Let $X$ be a stochastic process defined on a probability space $(\omega ,\mathcal F,P)$ endowed with a filtration $(\mathcal F)_{t \ge0}$ and let $T$ , $T^\prime$ be $\mathcal F_{t}-$stopping times. Then $X^T(t)=X(t\land T)$ is said to be the process $X$ stopped at $T$

I want prove follow statment :

1-$T$ and $T \land T'$ are $F_T-$measurable.

2-If $T \le T'$, then $F_T \subseteq F_T'$·

3-If $X$ is $F_t$ progressive, then $X(T)$ is $F_T$-measurable on ${T < \infty}$.

  1. Recall the definition:

$\mathcal{F}_T =\{A\in \mathcal{F} : \forall t>0, A\cap (T\leq t) \in \mathcal{F}_t\}$

T is $\mathcal{F}_T$ measurable just by definition. To see the other measurability, check the definition. We want to show that

$(T \wedge T') \cap (T\leq t) \in \mathcal{F}_t$ for all $t>0$.

But this means that at time $t$ you need to know the minimum of $T$ and $T'$ when $T$ has already occured. This is true - you can know that, so this event is $\mathcal{F}_T$ measurable. This is the way I'm used to thinking about stopping times, it is not super rigorous but it is more intuitive.

  1. You probably mean that $P(T \leq T') =1$. Then the inclusion follows from the above definition:

let $A \in \mathcal{F}_T$. Write $T' \leq t = [(T'\leq t)\cap (T \leq t)] \cup [(T'\leq t)\cap (T > t)]$. Check what this really means. Now use the definition and the fact that $(T \leq t) \in \mathcal{F}_{T'}$.

  1. I do not know what is a progressive process, sorry.

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