Stopping Time – Difference Between ??t and ?

Question

Let $\tau$ be a random variable, which is defined on the filtered probability space $(\Omega, \mathcal{F}, (\mathcal{F})_{t\in T}, P)$ with values in $T$. In most cases, $T=[0,\infty]$. Then $\tau$ is called a stopping time (with respect to the filtration $(\mathcal{F})_{t\in T}$), if the following condition holds:
$\{\tau\leq t\}\in (\mathcal{F})_t $ for all $t\in T$. Can someone explain a little bit what is the difference $\{\tau\leq t\}\in (\mathcal{F})_t $ and $\{\tau<t\}\in (\mathcal{F})_t $?
Can we replace $\{\tau\leq t\}\in (\mathcal{F})_t $ using $\{\tau<t\}\in (\mathcal{F})_t $ in the definition of stopping time?

Answer 1

Two definitions:
(ST1) for all $t \in [0,\infty]$, $\{\tau < t\} \in \mathcal{F}_t$
(ST2) for all $t \in [0,\infty]$, $\{\tau \le t\} \in \mathcal{F}_t$
It is true that (ST2) $\Longrightarrow$ (ST1). Indeed, assume (ST2). Given $t$, we have $$ \{\tau < t\} = \bigcup_{s < t, s \in \mathbb Q}\{\tau \le s\} \in \mathcal{F}_t . $$ Thus (ST1) holds.

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The other direction, (ST1) $\Longrightarrow$ (ST2) need not be true. It is true if the filtration $\big(\mathcal{F}_t\big)$ is right-continuous in the sense $$ \mathcal{F}_t = \bigcap_{s > t}\mathcal{F}_s . \tag{$*$}$$

Here is a counterexample in case $(*)$ fails. Fix a value $t_0$ such that there is an event $E \notin \mathcal{F}_{t_0}$ but for all $s>t_0$, $E \in \mathcal{F}_s$. Define $$ \tau(\omega) := \begin{cases} t_0, & \omega\in E \\ \infty, & \omega\notin E \end{cases} $$ Now $\{\tau \le t_0\} = E \notin \mathcal{F}_{t_0}$, so that (S2) fails. But for any $t$, consider $\{\tau < t\}$:
\begin{align} \text{if }t\le t_0,\quad\text{then}\quad &\{\tau < t\} = \varnothing \in \mathcal{F}_t, \\ \text{if }t > t_0,\quad\text{then}\quad &\{\tau < t\} = E \in \mathcal{F}_t. \end{align} Thus (S1) holds.

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Proof of (ST1) $\Longrightarrow$ (ST2) using $(*)$. Assume (ST1).

Fix $t$, then for all $s > t$, $$ \{\tau \le t\} \subseteq \{\tau < s \} \in \mathcal{F}_s $$ and therefore $$ \{\tau \le t\} \subseteq \bigcap\mathcal{F}_s = \mathcal{F}_t . $$ Thus (ST2) holds.