Three Conditions That Should Not Be Confused
Adaptedness, progressive measurability, predictability, and being a stopping time are related conditions, but they answer different questions. Adaptedness tests each fixed-time value against the information available at that time. Progressive measurability tests the process jointly over every time interval \([0,T]\). Predictability specifies a particular sigma-field of processes, generated by left-continuous adapted processes. A stopping time is a random time whose event of having occurred by time \(t\) is known at time \(t\). Confusing these tests can invalidate an argument before any stochastic-calculus calculation begins.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). No stochastic integral is needed here; in particular, the admissible class \(\mathcal{L}^2\) of progressively measurable integrands satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) is relevant as a reminder that adaptedness alone is not the integrand condition. The examples below isolate the measurability tests themselves.
What the Implications Actually Say
Progressive measurability implies adaptedness: apply the product-measurability condition on \([0,t]\), then take the section at time \(t\). Predictability also implies adaptedness, since the time-\(t\) sections of the generators of the predictable sigma-field belong to \(\mathcal{F}_t\); the collection of sets with this section property is a sigma-field. These implications do not say that adaptedness by itself supplies the joint measurability required for progressiveness.
A useful positive result goes in the other direction when path regularity is available: an adapted process with continuous paths is predictable. Here is a direct proof, which also shows why the value at time zero must be included.
Proof. For each integer \(n\ge1\), put \(\delta_n=2^{-n}\), set \(X^n_0=X_0\), and, for \(k=0,1,\ldots\), set \(X^n_t=X_{k\delta_n}\) when \(k\delta_n<t\le(k+1)\delta_n\). On each such interval the process is constant, and at a grid point its value is the value on the interval immediately to its left. Thus \(X^n\) is left-continuous on \((0,\infty)\). It is adapted because \(k\delta_n<t\) implies \(\mathcal{F}_{k\delta_n}\subseteq\mathcal{F}_t\); its value at zero is \(\mathcal{F}_0\)-measurable. Each \(X^n\) is therefore predictable by the definition of the predictable sigma-field. For every fixed \(t>0\), the grid point used in \(X^n_t\) lies in \([t-\delta_n,t)\), so it tends to \(t\). Path continuity gives \(X^n_t\to X_t\); at \(t=0\) equality holds for every \(n\). A pointwise limit of measurable functions is measurable, so \(X\) is predictable. \(\square\)
Worked Example: Adaptedness Without Progressiveness
This example shows why adaptedness alone is not a joint-measurability test. Let \(\Omega=\{\omega_0,\omega_1\}\), give each point probability \(1/2\), and let \(\mathcal{F}_t=\mathcal{F}\) be the full sigma-field for every \(t\). Choose a non-Borel set \(N\subset[0,1]\), and define \(X_t(\omega_0)=\mathbf{1}_N(t)\) for \(0\le t\le1\), with \(X_t(\omega_1)=0\); set \(X_t=0\) for \(t>1\).
At each fixed \(t\), \(X_t\) is a function on a finite probability space and is therefore \(\mathcal{F}_t\)-measurable. Thus \(X\) is adapted. But on \([0,1]\times\Omega\), the set where \(X=1\) is \(N\times\{\omega_0\}\). This set is not in \(\mathcal{B}([0,1])\otimes\mathcal{F}_1\): its section at \(\omega_0\) is \(N\), which is not Borel. Hence \(X\) is not progressive.
Check: the failure is not a failure of any fixed-time measurability test; it is the failure of joint product measurability. This deliberately pathological process is a warning against inferring progressiveness from adaptedness alone, not a typical model of a stochastic integrand.
Continuity Does Not Replace Adaptedness
The proposition uses both hypotheses. Continuous paths by themselves say nothing about when the information determining the process becomes available. Conversely, predictability does not require continuous paths: left-continuous adapted processes are predictable by definition, even if they jump immediately to the right of a time.
Worked Example: A Continuous Process That Is Not Predictable
Take \(\Omega=\{\omega_0,\omega_1\}\) with equal probabilities, \(A=\{\omega_1\}\), and set \(\mathcal{F}_t=\{\varnothing,\Omega\}\) for \(t<1\), while \(\mathcal{F}_t=2^\Omega\) for \(t\ge1\). Define \(X_t=\mathbf{1}_A\) for every \(t\). Each sample path is constant, hence continuous. At \(t=1/2\), however, \(X_{1/2}=\mathbf{1}_A\) is not measurable with respect to the trivial sigma-field \(\mathcal{F}_{1/2}\). So \(X\) is not adapted.
