Graduate real analysis
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 .
Definition 1.1.
Definition 1.2.
Let be a measure on a set . A set is measurable if, for every ,
(1.1) 
Remark 1.3.
Definition 1.4.
If is a measure on a set and , the restriction of to is defined as
Lemma 1.5.
Let be a measure on a set and . Then is a measure on and any measurable set is also measurable.
Proof.
The fact that is a measure follows immediately from the fact that is a measure. If is measurable, then for any ,
as required. ∎
Theorem 1.6.
Let be a measure on a set and let be the set of measurable subsets of .

If then and .

is countably additive on . That is, if are disjoint then

If then

If and then
Proof.
We first prove (1) for finite unions and intersections. If then for every ,
by subadditivity. Thus is measurable and induction gives finite unions. Taking complements gives finite intersections.
To prove (2) note that the inequality is given by subadditivity. For the other inequality, for each let be disjoint and for each let
which is measurable by (1). Note that
and that this union is disjoint. Therefore, since is measurable,
since the are all disjoint. Therefore, by induction, for each . Finally, for each , since is monotonic,
and so letting gives (2).
Finally, to prove (1) for countable unions, for each let
an increasing sequence, and let with . Since the are measurable,
using the fact that the are measurable in the second equality. Taking complements shows that countable intersections of measurable sets are measurable. ∎
Definition 1.7.
A collection of subsets of a set is a algebra if

;

;

.
Theorem 1.6 shows that the set of measurable sets is a algebra.
The Borel algebra of a topological space is the algebra generated by the open (respectively closed) subsets of . It will be denoted by and its elements called the Borel subsets of .
A measure for which all Borel sets are measurable is a Borel measure. It is Borel regular if for every there exists a Borel with .
Theorem 1.8 (Carathéodory criterion).
Let be a metric space and a measure on which is additive on separated sets. That is, whenever with
we have
Then is a Borel measure.
Proof.
Let be closed and with . We need to show
For each let
and
Since is closed,
Moreover, the with odd are pairwise separated so
and so the sum is convergent. Similarly the sum over even indices is convergent and so
Therefore
using the additivity on separated sets for the equality and countable subadditivity for the second inequality. ∎
Definition 1.9 (Carathéodory construction).
Let be a metric space, a set of subsets of and . For each and define
where the infimum is taken over all countable
such that
Finally, define .
For any , is a measure, as is . Theorem 1.8 shows that is a Borel measure on . Indeed, if then
If consists only of Borel sets then is Borel regular.
Remark 1.10.
The fact that whenever implies that
Definition 1.11.
We define some properties of a measure on a topological space .

is locally finite if every point in has a neighbourhood of finite measure.

is finite if there exist measurable with and .

is finite if .

A Borel regular measure is a Radon measure if

for all compact ,

for all Borel .

for all Borel .

Definition 1.12.
Let be a measure on a set . A property of points in holds almost everywhere (or a.e.) if the set of points for which the property doesn’t hold has measure zero.
Definition 1.13.
Let be sets, be a measure on and let . The push forward of under , written is defined by
Definition 1.14.
The Lebesgue measure on , denoted , is defined using the Carathéodory construction with the set of cubes and . Its measurable sets are called the Lebesgue measurable subsets of .
The following lemma is very useful.
Lemma 1.15.
Let be a finite measure on a set and let be a set of measurable subsets of . There exists disjoint such that any with
satisfies .
In particular, if each measurable subset of of positive measure contains an element of of positive measure then we can decompose almost all of into countably many disjoint elements of .
Proof.
We find the by induction. First let be countable and disjoint such that
Now let be the set of all that are countable, disjoint and disjoint from . Let be such that
Inductively, given countable, disjoint such that each and are disjoint for , let be the set of all that are countable, disjoint and disjoint from . Let be such that
We claim that any with
satisfies . If not, let be such that . Then and
a contradiction. ∎
1.1. Exercises
Exercise 1.1.
Usually in measure theory, a measure is defined as a countably additive function defined on a algebra. However, using our definition is simply a convenience rather than a restriction.
Indeed, suppose is a countably additive function defined on a algebra of with . Show that it can be extended to the power set of by
and that any is measurable. What about
Conversely, any measure is countably additive when restricted to any algebra of measurable sets.
Exercise 1.2.
Let be a set of subsets of a set . Show that is a algebra. Note that it is the smallest algebra of containing .
Exercise 1.3.
Show that the following sets are Borel subsets of : , , the set of points in whose first decimal is even.
Let . Show that the set of points where is continuous is a Borel set. What about the set of points where is differentiable?
Exercise 1.4.
Let be a set and . The Dirac measure at is defined as if , otherwise. Show that is a measure on . What are its measurable sets?
Define the counting measure on to be the cardinality (finite or ) of any subset of . Show that this is a measure. What are its measurable sets?
Exercise 1.5.
For a metric space and the dimensional Hausdorff measure on , denoted , is defined using the Carathéodory construction with the set of all sets and .

