Measure Theoretic Preliminaries

Measurable Spaces

Sigma Field

Definition (sigma-field).

Let be a given set, and be a collection of subsets in . is a -field (algebra) if:

: contains the entire set
If , then where : closed under complements
For any countable collection , we have : closed under countable unions

Here, note that the term 'sigma-' in mathematics usually implies 'countable union' or 'countably infinite'. Thus, -field means it is closed under the countable union, i.e., it contains every countable union of itself.

Furthermore, since countable set is isomorphic to , meaning that there exists an one-to-one correspondence between the two, we can understand -field as the set consists the countably infinite number of the subsets of , where we can index the each of the subsets in natural number bases.

Definition (measurable space).

For a -field on , elements in are called measurable sets, and the pair is a measurable space.

Note that it is more intuitive to understand the importance of ^32b063 Definition 1 (sigma-field) as a scope to narrow down the sets under our interest, more on the purpose of mathematical convenience.

Proposition (properties of sigma-field).

Let be a -field on . Then the following properties hold:

: contains empty set
For any finite collection , : closed under finite unions
For any countable collection , : closed under countable intersections
For any , : closed under set minus

Proof.The proof comes from ^32b063 Definition 1 (sigma-field) directly.

, and since is closed under complements, we have .
Let . Since and is closed under countable unions, we have .
Since , for every . Then by the closedness under countable unions and by the closedness under complements.
Let . Since and is closed under intersections, we have .
□

Proposition (intersection of sigma-field).

For a given set and -fields , is also a -field.

Proof.We check three properties of a ^32b063 Definition 1 (sigma-field) for :

Since for all , .
Let . Then for all . Thus, for all , therefore .
Let . Then for all . Thus, .
Therefore, is a -field. □

Definition (generated sigma-field).

Let be a collection of subsets of . The -field generated by is defined as the intersection of all -fields on that contain , and denoted as .

Remark (smallest sigma field).

is in the smallest -field containing .

Proof.Let be the countable collection of fields on . Now let be the index of containing . Then, is the smallest field containing by ^5a779e Proposition 4 (intersection of sigma-field). □

Borel Sigma-Field

Definition (Borel sigma-field).

Let be a topological space. The -field generated by the open sets in is called Borel -field and denoted as .

Remark (open sets as countable unions).

Any open set in can be represented as a countable union of open intervals.

Proof.If , thenand if , thenOther cases are trivial. □

Lemma (bases for Borel sigma-field).

For the following (), we have :

Proof.For , it is trivial from the definition.
For , since -field is closed under complements, , thus .
For , since , and -field is closed under complements, . Also, since and -field is closed under intersection, . Thus, .

Similarly, for , we have . □

Theorem (Borel sigma-field on euclidean space).

Let be the field generated by the open subsets of . Then where .

Proof.Remark that is the -field generated by the open sets in .

() We show that contains open sets in .

For an arbitrary open set and , there exists an open set that contains . Thus, any open set can be represented as a countable union of open sets in .
Note that this statement holds also for the case when includes or , since any open intervals can be represented as a countable union of open intervals by as shown in ^647970 Remark 8 (open sets as countable unions).

Therefore, as is the smallest -field containing open sets in , we have .

() Let be a collection of open sets in . We show that .

Remark that includes any open and closed sets in . For some , if is for some , then and if is or , then Thus, any element in can be represented as an intersection of open sets in . Since a -field is closed under countable intersection as a -field is closed under complements and countable unions, i.e.As is the smallest -field containing , we have . Combining the results, we have . □

Monotone Class Theorem

Definition (pi and lambda system).

Let be an arbitrary set and be a collection of subsets of .

is called a system if it is closed under intersection:
is called a system if the followings hold:
1. .
2. for any such that , we have .
3. for any sequence such that for all , we have .

Remark (sigma-field is pi-lambda system).

Any field is both a system and system.

Proof.Let be a field on .

(1) is a system since it is closed under intersection, as shown in third property of ^80c0c1 Proposition 3 (properties of sigma-field).

(2) is a system since:

by the first property of ^32b063 Definition 1 (sigma-field).
is closed under set minus by the fourth property of ^80c0c1 Proposition 3 (properties of sigma-field).
is closed under increasing union as it is closed under any unions by the last property of ^32b063 Definition 1 (sigma-field).

