Chapter 8

Integration of Functions of Several

Variables

ed
ct
In this chapter, we focus on the integration of bounded functions on bounded subsets of Rn .

te
8.1 Integrable Functions
P ro
We start with a simpler case n = 2.
ht

Definition 8.1. Let A Ď R2 be a bounded set. Define

a1 = inf x P R ˇ (x, y) P A for some y P R ,

␣ ˇ (
ig

b1 = sup x P R ˇ (x, y) P A for some y P R ,

␣ ˇ (
r
py

a2 = inf y P R ˇ (x, y) P A for some x P R ,

␣ ˇ (

b2 = sup y P R ˇ (x, y) P A for some x P R .

␣ ˇ (
Co

A collection of rectangles P is called a partition of A if there exists a partition Px of [a1 , b1 ]

and a partition Py of [a2 , b2 ],

Px = a1 = x0 ă x1 ă ¨ ¨ ¨ ă xn = b1 and Py = a2 = y0 ă y1 ă ¨ ¨ ¨ ă ym = b2 ,
␣ ( ␣ (

such that

P = ∆ij ˇ ∆ij = [xi , xi+1 ] ˆ [yj , yj+1 ] for i = 0, 1, ¨ ¨ ¨ , n ´ 1 and j = 0, 1, ¨ ¨ ¨ , m ´ 1 .

␣ ˇ (

The mesh size of the partition P and also called the norm of P, denoted by }P}, is defined
by
!b ˇ )
}P} = max (xi+1 ´ xi )2 + (yj+1 ´ yj )2 ˇ i = 0, 1, ¨ ¨ ¨ , n ´ 1, j = 0, 1, ¨ ¨ ¨ , m ´ 1 .
ˇ

275
276 CHAPTER 8. Integration

a
The number (xi+1 ´ xi )2 + (yj+1 ´ yj )2 is often denoted by diam(∆ij ), and is called the
diameter of ∆ij .

Similar to the integrability of f on a bounded subset of R, we have the following

Definition 8.2. Let A Ď R2 be a bounded set, and f : A Ñ R be a bounded function. For

any partition P = ∆ij ˇ ∆ij = [xi , xi+1 ] ˆ [yj , yj+1 ], i = 0, ¨ ¨ ¨ , n ´ 1, j = 0, ¨ ¨ ¨ , m ´ 1 , the
␣ ˇ (

upper sum and the lower sum of f with respect to the partition P, denoted by U (f, P)
and L(f, P) respectively, are numbers defined by

ed
ÿ A
U (f, P) = sup f (x, y)A(∆ij ) ,
0ďiďn´1 (x,y)P∆ij
0ďjďm´1

ct
ÿ A
L(f, P) = inf f (x, y)A(∆ij ) ,
(x,y)P∆ij

te
0ďiďn´1
0ďjďm´1

ro
where A(∆ij ) = (xi+1 ´ xi )(yj+1 ´ yj ) is the area of the rectangle ∆ij , and f is an extension
A

of f , called the extension of f by zero outside A, given by

P
"
A f (x) x P A ,
f (x) =
ht

0 x R A.
ig

The two numbers

r

ż
f (x, y)dA ” inf U (f, P) ˇ P is a partition of A
␣ ˇ (
py

A
Co

and ż
f (x, y)dA ” sup L(f, P) ˇ P is a partition of A
␣ ˇ (
A

are called the upper integral and lower integral of f over A, respectively. The function
ż ż
f is said to be Riemann (Darboux) integrable (over A) if f (x, y)dA = f (x, y)dA,
ż A A
and in this case, we express the upper and lower integral as f (x, y)dA, called the integral
A
of f over A.

In general, we can consider the integrability of a bounded function f defined on a

bounded set A Ď Rn as follows
§8.1 Integrable Functions 277

Definition 8.3. Let A Ď Rn be a bounded set. Define the numbers a1 , a2 , ¨ ¨ ¨ , an and

b1 , b2 , ¨ ¨ ¨ , bn by

ak = inf xk P R ˇ x = (x1 , ¨ ¨ ¨ , xn ) P A for some x1 , ¨ ¨ ¨ , xk´1 , xk+1 , ¨ ¨ ¨ , xn P R ,

␣ ˇ (

bk = sup xk P R ˇ x = (x1 , ¨ ¨ ¨ , xn ) P A for some x1 , ¨ ¨ ¨ , xk´1 , xk+1 , ¨ ¨ ¨ , xn P R .

␣ ˇ (

A collection of rectangles P is called a partition of A if there exists partitions P (k) of

(k) (k) (k)
[ak , bk ], k = 1, ¨ ¨ ¨ , n, P (k) = ak = x0 ă x1 ă ¨ ¨ ¨ ă xNk = bk , such that
␣ (

! ˇ
(1) (1) (2) (2) (n) (n+1)
P = ∆i1 i2 ¨¨¨in ˇ ∆i1 i2 ¨¨¨in = [xi1 , xi1 +1 ] ˆ [xi2 , xi2 +1 ] ˆ ¨ ¨ ¨ ˆ [xin , xin+1 ],
ˇ

ed
)
ik = 0, 1, ¨ ¨ ¨ , Nk ´ 1, k = 1, ¨ ¨ ¨ , n .

ct
The mesh size of the partition P, denoted by }P} and also called the norm of P, is defined

te
by #g +
f n ˇ
e (x(k) ´ x(k) )2 ˇˇ ik = 0, 1, ¨ ¨ ¨ , Nk ´ 1, k = 1, ¨ ¨ ¨ , n
fÿ ro
}P} = max ik +1 ik .
k=1
P
d
n
(k) (k)
The number (xik +1 ´ xik )2 is often denoted by diam(∆i1 i2 ¨¨¨in ), and is called the di-
ř
k=1
ht

ameter of the rectangle ∆i1 i2 ¨¨¨in .

ig

Definition 8.4. Let A Ď Rn be a bounded set, and f : A Ñ R be a bounded function. For

any partition
r
py

! ˇ
(1) (1) (2) (2) (n) (n+1)
P = ∆i1 i2 ¨¨¨in ˇ ∆i1 i2 ¨¨¨in = [xi1 , xi1 +1 ] ˆ [xi2 , xi2 +1 ] ˆ ¨ ¨ ¨ ˆ [xin , xin+1 ],
ˇ
Co

)
ik = 0, 1, ¨ ¨ ¨ , Nk ´ 1, k = 1, ¨ ¨ ¨ , n ,

the upper sum and the lower sum of f with respect to the partition P, denoted by
U (f, P) and L(f, P) respectively, are numbers defined by
ÿ A
U (f, P) = sup f (x, y)ν(∆) ,
∆PP (x,y)P∆
ÿ A
L(f, P) = inf f (x, y)ν(∆) ,
(x,y)P∆
∆PP

where ν(∆) is the volume of the rectangle ∆ given by

(1) (1) (2) (2) (n) (n)
ν(∆) = (xi1 +1 ´ xi1 )(xi2 +1 ´ xi2 ) ¨ ¨ ¨ (xin +1 ´ xin )
278 CHAPTER 8. Integration

(1) (1) (2) (2) (n) (n) A

if ∆ = [xi1 ´ xi1 +1 ] ˆ [xi2 ´ xi2 +1 ] ˆ ¨ ¨ ¨ ˆ [xin ´ xin +1 ], and f is the extension of f by
zero outside A given by "
A f (x) x P A ,
f (x) = (8.1.1)
0 x R A.
The two numbers
ż
f (x) dx ” inf U (f, P) ˇ P is a partition of A ,
␣ ˇ (
A

and ż
f (x) dx ” sup L(f, P) ˇ P is a partition of A
␣ ˇ (

ed
A

are called the upper integral and lower integral of f over A, respective. The function f

ct
ż ż
is said to be Riemann (Darboux) integrable (over A) if f (x) dx = f (x) dx, and
ż A A

te
in this case, we express the upper and lower integral as f (x) dx, called the integral of f
A
over A.
ro
Definition 8.5. A partition P 1 of a bounded set A Ď Rn is said to be a refinement of
P
another partition P of A if for any ∆1 P P 1 , there is ∆ P P such that ∆1 Ď ∆. A partition
P of a bounded set A Ď Rn is said to be the common refinement of another partitions
ht

P1 , P2 , ¨ ¨ ¨ , Pk of A if
ig

1. P is a refinement of Pj for all 1 ď j ď k.

r
py

2. If P 1 is a refinement of Pj for all 1 ď j ď k, then P 1 is also a refinement of P.

In other words, P is a common refinement of P1 , P2 , ¨ ¨ ¨ , Pk if it is the coarsest refinement.

Co

“+” “=”

Figure 8.1: The common refinement of two partitions

Qualitatively speaking, P is a common refinement of P1 , P2 , ¨ ¨ ¨ , Pk if for each j =

1, ¨ ¨ ¨ n, the j-th component cj of the vertex (c1 , ¨ ¨ ¨ , cn ) of each rectangle ∆ P P belongs to
(j)
Pi for some i = 1, ¨ ¨ ¨ , k.

Similar to Proposition 4.78 and Corollary 4.79, we have

§8.2 Conditions for Integrability 279

Proposition 8.6. Let A Ď Rn be a bounded subset, and f : A Ñ R be a bounded function.

If P and P 1 are partitions of A and P 1 is a refinement of P, then

L(f, P) ď L(f, P 1 ) ď U (f, P 1 ) ď U (f, P) .

Corollary 8.7. Let A Ď Rn be a bounded subset, and f : A Ñ R be a bounded function. If

P1 and P2 are partitions of A, then
ż ż
L(f, P1 ) ď f (x)dx ď f (x)dx ď U (f, P2 ) .
A A

8.2 Conditions for Integrability

ed
ct
In the following two sections, we discuss some equivalent conditions for Riemann integra-
bility of bounded functions (over bounded sets). We recall that in Section 4.7 we have

te
talked about two equivalent conditions for Riemann integrability: the Riemann condition
ro
(Proposition 4.80) and the Darboux theorem (Theorem 4.94). This section contributes to
the n-dimensional version of Riemann’s condition and Darboux theorem.
P
The proof of the following proposition is identical to the proof of Proposition 4.80.

Proposition 8.8 (Riemann’s condition). Let A Ď Rn be a bounded set, and f : A Ñ R be

ht

a bounded function. Then f is Riemann integrable over A if and only if

ig

@ ε ą 0, D a partition P of A Q U (f, P) ´ L(f, P) ă ε .

r
py

Definition 8.9. Let P = t∆1 , ∆2 , ¨ ¨ ¨ , ∆N u be a partition of a bounded set A Ď Rn . A

collection of N points tξ1 , ¨ ¨ ¨ , ξN u is called a sample set for the partition P if ξk P ∆k for
Co

all k = 1, ¨ ¨ ¨ , N . Points in a sample set are called sample points for the partition P.
Let A Ď Rn be a bounded set, and f : A Ñ R be a bounded function. A Riemann
sum of f for the the partition P = t∆1 , ∆2 , ¨ ¨ ¨ , ∆N u of A is a sum which takes the form
N
ÿ A
f (ξi )νn (∆k ) ,
k=1

where the set Ξ = tξ1 , ξ2 , ¨ ¨ ¨ , ξN u is a sample set for the partition P.

Similar to Theorem 4.94, the following theorem establishes the equivalence between
the Riemann condition and the Darboux integrals. The idea of the proof of the following
theorem are essentially identical to the proof of Theorem 4.94; however, the detail proof
requires a slight modification due to the fact that the dimension is bigger than one.
280 CHAPTER 8. Integration

Theorem 8.10 (Darboux). Let A Ď Rn be a bounded set, and f : A Ñ R be a bounded

A
function with extension f given by (8.1.1). Then f is Riemann integrable over A if and
only if there exists I P R such that for every given ε ą 0, there exists δ ą 0 such that if P
is a partition of A satisfying }P} ă δ, then any Riemann sums for the partition P belongs
to the interval (I ´ ε, I + ε). In other words, f is Riemann integrable over A if and only if
there exists I P R such that for every given ε ą 0, there exists δ ą 0 such that

ˇÿN ˇ
A
f (ξ )ν(∆ ) ´ I ˇăε (8.2.1)
ˇ ˇ
ˇ k k
k=1

ed
whenever P = t∆1 , ¨ ¨ ¨ , ∆N is a partition of A satisfying }P} ă δ and tξ1 , ξ2 , ¨ ¨ ¨ , ξN u is a
(

sample set for P.

ct
[ r r ]n
Proof. The boundedness of A guarantees that A Ď ´ , for some r ą 0. Let R =

te
[ r r ]n 2 2
´ , . ro
2 2

“ð” Suppose the right-hand side statement is true. Let ε ą 0 be given. Then there exists
P
δ ą 0 such that if P = t∆1 , ¨ ¨ ¨ , ∆N is a partition of A satisfying }P} ă δ, then for
(

all sets of sample points tξ1 , ¨ ¨ ¨ , ξN u for P, we must have

ht

N ˇ ε
ig

ˇÿ
A
f (ξk )ν(∆k ) ´ I ˇ ă .
ˇ ˇ
4
ˇ
k=1
r
py

Let P = t∆1 , ¨ ¨ ¨ , ∆N
(
be a partition of A with }P} ă δ. Choose two sample sets
tξ1 , ¨ ¨ ¨ , ξN u and tη1 , ¨ ¨ ¨ , ηN u for P such that
Co

A ε A A
(a) sup f (x) ´ ă f (ξk ) ď sup f (x);
xP∆k 4ν(R) xP∆k
A ε A A
(b) inf f (x) + ą f (ηk ) ě inf f (x).
xP∆k 4ν(R) xP∆k

Then
N
ÿ A
N
ÿ [ A ε ]
U (f, P) = sup f (x)ν(∆k ) ă f (ξk ) + ν(∆k )
4ν(R)
k=1 xP∆k k=1
N N
ÿ A ε ÿ ε ε ε
= f (ξk )ν(∆k ) + ν(∆k ) ă I + + = I +
4ν(R) 4 4 2
k=1 k=1
§8.2 Conditions for Integrability 281

and
N
ÿ A
N
ÿ [ A ε ]
L(f, P) = inf f (x)ν(∆k ) ą f (ηk ) ´ ν(∆k )
xP∆k 4ν(R)
k=1 k=1
N N
ÿ A ε ÿ ε ε ε
= f (ηk )ν(∆k ) ´ ν(∆k ) ą I ´ ´ = I ´ .
4ν(R) 4 4 2
k=1 k=1

ε ε
As a consequence, I ´ ă L(f, P) ď U (f, P) ă I + ; thus U (f, P) ´ L(f, P) ă ε.
2 2
ż
“ñ” Let I = f (x)dx, and ε ą 0 be given. Since f is Riemann integrable over A, there
A

ed
ε (i)
exists a partition P1 of A such that U (f, P1 ) ´ L(f, P1 ) ă . Suppose that P1 =
␣ (i) (i) 2
(i) (
y0 , y1 , ¨ ¨ ¨ , ymi for 1 ď i ď n. With M denoting the number m1 + m2 + ¨ ¨ ¨ + mn ,

ct
we define

te
ε
δ= ( A A ).
4rn´1 (M + n) sup f (R) ´ inf f (R) + 1
ro
Then δ ą 0. Our goal is to show that if P is a partition of A with }P} ă δ and
tξ1 , ¨ ¨ ¨ , ξN u is a set of sample points for P, then (8.2.1) holds.
P
Assume that P = t∆1 , ∆2 , ¨ ¨ ¨ , ∆N u is a given partition of A with }P} ă δ.
ht

