Proving Galmarino's Test

Question

Galmarino's Test gives a condition equivalent to being a stopping time. It says:

Let $X$ be a continuous stochastic process with index set $\mathbb{R}_+$ (i.e. each sample path is a continuous function of time). Let $\mathscr{F}$ be the filtration generated by $X$. Then a random time $T$ is a stopping time iff

for every pair of outcomes $\omega$ and $\omega'$, $T(\omega) = t$, $X_s(\omega) = X_s(\omega')$ for $s \leq t \implies T(\omega') = t$

The condition essentially says that the map $T$ restricted to $\{T \leq t\}$ factors through $(X_s)_{s \leq t}$, and I can prove that it is necessary by using a monotone class argument. I don't know how to prove the converse though.

Answer 1

Throughout my answer, $(X_t)_{t \geq 0}$ denotes a stochastic process with continuous sample paths on the canonical space, i.e.

$$X_t(\omega) := \omega(t), \qquad \omega \in \Omega := C([0,\infty)),$$

As usual we denote by

$$\mathcal{F}_t := \sigma(X_s; s \leq t)$$

the canonical filtation of $(X_t)_{t \geq 0}$ and set $\mathcal{F} := \sigma(X_s; s \geq 0)$. Moreover, we define $a_t: C([0,\infty)) \to C([0,\infty))$ by

$$a_t(\omega)(s) := \omega(s \wedge t), \qquad s \geq 0, \omega \in C[0,\infty).$$

For the proof of Galmarino's test we need an auxiliary result.

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Lemma: $$\mathcal{F}_t = a_t^{-1}(\mathcal{F}) \qquad \text{for all $t \geq 0$.}$$

Proof: First, we check that $a_t: (\Omega,\mathcal{F}_t) \to (\Omega,\mathcal{F})$ is measurable. Since $\mathcal{F} = \sigma(X_s; s \geq 0)$, this is equivalent to $$X_s \circ a_t: (\Omega,\mathcal{F}_t) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$$ being measurable for each $s \geq 0$. This, however, follows directly from the identity $$X_s \circ a_t(\omega) = (a_t(\omega))(s) = \omega(s \wedge t) = X_{s \wedge t}(\omega).$$ The measurability of $a_t$ yields $a_t^{-1}(\mathcal{F}) \subseteq \mathcal{F}_t$. To prove $\mathcal{F}_t \subseteq a_t^{-1}(\mathcal{F})$, it suffices to show that $\omega \mapsto X_s(\omega)$ is $a_t^{-1}(\mathcal{F})/\mathcal{B}(\mathbb{R})$-measurable for all $s \leq t$. Since

$$X_s(\omega) = \omega(s) = \omega(s \wedge t) = a_t(\omega)(s)= X_s(a_t(\omega))$$

for all $s \leq t$ we get

$$\{X_s \in B\} = \{X_s(a_t) \in B\} = \{a_t \in X_s^{-1}(B)\} \in a_t^{-1}(\mathcal{F})$$

for any $B \in \mathcal{B}(\mathbb{R})$. This shows that $X_s$ is indeed $a_t^{-1}(\mathcal{F})/\mathcal{B}(\mathbb{R})$-measurable for $s \leq t$.

Corollary 1: (Baby version of Galmarino's test) For any set $A \in \mathcal{F}$ the following statements are equivalent:

1. $A \in \mathcal{F}_t$
2. If $\omega \in A$, $\omega' \in \Omega$ are such that $X_s(\omega)=X_s(\omega')$ for all $s \leq t$, then $\omega' \in A$.

Proof: "1. $\implies$ 2.": Let $A \in \mathcal{F}_t$. By the above lemma, there exists $C \in \mathcal{F}$ such that $A=a_t^{-1}(C)$. Now if $\omega \in A$ and $\omega' \in \Omega$ are such that $X_s(\omega)=X_s(\omega')$ for all $s \leq t$, then $a_t(\omega)=a_t(\omega')$, and so $$1_A(\omega') = 1_{a_t^{-1}(C)}(\omega') = 1_C(a_t(\omega')) = 1_C(a_t(\omega))=1_A(\omega),$$ i.e. $\omega' \in A$.

