MA359 Lecture notes for Measure Theory - Of the Clux

MA359
Lecture notes for Measure Theory
Contents
1 Riemann Integral 2
1.1 Properties of the Riemann integral . . . . . . . . . . . . . . . . . 2
1.2 Some weak points of Riemann’s integral . . . . . . . . . . . . . . 5
1.2.1 The fundamental theorem of calculus . . . . . . . . . . . . 5
1.2.2 Convergence of functions. . . . . . . . . . . . . . . . . . . 6
1.2.3 Fourier coefficients. . . . . . . . . . . . . . . . . . . . . . . 6
2 Lebesgue’s list of properties for the integral 7
3 Outer Measure 9
3.1 The problem of measure of sets. . . . . . . . . . . . . . . . . . . . 9
3.2 General measure spaces . . . . . . . . . . . . . . . . . . . . . . . 18
4 Measurable functions 19
4.1 Approximation by step or simple functions . . . . . . . . . . . . . 21
4.2 Littlewood three principles . . . . . . . . . . . . . . . . . . . . . 23
5 Integration 25
5.1 Simple functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
5.2 Bounded functions supported on sets of finite measure. . . . . . . 27
5.3 Non-negative functions . . . . . . . . . . . . . . . . . . . . . . . . 29
5.3.1 Fatou’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . 30
5.4 Integrable functions . . . . . . . . . . . . . . . . . . . . . . . . . 33
5.4.1 The Lebesgue space L 1 . . . . . . . . . . . . . . . . . . . . 35
6 Lebesgue measure in R d . 37
6.1 Complex valued integrals . . . . . . . . . . . . . . . . . . . . . . 44
7 Lebesgue L p spaces 44
7.1 The space L 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
7.2 Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
7.3 The Lebesgue spaces L p . . . . . . . . . . . . . . . . . . . . . . . . 53
1

8 Abstract measures 56
8.1 Outer measures and Caratheodory measurable sets . . . . . . . . 56
8.2 Measurability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
8.3 Radon-Nikodym theorem . . . . . . . . . . . . . . . . . . . . . . 60
8.4 Signed measures . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
1 Riemann Integral
Since school we learned a lot about integration methods. However, these methods
were focused on the problem of how to integrate a given function rather
than discussing the nature of the integral itself.
In this section, we will deal with a definition of the integral due to Riemann.
Let P = {x 0 , x 1 , ..., x n } be a partition of [a, b]. A set of sampling points {t i },
for P, is a sequence of points in [a, b], such that t i ∈ [x i−1 , x i ]. We define
the mesh(P) to be the size of the biggest subinterval in P. Given a function
f : [a, b] → R define
S(f, P, {t i }) =
n∑
f(t i )(x i − x i−1 ).
i=1
Definition. A function f is called Riemann integrable if there exist a number
A, such that for every ǫ > 0, there exist δ > 0 such that if P is a partition of
[a, b] with mesh(P) < δ, then for every sampling {t i } we have
|S(f, P, {t i }) − A)| < ǫ.
In this case, we call A = ∫ b
f(x)dx the Riemann integral of f.
a
The value A is unique.
1.1 Properties of the Riemann integral
Assume f and g are Riemann integrable functions defined on [a, b].
• Linearity
∫ b
a
(αf + βg)(x)dx = α
• Positivity. If f(x) ≥ 0 then
∫ b
a
∫ b
a
f(x)dx ≥ 0
f(x)dx + β
In particular, if f(x) ≥ g(x) for all x ∈ [a, b] then
∫ b
a
f(x)dx ≥
∫ b
a
g(x)dx
∫ b
a
g(x)dx
2

• ∫ 1
1dx = 1.
0
Proposition 1. If f : [a, b] → R is continuous then f is Riemann integrable.
In fact, it is possible that a Riemann integrable function is discontinuous.
Nevertheless, the set of discontinuities can not be “too large” as we can see in
the following example.
Example 2 (Dirichlet’s function). Let f : [0, 1] → R be defined as
{
1 if x ∈ Q ∩ [0, 1]
f(x) =
0 if x ∈ [0, 1] \ Q .
The function f is not Riemann integrable. Since every open interval I ⊂ [0, 1]
contains rational and irrational numbers. Hence, given any partition P =
{x 0 , ..., x n }, we can construct samplings {r i } and {q i } with r i ∈ [x i−1 , x i ] ∩ Q
and q i ∈ [x i−1 , x i ] \Q. Then, the sampling set {r i } will render S(f, P, {r i }) = 1
whereas the sampling {t i } gives S(f, P, {q i }) = 0.
Riemann’s definition of the integral also imposes restrictions on the function
f, as we see in the following:
Theorem 3. If f : [a, b] → R is Riemann integrable, then f is bounded.
Proof. Consider δ > 0 such that
|S(f, P, {t i }) −
∫ b
a
f(x)dx| < 1 2
whenever mesh(P) < δ. Fix a partition P with mesh(P) < δ, and a set of
sampling points {t i } for P and let
and
M = max{|f(t 1 ), f(t 2 ), ..., f(t n )|}
△ = min{(x 1 − x 0 ), (x 2 − x 1 ), ..., (x n − x n−1 )} > 0.
Let x ∈ [a, b] and j be the smallest index such that x ∈ [x j−1, x j ]. Let T be the
sampling {t 1 , ..., t j−1 , x, t j+1 , ..., t n }. Note that
|f(x)(x j − x j−1 ) − f(t j )(x j − x j−1 )| = |S(f, P, T) − S(f, P, {t i })|
Moreover,
= |S(f, P, T) −
|S(f, P, T) − S(f, P, {t i })|
∫ b
f(x)dx +
∫ b
a
a
f(x)dx − S(f, P, {t i })|
3

Then
< |S(f, P, T) −
∫ b
f(x)dx| + |
∫ b
a
a
f(x)dx − S(f, P, {t i })| < 1.
|f(x)|(x j − x j−1 ) < |f(t j )|(x j − x j−1 ) + 1 ≤ M(x j − x j−1 ) + 1
which implies that
Thus, f is bounded.
|f(x)| < M +
1
(x j − x j−1 ) ≤ M + 1 △
Now, we will discuss another definition of the integral due to Darboux. In
fact, this definition is equivalent to Riemann’s definition. Let P = {x 0 , ..., x n }
be partition of [a, b] define
M i = sup{f(x)|x ∈ [x i−1 , x i ]}
and
So, we define
and
Put
m i = inf{f(x)|x ∈ [x i−1 , x i ]}.
n∑
U(f, P) = M i (x i − x i−1 )
i=1
n∑
L(f, P) = m i (x i − x i−1 ).
i=1
and
∫ b
f = inf{U(f, P)|P is a partition of [a, b]}
a
∫ b
a
f = sup{U(f, P)|P is a partition of [a, b]}
Whenever a function f : [a, b] → R is bounded the following inequality holds
∫ b
f ≤
a
Proposition 4. A function f : [a, b] → R is Riemann integrable if, and only if,
∫ b
f =
a
∫ b
a
∫ b
In this case, the common value is equal to ∫ b
a f(x)dx.
a
f
f.
4

Finally, let us conclude with another property of Riemann’s integral.
Proposition 5. If f : [a, b] → R is Riemann integrable, then |f| is Riemann
integrable and
|
∫ b
a
f(x)dx| ≤
∫ b
a
|f(x)|dx.
1.2 Some weak points of Riemann’s integral
Here we want to discuss some of the problems that arise with Riemann’s definition
of the integral.
1.2.1 The fundamental theorem of calculus
The fundamental theorem of calculus gives a link between differential calculus
and integral calculus. It consist of two parts:
Theorem 6 (The Fundamental Theorem of Calculus). Let f : [a, b] → R
be a Riemann integrable function.
i) Let F(x) be an antiderivative of f, then
∫ b
a
f(t)dt = F(b) − F(a).
ii) Define F(x) = ∫ x
a
f(t)dt. Then, F is continuous on [a, b]. Moreover, if f
is continuous at ζ ∈ [a, b], then F is differentiable at ζ and F ′ (ζ) = f(ζ).
The existence of continuous functions F that are nowhere differentiable, or
for which F ′ (x) exists at every x, but F ′ is not Riemann integrable, leads to the
problem of finding a general class of functions for which the theorem is valid.
The following example shows a function f which is differentiable at every
point in [0, 1] but whose derivative is not Riemann integrable.
Example 7. Let f : [0, 1] → R defined by
{
x 2 cos( π
f(x) =
x
) if 0 < x ≤ 1
2
0 if x = 0.
Then, f is differentiable on [0, 1] with derivative
{
f ′ 2xcos( π
(x) =
x
) + 2π 2 x sin( π x
) if 0 < x ≤ 1
2
0 if x = 0.
Note that f ′ is not bounded and hence f ′ is not Riemann integrable on [0, 1].
5

1.2.2 Convergence of functions.
Theorem 8. Let f, f k : [a, b] → R be functions for k ∈ N. Suppose that each
f k is Riemann integrable and that the sequence {f k } converges to f uniformly
on [a, b]. Then f is Riemann integrable over [a, b] and
lim
∫ b
a
f k (x)dx =
∫ b
a
f(x)dx.
Uniform convergence is quite strong condition, but pointwise convergence is
not enough in general, as the following example shows:
Example 9. Let {r i } be an enumeration of the rational numbers in [0, 1], define
{
1 if x = r i
δ ri (x) =
0 otherwise.
For every k, let f k (x) = ∑ k
i=1 δ r i
(x), then f k converges monotonically to f,
where f is Dirichlet’s function defined in Example 2.
Thus, it would be desirable to have an integral that satisfies weaker conditions
for convergence. However, it should be notice that the condition of
pointwise convergence alone is not sufficient to guarantee the equality of the
limits; we have to assume extra conditions, as we shall see later.
1.2.3 Fourier coefficients.
Let f : [−π, π] → R be Riemann integrable, then we can associate to f the
sequence {a n } of Fourier coefficients defined by
a n = 1
2π
∫ π
−π
f(x)e −inx dx.
Then f ∼ ∑ a n e inx and we have Parseval’s identity:
∑
|an | 2 = 1
2π
∫ π
−π
|f(x)| 2 dx.
Consider the space l 2 , of all sequences {a n } such that ∑ |a n | 2 < ∞. The
space l 2 is a complete metric space. On the other hand, the space of Riemann
integrable f is not complete. Moreover, for every sequence in l 2 we can define the
corresponding function. We would like to improve the definition of integrability,
so that include these functions. We also would like to describe the space of
functions arising from this construction.
6

2 Lebesgue’s list of properties for the integral
For any set E ⊂ R, let χ E (x) be the characteristic function of E defined by
{
1 if x ∈ E
χ E (x) =
0 otherwise.
We are looking for a definition of integrability that satisfies the following properties.
Given integrable functions f and g
1. ∫ b
a f(x)dx = ∫ b+h
f(x − h)dx
a+h
2. ∫ b
a f(x)dx + ∫ c
b f(x)dx + ∫ a
c f(x)dx = 0
3. ∫ b
a [f(x) + g(x)]dx = ∫ b
a f(x)dx + ∫ b
a g(x)dx
4. If f ≥ 0 and b > a, then ∫ b
a f(x)dx ≥ 0
5. ∫ 1
0 1 dx = 1
6. If {f} ∞ k=1 is a sequence, such that f k ր f monotonically, then
∫ b
a
f k (x)dx →
∫ b
a
f(x)dx
In other words, to define the integral we start with a list of properties we
want it to satisfy and then try to deduce a definition out of this list.
Theorem 10. Properties (1)-(5) imply:
(i) If f ≥ g and a < b, then ∫ b
a f(x)dx ≥ ∫ b
a g(x)dx
(ii) For every [a, b] then ∫ b
a 1dx = b − a
(iii) For every α, β ∈ R then
∫ b
a
(αf(x) + βg(x))dx = α
∫ b
a
f(x)dx + β
∫ b
a
g(x)dx
(iv) If f is Riemann integrable. The value ∫ b
f(x)dx coincides with the Riemann
integral of f on [a,
a
b].
Proof. Note that by setting g = −f in (3) we have ∫ b
a −f(x)d(x) = − ∫ b
a f(x)dx
(Proof of (i)). This is a consequence of (3) and for (4), since f > g implies
f − g > 0 and
∫ b
a
f(x)dx −
∫ b
a
f(x)dx =
∫ b
a
(f(x) − g(x))dx ≥ 0
7

(Proof of (ii)) First note that by (1) ∫ b
a 1dx = ∫ d
c
1dx whenever a − b = c − d.
Make r = a − b, so we want to check that ∫ r
0
1dx = r for every r > 0. By
induction and using (2) it is easy to see that ∫ n
1dx = n for every natural
0
number n. The same argument yields ∫ 1/n
0
1dx = 1/n and iterating (2) again,
we have ∫ q
1dx = q for every rational number q. For any real r let p, q rational
0
numbers with p < r < q. Since χ [0,p] < χ [0,r] < χ [0,q we have by (4) that
0 ≤
∫ q
0
1dx −
∫ r
0
1dx ≤
∫ q
0
1dx −
∫ p
Letting p and q approach to r, we have that for all r,
∫ r
0
1dx = r.
0
1dx = p − q.
(Proof of (iii)) By (3) and the observation at the beginning of the proof, it
is enough to prove that for every α ≤ 0, we have ∫ b
a αf(x)dx = α ∫ b
a f(x)dx.
Putting f = g in (3) and by iterating, we can see that ∫ b
a nf(x)dx = n ∫ b
a f(x)dx
for every natural number n. To prove it for all reals, we should repeat similar
arguments as the ones used in the proof of (ii), but we will omit them.
(Proof of (iv)) Let P = {x 0 , ..., x n } be any partition of the interval [a, b]. Remind
the definitions
M i = sup{f(x)|x ∈ [x i−1 , x i ]}
and
m i = inf{f(x)|x ∈ [x i−1 , x i ]}.
Then for every set of sampling points {t i }, we have
which implies
n∑
m i (x i − x i−1 ) ≤
i=1
∫ b
f ≤
a
∫ b
a
∫ b
a
f(x)dx ≤
f(x)dx ≤
n∑
M i (x i − x i−1 )
Thus, if f is Riemann integrable then ∫ b
f = ∫ b f and the middle integral must
a a
be equal to the Riemann integral of f.
By the theorem above, any integral satisfying properties (1)-(5) must coincide
with the Riemann integral whenever f is Riemann integrable. Now, let us
investigate what comes out of property (6).
Given f : R → R, ǫ > 0 and j ∈ Z, define
i=1
∫ b
a
f.
E j,ǫ = {x ∈ R|jǫ ≤ f(x) < (j + 1)ǫ}.
For [a, b], let A j,ǫ = E j,ǫ ∩ [a, b]. For every n let f n (x) = ∑ n
−n (j/n)χ A j,ǫ
, then,
by construction, we can check that f n ր f as n → ∞ in [a, b]. Then, by
8

property (6), we have ∫ b
a f n(x)dx → ∫ b
a
f(x)dx. Hence, in order to compute the
integral of f it suffices to know how to compute the integral of the characteristic
functions of the sets A j,ǫ . For every interval I = (c, d) ⊂ [a, b], we know that
∫ b
a χ I(x)dx = d−c, which is just the length of I. However, for general f, the sets
A j,ǫ above are not necessarily intervals. This leads to the problem of measure
of sets that will be discussed in the following section.
3 Outer Measure
3.1 The problem of measure of sets.
As we saw in last section, the problem of integration becomes the problem of
assign to every bounded set E ⊂ R a nonnegative number m(E) such that
satisfies the following properties:
1. (Translation invariance) For every h ∈ R define
then m(E + h) = m(E).
E + h = {x + h| for some x ∈ E},
2. (Additivity) If {E i } i∈σ is a finite or countable union of pairwise disjoint
sets, then
m( ⋃ E i ) = ∑ m(E i )
i∈σ i∈σ
3. m([0, 1]) = 1
For intervals we have already a function satisfying these properties: If I =
[a, b] the length of I given by l(I) = b − a satisfies properties (1)-(3) when
restricted to intervals. Note that, since the length of a point is zero, it is
irrelevant whether we include or not the endpoints in I. Our goal now, will be
to extend the concept of length to other sets rather than intervals.
In what follows, σ will always represent a countable set, that is either a finite
or countably infinite set.
Definition. Let E ⊂ R. The outer measure of E is defined by
m ∗ (E) = inf{ ∑ j∈σ
l(I j )}.
The infimum is taken over all countable coverings of E by open intervals {I j } j∈σ
For every open interval I, we have that l(I) > 0, then m ∗ (E) ≥ 0; also, for
every ǫ > 0, ∅ ⊂ (0, ǫ), then 0 ≤ m ∗ (∅) ≤ 0. Hence, we have m ∗ (∅) = 0.
Theorem 11. The outer measure m ∗ satisfies the following properties:
1. m ∗ is monotone. That is, whenever F ⊂ E, then m ∗ (F) ≤ m ∗ (E)
9

