This document defines key concepts in measure theory and integration, including σ-algebras, measurable spaces, measures, measure spaces, properties of measures (monotonicity, countable additivity), completion of measure spaces, and locally measurable sets. It provides examples of common measure spaces such as Lebesgue measure on the real line. The document establishes several important lemmas about measures, such as monotonicity, countable subadditivity, and limits of increasing/decreasing sequences.
Some forms of N-closed Maps in supra Topological spacesIOSR Journals
In this paper, we introduce the concept of N-closed maps and we obtain the basic properties and
their relationships with other forms of N-closed maps in supra topological spaces.
Intuitionistic First-Order Logic: Categorical semantics via the Curry-Howard ...Marco Benini
A novel approach to giving an interpretation of logic inside category theory. This work has been developed as part of my sabbatical Marie Curie fellowship in Leeds.
Presented at the Logic Seminar, School of Mathematics, University of Leeds (2012).
Some forms of N-closed Maps in supra Topological spacesIOSR Journals
In this paper, we introduce the concept of N-closed maps and we obtain the basic properties and
their relationships with other forms of N-closed maps in supra topological spaces.
Intuitionistic First-Order Logic: Categorical semantics via the Curry-Howard ...Marco Benini
A novel approach to giving an interpretation of logic inside category theory. This work has been developed as part of my sabbatical Marie Curie fellowship in Leeds.
Presented at the Logic Seminar, School of Mathematics, University of Leeds (2012).
Gamma Function mathematics and history.
Please send comments and suggestions for improvements to solo.hermelin@gmail.com. Thanks.
More presentations on different subjects can be found on my website at http://www.solohermelin.com.
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ON OPTIMALITY OF THE INDEX OF SUM, PRODUCT, MAXIMUM, AND MINIMUM OF FINITE BA...UniversitasGadjahMada
Chaatit, Mascioni, and Rosenthal de ned nite Baire index for a bounded real-valued function f on a separable metric space, denoted by i(f), and proved that for any bounded functions f and g of nite Baire index, i(h) i(f) + i(g), where h is any of the functions f + g, fg, f ˅g, f ^ g. In this paper, we prove that the result is optimal in the following sense : for each n; k < ω, there exist functions f; g such that i(f) = n, i(g) = k, and i(h) = i(f) + i(g).
This is my first PPT made for a college assignment on Measure Theory and Advanced Probability. I tried to cover the topic "Counting measure" under combinatorics from all dimensions according to my humble knowledge.
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The notion of proximal intersection property and diagonal property is introduced and used to establish some existence of the best proximity point for mappings satisfying contractive conditions.
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...Levi Shapiro
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
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• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
Introduction to AI for Nonprofits with Tapp NetworkTechSoup
Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
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Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
Normal Labour/ Stages of Labour/ Mechanism of LabourWasim Ak
Normal labor is also termed spontaneous labor, defined as the natural physiological process through which the fetus, placenta, and membranes are expelled from the uterus through the birth canal at term (37 to 42 weeks
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CHAPTER 1
Measure and Integration
1. σ- algebra and measure
Definition 1.1.1. Let X be a set. A subset A of the powerset P(X) of X is called a σ-
algebra if ∅ ∈ A , A is closed with respect to the formation of complement in X and A is
closed with respect to the formation of countable unions.
If A is a σ- algebra of subsets of X, then the pair (X, A ) is called a measurable space.
Let A be a σ- algebra of subsets of X. A subset A of X is called measurable if A ∈ A .
Note that {∅, X} and P(X) are the smallest and the largest σ- algebras of subsets of
X respectively.
Exercise 1.1.2. Show that the condition (3) in the definition 1.1.1 can be replaced by the
formation of countable intersections.
Definition 1.1.3. Let A be a σ- algebra of subsets of X. A map µ : A → [0, ∞] is called
a measure if
(i) µ(∅) = 0,
(ii) µ is countably additive, i.e., if {En} is a sequence of pairwise disjoint measurable
subsets of X, then µ( n En) = n µ(En)
If µ is a measure on a measurable space (X, A ), then the triplet (X, A , µ) is called a
measure space.
Examples 1.1.4.
(i) Let M be the σ- algebra of all measurable subsets of R, and let m be the Lebesgue
measure of R. Then (R, M , m) is a measure space.
(ii) Let A be the collection of all measurable subsets of [0, 1] and let m be the Lebesgue
measure on [0, 1]. Then ([0, 1], A , m) is a measure space.
(iii) Let B be the Borel σ algebra on R, and let m be the Lebesgue measure on R. Then
(R, B, m) is a measure space.
(iv) Let X be any set, and let A = {∅, X}. Let α > 0. Define µα : A → [0, ∞] by
µα(∅) = 0 and µα(X) = α. Then (X, A , µα) is a measure space.
(v) Let X be a any set. Let ν : P(X) → [0, ∞] be as follows. For a subset A of X put
ν(A) = ∞ is A is an infinite set and put ν(A) to be the number of elements in A. Then
i
3. P
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ii 1. MEASURE AND INTEGRATION
ν is a measure on X, called the counting measure (on X). The triplet (X, P(X), ν) is
a measure space.
(vi) Let X be an uncountable set. Let A = {E ⊂ X : either E or Ec
is countable}. Then
A is a σ- algebra of subsets of X. Let α > 0 Define µα : →[0, ∞] by µα(A) = 0 if A
is uncountable and µα(A) = α if A is countable. Then (X, A , µα) is a measure space.
(vii) Let X be a nonempty set, and let x ∈ X. Define δx : P(X) → [0, ∞] by δx(A) = 1 if
x ∈ A and δx(A) = 0 if x /∈ A. Then δx is a measure on (X, P(X)), called the Dirac
measure concentrated at x. Then (X, P(X), δx) is a measure space.
(viii) Let (X, A , µ) be a measure space, and let X0 be a measurable subset of X. Let
A0 = {U ⊂ X : U ∈ A , U ⊂ X0} = {U ∩ X0 : U ∈ A }. Then A0 is a σ- algebra of
subsets of X0. Define µ0 : A0 → [0, ∞] by µ0(E) = µ(E), E ∈ A0. In fact, µ0 = µ|A0
.
Then (X0, A0, µ0) is a measure space.
Lemma 1.1.5 (Monotonicity of a measure). Let (X, A , µ) be a measure space, and let E and
F be measurable subsets of X with F ⊂ E. Then µ(F) ≤ µ(E). Furthermore, if µ(F) < ∞,
then µ(E − F) = µ(E) − µ(F).
Proof. Clearly, E = F ∪ (E − F). As both E and F are measurable, E − F = E ∩ Fc
is
measurable. Since F and E − F are disjoint, µ(E) = µ(F) + µ(E − F) ≥ µ(F).
Let µ(F) < ∞. Since µ(E) = µ(F) + µ(E − F) and µ(F) < ∞, we have µ(E − F) =
µ(E) − µ(F).
Note that we cannot drop the condition that µ(F) < ∞ in the above lemma. For
example, let E = F = R in (R, M , m). Then m(E) = m(F) = ∞ and m(E−F) = m(∅) = 0
but m(E) − m(F) does not exist.
Lemma 1.1.6. Let (X, A , µ) be a measure space, and let {En} be a sequence of measurable
subsets of X. Then µ( n En) ≤ n µ(En).
Proof. Let F1 = E1 and for n > 1, let Fn = Fn − ( n−1
k=1 Fk). Then each Fn is measurable,
Fn ∩ Fm = ∅ if n = m and n Fn = n En. Also note that Fn ⊂ En for all n. Therefore
µ(Fn) ≤ µ(En) for all n. Now, µ( n En) = µ( n Fn) = n µ(Fn) ≤ n µ(En).
Lemma 1.1.7. Let {En} be an increasing sequence of measurable subsets of a measure space
(X, A , µ). Then µ( n En) = limn µ(En) = supn µ(En).
Proof. Since {En} is increasing, µ(En) ≤ µ(En+1) for all n. Also, µ(En) ≤ µ( n En).
