Path Regularity and Measurability in Time
Adaptedness says that the value \(X_t\) at each fixed time \(t\) is measurable with respect to the information \(\mathcal{F}_t\). By itself, this is a statement about separate time sections; it does not assert the joint measurability needed for many constructions involving an entire process. Progressive and predictable measurability supply stronger, process-level conditions. Path continuity or left continuity provides a useful route from adaptedness to these conditions, but the approximations must respect the direction of time.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\), with an increasing filtration, carrying a Brownian motion \(B_t\). The measurability results below do not require Brownian motion. A real-valued process \(X=(X_t)_{t\ge0}\) is adapted if \(X_t\) is \(\mathcal{F}_t\)-measurable for every \(t\). We take continuity or left continuity to hold for every path; the same statements apply to a chosen adapted version with that path property.
The half-open intervals in the predictable generators encode past information: on \((s,t]\), the event \(A\) is already known at time \(s\). The value at time \(0\) is handled separately by \(\mathcal{F}_0\). We now show that continuity and left continuity lead to the corresponding measurability properties.
Continuous Adapted Processes Are Progressive
Proof. Fix \(T>0\); the case \(T=0\) follows directly from adaptedness. For each integer \(n\ge1\), partition \([0,T]\) into \(2^n\) equal intervals of length \(\Delta_n=T2^{-n}\), with endpoints \(t_k=k\Delta_n\). Define \(X^n\) on \([0,T]\) by setting \(X^n_t=X_{t_k}\) for \(t_k\le t<t_{k+1}\), and \(X^n_T=X_T\). On any one of these intervals, the random value \(X_{t_k}\) is \(\mathcal{F}_{t_k}\)-measurable and therefore \(\mathcal{F}_T\)-measurable. The interval is Borel, so \(X^n\) is \(\mathcal{F}_T\otimes\mathcal{B}([0,T])\)-measurable. For every \((\omega,t)\), continuity of the path gives \(X^n_t(\omega)\to X_t(\omega)\): for \(t<T\), the left endpoint \(t_k\) of the interval containing \(t\) converges to \(t\), and at \(T\) the approximant equals \(X_T\). A pointwise limit of measurable real-valued functions is measurable. Thus \(X\) is measurable with respect to \(\mathcal{F}_T\otimes\mathcal{B}([0,T])\). Since \(T\) was arbitrary, \(X\) is progressive. \(\square\)
The proof uses a separate partition for each finite horizon. Its approximating value at time \(t\) comes from a time no later than \(t\), so it is compatible with the information available by the end of the horizon. The definition of progressive measurability asks for \(\mathcal{F}_T\)-measurability on each whole interval \([0,T]\); it does not require each approximant to be measurable with respect to \(\mathcal{F}_t\) at every individual \(t\).
Worked Example: A Continuous Adapted Process with an Initial Random Value
Let \(Y\) be a real-valued \(\mathcal{F}_0\)-measurable random variable and set \(X_t=Y+t\). For every \(t\), \(Y\) is \(\mathcal{F}_0\)-measurable and \(\mathcal{F}_0\subseteq\mathcal{F}_t\), while \(t\) is deterministic. Hence \(X_t\) is \(\mathcal{F}_t\)-measurable. Each path is continuous, so the theorem implies that \(X\) is progressive.
The approximation can also be checked directly on \([0,T]\). With \(\Delta_n=T2^{-n}\), set \(X^n_t=Y+k\Delta_n\) when \(k\Delta_n\le t<(k+1)\Delta_n\), and \(X^n_T=Y+T\). Each step is jointly measurable. For \(t<T\) in the \(k\)-th interval, \(X_t-X^n_t=t-k\Delta_n\), so \(0\le X_t-X^n_t<\Delta_n\); at \(T\) the difference is \(0\). Therefore \(\sup_{0\le t\le T}|X_t-X^n_t|\le\Delta_n\to0\). This also verifies convergence uniformly on the fixed horizon.
Worked Example: Brownian Motion Is Progressive
Take the filtration with respect to which \(B\) is adapted, and use its continuous paths. The theorem gives that \(B\) is progressive. For an explicit approximation on \([0,T]\), use the same partition and put \(B^n_t=B_{t_k}\) on \([t_k,t_{k+1})\), with \(B^n_T=B_T\). Each \(B_{t_k}\) is \(\mathcal{F}_{t_k}\)-measurable, hence \(\mathcal{F}_T\)-measurable, and each time interval is Borel. Thus every \(B^n\) is measurable with respect to \(\mathcal{F}_T\otimes\mathcal{B}([0,T])\).
For every continuous Brownian path, \(B^n_t\to B_t\) at each \(t\), so the limit is measurable on \(\Omega\times[0,T]\). As an endpoint check, at \(t=T\) the definition gives \(B^n_T=B_T\) exactly; there is no need to infer convergence at \(T\) from values just before it. Since this works for every finite \(T\), Brownian motion is progressive.
