Motivation: I want to prove that the existence of a $\sigma$-martingale implies NFLVR (No Free Lunch With Vanishing Risk). This comes from arbitrage theory in mathematical finance and was proved by Delbaen/Schachermayer.
We call $S$ a $\sigma$-martingale if there exists a local martingale $X$ and a predictable, $X$-integrable strictly positive process $\psi$ such that $S=\int\psi dX$. Then you can also define an equivalent $\sigma$-martingale measure in the obvious way.
Suppose $S$ is a semimartingale (under $P$) and it admits an equivalent $\sigma$-martingale measure, i.e. there exists a probability measure $Q\approx P$, s.t. $S$ is a $\sigma$-martingale (under $Q$). I know that there is predictable and $S$-integrable process $\vartheta$ such that $\int_0^t\vartheta_udS_u\ge -a$ for all $t\ge 0$ $P-$a.s. and an $a\ge 0$.
Clearly, by the equivalence of $Q$ and $P$, we have $\int_0^t\vartheta_udS_u\ge -a$ for all $t\ge 0$ $Q-$a.s. Since under $Q$ $S$ admits a $\sigma$-martingale we have:
$$-a\le\int_0^t\vartheta_udS_u=\int_0^t\vartheta_u\psi_udX_u$$
The thing which bothers me is: Is the integral $\int_0^t\vartheta_u\psi_udX_u$ well-defined? Obviously $\psi\vartheta$ is again predictable and $\psi$ is by definition $X$ integrable. But why is the product $\psi\vartheta$ $X$-integrable?