Bounded variation controls total oscillation. Absolute continuity is stronger: it says all variation is accumulated by integrating an ordinary derivative.
Total variation
Variation measures the total amount a function moves.
Definition. The total variation of \(f:[a,b]\to\mathbb R\) is \(V_a^b(f)=\sup_P\sum_i |f(x_i)-f(x_{i-1})|\), where the supremum is over finite partitions.
Proposition. \(f\) has bounded variation if and only if \(f=g-h\) for two increasing functions \(g,h\).
Proof. If \(f\) has bounded variation, set \(g(x)=V_a^x(f)\) and \(h(x)=g(x)-f(x)\). Both are increasing because variation over a larger interval increases and dominates increments of \(f\). Conversely, the variation of \(g-h\) is at most the variation of \(g\) plus the variation of \(h\), finite on \([a,b]\).
Proposition. If \(f\in BV([a,b])\), then \(f\) is bounded, has one-sided limits at every interior point, and has at most countably many discontinuities. Moreover, \[ |f(y)-f(x)|\le V_x^y(f),\qquad V_a^b(f+g)\le V_a^b(f)+V_a^b(g). \]
Proof. The inequalities follow directly from the supremum over partitions. The Jordan decomposition \(f=g-h\) expresses \(f\) as a difference of two increasing functions. Each increasing function is bounded on a compact interval, has one-sided limits, and has only countably many jump discontinuities; the same properties pass to their difference.
Absolute continuity
Absolute continuity is uniform smallness of image variation over collections of short intervals.
Definition. \(f\) is absolutely continuous on \([a,b]\) if for every \(\varepsilon>0\) there is \(\delta>0\) such that disjoint intervals \((a_k,b_k)\) with \(\sum_k(b_k-a_k)\lt \delta\) satisfy \(\sum_k|f(b_k)-f(a_k)|\lt \varepsilon\).
Proposition. Absolute continuity implies uniform continuity and bounded variation.
Proof. Uniform continuity follows by applying the definition to one interval. For bounded variation, choose \(\delta>0\) corresponding to \(\varepsilon=1\), and let \(P\) be any partition. Group its consecutive subintervals into at most \[ N\le 1+\left\lceil\frac{2(b-a)}{\delta}\right\rceil \] families, each having total length below \(\delta\). Absolute continuity bounds the sum of increments in each family by \(1\), so \(\sum_i|f(x_i)-f(x_{i-1})|\le N\), independently of \(P\). Taking the supremum gives bounded variation.
Fundamental characterization
Absolutely continuous functions are exactly indefinite integrals of \(L^1\) functions.
Theorem. \(f\) is absolutely continuous on \([a,b]\) if and only if there is \(g\in L^1([a,b])\) such that \(f(x)=f(a)+\int_a^x g(t)\,\mathrm{d}t\). In that case \(f'=g\) almost everywhere.
Proof. Suppose first that \(f(x)=f(a)+\int_a^xg\). Absolute continuity of the Lebesgue integral says that for every \(\varepsilon>0\) there is \(\delta>0\) such that \(m(E)<\delta\) implies \(\int_E|g|<\varepsilon\). For disjoint intervals with union \(E\), \[ \sum_k|f(b_k)-f(a_k)| \le\sum_k\int_{a_k}^{b_k}|g| =\int_E|g|<\varepsilon. \] Thus \(f\) is absolutely continuous, and the Lebesgue differentiation theorem gives \(f'=g\) almost everywhere.
Conversely, absolute continuity implies bounded variation, so increments define a finite signed measure \[ \nu((s,t])=f(t)-f(s). \] If \(m(E)=0\), cover \(E\) by intervals of arbitrarily small total length; absolute continuity makes the total variation \(|\nu|(E)\) vanish. Hence \(\nu\ll m\). Radon-Nikodym gives \(g\in L^1\) with \(\nu(E)=\int_Eg\,\mathrm{d}m\), so \[ f(x)-f(a)=\nu((a,x])=\int_a^xg(t)\,\mathrm{d}t. \] Differentiation of indefinite integrals identifies \(g=f'\) almost everywhere.
Singular bounded variation
Bounded variation allows movement on sets of Lebesgue measure zero.
Example (Cantor singular function). The Cantor function \(C:[0,1]\to[0,1]\) is continuous and increasing, maps the Cantor set onto \([0,1]\), and is constant on every complementary middle-third interval. Consequently \(C'=0\) almost everywhere, because the complement of the Cantor set has full measure. Yet \(C(1)-C(0)=1\), so \[ C(1)-C(0)\ne\int_0^1C'(x)\,\mathrm{d}x. \] It has bounded variation \(1\) but is not absolutely continuous.
Variation measures
Bounded-variation functions are distribution functions of signed measures.
Theorem (Lebesgue decomposition of a \(BV\) function). After choosing a right-continuous representative, every \(f\in BV([a,b])\) decomposes, up to an additive constant, as \[ f=f_{\mathrm{ac}}+f_{\mathrm{jump}}+f_{\mathrm{sc}}, \] where \(f_{\mathrm{ac}}\) is absolutely continuous, \(f_{\mathrm{jump}}\) is a sum of jumps, and \(f_{\mathrm{sc}}\) is continuous singular: its derivative is zero almost everywhere but it may be nonconstant.
Proof outline. Associate to \(f\) its finite signed Stieltjes measure \(\nu_f\). Lebesgue decomposition splits it as \(g\,m+\nu_s\), and the singular part splits uniquely into its atomic and nonatomic parts. Integrating these three measures from \(a\) to \(x\) produces \(f_{\mathrm{ac}}\), \(f_{\mathrm{jump}}\), and \(f_{\mathrm{sc}}\). Uniqueness follows from uniqueness of the measure decompositions.