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RA-6

7-12 hours

Advanced

Differentiation of Integrals & Maximal Control

Master the Hardy-Littlewood maximal function, Lebesgue differentiation theorem, Vitali covering lemma, and density theorems. Learn how to differentiate integrals and recover functions from their averages.

Learning Objectives

By the end of this course, you will be able to:

Understand the Hardy-Littlewood maximal function and its weak-type (1,1) inequality

Master the Lebesgue differentiation theorem for differentiating integrals

Apply the Vitali covering lemma to prove covering results

Understand density points and Lebesgue points

Prove that almost every point is a Lebesgue point for integrable functions

Apply differentiation theorems to recover functions from their integrals

Prerequisites

Before starting this course, you should be familiar with:

Lebesgue measure and integration
Convergence theorems
Basic properties of balls and cubes in ℝⁿ
Covering arguments and geometric measure theory basics

Core Concepts

Maximal functions and differentiation

Definition 6.1: Hardy-Littlewood Maximal Function

For $f \in L^1_{\text{loc}}(\mathbb{R}^n)$ , the Hardy-Littlewood maximal function is defined as:

Mf(x) = \sup_{r>0} \frac{1}{m(B(x,r))} \int_{B(x,r)} |f(y)| \, dy

where $B(x,r)$ denotes the open ball of radius $r$ centered at $x$ .

Definition 6.2: Lebesgue Point

A point $x$ is called a Lebesgue point of a locally integrable function $f$ if:

\lim_{r \to 0} \frac{1}{m(B(x,r))} \int_{B(x,r)} |f(y) - f(x)| \, dy = 0

Definition 6.3: Density Point

A point $x$ is called a density point of a measurable set $E$ if:

\lim_{r \to 0} \frac{m(E \cap B(x,r))}{m(B(x,r))} = 1

Theorem 6.1: Hardy-Littlewood Maximal Theorem

The maximal function $M$ satisfies:

Weak-type (1,1): For $f \in L^1$ and $\lambda > 0$ ,

m(\{x : Mf(x) > \lambda\}) \leq \frac{C_n}{\lambda} \|f\|_1

Strong-type (p,p): For

1 < p \leq \infty

and

f \in L^p

\|Mf\|_p \leq C_{n,p} \|f\|_p

where $C_n$ and $C_{n,p}$ are constants depending only on dimension and $p$ .

Proof of Theorem 6.1:

Proof:

The weak-type inequality is proved using the Vitali covering lemma. For each $x$ with $Mf(x) > \lambda$ , there exists a ball $B_x$ such that the average of $|f|$ over $B_x$ exceeds $\lambda$ .

Using Vitali's covering lemma, we extract a disjoint subcollection covering a large fraction of the set where $Mf > \lambda$ . The measure estimate follows.

The strong-type result for $p > 1$ follows from the weak-type (1,1) and the Marcinkiewicz interpolation theorem. ∎

∎

Theorem 6.2: Lebesgue Differentiation Theorem

Let $f \in L^1_{\text{loc}}(\mathbb{R}^n)$ . Then almost every $x$ is a Lebesgue point of $f$ . In particular,

\lim_{r \to 0} \frac{1}{m(B(x,r))} \int_{B(x,r)} f(y) \, dy = f(x)

for almost every $x$ .

Proof of Theorem 6.2:

Proof:

The proof uses the maximal function. For $\epsilon > 0$ , approximate $f$ by a continuous function $g$ such that $\|f - g\|_1 < \epsilon$ .

Since continuous functions satisfy the differentiation property, and the maximal function controls the error, we can show that the set of non-Lebesgue points has small measure.

Taking $\epsilon \to 0$ shows that almost every point is a Lebesgue point. ∎

∎

Theorem 6.3: Vitali Covering Lemma

Let $E$ be a measurable set in $\mathbb{R}^n$ and let $\mathcal{B}$ be a collection of balls such that every point of $E$ is contained in balls of $\mathcal{B}$ with arbitrarily small radius.

Then there exists a countable disjoint subcollection $\{B_i\}$ of $\mathcal{B}$ such that:

m\left(E \setminus \bigcup_i B_i\right) = 0

Proof of Theorem 6.3:

Proof:

The proof uses a greedy algorithm: at each step, select the largest ball that is disjoint from previously selected balls. The key is that if a ball is not selected, it must intersect a selected ball of comparable or larger size.

