MEASURE THEORY AND LEBESGUE INTEGRAL
15
Proof. Let E 2 M be such that µ(E) = 0, and f1 (x)  f2 (x)  · · · and limn!+1 fn (x) = f (x)
for x R2 c E. Setting
gn = c E fn andRg = c ERf , we observe that gn = fn a.e. and g = f a.e. so
R
that gn dµ = fn dµ, for all n and gdµ = f dµ, by Cor. 2.18.
We may apply the MCT to {gn } and obtain that
Z
Z
Z
Z
f dµ = g dµ = lim
gn dµ = lim
fn dµ
n!+1
n!+1
⇤
as we wish to show.
In the MCT the key assumption, besides the non-negativity of the functions, was the monotonicity of the sequence. Such monotonicity guaranteed the existence of the limits. In the
general case, we have the following.
Theorem 2.20. (Fatou’s Lemma.) Let {fn } ✓ L+ , then
Z ⇣
Z
⌘
lim inf fn dµ  lim inf fn dµ .
n!+1
n!+1
Proof. As in the proof of Thm. 2.6 (iv), if we set gk (x) = inf jk fj (x), for k = 1, 2, . . . , we have
that {gk } is an increasing sequence in L+ and limk gk = lim inf n!+1 fn . Moreover, gk  fk so
that
Z
Z
lim
gk dµ  lim inf fn dµ .
n!+1
k!+1
Therefore, by the MCT,
Z ⇣
Z ⇣
Z
Z
⌘
⌘
lim inf fn dµ =
lim gk dµ = lim
gk dµ  lim inf fn dµ ,
n!+1
k!+1
k!+1
n!+1
as we wished to show.
⇤
The following result follows at once from Fatou’s Lemma and Cor. 2.19.
Corollary 2.21. Let {fn } ✓ L+ and suppose fn ! f a.e. Then
Z
Z
f dµ  lim inf fn dµ .
n!+1
Proof. Let E 2 M be a set such that fn ! f in E and µ( c E) = 0. Set gn =
Then gn ! g, fn = gn a.e. and f = g a.e. By Fatou’s Lemma,
Z
Z
Z
Z
f dµ = g dµ  lim inf gn dµ = lim inf fn dµ . ⇤
n!+1
E fn
and g =
Ef.
n!+1
Remark 2.22. We observe that the previousR result cannot be improved to have equality even
if we assume that the limit of the sequence
fn dµ exists.
Indeed, consider the measure space (N, P(N), µ), where µ is the counting measure. Let sn
be the numerical sequence {sn,k } that is equal to 1 if k = n and equals 0 otherwise. Then the
sequence of functions on N {sn } converges to 0 pointwise, but
Z
+1
X
sn dµ =
sn,k = 1
N
for all n. Therefore,
0=
Z
N
k=1
s dµ  lim
Z
n!+1 N
sn dµ = 1 .
16
M. M. PELOSO
Remark 2.23. We also observe thatRthere exists a version of the MCT for decreasing sequences
+
{f
R n }, but one needs to assume that f1 dµ < +1. Precisely, if {fn } ✓ L , f1 f2 · · · , and
f1 dµ < +1, then
Z
Z
f dµ = lim
fn dµ .
n!+1
Proof. We first observe that if 0  g  f , then f g 0 and
Z
Z
Z
f g) dµ = f dµ
g dµ .
Next, since {fn } is monotone and non-negative, the limit f exists it is non-negative, and
fn
f for each n. Set gn = f1 fn . Then {gn } is a sequence of non-negative functions such
that gn  gn+1 for all n and converging to f1 f . Applying the MCT we obtain
Z
Z
Z ⇣
Z
Z
Z
⇣Z
f1 dµ
f dµ =
f1 f ) dµ = g dµ = lim
gn dµ = lim
f1 dµ
fn dµ .
n!+1
n!+1
R
Therefore, since f1 dµ < +1 we can subtractive from both sides to obtain
Z
Z
f dµ = lim
fn dµ ,
n!+1
⇤
as we wished to prove.
2.3. Integration of complex-valued functions. We now pass to consider generic real, or
complex, valued functions. If (X , M, µ) is a measure space, f : X ! C is measurable, and
u = Re f , v = Im f , we write
f = u+ u + i(v+ v ) .
Then, u± , v± : X ! [0, +1). If f is real-valued, we simply write f = f+
f .
Definition 2.24. Given a measure space (X , M, µ), given f : X ! C is measurable, we say
that f is absolutely integrable, or simply integrable, if
Z
|f | dµ < +1;
equivalently, if
0
Z
For f integrable we set
Z
Z
f dµ = u+ dµ
u± dµ,
Z
Z
v± dµ < +1 .
u dµ + i
In particular, if f is real-valued, we have
Z
Z
f dµ = f+ dµ
✓Z
v+ dµ
Z
f dµ .
