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 $a$

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.