A relationship between stopping times and their $\sigma$-algebras

54 Views Asked by At

I'm studying Stochast integration and stochastic differential equation by Protter and got confused by the solution of the very first exercise provided here.

So

Exercise 1. Let $S,T$ be stopping times, $S \leq T$ a.s. Show $\mathcal{F}_S \subset \mathcal{F}_T$

The provided solution starts like this:

$\forall A \subset \mathcal{F}_S, \forall t \geq 0, A \cap \{T \leq t\} = (A \cap \{S \leq t\}) \cap \{T \leq t\}$ since $\{T \leq t\} \subset \{S \leq t\}$

But $S \leq T$ almost surely. It means that there is a set $N \subset \mathcal{F}_0$ (usual hypotheses) with $\mathbf{P}(N) = 0$ such that $ \forall \omega \in N : S(\omega) > T(\omega)$ so $\exists t_0 : \{T \leq t_0\} \not\subset \{S \leq t_0\}$

Thus the start of the proof need to be corrected like that:

$\forall A \subset \mathcal{F}_S, \forall t \geq 0, A \cap \{T \leq t\} = (A \cap (\{S \leq t\} \cup N)) \cap \{T \leq t\}$ since $\{T \leq t\} \subset \{S \leq t\} \cup N$

The rest of the solution works if we replace $\{S \leq t\}$ by $\{S \leq t\} \cup N$:

Since $A \cap (\{S \leq t\} \cup N) \subset \mathcal{F}_t$ (since $N \subset \mathcal{F}_0 \subset \mathcal{F}_t$) and $\{T \leq t\} \subset \mathcal{F}_t$ we have $A \cap \{T \leq t\} \subset \mathcal{F}_t$ so $\mathcal{F}_S \subset \mathcal{F}_T$

Am I right or it is an unnecessary complication of the correct solution?