MEASURE THEORY AND LEBESGUE INTEGRAL
7
2. Abstract integration theory
In order to begin our approach to the theory of integration, we need to discuss the notion of
measurable functions.
2.1. Measurable functions. In analogy with the definition of continuous functions, as morphisms between topological spaces, we have the following definition.
Definition 2.1. Let (X , M), (Y, N ) be measurable spaces, and f : X ! Y be given. We say
that f is (M, N )-measurable if f 1 (E) 2 M for all E 2 N .
It is clear that if f : X ! Y is (M, N )-measurable and g : Y ! Z is (N , O)-measurable,
then g f : X ! Z is (M, O)-measurable. Also, observe that if N is a -algebra in P(Y),
then f 1 (E) : E 2 N is a -algebra in P(X ). In fact, [j f 1 (Ej ) = f 1 [j Ej and
c f 1 (E) = f 1 ( c E), and the conclusion follows.
Proposition 2.2. Let f : X ! Y be (M, N )-measurable and suppose that N is generated by E.
Then f is (M, N )-measurable if and only if f 1 (E) 2 M for all E 2 E.
Hence, if X , Y are topological spaces, and f : X ! Y continuous, then f is (BX , BY )measurable.
Proof. First of all, we observe that the implication “only if” is trivial. Conversely, suppose that
f 1 (E) 2 M for all E 2 E. It is easy to check that E ✓ Y : f 1 (E) 2 M is a -algebra in
P(Y) that contains E. Hence, it contains the -algebra generated by E, that is, it contains N
and f is (M, N )-measurable. The second conclusion now is obvious.
⇤
Definition 2.3. If (X , M) is a measurable space, then f : X ! R is said to be measurable if it
is (M, BR )-measurable.
It is however convenient to consider functions that take value in the extended reals, that is,
in R = R [ {±1} = [ 1, +1]. In this case, the -algebra of Borel sets BR is defined as
E ✓ R : E \ R 2 BR . Then we say that f : X ! R is measurable if it is (M, BR )-measurable.
Proposition 2.4. Let (X , M) be a measurable space, and let f : X ! R be given. Then the
following conditions are equivalent.
(i) f is measurable, that is, it is (M, BR )-measurable.
(ii) f
1
(iii) f
1
(iv) f
1
(v) f
1
[ 1, b) 2 M for every b 2 R.
[ 1, b] 2 M for every b 2 R.
(a, +1] 2 M for every a 2 R.
[a, +1] 2 M for every a 2 R.
Proof. If f : X ! R, the conclusions follow at once from Prop.’s 1.7 and 2.2. If f takes values
in [ 1, +1], the conclusions are a simple consequence of Def. 2.3.
⇤
The following simple result will be needed later on.
Lemma 2.5. Let (X , M) be a measurable space, and let f : X ! C be given. Then f is
measurable if and only if Re f and Im f are measurable.
8
M. M. PELOSO
Proof. We identify f with (f1 , f2 )f : X ! R2 . Observe that the coordinate maps ⇡j : R2 ! R,
j = 1, 2 are measurable. Since composition of measurable functions is measurable and f1 =
Re f = ⇡1 f , f2 = Im f = ⇡2 f , f measurable implies Re f, Im f are measurable.
Conversely, suppose Re f and Im f are measurable. By Prop. 1.8 we know that BC = BR2 =
BR ⌦ BR . Then, if E 2 BC , E = E1 ⇥ E2 , with E1 , E2 2 BR and f 1 (E) = f1 1 (E1 ) \ f2 1 (E2 )
that is in M by assumption.
⇤
We remark that when we require that f takes values in R we include the case of f having
finite values.
Theorem 2.6. Let (X , M) be a measurable space.
(i) If f, g : X ! R are measurable, then f + g, f g are measurable.
(ii) If fj : X ! R are measurable, j = 1, 2, . . . , then
g1 = sup fj ,
j
g2 = inf fj ,
g3 = lim sup fj ,
j
g4 = lim inf fj
j
j!+1
are measurable.
(iii) f, g : X ! R are measurable, then
max(f, g),
min(f, g)
are measurable.
