= Solution
Use the same <dyadic slope martingale> $G_n$ and dyadic <filtration> as in (c). Each $G_n$ is integrable, since it takes finitely many finite values. On each dyadic cell, the absolute slope is $2^n$ times the absolute endpoint increment. Consequently the hypothesis in the PDF is exactly
$$
\int_0^1|G_n(t)|\mathbf1_{\{|G_n(t)|\geq\lambda\}}dt=V_n(f,\lambda),
\qquad
\sup_n\int_0^1|G_n|\mathbf1_{\{|G_n|\geq\lambda\}}dt\longrightarrow0.
$$
Thus $(G_n)$ is <uniformly integrable>. It is also bounded in <L1 norm>: choose a finite $\lambda_0>0$ at which the supremum of the tails is finite, and use $\|G_n\|_1\leq\lambda_0+\sup_jV_j(f,\lambda_0)$. The <uniformly integrable martingale convergence theorem> supplies $g\in L^1([0,1])$ with $G_n\to g$ in <L1 norm>.
The functions $f_n(x)=f(0)+\int_0^xG_n(t)dt$ are again the dyadic <linear interpolations> of $f$. Since $f$ is continuous on a compact interval, it is <uniformly continuous>, and $\|f_n-f\|_\infty\leq\omega_f(2^{-n})\to0$, where $\omega_f$ is its <modulus of continuity>. On the other hand, the <integral> of $G_n-g$ is uniformly bounded in absolute value by $\|G_n-g\|_1\to0$. Hence the <dyadic slope-tail criterion for absolute continuity> gives
$$
\boxed{f(x)=f(0)+\int_0^xg(t)dt\quad\text{for every }x\in[0,1],\qquad g\in L^1([0,1]).}
$$
No boundedness of $g$ is asserted here; the tail condition permits integrable densities that are unbounded.
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