Why is $\mathbb R_{+}^{[0,\infty)}\cap C([0,\infty))$ measurable wrt $\mathcal{B}(\mathbb R)^{\otimes [0,\infty)}\cap C([0,\infty))$

Question

Why is $\mathbb R_{+}^{[0,\infty)}\cap C([0,\infty))$ measurable wrt $\mathcal{B}(\mathbb R)^{\otimes [0,\infty)}\cap C([0,\infty))$?

I have shown that $\mathbb R_{+}^{[0,\infty)}, C([0,\infty))\notin \mathcal{B}(\mathbb R)^{\otimes [0,\infty)}$ but I am stumped on showing

$\mathbb R_{+}^{[0,\infty)}\cap C([0,\infty))\in \mathcal{B}(\mathbb R)^{\otimes [0,\infty)}\cap C([0,\infty))$ since it does not seem intuitive at all. Any ideas?

Note $C([0,\infty))$ denotes the space of continuous functions from $[0,\infty) \to \mathbb R$

Answer 1

A continuous function $f:[0,\infty) \to \mathbb{R}$ is non-negative if, and only if, $$\forall t \in \mathbb{Q} \cap [0,\infty)\::\: f(t) \geq 0.$$ If we denote by $\pi_t$ the projection of $f \in \mathbb{R}^{[0,\infty)}$ onto the $t$-th coordinate, i.e. $\pi_t(f)=f(t)$, then this shows that

$$\mathbb{R}^{[0,\infty)}_+ \cap C([0,\infty)) = \bigcap_{t \in \mathbb{Q} \cap [0,\infty)} ( \{\pi_t \geq 0\} \cap C([0,\infty)). \tag{1}$$ By the definition of the trace $\sigma$-algebra, the preimage $\{\pi_t \geq 0\} \cap C([0,\infty))$ belongs to $\mathcal{B}(\mathbb{R})^{\otimes [0,\infty)} \cap C([0,\infty))$ for each $t \geq 0$. Since the intersection in $(1)$ is countable, we conclude that $\mathbb{R}^{[0,\infty)}_+ \cap C([0,\infty)) \in \mathcal{B}(\mathbb{R})^{\otimes [0,\infty)} \cap C([0,\infty))$.