Continuous and discrete

If $X$ and $Y$ are continuous, then in order for $f_{XY}$ to be their joint density function, we must have $P\{(X,Y) \in \mathbb{R}^2\} = \iint_{\mathbb{R}^2} f_{XY}(x,y)\,dx\,dy = 1. \tag{1}$ But what if the distribution of $X$ and $Y$ has support on a region of area 0? Consider the case when $X = Y$ . Then the support of the distribution in question is contained in the line $y = x$ and has area 0. In this case, $\iint_{\mathbb{R}^2} f_{XY}(x,y)\,dx\,dy = \lim_{b \rightarrow \infty} \int_{-b}^{b} \int_y^y f_{XY}(x,y)\, dx\, dy = 0,$ but this contradicts (1). Therefore, when $X$ and $Y$ are continuous, there does not necessarily exist a joint density function.

On the other hand, if $X$ and $Y$ are discrete, then in order for $p_{XY}$ to be their joint mass function, we must have, for all admissible values $(x,y)$ of $(X,Y)$ , $P\{X = x,Y = y\} = p_{XY}(x,y). \tag{2}$ Whereas in (1) the suitability of $f_{XY}$ depends on a relationship between $P$ and an integral of $f_{XY}$ , in (2) the suitability of $p_{XY}$ is based only on a relationship between $P$ and $p_{XY}$ itself . We simply take (2) as the definition of $p_{XY}$ , and then of course $p_{XY}$ is the desired mass function. Therefore, when $X$ and $Y$ are discrete, there must exist a joint mass function $p_{XY}$ .