Graduate real analysis

David Bate david.bate@warwick.ac.uk Zeeman Building, University of Warwick, Coventry CV4 7AL

Autumn 2021

Part I General measure theory

1. Measures

We wish to assign a value to the size of subsets of some given space, such as the length, area or volume of subsets of $R^{m}$ .

Definition 1.1.

A measure $μ$ on a set $X$ is a function

μ : {A : A \subset X} \to [0, \infty]

such that

$μ (\emptyset) = 0$ ;
$μ (A) \subset μ (B)$ whenever $A \subset B \subset X$ ;
$μ (⋃_{i \in N} A_{i}) \leq \sum_{i \in N} μ (A_{i})$ whenever $A_{1}, A_{2}, \dots \subset X$ .

A function satisfying (2) is said to be monotonic and a function satisfying (3) is said to be countably sub-additive.

Definition 1.2.

Let $μ$ be a measure on a set $X$ . A set $A \subset X$ is $μ$ -measurable if, for every $E \subset X$ ,

(1.1)

μ (E) = μ (E \cap A) + μ (E ∖ A) .

Remark 1.3.

Since a measure is countably sub-additive, it is sufficient to check the $\geq$ inequality in (1.1).
In particular, it suffices to check (1.1) for $E \subset X$ with $μ (E) < \infty$ .
If $A$ is $μ$ -measurable then so is $X ∖ A$ .
If $μ (A) = 0$ then $A$ is $μ$ -measurable.

Definition 1.4.

If $μ$ is a measure on a set $X$ and $S \subset X$ , the restriction of $μ$ to $A$ is defined as

μ |_{S} (A) := μ (S \cap A) .

Lemma 1.5.

Let $μ$ be a measure on a set $X$ and $S \subset X$ . Then $μ |_{S}$ is a measure on $X$ and any $μ$ -measurable set is also $μ |_{S}$ -measurable.

Proof.

The fact that $μ |_{S}$ is a measure follows immediately from the fact that $μ$ is a measure. If $A \subset X$ is $μ$ -measurable, then for any $E \subset X$ ,

	$μ \|_{S} (E) = μ (E \cap S)$	$= μ (E \cap S \cap A) + μ (E \cap S ∖ A)$
		$= μ \|_{S} (E \cap A) + μ \|_{S} (E ∖ A),$

as required. ∎

Theorem 1.6.

Let $μ$ be a measure on a set $X$ and let $M$ be the set of $μ$ -measurable subsets of $X$ .

If $A_{1}, A_{2}, \dots \in M$ then $⋃_{i \in N} A_{i} \in M$ and $⋂_{i \in N} A_{i} \in M$ .
$μ$ is countably additive on $M$ . That is, if $A_{1}, A_{2} \dots \in M$ are disjoint then

$μ (⋃ i \in N A_{i}) = \sum i \in N μ (A_{i}) .$
If $A_{1} \subset A_{2} \subset \dots \in M$ then

$μ (⋃ i \in N A_{i}) = lim i \to \infty μ (A_{i}) .$
If $A_{1} \supset A_{2} \supset \dots \in M$ and $μ (A_{1}) < \infty$ then

$μ (⋂ i \in N A_{i}) = lim i \to \infty μ (A_{i}) .$

Proof.

We first prove (1) for finite unions and intersections. If $A, B \in M$ then for every $E \subset X$ ,

	$μ (E)$	$= μ (E \cap A) + μ (E ∖ A)$
		$= μ (E \cap A) + μ ((E ∖ A) \cap B) + μ (E ∖ (A \cup B))$
		$\geq μ (E \cap (A \cup B)) + μ (E ∖ (A \cup B))$

by sub-additivity. Thus $A \cup B$ is $μ$ -measurable and induction gives finite unions. Taking complements gives finite intersections.

To prove (2) note that the inequality $\leq$ is given by sub-additivity. For the other inequality, for each $i \in N$ let $A_{i} \in M$ be disjoint and for each $j \in M$ let

B_{j} = j ⋃ i = 1 A_{i},

which is measurable by (1). Note that

B_{j} = B_{j - 1} \cup A_{j}

and that this union is disjoint. Therefore, since $A_{j}$ is $μ$ -measurable,

	$μ (B_{j})$	$= μ (B_{j} \cap A_{j}) + μ (B_{j} ∖ A_{j})$
		$= μ (A_{j}) + μ (B_{j - 1}),$

since the $A_{i}$ are all disjoint. Therefore, by induction, $μ (B_{j}) = \sum_{i = 1}^{j} μ (A_{i})$ for each $j \in N$ . Finally, for each $j \in N$ , since $μ$ is monotonic,

μ (⋃ i \in N A_{i}) \geq μ (B_{j}) = j \sum i = 1 μ (A_{i})

and so letting $j \to \infty$ gives (2).

(3) follows by applying (2) to the disjoint measurable sets $B_{j} = A_{j} ∖ A_{j - 1}$ .

(4) follows from (3) by setting $B_{j} = A_{1} ∖ A_{j}$ , so that

A_{1} = ⋂ i \in N A_{i} \cup ⋃ i \in N B_{i}

and the $B_{j}$ increase. By sub-additivity,

	$μ (A_{1})$	$\leq μ (⋂ i \in N A_{i}) + lim j \to \infty μ (B_{j})$
		$= μ (⋂ i \in N A_{i}) + lim j \to \infty μ (A_{1}) - μ (A_{j}),$

by applying (1) for finite unions. Since $μ (A_{1}) < \infty$ , (4) follows.

Finally, to prove (1) for countable unions, for each $j \in N$ let

B_{j} = j ⋃ i = 1 A_{i},

an increasing sequence, and let $E \subset X$ with $μ (E) < \infty$ . Since the $B_{j}$ are $μ$ -measurable,

	$μ (E)$	$= lim j \to \infty μ (E \cap B_{j}) + lim j \to \infty μ (E ∖ B_{j})$
		$= μ (E \cap ⋃ i \in N B_{i}) + μ (E ∖ ⋃ i \in N B_{i})$
		$= μ (E \cap ⋃ i \in N A_{i}) + μ (E ∖ ⋃ i \in N A_{i}),$

using the fact that the $B_{j}$ are $μ |_{E}$ -measurable in the second equality. Taking complements shows that countable intersections of measurable sets are measurable. ∎

Definition 1.7.

A collection $Σ$ of subsets of a set $X$ is a $σ$ -algebra if

$\emptyset \in Σ$ ;
$A \in Σ \Rightarrow X ∖ A \in Σ$ ;
$A_{1}, A_{2}, \dots \in Σ \Rightarrow ⋃_{i \in N} A_{i} \in Σ$ .

Theorem 1.6 shows that the set of $μ$ -measurable sets is a $σ$ -algebra.

For $Ω$ a set of subset of a set $X$ , the $σ$ -algebra generated by $Ω$ is

Σ (Ω) := ⋂ {Σ^{'} : Σ^{'} \supset Ω, Σ^{'} a σ -algebra} .

By creftype 1.2, it is a $σ$ -algebra.

The Borel $σ$ -algebra of a topological space $X$ is the $σ$ -algebra generated by the open (respectively closed) subsets of $X$ . It will be denoted by $B (X)$ and its elements called the Borel subsets of $X$ .

A measure for which all Borel sets are measurable is a Borel measure. It is Borel regular if for every $A \subset X$ there exists a Borel $B \supset A$ with $μ (B) = μ (A)$ .

Theorem 1.8 (Carathéodory criterion).

Let $(X, d)$ be a metric space and $μ$ a measure on $X$ which is additive on separated sets. That is, whenever $A, B \subset X$ with

inf {d (x, y) : x \in A, y \in B} > 0,

we have

μ (A \cup B) = μ (A) + μ (B) .

Then $μ$ is a Borel measure.

Proof.

Let $C \subset X$ be closed and $E \subset X$ with $μ (E) < \infty$ . We need to show

μ (E) \geq μ (E \cap C) + μ (E ∖ C) .

For each $j \in N$ let

E_{j} = {x \in E : \frac{1}{j + 1} < dist (x, C) \leq \frac{1}{j}}

and

E_{0} = {x \in E : dist (x, C) > 1} .

Since $C$ is closed,

E ∖ C = E_{0} \cup ⋃ j \in N E_{j} .

Moreover, the $E_{j}$ with $j$ odd are pairwise separated so

μ (E) \geq μ (⋃ j odd E_{j}) = \sum j odd μ (E_{j})

and so the sum is convergent. Similarly the sum over even indices is convergent and so

\sum j \geq n μ (E_{j}) \to 0 as n \to \infty .

Therefore

	$μ (E)$	$\geq μ (E \cap C \cup n ⋃ j = 0 E_{j})$
		$= μ (E \cap C) + μ (n ⋃ j = 0 E_{j})$
		$\geq μ (E \cap C) + μ (E ∖ C) - \sum j > n μ (E_{j})$
		$\to μ (E \cap C) + μ (E ∖ C),$

using the additivity on separated sets for the equality and countable sub-additivity for the second inequality. ∎

Definition 1.9 (Carathéodory construction).

Let $(X, d)$ be a metric space, $F$ a set of subsets of $X$ and $ζ : F \to [0, \infty]$ . For each $δ > 0$ and $A \subset X$ define

ψ_{δ} (A) = inf \sum S \in G ζ (S),

where the infimum is taken over all countable

G \subset {S \in F : diam (S)} < δ

such that

A \subset ⋃ S \in G S .

Finally, define $ψ (A) = {sup}_{δ > 0} ψ_{δ} (A)$ .

For any $δ > 0$ , $ψ_{δ}$ is a measure, as is $ψ$ . Theorem 1.8 shows that $ψ$ is a Borel measure on $X$ . Indeed, if $dist (A, B) > δ$ then

ψ_{δ} (A \cup B) \geq ψ_{δ} (A) + ψ_{δ} (B) .

If $F$ consists only of Borel sets then $ψ$ is Borel regular.

Remark 1.10.

The fact that $ψ_{δ^{'}} \leq ψ_{δ}$ whenever $δ^{'} \geq δ$ implies that

ψ (A) = lim δ \to 0 ψ_{δ} (A) .

Definition 1.11.

We define some properties of a measure $μ$ on a topological space $X$ .

$μ$ is locally finite if every point in $X$ has a neighbourhood of finite measure.
$μ$ is $σ$ -finite if there exist measurable $X_{i} \subset X$ with $μ (X_{i}) < \infty$ and $X = ⋃_{i \in N} X_{i}$ .
$μ$ is finite if $μ (X) < \infty$ .
A Borel regular measure $μ$ is a Radon measure if
1. $μ (K) < \infty$ for all compact $K \subset X$ ,
2. $μ (A) = sup {μ (K) : K \subset A compact}$ for all Borel $A \subset X$ .
3. $μ (A) = inf {μ (U) : U \supset A open}$ for all Borel $A \subset X$ .

Definition 1.12.

Let $μ$ be a measure on a set $X$ . A property of points in $X$ holds $μ$ almost everywhere (or $μ$ -a.e.) if the set of points for which the property doesn’t hold has $μ$ measure zero.

Definition 1.13.

Let $X, Y$ be sets, $μ$ be a measure on $X$ and let $f : X \to Y$ . The push forward of $μ$ under $f$ , written $f_{#} μ$ is defined by

f_{#} μ (S) = μ (f^{- 1} (S)) .

Definition 1.14.

The Lebesgue measure on $R^{n}$ , denoted $L^{n}$ , is defined using the Carathéodory construction with $F$ the set of cubes and $ζ (Q) = vol (Q)$ . Its measurable sets are called the Lebesgue measurable subsets of $R^{n}$ .

The following lemma is very useful.

Lemma 1.15.

Let $μ$ be a finite measure on a set $X$ and let $S$ be a set of $μ$ -measurable subsets of $X$ . There exists disjoint $S_{i} \in S$ such that any $S \in S$ with

S \subset X ∖ ⋃ i \in N S_{i}

satisfies $μ (S) = 0$ .

In particular, if each $μ$ -measurable subset of $X$ of positive measure contains an element of $S$ of positive measure then we can decompose almost all of $X$ into countably many disjoint elements of $S$ .

Proof.

We find the $S_{i}$ by induction. First let $S^{1} \subset S$ be countable and disjoint such that

μ (\cup S^{1}) \geq sup {μ (\cup S^{'}) : S^{'} \subset S countable and disjoint} - 1 / 1.

Now let $M_{2}$ be the set of all $S^{'} \subset S$ that are countable, disjoint and disjoint from $\cup S^{1}$ . Let $S^{2} \in M_{2}$ be such that

μ (\cup S^{2}) \geq sup {μ (\cup S^{'}) : S^{'} \in M_{2}} - 1 / 2.

Inductively, given countable, disjoint $S^{1}, \dots, S^{i - 1}$ such that each $S^{j}$ and $S^{k}$ are disjoint for $k < j$ , let $M_{i}$ be the set of all $S^{'} \subset S$ that are countable, disjoint and disjoint from $S^{1} \cup \dots \cup S^{i - 1}$ . Let $S^{i} \in M_{i}$ be such that

μ (\cup S^{i}) \geq sup {μ (\cup S^{'}) : S^{'} \in M_{i}} - 1 / i .

We claim that any $S \in S$ with

S \subset X ∖ ⋃ i \in N ⋃ S^{i}

satisfies $μ (S) = 0$ . If not, let $i \in N$ be such that $1 / i < μ (S)$ . Then $T := S^{i} \cup {S} \in M_{i}$ and

μ (\cup T) > sup {μ (\cup S^{'}) : S^{'} \in M_{i}} - 1 / i + 1 / i,

a contradiction. ∎

1.1. Exercises

Exercise 1.1.

Usually in measure theory, a measure is defined as a countably additive function defined on a $σ$ -algebra. However, using our definition is simply a convenience rather than a restriction.

Indeed, suppose $μ$ is a countably additive function defined on a $σ$ -algebra $Σ$ of $X$ with $μ (\emptyset) = 0$ . Show that it can be extended to the power set of $X$ by

¯ ¯ ¯ μ (A) = inf {μ (B) : A \subset B \in Σ}

and that any $B \in Σ$ is $μ$ -measurable. What about

μ - - (A) = sup {μ (B) : A \supset B \in Σ} ?

Conversely, any measure is countably additive when restricted to any $σ$ -algebra of measurable sets.

Exercise 1.2.

Let $Ω$ be a set of subsets of a set $X$ . Show that $Σ (Ω)$ is a $σ$ -algebra. Note that it is the smallest $σ$ -algebra of $X$ containing $Ω$ .

Exercise 1.3.

Show that the following sets are Borel subsets of $R$ : $Q$ , $[0, 1)$ , the set of points in $[0, 1]$ whose first decimal is even.

Let $f : [0, 1] \to [0, 1]$ . Show that the set of points where $f$ is continuous is a Borel set. What about the set of points where $f$ is differentiable?

Exercise 1.4.

Let $X$ be a set and $x \in X$ . The Dirac measure at $x$ is defined as $δ_{x} (A) = 1$ if $x \in A$ , $δ_{x} (A) = 0$ otherwise. Show that $δ_{x}$ is a measure on $X$ . What are its measurable sets?

Define the counting measure on $X$ to be the cardinality (finite or $\infty$ ) of any subset of $X$ . Show that this is a measure. What are its measurable sets?

Exercise 1.5.

For $(X, d)$ a metric space and $s \geq 0$ the $s$ -dimensional Hausdorff measure on $X$ , denoted $H^{s}$ , is defined using the Carathéodory construction with $F$ the set of all sets and $ζ (S) = diam (S)^{s}$ .

Show that $L^{n}$ and $H^{n}$ are non-zero, translation invariant and $n$ -homogenous measures. That is, for any $A \subset R^{n}$ , $x \in R^{n}$ and $t > 0$ , $L^{n} (A + x) = L^{n} (A)$ and $L^{n} (t A) = t^{n} L^{n} (A)$ (and similarly for $H^{n}$ ).
On $R^{n}$ show that there exists a $C > 0$ such that $H^{n} / C \leq L^{n} \leq C H^{n}$ .
Let $f : X \to Y$ be an $L$ -Lipschitz function between two metric spaces. Show that for any $s \geq 0$ and $A \subset X$ ,

$H^{s} (f (A)) \leq L^{s} H^{s} (A) .$
For any metric space $X$ , show that $H^{0}$ is the counting measure on $X$ .
For $0 \leq s < t < \infty$ , suppose that $H^{s} (A) < \infty$ . Show that $H^{t} (A) = 0$ . Hence there exists a single $0 \leq s \leq \infty$ for which $H^{t} (A) = 0$ for all $t > s$ and $H^{t} (A) = \infty$ for all $t < s$ . This $t$ is called the Hausdorff dimension of $A$ , denoted ${dim}_{H} A$ .

Exercise 1.6.

The Cantor set $K \subset [0, 1]$ is defined as follows. Let $K_{0} = [0, 1]$ and for each $i \in N$ let $K_{i}$ be obtained from deleting the “middle third” open interval from each of the intervals in $K_{i - 1}$ . That is, $K_{1} = [0, 1 / 3] \cup [2 / 3, 1]$ ,

K_{2} = [0, 1 / 9] \cup [2 / 9, 1 / 3] \cup [2 / 3, 7 / 9] \cup [8 / 9, 1],

etc. Define $K = ⋂_{i \in N} K_{i}$ . Note that $K$ is compact and hence Borel.

