If $X,Y$ are two random variable, is it possible that they have not a joint density ? If yes, what does it mean? Let $X,Y$ are two random variable (on the same probability space), is it possible that they have no density ? If yes, what does it mean ?
I'm asking this question, because my teacher always says : let $X,Y$ two random variable that have a joint density. So, I imagine that it can happen that they don't have a joint distribution.
 A: Let $X \sim \text{Unif}(0,1)$ and $Y\sim \text{Bernoulli}(1/2)$ be independent. These are defined on the common probability space $[0,1]\times \{0,1\}$, and they have a joint distribution. Namely if $A$ is a measurable subset of $[0,1]\times\{0,1\}$, then we can write $A=(A_0\times\{0\})\sqcup (A_1\times\{1\})$ for some measurable $A_0,A_1\subseteq [0,1]$.  Then we have
\begin{align*}
\mathbb{P}((X,Y) \in A) = \mu(A_0)\cdot \frac{1}{2}+\mu(A_1)\cdot \frac{1}{2}
\end{align*}
where $\mu$ is the Lebesgue measure on $[0,1]$.
Now the question is does $(X,Y)$ have a joint density? In other words, is there a function $f_{X,Y}(x,y):[0,1]\times\{0,1\}\to\mathbb{R}$ such that for any measurable $A \subseteq [0,1]\times\{0,1\}$ we have
$$\mathbb{P}((X,Y) \in A) = \int_A f_{X,Y}(x,y) \ dy \ dx?$$
In particular, for the set $A=[0,1]\times\{0\}$ we would need
$$\frac{1}{2}=\mathbb{P}((X,Y) \in A) = \int_0^1 \int_0^0f_{X,Y}(x,y) \ dy \ dx$$
but the right-hand-side is $0$ because $\int_0^0 f_{X,Y}(x,y) \ dy = 0$. So $X$ and $Y$ do not have a joint density.
