693
Internat. J. Math. & Math. Sci.
Vol. I0 No. 4 (1987) 693-706
THE LEBESGUE INTEGRAL AS A RIEMANN INTEGRAL
ENRIQUE A.
GONZALEZ-VELASCO
Department qf Mathematics
University of Lowell
Lowell, Massachusetts 01854 U.S.A.
IRecelved November 21, 1985 and in
revised form December
30, 19861
The object of this paper is to develop a very direct theory of the Lebesgue
ABSTRACT.
integral that is easily accessible to any audience familiar with the Riemann integral.
KEY WORDS AND PHRASES. Lebegue integration, Ricmann integration.
1980 AMS SUBJECT CLASSIFICATION CODE. 2801.
I.
INTRODUCTION.
The problem of making the theory of Lebesgue integration accessible to a wide
audience, including students of mathematics, science and engineering, is not easily
solved.
And yet, the Lebesgue integral is an indispensable tool in many branches of
mathematics and its applications, such as Fourier analysis and signal processing.
The inaccessibility of the theory is mainly due to the fact that the usual pre-
sentations, both those that are and those that are not based on measure theory, are
lengthy and indirect and rely on a number of new basic concepts that must be assimilated first.
For instance, in Royden
[I]
the general definition of the Lebesgue inte-
gral is preceded by definitions for simple functions, then for bounded functions over
a set of finite measure and then for nonnegative functions, and in each particular
case some amount of theory is elaborated.
In Apostol
[2]
the general definition is
preceded by definitions for step functions and then for upper functions, and again a
certain amount of theory is developed in
classic presentations represent a new
4ach
start,
particular case.
Furthermore, these
with all knowledge of Riemann integra-
tion to be abandoned in favor of the new Lebesgue theory.
[3]
To avoid this, McShane
has recently developed a unified theory of Riemann and Lebesgue integration that
is, hopefully, elementary enough for presentation to a wide audience. While sharing
McShane’s concern and agreeing that he succeeds in constructing an appealing unified
theory, this author believes that, regarding simplicity and accessibility, the answer
lies elsewhere.
There is an entirely different approach to
tegral, due to Mikusi6ski
[4],
te
definition of the Lebesgue in-
which is very direct and does not need the previous
development of either measure theory or the Riemann integral.
It is based, instead,
on a series representation of the function to be integrated, but, since this series
E.A. GONZALEZ-VELASCO
694
representation need not be unique, it requires a somewhat lengthy proof of consistency. In addition, it is not readily apparent how the definition of the Lebesgue
integral in any of the approaches mentioned so far can be immediately used for the
quick computation of simple integrals.
Our task is, then, to construct a simple theory that builds on concepts familiar
to any audience and that is based on a very direct and general definition, capable of
dealing with the computation of simple integrals. To this end, the best course of
action is to define the Lebesgue integral
s
a (possibly improper) Riemann integral.
This is not only brief but allows for complete generality at once and, moreover, we
know how to compute Riemann integrals (there is no need to develop a theory of impro-
per Riemann integrals as a prerequisite because, beyond the definition, all that is
needed is the fact that such an integral is a linear and increasing function of the
integrand).
Such a theory is presented in Rey Pastor
[5],
but only for functions
defined on a set of finite measure and his development makes use of the Riemann-
Stieltjes integral.
The main object of this paper is to reconstruct this theory from
the beginning (with one minor modification} for functions defined over sets of
arbitrary measure and, for additional simplicity, avoiding all use of the Riemann-
Stieltjes integral.
This has necessitated a complete set of new proofs and a collec-
3 and 4 and Lemmas 3, 4 and 5 below).
In addition, we have included the Levi monotone convergence theorem and have adopted
an entirely different approach to the Lebesgue dominated convergence theorem. Instead
of basing its proof on the additivity of the integral (not an easy fact to establish),
our approach highlights the role of uniform convergence. Lemmas 3 and 4 show that the
contribution of the dominated convergence is just to enable the uniform convergence to
tion of new auxiliary results (Propositions 2,
do the job via Egorov’s theorem and Proposition 4.
2.
THE LEBESGUE MEASURE AND MEASURABLE FUNCTIONS.
