How do we know the joint probability distribution measure is valid? Let $X,Y$ be $\mathbb{R}$-valued random variables on $(\Omega, \mathcal{F})$.  Then $(X,Y) : \Omega \to \mathbb{R}^2$ induces a joint probability distribution measure $\mu_{X,Y}: \mathcal{B} \otimes \mathcal{B} \to [0,1]$ given by
$$
\mu_{X,Y}(A) = P((X,Y) \in A).
$$
Now, if $A = A_1 \times A_2$ for $A_1, A_2 \in \mathcal{B}$ then I suppose $P((X,Y) \in A) = P(X \in A_1 \cap Y \in A_2)$, and since both $\{X \in A_1\}, \{Y \in A_2\} \in \mathcal{F}$ we get $\{X \in A_1\} \cap \{Y \in A_2\} \in \mathcal{F}$ and so this is a measurable set.
But what about general Borel subsets of $\mathbb{R}^2$?  Why would I expect $\{(X,Y) \in A\} \in \mathcal{F}$?
 A: Let $\mathcal{C} = \{ B \in \text{Borel of } \Bbb{R}^2 | \{ \omega \in \Omega | (X,Y)(\omega) \in B\} \in \mathcal{F} \}\}$.
Note that $\mathcal{C}$ is a $\sigma$-algebra.
1)$\mathbb{R}^2 \in \mathcal{C}$
2) $B \in \mathcal{C} \Rightarrow B^c \in \mathcal{C}$ 
3) $B_i \in \mathcal{C} \Rightarrow \cup_{i=1}^\infty B_i \in \mathcal{C}$
Now since the sets $ A_1 \times A_2$ belong to $\mathcal{C}$ then you conclude that $\mathcal{C} = \sigma(\mathcal{C}) = \mathcal{B}(\Bbb{R}^2)$ 
remark: the technique to prove 2) is the following. Assume $B \in \mathcal{C}$
$ (X,Y)(\omega) \in B^c \Leftrightarrow  (X,Y)(\omega) \notin B$
Therefore 
$$\{\omega \in \Omega | (X,Y)(\omega) \in B^c\} = \{\omega \in \Omega | (X,Y)(\omega) \in B\}^c$$
Since $\{\omega \in \Omega | (X,Y)(\omega) \in B\} \in \mathcal{F}$ and $\mathcal{F}$ is a $\sigma$-algebra, $$\{\omega \in \Omega | (X,Y)(\omega) \in B^c\} =\{\omega \in \Omega | (X,Y)(\omega) \in B\}^c \in \mathcal{F} \Rightarrow B^c \in \mathcal{C}$$
The proof of 3) is analogous
