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

Solution 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.

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. $\omega' \in A$.