Let P 1 be the common refinement of P and P1 . Write P 1 = t∆11 , ∆12 , ¨ ¨ ¨ , ∆1N 1 u and
(1) (2) (n) 1(1) 1(2) 1(n)
∆k = ∆k ˆ ∆k ˆ ¨ ¨ ¨ ˆ ∆k as well as ∆1k = ∆k ˆ ∆k ˆ ¨ ¨ ¨ ˆ ∆k . By the
ig

definition of the upper sum,

r

N
py

ÿ A
U (f, P) = sup f (x)ν(∆k )
k=1 xP∆k
Co

ÿ A
ÿ A
= sup f (x)ν(∆k ) + sup f (x)ν(∆k )
1ďkďN with xP∆k 1ďkďN with xP∆k
(i) (i) (i) (i)
y R∆ for all i, j y P∆ for some i, j
j k j k

and similarly,
ÿ A
ÿ A
U (f, P 1 ) = sup f (x)ν(∆1k ) + sup f (x)ν(∆1k ) .
1ďkďN 1 with xP∆k1 1ďkďN 1 with xP∆1k
(i) 1(i) (i) 1(i)
y R∆ for all i, j y P∆ for some i, j
j k j k

(i)
By the fact that ∆k P P 1 if yj R ∆k for all i, j, we must have
1(i)

ÿ ÿ
ν(∆k ) = ν(∆1k ) .
1ďkďN with 1ďkďN 1 with
(i) (i) (i) 1(i)
y P∆ for some i, j y P∆ for some i, j
j k j k
282 CHAPTER 8. Integration

The equality above further implies that

ÿ A
ÿ A
U (f, P)´U (f, P 1 ) = sup f (x)ν(∆k )´ sup f (x)ν(∆1k )
1ďkďN with xP∆k 1ďkďN 1 with xP∆k1
(i) (i) (i) 1(i)
y P∆ for some i, j y P∆ for some i, j
j k j k
( A A ) ÿ
ď sup f (R) ´ inf f (R) ν(∆k ) .
1ďkďN with
(i) (i)
y P∆ for some i, j
j k

Moreover, for each fixed i, j,

ď [ r r ]i´1 [ (i) (i) ] [ r r ]n´i
∆k Ď ´ , ˆ yj ´ δ, yj + δ ˆ ´ , ;
2 2 2 2

ed
(i) (i)
1ďkďN with yj P∆k

thus

ct
ÿ
ν(∆k ) ď 2δrn´1 @ 1 ď i ď n, 1 ď j ď mi .
(i) (i)

te
1ďkďN with yj P∆k

Therefore, ro
( A A )ÿ mi
n ÿ ÿ
U (f, P) ´ U (f, P 1 ) ď sup f (R) ´ inf f (R) ν(∆k )
P
i=1 j=0 1ďkďN with (i) (i)
yj P∆k
( A A )ÿ mi
n ÿ
ď sup f (R) ´ inf f (R) 2δrn´1
ht

i=1 j=0
n´1
( A R ) ε
(m1 + m2 + ¨ ¨ ¨ + mn + n) sup f (R) ´ inf f (A) ă ,
ig

ď 2δr
2
r

ε
and the fact that U (f, P1 ) ´ L(f, P1 ) ă shows that
py

2
U (f, P) ´ I ď U (f, P) ´ I + U (f, P1 ) ´ U (f, P1 )
Co

ď U (f, P) ´ L(f, P1 ) + U (f, P1 ) ´ U (f, P 1 ) ă ε .

Therefore, for any sample set tξ1 , ¨ ¨ ¨ , ξN u for P,

N
ÿ A
f (ξk )ν(∆k ) ď U (f, P) ă I + ε .
k=1

Similar argument can be used to show that

N
ÿ A
f (ξk )ν(∆k ) ě L(f, P) ą I ´ ε
k=1

which concludes the Theorem. ˝

§8.3 Lebesgue’s Theorem 283

In Section 5.1, we show that if a sequence of Riemann integrable functions tfk u8

k=1
converges to a function f uniformly on [a, b], then f is also Riemann integrable over [a, b] and
the integral of the limit function is the same as the limit of the integrals (of the sequences).
This theorem can also be established if the domain A under consideration is a bounded
subset of Rn . In fact, the same proof used to establish Theorem 5.16 can be applied to
conclude the following

Theorem 8.11. Let A Ď Rn be a bounded set, and fk : A Ñ R be a sequence of Riemann

integrable functions over A such that tfk u8
k=1 converges uniformly to f on A. Then f is
Riemann integrable over A, and

ed
ż ż
lim fk (x) dx = f (x) dx . (8.2.2)
kÑ8 A

ct
A

From now on, we will simply use fs to denote the zero extension of f when the

te
domain outside which the zero extension is made is clear.
ro
8.3 The Lebesgue Theorem
P
In this section, we talk about another equivalent condition of Riemann integrability, named
ht

the Lebesgue theorem. The Lebesque theorem provides a more practical way to check the
Riemann integrability in the development of theory. To understand the Lebesgue theorem,
ig

we need to talk about a new concept, sets of measure zero.

r
py

8.3.1 Volume and Sets of Measure Zero

Co

Definition 8.12. Let A Ď Rn be a bounded set, and 1A (or χA ) be the characteristic

function of A defined by
1 if x P A ,
"
1A (x) =
0 otherwise .
A is said to have volume if 1Aż is Riemann integrable (over A), and the volume of A,
denoted by ν(A), is the number 1A (x) dx. A is said to have volume zero or content
A
zero if ν(A) = 0.

Remark 8.13. Not all bounded set has volume.

Proposition 8.14. Let A Ď Rn be bounded. Then the following three statements are
equivalent.
284 CHAPTER 8. Integration

(a) A has volume zero;

(b) for every ε ą 0, there exists finite open rectangles S1 , ¨ ¨ ¨ , SN whose sides are parallel
to the coordinate axes such that
N
ď N
ÿ
AĎ Sk and ν(Sk ) ă ε (8.3.1)
k=1 k=1

(c) for every ε ą 0, there exists finite rectangles S1 , ¨ ¨ ¨ , SN such that (8.3.1) holds.

Proof. It suffices to show (a)ñ(b) and (c)ñ(a) since it is clear that (b)ñ(c).

ed
ż
“(a)ñ(b)” Since A has volume zero, 1A (x) dx = 0; thus for any given ε ą 0, there exists

ct
A
a partition P of A such that

te
ż
ε ε
U (1A , P) ă 1A (x) dx + = .
A
ro 2 2

1 if ∆ X A ‰ H ,
"
ε
Since sup 1A (x) = we must have . Now if ∆ P P
ř
ν(∆) ă
P
xP∆ 0 otherwise , ∆PP 2
∆XA‰H

and ∆XA ‰ H, we can find an open rectangle l such that ∆ Ď l and ν(l) ă 2ν(∆).
ht

N N
Let S1 , ¨ ¨ ¨ , SN be those open rectangles l. Then A Ď Sk and
Ť ř
ν(Sk ) ă ε.
k=1 k=1
ig

“(c)ñ(a)” W.L.O.G. we can assume that the ratio of the maximum length and minimum
r

length of sides of Sk is less than 2 for all k = 1, ¨ ¨ ¨ , N (otherwise we can divide Sk into
py

smaller rectangles so that each smaller rectangle satisfies this requirement). Then each
Sk can be covered by a closed rectangle lk whose sides are parallel to the coordinate
Co

? n
axes with the property that ν(lk ) ď 2n´1 n ν(Sk ). Let P be a partition of A such
that for each ∆ P P with ∆ X A ‰ H, ∆ Ď lk for some k = 1, ¨ ¨ ¨ , N . Then
N N
ÿ ÿ ? nÿ ? n
U (1A , P) = ν(∆) ď ν(lk ) ď 2n´1 n ν(Sk ) ă 2n´1 n ε ;
∆PP k=1 k=1
∆XA‰H

ż
thus the upper integral 1A (x) dx = 0. Since the lower integral cannot be negative,
A
ż ż
we must have 1A (x) dx = 1A (x) dx = 0 which shows that A has volume zero. ˝
A A

Example 8.15. Each point in Rn has volume zero.

§8.3 Lebesgue’s Theorem 285

Example 8.16. The Cantor set (defined in Exercise Problem 2.11) has volume zero.

Definition 8.17. A set A Ď Rn (not necessarily bounded) is said to have measure

zero（測度為零）or be a set of measure zero（零測度集）if for every ε ą 0, there exist
( 8 )
countable many rectangles S1 , S2 , ¨ ¨ ¨ such that tSk u8
k=1 is a cover of A that is, A Ď
Ť
Sk
k=1
8
and
ř
ν(Sk ) ă ε.
k=1

Example 8.18. The real line R ˆ t0u on R2 has measure zero: for any given ε ą 0, let
[ ´ε ε ]
Sk = [´k, k] ˆ k+3 , k+3 . Then
2 k 2 k

ed
8 8 8 8
ď ÿ ÿ 2ε ÿ ε ε
R ˆ t0u Ď Sk and ν(Sk ) = 2k ¨ = = ă ε.

ct
k=1 k=1 k=1
2k+3 k k=1
2k+1 2

te
Similarly, any hyperplane in Rn also has measure zero.
ro
Proposition 8.19. Let A Ď Rn be a set of measure zero. If B Ď A, then B also has
measure zero.
P
Modifying the proof of Proposition 8.14, we can also conclude the following
ht

Proposition 8.20. A set A Ď Rn has measure zero if and only if for every ε ą 0, there
ig

exist countable many open rectangles S1 , S2 , ¨ ¨ ¨ whose sides are parallel to the coordinate
8 8
axes such that A Ď Sk and
Ť ř
ν(Sk ) ă ε.
r
py

k=1 k=1

Remark 8.21. If a set A has volume zero, then it has measure zero.
Co

Proposition 8.22. Let K Ď Rn be a compact set of measure zero. Then K has volume
zero.

Proof. Let ε ą 0 be given. Then there are countable open rectangles S1 , S2 , ¨ ¨ ¨ such that
8
ď 8
ÿ
KĎ Sk and ν(Sk ) ă ε .
k=1 k=1

Since tSk u8
k=1 is an open cover of K, by the compactness of K there exists N ą 0 such that
N N 8
Sk , while ν(Sk ) ă ε. As a consequence, K has volume zero.
Ť ř ř
KĎ ν(Sk ) ď ˝
k=1 k=1 k=1

Since the boundary of a rectangle has measure zero, we also have the following
286 CHAPTER 8. Integration

Corollary 8.23. Let S Ď Rn be a bounded rectangle with positive volume. Then S is not a
set of measure zero.
8
Theorem 8.24. If A1 , A2 , ¨ ¨ ¨ are sets of measure zero in Rn , then Ak has measure
Ť
k=1
zero.

Proof. Let ε ą 0 be given. Since A1k s are sets of measure zero, there exist countable
␣ (k) (8
rectangles Sj j=1 , such that
8 8
ď (k)
ÿ (k) ε
Ak Ď Sj and ν(Sj ) ă @k P N.
2k+1

ed
j=1 j=1

(k)
Consider the collection consisting of all Sj ’s. Since there are countable many rectangles in

ct
this collection, we can label them as S1 , S2 , ¨ ¨ ¨ , and we have

te
8
ď 8 ď
ď 8 8
ď
(k)
Ak Ď Sj
ro = Sℓ
k=1 k=1 j=1 ℓ=1

and
P
8 8 ÿ
8 8
ÿ ÿ (k)
ÿ ε ε
ν(Sℓ ) = ν(Sj ) ď = ă ε.
k=1 k=1 j=1 k=1
2k+1 2
ht

8
Therefore, Ak has measure zero.
Ť
˝
ig

k=1

Corollary 8.25. The set of rational numbers in R has measure zero.

r
py

Theorem 8.26. Let A Ď Rn be bounded and B Ď Rm be a set of measure zero. Then A ˆ B

has measure zero in Rn+m .
Co

Proof. Let ε ą 0 be given. Since A is bounded, there exist a bounded rectangle R such that
k=1 Ď R such that
m
A Ď R. Since B has measure zero, there exist countable rectangles tSk u8
8 8
ď ÿ ε
BĎ Sk and νm (Sk ) ă .
k=1 k=1
ν(R)
8
Then A ˆ B Ď (R ˆ Sk ), and
Ť
k=1
8
ÿ 8
ÿ 8
ÿ
νn+m (R ˆ Sk ) = νn (R)νm (Sk ) = νn (R) νm (Sk ) ă ε .
k=1 k=1 k=1

Since R ˆ Sk is a rectangle for all k P N, we conclude that A ˆ B has measure zero. ˝

§8.3 Lebesgue’s Theorem 287

8.3.2 The Lebesgue Theorem

在之前我們提到了函數 Riemann 可積的兩個等價條件：Riemann’s condition 和 Darboux
定理。在這一節中，我們將引進函數是 Riemman 可積的另一個等價條件。這個等價條件
A
說的是一個函數 f 在 A 上是 Riemann 可積的若且唯若 f 的延拓 f （在函數可積分的定
義中有定義）的不連續點所構成的集合其測度為零。為了證明這個敘述，我們先對一個
函數的連續點做一個量化的刻劃。這個刻劃的方式，可以很容易用來檢驗一個函數在一
個點是否連續。
Definition 8.27. Let f : Rn Ñ R be a function. For any x P Rn , the oscillation of f at
x is the quantity

ed
ˇ ˇ
osc(f, x) ” inf sup ˇf (x1 ) ´ f (x2 )ˇ .
δą0 x1 ,x2 PD(x,δ)

ct
ˇ ˇ
我們注意到在上述定義中被取 infimum 的這個量 h(δ; x) ” sup ˇf (x1 ) ´ f (x2 )ˇ 是

te
x1 ,x2 PD(x,δ)
個 δ 的遞減函數（x 固定），而 osc(f, x) 則是 h(δ; x) 當 δ Ñ 0 時的極限。另外，我們也 ro
注意到說 h(δ; x) 也可以寫成 sup f (y) ´ inf f (y).
yPD(x,δ) yPD(x,δ)
以下的 Lemma 是關於如何檢驗一個函數在一個點是連續的。
P
Lemma 8.28. Let f : Rn Ñ R be a function, and x0 P Rn . Then f is continuous at x0 if
ht

and only if osc(f, x0 ) = 0.

Proof. “ñ” Let ε ą 0 be given. Since f is continuous at x0 ,
ig

ˇ ˇ ε
D δ ą 0 Q ˇf (x) ´ f (x0 )ˇ ă whenever x P D(x0 , δ).
r

3
py

In particular, for any x1 , x2 P D(x0 , δ),

ˇf (x1 ) ´ f (x2 )ˇ ď ˇf (x1 ) ´ f (x0 )ˇ + ˇf (x0 ) ´ f (x2 )ˇ ă 2ε ;
ˇ ˇ ˇ ˇ ˇ ˇ
Co

3
ˇf (x1 ) ´ f (x2 )ˇ ď 2ε which further implies that
ˇ ˇ
thus sup
x1 ,x2 PD(x0 ,δ) 3
2ε
0 ď osc(f, x0 ) ď ă ε.
3
Since ε is given arbitrarily, osc(f, x0 ) = 0.

