Relationship between algebras of sets

Question

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?

Answer 1

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 $S\times B$.