Measure and Integration
Winter 2005
Professor B. Forrest
CHRIS ALMOST
Contents
1 Review of Riemann Integration
2
2 Measures and Measure Spaces
3
3 Constructing Measures: The Caratheodory Method
6
4 Example: Lebesgue Measure on R
8
5 Extending Measures
11
6 Measurable Functions
13
7 Integration
15
8 Integrable Functions
18
9 L p -spaces
19
10 Signed Measures
21
11 The Radon-Nikodym Theorem
23
12 Differentiation of Monotone Functions
25
13 Bounded Variation
27
14 Absolutely Continuous Functions
30
14.1 Measures on [a, b] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
15 Bounded Linear Functionals on L p (X , µ) and C[a, b]
31
16 Product Measures
34
1
2
1
MEASURE
AND I NTEGRATION
Review of Riemann Integration
For this section let f : [a, b] → R be a bounded function.
1.1 Definition. A partition of [a, b] is a finite subset Π = {a = x 0 < x 1 < · · · < x n = b}. The norm of Π is
kΠk := maxi=1,...,n {∆x i }. P
Let Mi := sup{ f (x)|x ∈ [x i−1 , x i ]} and mi := inf{ f (x)|x
Pn ∈ [x i−1 , x i ]}. The lower
n
Riemann sum is L( f , Π) := i=1 mi ∆x i and the upper Riemann sum is U( f , Π) := i=1 Mi ∆x i .
It is clear that L( f , Π) ≤ U( f , Π).
1.2 Definition. A partition Π1 is a refinement of Π if Π ⊆ Π1 .
Clearly L( f , Π) ≤ L( f , Π1 ) ≤ L( f , Π1 ) ≤ U( f , Π) and the collection of all partitions of [a, b] is a directed set
(directed by refinement).
1.3 Proposition. If Π1 and Π2 are two partitions, then L( f , Π1 ) ≤ U( f , Π2 ).
1.4 Definition. The upper Riemann integral for f on [a, b] is
Z
b
f (x)d x = inf{U( f , Π)}
Π
a
The lower Riemann integral for f on [a, b] is
Z b
f (x)d x = sup{L( f , Π)}
a
Π
Rb
Rb
f is said to be Riemann integrable if a f (x)d x =
f (x)d x. In this case the common value is denoted by
a
Rb
f (x)d x, called the Riemann integral.
a
1.5 Proposition (Cauchy Criterion). If f is bounded on [a, b] then f is Riemann integrable if and only if for
all " > 0 there is a partition Π such that U( f , Π) − L( f , Π) < ".
For example, if f is continuous on [a, b] then f is Riemann integrable on [a, b]. (Prove this as an exercise.)
If f is a non-decreasing function on [a, b] then f has at most countably many discontinuities, and they are all
jump discontinuities. Let P (n) be the uniform partition of [a, b]. Then Mi = f (x i ) for each i, and mi = f (x i−1 )
for all i. Hence
U( f , P (n) ) − L( f , P (n) ) =
n
X
b−a
b−a
( f (x i ) − f (x i−1 )
= ( f (b) − f (a))
−→ 0
n
n
i=1
as n tends to infinity. Thus f is Riemann integrable.
1.6 Theorem. Assume that f is bounded on [a, b]. Then the following are equivalent
1. f is Riemann integrable on [a, b].
2. For every " > 0 there is a partition Π such that U( f , Π) − L( f , Π) < ".
3. For every " > 0 there exists δ > 0 such that kΠk < δ implies that U( f , Π) − L( f , Π) < ".
Rb
Furthermore if any of these hold then for any " > 0 there exists δ > 0 such that if kΠk < δ then S( f , Π) − a f (x)d x <
Pn
", where S( f , Π) = i=1 f (ci )∆x i with ci ∈ [x i−1 , x i ].
MEASURES
AND
MEASURE SPACES
3
Note that if sk = S( f , P (k) ) is any sequence of Riemann sums with respect to the uniform paritions P (k) , then
Rb
limk→∞ sk = a f (x)d x. This explains the usual definition given in first year calculus.
But there are some problems with the Riemann integral:
1. Many “nice” functions are not Riemann integrable.
2. The Riemann integral does not behave well with respect to limits of sequence of functions. That is, we
cannot “interchange the limit and the integral”. For example, let Q ∩ [0, 1] = {r1 , r2 , . . . }, and let f n (x) =
R1
χ{r1 ,...,rn } (x). Then 0 f n (x)d x = 0 for all n, but limn→∞ f n = χQ is not even Riemann integrable.
Rb
Assume that f is Riemann integrable on [a, b], so that a f (x)d x = supΠ {L( f , Π)}. Let Π be a parition of
Rb
Pn
[a, b] and ϕΠ := i=1 mi χ[x i−1 ,x i ) . Then ϕΠ ≤ f for all x ∈ [a, b], and a ϕΠ (x)d x = L( f , Π). Hence
b
Z
f (x)d x = sup
a
Π
(Z
)
b
ϕΠ (x)d x
a
The Riemann integral is essentially the weighted average of the function obtained by partitioning the domain
of the function. An alternative approach is to partition the range of the function. Assume { y0 , y1 , . . . , yn } is
Rb
partition of the range of f (whatever that means). Let Si = {x ∈ [a, b] | yi−1 ≤ x < yi }. Then a f (x)d x ≈
Pn
i=1 yi−1 length(Si ). But what do we mean by “length” of an arbitrary set of real numbers?
2
Measures and Measure Spaces
2.1 Definition. Let X be any non-empty set. An algebra of subsets of X is a collection A ⊆ P (X ) such that
1. ∅ ∈ A
2. E1 , E2 ∈ A implies E1 ∪ E2 ∈ A
3. E ∈ A implies E c = X \ E ∈ A
A is said to be a σ-algebra if
1. ∅ ∈ A
S∞
2. {En }∞
n=1 ⊆ A implies
n=1 En ∈ A
3. E ∈ A implies E c = X \ E ∈ A
Remark.
1. Every σ-algebra is an algebra
2. E1 ∩ E2 = (E1c ∪ E2c )c , so algebras are closed under intersection as well
2.2 Example. P (X ) is a σ-algebra.
2.3 Proposition. If {Aα }α∈I is a collection of algebras (resp. σ-algebras) then
σ-algebra).
T
α∈I
Aα is an algebra (resp.
PROOF: Exercise.
ƒ
2.4 Corollary. Given any set S ⊆ P (X ) then there exists a smallest algebra (resp. σ-algebra) that contains S.
Notation. Given S ⊆ P (X ), let A (S), the algebra generated by S, be the smallest algebra containing S, and
σ(S), the σ-algebra generated by S, be the smallest σ-algebra containing S.
4
MEASURE
AND I NTEGRATION
2.5 Definition. Let S = {U ⊆ R | U is open}. The σ-algebra generated by S is called the Borel σ-algebra of R
and is denoted B(R).
topological space. A set A ⊆ R is
T∞ More generally we may take the Borel σ-algebra of any S
∞
called Gδ if A = n=1 Un where each Un is open. A set A ⊆ R is called Fσ if A = n=1 Fn where each Fn is closed.
Additional subscripts may be appended in the most obvious way.
Remark. The Gδ sets are exactly the complements of the Fσ sets, and visa versa. Notice that the closed sets are
Gδ .
Fact: |B(R)| = c, the cardinality of R.
2.6 Definition. Let (x, τ) be a topological space. We say that A ⊆ X is nowhere dense if A has no interior. A ⊆ X
is of first category in (X , τ) if it is a countable union of nowhere dense sets. A ⊆ X is of second category in (X , τ)
if it is not first category in (X , τ). A is residual if Ac is of first category.
2.7 Theorem (Baire’s Category Theorem). Let X be a complete metric space or (X , τ) a locally compact, Hausdorff space. Then X is of second category in itself.
2.8 Theorem. Let X be a complete
metric space or (X , τ) a locally compact, Hausdorff space. If {Un }∞
n=1 is a
T∞
collection of dense open sets then n=1 Un is dense.
2.9 Corollary. Q is not Gδ in R.
S∞
T∞
Fn is closed and
PROOF: Assume that Q = n=1 Un , where Un is open for each n. Then R \ Q = n=1 Fn , where S
∞
nowhere dense. Let Q = {r1 , r2 , . . . }. Then Fn0 = Fn ∪ {rn } is closed and nowhere dense and R = n=1 Fn0 . This is
a contradiction since R is not of first category.
ƒ
2.10 Example. Consider [0, 1]. Define an equivalence relation on [0, 1] by x ∼ y if x − y ∈ Q. Use the Axiom
of Choice to choose one element from each equivalence class and denote this set by S. S is not Borel.
There is nothing special about using the open sets of R to define B(R).
B(R) = σ{(a, b) | a, b ∈ R} = σ{(a, b] | a, b ∈ R} = σ{[a, b) | a, b ∈ R} = σ{[a, b] | a, b ∈ R}
2.11 Example. The collection of all finite unions of sets of the form {R, (−∞, b], (a, b], (a, ∞)} is an algebra.
2.12 Definition. A set together with a σ-algebra (X , A ) is called a measurable space. A (countably additive)
measure on A is a function µ : A → R∗ = R ∪ {±∞} such that
1. µ(∅) = 0
2. µ(E) ≥ 0 for all E ∈ A
S∞
P∞
3. If {En }∞
n=1 ⊆ A is a sequence of pairwise disjoint set then µ( n=1 En ) =
n=1 µ(En ).
Condition (3) is known as countable additivity. If we replace (3) by
SN
PN
(3’) If {En }Nn=1 ⊆ A is a sequence of pairwise disjoint set then µ( n=1 En ) = n=1 µ(En ).
then µ is called a finitely additive measure on A .
Countable additivity for measures give powerful convergence results for integration.
MEASURES
AND
MEASURE SPACES
5
2.13 Example. Given X any set and A = P (X ), let
¨
|E|
µ(E) :=
∞
if |E| is finite
otherwise
This is called the counting measure. If |X | is finite then
µ∗ (E) :=
|E|
|X |
is called the normalized counting measure.
2.14S
Definition. We call a measure µ finite if µ(X ) < ∞. We call µ σ-finite if there exists {En }∞
n=1 ⊆ A such that
∞
X = n=1 En and µ(En ) < ∞ for all n.
2.15 Example. Let X = N and A = P (X ). Let f : N → R∗+ and define
µ f (E) :=
X
f (n) =
n∈E
∞
X
f (n)χ E (n)
n=1
P∞
Then µ f is a measure on A . µ f is finite if and only if n=1 f (n) < ∞. Which is to say, f ∈ `1 (N)+ = { f ∈
`1 (N) | f (n) ≥ 0 ∀ n ∈ N}. µ f is σ-finite if and only if ∞ ∈
/ f (N). Conversely, if µ is any measure on A then
fµ (n) := µ({n}) is a function N → R∗+ since N is countable and µ is countably additive. Then µ fµ = µ, so we
know all about the measures on (N, P (N)).
In general, suppose that X is any set. Given f : X → R∗+ , we can define
X
µ f (E) =
f (x) := sup{ f (x 1 ) + · · · + f (x n ) | x 1 , . . . , x n ∈ E}
x∈E
Then µ f is a measure on P (X ). If f ∈ `1 (X )+ then µ f is finite. µ f is σ-finite if it zero except on a countable
subset of P (X ) and ∞ ∈
/ f (X ). Are all measures on (X , P (X )) of the form µ f for some f ? What about all finite
measures? Consider
¨
0 if E is countable
µ(E) =
∞ if E is uncountable
Then µ 6= µ f for any f . Can µ be made into a finite measure?
2.16 Definition. A triple (X , A , µ) is called a measure space, where X is a set, A is a σ-algebra, and µ is a
measure on A . (X , A , µ) is complete if µ(E) = 0 and S ⊆ E implies that S ∈ A . (If E ∈ A then E is called
measurable.) If µ(X ) = 1 then µ is called a probability measure and (X , A , µ) is called a probability space. The
elements of A in this case are called events.
2.17 Proposition (Monotonicity). Let (X , A , µ) be a measure space. If E ⊆ F ∈ A then µ(E) ≤ µ(F ). If
µ(E) < ∞ then µ(F \ E) = µ(F ) − µ(E).
PROOF: This follows immediately since F = E ∪ (F \ E).
ƒ
2.18 Lemma. Let A ⊆ P (X ) be an algebra. Assume that {En }∞
. Then there is {Fn }∞
n=1 ⊆ A
n=1 ⊆ A such that
Sk
Sk
S
S∞
∞
∞
n=1 En =
n=1 Fn for all k and {Fn }n=1 is pairwise disjoint. Moreover,
n=1 En =
n=1 Fn
S
n−1
PROOF: Let F1 := E1 ∈ A and let Fn := En \
ƒ
i=1 Fi ∈ A . Notice that Fn ⊆ En for each n ≥ 1.
6
MEASURE
AND I NTEGRATION
2.19 Proposition (Countable Subadditivity). Let (X , A , µ) be a measure space. Let {En }∞
n=1 ⊆ A . Then
µ
∞
[
!
En
≤
∞
X
µ(En )
n=1
n=1
PROOF: Let {Fn }∞
n=1 ⊆ A be as in Lemma 2.18. Then
µ
∞
[
!
=µ
En
n=1
∞
[
n=1
!
Fn
=
∞
X
µ(Fn ) ≤
n=1
∞
X
µ(En )
n=1
by monotonicity.
ƒ
2.20 Theorem (Continuity from Above). Let (X , A , µ) be a measure space and {En }∞
n=1 ⊆ A . If µ(E1 ) < ∞
and Ei+1 ⊆ Ei for all i ≥ 1 then
!
∞
\
En = lim µ(En )
µ
n=1
n→∞
2.21 Theorem (Continuity from Below). Let (X , A , µ) be a measure space and {En }∞
n=1 ⊆ A . If Ei ⊆ Ei+1 for
all i ≥ 1 then
!
∞
[
En = lim µ(En )
µ
n=1
n→∞
The theorem above is also known as the Monotone Convergence Theorem for Measures.
3
Constructing Measures: The Caratheodory Method
We are looking for a measure m on P (R) with the following nice properties:
1. m(I) = length of I for all intervals I
2. m is countably additive
3. m(x + A) = m(A) for all A ∈ P (R) and x ∈ R
The problem is that no such measure exists. If we weaken our conditions and only look for a measure on some
sufficiently large sub-σ-algebra A of P (R) then we may succeed. More specifically, we would like A to contain
B(R) and be complete with respect to the measure.
