Why Measurability at a Random Time Needs Proof
For a fixed time \(n\), an adapted process \(X_n\) is measurable with respect to the information \(\mathcal{F}_n\) available by time \(n\). If the time is chosen randomly, it is not enough to say that \(X_\tau\) is one of the \(X_n\). Which \(X_n\) is used depends on the outcome, and the information available at the random time is itself a sigma-field that must be specified. The key is to separate outcomes according to the value of the stopping time.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) with a discrete-time filtration \((\mathcal{F}_n)_{n\ge0}\), meaning that each \(\mathcal{F}_n\) is a sigma-field and \(\mathcal{F}_n\subseteq\mathcal{F}_{n+1}\). A random time \(\tau:\Omega\to\{0,1,2,\ldots\}\cup\{\infty\}\) is a stopping time if \(\{\tau\le n\}\in\mathcal{F}_n\) for every \(n\ge0\). A real-valued process \((X_n)_{n\ge0}\) is adapted if \(X_n\) is \(\mathcal{F}_n\)-measurable for every \(n\).
First, this definition does give a sigma-field. The stopping-time condition ensures \(\{\tau\le n\}\in\mathcal{F}_n\), so \(\Omega\in\mathcal{F}_\tau\). If \(A\in\mathcal{F}_\tau\), then \(A^c\cap\{\tau\le n\}=\{\tau\le n\}\setminus(A\cap\{\tau\le n\})\in\mathcal{F}_n\), so \(A^c\in\mathcal{F}_\tau\). For a sequence \(A_1,A_2,\ldots\in\mathcal{F}_\tau\), \((\bigcup_r A_r)\cap\{\tau\le n\}=\bigcup_r(A_r\cap\{\tau\le n\})\in\mathcal{F}_n\); hence \(\bigcup_r A_r\in\mathcal{F}_\tau\).
The Integer-Time Decomposition
The stopping-time property makes each individual event \(\{\tau=n\}\) observable at time \(n\). For \(n\ge1\), \(\{\tau=n\}=\{\tau\le n\}\setminus\{\tau\le n-1\}\), which is in \(\mathcal{F}_n\) because \(\mathcal{F}_{n-1}\subseteq\mathcal{F}_n\). Also \(\{\tau=0\}=\{\tau\le0\}\in\mathcal{F}_0\).
Proof. Call the displayed union \(C\). For any fixed \(m\ge0\), the events \(\{\tau=n\}\) are disjoint, and \(C\cap\{\tau\le m\}=\bigcup_{n=0}^{m}(A_n\cap\{\tau=n\})\). For each \(n\le m\), \(A_n\in\mathcal{F}_n\subseteq\mathcal{F}_m\) and \(\{\tau=n\}\in\mathcal{F}_n\subseteq\mathcal{F}_m\). Thus every event in this finite union is in \(\mathcal{F}_m\), and so is the union. This holds for every \(m\), which is exactly the condition \(C\in\mathcal{F}_\tau\). \(\square\)
Proof. Use the stated extension by \(c\), and let \(B\) be any Borel subset of \(\mathbb{R}\). The event that the extended stopped value lies in \(B\) is \(\bigcup_{n\ge0}(\{X_n\in B\}\cap\{\tau=n\})\), together with \(\{\tau=\infty\}\) if \(c\in B\). Adaptedness gives \(\{X_n\in B\}\in\mathcal{F}_n\), so the proposition shows that the union over finite \(n\) belongs to \(\mathcal{F}_\tau\). Also \(\{\tau=\infty\}\in\mathcal{F}_\tau\), since its intersection with \(\{\tau\le m\}\) is empty for every \(m\). The event for the extended value is therefore in \(\mathcal{F}_\tau\) for every Borel \(B\), which proves measurability. If \(\tau<\infty\) almost surely, any two choices on \(\{\tau=\infty\}\) agree almost surely, giving the asserted version. \(\square\)
The proof is not claiming that the whole countable union is observable at any one fixed time. Instead, the definition of \(\mathcal{F}_\tau\) asks for the union after intersecting with \(\{\tau\le m\}\). That intersection involves only the finitely many indices \(0,\ldots,m\), all of which are observable by time \(m\).
