{ "id": "absolute-continuity", "title": "Absolute Continuity", "lang": "en", "source_lang": "en", "summary": "A property connecting controlled variation of functions, integration of derivatives, and measures admitting densities.", "category": "mathematics", "subcategory": "analysis", "type": "concept", "aliases": [ "Absolute continuity of functions", "Absolute continuity of measures" ], "generated_at": "2026-10-05", "model": "pro/gpt-6.1-sol", "url": "https://aiwiki.page/en/absolute-continuity/", "translations": { "zh-Hans": { "url": "https://aiwiki.page/zh/absolute-continuity/", "json": "https://aiwiki.page/zh/absolute-continuity.json" } }, "sections": [ { "heading": "", "text": "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)\n\nThe 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)" }, { "heading": "Definition for Functions on an Interval", "text": "Let \\(a0\\), there exists \\(\\delta>0\\) such that every finite family of pairwise disjoint open intervals \\((a_i,b_i)\\), with endpoints in \\([a,b]\\), satisfies\n\n\\[\\sum_{i=1}^{n}(b_i-a_i)<\\delta \\quad\\Longrightarrow\\quad \\sum_{i=1}^{n}|f(b_i)-f(a_i)|<\\varepsilon.\\]\n\nHere \\(n\\) is any positive integer and \\(a_i0\\), the estimate\n\n\\[|y^2-x^2|\\leq 2M|y-x|\\]\n\nmakes \\(\\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\n\n\\[|(x+h)^2-x^2|=2xh+h^2>1.\\]\n\nThus 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)" }, { "heading": "Comparison with Other Regularity Properties", "text": "On a compact interval \\([a,b]\\), the principal implications are\n\n\\[C^1([a,b]) \\subsetneq \\operatorname{Lip}([a,b]) \\subsetneq AC([a,b]) \\subsetneq UC([a,b]),\\]\n\nwhere 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\n\n\\[\\sum_i|f(b_i)-f(a_i)| \\le L\\sum_i(b_i-a_i).\\]\n\nThus \\(\\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)\n\nAbsolute continuity also implies bounded variation, meaning that\n\n\\[\\operatorname{Var}_{[a,b]}(f) =\\sup_{a=x_0<\\cdots0\\), \\(f'(x)=1/(2\\sqrt{x})\\), and\n\n\\[\\int_0^1\\frac{dt}{2\\sqrt{t}}=1, \\qquad \\sqrt{x}=\\int_0^x\\frac{dt}{2\\sqrt{t}}.\\]\n\nThe integral representation establishes absolute continuity, despite the unbounded derivative near \\(0\\). No finite Lipschitz constant can exist, since\n\n\\[\\frac{|f(x)-f(0)|}{|x-0|} =\\frac{1}{\\sqrt{x}}\\longrightarrow\\infty \\quad\\text{as }x\\downarrow0.\\]\n\nAbsolute continuity therefore permits this endpoint singularity; Lipschitz continuity requires a uniform bound on all secant slopes. (math.ualberta.ca)\n\nThe 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)\n\nThe 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)" }, { "heading": "Integral Characterization and the Fundamental Theorem of Calculus", "text": "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\n\n\\[f(x)=f(a)+\\int_a^x g(t)\\,dt \\qquad\\text{for every }x\\in[a,b].\\]\n\nHere 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)\n\nFor 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:\n\n\\[f(y)-f(x)=\\int_x^y f'(t)\\,dt \\qquad(a\\leq x\\leq y\\leq b).\\]\n\nConsequently, an absolutely continuous function whose derivative vanishes almost everywhere is constant. (math.ucdavis.edu)\n\nThe 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\n\n\\[m(E)<\\delta \\quad\\Longrightarrow\\quad \\int_E|g(t)|\\,dt<\\varepsilon,\\]\n\nwhere \\(m\\) denotes Lebesgue measure. This estimate does not require \\(g\\) to be bounded. Choose \\(M>0\\) sufficiently large that\n\n\\[\\int_{\\{|g|>M\\}}|g(t)|\\,dt<\\frac{\\varepsilon}{2}.