Thus is both a system and system. □

Lemma (pi-lambda system is sigma-field).

Let be a set and be a collection of subsets of . If is both a -system and a -system, then it is a -field on .

Proof.We will check the criteria for a -field in ^32b063 Definition 1 (sigma-field) one by one:

because is a -system.
Let . Since and , by the second property of a -system we have .
Let be an arbitrary collection of sets in . Now define as Then, is an increasing sequence of sets, where . If , trivially, . For any , because it is closed under complements by (2) above and finite intersections by the definition of a -system. Therefore, , and by the third property of -systems, .

Thus by definition, is a -field on . □

Lemma (intersection of lambda system).

For a given set and systems , is also a -system.

Proof.We check three properties of ^d386cb Definition 11 (pi and lambda system).

As for all , we have .
Suppose and . Then, for any , and for all . Thus, .
Suppose and for all . Then, for any , and for all . Thus, .

Therefore, is a system. □

Theorem (Dynkin's pi-lambda theorem).

Let be a system on and be a system on that contains . Then .

Proof.Define be the intersection of all systems containing : Thus is the smallest system containing . Our main goal is to show that is a field generated by .

Using ^1fb908 Lemma 13 (pi-lambda system is sigma-field) and ^21d26e Lemma 14 (intersection of lambda system), it suffices to show that is a system. Now let , then we need to show that .

(step 1) For a fixed , we show that is a system on , where We check three properties of ^d386cb Definition 11 (pi and lambda system):

since .
For any such that , we have where (as is a system and closed under intersection), , and as is a system. Thus .
Let be an increasing sequence. Then since is an increasing sequence and is a system. Thus .

Thus is a system.

(step 2) For any , we show that .

Let . Then for any , we have since is a system and closed under intersection. Then as , we have .
Thus for any , we have .

(step 3) For a fixed , as we have already seen in step 1, is also a system on , where Combining with the result in step 2, we have , and since is the smallest system containing , we have . Then by the definition of , we have Thus is a system, giving us that is a field generated by :

As shown in ^83bb5c Remark 12 (sigma-field is pi-lambda system), the field is also a -system, meaning that . Therefore, we have . This completes the proof. □

Measure

Definition (measure).

A set function is called a measure if for any countable collection of disjoint measurable sets , Additionally, we may add another condition for the lower bound: Furthermore, if , then it is a probability measure, denoted by .

Definition (measure space).

Let be a given set, be a -field on , and be a measure. The triple is called measure space. If is a probability measure, then is called probability space.

Theorem (properties of measure).

Let be a measure space. Then, the following properties hold:

If and , then : monotonicity
Let be any collection of sets, and , then : sub-additivity
If and for any , then : continuity from below
If and for any , then : continuity from above

Proof.

Assume and . Let . Then since and is closed under intersection by ^80c0c1 Proposition 3 (properties of sigma-field). Then and are disjoint, thus by ^c8f7a9 Definition 16 (measure), Since , we therefore have .
Since is an arbitrary collection of sets, we make another collection that is of disjoint sets and , where for each Now, let . Then by and , we have .
Then, as are disjoint sets, we have where the inequality holds since for all , and by the monotonicity of .
Construct a sequence of disjoint sets such that Then .
Thus, we have
Since for and , we have . Thus, Therefore,
Construct an increasing sequence as

Remark that , and . Then

Since is an increasing sequence,

As for all , and are disjoint, , thus

Finally, as , we have and thus This completes the proof. □

Proposition (sum of measures is measure).

Let be a sequence of measures on . Define as where the sum on the right-hand side is well defined due to the non-negativity of each . Then is actually a measure on :

Proof.We check ^c8f7a9 Definition 16 (measure):

since each for all .
For any disjoint , as are countably additive.

Therefore, is a measure on . □

Finite Measures

Definition (sigma-finite measure).

Let be a measure on . Then, we say

is finite if .
is finite if there exists a measurable and countable partition of () such that for any .
is finite if there exists a sequence of finite measures on such that .

Lemma (relationships between sigma-finiteness).

Let be a measure space. Then, the followings hold true:

If is finite, then it is finite.
If is finite, then it is finite.

