Let $M,N$ be continuous square-integrable martingales(this can probably be relaxed). Let $V$ be the total variation of $[M,N]$. Let $X,Y$ be progressively measurable stochastic processes.
By the Kunita Watanabe inequality we have that:
$\int_0^\infty|XY|dV\le(\int_0^\infty X^2d[M])^{0.5}(\int_0^\infty Y^2d[N])^{0.5}$, a.s.
Assume that we in some way know that the right hand side is finite almost surely. Then we have that:
$\int_0^\infty|XY|dV<\infty$ a.s.
My question is then, do we then have a meaningful value for $\int_0^\infty XYd[M,N]$ a.s.?, and why?
It does seem like the books I am looking at take this for granted but I am wondering why? What I have is that if since $[M,N]$ is of finite variation, then it can be written as $f_1-f_2$, where both functions are increasing. And they both generate measures $\mu_1, \mu_2$. The obvious way is to define the measure generated by $[M,N]$ as $\mu_1-\mu_2$(all of this is pathwise for each $\omega$). The problem I have is that I do not know that almost surely this is a well defined expression, that is, that almost surely I do not get $\infty-\infty$?
How do we know that that situation do not happen?