1
$\begingroup$

Let $\mathcal{N}$ and $\mathcal{M}$ be algebras of sets on $S$ and $T$ respectively. Let $\mathcal{N}\times\mathcal{M}$ the algebra generated by the rectangles in $S\times T$ (i,e the sets with the form $A\times B$ where $A\in\mathcal{N}$ and $B\in\mathcal{M}$). We denote by $\mathcal{N}\triangle\mathcal{M}$ to the $\sigma$-algebra generated by the algebra $\mathcal{N}\times\mathcal{M}$.

I want to show that, if $\mathcal{N'}$ and $\mathcal{M'}$ are the $\sigma$-algebras generated by $\mathcal{N}$ and $\mathcal{M}$ respectively then

$$ \mathcal{N}\triangle\mathcal{M}=\mathcal{N'}\triangle\mathcal{M'}. $$

It is easy to see that $\mathcal{N}\triangle\mathcal{M}\subseteq\mathcal{N'}\triangle\mathcal{M'}$. I have troubles with the other direction. My idea is to show that $\mathcal{N'}\times\mathcal{M'}\subseteq\mathcal{N}\triangle\mathcal{M}$ but I don't know how to get the last sentence. Can seomeone give me a hint?

$\endgroup$
1
$\begingroup$

Hint: $A\times B=(A\times T)\cap(S\times B)$

Let $A\in\mathcal N'$ and $B\in\mathcal M'$, then the same $\sigma$-construction which produces $A$, will produce $A\times T$, showing $A\times T\in \mathcal N\nabla\mathcal M$. Similarly for $S\times B$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.