Proof.(1) Suppose that . Now we construct a partition as then is a measurable where , thus by definition is finite.

(2) Suppose that is finite, i.e. there exists a measurable partition of such that for any . We first construct a sequence of measures and then show .

Now define as Then, we show that is a finite measure:

For any disjoint , as is a disjoint collection of sets, we have
.

Thus is a finite measures.

Now, for any , as is a disjoint collection of sets, we have As such, , and is -finite by definition.□

Theorem (uniqueness theorem on sigma-finite measure).

Suppose that and are measures on and , for a $\pi$-system -system . If and are both -finite on and they agree on , then they agree on .

Proof. First, let be such that , and define Since and ^83bb5c Remark 12 (sigma-field is pi-lambda system), is a -system that contains , hence it equals by ^aa7110 Theorem 15 (Dynkin's pi-lambda theorem).

By assumption, and are -finite. Then there exists a sequence such that , where . For each , As and , we have for all . Hence . □

Caratheodory's Extension Theorem

Definition (pre-measure).

A pre-measure on is defined as a function such that

null set:
finite additivity: for any finite disjoint set ,

Definition (outer-measure).

Definition (Lebesgue measure).

Definition (complete measure space).

Theorem (Caratheodory extension).

Let be a finite measure on the field of subsets of . Then has a unique extension to a measure on .

Almost Everywhere

Definition (almost everywhere).

Let be a measure space. We say that some property holds almost everywhere (a.e.) on if it holds true for any such that , and denote as .

Definition (equal almost everywhere).

Let are measurable functions. Then and are equal almost every on , or if

Measurable Function

Measurability

Definition (measurable function).

Let be a measurable space, and be a topological space. Then is -measurable if In particular, if , then is Borel measurable.

Proposition (domain of measurable function).

Let be a measurable space, and be a topological space. For a mapping , define Then is a -field in .

Proof.Remark that the inverse function preserves the set operations. Now we check the three conditions of ^32b063 Definition 1 (sigma-field) for .

, so .
Let , i.e. , then , thus .
Let , then for all . Then Thus .

Therefore, is a -field on . □

Theorem (measurability of function).

Suppose a function , for a measurable space . Let be a collection of subsets in such that . Then is -measurable if and only if for all .

Proof.() Since is -measurable, As , we have

() Let for all . We show that is -measurable.

First, let We show that contains . Since is the smallest -field containing and , this implies that is a -field:

, thus .
Let , then . Since is closed under complements, thus .
Let , then for all . Since is closed under countable unions, thus .

Therefore, is a -field and . This leads to Thus, is -measurable. □

In the most of the cases, we use the following corollary to check the measurability of the function.

Corollary (criteria for measurability).

By ^5ed7c8 Lemma 9 (bases for Borel sigma-field) and ^3cd757 Theorem 32 (measurability of function), for the function , the followings are equivalent:

is -measurable.
for any .
for any .
for any .
for any .

The proof is trivial.

Generated Sigma-Field

Definition (function-generated sigma-field).

Let be a measurable space and be a function. Then, is called as a field generated by , where

Remark (measurable function and sigma-field).

Let be a function and be a field generated by . Then we have the following results:

is a field on .
is measurable and is the smallest field on that makes measurable.
is measurable if and only if .

Proof.(1) We check three properties in ^32b063 Definition 1 (sigma-field).

because .
Let , i.e. such that . As is a field, by the closedness under complement. Then we have
Let be a countable collection. Then, there exist where for all . As is closed under countable union, . Thus we have Thus is a field on .

(2) Recall ^fc9748 Definition 30 (measurable function). Then, by ^aaba0f Definition 34 (function-generated sigma-field), meaning that is measurable. It follows trivially that is the smallest field that makes measurable.

(3) From (2), it can be shown directly. □

Exercise (2.1 family and function generated sigma field).

For a generated field , generates .

Proof.Here, we need to show that :

() Note that by ^aaba0f Definition 34 (function-generated sigma-field), and is the smallest field containing , thus .

() Assume , where . Then it can be represented as a countable unions, intersections, or complements using , since the inverse image preserves the set operations. As field is closed under those operations, we have . Thus . □

Measurability under Basic Operations

Theorem (measurability under various operations).