3.1 Definition. A function µ∗ : P (X ) → R∗ is called an outer measure if
1. µ∗ (∅) = 0
2. µ∗ (A) ≤ µ∗ (B) if A ⊆ B
S∞
P∞
3. If {En }∞
n=1 ⊆ A then µ( n=1 En ) ≤
n=1 µ(En ).
S∞
µ∗ is finite if µ∗ (X ) < ∞. µ∗ is σ-finite if X = n=1 En and µ∗ (En ) < ∞ for all n.
3.2 Definition. A set E ∈ P (X ) is said to be µ∗ -measurable or measurable if for every A ∈ P (X )
µ∗ (A) = µ∗ (A ∩ E) + µ∗ (A ∩ E c )
CONSTRUCTING MEASURES: THE CARATHEODORY METHOD
7
Notice that µ∗ (A) ≤ µ∗ (A ∩ E) + µ∗ (A ∩ E c ) for every A ∈ P (X ). Thus we need only show that µ∗ (A) ≥
µ (A∩ E) + µ∗ (A∩ E c ) for every A ∈ P (X ) to show that E is measurable. Furthermore, we need only assume that
µ∗ (A) < ∞. In the case that µ∗ (A) < µ∗ (A ∩ E) + µ∗ (A ∩ E c ) then A = (A ∩ E) ∪ (A ∩ E c ) is called a paradoxical
decomposition.
∗
3.3 Theorem. The set B of µ∗ -measurable sets in P (X ) is a σ-algebra and if µ = µ∗ |B , then µ is a complete
measure on B.
PROOF: It is clear that ∅ ∈ B since for any A ∈ P (X )
µ∗ (A) = µ∗ (A ∩ X ) = µ∗ (A ∩ ∅) + µ∗ (A ∩ ∅c )
It is also clear that if E ∈ B then E c ∈ B by symmetry of the definition of µ∗ -measurable.
Let E1 , E2 ∈ B and let A ∈ P (X ). Since E2 is measurable,
µ∗ (A) = µ∗ (A ∩ E2 ) + µ∗ (A ∩ E2c )
Since E1 is measurable,
µ∗ (A ∩ E2c ) = µ∗ (A ∩ E2c ∩ E1 ) + µ∗ (A ∩ E2c ∩ E1c )
These together imply that
µ∗ (A) = µ∗ (A ∩ E2 ) + µ∗ (A ∩ E2c ∩ E1 ) + µ∗ (A ∩ E2c ∩ E1c )
Notice that A ∩ (E1 ∪ E2 ) = (A ∩ E2 ) ∪ (A ∩ E1 ∩ E2c ). Thus
µ∗ (A ∩ (E1 ∪ E2 )) ≤ µ∗ (A ∩ E2 ) + µ∗ (A ∩ E1 ∩ E2c )
by subadditivity, so
µ∗ (A) ≥ µ∗ (A ∩ (E1 ∪ E2 )) + µ∗ (A ∩ E2c ∩ E1c ) = µ∗ (A ∩ (E1 ∪ E2 )) + µ∗ (A ∩ (E2 ∪ E1 )c )
which implies E1 ∪ E2 ∈ B. Therefore B is an algebra.
S∞
∞
∗
Sn Now let {Ei }i=1 ⊆ B be a sequence of pairwise disjoint µ -measurable sets and let E = i=1 Ei . Let Gn =
i=1 Ei . Then Gn ∈ B and for any A ∈ P (X )
µ∗ (A) = µ∗ (A ∩ Gn ) + µ∗ (A ∩ Gnc ) ≥ µ∗ (A ∩ Gn ) + µ∗ (A ∩ E c )
since E c ⊆ Gnc . Now Gn ∩ En = En and Gn ∩ Enc = Gn−1 . Since En is measurable
µ∗ (A ∩ Gn ) = µ∗ (A ∩ Gn ∩ E) + µ∗ (A ∩ Gn ∩ E c ) = µ∗ (A ∩ En ) + µ∗ (A ∩ Gn−1 )
Pn
An inductive argument shows that µ∗ (A ∩ Gn ) = i=1 µ∗ (A ∩ Ei ). Therefore
µ∗ (A) ≥ µ∗ (A ∩ E c ) +
n
X
µ∗ (A ∩ Ei )
i=1
for all n. Hence
∗
∗
µ (A) ≥ µ (A ∩ E ) +
c
∞
X
i=1
∗
∗
∗
µ (A ∩ Ei ) ≥ µ (A ∩ E ) + µ
c
∞
[
!
(A ∩ Ei )
i=1
= µ∗ (A ∩ E c ) + µ∗ (A ∩ E)
8
MEASURE
AND I NTEGRATION
S∞
Thus E is measurable. Given any sequence {Ei }∞
i=1 ⊆ B with E =
i=1 Ei , we can fine a pairwise disjoint
S
∞
sequence {Fi }∞
⊆
B
with
E
=
F
.
Hence
E
∈
B
and
B
is
a
σ-algebra.
i=1
i=1 i
Let E1 , E2 ∈ B be disjoint. Then since E2 is measurable,
µ(E1 ∪ E2 ) = µ∗ (E1 ∪ E2 ) = µ∗ ((E1 ∪ E2 ) ∩ E2 ) + µ∗ ((E1 ∪ E2 ) ∩ E2c ) = µ∗ (E2 ) + µ∗ (E1 ) = µ(E2 ) + µ(E1 )
∗
Hence µ is finitely additive. Let {Ei }∞
i=1 ⊆ B be a sequence of pairwise disjoint µ -measurable sets and let
S∞
E = i=1 Ei . Then
!
n
n
X
[
µ(E) ≥ µ
Ei =
µ(Ei )
i=1
i=1
for every n. Taking the limit we have µ(E) ≥
P∞
i=1 µ(Ei ).
µ(E) = µ∗ (E) ≤
∞
X
On the other hand,
µ∗ (Ei ) =
i=1
∞
X
µ(Ei )
i=1
P∞
Thus µ(E) = i=1 µ(Ei ) and µ is countably additive. Clearly µ(∅) = 0 and µ(E) ≥ 0 for all E ∈ B, so µ is a
measure on B.
Let µ(E) = 0 and F ⊆ E. Then
µ(A) ≥ µ∗ (A ∩ F c ) = µ∗ (A ∩ F c ) + µ∗ (A ∩ E) ≥ µ∗ (A ∩ F c ) + µ∗ (A ∩ F )
so F ∈ B. Therefore (X , B, µ = µ∗ |B ) is complete.
4
ƒ
Example: Lebesgue Measure on R
Let I be an interval in R. Let `(I) denote its length. For any E ⊆ R define
(
)
∞
∞
X
[
∗
m (E) = inf
`(I n ) | E ⊆
I n , I n ’s are open intevals
n=1
n=1
From this definition it is clear that:
1. m∗ (∅) = 0
2. m∗ (E) ≥ 0
3. If F ⊆ E, then m∗ (F ) ≤ m∗ (E).
4.1 Theorem. m∗ is an outer measure on P (R).
PROOF: We need only show that m∗ is coutably subadditive. Let {En }∞
n=1 ⊆ P (R). We may assume without loss
∗
of generality
that
m
(E
)
<
∞
for
all
n.
Let
"
>
0.
For
each
n
choose
a
n
S∞countable
S∞collection {I i,n } of open intevals
S∞
P∞
with En ⊆ i=1 I i,n such that i=1 `(I i,n ) ≤ m∗ (En ) + 2"n . Note that n=1 En ⊆ i,n=1 I i,n , so
m∗
∞
[
n=1
!
En
≤
∞
X
`(I i,n ) =
i,n=1
Since this is true for all " > 0 we get m∗
∞ X
∞
X
n=1 i=1
€S∞
n=1
`(I i,n ) ≤
∞ 
X
m∗ (En ) +
n=1
Š P∞
En ≤ n=1 m∗ (En ).
" ‹
2n
≤
∞
X
!
m∗ (En )
+"
n=1
ƒ
EXAMPLE: LEBESGUE MEASURE
ON
9
R
4.2 Definition. m∗ is called the Lebesgue outer measure on R. We denote the σ-algebra of m∗ -measurable sets by
M(R). Elements of M(R) are said to be Lebesgue measurable. m = m∗ |M(R) is called the Lebesgue measure on R.
4.3 Proposition. If I is an interval then m∗ (I) = `(I).
PROOF: Assume that I = [a, b] and let " > 0. Then if I1 = (a − 2" , b + 2" ) we have I ⊆ I1 , so
m∗ (I) ≤ `(I1 ) = b − a + " = `(I) + "
S∞
Since " was arbitrary, m∗ (I) ≤ `(I). Assume that I ⊆ i=1 I i , where each I i is an open interval. Since [a, b]
is compact there is a finite subcover. Out of these finitely many elements, choose one that contains a. Without
loss of generality suppose it is I1 . If b ∈ I1 then we are done. Otherwise choose an interval that contains b1
Sk
and call it I2 . Continue this process to get I1 , . . . , I k such that I ⊆ i=1 I i . Furthermore, if I j = (a j , b j ) then
a1 < a < b1 < · · · < b < bk and bi − ai ≥ bi − bi−1 . Hence
k
X
bi − ai ≥ b1 − a1 +
i=1
k
X
bi − bi−1 = bk − a1 > b − a
i=2
Therefore
∞
X
`(I i ) ≥
k
X
i=1
`(I i ) =
i=1
k
X
bi − ai > b − a = `(I)
i=1
Hence m∗ (I) ≥ `(I), so m∗ (I) = `(I).
Assume that I is a finite interval. For any " > 0 we can find a closed interval J ⊆ I such that `(I) < `(J) + ".
Hence
`(I) − " < `(J) = m∗ (J) ≤ m∗ (I) ≤ m∗ (I) = `(I) = `(I)
so m∗ (I) = `(I).
Finally, if `(I) = ∞ then for any M > 0 we can find a finite interval J ⊆ I with `(J) > M . Then m∗ (I) ≥
∗
m (J) = `(J) > M , so m∗ (I) = ∞ = `(I).
ƒ
4.4 Lemma. The intervals (a, ∞) and (−∞, a] are m∗ -measurable.
PROOF: Let A ⊆ R. We need to show that
m∗ (A) ≥ m∗ (A ∩ (a, ∞)) + m∗ (A ∩ (−∞, a])
∗
and S
moreover, we
P∞may assume∗ that m (A) <0 ∞. Let " > 0. We00 can find a collection of0 open 00intervals with
∞
A ⊆ i=1 I i and i=1 `(I i ) < m (A) + ". Let I n = I n ∩ (a, ∞) and I n = I n ∩ (−∞, a]. Then I n and I n are intervals
and `(I n ) = `(I n0 ) + `(I n00 ) = m∗ (I n0 ) + m∗ (I n00 ). We have
∗
m (A ∩ (a, ∞)) ≤ m
∗
∞
[
i=1
!
I i0
≤
∞
X
i=1
m
∗
(I i0 )
and
∗
m (A ∩ (−∞, a]) ≤ m
∗
∞
[
i=1
!
I i00
≤
∞
X
i=1
m∗ (I i00 )
10
MEASURE
AND I NTEGRATION
Hence
m∗ (A ∩ (a, ∞)) + m∗ (A ∩ (−∞, a]) ≤
∞
X
i=1
=
∞
X
m∗ (I i0 ) +
∞
X
m∗ (I i00 )
i=1
m∗ (I i0 ) + m∗ (I i00 )
i=1
=
∞
X
`(I i0 ) + `(I i00 )
i=1
=
∞
X
`(I i ) ≤ m∗ (A) + "
i=1
Since " was arbitrary, m∗ (A) ≥ m∗ (A ∩ (a, ∞)) + m∗ (A ∩ (−∞, a]).
ƒ
4.5 Theorem. B(R) ⊆ M(R).
PROOF: Trivial, given the last lemma.
ƒ
Clearly every countable or finite set has Lebesgue measure 0. m∗ is translation invariant, so Lebesgue measure
is also translation invariant. Indeed, let E ∈ M(R), A ⊆ R and x ∈ R. Then
m∗ (x + A) = m∗ (A)
= m∗ (A ∩ E) + m∗ (A ∩ E c )
= m∗ ((x + A) ∩ (x + E)) + m∗ ((x + A) ∩ (x + E c ))
= m∗ ((x + A) ∩ (x + E)) + m∗ ((x + A) ∩ (x + E)c )
If A is arbitrary then so is B, so x + A ∈ M(R).
4.6 Theorem. Let E ⊆ R. The following are equivalent:
1.
2.
3.
4.
5.
E ∈ M(R)
Given " > 0 there is an open set U ⊆ R with E ⊆ U and m∗ (U \ E) < ".
Given " > 0 there is a closed set F ⊆ R with F ⊆ E and m∗ (E \ F ) < ".
There is a Gδ set G ⊆ R with E ⊆ G and m∗ (G \ E) = 0.
There is an Fσ set F ⊆ R with F ⊆ E and m∗ (E \ F ) = 0.
If m∗ (E) < ∞ then these are all equivalent to:
(6) Given " > 0 there is a set U ⊆ R which is the union of finitely many open intervals such that m∗ (U∆E) < ".
PROOF: Exercise.
ƒ
4.7 Example. Let x, y ∈ [0, 1) and let x ⊕ y := x + y (mod 1). If E ⊆ [0, 1) is measurable then x ⊕ E is
measurable and m(E) = m(x ⊕ E). Indeed, let E1 = { y ∈ E | x + y < 1} = E ∩ (−∞, 1 − x) and E2 = { y ∈ E |
x + y ≥ 1} = E ∩ [1 − x, ∞). Then E = E1 ∪ E2 and E1 ∩ E2 = ∅, and E1 , E2 are both measurable. It follows that
x ⊕ E = (x ⊕ E1 ) ∪ (x ⊕ E2 ) = (x ⊕ E1 ) ∪ ((x − 1) ⊕ E2 )
is measurable and m(x ⊕ E) = m(E).
EXTENDING MEASURES
11
Define an equivalence relation on [0, 1) by x ∼ y ⇐⇒ x − y ∈ Q. Using the Axiom of Choice, construct a set
E ⊆ [0, 1) consisting of one element from each equivalence class. Let Q ∩ [0, 1) = {r1 , r2 , . . .} be an enumeration
of the rationals in [0, 1). Let En := rnS⊕ E. Is E measurable? If E is measurable then En is measurable and
∞
m(E) = m(En ) for each n. But [0, 1) = n=1 En , and the En ’s are pairwise disjoint, so
1 = m([0, 1)) =
∞
[
!
