"Brownian motion" without assuming continuity of path at origin of state space

Question

This question is inspired partly by this question Any reference on Brownian Motion continuity. In this post, the author asked if the following three axioms can define a Brownian motion without assuming the continuity axiom

"4.$W(t)$ is continuous with probability one. i.e. $\lim _{h\rightarrow 0}P(|W(t+h)-W(t)|>\epsilon )=0,\forall \epsilon>0, t\in S$" By assuming this, Brownian motion is a special case of Levy process.

1. $W(0) = 0$.
2. For all $0 \le t_1 \le t_2 \le t_3 \le t_4$, $W(t_2) - W(t_1)$ and $W(t_4) - W(t_3)$ are independent random variables.
3. For all $0 \le t_1 \le t_2$ , $W(t_2) - W(t_1)$ is normally distributed with mean 0 and variance $\sigma^2\,(t_2 - t_1)$.

OP

In fact, [Karlin&Taylor] defined Brownian motion to be a stochastic process satisfying 1,2,3 axioms with an additional stipulation

"4*.$W(t)$ is continuous at $t=0$"

And they derived continuity of Brownian path as a result using Karhunen–Loève representation Theorem at Sec 7.4. A possible relevant clue is that we always require the characteristic function $E(e^{Xt})$ to be continuous around origin in order to determine a random variables in distribution via characteristic functions. So I guess axiom 4* is a guarantee that some transform exists?

My question is that: If we only assume axiom 1,2,3 on a stochastic process as above, can we construct a stochastic process $W(t)$ that is not a Brownian motion (which is defined as a stochastic process with axiom 1,2,3,4 satisfied OR axiom 1,2,3,4* satisfied in [Karlin&Taylor])? OR Alternatively, is the continuity axiom redundant? (I do not think so but it does not seem very clear how I can construct a counter example to illustrate the point.)

After looking at @Bjørn Kjos-Hanssen's answer, I felt a more appropriate question to ask is that if there is a stochastic process that is not càdlàg and satisfies axioms 1,2,3.

[Karlin&Taylor]Karlin, S., and H. M. Taylor. "A first course in stochastic processes" Academic Press. New York (1975).

Answer 1

Yes, let $W$ be Brownian motion and let $V$ be the following modification: $V_t=W_t$ except that we pick a number $s\in [0,1]$ according to the uniform distribution, independently of $W$, and let $V_s=0$.

Then 1,2,3 are satisfied but the sample path of $V$ is almost surely discontinuous (at $s$).

To get almost sure discontinuity at 0, use $s_1,s_2,\dots$ in the unit interval, all $s_i$ independent of each other and of $W$, with say $$V_{s_i}=1\ne 0$$ for all $i$. Note that $S=\{s_i:i\ge 1\}$ is almost surely dense in the unit interval, but $S$ is random relative to $W$ so it will be disjoint from any countable set of $t$'s considered "in advance".

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Background: $W$ will be uniformly continuous on the rationals by axioms 1, 2, 3. But without axiom 4, the question whether the paths are continuous almost doesn't make sense -- the set $$\{f:f \text { is continuous}\} $$ is not measurable. So one solves this by redefining $W$ to be the unique continuous extension of the values of $W$ on the rationals.