Let be a measurable space and be a topological space.

Let be a continuous function, then is Borel measurable.
Let be an -measurable function and be a Borel measurable function. Then is -measurable.
Let be -measurable functions. Then, the following are all -measurable as well:
Let be an -measurable function, then for any is measurable.

Proof.

Assume that is continuous, thenthus is Borel measurable.
Let be an arbitrary Borel set. Then we have and since is Borel measurable, . Now let . Then,, as is -measurable, .
By 1 and 2, it is trivial to show and by letting them as a composition, By letting and continuous function, respectively.
Or, we can prove directly by the denseness of the real space as and Now consider the case when . For any , by the denseness of in , we have since is -measurable, for all , thus its countable union is also -measurable. Therefore, is -measurable.
Since is continuous function, by 1, is is Borel measurable. Thus by 2, is -measurable.

□

Theorem (measurable functions and measurable set).

Let be a measurable space, and suppose are -measurable functions. Then, the sets are all -measurable sets.

Proof.By the denseness of in , we have Since are measurable in , by ^91cf59 Corollary 33 (criteria for measurability), for any . Then by the closedness under union and intersection of ^80c0c1 Proposition 3 (properties of sigma-field), it follows that .

Likewise, because by ^6abc3b Theorem 37 (measurability under various operations).

In addition, and by the closedness under complement of ^32b063 Definition 1 (sigma-field).

Finally, by the closedness under intersection, and by the closedness under complement as well. □

Measurability under Limits

Theorem (measurability under limsup and liminf).

Let be a measurable space, and be a sequence of measurable functions. Then, the followings are all measurable:

Proof.Note that by definition, and

We first prove the measurability of supremum: for any , since are measurable functions, and is closed under countable intersection by (3) of ^80c0c1 Proposition 3 (properties of sigma-field), its supremum is measurable.
By (3) of ^6abc3b Theorem 37 (measurability under various operations), if is a sequence of measurable functions, is also measurable. Since the supremum is also measurable.
Using (1), for any , we have since is measurable, and is closed under countable intersection. Thus is measurable.
From (2), we have is measurable since is measurable functions.
□

Remark (limit, limsup, and liminf).

Let be a sequence in . Then exists if and only if

Proof.Remark that for a sequence , and

() Assume . We want to show that , i.e., First, let . Then, since , we haveAs for all and , we have Similarly, let . Then, since , we haveAs for all and , we have Therefore, by letting , we have for all , which equals . Thus, .

() Assume exists and its limit is . We want to show that .

As converges in , is bounded. Thus, its and exist.

By the assumption, for given arbitrarily, we have By taking supremum, we havewhich means .

Similarly, by taking infimum, we have Thus, . Therefore, . □

Corollary (measurability under limit).

Let be a measurable space, and be a sequence of measurable functions. If the sequence has a pointwise limit , its limit is measurable.

Proof.Let be a sequence of -measurable functions with a pointwise limit .
Then, for any , we havewhere the last two equalities come from ^cfd91c Remark 40 (limit, limsup, and liminf). Therefore, Since the latter two functions were shown to be measurable in ^2d4323 Theorem 39 (measurability under limsup and liminf), is measurable as well. □

Measurability of Positive and Negative Parts

Definition (positive and negative parts of function).

Let , then the positive and negative parts of are defined as Remark that the entire function can be decomposed as

Lemma (measurability of positive and negative parts).

Let be a measurable space, and . Then is measurable if and only if and are measurable.

Proof.
() Suppose is -measurable. The function defined as for any is -measurable, since for any .

Now we can write and . As the measurabilities under supremums and infimums are preserved by ^2d4323 Theorem 39 (measurability under limsup and liminf), and the both and are measurable, its positive and negative parts are also measurable.

() Suppose that are measurable. Then, for , and for , Therefore, for any , thus, is -measurable. □

Simple Functions

Measurable Simple Function

Proposition (measurability of indicator function).

For a set , an indicator function is defined asand is -measurable if and only if (i.e., is a measurable set).

Proof.() Let be -measurable, then ,

Thus, must be a measurable set.

() As is a measurable set, and . Trivially, and . □

Definition (simple function).

