Countable sets have zero probability under a density

If a random variable has a probability density function, each singleton has probability zero because its density has zero integral there. For a countable set $D=\{d_1,d_2,\dots \}$, remove duplicates to make the events $\{X=d_i\}$ disjoint. Countable additivity gives $\mathbb{P}(X\in D)={\sum }_i\mathbb{P}(X=d_i)=0$. The proof does not extend to uncountable unions merely because each constituent has zero probability.