MAT4300
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23/08-2010
Curriculum for MAT4300 is chapters 1-13 including 19 and parts of 17,18.
There are two main areas of study: measures and integration.
Measure
We wish to measure “the size” of sets
-lengths
-areas
-volumes
-mass
-electrical charges
-probabilities
Notationwise we have Ω: which is the set of all possible outcomes, and A
is a specific set and a subset of Ω. We have P (A) which is the probability
that A will happen. In the sense of a measure the probability is the size of
the set.
Integration
We introduce a new way to integrate, i.e finding the area/volume under the
curve. Consider the function z = f (x, y),
∞
X
αi · m(Ai ) ≈
i=−∞
Z
f dm
This is the basic idea of the Lebesgue-integral which provides a stronger
integration concept than Riemann. The Lebesgue integral is defined on any
space with a measure.
1
Set operations
A set is a collection of objects, usually from a predefined universe usually
denoted by Ω. A and B are subsets of Ω as shown in (i) in the image below,
and are written as A, B ⊆ Ω.
For objects or elements we have x ∈ A which means x is included in A,
as shown in (ii) and y 6∈ A. When a set A is a strict subset of B, A ⊂ B
as shown in (iii), this means ∀x ∈ A ⇒ x ∈ B: a ll the elements in A are
elements in B. If A ⊂ B and B ⊂ A, then A = B. This is a very useful trick.
Further set operations are
A ∪ B: the collection of all elements included in at least one of A and B. (i)
A ∩ B: the collection of all elements included in both A and B. (ii)
A\B = a ∈ A | a 6∈ B : the collections of elements in A that are not in B.
(iii)
Two important concepts for measure theory: for countable sequences
(which is mainly what one focuses on in measure theory):
[
A1 ∪ A2 ∪ . . . =
An
n∈N
A1 ∩ A2 ∩ . . . =
\
n∈N
2
An .
We have distributive laws for sets.
B ∪ A1 ∩ A2 ∩ . . . ∩ An ∩ . . . = (B ∪ A1 ) ∩ (B ∪ A2 ) ∩ . . . ∩ (B ∪ An ) ∩ . . .
and
B ∩ A1 ∪ A2 ∪ . . . ∪ An ∩ . . . = (B ∩ A1 ) ∪ (B ∩ A2 ) ∪ . . . ∪ (B ∩ An ) ∪ . . .
Proof of the second law
Left side: Assume x ∈ B ∩ (A1 ∪ . . .) ⇔ x ∈ B and x ∈ Ai for at least one i.
Right side: Assume x ∈ (B ∩ A1 ) ∪ . . . ⇔ x ∈ B ∩ Ai for at least one i ⇔
x ∈ B and x ∈ Ai for at least one i. DeMorgan’s Laws
c
A1 ∪ A2 ∪ . . . = Ac1 ∩ Ac2 ∩ . . .
c
A1 ∩ A2 ∩ . . . = Ac1 ∪ Ac2 ∪ . . .
Proved in exercise 2.3.
Cartesian Products
A × B = (a, b) : a ∈ A and b ∈ B
A1 × A2 × . . . × An = (a1 , a2 , . . . , an ) : a1 ∈ A1 , . . . , ai ∈ Ai , . . . an ∈ An
When the same set is multiplied n times, we have a shorthand notation:
n
A
| ×A×
{z. . . × A} = A .
n times
Functions: f : X 7→ Y
In this course we can think of functions as a rule that takes each element
x ∈ X and assigns a new element f (x) ∈ Y as shown in (i).
3
If A ⊂ X we define the image of A as f (A) = f (a) : a ∈ A as shown in
(ii).
If B ⊂ Y we define the inverse image by f −1 (B) = a ∈ X : f (a) ∈ B as
shown in the image.
Injective, Surjective and Bijective functions.
For f : X 7→ Y , we say f is injective if x1 6= x2 ⇒ f (x1 ) 6= f (x2 ). Two
different elements x1 , x2 will always give two different elements f (x1 ), f (x2 ).
We say that f : X 7→ Y is surjective if f (X) = Y , that is if the Y is
completely “filled up” by f (X). It is permissible for two points in X to map
to the same point in Y .
4
A function is bijective if it is both injective and surjective. We get a one-toone correspondence between X and Y .
Relation between functions and sets.
Result
(i) f −1 A1 ∩ A2 ∩ . . . = f −1 (A1 ) ∩ f −1 (A2 ) ∩ . . .
(ii) f −1 A1 ∪ A2 ∪ . . . = f −1 (A1 ) ∪ f −1 (A2 ) ∪ . . .
Inverse images commutes with intersections and unions.
Proof of (i)
x ∈ f −1 (A1 ∩ A2 ∩ . . .) ⇔
f (x) ∈ A1 ∩ A2 ∩ . . . ⇔
x ∈ f −1 (Ai ) ∀i ⇔ x ∈ f −1 (A1 ) ∩ f −1 (A2 ) ∩ . . .
This only holds for inverse images. Direct images does not commute over
intersections.
Result
(i) f (A1 ∪ A2 ∪ . . . = f (A1 ) ∪ f (A2 ) ∪ . . .
(i) f (A1 ∩ A2 ∩ . . . ⊆ f (A1 ) ∩ f (A2 ) ∩ . . .
Proof of (ii)
Since
A1 ∩ A2 ∩ . . . ⊆ Ai ∀i ⇔
f A1 ∩ A2 ∩ . . . ⊆ f (Ai )∀i ⇔
f A1 ∩ A2 ∩ . . . ⊆ f (A1 ) ∩ f (A2 ) ∩ . . .
Since the second last statement tells us that we have an inclusion in all the
sets, the third statement follows. Counterexample to equality for (ii)
We define X = {x1 , x2 } and Y = {y}, where we assume x1 6= x2 and we
consider the function f : X 7→ Y . The only possibility we have is f (x1 ) = y
and f (x2 ) = y. If we set A1 = {x1 } and A2 = {x2 } we get A1 ∩ A2 = ∅. Thus
f (A1 ∩ A2 ) = f (∅) = ∅. However, f (A1 ) ∩ f (A2 ) = {y} ∩ {y} = {y}.
The function is not injective since x1 6= x2 are mapped to the same point
y. If we assume f is injective we get an equality in (ii).
5
Countability
A set A is countable if there exists a list a1 , a2 , a3 , . . . that contain all the
elements in A.
Example
The natural numbers. As defined they make up a satisfactory list.
N = {1, 2, 3, . . .}
The integers. We wish to have a beginning to the list, so we rearrange them.
Z = {. . . , −2, −1, 0, 1, 2, . . .} =⇒ {0, 1, −1, 2, −2, 3, −3, . . .}
Result
If A and B are countable, then the product A × B is countable.
Proof
We define
A = {a1 , a2 , a3 , . . .} B = {b1 , b2 , b3 , . . .}
Then the product A × B = {(ai , bi ) : ai ∈ A, bi ∈ B}. We can write a list
based on the index numbers: first all elements such that the index numbers
sum to two, then three etc.
A × B = {(a1 , b1 ), (a2 , b1 ), (a1 , b2 ), (a3 , b1 ), (a2 , b2 ), (a1 , b3 ), . . .}
{z
} |
{z
}
| {z } |
sum: 2
sum: 3
sum: 4
Corollary
The set of rational numbers Q is a countable set.
Proof
Both Z and N are countable, so by our last result Z × N is countable.
Z × N = {(a1 , b1 ), (a2 , b1 ), (a1 , b2 ), . . .}
so the list of the rational numbers is as follows (all fractions are repeated
infinite times, but that doesn’t matter)
a1 a2 a1
Q=
, , ,...
b1 b1 b2
Non-countable sets can be R or even the interval (0, 1).
Result
The interval (0,1) is not countable.
Proof
6
Assume for contradiction that (0,1) is countable so that we can make en
exhaustive list.
(0, 1) := {x1 , x2 , x3 , . . .}.
All the xi are decimals, so we can write out the decimal expanions.
x1 =
x2 =
x3 =
..
.
0.d11 d12 d13 d14 . . .
0.d21 d22 d23 d24 . . .
0.d31 d32 d33 d34 . . .
..
.
xn = 0.dn1 dn2 dn3 dn4 . . .
..
..
.
.
Now we find a number not included on this list. We choose the decimal
number t.
1 dii =
6 1
t = 0.e1 e2 e3 e4 . . . where ei =
2 dii = 1
We stay away from 0’s and 9’s since they can produce the same number (like
how 0.999 . . . = 1). We are left with t ∈ (0, 1) which is not included in the
list. Contradiction! 7
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30/08-2010
Definition
If A is a σ-algebra on a set X and µ : A 7→ [0, ∞] is a function we call µ a
measure if
(i) µ(∅) = 0
(ii) Whenever {Ai }i∈N is a disjoint sequence of elements in A then
[ X
µ
Ai =
µ(Ai )
i∈N
i∈N
Theorem - Continuity of measures
Assume that µ is a measure on (X, A), then
(i) For any increasing (i.e Ai ⊂ Ai+1 ) sequence {Ai }, then
[ Ai = lim µ(Ai )
µ
i→∞
i∈N
(ii) For any decreasing sequence {Ai } of sets in A such that µ(A1 ) < ∞, we
have
\ Ai = lim µ(Ai )
µ
i→∞
i∈N
Proof
(i) We set B1 = A1 , B2 = A2 \A1 , B3 = A3 \A2 , etc. We then have a disjoint
sequence {Bi } of sets from {Ai } and
B1 ∪ B2 ∪ . . . ∪ Bn = A1 ∪ A2 ∪ . . . ∪ An (= An ) =⇒
B1 ∪ B2 ∪ . . . ∪ Bn ∪ . . . = A1 ∪ A2 ∪ . . . ∪ An ∪ . . . .
Hence
µ
[
i∈N
=µ
n
[ X
X
µ(B
)
=
lim
µ(Bi) = lim µ(B1 ∪. . . Bn ) = lim µ(An )
B
=
• i
i
i∈N
i∈N
n→∞
i=1
8
n→∞
→∞
(ii)
From (i) in the image we have a graphical representation of our situation. A1
is the first, finite set and A = ∩i∈N Ai . We also observe that A1 \An ր A1 \A.
By (i) of the Theorem, µ(A1 \A) = limn→∞ µ(A1 \An ). Hence µ(A1 ) −
µ(A) = µ(A1 ) − limn→∞ µ(An ) since µ(A1 ) < ∞. With a little algebra
limn→∞ µ(An ) = µ(A). Example
Showing that the condition µ(A1 ) < ∞ cannot be removed. The function
µ is the Lebesgue measure on R, or more precisely on the Borel set (the
measure generalising length/area/volume). For µ([a, b]) = b − a, the length
of the interval. For µ[n, ∞) = ∞ since the interval is infinitely long.
T If we define An = [n∞) with µ(An ) = ∞ and consider the sequence
i∈N An we eventually end up with the emptyset, so
\ µ
An = µ(∅) = 0.
i∈N
We see how important this condition is for (ii) in the theorem.