"2. $\implies$ 1.": It follows our assumption that we have $$\omega \in A \iff a_t(\omega) \in A, $$ and so $$1_A(\omega) = 1_A(a_t(\omega)) = 1_{a_t^{-1}(A)}(\omega)$$ for all $\omega \in \Omega$, i.e. $A = a_t^{-1}(A)$. It follows from the above lemma that $A \in \mathcal{F}_t$.

Corollary 2: (Galmarino's test) For any random time $T$ the following statements are equivalent:

1. $T$ is a stopping time, i.e. $\{T \leq t\} \in \mathcal{F}_t$ for all $t \geq 0$.
2. If $\omega, \omega' \in \Omega$ are such that $T(\omega)=t$ and $X_s(\omega)= X_s(\omega')$ for all $s \leq t$, then $T(\omega')=t$.

Proof: "1. $\implies$ 2." Since $$\{T = t\} = \{T \leq t\} \backslash \bigcup_{k \in \mathbb{N}} \{T \leq t-1/k\} \in \mathcal{F}_t,$$ it follows from Corollary 1 that for any $\omega \in \{T = t\}$ the implication $$\{\forall s \leq t=T(\omega): \, \, X_s(\omega) = X_s(\omega')\} \implies \omega' \in \{T = t\}$$ holds which proves the assertion.

"2. $\implies$ 1." Set $A := \{T \leq t\}$. If $\omega \in A$ and $\omega' \in \Omega$ are such that $X_s(\omega) = X_s(\omega')$ for all $s \leq t$, then it follows from our assumption that $T(\omega')=t$, and so $\omega' \in A$. Applying Corollary 1 yields $\{T \leq t\} = A \in \mathcal{F}_t$.

Remarks:

• Note that we haven't used the continuity of the sample paths; so, in fact, the claim does not hold only true for stochastic processes with continuous sample paths.
• In 2. we may replace $T(\omega)=t$ and $T(\omega')=t$ by $T(\omega) \leq t$ and $T(\omega') \leq t$, respectively.

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Thanks to @Shashi who helped a lot to improve this answer; (s)he come up with the idea to prove Galmarino's test using Corollary 1.

Answer 2

I think saz's proof is wrong in the last step when he claims $T\cdot I_{\{T\le t\}}$ is $\mathcal{F}_t$ measurable by Lemma 2. Also, his proof only works for real valued stochastic process. In fact, Galmarino Test works for process taking value in any measurable space $(E,\mathcal{E})$.

Galmarino Test: Let $(X_t)_{t\ge0}$ be a $(E,\mathcal{E})$ valued canonical process defined on the canonical probability space $(\Omega,\mathcal{F},P)$, that is, $\Omega=E^{[0,+\infty)}$, $\mathcal{F}=\mathcal{E}^{[0,+\infty)}$ and $X_t$ is the coordinate map on $E^{[0,+\infty)}$: for any $t\ge0$: $X_t(\omega)=\omega(t)\;\forall \omega\in \Omega$. Let $\mathcal{F}_t=\sigma(X_s,0\le s\le t)$ for $t\ge0$ be the natural filtration. Suppose a map $T$ from $\Omega$ to $[0,+\infty]$ is a random time. Then the following two statements are equivalent:

(a) $T$ is a $\mathcal{F}_t$ stopping time

(b) For any $\omega,\omega'\in\Omega$ and any $t\ge0$, $T(\omega)\le t$ and $X_s(\omega)=X_s(\omega')\forall s\le t$ imply $T(\omega)=T(\omega')$.

Proof: For each $t\ge0$, define $a_t$ to be a map from $\Omega$ to $\Omega$ such that for any $\omega\in\Omega(=E^{[0,+\infty)}$), $a_t(\omega)(s)=\omega(s\wedge t)$ for $s\ge0$.

Loosely speaking, $a_t(\omega)$ keeps the part of $\omega$ before $t$ and reset part of $\omega$ after $t$ to be $\omega(t)$. By the definition of $a_t$, for any $\omega,\omega'\in\Omega$, $a_t(\omega)=a_t(\omega')$ is equivalent to $\omega(s)=\omega'(s)\forall s\le t$, which is also equivalent to $X_s(\omega)=X_s(\omega)\forall s\le t$ (because $X_s$ is the coordinate map). Thus statement (b) is equivalent to the follow statement:

(b') For any $\omega,\omega'\in\Omega$ and any $t\ge0$, $T(\omega)\le t$ and $a_t(\omega)=a_t(\omega')$ imply $T(\omega)=T(\omega')$.

