Given $M$ a continuous local martingale, and $M^\text{*} = \sup_{0 \leq s \leq t} M_s$ its running maximum, we consider the finite variation integral
$$
I_T:= \int_0^T (M^\text{*}_s - M_s) \, \text{d}M^\text{*}_s
$$
I'm seeking to show **this integral is 0** a.s.. Intuitively, this is since $(M^\text{*}_s - M_s) \, \text{d}M^\text{*}_s$ is zero: if we're currently at our maximum then the bracket term zero, and otherwise $M^\text{*}_s$ is constant near $s$.

A little bit more rigorously, we have that is $$ I_T = \lim_{n \to \infty} \sum_{[s,t]\in \pi_n} (M^\text{*}_{t,s} - M_{t,s})M^\text{*}_{t,s} $$ (where the limit is in probability, and for a sequence of partitions $\pi_n$ with mesh tending to 0, and where we write $M_{t,s} = M_t - M_s$ and similarly for $M^\text{*}$) and that eventually we'll have "sufficiently small" (this is the part which I think needs more details) $t-s$. Then either we're reaching a new maximum, so that $M^\text{*}_t = M_t$ and $M^\text{*}_s = M_s$ and the difference in brackets is $0$. Or we're not reaching a new maximum, and so $M^\text{*}_{t,s} = 0$.

I think this isn't sufficient at the moment, since even if we could suppose $M$ is monotone on sufficiently small intervals we still need to choose our partitions $\pi_n$ uniformly in $\omega \in \Omega$ (for the convergence in probability).

I've also tried applying Itô's on $MM^\text{*}$, but it reduces this to a similar problem of reasoning for sufficiently small $t-s$ for a partition dependent on the outcome for the $\int M^\text{*} \, \text{d}M$ term.

(I've asked this question on MSE here, but gotten little traction. I thought Overflow might be more appropriate since I'm seeking to formalize and fill the gaps of the intuitive ideas I have - let me know if this isn't appropriate, for instance if the question particularly belongs on either site)