Show that $K$ is uncountable.
Let $s = log 2 / log 3$ . Show that $0 < H^{s} (K) < \infty$ .

In particular, $K$ is an uncountable subset of $R$ with $L^{1} (K) = 0$ .

Exercise 1.7.

Give an examples of $S \subset R^{2}$ with ${dim}_{H} S = 1$ for which $H^{1} |_{S}$ is:

finite,
$σ$ -finite but not finite,
not $σ$ -finite.

Exercise 1.8.

The fundamental properties of measures are those given in Theorem 1.6, in particular countable additivity. It is necessary for us to only require this to be true for measurable sets, as can be seen from the existence of non-measurable sets.

Define a Vitali set as follows. Consider the equivalence relation $\sim$ on $R$ defined by $x \sim y$ iff $x - y \in Q$ . By the density of $Q$ in $R$ , each equivalence class $V_{x}$ intersects $[0, 1]$ . Therefore, by the axiom of choice(!), we may construct a set $V \subset [0, 1]$ consisting of exactly one member of each equivalence class.

Show:

If $p \neq q$ are rational then $p + V$ and $q + V$ are disjoint.
$[0, 1] \subset ⋃ {q + V : q \in Q \cap [- 1, 1]} \subset [- 1, 2]$ .
Show that $L^{1} (V) \neq 0$ .
Deduce that $V$ is not Lebesgue measurable.

Exercise 1.9.

Show that the two extensions given in creftype 1.1 may not agree. For example, after extending Lebesgue measure (restricted to the Borel sets), what are the values of a Vitali set?

Exercise 1.10.

Let $μ$ be a finite Borel measure on a metric space $X$ . Prove that for every Borel $B \subset X$ ,

(1.2)

μ (B) = sup {μ (C) : C \subset B closed}

and

(1.3)

μ (B) = inf {μ (U) : U \supset B open} .

Property (1.2) is called inner regularity by closed sets and (1.3) is called outer regularity by open sets.

Hint: observe that it suffices to show that all Borel sets satisfy (1.2). Show that the set

{B \subset X : B and X ∖ B satisfy (???)}

is a $σ$ -algebra that contains all closed subsets of $X$ .

Show that a $σ$ -finite $μ$ is inner regular by closed sets. Show that a $σ$ -finite $μ$ is outer regular by open sets if there exist open sets $U_{i} \subset X$ with $μ (U_{i}) < \infty$ for all $i \in N$ and $X = ⋃_{i \in N} U_{i}$ . Give an example of a $σ$ -finite $μ$ that is not outer regular by open sets.

Exercise 1.11.

Let $X$ be a complete and separable metric space. Show that any finite Borel measure on $X$ is a Radon measure.

Hint: a metric space is compact if and only if it is complete and totally bounded.

2. Integration

Definition 2.1.

Let $μ$ be a measure on a set $X$ . A simple function is any function of the form

m \sum i = 1 a_{i} χ_{A_{i}},

where each $a_{i} \in R$ and the $A_{i} \subset X$ are disjoint $μ$ -measurable sets. We treat $0 \cdot \infty = 0$ .

Definition 2.2.

Let $μ$ be a measure on a set $X$ and let $f : X \to R^{+}$ . The (lower) integral of $f$ with respect to $μ$ is

\int f d μ := sup {m \sum i = 1 a_{i} μ (A_{i}) : s = m \sum i = 1 a_{i} χ_{A_{i}} \leq f, s simple} .

Definition 2.3.

Let $μ$ be a measure on a set $X$ . A function $f : X \to R$ is $μ$ -measurable if $f^{- 1} ((a, \infty))$ is $μ$ -measurable for every $a \in R$ .

For $f : X \to R$ measurable, let $f^{+} = max {f, 0}$ and $f^{-} = max {- f, 0}$ (both $μ$ -measurable), so that $f = f^{+} - f^{-}$ and $| f | = f^{+} + f^{-}$ . If one of $\int_{X} f^{+} d μ$ and $\int_{X} f^{-} d μ$ are finite, we say that $f$ is $μ$ -integrable and we define the integral of $f$ with respect to $μ$ to be

\int_{X} f d μ = \int_{X} f^{+} d μ - \int_{X} f^{-} d μ .

If only $\int_{X} f^{-} d μ < \infty$ (respectively $\int_{X} f^{+} d μ < \infty$ ) we write $\int_{X} f d μ = \infty$ (respectively $\int_{X} f d μ = - \infty$ ).

Let $X$ be a topological space. A function $f : X \to R$ is a Borel function if $f^{- 1} ((a, \infty))$ is a Borel set for every $a \in R$ .

There are some simple properties of the integral to check, such as linearity and monotonicity. See creftype 2.3.

Linear combinations of measurable functions are measurable, as are limits of measurable functions. See creftype 2.2.

Theorem 2.4 (Fatou’s lemma).

Let $μ$ be a measure on a set $X$ and $f_{k} : X \to [0, \infty]$ $μ$ -measurable. Then

\int_{X} liminf k \to \infty f_{k} d μ \leq liminf k \to \infty \int_{X} f_{k} d μ .

Proof.

Let

s = m \sum i = 1 a_{i} χ_{A_{i}}

be a simple function with

s \leq liminf f_{k} .

for each $x \in A_{i}$ and each $1 \leq i \leq m$ and let $0 < t < 1$ . For each $1 \leq i \leq m$ , the sets

G_{k, i} := {x \in A_{i} : f_{k} (x) \geq t a_{i} for all j \geq k}

monotonically increase to $A_{i}$ as $k$ increases. Therefore

\int f_{k} d μ \geq m \sum i = 1 t a_{i} μ (G_{k, i}) \to n \sum i = 1 t a_{i} μ (A_{i}) .

and hence

n \sum i = 1 t a_{i} μ (A_{i}) \leq liminf k \to \infty \int f_{k} d μ .

Since $0 < t < 1$ is arbitrary, the conclusion follows. ∎

Remark 2.5 (Reverse Fatou).

Suppose that there exists $g \geq 0$ with $\int g d μ < \infty$ and $f_{k} \leq g$ for all $k$ . Then

limsup k \to \infty f_{k} d μ \geq limsup k \to \infty \int_{X} f_{k} d μ .

Indeed, this follows by applying Fatou’s lemma to $g - f_{k}$ .

Theorem 2.6 (Monotone convergence theorem).

Let $μ$ be a measure on a set $X$ and $f_{k} : X \to [0, \infty]$ $μ$ -measurable. Suppose that for every $x \in X$ and all $k \in N$ , $f_{k + 1} (x) \geq f_{k} (x)$ . Then

Proof.

The monotonicity of the integral gives $\leq$ whilst Fatou’s lemma gives $\geq$ . ∎

Theorem 2.7.

Let $μ$ be a measure on $X$ and $f_{n} : X \to R$ $μ$ -measurable such that $f_{n} \to f$ pointwise. Suppose that there exists measurable $g : X \to [0, \infty]$ with $\int g d μ < \infty$ such that $| f_{n} (x) | \leq g (x)$ for all $x \in X$ . Then

\int f_{n} d μ \to \int f d μ .

Proof.

Observe that for all $n \in N$ , $| f - f_{n} | \leq 2 g$ and that $limsup | f - f_{n} | = 0$ . Then by the reverse Fatou lemma,

∣ ∣ ∣ \int f d μ - \int f_{n} d μ ∣ ∣ ∣ \leq \int | f - f_{n} | d μ \to 0.

∎

2.1. Exercises

Exercise 2.1.

For $μ$ a measure on a set $X$ , let $f : X \to R$ be measurable, respectively Borel. Show that the pre-image of any Borel $B \subset R$ is $μ$ -measurable, respectively Borel. Compare this to the definition of a continuous function.

Exercise 2.2.

Let $μ$ be a measure on $X$ and for each $i \in N$ let $f_{i} : X \to R$ be $μ$ -measurable. Show that the functions

limsup i \to \infty f_{i} and liminf i \to \infty f_{i}

are $μ$ -measurable.

Show that a linear combination of $μ$ -measurable functions is $μ$ -measurable. Show that a countable (pointwise) sum of $μ$ -measurable functions is $μ$ -measurable.

Exercise 2.3.

There are some simple properties of the integral to check:

If $f \leq g$ $μ$ -a.e. then

$\int f d μ \leq \int g d μ;$
The integral with respect to $μ$ is a linear operator;
If $S \subset X$ is $μ$ -measurable then

$\int_{X} f d μ = \int_{S} f d μ + \int_{X ∖ S} f d μ;$
$| \int f d μ | \leq \int | f | d μ$ ;
etc…

Exercise 2.4.

Show that $f : X \to R$ is $μ$ -measurable if and only if

μ (E) \geq μ (E \cap f^{- 1} ((- \infty, a))) + μ (E \cap f^{- 1} ((b, \infty)))

for every $E \subset X$ and $a < b \in Q$ .

Exercise 2.5.

State and prove a reverse monotone convergence theorem for monotonically decreasing sequences of functions.

Exercise 2.6.

Show that the Fatou lemma is false if the functions are not uniformly bounded below.

Show that the reverse Fatou lemma is false if the sequence is not bounded above by an integrable $g$ .

Show that the monotone convergence theorem is false if the sequence does not monotonically increase.

Exercise 2.7.

Let $X, Y$ be sets, $μ$ a measure on $X$ and $f : X \to Y$ . Show that $f_{#} μ$ is a measure on $Y$ . If $X, Y$ are topological spaces and $μ, f$ are Borel, show that $f_{#} μ$ is a Borel measure on $Y$ .

3. Some standard theorems

Theorem 3.1 (Egorov’s theorem).

Let $μ$ be a finite measure on a set $X$ and $f_{n} : X \to R$ a sequence of $μ$ -measurable functions such that $f_{n} \to f$ pointwise $μ$ -a.e. Then for every $ϵ > 0$ there exists a measurable $G \subset X$ with $μ (X ∖ G) < ϵ$ and $f_{n} \to f$ uniformly on $G$ .

Proof.

Fix $ϵ > 0$ , $k \in N$ and for each $n \in N$ let

B_{n, k} = {x \in X : | f_{n} (x) - f (x) | > 1 / k for some m \geq n} .

By assumption, the $B_{n, k}$ are measurable sets that monotonically decrease to a $μ$ -null set as $n \to \infty$ . Therefore, there exists $n \in N$ such that $μ (B_{n, k}) < ϵ 2^{- k}$ . Let

G_{k} = X ∖ B_{n, k} and G = ⋂ k \in N G_{k} .

Then $μ (X ∖ G) < ϵ$ and, for each $x \in G$ and each $k \in N$ , $G \subset G_{k}$ , so there exists $n \in N$ such that

| f (x) - f_{m} (x) | < 1 / k

for all $m \geq n$ . That is, $f_{m} \to f$ uniformly on $G$ , as required. ∎

Theorem 3.2 (Lusin’s theorem).

Let $μ$ be a finite Borel measure on a metric space $X$ and let $f : X \to R$ be $μ$ -measurable. Then for every $ϵ > 0$ there exists a closed $C \subset X$ with $μ (X ∖ C) < ϵ$ such that $f |_{C}$ is continuous.

Proof.

Fix $ϵ > 0$ and for each $i \in Z$ let

X_{i} = f^{- 1} ([i ϵ, (i + 1) ϵ)),

a collection of disjoint Borel sets which cover $X$ . Since $μ (X) < \infty$ , there exists $n \in N$ such that

μ (X ∖ n ⋃ i = 1 X_{i}) < ϵ .

Since $μ$ is Borel, for each $1 \leq i \leq n$ there exists a closed $C_{i} \subset X_{i}$ with $μ (X_{i} ∖ C_{i}) < ϵ / n$ .

For a moment fix $1 \leq i \leq n$ and let

D = ⋃ 1 \leq j \neq i \leq n C_{j} .

Since $D$ is closed, the sets $B (D, δ)$ monotonically decrease to $D$ as $δ \to 0$ , which is disjoint from $C_{i}$ . Therefore there exists $δ_{i} > 0$ such that

C_{i}^{'} := C_{i} ∖ B (D, δ_{i})

satisfies $μ (C_{i} ∖ C_{i}^{'}) < ϵ / n$ . Note that $C_{i}^{'}$ is closed.

Let $δ := {min}_{1 \leq i \leq n} δ_{i} > 0$ and $C_{ϵ} = ⋃_{i = 1}^{n} C_{i}^{'}$ , a closed set. For any $1 \leq i \neq j \leq n$ and $x \in C_{i}^{'}$ and $y \in C_{j}^{'}$ , $d (x, y) \geq δ$ . Therefore, if $x, y \in C_{ϵ}$ with $d (x, y) < δ$ , $| f (x) - f (y) | < ϵ$ . Repeat this for each $k \in N$ with $ϵ_{k} = 2^{- k} ϵ / 3$ and let $C = ⋂_{k \in N} C_{ϵ_{k}}$ , so that $μ (X ∖ C) < ϵ$ . Then $f$ is continuous on $C$ . ∎

Definition 3.3.

Let $μ, ν$ be measures on a set $X$ . We say $ν$ is absolutely continuous with respect to $μ$ , written $ν ≪ μ$ , if for every $S \subset X$ , $μ (S) = 0 ⟹ ν (S) = 0$ . We say that $ν$ is singular with respect to $μ$ , written $ν ⊥ μ$ if there exists $A \subset X$ with $μ (X ∖ A) = 0 = ν (A)$ .

Let $μ$ be a measure on a set $X$ and let $f : X \to R^{+}$ . The set valued function

ν (S) = \int_{S} f d μ

defines a measure on $X$ . Note also that $ν ≪ μ$ . The Radon-Nikodym theorem provides the converse to this statement.

Theorem 3.4 (Radon–Nikodym).

Let $μ, ν$ be finite measures on a set $X$ such that $ν ≪ μ$ . There exists a $μ$ and $ν$ -measurable $f : X \to R^{+}$ such that

ν (S) = \int_{S} f d μ

for all $μ$ and $ν$ -measurable $S \subset X$ . The function $f$ is called the Radon-Nikodym derivative of $ν$ with respect to $μ$ .

Proof.

Let $F$ be the set of all $μ$ and $ν$ -measurable $f : X \to R^{+}$ such that

\int_{S} f d μ \leq ν (S)

for all $μ$ and $ν$ -measurable $S \subset X$ . Note that $0 \in F$ and

(3.1)

f, g \in F \Rightarrow max {f, g} \in F .

Let

M = sup {\int f d μ : f \in F},

so that $0 \leq M \leq ν (X) < \infty$ , and let $f_{i} \in F$ be such that

\int f_{i} d μ \to M .

Eq. 3.1 implies that we may suppose the $f_{i}$ monotonically increase. Let $f : X \to R^{+}$ be the pointwise limit of the $f_{i}$ . Then $f$ is $μ$ and $ν$ -measurable and, by the monotone convergence theorem, $f \in F$ and $\int f d μ \geq M$ . Thus

(3.2)

\int f d μ = M .

We claim that $f$ satisfies the conclusion of the proposition. Indeed, suppose that $B \subset X$ is $μ$ -measurable but

ν (B) > \int_{B} f d μ

and let $ϵ > 0$ be such that

(3.3)

ν (B) > \int_{B} f + ϵ d μ .

Let $S$ be the collection of all $μ$ and $ν$ -measurable $S \subset B$ such that

ν (S) \leq \int_{S} f + ϵ d μ .

We claim that there exists a $μ$ and $ν$ -measurable $G \subset B$ of positive $μ$ -measure such that each $μ$ and $ν$ -measurable $G^{'} \subset G$ of positive $μ$ -measure is not contained in $S$ . Indeed, if not, then each $G \subset B$ of positive $μ$ -measure contains an element of $S$ of positive $μ$ -measure. Thus Lemma 1.15 gives a countable disjoint decomposition of $μ$ -almost all of $B$ into elements of $S$ . Since $ν ≪ μ$ this implies

ν (B) \leq \int_{B} f + ϵ d μ,

contradicting (3.3).

Note that $f + ϵ G \in F$ . Indeed, if $S \subset X$ is $μ$ -measurable,

	$\int_{S} f + ϵ G d μ$	$= \int_{S ∖ G} f + \int_{S \cap G} f + ϵ d μ$
		$\leq ν (S ∖ G) + ν (S \cap G)$
		$= ν (S) .$

On the other hand, since $μ (G) > 0$ ,

\int f + ϵ G d μ = M + ϵ μ (G) > M,

contradicting the definition of $M$ . ∎

Theorem 3.5 (Lebesgue decomposition theorem).

Let $μ, ν$ be finite measures on a set $X$ . There exists a $ν$ -measurable $A \subset X$ with $μ (X ∖ A) = 0$ such that, for all $S \subset A$ , $μ (S) = 0 \Rightarrow ν (S) = 0$ . That is, $ν = ν_{a c} + ν_{⊥}$ with $ν_{a c} ≪ μ$ and $ν_{⊥} ⊥ μ$ .

Proof.

Let $S$ be the set of all $ν$ -measurable $S \subset X$ with $μ (S) = 0$ . By Lemma 1.15, there exists $S_{i} \in S$ such that each $S \in S$ with

S \subset A := X ∖ ⋃ i \in N S_{i}

satisfies $ν (S) = 0$ . Since $μ (X ∖ A) = 0$ , this is the required decomposition. ∎

3.1. Exercises

Exercise 3.1.

Let $μ$ be a Borel measure on a metric space $X$ , $f : X \to R$ $μ$ -integrable and $ϵ > 0$ .