If µ(En0 ) = ∞ for some n0, then (µ(En) = ∞ for all n ≥ n0) clearly limn µ(En) = ∞
and µ( n En) = ∞. We are through in this case. Now assume that µ(En) < ∞ for all
n. Let F1 = E1 and for n > 1, let Fn = En − En−1. Then {Fn} is a sequence of pairwise
disjoint measurable subsets of X with n Fn = n En. Since µ(En) < ∞ for all n, we have
4. P
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1. σ- algebra and measure iii
µ(Fn) = µ(En) − µ(En−1) for all n > 1 and µ(F1) = µ(E1). Now
µ(∪nEn) = µ(∪nFn) =
n
µ(Fn) = lim
n
n
k=1
µ(Fk)
= lim
n
[µ(E1) +
n
k=2
(µ(Ek) − µ(Ek−1))]
= lim
n
µ(En).
Since the sequence {µ(En)} is increasing, limn µ(En) = supn µ(En).
Corollary 1.1.8. Let {En} be a sequence of measurable subsets of a measure space (X, A , µ).
Then µ( n En) = limn µ( n
k=1 Ek)
Proof. For each n set Fn = n
k=1 Ek. Then {Fn} is an increasing sequence of measurable
subsets of X and n Fn = n En. By above corollary we have
µ( n En) = µ( n Fn) = limn µ(Fn) = limn µ( n
k=1 Ek).
Lemma 1.1.9. Let {En} be a decreasing sequence of measurable subsets of a measure space
(X, A , µ). If µ(E1) < ∞, then µ( n En) = limn µ(En).
Proof. For each n, let Fn = E1 − En. Since {En} is a decreasing sequence, the sequence
{Fn} is an increasing sequence of measurable sets. As µ(E1) < ∞ (and hence µ(En) < ∞
for all n), we have µ(Fn) = µ(E1)−µ(En). It is also clear that n Fn = E1 −( n En). Now,
µ(E1) − µ(
n
En) = µ(E1 − (
n
En)) = µ(
n
Fn)
= lim
n
µ(Fn) = lim
n
(µ(E1) − µ(Fn)) = µ(E1) − lim
n
µ(En).
Hence µ( n En) = limn µ(En).
We cannot drop the condition that µ(E1) < ∞ in the above lemma. For example
consider En = [n, ∞) in (R, M , m). Then {En} is a decreasing sequence of measurable
subsets of R. Note that m(En) = ∞ for all n and so limn m(En) = ∞ while n En = ∅ gives
m( n En) = 0.
Definition 1.1.10. Let (X, A , µ) be a measure space. The measure µ is called a finite
measure (or the measure space (X, A , µ) is called a finite measure space) if µ(X) < ∞.
The measure µ is called a σ- finite measure (or the measure space (X, A , µ) is called a
σ- finite measure space) if X can be written has a countable union of measurable sets each
having finite measure, i.e., there is a sequence {En} of measurable subsets of X such that
µ(En) < ∞ for all n and n En = X.
A subset E of a measure space (X, A , µ) is said to have σ- finite measure if it can be
written as a countable union of measurable subsets of X each having finite measure.
5. P
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iv 1. MEASURE AND INTEGRATION
Exercise 1.1.11.
(i) If (X, A , µ) is a finite measure space, then it is a σ- finite measure space. Give an
example to show that the converse is not true.
(ii) Any measurable subset of finite measure space has a finite measure.
(iii) If E is a measurable subset of σ- finite measure space, then E is of σ- finite measure.
(iv) If E1, . . . , En are sets of finite measure in a measure space, then their union is a set of
finite measure.
(v) Countable union of sets of σ- finite measures is of σ- finite measure.
Definition 1.1.12. A measure space (X, A , µ) (or the measure µ) is called complete if A
contains all subsets of sets of measure zero.
Example 1.1.13.
(i) The measure space (R, M , m) is complete.
Let E ∈ M be such that m(E) = 0, and let F be a subset of E. Since F ⊂ E,
0 ≤ m∗
(F) ≤ m∗
(E) = m(E) = 0, i.e., m∗
(F) = 0. Let A be any subset of R. Then
A ∩ F ⊂ F gives m∗
(A ∩ F) = 0. Now A ∩ Fc
⊂ A gives m∗
(A ∩ Fc
) ≤ m∗
(A).
Therefore m∗
(A) ≥ m∗
(A ∩ F) = m∗
(A ∩ Fc
) + m∗
(A ∩ F). Hence F is measurable,
i.e., F ∈ M .
(ii) The measure space (R, B, m) is not a complete measure space as the Cantor set C has
measure 0 and it contains a subset which is not a Borel set (Construct such a set!!!).
Theorem 1.1.14 (Completion of a measure space). Let (X, A , µ) be a measure space. Then
there is a complete measure space (X, A0, µ0) such that
(i) A ⊂ A0,
(ii) µ0(E) = µ(E) for every E ∈ A ,
(iii) If E ∈ A0, then E = A ∪ B for some A ∈ A and B ⊂ C for some C ∈ A with
µ(C) = 0.
Proof. Let A0 = {E ⊂ X : E = A ∪ B, A ∈ A , B ⊂ C for some C ∈ A with µ(C) = 0}.
Clearly, A ⊂ A0 ( if E ∈ A , then E = E ∪∅ ∈ A0). Let E = a∪B ∈ A0. Then A ∈ A and
B ⊂ C for some C ∈ A with µ(C) = 0. Then Ec
= Ac
∩ Bc
= (Ac
∩ Cc
) ∪ (Ac
∩ (C − B)).
Clearly, Ac
∩ Cc
∈ A , (Ac
∩ (C − B) ⊂ C. Therefore Ec
∈ A0. Let {En} be a countable
collection of elements of A0. Then En = An ∪ Bn, where An ∈ A and Bn ⊂ Cn for some
Cn ∈ A with µ(Cn) = 0. Now n En = ( n An) ( n Bn). Clearly, n An ∈ A and
n Bn ⊂ n Cn, n Cn ∈ A and 0 ≤ µ( n Cn) ≤ n µ(Cn) = 0. Hence A0 is a σ- algebra
of subsets of X. Define µ0 on A0 as follows. If E = A ∪ B ∈ A0, then we put µ0(E) = µ(A).
Clearly, if E ∈ A , then µ0(E) = µ0(E ∪ ∅) = µ(E). By definition µ0(E) ≥ 0 for every
E ∈ A0. Let {En} be a sequence of pairwise disjoint elements of A0. Then En = An ∪ Bn,
6. P
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2. Measurable Functions v
where An ∈ A , Bn ⊂ Cn for some Cn ∈ A with µ(Cn) = 0. Since En’s are pairwise disjoint,
An’s are pairwise disjoint. Now
µ0(
n
En) = µ0((
n
An) (
n
Bn)) = µ(
n
An) =
n
µ(An) =
n
µ0(En).
Therefore µ0 is a measure on (X, A0). It remains to show that µ0 is complete. For that let
E = A ∪ B ∈ A0 with µ0(E) = 0 = µ(A), and let F ⊂ E. Then F = ∅ ((C A) F).
Obviously (C A) F ⊂ C A ∈ A and 0 ≤ µ(C A) ≤ µ(C) + µ(A) = 0. Hence
F ∈ A0. This finishes the proof.
Definition 1.1.15. Let (X, A , µ) be a measure space. A subset E of X is said to be locally
measurable if E ∩ A ∈ A for every A ∈ A with µ(A) < ∞.
A measure space (X, A , µ) is called saturated if every locally measurable subset of X
is measurable.
Note that every measurable set is locally measurable. The converse is not true (Example
???).
Lemma 1.1.16. Every σ- finite measure space is saturated.
Proof. Let (X, A , µ) be a σ- finite measure space. Then there is a sequence {En} of
measurable subsets of X such that n En = X and µ(En) < ∞ for all n. Let E be a locally
measurable subset of X. Then clearly, E ∩ En ∈ A as µ(En) < ∞. Now E = E ∩ X =
n(E ∩ En). Therefore E is a countable union of measurable subsets of X and hence it is
measurable. This proves that (X, A , µ) is saturated.
Since (R, M , m) and (R, B, m) are σ- finite measure spaces, they are saturated.
