Characterization of sets in $\mathcal F^X_t =\sigma(\{X_s: s\leq t\})$ where $(X_t)_t$ is a stochastic process

Question

I'm selfstudying stochastic processes. There is a question on the characterization of sets in $\mathcal F_t^X:=\sigma(\{X_s: s\leq t \})$.

Let $T \subset \mathbb R^+$ be the time index set. We consider $\Omega \subset E^T$ where $E$ is some (nice) set, e.g. $\mathbb R$ with the property that for eacht $t\in T$ and $\omega\in \Omega$ there is $\bar \omega \in \Omega$ such that $\bar \omega_s = \omega_{s\wedge t}$ for all $s\in T$ (I personally don't know why we need this). Let $\mathcal F=\mathcal E^T\cap \Omega$ where $\mathcal E^T$ is the $\sigma$-algebra on $E^T$. Let $X=(X_t)_{t\in T}$ be the canonical process on $(\Omega,\mathcal F)$.

The problem I have is to prove the following assertion:

Assertion. $A\in \mathcal F^X_t$ implies the following

1. $A\in\mathcal F_\infty^X$.
2. $\omega \in A$ and $X_s(\omega)=X_s(\omega')$ for all $s\in T$ with $s\leq t$ imply $\omega'\in A$.

Attempt.

Number (1) follows from $\mathcal F_t^X\subset\mathcal F_\infty^X$. But I'm having troubles with the second implication. I started like this, let $\mathcal C$ be a $\pi$-system that generates $\mathcal E$. Then in one of the notes they say $\mathcal F_t^X$ is generated by the following $\pi$-system $\mathcal C_t^X$, defined as \begin{align} \mathcal C_t^X = \{X_{t_1}^{-1}(C_1)\cap ...\cap X_{t_n}^{-1}(C_n) : t_1<t_2 <...<t_n\leq t, C_1,...,C_n\in\mathcal C, n=1,2,....\} \end{align}
Of course, if $A\in \mathcal C_t^X$, then $\omega \in A$ if and only if \begin{align*} X_{t_1}(\omega) \in C_1,....,X_{t_n}(\omega)\in C_n \end{align*} for some set $C_1, C_2,..$ etc. But then it is immediately clear that $\omega'\in A$. I can only do this in the case $A$ is in the $\pi$-system. How can I conclude that the assertion also holds if $A$ is not in the $\pi$-system?

Answer 1

For $t \in T$ define a mapping $a_t: \Omega \to \Omega$ by

$$a_t(\omega)(s) := \omega(t \wedge s), \qquad s \in T.$$

This mapping is well-defined because of the property of $\Omega \subseteq E^T$ which you mentioned in your question.

Claim: $\mathcal{F}_t^X \subseteq \mathcal{G}_t := a_t^{-1}(\mathcal{F}_{\infty}^X)$

Proof: It suffices to show that $X_s$ is $\mathcal{G}_t$-measurable for any $s \leq t$, $s \in T$. Since $$X_s(\omega) = \omega(s) = \omega(s \wedge t) = a_t(\omega)(s) = X_s(a_t(\omega))$$ for any $s \leq t$, $s \in T$, we have $$\{X_s \in B\} = \{X_s(a_t) \in B\} = \{a_t \in X_s^{-1}(B)\} \in a_t^{-1}(\mathcal{F}_{\infty}^X)$$ for any measurable set $B$. Hence, $X_s$ is $\mathcal{G}_t$-measurable, and this finishes the proof.

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Fix $A \in \mathcal{F}_t^X$. By the above result, there exists $C \in \mathcal{F}_{\infty}^X$ such that $A = a_t^{-1}(C)$. Now if $\omega \in A$ and $\omega' \in \Omega$ are such that $$\forall s \leq t, s \in T: \quad X_s(\omega) = X_s(\omega')$$ then $$\forall s \leq t, s \in T: \quad \omega(s) = \omega'(s),$$ and so $a_t(\omega)=a_t(\omega')$. Thus,

$$1_A(\omega') = 1_{a_t^{-1}(C)}(\omega') = 1_C(a_t(\omega')) = 1_C(a_t(\omega)) = 1_A(\omega)=1,$$

i.e. $\omega' \in A$.