uniqueness of Doob–Meyer decomposition

Question

If $X=(X_{n},n\ge1)$ is a square integrable martingale, set $$[X]_n=\sum\limits_{i=1}^n\Delta X_{i}^2,\langle X\rangle_n=\sum\limits_{i=1}^nE[\Delta X_{i}^2|\mathcal{F}_{i-1}],n\ge1$$where $X_{0}=0,\Delta X_{i}^2=(X_{i}-X_{i-1})^2$.Obviously ,$(X_{n}^2-\langle X\rangle_{n},n\ge 1)$ and $(X_{n}^2-[X]_{n},n\ge 1)$ are both different martingales. According to Doob–Meyer decomposition Theorem,the decomposition is unique, but the decomposition above have two distinctions.Why?

Answer 1

The uniqueness of $X_n=M_n+A_n$ holds with an additional predictable condition over $A$ : $A_n$ has to be $\mathcal{F}_{n-1}$ measurable.

Hence : $[X]_n$ is not predictable, but $<X>_n$ is.