Let be a measurable space. A simple function is a function that maps into a finite subset of , which can be represented as where are disjoint and for all . Furthermore, we denote the set of non-negative simple functions as

Proposition (measurability of simple function).

Let be a simple function on measurable space with canonical form Then is -measurable if and only if are pairwise disjoint.

Proof. Assume is measurable. then we have since , by ^fc9748 Definition 30 (measurable function).

Assume . Then, as , we have Now define for . If , then , while if , then and .
Therefore, and . This holds for any , so is -measurable. □

Lemma (operations on simple function).

Let be a measurable space, and be measurable simple functions. Then, and are also measurable simple functions.

Proof.Let and be given asand without loss of generality, assume . Now, define for any and . Since and are disjoint, we have where the last inclusion holds since and by the closedness under intersection from ^80c0c1 Proposition 3 (properties of sigma-field). Thus, is a measurable simple function.

For the product, we have and the last inclusion holds by the similar reason. □

Approximation by Simple Function

Theorem (approximation by simple functions).

Let be a measurable space, and a non-negative -measurable function. Then, there exists an increasing sequence of -measurable simple functions such that pointwise as .

We first construct a sequence of simple functions that rounds the value of function at the lower bound of the interval , and that ignores the extreme large values (). Thus is an increasing function that gets closer to the from below, as the interval gets smaller as increases.

Proof.Define Now define for . Then have canonical form as First, we show that is an non-decreasing sequence, i.e. for any :

if , then .
if , then by letting for some , we have
if , then by letting for some , we have which is larger or equal than .

Finally, we show that pointwise as . Choose any and any .

If , then let for any . Thus as .

On the other hand, suppose . Then, there exists a natural number such that , and for any , there exists a such that . As such, so that for any as . Therefore, by definition, . □

Corollary (operations on non-negative measurable function).

Let be a measurable space, and be non-negative measurable functions. Then, and are also non-negative measurable functions.

Proof.By ^d0d8b8 Theorem 48 (approximation by simple functions), there exist sequences such that and . As by ^e79b5c Lemma 47 (operations on simple function), , and the measurability is preserved under limits by ^58dafc Corollary 41 (measurability under limit), is also a non-negative measurable function.

Similarly, is also a non-negative measurable function as and . □

Corollary (generalization of approximation on simple function).

Let be any measurable function on . Then, there exists a sequence of measurable simple functions such that . Note that is not necessarily in .

Proof.By ^fce219 Lemma 43 (measurability of positive and negative parts), any measurable function can be decomposed into non-negative measurable functions where

Then, by ^d0d8b8 Theorem 48 (approximation by simple functions), there exist sequences such that , . Then, and is a measurable function, since measurability is preserved under linear transformation y ^6abc3b Theorem 37 (measurability under various operations). □

Abstract Integration

Integration of Simple Functions

Definition (integral of simple function).

Let be a measure space and be measurable simple function given as Then for any , the integration of with respect to over is defined as which takes values in .

Lemma (properties of integration on simple function).

Let be a measure space and be a non-negative measurable simple functions. Then, the followings hold true:

integration is well defined: for any , the mapping is a function.
monotonicity: if , then for any , .
linearity: for any and any , .

Proof.We first prove (2), then show it implies (1).

(2) Suppose and where where , and both are disjoint. Now, define for any and . Then, and are disjoint.

Then, by ^c8f7a9 Definition 16 (measure), we have and

Now, make sub-partitions for , as and note that for any , since for any , for any ,

Therefore, which proves (2).

(1) Suppose that has two representations as Now, label , . Then, by the result proved above, for any . Therefore, the integral of under over is invariant to the representation of , meaning that the mapping is a function for any . This proves (1).

(3) Suppose that where and let .

Because for any , is also a measurable simple function.

As above, define the disjoint sets for any , . Then by ^e79b5c Lemma 47 (operations on simple function), that we have Furthermore, using the proof of (2), This proves (3). □

Integration of Non-negative Functions

Definition (integral of non-negative function).

Let be a non-negative measurable function on . Then, for any , the integral of with respect to over is defined as where is the set of all measurable simple functions.

Theorem (properties of integral 1).

