Say I have two sigma algebras: $ \sigma(C) = \{E,\{\},\{1,2\},\{3,4,5\},\{1,2,3,4,5\},\{6\},\{3,4,5,6\},\{1,2,6\}\} $
where $E= \{1,2,3,4,5,6\}$
and $\sigma(Y) = \{E,\{\},\{1,3,5\},\{2,4,6\}\}$
How would one go about finding $\sigma(C) ⊗\sigma(Y) ?$
I know $\sigma(C) ⊗\sigma(Y) = \sigma(\{A\times B : A\in \sigma(C) , B\in \sigma(Y) \})$ is intuitively the smallest sigma algebra generated by the cartesian product.
Would this product algebra just give a set containing every tuple of sets from both sigma algebras eg: $(\{1,2\}\times\{1,3,5\})$?
Thanks.
Best Answer
Answer to your second question:
These sets will belong to the product $\sigma$-algebra, but not every set in it will be such a set.
Realize that the complement of e.g. $\{1,2\}\times\{1,3,5\}$ is not a set that can be written as $A\times B$.
As you know $\sigma$-algebras are closed under complementation.
Answer to your first question:
In this special case there are partitions $\mathcal P=\{\{1,2\},\{3,4,5\},\{6\}\}$ and $\mathcal Q:=\{\{1,3,5\},\{2,4,6\}\}$ on set $E$. A set is element of the $\sigma(C)$ if and only if it can be written as a finite union of elements of $\mathcal P$. This union is also allowed to be empty. Likewise a set is element of the $\sigma(Y)$ if and only if it can be written as a finite union of elements of $\mathcal Q$.
This partition induces a partition $\mathcal R:=\mathcal P\times\mathcal Q=\{A\times B\mid A\in\mathcal P,B\in\mathcal Q\}$ on set $E\times E$, and a set is element of $\sigma(C)\otimes\sigma(Y)$ if it can be written as a finite union of elements of $\mathcal R$.
Note: because set $E$ is finite the class of algebras on it coincides with the class of $\sigma$-algebras on it.