In stochastic process $\{X_t\}_{t\ge0}$ adapted to $\{\mathcal F_t\}_{t\ge0}$ where $\mathcal F_s\subset\mathcal F_t,\forall s<t$.
Many textbook say that $\{\mathcal F_t\}_{t\ge0}$ represents a increasing information but I don't quite understand. To simplify this question, I want to focus on random variable at first.
$X:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ is measurable. Someone says $\sigma(X)$ represents all information of $X$.
Here is my understanding:
Given $\sigma(X)$,what we know is "$\forall \omega_0\in\Omega,\forall a,b\in\mathbb R,\text{we can determine whether } \omega_0\in X^{-1}[a,b].$" i.e. we can determine whether $X(\omega_0)\in [a,b]$. So given $\omega_0$ we can obtian the value of $X(\omega_0)$ by adjust $a,b$. Is this right?
There is another concept "Stopping Time": If $\tau$ is a stopping time if $\{\tau<t\}\subset \mathcal F_t$.
My understanding is :
If we know $\mathcal F_{t_0}$ at time $t_0$, then we we can determine whether $\tau(\omega)<t_0$ for all $\omega\in\Omega$ but we can't determine the value of $\tau (\omega)$ for all $\omega$.
$A \in \sigma(X)$ means that we can decide for which $\omega \in \Omega$ the event $A$ happens if we know the value $X(\omega)$. For example, if we define
$$A := \{\omega \in \Omega; X^2(\omega)=1\},$$
then $A \in \sigma(X)$, since we can check whether $X^2(\omega)=1$ if we know $X(\omega)$. In contrast, if $Y$ is another (independent) random variable and we consider
$$A := \{\omega \in \Omega; X(\omega)+Y(\omega)=1\},$$
then $A \notin \sigma(X)$. Why? Because we have to know both $X(\omega)$ and $Y(\omega)$ in order to decide whether $X(\omega)+Y(\omega)=1$. Consequently, $A \in \sigma(X,Y) \supset \sigma(X)$.
For stopping times, the situation is quite similar. $\{\tau<t\} \in \mathcal{F}_t$ means that we can decide whether we stop before time $t$ (i.e. $\tau(\omega)<t$) if we know the sample path $s \mapsto X_s(\omega)$ up to time $t$. For example
$$\tau := \inf\{t \geq 0; X_t > a\}$$
is a stopping time whereas
$$\tau := \inf\{t \geq 0; X_{t+1}>a\}$$
is not a stopping time.