advanced35 min studyLesson 190 in Mathematics

Lebesgue differentiation and absolute continuity

Recover local values from averages and determine exactly when an integrable derivative reconstructs a function.

On this page

What you’ll understand

  • State the almost-everywhere conclusion of the Lebesgue differentiation theorem.
  • Use absolute continuity to justify the Lebesgue fundamental theorem of calculus.
  • Distinguish an integrable derivative from the stronger claim that its integral recovers every increment.

For a continuous function, averages over shorter intervals approach the value at their center. Lebesgue differentiation says a version of this remains true almost everywhere for locally integrable functions, even with very irregular pointwise behavior. The result reconnects integration with differentiation. A second concept, absolute continuity, identifies precisely which one-variable functions are reconstructed by integrating their almost-everywhere derivative. Neither statement says that every function differentiable almost everywhere is determined by that derivative.

Local averages find almost every representative value

Let f be locally integrable on the real line. Almost every x is a Lebesgue point: the mean of |f(t)−f(x)| over (x−r,x+r) tends to zero as r decreases to zero. Consequently the unsigned-error control forces the signed average of f to approach f(x). Local integrability is enough because each bounded interval has finite integral of |f|. The exceptional set may be nonempty, and modifying a representative on a null set can deliberately spoil the conclusion at selected points without changing its integrable equivalence class.

A step function illustrates an allowed exception. At a jump from zero to one, symmetric averages tend to one half, which need not equal the assigned value at the jump. Away from the jump, sufficiently small intervals lie on one side and averaging works immediately. The general theorem is much deeper: approximation by continuous functions is combined with a maximal-function estimate controlling where local averaging errors are large. A graph sampled at finitely many points cannot establish that almost-everywhere estimate for an arbitrary integrable function.

lim⁡r↓012r∫x−rx+r∣f(t)−f(x)∣ dt=0for a.e. x,f∈Lloc1(R)\lim_{r\downarrow0}\frac1{2r}\int_{x-r}^{x+r}|f(t)-f(x)|\,dt=0\quad\text{for a.e. }x,\qquad f\in L^1_{\mathrm{loc}}(\mathbb R)
The absolute-mean condition defines a Lebesgue point. It implies convergence of signed local averages, with an almost-everywhere rather than everywhere conclusion.↔ Scroll wide equations horizontally.

Absolute continuity controls many short intervals at once

A function F on [a,b] is absolutely continuous if for every epsilon>0 there is a delta>0 such that any finite family of pairwise disjoint subintervals with total length below delta has total absolute F-increment below epsilon. This is stronger than ordinary uniform continuity, which controls one small interval at a time without summing many increments. Lipschitz functions are absolutely continuous, but an absolutely continuous function can have an unbounded derivative. The condition restricts accumulated variation rather than imposing one global slope bound.

If g is integrable on [a,b], the function F(x)=F(a)+∫aˣ g(t)dt is absolutely continuous. To see why, bound the sum of absolute increments over disjoint intervals by the integral of |g| over their union. Integrability makes that integral small when the union has sufficiently small measure: truncate |g| at a large cutoff and control its remaining integral tail. Lebesgue differentiation then gives F′=g almost everywhere. The construction supplies both the continuity property and the correct derivative statement.

F∈AC([a,b])⟺F(x)=F(a)+∫axg(t) dt for some g∈L1([a,b]);F′=g a.e.F\in AC([a,b])\quad\Longleftrightarrow\quad F(x)=F(a)+\int_a^x g(t)\,dt\ \text{for some }g\in L^1([a,b]);\qquad F'=g\ \text{a.e.}
This is the Lebesgue fundamental theorem of calculus. For an absolutely continuous F, its almost-everywhere derivative is integrable and reconstructs every increment.↔ Scroll wide equations horizontally.

An almost-everywhere derivative can miss singular change

The converse direction is not merely that F has an integrable derivative almost everywhere. The Cantor function is continuous and increasing from zero to one, yet its derivative is zero almost everywhere. Integrating that derivative returns zero and misses the function's total increase. Its variation is concentrated on the null Cantor set, so it is not absolutely continuous. This example separates ordinary continuity, bounded variation, almost-everywhere differentiability, and absolute continuity. The singular component is a measure-theoretic phenomenon, not a numerical differentiation error.

For an absolutely continuous function, null-set changes to its derivative do not affect the reconstruction integral. Null-set changes to the function itself are different: changing a single value can destroy continuity and therefore destroy absolute continuity, even though the function represents the same L1 class. Thus the fundamental theorem concerns a particular continuous representative reconstructed at every point, while its derivative is specified only almost everywhere. Keeping those roles separate is important in weak derivatives, differential equations, and one-dimensional Sobolev spaces.

Absolute continuity without a bounded slope

Let F(x)=2xF(x)=2\sqrt{x} on [0,1]. Show it is absolutely continuous and determine where its classical derivative is finite.

  1. For x>0, differentiation gives F′(x)=x−1/2F'(x)=x^{-1/2}. The function g(t)=t⁻¹ᐟ² on (0,1] has finite integral two; its value at zero can be assigned arbitrarily for integration.

  2. Direct integration gives ∫₀ˣg(t)dt=2√x for every x in [0,1]. The integral-representation direction of the fundamental theorem therefore proves absolute continuity.

  3. The derivative is finite at every x>0 but becomes unbounded as x approaches zero. The right difference quotient at zero is 2/√h, so there is no finite right derivative there. This exceptional endpoint does not violate the almost-everywhere theorem.

The result

F is absolutely continuous despite an unbounded slope and a non-finite endpoint derivative. Integrable slope is sufficient; a uniform slope bound is unnecessary.

Explore the model

Interactive lab

Interactive lab · function

Derivative and accumulation in a smooth case

Use the coefficient a and position x controls for y=a x². Predict its tangent slope 2ax and its accumulated area a x³/3, then compare both readouts. Playback moves a trace marker along the curve; the selected endpoint, tangent, and area readouts stay fixed. It illustrates the smooth case of differentiation and integration; it does not model the square-root singularity, maximal functions, or the Cantor function.

Derivative and accumulation in a smooth caseThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe analytic quadratic obeys the fundamental theorem everywhere; general integrable data provide only an almost-everywhere derivative statement.

Check your understanding

Why does an integrable almost-everywhere derivative alone fail to reconstruct every continuous function?

Try it yourself

Let G(x)=x² on [0,1], except redefine G(1/2)=7. Does G still represent the same L1 element as x²? Is this representative absolutely continuous? Explain what integrating 2x reconstructs.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Separate equality almost everywhere from continuity of a specific representative.

Compare with the solution

The single altered point is Lebesgue-null, so G and x² define the same L1 element. This G is discontinuous at 1/2, hence not absolutely continuous. Its classical derivative still equals 2x away from that point, but integrating 2x from zero to x reconstructs x², the continuous absolutely continuous representative, not the isolated value seven. Derivative data specified almost everywhere cannot encode arbitrary point changes in a representative.

Take it with you

  • Locally integrable functions are recovered by local averages at almost every Lebesgue point.
  • Absolute continuity is equivalent to reconstruction from an L1 derivative on a compact interval.
  • Singular variation and isolated representative changes explain why weaker derivative claims are insufficient.

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