Generator of the $\sigma$-algebra on a product space 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 
 A: Let $J\subset \Gamma$ be finite and let $A_j\in \mathcal S$ for each $j\in \Gamma$. Then 
$$\prod_{j\in J} A_j\times \prod_{k\in J^c} S = \bigcap_{j\in J} X_j^{-1}(A_j)$$
So $M \subset \sigma(X_\gamma,\,  \gamma\in \Gamma)$ and therefore $\sigma(M)\subset \sigma(X_\gamma, \, \gamma\in \Gamma)$. 
On the other hand, each $X_\gamma$ is measurable w.r.t $\sigma(M)$, so we also get the other inclusion $\sigma(M)\supset \sigma(X_\gamma, \, \gamma\in \Gamma)$ and therefore
$$\sigma(M)= \sigma(X_\gamma, \, \gamma\in \Gamma)$$
A: Similar as in product topology we care about finite intersections, in measure theory we care about countable operations. But note one thing: let $\Gamma$ be infinite and $J\subset \Gamma$ be some infinite countable set. Let us numerate: $j_1,j_2,\dots, j_n,\dots \in J$.
If $\mathcal S^\Gamma$ is the product $\sigma$-algebra, and $A_j\in\mathcal S$ then any set of the form
$$
B_n = \prod\limits_{k=1}^nA_{j_n}\times \prod_{k\in \Gamma\setminus\{j_1,j_2,\dots,j_n\}}S\in \mathcal S^\Gamma
$$
for any finite $n$, hence
$$
\prod\limits_{k\in J}A_{j}\times \prod_{k\in J^c}S = \bigcap\limits_{n=1}^\infty B_n\in \mathcal S^\Gamma.
$$
So it means that there is no need to ask the product $\sigma$-algebra to be generated by a basis of products where there are possibly countably many sets which are not equal to $S$.
