 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-iii/6/a/iii/solution

{X = k}

are disjoint. By countable additivity and part ii,

P (X \in Z) = k \in Z \sum 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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