This allows us to cover most of $E$ with a controlled number of balls. ∎

∎

Corollary 6.1: Fundamental Theorem of Calculus for Lebesgue Integral

If $f \in L^1([a,b])$ and $F(x) = \int_a^x f(t) \, dt$ , then $F'(x) = f(x)$ almost everywhere.

Example 6.1: Differentiating an Integral

Problem: Let $f(x) = \chi_{[0,1]}(x)$ and $F(x) = \int_0^x f(t) \, dt$ . Find $F'(x)$ .

Solution:

We have $F(x) = \begin{cases} 0 & x < 0 \\ x & 0 \leq x \leq 1 \\ 1 & x > 1 \end{cases}$ .

By the Lebesgue differentiation theorem, $F'(x) = f(x)$ almost everywhere.

Indeed, $F'(x) = 1$ for $x \in (0,1)$ and $F'(x) = 0$ for $x \notin [0,1]$ , which equals $f(x)$ almost everywhere.

Theorem 6.4: Vitali Covering Lemma (Detailed Proof)

Let $E$ be a measurable set with $m(E) < \infty$ , and let $\mathcal{B}$ be a Vitali covering of $E$ (every point of $E$ is contained in balls of $\mathcal{B}$ with arbitrarily small radius).

Then for any $\epsilon > 0$ , there exists a finite disjoint subcollection $\{B_1, \ldots, B_N\}$ such that

m\left(E \setminus \bigcup_{i=1}^N B_i\right) < \epsilon

Proof of Theorem 6.4:

Proof:

Step 1: Use a greedy algorithm. Start with $B_1$ as any ball in $\mathcal{B}$ .

Step 2: Given $B_1, \ldots, B_k$ , if $E \subseteq \bigcup_{i=1}^k 5B_i$ (where $5B_i$ is the ball with the same center and 5 times the radius), we're done.

Step 3: Otherwise, choose $B_{k+1}$ to be a ball in $\mathcal{B}$ that is disjoint from $\bigcup_{i=1}^k B_i$ and has radius at least half the supremum of radii of such balls.

Step 4: Since $m(E) < \infty$ and the selected balls are disjoint, the process must terminate after finitely many steps.

Step 5: The key geometric fact is that if a ball $B$ is not selected, it must intersect some selected ball $B_i$ with $r(B) \leq 2r(B_i)$ , so $B \subseteq 5B_i$ .

This ensures that the union of the 5-times-dilated selected balls covers $E$ except for a set of small measure. ∎

∎

Theorem 6.5: Density Theorem (Complete Proof)

Let $E$ be a measurable set in $\mathbb{R}^n$ . Then for almost every $x \in E$ , we have

\lim_{r \to 0} \frac{m(E \cap B(x,r))}{m(B(x,r))} = 1

Such points are called density points of $E$ .

Proof of Theorem 6.5:

Proof:

Apply the Lebesgue differentiation theorem to the characteristic function $\chi_E$ .

For almost every $x \in E$ , we have

\lim_{r \to 0} \frac{1}{m(B(x,r))} \int_{B(x,r)} \chi_E = \chi_E(x) = 1

But $\int_{B(x,r)} \chi_E = m(E \cap B(x,r))$ , so the result follows. ∎

∎

Theorem 6.6: Lp Boundedness of the Maximal Function

For $1 < p < \infty$ , the Hardy-Littlewood maximal function $Mf$ satisfies

\|Mf\|_p \leq C_p \|f\|_p

where $C_p$ is a constant depending only on $p$ and the dimension.

Proof of Theorem 6.6:

Proof:

The proof uses the weak-type (1,1) bound (Theorem 6.2) and interpolation theory (Marcinkiewicz interpolation theorem).

For $p > 1$ , we can write $Mf = (Mf)^p \cdot (Mf)^{1-p}$ and use Hölder's inequality combined with the weak-type bound.

