I want to prove the following:
Let $A_0, A_1, ..$ be a martingale with respect to the sequence $B_0, B_1, ..$. then $(A_i)_{i\geq0}$ is also a martingale with respect to itself.
I have no idea how to approach this, but it seems like a pretty standard problem.
If $(A_n)$ is a martingale with respect to $(B_n)$, then, in particular, each random variable $A_n$ is measurable with respect to the sigma-algebra $\mathcal B_n=\sigma(B_i;i\leqslant n)$ since $A_n=E(A_{n+1}\mid\mathcal B_n)$, hence $\mathcal A_n=\sigma(A_i;i\leqslant n)$ is such that $\mathcal A_n\subseteq\mathcal B_n$. Then the hypothesis that $A_n=E(A_{n+1}\mid\mathcal B_n)$ and the tower property imply that $$E(A_{n+1}\mid\mathcal A_n)=E(E(A_{n+1}\mid\mathcal B_n)\mid\mathcal A_n)=E(A_n\mid\mathcal A_n)=A_n.$$ Thus, $(A_n)$ is also a martingale with respect to the filtration $(\mathcal A_n)$.