Advanced Calculus (II)

Advanced Calculus (II)

W^EN-C^HINGL^IEN

Department of Mathematics National Cheng Kung University

2009

Ch12: Integration on R

12.3: Iterate Integral

We call Z d

c

Z b a

f (x1, . . . ,xn)dxjdxk :=

Z d c

 Z b a

f (x1, . . . ,xn)dxj

 dxk

an iterated integral when the integrals on the right side exist.

Let R = [a, b] × [c, d ] be a two-dimensional rectangle and suppose that f : R →R is bounded. If f (x , ·) is integrable on [c, d ], then

(22) (L) Z Z

R

f dA ≤ (L) Z b

a

 Z d c

f (x , y )dy

 dx

≤ (U) Z b

a

 Z d c

f (x , y )dy



dx ≤ (U) Z Z

R

f dA.

Proof.

Let Rij = [xi−1,xi] × [yj−1,yj], where {x0, . . . ,xk} is a partition of [a, b] and {y0, . . . ,y`} is a partition of [c, d].

Then G = {Rij :i = 1, 2, . . . , k , j = 1, 2, . . . , `} is a grid on R.

Let R = [a, b] × [c, d ] be a two-dimensional rectangle and suppose that f : R →R is bounded. If f (x , ·) is integrable on [c, d ], then

(22) (L) Z Z

R

f dA ≤ (L) Z b

a

 Z d c

f (x , y )dy

 dx

≤ (U) Z b

a

 Z d c

f (x , y )dy



dx ≤ (U) Z Z

R

f dA.

Proof.

Let Rij = [xi−1,xi] × [yj−1,yj], where {x0, . . . ,xk} is a partition of [a, b] and {y0, . . . ,y`} is a partition of [c, d].

Then G = {Rij :i = 1, 2, . . . , k , j = 1, 2, . . . , `} is a grid on R.

Let R = [a, b] × [c, d ] be a two-dimensional rectangle and suppose that f : R →R is bounded. If f (x , ·) is integrable on [c, d ], then

(22) (L) Z Z

R

f dA ≤ (L) Z b

a

 Z d c

f (x , y )dy

 dx

≤ (U) Z b

a

 Z d c

f (x , y )dy



dx ≤ (U) Z Z

R

f dA.

Proof.

Let Rij = [xi−1,xi] × [yj−1,yj], where {x0, . . . ,xk} is a partition of [a, b] and {y0, . . . ,y`} is a partition of [c, d].

Then G = {Rij :i = 1, 2, . . . , k , j = 1, 2, . . . , `} is a grid on R.

Let ε > 0,choose G so that

(23) U(f , G)−ε < (U) Z Z

R

f dA,

and set

(24) Mij = sup

(x ,y )∈Rij

f (x , y ).

Since (U)Rb

a φ(x )dx =Pk

i=1(U)Rxi

xi−1φ(x )dx and

(U) Z b

a

(φ(x ) + ψ(x ))dx ≤ (U) Z b

a

φ(x )dx + (U) Z b

a

ψ(x )dx

Let ε > 0, choose G so that

(23) U(f , G)−ε < (U) Z Z

R

f dA,

and set

(24) Mij = sup

(x ,y )∈Rij

f (x , y ).

Since (U)Rb

a φ(x )dx =Pk

i=1(U)Rxi

xi−1φ(x )dx and (U)

Z b a

(φ(x ) + ψ(x ))dx ≤ (U) Z b

a

φ(x )dx + (U) Z b

a

ψ(x )dx

Let ε > 0, choose G so that

(23) U(f , G)−ε < (U) Z Z

R

f dA,

and set

(24) Mij = sup

(x ,y )∈Rij

f (x , y ).

Since (U)Rb

a φ(x )dx =Pk

i=1(U)Rxi

xi−1φ(x )dx and (U)

Z b a

(φ(x ) + ψ(x ))dx ≤ (U) Z b

a

φ(x )dx + (U) Z b

a

ψ(x )dx

Let ε > 0, choose G so that

(23) U(f , G)−ε < (U) Z Z

R

f dA,

and set

(24) Mij = sup

(x ,y )∈Rij

f (x , y ).