Let be a measure space, and be measurable non-negative functions. The following hold true:

integration is well defined: for any , the mapping is a function.
Monotonicity 1 : If , then for any , .
Monotonicity 2: If and , then .

Proof.(1) Note that the supremum in ^ab00c3 Definition 53 (integral of non-negative function) is well defined on : If the set is bounded above, then by the least upper bound property of real numbers, it has a supremum in , while if it is unbounded, then its supremum is . Furthermore, this value is unique due to the uniqueness of the supremum, which means that the mapping is a function.

(2) Choose any . Assume is a simple measurable function such that . Then, as , we have for any such . Thus is an upper bound of the set , and by the definition of the supremum,

(3) Choose such that . For any simple measurable function , by definition because for and measures are monotonic (^cdc44b Theorem 18 (properties of measure)).

Therefore, for any simple measurable function such that , so that is an upper bound of the set and thus This completes the proof. □

Theorem (properties of integral 2).

Let be a measure space, and be measurable non-negative functions. The following hold true:

For any , if for any , then .
For any and , .
If for some , then .
For any , .

Proof.(1) Let and suppose for any . Suppose is a measurable simple function such that . Then, for any .

Let . We show that this implies that for : RTA, suppose otherwise, for some . then letting , , a contradiction.

Therefore, This holds for any simple measurable function such that , so and as such

(2) Choose any and . If , then for any , so that by result (1).

Suppose on the other hand that . Then, for any simple measurable function such that , letting , we have is a simple measurable function such that , so

by ^b5d6e1 Lemma 52 (properties of integration on simple function).

Since the above result holds for any simple measurable function such that , so Since and is also a non-negative measurable function, applying the above result implies that or . Therefore,

(3) Suppose that for some . Then, for any simple measurable function such that , let . Because , we have for . This gives us which holds for any simple measurable function such that , so . Thus

(4) For any , first note that is a non-negative measurable function becausefor any .

() Let be a simple measurable function such that , and let , and as such and as such since .

This holds for any simple measurable function such that , so () To show the reverse inequality, suppose is a simple measurable function such that . Then, for any , . Now let for , and this means that for ; RTA, if there exists an such that , then for any , a contradiction.

Therefore,

This holds for any simple measurable function such that , so

Therefore we can conclude that . □

Monotone Convergence Theorem

Lemma (convergence of integral of simple function).

Let be a measure space, and be a simple -measurable function. Then, for any sequence of disjoint measurable sets , by letting , we have where the limit on the right exists in .

Proof.Let be the canonical form of . Then, we have We have the desired result. □

Theorem (monotone convergence theorem, MCT).

Let be a measure space, and a sequence of increasing non-negative -measurable functions. Then, the pointwise limit of exists and where the limit on the left hand side exists in .

Proof.First, define such that pointwise. Here, exists since is an increasing sequence in for any : If is bounded, then for some , while if is unbounded, then .

Now we show the equality .

() As is an increasing sequence, for any . Thus, by the monotonicity of integration (^c52afc Theorem 54 (properties of integral 1)), for any . Thus is an increasing sequence in bounded above by : where the limit on the left hand side exists in because is increasing.

() We show that . Let be a simple measurable function such that . Then by ^6abc3b Theorem 37 (measurability under various operations), is non-negative measurable functions for any .

Now define which is a measurable set by ^f6c8a0 Theorem 38 (measurable functions and measurable set). Also, as for all , we have i.e. an increasing sequence.

Now, we show that there exists some such that for all for any . If , then , and because , there exists an such that In other words, for any . On the other hand, if , then trivially for any as are non-negative.
Thus, we have .

By the monotonicity and linearity in ^b5d6e1 Lemma 52 (properties of integration on simple function), for any , we have Thus by ^a224ef Lemma 56 (convergence of integral of simple function), we have As this inequality holds for any simple function and for any , we have for any , and taking on both sides, we have Therefore, we have . □

Linearity of Integral

Corollary (linearity of integration).

Let be a measure space, and non-negative measurable functions. Then,

Proof.Let and be increasing sequences of simple measurable functions such that to and respectively.

Then by ^e79b5c Lemma 47 (operations on simple function), is a sequence of simple measurable functions increasing to the non-negative measurable function .