2. Measurable Functions
Lemma 1.2.1. Let (X, A ) be a measurable space, and let f be an extended real valued
function on X. Then the following conditions are equivalent.
(i) {x ∈ X : f(x) > α} ∈ A for every α ∈ R.
(ii) {x ∈ X : f(x) ≥ α} ∈ A for every α ∈ R.
(iii) {x ∈ X : f(x) < α} ∈ A for every α ∈ R.
(iv) {x ∈ X : f(x) ≤ α} ∈ A for every α ∈ R.
All the above conditions imply that the set {x ∈ X : f(x) = α} ∈ A for every α ∈ R.
Proof. (i) ⇒ (ii). Let α ∈ R. Then
{x ∈ X : f(x) ≥ α} =
n∈N
{x ∈ X : f(x) > α −
1
n
}.
7. P
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vi 1. MEASURE AND INTEGRATION
Since each set on the right side is measurable, it follows that {x ∈ X : f(x) ≥ α} is measur-
able.
(ii) ⇒ (iii) Let α ∈ R. Then {x ∈ X : f(x) < α} = {x ∈ X : f(x) ≥ α}c
. Therefore
{x ∈ X : f(x) < α} is measurable.
(iii) ⇒ (iv) Let α ∈ R. Then {x ∈ X : f(x) ≤ α} = n∈N{x ∈ X : f(x) < α + 1
n
}.
Therefore {x ∈ X : f(x) ≤ α} is measurable.
(iv) ⇒ (i) Let α ∈ R. Then {x ∈ X : f(x) > α} = {x ∈ X : f(x) ≤ α}c
. Therefore
{x ∈ X : f(x) > α} is measurable.
Since the all the above conditions are equivalent for any α ∈ R, the sets {x ∈ X : f(x) ≤
α} and {x ∈ X : f(x) ≥ α} are measurable. Therefore their intersection {x ∈ X : f(x) = α}
is measurable.
Definition 1.2.2. Let (X, A ) be a measurable space, and let f : X → [−∞, ∞]. Then f is
called measurable if {x ∈ X : f(x) > α} ∈ A for every α ∈ R.
Lemma 1.2.3. Let (X, A ) be a measurable space, and let E ⊂ X. Then E ∈ A if and only
if χE is measurable.
Proof. Assume that χE is measurable. Then the set {x ∈ X : χE(x) > 0} ∈ A . But
{x ∈ X : χE(x) > 0} = E. Therefore E is measurable.
Conversely, assume that E is measurable. Let α ∈ R. Then
{x ∈ X : χE(x) > α} =
∅ if α ≥ 1
E if 0 ≤ α < 1
X if α < 0
Hence χE is measurable.
Theorem 1.2.4. Let (X, A ) be a measurable space, and let f and g be measurable functions
on X. Let c ∈ R. Then
(i) cf is measurable.
(ii) f + c is measurable.
(iii) f+
, f−
and |f| are measurable.
(iv) max{f, g} and min{f, g} are measurable.
Definition 1.2.5. Let (X, A , µ) be a measure space. A property P is said to hold almost
everywhere [µ] on X if there is a measurable subset E of X with µ(E) = 0 such that P holds
on X − E.
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2. Measurable Functions vii
Lemma 1.2.6. Let (X, A , µ) be a complete measure space, and let f be measurable. Let
g : X → [−∞, ∞] be a map. If g = f a.e. [µ], then g is measurable.
Proof. Since f = g a.e. [µ], there is a measurable subset E of X with µ(E) = 0 such that
f = g on X −E. Let α ∈ R. Then {x ∈ X : g(x) > α} = ({x ∈ X : f(x) > α}∩(X −E))∪F
for some subset F of E. Since µ is complete, F is measurable. Hence g is measurable.
Theorem 1.2.7. Let (X, A , µ) be a measure space, and let f and g be measurable functions
on X which are finite a.e. [µ]. Then
(i) f ± g is measurable. (be very much careful in proving this.)
(ii) fg is measurable. (again be very much careful in proving this.)
(iii) f/g is measurable if g(x) = 0 for any x.
Lemma 1.2.8. Let (X, A ) be a measurable space, and let {fn} be a sequence of measurable
functions on X. Then supn fn, infn fn, lim supn fn, lim infn fn are measurable. In particular,
when fn → f (pointwise) on X, then f is measurable.
Proof. Let g = supn fn, and let h = infn fn. Let α ∈ R. Then {x ∈ X : g(x) ≤ α} =
n{x ∈ X : fn(x) ≤ α} and {x ∈ X : h(x) ≥ α} = n{x ∈ X : fn(x) ≥ α}. Therefore
both g and h are measurable. Let gk = supn≥k fn and hk = infn≥k fn. Then both gk and
hk are measurable for every k. Therefore lim supn fn = infk gk and lim infn fn = supk hk are
measurable. If a sequence {fn} converges to f, then f = lim supn fn = lim infn fn. Therefore
f is measurable.
Definition 1.2.9. Let X be a topological space. Then the smallest σ- algebra of subsets of
X containing all open subsets of X is called the Borel σ- algebra on X. Any element of the
Borel σ- algebra is called a Borel set.
Question 1.2.10. Let A be a σ- algebra of subsets of X. Does there exist a topology on
X whose Borel σ- algebra is A ?
In particular, we know that the set M of all measurable subsets of R is a σ- algebra.
Does there exist a topology on R whose Borel σ- algebra is M ? (There is such a topology
find it or construct it).
Consider X = [−∞, ∞]. The the standard topology on X is generated by a basis
{(a, b), (c, ∞], [−∞, d) : a, b, c, d ∈ R, a < b}. Also note that any open subset of [−∞, ∞]
can be written as a countable union of disjoint open sets of the form (a, b), (c, ∞], [−∞, d).
Theorem 1.2.11. Let (X, A ) be a measurable space, and let f be an extended real valued
map. Then the following are equivalent.
(i) f is measurable.
(ii) {x ∈ X : f(x) > r} is measurable for every r ∈ Q.
9. P
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viii 1. MEASURE AND INTEGRATION
(iii) f−1
(E) is measurable for every open subset E of [−∞, ∞].
Proof. (i) ⇒ (ii) is clear.
(ii) ⇒ (i) Let a ∈ R. Then
f−1
((a, ∞]) = {x ∈ X : f(x) > a} =
r∈Q,r>a
{x ∈ X : f(x) > r}.
Since each {x ∈ X : f(x) > r} is measurable and countable union of measurable sets is
measurable, {x ∈ X : f(x) > a} is measurable, i.e., f is measurable.
(iii) ⇒ (i) Assume that f−1
(E) is measurable for every open subset E of [−∞, ∞]. Let
α ∈ R. Then (α, ∞] is an open subset of [−∞, ∞]. Now f−1
(α, ∞] = {x ∈ X : f(x) > α}.
Hence f is measurable.
(i) ⇒ (iii) Assume that f is measurable. Let a ∈ R. Then f−1
(a, ∞] = {x ∈ X : f(x) > a}
and f−1
[−∞, a) = {x ∈ X : f(x) < a}. Since f is measurable, for any a ∈ R the sets
f−1
(a, ∞] and f−1
[−∞, a) are measurable. Let a, b ∈ R, then f−1
(a, b) = f−1
[−∞, b) ∩
f−1
(a, ∞], which is also measurable. Let E be an open subset of [−∞, ∞], then it follows
from the structure theorem of open subsets of [−∞, ∞] that E can be written as a countable
union of open intervals of the form [−∞, d), (a, b) and (c, ∞], i.e., E = n On, where On
takes one of the form [−∞, d), (a, b) and (c, ∞]. Now f−1
(E) = f−1
( n On) = n f−1
(On).
Since each f−1
(On) is measurable, it follows that f−1
(E) is measurable.
Definition 1.2.12. Let f be an extended real valued measurable function on a measurable
space (X, A ). Then for each α ∈ R the set Bα = {x ∈ X : f(x) < α} is measurable and it
satisfies Bα ⊂ Bα if α < α . The sets Bα ’s are called the ordinate sets of f.
Theorem 1.2.13. Let (X, A ) be a measurable space, and let D be a dense subset of R.
