in B$.
I wonder if, given some subset $F$ of $E$, if the set of $B$ not depending on what's happening in $F$ is the $\sigma$-algebra generated by the cylinders supported in $F^c$. The inclusion $\supseteq$ is easy and comes from the obvious fact that cylinders supported in $F^c$ do not depend on what's happening in $F$.
I am mostly interested in the case where $X_e$ is finite and $\mathcal{B}_e$ is the discrete $\sigma$-algebra, for every $e$ [EDIT] and when $F$ is finite[/EDIT].
Here is a proof of the other direction.
Choose $c_F\in \prod_{e\in F}X_e$. For any set $S\subseteq \prod_{e\in F}X_e$ define the modified set $\widehat S\subseteq \prod_{e\in F}X_e$ by $$(x|_F,x|_{E\setminus F})\in \widehat S\iff (c_F,x|_{E\setminus F}) \in S.$$ I hope the pair notation $(c_F, x|_{E\setminus F})$ is clear - I am splitting up the co-ordinates using $\prod_{e\in E}X_e = \prod_{e\in F}X_e\times \prod_{e\in E\setminus F}X_e$.
For arbitrary sets $B_n\subseteq \prod_{e\in E}X_e$ we have $$\bigcup_n \widehat{B_n}=\widehat{\bigcup_n B_n}\qquad\text{ and }\qquad\bigcap_n \widehat{B_n}=\widehat{\bigcap_n B_n}.$$ For any Borel set $B$, by transfinite induction we find that $\widehat{B}$ is in the $\sigma$-algebra generated by cylinders of the form $\widehat{C}$, which are necessarily supported in $F^c$. If $B$ doesn't depend on what's happening in $F$ then this gives $B=\widehat B$, but $\widehat B$ was constructed using cylinders supported in $F^c$ as required.
Here is another way of phrasing the same idea. Let $X_e'$ denote $X_e$ with the indiscrete $\sigma$-algebra, and consider the map $$f:\prod_{e\in F}X'_e\times\prod_{e\in E\setminus F}X_e\to\prod_{e\in E}X_e\\ (x|_F,x|_{E\setminus F})\mapsto(c_F,x|_{E\setminus F})$$
This $f$ is measurable (this can just be checked on cylinder sets), hence for any Borel set $B$, the preimage $f^{-1}(B)$ is Borel in $\prod_{e\in F}X'_e\times\prod_{e\in E\setminus F}X_e$. If $B$ doesn't depend on what's happening in $F$ then this gives $B=f^{-1}(B)$, but $f^{-1}(B)$ is constructed using cylinders supported in $F^c$ as required.