Functional representation of adapted jointly measurable stochastic processes

Question

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 $[0, \infty) \times E^{[0, \infty)}$)?

Answer 1

It is something of a folk theorem that every adapted measurable process (with values in a Polish space) admits a progressively measurable modification. This is in Dellacherie & Meyer's book, and a quick google search turned up a recent paper advertising a simpler proof. In short, the answer to your first question is yes, up to null sets: There exists a jointly measurable map $g$ as you describe, but the equality $Y_t = g(t,(X_u)_{u \le t})$ holds almost surely, for each $t$. As for your second question, if your process $Y$ has no continuity whatsoever, the domain of $g$ should probably be $[0,\infty) \times L^0([0,\infty);E)$, where $L^0([0,\infty);E)$ is the space of (equivalence classes of Lebesgue-a.e. equal) measurable functions $[0,\infty) \rightarrow E$. It just so happens that your particular function $g$ will satisfy $g(t,\omega) = g(t,\omega')$ whenever $\omega_u = \omega'_u$ for a.e. $u \le t$. This space $L^0$ is relatively pleasant in that it's Polish when topologized with convergence in measure, unlike $E^{[0,\infty)}$.