2. For any h ∈ R, m ∗ (E + h) = m ∗ (E)
3. For any interval I, m ∗ (I) = l(I)
4. m ∗ is countably subadditive; that is, if {E i } ∞ i=0 is a countable family of
sets in R then
∞⋃ ∞∑
m ∗ ( E i ) ≤ m ∗ (E i ).
i=0
Proof. For Part 1, it suffices to note that every cover of E covers F.
Part 2, the length of I = (a, b) is equal to the length of I +h = (a+h, b+h).
Every covering of E by intervals {I j } induces the covering {I j +h} of E+h with
∑ l(Ij ) = ∑ l(I j +h),which implies m ∗ (E+h) ≤ m ∗ (E). Reciprocally, if {I j } is
a covering of E +h, then {I j −h} is a covering of E with ∑ l(I j ) = ∑ l(I j −h),
so m ∗ (E) ≤ m ∗ (E + h). Hence, m ∗ (E) = m ∗ (E + h).
Let us prove now Part 3, since we can take I with or without a and b, let us
fix I = [a, b] for convenience, then I ⊂ (a − ǫ/2, b + ǫ/2), which implies
i=0
m ∗ (I) ≤ l((a − ǫ/2, b + ǫ/2)) = l(I) + ǫ.
The inequality holds for every ǫ, so m ∗ (I) ≤ l(I).
Conversely, let {I k } be any covering of I by open intervals, since I is compact,
there is a finite subcovering {I 1 , ..., I n } of I. We claim, that ∑ n
j=1 I j > l(I).
Since I is bounded, we can assume every I j to be bounded, then there
exist i 1 with I i1 = (a 1 , b 1 ), such that a 1 < a < b 1 . If b < b 1 , then I ⊂ I i1
and l(I i1 ) > l(I), otherwise there exists i 2 , such that I i2 = (a 2 , b 2 ), and a 2 <
b 1 < b 2 . The sequence {I j } is finite, then at some point there is a i k , with
I ik = (a k , b k ), satisfying a k < b k−1 < b k and b < b k . In this case, I ⊂ ∪ k j=1 I i j
and ∑ j = 1 k l(I ij ) > l(I) as we claimed. Then m ∗ (I) = l(I).
For the proof of Part 4, we may assume that m(E i ) < ∞ for every i, because
otherwise the equality holds. For every ǫ > 0, there exist {I j,k } covering of E j
such that
∑
l(Ij,k ) ≤ m ∗ (E j ) + ǫ
2 j ,
then E ⊂ ⋃ I j,k and
m(E) ≤ ∑ j,k
≤
∞∑
j=1
≤
(I j,k ) =
∞∑
j=1 k=1
(
m ∗ (E j ) + ǫ
2 j )
∞∑
m ∗ (E j ) + ǫ,
j=1
which implies m ∗ (E) ≤ ∑ ∞
j=1 m∗ (E j ).
∞∑
l(I j,k )
10

As a consequence of Part 4, if E is countable, then m ∗ (E) = 0. Hence, we
have:
Corollary 12. Every interval I is not countable.
The outer measure is additive for very especial cases:
Lemma 13. Let {C 1 , ..., C N } be a finite sequence of disjoint compact sets. If
C = ⋃ N
i=1 C i, then
N∑
m ∗ (C) = m ∗ (C i )
Proof. By subadditivity,
i=1
m ∗ (C) ≤ m ∗ (C 1 ) + ... + m ∗ (C N ).
To prove the other inequality, let δ = min{d(C i , C j )|i ≠ j}, choose a covering
{I j } of open intervals with
∑
l(Ij ) < m ∗ (C) + ǫ.
After subdividing assume that for all j with diam(I j ) < δ which implies that
each I j intersects at most one of C i . For every k, let
J k = { set of indexes j|I j ∩ C k }.
Then, E k ⊂ ⋃ j∈J k
I j and J k ∩ J k ′ = ∅ whenever k ≠ k ′ . Then
∑
m ∗ (C i ) ≤
N∑ ∑
k=1
j∈J k
l(I j ) ≤ ∑ l(I j ) ≤ m ∗ (C) + ǫ.
Lemma 14. Let {I j } j∈σ be a sequence of disjoint open and bounded intervals,
let E = ∪ j∈σ I j , then
m ∗ (E) = ∑ j∈σ
m ∗ (I j ).
Proof. Given ǫ > 0, let ˜Q j ⊂ I j be a closed interval such that l(I j ) ≤ l(Ĩj)+ǫ/2 j .
Now, the sets Q j are compact and disjoint. For a fixed N, by Lemma 13 and
Part 3 of Theorem 11 we have ⋃ N
j=1 Ĩj ⊂ E and, then
N⋃ N∑
m ∗ (E) ≥ m ∗ ( l(Ĩj) = m ∗ (Ĩj)
=
j=1
N∑
l(Ĩj) ≥
j=1
j=1
N∑
(l(I j ) − ǫ/2 j )
j=1
11

By taking the limit when N goes to ∞, we have
m ∗ (E) ≥
∞∑
l(I j ) − ǫ
j=1
for every ǫ > 0, hence m ∗ (E) ≥ ∑ ∞
j=1 l(I j), the other inequality holds by the
subadditivity of m ∗ .
The following example shows that, in general, the outer measure m ∗ is not
additive.
Example 15. Write x ∼ y if x ∼ y ∈ Q this is an equivalence relationship.
Since x ∼ x, x ∼ y implies y ∼ x, and whenever x ∼ y and y ∼ z then x ∼ z.
Let [x] denote an equivalence class of this relation. Then, [x] ∩ [y] ≠ ∅ implies
[x] = [y]. By the axiom of choice, we can construct a set N consisting of exactly
one element in each equivalence class of the given relation.
Let {r k } be an enumeration of Q ∩ [−1, 1] and N k = N + r k . Thus we have
[0, 1] ⊂
∞⋃
N k ⊂ [−1, 2],
k=1
which, by the monotonicity of m ∗ , implies 1 ≤ m ∗ ( ⋃ ∞
k=1 N k) ≤ 3.
Now N r ∩ N s ≠ ∅ whenever r ≠ s, then by translation invariance of m ∗ , we
have m ∗ (N k ) = m ∗ (N). So, ∑ ∞
k=1 m∗ (N k ) = ∑ ∞
k=1 m∗ (N) which can be either
0 or ∞, depending whether m ∗ (N) is either 0 or positive, respectively. In each
case, we have a contradiction.
Hence, m ∗ is not countable additive. The existence of sets like the one
constructed in Example 15 implies that a function satisfying the conditions of
the problem of measure of sets can not be defined in all sets. We come to the
definition of Lebesgue measurable sets, there are several equivalent definitions
we choose one that is intuitively convenient.
Definition. A set E ⊂ R is called Lebesgue measurable or simply measurable
if for any ǫ0 there exist an open set U containing E such that
m ∗ (U \ E) < ǫ.
In this case, we define the Lebesgue measure of E, or measure of E by
m(E) = m ∗ (E).
Lebesgue measure inherits all properties of outer measure. Now, we will
describe the family of measurable sets. By definition, it is immediate that every
open set in R is Lebesgue measurable. If m ∗ (E) = 0, then E is measurable,
since for every ǫ > 0 there is a covering of E by open intervals {I j } with
∑
l(Ij ) < m ∗ (E) + ǫ = ǫ.
12

Take U = ⋃ I j , then U \ E ⊂ U and m ∗ (U \ E) ≤ m ∗ (U) ≤ ǫ. The family of
measurable sets is also closed under countable unions.
Theorem 16. Let {E i } i∈σ be a sequence of measurable sets, then
E = ⋃ i∈σ
E i
is also measurable.
Proof. Given ǫ > 0, for every i ∈ σ there is an open set U i containing E i and
m ∗ (U \ E i ) < ǫ
let U = ⋃ i∈σ U i, then
U \ E ⊂
i∈σ(U ⋃ i \ E i )
hence,
m ∗ (U \ E) ≤ ∑ i∈σ
m ∗ (U i \ E i ) < ǫ.
Theorem 17. Closed sets are measurable.
Proof. Let F ⊂ R be a closed set. For every k ∈ N, let B k = [−k, k], then
F = ⋃ ∞
k=1 F ∩ B k. Thus, every closed set in R is countable union of compact
sets. By Theorem 16, it is enough to prove the theorem for compact sets.
Let K be a compact set in R, then there exist a bounded open interval I
such that l(I) = M. The set I \ K is open, so I \ K is a countable union of
disjoint open intervals {I j } j∈σ , moreover by Lemma 14, we have
m(I \ K) = ∑ j∈σ
l(I j ) < M.
Since the sum converges, there exist a finite set of intervals {I j1 , ..., I jN } such
that
N∑
l(I jk ) > m(I \ K) − ǫ/2,
k=1
for every k ∈ {1, ..., N}, let Q k be a closed interval, with Q k ⊂ I jk , such that
m(I jk \ Q k ) < ǫ
2N .
Notice that the set I jk \ Q k consist of two disjoint open intervals, and the set
Q = ⋃ N
k=1 Q k is compact.
So, W = I \ Q is an open set containing K and, from Lemma 14 again, it
follows
m(W \ K) = m( ⋃ N∑
I j \ Q) < m(I jk \ Q k ) + ǫ/2 < ǫ
j∈σ
k=1
13

As a consequence of Lemma 13, we have the following
Corollary 18. Let {C 1 , ..., C N } be a finite sequence of disjoint compact sets.
If C = ⋃ N
i=1 C i, then
N∑
m(C) = m(C i )
Proof. Every C i and C is closed, then m(C i ) = m ∗ (C i ) and m(C) = m ∗ (C).
i=1
Theorem 19. If E ⊂ R is measurable, then the complement E c in R is also
measurable.
Proof. For all n, there exist U n such that m ∗ (U n \ E) < 1/n. Now, Un c is closed
and measurable by Theorem 17. By Theorem 16, the set S = ⋃ ∞
n=1 Uc n is also
measurable. Since for every n, we have Un c ⊂ Ec , then ⋃ ∞
n=1 Uc n ⊂ Ec . Hence,
which implies for every n
m ∗ (E c \
(E c \
∞⋃
Un c ) ⊂ (U n \ E)
n=1
∞⋃
Un) c ≤ m ∗ (U n \ E) < 1/n.
n=1
Thus E c \ ⋃ ∞
n=1 Uc n has measure zero, and so is measurable. Now
∞⋃ ∞⋃
E c = (E c \ Un) c ∪ Un,
c
n=1
n=1
then E c is measurable because is union of measurable sets.
Theorem 20. Let {E i } i∈σ be a countable family of disjoint measurable sets. If
E = ⋃ i∈σ E i, then
m(E) = ∑ i∈σ
m(E i ).
Proof. First, assume that each E i is bounded. For every j, Ej c is measurable,
then for every ǫ > 0, there exist a closed set F j ⊂ E j with m(E j \ F j ) ≤ ǫ/2 j .
Because E j is bounded F j is compact. Fix N, so the sets F 1 , ..., F N are compact
and disjoint, by Corollary 18 we have
N⋃ N∑
m( F j ) = m(F j ),
j=1
j=1
also
N⋃
F j ⊂ E.
j=1
14

Then,
m(E) ≥
N∑ N∑
m(F j ) ≥ m(E j ) − ǫ,
j=1
j=1
by taking the limit when N goes to ∞, we have
m(E) ≥
∞∑
m(E j ) − ǫ.
j=1
Which holds for every ǫ > 0, then m(E) ≥ ∑ ∞
j=1 m(E j).
In the general case, let B k = [−k, k], then I k ⊂ I k+1 and R = ⋃ ∞
k=1 I k. Let
S 1 = I 1 and for every k let S k = I k \ I k−1 . For every j let
E j,k = E j ∩ S k
then
E = ⋃ j,k
E j,k ,
since E j,k is bounded we have
m(E) = ∑ j,k
m(E j,k ) = ∑ j
∑
m(E j,k ) = ∑ j
k
m(E j ).
Example 21 (A non measurable set). Let N be the set constructed in Example
15. We showed that for every rational numbers r, s ∈ [−1, 1], we have
N ∩ (N + r) =, thus m ∗ (N) = m ∗ (N + r) and
[0, 1] ⊂
⋃
r∈Q∩[−1,1]
(N + r) ⊂ [−1, 2].
By Theorem 20, N can not be a Lebesgue measurable set.
Let {E 1 , E 2 , ..., } be a sequence of measurable sets. We say that {E j } increases
to E if E k ⊂ E k+1 and E = ⋃ ∞
k=1 E k, then we write E j ր E. Similarly,
we say that {E j } decreases to E, if E k ⊃ E k+1 and E = ⋂ ∞
k=1 E k, in this case
we write E j ց E. We have
Corollary 22. Suppose {E 1 , E 2 , ...} is a sequence of measurable sets in R.
(i) If E j ր E then m(E) = lim
N→∞ m(E N)
(ii) If E j ց E and m(E k ) < ∞ for some k, then m(E) = lim
N→∞ m(E N)
15

Proof. Part (i). Let G 1 = E 1 , G 2 = E 2 \ E 1 and, recursively, G k = E k \ E k−1 .
Then, the sets G i are disjoint and measurable, also E = ⋃ ∞
i=1 G i so
m(E) =
∞∑
m(G k ) = lim
i=1
N→∞
i=1
= lim m( ⋃
N G k )
N→∞
i=1
= lim
N→∞ m(E N)
N∑
m(G i )
Proof of part (ii). Without loss of generality we can assume m(E 1 ) < ∞. For
every k define G k = E k \ E k+1 then
and E ∩ ⋃ ∞
k=1 G k which implies
Since m(E 1 ) < ∞ we have
E 1 = E ∪
m(E 1 ) = m(E) + lim
∞⋃
k=1
N→∞
k=1
G k
N∑
(m(E k ) − m(E k+1 ))
= m(E) − lim
N→∞ m(E N) + m(E 1 )
m(E) = lim
N→∞ m(E N).
We should remark here that without the condition m(E k ) < ∞ for some k,
the second statement is false. For instance, consider the sequence E k = [k, ∞).
What follows provides analytic and geometric insight into the nature of
measurable sets.
Theorem 23. Suppose that E is measurable in R. then for all ǫ > 0
(i) There exist an open set U, with E ⊂ U and m(U \ E) < ǫ.
(ii) There exist a closed set F, with F ⊂ E and m(E \ F) < ǫ.
(iii) If m(E) < ∞, there exist a compact set K, with K ⊂ E and m(E\K) < ǫ.
(iv) If m(E) < ∞, there exist F = ⋃ N
i=1 I j a finite union of intervals I j , such
that m(E △ F) < ǫ.
Let us remind that E △F is called the symmetric difference of E and F and
is defined by E △ F = (E \ F) ∪ (F \ E).
16

