It seems like the question stated [here](https://math.stackexchange.com/questions/982241/functional-representation-of-adapted-jointly-measurable-stochastic-processes) 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 $[0, \infty) \times E^{[0, \infty)}$)?



Functional representation of adapted jointly measurable stochastic processes
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](http://www.mat.ub.edu/EMIS/journals/EJP-ECP/article/download/2548/2548-13043-1-PB.pdf) 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 $E^{[0,\infty)}$.