(iv) If fj : X ! C are measurable, j = 1, 2, . . . , and g(x) = limj!+1 fj (x) exists, then g is
measurable.
Proof. (i) Observe that f + g = S F , where F (x) = (f (x), g(x)) and S : C ⇥ C ! C is the
sum-funciton, i.e. S(z + w) = z + w. Arguing as in Lemma 2.5 we see that F is measurable.
Since S is continuous is measurable, and then f + g is measurable. Replacing S by P , where
P (z, w) = zw, we obtain the measurability of f g.
(ii) Notice that
(a, +1] = x 2 X : sup fj (x) > a =
+1
[
g2 1 [ 1, b) = x 2 X : inf fj (x) < b =
+1
[
g1
1
j
while
j
The measurability of g1 and g2 follows from Prop. 2.4.
Next, we observe that3
fj
1
(a, +1] ,
fj
1
[ 1, b) .
j=1
j=1
g3 (x) = lim sup fj (x) = inf sup fj (x)
j!+1
k
j k
Then hk (x) = supj<k fj (x) are measurable, so is inf k hk (x) = g3 (x). The argument for g4 is
analogous, since
g4 (x) = lim inf fj (x) = sup inf fj (x) .
j!+1
k
Finally, (iii) and (iv) are trivial consequences of (ii)
j k
⇤
3We recall that, given a sequence {a }, lim sup a = s⇤ , where s⇤ is sup of the limit points of {a }. Then,
n
n
n n
s⇤ = inf k (supj k an ). Analogously, if we set s⇤ = lim inf n an , then s⇤ = supk (inf j k an ).
MEASURE THEORY AND LEBESGUE INTEGRAL
9
Corollary 2.7. If f : X ! R is measurable, then f+ = max(f, 0) and f = min(f, 0) are
measurable. If g : X ! C is measurable, then |g| and sgn g := g/|g| are measurable.
We point out that f = f+
f , |f | = f+ + f , and g = sgn g · |g|.
Proof. We only need to prove the statements for |g| and sgn g. But these are elementary and
we leave the details to the reader.
⇤
Definition 2.8. Let (X , M) be a measurable space, and let E 2 M. We define the characteristic
function of E as the function
(
1
if x 2 E
E (x) =
0
otherwise .
We call a simple function a finite linear combination with complex coefficients of characteristic
funcitons of measurable sets Ej
n
X
f (x) =
cj Ej (x) .
(2)
j=1
Observe that simple functions can characterized as the measurable functions whose range is
finite (that is, they attain at most finitely many values). We remark that the representation (2)
of a simple function f is not unique, but it becomes unique if we write
f (x) =
m
X
dj
Fj (x) .
j=1
where Fj = f 1 ({dj }), dj 6= 0, and d1 , . . . , dm = f X \ {0} is the range of f taken away the
value 0.
Notice also that finite sums and products of simple functions are again simple functions.
The main result of this section is that measurable can be suitable approximated with simple
functions, as the next result shows.
Theorem 2.9. Let (X , M) be a measurable space.
(1) Let f : X ! [0, +1] be a measurable funtion. Then there exists an increasing sequence of
non-negative measurable functions 0  s1  s2  · · ·  f such that sn (x) ! f (x) as n ! +1
and the convergence is uniform on all sets where f is bounded.
(2) Let f : X ! C be a measurable funtion. Then there exists a sequence of complex-valued
simple measurable functions {'n } such that 0  |'1 |  |'2 |  · · ·  |f | such that 'n (x) ! f (x)
as n ! +1 and the convergence is uniform on all sets where f is bounded.
Proof. (1) For n = 0, 1, 2, . . . and k integer, 0  k  22n
Fn = f
1
(2n , +1]
En,k = f
1 we define the sets
1
(k2
k2
n
n
, (k + 1)2
and set
sn (x) = 2
n
Fn (x)
+
2n 1
2X
k=0
En,k (x) .
n
] ,
10
M. M. PELOSO
Clearly {sn } are measurable and non-negative. In order to check that {sn } is increasing, notice
that
⇤
⇤
k2 n , (k + 1)2 n = 2k2 (n+1) , (2k + 2)2 (n+1)
⇤
⇤
= 2k2 (n+1) , (2k + 1)2 (n+1) [ (2k + 1)2 (n+1) , (2k + 2)2 (n+1) .