Proof. Part (i). Is just by definition.
Part (ii). Since E is measurable, then E c is measurable, and then there exist
an open set U, with E c ⊂ U and m(U \ E c ) < ǫ. Let F = U c , then F ⊂ E and
F is closed, also E \ F = U \ E c then m(E \ F) < ǫ.
Part (iii). Pick a closed set F, such that F ⊂ E and m(E \ F) < ǫ/2. For
each n, let D n = [−n, n] and K n = F ∩ D n , now
E \ K n ց E \ F
since m(E) < ∞ we have m(E \ K n ) → m(E \ F). For n large enough m(E \
K n ) < ǫ
Part (iv). Choose {Q j } ∞ j=1 a sequence of closed intervals with E ⊂ ⋃ ∞
j=1 Q j
and
∞∑
l(Q j ) ≤ m(E) + ǫ/2.
j=1
Since m(E) < ∞, the series converges and there exist N > 0 such that
∞∑
i=N+1
Then, take F = ⋃ N
j=1 Q j, we have
l(Q j ) < ǫ/2.
m(E △ F) = M(E \ F) + m(F \ E)
≤
≤ m(
∞∑
i=N+1
∞⋃
i=N+1
∞⋃
Q j ) + m( Q j \ E)
j=1
∞∑
l(Q j ) + ( l(Q j ) − m(E))
j=1
≤ ǫ/2 + ǫ/2 = ǫ
We say that a set G is a G δ set if it is a countable intersection of open sets
{U i } i∈σ . Similarly, we say that a set F is a F σ if is a countable union of closed
sets {F i } i∈σ .
We end our discussion of measurable sets with the following proposition:
Proposition 24. Let E be a subset of R, then the following are equivalent:
(i) E is measurable.
(ii) There exist G, a G δ set, such that m(E △ G) = 0
(iii) There exist F, a F σ set, such that m(E △ F) = 0
17

3.2 General measure spaces
Definition. Let X be any space. A family A, of subsets of X, is called a
σ-algebra if satisfies the following properties:
(i) ∅, X ∈ A
(ii) If E ∈ A then E c ∈ A
(iii) If {E j } j∈σ ⊂ A then ⋃ j∈σ E j ∈ A
Remark: As a consequence of De Morgan’s laws for sets:
( ⋃ E i ) c = ⋂
i∈σ i∈σ
E c i and ( ⋂ i∈σ
E i ) c = ⋃ i∈σ
E c i .
Every σ-algebra is also closed under intersections.
Definition. A measure , defined on a σ-algebra A, is a function µ : A → [0, ∞]
satisfying:
• µ(∅) = 0
• (Additivity). If {E i } i∈σ is a sequence of pairwise disjoint sets in A then
m( ⋃ E i ) = ∑ µ(E i ).
i∈σ i∈σ
If in addition, we have that µ(X) = 1, then µ is called a probability measure
in X. All together, a measure space is a triple (X, A, µ) consisting of the
underlying space X, a σ-algebra A of sets in X and a measure µ defined on A.
A consequence of Theorem 19 and Theorem 20 we have the following:
Proposition 25. The family of Lebesgue measurable sets in R is a σ-algebra
and the Lebesgue measure is a measure.
Another σ-algebra in R that plays a vital role in analysis is the Borel σ-
algebra B, which, by definition is the “smallest” σ-algebra containing all open
sets in R. Elements in B are called Borel sets. The term smallest means that,
if A is another σ-algebra containing all open sets in R, then B ⊂ A. In fact,
we could define B as the intersection of all such σ-algebras. Such intersection is
not empty, since it contains Lebesgue.
It arises the natural question: Is every Lebesgue measurable set a Borel set
The answer is no; there are sets of Lebesgue measure 0, and hence measurable,
that are not Borel.
From the point of view of Borel sets, Lebesgue sets arise as the completion
of Borel sets. That is, by adjoining to B all subsets of Borel sets of measure
zero. Later on, when we discuss product measure, we will see that the definition
of Lebesgue measurable sets and Borel sets generalize to R n .
18

4 Measurable functions
From now on, we will be considering extended functions f : R → [−∞, ∞] which
means that we allow f to take on the infinite values −∞ and ∞. In practice, f
will take on infinite values on at most a set of measure zero. We say that f is
finite valued if −∞ < f(x) < ∞ for all x ∈ R.
Definition. Let E be a measurable set, a function f : E → [−∞, ∞] is called
measurable if the set
is measurable for every a ∈ R.
To simplify notation, let us put
Using equations of the form
and
f −1 ([−∞, a)) = {x ∈ E : f(x) < a}
{x ∈ E : f(x) < a} = {f < a}
{f ≤ a} =
{f < a} =
∞⋂
{f < a + 1/k}
k=1
∞⋃
{f ≤ a + 1/k}
k=1
{f ≥ a} = {f < a} c .
One can check that, in the definition of measurable functions, we could equivalently
use either {f < a}, {f ≤ a}, {f > a} or {f ≥ a}. Also from these
equations it follows that if f is measurable, then −f is also measurable.
When f is finite valued, if f is measurable if, and only if, the set {a ≤ f ≤ b}
is measurable for every a, b ∈ R. From these observations it follows:
Proposition 26. Let f be a finite valued function. Then the following are
equivalent:
(i) f is measurable.
(ii) For every open set U ∈ R, the set f −1 (U) is measurable.
(iii) For every closed set F ∈ R the set f −1 (F) is measurable.
As a corollary, we have
Corollary 27. If f is continuous, then f is Lebesgue measurable. Moreover, if
f is measurable and finite valued, and g is a continuous function, then g ◦ f is
also measurable.
19

Notice that, in general, if g is continuous and f is measurable, then the
composition f ◦ g is not necessarily measurable.
Proposition 28. Suppose {f n } n∈σ is a sequence of measurable functions. Then
the following functions are all measurable:
sup f n (x), inf
n
n f n(x), limsup
n→∞
f n (x), liminf
n→∞ f n(x).
Proof. The proof is very similar for each case, let us prove that sup n f n is
measurable. To do this just note that {sup n f n > a} = ⋃ n {f n > a}, since if
we take x ∈ {sup n (x) > a}, then sup n f n (x) > a and, by definition of sup, this
happens if, and only if, there exist n such that f n (x) > a so x ∈ {f n > a}.
Since union of measurable sets is measurable. If every f n is measurable,
then {sup n f n > a} is measurable. The remaining functions can be proven to
be measurable by using the equations:
inf f n(x) = − sup{−f n (x)}
n
n
and
limsup f n (x) = inf {sup f n (x)}
n→∞
k
n≥k
liminf f n(x) = sup{ inf f n(x)}.
n→∞ n≥k
k
Corollary 29. If {f n } ∞ n=1 is a collection of measurable functions such that
lim n→∞ f n = f, then f is measurable.
Proof. Whenever the limit limf n exist, is equal to limsupf n and liminf f n .
Proposition 30. If f and g are measurable, then
(i) The integer powers f k are measurable for k ≥ 1.
(ii) If f and g are both finite valued, then f + g and fg are measurable.
Proof. Part (i). If k is odd then {f k > a} = {f > a 1/k }, if k is even then
{f k > a} = {f > a 1/k } ⋃ {f < −a 1/k }.
Part (ii). Notice that
{f + g > a} = ⋃ r∈Q{f > a − r} ∩ {g > r}
and
fg = 1/4[(f + g) 2 + (f − g) 2 ]
which implies that f + g and fg are measurable.
20

We say that f = g almost everywhere if {x|f(x) ≠ g(x)} has measure zero.
In general, we say that a property holds almost everywhere on a set X if the
property holds except for a set of measure zero in X.
Proposition 31. Assume that f = g almost everywhere, and f is measurable,
then g is measurable.
4.1 Approximation by step or simple functions
Let E be any set in R, then it is easy to check that the characteristic function
of E,
{
1 if x ∈ E
χ E (x) =
0 otherwise.
is measurable if, and only if, the set E is measurable in R. For Riemann integral,
we approximate a Riemann integrable function by step functions which are
function of the form
N∑
f =
k=1
there I k are intervals and a k are constants.
In this section, we will see that every measurable function can be approximated
by simple functions. That is, functions of the form:
f =
χ Ik
N∑
a k χ Ek ,
k=1
where the sets E k are measurable with finite measure. Since simple functions are
linear combinations of measurable functions, simple functions are measurable.
Moreover, by the Corollary to Proposition 28 every limit of simple functions is
a measurable function.
Theorem 32. Let f be a non negative measurable function on R. Then there
exist a sequence φ k , of step functions, such that for every k and x ∈ R φ k (x) ≤
φ k+1 (x) and
lim
k→∞ φ k(x) = f(x)
Proof. Let B k = [−k, k] the ball in R and consider the truncation of f:
⎧
⎪⎨ f(x) if x ∈ B k and f(x) ≤ k
F k (x) = k if x ∈ B k and f(x) ≥ k
⎪⎩
0 otherwise.
Then, clearly F k (x) converges to f(x) as k tends ∞. Now, define
E i,j = { i j < F k(x) ≤ i + 1 }
j
21

for i such that 0 ≤ i < kj and for every k, j let
Φ k,j (x) =
kj−1
∑
i=0
i
j χ E i,j
(x).
By definition, for each pair {k, j} the function Φ k,j is a simple function and
satisfies
0 ≤ F k (x) − Φ k,j (x) < 1 j ,
let φ k (x) = Φ k,k (x) then
0 ≤ F k (0) − φ k (x) < 1 k ,
and the sequence {φ k } satisfies the desired properties.
Note that the theorem allows f to take values on ∞. Now we drop the
non-negative assumption and allow f to take values on −∞:
Theorem 33. Let f be a measurable function, then there exist a sequence φ k
of simple functions, such that for every k and x ∈ R, |φ k (x)| < |φ k+1 (x)| and
Proof. Define
and
lim φ k(x) = f(x).
k→∞
f + (x) = max(f(x), 0)
f − (x) = max(−f(x), 0),
then f(x) = f + (x)−f − (x). The functions f + and f − are non negative measurable
functions. By Theorem 32, there are sequences {φ + k } and {φ− k
} of simple
functions such that for every k and every x
φ + k (x) ≤ φ+ k+1 (x) and φ− k (x) ≤ φ− k+1 (x)
Let φ k (x) = φ + k (x) − φ− k (x), then it is easy to check that φ k(x) → f(x) and
since |φ k (x)| = φ + k (x) + φ− k (x), we have that for every k and x ∈ R, |φ k(x)| <
|φ k+1 (x)|.
In fact, we can approximate every measurable function by step function.
However, the convergence is pointwise and almost everywhere:
Theorem 34. Suppose f is a measurable function on R, then there exist a
sequence of step functions {ψ k } such that
ψ k (x) → f(x) pointwise for a.e x ∈ R
22

Proof. By Theorem 33, it is enough to prove the theorem when f is the characteristic
function χ E of a measurable set E. Since E is measurable for every
ǫ there exist a finite sequence of intervals {I j } N j=1 such that m(E △ ⋃ I j ) < 2ǫ.
Then
N∑
χ E (x) = χ Ij (x)
j=1
except for a set of measure less than 2ǫ. So, for every k we can construct a step
function φ k such that
E k = {x|f(x) ≠ ψ k (x)}
satisfies m(E k ) < 2 −k . Let F k = ⋃ ∞
j=k+1 E j, then
m(F k ) ≤
∞∑
j=k+1
m(E j ) <
∞∑
j=k+1
2 −j = 2 −k .
If F = ⋂ ∞
k=1 F k, then m(F) = 0. Moreover, φ k (x) → f(x) as k → ∞ for all
x ∈ F c .
4.2 Littlewood three principles
The following statements, due to Littlewood, are very useful as an early guide
to the study of measure theory. The idea of them is that measure theoretical
objects are closed its topological counterparts.
Littlewood three principles
1. Every set is nearly a finite union of intervals.
2. Every measurable function is nearly continuous.
3. Every convergent sequence is nearly uniformly.
We saw already a precise formulation of the first principle in part (iv) of
Theorem 23. Which we write for convenience here:
If m(E) < ∞, there exist F = ⋃ N
i=1 I j a finite union of intervals I j , such
that m(E △ F) < ǫ.
The formal statement of the third principle translates into the following
theorem:
Theorem 35 (Egorov’s Theorem). Let {f k } ∞ k=1
be a sequence of measurable
functions defined on a measurable set E with m(E) < ∞. Assume that f k → f
almost everywhere on E. Then, for every ǫ > 0 there exist a closed set A ǫ ⊂ E
such that m(E \ A ǫ ) < ǫ and f k → f uniformly on A ǫ .
Proof. By restricting to a smaller set, we can assume that f k (x) → f(x) for all
x ∈ E. For every natural number n, let
E n k = {x ∈ E| |f j (x) − f(x)| < 1 n
for all j > k}
23

fix n and note that Ek n ⊂ En k+1 and E = ⋃ ∞
k=1 En k , then En k
∞. By Corollary 22 there exist K n such that
ր E as k tends to
and
m(E \ E n k n
) < 1
2 n
|f j (x) − f(x)| < 1 n
for all j > k n and x ∈ E n k n
. Choose N so that
and let Ãǫ = ⋂ n≥N En k n
so
∞∑
n=N
1
2 n < ǫ/2,
E \ Ãǫ = E ∩ Ãc ǫ
= E ∩ ( ⋂ E n k n
) c = E ∩ ⋃ (E n k n
) c
⊂ ⋃ (E ∩ (E n k n
) c ) = ⋃ (E \ E n k n
),
then
m(E \ Ãǫ) ≤ m( ⋃
∞∑
E \ Ek n n
) ≤ m(E \ Ek n n
) < ǫ/2.
n≥N
n=N
Let ǫ > 0 and choose n ≥ N such that 1/n < δ. If x ∈ Ãǫ, then x ∈ E n k n
and
|f j (x) − f(x)|1/n < δ for all j > k n .
Hence f k → f uniformly on Ãǫ. To finish the proof, notice that there exist a
closed set A ǫ ⊂ Ãǫ with m(Ãǫ) < ǫ/2, then we can check that f k → f uniformly
on A ǫ and m(E \ A ǫ ) < ǫ.
Now we are in position to state and proof a precise statement for Littlewood’s
second principle:
Theorem 36 (Lusin’s Theorem). Suppose that f is measurable and finite valued
on E, where E is a measurable set of finite measure. Then, for every ǫ > 0 there
exist a closed set F ǫ ⊂ E with m(E \ F ǫ ) < ǫ and such that f| Fǫ is continuous.
Remark: The theorem says that the function f restricted to the set F ǫ is
continuous, but not the stronger statement that the function f defined in E is
continuous at the points in F ǫ .
24

