Let $\Omega = \{\pm1\}^{\mathbb{Z}}$ with its product $\sigma$-algebra, and let $X = \ell^1(\mathbb{Z})$. For $\omega \in \Omega$ let

Then $X$ is separable and every $V_\omega$ belongs to the Grassmannian $\mathcal{G} X$, but there is no measurable map $W : \Omega \to \mathcal{G} X$ with $W(\omega) = V_\omega$ for every $\omega \in \Omega$.

Over the shift $\sigma(\omega)_n = \omega_{n+1}$ with the Bernoulli measure, the operators $\mathcal{L}_\omega x = \varphi_\omega(x)\, e_0$ form a cocycle of rank one with $\lambda_1 = 0$, fast space $E_1(\omega) = \operatorname{span}\{e_0\}$ and slow space $V_\omega$; so separability of $X$ alone does not make the slow spaces measurable. This example is due to J. A. Horan (PhD thesis, University of Victoria, 2020). The argument: distinct $V_\omega$ are at distance at least $1$ in $\mathcal{G} X$ and $V_\omega = V_{-\omega}$, so a measurable $W$ would make every set $S = -S$ measurable, while there are more such sets than measurable ones.

In Lean a sign $\omega_i \in \{\pm1\}$ is coded by a boolean, $true$ for $+1$.