Absolute Continuity

Absolute continuity is a mathematical property with closely related meanings for functions and measures. For a function on a real interval, it requires that the total change across any finite collection of disjoint subintervals becomes arbitrarily small when their combined length becomes sufficiently small. For measures, absolute continuity is a relation: a measure $\nu$ is absolutely continuous with respect to a measure $\mu$, written $\nu\ll\mu$, when every $\mu$-null measurable set is also $\nu$-null. These definitions connect through the measures associated with functions of bounded variation. (math.ucdavis.edu)

The concept provides a precise setting for the fundamental theorem of calculus with the Lebesgue integral: an absolutely continuous function on a compact interval can be recovered by integrating its derivative. In measure theory, the Radon–Nikodym theorem characterizes absolutely continuous measures by densities under appropriate hypotheses. These connections underpin the treatment of probability distributions, weak derivatives in Sobolev spaces, and curves taking values in metric spaces. (math.ucdavis.edu)

Definition for Functions on an Interval

Let $a<b$ be real numbers. A function $f:[a,b]\to\mathbb R$ is **absolutely continuous** if, for every $\varepsilon>0$, there exists $\delta>0$ such that every finite family of pairwise disjoint open intervals $(a_i,b_i)$, with endpoints in $[a,b]$, satisfies

Here $n$ is any positive integer and $a_i<b_i$. Equivalently, one may use closed subintervals with disjoint interiors: neighboring subintervals may share an endpoint, since the condition concerns their lengths and endpoint increments. The collection of all such real-valued functions is denoted $AC([a,b])$. (math.purdue.edu)

The order of the quantifiers is essential. Once $\varepsilon$ is fixed, a single $\delta$ must work for every admissible family, independently of its size and the locations of its intervals. It may depend on $f$, $\varepsilon$, and the ambient interval, but not on $n$ or the chosen endpoints. The hypothesis controls the total length, rather than merely requiring each interval to be short. Likewise, the conclusion controls the sum of the absolute increments, not the absolute value of their sum; increases and decreases therefore cannot cancel. Taking $n=1$ gives uniform continuity, but the full definition additionally controls increments accumulated over arbitrarily many disjoint intervals. (math.purdue.edu)

For functions taking values in the complex numbers, the same definition applies with complex modulus replacing real absolute value. Writing $f=u+iv$, this is equivalent to absolute continuity of both real-valued components $u$ and $v$. Indeed, each component increment is bounded by the corresponding complex increment; conversely,

Choosing the component tolerances as $\varepsilon/2$ and taking the smaller of their two admissible $\delta$'s proves the converse. (math.libretexts.org)

On a general interval $I\subseteq\mathbb R$, a function is locally absolutely continuous, written $f\in AC_{\mathrm{loc}}(I)$, if its restriction to every compact subinterval $[c,d]\subseteq I$ belongs to $AC([c,d])$. The admissible $\delta$ may then depend on $[c,d]$. By contrast, global absolute continuity on $I$, in the disjoint-interval sense, requires one $\delta$ for all finite families throughout $I$. These conditions coincide when $I$ itself is compact, but need not coincide on unbounded intervals. (sciencedirect.com)

For example, $f(x)=x^2$ is locally absolutely continuous on $\mathbb R$. On $[-M,M]$, with $M>0$, the estimate

makes $\delta=\varepsilon/(2M)$ sufficient for the defining inequality. Nevertheless, the global condition fails. Fix $\varepsilon=1$; given any $\delta>0$, set $h=\delta/2$ and choose $x>1/(2h)$. The single interval $(x,x+h)$ has length less than $\delta$, whereas

Thus the compact-interval bounds cannot be replaced by a location-independent bound on the whole real line. This follows directly from the distinction between the local and global definitions. (sciencedirect.com)

Comparison with Other Regularity Properties

On a compact interval $[a,b]$, the principal implications are

where the last space consists of uniformly continuous functions. For $f\in C^1([a,b])$, compactness bounds the derivative by $M=\max_{[a,b]}|f'|$; the mean value theorem then gives $|f(y)-f(x)|\le M|y-x|$. If $f$ is Lipschitz continuous with constant $L>0$, its increments over disjoint intervals satisfy

