Let $\{\mathcal{F}_n\}_n$ be a filtration and $A$ and $B$ two dependent events. Define the two following martingales:
$$X_n = \mathbb{E}[1_A\mid \mathcal{F}_n]\quad \mbox{ and } \quad Y_n = \mathbb{E}[1_B\mid \mathcal{F}_n]$$
Is it true that $X_nY_n$ is also a martingale with respect to the same filtration?
Using basic properties of conditional expectations we have $${\color{blue}{\mathbb{E}[}}X_nY_n\color{blue}\mid { \mathcal{F}_{n-1}]} = \color{blue}{\mathbb{E}\{} \mathbb{E}[1_A\mid \mathcal{F}_{n}]\mathbb{E}[1_B\mid \mathcal{F}_{n}]\color{blue}\mid { \mathcal{F}_{n-1}\}} = $$
$$=\color{blue}{\mathbb{E}\{} \mathbb{E}[\mathbb{E}[1_A\mid \mathcal{F}_{n}]1_B\mid \mathcal{F}_{n}]\color{blue}\mid {\mathcal{F}_{n-1}\}} = \mathbb{E}\{\mathbb{E}[1_A\mid \mathcal{F}_{n}]1_B\mid \mathcal{F}_{n-1}\} = E[X_n1_B\mid \mathcal{F}_{n-1}]$$
I'd like then to pull $X_n$ out of the expectation as $\mathbb{E}[1_A\mid \mathcal{F}_{n-1}]$ to prove the claim, but as $X_n$ and $1_B$ are not independent, I can't split this into a product of expectations.
I don't believe that any product of correlated martingales with respect to the same filtration is also a martingale, but I hope the fact that my martingale is exactly a contitional expectation might help.
Context: In an article with variables like these, the author said "Then using martingale convergence theorem the result follows", implicitely saying this product is also a martingale.
Thanks in advance.