Comment
One can turn the theorem around to prove that if A is a σ-algebra and
µ : A 7→ [0, ∞] satisfying
(i) µ(∅) = 0
(ii) µ(A ∪B)
·
= µ(A) + µ(B)
(iii) For any increasing sequence {An } of sets from A
[
An ) = lim µ(An )
µ(
n→∞
n∈N
then µ is a measure.
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Examples of measures
(i) Counting measure. Let X be finite, and A = P(X), the power set of X.
Define µ(A) = #A, which is the number of elements in set A.
#A
(ii) Normalised counting measure. As above except µ(A) = #X
, or the
(number of elements in A) divided by (the total number of elements in X),
and you get the proportion of elements in each set.
(iii) Dirac measure. Let (X, A) be arbitrary. Pick an element x0 ∈ X. The
Dirac measure of x0 is
1 x0 ∈ A
µx0 (A) =
0 x0 6∈ A
The physical interpretation is in a weightless room X where only the element
x0 has mass, which is 1, then all parts of the room excluding x0 will still be
weightless, whereas all parts of the room including x0 will have mass 1.
(iv) The n-dimensiona Lebesgue measure. If A is the Borel σ-algebra on Rn ,
then
µ [a1 , b1 ] × [a2 , b2 ] × · · · × [an , bn ] = (b1 − a1 )(b2 − a2 ) . . . (bn − an )
which can be length, area ,or volume depending on the dimension n. For
n = 2 this defines the area of a rectangle [a1 , b1 ] × [a2 , b2 ] = (b1 − a1 )(b2 − a2 ).
This is illustrated in the image:
10
(v) Coin tossing. For H, T for heads and tails, we have X : the set of all
infinite sequences of H’s and T ’s. We start by looking at how we should
ascribe to the measure where
1
µ HT
HHT
.
.
.
H
??
.
.
.
= n.
|
{z
}
2
n
For each independent outcome there is a 1/2 probability. For that exact
combination we thus get 1/2n . For probability measures we get the
probability we would expect.
The problem with these measures is what happens when we go from a simple
set to a σ-algebra and we have to deal with measures of infinite sets. This is
a problematic extension. This happens to be closely tied to the problem of
uniqueness.
Uniqueness of measures
Definition
Assume that X is a set. A collection D of subsets of X a Dynkin-system if
the following conditions are satisfied (similar as for σ-algebras).
(i) X ∈ D
(ii) If A ∈ D =⇒ Ac ∈ D
S
(iii) If {An } is a disjoint sequence of elements in D then • n∈N An ∈ D
Any σ-algebra is a Dynkin-system, but not the other way around; there are
Dynkin-systems that are not σ-algebras.
It is worth remembering that σ-algebras are the truly interesting object of
study. Dynkin-systems are just a tool used to prove things about σ-algebras.
Proposition
Assume G is any collection of subsets of X. Then there exists a smallest
Dynkin-system δ(G) containing G and G ⊂ δ(G) ⊂ σ(G). Sketch of proof
Show that
δ(G) =
\
F : F is a Dynkin system and G ⊂ F .
Show that the intersection of Dynkin systems is itself a Dynkin system. Recall
that
\
σ(G) =
F : F is a σ-algebra and G ⊂ F .
There are F ’s n the δ-intersection that are not used in the σ-intersection. It
is a ’bigger’-intersection, so it is a smaller set.
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Proposition
Assume that D is a Dynkin system that is stable/closed under finite
intersections. Then the D is a σ-algebra.
Stable/closed:
E, F ∈ D ⇒ E ∩ F ∈ D.
Proof
By definition, proposition (i) and (ii) are already satisfied. It is S
sufficient to
show that if {An } is a sequence of sets from D, then the union An ∈ D.
The trick is to turn the sequence {An } into a disjoint sequence {Bn } with
the same union.
Starting with the situation in a in the image. We define B1 = A1 , and then
B2 = A2 \A1 = A2 ∩ Ac1 . Writing this as an intersection, we already see that
B2 ∈ D. Next, for the set A3 we define B3 in such a way that we only include
the new information. B3 = A3 ∩ Ac2 ∩ Ac1 .
Continuing in this manner, we eventually get
Bn = An ∩ Acn−1 ∩ . . . ∩ Ac2 ∩ Ac1 .
The sequence {Bn } is disjoint (by construction), so we get
n
\
i=1
Bi =
n
[
Ai
and
i=1
[
i∈N
Ai =
[
Bn ∈ D
i∈N
This is quite straight forward, but it is quite hard to show in practice.
What we need is another result that states that if the generator set G is
closed under intersection, then so is δ(G). This is easier to use, but also much
harder to prove.
To make it easier we will introduce the supporting lemma.
Lemma
Assume D is a Dynkin system on X. For each set D ∈ D, the collection (also
called the trace of D)
DD = U ⊂ X : U ∩ D ∈ D
12
is also a Dynkin system.
Proof
Graphical illustration: we want to see that the gray area is a Dynkin system.
To do that, we must verify the three conditions for Dynkin systems.
(i) X ∈ D because X ∩ D = D and by definition D ∈ D, so X ∩ D = D ∈ D.
(ii) Assuming that U ∈ DD we must prove that U c ∈ DD , that is U c ∩D ∈ D.
First we note that (U c ∩ D)c = (U ∩ D) ∪D
· c . The first part is contained
in D by assumption, and D ∈ D ⇒ D c ∈ D by property (ii) for Dynkin
systems. By property (iii) of Dynkin systems, the disjoint union of elements
in D is again in D, so we know that (U c ∩ D)c ∈ D. Finally, by property (ii)
(U c ∩ D)c ∈ D implies U c ∩ D ∈ D.
(iii) Assume Ui i∈N is a disjoint sequence of sets from DD . We need to show
that
∪i∈N
Ui ∈ DD Since Ui ∈ DD we have Ui ∩ D ∈ D. This means that
Ui ∩ D is a disjoint sequence of elements in D, and since D is a Dynkin
system,
[
(Ui ∩ D) ∈ D.
i∈N
But
∪i∈N Ui ∩ D = ∪i∈N (Ui ∩ D) ∈ D, so we have
∪i∈N Ui ∈ DD
With this we have already proved the hard part of the theorem: If G is closed
under finite unions, then δ(G) = σ(G). This will be proved next time.
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06/09-2010
Recall: A Dynkin system is a collection of D of subsets of X such that:
X ∈D
(∆1 )
If D ∈ D =⇒ D c ∈ D
If {Dn } is a disjoint sequence of sets in D, then
(∆2 )
[
∈D
(∆3 )
n∈N
Proposition
A Dynkin system is a σ-algebra iff it is stable under finite intersections.
Proposition
If D is a Dynkin system and D ∈ D, then
DD = Q ⊂ X : Q ∩ D ∈ D
is a Dynkin system.
Theorem
If G is stable under finite intersections, then δ(G) = σ(G).
Proof
By previous proposition it suffices to prove that δ(G) is stable under finite
intersections.
Assume first that G ∈ G and consider the Dynkin system
DG = Q ⊂ X : Q ∩ G ∈ D
where D = δ(G). By previous proposition DG is a Dynkin system, and since
G is stable under finite intersections G ⊂ DG (and we will go on to check that
if Q ∈ G, then Q ∈ D).
Because if Q ∈ G then Q ∩ G ∈ G ⊂ D, because G is ∩-stable, and
D = δ(G).
But if DG is a Dynkin system containing G, then δ(G) ⊂ DG , for
δ(G) = D, since δ(G) is the smallest Dynkin system. This means that any
set D = D = δ(G) belongs to DG , that is D ∩ G ⊂ D. So far; for any G and
some D ∈ D then D ∩ G ∈ D.
Assume now that D, E ∈ D, and we shall prove that D ∩E ∈ D. Consider
DD = Q ⊂ X : Q ∩ D ∈ D .
14
By what we just proved, G ⊂ DD . Since δ(G) is the smallest Dynkin system
containing G, we have D = δ(G) ⊂ DD . This means that E ∈ DD and hence,
E ∩ D ∈ D. Proof II - Same proof repeated
It suffices to prove that if D, E ∈ δ(G), then D ∩ E ∈ δ(G). Let G ∈ G be
arbitrary, and let
DG = Q : Q ∩ G = D
where D = δ(G). Notice that by ∩-stability, G ⊂ DG . Any element in G ⊂ DG
(which can be verified).
Now choose Q ∈ G, then Q ∩ G ∈ G ⊂ D = δ(G). If the intersection is
in D, then Q ∈ DG . Since DG is a Dynkin system, D = δ(G) ⊆ DG , since
δ(G) is the smallest possible Dynkin system. This means that for any D ∈ D,
then we have D ∈ DG , and hence D ∩ G ∈ D. (An intersection between
an arbitrary element in G and D is still in D). However, this alone is not
enought to prove the theorem.
Second step in the proof: picking two elements in the Dynkin system:
D, E ∈ D, and we must prove that D ∩ E ∈ D.
DD = Q : Q ∩ D ∈ D .
By what we have above: G ∩ D ∈ D for all G ∈ G, hence G ⊂ DD and
moreover, D = δ(G) ⊂ DD . This means that E ∈ DD thus E ∩ D ∈ D. Theorem - Uniqueness of Measures
Assume that A = σ(G) where G satisfies the following conditions:
(i) G is ∩-stable.
(ii) G contains an exhausting sequence: there is an increasing sequence {Gi }i∈N
of sets in G such that ∪i∈N Gi = X.
If µ and ν are two measures such that µ(G) = ν(G), ∀G ∈ G, (the
measures agree on all sets in the generator), and µ(Gn ) = ν(Gn ) < ∞ for all
n ∈ N, then µ = ν; the measures are equal on all sets in the σ-algebra. (It is
always easier to check things in the generator than in the σ-algebra).
Proof
For each set Gi in the exhausting sequence, we define
Di = A ∈ A : µ(A ∩ Gi ) = ν(A ∩ Gi ) .
Assume we can prove that Di is a Dynkin system containing G as a subset:
Di ⊃ δ(G) = σ(G) = A.
15
where we have equality between the generators by proposition and ∩stability, and the σ-algebra equals A by assumption. Hence for all A ∈ A,
µ(A ∩ Gi ) = ν(A ∩ Gi ) for all i.
Since {A ∩ Gi } ր A, and by continuity of measures;
µ(A) = lim µ(A ∩ Gi ) = lim ν(A ∩ Gi ) = ν(A).
i→∞
i→∞
Since A was arbitrary, this is true for all A ∈ A.
We must now prove the assumption: that Di are Dynkin systems
containing G. If G ∈ G, then
µ(G ∩ Gi ) = ν(G ∩ Gi )
(where G ∩ Gi ∈ G by ∩-stability). Thus the inclusion is verified. We must
show this is a Dynkin system. We simply verify ∆1 − ∆3 .
(∆1 ) X ∈ Di by definition.
µ(X ∩ Gi ) = µ(Gi) = ν(Gi ) = ν(X ∩ Gi ).
(∆2 ) Assume that A ∈ Di , we must prove that Ac ∈ Di .
What we know is that A ∈ Di , which means that
µ(A ∩ Gi ) = ν(A ∩ Gi )
and what we need to show is that Ac ∈ Di which is
µ(Ac ∩ Gi ) = ν(Ac ∩ Gi ).
