Soumik Sarkar
Fall, 2010
Meaure Theory and Integration
Books
• Real Analysis (Princeton Lectures in Analysis III) by E.M. Stein and R.
Sakarachi
• Principles of Mathematical Analysis by W. Rudin (Chapter 11: The
Lebesgue Theory)
• Linear Operator Theory in Engineering and Science by Naylor and Sell
(Appendix D, especially D.10 approach)
• Real and Complex Analysis by W. Rudin (Advanced reference)
Topics
• Notion of Lebesgue Measure
• Measurable Functions
• Lebesgue Integration
• Convergence theorems
• Lp spaces
** The lectures will be given in a way that will complement the primary
course material (especially to Chapters: 1,3,4,5). Although some theorems
will be stated without proof, rough proof sketches will be provided.
1
1
Notion of Lebesgue Measure
** Ideas of open sets, closed sets, limit points etc. that are covered in class,
Chapter 1, will be used here.
Exterior Measure: A (closed) rectangle R in Rd is given by the product of
d one-dimensional closed and bounded intervals
R = [a1 , b1] × [a2 , b2] × · · · × [ad , bd ]
where, aj ≤ bj ∈ R for j = 1, 2, · · · , d.
So, lengths of the sides of R are b1 − a1 , · · · , bd − ad and volume of R, |R| =
(b1 − a1 ) · · · (bd − ad ). If b1 − a1 = · · · = bd − ad = l, then we call the rectangle
a cube Q ⊂ Rd and |Q| = ld .
Lemma 1.1. If a rectangle is the almost disjoint union of finitely many
PN
other rectangles, say R = ∪N
R
,
then
|R|
=
k=1 k
k=1 |Rk |.
The proof is intuitive. (See Stein and Sakarachi Lemma 1.1)
** A union of rectangles is said to be almost disjoint if the interiors of the
rectangles are disjoint.
Corollary 1.1. If R, R1, · · · , RN are rectangles and R ⊂ ∪N
k=1 Rk , then |R| ≤
PN
k=1 |Rk |.
Theorem 1.1. Every open set O ⊂ Rd , d ≥ 1, can be written as a countable
union of almost disjoint closed cubes.
Proof. (by construction, see Fig. 1)
• Procedure is repeated indefinitely
• Resulting collection of almost disjoint cubes, Q is countable.
• For a given x ∈ O, ∃ a cube of side length 2−N , N ∈ N such that it
contains x, and is entirely contained in O. This implies that union of all
cubes in Q covers O.
** Remember this construction is the primary notion of Lebesgue’s theory. It
will be used for many future proofs, including for the basic idea of Lebesgue
integration.
2
F – Fully Inside
O
P P
F F
F F
F F
O P F
O O P
O O P
O O P
P
O P
P F
P P
O
O
O
O
P
P
F
F
F
F
F
P
P – Partially Inside
O – Outside
O O
P O
P P
F P
F P
O
F P
F P
P O
O
O
O
P P
F F
F F
F F
O P F
O O P
O O P
O O P
P
O P
P F
P P
O
P
P
F
F
F
F
F
P
O O
P O
P P
F P
F P
…..
F P
F P
P O
Step 2
Step 1
Figure 1: Construction to show open set as a countable union of almost disjoint closed cubes
Definition 1.1. (Exterior Measure) Let E be any subset of Rd , then exterior
(outer) measure of E is
m∗ (E) = inf {
∞
X
j=1
|Qj |}
where the infimum is taken over all countable coverings of E ⊂ ∪∞
j=1 Qj by
closed cubes.
Properties:
1. m∗ (E) ∈ [0, ∞]
2. m∗ (φ) = 0; (Also note that external measure of a point is 0)
3. Monotonicity: If E1 ⊂ E2, then m∗ (E1) ≤ m∗ (E2). Every bounded
subset of Rd has finite exterior measure.
P∞
4. Countable sub-additivity: If E = ∪∞
j=1 Ej , then m∗ (E) ≤
j=1 m∗ (Ej ).
P
∞
Proof sketch: For m∗ (Ej ) < ∞, Ej ⊂ ∪∞
k=1 Qk,j and
k=1 |Qk,j | ≤
ǫ
∞
m∗ (Ej ) + 2j . then, E ⊂ ∪j,k=1Qk,j implies
m∗ (E) ≤
X
j,k
|Qk,j | =
∞ X
∞
X
j=1 k=1
∞
∞
X
X
ǫ
|Qk,j | ≤
[m∗ (Ej )+ j ] =
m∗ (Ej )+ǫ∀ǫ > 0
2
j=1
j=1
5. If E ⊂ Rd , then m∗ (E) = inf m∗ (O), where infimum is taken over all
open sets O containing E.
3
Proof sketch: m∗ (E) ≤ inf m∗ (O), by monotonicity and m∗ (E) ≥
inf m∗ (O) by Theorem 1.1 and Property 4. (See Stein and Sakarachi)
Note:
• To define m∗ (E), finite sums would not work (in general, volume of a finite
cover is larger than m∗ (E)). Again, uncountable cover is not tractable;
therefore, countably infinite cover is so important (brilliant contribution
by Lebesgue and others, used everywhere - Integration, Fourier series,
Probability theory etc.).