Show that there exists a simple function $s$ such that

$\int_{X} | f - s | d μ < ϵ .$
If $f$ is positive show that we may require $0 \leq s \leq f$ in the previous point.
Show that if $μ$ is finite, there exists $g \in C (X)$ with

$\int_{X} | f - g | d μ < ϵ .$
Show that the previous point may fail if $μ$ is only $σ$ -finite.

Exercise 3.2.

Prove the following variant of Lusin’s theorem for the case that $μ$ is not finite but $f$ is $μ$ -integrable: for every $ϵ > 0$ there exists a closed $C \subset X$ with

\int_{X ∖ C} | f | d μ < ϵ

such that $f |_{C}$ is continuous.

Exercise 3.3.

Let $μ, ν$ be measures on a set $X$ . Show that if $A \subset X$ is $μ + ν$ measurable then it is also $μ$ -measurable.

Exercise 3.4.

Prove the Radon-Nikodym and Lebesgue decomposition theorems for $σ$ -finite measures.

Exercise 3.5.

Let $μ, ν$ be finite measures on a set $X$ and suppose $ν ≪ μ$ . For any $μ, ν$ -measurable $g : X \to R^{+}$ show that

\int g d ν = \int g f d ν,

where $f$ is the Radon-Nikodym derivative of $ν$ with respect to $μ$ .

Exercise 3.6.

For a measure $μ$ on a set $X$ , we say a function $f : X \to R^{n}$ is $μ$ -measurable if each component of $f$ is $μ$ -measurable and define the integral of $f$ component by component.

An $n$ -dimensional vector valued measure is a function

ν : {A : A \subset X} \to R^{n}

for which there exists a measure $μ$ on $X$ and a $μ$ -measurable $f : X \to S^{n - 1}$ such that

ν (A) = \int_{A} f d μ

for each $A \subset X$ .

Show that if

$ν = \int f d μ = \int f^{'} d μ^{'}$

are two representations of a vector valued measure then $μ = μ^{'}$ (when restricted to the set of $μ$ -measurable sets), and hence $f = f^{'}$ $μ$ -a.e. We denote this unique measure by $| ν |$ . It is called the total variation of $ν$ .
Show that the set of all vector valued measures form a normed vector space when equipped with

$∥ ν ∥ = | ν | (X) .$
Show that this space is complete.

A signed measure is a 1-dimensional vector valued measure.

Exercise 3.7.

Let $Σ$ be a $σ$ -algebra on $X$ . The standard definition of a signed measure on $Σ$ is a countably additive function

μ : Σ \to R .

The Hahn decomposition theorem states that there exist disjoint $P, N \in Σ$ with $X = P \cup N$ such that:

For every $S \in Σ$ with $S \subset P$ , $μ (P) \geq 0$ and
For every $S \in Σ$ with $S \subset N$ , $μ (P) \leq 0$ .

That is, $μ |_{P}$ and $- μ |_{N}$ are (positive) measures.

Use the Lebesgue decomposition and Radon–Nikodym theorems to show that the two definitions of a signed measure agree.

4. The Daniell integral

Let $μ$ be a measure on a set $X$ and let $L$ be a set of real valued $μ$ -measurable functions on $X$ . The formula

T (f) := \int_{X} f d μ

defines an operator on $L$ . Moreover, it has the following two properties:

$T$ is monotonic: for all $f, g \in L$ with $g \leq f$ , $T (g) \leq f$ ;
$T$ is continuous with respect to monotone convergence: if $f_{i} \in C (X)$ monotonically increase to $f$ then $T (f_{i}) \to T (f)$ .

In the next theorem we see that these two properties completely characterise the Lebesgue integral. That is, we could equivalently develop a theory of integration (and hence measure) by beginning with operators on sets of functions.

Definition 4.1.

Let $X$ be a set. A lattice of functions on $X$ is a non-empty set $L$ of functions $X \to R$ which satisfies the following conditions: for any $c \in R^{+}$ and any $f, g \in L$ , $f + g$ , $c f$ , $inf {f, g}$ and $inf {f, c}$ all belong to $L$ and if $g \leq f$ then $f - g \in L$ too. Note that any vector space of functions closed under $inf$ is a lattice.

If $L$ is a lattice we let

L^{+} = {f \in L : f \geq 0} .

We say a $T : L \to R$ is

linear if, for all $f, g \in L$ and $a, b \in R^{+}$ ,

$T (a f + b g) = a T (f) + b T (g);$
monotonic if, for all $f, g \in L$ with $g \leq f$ ,

$T (g) \leq T (f);$
continuous with respect to monotone convergence if, for all $f_{i} \in L$ that monotonically increase to $f$ ,

$T (f_{i}) \to T (f) .$
bounded if for every $f \in L$ ,

$sup {T (g) : 0 \leq g \leq f} < \infty .$

A $T$ satisfying Items 3, 2 and 1 is called a monotone Daniell integral (it necessarily satisfies Item 4). A $T$ satisfying Items 3, 4 and 1 is called a Daniell integral.

Theorem 4.2.

Let $L$ be a lattice on $X$ and let $T : L \to R$ be a monotone Daniell integral. Then there exists a measure $μ$ on $X$ for which each $f \in L^{+}$ is $μ$ -measurable such that

(4.1)

T (f) = \int f d μ

for all $f \in L$ .

Proof.

First note that, for any $f \in L^{+}$ ,

T (f) \geq T (0 \cdot f) = 0.

For $A \subset X$ we say that a sequence $f_{i} \in L^{+}$ suits $A$ if the $f_{i}$ monotonically increase and

lim i \to \infty f_{i} (x) \geq 1 \forall x \in A .

Define

μ (A) = inf {lim i \to \infty T (f_{i}) : f_{i} suits A} .

Then $μ$ is a measure on $X$ . Indeed, $μ (\emptyset) = 0$ and $μ$ is monotonic. If $A_{j} \subset X$ for each $j \in N$ and $f_{i}^{j} \in L^{+}$ suit $A_{j}$ then

g_{i} := i \sum j = 1 f_{i}^{j}

suits $\cup_{j} A_{j}$ and

T (g_{i}) = i \sum j = 1 T (f_{i}^{j}) \leq \sum j \in N lim i \to \infty T (f_{i}^{j}) = \sum j \in N μ (A_{j}) .

Next we show that each $f \in L^{+}$ is $μ$ -measurable. By creftype 2.4, it suffices to show, for every $E \subset X$ and $0 \leq a < b \in R$ that, for

A := f^{- 1} ((- \infty, a)), B := f^{- 1} ((b, \infty)),

we have

μ (E) \geq μ (E \cap A) + μ (E \cap B) .

Suppose $g_{i}$ suit $E$ and let

h = \frac{inf {f, b} - inf {f, a}}{b - a}, k_{i} = inf {g_{i}, h} .

Then $0 \leq k_{i + 1} - k_{i} \leq g_{i + 1} - g_{i}$ for all $i$ and

h (x) = 1 whenever f (x) \geq b, h (x) = 0 whenever f (x) \leq a .

Then $k_{i}$ suit $B$ and $g_{i} - k_{i}$ suit $A$ . Therefore

lim i \to \infty T (g_{i}) = lim i \to \infty T (k_{i}) + T (g_{i} + k_{i}) \geq μ (B) + μ (A) .

Finally we show that

(4.2)

T (f) = \int f d μ f \in L^{+} .

First suppose that $A \subset X$ , $f_{i}$ suit $A$ and $g \in L^{+}$ satisfies $g \leq χ_{A}$ . Then $h_{i} = inf {f_{i}, g}$ monotonically increase to $g$ and so

T (g) = lim i \to \infty T (h_{i}) \leq lim i \to \infty T (f_{i}) .

Consequently,

(4.3)

T (g) \leq μ (A) .

Now fix $f \in L^{+}$ and for each $t \in R^{+}$ let $f_{t} = inf {f, t}$ . For a moment fix $ϵ > 0$ . Then, for each $k \in N$ ,

(4.4)

0 \leq f_{k ϵ} (x) - f_{(k - 1) ϵ} (x) \leq ϵ \forall x \in X,

(4.5)

f_{k ϵ} (x) - f_{(k - 1) ϵ} (x) = ϵ whenever f (x) \geq k ϵ

and

(4.6)

f_{k ϵ} (x) - f_{(k - 1) ϵ} (x) = 0 whenever f (x) \leq (k - 1) ϵ .

Note that, for any $k \in N$ , $(f_{k ϵ} - f_{(k - 1) ϵ}) / ϵ$ suits

{x : f (x) \geq k ϵ}

and so

(4.7)

T (f_{k ϵ} - f_{(k - 1) ϵ}) \geq ϵ μ ({x : f (x) \geq k ϵ}) .

By Eqs. 4.6 and 4.4 and Eq. 4.5 respectively,

	$ϵ μ ({x : f (x) \geq k ϵ})$	$\geq \int f_{(k + 1) ϵ} - f_{k ϵ} d μ$
(4.8)			$\geq ϵ μ ({x : f (x) \geq (k + 1) ϵ}) .$

Finally by (4.3),

(4.9)

ϵ μ ({x : f (x) \geq (k + 1) ϵ}) \geq T (f_{(k + 2) ϵ} - f_{(k + 1) ϵ})

Combining Eqs. 4.8, 4.9 and 4.7 gives

T (f_{k ϵ} - f_{(k - 1) ϵ}) \geq \int f_{(k + 1) ϵ} - f_{k ϵ} d μ \geq T (f_{(k + 2) ϵ} - f_{(k + 1) ϵ}) .

Summing from 1 to $i$ gives

T (f_{i ϵ}) \geq \int f_{(i + 1) ϵ} - f_{ϵ} d μ \geq T (f_{(i + 2) ϵ} - f_{2 ϵ}) .

Since $f_{i ϵ}$ monotonically increases to $f$ ,

T (f) \geq \int f - f_{ϵ} d μ \geq T (f - f_{ϵ}) .

Therefore, since $f_{ϵ}$ monotonically decreases to 0, $f - f_{ϵ}$ increases to $f$ and so this gives (4.2).

Note that, if $f \in L$ , then $f^{+}, f^{-} \in L^{+}$ and so

T (f) = T (f^{+}) - T (f^{-}) = \int f^{+} d μ - \int f^{-} d μ = \int f d μ .

∎

Observation 4.3.

For any measure $μ$ satisfying the conclusion of Theorem 4.2, the value of $μ ({f > t})$ , for $f \in L^{+}$ and $t > 0$ , is uniquely determined by the values of $T$ on $L^{+}$ .

Proof.

For any $t > 0$ and $0 < h < t$ , observe that the functions

g_{h} := \frac{inf {f, t + h} - inf {f, t}}{h}

converge pointwise to the characteristic function of $f^{- 1} ((t, \infty))$ and are bounded above by $2 f$ . Therefore, by (4.1) and the dominated convergence theorem,

μ ({x : f (x) > t}) = lim h \to 0 h^{- 1} T (inf {f, t + h} - inf {f, t}) .

∎

The Banach–Alaoglu theorem (see creftype 4.8) motivates us to consider representations of Borel measures by elements of the dual of a Banach space, namely of $C (X)$ . To use Theorem 4.2, we must upgrade the pointwise convergence in the hypotheses to uniform convergence in $C (X)$ . Recall that $C_{c} (X)$ is the set of all continuous functions on $X$ with compact support, and that pointwise monotonic convergence in $C_{c} (X)$ implies uniform convergence (see creftype 4.4). Therefore, for any monotonic $T \in C_{c} (X)^{'}$ , Theorem 4.2 produces a measure $μ$ that represents $T$ . If $X$ is a metric space, then all compact sets are measurable with respect to $μ$ . We next show, on locally compact spaces, how to obtain a Borel measure.

Lemma 4.4.

Let $X$ be a locally compact metric space and suppose that $τ$ is a finite measure on $X$ . There exists a unique Radon measure $μ$ on $X$ that agrees with $τ$ on $K (X)$ .

Proof.

Let $U$ be the set of open subsets of $X$ . For each $U \in U$ define

ν (U) = sup {τ (K) : K \subset U compact} .

Since $τ$ is monotone, for any $K \in K (X)$ , $ν (K^{o}) \leq τ (K)$ , for $K^{o}$ the interior of $K$ .

Since $X$ is locally compact, for any $U \in U$ and $K \in K (X)$ with $K \subset U$ , there exists $V \in U$ with $¯ ¯¯ ¯ V \in K (X)$ and

K \subset V \subset ¯ ¯¯ ¯ V \subset U .

Consequently,

(4.10)

ν (U) = sup {ν (V) : V \in U, V \subset U, ¯ ¯¯ ¯ V \in K (X)} .

Also, since $τ$ is a measure and $X$ is a metric space,

(4.11)

τ (K) = inf {τ (S) : K \subset S^{o}, S \in K (X)} .

For each $A \subset X$ define

μ (A) = inf {ν (U) : U \supset A open} .

Note that both $ν$ and $μ$ are monotone and give value 0 to the empty set. To see that $μ$ is a measure, let $A_{i} \subset X$ and let $U_{i} \supset A_{i}$ be open. We must show that

μ (⋃ i \in N A_{i}) \leq \sum i \in N μ (A_{i}) .

It suffices to show that

ν (⋃ i \in N U_{i}) \leq \sum i \in N ν (U_{i}) .

By (4.10), for any $ϵ > 0$ , there exists $V \subset \cup_{i} U_{i}$ with compact closure such that

ν (V) \geq ν (⋃ i \in N U_{i}) - ϵ

Since $V$ has compact closure, it is contained in the union of finitely many $U_{i}$ . Therefore, it suffices to show that $ν$ is finitely sub-additive.

Let $U, V \in U$ and $K \subset U \cup V$ be compact. Let $W = U \cap V$ and

K_{U} := {x \in K : d (x, U ∖ W) \leq d (x, V ∖ W)}

and

K_{V} := {x \in K : d (x, V ∖ W) \leq d (x, U ∖ W)} .

Then $K_{U}, K_{V}$ are closed subsets of $K$ with $K = K_{U} \cup K_{V}$ and $K_{U} \subset U$ and $K_{V} \subset V$ . Since $τ$ is finitely sub-additive,

τ (K) \leq τ (K_{U}) + τ (K_{V})

and so

ν (U \cup V) \leq ν (U) + ν (V) .

Therefore, $ν$ is finitely sub-additive by induction. As shown above, this implies that $μ$ is a measure.

To see that $μ$ is a Borel measure, if $A, B \subset X$ are separated, then there exist separated open sets $U \supset A$ , $V \supset B$ . Since $τ$ is finitely additive, $ν (U \cap V) = ν (U) + ν (V)$ and so $μ$ is additive on separated sets. By construction, $μ$ is Borel regular. Also, (4.10) shows that open sets are inner regular by compact sets. Combining this with outer regularity by open sets shows that $μ$ is a Radon measure.

To see that $μ$ agrees with $τ$ on $K (X)$ , note that for any $K \in K (X)$ and $U \in U$ with $K \subset U$ ,

τ (K) \leq ν (U) = μ (U)

and so $τ (K) \leq μ (K)$ . For the other inequality,

	$μ (K)$	$= inf {ν (U) : U \supset K open}$
		$\leq inf {ν (S^{o}) : S \in K (X), K \subset S^{o}}$
		$\leq inf {τ (S) : S \in K (X), K \subset S^{o}} = τ (K)$

by (4.11).

Finally, if $μ_{1}, μ_{2}$ are two Radon measures that agree with $τ$ on $K (X)$ , then for any open $U \subset X$ ,

	$μ_{1} (U)$	$= sup {μ_{1} (K) : K \subset U, K \in K (X)}$
		$= sup {μ_{2} (K) : K \subset U, K \in K (X)} = μ_{2} (U) .$

Since $μ_{1}, μ_{2}$ are both Borel regular, they must agree. ∎

Theorem 4.5 (Riesz representation theorem).

Let $X$ be a locally compact metric space and let $T \in C_{c} (X)^{'}$ be monotone. Then there exists a unique Radon measure $μ$ such that

(4.12)

T (f) = \int f d μ \forall f \in C_{c} (X) .

Proof.

Observe that $T$ is a monotone Daniell integral on $L = C_{c} (X)$ . By Theorem 4.2, there exists a measure $τ$ on $X$ for which (4.12) holds with $μ$ replaced by $τ$ . By Lemma 4.4, there exists a Radon measure $μ$ that agrees with $τ$ on $K (X)$ . For any $f \in C_{c} (X)$ , the value of

\int f d τ

is determined by the value of $τ$ on compact sets, and so

\int f d τ = \int f d μ,

so that (4.12) holds. creftype 4.3 implies that, on a locally compact space, the measure of any compact set is uniquely determined by (4.12). Thus any Radon measure satisfying (4.12) is uniquely determined. ∎

The Riesz representation theorem allows us to identify the set of finite Radon measures on a locally compact space $X$ with the set of monotonic elements of $C_{0} (X)^{'}$ . For $B$ a Banach space, a sequence $T_{n} \in B^{'}$ weak* converges to $T \in B^{'}$ if $T_{n} (x) \to T (x)$ for every $x \in B$ . In $C_{0} (X)^{'}$ , this translates to the following.

Definition 4.6.

Let $X$ be a locally compact metric space. A sequence $μ_{n}$ of finite Radon measures on $X$ weak* converges to a finite Radon measure $μ$ if, for every $f \in C_{c} (X)$ ,

\int_{X} f d μ_{n} \to \int_{X} f d μ .

