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Why is Brownian motion required to be almost surely continuous instead of merely continuous?

For example, this is stated as condition 2 in this article in section 1, Characterizations of the Wiener process. What is an example of a Brownian motion where there is a point at which the motion is not continuous?

user50229
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You are confusing the measure on path space with Lebesgue measure. The "almost everywhere" refers to the former: almost every individual path can be taken to be continuous everywhere. Indeed, the Wikipedia page you link to says that Brownian motion is "almost surely everywhere continuous".

In other words, if $\mathbb{P}$ is Wiener measure on a suitable measurable space $(\Omega, \mathcal{F})$, then there is a set $N \subset \Omega$ of $\mathbb{P}-$measure $0$ such that for all $\omega \notin N$, $t \rightarrow \omega(t)$ is continuous for all $t \in [0,1]$.

snar
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  • This is maybe not the best example, since if you define Brownian motion as the canonical process on the space of continuous paths, by definition it is continuous surely. This issue is more relevant if you define Brownian motion on some other probability space (often $\mathbb{R}^{[0,1]}$). – Nate Eldredge Dec 12 '14 at 17:58
  • @NateEldredge Thank you, that was sloppy. I've made a trivial modification and will wait for OP to respond. As an aside, can't one always redefine the measure such that every path is continuous? It could be an uncountable set, but contained within a set of measure 0. – snar Dec 12 '14 at 19:04