Absolute continuity
In mathematical analysis, absolute continuity is a property of real-valued functions defined on a closed interval [a, b] that strengthens the notion of uniform continuity by controlling the variation of the function over small disjoint subintervals in a uniform manner. Formally, a function f: [a, b] \to \mathbb{R} is absolutely continuous if for every \epsilon > 0, there exists \delta > 0 such that for any finite collection of pairwise disjoint subintervals [a_i, b_i] \subset [a, b] satisfying \sum_{i=1}^n (b_i - a_i) < \delta, it follows that \sum_{i=1}^n |f(b_i) - f(a_i)| < \epsilon.[1] This condition ensures that the function does not exhibit pathological oscillations or jumps that would prevent it from being recoverable from its derivative through integration.[2] Absolute continuity plays a central role in real analysis because it bridges classical calculus with Lebesgue integration. Every absolutely continuous function is uniformly continuous and of bounded variation, and conversely, continuous functions of bounded variation can be decomposed into an absolutely continuous part and a singular part.[1] Moreover, such functions are differentiable almost everywhere with respect to Lebesgue measure, and their derivative f' belongs to the Lebesgue space L^1[a, b].[2] A fundamental characterization states that f is absolutely continuous on [a, b] if and only if there exists an integrable function g \in L^1[a, b] such that f(x) = f(a) + \int_a^x g(t) \, dt for all x \in [a, b], with g = f' almost everywhere.[2] This equivalence establishes the validity of the Fundamental Theorem of Calculus in the Lebesgue setting and highlights absolute continuity as the precise condition under which indefinite integrals behave like antiderivatives.[1] Examples include all Lipschitz continuous functions, which form a proper subset, as well as non-Lipschitz cases like f(x) = \sqrt{x} on [0, 1].[1] In contrast, the Cantor function provides a continuous but singular (non-absolutely continuous) example, illustrating the distinction.[1] Beyond one dimension, the concept extends to measures, where a measure \mu is absolutely continuous with respect to Lebesgue measure if \mu(E) = 0 whenever the Lebesgue measure of E is zero, leading to the Radon-Nikodym theorem for representing such measures via densities.[3] This framework is essential in probability theory, functional analysis, and partial differential equations for studying regularity and integrability properties.[2]Absolute continuity of functions
Definition
A function f: [a, b] \to \mathbb{R} is absolutely continuous on the closed interval [a, b] if, for every \epsilon > 0, there exists a \delta > 0 such that for any finite collection of pairwise disjoint subintervals [a_i, b_i] \subset [a, b] (i.e., b_i \leq a_{i+1} for each i) satisfying \sum (b_i - a_i) < \delta, it holds that \sum |f(b_i) - f(a_i)| < \epsilon.[1] This condition strengthens uniform continuity by uniformly controlling the function's variation over small disjoint intervals. It parallels the notion for measures, where one measure vanishes on null sets of another, but here it applies directly to functions via the Lebesgue measure on intervals.[2]Equivalent characterizations
A function f: [a, b] \to \mathbb{R} is absolutely continuous if and only if there exists a Lebesgue integrable function g \in L^1[a, b] such that f(x) = f(a) + \int_a^x g(t) \, dt for all x \in [a, b].[4] Moreover, f is differentiable Lebesgue-almost everywhere, f' = g almost everywhere, and f' is integrable, so f is the indefinite Lebesgue integral of its derivative.[4] This equivalence follows from the fundamental theorem of calculus for Lebesgue integrals, which links the \varepsilon-\delta definition of absolute continuity to recovery via integration.[5] Absolute continuity also connects to bounded variation: a function f of bounded variation on [a, b] is absolutely continuous if and only if its total variation function V_f is absolutely continuous.[1] Since absolute continuity implies bounded variation (with total variation bounded by the \varepsilon-\delta condition applied uniformly), this provides a refinement within the class of bounded variation functions.[6] The link between the \varepsilon-\delta condition and differentiability almost everywhere relies on the Vitali covering lemma, which allows control of oscillations over fine covers to establish the existence of the derivative and its integrability.[7] Specifically, for an absolutely continuous f, the lemma helps show that the derivative exists almost everywhere by selecting disjoint intervals that approximate the behavior near points of differentiability.