Is this set negligible?

Question

Let $(W_t)_{t\ge 0}$ be a standard Brownian motion starting at zero. Let $f: [0,1]\to\mathbb R$ be a function that is righ-continuous with left limits. Set $$A:=\left\{\omega\in\Omega: \inf_{0\le t\le 1}\big(W_t(\omega)-f(t)\big)=0\right\}.$$ Is $A$ a negligible set? When $f$ is non-decreasing, it is shown by GJC20 in the comments below.

PS : My idea is as follows : Denote by $\tau:=\inf\{t\ge 0: W_t \le f(t)\}$ the hitting time. Then it holds that $$A= \Big(A\cap \{\tau<t\}\Big) \bigcup \Big(A\cap \{\tau=t\}\Big) = :A_1\cup A_2.$$ Clearly $A_2\subset \{W_1=f(1)\}$ is negligible. So it remains to prove so it is with $A_1$. As $$A_1\subset \{\tau<t \mbox{ and } W_{\tau}=f(\tau) \mbox{ and } \exists \epsilon>0 \mbox{ s.t. } W_s \ge f(s) \mbox { for } s\in [\tau+\epsilon, \epsilon] \}=:B,$$ so I wish to show $B$ is negligible. My feeling is to use the fact $f\approx f(\tau)$ uniformly on $[\tau,\tau+\epsilon]$ and the trajectory property of $W$, but I do not know how to make this rigorous.

Answer 1

Let $f(t)=-t^{1/3}$. (Or any negative function that goes to $0$ as $t\downarrow 0$, but slower than $-t^{1/2}\log\log(1/t)$.)

From the law of the iterated logarithm, there is positive probability that $W(t)\geq f(t)$ for all $t\in[0,1]$. In that case, $\inf_{0\leq t\leq 1} (W_t-f(t))=0$ (with the inf attained at $t=0$).