Preliminaries. We use two facts about any measure ΞΌ. Countable subadditivity: for events (Bmβ), ΞΌ(βmβBmβ)β€βmβΞΌ(Bmβ), by disjointifying (Dmβ=Bmβββl<mβBlβ), countable additivity, and monotonicity (ΞΌ(Dmβ)β€ΞΌ(Bmβ), since ΞΌ(Bmβ)=ΞΌ(Dmβ)+ΞΌ(BmββDmβ) 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 C1ββC2βββ― are events then P(βkβCkβ)=infkβP(Ckβ), since the complements increase to the complement of the intersection and P is finite.
First lemma. For every k, monotonicity and countable subadditivity give
P(mlimsupβAmβ)Β β€Β P(mβ₯kββAmβ)Β β€Β mβ₯kββP(Amβ),
and the right side tends to 0 as kββ, being the tail of a convergent series (its partial sums are bounded by the total sum minus the (kβ1)st partial sum). Hence P(limsupmβAmβ)=0.
Second lemma. First, complements of independent events are independent: replacing one event Aiβ by its complement Ξ©βAiβ preserves the product formula, since for any nonempty index set S containing i,
P((Ξ©βAiβ)β©jβSβ{i}ββAjβ)=P(jβSβ{i}ββAjβ)βP(jβSββAjβ)=(1βP(Aiβ))jβSβ{i}ββP(Ajβ),
by finite additivity and the product formula; iterating handles any number of complementations.
Next, 1βxβ€exp(βx) for every xβ₯0: the function h(x)=exp(βx)β(1βx) satisfies h(0)=0, and by claim 3 of Basic Properties of the Exponential Function together with the chain rule its derivative is hβ²(x)=1βexp(βx), which is β₯0 for xβ₯0 because exp(x)β₯1 there (claim 4) and hence exp(βx)=1/exp(x)β€1 (claim 2); by the mean value theorem, h is nondecreasing on [0,β), so h(x)β₯0.
Fix k and Kβ₯k. By independence of the complements, the product formula, the bound just proved, and the multiplicativity of exp (claim 1 of Basic Properties of the Exponential Function),
P(m=kβKβ(Ξ©βAmβ))=m=kβKβ(1βP(Amβ))Β β€Β m=kβKβexp(βP(Amβ))=exp(βm=kβKβP(Amβ)).
Since βmβP(Amβ)=β, the exponent tends to ββ as Kββ, so by claim 4 of Basic Properties of the Exponential Function the right side tends to 0; by continuity from above (the intersections decrease in K),
P(mβ₯kββ(Ξ©βAmβ))=0,henceP(mβ₯kββAmβ)=1.
The events βmβ₯kβAmβ decrease in k, so by continuity from above once more,
P(mlimsupβAmβ)=kinfβP(mβ₯kββAmβ)=1.β