How can $\mathcal{F}_{T}$ be a $\sigma$-algebra in this context of stopping time?

Question

I'm reading about stopping time in my lecture note:

Let $\left(\Omega, \mathcal{F},(\mathcal{F}_{n})_{n \in \mathbb{N}}, \mathbb P\right)$ be a filtered probability space.

In Remark 61, they said $X_{T}$ is $\mathcal{F}_{T}$-measurable. I can not understand whether $\mathcal{F}_{T}$ is a $\sigma$-algebra or it is a random variable. The latter case doesn't make sense to me, but the $T$ in $\mathcal{F}_{T}$ confuses me.

Could you please elaborate on this point? Many thanks!

Answer 1

Ah, I got it. My confusion arises because I misunderstand the definition of $\mathcal{F}_{T}$.

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Let $\left(\Omega, \mathcal{F},(\mathcal{F}_{n})_{n \in \mathbb{N}}, \mathbb P\right)$ be a filtered probability space. A random variable $T$ taking values in $I \cup\{\infty\}$ is a stopping time (or optional time) if $\{\omega: T(\omega)=n\} \in \mathcal{F}_{n}$ for all $n \in I$.

Let $T$ be a stopping time. The sigma field $\mathcal{F}_{T}$, sometimes called "the past before $T$" is $$\mathcal{F}_{T}=\left\{\Lambda \in \mathcal{F} \mid \forall n \in I:\Lambda \cap\{T=n\} \in \mathcal{F}_{n} \right\}$$