Two Different Measurability Questions
For a stochastic process, measurability can concern the time variable and the outcome variable together, or it can ask what is measurable at each fixed time. These are different questions. Joint measurability is a property of a map on the product of the time interval and the sample space. Adaptedness is a family of separate requirements, one at each time, relative to the information available then. Neither property alone generally gives the other.
Fix \(T>0\) and a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\), where \(\mathcal{F}_s\subseteq\mathcal{F}_t\subseteq\mathcal{F}\) whenever \(s\le t\). We consider real-valued processes \(X=(X_t)_{0\le t\le T}\). Joint measurability below means measurability with respect to the product sigma-field \(\mathcal{B}([0,T])\otimes\mathcal{F}\); in particular, the time coordinate uses the Borel sigma-field.
The distinction in the definitions matters. Joint measurability uses the full sigma-field \(\mathcal{F}\) for the outcome coordinate at every time. Adaptedness instead asks whether the particular value \(X_t\) can be measured using \(\mathcal{F}_t\). Thus joint measurability does not restrict the process to information available at time \(t\). Conversely, adaptedness does not impose any Borel regularity on how the values vary with \(t\).
What Joint Measurability Says About Fixed Times
Proof. First, if \(E\in\mathcal{B}([0,T])\otimes\mathcal{F}\), define its section at \(t\) by \(E_t=\{\omega\in\Omega:(t,\omega)\in E\}\). The collection of product-space sets \(E\) for which \(E_t\in\mathcal{F}\) is a sigma-field: taking complements takes complements of sections, and taking countable unions takes countable unions of sections. It contains every measurable rectangle \(A\times C\), because its section is \(C\) when \(t\in A\) and the empty set otherwise. It therefore contains the product sigma-field. Now fix any Borel set \(G\subseteq\mathbb{R}\). By joint measurability, \(E=\{(s,\omega):X_s(\omega)\in G\}\) belongs to the product sigma-field, so its section \(E_t=\{\omega:X_t(\omega)\in G\}\) belongs to \(\mathcal{F}\). This holds for every Borel \(G\), which is exactly \(\mathcal{F}\)-measurability of \(X_t\). The argument gives no reason for \(E_t\) to belong to the smaller sigma-field \(\mathcal{F}_t\). \(\square\)
Worked Example: Jointly Measurable but Not Adapted
Let \(\Omega=\{0,1\}\), let \(\mathcal{F}\) contain all subsets of \(\Omega\), and set \(\mathbb{P}(\{0\})=\mathbb{P}(\{1\})=\tfrac12\). On the time interval \([0,1]\), define \(\mathcal{F}_t=\{\varnothing,\Omega\}\) for \(t<1\), and \(\mathcal{F}_1=\mathcal{F}\). This is an increasing family of sigma-fields. Define \(X_t(\omega)=\omega\) at every time.
The inverse image of the Borel set \(\{1\}\) under the map \((t,\omega)\mapsto X_t(\omega)\) is \([0,1]\times\{1\}\). Both factors are measurable in their respective sigma-fields, so this rectangle belongs to \(\mathcal{B}([0,1])\otimes\mathcal{F}\). The map is jointly measurable: the inverse images of \(\{0\}\) and \(\{1\}\) are measurable rectangles, and these values exhaust the range.
At \(t=\tfrac12\), however, \(\{\omega:X_{1/2}(\omega)=1\}=\{1\}\), which is not in \(\mathcal{F}_{1/2}=\{\varnothing,\Omega\}\). Thus \(X_{1/2}\) is not \(\mathcal{F}_{1/2}\)-measurable and the process is not adapted. At \(t=1\), \(X_1\) is \(\mathcal{F}_1\)-measurable. The failure at even one time is enough to violate adaptedness.
Check: the event \(\{X_{1/2}=1\}\) has probability \(\tfrac12\), but probability alone is not the issue: it is absent from the time-\(\tfrac12\) sigma-field. Joint measurability uses \(\mathcal{F}\), which does contain that event.
Adaptedness Does Not Guarantee Joint Measurability
For a process that depends only on time, adaptedness is automatic: at each fixed time its value is a constant random variable. Yet an arbitrary function of time need not be Borel measurable. The next result makes this contrast precise for deterministic processes.
Proof. For each fixed \(t\), \(X_t\) is the constant random variable \(g(t)\). For any Borel set \(G\subseteq\mathbb{R}\), its inverse image is either \(\Omega\) or \(\varnothing\), both of which belong to every sigma-field \(\mathcal{F}_t\). Hence the process is adapted, without any condition on the function \(g\).
