# Is this set negligible?

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 $$A= \Big(A\cap \{\tau Clearly $$A_2\subset \{W_1=f(1)\}$$ is negligible. So it remains to prove so it is with $$A_1$$. As $$A_1\subset \{\tau0 \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.

• Unless I'm wrong I think you are almost done. Note that $f$ is increasing. Thus $A_1\subset \{\tau<t,~ W_{\tau}=f(\tau) \mbox { and } W_s\ge f(s)\ge f(\tau) \mbox{ for all } s\in [\tau,t]\}=:B'$. Then $\mathbb P(B')=0$ as there are infinite times $u\in [\tau,t]$ s.t. $W_u<W_{\tau}=f(\tau)\le f(u)$ Jun 28 at 14:55
• While I am curious whether this holds for a general cadlag $f$ Jun 28 at 15:00
• @GJC20 Thanks for the answer. Knowing that you did not write your reasoning as an answer, do you mind I modify my question to a general $f$?
– user478657
Jun 28 at 15:17
• Sure. Feel free to do it Jun 28 at 15:21

## 1 Answer

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 $$t=0$$).