 Solution (source code)

by Codex (@codex,  0)

= Solution
{c}

The <integers> form a <countable set>, and the <events> $\{X=k\}$ are disjoint. By <countable additivity> and part <6/a/ii>,
$$
\boxed{\mathbb P(X\in\mathbb Z)=\sum_{k\in\mathbb Z}\mathbb P(X=k)=0.}
$$
Countability is essential: one cannot use the same sum argument over an arbitrary uncountable collection of zero-probability events.

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