“ð” Let ε ą 0 be given. By the definition of infimum, there exists δ ą 0 such that
ˇ ˇ
sup ˇf (x1 ) ´ f (x2 )ˇ ă ε .
x1 ,x2 PD(x0 ,δ)
ˇ ˇ ˇ ˇ
In particular, ˇf (x) ´ f (x0 )ˇ ď sup ˇf (x1 ) ´ f (x2 )ˇ ă ε for all x P D(x0 , δ). ˝
x1 ,x2 PD(x0 ,δ)
288 CHAPTER 8. Integration

Lemma 8.29. Let f : Rn Ñ R be a function. Then for all ε ą 0, the set Dε = x P

␣

Rn ˇ osc(f, x) ě ε is closed.
ˇ (

Proof. Suppose that tyk u8k=1 Ď Dε and yk Ñ y. Then for any δ ą 0, there exists N ą 0 such
that yk P D(y, δ) for all k ě N . Since D(y, δ) is open, for each k ě N there exists δk ą 0
such that D(yk , δk ) Ď D(y, δ); thus we find that
ˇ ˇ ˇ ˇ
sup ˇf (x1 ) ´ f (x2 )ˇ ď sup ˇf (x1 ) ´ f (x2 )ˇ @k ě N .
x1 ,x2 PD(yk ,δk ) x1 ,x2 PD(y,δ)

The inequality above implies that osc(f, y) ě ε; thus y P Dε and Dε is closed. ˝

ed
Theorem 8.30 (Lebesgue). Let A Ď Rn be bounded, f : A Ñ R be a bounded function, and
A

ct
f be the extension of f by zero outside A; that is,

te
f (x) if x P A ,
"
A
f (x) =
0 otherwise . ro A
Then f is Riemann integrable over A if and only if the collection of discontinuity of f is a
P
set of measure zero.
ht

A A
Proof. Let D = x P Rn ˇ osc(f , x) ą 0 and Dε = x P Rn ˇ osc(f , x) ě ε . We remark
␣ ˇ ( ␣ ˇ (
8
here that D = D1 .
Ť
ig

k
k=1
r

“ñ” We show that D 1 has measure zero for all k P N (if so, then Theorem 8.24 implies
py

k
that D has measure zero).
Let k P N be fixed, and ε ą 0 be given. By Riemann’s condition there exists a partition
Co

P of A such that
ÿ[ A A
] ε
sup f (x) ´ inf f (x) ν(∆) ă .
xP∆ xP∆ k
∆PP

Define
(1) ˇ x P B∆ for some ∆ P P ,
␣ ˇ (
D1 ” x P D1
k k
(2) ˇ x P int(∆) for some ∆ P P .
␣ ˇ (
D1 ” x P D1
k k

(1) (2) (1)

Then D 1 = D 1 Y D 1 . We note that D 1 has measure zero since it is contained in
k k k k
(2)
B∆ while each B∆ has measure zero. Now we show that D 1 also has measure
Ť
∆PP k
§8.3 Lebesgue’s Theorem 289

(2)
∆ P P ˇ int(∆) X D 1 ‰ H . Then D 1 Ď
␣ ˇ (
zero. Let C = ∆ . Moreover, we
Ť
k k ∆PC
A 1 A
also note that if ∆ P C, sup f (x) ´ inf f (x) ě . In fact, if ∆ P C, there exists
xP∆ xP∆ k
y P int(∆) X D 1 ; thus choosing δ ą 0 such that D(y, δ) Ď int(∆),
k

A A ˇ A A ˇ A
ˇf (x1 ) ´ f A (x2 )ˇ
ˇ ˇ
sup f (x) ´ inf f (x) = sup ˇf (x1 ) ´ f (x2 )ˇ ě sup
xP∆ xP∆ x1 ,x2 P∆ x1 ,x2 PD(y,δ)
ˇ A A ˇ A 1
ě inf sup ˇf (x1 ) ´ f (x2 )ˇ = osc(f , y) ě .
δą0 x1 ,x2 PD(y,δ) k

As a consequence,

ed
1 ÿ ÿ[ A A
] ε
sup f (x) ´ inf f (x) ν(∆) = U (f, P) ´ L(f, P) ă

ct
ν(∆) ď
k xP∆ xP∆ k
∆PC ∆PP

te
(2)
which implies that ν(∆) ă ε. In other words, we establish that D 1 has measure
ř
∆PC
ro
zero. Therefore, D has measure zero for all k P N; thus D has measure zero.
1
k

k
P
“ð” Let R be a bounded closed rectangle with sides parallel to the coordinate axes and A
sĎ
ε
ht

int(R), and ε ą 0 be given. Define ε1 = , where }f }8 = sup |f (x)|.

2}f }8 + ν(R) + 1 xPA
ig

1. Since Dε1 is a subset of D, Proposition 8.19 implies that Dε1 has measure zero;
r

thus Proposition 8.20 provides open rectangles S1 , S2 , ¨ ¨ ¨ whose sides are parallel
py

8 8
to the coordinate axes such that Dε1 Ď Sk , and ν(Sk ) ă ε1 . In addition,
Ť ř
k=1 k=1
Co

we can assume that Sk Ď R for all k P N since Dε1 Ď R.

2. Since Dε1 Ď R is bounded, Lemma 8.29 implies that Dε1 is compact; thus Dε1 Ď
N
Sk for some N P N.
Ť
k=1

Ďk , and P1 be a partition of R satisfying

Let lk = S

(a) For each ∆ P P1 with ∆ X Dε1 ‰ H, ∆ Ď lk for some k = 1, ¨ ¨ ¨ , N .

(b) For each k = 1, ¨ ¨ ¨ , N , lk is the union of rectangles in P1 .

(c) Some collection of ∆ P P1 forms a partition P2 of A.

290 CHAPTER 8. Integration

R SN R SN

ñ
Dε1 Dε1
S1 S1
A A

Figure 8.2: Constructing partitions P1 and P2 from finite rectangles S1 , ¨ ¨ ¨ , SN

Rectangles in P1 fall into two families: C1 = ∆ P P1 ˇ ∆ Ď lk for some k = 1, ¨ ¨ ¨ , N ,

␣ ˇ (

and C2 = ∆ P P1 ˇ ∆ Ę lk for all k = 1, ¨ ¨ ¨ , N . By the definition of the oscillation

␣ ˇ (

ed
function, for x R Dε1 we let δx ą 0 be such that

ct
A A ˇ A
ˇf (x1 ) ´ f A (x2 )ˇ ă ε1 .
ˇ
sup f (y) ´ inf f (y) = sup
xPD(y,δy ) xPD(y,δy ) x1 ,x2 PD(x,δx )

te
(
Since K = ∆ is compact, there exists r ą 0 the Lebesgue number associated
Ť
∆PC2
with K and open cover
Ť )
ro
D(x, δx ) such that for each a P K, D(a, r) Ď D(y, δy ) for
xPK
P
some y P K. Let P 1 be a refinement of P1 such that }P 1 } ă r. Then if ∆1 P P 1 satisfies
that ∆1 Ď ∆ for some ∆ P C2 , we must have ∆1 Ď D(y, δy ) for some y P K; thus
ht

A A A A
sup f (x) ´ inf1 f (x) ď sup f (y) ´ inf f (y)
xP∆ xPD(y,δy )
ig

xP∆1 xPD(y,δy )
ˇ A
ˇf (x1 ) ´ f A (x2 )ˇ ă ε1
ˇ
= sup
r

x1 ,x2 PD(y,δy )
py

if ∆1 Ď ∆ for some ∆ P C2 . Let P = ∆1 P P 1 ˇ ∆1 Ď ∆ for some ∆ P P2 . Then P is

␣ ˇ (

a partition of A and
Co

( ÿ ÿ )( A A )
U (f, P) ´ L(f, P) = + sup f (x) ´ inf1 f (x) ν(∆1 )
xP∆1 xP∆
∆1 PP 1 ∆1 PP 1
∆1 Ď∆PC1 ∆1 Ď∆PC2
ÿ ÿ
ď 2}f }8 ν(∆1 ) + ε1 ν(∆1 )
∆1 PP 1 ∆1 PP 1
∆1 Ď∆PC1 ∆1 Ď∆PC2
ÿ
ď 2}f }8 ν(∆) + ε1 ν(R)
∆PPXC1
N
ÿ ( )
ď 2}f }8 ν(Sk ) + ε1 ν(R) ă 2}f }8 + ν(R) ε1 ď ε ;
k=1

thus f is Riemann integrable over A by Riemann’s condition. ˝

§8.3 Lebesgue’s Theorem 291

Example 8.31. Let A = Q X [0, 1], and f : A Ñ R be the constant function f ” 1. Then

1 if x P Q X [0, 1] ,
"
fs(x) =
0 otherwise .

The collection of points of discontinuity of fs is [0, 1] which, by Corollary 8.23, cannot be a

set of measure zero; thus f is not Riemann integrable.
Another way to see that f is not Riemann integrable is U (f, P) = 1 and L(f, P) = 0 for
all partitions P of A.

Corollary 8.32. A bounded set A Ď Rn has volume if and only if the boundary of A has

ed
measure zero.

ct
Proof. It suffices to show that the collection of discontinuities of the function 1A (which is
A
the same as 1Ď A ) is indeed BA.

te
1. If x0 R BA, then there exists δ ą 0 such that either D(x0 , δ) Ď A or D(x0 , δ) Ď AA ;
ro
thus 1A is continuous at x0 R BA since 1A (x) is constant for all x P D(x0 , δ).
P
2. On the other hand, if x0 P BA, then there exists xk P A, yk P AA such that xk Ñ x0 and
yk Ñ x0 as k Ñ 8. This implies that 1A cannot be continuous at x0 since 1A (xk ) = 1
ht

while 1A (yk ) = 0 for all k P N.

ig

As a consequence, the collection of discontinuity of 1A is exactly BA, and the corollary

r

follows from Lebesgue’s theorem. ˝

py

Corollary 8.33. Let A Ď Rn be a bounded set with volume. A bounded function f : A Ñ R

Co

with a finite or countable number of points of discontinuity is Riemann integrable over A.

Proof. We note that x P Rn ˇ osc(fs, x) ą 0 Ď BA Y x P A ˇ f is discontinuous at x . ˝

␣ ˇ ( ␣ ˇ (

Remark 8.34. In addition to the set inclusion listed in the proof of Corollary 8.33, we also
have
x P A ˇ f is discontinuous at x Ď x P Rn ˇ osc(fs, x) ą 0 .
␣ ˇ ( ␣ ˇ (

Therefore, if A Ď Rn is a bounded set with volume, then a bounded function f : A Ñ R is

Riemann integrable if and only if the collection of points of discontinuity of f has measure
zero.

Corollary 8.35. A bounded function is integrable over a compact set of measure zero.
292 CHAPTER 8. Integration

Proof. If f : K Ñ R is bounded, and K is a compact set of measure zero, then the collection
of discontinuities of fs is a subset of K. ˝

Corollary 8.36. Suppose that A, B Ď Rn are bounded sets with volume, and f : A Ñ R is
Riemann integrable over A. Then f is Riemann integrable over A X B.

Proof. By the inclusion

AXB A
, x) ą 0 Ď x P Rn ˇ osc(f , x) ą 0 ,
␣ ˇ ( ␣ ˇ (
x P int(A X B) ˇ osc(f

we find that

ed
AXB AXB
x P Rn ˇ osc(f
␣ ˇ ( ␣ ˇ (
, x) ą 0 Ď B(A X B) Y x P int(A X B) ˇ osc(f , x) ą 0

ct
A
Ď BA Y BB Y x P Rn ˇ osc(f , x) ą 0 .
␣ ˇ (

te
Since BA and BB both have measure zero, the integrability of f over A X B then follows
ro
from the integrability of f over A and the Lebesgue Theorem. ˝
P
Remark 8.37. Suppose that A Ď Rn is a bounded set of measure zero. Even if f : A Ñ R
is continuous, f might not be Riemann integrable. For example, the function f given in
ht

Example 8.31 is not Riemann integrable even though f is continuous on A.

ig

Remark 8.38. When f : A Ñ R is Riemann integrable over A, it is not necessary that A

has volume. For example, the zero function is Riemann integrable over A = Q X [0, 1] even
r
py

though A does not has volume.

Corollary 8.39 (Lebesgue’s Differentiation Theorem, a simple version). Let A Ď Rn be a

Co

bounded set with volume, and f : A Ñ R be bounded and Riemann integrable over A. Then
ż
1
lim f (x) dx = f (x0 ) (8.3.2)
rÑ0 ν(D(x0 , r) X A) D(x ,r)XA
0

for almost every x0 P A; that is, the equality above does not hold only for x0 from a set of
measure zero.

Proof. Let ε ą 0 be given, and suppose that fs, the zero extension of f outside A, is
continuous at x0 . Then there exists δ ą 0 such that

ˇfs(x) ´ fs(x0 )ˇ ă ε
ˇ ˇ
@ x P D(x0 , δ) X A .
2
§8.4 Properties Of The Integrals 293

Since BA has measure zero, by the fact that B(D(x0 , r) X A) Ď BD(x0 , r) Y BA we find that
B(D(x0 , r) X A) also has measure zero for all r ą 0. In other words, D(x0 , r) X A has volume.
Then if 0 ă r ă δ,
ż
ˇ 1 ˇ
f (x) dx ´ f (x0 )ˇ
ˇ ˇ
ν(D(x0 , r) X A) D(x0 ,r)XA
ˇ
( ) ˇˇ
ż
ˇ 1
=ˇ fs(x) ´ fs(x0 ) dxˇ
ˇ
ν(D(x0 , r) X A) D(x0 ,r)XA
ż
1 ˇ ˇ
ˇfs(x) ´ fs(x0 )ˇ dx
ď
ν(D(x0 , r) X A) D(x0 ,r)XA
ż
ε 1 ε

ed
ď 1 dx = ă ε .
2 ν(D(x0 , r) X A) D(x0 ,r)XA 2

ct
This implies that (8.3.2) holds for all x0 at which fs is continuous. The theorem then follows

te
from the Lebesgue theorem. ˝

8.4 Properties of the Integrals

P ro
Proposition 8.40. Let A Ď Rn be bounded, and f, g : A Ñ R be bounded. Then
ht

ż ż ż ż
(a) If B Ď A, then (f 1B )(x) dx = f (x) dx and (f 1B )(x) dx = f (x) dx.
A B A B
ig

ż ż ż ż ż ż
(b) f (x) dx + g(x) dx ď (f +g)(x) dx ď (f +g)(x) dx ď f (x) dx + g(x) dx.
r

A A A A A A
py

ż ż ż ż
(c) If c ě 0, then (cf )(x) dx = c f (x) dx and (cf )(x) dx = c f (x) dx.
Co

A A A A
ż ż ż ż
(d) If f ď g on A, then f (x) dx ď g(x) dx and f (x) dx ď g(x) dx.
A A A A
ż
(e) If A has volume zero, then f is Riemann integrable over A, and f (x) dx = 0.
A

Proof. We only prove (a), (b), (c) and (e) since (d) are trivial.