Show that and are nonzero, translation invariant and homogenous measures. That is, for any , and , and (and similarly for ).

On show that there exists a such that .

Let be an Lipschitz function between two metric spaces. Show that for any and ,

For any metric space , show that is the counting measure on .

For , suppose that . Show that . Hence there exists a single for which for all and for all . This is called the Hausdorff dimension of , denoted .
Exercise 1.6.
The Cantor set is defined as follows. Let and for each let be obtained from deleting the “middle third” open interval from each of the intervals in . That is, ,
etc. Define . Note that is compact and hence Borel.

Show that is uncountable.

Let . Show that .
In particular, is an uncountable subset of with .
Exercise 1.7.
Give an examples of with for which is:

finite,

finite but not finite,

not 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 nonmeasurable sets.
Define a Vitali set as follows. Consider the equivalence relation on defined by iff . By the density of in , each equivalence class intersects . Therefore, by the axiom of choice(!), we may construct a set consisting of exactly one member of each equivalence class.
Show:

If are rational then and are disjoint.

.

Show that .

Deduce that 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 be a finite Borel measure on a metric space . Prove that for every Borel ,
(1.2) 
and
(1.3) 
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
is a algebra that contains all closed subsets of .
Show that a finite is inner regular by closed sets. Show that a finite is outer regular by open sets if there exist open sets with for all and . Give an example of a finite that is not outer regular by open sets.
Exercise 1.11.
Let be a complete and separable metric space. Show that any finite Borel measure on 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 be a measure on a set . A simple function is any function of the form
where each and the are disjoint measurable sets. We treat .
Definition 2.2.
Let be a measure on a set and let . The (lower) integral of with respect to is
Definition 2.3.
Let be a measure on a set . A function is measurable if is measurable for every .
For measurable, let and (both measurable), so that and . If one of and are finite, we say that is integrable and we define the integral of with respect to to be
If only (respectively ) we write (respectively ).
Let be a topological space. A function is a Borel function if is a Borel set for every .
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 be a measure on a set and measurable. Then
Proof.
Let
be a simple function with
for each and each and let . For each , the sets
monotonically increase to as increases. Therefore
and hence
Since is arbitrary, the conclusion follows. ∎
Remark 2.5 (Reverse Fatou).
Suppose that there exists with and for all . Then
Indeed, this follows by applying Fatou’s lemma to .
Theorem 2.6 (Monotone convergence theorem).
Let be a measure on a set and measurable. Suppose that for every and all , . Then
Proof.
The monotonicity of the integral gives whilst Fatou’s lemma gives . ∎
Theorem 2.7.
Let be a measure on and measurable such that pointwise. Suppose that there exists measurable with such that for all . Then
Proof.
Observe that for all , and that . Then by the reverse Fatou lemma,
∎
2.1. Exercises
Exercise 2.1.
For a measure on a set , let be measurable, respectively Borel. Show that the preimage of any Borel is measurable, respectively Borel. Compare this to the definition of a continuous function.
Exercise 2.2.
Let be a measure on and for each let be measurable. Show that the functions
are measurable.
Show that a linear combination of measurable functions is measurable. Show that a countable (pointwise) sum of measurable functions is measurable.
Exercise 2.3.
There are some simple properties of the integral to check:

If a.e. then

The integral with respect to is a linear operator;

