Is this map measurable?

Question

Consider the map $\varphi:\mathbb{R}\times\Omega\times X\rightarrow X,(t,\omega,x)\rightarrow\varphi(t,\omega,x)$ is $(\mathcal{B}(\mathbb{R})\times\mathcal{F}\times\mathcal{C},\mathcal{C})$-measurable, where $\mathcal{F}$ is the $\sigma$-algebra on $\Omega$, $\mathcal{B}(\mathbb{R})$ is the Borel $\sigma$-algebra on $\mathbb{R}$, $X$ is a Polish space (one can regard it as $\mathbb{R}^n$), $\mathcal{C}$ is the $\sigma$-algebra on $X$. And also consider a random variable $v:\Omega\rightarrow X$.

My question is whether $\omega\mapsto\varphi(t,\omega,v(\omega))$ is $(\mathcal{F},\mathcal{C})$-measurable for any $t\in\mathbb{R}$.

My idea is that for any $t,\omega_0$, $x\mapsto\varphi(t,\omega_0,x)$ is $(\mathcal{C},\mathcal{C})$-measurable, and $v$ is $(\mathcal{F},\mathcal{C})$-measurable, hence $\omega\mapsto\varphi(t,\omega_0,v(\omega))$ is $(\mathcal{F},\mathcal{C})-$ measurable for any $t,\omega_0$. Does this mean that $\omega\mapsto\varphi(t,\omega,v(\omega))$ is $(\mathcal{F},\mathcal{C})$- measurable for any $t\in\mathbb{R}$?

Answer 1

Even more is true. The map $f: (t,\omega)\mapsto\varphi(t,\omega,v(\omega))$ is the composition $\varphi\circ g$ of the measurable maps $g:(t,\omega)\mapsto(t,\omega,v(\omega))$ and $\varphi$. As such, $f$ is $(\mathcal B(\Bbb R)\otimes\mathcal F,\mathcal C)$ measurable. In particular, for each fixed $t$ the partial map $\omega\mapsto f(t,\omega,v(\omega))$ is $(\mathcal F,\mathcal C)$ measurable.

The asserted measurability of $g$ follows because $\mathcal B(\Bbb R)\otimes\mathcal F\otimes\mathcal C$ is generated by events of the form $B\times C$ (where $B\in\mathcal B(\Bbb R)\otimes\mathcal F$ and $C\in \mathcal C$); for such an event, $g^{-1}(B\times C)=B\cap (\Bbb R\times v^{-1}(C))\in \mathcal B(\Bbb R)\otimes\mathcal F$ because $v$ is $(\mathcal F,\mathcal C)$ measurable.

Answer 2

Since $t$ is fixed in what you want to prove, forget about it and consider directly some measurable function $\psi:\Omega\times X\to X$. The map $$\tau:\omega\mapsto(\omega,v(\omega))$$ is measurable since its two components are. Therefore, the composition $$\psi\circ\tau:\omega\mapsto\psi(\omega,v(\omega))$$ is measurable.

You tried to use only the measurability of $x\mapsto\psi(\omega_0,x)$ for $\omega_0$ fixed, but this property of $\psi$ is not sufficient.