What is the name of the property for two random variables having disjoint supports? Let $X$ and $Y$ be two real-valued random variables.  There is a specific name for the property when the support for $X$ is disjoint from the support for $Y$.  I can't remember it, and I can't seem to find it in search results.  What is the name of that property?

More rigorously, let $R_X \subset \mathbb{R}$ be the values that $X$ will take with nonzero probability
$$
R_X = \arg \inf_{A \subset \mathbb{R}} \{ P(X \in A) = 1, A \, \text{closed} \}
$$
Similarly, let $R_Y \subset \mathbb{R}$ be the values that $Y$ will take with nonzero probability.
$$
R_Y = \arg \inf_{B \subset \mathbb{R}} \{ P(Y \in B) = 1, B \, \text{closed} \}
$$
Say $R_X \cap R_Y = \emptyset$.  What is this property called?
 A: After more research, the nearest property I can find is singularity of measure.  I don't think it's correct to say that random variables are themselves singular, but we can say it of their distributions.  I'll leave this answer unaccepted, just in case someone has a better answer.

Specifically, say the random variables $X, Y$ have the respective supports $R_X, R_Y$ disjoint as defined in the question, with distributions (or laws) $P^X, P^Y$.
Then $P^X, P^Y$ are singular.  Denoted $P^X \perp P^Y$
Proof:
By definition of support for probability distributions, we have that $P^X(R_X) = 1$.
Since we have assumed $R_X \cap R_Y = \emptyset$, we know that $R_Y \subset R_X^C$.
Also, $P^X(R_X^C) = 1 - P^X(R_X) = 1 - 1 = 0$.
Similarly, $P^Y(R_Y) = 1$ and $P^Y(R_Y^C) = 0$.
But $P^X(R_X^C) = 0 \implies P^X(R_Y) = 0$ by monotonicity of measure.
So we have that there exists $A,B \subset \mathbb{R}$ such that $A \cup B = \mathbb{R}$ and $P^X(A) = 0$ and $P^Y(B) = 0$.  So $P^X, P^Y$ are singular. $\square$

Note that, as per @whuber's comments, this only proves implication.  The converse is not true. As a counterexample, consider two random variables, one distributed uniformly over the closed unit interval, and the other distributed by Dirac delta centered in the midpoint:
$$
X \sim \text{Uniform}[0,1]\\
Y \sim \delta(\frac{1}{2})
$$
$P^X$ and $P^Y$ are mutually singular since $P^X(\{ \frac{1}{2} \}) = 0$ and $P^Y(\mathbb{R} \setminus \{ \frac{1}{2} \} ) = 0$, but the support of the former $R_X = [0,1]$ contains the support of the latter $R_Y = \{ \frac{1}{2} \}$.
