Show that stopping time $T$ is $\mathscr{F}_T$ measurable

Question

$T$ is a stopping time relative to a filtration if for every $t \ge 0$, $\{T \le t\} \in \mathscr{F}_t$.

For stopping time $T$ we define $\mathscr{F}_{T}$ as follows: Event $A$ is $\mathscr{F}_{T}$ measurable if and only if for any $t \ge 0$, $A \cap \{T \le t\} \in \mathscr{F}_{t}$.

Show that stopping time $T$ is $\mathscr{F}_T$ measurable.

I presume that means for any Borel set $B$, that $\{T \in B\} \cap \{T \le t \} \in \mathscr{F}_t$. I'm not sure how to do this or if this is even possible.

Answer 1

I assume it can be solved by just plugging in the definition. Indeed, $T$ is $\mathscr{F}_T$-measurable if and only if

$$ \{T\le t\} \in \mathscr{F}_T,\quad \forall t\ge 0,$$

which is true if and only if

$$\{ T\le \min(t, s)\} \in \mathscr{F}_s, \quad\forall t,s\ge 0.$$

However, by definition of filtration we know it is always the case that $\mathscr{F}_s \supset \mathscr{F}_{\min(t,s)}$. Hence the last statement is true.

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Moreover, observe that your $\mathscr{F}_T$ is well defined for any $T$, even if $T$ is not a stopping time.

Now suppose, on the contrary, that $T$ is not a stopping time, then $\{T\le t\} \notin \mathscr{F}_t$ for some $t \ge 0$. This means that for $s=t$, we have $\{T\le \min(s,t)\}\notin \mathscr{F}_s$, hence $T$ is not $\mathscr{F}_T$-measurable.

We conclude that $T$ is a stopping time if and only if $T$ is $\mathscr{F}_T$-measurable. Hence the latter can in fact serve as an alternative definition of $T$ being a stopping time.

Answer 2

It is enough to show that for any $a\in\mathbb{R}$, $\{T\leq a\}\in\mathcal{F}_T$ (why?)

Now, for any $t\geq0$, $$\{T\leq a\}\cap\{T\leq t\}=\{T\leq a\wedge t\}$$ If $a< 0$, then $\{T\leq a\wedge t\}=\emptyset\in\mathcal{F}_t$, while if $a\geq0$, $\{T\leq a\wedge t\}\in\mathcal{F}_{a\wedge t}\subset\mathcal{F}_t$.

Hence $\{T\leq a\}\in \mathcal{F}_T$ for all $a\in\mathbb{R}$.

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Here, $a\wedge t:=\min(a, t)$.