From $\sigma$-algebra on product space to $\sigma$-algebra on component space Given a $\sigma$-algebra on a Cartesian product of a collection of sets, do there always exist a $\sigma$-algebra on each set, so that their product $\sigma$-algebra on their Cartesian product will be the same as the given $\sigma$-algebra?  
If yes, how to construct the $\sigma$-algebras on the sets from the $\sigma$-algebra on their Cartesian product?
If no, when they exist, how to construct the component $\sigma$-algebras? 
Thanks and regards! Any reference is appreciated as well.
 A: Let $A=\{0,1\}$, $B=\{0,1\}$, $\mathbb{S}=\{\emptyset, \{(0,0),(1,1)\}, \{(1,0),(0,1)\}, A\times B\}$. Then $\mathbb{S}$ is a $\sigma$-algebra on $A\times B$. 
Are there $\sigma$-algebras on $A$ and on $B$ whose product is $\mathbb{S}$? Well, there aren't that many $\sigma$-algebras on the two-element set. There's the total $\sigma$-algebra, $\mathcal{P}(\{0,1\})$, and the $\sigma$-algebra $\{\emptyset,\{0,1\}\}$, and that's it (if it contains any proper subset, then it contains its complement, and it is the entire thing). Call these $\mathbb{S}_1$ and $\mathbb{S}_2$, respectively.
Is $\mathbb{S}$ the product of two of these? No.
It's not equal to $\mathbb{S}_1\times\mathbb{S}_i$, because this $\sigma$-algebra contains $\{1\}\times \{0,1\}$, but $\mathbb{S}$ does not. Symmetrically, it's not equal to $\mathbb{S}_i\times \mathbb{S}_1$.
So the only possibility would be that it is equal to 
$$\mathbb{S}_2\times\mathbb{S}_2 = \Bigl\{\emptyset,\{0,1\}\Bigr\} \times \Bigl\{\emptyset,\{0,1\}\Bigr\} = \Bigl\{ \emptyset, A\times B\Bigr\}$$
which is not equal to $\mathbb{S}$. 
So $\mathbb{S}$ is a $\sigma$-algebra on $A\times B$ which is not a product of a $\sigma$-algebra on $A$ and a $\sigma$-algebra on $B$. 
If it doesn't work for two, it has no hope of working for an infinite number of factors, so the potential distinction I suggested in the comments is in fact irrelevant.
Added. To answer the added question: suppose that there do exist $\sigma$-algebras $\mathbb{S}_i$ on $X_i$ such that $\mathbb{S}=\prod\mathbb{S}_i$. Any $A_i\in\mathbb{S}_i$ will necessarily satisfy that
$$A\times\prod_{j\neq i} X_j \in\mathbb{S}.$$
Conversely, suppose that $A\subseteq X_i$ is such that $A\times \prod_{j\neq i}X_j\in\mathbb{S}$. Looking at the projection from $\prod X_i$ to 
$X_i$, you get that $A\in\mathbb{S}_i$. The projection does not work; I'm fairly sure it is the case that such an $A$ will lie in $\mathbb{S}_i$, but I find myself unable to prove it right now. Edit: This gap was filled by Morning in his answer to this question.
Assuming this is the case, then $\mathbb{S}_i$ would consist exactly of the subsets of $X_i$ such that $A\times\prod_{j\neq i}X_j\in\mathbb{S}$. 
You can see how this fails in the example above. The collection of all $A\subseteq \{0,1\}$ such that $A\times\{0,1\}\in\mathbb{S}$ is just $\{\emptyset, \{0,1\}\}$, but the product of these does not exhaust the $\sigma$-algebra. 
If you define the $\mathbb{S}_i$ this way, then the product $\sigma$-algebra $\prod\mathbb{S}_i$ is always contained in $\mathbb{S}$, but need not be equal to it. 
A: Suppose $A$ and $B$ are each 2-element sets.  It's easy to create a measure on the 4-element set $A\times B$ which is not the product of measures on $A$ and $B$.  Or perhaps I misunderstood you?