Suppose that for each α ∈ D there is an associated Bα ∈ A such that Bα ⊂ Bα whenever
α < α . Then there is a unique measurable function f on X such that f ≤ α on Bα and
f ≥ α on Bc
α for every α ∈ D.
Proof. Define f : X → [−∞, ∞] as f(x) = inf{α ∈ D : x ∈ Bα} if x ∈ Bα for some
α ∈ D and f(x) = ∞ if x is not in any Bα. Let x ∈ Bα. Then inf{β ∈ D : x ∈ Bβ} ≤ α,
i.e., f(x) ≤ α. Therefore f ≤ α on Bα. Let x ∈ Bc
α. Suppose that f(x) = inf{β ∈ D :
x ∈ Bβ} < α. Then there is α ∈ D such that f(x) ≤ α < α and x ∈ Bα . Since α < α,
x ∈ Bα ⊂ Bα, which is a contradiction. Hence f ≥ α on Bc
α. Now we prove that f is
measurable. Let λ ∈ R. Since D is dense in R, there is a sequence {αn} with αn < λ for
all n such that αn → λ as n → ∞. Let x ∈ n Bαn . Then x ∈ Bαn for some n. Therefore
f(x) ≤ αn < λ, i.e., f(x) < λ. Let x ∈ X with f(x) = inf{α ∈ D : x ∈ Bα} < λ. Then
10. P
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3. Integration ix
there is β ∈ D such that f(x) ≤ β < λ and x ∈ Bβ. Since {αn} converges to λ, there is
n0 ∈ N such that β < αn0 < λ. Since β < αn0 and x ∈ Bβ, x ∈ Bαn0
⊂ n Bαn . Hence we
proved that {x ∈ X : f(x) < λ} = n Bαn . Since each Bαn is measurable, it follows that f
is measurable.
Let g be a measurable function on X such that g ≤ α on Bα and g ≥ α on Bc
α. Let
x ∈ X. If x is not in any Bα, then g(x) ≥ α for every α ∈ D. Since D is dense in R,
g(x) = ∞ = f(x). Let x ∈ Bα for some α. Then {α ∈ D : g(x) < α} ⊂ {α ∈ D : x ∈ Bα} ⊂
{α ∈ D : g(x) ≤ α}. Therefore inf{α ∈ D : g(x) < α} ≥ inf{α ∈ D : x ∈ Bα} ≥ inf{α ∈
D : g(x) ≤ α}, i.e., g(x) ≥ f(x) ≥ g(x). Hence f = g.
Theorem 1.2.14. Let (X, A , µ) be a measure space. Suppose that for each α in a dense
set D of real numbers, there is assigned a set Bα ∈ A such that µ(Bα − Bβ) = 0 for α < β.
Then there is a measurable function f such that f ≤ α a.e. on Bα and f ≥ α a.e. on Bc
α.
If g is any other function with this property, then g = f a.e.
Proof. (Verify the proof and write details)
Let C be a countable dense subset of D, let N = α,β∈C,α<β(Bα −Bβ). Then µ(N) = 0.
Let Bα = Bα ∪ N. If α, β ∈ C with α < β, then Bα − Bβ = (Bα − Bβ) − N = ∅. Thus
Bα ⊂ Bβ. By above theorem there is a unique measurable function f such that f ≤ α on
Bα and f ≥ α on B c
α .
Let α ∈ D. Let {γn} be a sequence in C with γn > α and limn γn = α. Then
Bα − Bγn
⊂ Bα − Bγn . Therefore P = n(Bα − Bγn
) has measure 0. Let A = n Bγn
. Then
f ≤ infn γn ≤ α on A and Bα − A = Bα ∩ ( n Bγn
)c
. Then f ≤ α a.e. on Bα and f ≥ α a.e.
on Bc
α.
Let g be an extended real function with g ≤ γ on Bγ and g ≥ γ on Bc
γ for each γ ∈ C.
Then g ≤ γ on Bγ and g ≥ γ on Bc
γ except for a set of measure 0, say Qγ. Thus Q = γ∈C Qγ
is a set of measure 0 and f = g on X − Q, i.e., f = g a.e. [µ].
3. Integration
Definition 1.3.1. A function s : X → R is called a simple function if it assumes finitely
many distinct values.
Let s : X → R be a simple function, and let α1, α2, . . . , αn be distinct values assumed
by s. Let Ai = {x ∈ X : s(x) = αi}, 1 ≤ i ≤ n. Then Ai’s are pairwise disjoint and
n
i=1 Ai = X. Thus s admits a canonical representation s = n
i=1 αiχAi
.
Lemma 1.3.2. Let (X, A ) be a measurable space, and let s = n
i=1 αiχAi
be a simple
function X. Then s is measurable if and only if each Ai is measurable.
Proof. Assume that s is measurable. Then for each i, the set {x ∈ X : s(x) = αi} is
measurable, i.e., Ai is measurable for each i.
11. P
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Conversely, assume that each Ai is measurable. Then each χAi
is measurable and hence
n
i=1 αiχAi
= s is measurable
Definition 1.3.3. Let (X, A , µ) be a measure space, and let s = n
i=1 αiχAi
be a non
negative measurable simple function on X. Then the Lebesgue integral of s over E ∈ A is
defined as
E
sdµ =
n
i=1
αiµ(Ai ∩ E).
By definition X
sdµ = n
i=1 αiµ(Ai ∩ X) = n
i=1 αiµ(Ai).
Remarks 1.3.4. Let (X, A , µ) be a measure space, and let s = n
i=1 αiχAi
be a non
negative measurable simple function on X.
(i) If E ∈ A , then
E
sdµ =
n
i=1
αiµ(Ai ∩ E) ≥ 0 as αi ≥ 0 and µ(Ai ∩ E) ≥ 0 for each i.
(ii) If E, F ∈ A and E ⊂ F, then E
sdµ ≤ F
sdµ.
Since Ai∩E ⊂ Ai∩F, we have µ(Ai∩E) ≤ µ(Ai∩F). Now E
sdµ = n
i=1 αµ(Ai∩E) ≤
n
i=1 αµ(Ai ∩ F) = F
sdµ.
(iii) If E1, E2, . . . , Em are pairwise disjoint measurable subsets of X, then ∪m
j=iEj
sdµ =
m
j=1 Ej
sdµ.
Since E1, E2, . . . , Em are pairwise disjoint measurable subsets of X and Ai is measur-
able, µ(( m
j=1 Ej) ∩ Ai) = m
j=1 µ(Ej ∩ Ai). Now
n
j=1 Ej
sdµ =
n
i=1
αiµ((
m
j=1
Ej) ∩ Ai) =
n
i=1
αi
m
j=1
µ(Ej ∩ Ai)
=
m
j=1
n
i=1
αiµ(Ej ∩ Ai) =
m
j=1 Ej
sdµ.
(iv) Let E ∈ A . If µ(E) = 0 or s = 0 a.e. [µ] on E, then E
sdµ = 0.
Assume that µ(E) = 0. Since E ∩ Ai ⊂ E, µ(E ∩ Ai) = 0 for all i. Hence E
sdµ =
n
i=1 αiµ(Ai ∩ E) = 0.
Now assume that s = 0 a.e. [µ] on E. Then there is a measurable subset F of E
with µ(F) = 0 such that s = 0 on E − F. Now E
sdµ = F
sdµ + E−F
sdµ = 0 as
µ(F) = 0 and s = 0 on E − F.
(v) If s ∈ A , then E
sdµ = X
sχEdµ.
12. P
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3. Integration xi
Here s = n
i=1 αiχAi
. Therefore sχE = n
i=1 αiχAi∩E.
Now X
sχE = n
i=1 αiµ(Ai ∩ E) = E
sdµ.
(vi) If α ≥ 0 and E ∈ A , then E
αsdµ = α E
sdµ.
Here s = n
i=1 αiχAi
. Therefore αs = n
i=1 ααiχAi
. So,
E
αsdµ =
n
i=1
ααiµ(Ai ∩ E) = α
n
i=1
αiµ(Ai ∩ E) = α
E
sdµ.