By the Banach-Alaoglu theorem (see creftype 4.8), the unit ball of $B^{'}$ is weak* compact. Since the weak* limit of a sequence of monotonic operators on $C_{0} (X)$ is monotonic, we have the following.

Theorem 4.7.

Let $X$ be a locally compact metric space and $μ_{n}$ a sequence of Radon measures with uniformly bounded total measures. There exists a finite Radon measure $μ$ on $X$ and subsequence $μ_{n_{k}}$ that weak* converges to $μ$ .

4.1. Exercises

Exercise 4.1.

Let $L$ be a lattice of functions on $X$ . For any $f \in L$ show that $f^{+}, f^{-} \in L^{+}$

Exercise 4.2.

Let $L$ be a lattice on $X$ and $T$ a monotone Daniell integral on $L$ . Give an example to show that there may be more than one measure satisfying (4.1) (recall creftype 1.9). Compare to creftype 4.3.

Exercise 4.3.

Let $L$ be a lattice on $X$ and $T$ a Daniell integral on $L$ . Define $T^{+}, T^{-}$ on $L^{+}$ by

T^{+} (f) = sup {T (g) : g \in L^{+}, g \leq f}

and

T^{-} (f) = - inf {T (g) : g \in L^{+} g \leq f} .

There show that $T^{+}, T^{-}$ are monotone Daniell integrals on $L$ .
If $f, g \in L^{+}$ with $g \leq f$ then $f \geq f - g \in L^{+}$ and so

$T (g) - T^{-} (f) \leq T (g) + T (f - g) \leq T (g) + T^{+} (f) .$
Deduce that $T = T^{+} - T^{-} .$

Exercise 4.4.

Let $K$ be a compact metric space and suppose that $f_{i} \in C (K)$ monotonically increase to $f \in C (K)$ . Show that $f_{i} \to f$ uniformly.

Exercise 4.5.

Let $c$ be the set of all $f : N \to R$ such that ${lim}_{j} f (x_{j})$ exists and define $T : c \to R$ by $T (f) = {lim}_{j} f (x_{j})$ . Let $C$ be the set of all bounded $f : N \to R$ . Equip $c$ and $C$ with the supremum norm.

Observe that $T$ is linear and continuous on $c$ and hence can be extended by the Hahn-Banach theorem to a linear and continuous element of $C$ . (Such an extension is called a Banach limit.)
Show that any finite Borel measure on $N$ is a convergent sum of Dirac masses.
Hence show that there is no Borel measure $μ$ on $R$ such that $T_{μ} = T$ .

Exercise 4.6.

Adapt the previous exercise to show that the Riesz representation theorem is false in non locally compact metric spaces.

Exercise 4.7.

Let $X$ be a metric space. Show that any $T \in C_{c} (X)^{'}$ is a Daniell integral.

Let let $T^{+}, T^{-} \in C_{c} (X)^{' +}$ be obtained from creftype 4.3.
Show that $∥ T ∥ = ∥ T^{+} ∥ + ∥ T^{-} ∥$ .
By Theorem 4.5, any $T \in C_{0} (X)$ can be identified with two measures $μ^{+}$ and $μ^{-}$ and hence with a signed measure (recall creftype 3.6). Show that this identification is an isometric isomorphism.

Exercise 4.8.

Let $B$ be a separable Banach space and $D$ a countable dense subset of $B$ . Suppose that $T_{n} \in B^{'}$ satisfy $∥ T_{n} ∥ \leq M$ for some $M > 0$ .

Show that there exists a subsequence $T_{n_{j}}$ and a $T \in B^{'}$ such that $T_{n_{j}} (d) \to T (d)$ for each $d \in D$ .
Deduce that, for any $x \in B$ , $T_{n_{j}} (x) \to T (x)$ .

That is, closed and bounded subsets of $B^{'}$ are weak* compact.

5. Fubini’s theorem

Definition 5.1.

Let $μ, ν$ be measures on sets $X, Y$ respectively and define the set of rectangles to be

R = {A \subset X : A μ -measurable} \otimes {B \subset Y : B ν -measurable} .

The product measure $μ \times ν$ on $X \times Y$ is defined by

μ \times ν (S) = inf \sum i \in N μ (A_{i}) ν (B_{i}),

where the infimum is taken over all countable collections of rectangles $A_{i} \times B_{i} \in R$ with

S \subset ⋃ i \in N A_{i} \times B_{i} .

This is a measure, see creftype 5.1.

Lemma 5.2.

Let $X, Y$ be sets and $μ, ν$ measures on $X, Y$ respectively. Then $μ \times ν$ is equivalently defined by the formula

μ \times ν (S) = inf \sum i \in N μ (A_{i}) ν (B_{i}),

where the infimum is taken over all disjoint countable collections of rectangles $A_{i} \times B_{i} \in R$ with

S \subset ⋃ i \in N A_{i} \times B_{i} .

Proof.

If $A, C$ are $μ$ -measurable and $B, D$ are $ν$ -measurable then

	$(A \times B) ∖ (C \times D)$	$= [(A ∖ C) \times B] \cup [(A \cap C) \times (B ∖ D)]$
		$:= A_{1} \times B_{1} \cup A_{2} \times B_{2}$

is a decomposition into disjoint rectangles. Since $A, C$ and $B, D$ are $μ$ and $ν$ measurable respectively,

	$μ (A_{1}) ν (B_{1}) + μ (A_{2}) ν (B_{2})$	$= [μ (A) - μ (A \cap C)] ν (B) + μ (A \cap C) [ν (B) - ν (B \cap D)]$
		$= μ (A) ν (B) - μ (A \cap C) ν (B \cap D)$
		$\leq μ (A) ν (B) .$

Thus the two formulae agree. ∎

Theorem 5.3 (Fubini’s theorem).

Let $X, Y$ be sets and $μ, ν$ $σ$ -finite measures on $X, Y$ respectively.

If $A$ is $μ$ -measurable and $B$ is $ν$ -measurable then $A \times B$ is $μ \times ν$ -measurable and

$μ \times ν (A \times B) = μ (A) ν (B) .$
If $S$ is $μ \times ν$ -measurable then

$S^{y} := {x \in X : (x, y) \in S}$

is $μ$ -measurable for $ν$ -a.e. $y \in Y$ , $y \mapsto μ (S^{y})$ is $ν$ -measurable;

$S_{x} := {y \in Y : (x, y) \in S}$

is $ν$ -measurable for $μ$ -a.e. $x \in X$ , $x \mapsto ν (S_{x})$ is $μ$ -measurable; and

$μ \times ν (S) = \int_{Y} μ (S^{y}) d ν (y) = \int_{X} ν (S_{x}) d μ (x) .$
If $f : X \times Y \to R^{+}$ is $μ \times ν$ -measurable or $f : X \times Y \to R$ is $μ \times ν$ integrable then

$\int_{X \times Y} f d μ \times ν = \int_{X} \int_{Y} f d ν (y) d μ (x) = \int_{Y} \int_{X} f d μ (x) d ν (y) .$

Proof.

Note that (3) follows from (2) by the monotone convergence theorem. We prove the theorem for finite $μ, ν$ .

To begin, let

U := {⋃ i \in N R_{i} : R_{i} \in R % pairwise disjoint} .

Let $P$ be the set of $S \subset X \times Y$ such that $y \mapsto μ (S^{y})$ is $ν$ -measurable and for any $S \in P$ define

ρ (S) := \int_{Y} μ (S^{y}) d ν .

Observe that $ρ (S)$ is monotonic in $S$ .

If $S = A \times B \in R$ then $y \mapsto μ (S^{y}) = μ (A) χ_{B}$ is $ν$ -measurable and $ρ (S) = μ (A) ν (B)$ . If $U \in U$ with

U = ⋃ i \in N A_{i} \times B_{i}

a disjoint union, then

y \mapsto μ (U^{y}) = \sum i \in N μ (A_{i}) χ_{B_{i}}

is a countable sum of $ν$ -measurable functions and so $U \in P$ . Moreover,

(5.1)

ρ (U) = \int_{Y} μ (U^{y}) d ν (y) = \sum i \in N μ (A_{i}) ν (B_{i}) .

Thus, for any $S \subset X \times Y$ ,

(5.2)

μ \times ν (S) = inf {ρ (U) : S \subset U \in U} .

To prove (1), let $A$ be $μ$ -measurable and $B$ be $ν$ -measurable. By definition, $ρ (A \times B) = μ (A) ν (B)$ and, since $ρ$ is monotonic, $ρ (A \times B) \leq ρ (U)$ whenever $A \times B \subset U \in U$ . Thus, by (5.2),

μ \times ν (A \times B) = μ (A) ν (B) .

Let $E \subset X \times Y$ and $E \subset U \in U$ . Observe

U \cap (A \times B) and U ∖ (A \times B)

are disjoint members of $U$ . Therefore, by (5.1) and (5.2),

	$ρ (U)$	$= ρ (U \cap (A \times B)) + ρ (V ∖ (A \times B))$
		$\geq μ \times ν (E \cap (A \times B)) + μ \times ν (E ∖ (A \times B)) .$

When taking the infimum over all such $U$ , the left hand side converges to $μ \times ν (E)$ , and so $A \times B$ is $μ \times ν$ -measurable. This also implies that all elements of $U$ are measurable.

Let $S \subset X \times Y$ and suppose that $U_{1}, U_{2}, \dots \in U$ are such that $ρ (U_{i}) \to μ \times ν (S)$ . Since the intersection of any two rectangles is a rectangle,

V_{i} := U_{i} \cap U_{i - 1} \cap \dots U_{1} \in U

for each $i \in N$ . Moreover, $ρ (V_{i})$ monotonically decreases to $μ \times ν (S)$ . Let $W = ⋂_{i \in N} V_{i} \supset S$ . Note that, for each $y \in Y$ , $μ (V_{i}^{y})$ monotonically decreases to $μ (W^{y})$ , so that $W \in P$ . Since $μ, ν$ are finite, the monotone convergence theorem implies that $ρ (V_{i}) \to ρ (W)$ and hence $μ \times ν (S) = ρ (W)$ . Since $μ \times ν (W) \geq μ \times ν (S)$ , the monotonicity of $ρ$ implies $μ \times ν (W) = ρ (W)$ .

To prove (2), if $S$ is $μ \times ν$ -measurable, then $μ \times ν (W ∖ S) = 0$ . By (5.2) there exists $Z \supset W ∖ U$ with $ρ (Z) = 0$ . That is, $μ (W^{y}) = μ (S^{y})$ for $ν$ -a.e. $y \in Y$ , and hence the first conclusion of (2). Moreover, $ρ (S) = ρ (W) = μ \times ν (S)$ , which concludes the proof. ∎

5.1. Exercises

Exercise 5.1.

Let $X, Y$ be sets and $μ, ν$ measures on $X, Y$ respectively. Prove that $μ \times ν$ is a measure on $X \times Y$ .

Exercise 5.2.

Let $X, Y$ be separable metric spaces. Show that

B (X \times Y) = Σ (B (X) \otimes B (Y)) .

Exercise 5.3.

Let $X, Y$ be separable metric spaces and $μ, ν$ finite Borel measures on $X, Y$ respectively. Show that $μ \times ν |_{B (X \times Y)}$ is the unique countably additive set function on $B (X \times Y)$ satisfying $μ (A \times B) = μ (A) \times ν (B)$ for all $A \times B \in B (X) \otimes B (Y)$ .

Exercise 5.4.

Prove Fubini’s theorem and Exercises 5.2 and 5.3 for $σ$ -finite measures $μ, ν$ .

Exercise 5.5.

Show that Theorem 5.3 (3) may fail if

$f : X \times Y \to R$ is measurable but not integrable; Hint: consider $μ, ν$ the counting measure on $N$ . Exploit the “identity”

$(1 - 1) + (1 - 1) + \dots = 1 + (- 1 + 1) + (- 1 + 1) + \dots$
$f : X \times Y \to R^{+}$ is measurable but $μ$ is not $σ$ -finite. Hint: consider $μ = L^{1}$ , $ν$ the counting measure on $R$ .

Prove Theorem 5.3 (3) for $f : X \times Y \to R$ integrable, even if $μ, ν$ are not $σ$ -finite.

Exercise 5.6.

Note in Theorem 5.3 (2) we must exclude a set of measure zero. Indeed, if $V$ is a Vitali set, note that $V \times {0} \subset R^{2}$ is $L^{2}$ -measurable.

Exercise 5.7.

Let $X$ be a separable metric space and $f : X \to [0, \infty)$ a Borel function. Prove that

\int_{X} f d μ = \int_{0}^{\infty} μ ({x \in X : f (x) \geq t}) d t .

Hint: consider

A = {(x, t) : f (x) \geq t} .

Exercise 5.8.

A measure $μ$ on a metric space $X$ is uniformly distributed if there exists a $g : (0, \infty) \to (0, \infty)$ such that $ν (B (x, r)) = g (r)$ for all $x \in X$ and $r > 0$ . Let $μ, ν$ be uniformly distributed Borel regular measures on a separable metric space $X$ (with functions $g$ and $h$ respectively). Let $U \subset X$ be open.

Observe that, for any $x \in U$ ,

$lim r \to 0 \frac{ν (U \cap B (x, r))}{h (r)} = 1$

for every $x \in U$ .
Deduce that

$μ (U) \leq liminf r \to 0 h (r)^{- 1} \int_{U} ν (U \cap B (x, r)) d μ (x) .$
Deduce that

$μ (U) \leq liminf r \to 0 h (r)^{- 1} \int_{U} μ (U \cap B (y, r)) d ν (y) = liminf r \to 0 \frac{g (r)}{h (r)} ν (U) .$

Deduce that $μ = c ν$ for some $c > 0$ .

6. Covering theorems

We will use $B (x, r)$ to denote the closed ball in a metric space $X$ centred at $x \in X$ with radius $r \geq 0$ . Since the centre and radius of a ball are not uniquely defined by its elements, formally by a “ball” we mean a pair $(x, r) \in X \times (0, \infty)$ , but in practice we mean the set of its elements.

Lemma 6.1 (Vitali covering lemma).

Let $X$ be a metric space and $B$ an arbitrary collection of closed balls of uniformly bounded radii. There exists a disjoint sub-collection $B^{'} \subset B$ such that any $B \in B$ intersects a ball $B^{'} \in B^{'}$ with

rad B^{'} \geq rad B / 2.

In particular,

⋃ B \in B^{'} 5 B \supset ⋃ B \in B B .

Here, $5 B$ denotes the ball with the same centre as $B$ and 5 times the radius.

Proof.

For each $n \in Z$ let

B_{n} = {B \in B : 2^{n} \leq rad B < 2^{n + 1}} .

Since the balls in $B$ have uniformly bounded radii, the $B_{n}$ are empty for all $n > N$ , for some $N \in N$ . Let $B_{N}^{'}$ be a maximal disjoint sub-collection of $B_{N}$ . That is, the elements of $B_{N}^{'}$ are disjoint elements of $B_{N}$ and if $B \in B_{N}$ , there exists a $B^{'} \in B_{N}^{'}$ with $B \cap B^{'} \neq \emptyset$ . (In general such a maximal collection exists by Zorn’s lemma. See also creftype 6.1.) Let $B_{N - 1}^{'}$ be a maximal collection such that $B_{N}^{'} \cup B_{N - 1}^{'}$ is a disjoint collection. Repeat this for each $i \in N$ , obtaining a maximal collection $B_{N - i}^{'}$ such that $B_{N}^{'} \cup \dots \cup B_{N - i}^{'}$ is a disjoint collection, and set $B^{'} = ⋃_{n \leq N} B_{n}^{'}$ .

Now suppose that $B \in B$ , say $B \in B_{n}$ . Then by construction there exists $B^{'} \in B_{m}^{'}$ for some $m \geq n$ with $B \cap B^{'} \neq \emptyset$ . In particular, $rad B^{'} \geq rad B / 2$ .

The final statement of the lemma follows from the triangle inequality. ∎

Definition 6.2.

Let $X$ be a metric space and $S \subset X$ . A Vitali cover of $S$ is a collection $B$ of closed balls such that, for each $x \in S$ and each $ϵ > 0$ , there exists a ball $B \in B$ with $rad B < ϵ$ and $x \in B$ .

Proposition 6.3.

Let $X$ be a metric space, $S \subset X$ and suppose that $B$ is a Vitali cover of $S$ . Then there exists a disjoint $B^{'} \subset B$ such that, for every finite $I \subset B^{'}$ ,

S ∖ ⋃ B \in I B \subset ⋃ B \in B^{'} ∖ I 5 B .

In particular, if $B^{'} = {B_{1}, B_{2}, \dots}$ is countable (for example, if $X$ is separable), then

S ∖ n ⋃ i = 1 B_{i} \subset ⋃ i > n 5 B_{i}

for each $n \in N$ .

Proof.

Note that we may suppose $B$ consists of balls with uniformly bounded radii. Let $B^{'}$ be a disjoint sub-collection of $B$ obtained from Lemma 6.1. If $I \subset B^{'}$ is finite then

C := ⋃ B \in I B

is closed. Therefore, if $x \in S ∖ C$ , since $B$ is a Vitali cover of $S$ , there exists $B \in B$ with $x \in B$ such that $B \cap C = \emptyset$ . However, $B$ must intersect some $B^{'} \in B^{'}$ with $rad B^{'} \geq rad B / 2$ , and so $x \in 5 B^{'}$ . That is, $x$ belongs to

⋃ B \in B^{'} ∖ I 5 B,

as required. ∎

Definition 6.4.