Since (U)Rb

a φ(x )dx =Pk

i=1(U)Rxi

xi−1φ(x )dx and (U)

Z b a

(φ(x ) + ψ(x ))dx ≤ (U) Z b

a

φ(x )dx + (U) Z b

a

ψ(x )dx

for any bounded function φ and ψ defined on [a, b] (see Exercise 7, p.116),we can write

(U) Z b

a

Z d c

f (x , y )dy

! dx

=

k

X

i=1

(U) Z x_i

x_i−1

`

X

j=1

Z y_j y_j−1

f (x , y )dy

! dx

≤

k

X

i=1

`

X

j=1

(U) Z xi

xi−1

Z yj

yj−1

f (x , y )dy

! dx

≤

k

X

i=1

`

X

j=1

Mij(xi− xi−1)(yj− yj−1) =U(f , G).

for any bounded function φ and ψ defined on [a, b] (see Exercise 7, p.116), we can write

(U) Z b

a

Z d c

f (x , y )dy

! dx

=

k

X

i=1

(U) Z x_i

x_i−1

`

X

j=1

Z y_j y_j−1

f (x , y )dy

! dx

≤

k

X

i=1

`

X

j=1

(U) Z xi

xi−1

Z yj

yj−1

f (x , y )dy

! dx

≤

k

X

i=1

`

X

j=1

Mij(xi− xi−1)(yj− yj−1) =U(f , G).

for any bounded function φ and ψ defined on [a, b] (see Exercise 7, p.116), we can write

(U) Z b

a

Z d c

f (x , y )dy

! dx

=

k

X

i=1

(U) Z x_i

x_i−1

`

X

j=1

Z y_j y_j−1

f (x , y )dy

! dx

≤

k

X

i=1

`

X

j=1

(U) Z xi

xi−1

Z yj

yj−1

f (x , y )dy

! dx

≤

k

X

i=1

`

X

j=1

Mij(xi− xi−1)(yj− yj−1) =U(f , G).

for any bounded function φ and ψ defined on [a, b] (see Exercise 7, p.116), we can write

(U) Z b

a

Z d c

f (x , y )dy

! dx

=

k

X

i=1

(U) Z x_i

x_i−1

`

X

j=1

Z y_j y_j−1

f (x , y )dy

! dx

≤

k

X

i=1

`

X

j=1

(U) Z xi

xi−1

Z yj

yj−1

f (x , y )dy

! dx

≤

k

X

i=1

`

X

j=1

Mij(xi− xi−1)(yj− yj−1) =U(f , G).

It follows from (23) that

(U) Z b

a

Z d c

f (x , y )dy

!

dx < (U) Z Z

R

f dA + ε.

Taking the limit of this inequality as ε → 0, we obtain

(U) Z b

a

Z d c

f (x , y )dy

!

dx ≤ (U) Z Z

R

f dA.

Similarly,

(L) Z b

a

Z d c

f (x , y )dy

!

dx ≥ (L) Z Z

R

f dA.

It follows from (23) that

(U) Z b

a

Z d c

f (x , y )dy

!

dx < (U) Z Z

R

f dA + ε.

Taking the limit of this inequality as ε → 0, we obtain

(U) Z b

a

Z d c

f (x , y )dy

!

dx ≤ (U) Z Z

R

f dA.

Similarly,

(L) Z b

a

Z d c

f (x , y )dy

!

dx ≥ (L) Z Z

R

f dA.

It follows from (23) that

(U) Z b

a

Z d c

f (x , y )dy

!

dx < (U) Z Z

R

f dA + ε.

Taking the limit of this inequality as ε → 0, we obtain

(U) Z b

a

Z d c

f (x , y )dy

!

dx ≤ (U) Z Z

R

f dA.

Similarly,

(L) Z b

a

Z d c

f (x , y )dy

!

dx ≥ (L) Z Z

R

f dA.

It follows from (23) that

(U) Z b

a

Z d c

f (x , y )dy

!

dx < (U) Z Z

R

f dA + ε.

Taking the limit of this inequality as ε → 0, we obtain

(U) Z b

a

Z d c

f (x , y )dy

!

dx ≤ (U) Z Z

R

f dA.

Similarly,

(L) Z b

a

Z d c

f (x , y )dy

!

dx ≥ (L) Z Z

R

f dA.

Let R = [a, b] × [c, d ] be a two-dimensional rectangle and let f : R →R. Suppose that f (x , ·) is integrable on [c, d ] for each x ∈ [a, b], that f (·, y ) is integrable on [a, b] for each y ∈ [c, d ], and that f is integrable on R (as a function of two variables). Then

Z Z

R

f dA = Z b

a

Z d c

f (x , y )dydx = Z d

c

Z b a

f (x , y )dxdy .

Note: These hypotheses hold if f is continuous on the rectangle [a, b] × [c, d ].