Claim 1: (b') is equivalent to the following statement:

(b'') For any $\omega,\omega'\in\Omega$ and any $t\ge0$, $T(\omega)\le t$ and $a_t(\omega)=a_t(\omega')$ imply $T(\omega')\le t$.

Let's first show (b') implies (b''). Suppose (b') holds. Let $\omega,\omega'\in\Omega$ and $t\ge0$ such that $T(\omega)\le t$ and $a_t(\omega)=a_t(\omega')$. By (b'), $T(\omega')=T(\omega)$. Since we already have $T(\omega)\le t$ so $T(\omega')\le t$.

Next, we show (b'') implies (b'). Suppose (b'') is true. Let $\omega,\omega'\in\Omega$ and $t\ge0$ such that $T(\omega)\le t$ and $a_t(\omega)=a_t(\omega')$. We need to show $T(\omega)=T(\omega')$.

Define $r=T(\omega)$ so $r\le t$ (becauase $T(\omega)\le t$). Now we have $$T(\omega)\le r\quad (1)\quad \mbox{ and }a_r(\omega)=a_r(\omega')\quad (2)$$ (1) is simply because of the definition of $r$. (2) is because $a_t(\omega)=a_t(\omega')$, that is, $\omega(s)=\omega'(s)\forall s\le t$. Since $r\le t$, so of course we also have $\omega(s)=\omega'(s)\forall s\le r$, which is (2). By (1)(2) and using (b''), we have $T(\omega')\le r$ which is $$T(\omega')\le T(\omega)\quad (*)$$ Define $r'=T(\omega')$. So $r'=T(\omega')\le T(\omega)\le t$. Similarly, we have $$T(\omega')\le r'\quad (3)\quad \mbox{ and }a_{r'}(\omega)=a_{r'}(\omega')\quad (4)$$ By (3)(4) and using (b'') (we need to exchange the role of $\omega$ and $\omega'$ when using (b'')), we have $T(\omega)\le r'$, which is $T(\omega)\le T(\omega')$. This combined with (*) shows $T(\omega)=T(\omega')$.

Claim 2: $a_t$ is $\mathcal{F}_t/\mathcal{F}$ measurable for each $t\ge0$.

Recall $X_t$ is the coordinate map so $\mathcal{F}=\mathcal{E}^{[0,+\infty)}=\sigma(X_t,t\ge0)$. Now let's fix $t\ge0$. $a_t$ takes value in $\Omega=E^{[0,+\infty)}$ so to show $a_t$ is $\mathcal{F}_t/\mathcal{E}^{[0,+\infty)}$ measurable, we only need to show each coordinate of its coordinate is $\mathcal{F}_t/\mathcal{E}$ measurable, that is, to show $X_s\circ a_t$ is $\mathcal{F}_t/\mathcal{E}$ measurable for each $s\ge0$.

For each $s\ge0$, $X_s\circ a_t(\omega)=a_t(\omega)(s)=\omega(s\wedge t)=X_{s\wedge t}(\omega)$ is $\mathcal{F}_t/\mathcal{E}$ measurable because $X_{s\wedge t}$ is $\mathcal{F}_{s\wedge t}/\mathcal{E}$ measurable and $\mathcal{F}_{s\wedge t}\subseteq\mathcal{F}_t$. Proof of claim 2 is done.

Claim 3: $\mathcal{F}_t=\{a_t^{-1}(A)|A\in\mathcal{F}\}$ for each $t\ge0$.

