Unfortunately I've trouble to see the following: If you work with stochastic process $X$ you often want to approximate this in the following sense, define:
$ X^{(n)}(s,\omega) = X\left(\frac{(k+1)t}{2^n},\omega\right)\cdot \mathbf1\left\{s\in \left(\frac{kt}{2^n},\frac{(k+1)t}{2^n}\right]\right\} $
for $t>0,n\ge 1, k=0,1,\dots, 2^n-1$ and $\mathbf1$ is the charateristic function of a set.
Now suppose $X$ has continuous trajectories. I don't see why
$ \sum_{k=0}^{2^n-1} X\left(\frac{(k+1)t}{2^n},\omega\right)\cdot \mathbf1\left\{s\in \left(\frac{kt}{2^n},\frac{(k+1)t}{2^n}\right]\right\} \to X_s(\omega)$
for all $(s,\omega) \in [0,t]\times\Omega$ as $n\to \infty$. Even more it's enough that the paths of $X$ are right continuous. I'm very thankful for your answer. This question comes up while reading Brownian Motion and Stochastic Calculus (Karatszas, Shreve).
hulik