In the image, the dark gray area is in Di by assumption, and we want to
show that the light gray area also is in Di .
Using A ∩ Gi ⊂ Gi < ∞:
µ(A ∩ Gi ) = µ Gi \(A ∩ Gi ) = µ(Gi ) − µ(A ∩ Gi )
c
ν(Gi ) − ν(A ∩ Gi ) = ν Gi \(A ∩ Gi ) = ν(Ac ∩ Gi ).
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(∆3 ) Assume that {An } is a disjoint sequence of sets in Ai . We need to prove
that ∪n∈N An ∈ Di .
What we know:
µ(An ∩ Gi ) = ν(An ∩ Gi )
for all n. We need to prove that
[
[
µ
An ∩ Gi = ν
An ∩ Gi .
n∈N
n∈N
Beginning on the left side.
[
[
Distrib.
µ
An ∩ Gi
An ∩ Gi == µ
=
|
{z
}
n∈N
n∈N
Disjoint
X
µ(An ∩ Gi ) =
ν
n∈N
ν(An ∩ Gi ) =
n∈N
n∈N
[
X
|
An ∩ Gi
{z }
Disjoint
=ν
[
n∈N
An ∩ Gi
Construction of Measures
We know how to measure rectangles and other areas, but we now want to
extend this to the Borel set or any other “big” σ-algebra.
Definition
A collection S of subsets of X is called a semi-ring if the following conditions
are satisfied.
(S1 ) ∅ ∈ S
(S2 ) S, T ∈ S =⇒ S ∩ T ∈ S.
(S3 ) If S, T ∈ S, then S\T = ∪· N
n=1 Sn where {Sn } is a disjoint family of sets
in S.
A graphical illustration of the third property. The universe is a half-open
rectangle, and S is all possible half open rectangles. For the difference S\T
we have three disjoint, half-open rectangles, S1 , S2 and S3 .
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The first step of the construction is to use µ to define an outer measure
µ∗ defined on all subsets A ⊂ X. For A ⊂ X, let
n o
[
CA =
Sn n∈N : Sn ∈ S and
Sn ⊃ A .
n∈N
For the set A in X, we cover it up with balls Sn as depicted in the image:
The area of any given ball is µ(Sn ). If we add up all the balls, we get
(measuring from the outside) a crude approximation:
X
µ(Sn ) ≥ µ∗ (A).
Define the outer measure µ∗ (A) by
o
nX
∗
µ(Sn ) : {Sn } ∈ CA .
µ (A) := inf
n∈N
This is the best we can do with the framework set out in the theorem. We
will come back and study the properties of outer measures.
Caratheodory’s Extension Theorem
Assume that S is a semi-ring of subsets of X and µ : S 7→ [0, ∞] (the positive,
extended real line, also called R̄+ ), is a function satifying
(i) µ(∅) = 0
(ii)
µ
[
• Sn
n∈N
!
=
X
µ(Sn )
n∈N
for all disjoint sequences {Sn } in S whose union happen to be in S.
Then; µ can be extended to a measure on the σ-algebra σ(S) generated by
S. If there is an exhausting sequence {Sn } of sets in S such that µ(Sn ) < ∞
for all n, then the extension is unique.
Will prove this next time.
18
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13/09-2010
Recall that a semiring of subsets is a collection S such that
(S1 ) ∅ ∈ S
(S2 ) S, T ∈ S =⇒ S ∩ T ∈ S.
(S3 ) S, T ∈ S ⇒ S\T is a finite disjoint union of sets in S:
S\T = S1 ∪S
· 2 ∪· . . . ∪S
· m.
We assume that µ : S 7→ [0, ∞] satisfies
(i) µ(∅) = 0.
(ii) Additivity (if the union happens to be disjoint. There is no requirement
that it must be)
[ X
µ • =
µ(Si )
i∈N
i∈N
for all families of {Si } of sets on S whose union is in S.
Our aim is to show that µ can be extended on σ(S) The outer measure µ∗
generated by µ is the function µ∗ : P (X) 7→ [0, ∞] defined by
Z X
[
∗
Si ⊃ A, Si ∈ S .
µ(Si ) |
µ (A) =
i∈N
i∈N
We want to prove that µ∗ restricted to σ(S) is a measure extending µ.
Lemma
(i)
µ(∅) = 0.
(ii)
A ⊆ B =⇒ µ∗ (A) ≤ µ∗ (B).
(iii) Sub-additivity.
µ∗
[
i∈N
X
Ai ≤
µ∗ (Ai )
i∈N
which holds for all sequences {Ai }.
19
Proof
With basis in the definition of µ∗ .
(i) Since ∅, ∅, ∅, . . . is a covering of ∅ we have
µ∗ (∅) ≤ µ(∅) + µ(∅) + . . . = 0 =⇒ µ∗ (∅) ≤ 0.
(ii)
∗
µ (A) = inf
nX
i∈N
inf
nX
o
µ Si ) | Si is a covering of A ≤
o
µ Si ) | Si is a covering of B = µ∗ (B).
i∈N
We have the inequality since A ⊆ B, so µ∗ (A) ≤ µ∗ (B).
(iii) (Useful argument). First we note that there is nothing to prove if we
have an infinite sum. It is therefore sufficient to prove the statement when
µ(Ai ) is finite.
Given ε > 0 we can for each Ai find a covering {Sn }n∈N such that
X
ε
µ(Sni ) ≤ µ∗ (Ai ) + i .
2
n∈N
(We can write it this way by the definition of the infimum). Note that
{Sni }i∈N,n∈N is a covering of ∪i∈N Ai , hence
[ X
X
XX
X
ε
µ∗ (Ai ) + ε
µ∗
Ai ≤
µ(Sni ) =
µ(Sni ) ≤
µ∗ (Ai ) + i =
2
n∈N
i∈N
i,n∈N
n∈N i∈N
n∈N
This is true for any ε > 0, so
[ X
µ∗
Ai ≤
µ∗ (Ai ).
i∈N
n∈N
Definition
S∪ consist of all subsets of X which can be written as a finite, disjoint union
of elements in S. Hence A ∈ S∪ ⇔ there exists disjoint set S1 , . . . , Sm ∈ S
such that A = ∪· m
i=1 Si .
Lemma
Assume that A, B ∈ S∪ , then
(i) A ∩ B ∈ S∪ .
(ii) A\B ∈ S∪ .
(iii) A ∪ B ∈ S∪ .
20
Proof
First observe that S∪ is closed under finite, disjoint unions, because if
A, B ∈ S∪ are disjoint, then A = ∪· m
· nj=1Tj so A ∪ B =
i=1 Si and B = ∪
S1 ∪· . . . Sm ∪T
· 1 . . . Tm ∈ S∪ . All S and T ’s are disjoint, so the union of A
and B is also disjoint.
(i) We have A = S1 ∪· . . . ∪S
· m and B = T1 ∪· . . . ∪T
· n.
A ∩ B = S1 ∪· . . . ∪S
· m T1 ∪· . . . ∪T
· n
To be in element in set, you must be in exactly one Si and on Tj .
[
= (Si ∩ Tj ), (S2 ) =⇒ Si ∩ Tj ∈ S.
i,j
Si ∩ Tj is disjoint from all other Si ’s and Tj ’s, so this is a A ∩ B can be broken
down into a disjoint union, and by the remark in this proof, A ∩ B ∈ S∪ .
(ii)
A\B = S1 ∪· . . . ∪S
· m \ T1 ∪· . . . ∪T
· n =
S1 ∪· . . . ∪S
· m
\
T1 ∪· . . . ∪T
· n
c
=
\ c
S1 ∪· . . . ∪S
· m
T1 ∩. . .∩Tnc = S1 ∩ T1c ∩. . .∩Tnc ∪· . . . ∪S
· m ∩ T1c ∩. . .∩Tnc
\
\
\
\
[\
=
S1 ∩Tjc ∪· . . . ∪·
Sm ∩Tjc =
S1 \Tj ∪· . . . ∪·
Sm \Tj = •
(Si \Tj )
j
j
j
j
i
j
Now Si \Tj ∈ S∪ by (S3 ). We have ∩j Si \Tj ∈ S∪ by (i) extended to several
sets and not just two. And finally we have ∪· i ∩j (Si \Tj ∈ S∪ by the remark
at the beginning of the proof.
(iii) For A ∪ B we divide it in three disjoint parts as illustrated in the image.
We have A ∪ B = (A\B) ∪(A
·
∩ B) ∪(B\A).
·
By parts (i) and (ii) all the
three separate parts are in S∪ , and by the remark at the beginning of the
proof, the disjoint union is also in S∪ , thus A ∪ B ∈ S∪ . 21
We would like to extend µ to a function µ on S∪ by A = S1 ∪· . . . ∪S
· m.
µ(A) = µ(S1 ) + . . . + µ(Sm ).
However, we may run into a problem if we have an alternative representation
for the set, like for instance A = T1 ∪· . . . ∪T
· n , in which case we would have
µ(A) = µ(T1 ) + . . . + µ(Tn ).
Suddenly µ(A) can seemingly have two different values.
Lemma
Assume A ∈ S∪ can be written as the disjoint union of sets in S∪ in two
different ways.
A = S1 ∪· . . . ∪S
· m
&
A = T1 ∪· . . . ∪T
· n.
Then
µ(S1 ) + . . . + µ(Sm ) = µ(T1 ) + . . . + µ(Tn ).
Proof
Note that Si = Si ∩ T1 ∪· . . . ∪· Si ∩ Tn . Since µ is additive on S:
µ(Si ) =
n
X
j=1
Hence
m
X
µ(Si ) =
i=1
µ(Si ∩ Tj ),
| {z }
m X
n
X
∈S∪
µ(Si ∩ Tj ).
i=1 j=1
By the exact same argument with the Si ’s and Tj ’s switched, we get
n
X
j=1
µ(Si) =
m
n X
X
µ(Tj ∩ Si )
j=1 i=1
and since both sums are equal to the same expression, we have equality. Definition
Define µ : S∪ 7→ [0, ∞] by
µ(A) = µ(S1 ) + . . . + µ(Sm )
whenever A = S1 ∪· . . . ∪S
· m . Observe that µ = µ on S.
∗
Note: It is µ which is the important measure. We merely introduce µ to
prove things about µ∗ .
22
Lemma
The measure µ is σ-additive on S∪ , that is if {Tk } is a disjoint sequence from
S∪ such that T = ∪k∈N Tk ∈ S∪ , then
X
µ(T ) =
µ(Tk )
k∈N
Proof
Since T ∈ S∪ , there are disjoint sets U1 , U2 , . . . , Uk such that T = ∪· ni=1 Uk .
We have a slight problem: the Ti that intersect with several Uj ’s. Solution:
we make a finer partition of Tk . Since Tk ∈ S∪ it can be decomposed as a
disjoint union of sets in S, and being clever we may assume that each set in
the decomposition is a subset of one of the U’s.
For T1 ∪· . . . Sn(1) , each set Si is found within exactly one Uj . We continue
T2 = Sn(1)+1 ∪· . . . ∪S
· n(2) and eventually Tk = Sn(k−1)+1 ∪· . . . ∪S
· n(k) .
