Running Maximum of Brownian motion is singularly continuous?

Question

Let $W_t$ be the standard Brownian motion, and define the running maximum of Brownian motion as $$M_t\doteq \max_{0\leq s\leq t} W_s.$$ Then $M_t$ is non-decreasing function. We can regard this non-decreasing function as a measure, and then by Radon-Nikodym theorem,$$M_t = \int_0^t \sigma(u) du + \nu$$ where $\nu$ concentrated on a Lebesgue negligible set. Can we say that $\sigma\equiv 0$? I want to prove that almost everywhere $M_t$ is constant, similar to the Cantor function.

Answer 1

$t>0$ is the growth point of $M_t$ iff $W_t\ge W_s$ for all $s\le t$. Therefore, $$ \mathrm{E}\bigl[\lambda(\{\text{growth points of $M$}\})\bigr] = \mathrm{E}\left[\int_0^\infty \mathbf1_{\forall s\le t: W_t\ge W_s}dt\right] = \int_0^\infty \mathrm{E}\left[\mathbf1_{\forall s\le t: W_t\ge W_s}\right]dt\\ = \int_0^\infty \mathrm{P}\bigl(\forall s\le t: W_t\ge W_s\bigr)dt = 0, $$ where the last equality follows e.g. from the law of iterated logarithm. Therefore, $\lambda(\{\text{growth points of $M$}\}) = 0$ a.s., as required.