65.14. A TECHNICAL INTEGRATION BY PARTS RESULT 2271

Thus

P(Akε) =∞

∑m=1

P(Akε ∩ ([τm = ∞]\ [τm−1 < ∞])) (65.14.28)

AlsoP(Akε ∩ ([τm = ∞]\ [τm−1 < ∞]))≤ P([τm = ∞]\ [τm−1 < ∞])

which is summable because these are disjoint sets. Hence one can apply the dominatedconvergence theorem in 65.14.28 and conclude

limk→∞

P(Akε) =∞

∑m=1

limk→∞

P(Akε ∩ ([τm = ∞]\ [τm−1 < ∞])) = 0