Def.
Let a filtration $\left\{\mathcal{F}_{t}:t\ge 0 \right\}$ on a probability space $(\Omega, \mathcal{F},\mathbf{P}).$
- An $\left\{\mathcal{F}_{t}\right\}-{\color{BLUE} {\text{stopping time}}}$ is a random variable $T$ on $\Omega$ taking values in $[0,\infty]$ and satisfying the condition $\left\{\omega\in\Omega:T(\omega){\color{Red} \le } t \right\}\in\mathcal{F_t}$ for each $t\ge 0.$
- An $\left\{\mathcal{F}_{t}\right\}-{\color{BLUE} {\text{wide (broad) stopping time}}}$ is a random variable $T$ on $\Omega$ taking values in $[0,\infty]$ and satisfying the condition $\left\{\omega\in\Omega:T(\omega) {\color{Red} < } t \right\}\in\mathcal{F_t}$ for each $t\ge 0.$
Let $\left\{\mathcal{F}_{t}\right\}_{t\geq 0}$ be a filtration.Show that
$\tau$ be a wide (broad) stopping time $\Longleftrightarrow$ there is a decreasing sequence of stopping times $(\tau_{n})_{n\in \mathbb{Z}^{+}}$,such that converges to $\tau.$
$\Longleftarrow.$ $\left\{\tau_{n}\le t \right\}\in\mathcal{F}_{t}$ , hence $\left\{\tau_{n}< t \right\}\in\mathcal{F}_{t}$ . Then $$\bigcap_{n\in\mathbb{Z}^{+}}\left\{\tau_{n}< t \right\}^{c}=\bigcap_{n\in\mathbb{Z}^{+}}\left\{\tau_{n}\ge t \right\}=\left\{\tau=\underset{n\in\mathbb{Z}^{+}}{\text{inf}}\tau_{n}\ge t\right\}\in \mathcal{F}_{t}\Longrightarrow\left\{\tau < t \right\}\in \mathcal{F}_{t} .$$ $\Longrightarrow.$ The proof in the opposite direction does not seem obvious.I don't know how to construct this monotonically decreasing sequence $(\tau_{n})_{n\in \mathbb{Z}^{+}}.$
Take $\tau_n := \tau + \frac{1}{n}$. Then
$ \{ \tau_n \leq t \} = \{ \tau \leq t - \frac{1}{n} \} = \bigcap \limits_{q \in \mathbb{Q} \cap (0,\frac{1}{n}) } \{ \tau < t - \frac{1}{n} + q \} \in \mathcal{F}_t $,
since all $\{ \tau < t - \frac{1}{n} + q \} \in \mathcal{F}_{t - \frac{1}{n} + q} \subset \mathcal{F}_t $.
So the $\tau_n$ are stopping times with $\tau_n = \tau + \frac{1}{n} \downarrow \tau$.