Lemma 1.3.5. Let s and t be non negative measurable simple functions on a measure space
(X, A , µ). Then X
(s + t)dµ = X
sdµ + X
tdµ.
Proof. Let s = n
i=1 αiχAi
and t = m
j=1 βjχBj
be the canonical representations of s and
t respectively. Then Ai’s are pairwise disjoint measurable sets, Bj’s are pairwise disjoint
measurable sets and n
i=1 Ai = m
j=1 Bj = X. For 1 ≤ i ≤ n and 1 ≤ j ≤ m, set Cij =
Ai ∩ Bj. Then each Cij is measurable and Cij’s are pairwise disjoint. Also i,j Cij =
i,j(Ai ∩ Bj) = n
i=1(Ai ∩ ( m
j=1 Bj)) = n
i=1 Ai = X. If x ∈ X. Then x is in exactly one
Cij and on Cij, (s + t)(x) = αi + βj. Therefore s + t is a non negative measurable simple
function and s + t = n
i=1
m
j=1(αi + βj)χCij
is the canonical representation of s + t. Now
X
(s + t)dµ =
n
i=1
m
j=1
(αi + βj)µ(χCij
)
=
n
i=1
m
j=1
αiµ(Ai ∩ Bj) +
n
i=1
m
j=1
βjµ(Ai ∩ Bj)
=
n
i=1
αi
m
j=1
µ(Ai ∩ Bj) +
m
j=1
βj
n
i=1
µ(Ai ∩ Bj)
=
n
i=1
αiµ(
m
j=1
(Ai ∩ Bj)) +
m
j=1
βjµ(
n
i=1
(Ai ∩ Bj))
=
n
i=1
αiµ(Ai ∩ (
m
j=1
Bj)) +
m
j=1
βjµ((
n
i=1
Ai) ∩ Bj)
=
n
i=1
αiµ(Ai) +
m
j=1
βjµ(Bj)
=
X
sdµ +
X
tdµ.
This completes the proof
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xii 1. MEASURE AND INTEGRATION
Lemma 1.3.6. Let (X, A , µ) be a measure space, and let s be a non negative measurable
simple function on X. Define ϕ on A by
ϕ(E) =
E
sdµ (E ∈ A ).
Then ϕ is a measure on (X, A ).
Proof. Obviously, ϕ(∅) = ∅
sdµ = 0 and ϕ(E) ≥ 0 for every E ∈ A . Let s = n
i=1 αiχAi
be the canonical representation of s. Let {En} be a sequence of pairwise disjoint measurable
subsets of X. Since En’s are pairwise disjoint and Ai’s are pairwise disjoint, Ai ∩ En are
pairwise disjoint. Now
ϕ(
j
Ej) =
j Ej
sdµ =
n
i=1
αiµ(Ai ∩ (
j
Ej))
=
n
i=1
αiµ(
j
(Ai ∩ Ej)) =
n
i=1
αi
j
µ(Ai ∩ Ej)
=
j
n
i=1
αiµ(Ai ∩ Ej) =
j Ej
sdµ =
j
ϕ(Ej).
Hence ϕ is a measure on (X, A ).
Theorem 1.3.7 (Lusin’s Theorem). Let (X, A ) be a measurable space, and let f be a non
negative measurable function X. Then there is an increasing sequence {sn} of non negative
measurable simple functions on X converging to f (pointwise) on X. Further, if f is bounded,
then {sn} converges to f uniformly on X.
Proof. Let n ∈ N. For i = 1, 2, . . . , n2n
, define Eni = {x ∈ x : i−1
2n ≤ f(x) < i
2n } and
Fn = {x ∈ X : f(x) ≥ n}. Since f is measurable each Eni and Fn are measurable. Note
that Eni’s and Fn are pairwise disjoint and their union is X. For n ∈ N, define
sn =
n2n
i=1
i − 1
2n
χEni
+ nχFn .
As each Eni and Fn are measurable, sn is a non negative measurable simple function.
First we prove that {sn} is increasing.For that let x ∈ X and n ∈ N.
Let f(x) ≥ n + 1. Then f(x) > n. Therefore sn+1(x) = n + 1 > n = sn(x).
Let n ≤ f(x) < n + 1. Then sn(x) = n. Since n ≤ f(x) < n + 1, there is i ∈
{n2n+1
, . . . , (n + 1)2n+1
} such that x ∈ E(n+1)i. Therefore sn+1(x) = i−1
2n+1 for some i ∈
{n2n+1
, . . . , (n + 1)2n+1
}. Since i ≥ n2n+1
, we have sn+1(x) = i−1
2n+1 ≥ n = sn(x).
14. P
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3. Integration xiii
Let f(x) < n. Then there is i ∈ {1, 2, . . . , n2n
} such that i−1
2n ≤ f(x) < i
2n , i.e.,
x ∈ Eni for some i ∈ {1, 2, . . . , n2n
}. Therefore sn(x) = i−1
2n . Since i−1
2n ≤ f(x) < i
2n ,
we have 2i−2
2n+1 ≤ f(x) < 2i
2n+1 . So, either 2i−2
2n+1 ≤ f(x) < 2i−1
2n+1 or 2i−1
2n+1 ≤ f(x) < 2i
2n+1 , i.e.,
x ∈ E(n+1)(2i−2) or x ∈ E(n+1)(2i−1). Therefore sn+1(x) = 2i−2
2n+1 or sn+1(x) = 2i−1
2n+1 . In any case,
sn+1(x) ≥ i−1
2n = sn(x).
Hence the sequence {sn} is increasing.
Now prove that if x ∈ X, then sn(x) → f(x) as n → ∞.
For that let x ∈ X. If f(x) = ∞, then sn(x) = n for all n and clearly sn(x) = n →
∞ = f(x) as n → ∞.
Let f(x) < ∞. Then there is k = k(x) > 0 such that f(x) < k(x). Let n ∈ N be such
that n > k, i.e., f(x) < n. Then x ∈ Eni for some i ∈ {1, 2, , . . . , n2n
}. Therefore sn(x) = i−1
2n
for some i ∈ {1, 2, , . . . , n2n
}. Now sn(x) = i−1
2n ≤ f(x) < i
2n gives |f(x) − sn(x)| =
f(x) − sn(x) ≤ i
2n − i−1
2n = 1
2n → 0 as n → ∞. Therefore limn sn(x) = f(x) for every x ∈ X.
Now assume that f is bounded. Then there is k > 0 such that f(x) < k for every
x ∈ X. As we have done in last paragraph, we get |sn(x) − f(x)| < 1
2n for all n ≥ k and for
all x ∈ X. Therefore supx∈X |sn(x) − f(x)| ≤ 1
2n . Hence {sn} converges to f uniformly on
X.
Example 1.3.8. Is it possible to approximate a non negative bounded measurable function
by a sequence of non negative measurable function vanishing outside a set of finite measure?
Consider the measure space (R, M , m). Let f(x) = 1 for all x ∈ R. Let {sn} be a
sequence of non negative measurable simple functions on R such that sn = 0 on R − En for
some measurable En with m(En) < ∞. Since m(En) < ∞, R − En = ∅. Let x ∈ R − En.
Then sn(x) = 0. Now 1 = |f(x) − sn(x)| ≤ supy∈R |sn(y) − f(y)|. Therefore {sn} does not
converge to f uniformly on X.
Definition 1.3.9. Let (X, A , µ) be a measure space, and let f be a non negative measurable
function on X. Then the Lebesgue integral of f over E ∈ A (with respect to µ) is defined
by
E
fdµ = sup
0≤s≤f
E
sdµ,
where s is a (non negative) measurable simple function.
Remarks 1.3.10. Let f and g be non negative measurable function on a measure space
(X, A , µ).
(i) If E ∈ A , then E
fdµ ≥ 0.
If s is any non negative measurable simple function, then E
sdµ ≥ 0. Therefore
E
fdµ = sup0≤s≤f E
sdµ ≥ 0.
15. P
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xiv 1. MEASURE AND INTEGRATION
(ii) If E, F ∈ A and E ⊂ F, then E
fdµ ≤ E
fdµ.
