Assume the filtration is right-continuous ($\mathcal{F}_{t+0} := \cap_{s>t}\mathcal{F_s} = \mathcal{F}_t$) and complete, then we have that $\tau(\omega)$ is a stopping time, if and only if $\{ \tau(\omega) < t\}$ $\in \mathcal{F}_t$, for all $t \geq 0$.

Here is the proof:

$\Rightarrow$: $\{ \tau(\omega) < t\}$ = $\cup_{n=1}^{\infty} \{ \tau(\omega)\leq t- 1/n\} \in \mathcal{F}_t;$

The "$\in$" is truth due to the definition of stopping time and the definition of sigma algebra; that is $\{ \tau(\omega)\leq t- 1/n\} \in \mathcal{F}_{t-1/n} \subset \mathcal{F}_t$ for all $n$, then the union is in $\mathcal{F_t}$.

$\Leftarrow$: $\{ \tau(\omega) \leq t\}$ = $\cap_{n=1}^{\infty} \{ \tau(\omega) < t+ 1/n\} \in \mathcal{F}_{t+0} = \mathcal{F}_t;$

One question is that:

The last equality follows from our right continuous assumption, but the why the "$\in$" is the case?

Another question is that:

Many books give the statement $\cap_{s>t}\mathcal{F_s} = \cap_{n=1}^{\infty}\mathcal{F}_{t+1/n}$, but they don't provide any proof of it, I've tried to prove it, but it doesn't make sense to me. can you give any thought about it?


1 Answer 1


Equivalently, $$\bigcap_{n=1}^{\infty} \{ \tau(\omega) < t+ 1/n\} \in \mathcal{F}_s$$ for every $s > t$, which is implied by, since $\{ \tau(\omega) < t+ 1/n\}$ is a decreasing sequence of events, that $\{ \tau(\omega) < t+ 1/n\} \in \mathcal{F}_s$ for all $n > N$ for some $N$. To see this just set $N = 1/(s-t)$.

For the second question, $$\bigcap_{s>t}\mathcal{F_s} \subseteq \bigcap_{n=1}^{\infty}\mathcal{F}_{t+1/n}$$ follows immediately from the fact that $t+1/n > t$ so the left intersection is over more sets.

The other direction is the archimedean property. Fix $s > t$. Then there is some positive integer $m$ such that $t+1/m < s$ showing $$\bigcap_{n=1}^{\infty}\mathcal{F}_{t+1/n} \subseteq \mathcal{F}_{t+1/m} \subseteq \mathcal{F}_s.$$ Now just intersect over all $s>t$.

  • $\begingroup$ i've been thinking this question for a long time, you save me, thanks! $\endgroup$
    – Wheea
    Commented Jul 9, 2020 at 3:53

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