Thus $\delta=\varepsilon/L$ establishes absolute continuity; constant functions require no restriction on $\delta$. Taking just one interval in the absolute-continuity definition establishes uniform continuity. The first inclusion is strict because $f(x)=|x|$ on $[-1,1]$ is Lipschitz with constant $1$, but is not differentiable at $0$. (math.ucdavis.edu)

Absolute continuity also implies bounded variation, meaning that

This follows directly from the definition. Choose the $\delta$ corresponding to $\varepsilon=1$, and divide $[a,b]$ into $m$ fixed intervals, each shorter than $\delta$. Refine any proposed partition by their endpoints. Within each fixed interval, the total increment of the refined partition is less than $1$, so the entire sum is less than $m$. Refinement cannot decrease a variation sum, giving a uniform bound for every original partition. Bounded variation alone does not imply continuity: the step function equal to $0$ for $x<1/2$ and $1$ for $x\ge1/2$ on $[0,1]$ has variation $1$, but a jump at $1/2$. (math.purdue.edu)

The function $f(x)=\sqrt{x}$ on $[0,1]$ separates absolute continuity from Lipschitz continuity. For $x>0$, $f'(x)=1/(2\sqrt{x})$, and

The integral representation establishes absolute continuity, despite the unbounded derivative near $0$. No finite Lipschitz constant can exist, since

Absolute continuity therefore permits this endpoint singularity; Lipschitz continuity requires a uniform bound on all secant slopes. (math.ualberta.ca)

The Cantor function $C:[0,1]\to[0,1]$ separates absolute continuity from both uniform continuity and continuous bounded variation. It is continuous and nondecreasing, with $C(0)=0$ and $C(1)=1$; consequently its total variation is $1$. It is constant on every complementary interval of the middle-thirds Cantor set. Since that set has Lebesgue measure zero, $C'=0$ almost everywhere, even though $C$ is not constant. (math.purdue.edu)

The failure of absolute continuity can be seen without a derivative criterion. At stage $n$ of the Cantor construction, the $2^n$ remaining intervals have total length $(2/3)^n$, while the sum of the increases of $C$ across them remains $1$: the removed gaps contribute no increase. Thus arbitrarily small total interval length can carry a fixed total increment, contradicting the definition with $\varepsilon=1/2$. Equivalently, integrating $C'=0$ yields $0$, not $C(1)-C(0)=1$; almost-everywhere differentiability and integrability of the derivative do not by themselves guarantee recovery of a function from its derivative. (math.ucdavis.edu)

Integral Characterization and the Fundamental Theorem of Calculus

A function $f:[a,b]\to\mathbb R$ is absolutely continuous if and only if there is a function $g\in [[lp-space|L^1([a,b])]]$ such that

Here the integral is the Lebesgue integral, and $g\in L^1$ means that $g$ is measurable and $\int_a^b|g(t)|\,dt<\infty$. Thus absolute continuity characterizes precisely the functions obtainable by integrating an integrable function and adding a constant. The equality holds at every point, including the endpoints, not merely almost everywhere. (math.ucdavis.edu)

For an absolutely continuous function, the classical derivative $f'$ exists almost everywhere in $(a,b)$, belongs to $L^1([a,b])$, and equals the representing function $g$ almost everywhere. Values assigned to $f'$ on the exceptional null set do not affect its integral. Conversely, the indefinite integral of any $L^1$ function has derivative equal to its integrand almost everywhere. Subtracting the representations at two points gives the Lebesgue form of the fundamental theorem of calculus:

Consequently, an absolutely continuous function whose derivative vanishes almost everywhere is constant. (math.ucdavis.edu)

The mechanism behind the integral-to-function direction is the absolute continuity of the integral. For $g\in L^1([a,b])$ and every $\varepsilon>0$, there is a $\delta>0$ such that every measurable $E\subseteq[a,b]$ satisfies

where $m$ denotes Lebesgue measure. This estimate does not require $g$ to be bounded. Choose $M>0$ sufficiently large that

