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Lecture 3: The Lebesgue Integral

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Course:
Theory of Probability I
Term:
Fall 2013
Instructor: Gordan Zitkovic

Lecture 3
The Lebesgue Integral
The construction of the integral
Unless expressly specified otherwise, we pick and fix a measure space
(S, S , µ) and assume that all functions under consideration are defined
there
...
1 (Simple functions)
...

The collection of all simple functions is denoted by LSimp,0 (more
precisely by LSimp,0 (S, S , µ)) and the family of non-negative simple
Simp,0
functions by L+

...
1)

k =1

for α1 ,
...
, An ∈ S
...

When the sets Ak , k = 1,
...
, αn )
...
The Lebesgue integral is very easy to define for non-negative
simple functions and this definition allows for further generalizations1 :
Definition 3
...
For f ∈
R
Simp,0
L+
we define the (Lebesgue) integral f dµ of f with respect to µ
by
Z

n

f dµ =

∑ αk µ( Ak ) ∈ [0, ∞],

k =1

where f =

∑nk=1

In fact, the progression of events you
will see in this section is typical for
measure theory: you start with indicator functions, move on to non-negative
simple functions, then to general nonnegative measurable functions, and finally to (not-necessarily-non-negative)
measurable functions
...

1

αk 1 Ak is a simple-function representation of f ,

Problem 3
...
Show that the Lebesgue integral is well-defined for simple functions, i
...
, that the value of the expression ∑nk=1 αk µ( Ak ) does
not depend on the choice of the simple-function representation of f
...
3
...
It is important to note that f dµ can equal +∞ even if f never
takes the value +∞
...
This is one of the reasons we start with non-negative
functions
...
On the other hand, such examples
cannot be constructed when µ is a finite measure
...


2
...
Indeed, if the simplefunction representation of f is given by f = ∑nk=1 1 Ak , for pairwise
disjoint A1 ,
...
When A1 ,
...
, n
...
Then
f = 1 A1 + 1 A2 = 1 A1 \C + 21C + 1 A2 \C ,
and so
Z

f dµ = µ( A1 \ C ) + µ( A2 \ C ) + 2µ(C ) = µ( A1 ) + µ( A2 ) + µ(C )
...
e
...
The
R
integral map f 7→ f dµ preserves this structure:
Simp,0

Problem 3
...
For f 1 , f 2 ∈ L+

and α1 , α2 ≥ 0 we have
R
R
1
...
(α1 f 1 + α2 f 2 ) dµ = α1 f 1 dµ + α2 f 2 dµ
...
In fact, at no extra cost we can
consider a slightly larger set consisting of all measurable [0, ∞]-valued


functions which we denote by L0+ [0, ∞]
...

2

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Lecture 3: The Lebesgue Integral

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Definition 3
...
For a
R


function f ∈ L0+ [0, ∞] , we define its Lebesgue integral f dµ by
Z

f dµ = sup

Z

g dµ : g ∈

Simp,0
L+
,


g ( x ) ≤ f ( x ), ∀ x ∈ S

∈ [0, ∞]
...
3
...
On the other hand, show
R
that f dµ = 0 for f of the form

∞, x ∈ A,
f ( x ) = ∞1 A ( x ) =
0, x 6= A,
whenever µ( A) = 0
...

A simple argument based on the monotonicity property of part 1
...
2 can be used to show that this is, indeed, the case
...


Finally, we are ready to define the integral for general measurable
functions
...
There exists a decomposition which is, in a sense,
minimal
...
The minimality we mentioned above is reflected in the fact that for each x ∈ S,
at most one of f + and f − is non-zero
...
5 (Integrable functions)
...

The collection of all integrable functions in L0 is denoted by L1
...
6 (The Lebesgue integral)
...


Remark 3
...

1
...
It is possible to
combine the two and define the Lebesgue integral for all functions
f ∈ L0 with f − ∈ L1
...


Note that no problems of the form ∞ − ∞ arise here, and also note
that, like L0+ , L0−1 is only a convex cone, and not a vector space
...

R
R
2
...

Problem 3
...
Show that the Lebesgue integral remains a monotone
operation in L0−1
...