Proof. Let f n be a sequence of step functions so that f n → f almost everywhere.
Then, for every n there exist E n such that m(E n ) < 1/2 n and f n is continuous
outside E n . By Egorov’s Theorem we may find a set A ǫ/3 ⊂ E on which f n → f
uniformly on A ǫ/3 and
m(E \ A ǫ/3 ) < ǫ/3.
Now let
where N is a number such that
F ′ = A ǫ/3 \ ⋃
∑
n≥N
n≥N
1
2 n < ǫ/3.
For every n, f n is continuous on F ′ , because f n converge uniformly to f on F ′ ,
the function f is continuous on F ′ . Now take a closed set F ⊂ F ′ with
E n ,
m(F ′ \ F) < ǫ/3.
Then F is the set we are looking for, since f is continuous in F, and
E \ F ⊂ (E \ A ǫ/3 ) ∪ (F ′ \ F) ∪ ( ⋃
E n )
n≥N
we have
m(E \ F) ≤ m(E \ A ǫ/3 ) + m(F ′ \ F) + m( ⋃
< ǫ/3 + ǫ/3 + ∑ n≥N
1
2 n < ǫ.
n≥N
E n )
5 Integration
In this section we will define Lebesgue integral for measurable functions on R.
In order to do so, we will proceed on several levels of generality, starting with
simple functions, then bounded functions supported on a set of finite measure,
non-negative functions, and finally, the general case of integrable functions.
5.1 Simple functions
We start by noting that simple functions φ = ∑ a i χ Ei have several representations;
for instance
χ [0,1] = χ [0,1/2) + χ [1/2,1] .
Nevertheless, we can speak of a canonical representation of any simple function:
25

Definition. Let φ = ∑ a i χ Ei a simple function, we say that φ is in its canonical
representation if, for all i, j with i ≠ j, we have that a i ≠ 0, a i ≠ a j and
E i ∩ E j = ∅. The canonical representation of φ is unique.
Given any simple function φ, to find the canonical representation of φ is
straightforward, since the function φ takes on finitely values c 1 , c 2 , ..., c n for
every k ∈ {1, ...n} let
F k = {x ∈ R|φ(x) = c k }
then φ = ∑ n
k=1 c kχ Fk is the canonical representation of φ. Let φ = ∑ n
k=1 a kχ Ek
be a simple function, define
∫ n∑
φ(x) = a k m(E k ).
R
k=1
From now on, whenever E is a measurable set and f a measurable function
define
∫ ∫
f(x)dx = f(x)χ E (x)dx.
R
E
Also we will adopt the following notations:
∫
∫ ∫
f(x)dm(x) = f(x)dm = f
R
Depending on whether we want to make emphasis on the variable x, the measure
m, or just the integral when no confusion is possible.
From the definition of simple functions, one can check the following Proposition,
which reflects Lebesgue’s list of properties for the integral of simple functions.
Proposition 37. For simple functions, the integral satisfies the following properties:
1) The integral is independent of the representation of φ.
2) (Linearity.) If φ and ψ are simple functions and a, b ∈ R, then
∫
∫ ∫
(aφ + bψ) = a φ + b psi.
3) (Additivity.) If E and F are disjoint measurable sets with m(E), m(F) <
∞, then ∫ ∫ ∫
φ = φ + φ.
F
E∩F
4) (Monotonicity) If φ and ψ are simple functions such that φ ≤ ψ then
∫ ∫
φ = ψ.
E
5) (Triangle inequality)
∫
|
∫
φ| ≤
|φ|.
26

5.2 Bounded functions supported on sets of finite measure.
We start with a definition.
Definition. Let f be a function, the support of f is defined by
supp(f) = {x|f(x) ≠ 0}
We say a measurable f is a bounded function supported on a set of finite measure,
if there is a measurable set E, with m(E) < ∞ and supp(f) ⊂ E.
Let f be a bounded function supported on a set of finite measure, a consequence
of Theorem 33 then there exist a sequence of simple functions {φ n } such
that supp(φ n ) ⊂ E, where m(E) < ∞ and φ n → f for all x. More precisely,
we have the following useful result:
Lemma 38. Let f be a bounded function supported on a set of finite measure
E. Let {φ n } be a sequence of simple functions such that there exist M > 0 such
that |φ n | < M and φ n (x) → f(x) a.e x ∈ E. Then,
1) The limit
exists.
∫
lim
n→∞
φ n
2) If f = 0 a.e, then
∫
lim
n→∞
φ n = 0.
Proof. Proof of (1). If the convergence were uniform, the result would be rather
clear. Nevertheless, we have Egorov’s theorem to fix the situation.
Since m(E) < ∞, for every ǫ > 0 there exist a closed set A ǫ ⊂ E such that
m(E \ A ǫ ) < ǫ and φ n → f uniformly on A ǫ . Let
∫
I n =
by the triangle inequality
∫
|I n − I m | ≤
E
φ n
|φ n (x) − φ m (x)|
∫
∫
= |φ n (x) − φ m (x)| + |φ n (x) − φ m (x)|
A ǫ E\A ǫ
∫
≤ |φ n (x) − φ m (x)| + 2Mm(E \ A ǫ )
A ǫ
∫
≤ |φ n (x) − φ m (x)| + 2Mǫ
A ǫ
27

Now, since the convergence in A ǫ is uniform there is N > 0 such that if n, m > N
then
|φ n (x) − φ m (x)| < ǫ, for all x ∈ A ǫ
which implies
|I n − I m | ≤ m(A ǫ )ǫ + 2Mǫ
= (m(A ǫ ) + 2M)ǫ
since A ǫ ⊂ E we have m(A ǫ ) < ∞, which implies that {I n } is a Cauchy sequence
in R, and then {I n } converges.
Proof (2). If f = 0 we repeat the argument to see that for n big enough
which implies that limI n = 0.
|I n | ≤ m(E)ǫ + Mǫ
Let f be a bounded function supported on a set of finite measure E. With
Lemma 38 at hand we can define the integral of f
∫
∫
f(x)dx = lim φ n (x)dx.
n→∞
Where φ n is a sequence of simple functions supported on E and converging to
f.
The second part of Lemma 38 also guarantees that this integral is well defined,
because if ψ n is another sequence of simple functions supported on E and
converging to f, then (φ n − ψ n ) → 0, hence ∫ (φ n − ψ n ) → 0 which implies
∫ ∫
lim
n→∞
φ n = lim
n→∞
Passing to the limit, from the properties of integrals of simple functions in
Proposition 37, we obtain the following:
Proposition 39. Let f and g be bounded functions over a set of finite measure.
Then the following properties hold
ψ n .
(1) (Linearity) For every a, b ∈ R
∫ ∫
(af + bg) = a
∫
f + b
g.
(2) (Additivity) For E and F disjoint measurable sets
∫ ∫ ∫
f = f + f.
E∪F E F
(3) (Monotonicity) If f ≤ g, then
∫
∫
f ≤
g.
28

(4) (Triangle inequality) The function |f| is also bounded supported on a set
of finite measure, and ∫ ∫
| f| ≤ |f|.
Now, let us prove a convergence theorem for bounded measurable functions:
Theorem 40 (Bounded Convergence Theorem). Suppose that {f n } is a sequence
of measurable function, such that for all n, |f n | < M, supp(f n ) ⊂ E
with m(E) < ∞ and
f n (x) → f(x) for a.e. x ∈ E.
Then f is measurable, bounded and supported on E almost everywhere. Moreover
∫
|f n − f| → 0
which implies
∫
∫
f n →
f
Proof. By passing to the limit, if follows from the assumptions that f is measurable,
|f| < M a.e, and m(supp(f) ∩ E c ) = 0. By the triangle inequality, it is
enough to prove
∫
|f n − f| → 0.
By Egorov’s theorem, for every ǫ > 0 there exist a closed set A ǫ ⊂ E such that
m(E \ A ǫ ) < ǫ and f n → f uniformly on A ǫ . Then, for every n > N
∫
∫
∫
|f n (x) − f(x)| ≤ |f n (x) − f(x)| + |f n (x) − f(x)|
A ǫ E\A ǫ
≤ ǫm(A ǫ ) + 2Mm(E \ A ǫ )
< ǫ(m(E) + 2M).
Which finishes the proof since ǫ is arbitrary and M(E) + 2M < ∞
5.3 Non-negative functions
Now, we will proceed to define the integral of non-negative measurable functions.
Let us remark that, in this case, we allow functions to take values on ∞. Let
f be a measurable, non-negative function. Let f be a non negative measurable
function, define ∫ ∫
f(x)dx = sup g(x)dx
g
where the supremum is taken over all measurable functions g, where g is bounded
and supported on a set of finite measure and 0 ≤ g ≤ g. This supremum is either
finite or infinite. When ∫
f(x)dx < ∞
29

we say that f is Lebesgue integrable (or just integrable for short).
Proposition 41. The integral of non-negative measurable functions satisfy:
(1) Linearity
(2) Additivity
(3) Let g be integrable and f such that 0 ≤ f ≤ g, then f is measurable and
integrable. I
(4) If f is integrable then f(x) < ∞ a.e. x.
(5) If ∫ f = 0 then f(x) = 0 a.e. x.
Proof. For part (1), let us check that ∫ f + g = ∫ f + ∫ g, let φ ≤ f and ψ ≤ g
be bounded measurable functions supported on a set of finite measure. Then
φ + ψ ≤ f + g
which implies
∫
∫
f +
∫
g ≤
f + g.
Now, let η ≤ f +g be a bounded measurable function supported on a set of finite
measure. Let η 1 = min{f(x), η(x)} and η 2 = η − η 1 . By definition, η = η 1 + η 2
and one can easily check that η 1 ≤ f and η 2 ≤ g. Hence
∫ ∫ ∫ ∫ ∫
η = η 1 + eta 2 ≤ f + g
which implies
∫
∫
(f + g) ≤
∫
f +
g.
To prove part (5), define E k = {x|f(x) ≥ k} and E ∞ = {x|f(x) = ∞}, then
E k ց E ∞ . Since f is integrable, for every k,
∫ ∫
∞ > f ≥ χ Ek f ≥ km(E k )
Hence m(E k ) → 0 as k → ∞, then m(E ∞ ) = 0.
5.3.1 Fatou’s Lemma
As in the case of bounded functions, we are concerned about how good the
definition of the integral behaves under limits. Suppose that we have a sequence
f n of non negative measurable functions, and that f n (x) → f(x) for almost every
x. Is it true that ∫ f n (x)dx → ∫ f(x)dx The answer is no as we can check on
the following example:
30

Example 42. Let f n (x) = nχ [0,1/n] , then f n → 0 a.e. ∫ f n = 1 but ∫ f = 0.
However, note that lim ∫ f n > ∫ f. This is true in general, and is the content
of Fatou’s Lemma.
Lemma 43 (Fatou’s Lemma). Suppose that {f n } is a sequence of non negative
measurable functions. If lim n→∞ f n (x) = f(x) for a.e. x. Then
∫ ∫
f ≤ liminf f n .
n→∞
Proof. Let g be a measurable function supported on a set E of finite measure,
and assume g is bounded by M and 0 ≤ g ≤ f. For every x, set
Then for every n,
g n (x) = min{g(x), f n (x)}.
|g n (x)| ≤ |g(x)| ≤ M
and g n (x) → g(x). By the Bounded Convergence Theorem
∫ ∫
g n → g
since g n < f n we have ∫ g n < ∫ f n which implies
∫ ∫
g ≤ liminf
f n
taking supreme over g:
∫
∫
f ≤ liminf
f n .
Fatou’s lemma is still true even if the limit f does not exist, in this case, we
replace f by liminf f n .
Corollary 44. Suppose f is a non negative measurable function, and {f n } is
a sequence of non negative measurable function with f n (x) ≤ f(x) and f n (x) →
f(x) a.e. x. Then
∫ ∫
lim f n = f.
n→∞
Proof. Since f n (x) ≤ f(x) for a.e x, we have ∫ f n ≤ ∫ f for all n. Hence, by
Fatou’s Lemma we have
∫ ∫ ∫ ∫
limsup f n ≤ f ≤ liminf f n ≤ limsup
f n
which implies
∫
lim
n→∞
∫
f n =
f.
31

In analogy with the symbols used for sets ր and ց, we write f n ր f if
f n (x) ≤ f n+1 (x) and f n (x) → f(x) for a.e. x. Similarly, we write f n ց f if
f n (x) ≥ f n+1 (x) and f n (x) → f(x) for a.e. x. As a special case of the previous
corollary we have the following:
Corollary 45 (Monotone Convergence Theorem). Suppose {f n } is a sequence
of measurable functions such that f n ≥ 0 and f n ր f. Then
∫ ∫
f n = f.
lim
n→∞
Monotone Convergence Theorem provides a useful criterion for convergence
of series:
Corollary 46. Consider a series ∑ ∞
k=1 a k(x) where for each k, a k is a non
negative measurable function. Then
∫
∑ ∞ ∞∑
∫
a k (x)dx = a k (x)dx
k=1
Moreover, if ∑ ∞
∫
k=1 ak (x)dx < ∞, then ∑ ∞
k=1 a k(x) converges for a.e. x.
Proof. Let f n (x) = ∑ n
k=1 a k(x) and f(x) = ∑ ∞
k=1 a k(x), then for each n, f n is
measurable non negative and f n ր f.
By the Monotone Convergence Theorem, lim ∫ f n = ∫ f, but
Then
Now, if ∑ ∞
k=1
k=1
∫
f n =
k=1
k=1
n∑
∫
a k (x)dx.
∞∑
∫ ∫
∑ ∞
a k (x)dx = a k (x)dx
k=1
∫
ak (x)dx < ∞, then ∑ ∞
k=1 a k(x) is integrable, hence
∞∑
a k (x) < ∞ for a.e. x.
k=1
Let {E k } be a sequence of measurable sets, define
limsup E k =
∞⋂
n=1 n=1
∞⋃
E k .
Corollary 47 (Borel-Cantelli Lemma). Let {E k } be a sequence of measurable
sets. If ∑ ∞
k=1 m(E k) < ∞, then m(limsupE k ) = 0.
32

Proof. Let a k (x) = χ Ek (x), then ∫ a k (x)dx = m(E k ). So
∞∑
∞∑
∫
m(E k ) = a k (x)dx < ∞,
k=1 k=1
then by the corollary above, ∑ ∞
k=1 a k(x) < ∞ a.e. Note that ∑ ∞
k=1 a k(x) = ∞,
if, and only if, x ∈ limsup E k , hence m(limsupE k ) = 0.
5.4 Integrable functions
Finally, we consider the definition of the integral for the general case, let f :
R → [−∞, ∞], we say that f is Lebesgue integrable (or just integrable) if |f| is
Lebesgue integrable (in the sense of non negative functions). Now, let f be an
integrable function. Define
f + (x) = max(f(x), 0), f − (x) = max(−f(x), 0)
Then f + , f − ≥ 0 and f = f + −f − . Also, note that f ± ≤ |f|. So f is integrable
if, and only if, f − and f + are integrable. We define
∫ ∫
f = f + − f −
Note that the integral of f does not depend on the decomposition of f into
non negative functions. Since, if f = g 1 − g 2 = f 1 − f 2 , where f 1 , f 2 , g 1 , g 2 ≥ 0
are integrable functions, then
g 1 + f 2 = f 1 + g 2
which implies
∫
∫
g 1 +
∫
f 2 =
∫
f 1 +
g 2 ,
so we have
∫
∫
g 1 −
∫
g 2 =
∫
f 1 −
f 2
Proposition 48. The integral of Lebesgue integrable functions is (1) linear, (2)
additive, (3) monotone, and satisfies the triangle inequality.
Theorem 49. Suppose f is integrable on R, then for every ǫ > 0.
(i) There exist a set B of finite measure such that
∫
B c |f| < ǫ
(ii) (Absolute continuity) There exist δ > 0, such that if m(E) < δ, then
∫
|f| < ǫ.
E
33