Splitting $E$ into the regions where $|g|\leq M$ and $|g|>M$ yields

Taking $\delta=\varepsilon/(2M)$ proves the assertion: integrability controls the large-value tail, while small measure controls the bounded part. (math.ucdavis.edu)

For a finite family of disjoint subintervals $(a_i,b_i)$, let $E=\bigcup_i(a_i,b_i)$. The integral representation gives

Since $m(E)=\sum_i(b_i-a_i)$, the small-set estimate supplies exactly the required absolute-continuity bound, independently of the number or placement of the intervals. (math.purdue.edu)

Almost-everywhere differentiability and integrability of the derivative, by themselves, do not imply the recovery formula. The Cantor function $C$ has $C'=0$ almost everywhere on $[0,1]$, yet $C(1)-C(0)=1$, whereas $\int_0^1C'(t)\,dt=0$. Its increase is therefore not recovered by integrating its almost-everywhere derivative. The characterization requires all three conditions together: existence of the derivative almost everywhere, its integrability, and recovery of every function increment by integration. (math.purdue.edu)

Algebraic Operations and Calculus Rules

On a compact interval $[a,b]$, the space $AC([a,b])$ is closed under linear combinations and pointwise multiplication. If $f,g\in AC([a,b])$ and $\alpha,\beta$ are constants, then $\alpha f+\beta g\in AC([a,b])$, with derivative $\alpha f'+\beta g'$ almost everywhere. Closure under multiplication uses boundedness: absolute continuity implies continuity, so $M_f=\sup_{[a,b]}|f|$ and $M_g=\sup_{[a,b]}|g|$ are finite. For $x,y\in[a,b]$,

Summing this estimate over disjoint subintervals establishes absolute continuity of $fg$. At every point where both derivatives exist, the ordinary product rule applies:

Thus the rule holds almost everywhere, without requiring either derivative to be continuous. (gyansanchay.csjmu.ac.in)

Quotients are also permitted when the denominator is bounded away from zero. Suppose $|g(x)|\ge c>0$ throughout $[a,b]$. The estimate

shows directly that $1/g$ is absolutely continuous; multiplication then gives $f/g\in AC([a,b])$. For a continuous denominator on a compact interval, being nowhere zero already guarantees such a constant $c$. The quotient rule holds almost everywhere:

Its right-hand side is integrable because $f,g$ are bounded, $f',g'$ are integrable, and $1/|g|^2\le c^{-2}$. (gyansanchay.csjmu.ac.in)

For composition, a sufficient preservation condition is Lipschitz continuity of the outer function. Let $f\in AC([a,b])$, and let $\Phi$ be Lipschitz with constant $L$ on an interval containing $f([a,b])$. Then

so $\Phi\circ f$ is absolutely continuous. This includes $|f|$, since $u\mapsto|u|$ is $1$-Lipschitz. Local Lipschitz continuity on an open interval containing the compact range of $f$ also suffices: one can restrict $\Phi$ to a compact interval on which it has a finite Lipschitz constant. (devel.isa-afp.org)

A straightforward sufficient hypothesis for the almost-everywhere chain rule is that $\Phi$ be continuously differentiable on an open interval containing $f([a,b])$. Its derivative is bounded on the compact range, and at every point where $f$ is differentiable,

The right-hand side belongs to $L^1([a,b])$. More generally, a Lipschitz outer function permits the classical rule wherever it is differentiable at $f(x)$. Lipschitz continuity alone does not make this pointwise expression meaningful everywhere: for $\Phi(u)=|u|$, $\Phi'(0)$ is undefined. Nevertheless, the composite remains absolutely continuous and has a derivative almost everywhere. (cs.cornell.edu)

Integration by parts follows by applying the fundamental theorem of calculus to the absolutely continuous product $fg$. Both terms in its derivative are Lebesgue integrable, since

Integrating the product rule and rearranging yields

Continuity and almost-everywhere differentiability alone do not suffice. Taking both factors to be the Cantor function on $[0,1]$ makes both derivative integrals zero, while the boundary term equals $1$; absolute continuity excludes precisely this unrecovered variation. (math.ucdavis.edu)

