Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, and $\{X_t:0\leq t\leq 1\}$ be a continuous stochastic process. We can then map $\Omega$ into Banach space $C([0,1])$ in a natural way, by sending $\omega$ to the continuous function $X(\omega)$.
My question is: how to show that this mapping is measurable when $C([0,1])$ is equipped with its Borel $\sigma$-algebra?
The hint says that $\|x\|\leq 1$ if and only if $|x(t)|\leq 1$ for all $t$ in a countable dense subset of $[0,1]$. But I have no idea how this hint is related to measurabilty of the mapping $\omega\mapsto X.(\omega)$.
Let $D\subseteq[0,1]$ be dense and countable.
Then according to the hint: $$\{|X|\leq 1\}=\bigcap_{t\in D}\{|X_t|\leq1\}$$showing that - if the $X_t$ are measurable - the set $\{|X|\leq1\}$ is measurable.
Of course more generally it can be proved that $\{|X-f|\leq r\}$ is measurable for all $r\geq0$ and $f\in C[0,1]$.
Proved is then that $X^{-1}(\mathcal V)\subseteq\mathcal F$ where $\mathcal V$ denotes the collection of closed balls in $C[0,1]$.
In general we have $X^{-1}(\sigma(\mathcal V))=\sigma(X^{-1}(\mathcal V))$ so that: $$X^{-1}(\sigma(\mathcal V))=\sigma(X^{-1}(\mathcal V))\subseteq\mathcal F$$
And here $\sigma(\mathcal V)$ can be recognized as the Borel $\sigma$-algebra on $C[0,1]$, so we are ready.