Question about a proof on the canonical process of Brownian motion. Here we have $\mathcal{C}_{(0)}$ is the space of continuous functions $[0,\infty)\to \mathbb{R}^d$ starting at $0$, with the metric of locally uniform convergence, 
$$\rho(w,v)=\sum_{n=1}^\infty (1\wedge \sup_{0\le t\le n}|w(t)-v(t)|)2^{-n}.$$
$B(v,r)$ is the ball centered at $v$ with radius $r$ in the metric $\rho$.
Let $(B_t)_{t\ge 0}$ be a Brownian motion on some probability space $(\Omega,\mathcal{A},\mathbb{P})$(we assume here that the BM has exclusively continuous paths) and consider the map 
$$\Psi:\Omega\to \mathcal{C}_{(0)},\;\omega\mapsto w:=\Psi(\omega):=B(\cdot,\omega).$$
We want to show that this map is $\mathcal{A}/\mathcal{B}(\mathcal{C}_{(0)})$ measurable. This is the proof given in the text.
Proof: The metric balls $B(v,r) \subset \mathcal{C}_{(0)}$  generate $ \mathcal{B}(\mathcal{C}_{(0)})$. From (4.1), that is, the map $h\mapsto \rho(h,g)$ for any $h,g\in \mathcal{C}_{(0)}$ is  $\mathcal{B}(\mathcal{C}_{(0)})$ measurable, we conclude that the map $\omega\mapsto \rho(B(\cdot,\omega),v)$ is $\mathcal{A}$ measurable; therefore 
$$\{\omega: \Psi(\omega)\in B(v,r)\}=\{\omega:\rho(B(\cdot,\omega),v)<r\}$$ shows that $\Psi$ is $\mathcal{A}/ \mathcal{B}(\mathcal{C}_{(0)})$ measurable.
However, I don't understand this proof. If we consider the map $F:\mathcal{C}_{(0)}\to \mathbb{R}$ defined as  $F(v)= \rho(v,w) $, then we need to show that for any $v\in \mathcal{C}_{(0)}$ and $r\ge 0$, $\{\omega:\Psi(\omega)\in B(v,r)\}\in \mathcal{A}$. I.e. we need to show that $(F\circ \Psi)^{-1}((-\infty,r))\in \mathcal{A}$. 
However, in the proof, he simply says that since the map $F$ is measurable with respect to $ \mathcal{B}(\mathcal{C}_{(0)})$ so we conclude that the map $\omega\mapsto \rho(B(\cdot,\omega),v)$ is $\mathcal{A}$ measurable and we have the result. I don't understand the reasoning of this proof. The map  $\omega\mapsto \rho(B(\cdot,\omega),v)$ itself is a composite of $F$ and $\Psi$ and we need to show that $\Psi$ is measurable, so how do we know that the composite is measurable to begin with? Doesn't the measurability of this map rely on the measurability of $\Psi$?
Below is (4.1).

 A: The key is that (4.1) doesn't quite say what you say it does. Discussing the measurability of $(v,w)\mapsto\rho(v,w)$ somewhat obscures the idea of this proof, so I'm going to do away with it and prove the result directly, although hopefully when I'm done you will be able to recognize that my proof is actually the same as the one in the textbook.
Let's start with what we know:


*

*$B(t,\cdot)$ is $(\mathcal A,\mathcal B(\mathbb R^d))$-measurable for all $t\ge0$;

*$B(\cdot,\omega)\in\mathcal C_{(0)}$ for all $\omega\in\Omega$.


That turns out to be all we need. From this and basic measure theory, we know that $\omega\mapsto|B(t,w)-v(t)|$ is measurable for any $t\ge0$, and hence
$$\omega\mapsto1\wedge\sup_{k\in\mathbb N}|B(t_k,\omega)-v(t_k)|$$
is measurable for any bounded sequence $\{t_k\}$ in $[0,\infty)$. In particular, we can take this sequence to be an enumeration of $[0,n]\cap\mathbb Q$, and so
$$\omega\mapsto\sum_{n=1}^\infty\left(1\wedge\sup_{t\in[0,n]\cap\mathbb Q}|B(t,\omega)-v(t)|\right)2^{-n}$$
is measurable. But since $B(\cdot,\omega)\in\mathcal C_{(0)}$, this last expression is equal to $\rho(B(\cdot,\omega),v)$. Hence
$$\{\omega:\Psi(\omega)\in B(v,r)\}=\{\omega:\rho(B(\cdot,\omega),v)<r)\}\in\mathcal A$$
which completes the proof.
