Let $X=(X_t)_{t\ge0}$ be a stochastic process on $\mathbb{R}$ and $W$ a Brownian motion, independent of $X$.
If $X$ is deterministic, then $X$ is adapted to the filtration generated by $W$.
Does the converse also hold, i.e., if $X$ is adapted to the filtration generated by $W$, is $X$ necessarily deterministic?
Yes, this is true. Because $X$ is adapted to the filtration generated by $W$, given any set of paths $A$ we have that $X^{-1}(A) \in \mathcal F^W$, so $X^{-1}(A) = W^{-1}(B)$ for some set $B$. Thus $$\mathbb{P}(X \in A) = \mathbb{P}(X \in A, W \in B) = \mathbb{P}(X \in A) \mathbb{P}(W \in B) = \mathbb{P}(X \in A)^2.$$ This implies $\mathbb{P}(X \in A) \in \{0,1\}$, i.e. $X$ is deterministic.