For any $A\in\mathcal{F}$, $a_t^{-1}(A)\in\mathcal{F}_t$ (claim 2), so $\{a_t^{-1}(A)|A\in\mathcal{F}\}\subseteq\mathcal{F}_t$. Now we only need to show $\mathcal{F}_t\subseteq\{a_t^{-1}(A)|A\in\mathcal{F}\}$. For any $B\in\mathcal{F}_t=\sigma(X_s,s\le t)$, there exist $t'\in [0,t]$ and $S\in\mathcal{E}$ such that $B=X_{t'}^{-1}(S)$. Now for any $\omega\in\Omega$, $X_{t'}(\omega)=\omega(t')=\omega(t'\wedge t)=a_t(\omega)(t')=X_{t'}(a_t(\omega))$. This shows $X_{t'}=X_{t'}\circ a_t$. Thus $B=X_{t'}^{-1}(S)=(X_{t'}\circ a_t)^{-1}(S)=a_t^{-1}(X_{t'}^{-1}(S))\in \{a_t^{-1}(A)|A\in\mathcal{F}\}$ where the last step is because $X_{t'}$ is $\mathcal{F}_{t'}/\mathcal{E}$ measurable so $X_{t'}^{-1}(S)\in\mathcal{F}_{t'}\subseteq\mathcal{F}$. Therefore $\mathcal{F}_t\subseteq\{a_t^{-1}(A)|A\in\mathcal{F}\}$. Claim 3 is proved.

Now we are ready to prove the theorem. Since (b) is equivalent to (b') which is also equivalent to (b''), now we only need to show (a) and (b'') are equivalent.

Step 1: Show (a) implies (b''). Suppose (a) is true. Let $\omega,\omega'\in\Omega$ and $t\ge0$ such that $T(\omega)\le t$ and $a_t(\omega)=a_t(\omega')$. We need to show $T(\omega')\le t$.

Since $T(\omega)\le t$ and by (a) and claim 3, $\omega\in\{T\le t\}\in\mathcal{F}_t=\{a_t^{-1}(A)|A\in\mathcal{F}\}$. Thus there exists $S\in\mathcal{F}$ such that $\{T\le t\}=a_t^{-1}(S)$. So $\omega\in a_t^{-1}(S)$. Thus $a_t(\omega)\in A$. Because $a_t(\omega)=a_t(\omega')$, $a_t(\omega')\in A$ which implies that $\omega'\in a_t^{-1}(A)=\{T\le t\}$ so $T(\omega')\le t$.

Step 2: Show (b'') implies (a). Suppose (b'') is true. We need to show $\{T\le t\}\in\mathcal{F}_t$ for any $t\ge0$. Now fix $t\ge0$. We claim $\{T\le t\}=a_t^{-1}(\{T\le t\})$.

To show this claim, we first show $\{T\le t\}\subseteq a_t^{-1}(\{T\le t\})$. For any $\omega\in\{T\le t\}$. So $T(\omega)\le t$. Define $\omega'=a_t(\omega)$ so $a_t(\omega')=a_t(a_t(\omega))=a_t(\omega)$ where in the last step we used the fact $a_t\circ a_t$ by the definition of $a_t$. Now we have $T(\omega)\le t$ and $a_t(\omega)=a_t(\omega')$ and thus by (b'') $T(\omega')\le t$ so $T(a_t(\omega))\le t$, thus $a_t(\omega)\in \{T\le t\}$ hence $\omega\in a_t^{-1}(\{T\le t\})$. Therefore $\{T\le t\}\subseteq a_t^{-1}(\{T\le t\})$ is proved. Next, we show $a_t^{-1}(\{T\le t\})\subseteq\{T\le t\}$. For any $\omega'\in a_t^{-1}(\{T\le t\})$, $a_t(\omega')\in\{T\le t\}$ thus $T(a_t(\omega'))\le t$. Define $\omega=a_t(\omega')$ so $T(\omega)\le t$ and $a_t(\omega)=a_t(a_t(\omega'))=a_t(\omega')$. By (b''), $T(\omega')\le t$. Hence $\omega'\in\{T\le t\}$. $a_t^{-1}(\{T\le t\})\subseteq\{T\le t\}$ is proved.

Therefore, we have proved $\{T\le t\}=a_t^{-1}(\{T\le t\})$. Now it's clear $\{T\le t\}=a_t^{-1}(\{T\le t\})\in\mathcal{F}_t$ because $\{T\le t\}\in\mathcal{F}$ ($T$ is a random time) and $a_t$ is $\mathcal{F}_t/\mathcal{F}$ measurable (claim 2).

Answer 3

Got a hint from http://wt.iam.uni-bonn.de/fileadmin/WT/Inhalt/people/Karl-Theodor_Sturm/Lectures/vorlesungWS0809/sheet1.pdf

Consider the problem in the canonical space. Define

$\alpha_t(\omega(\cdot)) = \omega(\cdot \wedge t)$

Using the monotone class theorem, we can show that the mapping:

$\alpha_t: (\Omega,{\mathcal F}_t^X) \to (\Omega,\mathcal{F})$

is measurable. And using the condition, we can show that

$ (T\le t) = \alpha_t^{-1}(T\le t)$

Therefore, $(T\le t)\in \mathcal{F}_t^X$.

