= Solution
Let $D_n=M_n-M_{n-1}$. Given $\mathcal F_{n-1}$, the new information at time $n$ is only the independent symmetric sign $\xi_n$. Thus
$$
D_n=C_n+B_n\xi_n
$$
for some $\mathcal F_{n-1}$-measurable $C_n,B_n$. Taking conditional expectation and using the martingale property gives $C_n=0$. More explicitly, one can take
$$
B_n=\mathbb E[D_n\xi_n\mid\mathcal F_{n-1}]
=\mathbb E[M_n\xi_n\mid\mathcal F_{n-1}],
$$
which is predictable. Therefore
$$
\boxed{M_n=M_0+\sum_{k=1}^nB_k\xi_k.}
$$
This is the <predictable representation in a Rademacher filtration>.
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