Absolute Continuity Implies Bounded Variation and Almost-Everywhere Differentiability

From Interval-Sum Control to Bounded Variation

Absolute continuity controls sums of increments over disjoint intervals whose total length is small. Bounded variation, by contrast, requires one finite bound for the sums of absolute increments over all partitions of the interval. The key step is to refine an arbitrary partition using a fixed grid: within each grid cell, absolute continuity controls the entire sum, and there are only finitely many cells.

Throughout, let \(a<b\) be finite real numbers, use Lebesgue measure \(m\) on \([a,b]\), and let \(f:[a,b]\to\mathbb{R}\). We use the interval-sum definition: \(f\) is absolutely continuous on \([a,b]\), written \(f\in AC[a,b]\), if for every \(\varepsilon>0\) there is a \(\delta>0\) such that every finite family of intervals \((u_j,v_j)\subseteq[a,b]\) with pairwise disjoint interiors and \(\sum_j(v_j-u_j)<\delta\) satisfies \(\sum_j|f(v_j)-f(u_j)|<\varepsilon\). Endpoints shared by intervals do not affect this condition.

Proof. Fix \(\varepsilon=1\) in the definition of absolute continuity and choose the resulting \(\delta>0\). Choose a positive integer \(N\) such that \((b-a)/N<\delta\), and fix the grid \(a=t_0<t_1<\cdots<t_N=b\), where \(t_k=a+k(b-a)/N\). Take any partition \(P\) of \([a,b]\), and refine it by adding all the grid points. Refinement cannot decrease the variation sum: if an interval \([x,y]\) is split at points \(x=z_0<z_1<\cdots<z_r=y\), then the triangle inequality gives \(|f(y)-f(x)|\le\sum_{\ell=1}^r|f(z_\ell)-f(z_{\ell-1})|\).

Group the intervals of the refined partition according to the grid cell \([t_{k-1},t_k]\) in which they lie. Within a given cell, their interiors are disjoint and their total length is \(t_k-t_{k-1}=(b-a)/N<\delta\). Absolute continuity therefore bounds the sum of their absolute increments by \(1\). Summing over the \(N\) cells shows that the refined variation sum is at most \(N\), so the original variation sum is at most \(N\) as well. This bound does not depend on \(P\). Taking the supremum over all partitions gives \(V_a^b(f)\le N<\infty\), as required. \(\square\)

The proof only needs a finite bound, not the sharpest possible one. In particular, it does not claim that a small total interval length makes the sum of increments small when that total length is large. The fixed grid makes the absolute-continuity condition applicable a finite number of times.

Almost-Everywhere Differentiability and Integrability

For an absolutely continuous function, the preceding theorem gives existence of the derivative almost everywhere. A further estimate shows that this derivative is in \(L^1(m)\), not merely finite almost everywhere. We give the estimate because it identifies a useful quantitative link between variation and the derivative.

Proof. The first result gives \(V=V_a^b(f)<\infty\), and the bounded-variation differentiability theorem gives a measurable derivative almost everywhere. Define \(v(x)=V_a^x(f)\) for \(a\le x\le b\). Variation is additive on adjacent intervals: \(V_a^y(f)=V_a^x(f)+V_x^y(f)\) for \(a\le x\le y\le b\). This follows by joining partitions of the two subintervals for one inequality and splitting any partition at \(x\) for the other. Since \(V_x^y(f)\ge |f(y)-f(x)|\), we have \(|f(y)-f(x)|\le v(y)-v(x)\). The function \(v\) is increasing, hence measurable, and \(v(a)=0,\ v(b)=V\).

For \(0<h<b-a\), define \(q_h(x)=|f(x+h)-f(x)|/h\) on \([a,b-h]\), and set \(q_h(x)=0\) on the rest of \([a,b]\). These functions are nonnegative and measurable because \(f\) is continuous; absolute continuity implies continuity. The increment bound gives

For the last inequality, \(0\le v(t)\le V\), so the first integral in parentheses is at most \(hV\), while the second is nonnegative. Now choose any sequence \(h_n>0\) with \(h_n\to0\) and \(h_n<b-a\). At every interior point where \(f'\) exists, \(q_{h_n}(x)\to |f'(x)|\); each such point eventually lies in \([a,b-h_n]\). Fatou's lemma, which states for nonnegative measurable functions that \(\int\liminf_n q_{h_n}\,dm\le\liminf_n\int q_{h_n}\,dm\), now yields \(\int_a^b|f'|\,dm\le V\). Thus the measurable derivative belongs to \(L^1(m)\). Values assigned at the null exceptional set do not change this integral. \(\square\)

The estimate is an upper bound; it does not assert that the variation equals the integral of the absolute derivative. The examples illustrate the conclusion and the bound, without requiring an integral representation of \(f\).

A Common Variation Mistake

A tempting but incorrect argument is: “For a sufficiently fine partition, every increment is less than \(\varepsilon\), so the variation sum is at most \(\varepsilon\).” The conclusion does not follow because the number of increments can be greater than one. For example, for \(f(x)=x\) on \([0,1]\), take \(\varepsilon=0.1\) and the partition \(x_i=i/20\), \(i=0,\ldots,20\). Each increment is \(1/20=0.05<0.1\), but the sum of all twenty increments is \(20(0.05)=1\), not less than \(0.1\).

The correction is to control the sum within each of a fixed finite number of cells, not to bound each increment separately. In the proof, each cell has length less than \(\delta\), so the sum of all increments belonging to that cell is controlled by absolute continuity. The resulting bound is the number of cells times the chosen tolerance, uniformly over every original partition.

Why the Conclusion Matters

Absolute continuity gives two useful conclusions here: a finite global bound on all partition variation sums, and an ordinary derivative defined almost everywhere whose absolute value has finite Lebesgue integral. The derivative conclusion does not say that \(f'\) is bounded, as the power-function example shows. Nor does this tutorial assert that \(f\) can be recovered from \(f'\) by integrating; that is a stronger conclusion requiring a separate result.