(a) Let ε ą 0 be given. By the definition of the lower integral, there exist partition PA of
A and PB of B such that
ż ÿ A
(f 1B )(x) dx ´ ε ă L(f 1B , PA ) = inf f 1B (x)ν(∆)
A xP∆
∆PPA
294 CHAPTER 8. Integration

and ż
ε ÿ B
f (x) dx ´ ă L(f, PB ) = inf f (x)ν(∆) .
B 2 ∆PP
xP∆
B

Let PA1 be a refinement of PA such that some collection of rectangles in PA1 forms a
B
partition of B. Denote this partition of B by PB1 . Since inf f (x) ď 0 if ∆ P PA1 zPB1 ,
xP∆
Proposition 8.6 implies that
ż ÿ A
(f 1B )(x) dx ´ ε ă L(f 1B , PA ) ď L(f 1B , PA1 ) = inf f 1B (x)ν(∆)
A 1
xP∆
∆PPA
( ÿ ÿ ) B
= + inf f (x)ν(∆)
xP∆

ed
1 zP 1
∆PPA 1
∆PPB
B
ÿ ż
B
ď inf f (x)ν(∆) = L(f, PB1 ) ď f (x) dx .

ct
1
xP∆ B
∆PPB

te
On the other hand, let P
rA be a partition of A such that PB Ď P
rA and
ro
ÿ ε
ν(∆) ď ,
2(M + 1)
∆PPA zPB , ∆XB‰H
P
r

where M ą 0 is an upper bound of |f |. Then

ht

ÿ B
ÿ B ε
inf f (x)ν(∆) ě ´M f (x)ν(∆) ě ´
xP∆ 2
∆PP
rA zPB , ∆XB‰H ∆PP
rA zPB , ∆XB‰H
ig

which further implies that

r

ż
py

ÿ A
(f 1B )(x) dx ě L(f 1B , P
rA ) = inf (f 1B (x)ν(∆)
A xP∆
∆PP
( ÿ )
rA
Co

ÿ ÿ B
= + + inf f (x)ν(∆)
xP∆
∆PPB ∆PP
rA zPB , ∆XB‰H ∆PP
rA zPB , ∆XB=H
ÿ ż
B
= L(f, PB ) + inf f (x)ν(∆) ą f (x) dx ´ ε .
xP∆ B
∆PP
rA zPB , ∆XB‰H

Therefore, we establish that

ż ż ż
f (x) dx ´ ε ă (f 1B )(x) dx ă f (x) dx + ε .
B A B
ż ż
Since ε ą 0 is given arbitrarily, we conclude that (f 1B )(x) dx = f (x) dx. Similar
A B
ż ż
argument can be applied to conclude that (f 1B )(x) dx = f (x) dx.
A B
§8.4 Properties Of The Integrals 295

(b) Let ε ą 0 be given. By the definition of the lower integral, there exist partitions P1
and P2 of A such that
ż ż
ε ε
f (x) dx ´ ă L(f, P1 ) and g(x) dx ´ ă L(g, P2 ) .
A 2 A 2

Let P be a common refinement of P1 and P2 . Then

ż ż
f (x) dx + g(x) dx ´ ε ă L(f, P1 ) + L(f, P2 ) ď L(f, P) + L(g, P)
A A
ÿ ÿ
= inf fs(x)ν(∆) + inf gs(x)ν(∆)
xP∆ xP∆

ed
∆PP ∆PP
ÿ ż
ď inf (fs + gs)(x)ν(∆) = L(f + g, P) ď (f + g)(x) dx .
xP∆

ct
∆PP A

Since ε ą 0 is given arbitrarily, we conclude that

te
ż ż ż ro
f (x) dx + g(x) dx ď (f + g)(x) dx .
A A A
P
ż ż ż
Similarly, we have (f + g)(x) dx ď f (x) dx + g(x) dx; thus (b) is established.
A A A
ht

(c) It suffices to show the case c = ´1. For each k P N, there exist partitions Pk and Qk
of A such that
ig

ż ż
1
´f (x) dx ´ ă L(´f, Pk ) ď ´f (x) dx
r

A k A
py

and ż ż
1
f (x) dx ď U (f, Qk ) ă f (x) dx + .
Co

A A k
Let Rk be the common refinement of Pk and Qk . Then
ż ż
1
´f (x) dx ´ ă L(´f, Pk ) ď L(´f, Rk ) ď ´f (x) dx
A k A

and ż ż
1
f (x) dx ď U (f, Rk ) ď U (f, Qk ) ă f (x) dx + .
A A k
which implies that
ż ż
lim U (f, Rk ) = f (x) dx and lim L(´f, Rk ) = ´f (x) dx
kÑ8 A kÑ8 A
296 CHAPTER 8. Integration

Since
A
ÿ ÿ A
L(´f, Rk ) = inf (´f ) (x)ν(∆) = ´ sup f (x)ν(∆) = U (f, Rk ) ,
xP∆
∆PRk ∆PRk xP∆
ż ż
we conclude that ´f (x) dx = ´ f (x) dx.
A A

(e) Since f is bounded on A, there exist M ą 0 such that ´M ď f (x) ď M for all x P A.
f
Therefore, ´1A ď ď 1A on A; thus (c) and (d) imply that
M
ż ż ż ż
f (x) 1
0= 1A (x) dx = 1A (x) dx ě dx = f (x) dx
A M M A

ed
A A
ż ż
which implies that f (x) dx ď 0. Similarly, ´f (x) dx ď 0 which further implies

ct
ż A A
that f (x) dx ě 0. Therefore, by Corollary 8.7 we conclude that

te
A ż ż
0ď f (x) dx ď f (x) dx ď 0
A
ro A ż
which implies that f is Riemann integrable over A and f (x) dx = 0. ˝
P
A
Remark 8.41. Let A Ď Rn be bounded and f, g : A Ñ R be bounded. Then (b) of
Proposition 8.40 also implies that
ht

ż ż ż ż ż ż
(f ´g)(x) dx ď f (x) dx´ g(x) dx and f (x) dx´ g(x) dx ď (f ´g)(x) dx .
ig

A A A A A A

Corollary 8.42. Let A, B Ď R be bounded such that A X B has volume zero, and f :
n
r
py

A Y B Ñ R be bounded. Then
ż ż ż ż ż ż
f (x) dx + f (x) dx ď f (x) dx ď f (x) dx ď f (x) dx + f (x) dx .
Co

A B AYB AYB A B

Proof. Note that f 1A + f 1B = f 1AYB + f 1AXB on A Y B. Therefore, (a), (b) of Proposition

8.40 and Remark 8.41 implies that
ż ż ż ż ż
f (x) dx + f (x) dx = (f 1A )(x) dx + (f 1B )(x) dx ď (f 1A + f 1B )(x) dx
A B AYB AYB AYB
( )
ż
= f 1AYB ´ (´f 1AXB ) (x) dx
żAYB ż
ď f 1AYB (x) dx ´ (´f 1AXB )(x) dx
AYB AYB
ż ż
= f (x) dx ´ (´f )(x) dx
AYB AXB
§8.4 Properties Of The Integrals 297

which, with the help of Proposition 8.40 (e), further implies that
ż ż ż
f (x) dx + f (x) dx ď f (x) dx .
A B AYB

The case of the upper integral can be proved in a similar fashion. ˝

Having established Proposition 8.40, it is easy to see the following theorem (except (c)).
The proof is left as an exercise.

Theorem 8.43. Let A Ď Rn be bounded, c P R, and f, g : A Ñ R be Riemann integrable.

Then

ed
ż ż ż
(a) f ˘ g is Riemann integrable, and (f ˘ g)(x) dx = f (x) dx ˘ g(x) dx.
A A A

ct
ż ż
(b) cf is Riemann integrable, and (cf )(x) dx = c f (x) dx.

te
A A
ˇż ˇ ż ro
(c) |f | is Riemann integrable, and ˇ f (x) dxˇ ď |f (x)|dx.
ˇ ˇ
A A
P
ż ż
(d) If f ď g, then f (x) dx ď g(x) dx.
A A
ht

ˇż ˇ
(e) If A has volume and |f | ď M , then ˇ f (x) dxˇ ď M ν(A).
ˇ ˇ
A
ig

Theorem 8.44. Let A Ď R be bounded, and f : A Ñ R be a bounded integrable function.

n
r

ż
1. If A has measure zero, then f (x) dx = 0.
py

A
ż ␣ ˇ (
2. If f (x) ě 0 for all x P A, and f (x) dx = 0, then the set x P A ˇ f (x) ‰ 0 has
Co

A
measure zero.

Proof. 1. We show that L(f, P) ď 0 ď U (f, P) for all partitions P of A. Let P =

␣ (
∆1 , ¨ ¨ ¨ , ∆N be a partition of A. By Corollary 8.23, ∆k X AA ‰ H for k = 1, ¨ ¨ ¨ , N ;
thus we must have inf fs(x) ď 0 and sup fs(x) ě 0. As a consequence, if P is a
xP∆k xP∆k
partition of A,
N
ÿ N
ÿ
L(f, P) = inf fs(x)ν(∆k ) ď 0 and U (f, P) = sup fs(x)ν(∆k ) ě 0 ;
xP∆k
k=1 k=1 xP∆k
ż ż ż
thus f (x) dx ď 0 ď f (x)dx. Since f is integrable over A, f (x) dx = 0.
A A A
298 CHAPTER 8. Integration

1(
. We claim that Ak has measure zero for all k P N.
␣ ˇ
2. Let Ak = x P A ˇ f (x) ě
k
ż
Let ε ą 0 be given. Since f (x)dx = 0, there exists a partition P of A such that
A ˇ
ε
U (f, P) ă . Let C = ∆ P P ˇ ∆ X Ak ‰ H . Then Ak Ď
␣ (
∆, and
Ť
k ∆PC

1 ÿ ÿ ÿ ε
ν(C) ď sup fs(x)ν(∆) ď sup fs(x)ν(∆) = U (f, P) ă
k ∆PC ∆PC xP∆ ∆PP xP∆
k

which implies that ν(∆) ă ε. Therefore, Ak has measure zero; thus Theorem 8.24
ř
∆PC
8

ed
implies that A = Ak also has measure zero.
Ť
˝
k=1

ct
Remark 8.45. Combining Corollary 8.35 and Theorem 8.44, we conclude that the integral
of a bounded function over a compact set of measure zero is zero.

te
Remark 8.46. Let A = Q X [0, 1] and f : A Ñ R be the constant function f ” 1. We have
ro
shown in Example 8.31 that f is not Riemann integrable. We note that A has no volume
P
since BA = [0, 1] which is not a set of measure zero. However, A has measure zero since it
consists of countable number of points.
ht

1. Since f is continuous on A, the condition that A has volume in Corollary 8.33 cannot
ig

be removed.
r

2. Since A has measure zero, the condition that f is Riemann integrable in Theorem 8.44
py

cannot be removed.
Co

Theorem 8.47 (Mean Value Theorem for Integrals). Let A be a subset of Rn such that A
has volume and is compact and connected. Suppose that f : A Ñ R is continuous, then there
exists x0 P A such that ż
f (x) dx = f (x0 )ν(A) .
A
ż
1
The quantity f (x) dx is called the average of f over A.
ν(A) A

Proof. Because of Theorem 8.44, it suffices to show the case that ν(A) ‰ 0. Let m =
min f (x) and M = max f (x). Then
xPA xPA

m1A (x) ď f (x) ď M 1A (x) ;

§8.5 Fubini’s Theorem 299

thus (b) and (d) of Theorem 8.43 imply that

ż ż ż
mν(A) = m1A (x) dx ď f (x) dx ď M 1A (x) dx = M ν(A) .
A A A

By the connectedness of A and continuity of f , Theorem 4.21 żand Theorem 3.38 implies
1
that f (A) = [m, M ]; thus by the fact that the quantity f (x) dx P [m, M ], there
ν(A) A
must be x0 P A such that ż
1
f (x0 ) = f (x) dx . ˝
ν(A) A

ed
Definition 8.48. Let A Ď Rn be a set and f : A Ñ R be a function. For B Ď A, the
restriction of f to B is the function f ˇB : A Ñ R given by f |B = f 1B . In other words,
ˇ

ct
f (x) if x P B ,
"

te
ˇ
f ˇB (x) =
0 if x P AzB .
ro
The following lemma is a direct consequence of Proposition 8.40 (a).
P
Lemma 8.49. Let A Ď Rn be bounded, and f : A Ñ R be a bounded function. Suppose that
ht

ˇ
B Ď A, and f ˇB is Riemann integrable over A. Then f is Riemann integrable over B, and
ig

ż ż
ˇ
f ˇB (x) dx = f (x) dx .
r

A B
py

Theorem 8.50. Let A, B be bounded subsets of Rn be such that A X B has measure zero,
and f : A Y B Ñ R be such that f ˇAXB , f ˇA and f ˇB are all Riemann integrable over A Y B.
ˇ ˇ ˇ
Co

Then f is integrable over A Y B, and

ż ż ż
f (x) dx = f (x) dx + f (x) dx .
AYB A B

Proof. Since 1AYB = 1A + 1B ´ 1AXB , we have

ˇ ˇ ˇ
f = f 1AYB = f ˇA + f ˇB ´ f ˇAXB ;

thus Theorem 8.43, Theorem 8.44 and Lemma 8.49 imply that
ż ż ż ż ż
ˇ ˇ
f (x) dx = f ˇA (x) dx + f ˇB (x) dx = f (x) dx + f (x) dx . ˝
AYB AYB AYB A B
300 CHAPTER 8. Integration

8.5 The Fubini Theorem

If f : [a, b] Ñ R is continuous, the fundamental theorem of Calculus (Theorem 4.90) can be
applied to computed the integral of f over [a, b]. In the following two sections, we focus on
how the integral of f over A Ď Rn , where n ě 2, can be computed if the integral exists. We
start with the special case n = 2.

Definition 8.51. Let S = [a, b] ˆ [c, d] be a rectangle in R2 , and f : S Ñ R be bounded.

For each fixed x P [a, b], the lower integral of the function f (x, ¨) : [c, d] Ñ R is denoted
żd żd
by f (x, y) dy, and the upper integral of f (x, ¨) : [c, d] Ñ R is denoted by f (x, y) dy.

ed
c c
If for each x P [a, b] the upper integral and the lower integral of f (x, ¨) : [c, d] Ñ R are
żd

ct
the same, we simply write f (x, y) dy for the integrals of f (x, ¨) over [c, d]. The integrals
żb żb c żb

te
f (x, y) dx, f (x, y) dx and f (x, y) dx are defined in a similar way.
a a a ro
Lemma 8.52. Let A = [a, b] ˆ [c, d] be a rectangle in R2 , and f : A Ñ R be bounded. Then
P
ż ż b(ż d ) ż b(ż d ) ż
f (x, y) dA ď f (x, y) dy dx ď f (x, y) dy dx ď f (x, y) dA (8.5.1)
A a c a c A
ht

and
ż d(ż b ) ż d(ż b )
ig

ż ż
f (x, y) dA ď f (x, y) dx dy ď f (x, y) dx dy ď f (x, y) dA . (8.5.2)
r

A c a c a A
py

Proof. It suffices to prove (8.5.1). Let ε ą 0 be given. Choose a partition

Co

P = ∆ij ˇ ∆ij = [xi , xi+1 ] ˆ [yj , yj+1 ] for i = 0, 1, ¨ ¨ ¨ , n ´ 1 and j = 0, 1, ¨ ¨ ¨ , m ´ 1

␣ ˇ (

ż
of A such that L(f, P) ą f (x, y) dA ´ ε. Using (4.7.3) and Remark 4.82, we find that
A

ż b(ż d ) n´1
ÿ ż xi+1 ( m´1
ÿ ż yj+1 )
f (x, y) dy dx = f (x, y) dy dx
a c i=0 xi j=0 yj
n´1
ÿ m´1
ÿ ż xi+1 ( ż yj+1 )
ě f (x, y) dy dx
i=0 j=0 xi yj
n´1
ÿ m´1
ÿ ż
ě inf f (x, y)ν(∆ij ) = L(f, P) ą f (x, y) dA ´ ε .
(x,y)P∆ij A
i=0 j=0
§8.5 Fubini’s Theorem 301

