# 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

-
Usually these results are stated with an a.s. (almost sure) precision rather than "for all $\omega$" – Ilya Feb 29 '12 at 11:28

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.
@ Ilya: If I want to approximate a left-continuous process, then I would do the following: $I_{n,k}(s)=\mathbf1_{[t_{k,n},t_{k+1,n})}(s)$, $t_{k,n}$ as above and $f_{n,k}=f(t_{k,n})$ is this right? – user20869 Feb 29 '12 at 12:50