Countable Limits Of Stopping Times

Why Countable Limits Need Care

For two stopping times, their minimum and maximum are again stopping times. Extending this fact to a countable family is not just a matter of replacing two events by a countable union or intersection: the events for an infimum involve strict inequalities, and their measurability may require right-continuity of the filtration. The supremum has a simpler event description and needs no such condition.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\). A stopping time \(\tau\) takes values in \([0,\infty]\) and satisfies \(\{\tau\le t\}\in\mathcal{F}_t\) for every \(t\ge0\). The arguments below concern stopping times and filtrations; they do not use a Brownian motion or stochastic integrals.

Definition: The filtration is right-continuous if \(\mathcal{F}_t=\bigcap_{s>t}\mathcal{F}_s\) for every \(t\ge0\). A countable family below is indexed by \(n\ge1\) and is nonempty. Its infimum and supremum are taken pointwise in \([0,\infty]\).

For an empty family, if one adopts the extended-value conventions \(\inf\varnothing=\infty\) and \(\sup\varnothing=0\), both are constant stopping times. The results of interest are for nonempty families, including finite families.

Countable Suprema

Theorem: If \((\tau_n)_{n\ge1}\) is any countable family of stopping times for a filtration \((\mathcal{F}_t)\), then \(\tau^*=\sup_{n\ge1}\tau_n\) is a stopping time. No right-continuity assumption is needed.

Proof. Fix \(t\ge0\). The supremum is at most \(t\) exactly when every member of the family is at most \(t\). Thus \(\{\tau^*\le t\}=\bigcap_{n\ge1}\{\tau_n\le t\}\). Each event in the intersection belongs to \(\mathcal{F}_t\), since each \(\tau_n\) is a stopping time. A sigma-field is closed under countable intersections, so \(\{\tau^*\le t\}\in\mathcal{F}_t\). This holds for every \(t\ge0\), proving that \(\tau^*\) is a stopping time. \(\square\)

The event formula is exact even when some \(\tau_n\) equal infinity: the supremum is at most a finite \(t\) precisely when no member exceeds \(t\). The countable intersection is therefore the direct route to measurability.

Worked Example: A Countable Family of Deterministic Times

Take a probability space with the constant filtration \(\mathcal{F}_t=\{\varnothing,\Omega\}\), and let \(\tau_n=2+1/n\) for every outcome. These deterministic random times are stopping times: for each \(t\), the event \(\{\tau_n\le t\}\) is either \(\varnothing\) or \(\Omega\).

The values decrease from \(\tau_1=3\) toward \(2\), so \(\inf_n\tau_n=2\) and \(\sup_n\tau_n=3\). At \(t=2.5\), \(\{\sup_n\tau_n\le2.5\}=\varnothing\), because \(\tau_1=3>2.5\). In contrast, \(\{\inf_n\tau_n\le2.5\}=\Omega\), because the infimum is \(2\). Both events belong to \(\mathcal{F}_{2.5}\).

There is a useful strict-inequality distinction here. The event \(\{\inf_n\tau_n<2.5\}\) is \(\Omega\): for example, \(\tau_3=2+\tfrac13<2.5\). The value \(\tau_2=2+\tfrac12=2.5\) does not itself satisfy the strict inequality, while \(\tau_1=3\) does not satisfy it either. This check agrees with the limiting value \(2\).

Countable Infima and Right-Continuity

For an infimum, the event \(\{\inf_n\tau_n\le t\}\) is generally not the intersection of the events \(\{\tau_n\le t\}\). An infimum can be at most \(t\) even when every individual value is greater than \(t\). The appropriate starting point is instead the strict event \(\{\inf_n\tau_n<u\}\).

Theorem: If \((\tau_n)_{n\ge1}\) is a countable family of stopping times and the filtration is right-continuous, then \(\tau_*=\inf_{n\ge1}\tau_n\) is a stopping time.

