Why Filtrations Are Augmented
A filtration records which events are observable by each time. Two regularity assumptions make arguments with that information more stable: completeness places all subsets of null events in the time-zero sigma-field, and right-continuity requires that information available at times just after \(t\) is already available at \(t\). Together they are called the usual conditions. They do not follow from the definition of a filtered probability space, so one must either assume them or construct a filtration satisfying them.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with filtration \((\mathcal{F}_t)_{t\ge0}\), where \(\mathcal{F}_s\subseteq\mathcal{F}_t\subseteq\mathcal{F}\) for \(s\le t\). A stopping time is a map \(\tau:\Omega\to[0,\infty]\) such that \(\{\tau\le t\}\in\mathcal{F}_t\) for every \(t\ge0\). All event inclusions and equalities below are setwise; claims about random variables that agree almost surely will be identified as such.
Completeness is relative to the chosen ambient probability space: if the ambient sigma-field has been completed, every subset of one of its null events must be in \(\mathcal{F}_0\). Right-continuity is a condition on the filtration’s time index, not a claim that sample paths are continuous.
Completing the Probability Space and Filtration
Let \(\mathcal{N}\) be the collection of all subsets of \(\mathbb{P}\)-null events in \(\mathcal{F}\). The completed ambient sigma-field is \(\mathcal{F}^{\mathbb{P}}=\sigma(\mathcal{F}\cup\mathcal{N})\). Its probability measure \(\mathbb{P}^{\mathbb{P}}\) extends \(\mathbb{P}\): for \(A\in\mathcal{F}^{\mathbb{P}}\), choose \(E\in\mathcal{F}\) with \(A\triangle E\subseteq N\) for some null \(N\in\mathcal{F}\), and set \(\mathbb{P}^{\mathbb{P}}(A)=\mathbb{P}(E)\). This is well-defined: if \(E'\) is another such choice, then \(E\triangle E'\) is contained in a null event, so \(\mathbb{P}(E)=\mathbb{P}(E')\).
Complete each time sigma-field by setting \(\mathcal{F}_t^c=\sigma(\mathcal{F}_t\cup\mathcal{N})\). This gives an increasing filtration: if \(s\le t\), then \(\mathcal{F}_s\subseteq\mathcal{F}_t\), and both completed sigma-fields contain \(\mathcal{N}\). Also, \(\mathcal{F}_0^c\) contains every subset of every ambient null event. Completion alone need not make a filtration right-continuous.
Worked Example: Completion Makes a Null-Set Modification Observable
Let \(\Omega=[0,1]\), let \(\mathcal{F}\) be the Borel sigma-field with Lebesgue probability, and set \(\mathcal{F}_t=\{\varnothing,\Omega\}\) for all \(t\ge0\). Consider \(N=\{0\}\), and define \(\tau(0)=0\) and \(\tau(x)=1\) for \(x\ne0\).
For \(0\le t<1\), \(\{\tau\le t\}=N\): at \(t=0\) this is \(\{0\}\), and for every \(0<t<1\) it remains \(\{0\}\). At \(t\ge1\), \(\{\tau\le t\}=\Omega\). The event \(N\) is Borel and null, but it is not in the raw trivial sigma-field \(\mathcal{F}_t\). Thus \(\tau\) is not a stopping time for that raw filtration.
After completion, \(N\in\mathcal{F}_t^c\) at every time, so all the displayed level events belong to the relevant sigma-fields and \(\tau\) is a stopping time. Check: the only changed event is the null set \(N\); completion adds it, while the event \(\Omega\) was already measurable. In particular, this example changes no probabilities, since \(\mathbb{P}(N)=0\), but it does change measurability.
Right-Continuous Augmentation
Starting from the completed filtration, define, for every finite \(t\ge0\), \(\mathcal{F}_t^u=\bigcap_{s>t}\mathcal{F}_s^c\). This is the right-continuous augmentation. The superscript \(u\) indicates the filtration obtained by imposing the usual conditions; it is not a new probability measure.
