Let $B_t$ the standard Brownian Motion. It's well known that Brownian Motion admits a continuous modification, call it $B_t$ as well.
I think we can use this to construct a discontinuous modification of the Brownian Motion as follow:
$$\tilde B_t(\omega)=B_t(\omega)\Bbb{1}_{\Bbb{R}\setminus{\Bbb{Q}}}+(B_t(\omega)+1)\Bbb{1}_{\Bbb{Q}}.$$
To check that is correct I need to prove that
- $(\omega,t)\mapsto \tilde B_t(\omega)$ is measurable
- That there exist $\tilde\Omega\in\mathcal{F}, P(\tilde\Omega)=1$ such that $g:t\mapsto \tilde B_t(\omega)$ is discontinuous.
I think that $2.$ comes from the fact that $\Bbb{Q}$ is dense as if $x\notin \Bbb{Q}$ we can find for all $a>0$ a rational $r$ such that $\vert x-r\vert<a$ so that $g(r)=B_t(\omega)+1$ and therefore $\vert g(x)-g(p)\vert=1.$ Which is strange is that I didn't use that $t\mapsto B_t(w)$ is continuous.
Is it correct ?
Not sure how to check measurability.
Of course I forgot to think that I must prove that is a modification (thanks to saz)
Regarding your construction: $(\tilde{B}_t)$ is a modification of $(B_t)$ if for all $t$, $\mathbb{P}[B_t=\tilde{B}_t]=1$. How high do you this is this probability for $t\in \mathbb{Q}$? Hint: How likely is it, that $1=0$?
You need a slightly more probabilistic approach. Consider $Y \sim U[0,1]$ (does it matter, which interval I pick?). Then consider $\tilde{B}_t=B_t+\mathbf{1}_{\{U\}}(t)$.
It is easy to show that this is measurable (sums of measurable sets are measurable, compositions of measurable maps are measurable). Also, it should be clear that $\tilde{B}_t$ cannot be continuous. Finally, show that $(\tilde{B}_t)$ is a modification of $(B_t)$ because - how likely is it that $U=t$ for a fixed $t$?