Answer 4

I think that the proofs by saz and MathGuy didn't need to use the continuity of $X$ because they didn't specify what $\sigma\{X_t: t\in\mathbb{R}_{+}\}$ is. If we take it to be the smallest $\sigma$-algebra containing rectangles of the form $\prod_{t \in \mathbb{R}_{+}}A_t$ where $A_t$ differs from $\mathbb{R}$ only on a finite number of $t$, as defined in planetmath, then we need the process to be continuous for Galmarino's test to apply. If $(\Omega, \mathscr{H})$ is a measurable space and $X: \Omega \rightarrow \mathbb{R}^{\mathbb{R}_+}$, this is the smallest $\sigma$-algebra such that all $X_t$ are measurable relative to $\mathscr{H}$ and $\sigma\{X_t: t\in\mathbb{R}_{+}\}$. Note that the rectangles form a $\pi$-system that generates the $\sigma$-algebra we'll use. The $A_t$'s that differ from $\mathbb{R}$ can be sets in the Borel $\sigma$-algebra on $\mathbb{R}$ denoted by $\mathscr{B}(\mathbb{R})$.

We start with the following proposition

Proposition 1: A random variable $V: \Omega \rightarrow \mathbb{R}_{+}$ is in $\sigma\{X_t: t\in\mathbb{R}_{+}\}$ if and only if there exists a countable sequence $(t_n)$ in $\mathbb{R}_{+}$ and a positive measurable function $f$ in $\prod_{n}\mathscr{B}(\mathbb{R})_{t_{n}}$ such that $$V = f(X_{t_{1}}, X_{t_{2}}, \dots)\tag{1}$$

Proof: Sufficiency of the condition follows from the measurability of $f$ which ensures that $V\in\sigma\{X_{t_n}: n \geq 1\}\subset\sigma\{X_t: t\in\mathbb{R}_{+}\}$. Necessity can be proved with the monotone class theorem for positive functions:

1. Let $\mathscr{M}_{+}$ be the collection of positive functions of the form (1)
2. Show that $\mathscr{M}_+$ is a monotone class
3. Show that $\mathscr{M}_+$ contains all indicators of products of the form $\prod_{t \in \mathbb{R}_{+}}A_t$ where $A_t$ differs from $\mathbb{R}$ only on a finite number of $t$
4. Since the products are a $\pi$-system that generates $\sigma\{X_t: t\in\mathbb{R}_{+}\}$, $\mathscr{M}_+$ includes all positive $V\in\sigma\{X_t: t\in\mathbb{R}_{+}\}$

Now we use proposition 1 to prove the necessity of the condition in Galmarino's test. Let $\Omega_t=\{\omega: T(\omega) \leq t\}$, since $T$ is a stopping time $\Omega_t\in\mathcal{F}_t\implies1_{\Omega_t}\in\mathcal{F}_t$. By proposition 1, there is a measurable function $f$ such that $$1_{\Omega_t}=f(X_{t_1}, X_{t_2}, \dots)\tag{2}$$ for some countable sequence $(t_n) \subset [0, t]$. In light of (2), for a pair of outcomes $\omega$ and $\omega'$ such that $T(\omega)=t$ and $X_s(\omega)=X_s(\omega')$ for $s\leq t$ we clearly have $1=1_{\Omega_t}(\omega) = 1_{\Omega_t}(\omega')$. In other words, $\omega'\in\Omega_t\implies T(\omega') \leq t$. Suppose $T(\omega') = t' < t$, by hypothesis $X_s(\omega')=X_s(\omega)$ for $s \leq t'$ so, in light of (2) again, $1=1_{\Omega_{t'}}(\omega') = 1_{\Omega_{t'}}(\omega)\implies T(\omega) \leq t'$, but this contradicts the hypothesis which leads us to conclude that $T(\omega')=t$.