If s is any non negative measurable simple function, then E
sdµ ≥ E
sdµ. Therefore
sup0≤s≤f E
sdµ ≤ sup0≤s≤f E
sdµ, i.e., E
fdµ ≤ F
fdµ.
(iii) Let E ∈ A . If µ(E) or f = 0 a.e. [µ] on E, then E
fdµ = 0.
Let µ(E) = 0. Let s be a non negative measurable simple function with s ≤ f. Since
µ(E) = 0, E
sdµ = 0. Therefore sup0≤s≤f E
sdµ = 0, i.e., E
fdµ = 0.
Suppose that f = 0 a.e. [µ] on E. Then there is a measurable subset F of E with
µ(F) = 0 such that f = 0 on E − F. Let s be any non negative measurable simple
function with s ≤ f. If x ∈ E −F, then 0 = f(x) ≥ s(x) ≥ 0, i.e., s(x) = 0. Therefore
s = 0 a.e. [µ] on E. Hence E
sdµ = 0. This implies that sup0≤s≤f E
sdµ = 0, i.e.,
E
fdµ = 0.
(iv) If E ∈ A and f ≤ g, then E
fdµ ≤ E
gdµ.
Let s be any non negative measurable simple function with s ≤ f. Then s ≤ g as
f ≤ g. Therefore E
sdµ ≤ E
gdµ. Therefore E
fdµ = sup0≤s≤f E
sdµ ≤ E
gdµ.
(v) If α ≥ 0 and E ∈ A , then E
αfdµ = α E
fdµ.
If α = 0, then it is clear. Let α > 0. Let s be any non negative measurable simple func-
tion. Then 0 ≤ s ≤ αf if and only if 0 ≤ s
α
≤ f and s is a simple function if and only
if αs is a simple fucntion. Now α E
fdµ = α sup0≤s≤f E
sdµ = sup0≤s≤f E
αsdµ =
sup0≤αs≤αf E
αsdµ = sup0≤t≤αf E
tdµ = E
αfdµ.
(vi) If E ∈ A , then X
fχEdµ = E
fdµ.
Let s be a non negative measurable simple function with s ≤ fχE. Then s ≤ f on E.
Therefore X
sdµ = E
sdµ ≤ E
fdµ. Since s is arbitrary X
fχEdµ ≤ E
fdµ. Let s
be a non negative measurable simple function with s ≤ f on E. Then sχE ≤ fχE on
X. Therefore E
sdµ = X
sχEdµ ≤ X
fχEdµ. Again, since s is arbitrary, we have
E
fdµ ≤ X
fχEdµ. Hence X
fχEdµ = E
fdµ.
Theorem 1.3.11 (Monotone Convergence Theorem). Let (X, A , µ) be a measure space, and
let {fn} be an increasing sequence of non negative measurable functions on X converging to a
function f (pointwise) on X. Then X
fdµ = lim
n→∞ X
fndµ. (In other words X
( lim
n→∞
fn)dµ =
lim
n→∞ X
fndµ.)
Proof. Since each fn is non negative and measurable, the limit function f is also non
negative and measurable and so X
fdµ exists. Since {fn} is an increasing sequence, f =
limn→∞ = supn fn. Therefore fn ≤ f for all n and hence X
fndµ ≤ X
fdµ. Since {fn} is
an increasing sequence, X
fndµ ≤ X
fn+1dµ for all n. Therefore { X
fndµ} is an increasing
16. P
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3. Integration xv
sequence of extended real numbers and hence limn→∞ X
fndµ (it may be infinity). As
X
fndµ ≤ X
fdµ for all n,
lim
n→∞
X
fndµ ≤
X
fdµ. (1.3.11.1)
Let 0 < c < 1. Let s be any non negative measurable simple function with s ≤ f.
Then clearly 0 ≤ cs(x) ≤ s(x) ≤ f(x) for every x ∈ X. For each n ∈ N, set En = {x ∈
X : fn(x) ≥ cs(x)}. As both fn and cs are measurable, the set En is measurable. The
sequence {En} is increasing because the sequence {fn} is increasing. Clearly, n En ⊂ X.
Let x ∈ X. If s(x) = 0, then fn(x) ≥ cs(x) = 0 for all n, i.e., x ∈ n En. Let s(x) > 0.
Then cs(x) < s(x) ≤ f(x). Since {fn(x)} converges to f(x), there is n0 ∈ N such that
cs(x) < fn0 (x) ≤ f(x). Therefore x ∈ En0 ⊂ n En. Hence n En = X. Now on En,
cs ≤ fn. Therefore
c
En
sdµ =
En
csdµ ≤
En
fndµ ≤
n En
fndµ =
X
fndµ. (1.3.11.2)
Define ϕ on A by ϕ(E) = E
sdµ, E ∈ A . Then ϕ is a measure on (X, A ). Since {En} is an
increasing sequence of measurable subsets of X, we have ϕ(X) = ϕ( n En) = limn→∞ ϕ(En),
i.e., X
sdµ = limn→∞ En
sdµ. Taking limit n → ∞ in equation (1.3.11.2) and applying this
equality we get c X
sdµ ≤ limn→∞ X
fndµ. Since the above is true for any non negative
measurable simple function, we get c X
fdµ ≤ limn→∞ X
fndµ. Since 0 < c < 1 is arbitrary,
taking c → 1, we get X
fdµ ≤ limn→∞ X
fndµ. The last inequality and the inequality in
equation (1.3.11.1) give the desired equality.
Theorem 1.3.12. Let f and g be non negative measurable functions on a measure space
(X, A , µ). Then X
(f + g)dµ = X
fdµ + X
gdµ.
Proof. By the Lusin’s theorem there exist increasing sequences {sn} and {tn} of non nega-
tive measurable simple functions converging to f and g respectively. Therefore by Monotone
Convergence Theorem X
fdµ = limn→∞ X
sndµ and X
gdµ = limn→∞ X
tndµ. Note that
the {sn + tn} is an increasing sequence of non negative measurable functions converging
to f + g. Again the application of Monotone Convergence Theorem give X
(f + g)dµ =
limn→∞ X
(sn + tn)dµ. Now
X
(f + g)dµ = lim
n→∞
X
(sn + tn)dµ
= lim
n→∞
X
sndµ + lim
n→∞
X
tndµ
17. P
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xvi 1. MEASURE AND INTEGRATION
=
X
fdµ +
X
gdµ.
Hence the proof.
Corollary 1.3.13. Let f1, f2, . . . , fn be non negative measurable functions on a measure
space (X, A , µ). Then X
( n
i=1 fi)dµ = n
i=1 X
fidµ.
Proof. Use the Principle of Mathematical Induction.
Theorem 1.3.14. Let {fn} be a sequence of non negative measurable functions on a measure
space (X, A , µ). Then X
( n fn)dµ = n X
fndµ.
Proof. For each n ∈ N, let gn = n
k=1 fk. Then {gn} is an increasing sequence of non
negative measurable functions converging to n fn. Therefore by Monotone Convergence
Theorem, X
( n fn)dµ = limn X
gndµ = limn X
( n
k=1 fk)dµ = limn
n
k=1 X
fkdµ =
n X
fndµ.
Theorem 1.3.15. Let f be a non negative measurable function on a measure space (X, A , µ).
Define ϕ on A by ϕ(E) = E
fdµ, E ∈ A . Then ϕ is a measure on (X, A ).
Proof. Clearly, ϕ(∅) = ∅
fdµ = 0 and ϕ(E) = E
fdµ ≥ 0 for every E ∈ A . Let {En}
be a sequence of pairwise disjoint measurable subsets of X. Observe that f = n fχEn on
n En. Also, we note that each fχEn is non negative and measurable. Therefore
ϕ(
n
En) =
En
fdµ =
En
(
n
fχEn )dµ
=
n
En
fχEn dµ =
n En
fdµ
=
n
ϕ(En).
Hence ϕ is a measure on (X, A ).
Theorem 1.3.16 (Fatou’s Lemma). Let {fn} be a sequence of non negative measurable
functions on a measure space (X, A , µ). Then X
(lim inf
n
fn)dµ ≤ lim inf
n X
fndµ.
