Given a probability space $(\Omega,\mathcal{F},P)$ and a process $X_{t}$ defined on it. We consider the natural Filtration generated by the process $\mathcal{F}_{t}=\sigma (X_{s}:s\leq t)$. Let $\tau$ be a stopping time. The corresponding $\sigma$-Algebra of the stopping time $\tau$ is given by
\begin{align}
F_{\tau}=\left\{A\in \bigcup_{t\geq 0} F_{t}: A\cap\{\tau \leq t\}\in \mathcal{F}_{t} \forall t\geq 0\right\}
\end{align}
Now i am considering the requirements under which
\begin{align}
\mathcal{F}_{\tau}=\sigma (X_{s\wedge \tau }:s\geq 0)
\end{align}
holds.
Shiryaev "Optimal Stopping Rules" Stochastic Modelling and Applied Probability 8 in Springer 2008
makes the assumptions, that $\Omega$ has to be sufficiently rich, that means for all $t\geq 0$ and $\omega \in \Omega$ there exists $\omega'\in\Omega$ such that
\begin{align}
X_{s}(\omega')=X_{s\wedge t}(\omega) \forall s\geq 0
\end{align}
holds; see Theorem 6 of Shiryaev. Then $\mathcal{F}_{\tau}=\sigma (X_{s\wedge \tau }:s\geq 0)$ should hold. Shiryaev shows this for arbitrary measurable space, and measurable process $X$ fullfilling this condition.
Also D. Stroock, S. Varadhan "Multidimensional Diffusion Processes" Lemma 1.3.3 proofs it, where here $\Theta$ is the space of continuous trajectories from $[0,\infty)$ to $\mathbb{R}^{d}$. In the proof Stroock also uses the assumption of the sufficiently richness of $\Omega$ under $X$, but he mentions it incidental, as if it would be clear for continuous $X$.
I guess, that this property also holds for cadlag processes $X$ i.e. fullfilling
\begin{align}
\mathcal{F}_{\tau}=\sigma (X_{s\wedge \tau }:s\geq 0)
\end{align}
But i dont know, how it is fullfilled.
Best Answer
The "saturation" property assumed by Shiryaev is indeed clear for $\Omega$ consisting of the space of cadlag paths from $[0,\infty)$ to $\Bbb R^d$, and $X_t(\omega)=\omega(t)$ for $t\ge0$ and $\omega\in\Omega$. For, given $t\ge 0$ and $\omega\in\Omega$, define $\omega'$ to be the path "$\omega$ stopped at time $t$"; that is, $\omega'(s):=\omega(s\wedge t)$ for $s\ge 0$. This path $\omega'$ is clearly an element of $\Omega$, and $X_{s\wedge t}(\omega) =\omega(s\wedge t) =\omega'(s)=X_s(\omega')$ for each $s\ge 0$.