Since ε ą 0 is given arbitrarily, we must have

ż b(ż d ) ż
f (x, y) dy dx ě f (x, y) dA .
a c A

ż b (ż d ) ż
Similarly, f (x, y) dy dx ď f (x, y) dA, so (8.5.1) is concluded. ˝
a c A

Theorem 8.53 (Fubini’s Theorem, the case n = 2). Let A = [a, b] ˆ [c, d] be a rectangle in
R2 , and f : A Ñ R be Riemann integrable. Then
żd żd

ed
1. the functions f (¨, y) dy and f (¨, y) dy are Riemann integrable over [a, b];
c c

ct
żb żb
2. the functions f (x, ¨)dx and f (x, ¨)dx are Riemann integrable over [c, d], and
a a

te
3. The integral of f over A is the same as the iterated integrals; that is, ro
ż ż b(ż d ) ż b(ż d )
f (x, y) dA = f (x, y) dy dx = f (x, y) dy dx
P
A a c a c
ż d(ż b ) ż d(ż b )
ht

= f (x, y) dx dy = f (x, y) dx dy .
c a c a
ig

żd
Proof. It suffices to prove that f (x, y) dy is Riemann integrable over [a, b] and
r

c
py

ż b(ż d ) ż
f (x, y) dy dx = f (x, y) dA . (8.5.3)
Co

a c A

ż b (ż d ) ż b (ż d )
Since f (x, y) dy dx ď f (x, y) dy dx , Lemma 8.52 implies that
a c a c

ż ż b(ż d ) ż b(ż d )
f (x, y) dA ď f (x, y) dy dx ď f (x, y) dy dx
A a c a c
ż b(ż d ) ż
ď f (x, y) dy dx ď f (x, y) dA .
a c A

żd
The integrability of f (x, y) dy and the validity of (8.5.3) are then concluded by the
c
integrability of f over A. ˝
302 CHAPTER 8. Integration

żbżd
Remark 8.54. To simplify the notation, sometimes we use f (x, y) dydx to denote
ż b (ż d ) a c
the iterated integral the iterated integral f (x, y) dy dx. Similar notation applies
a c żb żd
to the upper and the lower integrals. For example, we also have f (x, y) dydx =
a c
ż b (ż d )
f (x, y) dy dx.
a c

żd żd
Remark 8.55. For each x P [a, b], define φ(x) = f (x, y) dy and ψ(x) = f (x, y) dy.
c c
Then φ(x) ď ψ(x) for all x P [a, b], and the Fubini Theorem implies that

ed
[ ]
żb

ct
ψ(x) ´ φ(x) dx = 0 .
a

te
␣ ˇ (
By Theorem 8.44, the set x P [a, b] ˇ ψ(x) ´ φ(x) ‰ 0 has measure zero. In other words,
ro
except on a set of measure zero, f (x, ¨) is Riemann integrable over [c, d] if f is Riemann
integrable over [a, b] ˆ [c, d]. This property can be rephrased as that “f (x, ¨) is Riemann
P
integrable over [c, d] for almost every x P [a, b] if f is Riemann integrable over the rectangle
[a, b] ˆ [c, d]”. Similarly, f (¨, y) is Riemann integrable for almost every y P [c, d] if f is
ht

Riemann integrable over [a, b] ˆ [c, d].

ig

Remark 8.56. The integrability of f does not guarantee that f (x, ¨) or f (¨, y) is Riemann
r

integrable. In fact, there exists a function f : [0, 1] ˆ [0, 1] Ñ R such that f is Riemann
py

integrable, f (¨, y) is Riemann integrable for each y P [0, 1], but f (x, ¨) is not Riemann
integrable for infinitely many x P [0, 1]. For example, let
Co

$
& 0 if x = 0 or if x or y is irrational ,
f (x, y) = 1 q
% if x, y P Q and x = with (p, q) = 1 .
p p

Then

1. For each y P [0, 1], f (¨, y) żis continuous atżall irrational numbers. Therefore, f (¨, y) is
1 1
Riemann integrable, and f (x, y) dx = f (x, y) dx = 0.
0 0

ż1
2. For x = 0 or x R Q, f (x, ¨) is Riemann integrable, and f (x, y) dy = 0.
0
§8.5 Fubini’s Theorem 303

q
3. If x = with (p, q) = 1, f (x, ¨) is nowhere continuous in [0, 1]. In fact, for each
p
y0 P [0, 1],
1
lim f (x, y) = while lim f (x, y) = 0 ;
yÑy0
yPQ
p yÑy0
yRQ

thus the limit of f (x, y) as y Ñ y0 does not exist. Therefore, the Lebesgue theorem
implies that f (x, ¨) is not Riemann integrable if x P Q X (0, 1]. On the other hand, for
q
x = with (p, q) = 1 we have
p
ż1 ż1
1
f (x, y) dy = 0 and f (x, y) dy = .
p

ed
0 0

ż1 ż1

ct
4. Define φ(x) = f (x, y) dy and ψ(x) = f (x, y) dy. Then 2 and 3 imply that φ
0 0

te
ż1 ż1
and ψ are Riemann integrable over [0, 1], and φ(x)dx = ψ(x)dx = 0.
ro 0 0

5. For each a R Q X [0, 1] and b P [0, 1], f is continuous at (a, b). In fact, for any given
1
P
ε ą 0, there exists a prime number p such that ă ε. Let
p
ht

ℓ ˇˇ ˇ
!ˇ ˇ )
δ = min a ´ ˇ 0 ď ℓ ď k ď p, k P N, ℓ P N Y t0u .
ˇ
k
( )
ig

Then δ ą 0, and if (x, y) P D (a, b), δ X ([0, 1] ˆ [0, 1]), we have

r

ˇf (x, y) ´ f (a, b)ˇ = ˇf (x, y)ˇ ă 1 ă ε ,

py

ˇ ˇ ˇ ˇ
p
( )
Co

ℓ
where we use the fact that if (x, y) P D (a, b), δ and x P Q, then x = (in reduced
k
form) for some k ą p.
As a consequence, (a, b) P R2 ˇ fs is discontinuous at (a, b) Ď Q ˆ [0, 1]. Since
␣ ˇ (

Q ˆ [0, 1] is a countable union of measure zero sets, it has measure zero; thus f is
Riemann integrable by the Lebesgue theorem. The Fubini theorem then implies that
ż ż1ż1
f (x, y) dA = f (x, y) dxdy = 0 .
[0,1]ˆ[0,1] 0 0

Remark 8.57. The integrability of f (x, ¨) and f (¨, y) does not guarantee the integrability of
f . In fact, there exists a bounded function f : [0, 1] ˆ [0, 1] Ñ R such that f (x, ¨) and f (¨, y)
304 CHAPTER 8. Integration

are both Riemann integrable over [0, 1], but f is not Riemann integrable over [0, 1] ˆ [0, 1].
For example, let
& 1 if (x, y) = ( k , ℓ ), 0 ă k, ℓ ă 2n odd numbers, n P N ,
$

f (x, y) = 2n 2n
0 otherwise .
%

Then for each x P [0, 1], f (x, ¨) only has finite number of discontinuities; thus f (x, ¨) is
Riemann integrable, and ż1
f (x, y) dy = 0 .
0
ż1

ed
Similarly, f (¨, y) is Riemann integrable, and f (x, y) dx = 0. As a consequence,
0

ct
ż1ż1 ż1ż1
f (x, y) dydx = f (x, y) dxdy = 0 .

te
0 0 0 0

However, note that f is nowhere continuous on [0, 1] ˆ [0, 1]; thus the Lebesgue theorem
ro
implies that f is not Riemann integrable. One can also see this by the fact that U (f, P) = 1
and L(f, P) = 0 for all partition of [0, 1] ˆ [0, 1].
P
Corollary 8.58. 1. Let φ1 , φ2 : [a, b] Ñ R be continuous maps such that φ1 (x) ď φ2 (x)
ht

for all x P [a, b], A = (x, y) ˇ a ď x ď b, φ1 (x) ď y ď φ2 (x) , and f : A Ñ R be

␣ ˇ (

continuous. Then f is Riemann integrable over A, and

ig

ż ż b ( ż φ2 (x) )
r

f (x, y) dA = f (x, y) dy dx .
py

A a φ1 (x)

2. Let ψ1 , ψ2 : [c, d] Ñ R be continuous maps such that ψ1 (y) ď ψ2 (y) for all y P [c, d],
Co

A = (x, y) ˇ c ď y ď d, ψ1 (y) ď x ď ψ2 (y) , and f : A Ñ R be continuous. Then f is

␣ ˇ (

Riemann integrable over A, and

ż ż d ( ż ψ2 (y) )
f (x, y) dA = f (x, y) dx dy .
A c ψ1 (y)

Proof. It suffices to prove 1. First we show that f is Riemann integrable over A. By

ˇ ( )
Lebesgue’s theorem, it suffices to show that the set (x, y) P R2 ˇ osc fs, (x, y) ą 0 has
␣ (

measure zero, where fs is the extension of f by zero outside A. Nevertheless, we note that
ˇ ( ) [ ] [ ]
(x, y) P R2 ˇ osc fs, (x, y) ą 0 Ď tau ˆ φ1 (a), φ2 (a) Y tbu ˆ φ1 (b), φ2 (b) Y
␣ (
␣( )ˇ ( ␣( )ˇ (
Y x, φ1 (x) ˇ x P [a, b] Y x, φ2 (x) ˇ x P [a, b] .
§8.5 Fubini’s Theorem 305

[ ] [ ]
It is clear that tauˆ φ1 (a), φ2 (a) and tbuˆ φ1 (b), φ2 (b) have measure zero since they have
␣( )ˇ ( ␣( )ˇ (
volume zero. Now we claim that the sets x, φ1 (x) ˇ x P [a, b] and x, φ2 (x) ˇ x P [a, b]
also have measure zero.
Let ε ą 0 be given. Since φ1 is continuous on a compact set [a, b], φ1 is uniformly
continuous on [a, b]; thus there exists δ ą 0 such that
ˇ ˇ ε
ˇφ1 (x1 ) ´ φ1 (x2 )ˇ ă whenever |x1 ´ x2 | ă δ .
b´a
Let P = ta = x0 ă x1 ă ¨ ¨ ¨ ă xn´1 ă xn = bu be a partition of [a, b] such that |xi+1 ´xi | ă δ
[ ] [ ]
for all i = 0, ¨ ¨ ¨ , n ´ 1, and let ∆i = xi , xi+1 ˆ min φ1 (x), max φ1 (x) . Then

ed
xP[xi ,xi+1 ] xP[xi ,xi+1 ]

)ˇ ( n´1
ď

ct
␣(
x, φ1 (x) ˇ x P [a, b] Ď ∆i
i=0

te
and
n´1 n´1 n´1
ε ε ÿ
ÿ
ν(∆i ) ă
b
ÿ
´ a
(xi+1 ´ xi ) =
b
ro
´ a
(xi+1 ´ xi ) = ε .
i=0 i=0 i=0
)ˇ )ˇ
P
␣( ( ␣( (
Therefore, x, φ1 (x) ˇ x P [a, b] has volume zero; thus x, φ1 (x) ˇ x P [a, b] has measure
␣( )ˇ ( ␣
zero. Similarly, x, φ2 (x) ˇ x P [a, b] also has measure zero. By Theorem 8.24, (x, y) P
ˇ ( )
ht

R2 ˇ osc fs, (x, y) ą 0 has measure zero; thus f is Riemann integrable over A.
(

Let m = min φ1 (x), M = max φ2 (x), and S = [a, b] ˆ [m, M ]. Then A Ď S. By Lemma
ig

xP[a,b] xP[a,b]
8.49 and the Fubini Theorem,
r

ż b(ż M ) ż b ( ż φ2 (x) )
py

ż ż
f (x, y) dA = f (x, y) dA =
s f (x, y) dy dx =
s f (x, y) dy dx
A S a m a φ1 (x)
Co

which concludes 1. ˝

Example 8.59. Let A = (x, y) P R2 ˇ 0 ď x ď 1, x ď y ď 1 , and f : A Ñ R be given by

␣ ˇ (

f (x, y) = xy. Then Corollary 8.58 implies that

ż 1(ż 1 ) ż1(
x x3 )
ż ż 1 2 ˇy=1
xy ˇ 1 1 1
f (x, y) dA = xy dy dx = ˇ dx = ´ dx = ´ = .
A 0 x 0 2 y=x 0 2 2 4 8 8

On the other hand, since A = (x, y) P R2 ˇ 0 ď y ď 1, 0 ď x ď y , we can also evaluate the

␣ ˇ (

integral of f over A by
ż ż 1(ż y ) ż 1 2 ˇx=y ż1 3
x yˇ y 1
xy dA = xy dx dy = ˇ dy = dy = .
A 0 0 0 2 x=0 0 2 8
306 CHAPTER 8. Integration

?
Example 8.60. Let A = (x, y) P R2 ˇ 0 ď x ď 1, x ď y ď 1 , and f : A Ñ R be given by
␣ ˇ (
3
f (x, y) = ey . Then Corollary 8.58 implies that

ż ż 1(ż 1 )
3
f (x, y) dA = ?
ey dy dx .
A 0 x

Since we do not know how to compute the inner integral, we look for another way of finding
the integral. Observing that A = (x, y) P R2 ˇ 0 ď y ď 1, 0 ď x ď y 2 , we have
␣ ˇ (

ż ż 1 ( ż y2 ) ż1
1 y3 ˇˇy=1 e ´ 1

ed
y3 2 y3
f (x, y) dA = e dx dy = y e dy = e ˇ = .
A 0 0 0 3 y=0 3

ct
Now we prove the general Fubini Theorem.