To prove sufficiency of Galmarino's test condition, first we show the following proposition, $\Omega_t$ is as above

Proposition 2: Galmarino's test condition implies that for every pair of outcomes $\omega$ and $\omega'$ such that $X_s(\omega) = X_s(\omega')$ for all $s \leq t$ we have $1_{\Omega_t}(\omega)=1_{\Omega_t}(\omega')$

Proof: It's enough to show that $\omega \in \Omega_t \iff \omega' \in \Omega_t$. Assume $\omega \in \Omega_t$, so we have $T(\omega)=t'\leq t$ and since $X_s(\omega)=X_s(\omega')$ for all $s \leq t'$ Galmarino's condition applies and we have $T(\omega) = T(\omega') = t' \leq t \implies \omega' \in \Omega_t$. This shows that $\omega \in \Omega_t \implies \omega' \in \Omega_t$. To show the converse swap $\omega$ and $\omega'$ in the preceding argument.

Let $a_t: C[0, \infty) \rightarrow C[0, t]$ be a map that truncates continuous functions in $\mathbb{R}_+$ to $[0, t]$, that is $a_t(X)(s) = X_s$ for $s \leq t$.

From proposition 2 it's clear that $a_t(X(\omega)) = a_t(X(\omega')) \implies 1_{\Omega_t}(\omega)=1_{\Omega_t}(\omega')$, so there exists a function $g_t:C[0, t] \rightarrow \{0, 1\}$ such that $g_t(a_t(X(\omega)))=1_{\Omega_t}(\omega)$. In other words, $1_{\Omega_t}$ is determined by $\{X_s: s \leq t\}$.

Now, pick any dense sequence $(t_n)$ in $[0, t]$ and let $X' = (X_{t_1}, X_{t_2}, \dots)$. Because of the continuity of $X$, for every possible $X'$ there is a unique path of $X$ in $[0, t]$ that generates it, and so for any pair of events $\omega$ and $\omega'$ such that $X'(\omega) = X'(\omega')$ we have $a_t(X(\omega)) = a_t(X(\omega'))$. This implies the existence of a function $f: \prod_{n}\mathbb{R}_{t_{n}} \rightarrow \{0, 1\}$ where $f(X'(\omega)) = g_t(a_t(X(\omega)))=1_{\Omega_t}(\omega)$. By hypothesis $1_{\Omega_t}(\omega)$ is measurable which implies the measurability of $f(X'(\omega))$, and by proposition 1 $1_{\Omega_t} \in \mathcal{F}_t$ as needed.

Example

Let $T=\inf\{t \in \mathbb{R}_+: X_t \geq b\}$, it's clear that Galmarino's test applies to this random time so it must be a stopping time. However, we also need the continuity of $X$ for this to work in the minimial $\sigma$-algebra, without it we can't have a function as in (2) for $1_{\Omega_t}$. No matter what sequence $(t_n)$ we pick, there will be points we miss in $[0, t]$ that can flip the outcome of $1_{\Omega_t}$. On the other hand, if $X$ is continuous we can construct such a function. Let $\epsilon_i \rightarrow 0$ and $(t_n)$ as above, set $B_i=\prod_{s \in [0, t]}A_t$ where $A_t=(-\infty, b-\epsilon_i]$ for $t\in (t_n)$ and $A_t=\mathbb{R}$ otherwise. Note that ($B_i$) is an increasing sequence of rectangles in $\sigma\{X_s: s\in[0, t]\}$, and by the continuity of $X$ each $B_i$ only contains outcomes outside $\Omega_t$ so $a_t(X(\omega)) \in B_i \implies \omega \notin \Omega_t$. Furthermore, if $\omega \notin \Omega_t$ by the continuity of $X$ there has to be an $\epsilon \gt 0$ such that $X_s(\omega) \lt b-\epsilon$ for all $s \leq t$, which implies that for $i$ large enough $a_t(X(\omega)) \in B_i$. So we've shown that $a_t(X(\omega)) \in B_i \iff \omega \notin \Omega_t$. Finally, again owing to continuity there is a function $f: \prod_{n}\mathbb{R}_{t_{n}} \rightarrow \{0, 1\}$ $$ f(X_{t_1}(\omega), X_{t_2}(\omega), \dots) = \prod_{i} [1 - 1_{B_{i}}(a_t(X(\omega)))] $$ which evaluates to 1 when $\omega \in \Omega_t$ and 0 otherwise, so $1_{\Omega_t}=f(X_{t_1}, X_{t_2}, \dots)$.