This implies that
En,k = En+1,2k [ En+1,2k+1 .
Moreover,
n
2 ,2
n+1
so that
⇤
22(n+1)
[ 1
=
j2
Therefore, using (3) and (4),
sn (x) = 2
Fn (x)
2n 1
2X
+
k2
n
k=0
n
=2
2
Fn+1 (x)
n
Fn+1 (x)
+
⇣
+
+
2k2
+
2n 1
2X
(n+1)
=2
Fn+1 (x)
+2
n
En+1,2k+1 (x)
2k2
(n+1)
En+1,2k (x)
En+1,j (x)
j=22n+1
2
n+1
Fn+1 (x)
= sn+1 (x) .
+
X
(4)
⇣
+
⌘
En+1,2k (x)
2n 1
2X
+
(2k + 1)2
En+1,2k+1 (x)
(n+1)
⌘
En+1,2k+1 (x)
k=0
22(n+1)
X 1
22(n+1)
,
En+1,j .
k=0
n
⇤
j=22n+1
k=0
Fn \Fn+1 (x)
2n 1
2X
22(n+1)
[ 1
En+1,2k (x)
Fn \Fn+1 (x)
+
(n+1)
, (j + 1)2
j=22n+1
Fn \ Fn+1 =
n
(n+1)
(3)
+
22n+1
X 1
En+1,` (x)
`=0
1
j2
(n+1)
`2
(n+1)
En+1,j (x)
j=22n+1
+
22n+1
X 1
`2
(n+1)
En+1,` (x)
`=0
Hence, {sn } is monotone increasing. It is clear that sn (x)  f (x) for all x. Finally, notice that for
x2
/ Fn , 0  f (x) sn (x)  2 n . Then, sm ! f uniformly on each set c Fn = x : f (x)  2n .
On F = \n Fn , f (x) = +1 and sn (x) ! +1 for x 2 F . This proves (1).
(2) now follows easily from (1). If f : X ! C, we write f = u + iv, u = u+ u , v = v+ v .
±
Then, there exists increasing sequences {s±
n } and { n } of non-negative simple functions such
that
±
s±
and
n ! u± ,
n ! v± ,
as in (1). Then, setting
'n = s +
n
sn + i
+
n
n
MEASURE THEORY AND LEBESGUE INTEGRAL
11
we have that 'n ! u+ u + i(v+ v ) = f , pointwise, and uniformly on all sets where f is
bounded. Finally, it also holds that
⇥
⇤
⇥
⇤
2 1/2
2
2
+ 2
2 1/2
|'n | = (s+
sn )2 + ( n+
= (s+
n
n)
n ) + (sn ) + ( n ) + ( n )
⇥
⇤
+
2
2
2
2 1/2
 (s+
n+1 ) + (sn+1 ) + ( n+1 ) + ( n+1 )
 |'n+1 |
for all n, where we have used the obvious fact that g+ g = 0 for all g.
⇤
Definition 2.10. Given a measure space (X , M, µ), we say that a property (P) is valid µ-a.e.
(or, simply, a.e. if the measure µ is understood by the context), if there exists a set F 2 M,
with µ(F ) = 0 such that the property (P) holds on c F .
One convenience of working with complete measures is the following result.
Proposition 2.11. Let (X , M, µ) be a measure space The following implications are true if and
only if the measure µ is complete.
(1) If f is measurable and g = f µ-a.e., then g is measurable.
(2) If fn are measurable, n = 1, 2, . . . and fn ! f pointwise µ-a.e., then f is measurable.
Proof. Exercise.
⇤
On the other hand, we have the following result.
Proposition 2.12. Let (X , M, µ) be a measure space and let (X , M, µ) be its completion. If f
is M-measurable, then there exists g M-measurable function such that f = g µ-a.e.