For easy reference, and to specify how much Lebesgue measure theory is necessary
for our presentation, we collect in this section those results that will be used in
n
the rest of the article. For proofs see [I] or [6]. By a rectangle in
we shall
mean the product space of n bounded intervals, open, closed, or neither. The volume
of such a rectangle R, denoted by v(R), is the product of the lengths of its component intervals.
DEFINITION I. The outer measure of a set E C n is
m(E)
def
: inf
(2.1)
v(R.)
where the infimum is taken over all finite or countable collections of open rectan-
{Ri
R i.
It turns out that it is not always true that the outer measure of a finite union
of disjoint sets is the um of their outer measures. But this will always be true if
we restrict ourselves to sts of the following type.
DEFINITION 2. A set E
is called measurable if and only if for any set
S
m(S)
m(SE} + m{S/hEC), where E c is the complement of E in
If
E is measurable m{E) is called the Lebesgue measure of E.
gles
such that
n
n
Notice that
EC
n
n.
and the empty set are measurable with Lebesgue measures oo
695
LEBESGUE INTEGRAL AS A RIEMANN INTEGRAL
It is also clear that a set is measurable if and only it its
E and F are measurable with F C E then
and zero respectively.
complement is measurable and that if
m(F) < m(E).
THEOREM I.
(I)
(2)
O.
is a finite or countable collection of measurable sets then their
{Ei}
If
Ei
union
m(E)
is a finite or countable set then
E
If
is measurable and
m{U E
i)
<
m(E
E
If, in addition, the
i
are
i
disjoint then equality holds.
(3)
If
is a finite or countable collection of measurable sets then their
{El}
-
intersection
Ei
all
m(E
i)
E
and
i
(4)
then
m(f]E i)
as
i
--.
F
are measurable then E
F
C
m(E)
<
measurable. If, in addition,
E
and F
(5) Every half space in
is measurable.
(6)
If
n
n
Every rectangle in
m(E
If, in addition,
is measurable.
def
I)
<
{x
E
F}
x
m(E
then
Ei+1 C
and
for
i
is
m(E)
F)
E
m(F).
is measurable and its Lebesgue measure is its
volume.
(7)
Open and closed sets are measurable.
A function
y e
DEFINITION 3.
measurable and for any
f:E C
Bn
the set
B
THEOREM 2. The following statements
(I) For each y e
the set {x e
(2) For each y e
the set {x e
(3) For each y
the set {x e
(4) For each y e
the set {x e
EXERCISE I.
E
If
f(x) >
(2)
E
If
{fN}
E
3.
is measurable.
y}
< y}
is measurable.
f(x) >
f(x)
E
f(x)
E
f(x)
-
f:E
is measurable.
y}
B
is measurable.
is continuous then
f
is
be a measurable set.
are measurable then
f + g
is measurable.
is a sequence of measurable functions on
E
and if
fN--
on
f
then
f is measurable.
THEOREM 4. (Egorov’s theorem)
Let
be a sequence of measurable functions on
0
y}
E
E
is measurable and
cn
y}
is
is measurable.
are equivalent:
measurable.
THEOREM 3. Let E
(I) If f,g:E--
E
is measurable if and only if
{x e E
there is a subset
G
of
E
with
E C
E
n
be a measurable set and let
fN--
such that
m(G) < d and such that
- _
E.
on
f
fN
Go
THE LEBESGUE INTEGRAL.
For any
uniformly on
f
n
The Lebesgue approach to define the integral of a function f:E
is
to partition its range, not its domain as in the Riemann theory. To be specific,
consider the case of a measurable set E with m(E) <
and a measurable function
f:E--
whose values are between zero and
is a partition of
k,
then
sup
f
[0, M]
is said
k
i :I
and if we define
to be integrable on
m(Si)(Yi- Yi-1
inf
M > O.
If
{x
e E
.
S.
E
0
YO <
f(x) >
Y
Yi}
<
<
for
i
M
Yk
O,
if and only if
k
i:I
where the supremum and infimum are taken over all
m(Si-1)(Yi- Yi-1 )’
partitions of
[0, M],
(3.1)
and this
E.A. GONZALEZ-VELASCO
696
IE
common value will be denoted by
EXERCISE 2. Define a set E.
f.