Absolute Continuity of Measures

Let $\mu$ and $\nu$ be positive measures on the same measurable space $(X,\mathcal F)$. The measure $\nu$ is absolutely continuous with respect to $\mu$, written $\nu\ll\mu$, if

The direction matters: every $\mu$-null set must be $\nu$-null, not conversely. This definition requires neither finiteness nor sigma-finiteness and compares null sets rather than numerical sizes. For example, on the Borel subsets of $\mathbb R$, let $m$ be Lebesgue measure and define $\nu(E)=m(E\cap[0,1])$. Then $\nu\ll m$, whereas $m\not\ll\nu$, since $\nu((2,3))=0$ but $m((2,3))=1$. (sites.math.rutgers.edu)

Absolute continuity is reflexive and transitive: if $\rho\ll\nu$ and $\nu\ll\mu$, then a $\mu$-null set is first $\nu$-null and consequently $\rho$-null. Measures are mutually absolutely continuous, or equivalent, when both $\nu\ll\mu$ and $\mu\ll\nu$; the notation $\mu\sim\nu$ is commonly used. Equivalently, they have exactly the same measurable null sets. Thus equivalent measures give the same meaning to almost-everywhere assertions, although their values on sets need not agree. For instance, $m$ and $2m$ are equivalent. Equivalence is an equivalence relation on the measures defined on a fixed measurable space. (mat.univie.ac.at)

When $\nu(X)<\infty$, null-set absolute continuity is equivalent to the uniform small-set criterion

No finiteness assumption on $\mu$ is necessary. (math.ou.edu) The mechanism can be seen directly. If this criterion failed despite $\nu\ll\mu$, there would be an $\varepsilon_0>0$ and sets $E_n$ with $\mu(E_n)<2^{-n}$ and $\nu(E_n)\geq\varepsilon_0$. Set $F_n=\bigcup_{k\geq n}E_k$. Then $F_n$ decreases, $\mu(F_n)\leq2^{1-n}$, and $F=\bigcap_nF_n$ is $\mu$-null. But continuity from above for the finite measure $\nu$ gives $\nu(F)=\lim_n\nu(F_n)\geq\varepsilon_0$, contradicting absolute continuity. Conversely, applying the criterion to a $\mu$-null set makes its $\nu$-measure smaller than every positive $\varepsilon$. Finiteness cannot simply be dropped: the measure $\nu(E)=\int_E x^{-1}\,dx$ on $(0,1)$ is absolutely continuous with respect to Lebesgue measure, yet $\nu((0,t))=\infty$ for every $t>0$.

Two positive measures are mutually singular, written $\nu\perp\mu$, if there is an $S\in\mathcal F$ such that $\mu(S)=0$ and $\nu(X\setminus S)=0$. They are therefore concentrated on disjoint measurable parts of the space. A unit point mass $\delta_0$ and Lebesgue measure are singular, taking $S=\{0\}$. Singularity is stronger than merely failing absolute continuity: on $[0,1]$, $\nu=m+\delta_0$ is not absolutely continuous with respect to $m$, but it is not singular either, because it retains the Lebesgue component. Indeed, for every $m$-null set $S$, $\nu([0,1]\setminus S)\geq1$. A positive measure that is both absolutely continuous and singular relative to $\mu$ must be the zero measure. (sites.math.rutgers.edu)

For a finite signed measure $\nu$, the definition is expressed through its total variation measure, $|\nu|=\nu^++\nu^-$, where $\nu^+$ and $\nu^-$ are its positive and negative parts: $\nu\ll\mu$ means $|\nu|\ll\mu$. Equivalently, $\nu(E)=0$ for every $\mu$-null measurable set $E$. Cancellation does not invalidate this equivalence, since every measurable subset of a $\mu$-null set is also $\mu$-null. The small-set criterion consequently uses $|\nu|(E)<\varepsilon$, controlling both positive and negative variation. Singularity is likewise defined using total variation; for two finite signed measures it means $|\nu|\perp|\mu|$. (math.columbia.edu)