Proof. For each n ∈ N, let gn = inf{fn, fn+1, . . .} = infk≥n fk. Then {gn} is a sequence of
non negative measurable functions and gn ≤ fn for all n. Therefore X
gndµ ≤ X
fndµ for
all n. Since {gn} is increasing, we have limn gn = supn gn = lim infn fn. It follows from the
Monotone Convergence Theorem that
X
(lim inf
n
fn)dµ =
X
(lim
n
gn)dµ = lim
n
X
gndµ.
18. P
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3. Integration xvii
Since X
gndµ ≤ X
fndµ for all n, we get limn X
gndµ = lim infn X
gndµ ≤ lim infn X
fndµ.
Hence X
(lim inf
n
fn)dµ ≤ lim inf
n X
fndµ.
Theorem 1.3.17 (Beppo Levi’s Theorem). Let {fn} be a sequence of non negative measur-
able functions on a measure space (X, A , µ) converging to f (pointwise) on X. If fn ≤ f
for all n, then X
fdµ = limn X
fndµ.
Proof. Since {fn} converges to f and each fn is non negative and measurable, f is a non
negative measurable function on X. Here f = limn fn = lim infn fn. It follows from Fatous’
lemma that X
fdµ ≤ lim infn X
fndµ. Since fn ≤ f for all n, X
fndµ ≤ X
fdµ for all n.
Therefore lim supn x
fndµ ≤ x
fdµ. Now
X
fdµ ≤ lim inf
n
X
fndµ ≤ lim sup
n
X
fndµ ≤
X
fdµ.
Therefore the sequence { X
fndµ} convergent and it converges to X
fdµ, i.e., limn X
fndµ =
X
fdµ.
Example 1.3.18. Verify the Monotone Convergence Theorem, Fatou’s lemma and Beppo
Levi’s theorem for fn(x) = nx
1+nx
on [1, 7].
Lemma 1.3.19. Let f be a non negative measurable function on a measure space (X, A , µ).
If X
fdµ = 0, then f = 0 a.e. [µ] on X.
Proof. Let E = {x ∈ X : f(x) = 0} = {x ∈ X : f(x) > 0}. Then E is a measurable
subset of X. Let En = {x ∈ X : f(x) > 1
n
} for n ∈ N. Then each En is measurable. One
can verify easily that E = n En. If µ(E) > 0, then µ(EN ) > 0 for some N ∈ N. But
then 0 = X
fdµ ≥ EN
fdµ ≥ 1
N
µ(EN ) > 0, which is a contradiction. Hence µ(E) = 0, i.e.,
f = 0 a.e. [µ] on X.
Definition 1.3.20. A non negative measurable function on a measure space (X, A , µ) is
called integrable if X
fdµ < ∞.
Verify that f is integrable on (X, A , µ) if and only if E
fdµ < ∞ for every E ∈ A .
Now we define the integral of arbitrary measurable function (not necessarily non nega-
tive).
Definition 1.3.21. Let f be a measurable function on a measure space (X, A , µ). If at
least one of X
f+
dµ and X
f−
dµ is finite, then we define the Lebesgue integral of f by
X
fdµ = X
f+
dµ − X
f−
dµ. The function f is called integrable if both X
f+
dµ and
X
f−
dµ are finite.
Theorem 1.3.22. Let f be a measurable function on a measure space (X, A , µ). Then f is
integrable if and only if |f| is integrable.
19. P
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xviii 1. MEASURE AND INTEGRATION
Proof. Assume that f is integrable. Then both X
f+
dµ and X
f−
dµ are finite. Now
X
|f|dµ = X
(f+
+ f−
)dµ = X
f+
dµ + X
+f−
dµ < ∞. Therefore f is integrable. Con-
versely, assume that |f| is integrable. Then X
|f|dµ = X
f+
dµ + X
f−
dµ < ∞. Since
X
f+
dµ, X
f−
dµ ≤ X
f+
dµ + X
f−
dµ, it follows that f is integrable.
Lemma 1.3.23. Let f be a measurable function on a measure space (X, A , µ) such that
X
fdµ exists. Then X
fdµ ≤ X
|f|dµ.
Proof. If X
|f|dµ = ∞, then it is clear. Assume that X
|f|dµ < ∞, i.e., |f| is integrable.
Then f is integrable. Now X
fdµ = X
f+
dµ − X
f−
dµ ≤ X
f+
dµ + X
f−
dµ =
X
f+
dµ + X
f−
dµ = X
(f+
+ f−
)dµ = X
|f|dµ.
Lemma 1.3.24. Let f and g be integrable functions on a measure space (X, A , µ), and let
α ∈ R. Then
(i) f + g is integrable and X
(f + g)dµ = X
fdµ + X
gdµ.
(ii) αf is integrable and X
αfdµ = α X
fdµ.
Proof. Since f and g are measurable, f + g and αf are measurable. As both f and g are
integrable, both |f| and |g| are integrable.
(i) Now X
|f + g|dµ ≤ X
|f|dµ + X
|g|dµ < ∞. Therefore f + g is integrable. Let
h = f + g. Then h+
− h−
= f+
− f−
+ g+
− g−
gives h+
+ f−
+ g−
= h−
+ f+
= g+
.
Therefore X
h+
dµ + X
f−
dµ + X
g−
dµ = X
h−
dµ + X
f+
dµ + X
g+
dµ. As h, f and g
are integrable, all the numbers in above equation are finite. Therefore X
h+
dµ− X
h−
dµ =
X
f+
dµ − X
f−
dµ + X
g+
dµ − X
g−
dµ, i.e., X
(f + g)dµ = X
hdµ = X
fdµ + X
gdµ.
(ii) Since X
|αf|dµ = |α| X
|f|dµ, the function αf is integrable. If α = 0, then clearly
X
αfdµ = α X
fdµ.
Let α > 0. Then (αf)+
= αf+
and (αf)−
= αf−
. Now X
αfdµ = X
(αf)+
dµ −
X
(αf)−
dµ = X
αf+
dµ − X
αf−
dµ = α( X
f+
dµ − X
f−
dµ) = α X
fdµ.
Let α < 0. Then (αf)+
= (−α)f−
and (αf)−
= (−α)f+
. X
αfdµ = X
(αf)+
dµ −
X
(αf)−
dµ = X
(−α)f−
dµ − X
(−α)f+
dµ = (−α)( X
f−
dµ − X
f+
dµ) = α X
fdµ.
Corollary 1.3.25. If f1, f2, . . . , fn are integrable functions on a measure space (X, A , µ),
then X
( n
k=1 fk)dµ = n
k=1 X
fkdµ.
Proof. Use the Principle of Mathematical Induction.
Remarks 1.3.26.
(i) If f and g are measurable functions on a measure space, f is integrable and |g| ≤ |f|,
then g is integrable.
Here g is measurable and X
|g|dµ ≤ X
|f|dµ < ∞. Therefore g is integrable.
20. P
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3. Integration xix
(ii) Let f and g be integrable over E ∈ A . If f ≤ g a.e. [µ] on E, then E
fdµ ≤ E
gdµ.
Since f and g are integrable over E, g − f is integrable over E and E
(g − f)dµ =
E
gdµ − E
fdµ. Since f ≤ g a.e. [µ] on E, there is a measurable subset F of E with
µ(F) such that f ≤ g on E −F. Note that g−f is a non negative measurable function
on E. Therefore E
gdµ − E
fdµ = E
(g − f)dµ = F
(g − f)dµ + E−F
(g − f)dµ ≥ 0
as µ(F) = 0 and g − f ≥ 0 on E − F.
(iii) Let f be a measurable function on a measure space (X, A , µ), and let E ∈ A . If
µ(E) = 0 or f = 0 a.e. [µ] on E, then E
fdµ = 0.
Since µ(E) = 0 and f+
and f−
are non negative measurable functions, we get
E
f+
dµ = E
f−
dµ = 0. Therefore E
fdµ = 0.
If f = 0 a.e. [µ] on E, then f+
= 0 a.e. [µ] on E and f−
= 0 a.e. [µ] on E. Therefore
E
f+
dµ = E
f−
dµ = 0 and hence E
fdµ = 0.