{x
e E
for each
yi_
Prove
i.
that
"
"
and
k
i
m(Ei)Yi-1
k
i:I
m(Ei)Yi
Z {:Ik
(3.2)
m(Si)(Yi- Yi-1
m(Si-1 }(Y i
M]
f:[O,
Now, if we define a function
then
k
i:I
3
(3 3)
Yi-
f(y)
by
{x
m
f(x) > y}
e E
is clearly bounded and decreasing and, thus, Riemann integrable on
f
PROPOSITION I.
n
]R
E C
Let
IEf: RI
R
where the
PROOF.
m(E)<=o.
be a measurable set with
M
E
is integrable on
f:E
nonnegative measurable function
[0, M].
Every bounded,
and
(3.4)
f
0
indicates Riemann integral.
m(S i)
Notice that
i=kl m(Sl)(Yi
and then, since
f(yi
i=kl m(Si_1)(y
and
Yi-1
is decreasing,
f
(3.5)
Yi_1
i
are, respectively, the lower and upper Riemann sums of f with respect to the partiM of [0, M]. The result follows because f is
tion 0
< Y <
<
YO
Yk
[0, M],
Riemann integrable on
If
f
Q.E.D.
is nonpositlve instead of nonnegative we would like its integral on
equal the opposite of the integral of
IEf :-RI
M
0
R
0
M
RI
-f
-M_f(-y)
Mf(y)
This suggests that we define
to
that is, we want
-f,
-f
E
0
M
--f(-Y)
y < O,
if
(3.6)
dy
and then the entire prece-
ding discussion motivates the following general definitions.
DEFINITION 4.
table function.
n
E
Let
be a measurable set and let
The measure function of
m{
f(Y)
Notice that
f(yl
f(y2
values if
m(E)
subinterval of
{x
(-, O)
e E
y>O
f(x)< y}
if
y< 0
f
defined by
and
0
f
on
(O,oo),
and that
Yl
Y2
The measure function can have infinite
oo, but, if it does not, it is Riemann integrable on every bounded
(-, O) or (0,o}, and then the improper integrals
exist (they may be infinite).
or on
if
on each of these intervals.
R 0
and
(-oo, O)
y
f{x)>
be a measu-
f:E
is the function
(3.7)
-m
on
0
f
x e E
E
on
f
(0,=o)
(3.8)
’f
For convenience, if
f
has infinite values on
we define
-
RI o
f
or
R
Of:
oo-
(3.9)
697
LEBESGUE INTEGRAL AS A RIEMANN INTEGRAL
n
a measurable funcbe a measurable set, f:E-Let E C
the measure function of f on E. Then the (Lebesgue} integral of f
DEFINITION 5.
tion, and
E
on
f
is defined to be
-
R
f
E
(3.10)
f
if the right-hand side exists {that is if the two improper Riemann integrals over
(-4, O}
{0,}
exist}
If, in addition, the right-hand side is finite then f
is said to be {Lebesgue} integrable on E nd we write f e [{E}.
0
I/-- we have f(y}
0 < x < I} and f(x}
EXAMPLE I. If E
{x
-2
if y
y
and f(y)
I, and then
if 0 < y <
if y < O, f{y}
and
IE
EXAMPLE 2.
< 0,
y
f(y)
ix
E
If
0
if
f
and
0
iy_2
I
0 < x <
e
< y <
E f
Examples
R
dy
0
and
f(x}
f(y)
I/y
if
_
dy +
1
(3 11}
2
dy
y
I/x
f(y}
then
I.
0
if
We have
o
dy
(3 12}
and 2 show that the product of integrable functions need not be inte-
grable.
EXAMPLE 3.
f(x)
0
]R
0
E
and
f(x)
and
f{y)
f > 0
Show that if
I}
<
0 < x
is irrational, then
EXERCISE 3.
4.
{x
E
If
x
when
f
x
when
for all
y
0
m
then
0
{x
is rational and
E
f(x) #
e E
f
O}
O.
