= Solution
{c}
Each $Z_n$ is a bounded <Borel measurable function> of $X_n$, including the specified value at $X_n=1$. Therefore $(Z_n)$ is an <adapted process> for $(\mathcal F_n)$ and consists of <integrable random variables>.
Fix an <atom of a sigma-algebra> in $\mathcal F_n$ with positive <probability>, corresponding to the <dyadic interval> $[u,u+\delta)$, with $\delta=2^{-n}$. By the <uniform distribution>, each half has <conditional probability> $1/2$. The <conditional expectation> of $Z_{n+1}$ on that atom is therefore
$$
\begin{aligned}
\frac12\bigl[2^{n+1}(f(u+\delta/2)-f(u))
+2^{n+1}(f(u+\delta)-f(u+\delta/2))\bigr]
&=2^n(f(u+\delta)-f(u))\\
&=Z_n.
\end{aligned}
$$
The middle values cancel. On the zero-<probability> atom the <conditional expectation> can be chosen as zero, agreeing with our version of $Z_n$. Hence
$$
\boxed{\mathbb E[Z_{n+1}\mid\mathcal F_n]=Z_n\quad\text{almost surely},}
$$
so $(Z_n)$ is a <martingale>. This is the <dyadic slope martingale> associated with the <Lipschitz function> $f$.
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