Since predictability implies adaptedness, \(X\) cannot be predictable. The incorrect argument “the paths are continuous, so the process is predictable” omits adaptedness. The proposition applies only to a continuous adapted process.
For contrast, \(H_t=\mathbf{1}_{(1,2]}(t)\) is deterministic, adapted, and left-continuous: its value at \(1\) is \(0\), and its value at \(2\) is \(1\), matching its left limit there. Thus it is predictable, although it is discontinuous at \(1\) and \(2\). This example also fixes the endpoint convention: the interval includes \(2\), not \(1\).
A Random Time Is Not Automatically a Stopping Time
A random time need only be an \(\mathcal{F}\)-measurable map into \([0,\infty]\). For the stopping-time property, test the events \(\{\tau\le t\}\) against the corresponding \(\mathcal{F}_t\), for every \(t\). Measurability with respect to the final sigma-field \(\mathcal{F}\) is not enough. Under the right-continuity assumptions used in the rational-time test, that test can reduce the checking task; it does not change the defining event or remove the need for its filtration hypothesis.
Worked Example: A Future Event Used to Choose a Time
Use the two-point space and filtration from the preceding example, and let \(\tau=0\) on \(A\) and \(\tau=1\) on \(A^c\). This is an \(\mathcal{F}\)-measurable random time. At \(t=1/2\), however, \(\{\tau\le1/2\}=A\). Since \(A\) is neither \(\varnothing\) nor \(\Omega\), it is not in the trivial sigma-field \(\mathcal{F}_{1/2}\). Therefore \(\tau\) is not a stopping time.
The same check at \(t=0\) gives \(\{\tau\le0\}=A\notin\mathcal{F}_0\). The error would be to argue that \(\tau\) is a stopping time because it takes only the values \(0\) and \(1\), or because it is measurable at the terminal time. Neither fact checks the required event at the time it is tested.
A corrected choice in this filtration is \(\rho=1\) on \(A\) and \(\rho=2\) on \(A^c\). For \(0\le t<1\), \(\{\rho\le t\}=\varnothing\in\mathcal{F}_t\). For \(1\le t<2\), \(\{\rho\le t\}=A\in\mathcal{F}_t\), because these times are at least \(1\). For \(t\ge2\), \(\{\rho\le t\}=\Omega\in\mathcal{F}_t\). These cases cover every \(t\ge0\), so \(\rho\) is a stopping time.
Check: the middle case starts at \(t=1\), where \(A\) has become observable. The event tested is \(\{\rho\le t\}\), including equality at the endpoint.
A Practical Diagnostic
When an argument uses a process or a random time, write the relevant test before proceeding. The following table summarizes which facts answer which questions; an implication in the table is not a license to reverse its direction.
| Object or claim | Test to apply | Common invalid shortcut |
|---|---|---|
| Adapted process | For every fixed \(t\), verify \(X_t\) is \(\mathcal{F}_t\)-measurable. | Conclude joint measurability from fixed-time measurability. |
| Progressive process | For every \(T\), verify measurability on \([0,T]\times\Omega\) against \(\mathcal{B}([0,T])\otimes\mathcal{F}_T\). | Check only that each \(X_t\) is adapted. |
| Predictable process | Verify measurability for the sigma-field generated by adapted left-continuous processes. | Use continuity without first checking adaptedness. |
| Stopping time | For every \(t\), check \(\{\tau\le t\}\in\mathcal{F}_t\). | Check only that \(\tau\) is \(\mathcal{F}\)-measurable or discrete-valued. |
These distinctions matter in stochastic calculus because the admissibility of an integrand, the information available at a random time, and the sigma-field in which a process is measurable are separate hypotheses. A proof that checks one of them should not silently use it as a substitute for another. In particular, the continuity proposition is a reliable route from adaptedness to predictability, while the counterexamples show why the missing hypothesis cannot simply be dropped.
Check Your Understanding
For each question, state the sigma-field or event that must be checked, not merely the conclusion.
- Give the defining product-measurability condition for a process to be progressive on a fixed interval \([0,T]\).
- Why does the proposition that continuous adapted processes are predictable not imply that every continuous process is predictable?
- For \(H_t=\mathbf{1}_{(1,2]}(t)\), state its values at \(t=1\) and \(t=2\), and explain why it is left-continuous.
- In the two-point filtration example, which event at \(t=1/2\) disproves that \(\tau=0\) on \(A\), \(1\) on \(A^c\) is a stopping time?
- For the corrected time \(\rho=1\) on \(A\), \(2\) on \(A^c\), list the three time ranges and the corresponding events \(\{\rho\le t\}\).