\\]\n\nSplitting \\(E\\) into the regions where \\(|g|\\leq M\\) and \\(|g|>M\\) yields\n\n\\[\\int_E|g(t)|\\,dt \\leq M\\,m(E)+\\int_{\\{|g|>M\\}}|g(t)|\\,dt.\\]\n\nTaking \\(\\delta=\\varepsilon/(2M)\\) proves the assertion: integrability controls the large-value tail, while small measure controls the bounded part. (math.ucdavis.edu)\n\nFor a finite family of disjoint subintervals \\((a_i,b_i)\\), let \\(E=\\bigcup_i(a_i,b_i)\\). The integral representation gives\n\n\\[\\sum_i|f(b_i)-f(a_i)| \\leq\\sum_i\\int_{a_i}^{b_i}|g(t)|\\,dt =\\int_E|g(t)|\\,dt.\\]\n\nSince \\(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)\n\nAlmost-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)" }, { "heading": "Algebraic Operations and Calculus Rules", "text": "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]\\),\n\n\\[|f(y)g(y)-f(x)g(x)| \\le M_f|g(y)-g(x)|+M_g|f(y)-f(x)|.\\]\n\nSumming this estimate over disjoint subintervals establishes absolute continuity of \\(fg\\). At every point where both derivatives exist, the ordinary product rule applies:\n\n\\[(fg)'=f'g+fg'.\\]\n\nThus the rule holds almost everywhere, without requiring either derivative to be continuous. (gyansanchay.csjmu.ac.in)\n\nQuotients are also permitted when the denominator is bounded away from zero. Suppose \\(|g(x)|\\ge c>0\\) throughout \\([a,b]\\). The estimate\n\n\\[\\left|\\frac1{g(y)}-\\frac1{g(x)}\\right| \\le c^{-2}|g(y)-g(x)|\\]\n\nshows 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:\n\n\\[\\left(\\frac fg\\right)'=\\frac{f'g-fg'}{g^2}.\\]\n\nIts 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)\n\nFor 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\n\n\\[\\sum_i|\\Phi(f(b_i))-\\Phi(f(a_i))| \\le L\\sum_i|f(b_i)-f(a_i)|,\\]\n\nso \\(\\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)\n\nA 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,\n\n\\[(\\Phi\\circ f)'(x)=\\Phi'(f(x))f'(x).\\]\n\nThe 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)\n\nIntegration 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\n\n\\[\\int_a^b|f'g|\\le M_g\\int_a^b|f'|, \\qquad \\int_a^b|fg'|\\le M_f\\int_a^b|g'|.\\]\n\nIntegrating the product rule and rearranging yields\n\n\\[\\int_a^b f(x)g'(x)\\,dx =f(b)g(b)-f(a)g(a)-\\int_a^b f'(x)g(x)\\,dx.\\]\n\nContinuity 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)" }, { "heading": "Absolute Continuity of Measures", "text": "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\n\n\\[\\mu(E)=0\\quad\\Longrightarrow\\quad\\nu(E)=0 \\qquad(E\\in\\mathcal F).\\]\n\nThe 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)\n\nAbsolute 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)\n\nWhen \\(\\nu(X)<\\infty\\), null-set absolute continuity is equivalent to the uniform small-set criterion\n\n\\[\\forall\\varepsilon>0\\ \\exists\\delta>0\\ \\forall E\\in\\mathcal F: \\qquad \\mu(E)<\\delta\\ \\Longrightarrow\\ \\nu(E)<\\varepsilon.\\]\n\nNo 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\\).\n\nTwo 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)\n\nFor 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)" }, { "heading": "Densities and the Radon–Nikodym Theorem", "text": "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\n\n\\[\\nu(E)=\\int_E h\\,d\\mu \\qquad(E\\in\\Sigma).