Consider $M = (M_t)$ is a continuous square integrable local martingale and $$ \mathbb H ^2(M):= \left \{ \psi =(\psi_t)\ \text{is a real previsible process s.t.,} \forall t\geq 0, \ \mathbb E\left \{ \int _0 ^t \psi _s ~d \langle M \rangle _s \right \}< + \infty \ \text {a.e.} \right\}. $$ If $ \phi =(\phi_t) \in \mathbb H ^2(M)$ can we conclude that $I_t =\left(\int _0 ^ t \phi_s dM_s\right) $ is a martingale ?
My doubt comes from this problems also asked here.
Thank's in advance for your enlightenments.
As I said in my comment, no you cannot conclude that $I_t$ is a martingale solely from the fact that the integral is well defined, since $1 \in \mathbb{H}^2(M)$.
The result you want to complete your answer to the other question uses the following characterization of martingales (Prop 2.9): A local martingale $X_t$ is a true martingale if and only if the collection $\chi_t = \{ T : T \mbox{ is a stopping time and }T \leq t \}$ is uniformly integrable. A consequence of this is that if your local martingale is dominated by an integrable random variable, then it is a martingale. For the local martingale in that problem, $1$ is a dominating function.