Time Integral over the (positive) Innovations of a Stochastic Process Consider the Itô-Process
$$X(t) = X_{0} + \int_{0}^{t}\mu(s,X(s))\,ds + \int_{0}^{t}\sigma(s,X(s))\,dW(s)$$
where you can safely assume that the drift $\mu$ and the volatility $\sigma$ satisfy the necessary regularity conditions for this SDE to have a strong solution.
I am interested in any tools that would help me to make sense of / investigate an integral of the form
$$``\;Y(t)=\int_{0}^{t}\max[\,dX(s),0\,]\,ds\;``$$
i.e. the time integral of the ''positive innovations'' of $X$. In other words, $Y$ is the process that grows when $X$ grows (by the same magnitude), but is idle when $X$ shrinks.
Any comments, ideas and suggestions are highly appreciated. Thank you!
 A: One way to formalize the concept is to approximate $X(t)$ by a Markov jump process, then define the concept for this spatial approximation (where it makes perfect sense), and then take the limit. To construct this approximation, one approximates the infinitesimal generator $L_t$ of the SDE by a suitable spatial difference approximation  (see Chapter 2 of this reference for more detail).  Let $X^h(t)$ be the resulting Markov jump process where $h$ is a jump size parameter.  Note that this process only moves by jumps, and if the approximation is stable, the number of jumps in a finite time interval is a.s. bounded.  Define the positive innovation of $X(t)$ as:
$$
Y(t) = \lim_{h \to 0} \sum_{s \le t} \max( \Delta X^h(s), 0 ) 
$$ where $\Delta X^h(s)$ denotes the jump of $X^h$ at $s$ defined as $\Delta X^h(s) = X^h(s+) - X^h(s-)$.  This definition leads to a nontrivial $Y(t)$. Indeed, since $\max(a,0)=(a+|a|)/2 \ge a/2$, we have that
$$
 \sum_{s \le t} |\Delta X^h(s)| \ge  \sum_{s \le t} \max( \Delta X^h(s), 0 ) \ge \sum_{s \le t} \frac{1}{2} \Delta X^h(s) \tag{$*$}
$$  almost surely.  Note that the expected values of the random variables appearing in the upper and lower bounds are well-defined in the limit as $h \to 0$.  For example, the expected value of the lower bound converges to $\mathbb{E}_x \int_0^t \mu(s,X(s))/2 ds$, I think.
ADD
First, let us confirm numerically that $Y(t)$ is an integrable random variable.  For this purpose, suppose $X(t)$ is an OU process with unit drift/noise coefficients; and let $\overline{Y}^h(t)$,  $Y^h(t)$ and $\underline{Y}^h(t)$ denote the upper, middle, and lower bounds appearing in ($*$). The figures below plot these quantities for $t=1$ with initial condition $X(0)=1$.  

Why are these random variables integrable? The action of the infinitesimal generator of this particular approximation is given by:
$$
L^h f(x) = \frac{e^{-h x}}{2 h^2} \left( f(x+h) - f(x) \right) + 
\frac{e^{h x}}{2 h^2} \left( f(x-h) - f(x) \right)
$$
If $f \in C^4_b(\mathbb{R})$, a Taylor expansion about $h=0$ shows that $
L^h f(x) = L f(x) + O(h^2)
$ where $L f(x) = - x f'(x) + f''(x)/2$.  Moreover, for any $h>0$, this Markov jump process approximation is right continuous with left limits and is a process of finite variation with zero continuous part.  Thus, Ito's formula for this process reduces to a telescoping sum:
$$
f(X^h(t)) - f(X^h(0)) = \sum_{s \le t} \left( f(X^h(s)) - f(X^h(s-)) \right)
$$
Add to both sides of the equation $- \int_0^t L^h f(X^h(s)) ds$ to obtain
$$
f(X^h(t)) - f(X^h(0)) - \int_0^t L^h f(X^h(s)) ds = \sum_{s \le t} \left( f(X^h(s)) - f(X^h(s-)) \right) - \int_0^t L^h f(X^h(s)) ds
$$ where the LHS is a local martingale (and for suitable functions a true martingale).  Thus, we see that $- \int_0^t L^h f(X^h(s)) ds$ is the compensator for $\sum_{s \le t} \left( f(X^h(s)) - f(X^h(s-)) \right)$.  Now choose $f(x) = x^2$ and $f(x)=x$ to obtain upper and lower bounds on $Y^h(t)$.   