Let N(i) be the set of indices j suchSthat Sj ⊂ Ui : we collect all the Sj ’s
that belong in a specific Ui . Then Ui = • j∈N (i) Sj , and by addivity of µ, this
means
X
µ(Ui ) =
µ(Sj ).
j∈N (i)
µ(T ) =
k
X
µ(Ui ) =
µ(Ui ) =
i=1 i=1
i=1
On the other hand,
k
k X
X
µ(Tk ) =
n
X
X
µ(Sj ).
j∈N
(k)µ(Si ).
i=n(k−1)+1
We get
∞
X
k=1
hence
µ(Tk ) =
∞
X
n
X
(k)µ(Si ) =
k=1 i=n(k−1)+1
µ(T ) =
∞
X
k=1
23
X
µ(Si),
i∈N
µ(Tk )
The outer measure µ∗ is an extension of µ, but this is quite difficult to
prove. To do it we need µ. We start with a lemma.
Lemma
(i) If A ⊆ B, then µ(A) ≤ µ(B).
(ii) If {An } is a sequence from S∪ and ∪n∈N An ∈ S∪ , then
∞
X
µ(An )
µ ∪n An ≤
n=1
Proof
Consult text book. Similar to previous proofs.
Lemma
The measure µ∗ is an extension of µ, that is µ∗ (S) = µ(S) for all S ∈ S.
Proof
Since {S, ∅, ∅, . . .} is a covering of S,
µ∗ (S) ≤ µ(S) + µ(∅) + µ(∅) + . . . =⇒ µ∗ (S) ≤ µ(S).
∗
where we used the definition of µ∗ . We now need to prove that
Pµ (S) ≥ µ(S).
It is sufficient to prove that if {Si } is a covering of S, then
µ(Si ) ≥ µ(S).
µ(S) = µ(S) = µ
[
i∈N
a X
X
b X
µ(Si)
(S ∩ Si ≤
µ(S ∩ Si ) ≤
µ(Si ) =
i∈N
i∈N
i∈N
where a is from sub-additivity
and b from S ∩ Si ⊆ Si . With this we have
P
shown that µ(S) ≤ µ(Si ) for any covering {Si }, and thus µ(S) ≤ µ∗ (S).
We have shown the inequality both ways, which means µ∗ (S) = µ(S) for
all S ∈ S. 24
—————————————————————————
16/09-2010
We have S is a semiring, and we have the pre-measure µ : S :7→ [0, ∞], with
(i)
µ(∅) = 0
(ii)
µ
[
• Sn
k∈N
!
=
X
µ(Sn ).
k∈N
We defined the outer measure as:
X
µ∗ (A) = inf
µ(Si) : {Si } covers A}.
i
Fact: µ and µ∗ agree on S.
Definition
A∗ = A ⊂ X | µ∗ (Q ∩ A) + µ∗ (Q\A) = µ∗ (A) .
| {z }
Q∩Ac
∗
∗
A is a collection of µ -measurable sets.
We already have, by subadditivity,
µ∗ (Q ∩ A) + µ∗ (Q\A) ≥ µ∗ (Q)
so all we need to prove is,
µ∗ (Q ∩ A) + µ∗ (Q\A) ≤ µ∗ (Q).
Lemma
S ⊂ A∗ .
Proof
For any S ∈ S we have to prove
µ∗ (Q ∩ S) + µ∗ (Q\S) = µ∗ (Q)
∀Q ⊂ X.
Assume first Q = T ∈ S, then
∗
∗
)
+
µ
(T
\S)
=
µ(T
∩
S)
+
µ
S
∪
S
∪
.
.
.
∪
S
≤
µ∗ (T
∩
S
1
2
m
| {z }
S
25
µ(T ∩ S) +
X
µ∗ (Si ) = µ(T ∩ S) +
i
X
µ(Si ))µ(T ) = µ∗ (T ).
i
Now let Q be an arbitrary subset of X. Let {Tn } be any covering of Q,
!
[
Tn ∩ S + ....
µ∗ (Q ∩ S) + µ∗ (Q\S) ≤ µ∗
n∈N
(proof cuts off here).
Proposition
A∗ is a σ-algebra, and µ∗ restricted to A∗ is a measure.
Proof
To prove that A∗ is a σ-algebra, we use Dynkin systems. We shall prove
that A∗ is a Dynkin system closed under finite intersections. We are going
to verify the following 4 properties:
(∆1 ): ∅ ∈ A∗ .
(∆2 ): A ∈ A∗ implies Ac ∈ A∗ .
∩-stability: A, B ∈ A∗ implies A ∩ B ∈ A∗ A∗ .
(∆3 ): If {An } is a disjoint family from A∗ , then ∪· n An ∈ A∗ .
Verifying the first property.
(∆1 ):
µ∗ (Q ∩ ∅) + µ∗ (Q\∅) = µ∗ (∅) + µ∗ (Q) = 0 + µ∗ (Q) = µ∗ (Q).
(∆2 ): If A ∈ A∗ , then
µ∗ (Q ∩ A) + µ∗ (Q ∩ Ac ) = µ∗ (Q),
hence,
µ∗ (Q ∩ Ac ) + µ∗ (Q ∩ (Ac )c ) = µ∗ (Q ∩ Ac ) + µ∗ (Q ∩ A) = µ∗ (Q).
Thus Ac ∈ A∗ .
∩-stability. It is easier to prove that it is closed under unions, but this
is sufficient because you can use DeMorgan’s law and (∆2 ) to show that the
intersections are included in A∗ . (If complements are proved, we can always
use this trick).
26
Assume A, B ∈ A∗ . We want to prove that A ∪ B ∈ A∗ . Working with
the definition:
µ∗ Q ∩ (A ∪ B) + µ∗ Q ∩ (A ∪ B)c =
|
{z
}
Q\(A∪B)
Our goal is to prove that this sum is less than or equal to µ∗ (Q). All we
have to work with is that A, B ∈ A∗ . The second term is easy, we have 3
intersections which is okay. Intersections and unions combined are harder.
We write A ∪ B as the disjoint intersection A ∪(B\A).
·
µ∗ Q ∩ (A ∪(B\A))
·
+ µ∗ (Q ∩ Ac ∩ B c ) =
µ∗ (Q ∩ A) ∪(Q
·
∩ (B\A)) + µ∗ (Q ∩ Ac ∩ B c ) ≤
µ∗ (Q ∩ A) + µ∗ (Q ∩ (B\A)) + µ∗ (Q ∩ Ac ∩ B c ) =
µ∗ (Q ∩ A) + µ∗ (Q ∩ B ∩ Ac ) + µ∗ (Q ∩ Ac ∩ B c ) =
µ∗ (Q ∩ A) + µ∗ (Q ∩ Ac ∩ B) + µ∗ (Q ∩ Ac ∩ B c ) =
{z
}
|
B∈A∗
µ∗ (Q ∩ A) + µ∗ (Q ∩ Ac ) = µ∗ (Q)
by definition of A∗ and using that A ∈ A∗ .
All that
remains now is to prove ∆3 for A∗ , and after that you have to
∗
show µ A∗ , and then the proof of Caratheodory is completed.
Sublemma
If A1 , A2 ∈ A∗ are disjoint, then for any Q:
Proof
µ∗ (Q ∩ A1 ) + µ∗ (Q ∩ A2 ) = µ∗ Q ∩ (A1 ∪ A2 ) .
We have µ∗ Q ∩ (A1 ∪ A2 ) , and using that A1 ∈ A∗ and the definition of
A∗ , we et
µ∗ ∗ [Q ∩ (A1 ∪ A2 ) ∩ A1 + µ∗ [Q ∩ (A1 ∪ A2 )] ∩ Ac1 =
(this is the shaded set iin the image intersected with A1 ):
27
µ∗ (Q ∩ A1 ) + µ∗ (Q ∩ A2 ).
We can extend this by induction.
X ∗
µ∗ Q ∩ (A1 ∪ . . . ∪ An ) =
µ (Q ∩ Ai )
i
for all disjoint Ai ∈ A∗ .
-final parts of proof are missing-
28
—————————————————————————
20/09-2010
We recall that S is a semiring, which means:
S1 : ∅ ∈ S
S2 : S, T ∈ S =⇒ S ∩ T ∈ S
S3 : S, T ∈ S =⇒ S\T = ∪· N
i=1 Sn .
We also have the function µ : S 7→ [0, ∞] with
(i) µ(∅) = 0
(ii) µ ( ∪· ∞
n=1 Sn ) =
P∞
n=1
µ(Sn )
Caratheodory’s Ext. Thm
The function µ can be extended to a measure on σ(S).
How do we use this theorem to construct the Lebesgue-measure on Rn ?
Intuitively the Lebesgue measure is a Borel measure on Rn (rectangles/boxes) such that µ(R) = area(R) for each rectangular box with sides
parallel to the axes.
I 2 consists of the empty set plus all half open rectangles of the form
I = [a, b) × [c, d) (this way the rectagnles fit perfectly together).
We start with I 2 , and show that this is a semiring, and then we use
Caratheodory’s extension theorem and get a measure on the σ-algebra
generated by the semiring, which is the collection of all half-open sets: so
the σ-algebra is the Borel σ-algebra.
We verify that I 2 is indeed a semiring. We check the three properties.
S1 : ∅ ∈ I 2 (obvious).
S2 : S, T ∈ I 2 =⇒ S ∩ T ∈ I 2 . This is clear from the illustration,
where we can see that the intersection is a half open rectangle with the same
form as S and T (and if they are disjoint, then the intersection is the emptyset
which is in I 2 by S1 .
29
S3 : The third property is also clear from an illustration. We just need
to see that the difference of S and T can be composed by a finite amount of
half open rectangles.
So, S\T = ∪· ni=1 Si .
The second step is to show that I 2 generates the σ-algebra, that is σ(I 2 )
generates B(R). To show this, it suffices to show that all open sets are in
σ(I 2 ). The general idea: inside the box B(x, ε), you can find a box (half
open rectangle) where the corners are rational numbers: Ix . See image:
We do this for all x ∈ U, so
U=
[
Ix .
x∈U
This an unountable union, but we can make it countable: so there are only
a countable number of rectangles with rational corners.
Define
A function µ on I 2 is defined by
µ [a, b) × [c, d) = (b − a)(d − c).
30
To show that this is a true measure, we must berify the two properties,
(i) µ(∅) = 0.
S
P∞
(ii) µ ( • ∞
n=1 µ(Sn ).
n=1 Sn ) =
Showing the first property is easy, but the second one is difficult. Instead
of verifying it directly we replace it by (ii′ ): finite additivity: µ (A ∪B)
·
=
µ(A) + µ(B), and (iii′ ): For {An }\∅: some decreasing sequence in I 2 , then
limn→∞ µ(An ) = 0 for An ∈ I 2 .
With finite additivity, proving (iii′ ) is quite easy (and it is covered in the
book).
We also use Caratheodory’s theorem to construct product measures. Say we
have two measurable spaces (X, A, µ) and (Y, B, ν) and we want a measure
on the space X × Y such that
(µ × ν)(A × B) = µ(A)ν(B).