Proof. Each \(\mathcal{F}_t^u\) is an intersection of sigma-fields, so is a sigma-field contained in \(\mathcal{F}^{\mathbb{P}}\). If \(t\le v\), then every \(s>v\) also satisfies \(s>t\), so \(\bigcap_{s>t}\mathcal{F}_s^c\subseteq\bigcap_{s>v}\mathcal{F}_s^c\). Hence the family is increasing. Since \(\mathcal{F}_t^c\subseteq\mathcal{F}_s^c\) for every \(s>t\), \(\mathcal{F}_t^c\subseteq\mathcal{F}_t^u\). Every subset of every ambient null event belongs to each \(\mathcal{F}_s^c\), and therefore to \(\mathcal{F}_t^u\), including at \(t=0\). This proves completeness.
To prove right-continuity, first note that \(\mathcal{F}_t^u\subseteq\mathcal{F}_v^u\) for \(v>t\), so \(\mathcal{F}_t^u\subseteq\bigcap_{v>t}\mathcal{F}_v^u\). Conversely, take \(A\in\bigcap_{v>t}\mathcal{F}_v^u\). For any \(s>t\), choose \(v\) with \(t<v<s\). Since \(A\in\mathcal{F}_v^u=\bigcap_{r>v}\mathcal{F}_r^c\), and \(s>v\), we have \(A\in\mathcal{F}_s^c\). This holds for every \(s>t\), so \(A\in\mathcal{F}_t^u\). Thus \(\bigcap_{v>t}\mathcal{F}_v^u=\mathcal{F}_t^u\), as required. \(\square\)
Worked Example: Information Revealed Immediately After Time Zero
Let \(\Omega=\{a,b\}\), \(\mathcal{F}=2^\Omega\), and \(\mathbb{P}(\{a\})=\mathbb{P}(\{b\})=\tfrac12\). Set \(\mathcal{F}_0=\{\varnothing,\Omega\}\), and \(\mathcal{F}_t=2^\Omega\) for every \(t>0\). There are no nonempty null events, so completion does not change these sigma-fields.
At time zero, the right-continuous augmentation gives \(\mathcal{F}_0^u=\bigcap_{s>0}\mathcal{F}_s=2^\Omega\). At any \(t>0\), every sigma-field in the intersection is also \(2^\Omega\), so \(\mathcal{F}_t^u=2^\Omega\). Therefore the augmentation adds exactly the event \(\{a\}\) to the time-zero sigma-field.
Define \(\tau(a)=0\) and \(\tau(b)=1\). For \(t=0\), \(\{\tau\le t\}=\{a\}\); for \(0<t<1\), it is still \(\{a\}\); for \(t\ge1\), it is \(\Omega\). The first event is not in the original \(\mathcal{F}_0\), so \(\tau\) is not a stopping time for the original filtration. It is a stopping time for the augmented filtration. Check: at each \(t>0\), \(\{a\}\in\mathcal{F}_t\), and at \(t=0\) it belongs to \(\mathcal{F}_0^u\), matching the defining test at every time.
A Stopping-Time Test That Uses Right-Continuity
Proof. Suppose first that \(\tau\) is a stopping time. For \(t>0\), \(\{\tau<t\}=\bigcup_{q\in\mathbb{Q},\,0\le q<t}\{\tau\le q\}\). Indeed, if \(\tau(\omega)<t\), a rational \(q\) can be chosen with \(\tau(\omega)\le q<t\); the reverse inclusion follows from \(q<t\). Each event in the union belongs to \(\mathcal{G}_q\subseteq\mathcal{G}_t\), so the union does too. For \(t=0\), \(\{\tau<0\}=\varnothing\in\mathcal{G}_0\).