Proof. By replacing f by |f|, we can assume f ≥ 0. For every N ∈ N, let
E N = {x|f(x) ≤ N} and f N (x) = f(x)χ EN (x).
Proof of (i). By definition f N is non negative and measurable. Also f N (x) ≤
f N+1 (x) and lim N→∞ f N (x) = f(x). By the Monotone Convergence Theorem
∫ ∫
lim
N→∞
f N =
Thus, given ǫ > 0 there exist M > 0, such that
∫ ∫
f − fχ EM < ǫ.
f.
But 1 − χ EM = χ E c
M
, hence
∫
∫
fχ EM < ǫ = f < ǫ.
E c M
Proof of (ii). Again given ǫ > 0 there exist M > 0 such that
∫
f − f M < ǫ/2,
pick δ > 0 such that Mδ < ǫ/2. If m(E) < δ, we have
∫ ∫ ∫
f = f − f M +
E
E
f M
E
Since f M (x) < M for every x.
≤ ǫ/2 + Mm(E)
ǫ/2 + Mδ < ǫ.
Theorem 50 (Dominated Convergence Theorem). Suppose {f n } is a sequence
of measurable functions such that f n (x) → f(x) almost everywhere, and |f| < g
where g is an integrable function. Then
∫
|f n − f| → 0 as n → ∞,
which implies
∫
∫
f n →
f as n → ∞.
Proof. For each N ≤ 0, let E N = {x| |x| ≤ N, g(x) ≤ N}. Given ǫ > 0 there
exist M > 0 such that ∫
g < ǫ/3.
E c M
34

Then f n χ EM is measurable bounded and supported on a set of finite measure.
By the Bounded convergence theorem,
∫
|f n − f| < ǫ/3.
E M
Thus we have
∫
∫ ∫
|f n − f| = |f n − f| +
E M
∫
∫
≤ |f n − f| + 2
E M
≤ ǫ/3 + 2ǫ/3 = ǫ.
E c M
E c M
g
|f n − f|
5.4.1 The Lebesgue space L 1 .
Let L 1 = L 1 (R) be the space of integrable functions on R. under almost everywhere
equivalence. That is, two elements f, g ∈ L 1 (R) are equivalent if
f(x) = g(x) for almost every x. The properties of the integral imply that L 1 is
a vector space. Let f be an integrable function, the L 1 -norm of f, is given by
∫
‖f‖ 1 = |f|dx.
Proposition 51. The L 1 -norm ‖‖ 1 satisfies for all f, g in L 1 :
(i) For all a ∈ R, ‖af‖ = |a|‖f‖ 1 ,
(ii) ‖f + g‖ 1 ≤ ‖f‖ 1 + ‖g‖ 1 ,
(iii) ‖f‖ 1 = 0 if, and only if f = 0 a.e.,
(iv) d(f, g) = ‖f − g‖ 1 defines a metric L 1 (R).
Due to almost everywhere equivalence, the L 1 -norm is indeed a norm in L 1 .
On the other hand, the space L 1 is not quite a function space; since points
have measure zero, we can not evaluate elements in L 1 at a point. Nevertheless,
we keep the imprecise terminology that an element f ∈ L 1 (R) is an integrable
function, and every time we define an element in L 1 we keep in mind that the
definition holds almost everywhere.
Remind that a metric space (X, d) is complete if any Cauchy sequence {x i }
converges in X. In fact, for a Cauchy sequence to converge it is enough to have
a convergent subsequence:
Lemma 52. Let {x n } be a Cauchy sequence on a metric space X, if there exist
a subsequence x nk such that
Then {x n } converges to L.
R
x nk → L as k → ∞.
35

Proof. Let ǫ > 0, and let N > 0 such that for every n k , n, m > N we have
d(x nk , L) < ǫ/2 and d(x n , x m ) < ǫ/2. Then for every n > N
So, {x n } converges to L.
d(x n , L) ≤ d(x n , x nk ) + d(x nk , L) < ǫ
Now we prove that the metric defined in part (iv) of Proposition 51 is complete
in L 1 :
Theorem 53 (Riesz-Fisher). The Lebesgue space L 1 is a complete metric space.
Proof. Let {f n } be a Cauchy sequence in L 1 , by Lemma 52 to prove that {f n }
converges, it is enough to show that {f n } has a convergent subsequence. Since
{f n } is Cauchy, we can find a subsequence {f nk } satisfying
Define
and
Consider the partial sums
and
‖f nk+1 (x) − f nk (x)‖ 1 ≤ 1 for all k ≥ 1.
2k f(x) = f n1 +
g(x) = |f n1 | +
∞∑
(f nk+1 (x) − f nk (x))
k=1
∞∑
|f nk+1 (x) − f nk (x)|.
k=1
S K (f)(x) = f n1 +
S K (g)(x) = |f n1 | +
By the triangle inequality, for all k
‖S K (g)‖ 1 ≤ ‖f n1 ‖ 1 +
K∑
(f nk+1 (x) − f nk (x))
k=1
K∑
|f nk+1 (x) − f nk (x)|.
k=1
K∑
‖f nk+1 (x) − f nk (x)‖ 1 ≤ ‖f n1 ‖ 1 + 2,
k=1
thus the partial sums S K (g) are bounded and S K (g) ր g, by the Monotone
Convergence Theorem g is integrable. Since |f| ≤ g, f is also integrable. The
partial sums S K (f) telescopes to f nk so S K (f)(x) = f nk (x) → f(x) for almost
all x. To prove that f nk → f(x) in L 1 , note that
|f − S K (f)| ≤ (2g) for all K.
By the Dominated Convergence Theorem
‖f nk − f‖ 1 → 0
36

6 Lebesgue measure in R d .
So far, we have discussed measurability in dimension 1. All definitions carry
over higher dimension. Here we write down the relevant settings and differences.
Instead of intervals, in R d we will deal with rectangles of the form
R = (a 1 , b 1 ) × ... × (a d , b d ).
Where b j ≥ a j for all j. Note that all rectangles have sides parallel to the
coordinate axes. Instead of the length of an interval, we will deal with the
volume of a rectangle
|R| = vol(R) = (b 1 − a 1 ) · ... · (b d − a d )
In dimension 1, every open set is a countable union of disjoint open intervals.
In higher dimension this is not the case, however, in R d every open set is a
countable union of almost disjoint of open rectangles, meaning that we can cover
the open set except on the boundaries of the rectangle. Since the boundary of
rectangles have 0 volume, we do not care much for this issue.
Here are the definitions of Lebesgue outer measure in R d and Lebesgue
measurability:
Definition. Let E ⊂ R d . The outer measure of E is defined by
m ∗ (E) = inf{ ∑ j∈σ
|Q j |}.
The infimum is taken over all countable coverings of E by open rectangles
{Q j } j∈σ
Definition. A set E ⊂ R d is called Lebesgue measurable or simply measurable
if for any ǫ0 there exist an open set U ⊂ R d containing E such that
m ∗ (U \ E) < ǫ.
In this case, we define the Lebesgue measure of E, or measure of E by
m(E) = m ∗ (E).
The outer measure in R d and Lebesgue measurable sets have the same properties
than R. In this section, we will describe properties of Lebesgue measurable
sets in R d with respect to its factors.
Write R d = R d1 × R d2 with d = d 1 + d 2 and d 1 , d 2 ≥ 1. A point in R d has
the form (x, y) where x ∈ R d1 and y ∈ R d2 . Let f : R d → R be a function, the
y-slice of f corresponding to y ∈ R d2 is the function
f y : R d1 → R
given by
f y (x) = f(x, y)
37

Similarly, the x-slice of f corresponding to x ∈ R d1 is the function
defined by
f x : R d2 → R
f x (y) = f(x, y)
For measurable sets, let E be a measurable set in R d . For y ∈ R d2 and
x ∈ R d1 let
E y = {x ∈ R d1 |(x, y) ∈ E}
and
E x = {y ∈ R d2 |(x, y) ∈ E}
If f is measurable, it does not immediately follows f x or f y are measurable
for every x ∈ R d2 and y ∈ R d1 . For E measurable, it might happen that some
E y slice is not measurable.
Example 54. Let y ∈ R and E = N × y. Since E belongs to a set of measure
0 in R 2 , E is measurable in R 2 . But E y = N is not measurable on R. Fubini’s
theorem and its consequences deal with this kind of considerations.
Theorem 55 (Fubini’s Theorem). Suppose f(x, y) is integrable on R d = R d1 ×
R d2 , then for almost every y ∈ R d2
(1) The slice f y is integrable on R d1 .
(2) The function given by
is integrable in R d2 . Moreover,
(3) ∫
R d 2
∫
R d 1
∫
y ↦→
R d 1
f y (x)dx
∫
f y (x)dxdy = f(x)dx
R d
The theorem is symmetric on x and y. So we have that for almost every
x ∈ R d1 , f x is integrable on R d2 , the function x ↦→ ∫ R f x(y)dy is integrable on
d2
R d1 and
∫ ∫
∫ ∫
∫
f x (y)dydx = f y (x)dxdy = f(x)dx
R d 1 R d 2
R d 2
R d
Proof. Let F denote the class of integrable function on R d which satisfy all three
conditions of the theorem. The goal is to prove
R d 1
L 1 (R d ) ⊂ F.
We will achieve this goal in several steps.
Step 1 F is closed under linear combinations. Let {f k } N k=1 ⊂ F for all k
there exist A k ⊂ R d2 such that m(A k ) = 0 and f y k is integrable for all y ∈ Ac k .
38

Let A = ⋃ N
k=1 A k, then for all k and every y ∈ A c . f y k
is integrable, which
implies that ∑ N
k=1 a kf y k is integrable, for every linear combination of fy k . Also,
since the function
∫
y ↦→ f y k (x)dx
R d 1
is integrable, so it is every linear combination
∫
y ↦→
R d 1
k=1
and, by the properties of the integral,
∫
=
N ∑
R d k=1
∫
=
N∑
∫
a k
k=1
R d 2
∫
a k f k =
R d 2
N∑
a k f y k (x)dx,
∫
R d 1
k=1
N∑
a k f k
∫R d
k=1
R d 1
f y k (x)dxdy
N∑
a k f y k (x)dxdy.
So ∑ N
k=1 a kf k ∈ F.
Step 2 F is closed under limits.
Let {f k } be a sequence of elements in F, such that f k ր f or f k ց f. By
considering −f k if necessary we may assume f k ր f. Also, by replacing f k by
f k − f 1 , we may assume that f k ≥ 0. By the Monotone Convergence Theorem
∫ ∫
limk → ∞ = f,
R d R d
then f is integrable in R d . For all k, there exist A k ⊂ R d2 so that f y k
is integrable
on R d1 , whenever y ∈ A c k . Let A = ⋃ ∞
k=1 A k, then m(A) = 0, which implies
that for all y ∈ A c , the function f y k is integrable in , and since f y Rd1 k ր fy then
∫
∫
g k (y) = f y k (x)dx → g(y) = f y (x)dx,
R d 1
hence g is integrable in R d2 . Moreover
∫ ∫
g k (y)dy →
that is
which implies that f ∈ F.
R d 2
R d 2
∫ ∫
f k → f,
R d R d
R d 1
g(y)dy,
39

Step 3 For every G δ set E of finite measure or every E of measure cero,
χ E ∈ F. We separate this step into several cases. Let us consider first, the case
when E is an open rectangle, so E = R 1 × R 2 . Clearly, for every y ∈ R d2 , χ y E
is integrable and
∫
{
|R 1 | if y ∈ R 2
g(y) = χ E (x, y)dy =
0 otherwise
R d 1
hence g(y) = |R 1 |χ R2 which implies
∫
g(y)dy = |R 1 ||R 2 |.
But
R d 2
∫
χ = |R 1 ||R 2 |,
R d
so χ E ∈ F.
Suppose now that E ⊂ ∂(R 1 × R 2 ), thus ∫ R
χ d E = 0, it is clear that almost
every y-slice of χ E is integrable and
∫
g(y) = χ E (x, y)dx
R d 1
implies that g(y) = 0 for almost all y ∈ R d2 . Then
∫
g(y)dy = 0
R d 2
So, again χ E ∈ F.
To check that χ E ∈ F when E = ⋃ N
k=1 Q k is a finite union of disjoint
rectangles, note that χ E = ∑ χ Qk = ∑ χ intQk + ∑ χ ∂Qk where intQ k denotes
the interior of Q k . By the first step of the proof, and the previous cases χ E ∈ F.
When E is an open set of finite measure, then E = ⋃ ∞
k=1 Q k is almost a
disjoint union of rectangle Q k . Let f N = ∑ N
k=1 χ Q k
, then f k → χ E , and by the
previous case f k ∈ F. The second step of the proof implies χ E ∈ F..
Finally, let us check that χ E ∈ F when E is a G δ set of finite measure. By
definition there exist a countable sequence of open sets {Ũk} such that
E =
∞⋂
Ũ k ,
k=1
since m(E) < ∞, there exist an open set U 0 , which contains E and m(U 0 ) < ∞.
Let
U k = U 0 ∩ ⋂ j=1
Ũ j
then U k ց E and f k = χ Uk is a sequence such that f k ց χ E , by the previous
case χ Uk ∈ F and by the second step of the proof χ E ∈ F.
40

Assume that E is a set of measure 0. Then there exist a G δ set G, such that
E ⊂ G and m(G) = 0. Then χ G ∈ F and also E y ⊂ G y for all y. Then
∫ ∫
∫
χ G (x, y)dxdy = χ G = 0.
But this implies that
R d 2
R d 1
∫
R d 1
χ G (x, y)dx = 0
for almost all y. Then m(G y ) = 0 for a.e. y, since E y ⊂ G y we have m(E y ) = 0
for almost every y. Then ∫
χ Ey (x)dx = 0
and so
∫
R d 2
R d 1
∫
R d 1
χ E dxdy = 0.
Step 4 Conclusion. After all the cases of Step 3, let us check that χ E F,
when E is any measurable set of finite measure. In this case, there exist a G δ
set G, with E ⊂ G and m(G \ E) = 0, thus χ E = χ G − χ G\E , thus χ E is a
linear combination of elements in F. By the first step, χ E ∈ F. When f is
any integrable function, we can decompose f = f + − f − where f + and f − are
non negative functions. In each case, f + , and f − are limits of simple functions
over sets of finite measure. By the previous steps f ∈ F. So L 1 (R d ) ∈ F as we
wanted to show.
It is worth to note that Fubini’s theorem is valid when f is a non negative
measurable function:
Theorem 56. Suppose f(x, y) is integrable on R d = R d1 ×R d2 , then for almost
every y ∈ R d2
(1) The slice f y is integrable on R d1 .
(2) The function given by
is integrable in R d2 . Moreover,
(3) ∫
R d 2
∫
R d 1
∫
y ↦→
R d 1
f y (x)dx
∫
f y (x)dxdy = f(x)dx
R d
Here is how we apply Theorem 56, suppose we are given a measurable function
f and we want to compute ∫ R d f to justify the use of iterated integrals
we only apply it for |f|. If we found that ∫ |f| < ∞, then we get that f is
integrable.
41

