This is a homework question so no answers please
The problem is: Find a process $X_{t}$ s.t. $\forall t_{0}\geq 0$ and $\varepsilon>0$ we have
$lim_{n\to \infty}P(|X_{t_{0}}-X_{t_{n}}|>\varepsilon)=0$ (Stoc. continuous)
but has discontinuities with probability one.
Any mistakes:
For Brownian motion $B_{t}$ over [0,1] and uniform variable U over [0,1] independent of $B_{t}$, let $X_{t}=B_{t}1_{t\neq U}$.
It has a discontinuity with probability one because $P(B(U)\neq 0)=1-P(B(U)= 0)=1-\int_{0}^{1}P(B(u)=0)du=1- \int_{0}^{1}\int_{0}^{0}p(u,x,y)dydu=1$
It is Stochastically continuous as $P(U=t)=0$ and so $lim_{t\to t_{0}}P[|X_{t_{n}}-X_{t_{0}}|>\varepsilon]=lim_{t\to t_{0}}P[|B_{t}-B_{t_{0}}|>\varepsilon]=0$.
It fails to be continuous a.s. as $P[\exists N s.t. \forall n>N |X_{t_{n}}-X_{t_{0}}|>\varepsilon]$ uses an interval of times i.e. $[t_{N},t_{0}]$, where U has probability $t_{0}-t_{N}$ of being realized.