Why Test Only Rational Times?
A stopping time is defined through information available at every real time: by time \(t\), one must be able to determine whether the random time has occurred. At first this appears to require checking uncountably many events. The rational-time test reduces the task to a countable family. Its key ingredient is right-continuity of the filtration: information just after time \(t\), taken over times decreasing to \(t\), must already belong to \(\mathcal{F}_t\).
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\). A filtration is increasing, so \(\mathcal{F}_s\subseteq\mathcal{F}_t\) whenever \(s\le t\). It is right-continuous if \(\mathcal{F}_t=\bigcap_{s>t}\mathcal{F}_s\) for every \(t\ge0\). A random time here is an \(\mathcal{F}\)-measurable map \(\tau:\Omega\to[0,\infty]\). We use the stopping-time definition that \(\tau\) is a stopping time if \(\{\tau\le t\}\in\mathcal{F}_t\) for every \(t\ge0\).
The strict inequality in this test is essential to how the proof works. Knowing that \(\tau<q\) is observable at rational times just above \(t\) allows us to recover the event \(\tau\le t\) by intersection. It does not suffice to take rational times below \(t\): that would exclude the boundary case \(\tau=t\).
Proof of the Rational-Time Characterization
Proof. First suppose that \(\tau\) is a stopping time and fix a nonnegative rational \(q\). The following identity holds, including when \(\tau\) can equal \(0\) or \(\infty\):
If \(\tau<q\), there is a rational \(r\) with \(\tau\le r<q\), by density of the rationals; if \(\tau=\infty\), neither side contains that outcome. Conversely, \(\tau\le r<q\) implies \(\tau<q\). Each event in the union belongs to \(\mathcal{F}_r\) by the stopping-time property, and \(\mathcal{F}_r\subseteq\mathcal{F}_q\). The union is countable, so it belongs to \(\mathcal{F}_q\). When \(q=0\), the union is empty and \(\{\tau<0\}\) is empty as well.
Now suppose that \(\{\tau<q\}\in\mathcal{F}_q\) for every nonnegative rational \(q\), and fix \(t\ge0\). Choose rational numbers \(q_n>t\) decreasing to \(t\). Then
An outcome with \(\tau\le t\) lies in every event on the right. If \(\tau>t\), then some \(q_n\) is less than \(\tau\), so that outcome is excluded from the intersection. Each event \(\{\tau<q_n\}\) is in \(\mathcal{F}_{q_n}\); hence their intersection is in \(\bigcap_n\mathcal{F}_{q_n}\). This last sigma-field equals \(\mathcal{F}_t\). Indeed, \(\mathcal{F}_t\subseteq\mathcal{F}_{q_n}\) for every \(n\). Conversely, if \(A\in\bigcap_n\mathcal{F}_{q_n}\), then for any \(s>t\) there is \(n\) with \(q_n<s\), so \(A\in\mathcal{F}_{q_n}\subseteq\mathcal{F}_s\). Thus \(A\in\bigcap_{s>t}\mathcal{F}_s=\mathcal{F}_t\), using right-continuity. Therefore \(\{\tau\le t\}\in\mathcal{F}_t\) for every \(t\), and \(\tau\) is a stopping time. \(\square\)
This proof uses right-continuity precisely in the reverse implication. It does not assert that the test events themselves form a filtration or that one can replace right-continuity with a weaker assumption without further conditions.
Applying the Test
Worked Example: A Two-Valued Time Revealed at Time One
Let \(A\in\mathcal{F}_1\) and define \(\tau=2\) on \(A\) and \(\tau=3\) on \(A^c\). Check the rational-time condition, paying attention to strict endpoints.
For a nonnegative rational \(q\le2\), neither value of \(\tau\) is strictly less than \(q\), so \(\{\tau<q\}=\varnothing\in\mathcal{F}_q\). For \(2<q\le3\), exactly the outcomes in \(A\) have \(\tau<q\), including when \(q=3\), so \(\{\tau<q\}=A\). Here \(q>2>1\), and \(A\in\mathcal{F}_1\subseteq\mathcal{F}_q\). For \(q>3\), both possible values are strictly less than \(q\), so \(\{\tau<q\}=\Omega\in\mathcal{F}_q\). These cases cover every nonnegative rational \(q\). Thus the theorem gives that \(\tau\) is a stopping time.
Check: the stopping-time events are \(\{\tau\le t\}=\varnothing\) for \(t<2\), \(A\) for \(2\le t<3\), and \(\Omega\) for \(t\ge3\). In the middle range \(t\ge2>1\), so \(A\in\mathcal{F}_t\), as required. In particular, the endpoint \(t=2\) includes \(A\), whereas the strict test at \(q=2\) gives the empty event.
Worked Example: A Brownian Hitting Time
On a filtered probability space carrying a standard Brownian motion \(B_t\), let \(\mathcal{F}_t\) be its completed natural filtration, which is right-continuous, and define \(\tau=\inf\{s\ge0:B_s\ge1\}\), with \(\inf\varnothing=\infty\). Verify the rational-time test.
For every finite \(r\ge0\), continuity of the Brownian paths gives \(\{\tau\le r\}=\{\sup_{0\le s\le r}B_s\ge1\}\). The supremum over \([0,r]\) equals the supremum over rational times in \([0,r]\), together with the endpoint \(r\), by continuity and density. The resulting event is \(\mathcal{F}_r\)-measurable: it is determined by countably many \(B_s\) with \(s\le r\).