Z
◆
v dµ .
Remark 2.25. Notice that, if f is integrable, f = u + iv, then 0  u± , v±  |f | so that
Z
Z
Z
0  u± dµ,
v± dµ  |f | dµ < +1 .
MEASURE THEORY AND LEBESGUE INTEGRAL
17
R
R
Conversely, if 0  u± dµ, v± dµ < +1, then |f |  |u| + |v| = u+ + u + v+ + v , so that
Z
Z
Z
Z
Z
|f | dµ  u+ dµ + u dµ + v+ dµ + v dµ < +1 .
Moreover, the set of integrable functions is a complex vector space, as it is easy to check, and
for integrable f, g and a, b 2 C we have
Z
Z
Z
af + bg) dµ = a f dµ + b g dµ .
Proposition 2.26. The following properties hold true.
Z
Z
(i) If f is integrable, then
f dµ  |f |dµ .
Z
Z
(ii) If f, g are integrable, then
f dµ =
g dµ for every E 2 M if and only if |f
E
E
Z
a.e. if and only if
|f g| dµ = 0.
Proof. (i) If
g| = 0
R
f = 0 the result is trivial. Next, if f is real
Z
Z
Z
Z
Z
Z
f dµ =
f+ dµ
f dµ  f+ dµ + f dµ = |f | dµ .
R
If f isR complex-valued,
and f 6= 0, from Cor. 2.7 that sgn f, |f | are measurable and we set
R
↵ = | f dµ|/( f dµ). Then
Z
Z
Z
f dµ = ↵ f dµ = ↵f dµ
R
so that, in particular ↵f dµ is non-negative. Therefore,
Z
Z
Z
Z
Z
f dµ = Re ↵f dµ = Re(↵f ) dµ  |↵f | dµ = |f | dµ ,
since |↵| = 1.
R
(ii) If f, g 2 L1 (µ), from Cor. 2.18 we know
R that |f g| dµ = 0 if and only if |f
a.e., that is, if and only if f = g a.e. Now, if |f g| dµ = 0, then, for every E 2 M,
Z
Z
Z
Z
f dµ
g dµ 
g| dµ  |f g| dµ = 0 ,
E |f
E
so that
g| = 0
E
Z
f dµ =
E
Z
g dµ
E
for every E 2 M .
Finally, suppose that the above condition holds, and assume for simplicity that f and g are
real-valued. Consider theRfunction f g = m and its decomposition m = m+ m . Then, by
assumption we have that E mdµ = 0 for all E 2 M. In particular if we take E+ = x : m(x) 
0 , then we have
Z
Z
Z
0=
m dµ =
m+ dµ = m+ dµ ,
E+
E+
since m+ = 0 on c E+ . Thus, m+ = 0 a.e. and, with a similar argument we obtain also m = 0
a.e. Hence, m = 0 a.e., i.e. f = g a.e., as we wished to show.
⇤
18
M. M. PELOSO
From now on we will identify functions that di↵er only on a set of measure 0. Indeed, we
have the following definition.
Definition 2.27. Given a measure space (X , M, µ), we consider the equivalent relation ⇠ on
the space of integrable saying that f ⇠ g if f = g a.e. We define the space L1 (µ) as the space
of equivalent classes of integrable function modulo the relation ⇠ and we define the norm
Z
kf kL1 (µ) = |f | dµ .
Remark 2.28. We need to check that the above definition gives in fact a norm. It is clear that
kf kL1 (µ)
0 and that f = 0 in L1 (that is, f = 0 a.e.) implies kf kL1 (µ) = 0. Conversely, if
R
0 = kf kL1 (µ) = |f |dµ, then Cor. 2.18 implies that |f | = 0 a.e., ie. f = 0 a.e. and f = 0 in L1 .
The symmetry and triangular inequality of the norm follow easily.
In Thm. 2.31 we will show that L1 is a complete normed space, that is, a Banach space.
Therefore, we have a norm on L1 (µ), hence a metric, given by the expression
d(f, g) = kf
gkL1 (µ) .
Then, given {fn } ✓ L1 (µ) we say that fn ! f in L1 (µ) (or, simply in L1 if the measure µ is
understood) if
kfn f kL1 ! 0
as n ! +1 .
Theorem 2.29. (Dominated Convergence Theorem.) Let (X , M, µ) be a measure space
and {fn } ✓ L1 (µ). Suppose that fn ! f pointwise a.e. and that there exists a non-negative
g 2 L1 (µ) such that g |fn | for all n. Then, fn ! f in L1 (µ) and
Z
Z
f dµ = lim
fn dµ .
n!+1
Proof. Since |fn |  g for all n, since fn ! f pointwise a.e., it follows that f is measurable and,
by the comparison theorem, that |f |  g. Therefore, f 2 L1 (µ).