Because the integration of simple functions is linear by ^b5d6e1 Lemma 52 (properties of integration on simple function), for any . By ^c19a89 Theorem 57 (monotone convergence theorem, MCT), we can see that which completes the proof. □

Indefinite Integrals and Radon-Nikodym

Corollary (indefinite integrals).

Let be a measure space, and a non-negative measurable function. Then, the function defined as is a measure on . Moreover, for any measurable function , we have

Proof.Let be the sequence of simple measurable functions such that . Then, as the product of measurable simple function is also a measurable simple function (^e79b5c Lemma 47 (operations on simple function)), is a simple measurable function for any and for any .

First, we show that is measure satisfying ^c8f7a9 Definition 16 (measure):

by ^b702bf Theorem 55 (properties of integral 2) and .
Let be any sequence of disjoint measurable sets, and let . Then we have Thus, by definition, is a measure on .

Next, we show the equality .

Without loss of generality, let be a non-negative simple measurable function with canonical form Then, by the linearity of ^b5d6e1 Lemma 52 (properties of integration on simple function), Now let be an arbitrary non-negative measurable function. Letting be a sequence of simple measurable functions such that , then by ^47346b Corollary 49 (operations on non-negative measurable function), is a non-negative measurable functions increasing to . Thus, by repeated applications of ^c19a89 Theorem 57 (monotone convergence theorem, MCT), we finally have which completes the proof. □

Remark (Radon-Nikodym derivative).

Let and be a measures on . Then there exists a non-negative measurable function such that and is called the Radon-Nikodym derivative, or the density of with respect to .

Definition (absolutely continuous of measures).

For two measures on , we say is absolutely continuous with respect to , denoted as , if

Proposition (understanding absolute continuity).

For two measures on a measurable space , if and only if

Proof.() RTA: assume such that By letting , we have but , which gives us a contradiction.

() Assume for all , let . Then we have meaning . □

Theorem (Radon-Nikodym theorem).

Let and be $\sigma-$finite finite measures on . Then, , if and only if there exists a Radon-Nikodym derivative of with respect to .

Proof.Note that () part is trivial from ^a7bc19 Corollary 59 (indefinite integrals). Thus it remains to prove the converse part.

() Assume that there exists a Radon-Nikodym derivative of with respect to . Now suppose for some , . Then by the first of ^b702bf Theorem 55 (properties of integral 2), we have which completes the proof. □

Fatou's Lemma

Corollary (Fatou's Lemma).

Let be a measure space, and be a sequence of non-negative measurable functions. Then, for , we have

Proof.By the definition of the limit inferior, for any , we have where is a non-negative measurable function since is bounded below by for all and measurability is preserved under infimums by ^58dafc Corollary 41 (measurability under limit). Also, as we have .

Thus, is a sequence of non-negative measurable functions that increases to . By the ^c19a89 Theorem 57 (monotone convergence theorem, MCT), we have where the inequality holds by the monotonicity in ^c52afc Theorem 54 (properties of integral 1) and for any . □

Integration of Numerical Functions

Definition (integral of numerical function).

Let be a measure space and be a measurable function. We say is integrable if and define the integral over any as

Proposition (integrability and absolute function).

Let be a measure space and where Then is integrable if and only if is integrable.

Proof.() Suppose that is -integrable. Then, and take values in by definition. Then by ^d9bb31 Corollary 58 (linearity of integration), () Suppose . Then, by ^d9bb31 Corollary 58 (linearity of integration), as , Since and are integrals of non-negative functions and their sums are finite, and themselves are also finite. Therefore, is -integrable by definition.

Therefore, is -integrable if and only if is -integrable. □

Theorem (properties of integrability).

Let be a measure space, and numerical functions that are -measurable. Then, the following hold true:

If is -integrable, then for any is also -integrable and
Monotonicity: If are -integrable and , then
Linearity: For any , if are -integrable and real-valued, then is an -measurable and -integrable real-valued function such that
If is -integrable, then

Proof.(1) In ^b702bf Theorem 55 (properties of integral 2), we have already shown that if a function is non-negative measurable function, then . Therefore, using ^fda7b8 Definition 42 (positive and negative parts of function), we have (2) Suppose . Then, Both and are non-negative measurable functions, so by the monotonicity ^c52afc Theorem 54 (properties of integral 1) and ^d9bb31 Corollary 58 (linearity of integration), we have
As are -integrable, by ^252974 Definition 65 (integral of numerical function), each of the integrals are finite. Thus we have

(3) Since are measurable, by ^6abc3b Theorem 37 (measurability under various operations), is also measurable.

