It seems like the question stated here in MSE has no answer yet and seems therefore for me to be not of a basic question type. For this reason I move it to MO.

Let $X_t : \Omega \to E, \ t \geq 0$ be continuous-time stochastic process with (Polish) state space $E$ and canonical filtration $\mathcal{F}_t := \sigma(X_u \ | \ u \leq t)$. Let $Y_t : \Omega \to \mathbb{R}$ be a real-valued process that is adpated to $\mathcal{F}_t$.

From measure theory it is known that if a $\sigma$-Algebra $\mathcal{A}$ is generated by a real-valued map $h$, i.e. $\mathcal{A} = \sigma(h)$ then every real-valued map $f$ that is $\mathcal{A}$-measurable can be represented as measurable function of $h$, i.e. $f = g \circ h$ for some measurable map $g$.

Applying this to the stochastic process $X_t$ it follows that for each $t$ there exists a measurable map $g_t : E^{[0,t]} \to \mathbb{R}$ such that $Y_t = g_t((X_u)_{u \leq t})$.

I want to know, if there is some possibility to have only one map $g(t, (X_u)_{u \leq t})$ that is jointly measurable in $t \in \mathbb{R}$ and the sample paths $X_u$ up to time $t$ such that $Y_t = g(t, (X_u)_{u \leq t})$ for all $t$. Do I need $X_t$ to be jointly measurable and $Y_t$ to be progressively measurable? How can I formally choose a suitable domain for such a function $g$? Is it some fibre space related to $X$, e.g. a subspace of $[0, \infty) \times E^{[0, \infty)}$ that consists of all elements of the form $(t, (X_u)_{u \leq t})$ (i.e. some sort of measurable upper-diagonal of $[0, \infty) \times E^{[0, \infty)}$)?