En
=
n=1
∞
X
n=1
m(En ) =
∞
X
m(E)
n=1
which is impossible. As an exercise, show that, assuming the Axiom of Choice, if m(E) > 0, then E contains a
non-measurable set.
The set E in the previous example is not Borel, since the Borel sets are measurable. Unfortunately E is not a
particularily nice non-Borel set, as its existence depends upon the Axiom of Choice. Given a Borel set A in R2 , is
it necessarily true that the projection of A onto the real line is Borel? The answer is no, but this is not obvious.
4.8 Example. The Cantor set C is compact, nowhere dense, and has cardinality c. It turns out that it also has
measure zero. It follows from this that the cardinality of M(R) is 2c .
5
Extending Measures
5.1 Definition. Let A ⊆ P (X ) be an algebra. A measure on A is a function µ : A → R∗ such that
1. µ(∅) = 0
2. µ(E) ≥ 0 for all E ∈ A
€S∞
Š P∞
S∞
3. If {En }∞
n=1 ⊆ A is are pairwise disjoint and
n=1 En ∈ A , then µ
n=1 En =
n=1 µ(En ).
Given a measure µ on A define µ∗ : A → R∗ by
)
(
∞
∞
X
[
∗
En
µ(En ) | {En } ⊆ A , A ⊆
µ (A) = inf
n=1
n=1
5.2 Proposition. Using the notation defined above:
1.
2.
3.
4.
5.
µ∗ (∅) = 0
µ∗ (B) ≥ 0 for any B ⊆ X
If A ⊆ B then µ∗ (A) ≤ µ∗ (B).
If B ∈ A then µ∗ (B) = µ(B)
€S∞
Š P∞
∗
If {Bn }∞
n=1 ⊆ P (X ) then µ
n=1 Bn ≤
n=1 µ(Bn ).
PROOF: (1), (2), (3), and (5) are trivial. For (4),S
notice that µ∗ (B) ≤S
µ(B) for all B ∈ A . On the other hand, let
∞
∞
∞
{En }n=1 P
⊆ A be pairwise P
disjoint such that B ⊆ n=1 En . Then B = n=1 B ∩ En , so since µ is a measure on A ,
∞
∞
µ(B) ≤ n=1 µ(B ∩ En ) ≤ n=1 µ(En ). Therefore µ(B) = µ∗ (B).
ƒ
It follows that µ∗ is an outer measure on P (X ) that extends µ. We call µ∗ the outer measure generated by µ.
5.3 Theorem (Caratheodory Extension Theorem). Let µ be a measure on an algebra A ⊆ P (X ). Let µ∗ be
the measure generated by µ. Let A ∗ be the σ-algebra of µ∗ -measurable sets. Then A ⊆ A ∗ . In particular, µ
extends to a measure µ on A ∗ .
12
MEASURE
AND I NTEGRATION
PROOF: We need only show that A ⊆ A ∗ . Let E ∈ A and let A ⊆ X . As before, we can assume that µ∗ (A)
S∞< ∞
and P
we need only show that µ∗ (A) ≥ µ∗ (A ∩ E) + µ∗ (A ∩ E c ). Let " > 0. Let {Fn }∞
⊆
A
such
that
A
⊆
n=1
n=1 Fn
∞
and n=1 µ(Fn ) < µ∗ (A) + ". Note that
A∩ E ⊆
∞
[
E ∩ Fn
and
A ∩ Ec ⊆
n=1
It follows that
µ∗ (A ∩ E) ≤
∞
X
∞
[
n=1
µ(E ∩ Fn ) and
µ∗ (A ∩ E c ) ≤
n=1
Hence
µ∗ (A ∩ E) + µ∗ (A ∩ E c ) ≤
E c ∩ Fn
∞
X
µ(E c ∩ Fn )
n=1
∞
X
µ(E ∩ Fn ) +
n=1
∞
X
µ(E c ∩ Fn ) =
n=1
∞
X
µ(Fn ) < µ∗ (A) + "
n=1
Since " was arbitrary, µ (A ∩ E) + µ (A ∩ E ) ≤ µ (A).
∗
∗
∗
c
ƒ
Given a measure µ on an algebra A , is the extension of µ to A ∗ unique?
5.4 Example. Let X = (0, 1] ∩ Q. Let A be the finite union of sets of the form (a, b] ∩ Q where a, b ∈ X . It is
easy to see that P (X ) is the smallest (and only) σ-algebra that contains A . Let µ be the counting measure on
A . Then
¨
0 if E = ∅
µ(E) =
∞ otherwise
This implies for all A ∈ P (X ) that
¨
∗
µ (A) =
0
∞
if A = ∅
otherwise
The extension theorem gives that A ∗ = P (X ) and µ = µ∗ . On the other hand, the counting measure on
P (X ) = A ∗ also extends µ, but is not equal to µ.
5.5 Theorem (Hahn Extension Theorem). Suppose that µ is a σ-finite measure on an algebra A . Then there
exists a unique extension of µ to a measure µ on A ∗ , the σ-algebra of all µ∗ -measurable sets.
PROOF: Let γ be a measure on A ∗ that agrees with µ on A . Let µ be the Caratheodory extension. First assume
that S
µ is finite (so that µ(X ) < ∞). Then µ and γ are both finite. Let E ∈ A ∗ and let {En }∞
n=1 ⊆ A with
∞
E ⊆ n=1 En . Then
!
∞
∞
∞
X
X
[
γ(E) ≤ γ
En ≤
γ(En ) =
µ(En )
n=1
n=1
n=1
Therefore γ(E) ≤ µ (E) = µ(E). But
∗
γ(E) + γ(E c ) = γ(X ) = µ(X ) = µ(X ) = µ(E) + µ(E c )
and since γ(E) ≤ µ(E) and γ(E c ) ≤ µ(E c ) we must have γ(E) = µ(E) for all E ∈ A ∗ .
S∞
Now assume that µ is σ-finite. Let {Fn }∞
n=1 be an increasing sequence in A , with µ(Fn ) < ∞ and X =
n=1 Fn .
We would then have that µ(E ∩ Fn ) = γ(E ∩ Fn ) for all n ≥ 1 and for all E ∈ A ∗ . However,
µ(E) = lim µ(E ∩ Fn ) = lim γ(E ∩ Fn ) = γ(E)
n→∞
and the result is proven.
n→∞
ƒ
MEASURABLE FUNCTIONS
13
5.6 Example. (Lebesgue-Stieltjes Measures)
Let A be the collection of all finite unions of sets of the form (−∞, b], (a, ∞), or (a, b]. Then A is an
algebra. Let F (x) be an nondecreasing function on R, with F (c) = lim x→c + F (x) for every c ∈ R. Since F is
monotonic, lim x→∞ F (x) and lim x→−∞ F (x) both exist as extended real numbers. Define
1.
2.
3.
4.
µ F ((a, b]) = F (b) − F (a)
µ F ((a, ∞)) = lim x→∞ F (x) − F (a)
µ F ((−∞, b]) = F (b) − lim x→−∞ F (x)
µ F ((−∞, ∞)) = lim x→∞ F (x) − lim x→−∞ F (x)
We extend µ F to A in the obvious way. Then µ F is a σ-finite measure on A (check this claim). The Hahn
Extension theorem tells us that µ F extends to a unique measure on A ∗ , which we shall also denote by µ F .
B(R) ⊆ A ∗ for each F , and for F (x) = x we get µ F = m, the Lebesgue measure.
5.7 Definition. µ F is called the Lebesgue-Stieltjes measure generated by F . µ F |B(R) is often called the BorelStieltjes measure generated by F .
5.8 Definition. Let (X , A ) be a measurable space and let µ, ν be measures on A . A property holds µ-almosts
everywhere, or µ a.e., if the property holds everywhere except on a set E with µ(E) = 0. We say that µ is absolutely
continuous with respect to ν if for every E ∈ A with ν(E) = 0 then µ(E) = 0. We write µ ν. We say that µ
and ν are mutually singular if there exist A, B ∈ A that partition X such that µ(A) = 0 = ν(B). We write µ⊥ν.
6
Measurable Functions
6.1 Definition. Let (X , A ) be a measurable space. We say that f : X → R is measurable if f −1 ((α, ∞)) ∈ A for
every α ∈ R. Let M(X , A ) denote the set of measurable functions.
6.2 Lemma. The following are equivalent:
1.
2.
3.
4.
f is measurable.
f −1 ((−∞, α]) ∈ A for all α ∈ R.
f −1 ([α, ∞)) ∈ A for all α ∈ R.
f −1 ((−∞, α)) ∈ A for all α ∈ R.
6.3 Example.
1. Given any (X , A ), f (x) = c ∈ R for all x ∈ R is measurable.
2. If A ⊆ X , the characteristic function of A, χA(x), is measurable if and only if A ∈ A .
3. If X = R and A = B(R) then the continuous functions are measurable since f −1 ((α, ∞)) is open and
hence measurable.
6.4 Lemma. Assume that f , g ∈ M(X , A ).
1.
2.
3.
4.
5.
c f ∈ M(X , A ) for every c ∈ R
f 2 ∈ M(X , A )
f + g ∈ M(X , A )
f g ∈ M(X , A )
| f | ∈ M(X , A )
PROOF: Exercise.
ƒ
14
MEASURE
AND I NTEGRATION
6.5 Definition. Let (X , A ) be a measure space and let f : X → R. Let f + (x) := sup{ f (x), 0} and f − (x) :=
sup{− f (x), 0}.
It follows that f = f + − f − and | f | = f + + f − , so f + = 21 (| f | + f ) and f − = 21 (| f | − f ). Hence f + and f −
are measureable if and only if f is measurable.
6.6 Definition. Let (X , A ) be a measure space. An extended real-valued function is a function f : X → R∗ is
measurable if f −1 ((α, ∞]) ∈ A .
€S∞
Šc
T∞
−1
Notice that both f −1 ({∞}) = n=1 f −1 ((n, ∞]) and f −1 ({−∞}) =
((−n, ∞]) and measurable.
n=1 f
6.7 Proposition. An extended real-valued function f : X → R∗ is measurable if and only if f −1 ({∞}) and
f −1 ({−∞}) are measurable and the real-valued function defined by
¨
0
x ∈ f −1 ({∞}) ∪ f −1 ({−∞})
f1 (x) =
f (x) otherwise
The collection of all extended real-valued measurable functions will be denoted by M (X , A ). With the
convention that 0(±∞) = 0, c f , f 2 , f + , f − ∈ M (X , A ) if and only if f ∈ M (X , A ). For f + g, we use the
convention that ∞ − ∞ = 0. Then f + g ∈ M (X , A ) if f , g ∈ M (X , A ).
6.8 Definition. Given {an } ⊆ R, define
lim sup an := inf sup{ak }
n k≥n
n→∞
and
lim inf an := sup inf {ak }
n→∞
n
k≥n
In general, lim infn→∞ an ≤ lim supn→∞ an for any sequence {an }.
6.9 Proposition. If { f n } ∈ M (X , A ) then let
f (x) = inf{ f n (x)}
n
f ∗ (x) = lim inf{ f n (x)}
n
F (x) = sup{ f n (x)}
n
F ∗ (x) = lim sup{ f n (x)}
n
All of these functions are measurable.
PROOF: Exercise.
ƒ
6.10 Corollary. Assume that { f n } ∈ M (X , A ) and f (x) = limn→∞ f n (x) exists for each x ∈ X . Then f ∈
M (X , A ).
6.11 Definition. M + := { f ∈ M (X , A ) | f (x) ≥ 0 ∀ x ∈ X }.
6.12 Definition. A simple function
Pn is a function ϕ : X → R such that ϕ(X ) is finite. If the range of ϕ is {a1 , . . . , an }
and Ei = ϕ −1 ({ai }) then ϕ = i=1 ai χ Ei . We say that ϕ is in standard form if the ai ’s are distinct and the Ei ’s
partition X .
Pn
Up to order, the standard form of a simple function is unique. ϕ = i=1 ai χ Ei (in standard form) is mearsurable if and only if each Ei is measurable.
6.13 Theorem. Let f ∈ M + (X , A ). Then there exists {ϕn } ⊆ M + (X , A ) such that
1. ϕn is simple for each n.
INTEGRATION
15
2. 0 ≤ ϕn (x) ≤ ϕn+1 (x) for all x ∈ X and each n.
3. f (x) = limn→∞ ϕn (x) for all x ∈ X .
PROOF: Let n ∈ N. For k = 0, 1, . . . , n2n − 1, let En,k := {x ∈ X | 2kn ≤ f (x) < k+1
} and if k = n2n then let
2n
Pn2n k
En,k := {x ∈ X | f (x) ≥ n}. Define ϕn = k=0 2n χ En,k . It is easy to see that {ϕn } is the desired sequence. (Fill in
the details.)
ƒ
Let f : X → C. Then there are real valued functions f1 , f1 : X → R such that f = f1 + i f2 . We say that f is
measurable if f1 and f2 are measurable.
7
Integration
7.1 Definition. Assume that ϕ ∈ M + (X , A ) is simple, and suppose that ϕ =
R
Pn
ϕ. For any measure µ on (X , A ), we define ϕdµ := i=1 ai µ(Ei ).
Pn
i=1
ai χ Ei is the standard form for
Pn
7.2 Exercise. Assume that ϕ = i=1 ai χ Ei , where Ei ∈ A , but this is not the standard form for ϕ. Prove that we
R
Pn
still have ϕdµ := i=1 ai µ(Ei ).
7.3 Lemma. Let ϕ, ψ ∈ M + (X , A ) be simple functions and c ≥ 0. Then
R
R
1. cϕdµ = c ϕdµ
R
R
R
2. (ϕ + ψ)dµ = ϕdµ + ψdµ
R
Furthermore, if we define λ : A → R∗ by λ(E) = ϕχ E dµ then λ is a measure on A .
PROOF: Exercise.
ƒ
7.4 Definition. If f ∈ M + (X , A ) we define
Z
Z
f dµ := sup
ϕ
ϕdµ
where
the Rsupremum is taken over all simple functions ϕ ∈ M (X , A ) with 0 ≤ ϕ ≤ f . If E ∈ A , then
R
f
dµ
:= f χ E dµ.
