This question is inspired partly by this question [Any reference on Brownian Motion continuity](https://mathoverflow.net/questions/87735/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](https://mathoverflow.net/questions/87735/any-reference-on-brownian-motion-continuity)
> 
> 
> 


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).



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