BASIC PROPERTIES OF THE LEBESGUE INTEGRAL.
First we establish the basic properties of the measure function.
E
O.
f
and
instead of
f
LEMMA I.
Let
functions and let
(I}
f<_ g
(2)
cf
when we need to specify the domain
E
cn
be a measurable set,
c e
E
of
f.
f,g:E
let
We shall write
be measurable
Then
f g.
cf(y}
is measurable and
sgn(c}f(y/c}
for all y
O.
PROOF.
(I}
y < O,
{2)
y > O,
If
{x
{x
e E
g(x} < y}
e E
f(x) >
{x
e E
f(x} <
We shall consider the case
cf(y}
xe
y}
c < O,
m
m
{x
e E
x e E
E
g(x} > y}
f(y) <_ g(y}. If
-g(y} -f{y} f(y} g(y}.
y}
y > O,
and the others are similar
cf(x} > y}
f(x} <
y/c}
-f(y/c}
sgn(c}f(y/c}
(4.1}
Q.E.D.
LEMMA 2.
sets in
n
If
{Ei}
if
E
Ei
Ei
PROOF
m
{x
If
e E.
1
y > O,
f(x} >
-
is a finite or countable collection of disjoint measurable
y}
f:E
and if
{yl
m
i.
{x
E
is a measurable function, then
f(x} >
Similarly if
y}
y < 0
m{.Tix
Q E D
e E
i
f{x} >
y}
E
f
_
698
COROLLARY I.
E,F
If
E.A. GONZALEZ-VELASCO
n
C ]R
is a measurable function, then
PROOF.
E
f
E
(0,)
on
f
]R
n
f
E-F
Then
F
(-, O)
on
0
E,F
E
f
E
>_
>_ f on (-o, 0),
]R
sets, let f,g:E
f
be measurable
F
(0,)
on
> 0
(-, 0).
on
f
]R
f:E
and if
E
f
<_
Q.E.D.
be measurable
Then, if the integrals below exist,
c e
IE
|E f
g
f
f
Let
functions and let
(I)
E-F
and
THEOREM 5.
f f
E-F
E
F
f f+f
F)
FL)(E
F
F
are measurable sets with
F
E
on (0, } and
<_
g
f
and if, in addition,
0
then
L(E)
g
(E).
(3)
If
IEcfF C cE
then
(4)
If
and if
(2)
c
Ef
0
Assume that
min
(E).
cf
i(F)
f
def
fc
f > 0
and if, in addition,
f, e}
then
E
(proceed analogously if
c < 0
I
b
0
and taking the limit as
cf (y)
fc
E
f
then
c ---oo
as
b
0
-f(y/c)
dy
c
0b/c f(u)
(4.2)
du
cIO_f
gcf
-
Similarly for the integral from
By Corollary I,
IO
I
dy
By Lemma I(2),
c > 0).
ee,
b--
0
(3)
[(E)
f
[(E)
f
() is obvious from Lemma ().
PROOF.
(2)
and then
E
to
IO
(4.3)
Adding the two proves (2).
O.
I
F
F
E
(4.4)
If the outer integrals are finite, so are the inner integrals. Thus, f e L(E)
[(E). The last assertion follows from the last two inequalities on the right.
f
-
as
c
in
Bn,
c
(4)
E
oo,
Q.E.D.
THEOREM 6.
Let
EN
El,
E i and let
E
let
IEf
Furthermore, if
f
IE.
f
f
-
I"
f
E
(4 5)
f
be a finite collection of disjoint measurable sets
f:E
be a measurable function.
Then
f
(4.6)
has a constant value
IE
0
Z
c
fc
c.
on each
Ei,
then
cim(Ei)
(4,7)
PROOF. The first assertion follows from Lemma 2 and Definition 5.
E.
then
f(y)
otherwise.
m(E i)
when
c.
(here
m(E.)
can be
)
and
Now, if
Ei
f
(y)
c. >
0
Then
Elf
Similarly if
0.<
y <
c. < 0,
Q.E.D.
