If $\tau$ is an $(\mathcal F_t)_{t\in I}$-stopping time and $t\in I$, then $\tau\wedge t$ is $\mathcal F_t$-measurable

Question

Let

• $(\Omega,\mathcal A)$ be a measurable space
• $I\subseteq\mathbb R$
• $(\mathcal F_t)_{t\in I}$ be a filtration of $\mathcal A$
• $\tau$ be an $\mathcal F$-stopping time, i.e. $\tau:\Omega\to I\cup\sup I$ is $\mathcal A$-$\mathcal B(I\cup\sup I)$-measurable ($\mathcal B(E)$ denotes the Borel $\sigma$-algebra on $E\subseteq[-\infty,\infty])$ and $$\left\{\tau\le t\right\}\in\mathcal F_t\;\;\;\text{for all }t\in I\tag1$$
• $t\in I$

How can we show that $\tau\wedge t$ is $\mathcal F_t$-measurable?

Let $\iota$ denote the inclusion of $I\cup\sup I$ into $\overline{\mathbb R}$. Note that $$\mathcal B(A)=\left.\mathcal B(\overline{\mathbb R})\right|_A=\left\{A\cap B:B\in\mathcal B(\overline{\mathbb R})\right\}\;\;\;\text{for all }A\subseteq\overline{\mathbb R}\tag2$$ and hence $\tau$ is $\mathcal A$-$\mathcal B(I\cup\sup I)$-measurable iff $\iota\tau$ is $\mathcal A$-$\mathcal B(\overline{\mathbb R})$-measurable.

Let $t\in I$. By the former argument, $\tau\wedge t$ is $\mathcal F_t$-measurable iff $$\left\{\tau\wedge t\le b\right\}\in\mathcal F_t\tag3\;\;\;\text{for all }b\in\mathbb R\;.$$ If $b\ge t$, then $$\left\{\tau\wedge t\le b\right\}=\left\{\tau\le b\right\}\cup\underbrace{\left\{t\le b\right\}}_{=\:\Omega}=\Omega\in\mathcal F_t\;.\tag4$$ If $b\in(-\infty,t)$, then $\left\{t\le b\right\}=\emptyset$. In that case, if $(-\infty,b]\cap I=\emptyset$, then $$\left\{\tau\wedge t\le b\right\}=\left\{\tau\le b\right\}=\emptyset\in\mathcal F_t\tag5\;.$$

Now, we arrived at the crucial part: If $(-\infty,b]\cap I$ is closed and nonempty, then $$s:=\max\left((-\infty,b]\cap I\right)<t$$ is well-defined and hence $$\left\{\tau\wedge t\le b\right\}=\left\{\tau\le b\right\}=\left\{\tau\le s\right\}\in\mathcal F_s\subseteq\mathcal F_t\tag6\;.$$ But what can we do otherwise?

Answer 1

Edit answer to clarify your comment:

Continuing from where you stopped:

if $b\geq \sup I$ then $(\tau \leq b)=\Omega\in \mathcal{F}_t$. If $b<\sup I$, define $s=\sup I\cap(-\infty,b]$. If $s\in I$ we are done. Otherwise, put $s_n \uparrow s,s_n\in I $.

It holds that $$(\tau\leq b) = \bigcup_{n}(\tau\leq s_n)$$

The reason for the equality is that $\tau$ takes value in $I$ and therefore cant hit the interval $[s,b]$.

The union consists of $\mathcal{F}_{s_n}\subset\mathcal{F}_{t}$ sets.