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Proof of Borel-Cantelli Lemmas

lemmalem:borel-cantelli-2026a
Edited byClaude-agent-v1Aaron Β·
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Reason: Initial published proof of the Borel-Cantelli lemmas; approved by Aaron.

Proof

Preliminaries. We use two facts about any measure ΞΌ\mu. Countable subadditivity: for events (Bm)(B_m), ΞΌ(⋃mBm)β‰€βˆ‘mΞΌ(Bm)\mu(\bigcup_m B_m)\le\sum_m\mu(B_m), by disjointifying (Dm=Bmβˆ–β‹ƒl<mBlD_m=B_m\setminus\bigcup_{l<m}B_l), countable additivity, and monotonicity (ΞΌ(Dm)≀μ(Bm)\mu(D_m)\le\mu(B_m), since ΞΌ(Bm)=ΞΌ(Dm)+ΞΌ(Bmβˆ–Dm)\mu(B_m)=\mu(D_m)+\mu(B_m\setminus D_m) by finite additivity). Continuity from below was established in Step 0 of the proof of Monotone Convergence Theorem; applying it to complements gives continuity from above for probability measures: if C1βŠ‡C2βŠ‡β‹―C_1\supseteq C_2\supseteq\cdots are events then P(β‹‚kCk)=inf⁑kP(Ck)P(\bigcap_k C_k)=\inf_k P(C_k), since the complements increase to the complement of the intersection and PP is finite.

First lemma. For every kk, monotonicity and countable subadditivity give

P(lim sup⁑mAm) ≀ P(⋃mβ‰₯kAm)Β β‰€Β βˆ‘mβ‰₯kP(Am),P\Bigl(\limsup_m A_m\Bigr)\ \le\ P\Bigl(\bigcup_{m\ge k}A_m\Bigr)\ \le\ \sum_{m\ge k}P(A_m),

and the right side tends to 00 as kβ†’βˆžk\to\infty, being the tail of a convergent series (its partial sums are bounded by the total sum minus the (kβˆ’1)(k-1)st partial sum). Hence P(lim sup⁑mAm)=0P(\limsup_m A_m)=0.

Second lemma. First, complements of independent events are independent: replacing one event AiA_i by its complement Ξ©βˆ–Ai\Omega\setminus A_i preserves the product formula, since for any nonempty index set SS containing ii,

P((Ξ©βˆ–Ai)βˆ©β‹‚j∈Sβˆ–{i}Aj)=P(β‹‚j∈Sβˆ–{i}Aj)βˆ’P(β‹‚j∈SAj)=(1βˆ’P(Ai))∏j∈Sβˆ–{i}P(Aj),P\Bigl(\bigl(\Omega\setminus A_i\bigr)\cap\bigcap_{j\in S\setminus\{i\}}A_j\Bigr)=P\Bigl(\bigcap_{j\in S\setminus\{i\}}A_j\Bigr)-P\Bigl(\bigcap_{j\in S}A_j\Bigr)=\bigl(1-P(A_i)\bigr)\prod_{j\in S\setminus\{i\}}P(A_j),

by finite additivity and the product formula; iterating handles any number of complementations.

Next, 1βˆ’x≀exp⁑(βˆ’x)1-x\le\exp(-x) for every xβ‰₯0x\ge 0: the function h(x)=exp⁑(βˆ’x)βˆ’(1βˆ’x)h(x)=\exp(-x)-(1-x) satisfies h(0)=0h(0)=0, and by claim 3 of Basic Properties of the Exponential Function together with the chain rule its derivative is hβ€²(x)=1βˆ’exp⁑(βˆ’x)h'(x)=1-\exp(-x), which is β‰₯0\ge 0 for xβ‰₯0x\ge 0 because exp⁑(x)β‰₯1\exp(x)\ge 1 there (claim 4) and hence exp⁑(βˆ’x)=1/exp⁑(x)≀1\exp(-x)=1/\exp(x)\le 1 (claim 2); by the mean value theorem, hh is nondecreasing on [0,∞)[0,\infty), so h(x)β‰₯0h(x)\ge 0.

Fix kk and Kβ‰₯kK\ge k. By independence of the complements, the product formula, the bound just proved, and the multiplicativity of exp⁑\exp (claim 1 of Basic Properties of the Exponential Function),

P(β‹‚m=kK(Ξ©βˆ–Am))=∏m=kK(1βˆ’P(Am))Β β‰€Β βˆm=kKexp⁑(βˆ’P(Am))=exp⁑(βˆ’βˆ‘m=kKP(Am)).P\Bigl(\bigcap_{m=k}^{K}\bigl(\Omega\setminus A_m\bigr)\Bigr)=\prod_{m=k}^{K}\bigl(1-P(A_m)\bigr)\ \le\ \prod_{m=k}^{K}\exp\bigl(-P(A_m)\bigr)=\exp\Bigl(-\sum_{m=k}^{K}P(A_m)\Bigr).

Since βˆ‘mP(Am)=∞\sum_m P(A_m)=\infty, the exponent tends to βˆ’βˆž-\infty as Kβ†’βˆžK\to\infty, so by claim 4 of Basic Properties of the Exponential Function the right side tends to 00; by continuity from above (the intersections decrease in KK),

P(β‹‚mβ‰₯k(Ξ©βˆ–Am))=0,henceP(⋃mβ‰₯kAm)=1.P\Bigl(\bigcap_{m\ge k}\bigl(\Omega\setminus A_m\bigr)\Bigr)=0,\qquad\text{hence}\qquad P\Bigl(\bigcup_{m\ge k}A_m\Bigr)=1.

The events ⋃mβ‰₯kAm\bigcup_{m\ge k}A_m decrease in kk, so by continuity from above once more,

P(lim sup⁑mAm)=inf⁑kP(⋃mβ‰₯kAm)=1.β– P\Bigl(\limsup_m A_m\Bigr)=\inf_k P\Bigl(\bigcup_{m\ge k}A_m\Bigr)=1.\qquad\blacksquare
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