Proof. Consider the truncations
{
f(x, y) if |(x, y)| ≤ k and f(x, y) ≤ k
f k (x, y) =
0 otherwise.
Thus f k is integrable for every k, and f k ր f. By Fubini’s Theorem, for every
k there exist E k such that m(E k ) = 0 and f y k is integrable for every y ∈ Ec k .
Let E = ⋃ E k , so m(E) = 0 and for every y ∈ E c we have f y k ր fy where f y
is integrable. Moreover, by the Monotone Convergence Theorem
∫ ∫
f y k (x)dx → f y (x)dx
R d 1
R d 1
and
hence
∫ ∫
∫ ∫
f y k (x)dxdy → f y (x)dxdy
R d 2 R d 1
R d 2 R d 1
∫ ∫
f k → f
which implies by Fubini’s theorem on f k that
∫ ∫
∫
f y (x)dxdy =
R d 2 R d 1
f.
Corollary 57. Let E be a measurable set in R d1 × R d2 then
E y = {x ∈ R d1 |(x, y) ∈ E}
is measurable for almost every y ∈ R d2 , the function y ↦→ m(E y ) is measurable
and
∫
m(E) = m(E y )dy
Proof. The corollary is just Theorem 56 applied to χ E .
R d 2
The converse to this corollary is false as the following example shows.
Example 58. Consider E = [0, 1] × N, where N is the non-measurable set we
constructed in Example 15. Then E y is [0, 1] for y ∈ N or 0 if y ∈ N c . Hence
E y is measurable for every y ∈ R, but E is not measurable in R 2 .
Theorem 59. If E = E 1 × E 2 is measurable in R d and m(E 2 ) > 0 then E 1 is
measurable.
42

Proof. By the corollary above, for almost all y ∈ R d2 the function
χ y E 1×E 2
(x) = χ E1 (x) · χ E2 (y)
is a measurable function of x. Now, we claim that there exist y ∈ E 2 such that
χ y E 1×E 2
is measurable. If this claim is true, we have that χ y E 1×E 2
= χ E1 (x) is
measurable, which implies that E 1 is measurable. Let us proof the claim. Let
F ⊂ R d2 be the set of all y such that E y is measurable, then by assumption
m(F c ) = 0. Now
E 2 = (E 2 ∩ F) ∪ (E 2 ∩ F c ).
but
0 < m(E 2 ) = m(E 2 ∩ F) + m(E 2 ∩ F c ) = m(E 2 ∩ F)
since m(E ∩ F c ) = 0. Thus E 2 ∩ F ≠ ∅ which proves the claim.
Proposition 60. Let E 1 ⊂ R d1 and E 2 ⊂ R d2 be any subsets. Then
m ∗ (E 1 × E 2 ) ≤ m ∗ (E 1 )m ∗ (E 2 ).
If any E has outer measure 0, then m ∗ (E 1 × E 2 ) = 0.
Theorem 61. Suppose E 1 ⊂ R d1 and E 2 ⊂ R d2 are measurable subsets. Then
E = E 1 × E 2 is measurable in R d and
m(E) = m(E 1 ) · m(E 2 ).
Proof. It is enough to prove that E is measurable. Since E 1 and E 2 are measurable
there exist G δ sets, G 1 ⊂ R d1 and G 2 ⊂ R d2 , such that E i ⊂ G i and
m ∗ (E i ) = m(G i ), for i = 1, 2, but then G = G 1 × G 2 is G δ , hence measurable,
and
G 1 × G 2 \ (E 1 × E 2 ) ⊂ (G 1 \ E 1 ) × G 2 ∪ G 1 × (G 2 \ E 2 ).
Then m ∗ (G \ E) = 0, so E is measurable.
Corollary 62. Suppose f is a measurable function on R d1 , then ˜f(x, y) = f(x)
is measurable on R d .
Proof. For all a ∈ R, the set E 1 = {x ∈ R d1 |f(x) < α} is measurable in R d1 .
Then
{(x, y) ∈ R d : ˜f(x, y) < α} = E 1 × R
is measurable. Thus, ˜f is measurable.
Corollary 63. Suppose f(x) is a non-negative function on R d and let
A = {(x, y) ∈ R d × R : 0 ≤ y ≤ f(x)}.
Then f is measurable if and only if A is measurable in R d+1 . Moreover, if f is
measurable then ∫
R d f(x)dx = m(A).
43

Proof. If f is measurable on R d , then the previous proposition guarantees that
the function
F(x, y) = y − f(x)
is measurable, and the set A = {y > 0} ∩ {F ≤ 0} is measurable. For the
converse, assume A is measurable, then for every x ∈ R d
A x = {y ∈ R|(x, y) ∈ A} = [0, f(x)]
is measurable, and the function x ↦→ m(A x ) = f(x) is measurable. Also
∫
∫
∫
m(A) = χ A (x, y)dydx = m(A x )dx = f(x)dx.
R d 1
R d 1
Proposition 64. If f is measurable on R d , then ˜f(x, y) = f(x−y) is measurable
on R d × R d .
This proposition is particularly useful to define convolutions. Given f, g
measurable on R d , then f(x − y)g(y) is measurable on R 2d . If each function is
integrable then f(x−y)g(y) is also integrable. The convolution of two integrable
functions f, g is given by
∫
(f ∗ g)(x) = f(x − y)g(y)dy
R d
Which is a well defined function of x for almost all x.
6.1 Complex valued integrals
Let f : R d → C, then if Re(f) = u and Im(f) = v,
f(x) = u(x) + iv(x)
where u, v : R d → R. We say that f is Lebesgue integrable if an only if
u(x) and v(x) are both integrable. Note that f is integrable if and only if
|f(x)| = √ (u(x)) 2 + (v(x)) 2 is integrable. In case f is integrable we put
∫ ∫ ∫
f = u + i v.
R d R d R d
7 Lebesgue L p spaces
7.1 The space L 2
Let L 2 (E) ≡ L 2 (R d , C) be the set of complex valued functions f that satisfy
∫
R d |f(x)| 2 dx < ∞
44

The requirement that L 2 (R d ) consist of complex valued functions is just to
establish the definitions on its greater generality. However, all properties of
L 2 (R d ) that will be discussed here, apply for the space L 2 (R d , R) of real valued
functions satisfying ∫ E |f(x)|2 dx < ∞.
We can define the L 2 -norm on L 2 (R d ) for a function f by
(∫
‖f‖ 2 = |f(x)| 2 dx
R d
Again, as in the case of L 1 we will consider two functions f, g ∈ L 2 (R d ) to be
the same if f = g almost everywhere. An important property of L 2 (R d ), is that
is equipped with an inner product given by
∫
< f, g >= f(x)g(x)dx for all f, g ∈ L 2 (R d ).
R d
Here ¯z denotes complex conjugation. Note that
< f, f > 1 2 = ‖f‖2 .
) 1
2
Theorem 65. The space L 2 (R d ) has the following properties:
(1) L 2 (R d ) is a vector space over C.
(2) If f, g ∈ L 2 (R d ) then fḡ ∈ L 1 (R d ), and the Cauchy-Schwarz inequality
holds
| < f, g > | ≤ ‖f‖ 2 ‖g‖ 2 .
(3) If g ∈ L 2 (R d ) is fixed. The map f ↦→< f, g > is linear and < f, g >=
< g, f >.
(4) The triangle inequality holds
‖f + g‖ 2 ≤ ‖f‖ 2 + ‖g‖ 2 .
Proof. Proof part (1). If f, g ∈ L 2 (R d ) then
which implies
|f(x) + g(x)| ≤ 2 max{|f(x)|, |g(x)|}
|f(x) + g(x)| 2 ≤ 4(|f(x)| + |g(x)|) 2
integrating both sides:
∫
(∫ ∫ )
|f(x) + g(x)| 2 dx ≤ 4
R d |f(x)| 2 dx +
R d |g(x)| 2 dx
R d
so f + g ∈ L 2 (R d ). It is clear that for all λ ∈ C, λf ∈ L 2 (R d ) whenever
f ∈ L 2 (R d ).
45

Proof of part (2). Let us check that fḡ is integrable, for that we use the fact
that, A, B ≥ 0 then 2AB ≤ A 2 + B 2 so
|f||ḡ| ≤ 1 (
|f| 2 + |g| 2) ,
2
integrating
∫
|f||ḡ| ≤ 1 2
(∫
∫
|f| 2 +
|g| 2 )
< ∞,
thus fḡ is integrable. Let us prove Cauchy-Schwarz, if either f or g are such
that ‖f‖ 2 = 0 or ‖g‖ 2 = 0 then fḡ = 0 almost every where, so < f, g >= 0 and
the inequality holds. Assume first that ‖f‖ 2 = ‖g‖ 2 = 1, hence
∫
| < f, g > | ≤ |fḡ| ≤ 1 2
(
‖f‖
2
2 + ‖g‖ 2 )
2 = 1 = ‖f‖2 ‖g‖ 2 .
If neither f nor g have norm zero, in the general case consider ˜f =
˜g =
that is,
g
‖g‖ 2
, then we have ‖ ˜f‖ 2 = ‖˜g‖ 2 = 1, by the previous case
which implies
∣ ∣∣∣
∫
∣∫
∣∣∣
˜f(˜g)
∣ =
| < ˜f, ˜g > | ≤ 1,
|f||ḡ|
‖f‖ 2 ‖g‖ 2
∣ ∣∣∣
=
∣ ∣∣∣
∫
fḡ
∣ ≤ ‖f‖ 2‖g‖ 2
∣∫
1 ∣∣∣
‖f‖ 2 ‖g‖ 2
fḡ
∣ ≤ 1
f
‖f‖ 2
and
Proof part (3). This is a consequence of the linearity of the integral: For all
f 1 , f 2 ∈ L 2 (R d ) and α ∈ C
∫ ∫ ∫
< f 1 + f 2 , g >= (f 1 + f 2 )ḡ = f 1 ḡ + f 2 ḡ =< f 1 , g > + < f 2 , g >
also,
∫
< αf, g >=
∫
(αf)ḡ = α
fḡ = α < f, g >
and
∫
< f, g > =
∫
fḡ =
∫
fḡ =
¯fg =< g, f > .
Proof part (4).
‖f + g‖ 2 2 =< f + g, f + g >
= ‖f‖ 2 2+ < f, g > + < g, f > +‖g‖ 2 2
= ‖f‖ 2 2 + 2| < f, g > | + ‖g‖2 2
≤ ‖f‖ 2 2 + 2‖f‖ 2 ‖g‖ 2 + ‖g‖ 2 2
by the Cauchy-Schwarz inequality, hence
‖f + g‖ 2 2 ≤ (‖f‖ 2 + ‖g‖ 2 ) 2 .
46

As in the case of L 1 , the space L 2 (R d ) is a metric space with metric given
by d(f, g) = ‖f − g‖ 2 and Riesz-Fisher’s Theorem holds:
Theorem 66 (Riesz-Fisher). The space L 2 (R d ) is complete in its metric.
Proof. The proof is a modification of the argument in proof of Riesz-Fisher
theorem for L 1 . Let {f n } be a Cauchy sequence in L 2 , by Lemma 52 to prove that
{f n } converges, it is enough to show that {f n } has a convergent subsequence.
Since {f n } is Cauchy, we can find a subsequence {f nk } satisfying
Define
and
Consider the partial sums
and
‖f nk+1 (x) − f nk (x)‖ 2 ≤ 1 for all k ≥ 1.
2k f(x) = f n1 +
g(x) = |f n1 | +
∞∑
(f nk+1 (x) − f nk (x))
k=1
∞∑
|f nk+1 (x) − f nk (x)|.
k=1
S K (f)(x) = f n1 +
S K (g)(x) = |f n1 | +
By the triangle inequality, for all k
‖S K (g)‖ 2 ≤ ‖f n1 ‖ 2 +
K∑
(f nk+1 (x) − f nk (x))
k=1
K∑
|f nk+1 (x) − f nk (x)|.
k=1
K∑
‖f nk+1 (x) − f nk (x)‖ 2 ≤ ‖f n1 ‖ 2 + 2,
k=1
thus the partial sums S K (g) are bounded and S K (g) ր g, by the Monotone
Convergence Theorem g is integrable and belongs to L 2 (R d ). Since |f| ≤ g, f
is also integrable and
|f| 2 ≤ |g| 2
implies that the function f belongs to L 2 (R d ). The partial sums S K (f) telescopes
to f nk so S K (f)(x) = f nk (x) → f(x) for almost all x. To prove that
f nk → f(x) in L 2 (R d ), note that
|f − S K (f)| 2 ≤ (2g) 2 for all K,
then by the Dominated Convergence Theorem
‖f nk − f‖ 2 → 0
47

Definition. A metric vector space X is called separable, if it contains a countable
set L such that the set of all finite linear combinations of elements in L is
dense in X.
Theorem 67. The space L 2 (R d ) is separable.
Sketch of the proof. Let
and
Q × iQ = {z ∈ C|Re(x) ∈ Q, Im(z) ∈ Q}
Q = {All rectangles in R d with coordinates inQ × iQ}
Since Q × iQ and Q are countable, the set
A = {rχ R : r ∈ Q × iQ and R ∈ Q}
is countable. To finish the proof, one has to check that every simple function in
R d can be approximated by finite linear combination of elements in A.
7.2 Hilbert Spaces
Definition. A set H is called a Hilbert space if satisfies the following properties:
(1) H is a vector space over C (or R)
(2) H it is equipped with an inner product 〈·, ·〉 satisfying
(i) The map f ↦→ 〈f, g〉 is linear for every g fixed.
(ii) 〈f, g〉 = 〈g, f〉.
(iii) 〈f, f〉 ≥ 0 for all f ∈ H
(3) The inner product defines a norm ‖‖ given by
and ‖f‖ = 0 if and only if f = 0.
‖f‖ = 〈f, f〉 1 2 ,
(4) The Cauchy-Schwarz and triangle inequalities hold. Thus, for all f, g ∈ H
|〈f, g〉| ≤ ‖f‖‖g‖ and ‖f + g‖ ≤ ‖f‖ + ‖g‖
(5) H is complete in the metric d(f, g) = ‖f − g‖.
(6) H is separable.
As we have seen, the space L 2 (R d ) under almost everywhere equivalence is
a Hilbert space. Now we give a list of useful Hilbert spaces.
48

Example 68. If E is a measurable set in R d with m(E) > 0, the space
∫
L 2 (E) = {f supported on E| |f(x)| 2 dx < ∞}
is also a Hilbert space, with inner product given by
∫
〈f, g〉 = f(x)g(x)dx
and norm
(∫
‖f‖ 2 =
E
E
E
) 1
|f(x)| 2 2
dx .
Formally, we have to think L 2 (E) under the equivalence relation that f is equivalent
g if and only if f = g almost everywhere.
Example 69. A simple example is the finite dimensional complex Euclidean
space
C n = {(a 1 , ..., a n )|a k ∈ C}
the inner product is given by
and norm
〈a, b〉 =
‖a‖ =
n∑
a k¯bk
k=1
( n
∑
k=1
a 2 k
)1
2
.
Example 70. An infinite dimensional analog of C n is
l 2 (Z) = {(..., a −2 , a −2 , a 0 , a 1 , a 2 , ...)|a k ∈ C,
the inner product is given by
〈a, b〉 =
∞∑
k=−∞
a k¯bk
∞∑
k=−∞
|a k | 2 < ∞}
and norm
A variant is
‖a‖ =
( ∞
∑
k=−∞
a 2 k
l 2 (N) = {(a 1 , a 2 , ...)|a k ∈ C,
) 1
2
.
∞∑
|a k | 2 < ∞}
k=1
49