Worked Examples
Worked Example: A Capped First-Head Time
Let \(\Omega=\{H,T\}^{\mathbb{N}}\) be the space of infinite sequences of fair, independent coin tosses, with its product probability measure. Let \(\mathcal{F}_n\) record the first \(n\) tosses, with \(\mathcal{F}_0=\{\varnothing,\Omega\}\). Let \(X_n\) count heads among the first \(n\) tosses, with \(X_0=0\), and set \(\tau=1\) if the first toss is \(H\), and \(\tau=2\) if it is \(T\).
This is a stopping time: \(\{\tau\le0\}=\varnothing\), \(\{\tau\le1\}=\{\text{first toss is }H\}\in\mathcal{F}_1\), and \(\{\tau\le n\}=\Omega\in\mathcal{F}_n\) for \(n\ge2\). The process is adapted because \(X_n\) is determined by the first \(n\) tosses. If the first toss is \(H\), then \(X_\tau=X_1=1\). If it is \(T\), then \(\tau=2\), so \(X_\tau=X_2\). Thus \(X_\tau=0\) exactly when the first two tosses are \(TT\), and \(X_\tau=1\) for \(HH,HT,TH\) in the first two positions.
Check the event probabilities. Each of the four two-toss outcomes has probability \(\tfrac14\). Therefore \(\mathbb{P}(X_\tau=0)=\tfrac14\) and \(\mathbb{P}(X_\tau=1)=\tfrac34\); these sum to \(\tfrac14+\tfrac34=1\). For the stopped-field check, \(\{X_\tau=0\}\cap\{\tau\le1\}=\varnothing\), while \(\{X_\tau=0\}\cap\{\tau\le2\}=\{TT\}\in\mathcal{F}_2\). For \(m\ge2\), \(\{\tau\le m\}=\Omega\), and \(\{TT\}\in\mathcal{F}_m\). The level set is therefore in \(\mathcal{F}_\tau\), as the theorem says.
Worked Example: A Random-Walk Value at a Bounded Time
On the same infinite fair-coin space and filtration, define \(Y_n\) to be the number of heads in the first \(n\) tosses. Let \(T_2=\inf\{n\ge0:Y_n=2\}\), with \(\inf\varnothing=\infty\), and let \(\tau=T_2\wedge3\). This is a stopping time because \(\{\tau\le n\}=\{T_2\le n\}\in\mathcal{F}_n\) for \(n<3\), while \(\{\tau\le n\}=\Omega\in\mathcal{F}_n\) for \(n\ge3\). The process is adapted and \(\tau\le3\).
List the first three tosses; each of the eight strings has probability \(\tfrac18\). For \(HHH,HHT\), the process first reaches \(2\) at time \(2\), so \(Y_\tau=2\). For \(HTH,THH\), it reaches \(2\) at time \(3\), so again \(Y_\tau=2\). For \(HTT,THT,TTH\), it does not reach \(2\) by time \(3\) and \(Y_\tau=1\); for \(TTT\), \(Y_\tau=0\). Hence \(\mathbb{P}(Y_\tau=2)=\tfrac48=\tfrac12\), \(\mathbb{P}(Y_\tau=1)=\tfrac38\), and \(\mathbb{P}(Y_\tau=0)=\tfrac18\). The probabilities add to \(\tfrac48+\tfrac38+\tfrac18=1\).
Check the decomposition for the level set \(2\). It is \(\{Y_\tau=2\}=(\{Y_2=2\}\cap\{\tau=2\})\cup(\{Y_3=2\}\cap\{\tau=3\})\). The first event is the two-prefix \(HH\), with either third toss; the second consists of \(HTH\) and \(THH\). To check membership in \(\mathcal{F}_\tau\), intersect this event with \(\{\tau\le1\}\), obtaining \(\varnothing\); with \(\{\tau\le2\}\), obtaining the event \(HH\in\mathcal{F}_2\); and with \(\{\tau\le m\}\) for \(m\ge3\), obtaining the full level set, which is in \(\mathcal{F}_m\). Each required intersection is measurable at its corresponding time.
