# Why all previsible processes are also optional?

My doubt concerns a step on demonstration of the inclusion of the set of previsible processes in the set of optional processes.

The idea of the demonstration consists in:

Given a filtered probability space $(\Omega, \mathcal{F}, (\mathcal{F}_t), \mathbb{P} )$ ($t \in \mathbb{R}_+$ for example), lets take $X =(X_t)_{t\geq0}$ an $\mathcal{F}_t$-adapted and left-continuous process. Then we define the following sequence of processes: $$\forall n \in \mathbb{N}, \quad X_n (t, \omega)= \sum_{k\geq0} X_{\frac{k}{2^n }}(\omega) \mathbf{1}_{\left[\frac{k}{2^n }, \frac{k+1}{2^n }\right)} (t)$$

Clearly, we have that $X_n (t, \omega)\underset{n \rightarrow \infty} {\overset{w-a.e.}{\longrightarrow }} X(t, \omega)$ which is left-continuous by hypothesis. However, we conclude that it is also right-continuous as a limit of the sequence of right continuous processes $(X_n (t, \omega))_{n\in \mathbb{N}}$.

I have some problems with understanding the right continuity of $t\in\mathbb{R}_+\longmapsto X_n (t, \omega)$

Do you think it is proper to change without notice the intervals $[-,-]$ into intervals $[-,-)$ in your question, AFTER I posted my answer? –  Did Nov 5 '12 at 19:18
This is because your definition of $X_n$ is wrong. To get right-continuous processes $X_n$, consider $$X_n (t, \omega)= \sum_{k\geqslant0} X_{k/2^n}(\omega)\cdot\mathbf{1}_{k/2^n\leqslant t\lt(k+1)/2^n},$$ and/or see these notes.