The integral F(x)=integral from 0 to x of 2t dt equals x². Its derivative recovers 2x. This familiar calculation survives for arbitrary L1 integrands almost everywhere, but the converse requires more than continuity. A function can increase while its derivative is zero almost everywhere if its change is concentrated on a null set.
Predict
Predict whether a continuous increasing function with derivative zero almost everywhere must be constant. The Cantor distribution function is a counterexample. It is constant on every interval outside the Cantor set, yet its total increase is one. The property that rules out such singular accumulation is absolute continuity.
Experiment
Compare three functions on [0,1].
Function
Continuous
Absolutely continuous
Recovered from derivative integral
x²
Yes
Yes
Yes
Cantor distribution function
Yes
No
No
Indicator of [1/2,1]
No
No
No
For x², partition the interval into disjoint short pieces and sum absolute increments. Compare this with the integral of 2x over their union. For the Cantor function, the omitted null support carries the entire increase.
Understand
A function F on [a, b] is absolutely continuous if for every ε>0 there is δ>0 such that any finite collection of disjoint intervals with total length below δ has sum of absolute F increments below ε. If F is an indefinite integral of f in L1, this follows because the sum of absolute increments is bounded by the integral of abs (f) over the union, and L1 integrals are uniformly small on sets of sufficiently small measure.
At a Lebesgue point x of f, the difference quotient of F is the average of f on the interval from x to x+h. Its difference from f(x) is bounded by twice the centered mean oscillation on radius abs (h), which tends to zero. Hence F prime=f almost everywhere.
For the converse, first note that absolute continuity implies bounded variation. Choose a coarse partition into intervals shorter than a suitable δ. Refine any proposed partition by these coarse endpoints. On each coarse interval, the sum of absolute refined increments is bounded by the absolute-continuity condition; summing finitely many coarse bounds gives a bound independent of the proposed partition. The variation function V(x) is increasing, and both V+F and V−F are increasing. Thus F is the difference of two increasing functions. The same small-interval bound makes V continuous, so these increasing functions are continuous.
An increasing continuous function induces a finite measure whose value on (u, v] is its increment. Finite additivity is telescoping; continuity of the function gives continuity of these interval increments under shrinking endpoint gaps, which yields the premeasure condition and permits the extension theorem. Applying this to the variation decomposition constructs a finite signed Stieltjes measure for F. Absolute continuity of F forces its variation measure to vanish on length-null sets: cover a null set by open intervals of total length below δ; on any finite selection and any finite subdivisions, the sum of absolute increments is below ε. Take suprema over subdivisions and then countable unions. Let ε decrease to zero.
Radon–Nikodym now represents the signed increment measure by an L1 density g. Therefore F(x)−F(a)=integral from a to x of g. The already-proved differentiation result identifies g with F prime almost everywhere. This proves the fundamental theorem for absolutely continuous functions.
Apply
The interval extension step uses continuity, while the density step uses absolute continuity. Explain the role of each rather than merging them into a single regularity claim. For a monotone function, a singular component of its Stieltjes measure is exactly the obstruction to reconstructing all increments from a length density.
Continue
An almost-everywhere derivative alone does not control global changes. Absolute continuity connects local derivatives to all interval increments. This distinction is essential in weak differential equations, distribution functions and any argument that integrates a derivative to recover an original function.
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If F(x)=integral from 0 to x of 3t² dt on [0,1], compute F(1)−F(1/2).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For F(x)=2x on [0,3], what is its total variation?
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Prove an indefinite integral of an L1 function is absolutely continuous.
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Show F(x)=integral from a to x f has derivative f almost everywhere for f in L1.
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Outline the measure-theoretic proof that an absolutely continuous F equals F(a)+integral from a to x of an L1 derivative.
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Why does the Cantor function show that continuity and an almost-everywhere derivative are insufficient for the integral formula?
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