Definition of $\mathbb{P}((X, Y) \in A)$. I'm reading Larry Wasserman's "All of statistics", and I've come across a definition I can't "unpack".
Specifically the text defines $f(x, y)$ to be a PDF for the random variables $(X, Y)$, if:


*

*$f(x, y) \geq 0 $ for all $(x, y)$

*$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f(x, y)dxdy = 1 $ and

*For any set $A \subset \mathbb{R}\times\mathbb{R}, \mathbb{P}((X, Y) \in A) = \iint_{A}f(x, y)dxdy$.


Intuitively it all makes sense, but what exactly is $\mathbb{P}((X, Y) \in A)$?
Earlier in the text, we have $\mathbb{P}(X=x)$ defined to be $\mathbb{P}(X^{-1}(x))$, and this makes sense, since the pre-image of $X$ is the sample space.
This definition can be trivially extended to other arithmetic operators, i.e. $<$, $>$, $\leq$, etc and even set operations $\mathbb{P}(X \in A)$, as long as $A$ is a subset of $\mathbb{R}$.
I'm struggling to see how exactly this definition can be extended to $\mathbb{P}((X, Y) \in A)$. The text is not helpful in that regard.
 A: Assuming that you are working with a probability space $(\Omega,\mathcal A,\mathbb P)$, the expression $\mathbb P((X,Y)\in B)$  must actually be read as:$$\mathbb P(\{\omega\in\Omega\mid (X(\omega),Y(\omega))\in B\})$$
Also the set $\{\omega\in\Omega\mid (X(\omega),Y(\omega))\in B\}$ can be recognized as the preimage of $B$ under the function $\Omega\to\mathbb R^2$ prescribed by $\omega\mapsto(X(\omega),Y(\omega))$.
Here $B\subseteq\mathbb R^2$ is a Borel set, and $X,Y:\Omega\to\mathbb R$ are random variables.
A: $\mathbb{P}(X,Y)$ doesn't exist. $\mathbb{P}[(X,Y)\in A]$ is the probability of the two dimensional random variable (X,Y) landing in A. 
I am not quite sure why you don't understand $\mathbb{P}[(X,Y)\in A]$ if you understand $\mathbb{P}[X\in A]$.
So let us look at the one-dimensional case first. Say X is modelling a dice. So $\mathbb{P}(X=k)=1/6$ for $k\in \{1,...6\}$.
I can now define $A=\{2,4,6\}$. Then $\mathbb{P}[X\in A]=\mathbb{P}(\text{X is even})$. 
Or if you struggle with the formal definition: $\mathbb{P}[X\in A]=\mathbb{P}[X^{-1}(A)]=\mathbb{P}[\{\omega\in\Omega:X(\omega)\in\Omega\}]$.
Now look at the two dimensional case. Say (X,Y) are two dice. $\mathbb{P}[(X,Y)=(k,l)]=1/36$ for $k,l\in\{1,...,6\}$. I can now define $A=\{(k,l)\in\{1,...6\}^2:k+l=5\}$. Then $\mathbb{P}[(X,Y)\in A]=\mathbb{P}[\text{Sum of the two dice is 5}]$. 
Or if you struggle with the formal definition: $\mathbb{P}[(X,Y)\in A]=\mathbb{P}[(X,Y)^{-1}(A)]=\mathbb{P}[\{\omega\in\Omega:(X(\omega),Y(\omega))\in\Omega\}]$.
Of course this is the discrete case. But similarly if you can think of a continuous case for the one dimensional variable it shouldn't be too hard to find a example for the two dimensional continuous case. And discrete models are usually easier to grasp.
