# $\tau=s \mathbf{1}_{A^c}+t\mathbf{1}_A$, $A \in \mathcal F_s$ is a stopping time

Let $$(\mathcal F_t)_{t\in T}$$ be a filtration.

Consider $$s in T and $$A \in \mathcal F_s$$.

I want to show that $$\tau=s \mathbf{1}_{A^c}+t\mathbf{1}_A$$ is a stopping time.

My thoughts: In both cases, $$\omega \in A^c$$ and $$\omega \in A$$, we have that $$\tau(\omega)=const.$$, so $$\tau$$ is a stopping time in both cases (since it's constant). Is this a valid 'proof'? Feels wrong somehow.

In case this is wrong, I thought about the following: We have to show that $$\{\tau\leq u\}\in \mathcal F_u\ \forall u \in T$$.

$$\{\tau \leq s\}=A^c\in F_s$$

But how can I show that $$\{\tau \leq u\}\in F_u$$ for $$u>s$$?

The random variable is $$\tau(\omega)=s\mathbf{1}_{A^c}(\omega)+t\mathbf{1}_{A}(\omega),\,\omega \in \Omega$$ Now notice that $$\{\omega \in \Omega: \tau(\omega)\leq u\}=\begin{cases} \Omega&u\geq t\\ A^c& s\leq u As $$\mathcal{F}_s\subseteq \mathcal{F}_u$$, then $$A^c \in \mathcal{F}_u,\,\forall u >s$$. The empty set is in all $$\sigma$$-algebras, and the claim follows.
• I am confused, given $\omega \in \Omega$ is $\tau(\omega)$ a constant for every $s$ and $t$?
• If $\omega \in A$, then $\tau(\omega)=t$. If $\omega \in A^c$, then $\tau(\omega)=s$ @UBM Nov 21, 2021 at 18:44