I am working in a discrete setting. Consider any stochastic process $(X_n)_{n\in\mathbb N}$ with its natural filtration $(\mathcal F_n)_{n\in\mathbb N}$ and a stopping time $\tau$. We know that $$\mathcal F_\tau = \sigma(\tau,X_n^\tau, n\in\mathbb N).$$ For example in Filtration of stopping time equal to the natural filtration of the stopped process it is discussed if $$\mathcal F_\tau = \sigma(X_t^\tau, t\in\mathbb R_+)$$ holds in the continuous case. I am interested in the discrete analogon $$\mathcal F_\tau = \sigma(X_n^\tau, n\in\mathbb N).\tag{$\ast$}\label{1}$$
Shiryaev shows in his book "Optimal Stopping Rules" Stochastic Modelling and Applied Probability 8 in Springer 2008 that in the continuous case, the result is true if the probability space $\Omega$ is sufficiently rich, that means for all $t\ge0$ and $\omega\in\Omega$ there exists $\omega'\in\Omega$ such that $X_s(\omega')=X_{s\wedge t}(\omega)$ for all $s\ge0$ holds. The proof suggests that the analogue statement also holds in discrete time case.
I am looking for a counterexample, in which \eqref{1} does not hold. I tried to look at probability spaces consisting of only a few points but could not find any counterexample. For any suggestions, I am very thankful.