Ei
R ci
0 f
ci m(E i
(4.8)
699
LEBESGUE INTEGRAL AS A RIEMANN INTEGRAL
f(x}
b for all
such that a
EXERCISE 4. If there are constants a,b e
f e [(E).
bm(E) and m(E)<
f
x e E, then am(E)
B be measuset
let f,g:E
and
c
measurable
a
be
Let
E
EXERCISE 5.
of
E of measure
g on E except, possibly, on a subset
rable functions If f
E
-
Bn
IE =IEEE n
f
zero, then
THEOREM 7. Let
PROOF
{x
f
be a measurable
exists, then
y > 0
If
f(x) <-y}
E
B
f:E
Ill e L(E).
L(E)
f
Also,
IE
If
function
be a measurable set and let
C
f(x)l > Y)
{x e E
the identity
shows that
Ifl
R
dy :-/
is measurable and that
f(x) > y} U
{x e E
Ifl
f(Y)-f
(y)
Now,
0
/f(-Y)
.f(u)
0
R
du
p.f(y)
dy
so that
R
Ifl
0
0
R
(f(Y)- f(-Y))dy
0
f
f
The right-hand side is always well defined as a real number or oo
(-, 0}
and
f
b
a
b
fl
f e [(E}
a > 0
holds when
<_ O.
(0, ),
>_ 0 on
finite, that is
and
b
since
f
_
(4.10)
(4.11)
on
0
and it is finite if and only if both integrals are
e i(E}.
< O,
la
In either case, the inequality
a > 0 and
when
b
-,
+
bl
a
and when
and
Thus,
(4.12)
Q.E.D.
EXERCISE 6.
ble then
5.
Let
f e i(E)
R
Bn
be a rectangle in
IE RIE
f
and
If
f:R
is Riemann integra-
f
THE ADDITIVITY OF THE LEBESGUE INTEGRAL.
This is one of the week points of the Lebesgue theory and it will not be
straightforward to establish this fact.
We start by considering two particular cases:
those of piecewise continuous functions and nonnegative bounded functions.
Let
PROPOSITION 2.
E
cn
be a measurable set and let
rable functions each of which has a finite number of values.
IE
PROOF.
sets
F1,
If
f
F
N
on disjoint subsets
E
and if
GI,
G
c + C
G
on F
j
i
i
j"
using Theorem 6 we have
are
E
K
(5..I)
f+
g
g)
ci,
c
N
on disjoint sub-
.
has only a finite number of values
of
E,
Noticing that
(f
be measu-
IE IEg
(f + g)
has only a finite number of values
of
B
f,g:E
Then
then
F
i
" "
i
f + g
(ci
+
C
K
is the function whose values
is the disjoint union
j:1
CI,
Cj) m(Fif]G j)
jK=I
Fi G j
and
. .
E.A. GONZALEZ-VELASCO
700
Z i:IN
"
N
i =I
ci
Z j:1K
K
m(FiGj
K
ci m(Fi)
Cj
j=1
m(G
N
Cj
j:1
i :I
m(Gjf] F i
j
(5.2)
Q.E.D.
Let
PROPOSITION 3.
n
E c
IE(f
Let
PROOF.
M
[0, M].
tion
}
sf
mlF k)
by
sf(x}
Yi-1
IE
f +
IEg
[0, M], M > O,
Then define
x e
if
be non-
Then
+ g):
have values in
f
be a partition of
f,g :E--
be a measurable set and let
negatlve, bounded, measurable functions.
Fi_
(5.3)
and let
F.I
{x e E
Fi,
i
1,
0
< YW
YO < Yl <
f(x) > yi }_ and a funck. Noticing that
YO
and using Theorem 6 we have
0
k
IESf Z
Yi-1 m(Fi-1
i:i
iI Yi-1
mlF.))1
(mlFi-1)
k
i:I
(Yi
k
i:I
f(Yi)(Yl
IE
Fi)
Yi-1 m(Fi
Yi-I
(5.4)
f
and, as the norm of the partition approaches zero,
If
and
Sg is defined for g as sf was defined for f, applying Proposition 2 to sf
Sg, observing that sf + Sg <_ f + g, using Theorem 5(I}, and letting the norm
of the partition approach
zero, we
obtain
The first integral above is well defined because
f + g
and bounded.
