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CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-05
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Second Borel-Cantelli lemma under pairwise independence

Statement

Let (An)n∈N be pairwise independent events with ∑n=0∞P(An)=+∞. Then P(An i.o.)=1.

Facts & Assumptions

Given: Pairwise independent events (An)n∈N with ∑n=0∞P(An)=+∞.

[L1]

The frequency law gives ∑k=0n1Ak∑k=0nP(Ak)→1 almost surely. (Pairwise-independent Borel-Cantelli frequency law)

[L2]

The event {An i.o.} is the event that infinitely many of the An occur. (Limsup and the infinitely often event)

Proof

technique · direct
1.1givenL1

Let Sn:=∑k=0n1Ak and an:=∑k=0nP(Ak). The divergence hypothesis makes an→+∞, and [L1] gives Sn/an→1 almost surely. Therefore on a full-probability event there is N such that Sn≥an/2 for every n≥N, hence Sn→+∞.

2.1step 1.1L2∎

The partial counts Sn(ω) diverge to +∞ exactly when the event An occurs for infinitely many indices n. By [L2], this is precisely the event An i.o.. Since step 1.1 shows it has probability 1, the second Borel-Cantelli conclusion follows.

Depends on

Used by

Dependency tree · two levels

11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources