CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-05

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Second Borel-Cantelli lemma under pairwise independence

Statement

Let $(A_{n})_{n \in N}$ be pairwise independent events with $\sum_{n = 0}^{\infty} P (A_{n}) = + \infty .$ Then $P (A_{n} i . o .) = 1.$

Facts & Assumptions

Given: Pairwise independent events $(A_{n})_{n \in N}$ with $\sum_{n = 0}^{\infty} P (A_{n}) = + \infty$ .

[L1]

The frequency law gives $\frac{\sum_{k = 0}^{n} 1_{A_{k}}}{\sum_{k = 0}^{n} P (A_{k})} \to 1$ almost surely. (Pairwise-independent Borel-Cantelli frequency law)

[L2]

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

Proof

technique · direct

1.1

Let $S_{n} : = \sum_{k = 0}^{n} 1_{A_{k}}$ and $a_{n} : = \sum_{k = 0}^{n} P (A_{k})$ . The divergence hypothesis makes $a_{n} \to + \infty$ , and [L1] gives $S_{n} / a_{n} \to 1$ almost surely. Therefore on a full-probability event there is $N$ such that $S_{n} \geq a_{n} / 2$ for every $n \geq N$ , hence $S_{n} \to + \infty$ .

givenL1

2.1

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

step 1.1L2∎

Depends on

Used by

Nothing in the library uses this result yet.

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

Rick Durrett, Probability: Theory and Examples, 5th ed., Theorem 2.3.7 and Theorem 2.3.9 (standard reference, not scraped)