The key is that the weak-type bound gives control on the distribution function, which can be integrated to get Lp bounds. ∎

∎

Example 6.4: Applying Lebesgue Differentiation Theorem

Problem: Let $f(x) = |x|$ on $[-1, 1]$ . Verify that the Lebesgue differentiation theorem holds at $x = 0$ .

Solution:

For $x = 0$ , consider the average over $B(0, r) = (-r, r)$ :

\frac{1}{2r} \int_{-r}^r |t| \, dt = \frac{1}{2r} \cdot 2 \int_0^r t \, dt = \frac{1}{r} \cdot \frac{r^2}{2} = \frac{r}{2}

As $r \to 0$ , this tends to 0, which equals $f(0) = 0$ .

For $x \neq 0$ , $f$ is continuous, so the theorem holds trivially. Therefore, the Lebesgue differentiation theorem holds everywhere.

Example 6.5: Computing Density Points

Problem: Let $E = [0, 1] \cup [2, 3]$ . Find the density points of $E$ .

Solution:

For $x \in (0, 1)$ or $x \in (2, 3)$ , $x$ is in the interior of $E$ , so for small $r$ , $B(x, r) \subseteq E$ , giving density 1.

For $x \in (1, 2)$ or $x < 0$ or $x > 3$ , $x$ is in the complement, so for small $r$ , $B(x, r) \cap E = \emptyset$ , giving density 0.

At the boundary points $x = 0, 1, 2, 3$ , the density is $1/2$ (half the ball is in $E$ , half is not).

By the density theorem, almost every point of $E$ is a density point, which in this case means all interior points.

Example 6.6: Maximal Function Application

Problem: Estimate the maximal function of $f(x) = \chi_{[0,1]}(x)$ at $x = 0.5$ .

Solution:

For $r \geq 0.5$ , the ball $B(0.5, r)$ contains $[0,1]$ , so the average is $1/(2r)$ .

For $r < 0.5$ , the ball $B(0.5, r) = (0.5-r, 0.5+r)$ is contained in $[0,1]$ , so the average is 1.

Therefore, $Mf(0.5) = \sup_{r > 0} \frac{1}{2r} \int_{B(0.5, r)} f = \max(1, \sup_{r \geq 0.5} \frac{1}{2r}) = 1$ .

At $x = 1.5$ , for small $r$ , the average is 0, but for $r \geq 0.5$ , the ball intersects $[0,1]$ , giving $Mf(1.5) = 1/2$ .

Example 6.7: Derivative of an Absolutely Continuous Function

Problem: Let $F(x) = \int_0^x t^2 \, dt$ on $[0,1]$ . Show that $F$ is absolutely continuous and compute $F'(x)$ .

Solution:

Since $f(t) = t^2$ is in $L^1([0,1])$ , the function $F$ is absolutely continuous (this is a general property of indefinite integrals of L1 functions).

By the Lebesgue differentiation theorem (or the fundamental theorem of calculus), $F'(x) = f(x) = x^2$ almost everywhere.

In fact, since $f$ is continuous, $F'(x) = x^2$ for all $x$ .

We can verify: $F(x) = \frac{x^3}{3}$ , so $F'(x) = x^2$ .

Corollary 6.3: Local Properties of Differentiation

The Lebesgue differentiation theorem shows that differentiation is a local property:

If $f = g$ almost everywhere, then their derivatives (where they exist) are equal almost everywhere
The derivative at a point depends only on the behavior of $f$ in arbitrarily small neighborhoods
Null sets do not affect differentiation: if $f$ is modified on a null set, the derivative is unchanged

Remark 6.3: Historical Context of Differentiation Theorems

The differentiation theorems are fundamental results connecting integration and differentiation:

Lebesgue differentiation theorem (1904): Proved by Lebesgue, generalizing the fundamental theorem of calculus to Lebesgue integrals.
Hardy-Littlewood maximal function (1930): Introduced by G.H. Hardy and J.E. Littlewood as a tool for studying differentiation. The maximal function is now central to harmonic analysis.
Vitali covering lemma (1908): Proved by Giuseppe Vitali. It is a geometric tool essential for many results in measure theory and analysis.

These results form the foundation of modern analysis and are used constantly in partial differential equations, harmonic analysis, and geometric measure theory.