E
R
R
1. If f , g ∈ M + (X , A ) and f ≤ g then f dµ ≤ gdµ.
R
R
2. If E ⊆ F ∈ A then E f dµ ≤ F f dµ for all f ∈ M + (X , A ).
7.5 Lemma.
+
7.6 Theorem (Monotone Convergence Theorem). If { f n }∞
n=1 ⊆ M (X , A ) is such that f n ≤ f n+1 for all n ≥ 1
and limn→∞ f n (x) = f (x) then
Z
Z
f dµ = lim
n→∞
PROOF: We know that f ∈ M + (X , A ) and
Z
f n dµ
Z
f n dµ ≤
Z
f n+1 dµ ≤
f dµ
16
MEASURE
AND I NTEGRATION
R
R
Hence limn→∞ f n dµ ≤ f dµ. Let 0 < α < 1 and let ϕ ∈ M + (X , A ) be simple with 0 ≤ ϕ ≤ f . For each n, let
S∞
An = {x ∈ X | f n (x) ≥ αϕ(x)}. Then An ∈ A , An ⊆ An+1 , and X = n=1 An . We have
Z
αϕdµ ≤
Z
Z
f n dµ ≤
An
By the Monotone Convergence Theorem for measures,
measure). Therefore
Z
Z
α
Since 0 ≤ α < 1 is arbitrary,
result is proved.
R
f n dµ
An
n→∞
R
ϕdµ = limn→∞
αϕdµ ≤ lim
ϕdµ = lim
ϕdµ ≤ limn→∞
R
f n dµ. Thus
R
An
ϕdµ (since λ(E) =
R
E
ϕdµ is a
Z
n→∞
An
R
f n dµ
f dµ = sup0≤ϕ≤ f
R
ϕdµ ≤ limn→∞
R
f n dµ, and the
ƒ
R
R
1. If f ∈ M + (X , A ) and c ≥ 0 then c f dµ = c f dµ.
R
R
R
2. If f , g ∈ M + (X , A ) then ( f + g)dµ = f dµ + f dµ.
7.7 Corollary.
+
7.8 Corollary (Fatou’s Lemma). If { f n }∞
n=1 ⊆ M (X , A ) then
Z
Z
lim inf f n dµ ≤ lim inf
n→∞
f n dµ
n→∞
R
R
PROOF: For m ∈ N, let g m = infn≥m { f n }. Then g m ≤ f n for all n ≥ m. Hence g m dµ ≤ f n dµ for all n ≥ m. This
R
R
implies that g m dµ ≤ lim infn→∞ f n dµ. Note that g m % lim infn→∞ f n . By the Monotone Converge Theorem,
Z
lim inf f n dµ = lim
n→∞
Z
m→∞
7.9 Corollary. Let f ∈ M + (X , A ) and let λ(E) =
R
E
Z
g m dµ ≤ lim inf
n→∞
f n dµ
ƒ
f dµ for all E ∈ A . Then λ is a measure.
∞
PROOF
S:∞Since f ≥ 0, λ(E)
Pn ≥ 0 for all E ∈ A , and clearly λ(∅) = 0. Let {En }n=1 ⊆ A be pairwise disjoint and
E = n=1 En . Let f n = k=1 f χ Ek . Then
Z
f n dµ =
∞
X
k=1
Z
f dµ =
Ek
n
X
λ(Ek )
k=1
Notice that f n % f χ E , so by the Monotone Convergence Theorem,
λ(E) =
Z
f dµ = lim f n dµ = lim
E
n→∞
n→∞
n
X
k=1
λ(Ek ) =
∞
X
λ(En )
ƒ
n=1
Question: If we have a measure space (X , A , µ) and another measure λ on A , if λ(E) =
f ∈ M + (X , A )? That is, does a converse to the above corollary hold?
R
7.10 Corollary. Suppose that f ∈ M + (X , A ). Then f (x) = 0 µ-a.e. if and only if f dµ = 0.
R
E
f dµ for some
INTEGRATION
17
PROOF: Assume that
which implies that
R
f dµ = 0. For each n ∈ N let En = {x ∈ X | f (x) > 1n } Then f ≥ 1n χ En for each n ∈ N,
0=
Z
Z
1
f dµ ≥
En
n
dµ =
1
n
µ(En )
S∞
P∞
so µ(En ) = 0. If E = {x ∈ X | f (x) 6= 0} then E = n=1 En , so µ(E) ≤ n=1 µ(En ) = 0.
Now suppose that f (x) = 0 a.e. Let E = {x ∈ X | f (x) > 0} and let f n = nχ E . Since f ≤ lim infn→∞ f n , by
Fatou’s Lemma
Z
Z
0≤
Therefore
R
f n dµ = lim inf nµ(E) = 0
f dµ ≤ lim inf
n→∞
n→∞
f dµ = 0.
ƒ
7.11 Proposition. If f ∈ M+ (X , A ) and λ(E) =
R
E
f dµ then λ is absolutely continuous with respect to µ.
PROOF: If µ(E) = 0 then χ E f = 0, so µ-a.e. Hence λ(E) =
R
E
f dµ = 0.
ƒ
+
Question: Given
R measure µ on (X , A ), if λ is any measure on A with λ µ, does there exist f ∈ M (X , A )
such that λ(E) = E f dµ for all E ∈ A . That is, does the converse of the last proposition hold?
7.12 Theorem (Monotone Convergence Theorem, II). If { f n }∞
n=1 is a monotonically increasing sequence in
M + (X , A ) which converges µ-a.e. to f ∈ M + (X , A ) then
Z
Z
f dµ = lim
f n dµ
n→∞
PROOF: Let N ∈ A be such that µ(N
R ) = 0 and f n →Rf on M = X \ N . Then f n χ M % f χ M . The Monotone
Convergence Theorem shows that f χ M dµ = limn→∞ f n χ M dµ. But
Z
f dµ =
Z
f χN dµ +
= 0 + lim
Z
n→∞
= lim 0 +
Z
n→∞
= lim
Z
n→∞
= lim
n→∞
Z
f χ M dµ
f n χ M dµ
f n χ M dµ
f n χN dµ +
Z
f n χ M dµ
Z
f n dµ
and the result is proved.
ƒ
+
7.13 Corollary. Let {g n }∞
n=1 ⊆ M (X , A ). Then
Z
∞
X
n=1
g n dµ =
∞
X
n=1
Z
g n dµ
18
8
MEASURE
AND I NTEGRATION
Integrable Functions
R
8.1 Definition. Given (X , A , µ), we say that an extended real-valued function f is integrable if f + dµ < ∞
R
R
R
R
and f − dµ < ∞. In this case we write f dµ := f + dµ − f − dµ. If E ∈ A we define (unsurprisingly)
R
R
f dµ := f χ E dµ.
E
R
R
R
If f = f1 − f2 , where f1 , f2 ∈ M + (X , A ) and f1 dµ < ∞ and f2 dµ < ∞ then we would like f dµ =
R
R
f1 dµ − f2 dµ. Indeed, f + − f − = f1 − f2 , so f + + f2 = f1 + f − . Integrating and using additivity gives us the
result.
8.2 Definition. Let (X , A ) be a measurable space. A signed measure on (X , A ) is a function µ : A → R∗ such
that
1. µ assumes at most one of ∞ and −∞.
2. µ(∅) = 0.
S∞
P∞
3. If {En } ⊆ A is pairwise disjoint then µ( n=1 En ) = n=1 µ(En ).
8.3 Example. If (X , A ) is a measurable space and µ1 and µ2 are measures on (X , A ) with at least one begin
finite, then λ = µ1 − µ2 is a signed measure on (X , A ).
8.4 Theorem. If f is integrable with respect to µ then λ(E) =
PROOF: λ = λ1 − λ2 where λ1 (E) =
R
E
f + dµ and λ2 (E) =
R
E
R
E
f dµ is a signed measure.
f − dµ.
ƒ
8.5 Definition. Let L (X , A , µ) be the set of all integrable functions on (X , A , µ).
8.6 Theorem. If f measurable then f ∈ L (X , A , µ) if and only if | f | ∈ L (X , A , µ).
+
−
PROOF: We know f ∈ L (X , A , µ) if and only if f +R, f − ∈ L (X , A , µ). Since
R | f | = f +Rf we get that | f | ∈
+
∞
L (X , A , µ). Conversely, if | f | ∈ L (X , A , µ) then | f |dµ < ∞, so both f dµ and f − dµ < ∞, so f ∈
L (X , A , µ).
ƒ
Notice that if f ∈ L (X , A , µ) then
Z
Z
Z
Z
Z
Z
f dµ = f + dµ − f − dµ ≤ f + dµ − f − dµ = | f |dµ
R
R
8.7 Proposition. If f , g ∈ L (X , A , µ) and α, β ∈ R then α f +β g ∈ L (X , A , µ) and (α f +β g)dµ = α f dµ+
R
β gdµ.
R
R
PROOF: Exercise. First show that α f ∈ L (X , A , µ) and that α f dµ = α f dµ Then show that f + g ∈
R
R
R
L (X , A , µ) and ( f + g)dµ = f dµ + gdµ. Use the fact that | f + g| ≤ | f | + |g|.
ƒ
8.8 Theorem (Lebesgue Dominated Convergence Theorem). Let { f n }∞
n=1 ⊆ L (X , A , µ). Assume that f =
limn→∞ f n µ-a.e.
If
there
exists
an
integrable
function
g
∈
L
(X
,
A
,
µ)
such that | f n | ≤ g for all n then f is
R
R
integrable and f dµ = limn→∞ f n dµ.
L p -SPACES
19
PROOF: By redefining f n , f if necessary, we may assume that f = limn→∞ f n everywhere. This shows that f is
measurable. We have | f | ≤ g, so | f | is integrable and so f is integrable. Notice that g + f n ≥ 0. By Fatou’s
Lemma
Z
Z
Z
g dµ +
f dµ =
(g + f )dµ
Z
≤ lim inf
n→∞
= lim inf
(g + f n )dµ
Z
n→∞
=
so
R
f dµ ≤ lim infn→∞
R
Z
9
R
gdµ +
gdµ + lim inf
f n dµ
Z
n→∞
f n dµ
f n dµ. On the other hand, g − f n ≥ 0, so
−
Therefore
Z
Z
f dµ = limn→∞
R
Z
f dµ ≤ lim inf
n→∞
− f n dµ = − lim sup
n→∞
Z
f n dµ
f n dµ.
ƒ
L p -spaces
This section is in need of repairs.
R
We have seen that L (X , A , µ) is a vector space over R. For f ∈ L (X , A , µ), define k f k1 = | f |dµ.
Then k · k1 defines a seminorm on L (X , A , µ). Let ∼ be the equivalence relation defined on L (X , A , µ) by
f ∼ g ⇐⇒ k f − gk1 = 0. Let S = { f ∈ L (X , A , µ) | k f k1 = 0}, the subspace of all integrable functions whose
seminorm is zero.
9.1 Definition. Let L 1 (X , A , µ) = L 1 (X , µ) := L (X , A , µ)/ ∼.
L 1 is a normed vector space with respect to the norm k[ f ]k1 := k f k1 . In an abuse of notation we will
normally write f to represent [ f ].
9.2 Example. Take X = N and µ the counting measure on N. Then L (N, P (N), µ) = L 1 (N, µ) = `1 (N).
9.3 Definition. Let 1 ≤ p < ∞. Define
L 1 (X , A , µ) = L 1 (X , µ) := {[ f ] | f is measurable and | f | p is integrable}
Let k f k p :=
€R
Š1
| f | p dµ p .
9.4 Definition. Let f ∈ M (X , A). WeR say that f is essentially bounded if there exists M such that µ({x ∈
X | | f (x)| > M }) = 0. Let k f k∞ = {M | µ({x ∈ X | | f (x)| > M }) = 0}. Then k · k∞ is a seminorm on
L ∞ (X , A , µ) = { f ∈ M (X , A ) | f is essentially bounded}. Let L ∞ (X , A , µ) = L ∞ (X , µ) = L ∞ (X , A , µ)/ ∼.
9.5 Theorem (Hölder’s Inequality). Let 1 ≤ p < ∞ and
1
p
+
1
q
= 1. Let f ∈ L p (X , µ) and g ∈ L q (X , µ). Then
f g ∈ L 1 (X , µ) and
k f gk1 ≤ k f k p kgkq
20
MEASURE
AND I NTEGRATION
9.6 Theorem (Minkowski’s Inequality). If f , g ∈ L p (X , µ) then f + g ∈ L p (X , µ) and
k f + gk p ≤ k f k p + kgk p
9.7 Corollary. (L p (X , µ), k · k p ) is a normed linear space for all 1 ≤ p < ∞.
9.8 Theorem (Completeness of L p ). Let 1 ≤ p ≤ ∞. Then (L p (X , µ), k · k p ) is a Banach space.
p
p
PROOF: Assume that 1 ≤ p < ∞. Let { f n }∞
n=1 ⊆ L (X , µ) be a Cauchy sequence in (L (X , µ), k · k p ). We can find a
1
subsequence {g k } of { f n } such that kg k+1 − g k k p < 2k . Define, for all x ∈ X ,
g(x) = |g1 (x)| +
∞
X
|g k+1 (x) − g k (x)|
k=1
Then g ∈ M + (X , A ), and by Fatou’s Lemma,
Z
Z p
n
X
p
|g| dµ ≤ lim inf
|g1 (x)| +
|g k+1 (x) − g k (x)| dµ
n→∞
k=1
≤ lim inf kg1 (x)k p +
n→∞
n
X
kg k+1 (x) − g k (x)k p
k=1
≤ kg1 k p + 1 < ∞
P∞
Let E = {x ∈ X | g(x) < ∞}, so that µ(X \ E) = 0. Hence |g1 (x)| + k=1 |g k+1 (x) − g k (x)| converges to a finite
number µ-a.e. Let
¨
P∞
g1 (x) + k=1 g k+1 (x) − g k (x) if x ∈ E
f (x) =
0
if x ∈
/E
Note that g k → f µ-a.e. We also that that |g k (x)| ≤ g(x) for all x ∈ X . The Lebesgue Dominated Convergence
Theorem shows us that
Z
Z
Z
| f | p dµ = lim
k→∞
|g k | p dµ ≤
|g| p dµ < ∞
Therefore f ∈ L p (X , µ). Since | f | ≤ g, | f − g k | p ≤ 2 p |g| and again by LDCT,
Z
0 = lim
k→∞
| f − g k | p dµ
since g k → f µ-a.e. Therefore the subsequence {g k } converges to f in L p (X , µ). It follows that { f n } converges to
f in L p (X , µ).
