Probability that two singular random variables are equal to each other

Question

Let $X$ and $Y$ be independent random variables on $\mathbb{R}$. If they have an absolutely continuous distribution (w.r.t. the Lebesgue measure), we know that $P(X=Y)=0$, due to the fact that the event $X=Y$ is supported on a set with zero (Lebesgue) measure on $\mathbb{R}^2$. Now assume the variables have a discrete distribution, let's say $P(X=a_k)=P(Y=a_k)=p_k$, where $a_k\in\mathbb{R}$, $k$ runs through a countable set and $p_k\geqslant 0$ satisfy $\sum_k p_k=1$; in such a case $P(X=Y)=\sum_k p_k^2>0$. What happens if the two variables have a singular continuous distribution w.r.t. the Lebesgue measure, i.e. continuous but not absolutely continuous? Can we still say $P(X=Y)=0$?

Answer 1

$$\lim_{n\to \infty} \Pr(|X-Y|\leq 1/n)$$$$=\lim_{n\to \infty} \int_{-\infty}^{\infty}\mu_X(dx)\mu_Y\left[x-\frac{1}{n},x+\frac{1}{n}\right]=0$$ by dominated convergence.