Proof. The proof is indicative of a method that will be often used. Suppose first f is a characteristic function of an M-measurable set E = E [ N , with E 2 M, N ✓ F , µ(F ) = 0. Then
f = E = E µ-a.e. and the conclusion holds true in this
Pcase. In the same way we see that the
conclusion follows in the case of a simple function f = nj=1 cj E j .
In the general case, given f an M-measurable function, by Thm. 2.9 we can find a sequence
of M-measurable functions {'n } pointwise converging to f . For each n, let n be an Mmeasurable function, n = 'n except on a set Nn 2 M with µ(Nn ) = 0. Then, there exist
Fn 2 M, Fn ◆ Nn and µ(Fn ) = 0. Set F = [n Fn and define g = limn!+1 c F n . Then g is
M-measurable by the previous Prop. 2.11 (2), and g = f on c F , that is, µ-a.e.
⇤
2.2. Integration of non-negative functions. We begin the construction of the integral with
the simple functions.
Definition 2.13. Let (X , M, µ) be a measure space. We define
L+ = f : X ! [0, +1], M measurable .
(5)
P
n
If s 2 L+ is a simple function s = j=1 cj Ej we define the integral of s with respect to µ,
Z
n
X
s dµ =
cj µ(Ej ) ,
j=1
12
M. M. PELOSO
(with the standard convention that 0 · 1 = 0). More generally, if E 2 M we also define
Z
Z
n
X
s dµ =
cj µ(E \ Ej ) .
E s dµ =
E
j=1
We will use the following notation to denote the integral of s w.r.t. µ:
Z
Z
Z
Z
s dµ = s =
s dµ =
s(x) dµ(x)
X
X
R
and similarly in the case of E s dµ.
The following properties are easy to check.
Proposition 2.14. Let (X , M, µ) be a measure space and s,
on X . Then, the following
properties
Z
Z hold true.
be non-negative simple functions
(i) If c 0, then
cs dµ = c s dµ ;
Z
Z
Z
(ii) (s + ) dµ = s dµ +
dµ;
Z
Z
(iii) if s  , then
s dµ 
dµ;
Z
(iv) the map M 3 E 7!
s dµ is a measure.
E
We are now in the position to define the integral of non-negative measurable functions.
Definition 2.15. Let (X , M, µ) be a measure space. We define
If f 2 L+ we set
L+ = f : X ! [0, +1], M
Z
f dµ = sup
For any E 2 M we also define
nZ
Z
(6)
o
s dµ : 0  s  f, s simple .
f dµ =
E
measurable .
Z
f
E
(7)
dµ .
It follows at once that, if c
0 and f 2 L+ , then cf 2 L+ and
follows at once that if f, g 2 L+ and f  g, then
Z
Z
f dµ  g dµ .
R
cf dµ = c
R
f dµ. It also
(8)
The next result is a fundamental building block of this theory.
Theorem 2.16. (Monotone Convergence Theorem.) Let (X , M, µ) be a measure space
{fn } be given such that fn : X ! [0, +1], 0  f1  f2  · · · . Setting f (x) = limn!+1 fn (x),
we have
Z
Z
f dµ = lim
fn dµ .
n!+1
Notice that the conclusion can we written in the suggestive form
Z ⇣
Z
⌘
lim fn dµ = lim
fn dµ .
n!+1
n!+1
MEASURE THEORY AND LEBESGUE INTEGRAL
13
Proof. Since 0  f1 (x)  f2 (x)  · · · is a monotone sequence the limit function f exists, and it
is measurable by Thm. 2.6 (iv). Moreover, by the mononicity of the integral, property (8),
Z
Z
f1 dµ  f2 dµ  · · · ,
R
R
R
that is, the sequence
fn dµ is increasing, so that it has a limit and since fn dµ  f dµ for
all n4, by the comparison test
Z
Z
lim
fn dµ  f dµ .
n!+1
In order to prove the reverse inequality, let s be a simple function such that 0  s  f and
let 0 < ↵ < 1. Since limn!+1 fn (x) = f (x), setting
En = x : fn (x)
↵s(x) ,
we have that E1 ✓ E2 ✓ · · · , and [+1
n=1 En = X . Therefore, by the monotonicity of the integral,
Z
Z
Z
Z
↵
s dµ 
fn dµ =
f
dµ

fn dµ .