We just show that the product is a semiring and we must show that the
measure satisfies the two properties for measures.
Chapter 7 - Measurable Mappings
A measurable space (X, A) consists of a set X and a σ-algebra A of subsets
of X.
If we have twp measurable spaces (X, A) and (Y, B), then a map
(function) T : X 7→ Y is called A − B-measurable if T −1 (B) ∈ A for all
B ∈ B.
If this is true for any set B, the mapping is A − B-measurable. (A function
is continuous if the inverse of a open set is still open - a related concept).
Backwards images are more well-behaved than forward images.
Lemma
Assume that (X, A) and (Y, B) are measurable spaces, and B = σ(G), B is
generated by some set G. If
T −1 (G) ∈ A,
31
∀G ∈ G,
then the mapping is A − B-measurable.
We can check the generator instead of the σ-algebra: so this lemma makes
the potential work easier.
Proof
Let
C = C ∈ Y T −1 (C) ∈ A ,
the collection of sets in Y with an inverse image in the σ-algebra A. Note
that G ⊂ C, and that it suffices to prove that C is a σ-algebra, because B is
the smallest σ-algebra: B ⊆ C.
To prove C is a σ-algebra, we check the three conditions.
(i) Y ∈ C: T −1 (Y ) = X ∈ A.
(ii) Assume C ∈ C which implies that T −1 (C) ∈ A. We must show
that C c ∈ C
c
T −1 (C c ) = T −1 (C)
where we used that inverse
Since
c images commute with complements.
−1
−1
c
T (C) ∈ A, then T (C) ∈ A, since it is a σ-algebra, thus C ∈ C.
(iii) Assume {Cn } is a sequence of sets from C, we must prove that
∪n Cn ∈ C.
Since Cn ∈ C, T −1 (C) ∈ A. It follows that
!
[
[
Cn
T −1 (Cn ) = T −1
n∈N
n∈N
and since A is a σ-algebra
[
Cn ∈ C.
n∈N
Proposition
Let (X, A), (Y, B) and (Z, C) be three measurable spaces, and assume that
S : X 7→ Y and T : Y 7→ Z are measurable mappnigs (wrt the σ-algebras),
then the composition R = T ◦ S (or R(X) = T S(X) is a measurable map.
32
Proof
Start with a set C ∈ C, and if R−1 (C) ∈ A, then R is a measurable map.
But,
R−1 (C) = S −1 T −1 (C)
| {z }
∈B
{z
}
|
∈A
since both T and S are measurable functions. Assume now that (X, A) and (Y, B) are measurable spaces and T : X 7→ Y
is a measurable mapping. Assume also that µ is a measure on (X, A) and
introduce a function µT on B by
µT (B) = µ T −1 (B) .
| {z }
∈A
We measure the set B via X with the function µ.
Proposition
The mapping µT is a measure on (Y, B) and is called the image measure of
µ under T .
Proof
To prove something is a measure, we must show it satisfies the two defining
properties.
(i) µT (∅) = 0
(ii) For a disjoint sequence {Bn } in B, µT ( ∪· n∈N Bn ) =
Showing the first property.
since µ is a measure.
Def.
µT (∅) = µ T −1 (∅) = µ(∅) = 0
The second property.
µT
[
• Bn
n∈N
!
=µ T
−1
[
• Bn
n∈N
33
!!
=
P
n∈N
µT (Bn ).
µ
[
• T −1 (Bn )
n∈N
!
=
X
n∈N
X
µ T −1 (Bn ) =
µT (Bn )
n∈N
For a probability space (Ω, A, P ), and some measurable function X : Ω 7→ R
(which is a measurable map), this X is called a random variable. Now PX is
a measure on R such that
PX (A) = P X −1 (A) = P ω ∈ Ω X(ω) ∈ A ,
and this set is called the distribution of the random variable X. This is an
example of a setting where image construction is used.
Chapter 8 - Measurable Functions
Let (X, A) be any measurable space and let (R, B(R) be the real numbers
with the Borel σ-algebra. A measurable map f : X 7→ R is called a
measurable function.
Definition
The function f is measurable if f −1 (B) ∈ A for any Borel set B.
Lemma
A function f : X 7→ R is measurable if any one of the following conditions
are satisfied.
(i) f −1 (−∞, a] ∈ A ∀a ∈ R
(ii) f −1 (−∞, a) ∈ A ∀a ∈ R
(iii) f −1 (−∞, q] ∈ A ∀q ∈ Q
(iv) f −1 (−∞, q) ∈ A ∀q ∈ Q
In each case we get the Borel set. It is sufficient to check one of these.
Proof
In each case the sets (intervalls) in question generate the Borel σ-algebra.
Lemma
All continuous functions f : Rn 7→ R are B(Rn ) − B(R)-measurable.
Proof
If G ⊂ R is open, then f −1 (G) is also open, so f −1 (G) ∈ O(Rn ), and
O(Rn ) ⊂ B(Rn ). Since G is a generator for the Borel sets in R: B(R), this
shows that f is measurable. Remark: there are many important measures that are not continuous, so this
theorem has some restrictions.
34
Example
A measurable function that is not a continuous function. Let A ∈ A. Then
the indicator function IA , which is given by
1 x∈A
I=
0 x 6∈ A
is measurable. Confer with image:
For some intervall, we have the inverse image to the indicator function,
given by:

a≥1
 Xc
−1
I [−a, a] = A 0 ≤ a < 1

∅
a < 0.
All these sets are in the σ-algebra, thus the inverse image is in the σ-algebra,
so I is measurable.
35
—————————————————————————
27/09-2010
We introduce the extended real numbers.
R = R ∪ {−∞, ∞}.
In calculations we have the obvious extensions:
∞ + ∞ = ∞,
∞ + a = ∞,
∞ a>0
a·∞ =
−∞ a < 0
a∈R
In addition, we have the special property that only applies in measure theory:
0 · ∞ = 0.
Operations that are undefined are
±∞
.
±∞
∞ − ∞,
The Borel σ-algebra on R is defined as,
B(R) = A ⊆ R : A ∩ R ⊂ B(R) .
Lemma
A function f : X 7→ R is measurable if, and only if,
(i) {x : f (x) < α} is measurable.
(ii) {x : f (x) ≤ α} is measurable.
Simple Functions
A simple function on (X, A) is a function of the form
f (x) =
∞
X
ai IAi (x)
n=1
where A ∈ A for i = 1, . . . , n. A simple function is always measurable.
Lemma
Every simple function has a standard form representation.
36
Proof
Proved in the text book.
This is important since the standard form is unique, and we don’t measure
the same function with different size.
Observation
Any measurable function f : X 7→ R can be written as the difference between
two non-negative, measurable functions. We have f = f + − f − as shown in
the image:
where we define
f + = max{f (x), 0}
f − = − min{f (x), 0}
Each of these are measurable functions, which is easily seen from the
definition. Another useful equality is |f | = f + + f − .
Proposition
Any measurable function f : X 7→ R can be obtained as the limit of a
sequence of simple functions {un }, such that |un (x)| ≤ |u(x)| for all un and
all x. If f is non-negative we can choose the sequence {un } to be increasing.
Proof
It is enough to show that the theorem for non-negative f , because if f is not
non-negative we can treat f + and f − separately.
The general idea can be illustrated in an image.
37
We slice up the function using 2n parts on the y-axis, and construct a step
(n)
function. The set Ak is given the value 2kn (the minimal y-axis value). In
the next step we get 2n+1 parts, which means every part on the y-axis is split
in half. Since we used the minimal y-axis value in the previous step, the new
partition gets a higher value this time, and we get an increasing value as we
make the partition finer.
We define
un (x) =
n −1
2X
k=0
k
I (n) .
2 n Ak
As explained, this is an increasing function.
Every measurable function can be approximated by step functions (simple
functions). This concept will be very important for Lebesgue integration.
Proposition
Assume that {un } is a sequence of measurable functions. Then
sup un , inf un , limun = lim sup un , limu∈N = lim inf un
u∈N
u∈N
u∈N
u∈N
u∈N
and, if it exists limu∈N un are all measurable functions.
Proof
We begin with the ordinary, pointwise supremum supu∈N un . By the definition
of the supremum (for a fixed x):
sup un = sup un (x) : u ∈ N .
u∈N
We consider
{x : sup un (x) > α} =
u∈N
[
{x : un (x) > α}
n∈N
which is the same as writing
=
[
{x : un ≤ α}c .
n∈N
Now the set we have complemented in the last step is a collection of
measurable functions, so the set is measurable. But the complement of
a collection of measurable sets must also be measurable, so the sup is
measurable. The result for the infimum is proved in a similar way.
We now take a closer look at the lim sup. We need this concept for when
we have an oscillating function. For some oscillating sequence αn we get the
lim sup and lim inf as shown in the image:
38
and we note that when the sequence has a limit, i.e does not converge, the
lim sup and lim inf coincide. The definition is:
lim sup = inf sup αn .
n→∞
k→∞ n≥k
so we stop at k and look at the supremum from there and out.
Now, by the first part of the proof, we know that the supremum is
measurable, so the inner part of the lim sup is measurable, and we also know
the infimum is measurable, so the lim sup is measurable. The result for the
lim inf is similar.
If the limit exists, we also consider limn→∞ un , but as mentioned, this
coincides with both lim inf and lim sup, and since they are measurable so is
the limit. Proposition
If u and v are measurable functions, then so are u ± v, u · v max(u, v) and
min(u, v) whenever they exist/are defined.
Proof
If u and v are simple functions, so are the resulting functions, and they are
again measurable, since simple functions ⇒ measurable.
If u and v are general, measurable functions, we approximate them by
simple functions {un } and {vn }, and we get
u ± v = lim un ± vn ,
n→∞
max(u, v) = lim max(un , vn )
uv = lim un vn
n→∞
min(u, v) = lim min(un , vn ).
n→∞
n→∞
Since limits are measurable functions, so are all these functions. This kind of argument will be used a lot later.
Consider the case where we have the two measurable spaces (X, A) and
(Y, B), and some measurable mapping T : X 7→ Y . In addition we have a
function u : X 7→ R. Sometimes we want to have a function, w say, that
mimicks u and maps: w : Y 7→ R. Confer with image:
39
If such a function w exists, u is measurable wrt to the σ-algebra σ(T )
generated by T , because
u−1 (B) = (w ◦ T )−1 (B) = T −1 w −1 (B)
| {z }
∈B
|
{z
}
∈σ(T )
Factorization Lemma If (X, A) and (Y, B) are two measurable spaces,
and T : X 7→ Y is measurable wrt A and B, then for any σ(T )-measurable
function u : X 7→ R, there is a B-measurable function w from T 7→ R such
that u = w ◦ T .
Proof
Assume first that u is the indicator function u = IA . Since u is σ(T )measurable, then A ∈ σ(T ) and because A = T −1 (A) for some set B ∈ B,
but then we can take w = IB .
P
IfP
u is a simple function, u = ∞
n=1 ai IAn we use the same trick to get
w=
an IBn .
We have shown this is true for simple functions, and we approximate
those to more general functions.
