Doob's decomposition theorem states that, given any sequence of adapted, integrable functions $X_n$ (adapted to $\mathcal{F}_n$), we get an almost surely unique decomposition $X_n=A_n+Y_n$, where $Y_n$ is a martingale, $A_0=0$, and $A_n$ is $\mathcal{F}_{n-1}$-measurable ("previsible").
If we don't require the $A_n$ to be previsible, is the decomposition no longer unique? Can someone offer such an example?
If we don't require $A_n$ to be predictable, then we can set $A_n = X_n$ for all $n$. Then $Y_n = 0$ is trivially a Martingale. This gives us two Doob decompositions. The trivial one I just outlined, and the predictable Doob decomposition.