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Sep 26, 2011 at 21:44 comment added pde_bk However those other proofs mentioned above give stronger results as said by George and Michael in their posts. Thank everyone again!
Sep 26, 2011 at 21:41 comment added pde_bk After I got a relatively complete understanding(I think;-),let me comment a bit. The proof in spirit of BSteinhurst's post: Contradiction is got by working on $E[|W_s-W_t|^2]$($E[|W_s-W_t|]$ is not easy to work on). Certainly attention should be paid to classification of bounded and unbounded cases. There are other proofs, such as the one on page67 of Doob’s Stochastic Process and the one on page10 of Kallianpur’s Stochastic Filtering Theory, as well as Y.Sun's paper Michael pointed out in the answer. But seemingly they require mutual independence instead of only pair independence.
Sep 24, 2011 at 17:30 vote accept pde_bk
Sep 24, 2011 at 17:14 answer added Michael Greinecker timeline score: 5
Sep 24, 2011 at 17:14 comment added George Lowther @duomededuome: The process can't even be (jointly) measurable, which is a much weaker condition than continuity.
Sep 24, 2011 at 16:39 answer added BSteinhurst timeline score: 2
Sep 24, 2011 at 16:25 comment added pde_bk @Paul: Oksendal's Stochastic Differential Equation. chp3. Certainly he is talking about nontrivial cases.
Sep 24, 2011 at 16:15 comment added Paul Tupper Doesn't the process $W_t=0$ for all $t$ almost surely satisfy all the conditions and also have continuous paths? What book is this?
Sep 24, 2011 at 15:00 history asked pde_bk CC BY-SA 3.0