• One can replace cubes with rectangles or balls.
Definition 1.2. (Lebesgue Measurable Set) Let E be a subset of Rd . E is
called a Lebesgue measurable set if ∀ǫ > 0, ∃ an open set O s.t. E ⊂ O and
m∗ (O\E) ≤ ǫ.
Note, Property 5 holds for all sets, but this condition is more restrictive
(definition of infimum).
Let the collection of all such sets be L. If E ∈ L, then m(E) = m∗ (E), where
m is the Lebesgue measure, i.e., if E is Lebesgue measurable (or simply measurable), then the Lebesgue measure of E is its exterior measure. (Lebesgue
measure will be defined formally later)
Properties:
• Every open set in Rd is measurable (from definition)
• If m∗ (E) = 0, then E is measurable. In particular, if F is a subset of a
set of exterior measure 0, then F is measurable.
Proof: By Property 5 of exterior measure, ∀ǫ > 0, ∃ an open set O such
that E ⊂ O and m∗ (O) ≤ ǫ. Since (O − E) ⊂ O, we have m∗ (O\E) ≤ ǫ.
• A countable union of measurable sets is measurable, i.e., if Ej ∈ L, j ∈ N,
then ∪∞
j=1 Ej = E ∈ L.
Proof: Let, ∀j, Ej ⊂ Oj and m∗ (Oj − Ej ) ≤ 2ǫj . Now, O = ∪∞
j=1 Oj is
∞
open and also E ⊂ O. Moreover, (O − E) ⊂ ∪j=1(Oj − Ej ). Therefore,
by Monotonicity and Sub-additivity, we have
m∗ (O\E) ≤
∞
X
j=1
4
m∗ (Oj \Ej ) ≤ ǫ
E
Figure 2: Construction to prove that complement of measurable set is measurable
• Closed sets are measurable.
Proof sketch: Let F be a closed set and O be an open set with F ⊂ O
and m∗ (O) ≤ m∗ (F ) + ǫ. Open set (O\F ) = ∪∞
j=1 Qj , where Qj s are
almost disjoint cubes. Now it can be shown that,
∞
X
j=1
m∗ (Qj ) ≤ m∗ (O) − m∗ (F ) ≤ ǫ
and this implies, m∗ (O\F ) ≤
Stein and Sakarachi)
P∞
j=1 m∗ (Qj )
≤ ǫ. (For rigorous proof, see
• The complement of a measurable set is measurable.
Proof: (see Fig. 2) Let E ∈ L, ∀n ∈ N, ∃ an open set On with E ⊂ On
and m∗ (On \E) ≤ n1 . Since Onc is closed, we have Onc ∈ L, which implies
c
c
c
S = ∪∞
n=1 On ∈ L. Now, S ⊂ E and (E \S) ⊂ (On \E) ∀n. Therefore,
m∗ (E c\S) ≤ n1 ∀n ∈ N . Thus,
m∗ (E c\S) = 0 ⇒ E c \S ∈ L ⇒ E c = S ∪ (E c\S) ∈ L
(Note, S ∈ L and (E c\S) ∈ L)
• A countable intersection of measurable sets is measurable.
∞
c c
Proof: We observe, ∩∞
j=1 Ej = (∪j=1 Ej ) .
• If E1, E2, · · · are disjoint measurable sets and E = ∪∞
j=1 Ej , then m(E) =
P∞
j=1 m(Ej ).
5
Proof sketch: Prove first for bounded Ej s, then convert the sum ∪∞
j=1 Ej
to ∪∞
j=1 Ej,k where Ej,k is bounded. (See Stein and Sakarachi for the
detailed proof)
Definition 1.3. (σ-algebra) A σ-algebra of sets is a collection of subsets of
Rd that is closed under countable unions, countable intersections and complements.
Example: L is a σ-algebra. i) φ ∈ L; ii) If E ∈ L, then E c ∈ L; iii) If
E1, E2, E3, · · · ∈ L, then ∪∞
i=1 Ei ∈ L.
Definition 1.4. (Borel σ-algebra) Borel σ-algebra in Rd is by definition, the
smallest σ-algebra in Rd containing all open sets. Elements of Borel algebra
(BRd ) are called the Borel sets.
** Smallest: If S is a σ-algebra containing all open sets in Rd , then (BRd ) ⊂ S.
Hence, (BRd ) is the intersection of all σ-algebras that contain the open sets.
** Open sets are measurable. Hence, (BRd ) ⊆ L. But is (BRd ) = L? The
answer is NO!! In fact, completion of (BRd ) is L.
** Recall the definition of measurable sets and Property 5 of exterior measure. So, are there any non-measurable sets? The answer is YES. Non-trivial
example would be Vitali Set. See Stein and Sakarachi for construction of
such a set.
Definition 1.5. (Lebesgue measure) Let X be a set, A be a σ-algebra on X.
Then the Lebesgue measure µ on (X, A) is a function: µ : A → [0, ∞].