Densities and the Radon–Nikodym Theorem

Let $\mu$ and $\nu$ be sigma-finite positive measures on the same measurable space $(X,\Sigma)$: each admits a countable covering of $X$ by measurable sets of finite measure. The Radon–Nikodym theorem states that if $\nu\ll\mu$, there exists a nonnegative measurable function $h$, finite $\mu$-almost everywhere, such that

This function is the Radon–Nikodym derivative, written $h=d\nu/d\mu$. Conversely, a measure defined by such an integral is absolutely continuous with respect to $\mu$, because integration over a $\mu$-null set gives zero. The density is unique up to equality $\mu$-almost everywhere, not necessarily at every point. (sites.math.rutgers.edu)

Sigma-finiteness does not imply that $h$ belongs globally to the space $L^1(\mu)$. Indeed,

so $h\in L^1(\mu)$ exactly when $\nu$ is finite. For sigma-finite measures, one can partition $X$ into countably many measurable pieces on which both measures are finite; the density is integrable on each piece. As a direct example, if $\nu=\mu$ is Lebesgue measure on $\mathbb R$, then $d\nu/d\mu=1$, which is finite everywhere but not integrable over the whole line. (sites.math.rutgers.edu)

The defining identity extends from sets to functions through the change-of-measure formula:

It holds for every nonnegative measurable $\varphi$, allowing either side to be infinite, and for every real- or complex-valued $\nu$-integrable $\varphi$. In the latter case, $\int_X|\varphi|h\,d\mu<\infty$. The mechanism is to establish the identity first for indicator functions, extend it by linearity to nonnegative simple functions, and then use monotone convergence; positive and negative parts give the integrable real-valued case. (math.wustl.edu)

For three sigma-finite positive measures satisfying $\lambda\ll\nu\ll\mu$, the measure-theoretic chain rule is

The product can be defined as zero wherever $d\nu/d\mu=0$; changes to $d\lambda/d\nu$ on a $\nu$-null set do not affect the resulting $\mu$-almost-everywhere identity. If $\mu$ and $\nu$ are equivalent, meaning each is absolutely continuous with respect to the other, then $0<h<\infty$ almost everywhere for both measures, and

Values on the common exceptional null set may be chosen arbitrarily. (uab.edu)

A proof approach associated with John von Neumann uses Hilbert-space methods. For finite $\mu,\nu$, set $\rho=\mu+\nu$. The functional $T(f)=\int_X f\,d\nu$ is bounded on $L^2(\rho)$, since

The Riesz representation theorem supplies $g\in L^2(\rho)$ with $T(f)=\int_X fg\,d\rho$. Testing indicator functions gives $0\leq g\leq1$, $d\nu=g\,d\rho$, and $d\mu=(1-g)\,d\rho$. Absolute continuity forces $\{g=1\}$ to be $\rho$-null, so $h=g/(1-g)$ supplies the required density off that set. Applying this construction on a common countable finite-measure partition proves the sigma-finite version. (sites.math.rutgers.edu)

Stieltjes Measures, Singular Variation, and Null-Set Preservation

For a nondecreasing, right-continuous function $F:\mathbb R\to\mathbb R$, its Lebesgue–Stieltjes measure is the unique Borel measure $\mu_F$ satisfying

The interval increments define a premeasure, which extends to the Borel sigma-algebra by the Carathéodory construction. Right continuity is essential for the convention $(s,t]$; an arbitrary nondecreasing function must first be replaced by its right-continuous version. Adding a constant to $F$ leaves the measure unchanged. A jump becomes a point mass:

Thus $F(x)=x$ produces Lebesgue measure, whereas $F(x)=\mathbf 1_{[c,\infty)}(x)$ produces a unit mass at $c$. (math.ucdavis.edu)

On a compact interval $[a,b]$, take $F$ nondecreasing and right-continuous, and normalize its associated measure by $\mu_F(\{a\})=0$. Then $F$ is absolutely continuous exactly when $\mu_F\ll\lambda$, where $\lambda$ denotes Lebesgue measure on that interval. In the forward direction,

for every Borel set $E$. Conversely, if $\mu_F\ll\lambda$, its Radon–Nikodym density $h$ is nonnegative and integrable, and the interval identity gives