A Borel measure $μ$ on a metric space $X$ is a doubling measure if there exists a $C_{μ} \geq 1$ such that

0 < μ (2 B) \leq C_{μ} μ (B) < \infty

for all balls $B \subset X$ .

Remark 6.5.

Note that, for any $m \geq 2$ ,

μ (m B) \leq C_{μ}^{{log}_{2} m} μ (B) .

Lebesgue measure is a doubling measure.

Theorem 6.6 (Vitali covering theorem).

Let $μ$ be a doubling measure on a metric space $X$ and let $B$ be a Vitali cover of a set $S \subset X$ . There exists a countable disjoint $B^{'} \subset B$ such that

μ ⎛ ⎝ S ∖ ⋃ B \in B^{'} B ⎞ ⎠ = 0.

Proof.

First note that it suffices to prove the result for $S$ bounded, say $S$ is contained in some ball $~ B$ . We may also suppose that each $B \in B$ is a subset of $2 ~ B$ .

Let $B^{'}$ be a disjoint sub-collection of $B$ obtained from Proposition 6.3. Note that $B^{'}$ is countable. Indeed, for each $m \in N$ , at most $m μ (2 ~ B)$ balls $B \in B^{'}$ can satisfy $μ (B) > 1 / m$ .

Enumerate $B^{'} = {B_{1}, B_{2}, \dots}$ . Since the $B_{i}$ are disjoint subsets of $2 ~ B$ ,

\sum i > n μ (B_{i}) \to 0.

By the conclusion of Proposition 6.3,

S ∖ n ⋃ i = 1 B_{i} \subset ⋃ i > n 5 B_{i}

for each $n \in N$ . Since $μ$ is doubling, $μ (5 B_{i}) \leq C μ (B_{i})$ for each $i \in N$ and so

μ (S ∖ n ⋃ i = 1 B_{i}) \leq C \sum i > n μ (B_{i}) \to 0,

as required. ∎

Definition 6.7.

Let $μ$ be a Borel measure on a metric space $X$ and $f : X \to R$ $μ$ -measurable. Suppose that $0 < μ (B (x, r)) < \infty$ for all $x \in X$ and all $r > 0$ .

Define the Hardy–Littlewood maximal function of $f$ by

M f (x) = sup r > 0 \frac{1}{μ (B (x, r))} \int_{B (x, r)} | f | d μ .

By creftype 6.3, the maximal function is a Borel function.

Theorem 6.8 (Hardly–Littlewood maximal inequality).

Let $μ$ be a doubling measure on a metric space $X$ . There exists a $C > 0$ such that, for any $f : X \to R$ and $λ > 0$ ,

(6.1)

μ ({x : M f (x) > λ}) \leq \frac{C}{λ} \int | f | d μ .

Proof.

For $λ > 0$ , let

S = {x \in X : M f (x) > λ}

and, for each $R > 0$ , let $S_{R}$ be those $x \in X$ for which there exists $0 < r < R$ such that

(6.2)

\int_{B (x, r)} | f | d μ > λ μ (B (x, r)) .

Similarly to creftypecap 6.3, each $S_{R}$ is a Borel set. Moreover, the $S_{R}$ monotonically increase to $S$ . For a moment fix $R > 0$ . Let $B$ be the collection of balls $B (x, r)$ with $x \in S_{R}$ and $0 < r < R$ that satisfy (6.2) and let $B^{'}$ satisfy the conclusion of Lemma 6.1. Then

	$\int_{X} \| f \| d μ$	$\geq \sum B \in B^{'} \int_{B} \| f \| d μ$
		$> \sum B \in B^{'} λ μ (B)$
		$\geq \frac{1}{C_{μ}^{{log}_{2} 5}} \sum B \in B^{'} λ μ (5 B)$
		$\geq \frac{λ}{C_{μ}^{{log}_{2} 5}} μ (S_{R}) .$

In particular, (6.1) holds for $C = C_{μ}^{{log}_{2} 5}$ . ∎

Theorem 6.9 (Lebesgue differentiation theorem).

Let $μ$ be a doubling measure on a metric space $X$ and $f : X \to R$ with $\int f d μ < \infty$ . For $μ$ -a.e. $x \in X$ ,

\frac{1}{μ (B (x, r))} \int_{B (x, r)} | f - f (x) | d μ \to 0

as $r \to 0$ . Such an $x$ is called a Lebesgue point of $f$ .

Proof.

First note that the theorem is true if $f$ is continuous.

Fix $ϵ > 0$ and let $g : X \to R$ be continuous with

\int | f - g | d μ < ϵ

(such a $g$ exists by creftype 3.1). Let

B = {x \in X : | f (x) - g (x) | \geq \sqrt{ϵ}},

so that $μ (B) < \sqrt{ϵ}$ .

S = {x : M (f - g) > \sqrt{ϵ}},

then by Theorem 6.8,

μ (S) \leq \frac{C}{\sqrt{ϵ}} ∥ f - g ∥_{1} < C \sqrt{ϵ} .

Moreover, if $x \notin S$ ,

\frac{1}{μ (B (x, r))} \int_{B (x, r)} | f - g | d μ \leq \sqrt{ϵ}

for all $r > 0$ . In particular, since $g$ is continuous at $x$ ,

limsup r \to 0 \frac{1}{μ (B (x, r))} \int_{B (x, r)} | f - g (x) | d μ \leq \sqrt{ϵ} .

Therefore, if $x \notin B$ ,

limsup r \to 0 \frac{1}{μ (B (x, r))} \int_{B (x, r)} | f - f (x) | d μ \leq 2 \sqrt{ϵ} .

We are now done; repeat the above for a countable collection of $ϵ \to 0$ . The corresponding $B \cup S$ monotonically decrease to a set of measure zero. The set of $x \in X$ that does not belong to infinitely many of the $B \cup S$ has full measure, and for such an $x$ ,

lim r \to 0 \frac{1}{μ (B (x, r))} \int_{B (x, r)} | f - f (x) | d μ = 0.

∎

Corollary 6.10.

Let $μ$ be a doubling measure on a metric space $X$ and let $S \subset X$ be $μ$ -measurable with $μ (S) < \infty$ . Then

lim r \to 0 \frac{μ (S \cap B (x, r))}{μ (B (x, r))}

equals 1 for $μ$ -a.e. $x \in S$ and 0 for $μ$ -a.e. $x \notin S$ . Such an $x$ for which the limit equals 1 is called a density point of $S$ .

6.1. Exercises

Exercise 6.1.

Let $X$ be a separable metric space. Show that for any collection of balls, there exists a maximal disjoint sub-collection.

Exercise 6.2.

Show that the $5 r$ covering Lemma may not be true if the radii are not uniformly bounded.

Exercise 6.3.

Let $μ$ be a finite Borel measure on a metric space $(X, d)$ and let $x_{n} \to x \in X$ such that $d (x, x_{n})$ is a decreasing sequence. Let $U (y, r)$ denote the open ball centred on $y$ with radius $r$ .

Show that, for any $r > 0$ ,

$U (x, r) ∖ U (x_{n}, r)$

decreases to the empty set.
Deduce that $y \mapsto μ (U (y, r))$ is lower semi-continuous.
Give an example to show that $y \mapsto μ (U (y, r))$ may not be continuous.
Show that, for any $y \in X$ ,

$μ (B (y, r)) = lim Q ∋ q ↓ r μ (U (y, q)) .$
Deduce that, for any $r > 0$ , $y \mapsto μ (B (y, r))$ is a Borel function.
Show that the Hardy–Littlewood maximal function is equivalently defined by taking the supremum over all rational $r > 0$ .
Deduce that the Hardy–Littlewood maximal function is a Borel function.

Exercise 6.4.

Let $μ$ be a doubling measure on a metric space $X$ and let $S \subset X$ with $μ (S) < \infty$ . Suppose that there exists a $μ$ -measurable $S^{'} \supset S$ with $μ (S^{'}) = μ (S)$ . Show that

lim r \to 0 \frac{μ (S \cap B (x, r))}{μ (B (x, r))} = 1

for $μ$ -a.e. $x \in S$ .

Exercise 6.5.

Prove that on $R^{n}$ , $L^{n} = c H^{n}$ for some $c > 0$ . There are two ways to prove this (recall creftype 1.5 Items 1 and 2).

Let $μ, ν$ be two finite Borel measures on a set $X$ with $μ ≪ ν$ and suppose that $ν$ is doubling. Show that the Radon–Nikodym derivative of $μ$ with respect to $ν$ is given by

lim r \to 0 \frac{μ (B (x, r))}{ν (B (x, r))}

for $ν$ -a.e. $x \in X$ .

Exercise 6.6.

For $f = χ_{[0, 1]}$ , show that $M f$ does not have finite integral.

7. Differentiability of Lipschitz functions

The regularity of a Lipschitz function $f : R^{n} \to R$ is very interesting. Of course, Lipschitz functions are continuous, but they may not be differentiable everywhere. However, it is quite easy to convince yourself that they cannot be non-differentiable on quite a large set. The question to quantify how large the non-differentiability set of a Lipschitz function can be was one of the motivating questions of Lebesgue’s development of measure theory.

Definition 7.1.

Let $f : [a, b] \to R$ . The total variation of $f$ , $V f : [a, b] \to [0, \infty]$ is defined by

V f (x) = sup n \sum i = 1 | f (t_{i}) - f (t_{i - 1}) |

where the supremum ranges over all $a = t_{0} < t_{1} < \dots < t_{n} = b$ .

If $V f (b) < \infty$ , $f$ is said to have bounded variation (BV).

Definition 7.2.

A function $f : [a, b] \to R$ is absolutely continuous (AC) if for any $ϵ > 0$ there exists a $δ > 0$ such that, for any intervals $(a_{1}, b_{1}), (a_{2}, b_{2}) \dots \subset [a, b]$ with $\sum_{i} | b_{i} - a_{i} | < δ$ , we have $\sum_{i} | f (b_{i}) - f (a_{i}) | < ϵ$ .

Note that AC functions are BV, and Lipschitz functions are AC (see creftype 7.1). Also, if $f$ is BV then $V f$ and $V f - f$ are non-decreasing. If $f$ is AC then so are $V f$ and $V f - f$ , see creftype 7.2.

Theorem 7.3 (Lebesgue).

Let $f : [a, b] \to R$ be absolutely continuous. Then $f$ is differentiable $L^{1}$ almost everywhere. Moreover, for any $x > y \in [a, b]$ ,

f (x) - f (y) = \int_{y}^{x} f^{'} d x .

Proof.

By creftype 7.2 it suffices to assume that $f$ is non-decreasing. In this case define a measure $μ$ on $[a, b]$ using the Carathéodory construction with $F$ the set of compact intervals and $ζ ([c, d]) = f (d) - f (c)$ . This defines a finite Borel measure such that $μ ([c, d]) = f (d) - f (c)$ for all intervals $[c, d] \subset [a, b]$ . Indeed, for any $δ > 0$ , we may cover $[c, d]$ by finitely many intervals $[c, c_{1}], [c_{1}, c_{2}], \dots, [c_{k}, d]$ of width $δ^{'} \leq δ$ , showing

μ ([c, d]) \leq \sum f (c_{i + 1}) - f (c_{i}) = f (d) - f (c) .

The reverse inequality holds because $f$ is non-decreasing.

Note that $μ ≪ L^{1}$ . Indeed, given $ϵ > 0$ , let $δ > 0$ be given by the definition of $f$ being absolutely continuous. If $L^{1} (N) = 0$ , we may cover $N$ by countably many closed intervals $I_{i}$ such that $\sum_{i} L^{1} (I_{i}) < δ$ . In particular $\sum_{i} f (I_{i}) < ϵ$ and hence $μ (N) < ϵ$ . Therefore,

μ = \int \frac{d μ}{d L^{1}} d L^{1},

with $d μ / d L^{1} \in L^{1} (L^{1})$ .

By the Lebesgue differentiation theorem, for any Lebesgue point $x$ of $d μ / d L^{1}$ ,

	$lim t \to 0 \frac{f (x + t) - f (x)}{t}$	$= lim t \to 0 \frac{μ ([x + t, x])}{t}$
		$= lim t \to 0 \frac{1}{t} \int_{x}^{x + t} \frac{d μ}{d L^{1}} d L^{1}$
		$= \frac{d μ}{d L^{1}} (x) .$

∎

Theorem 7.4 (Rademacher).

Any Lipschitz $f : R^{n} \to R$ is differentiable $L^{n}$ almost everywhere.

Proof.

For notational simplicity, we prove the case $n = 2$ .

For each $y \in R$ , $x \mapsto f (x, y)$ is a Lipschitz function $R \to R$ and so is differentiable $L^{1}$ -a.e. That is, for every $y$ , $\partial_{1} f (x, y)$ exists for $L^{1}$ -a.e. $x$ . By Fubini’s theorem, $\partial_{1} f$ exists $L^{2}$ -a.e. Similarly, $\partial_{2} f$ exists almost everywhere too.

Fix $ϵ > 0$ . For $D \in Q^{2}$ and $j \in N$ let

X_{D, j} = {x : | f (x + h e_{i}) - f (x) - D_{i} h | < ϵ | h |, \forall 0 < | h | < 1 / j, i = 1, 2} .

These are Borel sets. Further, for $D \in Q^{2}$ , if

| \partial_{1} f (x) - D_{1} | < ϵ / 2 and | \partial_{2} f (x) - D_{2} | < ϵ / 2,

then $x \in X_{D, j}$ for sufficiently large $j$ . That is,

X^{ϵ} = ⋃ D \in Q ⋃ j \in N X_{D, j}

is a set of full measure.

Fix $D \in Q^{2}$ and $j \in N$ . Let $x$ be a density point of $X_{D, j}$ . Let $R > 0$ such that

L^{n} (B (x, r) \cap X_{D, j}) \geq (1 - ϵ^{n}) L^{n} (B (x, r))

for all $0 < r < R$ . In particular, for every $y \in B (x, r)$ there exists $y^{'} \in X_{D, j}$ with

∥ y - y^{'} ∥ < ϵ ∥ y - x ∥ .

Now let $r < min {R, 1 / j}$ and $∥ x - y ∥ < r$ . Set $h = y - x$ , $~ y = x + π_{1} y$ and $~ ~ y \in X_{D, j}$ with

∥ ~ y - ~ ~ y ∥ < ϵ ∥ x - ~ y ∥ \leq ϵ ∥ x - y ∥ .

Also let $y^{'}, y^{''}$ lie on the same vertical line as $~ ~ y$ such that $y^{'}, ~ y$ have the same vertical component as do $y^{''}, ~ ~ y$ . Then, since $x \in X_{D, j}$ ,

(7.1)

| f (~ y) - f (x) - D_{1} h_{1} | < ϵ | h_{1} | = ϵ ∥ x - ~ y ∥ \leq ϵ ∥ x - y ∥;

Since $f$ is Lipschitz,

(7.2)

| f (~ y) - f (y^{'}) | \leq L ∥ ~ y - y^{'} ∥ \leq L ϵ ∥ x - y ∥;

Since $~ ~ y \in X_{D, j}$ ,

(7.3)

| f (y^{''}) - f (y^{'}) - D_{2} h_{2} | \leq ϵ ∥ y^{'} - y^{''} ∥ \leq ϵ ∥ x - y ∥;

Since $f$ is Lipschitz,

(7.4)

| f (y^{''}) - f (y) | \leq L ∥ y^{''} - y ∥ = L ∥ y^{'} - ~ y ∥ \leq L ∥ ~ ~ y - ~ y ∥ \leq ϵ L ∥ x - y ∥ .

By combining Eqs. 7.4, 7.3, 7.2 and 7.1,

| f (y) - f (x) - D \cdot h | \leq 2 (1 + L) ϵ ∥ x - y ∥ .

This is true for all $y$ with $∥ x - y ∥ < r$ and for any density point $x$ of the full measure set $X^{ϵ}$ . That is, for $L^{n}$ -a.e $x$ . Taking a countable intersection over $ϵ \to 0$ concludes the proof. ∎

7.1. Exercises

Exercise 7.1.

Prove that Lipschitz functions are AC and that AC functions are BV.

Exercise 7.2.

Let $f : [a, b] \to R$ be BV. Show that $V f$ and $V f - f$ are non-decreasing. If $f$ is AC then show that $V f$ and $V f - f$ are AC.

Exercise 7.3.

Show that any monotonic $f : R \to R$ is continuous except at countably many points.

Exercise 7.4.

In this exercise we will show that monotonic functions are differentiable almost everywhere.

Let $f : [a, b] \to R$ be non-decreasing. For each $x \in (a, b)$ let

D - - f (x) = liminf h \to 0 \frac{f (x + h) - f (x)}{h} ¯ ¯¯¯ ¯ D f (x) = limsup h \to 0 \frac{f (x + h) - f (x)}{h} .

Observe that the set of $x \in (a, b)$ where $f$ is not differentiable at $x$ is the countable union, over $p < q \in Q$ , of the sets

B_{p, q} := {x \in (a, b) : D - - f (x) < p < q < ¯ ¯¯¯ ¯ D f (x)} .

We now fix $p < q \in Q$ .