(9)
En n
En
En
By Prop. R2.14 (iv), since
{En } is an increasing sequence of measurable sets whose union is X ,
R
limn!+1 En s dµ = s dµ, therefore, passing to the limit in (9) we obtain
Z
Z
↵ s dµ  lim
fn dµ .
n!+1
This holds for all 0 < ↵ < 1 so that
Z
s dµ  lim
n!+1
Z
fn dµ .
Passing to the supremum on the left hand side for all simple functions s with 0  s  f we
obtain
Z
Z
f dµ  lim
fn dµ ,
n!+1
⇤
and we are done.
Corollary 2.17. (1) Let f, g 2 L+ . Then
Z
Z
Z
(f + g) dµ = f dµ + g dµ .
(2) If fn 2 L+ , n = 1, 2, . . . , then
Z ⇣X
+1
n=1
⌘
fn dµ =
+1 Z
X
fn dµ
n=1
Notice that part (1) is just the statement of the additivity of the integral. In the present
theory, in order to obtain such property we had to recourse to the Monotone Convergence
Theorem (that we will abbreviate as MCT in what follows).
14
M. M. PELOSO
Proof. Let {sn }, { n } be sequences of non-negative simple functions, monotonically converging
to f and g, respectively. Then, {sn + n } is a sequence of non-negative simple functions,
monotonically converging to f + g. By the MCT,
Z
Z
Z
Z
Z
Z
(f + g) dµ = lim
(sn + n ) dµ = lim
sn dµ + lim
sigman dµ = f dµ + g dµ .
n!+1
n!+1
n!+1
This proves (1). In order to prove (2), notive that by (1) and induction we have that
Z ⇣X
N
N Z
⌘
X
fn dµ =
fn dµ ,
PN
n=1
n=1
+
for every N . Observe that
n=1 fn is a sequence of functions in L that converges monoP+1
tonically a n=1 fn . Hence, applying the MCT again we obtain
Z ⇣X
Z ⇣X
+1
N
N Z
+1 Z
⌘
⌘
X
X
fn dµ = lim
fn dµ = lim
fn dµ =
fn dµ ,
n=1
n!+1
n!+1
n=1
n=1
n=1
⇤
as we wished to show.
R
Corollary 2.18. Let f 2 L+ . Then f dµ = 0 if and only if f = 0 µ-a.e. Therefore, if
f, g 2 L+ and f = g a.e., then
Z
Z
f dµ = g dµ .
R
P
Proof. Assume first the f is simple, f = j cj Ej . Then f = 0 clearly implies f dµ = 0, while
R
P
if 0 = f dµ = j cj µ(Ej ), it must be either cj = 0 or µ(Ej ) = 0, for each j. In any case, f = 0
µ-a.e.
+
R Let now f 2 L and assume first f = 0 µ-a.e. If s is simple, 0  s  f R, then s = 0 µ-a.e.,
sdµ = R0, and passing to the supremum of such functions we obtain that f dµ = 0. Finally,
suppose f dµ = 0. Then
x : f (x) > 0 =
+1
[
x : f (x) > 1/n =:
n=1
Since E1 ✓ E2 ✓ · · · ,
µ
⇣
+1
[
En ,
n=1
x : f (x) > 0
⌘
= lim µ(En ) .
n!+1
If f is not equal to 0 a.e., then it must be µ(En ) > 0 for some n, but then
Z
Z
Z
1
1
f dµ
f dµ
dµ = µ(En ) > 0 ,
n
n
En
En
R
a contradiction. Then, f dµ = 0 implies f = 0 µ-a.e., and we are done.
⇤
The conclusion of the MCT holds true also if we just assume that {fn } is a sequence in L+
monotonically convergent a.e.
Corollary 2.19. Let {fn } is a sequence in L+ monotonically convergent a.e. to f 2 L+ , then
Z
Z
f dµ = lim
fn dµ .
n!+1
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
Remark 2.22. We observe that Rthe previous result cannot be improved to have equality even
if we assume that 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 .
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
n!+1
fn dµ .