It turns out, all infinite dimensional Hilbert spaces are l 2 in disguise
Orthogonality Let f, g ∈ H be elements in a Hilbert space, f and g are called
orthogonal or perpendicular if 〈f, g〉 = 0, and we write f ⊥ g. The first simple
observation is Pythagoras theorem for Hilbert spaces:
Theorem 71 (Pythagoras Theorem). If f ⊥ g then
‖f + g‖ 2 = ‖f‖ 2 + ‖g‖ 2 .
Proof. Note that 〈f, g〉 = 0 implies 〈g, f〉 = 0 and
‖f + g‖ 2 = ‖f‖ 2 + 〈f, g〉 + 〈g, f〉 + ‖g‖ 2 = ‖f‖ 2 + ‖g‖ 2 .
Definition. A finite or countable subset {e 1 , e 2 , ...} is called orthonormal if
{
1 if i = j
〈e i , e j 〉 =
0 if i ≠ j.
As a generalization of Pythagoras theorem we have:
Proposition 72. If {e k } ∞ k=1 is an orthonormal set in H and f = ∑ N
k=1 e k ∈ H
where N is a finite, then
N∑
‖f‖ 2 = |a k | 2 .
k=1
Given {e k } ∞ k=1 a natural problem is to determine if it spans all H. That is,
if the set of finite linear combinations of {e k } ∞ k=1
is dense in H. In this case we
say that {e k } ∞ k=1
is an orthonormal basis.
Example 73. Consider L 2 ([−π, π]) with normalized inner product
〈f, g〉 = 1 ∫ π
f(x)g(x)dx.
2π −π
The set {e inx } ∞ n=∞ is an orthonormal basis of L2 ([−π, π]). Given f ∈ L 2 ([−π, π]),
and a n = 〈f, e inx 〉 then f ∼ ∑ ∞
n=−∞ a ne inx .
We say that f n → f in norm if ‖f n − f‖ → 0.
Theorem 74. The following properties of an orthogonal set {e k } are equivalent:
(1) The set of finite linear combinations of elements in {e k } is dense in H.
(2) If f ∈ H and 〈f, e j 〉 = 0 for all j, then f = 0.
(3) If f ∈ H and S N (f) = ∑ N
k=1 a ke k , where a k = 〈f, e k 〉 and S N (f) → f as
N → ∞ in norm.
50

(4) If a k = 〈f, e k 〉, then ‖f‖ 2 = ∑ ∞
k=1 |a k| 2 .
Proof. (1)⇒ (2). There exist a sequence {g n } a sequence of finite linear combination
of e k with g n → f, such that ‖f − g n ‖ → 0, since 〈f, e j 〉 = 0 then
〈f, g n 〉 = 0 for all n. So
‖f‖ 2 = 〈f, f〉 = 〈f, f − g n 〉 = 0
for all n. By taking the limit we get ‖f‖ = 0 which implies f = 0.
(2)⇒ (3). For f ∈ H, let S N (f) = ∑ N
k=1 a ke k where a k = 〈f, e k 〉. Then
By Pythagoras theorem we have
thus
(f − S N (f)) ⊥ S N (f)
‖f‖ 2 = ‖f − S N (f)‖ 2 + ‖S N (f)‖ 2 = ‖f − S N (f)‖ 2 +
‖f‖ 2 ≥
N∑
|a k | 2
k=1
letting n → ∞, we obtain Bessel’s inequality
∞∑
‖f‖ 2 ≥ |a k | 2
So,
k=1
∞∑
|a k | 2
k=1
N∑
|a k | 2 ,
converges, which implies that {S N (f)} is a Cauchy sequence, because H is
complete, S N (f) → g for some g ∈ H. Fix j, for N large enough
〈f − S N (f), e j 〉 = a j − a j = 0
Since S N (f) → g, then 〈f − g, e j 〉 = 0 for all j, which by the assumption in (2)
implies f = g and
∞∑
f = a k e k
(3)⇒ (4). It is immediate that
‖f‖ 2 =
k=1
∞∑
|a k | 2 .
k=1
This equation is known as Parseval’s identity.
(4)⇒ (1). By definition S N (f) is a finite linear combination of {e k } and by the
assumption in (4),
‖f − S N (f)‖ → 0.
k=1
51

Let us remark that Bessel’s inequality holds for all orthogonal families, in
contrast Parseval’s identity holds only when the family is a basis.
Proposition 75. Any Hilbert space has an orthonormal basis.
Riesz Representation Theorem.
Definition. A bounded linear functional in H, is a linear function φ : H → R
such that there exist A > 0, such that for all f ∈ H
|φ(f)| ≤ A‖f‖
Example 76. If g ∈ L 2 (R d ) is fixed then
∫
φ g (f) =< f, g >=
R d }fḡ
is a bounded linear functional. Note that φ g is well defined, that is if g ′ = g a.e.,
then φ g = φ g ′. The function φ is bounded by the Cauchy-Schwarz inequality:
∫
|φ g (f)| ≤ |fḡ|dµ ≤ ‖f‖ 2 ‖ḡ‖ 2
R d
and taking A = ‖ḡ‖ 2 .
The following theorem states that all bounded linear functional in a H are
of this form.
Theorem 77 (Riesz Representation Theorem). Let φ be a continuous linear
functional on a Hilbert space H. Then there exist a unique g ∈ H such that
for all f ∈ H.
φ(f) = 〈f, g〉
Proof. Consider the subspace of H defined by
S = {f ∈ H|φ(f) = 0}.
Since φ is continuous, S is a closed subspace of H. The space S is called the
null-space of φ. If S = H, then φ = 0 and we can take g = 0. Otherwise, S ⊥ ≠ 0
so there exist h ∈ S ⊥ with ‖h‖ = 1, let g = φ(h)h and for every f ∈ H let
u = φ(f)h − φ(h)f, then φ(u) = 0 which implies that u ∈ S and 〈u, h〉 = 0.
Then we have
that is,
0 = 〈φ(f)h − φ(h)f, h〉 = φ(f)〈h, h〉 − 〈f, φ(h)h〉
as we wanted to show.
φ(f) = 〈f, g〉,
52

7.3 The Lebesgue spaces L p .
For p ≥ 1 we sat that f ∈ L p (R d ) (resp. L p (E), for a measurable set E in R d ),
if |f| p is integrable over R d (resp. E). Again, we will consider two functions, f
and g to be the same if f = g a.e.
Definition. For each p ≥ 1, and p < ∞ we define
∫
L p (E) = {f measurable on E| |f| p < ∞}
the L p -norm of a function f ∈ L p (E) is given by
(∫ ) 1
‖f‖ p = |f| p p
(Case p = ∞) Let f : E → [−∞, ∞] be a measurable function, then
esssup f = inf{c||f| < c a.e. }
Definition. A measurable function f : E → [−∞, ∞] is called essentially
bounded if
esssup |f| < ∞.
The space of essentially bounded functions on E is denoted by L ∞ (E).
It is easy to check that L ∞ (E) is a vector space with norm
E
‖f‖ ∞ = esssup |f|.
Here again, two functions f and g are considered the same if they are equal
almost everywhere.
Let us check that for all p ≥ 1, the Lebesgue space L p is a vector space.
Remind that if f and g are measurable, then cf and f + g are measurable.
Moreover, since |cf(x)| p = |c| p |f(x)|, we have
(∫
‖cf(x)‖ p =
E
|cf(x)| p ) 1
p
= |c|
(∫
E
E
|f(x)| p )
= |c|‖f‖ p
Thus cf ∈ L p (E) if, and only if, f ∈ L p (E). Now note that
|f(x) + g(x)| p ≤ 2 p max{|f(x)| p , |g(x)| p }
which implies ‖f + g‖ p < ∞ if, and only if, ‖f‖ p < ∞ and ‖g(x)‖ p < ∞.
Clearly, ‖f‖ p = 0 if, and only if, |f(x)| = 0 a.e., that is f(x) = 0 a.e.
For 1 ≤ p < ∞ triangle inequality in L p (E) is by no means obvious. However,
the case p = ∞ is rather easy:
implies
|f + g| ≤ |f| + |g|
‖f + g‖ ∞ ≤ ‖f‖ ∞ + ‖g‖ ∞ .
Besides ‖f‖ ∞ = 0, if and only if f = 0 a.e.
53

Lemma 78. For any non negative real numbers x, y and all α, β ∈ (0, 1) such
that α + β = 1, we have the following inequality
x α y β ≤ αx + βy
Proof. If x = 0, the inequality is obvious. Let x > 0 and let f(t) = (1 − β) +
βt − t β , for t ≥ 0 and β as given. Then
so we have
and
f ′ (t) = (1 − β) − βt β−1
f ′ (t) < 0 for t ∈ (0, 1)
f ′ (t) > 0 for t ∈ (1, ∞)
which implies that f(1) is the only minimum of f on [0, ∞], hence f(t) ≥ 0 for
all t ≥ 0. Let t = y x , then (1 − β) + β y ( y
) β
x − ≥ 0,
x
(1 − β) + β y x ≥ ( y
x) β
,
(1 − β)x + βy ≥ y β x 1−β ,
αx + βy ≥ x α y β .
Theorem 79 (Hölder’s Inequality). If 1 p + 1 q = 1 and p > 1, then for f ∈ Lp (E)
and g ∈ L q (E) we have fg ∈ L 1 (E) and ‖fg‖ 1 ≤ ‖f‖ p ‖q‖ q , that is
∫
E
(∫
|fg| ≤
E
) 1 (∫ ) 1
|f| p p
|g| q q
E
Proof. If ‖f‖ p = 0 or ‖g‖ q = 0 then |fg| = 0 a.e., and the inequality follows,
hence we can assume that ‖f‖ p ≠ 0 and ‖g‖ q ≠ 0. Let us consider first the case
where ‖f‖ p = ‖g‖ q = 1, so we want to show
∫
|fg| ≤ 1.
Take α = 1 p , β = 1 q , x = |f|p , y = |g| q on the previous Lemma so we have
|fg| = x 1 p y
1
q ≤
1
p |f|p + 1 q |g|q .
Integrating both sides
∫
|fg| ≤ 1 ∫
p
|f| p + 1 q
∫
|g| q = 1 p + 1 q = 1
54

as we wanted to show. For the general case let
˜f =
f
‖f‖ p
and ˜g =
then ‖ ˜f‖ p = ‖˜g‖ q = 1 by the previous case
g
‖g‖ p
‖ ˜f˜g‖ 1 ≤ ‖ ˜f‖ p ‖˜g‖ q ≤ 1,
hence
‖fg‖ 1
‖f‖ p ‖g‖ q
≤ ‖f‖ p‖g‖ q
‖f‖ p ‖g‖ q
‖fg‖ 1 ≤ ‖f‖ p ‖g‖ q .
Now we prove the triangle inequality for L p , which is called Minkowski’s
inequality:
Theorem 80 (Minkowski’s Inequality). Let f, g ∈ L p where p ≥ 1, then
Proof. First, note that
‖f + g‖ p ≤ ‖f‖ p + ‖g‖ p .
|f + g| p ≤ |f + g||f + g| p−1 ≤ |f||f + g| p−1 + |g||f + g| p−1 .
Let q be such that 1 p + 1 q = 1, by definition, f +g ∈ Lp implies that |f +g| p ∈ L 1 ,
since p = q(p − 1) then |f + g| p−1 ∈ L q by Hölder’s inequality
∫
|f||f + g| p−1 ≤ ‖f‖ p
(∫
|f + g| (p−1)q ) 1
q
≤ ‖f‖p ‖f + g‖ p q p . (1)
Similarly,
∫
|g||f + g| p−1 ≤ ‖g‖ p ‖f + g‖ p q p . (2)
Then combining previous inequalities, we obtain
∫ ∫
∫
‖f + g‖ p p = |f + g| p ≤ |f||f + g| p−1 +
|g||f + g| p−1
≤ (‖f‖ p + ‖g‖ p ) ‖f + g‖ p q .
If ‖f + g‖ p = 0, Minkowski’s inequality is valid. Assume ‖f + g‖ p ≠ 0 so we
can divide each side by |f + g‖ p q
p to get
‖f + g‖ p− p q
p = ‖f + g‖ p ≤ ‖f‖ p + ‖g‖ p .
55

For all p, the L p norm ‖‖ p defines a metric, given by d(f, g) = ‖f − g‖ p , as
in the cases for p = 1 and p = 2 we have:
Theorem 81 (Riesz-Fisher). For all p with 1 ≤ p ≤ ∞, the Lebesgue space
L p (E) is complete on its metric.
Proof. The case p < ∞ uses the same argument as for p = 1 or p = 2, so we do
not repeat it. The case p = ∞ is much simpler, let {f n } be a Cauchy sequence
in L ∞ (E), we want to show that {f n } converges. For each n, there exist A n
with m(A n ) = 0 and such that
|f n (x)| ≤ ‖f n ‖ ∞ for all x ∈ E \ A n
Let A = ⋃ A n so m(A) = 0, then we can assume that
and
|f n (x)| ≤ ‖f n ‖ ∞ for all x ∈ E \ A and all n
|f n (x) − f m (x)| ≤ ‖f n − f m ‖ ∞ for all x ∈ E \ A and all n, m.
Since {f n } is a Cauchy sequence in L ∞ , then the sequence {f n (x)} is uniformly
convergent in E \ A, define
{
limf n (x) x ∈ E \ A
f(x) =
0 x ∈ A.
Then, f is measurable and ‖f n − f‖ ∞ → 0.
8 Abstract measures
8.1 Outer measures and Caratheodory measurable sets
Let X be any set, an exterior measure µ ∗ on X is a function µ ∗ : P → [0, ∞]
satisfying
(1) µ ∗ (∅) = 0
(2) If E 1 ⊂ E 2 , then µ ∗ (E 1 ) ≤ µ ∗ (E 2 ),
(3) If {E n } is a countable family of sets, then
∞⋃
∞∑
µ ∗ ( E n ) ≤ µ ∗ (E n )
n=1
n=1
Example 82. Lebesgue outer measure m ∗ on R d .
56

8.2 Measurability
A set E ⊂ X is called Caratheodory measurable (or just measurable) if for all
A ⊂ X one has
µ ∗ (A) = µ ∗ (E ∩ A) + µ ∗ (A ∩ E c ). (3)
Because µ ∗ is subadditive, in order to prove equation (3) for a set E, we just
need to check
µ ∗ (A) ≥ µ ∗ (E ∩ A) + µ ∗ (A ∩ E c ).
Proposition 83. Given an exterior measure µ ∗ on a set X, the collection M
of Caratheodory measurable sets forms a σ-algebra. Moreover, µ = µ ∗ restricted
to the sets in M is a measure on M.
Proof. For all A ⊂ X,
µ ∗ (∅ ∩ A) + µ ∗ (∅ c ∩ A)
= µ ∗ (∅ ∩ A) + µ ∗ (X ∩ A) = µ ∗ (A),
hence ∅, X ∈ M. By the symmetry of equation (3), it is clear that E ∈ M if,
and only if E c ∈ M. Now, let us check that M is closed under finite unions. If
E 1 , E 2 ∈ M, and A ⊂ X, then
µ ∗ (A) = µ ∗ (E 2 ∩ A) + µ(E c 2 ∩ A)
= µ ∗ (E 1 ∩ E 2 ∩ A) + µ ∗ (E c 1 ∩ E 2 ∩ A)
+µ ∗ (E 1 ∩ E c 2 ∩ A) + µ∗ (E c 1 ∩ Ec 2 ∩ A)
≥ µ ∗ ((E 1 ∪ E 2 ) ∩ A) + µ ∗ ((E 1 ∪ E 2 ) c ∩ A)
where the second equality is equation (3) apply to E 2 and E 1 ∩ A, and the last
inequality uses
(E 1 ∪ E 2 ) ∩ A = (E 1 ∩ E 2 ∩ A) ∪ (E c 1 ∩ E 2 ∩ A) ∪ (E 1 ∩ E c 2 ∩ A)
sudadditivity of µ ∗ , and E c 1 ∩Ec 2 = (E 1∪E 2 ) c . All together implies that E 1 ∪E 2 ∈
M, if in addition we assume E 1 ∩ E 2 = ∅, we have
µ(E 1 ∪ E 2 ) = µ ∗ (E 1 ∪ E 2 ) = µ ∗ (E 1 ∩ (E 1 ∪ E 2 ) + µ ∗ (E c 1 ∩ (E 1 ∪ E 2 ))
= µ ∗ (E 1 ) + µ ∗ (E 2 ) = µ(E 1 ) + µ(E 2 ).
Hence, µ is additive for finite union of disjoint sets in M. Now we have to
check that M is closed under countable unions, and µ is additive for unions of
countable disjoint sets.
Since M is closed under finite unions and complements, every set E that is
countable union of sets in M, is also a countable union of disjoint sets. Hence
it is enough to check that M is closed under countable union of disjoint sets in
M. Let {E k } be a countable collection of disjoint sets in M and define
G n =
n⋃
∞⋃
E k , and G = E k .
k=1
57
k=1