Conversely, suppose \(\{\tau<t\}\in\mathcal{G}_t\) for every \(t\ge0\). For fixed \(t\ge0\), choose \(t_n=t+1/n\). Then \(\{\tau\le t\}=\bigcap_{n\ge1}\{\tau<t_n\}\): if \(\tau\le t\), it is strictly less than every \(t_n\); if \(\tau>t\), some \(1/n\) is smaller than \(\tau-t\), so \(\tau\not<t_n\). The intersection belongs to \(\bigcap_{n\ge1}\mathcal{G}_{t+1/n}=\mathcal{G}_t\). The last equality follows from right-continuity and the fact that \(t+1/n\) decreases to \(t\): membership in all these sigma-fields implies membership in every \(\mathcal{G}_s\), \(s>t\), by choosing \(n\) with \(t+1/n<s\) and using the increasing property. Thus \(\{\tau\le t\}\in\mathcal{G}_t\) for every \(t\), proving that \(\tau\) is a stopping time. \(\square\)
Worked Example: Why the Strict-Time Test Can Fail Without Right-Continuity
Use the two-point space and filtration from the preceding example, before augmentation, with \(\tau(a)=0\), \(\tau(b)=1\). If \(t=0\), then \(\{\tau<t\}=\varnothing\in\mathcal{F}_0\). If \(0<t\le1\), then \(\{\tau<t\}=\{a\}\in\mathcal{F}_t\). If \(t>1\), then \(\{\tau<t\}=\Omega\in\mathcal{F}_t\). Thus the strict-time condition holds at every \(t\).
However, at \(t=0\), \(\{\tau\le0\}=\{a\}\notin\mathcal{F}_0\), so \(\tau\) is not a stopping time. The filtration is not right-continuous because \(\bigcap_{s>0}\mathcal{F}_s=2^\Omega\ne\mathcal{F}_0\). Check: after right-continuous augmentation, \(\mathcal{F}_0^u=2^\Omega\), and the missing level event \(\{\tau\le0\}\) becomes measurable, exactly as the theorem predicts.
Null Modifications and Common Pitfalls
Completeness also handles changes to a random time on a null set. Suppose \(\tau\) is a stopping time and \(\tau'=\tau\) outside a null event \(N\). If the filtration is complete, then \(N\in\mathcal{F}_0\subseteq\mathcal{F}_t\), and for every \(t\), \(\{\tau'\le t\}\triangle\{\tau\le t\}\subseteq N\). Since \(\{\tau\le t\}\in\mathcal{F}_t\) and every subset of \(N\) belongs to \(\mathcal{F}_t\), the symmetric difference is measurable there, and hence so is \(\{\tau'\le t\}\). This is a setwise stopping-time statement, not merely an almost-sure identification.
A common mistake is to replace \(\{\tau\le t\}\) by \(\{\tau<t\}\) without checking right-continuity. In the two-point example, the strict event at time zero is empty, but the non-strict event is \(\{a\}\), which is not in the original time-zero sigma-field. The incorrect calculation “\(\{\tau<0\}=\varnothing\), so \(\{\tau\le0\}\in\mathcal{F}_0\)” changes the event being tested and is invalid. Right-continuity justifies the passage through the decreasing times \(t+1/n\); completeness justifies adding null modifications. These are distinct properties with distinct roles.
| Assumption | What it supplies | What it does not supply by itself |
|---|---|---|
| Completeness | All subsets of ambient null events are measurable from time zero | Right-continuity |
| Right-continuity | Information in every \(\mathcal{F}_s\), \(s>t\), is in \(\mathcal{F}_t\) | Completeness |
| Usual conditions | Both properties, supporting standard stopping-time arguments | A license to omit the event-measurability check |
Check Your Understanding
For each question, distinguish explicitly between the raw filtration, its completion, and its right-continuous augmentation where relevant.
- State the defining event condition for a stopping time and explain which event the strict-time criterion tests instead.
- Why does completing each \(\mathcal{F}_t\) not necessarily make the filtration right-continuous?
- In the two-point example, compute \(\mathcal{F}_0^u\) directly from \(\bigcap_{s>0}\mathcal{F}_s\).
- Identify the step in the proof of the strict-time criterion that requires right-continuity.
- If two random times agree outside a null event, what role does completeness play in preserving the stopping-time property?