The integral representation therefore establishes absolute continuity of $F$. For continuous functions of bounded variation that are not monotone, the analogous construction gives a signed Stieltjes measure; absolute continuity of the function is equivalent to absolute continuity of that signed measure, or of its total variation. (terrytao.wordpress.com)

The Lebesgue decomposition theorem, followed by separation of the atoms, gives every finite positive Borel measure on $\mathbb R$ a unique decomposition

Here $c_k=\mu(\{x_k\})>0$, and the atoms form an at most countable set with $\sum_k c_k<\infty$. The singular-continuous component $\mu_{\mathrm{sc}}$ is concentrated on a Lebesgue null set but assigns zero mass to every singleton. Consequently, the corresponding nondecreasing function separates, up to an additive constant, into an absolutely continuous contribution, a jump contribution, and a continuous singular contribution. The last two are singular variation: they contribute interval increments without contributing to the classical derivative almost everywhere. (terrytao.wordpress.com)

The Cantor function $C$, extended constantly outside $[0,1]$, illustrates singular-continuous variation. Its Stieltjes measure has total mass $1$, is concentrated on the Cantor set $K$, and has no atoms because $C$ is continuous. Nevertheless, $\lambda(K)=0$ and $C'=0$ almost everywhere, so integration of the classical derivative recovers none of the increase from $C(0)=0$ to $C(1)=1$. At the level of images, the same obstruction appears as $C(K)=[0,1]$: a null set is mapped onto a set of measure $1$. (sites.math.rutgers.edu)

The Banach–Zarecki theorem characterizes this obstruction for arbitrary real-valued functions on $[a,b]$: $f$ is absolutely continuous if and only if it is continuous, has bounded variation, and satisfies Lusin’s property $N$. This property means

for every $E\subseteq[a,b]$, with outer measure $\lambda^*$ avoiding any preliminary measurability assumption on the image. Continuity excludes jumps, bounded variation controls the total accumulated change, and property $N$ excludes the remaining singular-continuous variation. The Cantor function meets the first two conditions but fails the third precisely on $K$. (arxiv.org)

Absolutely Continuous Probability Distributions

A probability law $\mu$ on $\mathbb R$ is absolutely continuous with respect to Lebesgue measure $\lambda$ if $\lambda(A)=0$ implies $\mu(A)=0$ for every Borel set $A$. Equivalently, it has a probability density function: a nonnegative measurable function $p=d\mu/d\lambda$ satisfying

The density is unique up to equality Lebesgue-almost everywhere. Its value at an individual point is not that point’s probability; indeed, every singleton has probability zero. Probabilities are obtained by integrating over sets, rather than by evaluating $p$. (ocw.mit.edu)

The cumulative distribution function $F(x)=\mu((-\infty,x])$ consequently satisfies

Thus $F$ is absolutely continuous and $F'(x)=p(x)$ almost everywhere. Conversely, an absolutely continuous cumulative distribution function determines a density through its almost-everywhere derivative. For example, the uniform distribution on $[0,1]$ has density $p=\mathbf 1_{[0,1]}$, and its cumulative distribution function equals $0$ below $0$, $x$ on $[0,1]$, and $1$ above $1$. Differentiability almost everywhere alone is insufficient: the derivative must recover all the probability through integration. (services.math.duke.edu)

A discrete law concentrates its probability on a countable set and therefore cannot be absolutely continuous with respect to Lebesgue measure. However, excluding point masses does not ensure absolute continuity. For any cumulative distribution function,

so continuity of $F$ is equivalent to the absence of atoms, not to the existence of a Lebesgue density. The Cantor distribution gives a concrete counterexample. Let $B_1,B_2,\ldots$ be independent random variables taking $0$ and $1$, each with probability $1/2$, and put

