Why is a stochastic integral w.r.t a martingale always a local martingale?

Question

In my course on stochastic calculus, the professor mentioned that stochastic integral w.r.t a martingale always a local martingale? How can I rigorously show this? I know that when integrating wrt to Brownian Motion, we need the integrand to be in $\Lambda^2_{loc}$ for it to be a local martingale but I cannot show why the above statement follows from what I know. Can someone help me here? Thanks

Answer 1

Given any $a>0$ any local martingale can be decomposed into a local martingale which is of finite variation and a local martingale (starting at zero) with jumps bounded by $a$.

A local martingale $M$ whose jumps are bounded and with $M_0 \in L^2$ belongs to $H^2_{loc}$.

For the part $M$ with finite variation you can use the fact that if $H$ is a locally bounded predictable process, then the compensator $(H \cdot M)^p$ of the process $H \cdot M$ is $H \cdot M^p$ (and $M^p$- i.e the compensator of $M$ is obviously zero).

Answer 2

you can find the proof of the result in Jacod & Shirjaev (Lemma 4.24 of chapter 1); $H^2_{loc}$ denotes the set of all locally square integrable martingales.