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I am confused on the definition of continuous/discrete random variables defined on the same probability space. Consider the random variables $X,Y$ defined on the same probability space $(\Omega, \mathcal{F}, P)$.

Suppose we are told that $X:\Omega\rightarrow \mathcal{X}\subseteq \mathbb{R}$ is continuous. Hence, by definition, $\mathcal{X}$ is uncountable and $P(X=x)=P(\{\omega \in \Omega \text{ s.t. } X(\omega)=x\})=0$ for any $x\in \mathcal{X}$.

Suppose instead that $Y:\Omega\rightarrow \mathcal{Y}\subseteq \mathbb{R}$ is discrete. Hence, by definition, $\mathcal{Y}$ is finite and $P(Y=y)=P(\{\omega \in \Omega \text{ s.t. } Y(\omega)=y\})>0$ for any $y\in \mathcal{Y}$ [assuming for simplicity that all points in $\mathcal{Y}$ are mass points].

Since $X,Y$ are defined on the same probability space, does the fact that $X$ is continuous and $Y$ is discrete imply that $$ \{\omega \in \Omega \text{ s.t. } Y(\omega)=y \}\cap \{ \omega \in \Omega \text{ s.t. } X(\omega)=x\}=\emptyset $$ for any $x\in \mathcal{X}$ and $y\in \mathcal{Y}$? Or do we have other implications on the probability measure $P$?

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Consider $\Omega = [-1, 1]$, $X(t) = t$, and set the codomain of $Y$ to $\{0, 1\}$, with $$ Y(t) = \begin{cases} 0 & t \le 0 \\ 1 & \text{else} \end{cases}. $$ Let $x = 0$ and $y = 0$ in your last paragraph. The first set is $\{0\}$ and the second is $[-1, 0]$, and their intersection is $\{0\}$.

So the answer is "no" to that main question.

It is true that the measure of that intersection is always zero (assuming the intersection is measurable, which I think it always is, but I'm not good at this stuff...), but that's because it's a subset of the right set, which has measure zero to start with.

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