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?



What is the difference $\{\tau\leq t\}\in (\mathcal{F})_t $ and $\{\tau<t\}\in (\mathcal{F})_t $ in the definition of stopping time?
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.