Remark 6.4: Applications in Harmonic Analysis

The maximal function and differentiation theorems are central to harmonic analysis:

Pointwise convergence: The maximal function is used to prove pointwise convergence of Fourier series and other approximation methods.
Singular integrals: The Lp boundedness of the maximal function is a model for studying more general singular integral operators.
Calderón-Zygmund theory: The techniques developed for the maximal function extend to a wide class of operators in harmonic analysis.
Weighted inequalities: The study of weighted Lp spaces (Ap weights) grew out of questions about the maximal function.

These applications show why the differentiation theorems are not just about calculus, but are fundamental tools in modern analysis.

Example 6.8: Maximal Function: Lp Estimates

Problem: Show that for $f \in L^p$ with $1 < p \leq \infty$ , the maximal function $Mf$ satisfies $\|Mf\|_p \leq C_p \|f\|_p$ for some constant $C_p$ .

Solution:

This is Theorem 6.6. The proof uses the weak-type (1,1) bound (Theorem 6.2) and Marcinkiewicz interpolation.

For $p = \infty$ , we have $Mf(x) \leq \|f\|_\infty$ almost everywhere, so $\|Mf\|_\infty \leq \|f\|_\infty$ .

For $1 < p < \infty$ , the constant $C_p$ depends on $p$ and the dimension, and grows like $\frac{p}{p-1}$ as $p \to 1$ .

Example 6.9: Density Points: Application to Sets

Problem: Let $E$ be a measurable set with $m(E) > 0$ . Show that almost every point of $E$ is a density point.

Solution:

By Theorem 6.5 (the density theorem), for almost every $x \in E$ , we have

\lim_{r \to 0} \frac{m(E \cap B(x,r))}{m(B(x,r))} = 1

This means that almost every point of $E$ is "fully surrounded" by $E$ in a measure-theoretic sense, even if $E$ is topologically "thin".

This is a remarkable fact: even sets with complicated boundaries have most of their points as density points.

Corollary 6.4: Differentiation of Convolutions

If $f \in L^1$ and $g$ is a smooth function with compact support, then the convolution $f * g$ is differentiable and

(f * g)'(x) = (f * g')(x)

This follows from the Lebesgue differentiation theorem and properties of convolution.

Remark 6.5: Maximal Functions in Higher Dimensions

The Hardy-Littlewood maximal function extends naturally to $\mathbb{R}^n$ :

Mf(x) = \sup_{r > 0} \frac{1}{m(B(x,r))} \int_{B(x,r)} |f(y)| \, dy

The weak-type (1,1) bound and Lp boundedness extend to higher dimensions with constants depending on the dimension.

The maximal function is a fundamental tool in harmonic analysis, used to prove pointwise convergence results for Fourier series, singular integrals, and other operators.

Example 6.10: Lebesgue Points: Detailed Calculation

Problem: Show that $x = 0$ is not a Lebesgue point for the function $f(x) = |x|^{-1/2}$ on $[-1,1]$ (with $f(0) = 0$ ).

Solution:

For small $r$ , consider the average:

\frac{1}{2r} \int_{-r}^r |f(x) - f(0)| \, dx = \frac{1}{2r} \int_{-r}^r |x|^{-1/2} \, dx = \frac{1}{r} \int_0^r x^{-1/2} \, dx = \frac{2\sqrt{r}}{r} = \frac{2}{\sqrt{r}}

As $r \to 0$ , this tends to infinity, so $x = 0$ is not a Lebesgue point.

This is because $f$ has a singularity at 0 that is too strong for the function to be "well-behaved" in the average sense.

Example 6.11: Maximal Function: Controlling Averages

Problem: Show that for $f \in L^1$ , the maximal function $Mf$ controls the local averages in the sense that if $Mf(x) < \infty$ , then $x$ is a Lebesgue point of $f$ .

Solution:

This is a key property of the maximal function. If $Mf(x) < \infty$ , then the averages $\frac{1}{m(B(x,r))} \int_{B(x,r)} |f|$ are bounded as $r \to 0$ .

This boundedness, combined with the Lebesgue differentiation theorem, ensures that the limit exists and equals $f(x)$ almost everywhere where $Mf(x) < \infty$ .