Then $Y$ lies in the middle-thirds Cantor set $C$, which has Lebesgue measure zero, with probability one. Each of the $2^n$ remaining construction intervals at stage $n$ receives probability $2^{-n}$, implying that no individual point has positive probability. Its cumulative distribution function is continuous but has derivative zero almost everywhere: this law is singular-continuous, rather than discrete or absolutely continuous. (sites.math.duke.edu)

Absolute continuity always depends on the reference measure. If a law is supported on a countable set $S$, counting measure $c$ on $S$ gives it a density

This is precisely its probability mass function, and integration becomes summation: $\mu(A)=\sum_{s\in A\cap S}\mu(\{s\})$. Accordingly, the statement that a discrete law “has no density” means that it has no density relative to Lebesgue measure; it does not preclude densities relative to other measures. (stat.yale.edu)

More generally, let $P$ and $Q$ be probability measures on the same measurable space, with $Q\ll P$. Their Radon–Nikodym derivative $Z=dQ/dP$ is a nonnegative measurable weight satisfying

The change-of-measure identity is

It holds for nonnegative measurable $X$, allowing infinite expectations, and for signed $X$ whenever $\mathbb E_Q[|X|]=\mathbb E_P[|X|Z]<\infty$. Thus $Z$ reweights probabilities and expectations without assigning positive probability to any $P$-null event. It need not be positive everywhere: $Q\ll P$ does not require $P\ll Q$. (ocw.mit.edu)

Sobolev Spaces and Differential Equations

On a bounded interval $(a,b)$, the Sobolev space $W^{1,1}(a,b)$ consists of functions $u\in L^1(a,b)$ with a weak derivative $g\in L^1(a,b)$: for every smooth, compactly supported test function $\varphi$,

Every such $u$ has an absolutely continuous representative $\widetilde u$ on $[a,b]$, with

The mechanism is that subtracting the indefinite integral of $g$ leaves a function with zero weak derivative, which is constant almost everywhere. Conversely, integration by parts shows that an absolutely continuous function belongs to $W^{1,1}(a,b)$, with weak derivative equal to its classical derivative almost everywhere. (math.ucdavis.edu)

Strictly speaking, an element of $W^{1,1}(a,b)$ is an equivalence class under equality almost everywhere, not a function with prescribed values at every point. Altering a value at one interior point leaves the Sobolev element unchanged but can destroy continuity. Its continuous representative is unique: two continuous functions agreeing almost everywhere agree everywhere. Consequently, $\widetilde u(a)$ and $\widetilde u(b)$ are well-defined endpoint values, or traces, determined by the class rather than by arbitrary endpoint assignments. (math.ucdavis.edu)

For every $1\leq p\leq\infty$, the corresponding characterization is

Boundedness of the interval ensures that $L^p(a,b)\subset L^1(a,b)$, while an absolutely continuous representative is bounded and hence belongs to $L^p(a,b)$. For $1<p<\infty$, Hölder’s inequality applied to the integral of $\widetilde u'$ gives

Thus the representative is Hölder continuous with exponent $1-1/p$. For $p=\infty$, the estimate becomes

yielding Lipschitz continuity; conversely, every Lipschitz function on $[a,b]$ defines an element of $W^{1,\infty}(a,b)$. (people.cas.uab.edu)

In higher dimensions, the correspondence takes the form of absolute continuity on lines. If $\Omega\subset\mathbb R^n$ is open and $u\in W^{1,p}(\Omega)$, a suitable representative restricts to an absolutely continuous function on every compact subinterval of almost every line parallel to each coordinate axis. Here “almost every” refers to $(n-1)$-dimensional Lebesgue measure on the transverse coordinates. Along those lines, its classical derivative agrees almost everywhere with the corresponding weak partial derivative. This does not assert continuity throughout $\Omega$, or absolute continuity on every line. Conversely, this linewise property, together with $u\in L^p(\Omega)$ and linewise derivatives in $L^p(\Omega)$, characterizes $W^{1,p}(\Omega)$. (math.ucv.ro)

For an ordinary differential equation, absolute continuity permits an initial-value problem to be interpreted through

