Let $(\Omega ,\mathcal F,\mathbb P)$ a probability space. I'm trying to understand better $\sigma -$algebra in probability, in particular the $\sigma -$algebra $$\mathcal F_T=\{A\in \mathcal F\mid A\cap \{T\leq t\}\in \mathcal F_t, \forall t\geq 0\},$$ where $T$ is a stopping time and $(\mathcal F_t)$ is a filtration.
Let $(X_n)_{n\in\mathbb N}$ independant Bernoulli r.v. Set $$S=\inf\left\{n\in\mathbb N\mid \sum_{k=1}^nX_i\geq 1\right \}$$ and $$T=\inf\left\{n\in\mathbb N\mid \sum_{k=1}^n X_k\geq 2\right \}.$$
It's written that $S\in \mathcal F_T$ but $S\notin \sigma (T)$, and I don't really understand why. The fact that $S\in \mathcal F_T$ I can imagine that it come from the fact that if $T$ occur, then $S$ occurred. It's what I read (even if I don't really understand in what this is related to $\mathcal F_T$ I totally agree with the last sentence). But Why isn't it in $\sigma (T)$ ? Because if we know $T$, then we know $S$, no ?
$\mathcal{F}_T$ is basically everything that you know by knowing the process up to time $T$. So $S\in\mathcal{F}_T$ since the stopping time $S$ has to happened before $T$.
On the other hand, $\sigma(T)$ "is" what knowing $T$ alone gives you, in other words, measurable sets recovered from deterministic functions of $T$. Now knowing $T$ doesn't exactly tell you what $S$ is, since we could have $S=T-1,T-2,\dots,1$ and there is no way to recover $S$ from $T$ alone, which is why $S\notin\sigma(T)$.