Dyadic slope martingale ($G_n(t)=2^n(f((k+1)2^{-n})-f(k{2}^{-n}))$)

For a real continuous function $f$ on $[0,1]$, let $G_n$ be its secant slope on each length-$2^{-n}$ dyadic cell. Regard $([0,1],\mathcal{B},dt)$ as a probability space and use the filtration of half-open dyadic cells with the endpoint $\{1\}$ as a separate null cell. The mean of the two child slopes equals the parent slope, so $(G_n)$ is a martingale. Its integral gives the linear interpolation of $f$ on that grid. If $f$ is Lipschitz continuous, these slopes are bounded by its Lipschitz constant; the Lp martingale convergence theorem then supplies a bounded integral density for $f$.

For an $L$-Lipschitz function $f$ on $[0,1]$, its slopes on the dyadic partition, evaluated at a uniform distribution input, form a bounded martingale for the dyadic filtration. The uniformly integrable martingale convergence theorem yields a bounded terminal random variable $g(X)$, with $g$ chosen as a Borel measurable function by the Doob-Dynkin lemma. The terminal conditional expectation identity says that each slope is the average of $g$ over the corresponding dyadic interval. Integrating over adjacent dyadic intervals and taking a telescoping sum proves the displayed representation first at dyadic endpoints. The Lipschitz condition for $f$ and the bound $|g|\le L$ extend it to every endpoint by continuity. This proves absolute continuity of a function without presupposing a derivative.