Showing that $P(X\in A,Y\in B)=P(X\in A)P(Y\in B)$ closed under increased limits and difference.

Question

I need to showing that the collection of sets $A \in M$ with $$P(X\in A,Y\in B)=P(X\in A)P(Y\in B)$$ with a set $B\in N$ with with $M,N$ being $\sigma-$Algebras, is closed under increased limits and difference. I think that I have problems understanding the concept of $P(X\in A,Y\in B)$, which I know is the Probability of the set of $\omega$ with $X(\omega) \in A \wedge Y(\omega)\in B$. But I am unsure how to get the result for differences. Let $A\subset B$: $$ P\{(X\in (B\backslash A),Y\in B)\}=P\{((X\in B)\backslash (X\in A)),Y\in B)\}=P\{(X\in B,Y\in B)\backslash (X\in A,Y\in B)\}=P\{(X\in B,Y\in B)\}-P\{(X\in A,Y\in B)\}=(P(B)-P(A))P(Y)=P(B\backslash A)P(B) $$ Is this correct?

Answer 1

Your attempt contains the right idea, but also contains some mistakes and typos. First, note that you can't use $B$ as that is already used as a fixed constant. So suppose $A \subseteq C$ instead. Then

$$P(X \in C \setminus A, Y \in B) = P(\{X \in C, Y \in B\}\setminus \{X \in A, Y \in B\})$$ $$P(X \in C, Y \in B)- P(X \in A, Y \in B)$$ $$P(X \in C)P(Y \in B) - P(X \in A)P(Y \in B)$$ $$= (P(X \in C)-P(X \in A))P(Y \in B)$$ $$= P(X \in C\setminus A)P(Y \in B)$$

and we are done.