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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 a continuous function $f$ on $[0,1]$, let $G_n$ be its dyadic slope martingale. Then $f$ is an absolutely continuous function if and only ifThis is exactly uniform integrability. The uniformly integrable martingale convergence theorem gives convergence in L1 $G_n\to g$, while their integrated linear interpolations converge uniformly to $f$. Hence $f(x)=f(0)+{\int }_0^{x}g$. Conversely, if $f$ has density $g\in L^1$, its slopes are $\mathbb{E}[g\mid {\mathcal{F}}_n]$, and the uniform integrability of conditional expectations proves the criterion. The dyadic tail integral is also the sum of the absolute endpoint increments in cells whose slope is at least $\lambda$ in magnitude.

Between two prescribed values, linear interpolation assigns $(1-t)f(a)+tf(b)$ at $(1-t)a+tb$, for $0\le t\le 1$. Joining successive samples produces a piecewise-linear function. For a continuous function on a compact interval, interpolants along grids with mesh tending to zero converge in the uniform norm; their error is bounded by the modulus of continuity at the mesh size.

Put $M_k=S_k-mk$ as above. Centering the linear interpolation also interpolates the centered values: for $u\in [0,1]$,The absolute value of this convex combination is at most the larger endpoint absolute value. For every fixed $A>0$, part (c) therefore givesHence there is uniform convergence on compacts in probability to the deterministic path $h(t)=mt$, by Markov inequality. Equip $C([0,\infty ),\mathbb{R})$ with its standard compact-open topology, metrized byFor each finite number of terms their suprema converge to zero in probability, and the remaining tail is bounded deterministically by ${\sum }_{j>J}{2}^{-j}$. Thus $d(S^{(N)},h)\to 0$ in probability. Convergence in probability to a deterministic point implies weak convergence of probability measures, so the fluid limit isHere ${\delta }_h$ is the Dirac measure concentrated on that continuous path. The topology is locally uniform convergence; no assertion of uniform convergence over the entire unbounded half-line is needed. The interpolation is intended for integers $k\ge 0$, including the initial interval $[0,1]$. If the PDF's ${\mathbb{Z}}^{+}$ is interpreted as strictly positive integers, that initial piece is omitted from the displayed definition and must be supplied by the same formula using $S_0=0$.

Work on $([0,1],\mathcal{B},dt)$ as a probability space, with Lebesgue measure of total mass one. Let ${\mathcal{F}}_n$ be the filtration generated by the level-$n$ half-open dyadic cells together with the separate null cell $\{1\}$. Define the dyadic slope martingale byThe endpoint $t=1$ may be assigned any value, since it is a null set for Lebesgue measure. The two child slopes average to their parent slope, by telescoping the two increments. Therefore $\mathbb{E}[G_{n+1}\mid {\mathcal{F}}_n]=G_n$ almost everywhere, so this is a martingale. The Lipschitz condition gives $|G_n|\le K$ everywhere except possibly at the freely chosen endpoint, where we take zero.

Apply the Lp martingale convergence theorem with $p=2$. Its limit $g$ has $|g|\le K$ almost everywhere and $G_n\to g$ in L1 norm as well, by Cauchy-Schwarz inequality. Choose a measurable representative of $g$ and set it to zero on any exceptional null set; it is then a bounded measurable function on the entire interval.

Set $f_n(x)=f(0)+{\int }_0^x{G}_n(t)dt$. Telescoping at the grid points shows that $f_n$ is the linear interpolation of $f$ on the dyadic grid. The Lipschitz condition gives $\Vert f_n-f{\Vert }_{\infty }\le K{2}^{-n}\to 0$. AlsoThe two uniform limits coincide, giving the absolutely continuous function representationThe chosen bound holds for every $t$ after the null-set modification; the integral identity holds for every $x$ simultaneously.

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 exactlyThus $(G_n)$ 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 $\Vert G_n{\Vert }_1\le {\lambda }_0+{\sup }_j{V}_j(f,{\lambda }_0)$. The uniformly integrable martingale convergence theorem supplies $g\in L^1([0,1])$ with $G_n\to g$ in L1 norm.

The functions $f_n(x)=f(0)+{\int }_0^x{G}_n(t)dt$ are again the dyadic linear interpolations of $f$. Since $f$ is continuous on a compact interval, it is uniformly continuous, and $\Vert f_n-f{\Vert }_{\infty }\le {\omega }_f(2^{-n})\to 0$, 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 $\Vert G_n-g{\Vert }_1\to 0$. Hence the dyadic slope-tail criterion for absolute continuity givesNo boundedness of $g$ is asserted here; the tail condition permits integrable densities that are unbounded.

Fix an exponent in the range from (a), and work on the single probability-one event where $K_{\alpha }<\infty$ and all dyadic values are finite. We give the dyadic increment chaining argument. For dyadic $s<t$, put $\delta =t-s$ and choose $n\ge 0$ with $2^{-n}\le \delta <2^{1-n}$. Let $s_j=2^{-j}\lfloor 2^{j}s\rfloor$ and $t_j=2^{-j}\lfloor 2^{j}t\rfloor$. The level-$n$ approximations are at most two grid steps apart, so their difference is bounded by $2A_n$. At each subsequent level an approximation either stays fixed or moves by one adjacent level increment. Because $s,t$ are dyadic, these approximations eventually equal $s,t$. ThusThe dyadic rationals are dense in $[0,1]$, so this uniform bound gives a unique continuous extension $X$ to the entire interval, withAll these equalities hold on that one event, not merely one event per dyadic time. Define $X$ to be the zero path on its null complement. Each $X_t$ is measurable as the limit of the random variables at deterministic left dyadic approximations. Alternatively, their measurable linear interpolations converge uniformly to $X$, which also shows measurability as a random element of $C([0,1])$. This is the continuous extension version of a continuous modification.