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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 $\tau_n$ ?

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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 $\tau = \infty$.

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    $\begingroup$ 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? $\endgroup$ Feb 10, 2023 at 18:26
  • $\begingroup$ @AlmostSureUser Yes, that's exactly right. $\endgroup$ Feb 10, 2023 at 19:21

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