Observing that |fn f |  2g, we apply Fatou’s Lemma to the sequence {2g |fn f |} and
obtain
Z
Z
Z
⇣ Z
⌘
2g dµ  lim inf 2g |fn f | dµ = 2g dµ + lim inf
|fn f | dµ
n!+1
n!+1
Z
Z
= 2g dµ lim sup |fn f | dµ .
Since
R
n!+1
2g dµ < +1, we can subtract it from both sides of the inequality and obtain that
Z
lim sup |fn f | dµ  0 .
n!+1
This implis that limn!+1
R
|fn f | dµ = 0, that is, fn ! f in L1 (µ). Finally,
Z
Z
Z
fn dµ
f dµ  |fn f | dµ ! 0
and the last conclusion follows.
⇤
MEASURE THEORY AND LEBESGUE INTEGRAL
19
2.4. The space L1 (µ). Our next goal is to show that L1 (µ) is a Banach space, that is it is
complete in its norm. We recall that a normed space (X, k · kX ) is complete if P
it is complete as a
metric space w.r.t the metric d(f, g) = kf gkX . We also recall that aP
series n fn of elements
in X is said to be absolutely convergent in X if the numerical series n kfn kX is convergent.
We have the following result.
Theorem 2.30. P
A normed space (X, k · kX ) is a Banach space if and only if every absolutely
convergent series n fn is convergent in X.
P
Proof. Suppose first X is a Banach space and that n fn is an absolutely convergent series. Let
P
{sN } be the sequence of partial sums, that is, sN = +1
n=1 fn . Then, given " > 0, there exists
n" such that for n" < M < N ,
d(sN , sM ) = ksN
s M kX =
N
X
fn
n=M +1
X

N
X
n=M +1
kfn kX < " .
Hence, {sN } is a Cauchy sequence in X, that converges, since X is complete.
Conversely, suppose every absolutely convergent series is convergent in X. Let {gn } be a
Cauchy sequence in X. Hence, for every k = 1, 2, . . . , there exists an integer Nk such that if
m, n Nk , d(gn , gm ) = kgn gm kX < 2 k . Notice that we may assume that
Nk+1 > Nk for all
P+1
k
k. Then, the subsequence {gNk } is such that kgNk+1 gNk kX < 2 . Then, k=1 kgNk+1 gNk kX
P
converges, so that by assumption, +1
gNk converges to an element g 2 X. But, the
k=1 gNk+1
P+1
series k=1 gNk+1 gNk is telescopic and the partial sum
M
X
gNk+1
gNk = gNM +1
g N1 ,
k=1
so that gNk gN1 ! g as k ! +1. Hence, the original sequence must converge too, and the
conclusion follows.
⇤
Theorem 2.31. Let {fn } be a sequence of functions in L1 (µ) such that
P
1
Then, there exists f 2 L1 (µ) such +1
n=1 fn converges to f in L (µ) and
Z
+1 Z
X
f dµ =
fn dµ .
P+1
n=1 kf kL1 (µ)
< +1.
n=1
As a consequence,
L1 (µ)
is a Banach space. Finally, the simple functions are dense in L1 (µ).
Proof. By Cor. 2.17 (ii) we know that
Z X
+1
XZ
X
|fn |dµ =
|fn |dµ 
kfn kL1 (µ) < +1 .
P+1
n
n
n=1
L1 (µ),
Then, setting g = n=1 |fn |, we have g 2
and hence it is finite a.e. This implies that
P+1
n=1 fn converges but on a set
P of measure 0, call f such limit. We apply the DCT to the
sequence of partial sums sn = nk=1 fk . Clearly, sn ! f pointwise, and also
|sn | =
n
X
k=1
fk 
n
X
k=1
|fk | = g .
20
M. M. PELOSO
R
R
Hence, the DCT applies to give that sn ! f in L1 (µ) and limn!+1 sn dµ = f dµ, that is
Z
Z ⇣X
n
n Z
+1 Z
⌘
X
X
f dµ = lim
fk dµ = lim
fk dµ =
fn dµ .
n!+1
n!+1
k=1
n=1
k=1
L1 (µ),
In particular, every absolutely convergent series also converges in
hence L1 (µ) is complete.
Finally, let f 2 L1 (µ). Then, |f | is finite a.e. and the sequence of simple functions {'n } in
Thm. 2.9 (2), is such that 'n ! f pointwise, and |'n |  |f |. Therefore, we can apply the DCT
to obtain that 'n ! f in L1 (µ).
⇤
We conclude this part by comparing L1 (µ) with L1 (µ). If f 2 L1 (µ) and g is given by by Prop.