Sf
F.1
Fi
Similarly, if we define a function
and if
S
g
is defined for
g
as
Sf
is measurable, nonnegative
f
is for
by
Sf(x)
Yi
if
x
f, a similar argument would
yield
IE
(f + g)
IE
f +
IEg
(5.7)
and the conclusion follows from the last two inequalities, Q.E.D.
THEOREM 8.
L(E)
Let
E c
n
be a measurable set and let
f,g
[(E}.
Then
f + g
and
IE(f
+
g):
IEf IEg
+
(5.8)
701
LEBESGUE INTEGRAL AS A RIEMANN INTEGRAL
mln
f
Assume first that
PROOF.
{f, M)
g
and, similarly,
M
M > 0
are nonnegative and for
and
g
and
{f
g}M
Since
fM
g
M
fM
define
f + g,
Proposi-
tlon 3 gives
IE gM
fM +
E
(fM
E
g
M)
<
E
g)
(5.9)
(f + g)
(5.10)
(f
and then Theorem 5{4} implies that
as
IE
fM +
E
g
M
E’ f+I E g
IE
(f +
fM gM
g)M
g)
(f
It is not difficult to see that
M--*
E
M
IE
fM
fM
+ g
M
and then
E
E
Hence
as
M--o
It follows that
E
IEf IEg
(f + g)
E
+,
g e [(E).
f
and, since the right-hand side is finite,
{x
(5.13)
+
{x
If, instead, f > 0 and g < O, define E
E
g < 0}. Since f + g, -g and f
f
E
f + g
f + g + (-g)
>_ O}
and
E
are nonnegative on
the preceding proof shows that
IE+f
Similarly,
E
+(f
+ g
f
and
E-
-:)
-(f + g),
+
E
+
(-g):
E
+(f
-(f + g)
-g
I_
E
f
+
g)
(5.14)
+g
E
are nonnegative on
E
and
_.
+gl +
5.5)
E
Subtracting the last equation from the previous one and using Theorem 6 proves the
desired result in this case.
The two cases proved above imply the truth of the assertion in the general case
after decomposing
and
6.
g
E
into the union of four disjoint subsets, on each of which
f
do not change sign, Q.E.D.
THE CONVERGENCE THEOREMS.
IE fN IE
f when fN
f"
In this section we study the validity of the limit
In the case of Riemann integration the uniform convergence of this sequence is a
sufficient condition. We start by proving a similar result for the Lebesgue integral.
and let {fN}
be a measurable set with m(S) < o
PROPOSITION 4. Let S
be a sequence of measurable functions such that
[(S) and
f e [(S)
n
fN
uniformly on
S.
ISfN-- IS
PROOF.
Given
fN
Then
e
>
0
(6.1)
f
there is a
K e
+
such that
f
6
fN
f
’
and
E.A. GONZALEZ-VELASCO
702
then
For
N > K.
for
fN0 andf+e’N > K,
f-
y >
f{y
+ )
m
{x
S
m
{x
e S
f(x) > y + }
> y}
f(x)
and then
f
0
For
y >
and
oo f
em(S) <_ i
N
oo0
gf (y + e) dy
_
(6.2)
(6.3)
i
> K,
f+(Y)
fN(Y
m{x e S
f(x) + e >y}
m {x e S
f(x)
>
y
e}
(6.4)
}
f{y
and then
0
fN <-
: fN
m(S)
e m(S) +
Thus, given
e
> 0
Z+
K
there is a
N > K
for
and, since
e
0
as
=o.
N
.
N
0
Pf
(6.6)
m(S)
is arbitrary,
0
0
as
(6.5)
f
such that
I,fN
-m(S)
0
(6.7)
gf
Similarly,
RI
O
I
pfN
0
(6.8)
The last two limits and Definition 5 establish the desired convergence,
Q.E.D.
The following theorem is usually cited as one of the triumphs of the Lebesgue
theory because it replaces the uniform convergence of
condition that the
fN
be
unifdmly
bounded.
{fN
by the less restrictive
It also allows the domain of these
functions to have arbitrary Lebesgue measure.
