Let $C([0,\infty))$ be the space of all continuous functions from $[0,\infty)$ to $\mathbb{R}$ and let $\mathcal{B}(C([0,\infty)))$ be the Borel $\sigma$-alegbra generated by the metric
$$d(f,g) = \sum_{n=1}^\infty \frac{1}{2^n} \max(1, |f-g|_n)$$ $$ |f-g|_n = \sup_{x \in [0,n]} |f(x)-g(x)| $$
(This metric is one that makes $C([0,\infty)), \mathcal{B}(C([0,\infty)))$ a polish space.)
I am struggling with how to show that the event $A$ defined as $$ A = \left\{ f \in C([0,\infty)) : \lim\limits_{t \to \infty} \frac{f(t)}{t} = 0 \right\} $$ is contained in $\mathcal{B}(C([0,\infty)))$. I think this set is an open set? As it seems like any function ``close enough" to $f$ such that $\lim\limits_{t \to \infty} \frac{f(t)}{t} = 0$ should also go to zero when divided by $t$ but I am struggling to prove that using the metric. I also don't know if there is an easier way to get this.
For context, this question was motivated by this question: Let $\{B_t: t \geq 0\}$ be a standard Brownian Motion. Fix $\mu \in \mathbb{R}$ and define $W_t = B_t + \mu t$ for $t \geq 0$ to be Brownian Motion with drift. Let $P_{\mu} $ denote the Law $W_t$ induces on $C([0,\infty])$. Prove there exists a measurable subset $A \in \mathcal{B}(C([0,\infty]))$ such that $P_{0}(A) =1$ and $P_\mu(A) = 0$ for all $\mu \neq 0$.
The event $A$ from above suffices as long as its measurable (which I am almost certain it is) as then you can leverage the fact that $P( \lim\limits_{t \to \infty} \frac{B_t}{t} = 0) = 1$.
For $\epsilon>0$ and $n\in \mathbb N$ we put $$A_{n,\epsilon} = \{f\in C([0,\infty)): t^{-1}|f_t|<\epsilon,\text{ for all }t\geq n\}$$ Then $$A = \bigcap_{\substack{\epsilon \in \mathbb Q \\ \epsilon >0}}\bigcup_{n\in \mathbb N}A_{n,\epsilon}$$ By continuity, whether or not $f\in A_{n,\epsilon}$ can be ascertained by the values of $f$ on $\mathbb Q_+$. Thus $A_{n,\epsilon} \in \mathcal B(C([0,\infty)))$, and consequently so is $A$. (Here we are using that the Borel $\sigma$-algebra on $C([0,\infty))$ generated by the topology of uniform convergence on compacts agrees with the cylindrical one.)