Let

$B = {[x, x + h] : f (x + h) - f (x) < p h} .$

Note that $B$ satisfies the hypotheses of the Vitali covering theorem (recall Theorem 6.6). Let $B^{'}$ be a disjoint sub-cover obtained from the Vitali covering theorem with respect to Lebesgue measure and let $S = \cup B^{'}$ . Prove that

$L^{1} (f (B_{p, q} \cap S)) \leq p L^{1} (B_{p, q} \cap S) .$

Note: this is the step where we require $f$ to be monotonic.
Similarly, prove that $L^{1} (f (B_{p, q} \cap S)) \geq q L^{1} (B_{p, q} \cap S)$ .
Deduce that $f$ is differentiable almost everywhere.
Deduce that a BV function is differentiable almost everywhere.

However, BV functions do not satisfy the fundamental theorem of calculus:

Exercise 7.5.

Recall the definition of the Cantor set from creftype 1.6. Define the Cantor function $f : [0, 1] \to [0, 1]$ as follows. For each $n \in N$ , define $f_{n} : [0, 1] \to [0, 1]$ by

f (x) = {(\frac{3}{2})}^{n} L^{1} ([0, x] \cap C_{n}) .

Show that the $f_{n}$ converge uniformly on $[0, 1]$ to a monotonic, continuous function $f$ . For each $x \in [0, 1] ∖ C$ , show that $f^{'} (x) = 0$ .

Thus, $f$ is monotonic and hence BV, has derivative $0$ almost everywhere, but does not satisfy the fundamental theorem of calculus.

Exercise 7.6.

In lectures we proved that the derivative of any AC function is an absolutely continuous measure. Prove the converse: for any finite, absolutely continuous measure $μ$ on $[0, \infty)$ , show that

defines an absolutely continuous function.

Up to now, we have considered points where functions are differentiable. We now consider points of non-differentiability (which are much more interesting).

Exercise 7.7.

Show that the Cantor function is not differentiable at any point of the Cantor set.

Exercise 7.8.

Let $N \subset [0, 1]$ satisfy $L^{1} (N) = 0$ .

For each $n \in N$ , iteratively construct a countable collection of open intervals $O_{n}$ such that, for each $n \in N$ ,
- $N$ is contained in the union of $O_{n}$ ;
- for every $I \in O_{n}$ there exists $J \in O_{n - 1}$ with $I \subset J$ ;
- for each $I \in O_{n - 1}$ ,
  
  $L^{1} (I \cap \cup {J : J \in O_{n}}) < 2^{- n} | I | .$
Let

$S = ⋂ n \in N ⋃ m > n \cup {J : J \in O_{m}},$

the “limsup” of the $O_{n}$ ( $S$ is the set of points that are contained in infinitely many intervals from the $O_{n}$ ). In particular, $S \supset N$ .

For each $x \in [0, 1] ∖ S$ , let $N (x)$ be the largest $n$ for which there exists $I \in O_{n}$ with $x \in I$ . Define $P (x) = 1$ if $N (x)$ is even, $P (x) = 0$ otherwise. Finally, for each $x \in [0, 1]$ define

$f (x) = L^{1} ({t \in [0, x] : P (t) = 1}) .$

Show that $f$ is Lipschitz, monotonic, and not differentiable at any point of $N$ . Hint: show that $D - - f (x) = 0$ and $¯ ¯¯¯ ¯ D f (x) = 1$ for each $x \in N$ .

Part II Some topics in Geometric Measure Theory

8. Hausdorff measure and densities

Recall the definition of Hausdorff measure from creftype 1.5.

We are interested in the measure $H^{s} |_{S}$ , for $S \subset X$ some $H^{s}$ -measurable set with $H^{s} (S) < \infty$ . In particular, we require some counterpart to the Lebesgue density theorem, but, of course, $H^{s} |_{X}$ may not be locally finite.

Definition 8.1.

Let $X$ be a metric space, $A \subset X$ and $s \geq 0$ . The upper and lower Hausdorff densities of $A$ are

Θ^{*, s} (A, x) = limsup r \to 0 \frac{H^{s} (A \cap B (x, r))}{(2 r)^{s}}

and

Θ_{*}^{s} (A, x) = liminf r \to 0 \frac{H^{s} (A \cap B (x, r))}{(2 r)^{s}} .

Lemma 8.2.

Let $X$ be a metric space, $s \geq 0$ and $A \subset X$ with $H^{s} (A) < \infty$ . Then

2^{- s} \leq Θ^{*, s} (A, x) \leq 1

for $H^{s}$ -a.e. $x \in A$ .

Proof.

The set of $x \in A$ with $Θ^{*, s} (A, x) < 2^{- s}$ is a countable union countable of the sets

S_{δ} := {x \in A : H^{s} (A \cap B (x, r)) < (1 - δ) r^{s} \forall 0 < r < δ} .

Thus, for the first inequality, it suffices to show that $H^{s} (S_{δ}) = 0$ for all $δ > 0$ .

Fix $δ, ϵ > 0$ . We may cover $S_{δ}$ by sets $E_{1}, E_{2}, \dots$ such that, for each $i \in N$ , $diam E_{i} < ϵ$ , $S_{δ} \cap E_{i} \neq \emptyset$ and

\sum i \in N diam E_{i}^{s} \leq H^{s} (S_{δ}) + ϵ .

For each $i \in N$ let $x_{i} \in S_{δ} \cap E_{i}$ and set $r_{i} = diam E_{i}$ . Then

H^{s} (S_{δ}) \leq \sum i \in N H^{s} (S_{δ} \cap E_{i}) \leq \sum i \in N H^{s} (A \cap B (x_{i}, r_{i})) \leq (1 - δ) \sum i \in N diam E_{i}^{s} \leq (1 - δ) (H^{s} (S_{δ}) + ϵ) .

Since $ϵ > 0$ is arbitrary and $δ > 0$ , this implies $H^{s} (S_{δ}) = 0$ , as required.

For the second inequality, since $H^{s}$ is Borel regular (see creftype 8.2), it suffices to assume that $A$ is Borel. As before, given $δ > 0$ , it suffices to prove that

S := {x \in A : Θ^{*, s} (A, x) > 1 + δ}

satisfies $H^{s} (S) = 0$ . Fix $ϵ > 0$ and let $U \supset S$ be open with

H^{s} (A \cap U) \leq H^{s} (S) + ϵ

(which exists by the outer regularity of the measure $H^{s} |_{A}$ ). Let $B_{ϵ}$ be the collection of balls $B$ centred at a point of $S$ with $rad B < ϵ$ such that $B \subset U$ and

(8.1)

H^{s} (A \cap B) > (1 + δ) (2 rad B)^{s} .

This is a Vitali cover of $S$ . Let $B_{ϵ}^{'}$ be obtained from Proposition 6.3.

Since $H^{s} (S) < \infty$ , $S$ is separable (see creftype 8.4) and so $B_{ϵ}^{'} = {B_{1}, B_{2}, \dots}$ is countable and the conclusion of Proposition 6.3 states that

S ∖ ⋃ i \in N B_{i} \subset ⋃ i > n 5 B_{i}

for each $n \in N$ . Since $diam B_{i} < ϵ$ for each $i \in N$ , the $B_{i}$ and $5 B_{i}$ may be used to estimate $H_{10 ϵ}^{s} (S)$ . For each $n \in N$ we obtain

	$H_{10 ϵ}^{s} (S)$	$\leq \sum i \in N (2 rad B_{i})^{s} + \sum i > n (10 rad B_{i})^{s}$
		$\leq \sum i \in N \frac{H^{s} (A \cap B_{i})}{1 + δ} + 5^{s} \sum i > n \frac{H^{s} (A \cap B_{i})}{1 + δ}$

where the second inequality follows by (8.1). Since the $B_{i}$ are disjoint and $H^{s} (A) < \infty$ , the second term converges to 0 as $n \to \infty$ . Since the $B_{i}$ are subsets of $U$ we obtain

H_{10 ϵ}^{s} (S) \leq \frac{H^{s} (A \cap U)}{1 + δ} \leq \frac{H^{s} (S) + ϵ}{1 + δ} .

Since $ϵ > 0$ is arbitrary, this implies $H^{s} (S) \leq H^{s} (S) / (1 + δ)$ and hence $H^{s} (S) = 0$ , as required. ∎

Lemma 8.3.

Let $X$ be a metric space, $s \geq 0$ and let $A \subset X$ be $H^{s}$ -measurable with $H^{s} (A) < \infty$ . Then

Θ^{*, s} (A, x) = 0

for $H^{s}$ -a.e. $x \notin A$ .

Proof.

It suffices to show that, for $t > 0$ , the set

S = {x \in X ∖ A : Θ^{*, n} (A, x) > t}

satisfies $H^{s} (S) = 0$ . Fix $ϵ > 0$ . Since $A$ is $H^{s}$ -measurable, $H^{s} |_{A}$ is Borel regular. Therefore, since $H^{s} |_{A} (S) = 0$ , there exists an open $U \supset S$ with

H^{s} (A \cap U) = H^{s} |_{A} (U) < ϵ .

For each $x \in S$ and $δ > 0$ there exists a ball $B$ centred on $x$ with $rad B < δ$ such that

\frac{H^{s} (A \cap B)}{(2 rad B)^{s}} > t .

By Lemma 6.1 there exists a disjoint collection $B$ of such balls such that

S \subset ⋃ B \in B 5 B .

Since $H^{s} (A) < \infty$ , $A$ is separable and each of these balls contains a point of $A$ , $B$ is countable. Therefore

t H_{5 δ}^{s} (S) \leq t \sum B \in B (2 rad 5 B)^{s} < 5^{s} \sum B \in B H^{s} (A \cap B) \leq 5^{s} H^{s} (A \cap U) < 5^{s} ϵ .

Since $δ, ϵ > 0$ are arbitrary, this completes the proof. ∎

8.1. Exercises

Exercise 8.1.

Let $V \subset [0, 1]$ be a Vitali set as constructed in creftype 1.8.

Show that, for any Borel $B \subset V$ , $L^{1} (B) = 0$ .
Deduce that $L^{1} ([0, 1] ∖ V) = 1$ and hence, if $C$ is a Borel set with

$[0, 1] ∖ V \subset C \subset [0, 1],$

then $L^{1} (C) = 1$ .
Hence show that $L^{1} (V \cap C) = L^{1} (V) > 0$ .

Note however that we cannot deduce the value of $L^{1} (V)$ from our construction in creftype 1.8. Indeed, for any $ϵ > 0$ , that construction may produce a $V \subset [0, ϵ]$ .

Exercise 8.2.

Let $X$ be a metric space and $s \geq 0$ .

Show that $H^{s}$ is Borel regular. Hint: first show that in the definition of $H^{s}$ , we may take $F$ to be the collection of closed sets.
We are usually interested in $H^{s} |_{A}$ for some $A \subset X$ . Show that for any $A \subset X$ , $H^{s} |_{A}$ is a Borel measure.
Now assume that $A \subset X$ is $H^{s}$ -measurable with $H^{s} (A) < \infty$ . Show that $H^{s} |_{A}$ is Borel regular. Hint: show that there exist Borel sets $B \supset A \supset B^{'}$ with $H^{s} (B ∖ B^{'}) = 0$ .
Show that $H^{s} |_{A}$ may not be Borel regular if $A$ is not $H^{s}$ measurable. Hint: consider creftype 8.1.

Exercise 8.3.

In this exercise we construct the four corner Cantor set. Let $K_{0} = [0, 1]^{2}$ . Let $K_{1}$ be the “four corners” of $K_{0}$ of side length 1/4. That is

K_{1} = [0, 1 / 4]^{2} \cup [3 / 4, 1]^{2} \cup [0, 1 / 4] \times [3 / 4, 1] \cup [3 / 4, 1] \cup [0, 1 / 4] .

Inductively, $K_{n}$ is constructed by taking the four corners of side length $1 / 4^{n}$ of all the squares of $K_{n - 1}$ . Finally let $K = ⋂_{n \in N} K_{n}$ , a compact set. Show that $0 < H^{1} (K) < \infty$ .

Exercise 8.4.

For $s \geq 0$ let $X$ be a metric space with $H^{s} (X) < \infty$ . Show that $X$ is separable.

Exercise 8.5.

Show that Lemmas 8.3 and 8.2 may be false if $A$ has only $σ$ -finite $H^{s}$ measure.

9. Rectifiable sets and approximate tangent planes

Rectifiable sets are the measure theoretic counterpart to manifolds.

Definition 9.1.

A $H^{n}$ -measurable set $E \subset R^{m}$ is $n$ -rectifiable if there exist Lipschitz $f_{i} : R^{n} \to R^{m}$ such that

H^{n} (E ∖ ⋃ i \in N f_{i} (R^{n})) = 0.

We will show that $n$ -rectifiable sets possess a unique approximate $n$ -dimensional tangent plane at almost every point.

Given $V \in G (m, n)$ , $a \in R^{n}$ and $0 < s < 1$ define the cone around $V$ centred at $a$ with aperture $s$ as

C (a, V, s) = {x \in R^{n} : dist (x - a, V) < s ∥ x - a ∥} .

Definition 9.2.

Let $A \subset R^{m}$ and $a \in A$ . A $V \in G (m, n)$ is an approximate tangent plane to $A$ at $a$ if

(9.1)

limsup r \to 0 \frac{H^{n} (A \cap B (a, r))}{r^{n}} > 0

and, for every $0 < s < 1$ ,

(9.2)

lim r \to 0 \frac{H^{n} (A \cap B (a, r) ∖ C (a, V, s))}{r^{n}} = 0.

Rademacher’s theorem gives a candidate for the approximate tangent plane to a rectifiable set. There are three steps required to prove that the derivative is indeed an approximate tangent plane: show that the derivative has full rank at almost every point; prove the density condition (9.1); and show that the sets from other parametrisations of the rectifiable set do not destroy the approximation by a tangent plane at almost every point.

The second and third steps follow from the results of the previous section. For the first step we use the following.

Lemma 9.3 (Easy Sard’s theorem).

If $f : R^{n} \to R^{m}$ is Lipschitz then

H^{n} ({f (x) : rank D f (x) < n}) = 0.

Proof.

Let $L = Lip f$ . Fix $0 < R < \infty$ , $δ, ϵ > 0$ and let

A = {x \in B (x, R) : rank D f (x) < n} .

For $x \in A$ let

W_{x} = f (x) + D f (x) (R^{n}) .

Then for sufficiently small $0 < r_{x} < δ$ ,

f (B (x, r_{x})) \subset B (f (x), L r_{x}) \cap {y : dist (y, W_{x}) < ϵ r_{x}} .

Since $rank D f (x) < n$ , the set on the right hand side can be covered by $(L / ϵ)^{n - 1}$ cubes of side length $ϵ r_{x}$ .

Since $A$ is covered by balls of the form $B (x, r_{x} / 5)$ , there exists a disjoint collection of balls $B (x_{i}, r_{i} / 5)$ such that $A$ is covered by the union of the $B (x_{i}, r_{i})$ . Then

f (A) \subset f (⋃ i \in N B (x_{i}, r_{i})) \subset ⋃ i \in N f (B (x_{i}, r_{i})) .

By the previous argument, each factor of the right hand side is covered by $(L / ϵ)^{n - 1}$ cubes of side length $ϵ r_{i}$ . Thus

H_{2 δ}^{n} (f (A)) \leq \sum i \in N {(\frac{L}{ϵ})}^{n - 1} (ϵ r_{i})^{n - 1} = L^{n - 1} ϵ \sum i \in N r_{i}^{n} .

However, the $B (x_{i}, r_{i} / 5)$ are disjoint subsets of $B (0, R + δ) \subset R^{n}$ and so

\sum i \in N {(\frac{r_{i}}{5})}^{n} \leq (R + δ)^{n} .

Since $ϵ > 0$ is arbitrary, this implies that $H_{2 δ}^{n} (f (A)) = 0$ and hence $H^{n} (f (A)) = 0$ . Taking a countable union over $R \to \infty$ completes the proof. ∎

Lemma 9.4.

Let $f : R^{n} \to R^{m}$ be Lipschitz and $S \subset R^{n}$ . Suppose that there exists a $δ > 0$ such that, for each $x, y \in S$ ,

∥ f (x) - f (y) ∥ \geq δ ∥ x - y ∥ .

Then $f (S)$ has a unique approximate tangent plane at $H^{n}$ almost every point.

Proof.

By Lemma 9.3, we may suppose $rank D f (x) = n$ for every $x \in S$ . By Lemma 8.2, we may suppose $Θ^{*, n} (E, f (x)) > 0$ for every $x \in S$ . Fix $x \in S$ and $0 < s < 1$ . There exists $ϵ > 0$ such that

∥ f (y) - f (x) - D f (x) (y - x) ∥ < s ∥ y - x ∥

for all $y \in B (x, ϵ) \cap S$ . Moreover, if $y \in S ∖ B (x, ϵ)$ then

∥ f (y) - f (x) ∥ \geq δ ϵ .

That is, if $a = f (x)$ and $b = f (y)$ with $∥ a - b ∥ \leq δ ϵ$ and $V = a + D f (x) (R^{n})$ ,

dist (b - a, V) < s ∥ y - x ∥ \leq s ∥ b - a ∥ / δ .

Therefore, $V$ is an approximate tangent plane to $f (S)$ at $a$ .