Properties:
• µ(φ) = 0
• If Ej ∈ A for j ∈ N and are disjoint, then
µ(∪∞
j=1 Ej )
=
∞
X
µ(Ej )
j=1
→ (X, A) is called a measurable space.
→ (X, A, µ) is called a measure space.
** Finite Measure: (X, A, µ) is called a finite measure space if µ(X) < ∞
(and µ is called a finite measure).
6
** σ-finite measure: (X, A, µ) is called a σ-finite measure space, if X is
the countable union of measurable sets of finite measure, i.e., if X = ∪∞
j=1 Xj
and Xj ∈ A and µ(Xj ) < ∞ (and µ is called a σ-finite measure).
** Counting measure: Counting measure of any finite set is the number
of elements in the set and counting measure of an infinite set is ∞.
Examples:
1. X = R and µ is Lebesgue measure. Therefore, X = ∪k∈Z [k, k + 1] and
µ[k, k + 1] = 1. Hence, µ is σ-finite; but not finite as µ(X) = ∞. Also,
counting measure of X is ∞.
2. X = N and µ is Counting measure, with σ-algebra, A = P(N). We
have, N = ∪n∈N {n}. Hence, µ is σ-finite as µ({n}) = 1; but not finite as
µ(X) = ∞.
3. Probability Space: (Γ, G, P )
• Γ is the sample space, i.e., set of all outcomes (the Set)
• G is the event space, i.e., collection of all sets of outcomes to which
we can assign probability (the σ-algebra)
• P is the Probability measure, i.e., a function P : G → [0, 1], satisfying
the following axioms: (i) P (φ)
0, P (Γ) = 1, (ii) If E1 , E2, · · · ∈ G
P=
∞
∞
are disjoint then P (∪1 Ei) = 1 P (Ei) (the Measure)
Example 1: (Double Coin Flip) Flip a coin twice in succession.
Γ = {(H, H), (H, T ), (T, H), (T, T )}
G = P(Γ) = {φ, {(H, H)}, · · · , {(H, H), (H, T )}, · · · , Γ} (the entire Power
set)
Assuming fairness of the coin, what is the probability that the second
coin flip comes up tails?
P ({(H, T ), (T, T )}) = P (H, T ) + P (T, T ) = 0.5
Example 2: (Gaussian Random Variable) Generate y ∈ R from the following probability density function:
1
2
2
e−(y−µ) /2σ
p(y) = √
2πσ 2
Now, Γ = R, but G =
6 P(R), instead G = B(R), where B(R) is the Borel
σ-algebra on R, basically excludes non-measurable sets (non-nice sets on
which we can not assign probability)
7
What is the probability that the generated random number y is less that
µ?
Rµ
P (y < µ) = −∞ p(y)dy = 1/2
2
Measurable Functions
Definition 2.1. (Measurable functions) A function f defined on a measurable
subset E of Rd is measurable if ∀a ∈ R, the set
f −1([−∞, a)) = {x ∈ E : f (x) < a}
is measurable.
** −∞ can be excluded to define finite valued measurable functions.
Equivalent definitions: (by using properties of measurable sets)
• f is measurable iff {x ∈ E : f (x) ≤ a} ∈ L ∀a ∈ R.
Proof. ⇒ We can write the following:
1
{x : f (x) ≤ a} = ∩∞
k=1 {x : f (x) < a + }
k
Also, countable intersection of measurable sets is measurable. Done!!
⇐ We can write the following:
1
{x : f (x) < a} = ∪∞
k=1 {x : f (x) ≤ a − }
k
Also, countable union of measurable sets is measurable. Done!!
• f is measurable iff {x ∈ E : f (x) > a} ∈ L ∀a ∈ R.
Proof. Note, the set {x ∈ E : f (x) > a} is complement of the set {x ∈ E :
f (x) ≤ a}. Hence, f is measurable iff {x ∈ E : f (x) ≤ a} is measurable
∀a ∈ R.
Corollary 2.1. If f is measurable, then −f is measurable.
• For finite valued functions, f is measurable iff {x : a < f (x) < b} is
measurable ∀a, b ∈ R. By using previous arguments, f is measurable
iff f −1((a, b]) ∈ L; iff f −1([a, b)) ∈ L; iff f −1([a, b]) ∈ L
8
Formally, a finite valued function f is measurable iff f −1(O) is measurable
for every open set O, and iff f −1(F ) is measurable for every closed set F .
** If f −1(∞), f −1(−∞) are measurable sets, the definition can be extended for extended valued functions.
Properties:
1. If f is continuous on Rd , then f is measurable. If f is measurable and
finite valued and φ is continuous, then φ ◦ f is measurable.
Proof. Recall the topological definition of continuity ⇒ If inverse image
of every open set is open, then the function is continuous. Hence, f is
measurable if it is continuous on Rd .
Let φ is continuous, then φ−1((−∞, a)) is an open set, say O. This
implies:
(φ ◦ f )−1((−∞, a)) = f −1(O)
But, f is measurable, so f −1(O) is a measurable set. Hence, φ ◦ f is
measurable.
2. Let {fn}∞
n=1 be a sequence of measurable functions. Then supn fn (x),
limn→∞ sup fn (x), inf n fn(x), limn→∞ inf fn(x) are measurable.
