= 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> $D=\{d_1,d_2,\ldots\}$, 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.
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