If u is a general σ(T )-measurable function, we approximate by a sequence
of simple functions, such that un (x) → u(x). We know that for each n there
is a simple function w such that wn ◦ T = un .
We let w = lim sup un , since the limit may not exist, but the lim sup
always does.
w(T (x)) = sup lim wn (T (x)) = sup lim un (x) = lim un (x) = u(x).
n→∞
n→∞
n→∞
Where we used the construction of wn and the fact that un converges.
This is also a much used trick we will see again.
40
—————————————————————————
30/09-2010
Chapter 9 - Integration -Aim of this lecture R
We are going to define integrals of the form f dµ for all measures µ and
”all” measurable functions f .
We do this in steps.
(i) f is the indicator function.
(ii) f is a positive, simple function.
(iii) f is a positive, measurable function.
(iv) f is an integrable functions. (Not just for measurable functions,
since we can’t integrate all measurable functions).
•(i) When f is the indicator function, f = IA for some measurable A, we
have the case as shown in the image:
Thinking in terms of classical integrals, we would want the volume of the
”box” indicated in the image. The volume of this is simply the area of A times
the heigth, which is 1. If we use the Lebesgue measure, area(A) = µ(A). For
the indicator function we define,
Z
f dµ = µ(A).
Pn
•(ii) We now assume that f =
i=0 ai IAi , that is a simple function on
standard form. Looking at the image:
41
and using the same reasoning as for the indicator function (i), we would like
to define this as
Z
n
X
f dµ =
ai µ(Ai).
i=0
However, before we make this a formal definition, there is a potential problem
we must adress. What if there
P is another standard representation (they are,
after all, not unique) f = nj=0 bj IBj , in which case we would have
Z
f dµ =
m
X
bj µ(Bj ).
j=0
If this is different from the first integral, we have a problem, since we
have inconsistency. We have to show that the volume of the function f is
independent of the standard representation.
Lemma
P
P
Assume that ni=0 ai IAi and nj=0 bi IBj are standard representations of the
same function f . Then,
n
X
ai µ(Ai ) =
m
X
bj µ(Bj ).
j=0
i=0
Proof
What we have with these standard representations, is two ways of
partitioning the set X, as illustrated in the image. We make a new, finer,
partition and show that the volumes of the different partitions coincide.
Since B1 , . . . , Bm is a partition of X, we can write
Ai = Ai ∩ B0 ∪· Ai ∩ B1 ∪· . . . ∪· Ai ∩ Bm
Since X = B0 ∪B
· 1 ∪· . . . ∪B
· m , and Ai = Ai ∩ X. This means we can write
µ(Ai ) =
m
X
µ(Ai ∩ Bj ).
j=0
42
Using this, we can rewrite the proposed definition of the integral (volume):
n
X
ai µ(Ai ) =
i=1
n
X
i=0
ai
m
X
µ(Ai ∩ Bj ) =
j=0
n X
m
X
ai µ(Ai ∩ Bj ).
i=0 j=0
Using the exact same approach we get a similar expression for the other
representation:
m
m X
n
X
X
bj µ(Bj ) =
bj µ(Ai ∩ Bj ).
j=0
j=0 i=0
The order of the sums are different, but that doesn’t matter since they are
finite sums. The main difference here is ai and bj . From the definitions of
simple functions, ai is the value of f at Ai , and bj is the value of f on Bj . If
we are on the intersection Ai ∩ Bj then f can only have one value so ai = bj .
If they are disjoint we get µ(∅) = 0 so the values don’t matter. This verifies
the uniqueness of the sums.
m
X
j=0
bj µ(Bj ) =
n
m X
X
bj µ(Ai ∩ Bj ) =
n X
m
X
ai µ(Ai ∩ Bj ) =
i=0 j=0
j=0 i=0
n
X
ai µ(Ai ) i=1
Definition
If f is a positive, simple function with standard representation
f=
n
X
ai I( Ai )
i=1
then we define the integral of f wrt the underlying measure µ to be:
Z
n
X
f dµ =
ai µ(Ai).
i=1
Proposition
Assume f, g are positive, simple functions. Then
R
R
(i) af dµ = a f dµ for all real a > 0.
R
R
R
(ii) (f + g)dµ = f dµ + gdµ.
R
R
(iii) If f ≤ g, then f dµ ≤ gdµ.
Proof
(i) By the definition of simple functions:
Z
Z
n
n
X
X
af dµ =
a · ai µ(Ai ) = a
ai µ(Ai ) = a f dµ.
i=0
i=0
43
Pn
Pm
(ii) Assume that f =
a
I
and
g
=
i
A
i
i=0
j=0 bj IBj are standard
representations. Then, the right side of the proposition is:
Z
f dµ +
Z
gdµ =
n
X
ai µAi +
i=0
m
X
bj µBj .
j=0
We have to show that the left side is equal to this. To compute the integral
of f + g, we need a standard representation of f + g. Observe that the sets
Ai ∩ Bj gives us a partition of X (just like in the previous proof), and on
each Ai ∩ Bj , f has aj and g has bj , so f + g has a constant value on that
set: consequently we have a standard representation of f + g:
X
f +g =
(ai + bj )IAi ∩Bj .
i,j
By our definition:
Z
(f + g)dµ =
X
(ai + bj )µ(Ai ∩ Bj )
i,j
and we split it up
X
ai µ(Ai ∩Bj )+
i,j
X
bj µ(Ai∩Bj ) =
i,j
=
n
X
i=0
ai µ(Ai ) +
n
X
i=0
m
X
ai
m
X
j=0
bj µ(Bj ) =
j=0
µ(Ai ∩Bj )+
m
X
bj
j=0
Z
f dµ +
Z
n
X
µ(Ai ∩Bj )
i=0
gdµ.
(iii) This is easily proved using a simple trick. g = f + (g − f ). Since g ≥ f ,
g − f is positive.
Z
Z
Z
Z
Z
(ii)
gdµ =
f + (g − f ) dµ =
f dµ + (g − f )dµ ≥ f dµ.
|
{z
}
≥0
44
We are now ready for the third step: integration of positive, measurable
functions.
Definition
R
If f is a positive, measurable function, define the integral f dµ of f wrt µ
by
Z
o
nZ
f dµ = sup
gdµ g ≤ f, g positive, simple funcion .
So for some function f , we approximate with the largest possible step
function g that is smaller than or equal to f . In a sense this is similar like
the Riemann integral (except we can do the same over far more complex
funtions).
Remark: We now have two seperate definitions for the integral, but this is
okay since the definitions agree. In the text book they have worked their
way around this by designating the first integral definition by Iµ . (For this
definition you use that f itself is one of the g’s and use proposition (iii)).
—————————————————————————
04/10-2010
Positive, simple functions
f=
X
ai IAi ,
Z
f dµ =
X
ai µ(Ai ).
Positive, measurable functions
Z
Z
gdµ = sup
f dµ | f ≤ g, f’s are positive, simple functions
Question: If fn → f , will
R
fn dµ →
R
f dµ?
45
Monotone Convergence Theorem (Beppo-Levi’s Theorem)
Assume that {un } is an increasing sequence of positive, measurable functions.
Then the limit
Z
Z
un dµ =
lim un dµ.
lim
n→∞
n→∞
(If we have an increasing sequence of positive functions, you can interchange
limits and integral signs).
Proof —————————
Since fn ≤ limn→∞ fn , we have by a previous result, (if a function is greater
or equal, so is its integral)
Z
Z
fn dµ ≤
lim fn dµ,
∀n.
n→∞
hence, since this is true for all n,
Z
Z
fn dµ ≤
lim fn dµ.
lim
n→∞
n→∞
This was the easy inequality. We now want to show the opposite inequality,
Z
Z
fn dµ ≥
lim fn dµ,
lim
n→∞
n→∞
but to show this it suffices to show that
Z
Z
fn dµ ≥ gdµ,
lim
n→∞
for any simple function g ≤ limn→∞ fn , or alternatively, limn→∞ fn =
Psup g,
and if it holds for all such g it also holds for the sup. Assume that g =
ai IAi
is a positive, simple function and g ≤ limn→∞ fn . We let 0 < α < 1 (which
we think as something close to 1) and consider the set
Bnα = {x ∈ X | αg(x) ≤ fn (x)} .
Note that {Bnα }n∈N ր X as n gets larger. The set Bnα gets larger since the fn
gets larger, and when they do the set of points x that satisfy the inequality
grows. This means that
Z
Z
Z
Z
fn dµ.
αg · IBnα dµ ≤ fn · IBnα dµ ≤ fn dµ ≤ lim
n→∞
or, in short,
Z
αg · IBnα dµ ≤ lim
n→∞
46
Z
fn dµ.
Now we consider what happens as n → ∞ and α → 1. Note that
Z
Z X
Z X
αgIBnα dµ = α
ai IAi IBnα dµ = α
ai IAi ∩Bn∞ dµ
and by the definition of simple functions,
=α
X
ai µ(Ai ∩
n→∞
Bnα ) −→
α
X
ai µ(Ai ) = α
Z
gdµ.
Where we used that as n → ∞, Bnα → X, so Ai ∩ Bnα → Ai ∩ X = Ai . We
thus have that
Z
Z
αgIBnα dµ ≤ lim
fn dµ
∀α, n
n→∞
and since this is true for all n, we can take the limit, so Bnα and we are left
with
Z
Z
α gdµ ≤ lim
fn dµ
∀α
n→∞
Again, by similar reasoning, this is true for all α, so we can make it come
arbitrarily close to 1, or in effect let α → 1, and get
Z
Z
gdµ ≤ lim
fn dµ.
n→∞
A useful converse result to the theorem.
Corollary
Assume that f is a positive, measurable function and that fn is any increasing
sequence of simple functions converging to f . Then
Z
Z
f dµ = lim
fn dµ.
n→∞
Proof : Special case of the theorem. By the definition
Z
Z
gdµ | g simple, positive functions .
f dµ = sup
R
R
We only require that gn ր f , and then we can write f dµ = limn→∞ gn dµ.
Since we can freely choose such a sequence, this gives us an extra degree of
freedom when working with converging integrals.
47
Proposition
Assume that f, g are positive, measurable functions. Then
R
R
(i) αf dµ = α f dµ.
R
R
R
(ii) (f + g)dµ = f dµ + gdµ.
R
R
(iii) If f ≤ g, then f dµ ≤ gdµ.
Proof
(i), (iii): Follows immediately from the definition.
(ii) Let un and vn be increasing sequences of step functions converging to f
and g. (We know such sequences exist). By the corollary,
Z
Z
Z
Z
vn dµ.
f dµ = lim
un dµ
gdµ = lim
n→∞
n→∞
Also, since un + vn ր f + g, we also have (by the same corollary)
Z
Z
(f + g)dµ = lim (un + vn )dµ.
n→∞
Hence,
Z
(f + g)dµ = lim
n→∞
= lim
n→∞
Z
Z
(un + vn )dµ = lim
n→∞
un dµ + lim
n→∞
Z
vn dµ =
Z
Z
un dµ +
f dµ +
Z
Z
vn dµ
gdµ.
