Show that $\{ S < T\} \in \mathcal{F}_{S}$ and $\mathcal{F}_{T}$ where $S,T$ are stopping times

Question

Let $S,T$ be stopping times. Show that $\{ S < T\} \in \mathcal{F}_{S}$ and $\mathcal{F}_{T}$

Let's evaluate $\{ S < T\}\cap \{ T< t\}$

My idea:

$\{ S < T\}\cap \{ T< t\}=(\bigcup\limits_{r \in \mathbb Q, r < T}\{ S < r\})\cap \{ T< t\}$

But I am not sure whether I can just choose and $r < T$ since $T$ is not necessarily constant

Answer 1

If $\omega \in \Omega$ is such that $S(\omega)<T(\omega)$ then there exists a rational number $r \in \mathbb{Q} \cap [0,\infty)$ such that $S(\omega)<r<T(\omega)$. This gives

$$\{S<T\} = \bigcup_{r \in \mathbb{Q} \cap [0,\infty)} \{S<r\} \cap \{r<T\}.$$

Intersecting both sides with $\{T<t\}$, we get

\begin{align*} \{S<T\} \cap \{T<t\} &= \bigcup_{r \in \mathbb{Q} \cap [0,\infty)} \{S<r\} \cap \{r<T<t\} \\ &= \bigcup_{r \in \mathbb{Q} \cap [0,t)} \underbrace{\{S<r\}}_{\in \mathcal{F}_r \subseteq \mathcal{F}_t} \cap \underbrace{\{r<T<t\}}_{\in \mathcal{F}_t} \in \mathcal{F}_t. \end{align*}