Consider a stochastic process $X: [0, \infty) \times \Omega \to \mathbb{R}^d$
We define $\mathcal{F}_t = \sigma(X(s), 0 \leq s \leq t)$ The sigma algebra generated by the sets $\{\omega: X(s,\omega) \in \Gamma \}$ for $\Gamma \in \mathcal{B}(\mathbb{R}^d)$.
Now a stopping time is a function $\tau: \Omega \to [0,\infty] $ such that $[\tau \leq t]=\{\omega:\tau(\omega) \leq t\} \in \mathcal{F}_t $ for all $t$.
Let $\mathcal{F} = \sigma(X(s), 0 \leq s <\infty)$. We consider now the sigma algebra given by $$\mathcal{F}_\tau = \{A \in \mathcal{F}: A \cap [\tau \leq t] \in \mathcal{F}_t , \forall t \}.$$
We would like to prove that $$\mathcal{F}_\tau = \sigma (X(s \wedge \tau): 0 \leq s < \infty )$$
It is clear that $$\sigma (X(s \wedge \tau): 0 \leq s < \infty ) \subset \{A \in \mathcal{F}: A \cap [\tau \leq t] \in \mathcal{F}_t , \forall t \}$$
but to the other inclusion I am trying to follow Stroock and Varadhan (Diffusion processes with continuous coefficients I - 1969 - appendix - pg 394) they say that:
Claim if $\xi $ is $\mathcal{F}_\tau$ measurable then there is a measurable function $f: (\mathbb{R}^d)^{\infty} \to \mathbb{R}$ and a sequence $t_1 < \ldots <t_n <\ldots$ such that $$\xi(\omega) = f(X(t_1,\omega),\ldots,X(t_n,\omega), \ldots).$$
The question is to show the above claim.
Edit: Just to give the originals here below:
In pg 394 of Diffusion processes with continuous coefficients I one reads
but in Dynkin we read
It is not clear we can use Dynkin 1.5 since we don't know that $\mathcal{F}_\tau$ is generated by a family of times $t \ in T$ that index it as in Dynkin's case