Corollary 2.2. If lim n → ∞fn (x) = f (x), then f is measurable.
Background: Introduction to sequence of functions can be found in
Chapter 7, Principles of Mathematical Analysis, 3rd Ed.
Supremum or Infimum of a function can be defined similarly as for a
sequence. For example:
{x : sup fn (x) > a} = ∪∞
n=1{x : fn (x) > a}
n
Proof. ⇒ Suppose, fn (x) > a for some n, then supn fn(x) > a. So,
∪n∈N {x : fn(x) > a} ⊆ {x : sup fn (x) > a}
⇐ If supn fn(x) > a, then ∃k s.t. fk (x) > a, i.e., x ∈ {y : fk (y) > a}
implies
∪n∈N {x : fn(x) > a} ⊇ {x : sup fn (x) > a}
9
** Similarly, {x : inf n fn(x) > a} = ∩∞
n=1 {x : fn (x) > a}
** inf n fn (x) = − supn (−fn(x)) (understand intuitively)
** What is limn→∞ sup fn (x)?
Let,
h1 (x) = sup{fk (x)|k ≥ 1}
h2 (x) = sup{fk (x)|k ≥ 2}
.
.
Hence, h1 (x) ≥ h2 (x) ≥ h3 (x) ≥ · · · . Therefore,
lim sup fn (x) = lim hn (x) = inf hn (x)
n→∞
n→∞
n
Or in other words,
∞
lim sup fn (x) = inf sup fk (x) = ∩∞
n=1 ∪k=n fk (x)
n→∞
n k≥n
∞
** Similarly, limn→∞ inf fn (x) = supn inf k≥n fk (x) = ∪∞
n=1 ∩k=n fk (x).
** For any sequence xn, limn→∞ sup xn ≥ limn→∞ inf xn ; and, we have
limn→∞ xn = y iff limn→∞ sup xn = limn→∞ inf xn = y.
Proof. Let, yn = inf k≥n xk ; zn = supk≥n xk .
Hence, limn→∞ inf xn = supn yn ; limn→∞ sup xn = inf n zn .
Note, ∀n yn ≤ zn . Also, if n ≤ m, then yn ≤ ym and zn ≥ zm .
Therefore, ∀m, n zm ≥ yn and this implies that
limn→∞ sup xn ≥ limn→∞ inf xn .
⇒ Let limn→∞ xn = y. Then, ∀ǫ > 0, ∃ an N s.t. |xn − y| < ǫ ∀n ≥ N .
Hence, supk≥n xk ≤ y+ǫ that implies limn→∞ sup xn ≤ y+ǫ or, limn→∞ sup xn ≤
y. Similarly, limn→∞ inf xn ≥ y.
These imply limn→∞ sup xn ≤ limn→∞ inf xn. But, we proved otherwise
earlier. Therefore, limn→∞ sup xn = limn→∞ inf xn .
⇐ Let limn→∞ sup xn = limn→∞ inf xn = y.
Suppose, limn→∞ xn 6= y, i.e., ∃ǫ > 0 s.t. ∀N , there is n ≥ N s.t.
|xn − y| ≥ ǫ, i.e., xn ≥ y + ǫ or xn ≤ y − ǫ and there are infinitely many
n s for which these relations are possible.
Now, xn ≥ y +ǫ for infinitely many n s implies limn→∞ sup xn ≥ y +ǫ. On
the other hand, xn ≤ y −ǫ for infinitely many n s implies limn→∞ inf xn ≤
y − ǫ. But these are contradictions.
10
Proof. (of Prop. 2) Let Lf be the collection of measurable functions. We
can write,
{x : sup fn (x) > a} = ∪∞
n=1{x : fn (x) > a}
n
Now, {x : fn (x) > a} ∈ Lf ∀n and countable union of measurable
functions is measurable.
→ We can write inf n fn (x) = − supn (−fn(x)) and we know that −fn (x) ∈
Lf and supn fn(x) ∈ Lf . Therefore, inf n fn(x) ∈ Lf .
→ We can write limn→∞ sup fn (x) = inf n supk≥n fk (x). We already proved
that both supn fn (x) and inf n fn (x) are elements of Lf , when fn (x) ∈ Lf .
Therefore, limn→∞ sup fn(x) ∈ Lf . Similarly, limn→∞ inf fn (x) ∈ Lf .
→ If limn→∞ fn(x) = f (x), then,
lim sup fn (x) = lim inf fn (x) = f (x)
n→∞
n→∞
So, f (x) ∈ Lf .
3. If f and g are measurable, then (i) f k , k ∈ N are measurable; (ii) f + g
and f g are measurable if both f and g are finite valued.
Proof. (i) For k odd,
{x : f k (x) > a} = {x : f (x) > a1/k }
For k even,
{x : f k (x) > a} = {x : f (x) > a1/k } ∪ {x : f (x) < −a1/k }
So, f k ∈ Lf .
(ii) We can write,
{x : f (x) + g(x) > a} = ∪r∈Q {x : f (x) > a − r} ∩ {x : g(x) > r}
where, Q is the set of rationals. Both terms on the r.h.s are elements of
Lf , hence, f + g is measurable. Finally,
1
f g = [(f + g)2 + (f − g)2 ]
4
We use the previous results to prove that f g ∈ Lf .