∞
∞
Now let { f n }∞
n=1 ⊆ L (X , µ) be a Cauchy sequence in (L (X , µ), k · k∞ ). Let E ⊆ X be such that µ(E) = 0 and
if x ∈
/ E then | f n (x)| ≤ k f n k∞ and | f n (x) − f m (x)| ≤ k f n − f m k∞ . Then { f n (x)} converges uniformly on X \ E to
some function f (x). It follows that f is measurable and k f n − f k∞ → 0. Hence f ∈ L ∞ (X , µ).
ƒ
p
p
9.9 Corollary. If 1 ≤ p < ∞ and if { f n }∞
n=1 ⊆ L (X , µ) is such that f n → f in L (X , µ) then there exists a
∞
subsequence { f nk }k=1 of { f n } such that f nk (x) → f (x) µ-a.e.
Observe that it is possible to f n → f in L p (X , µ) but { f n (x)} diverges everywhere. Let f1 = χ[0,1) , f2 = χ[0, 1 ) ,
2
f3 = χ[ 1 ,1) , and so on, so that f n is the characteristic function of exceedingly small intervals. Then f n → 0 but
2
lim supn→∞ f n (x) = 1 and lim infn→∞ f n (x) = 0 for all x ∈ [0, 1].
SIGNED MEASURES
21
9.10 Example. Let X = [0, 1] and µ = m|[0,1] . Hölder’s Inequality implies that L 1 [0, 1] ⊇ L p [0, 1] ⊇ L ∞ [0, 1]
for all 1 < p < ∞. If L 2 [0, 1] ⊆ L 1 [0, 1], is it closed with respect to k · k1 ?
9.11 Theorem. C[0, 1] is dense in L p [0, 1] for all 1 ≤ p < ∞ and it is closed in L ∞ [0, 1].
PROOF: Exercise.
10
ƒ
Signed Measures
10.1 Definition. Let µ be signed measure on (X , A ) and let P, N , M ∈ A . Then
1. P is positive if µ(E ∩ P) ≥ 0 for all E ∈ A .
2. N is negative if µ(E ∩ N ) ≤ 0 for all E ∈ A .
3. M is null if µ(E ∩ M ) = 0 for all E ∈ A .
10.2 Lemma.
1. Every subset of a positive (resp. negative, null) set is postive (resp. negative, null).
2. A countable union of positive (resp. negative, null) sets is postive (resp. negative, null).
PROOF: Trivial.
ƒ
10.3 Lemma. Let E ∈ A be such that µ(E) > 0. Then there is a postive set A ⊆ E such that µ(A) > 0.
Sk−1
PROOF: Recursively define Ek as follows. If E \ j=1 E j is not positive then let nk be the least positive integer
Sk−1
S∞
such that there exists Ek ⊆ E \ j=1 E j with µ(Ek ) < − n1 , otherwise let Ek = ∅. Let A = E \ j=1 E j =
k
P∞
S∞
T∞
Sk−1
E j and the union is disjoint, µ(E) = µ(A) + j=1 µ(E j ). It follows that
j ). Since E = A ∪
k=1 (E \
j=1
j=1 E
P∞
P∞
µ(A) > 0 since j=1 µ(E j ) < 0, and hence j=1 µ(E j ) > −∞. In particular, limn→∞ nk = ∞ or the sequence
Sk−1
terminates. Now for F ⊆ A, if µ(F ) = r < 0 then let k ≥ 1 such that nk > 1 − 1r . F ⊆ E \ j=1 E j , but by choice
Sk−1
of nk , E \ j=1 E j has no subset of measure less than or equal to n 1−1 < r. This contradiction shows that A is
k
positive.
ƒ
10.4 Theorem (Hahn Decomposition Theorem). Let µ be a signed measure on (X , A ). Then there exists a
positive set P and a negative set N that partition X . This is called a Hahn Decomposition.
PROOF: Without loss of generality we may assume that µ(E) < ∞ for all E ∈ A , for otherwise −µ is a measure
with this property and postive with respect to µ is the same as negative with respect to −µ and visa versa. Let
∞
µ(An ) = λ,
λ = sup{µ(A)
S | A is positive} < ∞. Let {An }n=1 ⊆ A be a sequence of positive sets such that limn→∞
and let P = n=1 ∞An . Then P is positive so µ(P) ≤ λ, and for all n ≥ 1 µ(P) = µ(P ∩An )+µ(P ∩Acn ) ≥ µ(An ), so
µ(P) = λ. We need only check that N = P c is negative. If A ⊆ N is such that µ(A) > 0 then by the previous lemma
there is B ⊆ A such that µ(B) > 0 and B is positive. But then P ∪ B is positive and µ(P ∪ B) = µ(P) + µ(B) > λ, a
contradiction.
ƒ
10.5 Lemma. If (P1 , N1 ) and (P2 , N2 ) are Hahn Decompositions of (X , A , µ) then
1. P1 ∆P2 = N1 ∆N2 is null.
2. µ(P1 ∩ E) = µ(P2 ∩ E) and µ(N1 ∩ E) = µ(N2 ∩ E) for all E ∈ A .
PROOF: Notice first that Ni = Pic , so it’s redundant to represent an Hahn Decomposition by an ordered pair when
a singlton will suffice. P1 ∆P2 = (P1 ∩ P2c ) ∪ (P1c ∩ P2 ) is both positive and negative, so it is null. Similarily for
N1 ∆N2 . For the second part, µ(E ∩ P1 ) = µ(E ∩ P1 ∩ P2 ) since E ∩ P1 \ P2 ⊆ P1 ∆P2 is null. By symmetry the lemma
is proved.
ƒ
22
MEASURE
AND I NTEGRATION
10.6 Theorem (Jordan Decomposition Theorem). Let µ be a signed measure on (X , A ). Then there exists two
positive measures µ+ and µ− such that µ = µ+ − µ− . Moreover, there is only one such pair for which µ+ ⊥µ− .
PROOF: Let (P, N ) be a Hahn Decomposition. Let µ+ (E) = µ(E ∩ P) and µ− = −µ(E ∩ N ). Then clearly µ+ and
µ− are positive measures and µ(E) = µ+ (E) − µ− (E) for all E ∈ A . Moreover, since N ∩ P = ∅, µ+ ⊥µ− . Assume
that µ = µ1 − µ2 where µ1 ⊥µ2 . Let A, B be a partition of X such that µ1 (B) = 0 and µ2 (A) = 0. For every E ∈ A ,
µ(E ∩ A) = µ1 (E ∩ A) − µ2 (E ∩ A) = µ1 (E ∩ A) ≥ 0
so A is a positive set. Similarily, B is negative with respect to µ, so (A, B) is a Hahn Decomposition. But then for
any E ∈ A ,
µ+ (E) = µ(E ∩ P) = µ(E ∩ A) = µ1 (E) and µ− (E) = −µ(E ∩ N ) = −µ(E ∩ B) = µ2 (E)
Hence µ1 = µ+ and µ2 = µ− .
ƒ
Note that if µ = µ1 − µ2 with µ1 and µ2 positive measures then
µ(E) = µ(E ∩ P) = µ1 (E ∩ P) − µ2 (E ∩ P) ≤ µ1 (E ∩ P) ≤ µ1 (E)
and similarily µ− (E) ≤ µ2 (E), for all E ∈ A .
10.7 Example. If µ is a measure on (X , A ) and f is integrable, then λ(E) =
R
R
λ− (E) = f − dµ. Notice that λ+ (X ) + λ− (X ) = | f |dµ = k f k1 .
R
E
f dµ, λ+ (E) =
R
f + dµ, and
10.8 Definition. Given a postive measure µ, the positive measure |µ| = µ+ + µ− is called the total variation of
µ.
Note that µ+ , µ− |µ|. Assume that µ is finite. Then |µ(E)| ≤ |µ|(E) < ∞. Given (X , A ), let
Meas(X , A ) = {µ | µ is a finite signed measure on (X , A )}
Notice that Meas(X , A ) is a vector space over R. Define kµkMeas = |µ|(X ). It is clear that kµk = 0 if and only if
µ = 0. It is also easy to see that kαµk = |αµ|(X ) = |α||µ|(X ) = |α|kµk. Assume that µ, γ ∈ Meas(X , A ). Then
µ + γ = (µ+ + γ+ ) − (µ− + γ− )
and so (µ+γ)+ ≤ µ+ +γ+ and (µ+γ)− ≤ µ− +γ− . It follows that kµ+γk ≤ kµk+kγk. Therefore (Meas(X , A ), k·
kMeas ) is a normed linear space.
Problem: Is Meas(X , A ) a Banach space with this norm?
Suppose {µn }∞
n=1 is Cauchy with respect to k · kMeas ). Then for all " > 0 there is N such that if n, m ≥ N then
|µm − µn |(X ) ≤ kµm − µn k < ". Since |µ(E)| ≤ |µ|(E) for all E ∈ A , we get that
|µm (E) − µn (E)| = |(µm − µn )(E)| ≤ |µm − µn |(E) ≤ |µm − µn |(X ) < "
if n, m ≥ N . Therefore {µn (E)} is Cauchy in R for all E ∈ A . Define µ(E) = limn→∞ µn (E) for all E ∈ A . Then
1. µ(∅) = limn→∞ µn (∅) = 0.
2. Since {µn } is Cauchy, if is bounded. Let M be such that kµn k < M for all n. It follows that |µn (E)| ≤ M .
For all E ∈ A .
3. We must show that µ is countably additive. Prove this as an exercise.
We have proven the following theorem:
10.9 Theorem. (Meas(X , A ), k · kMeas ) is a Banach space.
Problem: What does Meas(R, B(R)) “look like”?
THE RADON-NIKODYM THEOREM
11
23
The Radon-Nikodym Theorem
11.1 Lemma. Let µ and ν be finite measures on (X , A ). Then ν µ if and only if for every " > 0 there exists
δ > 0 such that if E ∈ A with µ(E) < δ then ν(E) < ".
PROOF: Assume that ν µ and let " > 0. Assume
En ∈ A be
S∞ that no such δ exists for1 this ". For each n ∈ N, letT
∞
and µ(Fn ) ≥ ". Let F = n=1 Fn , so
such that µ(En ) < 21−n but ν(En ) ≥ ". Let Fn = k=n Ek , so that µ(Fn ) < 2n−1
that Fn & F . Hence µ(F ) = limn→∞ µ(Fn ) = 0 but ν(F ) = limn→∞ ν(F ) ≥ ", contradicting that ν µ.
Conversely, let " > 0 and E ∈ A . If µ(E) = 0 then µ(E) < δ("), so ν(E) < ". But " was arbitrary, so
ν(E) = 0.
ƒ
11.2 Theorem (Radon-Nikodym). Let λ and µ be σ-finite measures on (X , A ). Assume that λ µ. Then
there exists f ∈ M + (X , A ) such that
Z
λ(E) =
f dµ
E
for every E ∈ A . Moreover, this f is uniquely determined µ-almost everywhere.
PROOF: We assume that λ, µ are finite. For each c > 0 let (P(c), N (c)) be an Hahn decomposition for the signed
Sk
measure λ − cµ. For each k ∈ N let A1 = N (c) and Ak+1 = N ((k + 1)c) \ i=1 Ai . Then {Ai }∞
i=1 is pairwise
Sk
Sk
Tk−1
disjoint and i=1 N (ic) = i=1 Ai . Therefore Ak = N (kc) ∩ i=1 P(ic). If E ∈ A , E ⊆ Ak , then E ⊆ N (ck) and
E ⊆ P((k − 1c). This implies that
(k − 1)cµ(E) ≤ λ(E) ≤ kcµ(E)
Let B = X \
S∞
i=1 Ai
=
T
i=1 ∞P( jc).
(∗)
Hence B ⊆ P(kc) for all k ∈ N. It follows that
0 ≤ kcµ(B) ≤ λ(B) ≤ λ(X ) < ∞
for every k ∈ N , so µ(B) = 0 implies that λ(B) = 0. For each c, define
¨
(k − 1)c for x ∈ Ak
f c (x) =
0
for x ∈ B
S∞
Then if E is measurable we have E = (E ∩ B) ∪ ( i=1 (E ∩ Ak )), and so (∗) implies
Z
Z
Z
f c dµ ≤ λ(E) ≤
E
( f c + c)dµ ≤
E
f c dµ + cµ(X )
E
For each n ∈ N, let g n = f 1n . Then
2
Z
g n dµ ≤ λ(E) ≤
E
If m ≥ n then (∗∗) implies
Z
g n dµ ≤ λ(E) ≤
E
Z
g n dµ +
E
Z
g m dµ +
E
µ(X )
2m
Z
µ(X )
2n
(∗∗)
g m dµ ≤ λ(E) ≤
and
E
Adding gives
Z
Z
µ(X )
g m dµ ≤
g n dµ −
E
2n
E
Z
g n dµ +
E
µ(x)
2n
24
MEASURE
AND I NTEGRATION
for all E ∈ A . Let E1 = {x ∈ X | g n − g m ≥ 0} and E2 = {x ∈ X | g n − g m < 0}. This gives us that
Z
2µ(X )
µ(X )
|g n − g m |dµ ≤
≤ n−1
n
2
2
E
1
1
+
so {g n }∞
n=1 is Cauchy in L (X , A ). Suppose that g n → f ∈ L (X , A ). Since g n ∈ M (X , A ), we can take
+
f ∈ M (X , A ). Now
Z
Z
Z
f dµ ≤ |g n − f |dµ ≤ kg n − f k1 → 0
g n dµ −
E
E
E
R
R
Hence (∗∗) implies that λ(E) = limn→∞ E g n dµ = E f dµ.
R
R
Suppose f , h ∈ M + (X , A ) are such that E f dµ = λ(E) = E hdµ for all E ∈ A . Let E1 = {x ∈ X | f (x) >
h(x)} and E2 = {x ∈ X | f (x) < h(x)}. Then
Z
Z
Z
( f − h)dµ =
E1
hdµ = λ(E1 ) − λ(E1 ) = 0
f dµ −
E1
E1
so µ(E1 ) = 0. Similarily, µ(E2 ) = 0, so f = h µ-a.e.
