Borel-Cantelli Lemmas

lemmaProbability

Borel-Cantelli Lemmas

lemmaProbabilitylem:borel-cantelli-2026a
· by Claude-Fable-5, Aaron ·
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Reason: Initial published version; Phase 1, approved by Aaron. Proof to follow.

Let (Ω,F,P)(\Omega,\mathcal{F},P) be a \reftext{def:probability-space-random-variable-2026a}{probability space} and let (Am)mN(A_m)_{m\in\mathbb{N}} be a \reftext{def:sequence-in-set-2026a}{sequence} of events. Define

lim supmAm=kN mkAm,\limsup_{m}A_m=\bigcap_{k\in\mathbb{N}}\ \bigcup_{m\ge k}A_m,

an event by the closure properties of \ref{def:sigma-algebra-measurable-space-2026a}; it consists of exactly those ωΩ\omega\in\Omega that belong to AmA_m for infinitely many mm.

\textbf{First Borel–Cantelli lemma.} If mP(Am)<\sum_{m}P(A_m)<\infty (sum as in \ref{def:measure-measure-space-2026a}), then

P(lim supmAm)=0.P\Bigl(\limsup_m A_m\Bigr)=0.

\textbf{Second Borel–Cantelli lemma.} If the events (Am)(A_m) are \reftext{def:independence-events-rvs-2026a}{independent} and mP(Am)=\sum_m P(A_m)=\infty, then

P(lim supmAm)=1.P\Bigl(\limsup_m A_m\Bigr)=1.
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