First properties of the integral
The wider the generality to which a definition applies, the harder it is
to prove theorems about it
...
In fact, it seems that the easiest route towards linearity
is through two important results: an approximation theorem and a
convergence theorem
...
5
...
if f 1 ( x ) ≤ f 2 ( x ) for all x ∈ S then f 1 dµ ≤ f 2 dµ
...
α f dµ = α f dµ
...
8 (Monotone convergence theorem)
...
for all x ∈ S
...


Proof
...
5 above implies
R
R
R
that the sequence f n dµ is non-decreasing and that f n dµ ≤ f dµ
...
To show the opposite inequality, we
R
Simp,0
deal with the case f dµ < ∞ and pick ε > 0 and g ∈ L+
with
R
R
R
g( x ) ≤ f ( x ), for all x ∈ S and g dµ ≥ f dµ − ε (the case f dµ = ∞
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Lecture 3: The Lebesgue Integral

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is similar and left to the reader)
...

By the increase of the sequence { f n }n∈N , the sets { An }n∈N also increase
...
By non-negativity of f n and
monotonicity,
Z

Z

f n dµ ≥

and so
sup

Z

n

f n 1 An dµ ≥ c

f n dµ ≥ c sup
n

Z

Z

g1 An dµ,

g1 An dµ
...
Then
Z

g1 An dµ =

Z

k

k

i =1

i =1

∑ αi 1Bi ∩ An dµ = ∑ αi µ( Bi ∩ An )
...
, k, and the continuity
of measure implies that µ( An ∩ Bi ) % µ( Bi )
...


i =1

Consequently,
lim
n

Z

f n dµ = sup
n

Z

f n dµ ≥ c

Z

g dµ, for all c ∈ (0, 1),

and the proof is completed when we let c → 1
...
9
...
The monotone convergence theorem is really about the robustness
of the Lebesgue integral
...

2
...
Take, for example S = [0, 1], S = B([0, 1]),
and µ = λ (the Lebesgue measure), and define
f n = n1(0,n−1 ] , for n ∈ N
...
On the other hand
Z

1
x

and x > 0
...

n

We will see later that the while the equality of the limit of the integrals and the integral of the limit will not hold in general, they
will always be ordered in a specific way, if the functions { f n }n∈N
are non-negative (that will be the content of Fatou’s lemma below)
...
10 (Approximation by simple functions)
...
gn ( x ) ≤ gn+1 ( x ), for all n ∈ N and all x ∈ S,
2
...
f ( x ) = limn gn ( x ), for all x ∈ S, and
4
...

Proof
...
, n2n be a collection of subsets of S
given by


Ank = { k2−n1 ≤ f < 2kn } = f −1 [ k2−n1 , 2kn ) , k = 1,
...

Note that the sets Ank , k = 1,
...
, n2n
...


The statements 2
...
, and 4
...

Finally, we leave it to the reader to check the simple fact that { gn }n∈N
is non-decreasing
...
6
...


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Lecture 3: The Lebesgue Integral

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Proposition 3
...



For f 1 , f 2 ∈ L0+ [0, ∞] and α1 , α2 ≥ 0 we have
Z

(α1 f 1 + α2 f 2 ) dµ = α1

Z

f 1 dµ + α2

Z

f 2 dµ
...
Thanks to Problem 3
...
Let { gn1 }n∈N and { gn2 }n∈N be sequences in L+
which
1
2
approximate f and f in the sense of Proposition 3
...
The sequence
{ gn }n∈N given by gn = gn1 + gn2 , n ∈ N, has the following properties:
Simp,0

• gn ∈ L +

for n ∈ N,

• gn ( x ) is a nondecreasing sequence for each x ∈ S,
• gn ( x ) → f 1 ( x ) + f 2 ( x ), for all x ∈ S
...
8) to
conclude that
Z

Z
Z
Z
( f 1 + f 2 ) dµ = lim ( gn1 + gn2 ) dµ = lim
gn1 dµ + gn2 dµ
n

=

Z

n

f 1 dµ +

Z

f 2 dµ
...
12 (Countable additivity of the integral)
...
Then
Z

∑

n ∈N

f n dµ =

∑

Z

f n dµ
...
Apply the monotone convergence theorem to the partial sums
gn = f 1 + · · · + f n , and use linearity of integration
...
We leave it to the reader to
prove all the statements in the following problem:
Problem 3
...
The family L1 of integrable functions has the following
properties:
R
1
...
L1 is a vector space,
R
 R
3
...

R
R
R
4
...


