Let $X\sim P$ and $Y\sim Q$ be independent random variables with their subsets of atoms $A_P$ and $A_Q$ respectively.
What is the probability that $X=Y$?
I am aware that there can be at most countably many atoms. Intuitively, I would say that $$ \mathbb P\{X=Y\} = \sum _{x\in A_P\cap A_Q} P\{x\}Q\{x\}. $$
So if either variable has no atoms, then this probability is immediately zero.
Attempted to solve special case $P=Q$ with $$ \mathbb P\{X=Y\} = EI_{\{X=Y\}} =: EI_C + EI_D, $$ where $D = \{\omega \mid X(\omega) = Y(\omega) \in A_P\}$ and $C = \{\omega \mid X(\omega) = Y(\omega), X(\omega)\notin A_P\}$. For the atoms it would then work out as $$ EI_D = \mathbb P\left\{\bigcup _{x\in A_P} \{X=x\}\cap \{Y=x\} \right\} = \sum _{x\in A_P} \mathbb P\{X=x\}\mathbb P\{Y=x\} = \sum _{x\in A_P} P\{x\}^2, $$ but I don't understand how to evaluate $EI_C$ or if and why it should be zero.
My way to see this is to use $(X,Y)\sim P\otimes Q$ which charaterizes the independence and Fubini's theorem. For the diagonal $\Delta=\{(x,y)\in\mathbb R^2: x=y\}$ one gets $$\mathbb P(\{X=Y\}=P\otimes Q(\Delta) =\int\int I_\Delta(x,y)dP(x)dQ(y) =\int\mathbb P(\{X=y\})dQ(y).$$ There are at most countably many $y\in\mathbb R$ with $\mathbb P(\{X=y\})\neq 0$ so that the integral becomes the sum over these probabilities weighted with $Q(y)$, i.e., $$\mathbb P(\{X=Y\})=\sum \mathbb P(\{X=y\})\mathbb P(\{Y=y\}).$$