I know that the set $S$ where a standard Brownian motion $M:=B[\mathbb{R}]$ attains a strict local minimum is a.s. dense in $\mathbb{R}$. For every point $s \in S$, consider the interval $(s,t)$ such that $B(y) > B(s)$ for all $y \in (s,t)$, and such that $B(t) = B(s)$ (it exists because $M$ is a.s. diverging to $+\infty$ and $-\infty$). Call the set of such intervals $I$.
My question is, is it true that the complement of the union of all the intervals in $I$ is a.s. of Lebesgue measure $0$?
Somehow, it seems to be difficult as, with probability $1$, this complement will not contain any interval.
I'd be very happy if someone could give me a vague hint or a reference.