2.12,Rthen g = Rf µ-a.e. and g 2 L1 (µ). Conversely, if g 2 L1 (µ), then g is also M-measurable
and |g|dµ = |g|dµ. Then g 2 L1 (µ). Therefore, we have shown that the spaces L1 (µ) and
L1 (µ) can be identified and we will do in what follows.
3. The Lebesgue measure in R
Having established the definition and main properties of abstract integrals, we now go back
to the construction of measures and in particular of the Lebesgue measure and of other Borel
measures in R, and eventually in Rn .
3.1. Outer measures. Given a set X , an outer measure on X is a set function µ⇤ defined on
the set of parts of X , P(X ) such that
(i) µ⇤ (;) = 0;
(ii) if A ✓ B ✓ X , then µ⇤ (A)  µ⇤ (B);
P+1 ⇤
(iii) µ⇤ [+1
j=1 µ (Aj ), for all {Aj } ✓ X .
j=1 Aj 
The reason for this definition is that outer measures arise naturally when one has a family of
elementary sets and a notion of measure for such sets.
Proposition 3.1. Let E be a collection of sets in P(X ) such that ;, X 2 E, and let ⇢ : E !
[0, +1] such that ⇢(;) = 0 be given. Define
⇤
µ (A) = inf
+1
nX
j=1
Then
µ⇤
⇢(Ej ) : A ✓
+1
[
j=1
o
Ej , E j 2 E .
is an outer measure.
The key example to keep in mind is when E is the collection of (all, or a subclass of) intervals
I in R, and ⇢(I) is the length of I, which could be +1 if I is a ray.
Proof. First of all we observe that the definition makes sense, since for any A ✓ X there exists a
covering of A with sets in E, e.g. {X , ;, ;, . . . }, where ;, X 2 E. Also, it is clear that µ⇤ (;) = 0,
since ; can be covered by {;, ;, . . . }. Property (ii) in the definition of an outer measure is
satisfied since, if A ✓ B, and {Ej } is a covering of B with sets in E, {Ej } is also a covering
of A. Thus, the infimum in the definition of µ⇤ (A) is taken with respect a larger collection of
converings than in the case of µ⇤ (B); hence (ii) holds.
MEASURE THEORY AND LEBESGUE INTEGRAL
21
Finally, let {Aj } ✓ P(X ). Fix " > 0. Then, by the properties of the infimum of real numbers,
for each j, there exists a covering {Ej,k } of sets in E of Aj such that
+1
X
⇢(Ej,k ) < µ⇤ (Aj ) + "2
j
.
k=1
Therefore, {Ej,k } for j, k = 1, 2, . . . is a covering of [+1
j=1 Aj by sets in E such that
µ
⇤
⇣ +1
[
⌘
Aj = inf
j=1
nX
n

=
+1
X
⇢(En ) : En 2 E,
µ⇤ (Aj ) + "2
+1
[
j=1
Aj ✓
+1
[
n=1
o
En 
+1
X
⇢(Ek,j )
j,k=1
j
j=1
+1
X
µ⇤ (Aj ) + " .
j=1
⇤
Since " > 0 was arbitrary, (iii) follows.
The key step to pass from an outer measure to an actual measure is based on the following
definition.
Definition 3.2. Let µ⇤ be an outer maesure on a set X . We say that a set E ✓ X is µ⇤ maesurable if
µ⇤ (A) = µ⇤ (A \ E) + µ⇤ (A \ c E) ,
for all A ✓ X .
Observe that we always have the inequality
µ⇤ (A)  µ⇤ (A \ E) + µ⇤ (A \ c E) ,
by subadditivity. Then, in order to prove the µ⇤ -maesurability of a set E, we only need to prove
the reverse inequality
µ⇤ (A) µ⇤ (A \ E) + µ⇤ (A \ c E) ,
for all sets A ✓ X . This inequality is trivial if µ⇤ (A) = +1. Therefore, E ✓ X is µ⇤ -maesurable
if and only if
µ⇤ (A)
µ⇤ (A \ E) + µ⇤ (A \ c E)
for all A with µ⇤ (A) < +1 .
(10)
The key step to pass from an outer measure to an actual measure is the following result,
known as Carathéodory’s theorem.
Theorem 3.3. (Carathéodory’s Theorem.) Let X be a set and µ⇤ an outer measure on X .
Let M be the collection of µ⇤ -maesurable sets, and µ the restriction of µ⇤ to M. Then M is a
-algebra, and µ is a complete measure on M.
Proof. We first observe that E 2 M implies that also c E 2 M since the condition (10) is
symmetric in E and c E. In order to show that M is a -algebra, it suffices to show that it is
closed under countable unions of disjoint sets – see the comment regarding formula (1). To this
end, we first show that M is closed under finite unions. Let E, F 2 M and let A ✓ X . Then