What is the difference $\{\tau\leq t\}\in (\mathcal{F})_t $ and $\{\tauLet  $\tau$ 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?
 A: You find explanation for these two concepts of stopping times f.i. in Hackenbroch/Thalmaier, Stochastische Analysis (1994), ch. 3.2 (Stopzeiten). Here a stopping time is a map $\tau : \Omega \to \overline{\mathbb{R}_+}$ with $\{\tau \leq t\} \in \cal{F}_t$ for all $t$ and a stopping time in extended sense, if $\{\tau \leq t\} \in \cal{F}_{t+}$, where $\cal{F}_{t+} := \bigcap_{s > t} \cal{F}_s$. In Bem. 3.10 it is shown that this holds iff $\{\tau < t\} \in \cal{F}_t$ for all $t$. Often the filtration $(\cal{F}_t)_{t \geq 0}$ has the property that $\cal{F}_t = \cal{F}_{t+}$ (by assumption) and then both types of stopping times are the same.
This is only one place where you can find information about these types of stopping times. A further good starting point are the books of Dellacherie and Mayer, Probabilities and Potential.
A: 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.

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.

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.
