Let $\Gamma$ be an arbitrary index set and $(S,\mathcal{S})$ a measure space. I want to study the product space $S^\Gamma$. For that we define
$X_\gamma : S^\Gamma \to S, \hspace{12pt}\omega=(\omega_\gamma)_{\gamma \in \Gamma} \mapsto X_\gamma(\omega):=\omega_\gamma$
where $X_\gamma$ is the coordinate map on the $\gamma$-th coordinate. Moreover, we define
$\mathcal{S}^\Gamma=\sigma(X_\gamma,\gamma \in \Gamma):=\sigma(\{\{X_\gamma \in A_\gamma\};\gamma\in \Gamma,A_\gamma\in \mathcal{S}\})$
this should be the smallest $\sigma$-algebra, such that all $X_\gamma$'s are measurable. Now my question, why is the following set a generator of this $\sigma$-algebra?
$M:=\{\omega\in S^\Gamma;\omega_j\in A_j,j\in J\}=\prod_{j\in J}A_j\times \prod_{k\in J^c}S$
for all $J\subset \Gamma$ which are finite and $A_j\in \mathcal{S}$. Why is this true? It would be appreciated if someone could give me a proof or post a reference.
As I saw the statement, I had to think about the product topology. This can be described in a similar way (using a basis). However, in topology one has often that a property is true just for finite intersections or similar things. In measure theory this is not the case. So I do not see where the finiteness come in. Thank you for your help.
cheers
math