I am stuck with the following doubt (for which I cannot find a proof, or a counter example). Let us have $(\mathcal{F}_t)_{t \geq0}$ as a filtration on $(\Omega,\mathcal{F}, \mathbb{P})$. Let us have a process $X = (X_t)_{t \geq0}$
Is it true that: $$\text{if} \ \ \ \ \ \mathbb{E}\left[f( X_t) | \mathcal{F}_s \right] = a$$ $$\text{and} \ \ \ \ \ \mathbb{E}\left[f( X_t) \right] = a$$ $$\text{then} \ \ \ \ \ \ X_t \ \ \text{ is independent of} \ \ \ \ \mathcal{F}_s ?$$ where $f()$ is a deterministic, continuous function and $a \in \mathbb{R}$.
If this is true for every deterministic continuous function $f$, then you do have $X_t$ is independent of $\mathcal F_s$. The reason is that you can approximate the indicator function of a Borel set $B$ by continuous functions to show that it holds for indicator functions, then using the definition of conditional expectation we have for every $A \in \mathcal F_s$ that \begin{align*} \mathbb{E}[1_A 1_{X_t \in B}] = \mathbb{E}[1_A \mathbb{E}[1_{X_t \in B}|\mathcal F_s]] = \mathbb{E}[1_A]\mathbb{E}[1_{X_t \in B}] = P(A)P(X_t \in B). \end{align*} Since $\sigma(X)$ is generated by sets of the form $\{X_t \in B\}$, this shows $X_t$ is independent of $\mathcal F_s$.
If you just know it is true for a particular deterministic continuous function $f$, you can't say anything. For example, the constant function $f(x)=a$ for all $x$ satisfies those conditions for any $X_t$ and $\mathcal F_s$.