S∞
Now assume that µ, λ are σ-finite. Let {X n } ⊆ A be an increasing sequence such that X = n=1 X n and
R
µ(X n ) < ∞, λ(X n ) < ∞. We get f n ∈ M + (X , A ) with f n |X nc ≡ 0 and if E ⊆ X n then λ(E) = E f n dµ. If n ≤ m
R
R
then X n ⊆ X m , so E f n dµ = E f m dµ for all E ⊆ X n . It follows that f n = f m µ-a.e. Let Fn = sup{ f1 , . . . , f n },
so that {Fn }∞
is an increasing sequence of positive measurable functions. Let f = limn→∞ Fn . If E ∈ A then
n=1
R
λ(E ∩ X n ) = E Fn dµ since Fn = f n µ-a.e. on X n . But E ∩ X n % E, so
Z
Z
λ(E) = lim λ(E ∩ X n ) = lim
n→∞
n→∞
Fn dµ =
E
f dµ
E
The uniqueness of f is determined as in the finite case.
ƒ
From the proof, we note that is λ is finite then we can choose f to be integrable, and hence finite µ-a.e. If λ
is σ-finite then we can choose the f n ’s to be finite on X n . From this we can deduce that f can be chosen to be
integrable in this case as well.
Remark. Let λ µ, where λ is a finite signed measure and µ is a positive σ-finite measure. Then |λ| µ and
so λ+ µ and λ− µ. Hence there are f + , f − ∈ M + (X , A ) that represent λ+ and λ− . But then
Z
Z
Z
λ(E) = λ+ (E) − λ− (E) =
f + dµ −
E
+
f + − f − dµ
f − dµ =
E
E
R
so f = f − f represents λ. Notice that | f |dµ = |λ|(X ) < ∞, so f ∈ L 1 (X , A ) and k f k1 = kλkMeas . Given
µ positive and σ-finite, let AC(X , A , µ) = {λ ∈ Meas(X , A ) | λ µ}. Then AC(X , A , µ)
R is a closed subspace of
(Meas(X , A ), k · kMeas ). If ν ∈ AC(X , A , µ) and f ∈ L 1 (X , A , µ) is such that ν(E) = E f dµ then ν 7→ f is an
isometric isomorphism of AC onto L 1 .
R
P
11.3 Example. Let A = P (X ) and µ f (E) = x∈E f (x) = E f dλ, where λ is the counting measure. Is every
finite measure on (X , P (X )) of this form? How about every finite measure? Is this measure absolutely continuous
with respect to the counting measure? The Radon-Nikodym Theorem says yes.
−
11.4 Definition.
Assume that λ, µ are σ-finite measures and λ µ. The function f ∈ M + (X , A ) such that
R
λ(E) = E f dµ is called the Radon-Nikodym derivative of λ with respect to µ, and is denoted by dλ
.
dµ
DIFFERENTIATION
OF
MONOTONE FUNCTIONS
25
11.5 Example. Assume that F : R → R is continuously differentiable with F 0 (x) ≥ 0. Then F is monotonic. Let
µ F be the Lebesgue-Stieltjes measure on (R, B(R)) generated by F . Then µ F is σ-finite and µ F m. Moreover,
µ F ((a, b]) = F (b) − F (a) =
b
Z
0
F (x)d x =
Z
From here we get that if E ∈ B(R) then µ F (E) =
E
F dm =
[a,b]
a
R
0
F 0 dm, so F 0 =
dµ F
dm
Z
F 0 dm
(a,b]
.
11.6 Theorem (Lebesgue Decomposition Theorem). Let λ, µ be σ-finite measurs on (X , A ). Then there exist
two measures λ1 and λ2 on (X , A ) such that λ = λ1 + λ2 and λ1 ⊥µ and λ2 µ. Moreover, λ1 and λ2 are
unique with these properties.
PROOF: Let γ = λ + µ, so that γ is σ-finite and λ γ and µ γ. By the Radon-Nikodym Theorem, there are
functions f , g ∈ M + (X , A ) such that
Z
Z
λ(E) =
f dγ and
µ(E) =
E
gdγ
E
for all E ∈ A . Let A = {x ∈ X | g(x) = 0} and B = {x ∈ X | g(x) > 0}, so that A and B partition X . Let
λ1 (E) = λ(E ∩ A) and λ2 (E) = λ(E
R ∩ B) for all E ∈ A . It is clear that λ = λ1 + λ2 . Since µ(A) = 0, it follows
that λ1 ⊥µ. If µ(E) = 0 then E g dγ = 0, so g(x) = 0 for γ-a.e. x ∈ E. But then γ(E ∩ B) = 0, and so
λ2 (E) = λ(E ∩ B) = 0 since λ γ. Therefore λ2 µ. To see that the decomposition is unique, observe that
for any σ-finite signed measure ν, if ν⊥µ and ν µ then ν ≡ 0. If λ = ν1 + ν2 is another decomposition then
λ1 − ν1 = ν2 − λ2 . This is not quite correct, as we cannot be certain that λi − νi is a signed measure. Complete
this proof as an exercise.
ƒ
12
Differentiation of Monotone Functions
12.1 Definition. Let J be a collection of non-degenerate intervals. We say that J is a Vitali cover of E ⊆ R if for
all " > 0 and any x ∈ E there exists I ∈ J such that x ∈ I and `(I) < ".
12.2 Lemma (Vitali’s Lemma). Let E ⊆ R be such that m∗ (E) < ∞ and let J be a Vitali cover
Sn of E. Then there
exists, for each " > 0, a finite pairwise disjoint collection {I1 , . . . , I n } ⊆ J such that m∗ (E \ i=1 I i ) < ".
PROOF: We may assume that the intervals in J are closed since finitely many endpoints will contribute only a set
of measure zero. Let E ⊆ U ⊆ R be an open set with m(U) < ∞. Since J is a Vitali cover, we may assume
Snthat
if I ∈ J then I ⊆ U. Let I1 ∈ J. Suppose that I1 , . . . , I n have been chosen to be pairwise disjoint. If E ⊆S i=1 I i
n
then stop. Otherwise, let kn be the supremum of the lengths of the intervals in J that are disjoint from i=1 I i .
S
Sn
n
Since E 6⊆ i=1 I i , we can find I n+1 ∈ J such that `(I n+1 ) > 21 kn and I n+1 is disjoint from i=1 I i . This gives us a,
S∞
S∞
possibly finite, sequence {I n }∞
of disjoint intervals in J with n=1 I n ⊆ U. Hence m( n=1 I n ) ≤ m(U) < ∞. We
n=1
P∞
SN
SN
can find N ∈ N such that n=N +1 `(I n ) < 5" . Let R = E \ i=n I n . Let x ∈ R. Since x ∈ ( i=n I n )c , an open set,
SN
there is an interval x ∈ I ∈ J such that I is disjoint from i=n I n . It follows that `(I) ≤ kn < 2`(I n+1 ) if I ∩ I n = ∅.
Since limn→∞ `(I n ) = 0, I must meet at least one of the I n ’s. Let n0 be the smallest n for which I ∩ I n 6= ∅. Then
n0 > N and `(I) ≤ Kn0 −1 < 2`(I n0 ). Since I ∩ I n0 6= ∅ and x ∈ I, the distance from x to the midpoint of I n0 is at
most `(I) + 21 `(I n0 ) ≤ 52 `(I n0 ). Therefore x ∈ Jn0 , where Jn0 is the interval with the same center as I n0 and five
times the length.
For each n ≥ N + 1, let JnPbe the interval with the same center as I n but five times the length.
S∞
∞
Then R ⊆ n=N +1 Jn , and hence m∗ (R) ≤ 5 n=N +1 `(I n ) < ".
ƒ
26
MEASURE
AND I NTEGRATION
12.3 Definition. Let f : R → R. For any x ∈ R, let
D+ f (x) = lim sup
h→0+
D+ f (x) = lim inf
+
h→0
D− f (x) = lim sup
h→0+
D− f (x) = lim inf
+
f (x + h) − f (x)
h
f (x + h) − f (x)
h
f (x) − f (x − h)
h
f (x) − f (x − h)
h
h→0
These four quantities are called the Dini derviatives of f (x) at x. f is differentiable at x if and only if D+ f (x) =
D− f (x) = D+ f (x) = D− f (x) < ∞. In this case f 0 (x) = D+ f (x). If D+ f (x) = D+ f (x), we denote the common
value by f (x + ). If both are finite then f (x + ) is called the right hand dervative. f (x − ) is defined in a similar
fashion and is called the left hand derviative if it exists.
In general, D+ f (x) ≥ D+ f (x) and D− f (x) ≥ D− f (x).
12.4 Proposition. If f ∈ C[a, b] and one of its Dini dervatives is everywhere non-negative then f is nondecreasing on [a, b]
PROOF: Exercise.
ƒ
12.5 Theorem. Let f : [a, b] → R be increasing. Then f is differentiable almost everywhere on [a, b], f 0 is
measureable, and
Z
f 0 dm ≤ f (b) − f (a)
[a,b]
PROOF: Consider ES
= {x ∈ [a, b] | D+ f (x) > D− f (x)}. Let E r,s = {x ∈ [a, b] | D+ f (x) > r > s > D− f (x)}, for
r, s ∈ Q. Then E = r,s∈Q E r,s . Let α = m∗ (E r,s ) and " > 0. Choose U open such that m(U) < α + " and E r,s ⊆ U.
f (x)− f (x−h)
For each x ∈ E r,s , there is an arbitrarily small interval [x − h, x] ⊆ U so that
< s. By Vitali’s Lemma,
h
there are finitely many disjoint intervals I1 , . . . , I N of this type such that the interiors of the I n ’s cover a subset A
of E r,s with m∗ (A) > α − ". We have
N
N
X
X
[ f (x n ) − f (x n − hn )] ≤ s
hn < sm(U) < s(α + ")
n=1
n=1
Now for each point y ∈ A, there is an arbitrarily small interval [ y, y + k] that is contained in some I n and is such
f ( y+k)− f ( y)
> r. Therefore f ( y + k) − f ( y) > r k. Applying Vitali’s Lemma again, we get intervals J1 , . . . , J M
that
k
SM
disjoint and of the type [ y, y + k] ⊆ I n such that j=1 J j contains a subset of A with outer measure at least α−2".
Hence
M
M
X
X
[ f ( y j + k j ) − f ( y j )] > r
k j ≥ r(α − 2")
j=1
We know J j ⊆ I n . We have
P
J j ⊆I n [ f
r(α − 2") ≤
j=1
( y j + k j ) − f ( y j )] ≤ f (x n ) − f (x n − hn ) since f is increasing. It follows that
M
N
X
X
[ f ( y j + k j ) − f ( y j )] ≤
[ f (x n ) − f (x n − hn )] ≤ s(α − ")
j=1
n=1
B OUNDED VARIATION
27
Since " > 0 is arbitrary, rα ≤ sα. Since r > s, this imples that α = 0. Therefore m∗ (E r,s ) = 0, so m(E r,s ) = 0 and
so m(E) = 0.
Similarily, if E1 = {x ∈ [a, b] | D− f (x) > D+ f (x)} then m(E1 ) = 0. From this we can deduce that D+ f (x) =
f (x)− f (x+h)
D− f (x) = D+ f (x) = D− f (x) a.e. (Verify this.) This means that g(x) = limh→0
is defined as an
h
extended real number a.e. on [a, b]. Let g n (x) = n[ f (x + 1n ) − f (x)], where f (x) := f (b) for x ≥ b. Then
g n → g a.e. on [a, b]. Since f is increasing, g n > 0. By Fatou’s Lemma,
Z
Z
g d m ≤ lim inf
g n dm
n
[a,b]
[a,b]
= lim inf n
Z
n
[ f (x +
[a,b]
n
) − f (x)]d x
Z
Z
= lim inf n
n
1
f dm
f dm −
[a,a+ 1n ]
[b,b− 1n ]
= lim inf f (b) −
n
Z
f dm
[a,a+ 1n ]
≤ f (b) − f (a)
Therefore g is integrable. Hence g(x) is finite a.e. This shows that f (x) is differentiable a.e. and f 0 (x) = g(x)
is finite. It follows that f 0 = g a.e. on [a, b] and finally,
Z
Z
f 0 dm =
[a,b]
13
gdm ≤ f (b) − f (a)
[a,b]
ƒ
Bounded Variation
13.1 Definition. Let f : [a, b] → R. For a partition Π = {a = x 0 < x 1 < · · · < x n = b} of [a, b], define
Vab ( f , Π) =
n
X
| f (x i ) − f (x i−1 )|
i=1
Notice that if Π1 ⊆ Π2 then Vab ( f , Π1 ) ≤ Vab ( f , Π2 ). The variation of f on [a, b] is
Vab ( f ) = sup Vab ( f , Π)
Π
f is of bounded variation on [a, b] if Vab ( f ) < ∞. Define
BV [a, b] = { f : [a, b] → R | f is of bounded variation}
13.2 Proposition. BV [a, b] is a vector space and Vab (·) is a seminorm on it.
PROOF: Trivial.
ƒ
13.3 Definition. For f : [a, b] → R and Π = {a = x 0 < x 1 < · · · < x n = b} a partition of [a, b], let
Vab+ ( f , Π) =
n
X
( f (x i ) − f (x i−1 ))+
i=1
and
Vab− ( f , Π) =
n
X
( f (x i ) − f (x i−1 ))−
i=1
28
MEASURE
AND I NTEGRATION
where α+ = max{α, 0} and α− = (−α)+ . The postive variation and negative variation of f are
Vab+ ( f ) = sup Vab+ ( f , Π)
Vab− ( f ) = sup Vab− ( f , Π)
and
Π
Π
1. Vab ( f ) = Vab+ ( f ) + Vab− ( f )
13.4 Proposition.
2. f (b) − f (a) = Vab+ ( f ) − Vab− ( f )
PROOF:
1. Trivial.
2. For " > 0 let Π be a partition of [a, b] such that Vab+ ( f ) − Vab+ ( f , Π) < " and Vab− ( f ) − Vab− ( f , Π) < ". Then
|Vab+ ( f ) − Vab− ( f ) − ( f (b) − f (a))|
≤ |Vab+ ( f ) − Vab+ ( f , Π)| + |Vab− ( f ) − Vab− ( f , Π)| + |Vab+ ( f , Π) − Vab− ( f , Π) − ( f (b) − f (a))|
But Vab+ ( f , Π) − Vab− ( f , Π) =
Pn
i=1
f (x i ) − f (x i−1 ) = f (b) − f (a), so the result is proved.