Provided $s\mapsto f(s,x(s))$ is integrable, this identity is equivalent to $x$ being absolutely continuous, satisfying $x(t_0)=x_0$, and obeying $x'(t)=f(t,x(t))$ almost everywhere. The Carathéodory formulation accommodates right-hand sides measurable in time and continuous in the state for almost every time, with locally integrable bounds. (karlin.mff.cuni.cz) For example, Gerald Teschl gives $x'=\operatorname{sgn}(t)x$, $x(0)=1$: the solution $x(t)=e^{|t|}$ is absolutely continuous on compact intervals and satisfies the equation almost everywhere, although it is not differentiable at $t=0$. (schecter.math.ncsu.edu)

Absolutely Continuous Curves in Metric Spaces

Let $(X,d)$ be a metric space and $\gamma:[a,b]\to X$. The curve is absolutely continuous if, for every $\varepsilon>0$, there is a $\delta>0$ such that every finite family of subintervals $(a_i,b_i)\subseteq[a,b]$ with pairwise disjoint interiors satisfies

This replaces absolute values of scalar increments by distances, without requiring addition or subtraction in $X$. On a compact parameter interval, it is equivalent to the existence of a nonnegative $g\in L^1(a,b)$ such that

The integral bound implies the disjoint-interval condition because the integral of $g$ over a set of sufficiently small Lebesgue measure is small. Conversely, the disjoint-interval condition makes the curve’s accumulated variation an absolutely continuous scalar function; its derivative supplies such a control $g$. (arxiv.org)

The intrinsic speed is the metric derivative, defined wherever the following limit exists:

For an absolutely continuous curve, this limit exists almost everywhere, belongs to $L^1(a,b)$, and itself satisfies the integral bound. Moreover, it is the smallest admissible control: $|\dot\gamma|\leq g$ almost everywhere for every $g$ satisfying that bound. Thus metric differentiation yields a scalar speed even when the target space has no tangent vectors. The existence and minimality theorem is presented, for example, in Luigi Ambrosio and Giuseppe Savaré’s 2007 lectures on gradient flows of probability measures. (web.math.ucsb.edu)

The length on any subinterval is exactly the integral of this speed:

Consequently, absolutely continuous curves have finite length. The formula counts all motion along the curve, not merely the distance between its endpoints, which can be strictly smaller. Absolute continuity is a property of the parametrized curve: finite length and continuity alone do not ensure it under an arbitrary parametrization. (research-collection.ethz.ch)

For $1\leq p\leq\infty$, the notation $AC^p([a,b];X)$ means that the integral bound holds with a control in the space $L^p$, equivalently that the curve is absolutely continuous and $|\dot\gamma|\in L^p$. Thus $AC^1=AC$. For $1<p<\infty$, Hölder’s inequality gives

For $p=\infty$, the bound becomes $d(\gamma(s),\gamma(t))\leq\|\,|\dot\gamma|\,\|_\infty|t-s|$, so $AC^\infty$ is precisely the class of Lipschitz curves. (web.math.ucsb.edu)

On a smooth Riemannian manifold, using its Riemannian distance recovers the usual notion of absolutely continuous curves, with metric speed equal almost everywhere to the Riemannian norm of the velocity. The same definitions apply to paths of probability measures equipped with a Wasserstein distance, such as $(\mathcal P_2(\mathbb R^n),W_2)$. Here regularity concerns the parameterized path, not whether its individual measures have densities. For example, the path $\mu_t=\delta_t$ on $[0,1]$ satisfies $W_2(\delta_s,\delta_t)=|s-t|$: it is Lipschitz with speed $1$, although every $\mu_t$ is singular relative to Lebesgue measure. Conversely, the constant path at a singular measure has speed zero. These examples follow directly from the transport definition of $W_2$; neither continuity nor absolute continuity of a measure-valued path implies absolute continuity of its constituent measures relative to a reference measure. (cvgmt.sns.it)

References

1. 2: More on L-Integrals and Absolute Continuitymath.libretexts.org
2. Notes on Partial Differential Equationsmath.ucdavis.edu
3. Modern Real Analysismath.purdue.edu
4. Honours Real Variables Imath.ualberta.ca
5. Notes on Partial Differential Equationsmath.ucdavis.edu
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