Let $\Omega = C(\mathbb{R}_+,\mathbb{R})$ be the space of continuous functions from $[0,\infty)$ to $\mathbb{R}$. The $\sigma$-field $\mathscr{A}$ on $\Omega$ is defined to be the smallest $\sigma$-field such that the coordinate map $\Phi_t: \omega \mapsto \omega(t)$ is measurable for all $t \ge 0$.
Now I want to show that the function $\Phi: \Omega \to \mathbb{R}$ $$ \Phi = \sup_{t\le s} \Phi_t$$ is measurable.
One way is to observe that $$\Phi(\omega) = \sup_{t\le s} \Phi_t(\omega) = \sup_{t\le s}\omega(t) = \sup_{t\le s \ \text{and } t \in \mathbb{Q}} \omega(t)$$ by the continuity of $\omega$. Hence, as a supremum of a countable collection of measurable functions, $\Phi$ is measurable.
I was told that another way is to use the following fact:
The $\sigma$-field $\mathscr{A}$ coincides with the Borel $\sigma$-field when $\Omega$ is endowed with the topology of compact convergence (i.e., uniform convergence on every compact set).
But I can't see how this fact can be used to prove the measurability of $\Phi$.