Stopping Time Subset Proof My probability textbook has a really crappy proof for the following result.  Suppose $S$ and $T$ are stopping times, with $S(\omega) \le T(\omega)$ for all $\omega$.  Prove that $\mathcal{F}_S \subset \mathcal{F}_T$.  Can anyone give me a direct proof for this?
Take $A \in \mathcal{F}_S$.  Then $A \cap \{S = n\} \in \mathcal{F}_n$ for all $n$.  Since $\mathcal{F}_m \subset \mathcal{F}_n$ for all $m \le n$, we also have $A \cap \{S \le n\} \in \mathcal{F}_n$, we know $A \cap \{T = n\} = A \cap \{S \le n\} \cap \{T = n\} \in \mathcal{F}_n$.  Therefore, $A \in \mathcal{F}_T$ as well.
Starting from a probability space $(\Omega,\mathcal{F},P)$ with a filtration $\{\mathcal{F}_t\}$ we have the following definitions:

$$\mathcal{F}_S:=\{A\in\mathcal{F}:A\cap\{S\le t\}\in\mathcal{F}_t, \forall t\ge 0\}$$

The proof $\mathcal{F}_S\subset\mathcal{F}_T$ is straightforward. Let $A\in\mathcal{F}_S$ and $t\ge 0$. All we need to show is $A\cap\{T\le t\}\in\mathcal{F}_t,\forall t\ge 0$:

Since $T$ is a stopping time we know $\{T\le t\}\in\mathcal{F}_t$,$\forall t$. Hence

$$A\cap\{T\le t\}=A\cap \{S\le t\}\cap \{T\le t\}\in\mathcal{F}_t$$

Since both, $A\cap \{S\le t\}$ and $\{T\le t\}$ are elements of $\mathcal{F}_t$, so is their intersection. The equality is true since $\{S\le t\}\subset\{T\le t\}$.