Theorem 8.61 (Fubini’s Theorem). Let A Ď Rn and B Ď Rm be rectangles, and f :

te
ro
A ˆ B Ñ R be bounded. For x P Rn and y P Rm , write z = (x, y). Then
P
ż ż (ż ) ż (ż ) ż
f (z) dz ď f (x, y)dy dx ď f (x, y)dy dx ď f (z) dz (8.5.4)
ht

AˆB A B A B AˆB
ig

and
r

ż (ż ) ż (ż )
py

ż ż
f (z) dz ď f (x, y)dx dy ď f (x, y)dx dy ď f (z) dz . (8.5.5)
AˆB B A B A AˆB
Co

In particular, if f : A ˆ B Ñ R is Riemann integrable, then

ż ż (ż ) ż (ż )
f (z) dz = f (x, y)dy dx = f (x, y)dy dx
AˆB A B A B
ż (ż ) ż (ż )
= f (x, y)dx dy = f (x, y)dx dy .
B A B A

Proof. It sufficesż to prove (8.5.4). Let ε ą 0 be given. Choose a partition P of A ˆ B such

that L(f, P) ą f (z) dz ´ ε. Since P is a partition of A ˆ B, there exist partition Px
AˆB
of A and partition Py of B such that P = ∆ = R ˆ S ˇ R P Px , S P Py . By Proposition
␣ ˇ (
§8.5 Fubini’s Theorem 307

8.40 and Corollary 8.42, we find that

ż (ż ) ż (ż )
f (x, y) dy dx = Ť 1A (x) Ť f (x, y)1B (y) dy dx
A B RPPx R SPPy S
ÿ ż ( ÿ ż AˆB
)
ě f (x, y) dy dx
R PPx R S PPy S
ÿ ÿ ż (ż AˆB
)
ě f (x, y) dy dx
R PPx S PPy R S
ÿ AˆB
ě inf f (x, y)νm (S)νn (R)
(x,y)PRˆS
R PPx ,S PPy

ed
ÿ ż
AˆB
= inf f (x, y)νn+m (∆) = L(f, P) ą f (z)dz ´ ε .
(x,y)P∆ AˆB

ct
∆PP

Since ε ą 0 is given arbitrarily, we conclude that

te
ż ż (ż ) ro
f (z) dz ď f (x, y)dx dy .
AˆB B A

ż (ż )
P
ż
Similarly, f (x, y)dy dx ď f (z) dz; thus (8.5.4) is concluded. ˝
A B AˆB
ht

Corollary 8.62. Let S Ď Rn be a bounded set with volume, φ1 , φ2 : S Ñ R be continuous

ig

maps such that φ1 (x) ď φ2 (x) for all x P S, A = (x, y) P Rn ˆR ˇ x P S, φ1 (x) ď y ď φ2 (x) ,
␣ ˇ (

and f : A Ñ R be continuous. Then f is Riemann integrable over A, and

r
py

ż ż ( ż φ2 (x) )
f (x, y) d(x, y) = f (x, y) dy dx . (8.5.6)
A S φ1 (x)
Co

Proof. Since BA has measure zero, and f is continuous on A, Corollary 8.33 implies that f is
Riemann integrable over A. Let m = min φ1 (x) and M = max φ2 (x). Then A Ď S ˆ[m, M ];
xPS xPS
thus Theorem 8.50 and the Fubini Theorem imply that
ż ż ż (ż M )
A A
f (x, y) d(x, y) = f (x, y) d(x, y) = f (x, y) dy dx
A Sˆ[m,M ] S m
ż (ż M )
A
= f (x, y) dy dx .
S m

A
Noting that [m, M ] has a boundary of volume zero in R, and for each x P S, f (x, ¨) is
A
continuous except perhaps at y = φ1 (x) and y = φ2 (x), Corollary 8.33 implies that f (x, ¨) is
308 CHAPTER 8. Integration

żM żM
A A
Riemann integrable over [m, M ] for each x P S. Therefore, f (x, y) dy = f (x, y) dy
m m
which further implies that
ż ż (ż M )
A
f (x, y) d(x, y) = f (x, y) dy dx . (8.5.7)
A S m
A
For each fixed x P S, let Ax = y P R ˇ φ1 (x) ď y ď φ2 (x) . Then f (x, y) = f (x, y)1Ax (y)
␣ ˇ (
A
for all (x, y) P S ˆ [m, M ] or equivalently, f (x, ¨) = f (x, ¨)|Ax for all x P S; thus Proposition
8.40 (a) implies that
żM ż ż φ2 (x)
A
f (x, y) dy = f (x, y) dy = f (x, y) dy @x P S . (8.5.8)

ed
m Ax φ1 (x)

Combining (8.5.7) and (8.5.8), we conclude (8.5.6). ˝

ct
Example 8.63. Let A Ď R3 be the set (x1 , x2 , x3 ) P R3 ˇ x1 ě 0, x2 ě 0, x3 ě 0, and x1 +
␣ ˇ

te
x2 + x3 ď 1 , and f : A Ñ R be given by f (x1 , x2 , x3 ) = (x1 + x2 + x3 )2 . Let S =
(

[0, 1] ˆ [0, 1] ˆ [0, 1], and fs : R3 Ñ R be the extension of f by zero outside A. Then
ro
Corollary 8.33 implies that f is Riemann integrable (since BA has measure zero). Write
P
x p2 = (x1 , x3 ) and x
p1 = (x2 , x3 ), x p3 = (x1 , x2 ). Lemma 8.49 implies that
ż ż
f (x)dx = fs(x)dx ,
ht

A S

and Theorem 8.61 implies that

ig

ż ż (ż )
f (x)dx = fs(p x3 dx3 .
x3 , x3 )dp
r

s
py

S [0,1] [0,1]ˆ[0,1]

= (x1 , x2 ) P R2 ˇ x1 ě 0, x2 ě 0, x1 + x2 ď 1 ´ x3 . Then for each x3 P [0, 1],

␣ ˇ (
Let Ax3
Co

ż ż ż 1´x3 ( ż 1´x3 ´x2 )

f (p
s x3 , x3 )dp
x3 = f (p
x3 , x3 )dp
x3 = f (x1 , x2 , x3 )dx1 dx2 .
[0,1]ˆ[0,1] Ax3 0 0

Computing the iterated integral, we find that

ż ż 1 [ ż 1´x3 ( ż 1´x3 ´x2 ) ]
2
f (x)dx = (x1 + x2 + x3 ) dx1 dx2 dx3
A 0 0 0
ż 1 [ ż 1´x3 ]
(x1 + x2 + x3 )3 ˇˇx1 =1´x3 ´x2
= dx2 dx3
3
ˇ
x1 =0
ż01 [ ż01´x3 ( 3) ]
1 (x2 + x3 )
= ´ dx2 dx3
3 3
ż01 ( 0 4)
1 x3 x3 1 1 1 15 ´ 10 + 1 1
= ´ + dx3 = ´ + = = .
0 4 3 12 4 6 60 60 10
§8.6 Change of Variables Formula 309

Example 8.64. In this example we compute the volume of the n-dimensional unit ball ωn .
By the Fubini theorem,
ż 1 ż ?1´x21 ż ?1´x21 ´¨¨¨´x2n´1
ωn = ? ¨¨¨ ? dxn ¨ ¨ ¨ dx1 .
´1 ´ 1´x21 ´ 1´x21 ´¨¨¨´x2n´1

ż ?1´x2 ż ?1´x2 ´¨¨¨´x2

1 1 n´1 n´1
Note that the integral ? ¨¨¨ ? dxn ¨ ¨ ¨ dx2 is in fact ωn´1 (1´x21 ) 2 ; thus
´ 1´x21 ´ 1´x21 ´¨¨¨´x2n´1
ż1 ż π
n´1 2
2
ωn = ωn´1 (1 ´ x ) 2 dx = 2 ωn´1 cosn θdθ . (8.5.9)
´1 0

ed
Integrating by parts,
żπ żπ ˇθ= π żπ

ct
2 2 2
n n´1 n´1 ˇ 2
cos θ dθ = cos θ d(sin θ) = cos θ sin θˇ + (n ´ 1) cosn´2 θ sin2 θ dθ
0 0 θ=0 0

te
żπ
2
= (n ´ 1) cosn´2 θ(1 ´ cos2 θ) dθ
0
ro
which implies that π π
P
ż ż
2
n n´1 2
cos θ dθ = cosn´2 θ dθ .
0 n 0
ht

As a consequence,
$ ż π
(n ´ 1)(n ´ 3) ¨ ¨ ¨ 2
ig

2
ż π
’
’ cos θ dθ if n is odd ,
2
n
& n(n ´ 2) ¨ ¨ ¨ 3 0
cos θ dθ =
r

ż π
0 ’ (n ´ 1)(n ´ 3) ¨ ¨ ¨ 1 2
if n is even ;
py

dθ
’
%
n(n ´ 2) ¨ ¨ ¨ 2 0

n´2 2ω
Co

thus the recursive formula (8.5.9) implies that ωn = π . Further computations shows
n
that $ n´1
(2π) 2
if n is odd ,
’
& n(n ´ 2) ¨ ¨ ¨ 3 ω1
’
’
ωn = n´2
’ (2π) 2
ω2 if n is even .
’
’
%
n(n ´ 2) ¨ ¨ ¨ 4
ż8
Let Γ be the Gamma function defined by Γ(t) = xt´1 e´x dx for t ą 0. Then Γ(x + 1) =
(1) ? 0
xΓ(x) for all x ą 0, Γ(1) = 1 and Γ = π. By the fact that ω1 = 2 and ω2 = π, we can
2
express ωn as n
π2
ωn = ( n+2 ) .
Γ 2
310 CHAPTER 8. Integration

8.6 Change of Variables Formula

Fubini theorem can be used to find the integral of a (Riemann integrable) function over a
rectangular domain if the iterated integrals can be evaluated. However, like the integral of
a function of one variable, in many cases we need to make use of several change of variables
in order to transform the integral to another integral that can be easily evaluated. In this
section, we establish the change of variables formula for the integral of functions of several
variables.

Theorem 8.65 (Change of Variables Formula). Let U Ď Rn be an open bounded set, and

ed
g : U Ñ Rn be an one-to-one C 1 mapping with C 1 inverse; that is, g ´1 : g(U) Ñ U is
also continuously differentiable. Assume that the Jacobian of g, Jg = det([Dg]), does not

ct
vanish in U, and E ĂĂU has volume. Then g(E) has volume. Moreover, if f : g(E) Ñ R is

te
bounded and integrable, then (f ˝ g)Jg is integrable over E, and
ż ż roż
f (y) dy = (f ˝ g)(x)ˇJg (x)ˇ dx = (f ˝ g)(x)ˇ 1
ˇ ˇ ˇ B(g , ¨ ¨ ¨ , gn ) ˇ
ˇ ˇ
ˇ dx .
g(E) E E B(x1 , ¨ ¨ ¨ , xn )
P
Remark 8.66. The condition that g has to be defined on a larger open set U can be
removed. In other words, E Ď U has volume is enough for the change of variable formula
ht

to hold; however, we will not prove this more generalized version here.
ig

The proof of the change of variables formula is separated into several steps, and we list
r

each step as a lemma.

py

First, we show that the map g in Theorem 8.65 has the property that g ´1 (Z) has measure
zero (or volume zero) if Z itself has measure zero (or volume zero). This establishes that if
Co

A and B are not overlapping; that is, ν(A X B) = 0, then ν(g ´1 (A X B)) = 0.

Lemma 8.67. Let U Ď Rn be an open set, and ϕ : U Ñ Rn be Lipschitz continuous; that

is, there exists L ą 0 such that }ϕ(x) ´ ϕ(y)}Rn ď L}x ´ y}Rn for all x, y P U. Suppose
that Z Ď U is a set of measure zero (or a set of volume zero) and Zs Ď U. Then ϕ(A) has
measure zero (or volume zero).

Proof. We prove the case that Z has measure zero, and the proof for the case that Z has
volume zero is obtained by changing the countable sum/union to finite sum/union.
First we note that if S Ď U is a rectangle on which the ratio of the maximum length and
minimum length of sides is less than 2, then ϕ(S) Ď R for some n-cube R with side of length
§8.6 Change of Variables Formula 311

? ?
L nδ, where δ is the maximum length of sides of S. Therefore, ν(ϕ(S)) ď (2 nL)n ν(S).
Let ε ą 0 be given. Since Z has measure zero, there exists countable rectangles S1 , S2 , ¨ ¨ ¨
8 8 ε
such that Z Ď Sk and . Moreover, as in the proof of Proposition
Ť ř
ν(Sk ) ă ? n
k=1 k=1 (2 nL)
8.14 we can also assume that the ratio of the maximum length and minimum length of sides
8 8
of Sk is less than 2 for all k P N; thus ϕ(Z) Ď Rk and ν(Rk ) ă ε for some rectangles
Ť ř
k=1 k=1
Rk ’s. ˝

Next, we prove that it suffices to show the change of variables formula for the case that
f is a constant and E is the pre-image of closed rectangle under g in order to establish the

ed
full result.

ct
Lemma 8.68. Let U Ď Rn be an open bounded set, and g : U Ñ Rn be an one-to-one C 1

te
mapping that has a C 1 inverse. Assume that the Jacobian of g, Jg = det([Dg]), does not
vanish in U, and
ż
ro
for all closed rectangle R Ď g(U) . (8.6.1)
P
ν(R) = |Jg (x)|dx
g ´1 (R)
ht

Then if E ĂĂU has volume and f : g(E) Ñ R is bounded and integrable, then (f ˝ g)|Jg | is
Riemann integrable over E, and
ig

ż ż
f (y) dy = (f ˝ g)(x)|Jg (x)|dx .
r
py

g(E) E

Proof. Consider the extensions of f and (f ˝ g)|Jg | given by

Co

f (x) if x P g(E) , (f ˝ g)(x)|Jg |(x) if x P E ,

" "
g(E) E
f (x) = and (f ˝ g)|Jg | (x) =
0 if x P g(E) ,
A
0 if x P E A .
ˇ g(E)
By the integrability of f over g(E), the set y P Rn ˇ f
␣ (
is discontinuous at y has measure
zero. Since
E
x P Rn ˇ (f ˝ g)|Jg | is discontinuous at x
␣ ˇ (
␣ ˇ (
Ď BE Y x P int(E) ˇ f is discontinuous at g(x)
␣ ˇ (
= BE Y y P g(int(E)) ˇ f is discontinuous at y
ˇ g(E)
Ď BE Y y P Rn ˇ f
␣ (
is discontinuous at y ,
312 CHAPTER 8. Integration

E
we conclude that x P Rn ˇ (f ˝ g)|Jg | is discontinuous at x has measure zero. Therefore,
␣ ˇ (

(f ˝ g)|Jg | is Riemann integrable over E. On the other hand, by the fact that

g(E) E E
(f ˝ g)|Jg | = (f ˝ g) |Jg | = (f ˝ g)|Jg | on U,

g(E)
the Lebesgue theorem also implies that (f ˝ g)|Jg | is Riemann integrable over F if E Ď
F Ď U since
F
E E
(f ˝ g)|Jg | = (f ˝ g)|Jg | @F Ě E .

ed
Moreover, it follows from Lemma 8.49 that

ct
ż ż
g(E)
(f ˝ g)(x)|Jg (x)|dx = (f ˝ g)(x)|Jg (x)|dx @E Ď F Ď U . (8.6.2)

te
F E

ro
Since the Jacobian of g does not vanish in U, Remark 7.2 implies that g is an open
mapping; thus g(U) is open. By the fact that g(E) s is compact, there exists an open set V
P
in Rn such that g(E)s Ď V ĂĂg(U). It then follows from g ´1 P C 1 (g(U)) and V s Ď U that
there exists L ą 0 such that ›g (y1 ) ´ g ´1 (y2 )›Rn ď L}y1 ´ y2 }Rn for all y1 , y2 P V. In other
› ´1 ›
ht

words, g ´1 is Lipschitz on V, and Lemma 8.67 implies that g ´1 (Z) has volume zero if Z Ď V
ig

has volume zero.

