Proving $\mathbb{P}(X=Y)=0$ for independent random variables If I assume that $\mathbb{P}(X=x)=\mathbb{P}(Y=x)=0$ for all $x\in\mathbb{R}$ with $X$ and $Y$ independent random variables on the same probability space, does it follow that $\mathbb{P}(X=Y)=0$?
My reasoning went as follows: $\mathbb{P}(X=Y)=\sum_{x\in\mathbb{R}}\mathbb{P}(X=x \space\cap\space Y=x)=\sum_{x\in\mathbb{R}}\mathbb{P}(X=x)\mathbb{P}(Y=x)=\sum_{x\in\mathbb{R}}0=0$. However, I'm not sure that every concept that I have used here is applicable in the most general setting. Could anyone verify whether this proof is correct for all probability spaces?
 A: Let $X,Y$ be real-random variables such that $\forall t \in \mathbb R$ we have $\mathbb P(X=t) = \mathbb P(Y=t) = 0$.
Firstly: $X,Y$ need not be continuous random variable in the sense of having density with respect to Lebesgue measure! Condition $\mathbb P(X=t)=0$ for any $t \in \mathbb R$ only ensures that the CDF of $X$ $($and $Y)$ is a continuous function $($since $F_X(t) - F_X(t-) = \mathbb P(X=t) = 0 )$. There are examples ( look for Cantor distribution ) of distribution with continuous CDF but without density function.
Having said that, let's prove the statement. It is worth mentioning that you only need one of those variables to have continuous CDF for it to work.
Since $X,Y$ are independent, the distribution function of rv $(X,Y)$ is $\mu_X \otimes \mu_Y$.
We have $\mathbb P(X=Y) = \mathbb P( (X,Y) \in \Delta )$, where $\Delta = \{(x,y) \in \mathbb R^2 : x=y \}$.
Then:
$$ \mathbb P(X=Y) = \int_\Delta d(\mu_X \otimes \mu_Y) = \int_\mathbb R \int_{ \{x=y \}} d\mu_X(x)d\mu_Y(y) = \int_\mathbb{R} \mathbb P(X=y) d\mu_Y(y) = 0 $$
A: Suppose that $X$ and $Y$ are discrete random variables. If $\mathbb{P}(X=x)=\mathbb{P}(Y=x)=0$ for all $x\in\mathbb{R}$, it violates the basic property of the probability density function that it should sum (or integrate to) to 1. Therefore, $X$ and $Y$ are continuous random variables. 
For continuous random variables, the condition $\mathbb{P}(X=x)=\mathbb{P}(Y=x)=0$ for all $x\in\mathbb{R}$ always holds. Therefore, that part does not add any information. However, we note that $\mathbb{P}(X=Y)=\mathbb{P}(X-Y=0)$. Since $X-Y$ is a continuous random variable, the probability is zero.
