# Is this map measurable?

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 $$t\in\mathbb{R}$$?

• What is $\phi (t,\omega)$? Commented Dec 28, 2023 at 8:56
• $\varphi(t,\omega)$ is a map from $X$ to $X$. It is $\mathcal{C},\mathcal{C}$ measurable Commented Dec 28, 2023 at 9:43

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.
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.