Suppose we have an increasing, non-negative sequence of random variables $0 = A_0 \leq A_1 \leq ... \leq A_n \leq A_{n+1} \leq ...$ with all $A_n$ integrable. The question I am attempting to solve states that $$\{A_n\}_{n \in \mathbb{N}} \text{ is predictable} \quad \iff \quad E(M_nA_n) = E\left(\sum_{k=1}^n M_{k-1}(A_k-A_{k-1}) \right) $$
for every bounded martingale $M_n$.
One of the directions is ok/easy, but I cannot do the converse:
$(\implies)$ If $A_n$ is predictable, then for each bounded martingale $M$ we have $$E\left(\sum_{k=1}^n M_{k-1}(A_k-A_{k-1}) \right) = \sum_{k=1}^n E\left( E(M_n|\mathcal{F}_{k-1})(A_k-A_{k-1})\right) \\ = \sum_{k=1}^n E\left( E(M_n(A_k-A_{k-1})|\mathcal{F}_{k-1})\right) = E\left( M_n\sum_{k=1}^n(A_k-A_{k-1})\right) \\ = E(M_nA_n)$$
where the second equality follows from the $\mathcal{F}_{k-1}$ measurability of the $A_k$ and the last equality follows as the sum telescopes and the first term $A_0 = 0$.
$(\impliedby)$ For this direction, I am very unsure about how to proceed. I know that if the expectations are equal then we have $$E\left(\sum_{k=1}^n M_{k-1}(A_k-A_{k-1}) \right) = E\left(\sum_{k=1}^n M_{k}(A_k-A_{k-1}) \right) $$
since $E(M_nA_n) = E\left(M_n \sum_{k=1}^n (A_k - A_{k-1}) \right) = \sum_{k=1}^n E\left(E(M_n (A_k - A_{k-1}) | \mathcal{F}_k) \right)$, but I can't figure anything from here. I figured maybe fixing $N \in \mathbb{N}$ and considering the event $E_N = \{A_{N+1} > E(A_{N+1} | \mathcal{F}_N\}$ with associated martingale $M_n \equiv E(1_{E_N}|\mathcal{F}_n)$ might yield something useful but I can't come up with something from there either.
I'd be grateful for any help! Thanks for taking the time to read.
I have a solution based on $E_N$ as defined in the question but I'm not sure if it's enough to make the conclusion because I obtain that $A_{N+1} = E(A_{N+1}|\mathcal{F}_N)$ almost surely - thus I'm not sure if I can conclude that $A_{N+1}$ is indeed $\mathcal{F}_N$ measurable.
(1) Notice that $M_{N+1} = 1_E$ as $E$ is $\mathcal{F}_{N+1}$ measurable and that $$E(M_{N+1}A_{N+1}) - E(M_NA_N) = E(M_N A_{N+1}) - E(M_N A_N)$$ by the hypothesis we are assuming.
(2) Also since $Y \equiv E(A_{N+1}|\mathcal{F}_N)$ is trivially $\mathcal{F}_N$ measurable, we have $E(Y(M_{N+1}-M_N)) = 0$
Combining (1) and (2) yields $$E(1_{E_N} (A_{N+1} - E(A_{N+1}|\mathcal{F}_N))) - \underbrace{E(M_N(A_{N+1} - E(A_{N+1}|\mathcal{F}_N)))}_{=0} = 0$$
As $1_{E_N} (A_{N+1} - E(A_{N+1}|\mathcal{F}_N)) \ge 0$ by definition of the set $E_N$, this quantity must be $=0$ almost surely. Hence $A_{N+1} \leq E(A_{N+1}|\mathcal{F}_N)$ almost surely. Replacing $E_N$ with the reversed strict inequality yields that $$ A_{N+1} = E(A_{N+1}|\mathcal{F}_N) \quad \text{almost surely for every } N \in \mathbb{N}$$
Is this enough to conclude? i.e. is almost sure equivalence to an $\mathcal{F}_N$ measurable random variable imply that $A_{N+1}$ is $\mathcal{F}_N$ measurable? I know it is in the case where the filtration is complete, but is this necessarily the case in general?