This approximate tangent plane is unique at any density point $x$ of $S$ . Indeed, if $V^{'} \neq V$ , let $v \in V ∖ V^{'}$ and let $0 < s < 1$ be such that $C (f (x), R v, s) \cap C (f (x), V^{'}, s) = {0}$ . Since $rank D f (x) = n$ and $x$ is a density point of $S$ , for sufficiently small $r > 0$ there exists $y \in S \cap B (x, r)$ such that $B (f (y), s r) \cap C (f (x), V^{'}, s) = \emptyset$ and

\frac{H^{n} (B (f (y), s r) \cap S)}{r^{n}} \geq δ s .

In particular, $V^{'}$ is not an approximate tangent to $f (S^{'})$ at $x$ . ∎

Theorem 9.5.

Let $E \subset R^{m}$ be $n$ -rectifiable with $H^{n} (E) < \infty$ . Then for $H^{n}$ -a.e. $x \in E$ , $E$ has a unique approximate tangent plane at $x$ .

Proof.

Let $f : R^{n} \to R^{m}$ be one of the Lipschitz functions as in the definition of a rectifiable set and let $S = f^{- 1} (E)$ . It suffices to prove that $E$ has a unique approximate tangent plane at $f (x)$ for $L^{n}$ -a.e. $x \in S$ .

By Lemma 9.3, we may suppose that $dim D f (R^{n}) = n$ for all $x \in S$ . Fix such an $x$ and let $0 < ϵ < ∥ D f (x)^{- 1} ∥ / 2$ . There exists $δ > 0$ such that

∥ f (y) - f (x) - D f (x) (y - x) ∥ < ϵ ∥ y - x ∥

for all $y \in B (x, δ) \cap S$ . In particular, by the triangle inequality,

∥ f (y) - f (x) ∥ > ϵ ∥ y - x ∥ / 2.

Therefore, the sets

S_{ϵ} := {x \in G : ∥ f (y) - f (x) ∥ > ϵ ∥ y - x ∥ \forall y \in B (x, ϵ)}

are Borel and monotonically increase to $S$ as $ϵ \to 0$ . Therefore it suffices to prove the result for $L^{n}$ -a.e. $x$ in some fixed $S_{ϵ}$ . Cover $S_{ϵ}$ by finitely many balls $B_{1}, B_{2}, \dots, B_{N}$ of radius $ϵ$ . It suffices to prove the result for $L^{n}$ -a.e. $x$ in some fixed $S^{'} := S_{η} \cap B_{i}$ .

However, $S^{'}$ satisfies the hypotheses of Lemma 9.4 and so $f (S^{'})$ has a unique approximate tangent at $H^{n}$ almost every point. To see that this tangent is a unique approximate tangent to $E$ at $H^{n}$ almost every point, we simply use Lemma 8.3: for $H^{n}$ -a.e. $x \in f (S^{'})$ , $Θ^{*, n} (E ∖ f (S^{'}), x) = 0$ . ∎

In Theorem 10.10 we will see that the converse to Theorem 9.5 holds.

For $V \in G (n, m)$ , write $π_{V}$ for the orthogonal projection onto $V$ and equip $G (n, m)$ with the metric $d (V, W) = ∥ π_{V} - π_{W} ∥$ . We will consider $L^{n}$ on an element of $G (n, m)$ .

Lemma 9.6.

Let $f : R^{m} \subset R^{n} \to R^{m}$ be Lipschitz and let $S \subset R^{n}$ satisfy $L^{n} (S) > 0$ . For $ϵ > 0$ suppose that there exists an invertible linear $L : R^{n} \to R^{m}$ such that, for all $x, y \in S$ ,

∥ f (x) - f (y) - L (x - y) ∥ < \frac{ϵ}{∥ L^{- 1} ∥} ∥ x - y ∥ .

Then for any $V \in G (n, m)$ with $∥ (π_{V} |_{L (R^{n})})^{- 1} ∥^{- 1} \geq 2 ϵ$ , $L^{n} (π_{V} (f (S))) > 0$ .

Proof.

For any $V \in G (n, m)$ ,

∥ π_{V} (f (x) - f (y)) - π_{V} (L (x - y)) ∥ < ϵ ∥ L^{- 1} ∥^{- 1} ∥ x - y ∥

and so, if $∥ (π_{V} |_{L (R^{n})})^{- 1} ∥^{- 1} \geq 2 ϵ$ ,

	$∥ π_{V} (f (x) - f (y)) ∥$	$\geq ∥ π_{V} (L (x - y)) ∥ - ϵ ∥ L^{- 1} ∥^{- 1} ∥ x - y ∥$
		$\geq ∥ (π_{V} \|_{L (R^{n})})^{- 1} ∥^{- 1} ∥ L (x - y) ∥ - ϵ ∥ L (x - y) ∥$
		$\geq ϵ ∥ L (x - y) ∥$
		$\geq ϵ ∥ L^{- 1} ∥ ∥ x - y ∥ .$

Thus $π_{V} \circ f$ has Lipschitz inverse on $S$ and hence $L^{n} (π_{V} (f (S))) > 0$ by creftype 1.5. ∎

Corollary 9.7.

Let $E \subset R^{m}$ be $n$ -rectifiable with $H^{n} (E) > 0$ . Then there exists $W \in G (m - n, m)$ such that $π_{V} (E) > 0$ for all $V \in G (n, m)$ with $V \cap W = {0}$ .

Remark 9.8.

The set of $V$ that satisfy the conclusion of Corollary 9.7 is very large; try some examples in reasonable dimensions.

Proof.

Since $E \subset R^{m}$ is rectifiable with $H^{n} (E) > 0$ , there exists a Lipschitz $f : R^{n} \to R^{m}$ with $H^{n} (E \cap f (R^{n})) > 0$ . In particular, $S := f^{- 1} (E)$ satisfies $L^{n} (S) > 0$ . By Lemma 9.3, for $L^{n}$ -a.e. $x \in S$ , $D f (x)$ is injective.

Fix $ϵ > 0$ For $M > 0$ , the set of invertible $L : R^{n} \to R^{m}$ with $∥ L^{- 1} ∥ < M$ may be covered by countably many sets of diameter $ϵ / M$ . Varying $M \in N$ , we see that $L^{n}$ almost all of $S$ is covered by countably many sets of the form

{x \in S : ∥ D f (x) - L ∥ < ϵ / 2 ∥ L^{- 1} ∥} .

Moreover, each of these sets may be covered by countably many sets of the form

{x \in S : ∥ f (x) - f (y) - L (x - y) ∥ < ϵ ∥ x - y ∥ / ∥ L^{- 1} ∥ \forall y \in B (x, ϵ)} .

Finally, these sets may be covered by countably many sets of diameter $ϵ$ . Therefore, for each $j \in N$ there exists $S_{j} \subset S$ and invertible $L_{j} : R^{n} \to R^{m}$ such that, for all $x, y \in S_{j}^{ϵ}$ ,

∥ f (x) - f (y) - L_{j} (x - y) ∥ < ϵ ∥ x - y ∥ / ∥ L_{j} ∥^{- 1},

and $L^{n} (S ∖ ⋃_{j \in N} S_{j}) = 0$ .

Since $L^{n} (S) > 0$ , there exists $j \in N$ with $L^{n} (S_{j}) > 0$ . Then $S_{j}$ satisfies the hypotheses of Lemma 9.6 and so $L^{n} (π_{V} (S)) \geq L^{n} (π_{V} (S_{j})) > 0$ for all $V \in G (n, m)$ with $∥ (π_{V} |_{L_{j} (R^{n})})^{- 1} ∥^{- 1} \geq 2 ϵ$ . Let $L_{ϵ} = L_{j}$ . Repeat this for each $i \in N$ with $ϵ = 1 / i$ . The set $G (n, m)$ is compact and so we may suppose that $L_{1 / i} (R^{n}) \to W \in G (n, m)$ . The only $V \in G (n, m)$ for which $L^{n} (π_{V} (S)) = 0$ satisfy $∥ (π_{V} |_{L_{1 / i} (R^{n})})^{- 1} ∥^{- 1} < 2 / i$ and hence $∥ (π_{V} |_{W})^{- 1} ∥^{- 1} < 2 / i$ for each $i \in N$ . That is, $V \cap W^{⊥} \neq {0}$ as required. ∎

9.1. Exercises

Exercise 9.1.

Let $X$ be a metric space, $Y \subset X$ and $f : Y \to R$ $L$ -Lipschitz. Define $~ f : X \to R$ by

~ f (x) = sup {f (y) - L d (x, y) : y \in Y} .

Show that $~ f$ is an $L$ -Lipschitz extension of $f$ to $X$ . This is called the McShane–Whitney extension theorem
If $f : Y \to R^{n}$ is $L$ -Lipschitz, show that there is a $\sqrt{n} L$ -Lipschitz extension of $f$ to $X$ .
The following example shows that the vector valued extension cannot have the same Lipschitz constant in general: Let

$Y = {(- 1, 1), (1, - 1), (1, 1)} \subset ℓ_{\infty}^{2}$

and define

$f (- 1, 1) = (- 1, 0), f (1, - 1) = (1, 0), f (1, 1) = (0, \sqrt{3}) .$

Show that $f$ is 1-Lipschitz but has no 1-Lipschitz extension to $Y \cup {(0, 0)}$ .
However, the Kirszbraun extension theorem states that any Lipschitz map between any two Hilbert spaces may be extended whilst preserving the Lipschitz constant.

Exercise 9.2.

Let $E \subset R^{m}$ be $n$ -rectifiable. Show that $H^{n} |_{E}$ is $σ$ -finite.
Show that Theorem 9.5 may not be true if $E$ does not satisfy $H^{n} (E) < \infty$ .

10. Purely unrectifiable sets

Definition 10.1.

A $H^{n}$ -measurable set $S \subset R^{m}$ is purely $n$ -unrectifiable if, for all $n$ -rectifiable $E \subset R^{n}$ , $H^{n} (S \cap E) = 0$ .

Lemma 10.2.

The four corner Cantor set $K \subset R^{2}$ is purely 1-unrectifiable.

Proof.

Observe that the coordinate projections of $K$ have $L^{1}$ measure zero (see creftype 10.1). If there existed a rectifiable $γ \subset R^{2}$ with $H^{1} (γ \cap K) > 0$ , then $γ \cap K$ is a 1-rectifiable set of positive measure and hence, by Corollary 9.7, one of the coordinate projections must have positive measure, a contradiction.

For a second proof see creftype 10.2. ∎

Lemma 10.3.

Let $A \subset R^{m}$ be $H^{m}$ -measurable with $H^{m} (A) < \infty$ . There exists a decomposition $A = E \cup S$ with $E$ $n$ -rectifiable and $S$ purely $n$ -unrectifiable.

Proof.

Let

t = sup {H^{n} (E) : E \subset A, n -rectifiable} .

Since $H^{n} (A) < \infty$ , $t < \infty$ . Let $E_{i} \subset A$ be $n$ -rectifiable with $H^{n} (E_{i}) \to t$ . Then $E = ⋃_{i \in N} E_{i}$ is $n$ -rectifiable and is contained in $A$ . Therefore

t \geq H^{n} (E) \geq H^{n} (E_{i}) \to t

and so $H^{n} (E) = t$ . Then $S = A ∖ E$ is purely $n$ -unrectifiable. Indeed, if $f : R^{n} \to R^{m}$ is Lipschitz, $E^{'} = (A ∖ E) \cap f (R^{n})$ , is $H^{m}$ -measurable and so

t \geq H^{n} (E \cup E^{'}) = H^{n} (E) + H^{n} (E^{'}) = t + H^{n} (E^{'}) .

∎

We now state a very important theorem on the structure of purely unrectifiable sets. It requires the notion of a natural measure $γ_{n, m}$ on $G (n, m)$ that is invariant under the action of $S O (m)$ . There are several ways to construct this measure. The simplest is to consider $G (n, m)$ as a (compact) metric space equipped with the metric

d (V, W) = ∥ π_{V} - π_{W} ∥ .

Then $γ_{n, m}$ is given by (a scalar multiple of) $H^{n (m - n)}$ . We will not discuss the specific details of this measure. When $n = 1$ , we may identify $G (1, m)$ with $S^{m - 1}$ . In this case, $γ_{1, m}$ is simply $H^{m - 1}$ .

Theorem 10.4 (Besicovitch–Federer projection theorem).

Let $S \subset R^{m}$ be purely $n$ -unrectifiable with $H^{n} (S) < \infty$ . Then, for $γ_{n, m}$ -a.e. $V \in G (n, m)$ ,

L^{n} (π_{V} (S)) = 0.

Conversely, if $E \subset R^{m}$ is purely $n$ -unrectifiable with $H^{n} (E) > 0$ , for $γ_{n, m}$ -a.e. $V \in G (n, m)$ ,

L^{n} (π_{V} (E)) > 0.

Remark 10.5.

The converse statement is given by Corollary 9.7.

We will prove the projection theorem for $n = 1$ and $m = 2$ , which was proved by Besicovitch. First we prove some preliminary geometric properties of purely unrectifiable sets.

Lemma 10.6.

Let $E \subset R^{m}$ , $V \in G (m - n, m)$ and $0 < s < 1$ . Suppose that, for every $x \in E$ ,

E \cap C (x, V, s) \cap B (x, r) = \emptyset .

Then $E$ is $n$ -rectifiable.

Proof.

Since $E$ may be divided into countably many sets of diameter at most $r$ , we may suppose $diam E \leq r$ . In this case, $π_{V^{⊥}}$ has Lipschitz inverse on $E$ . Indeed, if $x, y \in E$ then $y \notin C (x, V, s)$ and so

∥ π_{V^{⊥}} (x - y) ∥ = dist (x - y, V) \geq s ∥ x - y ∥ .

Therefore, $E$ is covered by a Lipschitz image of $R^{n}$ . ∎

Lemma 10.7.

Let $S \subset R^{m}$ be purely $n$ -unrectifiable, $V \in G (m - n, m)$ , $0 < s < 1$ and $0 < δ, λ < \infty$ . If

(10.1)

H^{n} (S \cap C (x, V, s) \cap B (x, r)) \leq λ r^{n} s^{n}

for every $x \in S$ and $0 < r < δ$ then

H^{n} (S \cap B (a, δ / 6)) \leq 2 \cdot 20^{n} λ δ^{n}

for every $a \in S$ .

Remark 10.8.

Note that this is certainly not true for a rectifiable set $E$ ; the first cone may be empty for every $x \in E$ .

Proof.

For a fixed $a \in S$ , we may suppose $S \subset B (a, δ / 6)$ . By Lemma 10.6, we may suppose that

S \cap C (x, V, s / 4) \neq \emptyset

for every $x \in S$ . For every $x \in S$ let

h (x) = sup {| x - y | : y \in S \cap C (x, V, s / 4)},

so that $0 < h (x) < δ / 3$ for all $x \in S$ . Pick $x^{*} \in S \cap C (x, V, s / 4)$ with $| x - x^{*} | \geq 3 h (x) / 4$ and let $C_{x}$ be the cylinder

C_{x} = π_{V^{⊥}}^{- 1} (π_{V^{⊥}} (B (x, s h (x) / 4))) .

We claim that

(10.2)

C_{x} \cap S \subset C (x, V, s) \cap B (x, 2 h (x)) \cup C (x^{*}, V, s) \cap B (x^{*}, 2 h (x)) .

(Draw a picture!) Suppose $z \in C_{x} \cap S$ does not belong to the first set. Then

	$s ∥ x^{*} - z ∥$	$\leq ∥ π_{V^{⊥}} (x^{*} - z) \|$
		$\leq ∥ π_{V^{⊥}} (x^{*} - x) ∥ + ∥ π_{V^{⊥}} (x - z) ∥$
		$\leq s \| x^{*} - x \| / 4 + s h (x) / 4$
		$\leq s h (x) / 2,$

where the penultimate inequality follows because $x^{*} \in C (x, V, s / 4)$ and $z \in C_{x}$ . Therefore

	$∥ x - z ∥$	$\geq ∥ x - x^{} ∥ - ∥ x^{} - z ∥$
		$> 3 h (x) / 4 - h (x) / 2$
		$\geq ∥ π_{V^{⊥}} (x - z) ∥ / s .$

That is, $z$ belongs to the first set in (10.2).

By (10.2) and (10.1),

H^{1} (S \cap C_{x}) \leq 2 λ (2 h (x) s)^{n} .

We apply Lemma 6.1 to the balls

π_{V^{⊥}} (B (x, s h (x) / 20))

with $x \in S$ . This gives countably many $x_{i} \in S$ , for which these balls are disjoint, and

S \subset ⋃ i \in N C_{x_{i}} .

Therefore

	$H^{n} (S)$	$\leq \sum i \in N H^{n} (C_{x_{i}})$
		$\leq 2 λ 2^{n} \sum i \in N (s h (x_{i}))^{n}$
		$= 2 λ 2^{n} 20^{n} \sum i \in N {(\frac{s h (x_{i})}{20})}^{n} .$

But, the $π_{V^{⊥}} (B (x_{i}, s h (x_{i}) / 20))$ are disjoint subsets of $B (π_{V^{⊥}} (a), δ / 2) \subset V^{⊥}$ and so the final sum is bounded above by $(δ / 2)^{n}$ . ∎

Corollary 10.9.

If $S \subset R^{m}$ is purely $n$ -unrectifiable with $H^{n} (S) < \infty$ then for every $V \in G (m - n, m)$ , every $0 < s < 1$ and $H^{n}$ -a.e. $x \in S$ ,

Θ^{*, n} (S \cap C (a, V, s), a) \geq 240^{- n - 1} s^{n} .

