Let $M_t$ be a $\mathcal F_t$-local martingale. What would be the weakest condition we could put on $M$ in order for it to satsify $E[M_T \mid \mathcal F_t]=M_t$ for $t<T<\infty$?
Or in other words, can one exploit the fact that a local martingale is driftless to satisfy that equation without requiring it to be a true martingale?
I know that one possible condition is $\sup E[[M]_t]<\infty$ where $[.]_t$ is its quadratic variation. But this is indeed a condition for $M$ to be a true martingale.
I feel that this question probably does not make sense, because in order to satisfy $E[M_T \mid \mathcal F_t] = M_t$ it seems that we must also impose $E[|M_t|]<\infty$ for all $t$. But then it satisfies all criteria for being true martingale again.
Does this mean that the equation holds for a local martingale if and only if it is also a true martingale?
To add some context for where I started to consider this problem: I have a local martingale $M_t$ and would like it to satisfy $$ \mathbb E \left[\int^T_t dM_s \mid \mathcal F_t \right] = 0 $$ The question is then what condition one needs to impose in order for this to hold, if it is at all possible to do it without imposing true martingality.
The requirement $E[|M_t|]<\infty$ for $t\le T$ is not sufficient (I can provide a counterexample if you want). The simplest sufficient condition I know is that the family $\{M_t,t\in[0,T]\}$ is uniformly integrable.