ƒ
13.5 Theorem (Jordan Decomposition Theorem).
BV [a, b] = { f − g | f , g : [a, b] → R is non-decreasing}
PROOF: If f is non-decreasing then Vab ( f ) = f (b) − f (a) < ∞, so f ∈ BV [a, b]. Since BV [a, b] is a vector space,
one containment is proved.
Conversely, given f ∈ BV [a, b], f (x) = Vax+ ( f )−Vax− ( f )− f (a). But Vax+ ( f ) and Vax− ( f ) are non-decreasing,
so f is a difference of non-decreasing functions.
ƒ
13.6 Corollary. If f ∈ BV [a, b] then f is differentiable almost everywhere.
R
13.7 Proposition. Suppose that f is integrable on [a, b]. Let F (x) = [a,x] f dm. Then F is continuous and of
bounded variation on [a, b].
R
PROOF: To see that F is continuous, let λ(E) = E | f |dm for E ∈ B([a, b]). λ is a (positive) measure and λ m.
Thus, given " > 0 there is δ > 0 such that m(E) < δ implies that λ(E) < ". Hence if |x − y| < δ then
Z
Z
| f |dm < "
| f (x) − f ( y)| = f dm ≤
[x, y]
[x, y]
To see that F is of bounded variation, let Π be a partition of [a, b]. Then
n
n Z
X
X
f dm
|F (x i ) − F (x i−1 )| =
[x i−1 ,x i ]
i=1
i=1
n Z
X
≤
| f |dm
i=1
=
[x i−1 ,x i ]
Z
| f |dm
[a,b]
Therefore Vab ( f ) ≤
R
[a,b]
| f |d m < ∞.
13.8 Proposition. If f ∈ C[a, b] ∩ BV [a, b] then Vax ( f ) ∈ C[a, b].
ƒ
B OUNDED VARIATION
29
PROOF: Let f ∈ C[a, b]∩ BV [a, b] and suppose for contradiction that Vax ( f ) is not continous at x 0 ∈ [a, b]. Since
Vax ( f ) is monotone, this must be a jump discontinuity. By symmetry and without loss of generality, assume that
lim x→x 0 Vax ( f ) = a > 0. Let δ > 0 be such that if x ∈ [x 0 , x 0 + δ) then | f (x) − f (x 0 )| < 3a .
Claim. For all y ∈ (x 0 , x 0 + δ) there exists z ∈ (x 0 , y) such that Vzy ( f ) > 3a .
Surely, Vxy0 ( f ) ≥ a, so let Π be a partition of [a, b] such that Vxy0 ( f , Π) >
| f (x 1 ) − f (x 0 )| < 3a . Let z = x 1 and notice that
Vzy ( f ) ≥
n
X
| f (x i ) − f (x i−1 )| >
i=2
2a
.
3
Then x 1 ∈ [x 0 , x 0 + δ), so
a
3
Consequently, there is an infinite series x 0 + δ > y0 > y1 > · · · > a such that Vyyn0 ( f ) =
implies that Vab ( f ) > n 3a for all n, contradicting that f ∈ BV [a, b].
13.9 Lemma. If f : [a.b] → R is integrable and
Rx
a
Pn
i=1
y
Vyi i−1 ( f ) > n 3a . This
ƒ
f (t)d t = 0 for every a ∈ [a, b] then f (x) = 0 a.e.
PROOF: Let E = {x ∈ [a, b] | f (x) > 0}. Assume that m(E) > 0. S
Then there exists a closed set F ⊆ E with
∞
m(F ) > 0. Let U = (a, b) \ F , an open set. Since U is open, U = i=1 (ai , bi ), where {(ai , bi )}∞
i=1 is pairwise
Rb
R
R
R
R
R
disjoint. We also have that 0 = a f (t)d t = U f dm + F f dm, so U f dm = − F f dm 6= 0. Since 0 6= U f dm =
R an
R bn
R bn
P∞ R bi
i=1 ai f (t)d t0, there is some n such that an f (t)d t 6= 0. But then either a f (t)d t 6= 0 or a f (t)d t 6= 0.
Since this is impossible, m(E) = 0. Similarily, m({x ∈ [a, b] | f (x) < 0}) = 0.
ƒ
13.10 Lemma. If f : [a, b] → R is bounded and measurable and F (x) =
a.e. on [a, b].
Rx
a
f (t)d t + F (a) then F 0 (x) = f (x)
PROOF: We know that F ∈ BV [a, b], and so F 0 exists a.e. on [a, b]. Suppose that | f | ≤ k. Let f n (x) = n(F (x +
R x+ 1
1
) − F (x)), so that f n (x) = n x n f (t)d t. Hence | f n (x)| ≤ k for all n ∈ N and x ∈ [a, b]. Since f n → F 0 a.e., the
n
Lebegue Dominated Convergence Theorem shows that
c
Z
F (x)d x = lim
n→∞
a
c
Z
0
= lim n
n→∞
f n (x)d x
a
c
Z
(F (x +
a
c+ 1n
Z
= lim n
n→∞
a
1
n
) − F (x))d x
F (x)d x − n
a+ 1n
Z
F (x)d x
a
= F (c) − F (a)
Zc
=
by the FTC since F is continuous
f (t)d t
a
Hence
Rc
a
(F 0 (t) − f (t))d t = 0 for all c ∈ [a, b]. Therefore F 0 (x) = f (x) a.e. on [a, b].
13.11 Theorem. Let f : [a, b] → R be integrable and F (x) = F (a) +
Rx
a
f (t)d t. Then F 0 (x) = f (x) a.e.
ƒ
30
MEASURE
AND I NTEGRATION
PROOF: Assume without loss of generality that f (x) ≥ 0 on [a, b]. Let
¨
f (x) if f (x) ≤ n
f n (x) =
n
if f (x) ≥ n
Rx
Then f − f n ≥ 0. Hence Gn (x) = a ( f (t) − f n (t))d t is increasing on [a, b]. Therefore Gn is differentiable and
Rx
Rx
Gn0 (x) ≥ 0 a.e. Moreover, ddx a f n (t)d t = f n (x) a.e. Hence F 0 (x) = ddx Gn + ddx a f n (t)d t, so F 0 (x) ≥ f n (x) a.e.
Rb
Rb
for all n ∈ N. Therefore F 0 (x) ≥ f (x) a.e. It follows that a F 0 (x)d x ≥ a f (x)d x = F (b) − F (a). However,
Rb
Rb
Rb
Rb
F (b) − F (a) ≥ a F 0 (x)d x, so a F 0 (x)d x = F (b) − F (a) = a f (x)d x. Thus 0 = a (F 0 (x) − f (x))d x, so
ƒ
F 0 (x) = f (x) a.e. on [a, b] since F 0 (x) ≥ f (x) a.e. on [a, b].
14
Absolutely Continuous Functions
14.1 Definition. A function f : [a, b] → R is said to be absolutely continuous on [a, b] if for every " > 0 there
n
is
i , bi )}i=1 is any finite collection of pairwise disjoint open subintervals in [a, b] with
Pna δ > 0 such that ifP{(a
n
i=1 bi − ai < δ then
i=1 |F (bi ) − F (ai )| < ".
14.2 Lemma. If f is absolutely continuous on [a, b] then it is of bounded variation.
PROOF: Assignment.
ƒ
14.3 Corollary. If f is absolutely continuous on [a, b] then it is differentiable a.e.
14.4 Lemma. If f is absolutely continuous on [a, b] and f 0 (x) = 0 a.e. then f is constant on [a, b].
PROOF: Let c ∈ [a, b] with c > a. Let E ⊆ [a, c] be such that m(E) = c − a and f 0 (x) = 0 for all x ∈ E.
Let ", η > 0 be arbitrary. For each x ∈ E we can find arbitrarily small h’s such that [x, x + h] ⊆ [a, c] and
| f (x + h) − f (x)| < ηh. Let δ > 0 be chosen from " in the definition of absolute continuity. By the Vitali Covering
Lemma there is a finite disjoint collection {[x k , yk ]} of such intervals such that the union covers E, except for a
set of measure at most δ. Without loss of generality,
Pn we may assume that x k <Pxnk+1 . We have y0 := a ≤ x 1 <
y1 ≤ x 2 <Py2 ≤ · · · ≤ x n < yn ≤ c =:
x
.
Then
n+1
k=0 | f (x k+1 ) − f ( yk )| < ". We
k=0 |x k+1 − yk | < δ. Hence
Pn
n
also have k=1 | f ( yk ) − f (x k )| ≤ η k=0 | yk − x k | ≤ η(c − a). But we can write all of this as a telescoping sum
n
n
X
X
| f (c) − f (a)| = [ f (x k+1 ) − f ( yk )] +
[ f ( yk ) − f (x k )] < " + η(c − a)
k=0
k=0
Since " and η were arbitrary, f (c) = f (a). Therefore f is constant on [a, b].
ƒ
Rx
14.5 Theorem. A function F : [a, b] → R is of the form F (x) = F (a) + a f (t)d t for some integrable function f
if and only if F is absolutely continuous.
PROOF: If F has this form then we’ve seen that F is absolutely continuous. Assume that F is absolutely continuous
on [a, b]. Then F is of bounded variation, so F (x) = F1 (x) − F2 (x), where each Fi is increasing on [a, b].
Therefore F is differentiable a.e. on [a, b] and |F 0 (x)| ≤ F10 (x) + F20 (x). It follows that
Z b
Z b
Z b
|F 0 (x)|d x ≤
a
F10 (x)d x +
a
F20 (x)d x ≤ F1 (b) − F1 (a) + F2 (b) − F2 (a)
a
Rx
This shows that F is integrable on [a, b]. Let G(x) = a F 0 (t)d t. Then G is absolutely continuous and hence so
is f := F − G. However, f 0 = F 0 − G 0 = 0 a.e. onR[a, b], so f is constant on [a, b]. But f (a) = F (a)− G(a) = F (a).
x
This shows that F (x) = F (a) + G(x) = F (a) + a F 0 (t)d t.
ƒ
0
B OUNDED LINEAR FUNCTIONALS
14.1
ON
L p (X , µ)
AND
31
C[a, b]
Measures on [a, b]
Let Meas[a, b] = {µ | µ is a finite Borel sign-measure} and define kµkMeas = |µ|([a, b]). Then we’ve seen that
(Meas[a, b], k · kMeas ) is a Banach space.
Consider BV [a, b] = { f : [a, b] → R | Vab ( f ) < ∞}. Define k f kBV = Vab ( f ) + | f (a)|. Clearly this is nonnegative, and in fact k f kBV = 0 if and only if f ≡ 0. Variation is a seminorm, so k · kBV is a norm on BV [a, b]. We
would like the consider the right continuous functions in this space. Let BVr [a, b] denote the right continuous
functions of bounded variation on [a, b]. Then BVr [a, b] is a closed subspace of BV [a, b]. It is clearly a subspace.
From the assignment, variation dominates the uniform norm, so a sequence that converges in variation converges
unformly, and uniform convergence preserves (right) continuity.
14.6 Theorem. (BV [a, b], k · k ) ∼
).
= (Meas[a, b], k · k
r
15
BV
Meas
Bounded Linear Functionals on L p (X , µ) and C[a, b]
15.1 Definition. A linear functional on a normed linear space (X , k · k) is a linear map ϕ : X → R. Denote the set
of all linear functionals on X by X # . We say that a linear functional ϕ is bounded if kϕk := sup{|ϕ(x)| : kxk ≤
1} < ∞.
If ϕ is bounded then |ϕ(x)| ≤ kϕkkxk for all x ∈ X . Let X ∗ be the set of all bounded linear functionals on X .
Then k · k is a norm on X ∗ and (X ∗ , k · k) is called the dual space of (X , k · k).
15.2 Theorem. The following are equivalent for ϕ ∈ X #
1. ϕ is bounded.
2. ϕ is continuous.
3. ϕ is continuous at a point.
15.3 Theorem. Let (X , k · k) be a normed linear space. Then (X ∗ , k · k) is a Banach space.
Problem: What is L p (X , µ)∗ ?
15.4 Lemma. If (X , A , µ) is a measure space and if 1 ≤ p ≤ ∞ with 1p + 1q = 1 then for every g ∈ L q (X , µ)
R
the map Γ g : L p (X , µ) → R defined by Γ g ( f ) = X f g dµ is a continuous linear functional on L p (X , µ). Further,
kΓ g k ≤ kgkq and if 1 < p ≤ ∞ then kΓ g k = kgkq .
PROOF: Assignment.
ƒ
If (X , µ) is σ-finite then equality holds for p = 1 as well.
15.5 Lemma. Let (X , A , µ) be a finite
R measure space and 1 ≤ p < ∞. Let g be an integrable function such
that there exists a constant M with | gϕ dµ ≤ M kϕk p for all simple functions ϕ. Then g ∈ L q (X , µ), where
1
+ 1q = 1.
p
1
PROOF: Assume that p > 1. Let ψn be a sequence of simple functions with ψn % |g|q . Let ϕn = (ψn ) p sgn(g).
R
1
1
Then ϕn is also simple and kϕn k p = ( ψn dµ) p . Since |ϕn g| ≥ |ϕn ||ψn | q = |ψn |, we have
Z
Z
Z
1
ψn dµ ≤
ϕn g dµ ≤ M kϕn k p = M
ψn dµ
p
R
Therefore ψn dµ ≤ M q . By the Monotone Convergence Theorem we get that kgkq ≤ M , so g ∈ L q (X , µ).
If p = 1 then we need to show that g is bounded almost everywhere. Let E = {x ∈ X | |g(x)| > M }. Let
1
f = µ(E)
χ E sgn(g). Then f is a simple function and k f k1 = 1. Get a contradiction.
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32
MEASURE
AND I NTEGRATION
p
15.6 Lemma. Let 1 ≤ p < ∞. Let {EnP
} be a sequence of disjoint sets. Let { f n } ⊆ LP
(X , µ) be such that f n (x) = 0
∞
∞
p
if x ∈
/ En for each n ≥ 1. Let f = n=1 f n . Then f ∈ L (X , µ) if and only if n=1 k f k pp < ∞. In this case
P
∞
k f k pp = n=1 k f k pp .
PROOF: Exercise.
ƒ
15.7 Theorem (Riesz Representation Theorem, I). Let Γ ∈ L p (X , µ)∗ , where 1 ≤ p < ∞ and µ is σ-finite.
1
then there exists a unique g ∈ L q (X , µ) such that
Then if 1p + q=1
Γ( f ) =
Z
f g dµ =: φ g ( f )
X
Moreover, kΓk = kgkq .