Note that there exists δ ą 0 such that d(x, y) ą δ for all x P g(E) and y P V A . Let P be
r
py

a partition of g(E) such that }P} ă δ; that is, diam(∆) ă δ for all ∆ P P. Then ∆ Ď V if
g(E) g(E)
∆ P P and ∆ X g(E) ‰ H. Since inf f (y) = inf
´1
(f ˝ g)(x) if ∆ Ď U, using (8.6.1)
yP∆ xPg (∆)
Co

we find that

ÿ g(E)
ÿ g(E)
L(f, P) = inf f (y)ν(∆) = inf
´1
(f ˝ g)(x)ν(∆)
yP∆ xPg (∆)
∆PP ∆PP
∆Xg(E)‰H ∆Xg(E)‰H
ÿ ż
g(E)
= inf (f ˝ g)(x) |Jg (x)|dx
∆PP
xPg ´1 (∆) g ´1 (∆)
∆Xg(E)‰H
ÿ ż
g(E)
ď (f ˝ g)(x)|Jg (x)|dx .
∆PP g ´1 (∆)
∆Xg(E)‰H

Since each “face” of the rectangle ∆ P P has volume zero, Lemma 8.67 implies that g ´1 (∆)X
§8.6 Change of Variables Formula 313

g ´1 (∆1 ) has volume zero if ∆ X ∆1 has volume zero. Therefore, Corollary 8.42 shows that
ż
g(E)
L(f, P) ď Ť (f ˝ g)(x)|Jg (x)|dx
∆PP,∆Xg(E)‰H g ´1 (∆)
ż
g(E)
= (f ˝ g)(x)|Jg (x)|dx
g ´1 ( ∆PP,∆Xg(E)‰H ∆)
Ť
ż
g(E)
= (f ˝ g)(x)|Jg (x)|dx
g ´1 ( ∆PP,∆Xg(E)‰H ∆)
Ť
ż
= (f ˝ g)(x)|Jg (x)| dx ,
E
g(E)

ed
where we have used the integrability of (f ˝ g)|Jg | over the set g ´1 (
Ť
∆PP,∆Xg(E)‰H ∆)
(since this set is a super set of E) and (8.6.2) to conclude the last two equalities.

ct
g(E) g(E)
Similarly, by the fact that sup f (y) = sup (f ˝ g)(x) if ∆ Ď U, we obtain that
yP∆ xPg ´1 (∆)

te
ÿ ż
g(E)
U (f, P) = sup (f ˝ g)(x) |Jg (x)|dx
ro
xPg ´1 (∆) g ´1 (∆)
∆PP
∆Xg(E)‰H
ż ż
P
ÿ g(E) g(E)
ě (f ˝ g)(x)|Jg (x)|dx = (f ˝ g)(x)|Jg (x)|dx
g ´1 (∆) g ´1 ( ∆PP,∆Xg(E)‰H ∆)
Ť
∆PP
∆Xg(E)‰H
ht

ż
= (f ˝ g)(x)|Jg (x)| dx .
E
ig

ż ż
The integrability of f over g(E) then implies that f (y) dy = (f ˝ g)(x)|Jg (x)|dx. ˝
r

g(E) E
py

Since the differentiability of g implies that locally g is very closed to an affine map; that
is, g(x) « Lx+c for some L P B(Rn , Rn ) and c P Rn (in fact, g(x) « g(x0 )+(Dg)(x0 )(x´x0 )
Co

in a neighborhood of x0 ), our next step is to establish (8.6.1) first for the case that g is an
affine map. Since the volume of a set remains unchanged under translation, W.L.O.G. we
can assume that g is linear.
Lemma 8.69. Let g P B(Rn , Rn ), and A Ď Rn be a set that has volume. Then g(A) has
volume, and
( )
ż ż
ν g(A) = 1dy = |Jg (x)|dx . (8.6.3)
g(A) A

Remark 8.70. If g P B(Rn , Rn ), then g(x) = Lx for some n ˆ n matrix. In this case
Jg (x) = det(L) for all x P A; thus (8.6.3) is the same as that
( )
ż ż
ν L(A) = 1dy = | det(L)|dx = | det(L)|ν(A) . (8.6.4)
L(A) A
314 CHAPTER 8. Integration

Therefore, in the following we prove (8.6.4) instead of (8.6.3).

Proof of Lemma 8.69. Since any n ˆ n matrices can be expressed as the product of ele-
mentary matrices, it suffices to prove the validity of the lemma for the case that L is an
elementary matrix.
Suppose first that A = [a1 , b1 ] ˆ ¨ ¨ ¨ ˆ [an , bn ] is a rectangle.
1. If L is an elementary matrix of the type
 
1 0 ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ 0
 . .. 
 0 .. 0 . 
 . . .. 
 .. . . 1 . . . . 
 

ed
 .. . 
 . . 
.
 0 0 0 1  Ð the i0 -th row
 .. .. .. .. 

ct
 . . 1 . . 
 . .. 
L=  .. 0
. .. 0 .  

te
 . . . . 
 .. .. 1 .. .. 
 
 .. .. 
 . 1 0 0 0
ro .  Ð the j0 -th row
 
 ... ... . .
 0 . 
P
1
 .. 
 0 0 . 0 
ht

0 ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ 0 1

Ò Ò
the i0 -th column the j0 -th column
ig

then
r

L(A) = [a1 , b1 ] ˆ ¨ ¨ ¨ ˆ [ai0 ´1 , bi0 ´1 ] ˆ [aj0 , bj0 ] ˆ [ai0 +1 , bi0 +1 ] ˆ ¨ ¨ ¨ ˆ

py

ˆ[aj0 ´1 , bj0 ´1 ] ˆ [ai0 , bi0 ] ˆ [aj0 +1 , bj0 + 1] ˆ ¨ ¨ ¨ ˆ [an , bn ] ;

( )
thus ν L(A) = ν(A) = | det(L)|ν(A).
Co

2. If L is an elementary matrix of the type

 
1 0 ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ 0
 .. 
 0 1 0 . 
 . 
..
 .. . . . . . . ..
. 
 
.
 . .. 
 .. 0 1 0 . 
 
 .. .. 
L= . 0 c 0 .  Ð the k0 -th row
 
 .. ..
 . 0 1 0 .
 . ... ... . . . .. 
 .. .
 
 .. 
 . 0 1 0 
0 ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ 0 1
§8.6 Change of Variables Formula 315

then
L(A) = [a1 , b1 ] ˆ ¨ ¨ ¨ ˆ [ak0 ´1 , bk0 ´1 ] ˆ [cak0 , cbk0 ] ˆ [ak0 +1 , bk0 +1 ] ˆ ¨ ¨ ¨ ˆ [an , bn ]
if c ě 0 or
L(A) = [a1 , b1 ] ˆ ¨ ¨ ¨ ˆ [ak0 ´1 , bk0 ´1 ] ˆ [cbk0 , cak0 ] ˆ [ak0 +1 , bk0 +1 ] ˆ ¨ ¨ ¨ ˆ [an , bn ]
( )
if c ă 0. In either case, ν L(A) = |c|ν(A) = | det(L)|ν(A).
3. If L is an elementary matrix of the type
 
1 0 ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ 0
 0 1 0 0 
 

ed
 .. . . . . ... 
 . . . c 0  Ð the i0 -th row
 . . ... ... 
 .. .. 0 

ct
 
 . .. 
L=  .
. 0 1 0 . 

te
 .. .. .. .. .. 
 . . . . . 
 . 
 .. .. .. . . .. 

ro . . . . 
 . 
 .. 0 1 0 
P
0 ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ ¨¨¨ 0 1
Ò
ht

the j0 -th column

then L(A) is a parallelepiped
ig

L(A) = (x1 , ¨ ¨ ¨ , xi0 ´1 , xi0 + cxj0 , xi0 +1 , ¨ ¨ ¨ , xn ) P Rn ˇ xi P [ai , bi ] @ 1 ď i ď n

␣ ˇ (
r
py

= (x1 , ¨ ¨ ¨ , xi0 ´1 , yi0 , xi0 +1 , ¨ ¨ ¨ , xn ) P Rn ˇ ai0 + cxj0 ď yi0 ď bi0 + cxj0 ,

␣ ˇ
(
xi P [ai , bi ] @ i ‰ i0 ;
Co

xi0 axis xi0 axis

xj0 axis xj0 axis

(x1 , ¨ ¨ ¨ , xi0 ´1 , xi0 +1 , ¨ ¨ ¨ , xn ) hyper-space (x1 , ¨ ¨ ¨ , xi0 ´1 , xi0 +1 , ¨ ¨ ¨ , xn ) hyper-space

Figure 8.3: The image of a rectangle under a linear map induced by the elementary matrix
of the third type
316 CHAPTER 8. Integration

thus the Fubini theorem (or Corollary 8.62) implies that

ż ( ż bi0 +cxj0 )
ν(L(A)) = 1dyi0 dp
xi0 = ν(A) .
[a1 ,b1 ]ˆ¨¨¨ˆ[ai0 ´1 ,bi0 ´1 ]ˆ[ai0 +1 ,bi0 +1 ]ˆ¨¨¨ˆ[an ,bn ] ai0 +cxj0
) (
On the other hand, | det(L)| = 1, so ν L(A) = | det(L)|ν(A) is validated.

Therefore, (8.6.4) holds if A is a rectangle and L is an elementary matrix. An immediate

consequence of this observation is that if Z is a set of measure zero, so is L(Z).
Now suppose that A is an arbitrary set with volume, and L is an elementary matrix.

1. If det(L) = 0, L must be an elementary matrix of the second type (with c = 0), and

ed
in this case,

ct
L(A) Ď [´r, r] ˆ ¨ ¨ ¨ ˆ [´r, r] ˆ lo
[´ε, n ˆ ¨ ¨ ¨ ˆ [´r, r]
omooε]
the k0 -th slot

te
for some r ą 0 sufficiently large and arbitrary ε ą 0. Therefore, L(A) has volume
( )
zero; thus L(A) has volume and ν L(A) = | det(L)|ν(A).
P ro
2. Suppose that det(L) ‰ 0. Let ε ą 0 be given. Since A has volume, by Riemann’s
condition there exists a partition of A such that
ht

ε
U (1A , P) ´ L(1A , P) ă .
| det(L)|
ig

Note that the inequality above also implies that

r

ε ε
py

U (1A , P) ´ ν(A) ă and ν(A) ´ L(1A , P) ă .

| det(L)| | det(L)|
Co

Let C1 = ∆ P P ˇ ∆ X A ‰ H and C2 = ∆ P P ˇ ∆ Ď A , and define R1 =

␣ ˇ ( ␣ ˇ ( Ť
∆
∆PC 1
and R2 = ∆. Then R2 Ď A Ď R1 . Moreover,
Ť
∆PC2
( ) ÿ ÿ
ν L(R1 ) = ν(L(∆)) = | det(L)|ν(∆) = | det(L)|U (1A , P)
∆PC1 ∆PC1

ă | det(L)|ν(A) + ε

and
( ) ÿ ÿ
ν L(R2 ) = ν(L(∆)) = | det(L)|ν(∆) = | det(L)|L(1A , P)
∆PC2 ∆PC2

ą | det(L)|ν(A) ´ ε .
§8.6 Change of Variables Formula 317

As a consequence, by the fact that L(R2 ) Ď L(A) Ď L(R1 ) we conclude that

ˇż ( ) ( ) ( )
ż ˇ
1dx ´ 1dxˇ ď ν L(R1 ) ´ ν L(R2 ) = | det(L)| U (1A , P) ´ L(1A , P) ă ε .
ˇ ˇ
ˇ
L(A) L(A)
ż ż
Since ε ą 0 is arbitrary, we find that 1dx = 1dx which implies that 1L(A) is
L(A) L(A)
Riemann integrable, or equivalently, L(A) has volume.
On the other hand, observing that
( ) ( ) ( )
| det(L)|ν(A) ´ ε ă ν L(R2 ) ď ν L(A) ď ν L(R1 ) ă | det(L)|ν(A) + ε ,

ed
( )
we conclude that ν L(A) = | det(L)|ν(A) again because ε ą 0 is arbitrary. ˝

ct
Lemma 8.71. Let U Ď Rn be an open bounded set, and g : U Ñ Rn be an one-to-one C 1

te
mapping that has a C 1 inverse. Assume that the Jacobian of g, Jg = det([Dg]), does not
vanish in U. Then
ż
P ro
ν(R) = |Jg (x)|dx for all closed rectangle R Ď g(U) . (8.6.1)
g ´1 (R)
ht

Proof. First, we note that by the the compactness of R, there exist m ą 0 and Λ ą 0 such
that
ig

› ›
|Jg (x)| ě m and ›(Dg)(x)›B(Rn .Rn ) ď Λ @ x P g ´1 (R) .
r

Let 0 ă ε ă 1 be given. By the compactness of g ´1 (R), (Theorem 4.52 implies that)

py

Jg : g ´1 (R) Ñ R is uniformly continuous; thus there exists δ1 ą 0 such that

Co

ˇ ˇ
ˇJg (x1 ) ´ Jg (x2 )ˇ ă mε if }x1 ´ x2 }Rn ă δ1 and x1 , x2 P g ´1 (R) .

Since g ´1 is of class C 1 , the continuity of g ´1 and Corollary 6.36 guarantee that there exists
δ ą 0 such that if }y1 ´ y2 }Rn ă δ and y1 , y2 P R, we have
› ´1 ›
›g (y1 ) ´ g ´1 (y2 )› n ă δ1
R

and
› ´1 ›
›g (y2 ) ´ g ´1 (y1 ) ´ (Dg ´1 )(y1 )(y2 ´ y1 )› ε
Rn
ď ? }y1 ´ y2 }Rn .
2 nΛ
Let P be a partition of R with mesh size }P} ă δ and the ratio of the maximum length
and minimum length of sides of each ∆ is less than 2. For ∆ P P, let c∆ denote the center
318 CHAPTER 8. Integration

( )
of ∆ for ∆ P P, and define A∆ = (Dg)(g ´1 (c∆ )) as well as h∆ (x) = A∆ x ´ g ´1 (c∆ ) + c∆ .
Then
ˇ ˇ
ˇJg (g ´1 (y)) ´ Jg (g ´1 (c∆ ))ˇ ă mε @y P ∆.

Moreover, the inverse function theorem (Theorem 7.1) implies that A´1
∆ = (Dg )(c∆ ); thus
´1

for y P ∆,
› › › ( )›
›(h ˝ g ´1 )(y) ´ y › n = ›A∆ g ´1 (y) ´ g ´1 (c∆ ) ´ (Dg ´1 )(c∆ )(y ´ c∆ ) › n
R R
› ´1 ´1 ´1
›
ď }A∆ }B(Rn ,Rn ) g (y) ´ g (c∆ ) ´ (Dg )(c∆ )(y ´ c∆ )›
›
Rn
´1
ε}(Dg)(g (c∆ ))}B(Rn ,Rn ) ε
}y ´ c∆ }Rn ď ? diam(∆) .

ed
ď ?
2 nΛ 4 n

The inequality above implies that for all ∆ P P,

ct
( )

te
(1 ´ ε)n ν(∆) ď ν (h∆ ˝ g ´1 )(∆) ď (1 + ε)n ν(∆) .
ro
Since Jh∆ = det(A∆ ) = Jg (g ´1 (c∆ )), Lemma 8.69 or (8.6.4) provides that
P
( ) ( ) ( )
ż ż
ˇ ˇ
ˇJg (x)ˇ dx ď |Jg (g ´1 (c∆ ))| + mε dx = |Jg (g ´1 (c∆ ))| + mε ν g ´1 (∆)
g ´1 (∆) g ´1 (∆)
( )
ht

( ) ν (h ∆ ˝ g) ´1
(∆)
= |Jg (g ´1 (c∆ ))| + mε ď (1 + ε)n+1 ν(∆) .
|Jg (g ´1 (c∆ ))|
ig

A similar argument provides a lower bounded of the left-hand side, and we conclude that
r
py

ż
n+1
(1 ´ ε) ν(∆) ď |Jg (x)|dx ď (1 + ε)n+1 ν(∆) @∆ P P .
g ´1 (∆)
Co

Summing over all ∆ P P, we find that

ÿż
n+1
(1 ´ ε) ν(R) ď |Jg (x)|dx ď (1 + ε)n+1 ν(R) .
∆PP g ´1 (∆)

ř ż ż
Identity (8.6.1) is then concluded since |Jg (x)|dx = |Jg (x)|dx and ε P (0, 1)
∆PP g ´1 (∆) g ´1 (R)
is arbitrary. ˝

Example 8.72. Let A be the triangular region with vertices (0, 0), (4, 0), (4, 2), and f :
A Ñ R be given by
a
f (x, y) = y x ´ 2y .
§8.6 Change of Variables Formula 319

( u ´ v)
Let (u, v) = (x, x ´ 2y). Then (x, y) = g(u, v) = u, ; thus
2
ˇ ˇ
ˇ1 0 ˇ 1
Jg (u, v) = ˇ 1 1ˇ = ´ .
ˇ ˇ
ˇ ´ ˇ 2
2 2