Suppose next that \(g\) is Borel measurable. For each Borel set \(G\subseteq\mathbb{R}\), the joint inverse image is \(g^{-1}(G)\times\Omega\), which belongs to \(\mathcal{B}([0,T])\otimes\mathcal{F}\). Thus \(X\) is jointly measurable. Conversely, suppose \(X\) is jointly measurable. Choose any \(\omega_0\in\Omega\), which is possible because \(\Omega\) is nonempty. The section-at-\(\omega_0\) argument for product sigma-fields shows that \(\{t:X_t(\omega_0)\in G\}\) is Borel for every Borel \(G\). Since \(X_t(\omega_0)=g(t)\), this set is \(g^{-1}(G)\). Therefore \(g\) is Borel measurable, proving the reverse implication as well. \(\square\)
Worked Example: Adapted but Not Jointly Measurable
Take \([0,1]\) as the time interval, and choose a subset \(D\subseteq[0,1]\) that is not Borel. Such subsets exist. Let \(\Omega=\{0,1\}\) with \(\mathcal{F}\) the set of all subsets, and take \(\mathcal{F}_t=\mathcal{F}\) at every time. Define \(X_t(\omega)=\mathbf{1}_D(t)\), where \(\mathbf{1}_D(t)\) is \(1\) if \(t\in D\) and \(0\) otherwise.
For every fixed \(t\), \(X_t\) is either the constant random variable \(0\) or the constant random variable \(1\). Both are measurable with respect to \(\mathcal{F}_t\), so the process is adapted at every time.
To test joint measurability, take the Borel target set \(\{1\}\). Its inverse image is \(D\times\Omega\). If the process were jointly measurable, the section of this set at \(\omega=0\) would be Borel in time. That section is exactly \(D\), contradicting the choice of \(D\). Therefore \(X\) is not jointly measurable.
Check: the time section at any fixed \(t\) is measurable, but the set of times at which the value is \(1\) is non-Borel. Measurability of all the random variables \(X_t\) does not fix this failure in the time coordinate.
A Process with Both Properties
Worked Example: Jointly Measurable and Adapted
Let \(\Omega=\{0,1\}\), let \(\mathcal{F}\) contain all subsets, and assign probability \(\tfrac12\) to each outcome. On \([0,1]\), let \(\mathcal{F}_t=\{\varnothing,\Omega\}\) for \(t<\tfrac12\), and let \(\mathcal{F}_t=\mathcal{F}\) for \(t\ge\tfrac12\). Write \(A=\{1\}\) and define \(X_t(\omega)=\mathbf{1}_{[1/2,1]}(t)\mathbf{1}_A(\omega)\).
For joint measurability, the inverse image of \(\{1\}\) is \([1/2,1]\times A\), a measurable rectangle. The process only takes the values zero and one, so the inverse image of any Borel set is one of the empty set, the full product space, this rectangle, or its complement. Each is product measurable.
For adaptedness, if \(t<\tfrac12\), then \(X_t=0\), which is measurable with respect to the trivial sigma-field. If \(t\ge\tfrac12\), then \(\{X_t=1\}=A\in\mathcal{F}_t\); the inverse image of every Borel set is consequently in \(\mathcal{F}_t\). Thus \(X\) is adapted at all times as well as jointly measurable.
Check: before time \(\tfrac12\), the process is identically zero, so it does not require knowledge of \(A\). At and after time \(\tfrac12\), \(A\) belongs to the available sigma-field. These two cases verify adaptedness at the boundary time as well as on either side.
How to Keep the Properties Separate
| Property | What is tested | Measurability used |
|---|---|---|
| Joint measurability | The map \((t,\omega)\mapsto X_t(\omega)\) | \(\mathcal{B}([0,T])\otimes\mathcal{F}\) |
| Adaptedness | The random variable \(X_t\), separately at each \(t\) | \(\mathcal{F}_t\) |
A useful first check is to ask which sigma-field appears in the claim. If a statement concerns a map of both time and outcome, test inverse images in the product sigma-field. If it concerns information at a specified time, test inverse images under \(X_t\) in \(\mathcal{F}_t\). The proposition above supplies one implication: joint measurability gives \(\mathcal{F}\)-measurable time sections. It does not give the stronger, time-dependent requirement in adaptedness.
Common mistake. A tempting but incorrect argument is: “For each fixed \(t\), \(X_t\) is measurable, therefore \((t,\omega)\mapsto X_t(\omega)\) is jointly measurable.” This confuses separate measurability in the outcome variable with measurability on a product space. In the deterministic example, each \(X_t\) is a measurable constant, but the inverse image of \(\{1\}\) is \(D\times\Omega\), whose time section is the non-Borel set \(D\). The corrected conclusion is that adaptedness verifies the fixed-time tests only; a separate argument is needed for joint measurability.
Check Your Understanding
For each question, identify which sigma-field or product sigma-field must contain the relevant inverse image.
- In the first worked example, why does joint measurability hold even though adaptedness fails at \(t=\tfrac12\)?
- For a deterministic process \(X_t(\omega)=g(t)\), which condition on \(g\) is equivalent to joint measurability?
- In the non-Borel-set example, what is the inverse image of \(\{1\}\), and which section shows that it is not product measurable?
- Can an adapted process fail to be \(\mathcal{F}\)-measurable at a fixed time? Explain using \(\mathcal{F}_t\subseteq\mathcal{F}\) and the definitions.
- In the final worked example, why must the case \(t=\tfrac12\) be checked separately from \(t<\tfrac12\)?