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Lecture 3: The Lebesgue Integral

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We conclude the present section with two results, which, together
with the monotone convergence theorem, play the central role in the
Lebesgue integration theory
...
13 (Fatou’s lemma)
...

Then
Z
Z
lim inf f n dµ ≤ lim inf f n dµ
...
Set gn ( x ) = infk≥n f k ( x ), so that gn ∈ L0+ [0, ∞] and gn ( x ) is a
non-decreasing sequence for each x ∈ S
...
Its conclusion is not as strong as that of the
monotone convergence theorem, but it
proves to be very useful in various settings because it Rgives an upper bound
(namely lim infn f n dµ) on the integral
of the non-negative function lim inf f n
...

n

On the other hand, gn ( x ) ≤ f k ( x ) for all k ≥ n, and so
Z

gn dµ ≤ inf

k≥n

Z

f k dµ
...


Remark 3
...

1
...
You can use the
sequence { f n }n∈N of Remark 3
...

2
...
This requirement is necessary - to see that, simply consider the sequence {− f n }n∈N , where
{ f n }n∈N is the sequence of Remark 3
...

Theorem 3
...
Let { f n }n∈N be a
sequence in L0 with the property that there exists g ∈ L1 such that | f n ( x )| ≤
g( x ), for all x ∈ X and all n ∈ N
...


Proof
...
Since f n+ ≤ g, f n− ≤ g and g ∈ L1 ,
we immediately have f n ∈ L1 , for all n ∈ N
...
The price to be
paid is the uniform boundedness requirement
...
Still, it is an unexpectedly useful
theorem
...

Clearly g( x ) + f n ( x ) ≥ 0 for all n ∈ N and all x ∈ S, so we can
apply Fatou’s lemma to get
Z

g dµ + lim inf

Z

n

f n dµ = lim inf

Z

( g + f n ) dµ ≥

n

=

Z

( g + f ) dµ =

Z

Z

lim inf( g + f n ) dµ
n

g dµ +

Z

f dµ
...


Therefore
lim sup
n

and, consequently,

R

Z

f n dµ ≤

Z

f dµ = limn

f dµ ≤ lim inf

Z

n

R

f n dµ,

f n dµ
...
16
...
To make this statement
precise, we need to introduce some language:
Definition 3
...
Let (S, S , µ) be a measure space
...
N ∈ S is said to be a null set if µ( N ) = 0
...
A function f : S → R
set N such that f ( x ) = 0 for x ∈ N c
...
Two functions f , g are said to be equal almost everywhere - denoted
by f = g, a
...
- if f − g is a null function, i
...
, if there exists a null
set N such that f ( x ) = g( x ) for all x ∈ N c
...
18
...
In addition to almost-everywhere equality, one can talk about the
almost-everywhere version of any relation between functions which
can be defined on points
...
e
...


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Lecture 3: The Lebesgue Integral

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2
...
e
...
e
...
e
...
e
...

3
...
g
...
Indeed,
there is no guarantee that it is measurable at all
...
Any such (measurable)
null set is often referred to as the exceptional set
...
8
...
The almost-everywhere equality is an equivalence relation between
functions
...
The family { A ∈ S : µ( A) = 0 or µ( Ac ) = 0} is a σ-algebra (the
so-called µ-trivial σ-algebra)
...
19
...
e,
...
Then
Z

f dµ =

Z

g dµ
...
Let N be an exceptional set for f = g, a
...
, i
...
, f = g on N c
R
R
and µ( N ) = 0
...
On
R
the other hand f 1 N ≤ ∞1 N and ∞1 N dµ = 0, so, by monotonicity,
R
R
f 1 N dµ = 0
...
It remains to use the additivity
of integration to conclude that
Z

f dµ =

=

Z
Z

f 1 N c dµ +
g1 N c dµ +

Z
Z

f 1 N dµ
g1 N dµ =

Z

g dµ
...
19
also holds:
Problem 3
...
If f ∈ L0+ and

R

f dµ = 0, show that f = 0, a
...


Hint: What is the negation of the statement “ f = 0
...
e
...
A slightly weaker notion of convergence is
required, though:

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Lecture 3: The Lebesgue Integral

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Definition 3
...
A sequence of functions { f n }n∈N is said to converge
almost everywhere to the function f , if there exists a null set N such
that
f n ( x ) → f ( x ) for all x ∈ N c
...
21
...

