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 $t\in\mathbb{R}_+\longmapsto X_n (t, \omega)$

Could someone help me please?

  • $\begingroup$ Do you think it is proper to change without notice the intervals $[-,-]$ into intervals $[-,-)$ in your question, AFTER I posted my answer? $\endgroup$ – Did Nov 5 '12 at 19:18
  • $\begingroup$ Why would it be a problem ? I thank you for your remark. You are perfectly right. I didn't thank you before beacause I was having dinner. I expect you understand. Thank you again! Now, I'm shamed for having posted this question because there is nothing to be understended here, all thing are pretty clear in the solutiion. Lack of attention! =) $\endgroup$ – Paul Nov 5 '12 at 20:00
  • $\begingroup$ It is a problem because it makes irrelevant every answer already posted. $\endgroup$ – Did Nov 5 '12 at 20:43

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.

  • $\begingroup$ Thank you for the notes also! $\endgroup$ – Paul Nov 5 '12 at 20:06

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