Proof. First fix \(u>0\). Pointwise, \(\inf_n\tau_n<u\) holds exactly when \(\tau_n<u\) for at least one \(n\). For each stopping time, \(\{\tau_n<u\}=\bigcup_{q\in\mathbb{Q},\,0\le q<u}\{\tau_n\le q\}\). Indeed, if \(\tau_n<u\), a rational \(q\) can be chosen with \(\tau_n\le q<u\); the reverse implication follows from \(q<u\). Every event \(\{\tau_n\le q\}\) is in \(\mathcal{F}_q\subseteq\mathcal{F}_u\). Taking the indicated countable unions shows that \(\{\tau_n<u\}\in\mathcal{F}_u\). Taking the union over \(n\) then gives \(\{\tau_*<u\}\in\mathcal{F}_u\).

Now fix \(t\ge0\). For extended nonnegative values, \(\tau_*\le t\) if and only if \(\tau_*<t+1/k\) for every integer \(k\ge1\). Therefore \(\{\tau_*\le t\}=\bigcap_{k\ge1}\{\tau_*<t+1/k\}\). Each event on the right belongs to \(\mathcal{F}_{t+1/k}\), so the intersection belongs to \(\bigcap_{k\ge1}\mathcal{F}_{t+1/k}\). This last intersection equals \(\mathcal{F}_t\): right-continuity gives one inclusion, and for any \(s>t\) some \(k\) satisfies \(t+1/k<s\), so \(\mathcal{F}_{t+1/k}\subseteq\mathcal{F}_s\), giving the other inclusion. Thus \(\{\tau_*\le t\}\in\mathcal{F}_t\) for every \(t\), as required. \(\square\)

The proof also explains the role of the hypothesis. Strict sublevel events are measurable at times just after \(t\); right-continuity is what brings their intersection back into \(\mathcal{F}_t\). Without that property, the infimum need not be a stopping time.

Increasing and Decreasing Sequences

Corollary: If \((\tau_n)\) is a pointwise increasing sequence of stopping times, then its limit \(\lim_{n\to\infty}\tau_n=\sup_n\tau_n\) is a stopping time for any filtration. If \((\tau_n)\) is a pointwise decreasing sequence of stopping times, then its limit \(\lim_{n\to\infty}\tau_n=\inf_n\tau_n\) is a stopping time when the filtration is right-continuous.

Proof. For an increasing sequence, its pointwise limit in \([0,\infty]\) is its supremum, so the countable-supremum theorem applies. For a decreasing sequence, its pointwise limit is its infimum, so the countable-infimum theorem applies under right-continuity. These arguments include the cases where the limit is \(0\) or \(\infty\). \(\square\)

Worked Example: Limits of Increasing and Decreasing Sequences

Let \(\Omega=\{a,b\}\) and \(\mathcal{F}_t=2^\Omega\) for every \(t\ge0\). Every \([0,\infty]\)-valued random variable on this space is a stopping time, since every event is in every \(\mathcal{F}_t\).

Define an increasing sequence by \(\alpha_n(a)=1-\tfrac1{n+1}\) and \(\alpha_n(b)=2-\tfrac1{n+1}\). At \(a\), the first values are \(\alpha_1(a)=\tfrac12\) and \(\alpha_2(a)=\tfrac23\), and the limit is \(1\). At \(b\), the first values are \(\alpha_1(b)=\tfrac32\) and \(\alpha_2(b)=\tfrac53\), and the limit is \(2\). Hence \(\sup_n\alpha_n\) takes values \(1\) at \(a\) and \(2\) at \(b\). At \(t=1\), its event \(\{\sup_n\alpha_n\le1\}\) is \(\{a\}\), which belongs to \(\mathcal{F}_1\).

