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](https://math.stackexchange.com/questions/4716163/integral-of-m-text-m-the-difference-of-running-maximum-and-current-value-w), 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)



Integral of $M^\text{*} - M$ with respect to $M^\text{*}$ is zero for $M^\text{*}$ the running maximum of $M$ a continuous local martingale
The integral with regard do $\mathrm{d}M^\*$ is a pathwise Stieltjès integral, so the question is an analysis problem.


Let $f : \mathbb{R}\_+ \to \mathbb{R}$ be any continuous function, $F$ its current maximum, and $\mu$ the Stieltjès measure associated to $F$


One checks that $F$ is also continuous, so $O := \{s \in \mathbb{R}\_+ : F(s)-f(s)>0\}$ is open subset in $\mathbb{R\_+}$ and contained in $\mathbb{R\_+}^\*$ since $F(0)=f(0)$. Hence it is an at most countable union of disjoint open intervals. On each one of these open intervals, $F$ remains constant, so the $\mu$-measure of this interval is $0$. As a result $\mu(O)=0$, so $F-f$ is null $\mu$-almost everywhere and the integral
$\int(F-f) \mathrm{d}\mu$ is $0$.


