Cfrgtkky

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

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 $tin T$ and $omegain Omega$ there is $bar omega in Omega$ such that $bar omega_s = omega_{swedge t}$ for all $sin T$ (I personally don't know why we need this). Let $mathcal F=mathcal E^Tcap Omega$ where $mathcal E^T$ is the $sigma$-algebra on $E^T$. Let $X=(X_t)_{tin T}$ be the canonical process on $(Omega,mathcal F)$.

The problem I have is to prove the following assertion:

Assertion. $Ain mathcal F^X_t$ implies the following

1. 
   $Ainmathcal F_infty^X$.


2. 
   $omega in A$ and $X_s(omega)=X_s(omega')$ for all $sin T$ with $sleq t$ imply $omega'in A$.

Attempt.

Number (1) follows from $mathcal F_t^Xsubsetmathcal F_infty^X$. 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_nleq t, C_1,...,C_ninmathcal C, n=1,2,....}
end{align}

Of course, if $Ain mathcal C_t^X$, 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?

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. $omega' in A$.