4. Let f is measurable and f (x) = g(x) a.e., then g is measurable.
11
Proof. If µ{x ∈ E : f (x) 6= g(x)} = 0, then we say, f (x) = g(x) a.e.
f is measurable then {x : f (x) > a} is a measurable set, but {x : g(x) >
a} and {x : f (x) > a} differs by a set of measure zero. Hence, g is also
measurable.
Relaxations:
** Let limn→∞ fn(x) = f (x) a.e. where fn is a sequence of measurable functions (n ∈ N), then f is measurable.
** Even if f and g are finite valued almost everywhere, f + g and f g are
measurable as, f + g and f g are only defined on the intersection of domains
of f and g and union of two sets of measure zero has measure zero.
3
Lebesgue Integration
Riemann vs. Lebesgue Integral
• Let f be a bounded real valued function defined on compact interval
[a, b]. Then f is Riemann integrable if and only if D = {x ∈ [a, b] :
f is not continuous at x} has zero Lebesgue measure.
• If f is a bounded and continuous function that is Riemann integrable,
then its Lebesgue integral coincides with Riemann integral.
Case Study 1: Let us consider the following function for integration in the
real interval [a, b]:
q(t) = 1 when t ∈ Q
q(t) = 0 otherwise
In the sense of Riemann, the upper integral is (b − a) and the lower integral
is 0. Hence, the Riemann integral does not exist. Note, this function is no
where continuous.
Case Study 2: Let us consider a Cantor-like function. Following Fig. 3 ,
let, fn = F1 · F2 · · · Fn with l1 + 2l2 + · · · + 2k−1lk < 1 and Fk is continuous
∀k ∈ N. Now,
• 0 ≤ fn (x) ≤ 1 ∀x
• fn ≥ fn+1
12
F1
1
0
1
l1
1
F2
0
l2
l2
1
Figure 3: Construction of Cantor-like function for Case Study 2
• Let limn→∞ fn = f ; where f is no where continuous in C, where C = ∩Ck ,
Ck s are domains after removal of lk s.
R1
Hence, although fn s are Riemann integrable, f is not. So, 0 f (x)dx =? and
R1
R1
is 0 f (x)dx = limn→∞ 0 fn (x)dx ? i.e., can we switch limit and integral?
Furthermore, if in a sequence fn s are continuous or differentiable, is that
true for limit functions? Concepts of Lebesgue integration and corresponding
convergence theorems will help us answer these questions.
Definition 3.1. (Characteristic function) Characteristic function χE of a set
E is defined as
1
if x ∈ E
χE (x) =
0
if x ∈
/E
Definition 3.2. (Simple function) A function f is known as a simple function, if it can be written as
f=
N
X
ak χEk
k=1
where, N is finite, ak ∈ R are constants and each Ek is a measurable set with
µ(Ek ) < ∞.
Step 1: Let (X, M, µ) be a measure space and f be a (non-negative) simple
P
function, i.e., f = N
i=1 ai χEi with Ei ∈ M and ai ≥ 0.
13
f (x)
f (x)
I
x
f 1(I)
x
Figure 4: Illustrations of Riemann and Lebesgue Integral, note, we partition the ordinate for Lebesgue
integral instead of abscissa as done in Riemann Integral
The Lebesgue integration of f is defined as
Z
N
X
f dµ =
ai µ(Ei )
X
i=1
Note that for µ(Ei) to exist ∀i, f has to be a measurable function, i.e.,
f −1(I) ∈ M.
Properties:
1. Linearity: If ϕ and ψ are simple functions and a, b ∈ R, then (from
definition)
Z
Z
Z
(aϕ + bψ) = a ϕ + b ψ
2. Additivity: If E and F are disjoint subsets of Rd with finite measure,
then
Z
Z
Z
ϕ=
ϕ+
ϕ
E∪F
E
F
Note that χE∪F = χE + χF .
R
R
3. Monotonicity: If ϕ ≤ ψ are simple functions, then ϕ ≤ ψ.
R
Let, η = ψ − ϕ, then η ≥ 0 which implies η ≥ 0.
R
R
Also, | ϕ| ≤ |ϕ|
R
4. Let ν(A) = A f dµ, then ν : M 7→ [0, ∞] is a measure. (prove it !!)
14
Step 2: Non-negative measurable functions f . Lets define the space of such
functions as
L+ = {f : X → [0, ∞]| f is M-measurable}
the (extended) Lebesgue integral of f is defined as,
Z
Z
f (x)dx = sup g(x)dx
g
where, the supremum is taken over all measurable functions g s.t. 0 ≤ g ≤ f ,
and where g is bounded and supported on a set of finite measure. This is
a general enough definition and it can be understood
that the integral will
R
either be finite or infinite. When finite, i.e., f (x)dx < ∞, f is called
Lebesgue integrable or simply integrable. The reason of involving g is a bit
technical that is discussed below.
