# Hitting time of continuous process is predictable

I find some difficulties in understanding a specific point of a proof in the George Lowther's Blog: Theorem 2 at this link. Here's the theorem (it is implicitly assumed that the filtration is complete and right-continous) and the point of the proof which is not clear to me.

Theorem Let $$X$$ be a continuous adapted process and $$K$$ a real constant. Then the hitting time $$\tau(\omega)=\inf\{t\geq 0| X_t(\omega)\geq K\}$$ is a predictable stopping time.

Proof. Consider the sequence $$\tau_n=\inf\{t\geq 0|X_t\geq K-\frac{1}{n}\}$$. By Right-continuity (and the fact that the filtration is complete and right-continuous) we can apply the Debut theorem so that $$\tau$$ as well as $$\tau_n$$ are stopping times. We clearly have that $$\tau_n\leq \tau$$ and since $$X$$ is left-continuous we also have $$X_{\tau_n}=K-\frac{1}{n}\neq K=X_{\tau}$$, so that $$\tau_n<\tau$$ on $$\tau>0$$. On $$\tau<\infty$$ the sequence $$\tau_n$$ is bounded from above by $$\tau$$ and, being non-decreasing, has a limit $$\sigma$$, i.e. $$\tau_n\uparrow\sigma$$. By left-continuity we get $$X_{\sigma}=K$$. Now I would say that we have finished and the sequence $$\tau_n$$ is an announcing sequence for $$\tau$$. In the blog it is however reported that $$n\wedge\tau_n$$ is an announcing sequence, I wonder what I am missing: why do we need to consider the minimum between $$n$$ and $$\tau_n$$ ?

If you don't take $$\tau_n \wedge n$$, there might be positive probability $$\tau_n = \tau = \infty$$. For an announcing sequence, we require $$\tau_n < \tau$$, so this is to rule out $$\tau_n = \tau$$ in the case where $$\tau = \infty$$.
• Thanks. So the reasoning could be the following. Define $A_n=\{\tau_n=+\infty\}$. If $\omega\in A_{n}$ we have $\tau_n(\omega)=+\infty$ and $\tau(\omega)=+\infty>0$ so that $n\wedge \tau_n(\omega)=n<+\infty=\tau(\omega)$ and the requirement $\tau_n<\tau$ whenever $\tau>0$ is respected. Correct? Feb 10, 2023 at 18:26