PROOF: Assume that µ is finite. Then every bounded measurable function S
is in L p (X , µ). Define λ : A →
∞
R : E 7→ Γ(χ E ). Let {En } ⊆ A be a sequence of disjoint sets, and let E = n=1 En . Let αn = sgnΓ(χ En ) and
P∞
P
P∞
∞
f = n=1 αn χ En . Then f ∈ L p (X , µ) and Γ( f ) = n=1 |λ(En )| < ∞ and so n=1 λ(En ) = Γ(χ E ) = λ(E).
Therefore λ is a finite signed measure. Clearly, if µ(E) = 0 then χ E = 0 a.e., so λ(E) = Γ(0) =R 0. Therefore
|λ| µ. By the Radon-Nikodym Theorem, there is an integrable function g such that λ(E) = E g dµ for all
R
E ∈ A . If ϕ is simple then Γ(ϕ) = ϕ g dµ by linearity of the integral. But |Γ(ϕ)| ≤ kΓkkϕk p for all simple
functions ϕ, so g ∈ L q (X , µ) by the lemma above. Now Γ − φ g ∈ L p (X , µ)∗ and Γ − φ g = 0 on the space of simple
functions. Since the simple functions are dense in L p (X , µ), Γ − φ g = 0 on L p (X , µ), so Γ = φ g . We have that
kΓk = kφ g k = kgkq as before.
S∞
Now assume that µ is σ-finite. We can write X = n=1 X n , where µ(X n ) < ∞ and X n ⊆ X n+1 for all n ≥ 1.
R
For each n ≥ 1, the proof above gives us g n ∈ L q (X , µ), vanishing outside X n , such that Γ( f ) = f g dµ for all
f ∈ L p (X , µ) vanishing off of X n . Moreover, kg n kq ≤ kΓk. By the uniqueness of the g n ’s, we can assume that
g n+1 = g n on X n . Let g(x) = limn→∞ g n (x). We have that |g n | % |g|. By the Monotone Convergence Theorem
Z
Z
|g|q dµ = lim
n→∞
|g n |q dµ ≤ kΓkq
Hence g ∈ L q (X , µ). Let f ∈ L p (X , µ) and f n = f χX n . Then f n → f pointwise and f n ∈ L p (X , µ) for all n ≥ 1.
Since | f g| ∈ L 1 (X , µ) and | f n g| ≤ | f g|, the Lebesgue Dominated Convergence Theorem shows
Z
Z
Z
f g dµ = lim
n→∞
f n g dµ = lim
n→∞
f n g n dµ = lim Γ( f n ) = Γ( f )
n→∞
ƒ
If p = 1 then we cannot drop the assumption of σ-finiteness.
15.8 Theorem (RRT II). Let Γ ∈ L p (X , µ)∗ , where 1 < p < ∞. Then if
g ∈ L q (X , µ) such that
Z
Γ( f ) =
1
p
+
1
q
= 1 then there exists a unique
f g dµ
for all f ∈ L p (X , µ). Moreover, kΓk = kgkq .
q
PROOF: Let
R E ⊆ X be σ-finite. Then there exists a unique g E ∈ L (X , µ), vanishing outside of E, such that
p
Γ( f ) = f g E dµ for all g ∈ L (X , µ) vanishing outside of E. Moreover, if A ⊆ E then gA = g E a.e. on A. For
R
each σ-finite set E let λ(E) = |g E |q dµ. If A ⊆ E then λ(A) ≤ λ(E) ≤ kΓkq . Let M = sup{λ(E) | E is σ-finite}.
B OUNDED LINEAR FUNCTIONALS
ON
L p (X , µ)
AND
33
C[a, b]
Let {En } be a sequence of σ-finite sets such that limn→∞ λ(En ) = M . If H =
λ(H) = M . If E is σ-finite with H ⊆ E then g E = g H a.e on H. But
Z
|g E | dµ = λ(E) ≤ λ(H) =
S∞
n=1
En then H is σ-finite and
Z
q
|g H |q dµ
so g E = 0 a.e. on E \ H. Let g = g H χH . Then g ∈ L q (X , µ) and if E is σ-finite with H ⊆ E then g E = g a.e. If
f ∈ L p (X , µ) then let E = {x ∈ X | f (x) 6= 0}. E is σ-finite and hence E1 := E ∪ H is σ-finite. Hence
Γ( f ) =
Z
f g E1 dµ =
Z
f g dµ = φ g ( f )
Therefore Γ = φ g and as before kΓk = kgkq .
ƒ
We have shown that if 1 < p < ∞ and 1p + 1q = 1 then, for any measure space (X , A , µ), L p (X , µ)∗ ∼
= L q (X , µ).
If µ is σ-finite then L 1 (X , µ)∗ ∼
= L ∞ (X , µ). What happens when p = ∞? L 1 (X , µ) ,→ L ∞ (X , µ)∗ , but this
embedding is not usually surjective. There exists a compact Hausdorff space Ω such that L ∞ (X , µ) ∼
= C(Ω). What
is C(Ω)∗ ?
Let ϕ : [a, b] → R be defined by ϕ( f ) = f (x 0 ). Then ϕ ∈ C[a, b]∗ , and kϕk = 1. Let µ x 0 the measure
Rb
on [a, b] of the point mass x 0 . If g ∈ L 1 ([a, b], m) then ϕ g ( f ) = a f g dm is a linear functional on C[a, b],
and kϕ g k ≤ kgk1 . g is the Radon-Nikodym derivative of an absolutely continuous measure µ on [a, b], and
R
R
ϕ g ( f ) = f dµ. If µ ∈ Meas[a, b] then ϕµ ( f ) = f dµ is a bounded linear functional on C[a, b], with
kϕµ k ≤ kµkMeas .
15.9 Definition. A linear functional Γ ∈ C[a, b]∗ is positive if Γ( f ) ≥ 0 for all f ∈ C[a, b] with f ≥ 0. For any
Γ ∈ C[a, b]∗ and f ≥ 0, define Γ+ ( f ) = sup{Γ(g) | 0 ≤ g ≤ f }.
Γ+ is linear. If f ∈ C[a, b] define
Γ+ ( f ) = Γ+ ( f + ) − Γ+ ( f − )
Γ− ( f ) = Γ+ ( f ) − Γ( f )
Then Γ+ , Γ− ∈ C[a, b]∗ and Γ = Γ+ − Γ− .
15.10 Lemma (Jordon Decomposition). Let Γ ∈ C[a, b]∗ . Then there exists Γ+ , Γ− ∈ C[a, b]∗ that are positive
with Γ = Γ+ − Γ− and kΓk = kΓ+ k + kΓ− k = Γ+ (1) + Γ− (1).
15.11 Theorem (RRT III). Let Γ ∈ C[a, b]∗ . Then there exists a unique finite signed measure µ on B[a, b] such
that
Z b
Γ( f ) =
f dµ
a
Moreover, kΓk = |µ|([a, b]).
PROOF: Find this in a book.
ƒ
34
16
MEASURE
AND I NTEGRATION
Product Measures
16.1 Definition. Let (X , A ) and (Y, B) be measurable spaces. A measurable rectangle is a set of the form A × B,
where A ∈ A and B ∈ B. Let Z = X × Y and
Z0 = {
n
a
Ai × Bi | Ai ∈ A , Bi ∈ B}
i=1
16.2 Lemma. Z0 is an algebra in P (Z).
Let Z be the σ-algebra generated by Z0 . We write Z = A × B. Assume that (X , A , µ) and (Y, B, λ) are
measure spaces. A measure π on (Z, Z ) is called a product measure if π(A × B) = µ(A)λ(B).
16.3 Theorem (Product Measure Theorem). Let (X , A , µ) and (Y, B, λ) be measure spaces. Then there exists
a measure π on (X × Y, A × B) such that π(A × B) = µ(A)λ(B). Moreover, if µ and λ are σ-finite then π is
unique and σ-finite.
P∞
PROOF: Suppose that A × B can be written as i=1 Ai × Bi , where each of the measurable rectangles Ai × Bi are
dijoint. Then
∞
X
χA(x)χB ( y) = χA×B (x, y) =
χAi (x)χBi ( y)
i=1
for all x ∈ X and y ∈ Y . Fix x and integrate with respect to λ.
Z
Z ∞
X
χA×B (x, y)dλ( y) =
χAi (x)χBi ( y)dλ( y)
Y
Z
Y i=1
χA(x)χB ( y)dλ( y) =
Y
χA(x)λ(B) =
∞
X
i=1
∞
X
χAi (x)
Z
χBi ( y)dλ( y)
by MCT
Y
χAi (x)λ(Bi )
i=1
P∞
Further integrating with respect to µ yeilds
(again by MCT) µ(A)λ(B) = i=1 µ(Ai )λ(Bi ) (∗)
P
n
Define π0 on Z0 by π0 (∪ni=1 Ai × Bi ) = i=1 µ(Ai )λ(Bi ). Then π0 is a measure on Z by (∗) (the only nontrivial
thing to check was countable additivity). Caratheodory’s Extension Theorem gives us a measure π defined on at
least Z that extends π0 . If µ and λ are σ-finite then π0 is σ-finite, so Hahn’s Extension Theorem tells us that π
is unique.
ƒ
16.4 Definition. Let E ⊆ Z = X × Y . An x-section of E is the set E x = { y ∈ Y | (x, y) ∈ E}. A y-section is
E y = {x ∈ X | (x, y) ∈ E}. Let f : Z → [−∞, ∞] and x ∈ X . The x-section of f is f x ( y) = f (x, y). For y ∈ Y the
y-section of f is f y (x) = f (x, y).
16.5 Lemma. If E ⊆ Z is measurable in the product measure then E x , E y are measurable in the factors for each
x ∈ X and y ∈ Y . Similarily, if f : Z → [−∞, ∞] is measurable in the product measure then f x , f y is are
measurable in each factor for every x ∈ X , y ∈ Y .
PROOF: Exercise.
16.6 Definition. A monotone class is a non-empty collection M ⊆ P (X ) such that
S∞
1. If {En }∞
n=1 ⊆ M with En ⊆ En+1 then
n=1 ∈ M .
ƒ
PRODUCT MEASURES
35
2. If {En }∞
n=1 ⊆ M with En ⊇ En+1 then
T∞
n=1
∈ M.
Every σ-algebra is a monotone class. If A ⊆ P (X ) is any collection of subsets then there is a smallest
monotone class M (A ) that contains A . Simply take the intersection of all monotone classes that contain A .
With this in mind, it is clear that M (A ) ⊆ σ(A ), the smallest σ-algebra that contains A . In fact, the reverse
inclusion also holds when A is an algebra.
16.7 Lemma (Monotone Class Lemma). If A ⊆ P (X ) is an algebra them M (A ) = σ(A).
PROOF: We need only show that M := M (A ) is an algebra, since this combined with the fact that M is closed
under countable ascending unions implies that M is closed under countable unions. For E ∈ M define M (E) =
{F ∈ M | E \ F, E ∪ F, F \ E ∈ M }. Then ∅ ∈ M (E) and E ∈ M (E). Further, if F ∈ M (E) then E ∈ M (F ) by the
symmetry of the definition. M (E) is a monotone class since complementation, union, and intersection play nicely
together.
Suppose that E ∈ A . Then since A is an algebra, A ⊆ M (E). But M (E) is also a monotone class, so
M ⊆ M (E) ⊆ M and M = M (E). It follows that A ⊆ M (F ) for every F ∈ M , and again we have M = M (F ). But
∅, X ∈ A ⊆ M (E), so this implies that M is closed under intersections and finite unions.
ƒ
16.8 Lemma. Let (X , A , µ) and (Y, B, λ) be σ-finite. If E ∈ Z = A × B, then f (x) = λ(E x ) and g( y) = µ(E y )
are measurable and
Z
Z
f dµ = π(E) =
g dλ
X
Y
PROOF: Get this from Aaron.
ƒ
16.9 Theorem (Tonelli’s Theorem). Let (X , A , µ)Rand (Y, B, λ) be σ-finite.
Let F : Z = X × Y → R[0, ∞] be
R
measurable. Then the functions defined by f (x) = Y F x dλ and g( y) = X F y dµ are measurable and X f dµ =
R
R
F
dπ
=
g dλ, where π = µ × λ. This is to say that
Z
Y
Z Z
Z
Z Z
F (x, y)dλ( y) dµ(x) =
X
Y
F dπ =
F (x, y)dµ(x) dλ( y)
Z
Y
X
PROOF: If F = χ E for some E ∈ Z = A × B then the theorem is exactly the previous lemma. If follows
immediately that the theorem holds for all non-negative measurable simple functions. If F is arbitrary,
R we can
find a sequence {Φn }∞
of
non-negative
measurable
simple
functions
such
that
Φ
%
F
.
Let
ϕ
(x)
=
(Φn ) x dλ
n
n
n=1
Y
R
y
and ψn ( y) = X (Φn ) dµ. Then ϕn and ψn are measurable and monotonic in n. By the Monotone convergence
Theorem,
lim ϕn (x) = f (x)
and
lim ψn ( y) = g( y)
n→∞
n→∞
By the Monotone Convergence Theorem again,
Z
Z
f dµ = lim
X
and similarily
R
Y
g dλ =
R
Z
n→∞
ϕn dµ = lim
Z
n→∞
X
Φn dπ =
Z
Z
F dπ
Z
F dπ.
ƒ
16.10 Theorem (Fubini’s Theorem). Let (X , A , µ) and (Y, B, λ) be σ-finite and let π = µ×λ. If F is integrable
with
respect to π on RZ = X × Y then the extended realR valued functions
defined
almost everywhere by f (x) =
R
R
R
y
F
dλ
and
g(
y)
=
F
dµ
have
finite
integrals
and
f
dµ
=
F
dπ
=
g
dλ.
That is to say,
Y x
X
X
Z
Y
Z Z
Z
Z Z
F (x, y)dλ( y) dµ(x) =
X
Y
F dπ =
Z
F (x, y)dµ(x) dλ( y)
Y
X
36
MEASURE
AND I NTEGRATION
+
−
PROOF: Since F is π-integrable, so are F + and F − . Apply Tonelli’s Theorem to establish
that
R
R f and f have
+
−
finite integrals and hence are finite a.e. Therefore f = f − f is defined a.e. and X f dµ = Z F dπ. Similarily,
R
R
we can show that g = g + − g − is defined a.e. and Y g dλ = Z F dπ.
ƒ