Definition 3.3. A function g is called a bounded function, supported on a
finite measure, if
• g is bounded
• The support of a measurable function f is defined to be the set of all
points where f is not zero, i.e.,
supp(f ) = {x : f (x) 6= 0}
Also, f is said to be supported on a set E, if f (x) = 0 ∀x ∈
/ E, and as f is
measurable, so is its support. In the present case, µ(supp(g)) < ∞
Theorem 3.1. Let f be a bounded function (bounded by M) and is supported
on a set E, then there exists a sequence {ϕn } of simple functions with each
ϕn bounded by M and supported on E, and such that ϕn(x) → f (x) ∀x.
(** The result also holds for infinite valued functions)
Intuitive proof sketch:(See Fig. 5)
ϕk (x) ≤ ϕk+1(x);
lim ϕk (x) = f (x) ∀x
k→∞
Theorem 3.2. Let f be a bounded function supported on a set E of finite
measure. If {ϕn }∞
n=1 is any sequence of simple functions bounded by M,
supported on E, and with ϕn (x) → f (x) for a.e. x, then:
• limn→∞ ϕn exists
15
f (x)
f (x)
! k 1 ( x)
! k (x)
x
Figure 5: Basic idea behind the concept of approximating a bounded function by a sequence of simple
functions at the limiting point
• If f = 0 a.e. then limn→∞ ϕn = 0
See Stein & Sakarachi for proof.
Finally, if g is a bounded function that is supported on a set of finite measure,
its Lebesgue Integral is defined as,
Z
Z
ϕn (x)dx
g(x)dx = lim
n→∞
where, {ϕn } is any sequence of simple functions such that |ϕn | ≤ M, each ϕn
is supported on the support of f and
lim ϕn (x) = f (x) a.e. x
R
** It can be proved that, f is independent of the limiting sequence {ϕn}
used and
Z
Z
f (x)dx = sup g(x)dx
n→∞
g
where, f is measurable, non-negative but not necessarily bounded, i.e., f can
take the value ∞ on a set of finite measure.
Note, the supremum is taken over all measurable functions g, s.t. 0 ≤ g ≤ f ,
and where g is bounded and supported on a set of finite measure. The
implications of this definition are:
→ If f is bounded too, then the situation is trivial.
→ If fR is ∞ over a set of measure zero, still g can be equal to f a.e.; in that
case, f can still be finite.
→ But if f is ∞ over a set of finite measure, then
g can never actually model
R
f (in almost everywhere sense). Hence, supg g(x)dx has to be ∞.
16
Properties:
R
R
R
1. Linearity: For f, g ≥ 0 and a, b ∈ R+ , we have (af +bg) = a f +b g.
d
2. Additivity:
R
R E, F
R are disjoint subsets of R , and f ≥ 0, then we have
E∪F f = E f + F g.
R
R
3. Monotonicity: If 0 ≤ f ≤ g, then f ≤ g.
4. If h is integrable and 0 ≤ f ≤ h, then f is integrable.
5. If f is integrable, then f (x) < ∞ a.e.
Proof Rsketch:R Let Ek = {x : f (x) ≥ k} and E∞ = {x : f (x) = ∞}.
Then, Rf ≥ χEk f ≥ kµ(Ek ). This implies µ(Ek ) → 0 as k → ∞
because f < ∞, i.e., Ek ց E∞ . This implies µ(E∞ ) = 0.
R
6. If f = 0, then f (x) = 0 a.e.
** Proof uses the Bounded Convergence Theorem (BCT).
Theorem 3.3. (Bounded Convergence Theorem) Suppose {fn } is a sequence
of measurable functions that are all bounded by M, supported on a set E of
finite measure and fn(x) → f (x) a.e. as n → ∞. Then f is measurable,
bounded, supported on E for a.e. x and
Z
|fn − f | → 0 as n → ∞
R
R
Moreover, as | f | ≤ |f |, we can say,
Z
Z
fn → f as n → ∞
** See Stein & Sakarachi for proof.
As an observation, we can deduce from the above
R that if f ≥ 0 is bounded
and supported on a set of finite measure E and f = 0, then f = 0 a.e.
By using the above definition of Lebesgue integral, we can now evaluate the
following integral:
Z
1Q dx = µ(Q ∩ [0, 1]) = 0
[0,1]
Let us now look at another related example:
Example 3.1. Let, for n ∈ N
1
n
for 0 < x < n
0
otherwise
R
Therefore, f (x) = limn→∞ fn (x) = 0 a.e. However, fn(x) = 1 ∀n
fn (x) =
17
The general result is given by Fatou’s Lemma.
Theorem 3.4. (Fatou’s Lemma) Suppose {fn} is a sequence of measurable
functions with fn ≥ 0. If limn→∞ fn(x) = f (x) for a.e. x, then
Z
Z
Z
Z
f ≤ lim inf fn or
lim inf fn ≤ lim inf fn
n→∞
n→∞
n→∞
Proof. Suppose, 0 ≤ g ≤ f , where g is bounded and supported on a set
E of finite measure. Let, gn (x) = min(g(x), fn(x)). Then gn is measurable,
R
R
bounded, supported on E and gn (x) → g(x) a.e. Hence, by BCT, gn → g.
By construction, we also have gn ≤ fn .
