In a probability space $\left(\Omega,(\mathcal{F}_t)_{t=0,..,T},\mathcal{F},\mathbb{P}\right)$ let $\tau$ be a stopping time. Consider the definition of "stopped" filtration as
$$ \mathcal{F}_{\tau} = \left\{F\in\mathcal{F}\mid F\cap\left\{\tau\leq t\right\}\subset\mathcal{F}_t,t=0,...,T\right\}. $$
Let $X_t$ be and adapted stochastic process.
Theorem.
$X_{\tau}$ is $\mathcal{F}_{\tau}$ measurable.
I didn't completely understood the proof of the Theorem as provided by Pascucci (PDE and Martingale Methods in Option Pricing). The proof proceed as follows.
First notice that (obvious):
$$ X_{\tau} = \sum_{t=0}^T X_t\,\mathbb{1}_{\left\{\tau=t\right\}}. $$
hence it is enough to prove that $X_t\,\mathbb{1}_{\left\{\tau=t\right\}}$ is $\mathcal{F}_{\tau}$ measurable for all $t$. So we have to prove that
$$ \left\{X_t\,\mathbb{1}_{\left\{\tau=t\right\}}\in H\right\}\in\mathcal{F}_{\tau},\quad H\in\mathcal{B},\quad t=0,...,T. $$
Here it comes the problem. The author distinguishes between two possible cases for the Borel set $H$: either $0\notin H$ or $H=\left\{0\right\}$.
Didn't we miss something? I would divide the possible cases as 1) $0\notin H$ or 2) $0\in H$, of course $0\in H$ is more general than $H=\left\{0\right\}$.
Ok, I think I got it, it was quite simple, I am attaching the screenshot of the proof.