If is measurable then

;

etc…
Exercise 2.4.
Show that is measurable if and only if
for every and .
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 .
Show that the monotone convergence theorem is false if the sequence does not monotonically increase.
Exercise 2.7.
Let be sets, a measure on and . Show that is a measure on . If are topological spaces and are Borel, show that is a Borel measure on .
3. Some standard theorems
Theorem 3.1 (Egorov’s theorem).
Let be a finite measure on a set and a sequence of measurable functions such that pointwise a.e. Then for every there exists a measurable with and uniformly on .
Proof.
Fix , and for each let
By assumption, the are measurable sets that monotonically decrease to a null set as . Therefore, there exists such that . Let
Then and, for each and each , , so there exists such that
for all . That is, uniformly on , as required. ∎
Theorem 3.2 (Lusin’s theorem).
Let be a finite Borel measure on a metric space and let be measurable. Then for every there exists a closed with such that is continuous.
Proof.
Fix and for each let
a collection of disjoint Borel sets which cover . Since , there exists such that
Since is Borel, for each there exists a closed with .
For a moment fix and let
Since is closed, the sets monotonically decrease to as , which is disjoint from . Therefore there exists such that
satisfies . Note that is closed.
Let and , a closed set. For any and and , . Therefore, if with , . Repeat this for each with and let , so that . Then is continuous on . ∎
Definition 3.3.
Let be measures on a set . We say is absolutely continuous with respect to , written , if for every , . We say that is singular with respect to , written if there exists with .
Let be a measure on a set and let . The set valued function
defines a measure on . Note also that . The RadonNikodym theorem provides the converse to this statement.
Theorem 3.4 (Radon–Nikodym).
Let be finite measures on a set such that . There exists a and measurable such that
for all and measurable . The function is called the RadonNikodym derivative of with respect to .
Proof.
Let be the set of all and measurable such that
for all and measurable . Note that and
(3.1) 
Let
so that , and let be such that
Eq. 3.1 implies that we may suppose the monotonically increase. Let be the pointwise limit of the . Then is and measurable and, by the monotone convergence theorem, and . Thus
(3.2) 
We claim that satisfies the conclusion of the proposition. Indeed, suppose that is measurable but
and let be such that
(3.3) 
Let be the collection of all and measurable such that
We claim that there exists a and measurable of positive measure such that each and measurable of positive measure is not contained in . Indeed, if not, then each of positive measure contains an element of of positive measure. Thus Lemma 1.15 gives a countable disjoint decomposition of almost all of into elements of . Since this implies
contradicting (3.3).
Note that . Indeed, if is measurable,
On the other hand, since ,
contradicting the definition of . ∎
Theorem 3.5 (Lebesgue decomposition theorem).
Let be finite measures on a set . There exists a measurable with such that, for all , . That is, with and .
Proof.
Let be the set of all measurable with . By Lemma 1.15, there exists such that each with
satisfies . Since , this is the required decomposition. ∎
3.1. Exercises
Exercise 3.1.
Let be a Borel measure on a metric space , integrable and .

Show that there exists a simple function such that

If is positive show that we may require in the previous point.

Show that if is finite, there exists with

Show that the previous point may fail if is only finite.
Exercise 3.2.
Prove the following variant of Lusin’s theorem for the case that is not finite but is integrable: for every there exists a closed with
such that is continuous.
Exercise 3.3.
Let be measures on a set . Show that if is measurable then it is also measurable.
Exercise 3.4.
Prove the RadonNikodym and Lebesgue decomposition theorems for finite measures.
Exercise 3.5.
Let be finite measures on a set and suppose . For any measurable show that
where is the RadonNikodym derivative of with respect to .
Exercise 3.6.
For a measure on a set , we say a function is measurable if each component of is measurable and define the integral of component by component.
An dimensional vector valued measure is a function
for which there exists a measure on and a measurable such that
for each .

Show that if
are two representations of a vector valued measure then (when restricted to the set of measurable sets), and hence a.e. We denote this unique measure by . It is called the total variation of .

Show that the set of all vector valued measures form a normed vector space when equipped with

Show that this space is complete.
A signed measure is a 1dimensional vector valued measure.
Exercise 3.7.
Let be a algebra on . The standard definition of a signed measure on is a countably additive function
The Hahn decomposition theorem states that there exist disjoint with such that:

For every with , and

For every with , .
That is, and 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 be a measure on a set and let be a set of real valued measurable functions on . The formula
defines an operator on . Moreover, it has the following two properties:

is monotonic: for all with , ;

is continuous with respect to monotone convergence: if monotonically increase to then .
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 be a set. A lattice of functions on is a nonempty set of functions which satisfies the following conditions: for any and any , , , and all belong to and if then too. Note that any vector space of functions closed under is a lattice.
If is a lattice we let
We say a is

linear if, for all and