Since G n ∈ M then
µ ∗ (G n ∩ A) = µ ∗ (E n ∩ (G n ∩ A)) + µ ∗ (En c ∩ (G n ∩ A))
= µ ∗ (E n ∩ A) + µ ∗ (G n−1 ∩ A) =
but G c ⊂ G c n thus
µ ∗ (A) = µ ∗ (G n ∩ A) + µ ∗ (G c n ∩ A) ≥
letting n → ∞ we obtain
n∑
µ ∗ (E k ∩ A)
k=1
n∑
µ ∗ (E k ∩ A) + µ ∗ (G c ∩ A)
k=1
∞∑
µ ∗ (A) ≥ µ ∗ (E k ∩ A) + µ ∗ (G c ∩ A)
k=1
≥ µ ∗ (G ∩ A) + µ ∗ (G c ∩ A) ≥ µ ∗ (A)
that is G ∈ M. To check that µ is additive take A = G, then
∞∑
∞∑
µ ∗ (G) = µ ∗ (E k ∩ G) + µ ∗ (G c ∩ G) = µ ∗ (E k ).
k=1
k=1
Let E ⊂ X be a measurable set such that µ ∗ (E) = 0, then for all A ⊂ X,
A ∩ E ⊂ E and µ ∗ (A ∩ E) = 0 so
µ ∗ (A ∩ E) + µ ∗ (A ∩ E c ) ≤ µ ∗ (A)
which implies that E ∈ M. Since, the exterior measure µ ∗ is monotone whenever
F ⊂ E, µ ∗ (F) = 0 and then F M. Then, the measure µ restricted to
Caratheodory measurable sets is complete. Not all measures µ on σ-algebras A
are complete. Since, it is possible that subsets of sets of µ-measure 0 in A may
not belong to A:
Example 84. The Borel σ-algebra B in R is the minimal σ-algebra containing
all open sets in R. The Lebesgue measure m restricted to the Borel σ-algebra is
not complete. There are sets of Lebesgue measure 0 that are not Borel.
Now we are going to prove that the measurability notions of Caratheodory
and Lebesgue are equivalent. For convenience, let us remind that a set E ⊂ R d
is Lebesgue measurable if for every ǫ > 0 there exist an open set U, with E ⊂ U
and such that
m ∗ (U \ E) < ǫ.
This is equivalent to the existence of a G δ set G, such that m ∗ (G△E) = 0
58

Theorem 85. Let E ∈ R d , then E is Lebesgue measurable if and only if E is
Caratheodory measurable with respect to m ∗ .
Proof. Let E be Lebesgue measurable and let A be any subset of R d . Then
there exist a G δ set G such that m ∗ (A) = m ∗ (G) then
m ∗ (E ∩ A) + m ∗ (E c ∩ A) ≤ m ∗ (E ∩ G) + m ∗ (E c ∩ G)
= m(E ∩ G) + m(E c ∩ G) = m(G) = m ∗ (A).
Hence E is Caratheodory measurable.
Now, assume E is a Caratheodory measurable set, since m∗ is σ-finite on R d ,
we may assume m ∗ (E) < ∞. Take a G δ such that E ⊂ G, with m ∗ (E) = m ∗ (G),
m ∗ (G) = m ∗ (E ∩ G) + m ∗ (E c ∩ G)
= m ∗ (E) + m ∗ (E c ∩ G)
so m ∗ (G \ E) = 0 which implies that E is Lebesgue measurable.
Let (X, A, µ) a measure space, where X is the underlying set, A a σ-algebra.
Assume that the measure µ is σ-finite, that is there exist a countable collection
{E k } ⊂ A with µ(E k ) < ∞ for all k, and X = ∪E k .
A function f : X → [−∞, ∞] is called µ-measurable (or just measurable) if
for all a ∈ R
f −1 ([−∞, ∞]) = {x ∈ X|f(x) < a} ∈ A.
Then we can define the integral
∫ ( ∫
f(x)dx =
X
X
∫ ∫
fdµ = fdµ =
in the same way we defined the Lebesgue integral. For instance, if φ = ∑ a k χ Ek
is a simple function, define
∫
φdµ = ∑ a k µ(E k ),
X
then generalize to bounded functions, non negative and finally, we say that a
µ-measurable function f is integrable if
∫
fdµ < ∞
X
)
f
Moreover, this definition of integral satisfies the following properties:
Proposition 86. Let {f n } be a sequence of µ-measurable functions.
(1) Fatou’s Lemma. If for every n, f n is non negative
∫
∫
liminf f n dµ ≤ liminf f n dµ.
59

(2) Monotone Convergence Theorem. If all f n are non negative, and f n ր f,
then
∫ ∫
f n dµ = fdµ
lim
n→∞
(3) Dominated Convergence Theorem. If f n → f a.e., and |f n | < g where g
is a µ-integrable function, then
∫
|f n − f|dµ → 0
which implies
∫
∫
f n dµ →
fdµ.
For (X, A, µ), we can define L p (X, A, µ) = {f µ − measurable | ∫ X |f|p dµ <
∞}. The L 2 space is a Hilbert space, and L p are vector spaces with a norm
given by
(∫ ) 1
‖f‖ p = |f| d p
dµ .
X
All theorems we proved for L p (R d ) and L 2 (R d ) carry on to the spaces L p (X, A, µ).
8.3 Radon-Nikodym theorem
Definition. Let µ, ν be two σ-finite measures on (X, A), we say that µ is absolutely
continuous with respect to ν, and write µ ≺≺ ν if ν(E) = 0 implies
µ(E) = 0.
Example 87. Let f be a non negative ν-measurable function, define
∫
µ(E) = fdν
for all E ∈ A. Then µ ≺≺ ν.
Example 88. Let m be the Lebesgue measure on R, and Q = {q n } some enumeration
of the rational numbers. Define
µ(E) =
∞∑
n=1
E
1
2 nχ E(q n ),
then µ is not absolutely continuous with respect to m.
Theorem 89 (Radon-Nikodym). Let µ, ν be two σ-finite measures on (X, A).
If µ ≺≺ ν, then there exists a non negative measurable function g : X → R such
that for every µ-integrable function f,
∫ ∫
fdµ = fgdν.
Moreover, g is unique up to null sets.
X
X
60

Definition. In this case we write gdν = dµ or g = dµ
dν
, the Radon-Nikodym
derivative of µ with respect to ν.
Proof. Since µ and ν are σ-finite, it is enough to consider the case µ(X), ν(X) <
∞, once this case is settle the general case follows by an argument over a countable
partition of the space.
Define λ = µ + ν and let φ : L 2 (X, A, λ) → R be the functional
∫
φ(f) = fdµ.
This φ is well defined, if f 1 = f 2 λ-a.e. Then
0 = λ({f 1 ≠ f 2 }) = µ({f 1 ≠ f 2 }) + ν({f 1 ≠ f 2 })
X
≥ µ({f 1 ≠ f 2 })
then f 1 = f 2 µ-a.e. Which implies
∫ ∫
φ(f 1 ) = f 1 dµ = f 2 dµ = φ(f 2 ).
Clearly φ is linear, and by Cauchy-Schwarz inequality in L 2 (X, A, λ), φ is also
bounded:
∫ ∫
|φ(f)| ≤ |f|dµ ≤ |f|dλ
√ ∫ ∫
≤ 1 dλ√ 2 |f| 2 dλ ≤ √ λ(X) · ‖f‖ L2 (X,λ)
By the Riesz Representation Theorem there exist h ∈ L 2 (X, A, λ) such that for
all f
∫
φ(f) = fhdλ
then
∫
∫
fdµ =
∫
fhdλ =
∫
fhdµ +
fhdν,
and
∫
∫
f(1 − h)dµ =
fhdν
so, we have (1 − h)dµ = hdν which suggest
dµ = h
1 − h dν.
But we have to prove it rigorously:
First, we claim that 0 ≤ h ≤ 1 for λ-a.e. To prove the claim, note that for
all E ∈ A, χ E ∈ L 2 (X, A, λ). We also have
∫
µ(E) = φ(χ E ) = hdλ
61
E

and
Now
then
∫
ν(E) = λ(E) − µ(E) =
∫
0 ≥
{h1}
then ν({h > 1}) = 0 and since µ ≺≺ ν, we have µ({h > 1}) = 0, thus
λ({h > 1}) = 0 which proves our claim.
Note that ∫
∫
χ E · (1 − h)dµ = χ E hdν,
by linearity we also have
∫
∫
φ · (1 − h)dµ =
φhdν,
for all simple functions φ. If f is non negative, choose a sequence of non negative
simple functions φ k ր f, then we also have 0 ≤ φ k (1 − h) ր f(1 − h) and
0 ≤ φ k h ր fh. By the Monotone Convergence Theorem (applied twice)
∫
∫
f(1 − h)dµ = lim
n→∞
φ n (1 − h)dµ
∫ ∫
= lim
n→∞
φ n hdν = fhdν
for all functions f non negative and measurable. But if f ≥ 0 then f
1−h ≥ 0 by
the claim before. By replacing f
1−h
instead of f in the equation above, we have
∫ ( ) ∫ ( )
f
fh
(1 − h)dµ = dν
1 − h
1 − h
that is
So, g = h
1−h
∫ ∫
h
fdµ = f
1 − h dν.
is the function we were looking for.
62

8.4 Signed measures
Definition. Let (X, A) be a measurable space. Where X is the underlying set
and A is a σ-algebra. A signed measure µ is a set function defined on A,
µ : A → [−∞, ∞] satisfying:
(1) µ attains at must one of the values ±∞,
(2) µ(∅) = 0
(3) (Additivity) If E = ⊔ ∞
n=1 E n is a disjoint union of sets {E n } in A, then
m(E) =
∞∑
n=1
E n
The triple (X, A, µ) is called a signed measure space.
We will need the following definition for some measurable sets in A:
Definition. Let (X, A, µ) be a signed measure space.
(1) A set E ∈ A is called null if for every E ′ ⊆ E measurable in A, m(E ′ ) = 0.
(2) A set E ∈ A is called positive if for every E ′ ⊆ E measurable in A,
m(E ′ ) ≥ 0.
(3) A set E ∈ A is called negative if for every E ′ ⊆ E measurable in A,
m(E ′ ) ≤ 0.
Lemma 90. Let (X, A, µ) be a signed measure space. Let E ∈ A such that
µ(E) > 0, then there exist a positive set E ′ ⊂ E and such that µ(E ′ ) > 0.
Proof. If E is not positive, there is a subset if E with negative measure. Let n 1
be the smallest positive number such that there exist E 1 ⊂ E and µ(E 1 ) < − 1 n 1
.
So
µ(E \ E 1 ) = µ(E) − µ(E 1 ) > µ(E) ≥ 0.
Again, if E \ E 1 is positive then we are done, otherwise let n 2 be the smallest
positive number such that there exist E 2 ⊂ E \ E 1 and µ(E 2 ) < − 1 n 2
.
Repeating the argument we obtain a sequence of disjoints sets {E k }, and a
sequence of numbers {n k } such that µ(E k ) ≤ − 1
n k
. Let F = ⋃ ∞
k=1 E k, so
∞∑
∞∑ 1
µ(F) = µ(E k ) < − ≤ 0
n k
k=1
Hence 1
n k
→ 0. Let G ⊂ E \ F be a measurable set in A, if µ(G) < 0, then
there exist k, such that µ(G) < − 1
(n k −1) , but G ⊂ E \F \E \ {E 1 ∪... ∪E k } and
n k − 1 < n k contradicting the definition of n k , Therefore µ(G) ≥ 0 and then
µ(E \ F) = µ(E) − µ(F) > 0.
k=1
63

Theorem 91 (Hahn’s Decomposition Theorem). If µ is a signed measure on
A, there are sets P, and N with X = P ∪N and P ∩N = ∅ where P is positive
and N is negative.
Proof. The class of positive sets P since ∅ ∈ P. Then, we can define
α = sup{µ(A)|A ∈ P}
By replacing −µ instead of µ, we can assume that µ does not attain the value
∞ and α < ∞. By definition, there exist a sequence {A n } of positive sets
such that µ(A n ) → ∞. The union of two positive sets is positive, thus we
can choose A n to be strictly increasing. Let P = ⋃ A n . Then A n ր P and
µ(P) = limµ(A n ) = α, also P is positive, since for every E ∈ A
µ(E ∩ P) = µ(E ∩ ⋃ A n ) = µ( ⋃ (A n ∩ E)) = limµ(A n ∩ E) ≥ 0
Let N = X \P, we claim that N is negative, otherwise there is a set E ⊂ N with
µ(E) > 0, by the previous Lemma, there exist S ⊂ E positive with µ(S) > 0,
then
µ(P ∪ S) = µ(P) + µ(S) > α
contradicting the definition of α.
Lemma 92. If X = P 1 ∪N 1 = P 2 ∪N 2 are two Hanh decompositions for µ then
µ(P 1 △ P 2 ) = 0 and µ(N 1 △ N 2 ) = 0.
Proof. We only prove that µ(P 1 △P 2 ) = 0. Since P 1 \P 2 is positive and P 1 \P 2 ⊂
N 2 , then µ(P 1 \P 2 ) = 0. Similarly µ(P 2 \P 1 ) = 0, and P 1 △P 2 = P 1 \P 2 ∪P 2 \P 1
implies the equation.
Definition. Let µ, ν be two measures on (X, A), we say that µ and ν are mutually
singular, and write µ⊥ν, if there are disjoint sets X µ and X ν in A such
that
X = X µ ∪ X ν , and µ(X µ ) = µ(X ν ) = 0
Theorem 93 (Jordan’s decomposition Theorem). Let (X, A, µ) be a signed
measure. Then there exist a decomposition µ = µ + − µ − , where µ + and µ − are
proper measures and µ + ⊥µ − .
Proof. Let X = P ∪ N be a Hanh decomposition of X, and for every E ⊂ A,
let
µ + (E) = µ(E ∩ P) and µ − (E) = −µ(E ∩ N).
The reader can check that µ + and µ − are proper measures, µ = µ + − µ − and,
µ + ⊥µ − .
64