Worked Example: Allowing the Stopping Time to Be Infinite
Let \(\Omega=\{0,1\}^{\mathbb{N}}\) with independent coordinates \(Z_1,Z_2,\ldots\), each equal to \(1\) with probability \(\tfrac12\), and let \(\mathcal{F}_n\) record the first \(n\) coordinates. Define \(X_n=Z_1+\cdots+Z_n\), with \(X_0=0\), and \(\tau=\inf\{n\ge1:Z_n=1\}\). For each \(n\), \(\{\tau\le n\}=\bigcup_{k=1}^n\{Z_k=1\}\in\mathcal{F}_n\), so \(\tau\) is a stopping time. The process is adapted.
On \(\{\tau<\infty\}\), exactly one of the first \(\tau\) coordinates is \(1\), so \(X_\tau=1\). On \(\{\tau=\infty\}\), define the extended stopped value to be \(c=0\). For the Borel set \(B=\{1\}\), the level event is \(\{\tau<\infty\}\). For every \(m\), its intersection with \(\{\tau\le m\}\) is \(\{\tau\le m\}\in\mathcal{F}_m\), verifying the stopped-field condition directly. Also, \(\mathbb{P}(\tau=\infty)=0\): the event that the first \(m\) coordinates are all zero has probability \(2^{-m}\), and \(\{\tau=\infty\}\) is contained in that event for every \(m\). Since \(2^{-m}\to0\), its probability is zero. The choice \(c=0\) thus does not change the stopped value almost surely, but it makes the variable defined on all of \(\Omega\).
Bounded Times and a Common Pitfall
Proof. By the theorem, \(X_\tau\) is \(\mathcal{F}_\tau\)-measurable. If \(A\in\mathcal{F}_\tau\), then \(\tau\le N\) on all outcomes, so \(A=A\cap\{\tau\le N\}\in\mathcal{F}_N\) by the definition of \(\mathcal{F}_\tau\). Thus \(\mathcal{F}_\tau\subseteq\mathcal{F}_N\), and in particular \(X_\tau\) is \(\mathcal{F}_N\)-measurable. \(\square\)
Common mistake: treating a countable union as measurable at one fixed time. In the bounded random-walk example, \(\{Y_\tau=2\}\) includes \(HTH\) but not \(HTT\). Those two outcomes have the same first two tosses, so this level event is not in \(\mathcal{F}_2\). It would be incorrect to say that every piece \(\{Y_n=2\}\cap\{\tau=n\}\) is in \(\mathcal{F}_n\), and therefore their infinite union is in \(\mathcal{F}_2\). The pieces are measurable at different times. The correction is to use the definition: for each \(m\), intersect with \(\{\tau\le m\}\), leaving only indices \(n\le m\). In this example, the intersection is empty for \(m=1\), is the \(HH\) event for \(m=2\), and is the full level event for \(m\ge3\).
Adaptedness is essential as well. On the infinite fair-coin space, let \(\mathcal{F}_0=\{\varnothing,\Omega\}\), let \(\tau=0\), and set \(X_0=\mathbf{1}_{\{Z_1=1\}}\). Then \(\mathcal{F}_\tau=\mathcal{F}_0\), but \(\{X_\tau=1\}=\{Z_1=1\}\) is not in \(\mathcal{F}_0\): it is neither empty nor all of \(\Omega\). Here \(X_0\) is not \(\mathcal{F}_0\)-measurable, so the adaptedness hypothesis fails. The measurability theorem concerns the value at a stopping time; by itself it does not assert any martingale identity or expectation formula.
Check Your Understanding
For each question, identify the relevant information at the random time and use the stopped-field definition where appropriate.
- Why does \(\{\tau=n\}\in\mathcal{F}_n\) follow from the stopping-time condition when \(n\ge1\)?
- If \(A_n\in\mathcal{F}_n\), what finite union describes \((\bigcup_{n\ge0}A_n\cap\{\tau=n\})\cap\{\tau\le m\}\)?
- For a stopping time bounded by \(N\), explain why every event in \(\mathcal{F}_\tau\) is in \(\mathcal{F}_N\).
- In the capped first-head example, what are \(\mathbb{P}(X_\tau=0)\) and \(\mathbb{P}(X_\tau=1)\)? Check that they sum to \(1\).
- Why does assigning a fixed value on \(\{\tau=\infty\}\) produce a well-defined measurable stopped value even when \(\tau\) may be infinite?