Left-Continuous Adapted Processes Are Predictable
Proof. Let \(X\) be left-continuous and adapted. For each integer \(n\ge1\), set \(\delta_n=2^{-n}\) and define \(X^n_0=X_0\), while \(X^n_t=X_{k\delta_n}\) on each interval \((k\delta_n,(k+1)\delta_n]\), for \(k=0,1,2,\ldots\). To verify predictability, first take an interval \((k\delta_n,(k+1)\delta_n]\). For any Borel set \(C\subseteq\mathbb{R}\), its inverse image under the process that equals \(X_{k\delta_n}\) on this interval is \(\{\omega:X_{k\delta_n}(\omega)\in C\}\times(k\delta_n,(k+1)\delta_n]\). The event in the first factor belongs to \(\mathcal{F}_{k\delta_n}\) by adaptedness, so this rectangle is a predictable generator. Taking the countable union over \(k\), and adding the time-zero process \(X_0\), which is measurable on \(\{0\}\times\Omega\) by \(\mathcal{F}_0\)-measurability, shows that \(X^n\) is predictable. For \(t>0\), the left endpoint \(k\delta_n\) of the interval containing \(t\) is strictly less than or equal to \(t\) and converges to \(t\) from below as \(n\to\infty\). Left continuity gives \(X^n_t\to X_t\); at \(t=0\), equality holds for every \(n\). Hence \(X\), a pointwise limit of predictable processes, is predictable. \(\square\)
The intervals are deliberately \((k\delta_n,(k+1)\delta_n]\), not intervals that include their left endpoint. This convention makes the value on each interval known at its left endpoint, as required by the predictable generators. At a grid point \(t>0\), the approximating value comes from the preceding grid point, which still converges to \(X_t\) by left continuity.
Worked Example: A Random Left-Continuous Step Process
Let \(0\le a<b\), let \(A\in\mathcal{F}_a\), and define \(H_t=\mathbf{1}_A\mathbf{1}_{(a,b]}(t)\). For \(t\le a\), \(H_t=0\); for \(a<t\le b\), \(H_t=\mathbf{1}_A\); and for \(t>b\), \(H_t=0\). At \(a\), the value is \(0\), matching the limit from the left. At \(b\), the value is \(\mathbf{1}_A\), also matching the limit from the left. Thus every path is left-continuous, including when \(a=0\), where only right-hand times are relevant to left continuity.
The process is adapted: before or at \(a\) its value is constant, and for \(t>a\) the event \(A\) belongs to \(\mathcal{F}_a\subseteq\mathcal{F}_t\). Predictability is visible directly as well. For any Borel set \(C\), the inverse image \(H^{-1}(C)\) is a union of sets formed from \((a,b]\times A\), \((a,b]\times A^c\), and time intervals on which \(H\) is constantly zero. The first two are predictable generators because \(A,A^c\in\mathcal{F}_a\); the constant parts are measurable too. In particular, the value at \(b\) belongs to the interval \((a,b]\), as required by the left-continuous path.
Progressive Step Processes and the Direction of Approximation
A process can be progressive without being left-continuous. For example, fix \(s>0\), take \(A\in\mathcal{F}_s\), and set \(J_t=\mathbf{1}_A\mathbf{1}_{[s,\infty)}(t)\). It is adapted: before \(s\) it is zero, and at and after \(s\) the event \(A\) is known. For any horizon \(u\), if \(u<s\) its restriction is zero; if \(u\ge s\), the only nonzero part is \([s,u]\times A\), which is measurable in \(\mathcal{F}_u\otimes\mathcal{B}([0,u])\). Thus \(J\) is progressive. On \(A\), however, its value at \(s\) is \(1\), while its left limit is \(0\), so its paths need not be left-continuous.
This example highlights a distinction in the results: progressive measurability is obtained from continuity and adaptedness by approximating with values at earlier times, whereas predictability of a left-continuous process is obtained by step processes whose value on \((s,t]\) is already \(\mathcal{F}_s\)-measurable. In particular, every continuous adapted process is also left-continuous, so it is predictable as well as progressive.
Common Pitfall: Using Future Values
A tempting approximation for proving predictability is to put \(X_{(k+1)\delta_n}\) on \((k\delta_n,(k+1)\delta_n]\). Although this may converge to \(X_t\) for a continuous process, the coefficient is measurable with respect to \(\mathcal{F}_{(k+1)\delta_n}\), not necessarily \(\mathcal{F}_{k\delta_n}\). Thus the resulting rectangle need not be one of the predictable generators. For instance, with \(X=B\), the value \(B_{(k+1)\delta_n}\) uses an increment after \(k\delta_n\); it is not justified to treat that value as known at \(k\delta_n\). The correction is to use \(X_{k\delta_n}\) on \((k\delta_n,(k+1)\delta_n]\). Left continuity still gives convergence, and the coefficient now uses only information available at the left endpoint.
The time-zero convention is another point to check. The interval generators \((s,t]\times A\) do not include time \(0\). A predictable step approximation must therefore specify its value at \(0\) separately and use an \(\mathcal{F}_0\)-measurable value there. In the theorem, that value is \(X_0\), which is \(\mathcal{F}_0\)-measurable by adaptedness.
Check Your Understanding
For each question, identify which measurability definition or predictable generator is being used.
- In the progressive measurability proof, why is \(X_{t_k}\) measurable with respect to \(\mathcal{F}_T\) whenever \(t_k\le T\)?
- For a left-continuous adapted process, which value approximates \(X_t\) when \(t\) lies in \((k\delta_n,(k+1)\delta_n]\), and why does it converge?
- Verify directly that \(\mathbf{1}_A\mathbf{1}_{(a,b]}\) is adapted when \(A\in\mathcal{F}_a\).
- Why does an approximation using \(X_{(k+1)\delta_n}\) on \((k\delta_n,(k+1)\delta_n]\) not automatically prove predictability?
- Does every continuous adapted process satisfy both conclusions of the two theorems? Explain the role of continuity in your answer.