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.

OP

- $W(0) = 0$.
- 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.
- For all $0 \le t_1 \le t_2$ , $W(t_2) - W(t_1)$ is normally distributed with mean 0 and variance $\sigma^2\,(t_2 - t_1)$.

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).