 Solution

ID: past-exam-of-the-integration-course-of-the-university-of-oxford/2021/3/b/i/solution

The points belonging to infinitely many

E_{n}

G = ⋂_{N \geq 1} ⋃_{n \geq N} E_{n} .

For every

N

0 \leq m (G) \leq \sum_{n \geq N} m (E_{n}) \leq \sum_{n \geq N} 2^{- n} = 2^{1 - N} ⟶ 0.

Thus

m (G) = 0

. Outside this null set, each point belongs to only finitely many

E_{n}

, so its indicator functions are eventually zero. Therefore

χ_{E_{n}} (x) \to 0 almost everywhere

. This is the summable-measure argument behind the first Borel-Cantelli lemma; no independence assumption is needed.

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