Fatou’s Lemma
If {un } is a sequence of positive functions, then
Z
Z
lim inf un dµ ≤ lim inf un dµ.
n→∞
n→∞
(We have weaker conditions in this case, so we don’t know if the limit exists:
so we take the inf in addition to the lim). We only have an inequality, and
show a counterexample to the equality.
Counterexample
We define the function
1 |x| ≥ n
un (x) =
0 |x| < n
which has a graph as shown in the following image.
48
In this particular case the limit does exist, but lim inf un (x)
n→∞
=
limn→∞ un (x) = 0 since when n becomes
infinitely large, the interval which
R
is 0 also becomes infinitely long, so lim inf un (x)dµ = 0. But, if we look at
n→∞
R
any function in the sequence, un (x)dµ = ∞, since we are looking at the
area under the graph, and for, say n = 3, we have infinite areas under the
graph for
R n > |3|. Since this is true for at least one element in the sequence,
lim inf un (x)dµ = ∞, and we have the counterexample.
n→∞
Proof (of Fatou’s Lemma)
By the definition of the lim inf,
lim inf un = lim inf un
n→∞
n→∞ k≥n
where inf k≥n un is an increasing sequence, since we are taking the infimum of
a smaller and smaller set. We set fk (x) = inf k≥n un , and have that fk (x) is an
increasing sequence, in which case we can apply the monotone convergence
theorem to {fn (x)} and get
Z
Z
lim fn dµ = lim
fn dµ.
n→∞
n→∞
But, limn→∞ fk (x) = lim inf un , so we can rewrite this as
n→∞
Z
lim inf un dµ = lim
n→∞
n→∞
Z
fn dµ.
(Using that fn (x) ≤ un (x), and that lim inf = lim when the limit exists):
Z
Z
= lim inf f dµ ≤ lim inf un dµ.
n→∞
n→∞
Fatou’s lemma is weak due to the inequality, but at the same time strong
since the conditions are so weak. It is typically used as a first step, opening
up the usage for other theorems.
49
Chapter 10 - Integrable Functions
Recall that we can split f as f + − f − .
Definition
A measurable function f : X 7→ R is called integrable if
Z
Z
+
f dµ < ∞ and
f − dµ < ∞,
and we define the integral
Z
Z
Z
+
f dµ := f dµ − f − dµ.
(We assume they are finite to avoid ∞ − ∞ in this definition. Some books
allow one of them to be infinite which allows ±∞ integrals, but we will not).
The set of integrable functions is denoted L1 (µ) and will be a key concept
for a while.
Proposition
This gives us an easy way to verify integrability. The following are equivalent
for a measurable function f . (Propery (iv) is a useful trick).
R
(i) f is integrable or ( f dµ is finite).
R
R
(ii) f + and f − are integrable (or f + dµ and f − dµ are finite).
R
(iii) |f | is integrable (or |f |dµ is finite).
(iv) There exists a function ω ∈ L1 (µ) such that |f | ≤ ω.
Proof
Suffices to show (i)⇒(ii)⇒(iii)⇒(iv)⇒(i).
(i)⇒(ii): This is just a slight rephrasing of the definition.
(ii)⇒(iii): We know that
+
−
1
f , f ∈ L (µ) =⇒
Z
+
f dµ,
Z
f − dµ < ∞.
Since |f | = f + + f − |, we have
Z
Z
Z
+
|f |dµ = f dµ + f − dµ < ∞,
| {z } | {z }
<∞
<∞
so |f | is a positive function with finite integral, thus f ∈ L1 (µ).
50
(iii)⇒(iv): We simply choose ω = |f | and since |f | ≤ |f | is true the
implication follows.
(iv)⇒(i): We have
Z
+
f dµ ≤
Z
|f |dµ ≤
Z
ωdµ < ∞
since, by assumption, ω is integrable.
Z
Z
Z
−
f dµ ≤ |f |dµ ≤ ωdµ < ∞
Both f + and f − are finite, so f ∈ L1 (µ). Lemma
Assume f is a measurable function, and that h and g are positive, integrable
functions such that f = h − g. Then f is integrable and
Z
Z
Z
f dµ = hdµ − gdµ.
Proof
We have |f | ≤ |h| + |g| = h + g (by the triangleRinequality and
are
R since they
R
positive), and we know that h+g ∈ L1 (µ) since (h+g)dµ = hdµ+ gdµ <
∞ (by assumption they are integrable). By (iv) of the previous proposition,
f is integrable. It remains to show that f can be written as h − g. Combining
f = h − g and f = f + − f − , we get
f + − f − ⇒ f + + g = f − + h,
hence,
Z
Z
+
f dµ +
+
Z
gdµ =
Z
Z
Z
−
f dµ +
−
Z
Z
hdµ
f dµ + gdµ = f dµ + hdµ
Z
Z
Z
Z
Z
Z
Z
+
−
f dµ − f dµ = hdµ − gdµ =⇒
f dµ = hdµ − gdµ. Theorem
Assume that f, g ∈ L1 (µ). Then
(i) αf ∈ L1 (µ) for all α ∈ R and
Z
Z
αf dµ = α f dµ.
51
(ii) f + g ∈ L1 (µ) and
Z
Z
Z
(f + g)dµ = f dµ + gdµ.
(iii) max(f, g), min(f, g) ∈ L1 (µ).
R
R
(iv) f ≤ g =⇒
f dµ ≤ gdµ
(v) Rather intuitive result:
Z
Z
f dµ ≤ |f |dµ.
(In the first integral we can have positive and negative parts of the function
that cancel each other out, while on the right side we take the absolute value
of the function before we integrate).
Proof - Only (ii), the rest is left to the student.
(ii) Let f = f + − f − and g = g + − g − , then
(f + g) = f + + g + + f − + g −
| {z }
| {z }
pos. in L1 (µ)
pos. in L1 (µ)
By assumption each part is finite, so we can apply the previous lemma with
h = (f + + g + ) and g ∗ = (f − + g − ). We get f + g ∈ L1 (µ) and
Z
Z
Z
+
+
(f + g)dµ = (f + g )dµ − (f − + g − )dµ =
Z
Z
Z
+
(i) Since α ∈ R and not R, α < ∞ so
R
−
Z
f dµ + g dµ − f dµ − g − dµ =
Z
Z
Z
Z
Z
Z
+
−
+
−
f dµ − f dµ + g dµ − g dµ = f dµ + gdµ
+
αf dµ < ∞ and thus integrable. By
definition of the integral,
Z
Z
Z
Z
Z
Z
+
−
+
−
αf dµ = αf dµ − αf dµ = α
f dµ − f dµ = α f dµ.
(iii) Both max(f, g) and min(f, g) return one of the functions f, g which are
integrable by assumption.
52
(iv) We use the trick, g = f + (g − f ), and write out the integral.
Z
Z
Z
Z
Z
(ii)
gdµ = f + (g − f )dµ =
f dµ + (g − f )dµ ≥ f dµ.
R
since g ≥ f , we have (g − f ) ≥ 0 and then (g − f )dµ ≥ 0 and the inequality
is apparent.
(v) Since |f | = f + + f − , we can write
Z
Z
Z
Z
(ii)
+
−
+
|f |dµ = f + f dµ =
f dµ + f − dµ
whereas
Z
Z
Z
Z
Z
f dµ = f + dµ − f − dµ ≤ f + dµ + f − dµ
and now we can remove the absolute value, since f + and f − are positive.
Using the first step we conclude the proof.
Z
Z
Z
Z
(ii)
+
−
+
−
= f dµ + f dµ =
f + f dµ = |f |dµ.
53
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07/10-2010
We have previously defined
+
Z
f <∞
1
f dµ, f ∈ L (µ) ⇐⇒
f− < ∞
Definition
If A ∈ A, we define
Z
Z
f dµ =
A
IA · f dµ.
We only consider the the function f over the set A and set it to 0 everywhere
else. Graphical illustration.
Note that if f ∈ L1 (µ), then IA f ∈ L1 (µ), simply because |IA | ≤ |f |
(bigger without being restricted to the set A, and by assumption f ∈ L1 (µ)
and then we can apply previous result with f = ω).
Proposition
If f is a positive, measurable function, then
Z
ν(A) =
f dµ
A
is a measure on A.
Proof
Since ν(A) ≥ 0 ∀A ∈ A, all we need to show are the two axioms for measures.
(i) ν(∅) = 0.
(ii)
ν
[
• Ai
i∈N
!
=
for all disjoint sequences of sets from A.
54
X
i∈N
ν(Ai )
(i) Observe that
ν(∅) =
Z
f dµ =
∅
Z
I∅ f dµ =
Z
0dµ = 0.
(The zero function, 0 is a simple function).
(ii) Assume that we have a disjoint sequence of measurable sets{Ai }. Note
that
! Z
Z
[
ν • Ai =
f dµ = I∪· ∞
f dµ
(1)
i Ai
∪· ∞
i Ai
i∈N
X
i∈N
ν(Ai ) = lim
n→∞
n
X
ν(Ai ) = lim
i=1
n→∞
n Z
X
IAi f dµ = lim
n→∞
i=1
Z X
n
i=1
IAi f dµ = lim
n→∞
Z
I∪· ni Ai f dµ
(2)
We must show that (1) and (2) are equal.
Note that
I∪ni Ai f ր I∪∞
f
i Ai
as n → ∞, since, if we call define the left side as fn , we see that f1 = f (A1 ),
f2 = f (A1 ) + f (A2 ) etc, so we have an increasing sequence! By the monotone
convergence theorem, we can interchange limits and integrals, so
Z
Z
Z
I∪· ni Ai f dµ =
lim I∪· ni Ai f dµ = I∪· ∞
f dµ
lim
i Ai
n→∞
n→∞
and using (1) and (2) with this, we get that
!
[
X
ν • Ai =
ν(Ai )
i∈N
i∈N
(You need to apply some theorem when you are faced with interchanging
limits and integrals, so you must look close at the elements you are working
with and see if you can use a theorem).
Concept of ’Almost Everywhere’
In measure theory we can’t distringuish between two sets that only differ on
a null set.
Example
If f, g : X 7→ R, we say that f and g are equal almost everywhere if there is
a set N of measure zero such that
{x : f (x) 6= g(x)} ⊂ N.
55
In general a property holds almost everywhere (abbreviated a.e) if the set
where it fails is contained in a set of measure zero.
Example
fn → f a.e. What does this mean? It means the set
{x : fn (x) 6→ f (x)}
is contained in a null set.
Proposition
For u ∈ L1 (µ)
Z
|u|dµ ⇐⇒ u = 0 a.e
Proof
Assume that u = 0 a.e, which
P means that the set {x : ux) 6= 0} has measure
zero (def of ’a.e’). If f =
ai IAi is a simple function on standard form less
than |u|, then f must be 0 a.e, and hence
Z
X
f dµ =
ai µ(Ai) = 0.
Whenever ai 6= 0 we are on the null set which means µ(Ai ) = 0. So by the
definition of the integral,
Z
Z
|u|dµ = sup
f dµ : f ≤ |u|, f positive, simple funcion .