We just defined fn as supremum over all functions of the type gn , that is in
turn depends on Rthe choice
R of g (so necessarily supremum over all choices of
g). This implies gn ≤ fn. Therefore,
Z
Z
lim inf gn ≤ lim inf fn
n→∞
n→∞
By BCT, we can push the lim inf inside the integral for the term on left hand
side, but we can’t do the same for the right hand side term. Therefore,
Z
Z
lim inf gn ≤ lim inf fn
n→∞
n→∞
Z
Z
or,
lim gn ≤ lim inf fn
n→∞
n→∞
Z
Z
or, g ≤ lim inf fn
n→∞
But, g is any function satisfying 0 ≤ g ≤ f and it is bounded,
measurable,
R
R
supported
on
a
finite
measure
set.
Also
our
definition
of
f
says
that
f=
R
R
supg g. So, the above inequality will hold for f as well, i.e.,
Z
Z
f ≤ lim inf fn
n→∞
The fact of the matter is whenever
be equal to ∞ as well.
R
f = supg
R
g = ∞, limn→∞ inf
R
fn will
Theorem 3.5. (Monotone Convergence Theorem) Let {fn } be a sequence of
non-negative measurable functions s.t. fn ≤ fn+1 ∀n ∈ N, then,
Z
Z
lim
fn =
lim fn
n→∞
n→∞
18
Proof. We have
R
fn ≤
R
f ∀n ∈ N, where f = limn→∞ fn. Hence,
Z
Z
lim sup fn ≤ f
n→∞
Combining this inequality with that of Fatou’s lemma, we have,
Z
Z
Z
lim
fn = f =
lim fn
n→∞
n→∞
Step 3: Let f : X → R̄ be M-measurable, then we split up f as f = f + −f −,
where f +(x) = max(f (x), 0); f −(x) = max(−f (x), 0). Note, both f +, f −
are non-negative and measurable,
also, |f | ≥ f ±. We define f to be Lebesgue
R
integrable or (∈ L1) iff |f | < ∞. So, f +, f − are integrable whenever f is.
Finally, the Lebesgue integral of f is
Z
Z
Z
+
f = f − f−
Again, this integral is linear, additive, monotonic and satisfies |
Also, L1 is a real vector space.
R
f| ≤
R
|f |.
When f : X → C,R f is measurable
ifR both Re(f ) and Im(f ) are measurR
able and we have, f = Re(f ) + i Im(f ). Also, Re(f ) and Im(f ) are
integrable whenever f is.
Theorem 3.6. (Dominated Convergence Theorem) Let {fn} be a sequence
of complex valued measurableR functions such
R that fn (x) → f (x) a.e. and
1
|fn | ≤ g, where g ∈ L , then f = limn→∞ fn .
Proof. (Special case) Assume fn : X → R̄, f : X → R̄ and |fn | ≤ g.
So, −g ≤ fn ≤ g which implies g + fn ≥ 0 and g − fn ≥ 0. Now,
Z
Z
Z
Z
Z
g+ f =
(g + f ) = lim(g + fn ) = lim inf(g + fn)
Z
Z
Z
≤ lim inf (g + fn) = lim inf[ g + fn]
Note the application of Fatou’s lemma above.
Fact: If an → a, then lim inf(an + bn) = lim an + lim inf bn.
R
R
R
R
Therefore, g + f ≤ g + lim inf fn which implies (as g ∈ L1)
Z
Z
f ≤ lim inf fn
∗
19
Also, for every sequence, lim sup fn ≥ lim inf fn. ∗∗
Similarly,
Z
Z
(g − f ) ≤ lim inf (g − fn )
Z
Z
Z
or, − f ≤ lim inf −fn = − lim sup fn
Z
Z
or, f ≥ lim sup fn
∗ ∗∗
By ∗, ∗∗ and ∗ ∗ ∗, we have
or,
4
Z
Z
f = lim
Z
fn
lim fn = lim
Z
fn
n→∞
n→∞
n→∞
Lp Spaces
Basic definitions related to Lp spaces are omitted here. In class material
(Chapter-3), completion of lp spaces has been shown. In this advanced material, completion of Lp spaces has been proved along with necessary background. Although the proof is somewhat lengthy, it is worth going through
as it uses some of the important concepts introduced before including that
of the convergence theorems.
Theorem 4.1. (Completeness of Lp spaces) If 1 ≤ p ≤ ∞, then Lp space is
a Banach space.
Background: Let {fn } be a sequence in Lp , if limn→∞ ||fn − f ||p = 0 and
f ∈ Lp, we say that {fn} converges to f in Lp .
Completeness: If ∀ǫ > 0, ∃N ∈ N, s.t. ||fn − fm ||p < ǫ ∀n, m > N , we call
{fn} a Cauchy sequence in Lp . Furthermore, if every Cauchy sequence in Lp
converges to an element of Lp, we call Lp a complete metric space.
Summability: A sequence {fn } in a normed linear space is said to be
P
summable to a sum s, if s is also in the space and ||s − ni=1 fi|| → 0.
P
In this case, we write s = ∞
i=1 fi . Furthermore, we call {fn } absolutely
P∞
summable if n=1 ||fn || < ∞.
