= Solution
{c}
Choose $Z_\infty$ as the pointwise <limit of a sequence> of $(Z_n)$ where it exists, and as zero otherwise. The convergence <event> and every $Z_n$ are $\sigma(X)$-<measurable>, so this is an exactly $\sigma(X)$-<measurable> version, bounded by $L$. Part (a)(ii), the <Doob-Dynkin lemma>, supplies a <Borel measurable function> $g$ on $[0,1]$ such that $Z_\infty=g(X)$. Applying <scalar clipping> to $g$ with bounds $[-L,L]$ preserves that equality and ensures $|g(u)|\leq L$ everywhere.
Part (c)(ii) and the <conditional expectation> formula from part (b)(iv) now give
$$
\boxed{Z_n=\mathbb E[g(X)\mid\mathcal F_n]
=2^n\int_{X_n}^{X_n+2^{-n}}g(u)\,du\quad\text{almost surely}.}
$$
Since the indices $n$ form a <countable> family, these identities may be taken on one common full-<probability> <event>.
For fixed $n,k$, both the slope and the displayed average are constant on the atom $A_{n,k}$, whose <probability> is $2^{-n}>0$. An equality holding <almost surely> of these constants forces the deterministic equality
$$
f((k+1)2^{-n})-f(k2^{-n})
=\int_{k2^{-n}}^{(k+1)2^{-n}}g(u)\,du.
$$
Sum over $0\leq k<j$. The <telescoping sum> gives
$$
f(j2^{-n})-f(0)=\int_0^{j2^{-n}}g(u)\,du,
\qquad 0\leq j\leq2^n.
$$
Thus the desired identity holds at every dyadic endpoint, including $1$, without evaluating the undefined slope at $X=1$.
Finally, given $x\in[0,1]$, take dyadic endpoints $x_n\to x$. The <Lipschitz condition> implies $|f(x_n)-f(x)|\leq L|x_n-x|$, and the bound on the <Borel measurable function> $g$ gives
$$
\left|\int_0^{x_n}g(u)\,du-\int_0^xg(u)\,du\right|\leq L|x_n-x|.
$$
Passing to the <limit of a sequence> proves
$$
\boxed{f(x)-f(0)=\int_0^xg(u)\,du\quad\text{for every }x\in[0,1].}
$$
This is a <martingale proof of the integral representation of a Lipschitz function>. It constructs a bounded <Lebesgue integrable function> representing the increments of $f$, and in particular proves <absolute continuity> of $f$. No prior assumption that $f$ has a <derivative> is needed.
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