For each nonnegative rational \(q\), the deterministic identity used in the theorem gives
Each event in this countable union is in \(\mathcal{F}_r\subseteq\mathcal{F}_q\); if \(q=0\), the event is empty. Hence \(\{\tau<q\}\in\mathcal{F}_q\) for every such \(q\), and the rational-time theorem confirms that \(\tau\) is a stopping time.
Check: at \(r=0\), \(B_0=0<1\), so \(\{\tau\le0\}=\varnothing\), consistent with \(\{\sup_{0\le s\le0}B_s\ge1\}=\varnothing\). The supremum event uses the non-strict threshold \(\ge1\), because touching level \(1\) counts as hitting it.
Why Right-Continuity Cannot Be Dropped
The reverse implication can fail without right-continuity, even if every rational-time test event is measurable at its designated time. Here is a finite-space example that isolates the issue.
Proof. Take \(\Omega=\{0,1\}\), \(\mathcal{F}=2^\Omega\), and \(A=\{1\}\). Define \(\mathcal{F}_t=\{\varnothing,\Omega\}\) for \(0\le t\le1\), and \(\mathcal{F}_t=2^\Omega\) for \(t>1\). This is an increasing filtration, but it is not right-continuous at \(t=1\), since \(\bigcap_{s>1}\mathcal{F}_s=2^\Omega\ne\mathcal{F}_1\). Set \(\tau=1\) on \(A\) and \(\tau=2\) on \(A^c\). For rational \(q\le1\), \(\{\tau<q\}=\varnothing\); for \(1<q\le2\), it is \(A\), which belongs to \(\mathcal{F}_q=2^\Omega\); and for \(q>2\), it is \(\Omega\). Thus every rational-time test event is measurable in \(\mathcal{F}_q\). But \(\{\tau\le1\}=A\notin\mathcal{F}_1\), so \(\tau\) is not a stopping time. \(\square\)
Here the test reveals \(A\) at times strictly after \(1\), but the filtration does not require that information to be available at time \(1\). Right-continuity is exactly what rules out this gap when the rational times decrease to \(1\).
A Future-Dependent Time That Fails
Worked Example: A Test Event Depending on a Future Increment
Again use the completed natural filtration of a standard Brownian motion \(B_t\). Set \(\tau=2\) on \(\{B_3>0\}\) and \(\tau=4\) on \(\{B_3\le0\}\). Show that this is not a stopping time.
The rational time \(q=\tfrac52\) gives \(\{\tau<\tfrac52\}=\{B_3>0\}\). Write \(Y=B_3-B_{5/2}\). The Brownian increment \(Y\) is independent of \(\mathcal{F}_{5/2}\) and has distribution \(N(0,\tfrac12)\). Therefore
Since \(B_{5/2}\) is finite almost surely and the standard normal CDF satisfies \(0<\Phi(x)<1\) for every finite real \(x\), this conditional expectation is strictly between \(0\) and \(1\) almost surely. If \(\{B_3>0\}\) belonged to \(\mathcal{F}_{5/2}\), its conditional expectation given \(\mathcal{F}_{5/2}\) would equal its indicator, which takes only the values \(0\) and \(1\). This is a contradiction. Thus the rational-time condition fails at \(q=\tfrac52\), so \(\tau\) is not a stopping time.
Check: the calculation uses the increment variance \(3-\tfrac52=\tfrac12\), not \(3\). The event \(B_3>0\) depends on information arriving after time \(\tfrac52\); its conditional probability at that earlier time is neither zero nor one.
Common Mistake: Taking Rational Times from Below
A tempting but incorrect identity is \(\{\tau\le t\}=\bigcap_{q<t,\ q\in\mathbb{Q}}\{\tau<q\}\). For the deterministic time \(\tau=t\), the left side is \(\Omega\), but every event on the right is empty because \(t<q\) is false for \(q<t\). The correct intersection uses rational times strictly above \(t\): choose \(q_n\downarrow t\), then \(\{\tau\le t\}=\bigcap_n\{\tau<q_n\}\). This choice both preserves the boundary case \(\tau=t\) and places each event in a sigma-field just after \(t\), where right-continuity applies.
The practical advantage is that a countable collection of measurable events can determine whether a random time is a stopping time. In applications, the main work is often identifying each event \(\{\tau<q\}\) correctly and locating the information it uses. Strict versus non-strict inequalities matter at endpoints, and right-continuity matters when moving back from times just after \(t\) to time \(t\) itself.
Check Your Understanding
For each question, specify the relevant strict or non-strict event and the sigma-field in which it must lie.
- In the theorem’s reverse direction, why must the rational sequence satisfy \(q_n>t\) and \(q_n\downarrow t\)?
- For the two-valued time in the first example, what is \(\{\tau<q\}\) when \(q=3\), and what is \(\{\tau\le t\}\) when \(t=3\)?
- For the Brownian hitting time, why is \(\{\tau\le r\}\) expressed using \(\sup_{0\le s\le r}B_s\ge1\), rather than a strict inequality?
- In the finite-space counterexample, which event fails to belong to the filtration at its required time?
- For the future-dependent time, why does an \(\mathcal{F}_{5/2}\)-measurable event have conditional expectation equal to its indicator?