20
Fact: For a sequence of real numbers, absolute summability implies summability. However, it is not true in general for sequence of elements in a normed
linear space.
Proposition 4.1. A normed linear space X is complete iff every absolutely
summable series is summable.
Proof. ⇒ Let X be complete and {fn } be an absolutely summable sequence
P
of elements of X. Since, ∞
n=1 ||fn || = M < ∞, for each ǫ > 0 we have:
∃N ∈ N s.t.
∞
X
n=N
||fn || < ǫ
Pn
Let sn =
i=1 fi be the partial sum of the sequence {fn } (Note that the
equality for n = ∞ was defined in norm sense). Then for n ≥ m ≥ N , we
have
n
∞
∞
X
X
X
||fi || < ǫ
||sn − sm || = ||
fi || ≤
||fi || ≤
i=m
i=m
i=N
Hence, sequence {sn } is a Cauchy sequence and must converge to an element
s in X as X is complete.
⇐ Let {fn} be a Cauchy sequence in X. For each integer k, there is an
integer nk s.t. ||fn − fm|| < 2−k ∀n, m > nk and also we may choose nk s s.t.
n1 < n2 < n3 < · · · . Then {fnk } is a subsequence of {fn} for k = 1, 2, · · · , ∞
and we set
g1 = fn1
gk = fnk − fnk−1 for k > 1
P
We obtain a sequence {gk } whose k th partial sum ki=1 gi = fnk .
For k > 1, we have ||gk || ≤ 2−(k−1). Therefore,
∞
X
k=1
||gk || ≤ ||g1 || +
∞
X
k=2
2−k+1 = ||g1 || + 1
Thus the sequence
P∞ {gk } is absolutely summable and hence there is an element
f in X, s.t. k=1 gk = f . In other words, subsequence {fnk } converges to f .
Now it remains to be shown that f = limn→∞ fn.
{fn} is a cauchy sequence, hence given ǫ > 0, ∃ N s.t. ||fn − fm || <
ǫ/2 ∀n, m > N . Since, fnk → f , ∃ a K s.t. ∀k > K, we have ||fnk − f || < ǫ/2.
21
Let us consider k to be so large that k > K and nk > N . In that case, for
n > N we have
||fn − f || ≤ ||fn − fnk || + ||fnk − f || < ǫ/2 + ǫ/2 = ǫ
Therefore, ∀n > N , we have ||fn − f || < ǫ that implies fn → f .
Proof. (Lp space completeness)
1≤p<∞
Considering the above proposition, we just need to show that each absolutely
summable series in Lp is summable to some element in Lp.
P
Let {fn}P
be a sequence in Lp and ∞
n=1 ||fn || = M < ∞. Let us also define,
n
gn (x) = k=1 |fk (x)|. Applying Minkowski inequality, we have
||gn || ≤
or,
Z
n
X
k=1
||fk || ≤ M
gnp ≤ M p
Now for each x, gn (x) is an increasing sequence of (extended) real numbers
and hence converges to an (extended) real number g(x), where g(x) is measurable; also gn ≥ 0. Therefore, by applying the Monotone Convergence
theorem, we have
Z
Z
g p = lim
gnp ≤ M p
n→∞
1
or, g p ∈ L and g(x) < ∞ a.e.
For each x, s.t. g(x) < ∞, the sequence {fk (x)} is an absolutely summable
series of real numbers and hence must be summable to a real number s(x)
and we set s(x) = 0 for those x, where g(x) = ∞. Thus,
sn =
n
X
k=1
fk and sn → s a.e.
Hence, s is measurable.
Now, |sn (x)| ≤ g(x) implies |s(x)| ≤ g(x). Also, s ∈ Lp since g p ∈ L1.
Therefore, we have
|sn (x) − s(x)| ≤ |sn (x)| + |s(x)| ≤ 2g(x)
or,|sn (x) − s(x)|p ≤ 2pg p (x)
22
Now, 2pg p ∈ L1 and this will be our dominating function for application of
the Dominated Convergence theorem in the following:
Z
Z
p
p
lim ||sn − s|| = lim
|sn − s| =
lim |sn − s|p = 0
n→∞
n→∞
n→∞
as sn → s a.e.
Hence, the absolutely summable sequence {fn} in Lp is summable to an
element s in Lp . Therefore, Lp is complete for 1 ≤ p < ∞.
p=∞
Let {fn} be a Cauchy sequence in L∞ and let Ak and Bm,n be the sets
where, |fk (x)| > ||fk ||∞ and |fn (x) − fm (x)| > ||fn − fm ||∞ respectively. Let,
E = ∪k,m,n (Ak ∪ Bm,n ) for k, m, n = 1, 2, 3, · · · . With all these definitions, we
have µ(E) = 0.
On the set E c sequence {fn} converges uniformly to a bounded function f
and let, f (x) = 0 ∀x ∈ E. Then we have
f ∈ L∞ and ||fn − f ||∞ → 0 as n → ∞
Note,
||f ||∞ = inf{c ≥ 0 : |f (x)| ≤ c a.e. x}
Hence, L∞ is complete.
23