I would like to pose a question to you, and I will appreciate any hint or partial solution for my question.
Let $\left(\Omega, \mathcal{F}\right)$ be a measurable space, and let $\left(\Omega \times \Omega, \mathcal{F} \times \mathcal{F}\right)$ be the product space of $\left(\Omega, \mathcal{F}\right)$ with itself.
Let $\mathcal{C}:= \left\{ A \times A : A \in \mathcal{F} \right\}$.
How can I show that $\mathcal{C}$ is not a sigma-algebra, but it is closed under finite intersections?
Thanks
On
Wikipedia says that $(A \times C) \cap (B \times D) = (A \cap B) \times (C \cap D)$ from which it follows that it is closed with respect to finite intersection.
That it is not a $\sigma$-algebra stems from the fact that it is not necessarily closed with respect to unions. After all how in general can you write $(A \times B) \cup (C \times D)$ as $U \times V$?
As soon as $\mathcal F\ne\{\varnothing,\Omega\}$, choose $A$ in $\mathcal F$, $A\ne\varnothing$, $A\ne\Omega$, and consider $$ B=(A\times A)\cup(A^c\times A^c). $$ Then $B$ is not in $\mathcal C$ while $A\times A$ and $A^c\times A^c$ are, hence $\mathcal C$ is not a sigma-algebra.
To show that $B$ is not in $\mathcal C$, consider that, for every $C$, if $(x,x)$ and $(y,y)$ are in $C\times C$ then $x$ and $y$ are in $C$ hence $(x,y)$ is in $C\times C$. But, if $x$ is in $A$ and $y$ in $A^c$ then $(x,x)$ and $(y,y)$ are in $B$ while $(x,y)$ is not.