THEOREM 9.
(The Lebesgue dominated convergence theorem)
measurable set and let
{fN }
Let
E Cn
be a sequence of integrable functions on
verges to an integrable function
f
on
E.
If there is a function
g
E
be a
that con-
[(E)
such
703
LEBESGUE INTEGRAL AS A RIEMANN INTEGRAL
fNl
that
<_ g
.+
N e
for all
then
1E
IEfN
as
(6.9}
f
N --oo,
f
and
fN
This theorem is sometimes stated with the
measurable rather than
integrable, but, in fact, these functions must be integrable.
because
g
fN
and
f
[(E)
6
-g <
(E)
-g
fN <
g
IN
-g
and
-g
g
fN
-g < f < g"
For the
g"
this is so
fN
For
f
because
The mechanism through which the uniform boundedness by the integrable function
g
operates in this theorem can be better understood by means of the next two lemmas,
whose proof is temporarily postponed.
LEMMA 3.
Let
there is a subset
n
E
F
be a measurable set and let
E
of
m(F) < oo
with
IIE_Fg
LEMMA 4.
subset of F.
Let F c
For any
]R
n
e > 0
(6.10)
> 0
there is a
IIGgl
For any
<
be a measurable set, let
> 0
g e [(E).
such that
g e (F)
such that if
G
and let
m(G) <
be a
then
(6.11)
<e
Because of Lemma 3, the integrals of all functions involved in Theorem 9 are
F of finite measure and, because of Lemma 4, the same is
true in a sufficiently small subset G of F. This is significant because, according
f uniformly
to Egorov’s theorem (Theorem 4), the set G can be chosen so that
negligeable outside a set
fN
on
F
G.
And now, as in the case of Riemann integration, uniform convergence is
seen to be the main driving force in the proof of Theorem 9 through Proposition 4.
PROOF OF THEOREM 9. Given
Lemma 4 and, for this choice of
,
> 0
let
let
triangle inequality and Theorems 6 and
G
Lemma 3, let
be as in Egorov’s theorem.
F
By Proposition 4,
smaller than
,
N
By Definition 5, given
F-G
+ 2
IE_F
g
fl 21Gg 21E-Fg
+
+
can be chosen so large that the first term on the right is
and, since
PROOF OF LEMMA 3.
IF-G fN
Using the
7 we obtain
Ifl
<--
be as in
be as in
e
is arbitrary, the result follows, Q.E.D.
In view of Theorem 7 we need only consider
> 0 there is an a > 0 such that
la gE <
0
the case
g > O.
(6.13)
E.A. GONZALEZ-VELASCO
704
If we define
F
E
But
gE-F(y)
0
F
{x
e E
{x
E
0
y
if
g(x) > a}
a},
g(x)
a.
we have
m(F)
so that
gE-F(y)
ggE(a)E < ee since
g (y) if y <
g e L(E).
a
and
Then
R
E_Fg
0
gE-F <_
a
0
R
gE
(6.14)
<e
Q.E.D.
PROOF OF LEMMA 4.
e
Given
If
m(G7 <
,
a > 0
there is an
> 0
G C F
gG
we have
_
In view of Theorem 7 we need only consider the case
m(G7
g
O.
such that
and
(6.15)
gG Mg-F
Then, if we choose
e/2a
and if
we obtain
IG
RI aO ggG RIGa
g
g
F
R[
a g
am(G}
< am(G)
(6.16)
<
+
Q.E.D.
In some applications the sequence
may not be uniformly bounded by an inte-
{fN }
Such a case frequently occurs when the
grable function.
are the partial sums of
fN
In this type of situation Theorem 11 below is useful,
a series of positive functions.
but in order to prove it we need some preparation.
First we observe that the Lebesgue
integral of a nonnegative function can be characterized in terms of those of bounded
integrable functions.
LEMMA 5.
E Cn
Let
measurable function.
be a measurable set and let
IE
If
{E),
f e
given
0
If we define
y < b
and
g
g(y}
(6.177
min
0
{f, b}
if
e > 0
b
IEf RIO f
<_
g
then
y
>_
there is a
b.