First, we show that is integrable. We divide the case by the sign of .
If , as and , both positive and negative parts of are finite: thus by ^ab00c3 Definition 53 (integral of non-negative function), is -integrable, and is computed by On the other hand, for , we have and , and still both positive and negative parts of are finite and is integrable. Similarly, it is computed as Thus we can conclude that for any , is integrable and .

Now denote . Since and there integrations are both finite, is integrable by ^252974 Definition 65 (integral of numerical function). Note that and by rearranging the term, we get where the both sides of the equation is composed of non-negative measurable functions. Therefore, by ^d9bb31 Corollary 58 (linearity of integration), we have which gives us

(4) As by ^252974 Definition 65 (integral of numerical function), we have which completes the proof. □

Dominated Convergence Theorem

Theorem (dominated convergence theorem).

Let be a sequence of measurable functions, and let and be measurable functions such that

a.e.
a.e.
.

Then,

Proof.By assumption, we have a.e. , hence the followings hold: () It follows from ^cb5248 Corollary 64 (Fatou's Lemma) and ^d9bb31 Corollary 58 (linearity of integration) that therefore,

() Similarly, it follows that thus

Together, we have which concludes the proof. □

Integration of Complex Functions

Measurability for Complex Functions

Integrability of Complex Functions

Product Measures

Product Sigma-Field

Definition (rectangles).

Let be measurable spaces. The set of all measurable rectangles is defined as Then is a collection of subsets of the product space .

Remark (rectangle is pi-system).

is a $\pi-$system system on .

Proof.Consider any , then we have
where we used the fact that algebras are closed under finite intersections.□

Definition (product sigma-field).

Let be two measurable spaces. The product field on is defined as where is a measurable rectangles.

Remark (set product is not product sigma-field).

Note that as may not be closed on complement or unions. For instance, consider or .

Remark (product borel sigma-field).

Let be the Borel $\sigma-$field Borel field on real space. Then, the product Borel field satisfies and in general, we have

Product Measures

Theorem (uniqueness of product measure).

Let and be two measure spaces, where are $\sigma-$finite finite. Then, there exists a unique measure on such that Here, is called the product measure, denoted as .
In general, for finite measure spaces , the unique product measure on is

Proof.Here, we only prove for the case when two measure spaces. The general case is directly followed using induction from there.

(finiteness) First we show that any such measure must be -finite. Since and are -finite, there exist and such that , , and are finite for all .
Now let for all and . Then . Thus, we have implying that is -finite.

(uniqueness) Suppose there are two measures and satisfying the condition. By ^aab6e6 Remark 70 (rectangle is pi-system), is a -system. and are both -finite and agree on . By the ^9d6b39 Theorem 22 (uniqueness theorem on sigma-finite measure), they agree on the generated -field , i.e., , which means such measure must be unique.

(existence) For any , let where as introduced previously. Then is a measure. Then for any , , Hence such a measure exists. □

Fubini's Theorem

Theorem (Fubini theorem).

Let and be two measurable spaces where and are -finite measures. Let be the product measure on . Let be a nonnegative measurable function. Then the following holds: Furthermore, this be extended to integrable functions with respect to the product measure , i.e. .

Proof.Here, we first prove the case when is indicator function, and then prove step by step to show the case when is the real-valued function.

(Step 1) Suppose for some , where for and . Then, define and similarly, we also have (Step 2) Let where . Then by linearity of ^b5d6e1 Lemma 52 (properties of integration on simple function), we have (Step 3) Let , then by ^d0d8b8 Theorem 48 (approximation by simple functions), there exist a sequence of simple functions such that . Then by ^c19a89 Theorem 57 (monotone convergence theorem, MCT), we have

(Step 4) Let be a real-valued function, where . Then by ^d9bb31 Corollary 58 (linearity of integration), we have This completes the proof. □