In particular, if f and g are integrable functions on X with f = g a.e. [µ] on X, then
X
fdµ = X
gdµ.
Theorem 1.3.27 (Lebesgue Dominated Convergence Theorem). Let {fn} be a sequence of
measurable functions on a measure space (X, A , µ) converging to a function f (pointwise)
on X. Let E ∈ A . If g is integrable over E and |fn| ≤ g on E for all n, then E
fdµ =
limn E
fndµ.
Proof. Since {fn} converges to f and each fn is measurable, the function f is measurable.
As |fn| ≤ g on E for all n and fn → f as n → ∞, we have |f| ≤ |g| on E. Since g is integrable
over E, each f is integrable over E and E
|f|dµ ≤ E
gdµ. Consider the sequences {g + fn}
and {g − fn}. Then both are sequences of non negative measurable functions converging to
g + f and g − f respectively. Applying Fatou’s lemma we get
E
lim
n
(g + fn)dµ =
E
lim inf
n
(g + fn)dµ ≤ lim inf
n
E
(g + fn)dµ,
it means
E
gdµ +
E
fdµ ≤ lim inf
n
E
gdµ + lim inf
n
E
fndµ =
E
gdµ + lim inf
n
E
fndµ.
Therefore E
fdµ ≤ lim infn E
fndµ. Applying Fatou’s lemma to the sequence {g − fn} we
obtain
E
gdµ −
E
fdµ ≤ lim inf
n
E
gdµ + lim inf
n
(−
E
fndµ) =
E
gdµ − lim sup
n
E
fndµ.
Therefore E
fdµ ≥ lim supn E
fndµ. Hence E
fdµ ≤ lim infn E
fndµ ≤ lim supn E
fndµ ≤
E
fdµ. It means that the sequence { E
fndµ} is convergent and it converges to E
fdµ, i.e.,
limn E
fndµ = E
fdµ.
21. P
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xx 1. MEASURE AND INTEGRATION
Corollary 1.3.28 (Bounded Convergence Theorem). Let {fn} be a sequence of measurable
functions on a measure space (X, A , µ) converging to a function f (pointwise) on X, and
let E ∈ A with µ(E) < ∞. If there is M > 0 such that |fn| ≤ M on E for all n, then
E
fdµ = limn E
fndµ.
Proof. Define g(x) = M for x ∈ E. Then g is integrable over E and it follows from the
last theorem that E
fdµ = limn E
fndµ.
Examples 1.3.29.
(i) Let µ1, µ2, . . . , µk be measures on (x, A ), and let α1, α2, . . . , αk be nonnegative real
numbers. Then α1µ1 + α2µ2 + · · · + αkµk is a measure on (X, A ).
Let µ = α1µ1 + α2µ2 + · · · + αkµk. Clearly µ(∅) = 0 and µ(E) ≥ 0 for every E ∈ A .
Let {En} be a sequence of pairwise disjoint measurable subsets of X. Then
µ(
n
En) = α1µ1(
n
En) + α2µ2(
n
En) + · · · + αkµk(
n
En)
=
n
α1µ1(En) +
n
α2µ2(En) + · · · +
n
αkµk(En)
=
n
(α1µ1(En) + α2µ2(En) + · · · + αkµk(En)) (why?)
=
n
µ(En)
Therefore µ is a measure.
(ii) Let µ and η be measures on a measurable space (X, A ). Is ν = max{µ, η} a measure
on (X, A )?
Consider the measure space (R, M ). Let m be the Lebesgue measure on R, and let
δ0 be the point mass measure at 0. Then ν([0, 1]) = 1, ν([0, 1
2
]) = 1 and ν((1
2
, 1]) = 1
2
.
Hence ν([0, 1
2
]) + ν((1
2
, 1]) = ν([0, 1]). Therefore ν is not a measure on (R, M ).
(iii) Let (X, A ) be a measurable space, and let f : X → [−∞, ∞] be a map. Then f is
measurable if and only if f−1
({−∞}), f−1
({∞}) are measurable and that f−1
(E) is
measurable subset for every Borel subset E of R.
First assume that f−1
({−∞}), f−1
({∞}) are measurable and that f−1
(E) is measur-
able subset for every Borel subset E of R. Let α ∈ R. Then {x ∈ X : f(x) > α} =
f−1
((α, ∞]) = f−1
((α, ∞)) ∪ f−1
({∞}). As f−1
({∞}) is measurable and f−1
((α, ∞))
is measurable (as (α, ∞) is a Borel set), the set {x ∈ X : f(x) > α} is measurable.
Hence f is measurable.
Assume that f is measurable, then f−1
({−∞}) = n{x ∈ X : f(x) < −n} and
f−1
({∞}) = n{x ∈ X : f(x) > n} are measurable. Let a, b ∈ R. Then f−1
((a, ∞)) =
{x ∈ X : f(x) > a} − f−1
({∞}), f−1
((−∞, b)) = {x ∈ X : f(x) < b} − f−1
({−∞})
22. P
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3. Integration xxi
and f−1
((a, b)) = f−1
((a, ∞))∩f−1
((−∞, b)). It follows that f−1
((a, ∞)), f−1
((−∞, b))
and f−1
((a, b)) are measurable. Let C = {E ⊂ R : f−1
(E)} is measurable. Then C is
a σ- algebra of subsets of R. As f−1
((a, ∞)), f−1
((−∞, b)) and f−1
((a, b)) are mea-
surable, (a, ∞), (−∞, b), (a, b) ∈ C . Since any open subset of R is a countable union of
intervals of the form (a, ∞), (−∞, b) and (a, b), it follows that C contains every open
subset of R. Since B is the smallest σ- algebra containing all open subsets of R, it
follows that B ⊂ C . Let E be a Borel set, i.e., E ∈ B ⊂ C . Then by definition of C ,
f−1
(E) is measurable.
(iv) Let (X, A , µ) be a measure space which is not complete. If f = g a.e. [µ] on X and
f is measurable, show that g need not be measurable.
Since (X, A , µ) is not complete, there is a measurable subset E of X with µ(E) = 0
and E has a nonmeasurable subset say F. Define f = 0 and g = χF . Then f is
measurable and g is not measurable as F is not measurable. If x ∈ X − E, then
f(x) = g(x). Therefore f = g on X − E and µ(E) = 0. Hence f = g a.e. [µ] on X.
(v) Consider a measurable space (X, P(X)). Let η be a counting measure and let δx0 be
a Dirac measure at x0 ∈ X. Let f, g : X → [−∞, ∞].
(a) Show that f = g a.e. [δx0 ] if and only if f(x0) = g(x0).
(b) Show that f = g a.e. [η] if and only if f(x) = g(x) for every x ∈ X.
(a) Assume that f = g a.e. [δx0 ]. Then there is a measurable subset E of X with
δx0 (E) = 0 and f = g on X − E. Since δx0 (E) = 0, x0 /∈ E. Hence f(x0) = g(x0).
Conversely, assume that f(x0) = g(x0). Then X − {x0} is a measurable subset of X
with δx0 (X − {x0}) = 0. Therefore f = g a.e. [δx0 ].
(b) Assume that f = g a.e. [η]. Then there is a measurable subset E of X with
η(E) = 0 and f = g on X − E. Since η(E) = 0, E = ∅. Hence f(x) = g(x) for every
x ∈ X. Conversely, assume that f(x) = g(x) for every x ∈ X. Then clearly f = g a.e.
[η].
(vi) Let f and g be nonnegative measurable functions on a measure space (X, A , µ) with
g ≤ f. Show that f = g a.e. [µ] if and only of X
gdµ = X
fdµ.
We may write f = (f −g)+g. Note that both f −g and g are nonnegative measurable
functions therefore X
fdµ = X
(f − g)dµ + X
gdµ. We know that for a nonnegative
measurable function h, h = 0 a.e. [µ] if and only if X
hdµ = 0. Hence f = g a.e. [µ],
i.e., f − g = 0 a.e. [µ] if and only if X
(f − g)dµ = 0 if and only if X
fdµ = X
gdµ.