If
f
[(E)
>
0
g
0
f.
such that
(6.18)
0
g
f,
g(y)
f(y7
if
O<
b
O
(6.19)
f
and
0
cases.
b
such that
< e
is bounded,
b
g
g e [(E)
g e L(E7
We have
IE RIO g RI
so that
be a nonnegative
suPlEg
f
where the supremum is taken over all bounded functions
PROOF.
f:E--
Then
define
E
IE
{x
IEg
f
E
(6.207
<
f(x) > a}
for each
a > 0
and consider two
705
LEBESGUE INTEGRAL AS A RIEMANN INTEGRAL
m(E
{I}
<
0 < a
a
and
b
is bounded,
g{yl
0
(2)
m(E
m(RE
g
by
m(Ea}
b
a
then
g
for some
a > O.
Given
g(x)
If we define
a
is bounded,
g
0
f
a
m(Ea
M > 0
let
f(y}
a
a
if
_
then
g
y < b
(6.22)
+ M
R
RE a
x e
if
E
(6.21}
and that
> a
f
E
on
0
g{y)
[{E}
g
g
and
a
y < a,
0
if
a
m(E
E
on
It follows that
b.
> M/a.
{f, b}
min
g(y}
f,
y
o
a
such that
a,b e
b
0
that
there are
la f > M
g
if
0
M > 0
For any
> O.
for all a
oo
g:E
If we define
and
__
<
be a rectangle so large
and
g(xl
0
otherwise,
and
IE IRfE
g
am(Rf]E
a
> M
a
a
(6.23}
Q.E.D.
THEOREM 10. (Fatou’s lena} Let E :21R n
sequence of nonnegative measurable functions on
>
0
N > M.
for
PROOF.
[(E},
_
/
M e 2Z
there is an
0
> O,
g
f
IE fN
gN
{x
gN
> y}
e E
fN
:gNIx}
min
{g,
and let
fN
{fN)
be a
Then for any
f"
(6.241
+ e
Lemma 5 implies that there is a bounded function
g
and
fIf we define
E
such that
f
For any
such that
be a measurable set, let
fN}
IE g<-e
(6.25}
2
N
for each
{x e E :g{xl > y}h{x
2Z
e E
/
then
:fN(x}
g
> y}
is measurable because
for each
y
> O,
and
implies that
Also, since
gN
-
IEgN
g
on
E
and
IEfN
(6.26}
g,
IgNl
the Lebesgue dominated convergence theo-
rem implies that
IEg
as
N
e.
Therefore, there is an
Eg
for
N > M,
(6.27}
M >0
E
fN
such that
{6.28}
2
or else we would have
{6.29}
for arbitrarily large
N,
contradicting the limit above.
We conclude that
E.A. GONZALEZ-VELASCO
706
E
N > M,
for
M < N,
for
f
E N
(6 30)
+
(The Levi monotone convergence theorem)
{fN }
and let
fN
E c
Let
be a sequence of nonnegative functions on
E
n
be a measura-
such that
fM fN
Then
f"
E
as
e
Eg+
O.E.D.
THEOREM 11.
ble set, let
f<
fN
{6.31)
f
E
oo.
N
PROOF
fM fN
for
M < N
fN <-
f
for all
N
(6.32)
for all
N
/
By Fatou’s lemma, given
f <
N
for
large enough.
E
These two inequalities imply that
IE
as
N
oo,
(6.33)
fN
f
(6.34)
Q.E.D.
REFERENCES
I.
2.
3.
4.
5.
6.
ROYDEN, H.L. Real Analysis, 2nd. ed., Macmillan, New York, 1968
APOSTOL, T.M. Mathematical Analysis, 2nd. ed., Addison-Wesley, Reading, Mass.,
1974.
MCSHANE, E.J. ified Integration, Academic Press, Orlando, Fla., 1983.
MIKUSINSKI, J. The Bochner Integral, Birkhuser Verlag, Boston, 1978.
REY PASTOR, J. Elementos de la Teorla de Funciones, 3rd. ed., Ibero-Americana,
Madrid, 1953.
KESTELMAN, H. Modern Theories of .Integration, Oxford University Press, Oxford,
1937.