Define E as the triangle with vertices (0, 0), (4, 0), (4, 4). Then A = g(E).

v y

ed
E

ct
A
u x

te
Figure 8.4: The image of E under g
ro
Therefore,
P
( )
ż ż ż
1
f (x, y)d(x, y) = f (x, y)d(x, y) = f g(u, v) d(u, v)
ht

A g(E) 2 E
1 4 [ 2 3 2 5 ]ˇˇv=u
ż4żu
?
ż
1
= (u ´ v) vdvdu = uv 2 ´ v 2 ˇ du
ig

4 0 0 4 0 3 5 v=0
( )
ż4
1 2 2 5 1 2 7ˇ u=4 256
r

ˇ
= ´ u 2 du = ˆ u2 ˇ = .
py

4 0 3 5 15 7 u=0 105
ż1
Co

Example 8.73. Suppose that f : [0, 1] Ñ R is Riemann integrable and (1´x)f (x) dx = 5
0
(note that the function g(x) = (1 ´ x)f (x) is Riemann integrable over [0, 1] because of the
ż1żx
Lebesgue theorem). We would like to evaluate the iterated integral f (x ´ y) dydx.
0 0
It is nature to consider the change of variables (u, v) = (x ´ y, x) or (u, v) = (x ´ y, y).
Suppose the later case. Then (x, y) = g(u, v) = (u + v, v); thus
ˇ ˇ
ˇ1 1 ˇ
Jg (u, v) = ˇˇ ˇ = ´1 .
1 0ˇ

Moreover, the region of integration is the triangle A with vertices (0, 0), (1, 0), (1, 1), and
three sides y = 0, x = 1, x = y correspond to u = 0, u + v = 1 and v = 0. Therefore, if
320 CHAPTER 8. Integration

E denotes the triangle enclosed by u = 0, v = 0 and u + v = 1 on the (u, v)-plane, then

g(E) = A, and
ż1żx ż ż
f (x ´ y) dydx = f (x ´ y)d(x, y) = f (x ´ y)d(x, y)
0 0 A g(E)
( )
ż ż 1 ż 1´u
= f g1 (u, v) ´ g2 (u, v) |Jg (u, v)|d(u, v) = f (u) dvdu
żE1 0 0

= (1 ´ u)f (u) du = 5 .
0

Example 8.74 (Polar coordinates). In R2 , when the domain over which the integral is taken

ed
is a disk D, a particular type of change of variables is sometimes very useful for the purpose
of evaluating the integral. Let (x, y) = (x0 + r cos θ, y0 + r sin θ) ” ψ(r, θ), where (x0 , y0 ) is

ct
the center of D under consideration. If the radius of D is R, then D, up to removing a line
segment with length R, is the image of (0, R) ˆ (0, 2π) under ψ. Note that the Jacobian of

te
ψ is ˇ ˇ ro
ˇ Bψ1 Bψ1 ˇ ˇ ˇ
ˇ ˇ ˇcos θ ´r sin θˇ
Jψ (r, θ) = ˇ Br Bθ
ˇ=ˇ ˇ = r.
ˇ ˇ ˇ ˇ
P
ˇ Bψ2 Bψ2 ˇ ˇ sin θ r cos θ ˇ
ˇ ˇ
Br Bθ
Therefore, if f : D Ñ R is Riemann integrable, then
ht

ż ż ż
ˇ ˇ
f (x, y) d(x, y) = f (x, y) d(x, y) = (f ˝ ψ)(r, θ)ˇJψ (r, θ)ˇ d(r, θ)
ig

D ψ((0,R)ˆ(0,2π)) (0,R)ˆ(0,2π)
ż
r

= f (x0 + r cos θ, y0 + r sin θ) r d(r, θ) .

py

(0,R)ˆ(0,2π)

Example 8.75 (Cylindrical coordinates). In R3 , when the domain over which the integral
Co

is taken is a cylinder C; that is, C = D ˆ [a, b] for some disk D and ´8 ă a ă b ă R, then
the change of variables

ψ(r, θ, z) = (x0 + r cos θ, y0 + r sin θ, z) 0 ă r ă R , 0 ă θ ă 2π , a ď z ď b ,

where (x0 , y0 ) is the center of D and R is the radisu of D, is sometimes very useful for
evaluating the integral. Since the Jacobian of ψ is
ˇ ˇ
ˇ Bψ1 Bψ1 Bψ1 ˇ
Bz ˇ ˇcos θ ´r sin θ 0ˇ
ˇ ˇ ˇ ˇ
ˇ Br Bθ
ˇ ˇ ˇ ˇ
ˇ Bψ Bψ2 Bψ2 ˇ ˇ
Jψ (r, θ, z) = ˇ 2 ˇ = ˇ sin θ r cos θ 0ˇ = r ,
ˇ
ˇ Br Bθ Bz ˇ ˇ ˇ
ˇ ˇ ˇ 0 0 1 ˇ
ˇ Bψ3 Bψ3 Bψ3 ˇ
ˇ ˇ
Br Bθ Bz
§8.7 Exercises 321

we must have
ż ż
f (x, y, z) d(x, y, z) = f (x, y, z) d(x, y, z)
C ψ((0,R)ˆ(0,2π)ˆ[a,b])
ż
ˇ ˇ
= (f ˝ ψ)(r, θ, z)ˇJψ (r, θ, z)ˇ d(r, θ, z)
(0,R)ˆ(0,2π)ˆ[a,b]
ż
= f (x0 + r cos θ, y0 + r sin θ, z) r d(r, θ, z) .
(0,R)ˆ(0,2π)ˆ[a,b]

Example 8.76 (Spherical coordinates). In R3 , when the domain over which the integral is
taken is a ball B, the change of variables

ed
ψ(ρ, θ, ϕ) = (x0 + ρ cos θ sin ϕ, y0 + ρ sin θ sin ϕ, z0 + ρ cos ϕ) 0 ă ρ ă R, 0 ă θ ă 2π, 0 ă ϕ ă π,

ct
te
where (x0 , y0 , z0 ) is the center of B and R is the radius of B, is often used to evaluate the
integral a function over B. Since the Jacobian of ψ is ro
ˇ ˇ
ˇ Bψ1 Bψ1 Bψ1 ˇ
P
ˇ ˇ
ˇ Bρ Bθ Bϕ ˇ ˇˇcos θ sin ϕ ´ρ sin θ sin ϕ ρ cos θ cos ϕˇˇ
ˇ ˇ
ˇ Bψ2 Bψ2 Bψ2 ˇ ˇˇ ˇ
Jψ (ρ, θ, ϕ) = ˇˇ ˇ = ˇ sin θ sin ϕ ρ cos θ sin ϕ ρ sin θ cos ϕ ˇˇ
ht

ˇ Bρ Bθ Bϕ ˇˇ ˇ ˇ
ˇ Bψ3 Bψ3 Bψ3 ˇ ˇ cos ϕ 0 ´ρ sin ϕ ˇ
ˇ ˇ
ig

ˇ Bρ Bθ Bϕ ˇ
= ´ρ2 cos2 θ sin3 ϕ ´ ρ2 sin2 θ sin ϕ cos2 ϕ ´ ρ2 cos2 θ sin ϕ cos2 ϕ ´ ρ2 sin2 θ sin3 ϕ
r
py

= ´ρ2 sin3 ϕ ´ ρ2 sin ϕ cos2 ϕ = ´ρ2 sin ϕ ,

Co

if the radius of B is R, we must have

ż ż
f (x, y, z) d(x, y, z) = f (x, y, z) d(x, y, z)
B ψ((0,R)ˆ(0,2π)ˆ(0,π))
ż
ˇ ˇ
= (f ˝ ψ)(ρ, θ, ϕ)ˇJψ (ρ, θ, ϕ)ˇ d(ρ, θ, ϕ)
(0,R)ˆ(0,2π)ˆ(0,π)
ż
= f (x0 + ρ cos θ sin ϕ, y0 + ρ sin θ sin ϕ, z0 + ρ cos ϕ) ρ2 sin ϕ d(r, θ, z) .
(0,R)ˆ(0,2π)ˆ(0,π)

8.7 Exercises
§8.2 Conditions for Integrability
322 CHAPTER 8. Integration

Problem 8.1. Let f : [0, 1] ˆ [0, 1] Ñ R be a bounded function such that f (x, y) ď
f (x, z) if y ă z and f (x, y) ď f (t, z) if x ă t. In other words, f (x, ¨) and f (¨, y) are both
non-decreasing functions for fixed x, y P [0, 1]. Show that f is Riemann integrable over
[0, 1] ˆ [0, 1].

Problem 8.2. Let A Ď Rn be a bounded set, and fk : A Ñ R be a sequence of Riemann

integrable functions which converges uniformly to f on A. Show that f is Riemann integrable
over A, and ż ż ż
lim fk (x) dx = lim fk (x) dx = f (x) dx .
kÑ8 A A kÑ8 A

ed
§8.3 Lebesgue’s Theorem

ct
Problem 8.3. Complete the following.

te
1. Show that if A is a set of volume zero, then A has measure zero. Is it true that if A
ro
has measure zero, then A also has volume zero?
P
2. Let a, b P R and a ă b. Show that the interval [a, b] does not have measure zero (in
R).
ht

3. Let A Ď [a, b] be a set of measure zero (in R). Show that [a, b]zA does not have
ig

measure zero (in R).

r
py

4. Show that the Cantor set (defined in Exercise Problem 2.11) has volume zero.
8 (1 1 ) 8 (1 1 1 1)
Problem 8.4. Let A = ´ k , + k be a subset of R. Does A have
Ť Ť
D , k =
Co

k=1 k 2 k=1 k 2 k 2
volume?

Problem 8.5. Let f : [a, b] Ñ R be bounded and Riemann integrable. Show that the graph
of f has volume zero by considering the difference of the upper and lower sums of f .

Problem 8.6. Let A Ď żRn be an open bounded set with volume, and f : A Ñ R be
continuous. Show that if f (x) dx = 0 for all subsets B Ď A with volume, then f = 0.
B

Problem 8.7. Prove the following statements.

1
1. The function f (x) = sin is Riemann integrable over (0, 1).
x
§8.7 Exercises 323

2. Let f : [0, 1] Ñ R be given by

$
& 1 if x = q P Q, (p, q) = 1 ,
f (x) = p p
% 0 if x is irrational.

ż1
Then f is Riemann integrable over [0, 1]. Find f (x)dx as well.
0

3. Let A Ď Rn be a bounded set, and f : A Ñ R is Riemann integrable. Then f k （f 的

k 次方）is integrable for all k P N.

ed
Problem 8.8. Suppose that f : [a, b] Ñ R is Riemann integrable, and the set
␣
x P
ˇ ( żb
[a, b] f (x) ‰ 0 has measure zero. Show that f (x) dx = 0.

ct
ˇ
a

§8.5 Fubini’s Theorem

te
ż1żx
2
Problem 8.9. Evaluate the iterated integral (2y ´ y 2 ) 3 dydx.
ro 0 0

Problem 8.10. Let A = [a, b] ˆ [c, d] be a rectangle in R2 , and f : A Ñ R be Riemann

P
integrable. Show that the sets
ht

! ˇ żd żs d ) ! ˇ żb żs b )
x P [a, b] ˇ f (x, y)dy ‰ f (x, y)dy and y P [c, d] ˇ f (x, y)dx ‰ f (x, y)dx
ˇ ˇ
ig

c c a a

have measure zero (in R1 ).

r
py

Problem 8.11. Define a set S Ď [0, 1] ˆ [0, 1] by

Co

!(
p k)
ˇ )
S= , P [0, 1] ˆ [0, 1] ˇ m, p, k P N , gcd(m, p) = 1 and 1 ď k ď m ´ 1 .
ˇ
m m
Show that ż 1(ż 1 ) ż 1(ż 1 )
1S (x, y) dy dx = 1S (x, y) dx dy = 0
0 0 0 0

but 1S is not Riemann integrable over [0, 1] ˆ [0, 1].

Problem 8.12. Let f : [0, 1] ˆ [0, 1] Ñ R be given by

22n if (x, y) P [2´n , 2´n+1 ) ˆ [2´n , 2´n+1 ), n P N ,

$
’
&
f (x, y) = ´22n+1 if (x, y) P [2´n , 2´n+1 ) ˆ [2´n´1 , 2´n ), n P N ,
’
otherwise .
%
0
324 CHAPTER 8. Integration

ż1
1. Show that f (x, y) dx = 0 for all y P [0, 1).
0
ż1 [ 1)
2. Show that f (x, y) dy = 0 for all x P 0, .
0 2
ż1ż1 ż1ż1
3. Justify if the iterated (improper) integrals f (x, y)dxdy and f (x, y) dydx
0 0 0 0
are identical.

Problem 8.13.
ż 1 ( ż ex )
1. Draw the region corresponding to the integral (x + y) dy dx and evaluate.

ed
0 1

2. Change the order of integration of the integral in 1 and check if the answer is unaltered.

ct
§8.6 Change of Variables Formula

te
Problem 8.14. Prove Theorem 4.95 using Theorem 8.65. ro
Problem 8.15. Find the volume of the set (x, y, z) P R3 ˇ 0 ď x2 + y 2 + xy ď z 2 ď 4 .
␣ ˇ (
P
Problem 8.16. Suppose that U Ď Rn is an nonempty open set, and f : U Ñ R is of class
C 1 such that Jf (x) ‰ 0 for all x P U. Show that
ht

( )
ν f (D(x0 , r))
lim ( ) = Jf (x0 ) @ x0 P U .
ig

rÑ0+ ν D(x0 , r)

(2 1)
r

Problem 8.17. 1. Let A be the parallelogram with vertices (0, 0), ,´ , (1, 0) and
py

3 3
(1 1)
, . Evaluate the integral
3 3
Co

ż
? a
x ´ y x + 2y dA .
A

2. Let A be the parallelogram bounded by lines x = 3y, x = 1 + 3y, y = ´2x and

y = 1 ´ 2x. Evaluate the integral
ż a
3
2x2 ´ 5xy ´ 3y 2 dA .
A

3. Let A be the trapezoid with vertices (1, 1), (2, 2), (2, 0) and (4, 0). Evaluate the
integral ż
e(y´x)/(y+x) dA .
A
§8.7 Exercises 325

Problem 8.18 (True or False). Determine whether the following statements are true or
false. If it is true, prove it. Otherwise, give a counter-example.

1. Let A Ď Rn be bounded, and f : A Ñ R be Riemann integrable. If P be a partition

of A, and m ď f (x) ď M for all x P A. Then mν(A) ď L(f, P) ď U (f, P) ď M ν(A).

2. Let A Ď Rn be a set of measure zero. If AzA

s is countable, then A has volume zero.

3. Let A Ď Rn be a closed rectangle and f, g : B Ñ R be Riemann integrable functions.

If there exists aż set Z Ď A żsuch that Z has measure zero and g(x) = f (x) for all
x P AzZ, then

ed
f (x) dx = g(x) dx.
A A

4. Let A Ď Rn be a closed rectangle. Suppose that f and g are two bounded real-valued

ct
functions defined on A such that f is continuous and g = f except on a set of measure

te
zero, then f and g are both Riemann integrable over A.
ro
5. Let A, B Ď R be bounded, and f : A Ñ R and g : f (A) Ñ R be Riemann integrable.
Then g ˝ f is Riemann integrable over A.
P
6.
ht
r ig
py
Co
 Co
py
rig

326
ht
P ro
te
ct
ed