Assume
R
|u|dµ = 0. Note that in particular,
An = {x : |u(x)| ≥
1
}
n
1
has measure 0. (Because
R 1 n IAn is a1 simple function less than the measurable
function |u(x)|, and n IAn dµ = n µ(An )). But then the set
{x : |u(x)| =
6 0} =
[
An ,
n∈N
but this is a union of null sets, which is again a null set. By subadditivity,
X
µ{x : |u(x)| =
6 0} ≤
µ(An ) = 0.
| {z }
n∈N
56
0
Corollary
If f ∈ L1 (µ) and µ(N) = 0, then
Z
f dµ = 0.
N
(The integral over a null set is 0).
Proof
Z
f dµ =
N
Z
IN f dµ ≤
Z
Prop.
|In f |dµ = 0.
Corollary
If u, v ∈ L1 (µ) and u = v a.e, then
Z
Z
udµ = vdµ.
Proof
Z
Z
Z
Z
udµ − vdµ = (u − v)dµ ≤ |u − v|dµ Prop.
= 0
since |u − v| =
6 0 on a set of measure 0, or alternatively, |u − v| = 0 a.e. 57
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11/10-2010
(X, A, µ) is a measure space.
Proposition
If f, g are measurable functions and
Z
Z
f dµ =
gdµ,
A
∀A ∈ A
A
then f = g a.e.
Proof
Assume for a contradiction this is not the case. Then,
0 < µ {x : f (x) 6= g(x)} = µ {x : f (x) > g(x)} + µ {x : g(x) > f (x)}
(the first set is the union of the two on the right, and since they are disjoint
we can add their measures). Since these sets are strictly greater than 0 one of
them must have positive measure, say the one for f (x) > g(x). (The following
reasoning can be used on the other case, where we just exchange f and g).
This set can be rewritten:
[
1
{x : f (x) > g(x)} =
x : f (x) > g(x) +
.
n
n∈N
At least one of the sets on the right side must have a positive measure.
1
x : f (x) > g(x) +
> 0 for some n.
n
But the, the indicator function multiplied by n1 , (to shorten the notation we
define θ = {x : f (x) > g(x) + n1 }: n1 Iθ is a simple function majorized by
f (x) − g(x) > n1 : since f (x) > g(x) + n1 we have f (x) − g(x) > n1 . This means,
Z
1
1
> 0.
[f (x) − g(x)] dµ ≥ µ x : f (x) > g(x) +
n
n
θ
Hence, since both f and g are measurable functions which means we can split
the integral,
Z
Z
Z
Z
f dµ − gdµ > 0 =⇒
f dµ > gdµ.
θ
θ
θ
θ
which contradicts the assumption theorem. (This theorem also applies for a
sub-σ-algebra C ⊂ A). 58
Chapter 11
Basic question: When can we interchange limits and integrals?
Z
Z
lim un dµ = lim
un dµ.
n→∞
n→∞
Fatou’s Lemma
Z
lim inf un dµ = lim inf
n→∞
n→∞
Z
un dµ
(un positive)
This lemma is an example of when we can change integrals and limits.
Generally the measurability of the limit is never the problem, only that a
function can be too ”big”. In the Riemann theory it can be both too big and
too irregular.
Monotone Convergence Theorem (Generalized)
Assume {un } is an increasing sequence of functions
in L1 (µ), and let u =
R
limn→∞ un . Then u ∈ L1 (µ) if and only if limn→∞ un dµ < ∞ and in that
case
Z
Z
Z
Z
udµ = lim
un dµ =⇒
lim udµ = lim
un dµ.
n→∞
n→∞
n→∞
(Compared to the first version, Beppo-Levi, we do not assume the functionals
are positive, only that the sequence is increasing).
Proof
We reduce this to the Beppo-Levi case. Note that {un − u1 } is an increasing
sequence of positive functions converging to u − u1 . Hence, by Beppo Levi’s
theorem,
Z
Z
Z
Z
Z
B.L
(u−u1 )dµ =
lim (un −u1 ) = lim (un −u1 )dµ = lim
un dµ− u1 dµ.
n→∞
R
n→∞
n→∞
(*)
un dµ = ∞, then u cannot be integrable, so we have
Note that if limn→∞
proved the sufficiency.
R
R
If limn→∞ un dµ < ∞, then (u − u1 )dµ < ∞, and since it is a finite,
positive function it is measurable, so u−u1 ∈ L1 (µ). But then u = (u−u1 )+u1
are both two measurable functions, and their sum is also measurable which
means u ∈ L1 (µ), which proves the necessity.
Returning now to (*), where we verified
Z
Z
Z
(u − u1) = lim
un dµ − u1 dµ =⇒
n→∞
59
we can now use that we know the terms in the integral are finite and
measurable. Since we know they are finite we can cancel them.
Z
Z
Z
Z
Z
Z
un dµ un dµ−u1 dµ = udµ = lim
udµ−u1 dµ = lim
n→∞
n→∞
Note: we cannot cancel out terms unless we know they are finite. If we do we
may reach contradictions, like for instance
∞ − a = ∞ − b,
which is true since both sides are infinite. But if we cancel the infinity terms
we are left with
−a = −b
which is incorrect.
We now look at a very important convergence theorem, and the one which
is most used in practice.
Lebesgue’s Dominated Convergence Theorem
Assume that {un } is a sequence of functions in L1 (µ) converging a.e to a
function u. Assume further that there is a function w ∈ L1 (µ) such that
|un | ≤ w for all n, then
Z
Z
Z
Z
un dµ
un dµ =⇒
lim un dµ = lim
udµ = lim
n→∞
n→∞
Also,
lim
n→∞
Z
n→∞
|u − un |dµ = 0
(which is a slightly stronger L1 (µ)-convergence.
Before we prove this, we take a look at why we need dominated convergence.
Consider the function given by
1 if x ∈ [n, n + 1)
un (x) =
0
otherwise.
The graph to this function is something like this:
60
R
We note that limn→∞ dµ = 0, because sooner or later we will pass n, and
from there on out we always have 0. On the other hand, for every n, we have
∈ un dµ = 1, since we have an interval of size 1 where the function is 1. So
Z
Z
Z
lim un (x)dµ = u(x)dµ = 0 but
un dµ = 1, ∀n.
n→∞
In this case the usual convergence theorems don’t work, but we could use
DCT.
Proof
Note that 2w − |u − un | ≥ 0. Since |un | ≤ w and |u| ≤ |w| and w ∈ L1 (µ),
then u ∈ L1 (µ), or u is integrable. This means that
Z
Z
Z
Z
udµ − un dµ = (u − un )dµ ≤ |u − un |dµ.
R
It suffices to prove that limn→∞ |u − un |dµ = 0. We use Fatous’ lemma:
Z
Z
F.L
2wdµ = lim inf (2w − |u − un |)dµ ≤
n→∞
lim inf
n→∞
Z
(2w − |u − un |)dµ =
Z
2wdµ − lim sup
n→∞
Z
|u − un |dµ.
where we in the last step used that both terms are measurable so we can
split up the integral, 2w has no n-dependance so we can remove the lim inf,
and the lim inf becomes − lim sup. We have thus established,
Z
Z
Z
2wdµ ≤ 2wdµ − lim sup |u − un |dµ.
n→∞
and since the integral of absolute values must be positive, the only possibility
for this inequality is that we have equality and that the integral we are
subtracting is 0. So, we have verified that
Z
lim sup |u − un |dµ = 0.
n→∞
Since the integral must be positive, and the lim sup is 0, then so is the lim inf
and when they agree we have convergence, so
Z
|u − un |dµ = 0
lim
n→∞
Applications
We see some applications of the convergence theorem.
61
Proposition
Let u : (a, b) × X 7→ R (the function maps from an intervall and a set in the
measurable space X to the real numbers) satisfies
(i) x 7→ u(t, x) is L1 (µ) for all t ∈ (a, b).
(ii) t 7→ u(t, x) is continuous for all x.
(iii) There is a w ∈ L1 (µ) such that |u(t, x)| ≤ w(x) for all t and x.
Then the function v(t), defined as
v(t) =
is continuous.
Z
u(t, x)dµ(x)
Proof
It is sufficient to show that v(tn ) → v(t) for all sequences tn → t, i.e we want
to verify
Z
Z
?
u(tn , x) dµ(x) −→
| {z }
un (x)
u(t, x) dµ(x)
| {z }
u(x)
where we use the underbraced versions since t is fixed. Now we have
the conditions in the dominated convergence theorem, especially due to
assumption (iii), so we know that
Z
Z
un dµ(x) = u(x)dµ(x)
lim
n→∞
Proposition
Let u : (a, b) × X 7→ R satisfies
(i) x 7→ u(t, x) is in L1 (µ) for all t.
(ii) t 7→ u(t, x) is continuosuly differentiable at all points (a, b).
| ≤ w(x).
(iii) There is a function w ∈ L1 (µ) such that | ∂u
∂t
Then the function
Z
v(t) = u(t, x)dµ(x)
is differentiable and (we differentiate under the integral sign)
Z
∂u
′
(t, x)dµ(x).
v (t) =
∂t
So the proposition basically tells us that
Z
Z
d
∂u
u(t, x)dx =
(t, x)dµ(x)
dt
∂t
62
Proof
The conclusion means that we have
Z
v(s) − v(t)
∂u
lim
=
(t, x)dt.
s→t
s−t
∂t
We look at sequences of functions converging to, and not the limit-definition
of the derivative, so it suffices to prove that
Z
∂u
v(tn ) − v(t)
lim
=
(t, x)dµ(x)
n→∞
tn − t
∂t
for all sequences tn → t. Using the definition of v(t), we get
R
R
u(tn , x)dµ(x) − u(t, x)dµ(x)
u(tn , x) − u(t, x)
v(tn ) − v(t)
= lim
= lim
dµ(x).
lim
n→∞
n→∞
n→∞
tn − t
tn − t
tn − t
From calculus/real analysis we see we can use the mean value theorem, so
for some intermediate point cn we have that this equals ∂u
(c , x), and when
∂t n
∂u
cn → t we get ∂t (t, x) which, by assumption, is bounded by w. We apply the
Lebesuge dominated convergence theorem, and get
Z
∂u
(t, x)dµ(x)
=
∂t
Lebesgue vs Riemann
We consider the classical example of the function that is not Riemann
integrable but Lebesgue integrable. For f : [0, 1] 7→ R, we define
1 if x is rational
.
f (x) =
0
otherwise
The upper and lower step functions in the Riemann integral always converge
to 1 and 0 respectively, so it is not Riemann integrable. It is Lebesgue
integrable since it is 0Ra.e (the rational numbers are counable, and countable
sets are null sets), so [0,1] f dµ = 0.
In a more general version,
1 if x can be written on the form m
for k ≤ n
k
fn (x) =
.
0
otherwise
In this case every fn (x) is Riemann integrable (since we can partition them
as step functions over the rational points), but as fn → f , f is not Riemann
integrable as we saw over. When we take the limits of functionals that are
Riemann integrable, we get problems with the limit, which is never the case
with Lebesgue integrals. In view of how important limits are as fundamental
analysis wholly depend on them, this becomes on of the most important
reasons to why we discard Riemann integrals and use Lebesgue integrals
instead.
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18/10-2010
64