Does $\mathbb P\{X\in A, Y\in B\}$ mean $\mathbb P(\{X\in A\}\cap \{Y\in B\})$? Let $X,Y:\Omega \to \mathbb R$ two randoms variables on $(\Omega ,\mathcal F,\mathbb P)$. Now, let consider $(X,Y)$ on $(\Omega ^2, \mathcal F^2, \mathbb P\times \mathbb P)$. I am a bit confuse with $\mathbb P\times \mathbb P=:\mathbb P_2$. Now, I always thought that $\mathbb P_2\{X\in A, Y\in B\}$ was to denote $\mathbb P(\{X\in A\}\cap \{Y\in B\})$. But $\{X\in A, Y\in B\}\in \mathcal F^2$, whereas $\{X\in A\}\cap \{Y\in B\}$ is in $\mathcal F$, no ? So I'm a bit confuse with this. 
 A: Yes, in probability theory
$$
\mathbb{P}(A,B,C,\dots) 
$$
traditionally means the same as 
$$
\mathbb{P}(A\cap B\cap C\cap\cdots).
$$
There are also some other traditional conventions: using 
$\{X \in A\} $ for $\{\omega\in \Omega \mid X(\omega) \in A\}$, $\mathbb{P}(X\in A)$ for $\mathbb{P}(\{X\in A\})$ etc. 
A: If $X,Y$ are random variables on probability space $(\Omega,\mathcal F,\mathbb P)$ then $(X,Y)$ is a notation for the function $\Omega\to\mathbb R^2$ that is prescribed by:$$\omega\mapsto(X(\omega),Y(\omega))$$
In that context $\{X\in A,Y\in B\}$ is a notation for the set $\{\omega\in\Omega\mid X(\omega)\in A\wedge Y(\omega)\in B\}$.
It is linked with the fact that $X$ and $Y$ have a common codomain.
Further $\mathbb P(X\in A,Y\in B)$ is an abbreviation of $\mathbb P(\{\omega\in\Omega\mid X(\omega)\in A\wedge Y(\omega)\in B\})$.

But next to that $X$ and $Y$ also induce a function $\Omega^2\to\mathbb R^2$ by:$$(\omega,\omega')\mapsto(X(\omega),Y(\omega'))$$for which I would rather use the notation $X\times Y$.
In that context we have sets like $\{X\times Y\in C\}\subseteq\Omega^2$ where $C\subseteq\mathbb R^2$. 
The special case where $C=A\times B$ then causes mainly the confusion because we are tempted to write: $$\{X\times Y\in A\times B\}=\{X\in A,Y\in B\}$$

There is even more, since $X,Y$ also have a common domain, leading to a function: $$\Omega\sqcup\Omega=\Omega\times\{1\}\cup\Omega\times\{2\}\to \mathbb R$$
A notation for it is $[X,Y]$ and the function is prescribed by $(\omega,1)\mapsto X(\omega)$ and $(\omega,2)\mapsto Y(\omega)$. 
