I am interested in a proof of the following statement which seems intuitive, but is somehow really tricky:

Let $X$ be a stochastic process and let $(\mathcal{F}(t) : t \geq 0)$ be the filtration that it generates (unaugmented). Let $T$ be a bounded stopping time. Then we have $\mathcal{F}(T) = \sigma(X(T \wedge t) : t \geq 0)$

I have a proof at hand (Bain and Crisan, Fundamentals of Stochastic Filtering, page 309), but in my opinion there is a major gap. I will try to explain the idea of proof.

Let $V$ be the space of functions $[0,\infty) \rightarrow \mathbb{R}$ equipped with the sigma algebra generated by the cylinder sets. Consider the canonical map $X^T:\Omega \rightarrow V$ which maps $\omega$ to the trajectory $t \mapsto X(t \wedge T(\omega),\omega)$. Then we have $\sigma(X(T \wedge t) : t \geq 0) = \sigma(X^T)$.

The difficult part is $\subseteq$. Let $A \in \mathcal{F}(T)$. We want to find a measurable map $g:V \rightarrow \mathbb{R}$ such that $1_A = g \circ X^T$, then we're done. It is now straightforward to show that $1_A$ is constant on sets where the sample paths of $X^T$ are constant. (To be more precise: for $\rho \in \Omega$ consider the set $\mathcal{M}(\rho) = \lbrace \omega : X(\omega,t) = X(\rho,t), 0 \leq t \leq T(\rho) \rbrace$. Then $T$ and $1_A$ are constant on every set of this form).

The problem is: this is not sufficient! It suffices to construct a map $g$ such that $1_A = g \circ X^T$, but how we can we know that $g$ is measurable? This is where the proof of Bain and Crisan comes up short IMO.

I can show this result only under the assumption that the map $X:\Omega \rightarrow V$ be surjective: Since $A \in \mathcal{F}(\infty)$, we have a measurable map $g$ such that $1_A = g \circ X$. Let $\rho \in \Omega$. Then $T$ and $1_A$ are constant on $\mathcal{M}(\rho)$. Therefore, $g$ must be constant on the image of $\mathcal{M}(\rho)$ under $X$. Because $X$ is assumed to be surjective, this image contains the function $X^T(\rho)$. Hence, $g \circ X = g \circ X^T$, and we are done.

I think that this result could be a little bit deeper. I have seen two proofs of this for the special case that $X$ is the coordinate process on $C[0,\infty)$, one is given in the book of Karatzas & Shreve, Lemma 5.4.18. The fact that Karatzas proves this late in the book only in this special case somehow makes me think that the general case is not so easy.

I would really appreciate any comment or other reference for this result.

Foundations of Modern Probability(exercise 6 of chapter 7, page 138 of the 2nd ed.) giving something in this direction. He assumes however that the process is progressive. The general result you mention would indeed be very nice but I also have doubts wether it holds in that generality. $\endgroup$ – Jochen Wengenroth Jul 9 '12 at 7:18necessaryconditions in A.22. and A.23. which they verify in thesufficiencypart of the proof -- of course, this is not conclusive. The way you describe the argument shows the difficulty if one does not have any regularity: Measurability of the function $g: \mathbb{R}^{[0,\infty)} \to \mathbb{R}$ is a very restrictive condition because $g$ must only depend on countably many variables. $\endgroup$ – Jochen Wengenroth Jul 9 '12 at 14:38