= Solution
Take $\Omega=[0,1)$ with its <Borel sigma-algebra> and <Lebesgue measure> of total mass one. Let $\mathcal F_n$ be the finite <sigma-algebra> generated by the half-open <dyadic intervals> of length $2^{-n}$. Set $X_0=f(1)-f(0)$ if the indexing begins at one. Each $X_n$ is <adapted> to this <filtration>, and if $L$ is a <Lipschitz constant> of $f$, then $|X_n|\leq L$, so every coordinate is integrable.
On a parent interval $[u,u+2^{-n})$, <conditional expectation> averages the two equally likely child values. Their average is
$$
\frac12\,2^{n+1}\bigl(f(u+2^{-n-1})-f(u)+f(u+2^{-n})-f(u+2^{-n-1})\bigr)
=2^n\bigl(f(u+2^{-n})-f(u)\bigr)=X_n.
$$
Thus \b[$\mathbb E[X_{n+1}\mid\mathcal F_n]=X_n$], establishing the <dyadic slope martingale>. One can equivalently include $1$ as a separate null atom without changing any <probability law>.
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