Proof.

For a fixed $V, s$ , this is immediate from the fact that $Θ^{*, n} (S, a) \geq 2^{- n}$ almost everywhere. To obtain the conclusion for all $V, s$ , note that the conclusion is determined by a countable dense set of $V, s$ . ∎

Theorem 10.10.

Let $E \subset R^{m}$ satisfy $H^{n} (E) < \infty$ and suppose that, for $H^{n}$ -a.e. $x \in E$ , $E$ has a unique approximate tangent plane at $x$ . Then $E$ is $n$ -rectifiable.

Proof.

By Lemma 10.3, there exists a decomposition $E = E^{'} \cup S$ , where $E^{'}$ is $n$ -rectifiable and $S$ is purely $n$ -unrectifiable. We must show that $H^{n} (S) = 0$ . Note that, by applying Lemma 8.3 to $E^{'}$ , we see that the approximate tangent plane to $E$ at $x \in S$ is also an approximate tangent plane to $S$ for $H^{n}$ -a.e. $x$ .

It suffices to show, for a fixed $W \in G (n, m)$ , that the set $S^{'}$ of $x \in S$ whose approximate tangent plane $V_{x}$ lies in $B (W, δ)$ has measure zero. Suppose not. Then, for any $λ > 0$ , there exists an $R > 0$ such that the set $S^{''}$ of those $x \in S^{'}$ with

sup 0 < r < R \frac{H^{n} (S \cap B (a, r) ∖ C (a, V_{a}, 1 / 3))}{r^{n}} < λ 3^{- n}

has positive measure. Fix an $x \in S^{''}$ . Since $∥ π_{V_{x}} - π_{W} ∥ \leq 1 / 3$ , for every $0 < r < R$ we have

C (x, W^{⊥}, 1 / 3) \cap B (x, r) \subset B (x, r) ∖ C (x, V_{x}, 1 / 3) .

Thus, for $x \in S^{''}$ and $0 < r < R$ ,

H^{n} (S^{'} \cap C (x, W^{⊥}, 1 / 3) \cap B (x, r)) < λ 3^{- n} r^{n} .

If $λ < 240^{- m - 1}$ , Corollary 10.9 implies $H^{n} (S^{''}) = 0$ , a contradiction. ∎

We now prove the Besicovitch projection theorem [1]

Theorem 10.11.

Let $S \subset R^{2}$ be purely 1-unrectifiable with $H^{1} (S) < \infty$ . Then for $H^{1}$ -a.e. $e \in S^{1}$ ,

L^{1} (π_{e} (S)) = 0.

We follow the presentation of Orponen [2].

From Corollary 10.9, we see that a purely unrectifiable set has many radiating out of almost every point in all directions at almost every point. We now precisely describe two ways in which this can occur.

Notation 10.12.

Let $S \subset R^{2}$ and $x \in S$ . For $e \in S^{1}$ let $l_{e} (x)$ be the half line $x + [0, \infty) e$ and for $I \subset S^{1}$ , let $C (I, x)$ be the cone $⋃_{e \in I} l_{e} (x)$ . For $r > 0$ let $H_{x} (r)$ be those $e \in S^{1}$ for which

| K \cap l_{e} (x) \cap B (x, r) | \geq 2.

That is, $S \cap l_{e} (x) \cap B (x, r)$ contains another point of $K$ . Also let $H_{x} = ⋂_{r > 0} H_{x} (r)$ , the directions that contain other points of $S$ arbitrarily close to $x$ . For $e \in S^{1}$ , we let $H_{e}$ be those $x \in S$ for which $e \in H_{x}$ .

For $R, M, ϵ > 0$ let $D_{x} (R, M, ϵ)$ be those $e \in S^{1}$ for which there exists $0 < r < R$ and an interval $I \subset S^{1}$ with $e \in I$ and $0 < H^{1} (I) < ϵ$ such that

\frac{H^{1} (S \cap C (x, I) \cap B (x, r))}{r} \geq M H^{1} (I) .

That is, the density of $S$ in the cone $C (x, I)$ at scale $r$ is very high, compared to the length of $I$ . Also let $D_{x} = ⋂_{R, M, ϵ > 0} D_{x} (R, M, ϵ)$ . For $e \in S^{1}$ , we let $D_{e}$ be those $x \in S$ for which $e \in D_{x}$ .

The main step in proving Theorem 10.11 is the following.

Proposition 10.13.

Let $S \subset R^{2}$ be purely 1-unrectifiable with $H^{1} (S) < \infty$ . For $H^{1}$ -a.e. $x \in S$ , $H^{1} (S ∖ H_{x} \cup D_{x}) = 0$ .

Before proving Proposition 10.13, we will demonstrate how it is used to prove Theorem 10.11.

Lemma 10.14 (Special case of the coarea formula).

For any $e \in S^{1}$ and any compact $K \subset R^{2}$ ,

\int_{R} card (K \cap l_{e} (t)) d t \leq H^{1} (K) .

In particular, if $H^{1} (K) < \infty$ then for any $e \in S^{1}$ , $L^{1} (π_{e^{⊥}} (H_{e})) = 0$ .

Proof.

Since $K$ is compact,

f (t) = card (K \cap l_{e} (t))

is a Borel function. Indeed, if, for $δ > 0$ ,

f_{δ} (t) = max {n \in N : \exists x_{1}, \dots, x_{n} \in K \cap l_{e} (t) with ∥ x_{i} - x_{j} ∥ \geq δ \forall 1 \leq i \neq j \leq n}

then $f_{δ}$ monotonically increases to $f$ as $δ \to 0$ . Since $K$ is compact, the $f_{δ}$ are lower semi-continuous. Therefore, by the monotone convergence theorem, it suffices to bound the integral of each $f_{δ}$ .

Fix $δ > 0$ and cover $K$ by sets $E_{1}, E_{2}, \dots$ with $diam E_{i} < δ$ such that

\sum i \in N diam E_{i} \leq H^{1} (K) + δ .

Note that

f_{δ} (t) \leq card ({i : E_{i} \cap l_{e} (t) \neq \emptyset}) .

Therefore

	$\int_{R} f_{δ} d L^{1}$	$\leq \int_{R} \sum i \in N χ_{{(i, t) : E_{i} \cap l_{e} (t) \neq \emptyset}}$
		$= \sum i \in N \int_{R} χ_{{(i, t) : E_{i} \cap l_{e} (t) \neq \emptyset}}$
		$\leq \sum i \in N diam E_{i}$
		$\leq H^{1} (K) + δ,$

as required. ∎

Lemma 10.15.

Let $S \subset R^{2}$ be $H^{1}$ -measurable with $H^{1} (S) < \infty$ . Then for any $e \in S^{1}$ , $L^{1} (π_{e^{⊥}} (D_{e})) = 0$ .

Proof.

Fix $e \in S^{1}$ . For any $M \in N$ and $t \in π_{e^{⊥}} (D_{e})$ there exists an $x \in D_{e}$ , $r_{x} > 0$ and an interval $e \in I_{x} \subset S^{1}$ with $diam I < 1 / 10$ such that

H^{1} (S \cap C (x, I) \cap B (x, r)) \geq M H^{1} (r I) .

Apply Lemma 6.1 to the intervals $J_{x} = π_{e^{⊥}} (C (x, I_{x}) \cap B (x, r_{x}))$ to obtain a disjoint collection $J_{x_{1}}, J_{x_{2}}, \dots \subset R$ such that

D_{e} \subset ⋃ i \in N 5 J_{x_{i}} .

Therefore

	$L^{1} (D_{e})$	$\leq \sum i \in N 5 L^{1} (J_{x_{i}})$
		$\leq 5 \sum i \in N H^{1} (r I_{x_{i}})$
		$\leq \frac{5}{M} \sum i \in N H^{1} (S \cap C (x_{i}, I_{x_{i}}) \cap B (x_{i}, r_{x_{i}}))$
		$\leq \frac{5}{M} H^{1} (S),$

where the final inequality follows from the disjointness of the sets $C (x_{i}, I_{x_{i}}) \cap B (x_{i}, r_{i})$ , $i \in N$ . Since this is true for all $M \in N$ , $L^{1} (D_{e}) = 0$ . ∎

Proof of Theorem 10.11 using Proposition 10.13.

By the inner regularity of $H^{1}$ , it suffices to prove the result for compact $S$ . By definition, we have

{(x, e) \in S \times S^{1} : e \notin H_{x} \cup D_{x}} = {(x, e) : x \notin H_{e} \cup D_{e}} .

Proposition 10.13 implies that the left hand expression has $H^{1} \times H^{1}$ -measure zero and so Fubini’s theorem implies that, for $H^{1}$ -a.e. $e \in S^{1}$ , $H^{1} (S ∖ H_{e} \cup D_{e}) = 0$ . Therefore, by Lemmas 10.14 and 10.15, $π_{e^{⊥}} (S) = 0$ for $H^{1}$ -a.e. $e \in S^{1}$ . ∎

Proof of Proposition 10.13.

Fix $R, M, ϵ > 0$ and $x \in S$ which satisfies the conclusion of Corollary 10.9. That is,

(10.3)

Θ^{*, 1} (S \cap C (x, I), x) \geq c_{0} H^{1} (I)

for every interval $I \subset S^{1}$ . It suffices to show that $H^{1} (S^{1} ∖ H_{x} (R) \cup D_{x} (R, M, ϵ)) = 0$ . In fact, we will show that, for any $e \in S^{1}$ ,

Θ^{*, 1} (H_{x} (R), e) > 0 or Θ_{*}^{1} (D_{x} (R, M, ϵ), e) > 0,

from which the result follows by the Lebesgue density theorem.

To this end, fix $e \in S^{1}$ with $Θ^{*, 1} (H_{x} (R), e) = 0$ . Then for all sufficiently small intervals $I$ with $e \in I \subset S^{1}$ ,

(10.4)

H^{1} (H_{x} (R) \cap I) < c_{0} H^{1} (I) / 4 M .

Fix such an $I$ . By Eq. 10.3, there exists $r < R$ with

(10.5)

H^{1} (S \cap C (x, I) \cap B (x, r)) \geq c_{0} r H^{1} (I) .

Note that (10.4) implies

(10.6)

H^{1} (H_{x} (r) \cap I) < c_{0} H^{1} (I) / 4 M .

We will prove that

(10.7)

H^{1} (D_{x} (R, M, ϵ) \cap I) \geq c_{0} H^{1} (I) / 4 M .

Since $I$ is any sufficiently small interval containing $e$ , this implies

Θ_{*}^{1} (D_{x} (R, M, ϵ), e) \geq c_{0} / 4 M > 0

as required.

By (10.6) we may cover $H_{x} (r) \cap I$ by disjoint intervals $I_{1}, I_{2}, \dots$ with

\sum i \in N H^{1} (I_{i}) < c_{0} H^{1} (I) / 4 M

(indeed, the disjointness of the intervals is shown in creftype 10.6). By the definition of $H_{x} (r)$ , we know that

(10.8)

S \cap C (x, I) \cap B (x, r) \subset ⋃ i \in N S \cap C (x, I_{i}) \cap B (x, r) .

Let $G$ be those $i \in N$ with

(10.9)

\frac{H^{1} (S \cap C (x, I_{i}) \cap B (x, r))}{r} \geq M H^{1} (I_{i}) .

Note that by (10.5) and (10.8),

	$\sum i \in G \frac{H^{1} (S \cap C (x, I_{i}) \cap B (x, r))}{r}$	$\geq \frac{H^{1} (S \cap C (x, I) \cap B (x, r))}{r} - \frac{c_{0} H^{1} (I)}{4}$
		$\geq \frac{3 c_{0} H^{1} (I)}{4} .$

That is, the cones associated to $G := ⋃_{i \in G} I_{i}$ cover a large proportion of $S \cap C (x, I) \cap B (x, r)$ . Moreover, $G \subset D_{x} (R, M, ϵ)$ , because if $ξ \in G$ then $ξ \in I_{i}$ for some $i \in G$ which satisfies the definition of $D_{x} (R, M, ϵ)$ .

Thus, to show (10.7), it would be enough to bound $H^{1} (G)$ from below by a multiple of $H^{1} (I)$ . But this is not necessarily true: the intervals that form $G$ could be extremely thin compared to $I$ . To accommodate this, we enlarge the intervals $I_{i}$ with $i \in G$ as follows. For each $i \in G$ enlarge $I_{i}$ until (10.9) becomes an equality or until $I_{i}$ intersects another $I_{j}$ . If the first possibility occurs then we still have $I_{i} \subset D_{x} (R, M, ϵ)$ . If the second possibility occurs then we merge the two intervals; since both sides of (10.9) are linear in $I$ , and the boundary of each $C (x, I_{i})$ contains no points of $S$ , (10.9) remains true after the merge. By the same reasoning, both sides of (10.9) are continuous under expanding $I$ and consequently one of the two possibilities must occur.

This results in a disjoint collection of intervals ${~ I}_{i}$ for which (10.9) is an equality. Moreover,

G = ⋃ i \in G I_{i} \subset ⋃ {~ I}_{i}

and, by construction, each ${~ I}_{i} \in D_{x} (R, M, ϵ)$ . Therefore

	$H^{1} (D_{x} (r_{0}, ϵ, M) \cap I)$	$\geq \sum H^{1} ({~ I}_{i})$
		$= \sum \frac{H^{1} (S \cap B (x, r) \cap C (x, {~ I}_{i}))}{r M}$
		$\geq \sum i \in G \frac{H^{1} (S \cap B (x, r) \cap C (x, I_{i}))}{r M}$
		$\geq \frac{3 c_{0} H^{1} (I)}{4 M} .$

∎

10.1. Exercises

Exercise 10.1.

Prove that the coordinate projections of the four corner Cantor set $K \subset R^{2}$ have Lebesgue measure zero.

Exercise 10.2.

A second proof that the four corner Cantor set is purely 1 unrectifiable.

Suppose that $f : R \to R^{2}$ satisfies $H^{1} (f (R) \cap K) > 0$ .

Prove that there exists $x \in R$ that is a density point of $f^{- 1} (K)$ such that $f^{'} (x) \neq 0$ .
Therefore, for sufficiently small $r$ , $f (x - r, x + r)$ is approximated by a line segment of length $2 r f^{'} (x)$ that is mostly contained in $K$ . Derive a contradiction.

Exercise 10.3.

Prove that the decomposition given in Lemma 10.3 is unique up to $H^{n}$ -null sets.

Exercise 10.4.

Think about Propositions 10.13 and 10.9 in regard to the four corner Cantor set.

Exercise 10.5.

Let $K \subset R^{2}$ be compact. Show that

{(x, e) : e \in H_{x}} and {(x, e) : e \in D_{x}}

are Borel subsets of $K \times S^{1}$ .

Exercise 10.6.

Show that in the definitions of $L^{n}$ and $H^{n}$ we may suppose the covering intervals, respectively sets, may be chosen to be disjoint.

Exercise 10.7 (Open problem).

For $k \geq 2$ , does there exist a compact purely 1-unrectifiable $S \subset R^{2}$ with $H^{1} (S) > 0$ that intersects every line in at most $k$ points?

References

[1] A. S. Besicovitch (1939) On the fundamental geometrical properties of linearly measurable plane sets of points (III). Math. Ann. 116 (1), pp. 349–357. External Links: ISSN 0025-5831, Document, Link, MathReview Entry Cited by: §10.
[2] T. Orponen (2018) Geometric measure theory. Online lecture notes. Cited by: §10.

Graduate real analysis

Contents

Part I General measure theory

1. Measures

Definition 1.1.

Definition 1.2.

Remark 1.3.

Definition 1.4.

Lemma 1.5.

Proof.

Theorem 1.6.

Proof.

Definition 1.7.

Theorem 1.8 (Carathéodory criterion).

Proof.

Definition 1.9 (Carathéodory construction).

Remark 1.10.

Definition 1.11.

Definition 1.12.

Definition 1.13.

Definition 1.14.

Lemma 1.15.

Proof.

1.1. Exercises

Exercise 1.1.

Exercise 1.2.

Exercise 1.3.

Exercise 1.4.

Exercise 1.5.

Exercise 1.6.

Exercise 1.7.

Exercise 1.8.

Exercise 1.9.

Exercise 1.10.

Exercise 1.11.

2. Integration

Definition 2.1.

Definition 2.2.

Definition 2.3.

Theorem 2.4 (Fatou’s lemma).

Proof.

Remark 2.5 (Reverse Fatou).

Theorem 2.6 (Monotone convergence theorem).

Proof.

Theorem 2.7.

Proof.

2.1. Exercises

Exercise 2.1.

Exercise 2.2.

Exercise 2.3.

Exercise 2.4.

Exercise 2.5.

Exercise 2.6.

Exercise 2.7.

3. Some standard theorems

Theorem 3.1 (Egorov’s theorem).

Proof.

Theorem 3.2 (Lusin’s theorem).

Proof.

Definition 3.3.

Theorem 3.4 (Radon–Nikodym).

Proof.

Theorem 3.5 (Lebesgue decomposition theorem).

Proof.

3.1. Exercises

Exercise 3.1.

Exercise 3.2.

Exercise 3.3.

Exercise 3.4.

Exercise 3.5.

Exercise 3.6.

Exercise 3.7.

4. The Daniell integral

Definition 4.1.

Theorem 4.2.

Proof.

Observation 4.3.

Proof.

Lemma 4.4.

Proof.