A property of natural filtrations Let $(X_t)_{t \in [0,\infty)}$ be a stochastic process, and let $\mathcal{F}_t$ be the natural filtration generated by $X_t$; i.e. $$\mathcal{F}_t=\sigma(X_s:0\leq s \leq t).$$ Now we fix a $t \in [0,\infty)$. Let $A \in \mathcal{F}_t$ and let $\omega \in A$. Suppose that $\omega'\in \Omega$ satisfies $$X_s(\omega)=X_s(\omega')$$ for any $s \in [0,t]$. How can we show that $\omega'\in A$?
 A: There is a result (due to Doob, I think), following from the monotone class theorem, that if $A\in\mathcal F_t$ then there is a sequence $(t_n)\subset[0,t]$ and an element $B\in\mathcal B^{\Bbb N}$ such that
$$
A=\{\omega: (X_{t_1}(\omega), X_{t_2}(\omega),\ldots)\in B\}.
$$
From this it is clear that if $\omega\in A$ and $X_s(\omega') =X_s(\omega)$ for all $s\in[0,t]$ (or just for each $s=t_n$, $n\in\Bbb N$) then $\omega'\in A$ as well.
A: Let $(\Omega, \mathcal{F}, P)$ be the underlying probability space. We say that $A \in \mathcal {F}$ satisfies the property $\mathcal P$ if and only if for any $\omega \in A$ and $\omega' \in \Omega$ with $X_s(\omega')=X_s(\omega) \quad \forall s \in [0,t]$, we have $\omega'\in A$. Now we use the principle of good sets as suggested by @user159517.
Define $\mathcal A=\{A \in \mathcal {F}: A \text{ satisfies property } \mathcal P\}$. We need to show two statements:

*

*$\mathcal A$ is a $\sigma$-algebra;

*$\bigcup_{s \in [0,t]}\sigma(X_s) \subset \mathcal A$.

It follows that $\mathcal F \subset\sigma(\mathcal A)=\mathcal A\subset \mathcal F$ and hence $\mathcal A=\mathcal F$.
Argument for 1:
Trivially $\emptyset \in \mathcal A$. Let $A \in \mathcal A$ and suppose for the sake of contradiction that $A^c \notin \mathcal A$. Then there exists $\omega \in A^c$ and $\omega' \in (A^c)^c=A$ such that $X_s(\omega)=X_s(\omega')$ for every $s\in [0,t]$. This is a contradiction to $A \in \mathcal A$. The $\sigma-$additivity is trivially true.
Argument for 2:
Let $A \in \bigcup_{s\in[0,t]}\sigma(X_s)$. Then $A \in \sigma(X_s)$ for some $s\in [0,t]$. Hence, $A=X_s^{-1}(B)$ for some Borel set $B$. Suppose $\omega \in A$, then $$\omega'\in X_s^{-1}(X_s(\omega'))=X_s^{-1}(X_s(\omega))\subset X_s^{-1}(B)=A.$$
