# Why is a constant always a stopping time?

In my book, it states that any constant, including, $$\infty$$, is a stopping time. But why is this? I thought the reason might be that a constant is measurable with regards to the trivial sigma-algebra, i.e. $$\{\emptyset, \Omega\}$$ and since $$\emptyset \in \mathcal{F}_n$$ the results follow? Or what is the reason?

Thanks.

• We're going to need more context here. Commented Jun 18, 2022 at 15:48

Let $$\tau=k,\,k \in \mathbb{N}\cup \{\infty\}$$. Recalling the definition, $$\tau$$ is a $$\mathscr{F}_n$$-stopping time if $$\{\tau\leq n\}\in \mathscr{F}_n,\,\forall n \in \mathbb{N}$$. Now if $$k<\infty$$ we have $$\{\tau\leq n\}=\begin{cases}\Omega&n\geq k\\ \emptyset &n while if $$k=\infty$$, $$\{\tau\leq n\}=\emptyset$$ for all $$n$$ so $$\tau$$ is always a stopping time.
• Do I understand it correctly: $\{\tau \leqslant n\} = \{\tau(\omega) \leqslant n\}$ is the set of instances where the stopping time $\tau$ is less or equal than $n$? And if $k<\infty$ is $\tau < n$ always true if $n\geqslant k$ and never true if $n<k$? Commented Jan 17, 2023 at 18:03
• @HelloWorld By definition $\{\tau \leq n\}=\{\omega: \tau(\omega) \leq n\}$. Since $\tau(\omega)=k$ for all $\omega$, we have that $\{\omega:\tau(\omega)\leq n\}=\Omega$ for all $n\geq k$ (so, always) and $\{\omega:\tau(\omega)\leq n\}=\emptyset$ for all $n<k$ (so, never). Commented Jan 17, 2023 at 18:24
• Thanks! Yes I meant the $\{\omega...\}$ not $\{\tau(\omega)\}...$ ;) Commented Jan 17, 2023 at 20:19