in my research, I need to deal with a stochastic integral with respect to a compensated poisson process, namely, $ \int_0^t f(X_t) dM_t,$ where $M(t) = N(t) - \int_0^t \lambda(s)ds$.
The integrand $f$ is bounded, and my question is do I still need $f$ to be predictable in order to make sure such integral is again a martingale.
Yes, you do need the integrand $f(X_t)$ to be predictable. If it is merely adapted you may not get a martingale.
Intuitively, the stochastic integral $\int Y_t\,dM_t$ tells you the profit from a stock trading strategy. $M_t$ is the share price at time $t$, and $Y_t$ is the number of shares you hold at time $t$. Requiring that $Y$ be predictable means that in determining your holdings at time $t$, you can only use information about the stock prior to time $t$. If $Y$ is adapted, you may also use the stock price at time $t$. But since our compensated Poisson process $M_t$ moves by jumping up and drifting down, if $Y$ is adapted, you can buy shares at the instant the price jumps up to turn a quick profit, and sell them a short time later to lock it in. That strategy gives you a guaranteed profit, so it can't be a martingale.
Formally, consider a time-homogeneous Poisson process ($\lambda(t) = 1$) and let $T$ be the time of the first jump: $T = \inf\{t : M_t > M_{t-}\}$. $T$ is a stopping time, so if we let $X_t = 1_{\{T \le t < T+\epsilon\}}$ where $\epsilon < 1$, we can check that $X_t$ is adapted (but not predictable). Thus our strategy is "buy 1 share when the stock jumps for the first time, sell it $\epsilon$ seconds later".
Let $I_t = \int_0^t X_t\,dM_t$ be the stochastic integral. You can check that $I_t \ge 0$, and for any $t > 0$, we have $I_t > 0$ with positive probability ($I_t$ is positive, indeed at least $1-\epsilon$, on the event that $M_t$ has made a jump by time $t$). So $I_t$ certainly is not a martingale.
