By Daniel Revuz, Marc Yor

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**Example text**

Forapoints (resp. t) in D, there is an increasing sequence (sm) (resp. (tm)) of points in D such that sm (resp. tm) is in Dm for each m, Sm :::;; s (tm :::;; t) and sm = s (tm = t) from some m on. lf ls- tl :::;; 2-m, then either Sm = tm or (sm, tm) E Am and 26 I. '"- x,'" + i=m L (X,,- x,i+,) i=m where the series are actually finite sums. It follows that 00 IX. - X,l :::;; Km +2 L m+l 00 Ki :::;; 2 L Ki. l/lt- slcr; s, t e D, s-:/- t}, we have M .. l; s, t e D, s -:1- t} L 2icrKi. 00 i=O For y ~ 1 and a < ejy, we get, with J' = 2cr+t J, IIMcrlly:::;; J' L 2icr11Kill 1 :::;; J' L 2i(cr-(•/Y)) < 00.

S. zero Lebesgue measure. 18) Exercise. Let B be the standardlinear BM. s 1 B•. 7) Chap. 111. 19) Exercise. Rd with llx II = process (x,X,) isalinear BM. 1, the § 1. Examples of Stochastic Processes. Brownian Motion 23 2°) Prove that the converse is false. 20) Exercise (Polar functions and points). A continuous function f from ~+ into ~ 2 is said to be polar for BM 2 if for any x E ~ 2 , P[Fx] = 0 where f'x = {3t > 0: B, + x = f(t)} (the measurability of the set f'x will follow from results in Sect.

1 since (> is arbitrary. We shall now prove the reverse inequality. Again, we pick (> E ]0, 1[ and 6 > 0 such that (1 + 6)(1 - (>) > 1 + b. Let K be the set ofintegers k suchthat 0 < k = j - i ~ 2n~ with 0 ~ i < j ~ 2n; using the inequality 1oo exp( -b 2/2)db < 1oo exp( -b 2j2)(bja)db = a- 1exp( -a2/2), and setting L = P[maxkex(IBi 2-n- B; 2-nl/h(k2-n)) L ~ L: ~- keK ~ D 2 fo L 1 00 (1+e)h(k2-") ~ 1 + 6], we have (kTnf 112 exp(- x 2j2(kTn)) dx (log(k- 12nW 1i 2 exp( -(1 keK + 6) 2 log(k- 12n)) where Dis a constant which may vary from line to line.