I was reading this interesting discussion: Is the condition "sample paths are continuous" an appropriate part of the "characterization" of the Wiener process? and feel some follow up questions might be interesting.

Basically, all the other conditions in a standard Brownian motion definition specify the "law", the distribution. Path continuity however, is something “extra”, it surely makes our life much easier, but somehow, I feel it is not “essential”, and I wonder whether properties of a standard Brownian motion also (in some way) hold for Brownian motion without path continuity assumption (Let’s call such process the Pre-Brownian motion below).

For example, in one link of the post quoted above, Byron has constructed an elegant counter example to show that, without the explicit path continuity assumption, a Pre-Brownian motion may have no continuous sample paths at all. But we also know that, a Pre-Brownian motion is a Martingale, therefore, almost surely all sample paths have left/right limits everywhere (I hope I get it right here…).

So a natural question will be:

Is it true that, the left and right limits are all the same a.s. for a Pre-Brownian motion?

For another question, let

$M_t(\omega) = \sup_{s\in [0\ t]} B_s(\omega)$.

For standard B.M., we can derive the distribution for $M_t$. But for a pre-Brownian motion, $M_t$ might not even be measurable. But how about a modified definition, an “essential supremum”, defined as:

$\tilde M_t(\omega) = \sup\{a | \mu(\{s | B_s(\omega)>a, s\in [0,t]\})>0\}$

where $\mu$ is the standard Borel measure. Will such a definition of $\tilde M_t$ measurable? Has a same distribution as $M_t$ in a standard Brownian motion?

We may also construct similar questions for hitting times for a Pre-Brownian Motion.

[EDIT] John answered the first question below.
For the second question about running maximum, I am still struggling with a reasonable defination. To avoid taking superimum over uncountable number of points in time (which always leads to question of measurebility), how about let's define the pre-Brownian motion on $\Bbb Q^+$ only? Now $M_t$ is measurable, and we can talk about its distribution. Will it be the same as in standard B.M.?

  • $\begingroup$ The very definition of your $\tilde M_t(\omega)$ is problematic, in that $s\mapsto B_s(\omega)$ need not be measurable. $\endgroup$ Nov 20, 2015 at 19:24
  • $\begingroup$ You are right, perhaps we can try to use "outer measure" instead of Borel measure? but then I am not confident $\tilde M$ is measurable. Maybe it is hopeless here, because without some kind of path continuity, the sup operation consists of essentially uncountable number of operations. $\endgroup$
    – Jay.H
    Nov 20, 2015 at 19:40

1 Answer 1


Concerning your first question, you overstate the case for right/left limits slightly. Let's suppose that $B=(B_t)_{t\ge 0}$ is a pre-Brownian motion defined on a probability space $(\Omega,\mathcal F,\Bbb P)$. Then $B$ is indeed a martingale (with respect to its natural filtration). From this we know (thanks to Doob) that there exists $G\in\mathcal F$ with $\Bbb P(G)=1$ such that, for all $\omega\in G$, $$ X_t(\omega):=\lim_{q\to t, q>t,q\in\Bbb Q}B_q(\omega)\qquad\hbox{ exists }\forall t\ge 0. $$ (Let's define $X_t(\omega):=0$ for all $t\ge 0$ if $\omega\in G^c$.) The process $X$ so defined satisfies $$ \Bbb P[X_t=B_t]=1,\qquad\forall t\ge 0. $$ In particular, $X$ is also a pre-Brownian motion. Finally, $$ \Bbb P[\{\omega: t\mapsto X_t(\omega) \hbox{ is continuous on }[0,\infty)\}]=1, $$ so that $X$ is a Brownian motion.

The answer to your (revised) second question is YES, because for a standard Brownian motion $X$, $M_t^X:=\max_{0\le s\le t}X_s=\sup_{0\le q\le t,q\in\Bbb Q} X_q$ by path continuity, and the stochastic processes $\{X_q:q\ge 0,q\in\Bbb Q\}$ and $\{B_q:q\ge 0,q\in\Bbb Q\}$ (for $B$ the pre-Brownian motion) have the same distribution.

  • $\begingroup$ Thanks, I agree this addresses my first question. I added some thoughts about the running maximum, can you also take a look? $\endgroup$
    – Jay.H
    Nov 20, 2015 at 22:41

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