Since $Mf$ is finite almost everywhere (by the weak-type bound), almost every point is a Lebesgue point.

Lemma 6.2: Covering by Disjoint Balls

Given a finite collection of balls in $\mathbb{R}^n$ , there exists a disjoint subcollection whose union covers at least $1/3^n$ of the total measure of the original collection.

This is a simplified version of the Vitali covering lemma, useful for many geometric arguments.

Proof of Lemma 6.2:

Proof:

Use a greedy algorithm: select the largest ball, remove all balls that intersect it, and repeat.

The key observation is that if a ball is not selected, it must intersect a selected ball, and the selected ball is at least as large, so the 3-times-dilated selected ball covers the non-selected one. ∎

∎

Remark 6.6: Differentiation and Integration: The Complete Picture

The Lebesgue differentiation theorem completes the fundamental theorem of calculus:

Differentiation of integrals: If $F(x) = \int_a^x f$ with $f \in L^1$ , then $F' = f$ almost everywhere.
Integration of derivatives: If $F$ is absolutely continuous, then $F(x) = F(a) + \int_a^x F'$ .
Connection: These two results together show that differentiation and integration are inverse operations (almost everywhere) for absolutely continuous functions.

This is the measure-theoretic version of the fundamental theorem of calculus, more general than the classical version.

Remark 6.1: Key Insights

Key takeaways:

The maximal function controls local averages and is a key tool in analysis
The Lebesgue differentiation theorem generalizes the fundamental theorem of calculus
Almost every point is a Lebesgue point for integrable functions
Vitali covering lemma is essential for geometric measure theory
Density points characterize the 'interior' of measurable sets

Practice Quiz

Differentiation of Integrals & Maximal Control

Questions

Correct

Accuracy

The Hardy-Littlewood maximal function

Mf

is defined as:

Medium

Not attempted

The Lebesgue differentiation theorem states that for

f \in L^1_{\text{loc}}

, almost every

x

is a Lebesgue point, meaning:

Medium

Not attempted

The Hardy-Littlewood maximal theorem states that

M

is:

Hard

Not attempted

The Vitali covering lemma allows us to:

Medium

Not attempted

F(x) = \int_a^x f(t) \, dt

for

f \in L^1

, then almost everywhere:

Medium

Not attempted

A point

x

is a density point of a measurable set

E

if:

Easy

Not attempted

The weak-type (1,1) inequality for the maximal function states:

Hard

Not attempted

The Lebesgue differentiation theorem generalizes:

Easy

Not attempted

f \in L^1

, then the set of non-Lebesgue points has:

Medium

Not attempted

The maximal function is useful because:

Medium

Not attempted

Frequently Asked Questions

What is the Hardy-Littlewood maximal function?

The maximal function $Mf(x)$ is the supremum over all balls centered at $x$ of the average of $|f|$ over those balls. It measures the 'worst-case' local average of a function and is a key tool in analysis.

Why is the maximal function not bounded on L1?

The maximal function can be infinite on sets of positive measure even for L1 functions. However, it satisfies a weak-type (1,1) inequality, which is sufficient for many applications.

What does the Lebesgue differentiation theorem tell us?

It tells us that for locally integrable functions, almost every point is a Lebesgue point, meaning the average value over small balls converges to the function value. This generalizes the fundamental theorem of calculus.

What is a Vitali covering?

A Vitali covering is a collection of balls (or sets) such that every point is contained in arbitrarily small sets from the collection. The Vitali covering lemma allows extracting a disjoint subcollection.

How does the differentiation theorem relate to the fundamental theorem of calculus?

The Lebesgue differentiation theorem shows that if $F(x) = \int_a^x f(t) \, dt$, then $F'(x) = f(x)$ almost everywhere. This is the measure-theoretic version of the fundamental theorem of calculus.

What is a density point?

A density point of a set $E$ is a point $x$ where $E$ occupies nearly all of small balls around $x$ (the density is 1). Almost every point of a measurable set is a density point.

Why are covering lemmas important?

Covering lemmas (like Vitali's) allow us to control measure-theoretic arguments by selecting well-behaved subcollections from coverings. They are essential tools in geometric measure theory and differentiation theory.