For a decreasing sequence, let \(\beta_n(a)=1+\tfrac1n\) and \(\beta_n(b)=2+\tfrac1n\). Its first values are \(\beta_1(a)=2\), \(\beta_2(a)=\tfrac32\), \(\beta_1(b)=3\), and \(\beta_2(b)=\tfrac52\). The limits are \(1\) at \(a\) and \(2\) at \(b\). Thus \(\{\inf_n\beta_n\le1\}=\{a\}\), again an event in \(\mathcal{F}_1\). By contrast, every \(\beta_n(a)>1\), so \(\bigcap_n\{\beta_n\le1\}=\varnothing\). This verifies concretely why the infimum event cannot be replaced by that intersection.

Why Right-Continuity Cannot Be Dropped

Worked Example: An Infimum That Is Not a Stopping Time

Let \(\Omega=\{a,b\}\), with \(\mathcal{F}=2^\Omega\), and define \(\mathcal{F}_0=\{\varnothing,\Omega\}\) and \(\mathcal{F}_t=2^\Omega\) for every \(t>0\). This is an increasing filtration, but it is not right-continuous at \(0\), since \(\bigcap_{s>0}\mathcal{F}_s=2^\Omega\ne\mathcal{F}_0\).

Set \(\tau_n(a)=1/n\) and \(\tau_n(b)=1\). Each \(\tau_n\) is a stopping time. At \(t=0\), \(\{\tau_n\le0\}=\varnothing\in\mathcal{F}_0\). For \(0<t<1\), the event is \(\{a\}\) if \(1/n\le t\), and is empty if \(1/n>t\); either event belongs to \(\mathcal{F}_t=2^\Omega\). For \(t\ge1\), the event is \(\Omega\).

The pointwise infimum is \(0\) at \(a\) and \(1\) at \(b\). For \(0<t<1\), \(\{\inf_n\tau_n\le t\}=\{a\}\), which belongs to \(\mathcal{F}_t\); for \(t\ge1\), the event is \(\Omega\). But at \(t=0\), the event is \(\{a\}\notin\mathcal{F}_0\). Therefore the infimum is not a stopping time. This example pinpoints the missing measurability at the boundary time \(0\).

Common Pitfall: Confusing Infimum Events

A common incorrect calculation is \(\{\inf_n\tau_n\le t\}=\bigcap_n\{\tau_n\le t\}\). The right-hand side says that every member is at most \(t\), which describes the supremum event, not the infimum event. For the decreasing sequence \(\beta_n\) above, at \(t=1\) the infimum event is \(\{a\}\), whereas the intersection of the individual events is empty.

Another useful distinction is between strict and non-strict inequalities. For \(\tau_n=2+1/n\) and \(u=2.5\), the value at \(n=2\) equals \(u\), so it does not belong to \(\{\tau_n<u\}\). The value at \(n=3\) is \(2+\tfrac13<2.5\), so the union of strict events does contain the outcome. The proof of the infimum theorem uses this union of strict events and then right-continuity; it does not assume the infimum is attained by some \(n\).

In applications, the distinction matters whenever stopping rules are formed by taking limits of sequences. Increasing limits are controlled directly by countable intersections of \(\{\tau_n\le t\}\). Decreasing limits require the additional right-continuity argument, so that the limiting event is measurable at the time being tested.

Check Your Understanding

For each question, keep track of whether the relevant event uses a strict or a non-strict inequality.

Why is a countable supremum of stopping times a stopping time even if the filtration is not right-continuous?
Which event identity starts the proof for an infimum: an intersection of events \(\{\tau_n\le t\}\), or a union of events \(\{\tau_n<u\}\)? Explain the difference.
For the family \(\tau_n=2+1/n\), compute \(\inf_n\tau_n\) and \(\sup_n\tau_n\), and test their sublevel events at \(t=2.5\).
For the sequence \(\beta_n(a)=1+1/n\), \(\beta_n(b)=2+1/n\), what are \(\{\inf_n\beta_n\le1\}\) and \(\bigcap_n\{\beta_n\le1\}\)?
In the two-point counterexample, at which time does the infimum fail the stopping-time condition, and which filtration property fails there?