Pointwise convergence almost surely of an Approximation sequence 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 
 A: Well, this problem is rather from analysis since the only probabilistic fact is used here is that for a.a. $\omega\in \Omega$ the function $s\mapsto X(s,\omega)$ is right-continuous in $s$. For the simplicity of notation, let us put the function $f:[0,t]\to \mathbb R$ to be right-continuous:
$$
\lim\limits_{h\downarrow 0}f(s+h) = f(s)\quad\text{ for all }s\in[0,t) 
$$
and define 
$$
\begin{cases}
t_{k,n} &= \frac{kt}{2^n},
\\
\\
f_{k,n} &= f\left(t_{k+1,n}\right),
\\
\\
I_{k,n}(s) &= 1_{(t_{k,n},t_{k+1,n}]}(s)
\end{cases}
$$ for $n\in\mathbb N $ and $k = 0,1,2,\dots,2^{n}-1$. Let us show that for any $s\in [0,t)$ we have
$$
\lim\limits_{n\to\infty}\sum\limits_{k=0}^{2^n-1}f_{k,n}I_{k,n}(s) = f(s).
$$
Fix $s$ and let $\hat k(n) = \{k:I_{k,n}(s) = 1\}$ be the index of the interval of the partition where $s$ falls. Then:
$$
\lim\limits_{n\to\infty}\sum\limits_{k=0}^{2^n-1}f_{k,n}I_{k,n}(s) = f_{\hat k(n),n} = f\left(t_{\hat k(n)+1}\right)
$$
and $s<t_{\hat k(n)+1}$ toghether with the fact that $t_{\hat k(n)+1}-s\leq 2^{-n}t$ imply that $t_{\hat k(n)+1}\downarrow s$ with $n\to\infty$. As a result, $\lim\limits_{n\to\infty}f\left(t_{\hat k(n)+1}\right) = f(s)$ by right-continuiuty. Finally, if $s =t$ then 
$$
\lim\limits_{n\to\infty}\sum\limits_{k=0}^{2^n-1}f_{k,n}I_{k,n}(t) = f(t)
$$
and there is nothing to prove.
