Doob Meyer decomposition for Super-martingales Let $Z$ be a _super-martingale_ with usual Doob-Meyer decomposition: $Z=M-A$. Is it true that : $A\leq M$ and therefore: $\mathbb{E}[A^2]\leq \mathbb{E}[M^2]$ ?

No, this is in general not correct.

Let $(M_t)_{t \geq 0}$ be a martingale and $A:[0,\infty) \to [0,\infty)$ an increasing deterministic function. Then $Z(t) := M_t-A(t)$ is a supermartingale. This means in particular that we can in general not expect $A \leq M$.

Consider e.g. a one-dimensional Brownian motion $(M_t)_{t \geq 0}$ and $A(t) := t^{1/4}$. Then neither the inequality $A(t) \leq M(t)$ nor $\mathbb{E}(A_t^2) \leq \mathbb{E}(M_t^2)$, $t \geq 1$, are satisfied.