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


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 $E^{[0,\infty)}$.

  • $\begingroup$ Thank you! I'm not an expert in this area, but just to recapitulate: For the domain for $g$ you use that the path $X^\omega\in M([0, \infty), E)$ the set of measurable maps. By fixing the Lebesgue measure $\lambda$ on $[0, \infty)$ one can define the topology of convergence in measure on $M$ which results in $L^0$. This one is Polish, by going from $\lambda$ to an equivalent finite measure. $g$ is then defined as $g(t,f) := g_t(f|_{[0,t]})$ for $f \in L^0$. Why is $g$ jointly measurable on its whole domain? Is there something special in the choice of $\lambda$ for generating a topology on $M$? $\endgroup$ – yadaddy Oct 31 '14 at 13:18

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