Proposition 3
...
Let { f n }n∈N be a sequence in L0+ [0, ∞] with the property that
f n ≤ f n+1 a
...
, for all n ∈ N
...
e
...
There are “∞ + 1 a
...
-statements” we need to deal with: one
for each n ∈ N in f n ≤ f n+1 , a
...
, and an extra one when we assume that f n → f , a
...
Each of them comes with an exceptional set;
more precisely, let { An }n∈N be such that f n ( x ) ≤ f n+1 ( x ) for x ∈ Acn
and let B be such that f n ( x ) → f ( x ) for x ∈ Bc
...
Moreover, consider the
functions f˜, { f˜n }n∈N defined by f˜ = f 1 Ac , f˜n = f n 1 Ac
...
Therefore, the monotone convergence theR
R
orem (Theorem 3
...

R
R
Finally, Proposition 3
...

Problem 3
...
State and prove a version of the dominated convergence
theorem where the almost-everywhere convergence is used
...
23
...
If a sequence { f n }n∈N of measurable functions converges to the function f
everywhere, then f is necessarily a measurable function (see Proposition 1
...
However, if f n → f only almost everywhere, there is no
guarantee that f is measurable
...


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Lecture 3: The Lebesgue Integral

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Additional Problems
Problem 3
...
Prove the following result, known as the monotone-class theorem (remember that an % a means
that an is a non-decreasing sequence and an → a)

Hint: Use Theorems 3
...
12

Let H be a class of bounded functions from S into R satisfying the
following conditions
1
...
the constant function 1 is in H, and
3
...

Then, if H contains the indicator 1 A of every set A in some π-system
P , then H necessarily contains every bounded σ(P )-measurable
function on S
...
12 (A form of continuity for Lebesgue integration)
...
Show that
for each ε > 0 there exists δ > 0 such that if A ∈ S and µ( A) < δ, then
R

 f dµ < ε
...
13 (Sums as integrals)
...

1
...


n =1

2
...

3
...


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Lecture 3: The Lebesgue Integral

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Problem 3
...
Let (S, S , µ) be a finite
measure space
...


n ∈N

Problem 3
...
Let (S, S , µ) be a measure space,
R
and suppose f ∈ L1+ is such that f dµ = c > 0
...




n log 1 + ( f /n)α dµ

n

exists in [0, ∞] for each α > 0 and compute its value
...
16 (Integrals converge but the functions don’t
...

Problem 3
...
or they do, but are not dominated)
...

Problem 3
...
Let
S 6= ∅ be a set and let f : S → R be a function
...

Problem 3
...
Let (S, S , µ) and ( T, T , ν)
be two measurable spaces, and let F : S → T be a measurable function
with the property that ν = F∗ µ (i
...
, ν is the push-forward of µ through
F)
...


Problem 3
...
A finite collection ∆ = {t0 ,
...
The set of all partitions of [ a, b] is denoted by
P([ a, b])
...
, tn } ∈
P([ a, b]), we define its upper and lower Darboux sums U ( f , ∆) and
L( f , ∆) by
!
U ( f , ∆) =

n

∑

k =1

sup

f (t)

( t k − t k −1 )

t∈(tk−1 ,tk ]

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Lecture 3: The Lebesgue Integral

and
L( f , ∆) =

14 of 14

!

n

∑

k =1

inf

t∈(tk−1 ,tk ]

( t k − t k −1 )
...


In that case the common value of the supremum and the infimum
above is called the Riemann integral of the function f - denoted by
Rb
( R) a f ( x ) dx
...
Suppose that a bounded Borel-measurable function f : [ a, b] → R is
Riemann-integrable
...


2
...

3
...

4
...

Show that
• all monotone functions are Riemann-integrable,
• f ◦ g is Riemann integrable if f : [c, d] → R is Riemann integrable
and g : [ a, b] → [c, d] is continuous,
• products of Riemann-integrable functions are Riemann-integrable
...
Let ([ a, b], B([ a, b])∗ , λ∗ ) denote the completion of ([ a, b], B([ a, b]), λ)
...

Hint: Pick a sequence {∆n }n∈N in P([ a, b]) so that ∆n ⊆ ∆n+1 and U ( f , ∆n ) −
L( f , ∆n ) → 0
...
Conclude that f agrees with a Borel measurable function on a complement of a subset of the set { f 6= f } which has Lebesgue
measure 0

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