I'm interested in proving that any optional process is adapted. Here optional and adapted are used in the conventional sense but the precise definitions can be found here.
In the proof I'm actually using I'm confuse with two statements:
- If $X$ is optional then $X^n:=X 1_{|X|<n}$ is also optional
- Noting $\mathcal{O}_b$ for the set of all the optional bounded process and $\mathcal{M}:=\{$ all bounded RCLL process $\}$ we have $\mathcal{O}_b=\sigma(\mathcal{M})$ or at least $\mathcal{O}_b \subset \sigma(\mathcal{M})$
Of course all the objects are defined on the same probability space $(\Omega,\mathcal{F},\mathbb{P} )$
I'm stuck in proving that both this statements are valid and I will greatly appreciate any help :)
Thanks by advance,
Arthur
First observe that $|X|$ is optional provided that $X$ is: This follows from $$ \{|X|< b\}=\{X<b\}\cap\{X\ge 0\}\cup\{X>-b\}\cap\{X<0\}\,. $$
For $b> 0\,,$ $$ \{X^n< b\}=\{X< b\}\cup\{|X|\ge n\}\in{\cal O}\,. $$ For $b\le 0\,,$ $$ \{X^n< b\}=\{X< b\}\cap\{|X|<n\}\in{\cal O}\,. $$ Since the Borel $\sigma$-albgebra in $\mathbb R$ is generated by the intervals $(-\infty,b)$ this shows that $X^n$ is optional.
You are confusing the set of all optional bounded processes with the $\sigma$-algebra ${\cal O}_b$ they generate. You denoted by ${\cal M}$ the space of all RCLL (right continuous having left limits) processes which are bounded. Therefore, by definition in the link you provided, the optional $\sigma$-algebra ${\cal O}_b$ is equal to $\sigma({\cal M})\,.$
A priori, ${\cal O}_b\subset{\cal O}$ because the set of bounded optional processes is smaller than the set of optional processes. However, by point 1., every optional process $X$ is the limit of a sequence of bounded optional processes $X^n\,.$ Therefore, $X$ is measurable w.r.t. ${\cal O}_b$ which shows ${\cal O}\subset{\cal O}_b\,.$ To summarize: ${\cal O}_b={\cal O}=\sigma({\cal M})\,.$