[8] Absolute continuity implies differentiability almost everywhere, but the converse fails: there exist functions differentiable almost everywhere whose derivatives are integrable yet fail absolute continuity.[5] The Cantor function provides such a counterexample, as it is continuous and increasing with derivative zero almost everywhere, but it maps a set of Lebesgue measure zero (the Cantor set) to a set of positive measure, violating the null-set preservation required for absolute continuity.[9] The Banach–Zarecki theorem offers another characterization: a function f: [a, b] \to \mathbb{R} is absolutely continuous if and only if it is continuous, of bounded variation, and maps Lebesgue-null sets to Lebesgue-null sets (Lusin's condition (N)).[4] Equivalently, f is continuous, differentiable almost everywhere with f' \in L^1[a, b], and satisfies condition (N).[10] This theorem interconnects the measure-theoretic, variational, and analytic perspectives on absolute continuity.[4]Basic properties
Absolutely continuous functions possess several key properties that distinguish them within the class of continuous functions. First, every absolutely continuous function is uniformly continuous on [a, b], as the \varepsilon-\delta condition implies the standard uniform continuity criterion by considering single intervals.[1] Second, absolutely continuous functions are of bounded variation. The total variation is controlled by the \varepsilon-\delta property, ensuring that the supremum of sums of absolute differences over partitions is finite. Moreover, any function of bounded variation can be decomposed into an absolutely continuous part and a singular part via the Lebesgue decomposition.[2] Third, an absolutely continuous function f is differentiable almost everywhere with respect to Lebesgue measure, and its derivative f' belongs to L^1[a, b]. Furthermore, f(x) = f(a) + \int_a^x f'(t) \, dt for all x \in [a, b], validating the fundamental theorem of calculus in the Lebesgue sense.[4] Additionally, absolutely continuous functions satisfy Lusin's condition (N): they map Lebesgue-null sets to Lebesgue-null sets. This property ensures that the function preserves the notion of measure zero in its range.[5]Examples and counterexamples
A canonical class of absolutely continuous functions consists of Lipschitz continuous functions on a closed interval [a, b]. For such a function f satisfying |f(x) - f(y)| \leq K |x - y| for some constant K > 0 and all x, y \in [a, b], absolute continuity follows directly: given \epsilon > 0, choose \delta = \epsilon / K, so that for any finite collection of disjoint subintervals with total length less than \delta, the sum of |f(x_i) - f(y_i)| is less than \epsilon.[2] Another fundamental class comprises indefinite integrals of integrable functions. Specifically, if g \in L^1[a, b], then F(x) = \int_a^x g(t) \, dt is absolutely continuous on [a, b], as it satisfies the \epsilon-\delta condition via the absolute continuity of the Lebesgue integral.[11] A concrete example is f(x) = \sqrt{x} on [0, 1], which can be expressed as f(x) = \int_0^x \frac{1}{2\sqrt{t}} \, dt; here, \frac{1}{2\sqrt{t}} belongs to L^1[0, 1] since \int_0^1 t^{-1/2} \, dt = 2, confirming absolute continuity. This function is not Lipschitz continuous, however, because its derivative \frac{1}{2\sqrt{x}} is unbounded near x = 0.[3] Counterexamples illustrate the distinction from mere continuity. The Cantor function (or devil's staircase), defined on [0, 1], is continuous and non-decreasing, mapping the Cantor set (of measure zero) onto an interval of positive length, but it is constant almost everywhere and has derivative zero almost everywhere while increasing overall from 0 to 1; thus, it fails absolute continuity.[12] The Heaviside step function H(x), defined as H(x) = 0 for x < 0 and H(x) = 1 for x \geq 0, is discontinuous at x = 0. Since absolute continuity implies uniform continuity (and hence continuity), H cannot be absolutely continuous on any interval containing 0.[13] The Weierstrass function w(x) = \sum_{n=0}^\infty a^n \cos(b^n \pi x) (with $0 < a < 1, ab > 1 + \frac{3\pi}{2}) is continuous on \mathbb{R} but differentiable nowhere. Absolutely continuous functions are differentiable almost everywhere, so w is not absolutely continuous.[14][11] The following table compares these properties for representative functions on [0, 1]:| Function | Uniformly Continuous | Lipschitz Continuous | Absolutely Continuous |
|---|---|---|---|
| Constant function f(x) = c | Yes | Yes | Yes |
| Linear function f(x) = x | Yes | Yes | Yes |
| f(x) = \sqrt{x} | Yes | No | Yes |
| Cantor function | Yes | No | No |
| Heaviside function H(x) (adjusted to [0,1]) | No | No | No |
| Weierstrass function | Yes | No | No |