\\]\n\nThis 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)\n\nSigma-finiteness does not imply that \\(h\\) belongs globally to the space L^1(\\mu). Indeed,\n\n\\[\\int_X h\\,d\\mu=\\nu(X),\\]\n\nso \\(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)\n\nThe defining identity extends from sets to functions through the change-of-measure formula:\n\n\\[\\int_X \\varphi\\,d\\nu = \\int_X \\varphi h\\,d\\mu.\\]\n\nIt 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)\n\nFor three sigma-finite positive measures satisfying \\(\\lambda\\ll\\nu\\ll\\mu\\), the measure-theoretic chain rule is\n\n\\[\\frac{d\\lambda}{d\\mu} = \\frac{d\\lambda}{d\\nu}\\frac{d\\nu}{d\\mu} \\qquad\\mu\\text{-almost everywhere}.\\]\n\nThe 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 \\(00\\), 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)\n\nThe 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)\n\nThe 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\n\n\\[\\lambda(E)=0\\quad\\Longrightarrow\\quad\\lambda^*(f(E))=0\\]\n\nfor 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)" }, { "heading": "Absolutely Continuous Probability Distributions", "text": "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\n\n\\[\\mu(A)=\\int_A p(x)\\,dx,\\qquad \\int_{\\mathbb R}p(x)\\,dx=1.\\]\n\nThe 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)\n\nThe cumulative distribution function \\(F(x)=\\mu((-\\infty,x])\\) consequently satisfies\n\n\\[F(x)=\\int_{-\\infty}^{x}p(t)\\,dt,\\qquad F(b)-F(a)=\\int_a^b p(t)\\,dt.\\]\n\nThus \\(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)\n\nA 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,\n\n\\[\\mu(\\{x\\})=F(x)-F(x-),\\]\n\nso 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\n\n\\[Y=\\sum_{n=1}^{\\infty}\\frac{2B_n}{3^n}.\\]\n\nThen \\(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)\n\nAbsolute 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\n\n\\[\\frac{d\\mu}{dc}(s)=\\mu(\\{s\\}).\\]\n\nThis 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)\n\nMore 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\n\n\\[Q(A)=\\mathbb E_P[\\mathbf 1_A Z],\\qquad \\mathbb E_P[Z]=1.\\]\n\nThe change-of-measure identity is\n\n\\[\\mathbb E_Q[X]=\\mathbb E_P[XZ].\\]\n\nIt 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)" }, { "heading": "Sobolev Spaces and Differential Equations", "text": "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\\),\n\n\\[\\int_a^b u(x)\\varphi'(x)\\,dx =-\\int_a^b g(x)\\varphi(x)\\,dx.\\]\n\nEvery such \\(u\\) has an absolutely continuous representative \\(\\widetilde u\\) on \\([a,b]\\), with\n\n\\[\\widetilde u(x)=c+\\int_a^x g(t)\\,dt, \\qquad \\widetilde u'=g\\quad\\text{almost everywhere}.\\]\n\nThe 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)\n\nStrictly 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)\n\nFor every \\(1\\leq p\\leq\\infty\\), the corresponding characterization is\n\n\\[u\\in W^{1,p}(a,b) \\quad\\Longleftrightarrow\\quad u=\\widetilde u\\ \\text{almost everywhere},\\qquad \\widetilde u\\in AC([a,b]),\\qquad \\widetilde u'\\in L^p(a,b).\\]\n\nBoundedness 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 \\(10\\), there is a \\(\\delta>0\\) such that every finite family of subintervals \\((a_i,b_i)\\subseteq[a,b]\\) with pairwise disjoint interiors satisfies\n\n\\[\\sum_i(b_i-a_i)<\\delta \\quad\\Longrightarrow\\quad \\sum_i d\\bigl(\\gamma(b_i),\\gamma(a_i)\\bigr)<\\varepsilon.\\]\n\nThis 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\n\n\\[d\\bigl(\\gamma(s),\\gamma(t)\\bigr)\\leq\\int_s^t g(r)\\,dr \\qquad(a\\leq s