For each x ∈ [a, b],set g(x ) =Rd

c f (x , y )dy . Since f is integrable on R, Lemma 12.30 implies

Z Z

R

f dA= (U) Z b

a

g(x )dx = (L) Z b

a

g(x )dx .

Hence, g is integrable on [a, b] and the first identity in (25) holds. Reversing the roles of x and y , we obtain

Z Z

R

f dA = Z d

c

Z b a

f (x , y )dx dy .

Hence, the second identity in (25) holds.

For each x ∈ [a, b], set g(x ) =Rd

c f (x , y )dy .Since f is integrable on R, Lemma 12.30 implies

Z Z

R

f dA = (U) Z b

a

g(x )dx =(L) Z b

a

g(x )dx .

Hence, g is integrable on [a, b] and the first identity in (25) holds. Reversing the roles of x and y , we obtain

Z Z

R

f dA = Z d

c

Z b a

f (x , y )dx dy .

Hence, the second identity in (25) holds.

For each x ∈ [a, b], set g(x ) =Rd

c f (x , y )dy . Since f is integrable on R, Lemma 12.30 implies

Z Z

R

f dA= (U) Z b

a

g(x )dx = (L) Z b

a

g(x )dx .

Hence, g is integrable on [a, b] and the first identity in (25) holds. Reversing the roles of x and y , we obtain

Z Z

R

f dA = Z d

c

Z b a

f (x , y )dx dy .

Hence, the second identity in (25) holds.

For each x ∈ [a, b], set g(x ) =Rd

c f (x , y )dy . Since f is integrable on R, Lemma 12.30 implies

Z Z

R

f dA = (U) Z b

a

g(x )dx =(L) Z b

a

g(x )dx .

Hence, g is integrable on [a, b] and the first identity in (25) holds.Reversing the roles of x and y , we obtain

Z Z

R

f dA = Z d

c

Z b a

f (x , y )dx dy .

Hence, the second identity in (25) holds.

For each x ∈ [a, b], set g(x ) =Rd

c f (x , y )dy . Since f is integrable on R, Lemma 12.30 implies

Z Z

R

f dA = (U) Z b

a

g(x )dx = (L) Z b

a

g(x )dx .

Hence, g is integrable on [a, b] and the first identity in (25) holds. Reversing the roles of x and y ,we obtain

Z Z

R

f dA = Z d

c

Z b a

f (x , y )dx dy .

Hence, the second identity in (25) holds.

For each x ∈ [a, b], set g(x ) =Rd

c f (x , y )dy . Since f is integrable on R, Lemma 12.30 implies

Z Z

R

f dA = (U) Z b

a

g(x )dx = (L) Z b

a

g(x )dx .

Hence, g is integrable on [a, b] and the first identity in (25) holds.Reversing the roles of x and y , we obtain

Z Z

R

f dA = Z d

c

Z b a

f (x , y )dx dy .

Hence, the second identity in (25) holds.

For each x ∈ [a, b], set g(x ) =Rd

c f (x , y )dy . Since f is integrable on R, Lemma 12.30 implies

Z Z

R

f dA = (U) Z b

a

g(x )dx = (L) Z b

a

g(x )dx .

Hence, g is integrable on [a, b] and the first identity in (25) holds. Reversing the roles of x and y ,we obtain

Z Z

R

f dA = Z d

c

Z b a

f (x , y )dx dy .

Hence, the second identity in (25) holds.

For each x ∈ [a, b], set g(x ) =Rd

c f (x , y )dy . Since f is integrable on R, Lemma 12.30 implies

Z Z

R

f dA = (U) Z b

a

g(x )dx = (L) Z b

a

g(x )dx .

Hence, g is integrable on [a, b] and the first identity in (25) holds. Reversing the roles of x and y , we obtain

Z Z

R

f dA = Z d

c

Z b a

f (x , y )dx dy .

Hence, the second identity in (25) holds.

For each x ∈ [a, b], set g(x ) =Rd

c f (x , y )dy . Since f is integrable on R, Lemma 12.30 implies

Z Z

R

f dA = (U) Z b

a

g(x )dx = (L) Z b

a

g(x )dx .

Hence, g is integrable on [a, b] and the first identity in (25) holds. Reversing the roles of x and y , we obtain

Z Z

R

f dA = Z d

c

Z b a

f (x , y )dx dy .

Hence, the second identity in (25) holds.

Find

Z 1 0

Z 1 0

y³e^xy2dydx .

There exists a function f :R²→ R such that f (x, ·) and f (·, y ) are both integrable on [0, 1], but iterated integrals are not equal.

Proof.

Set

f (x , y ) =







2²ⁿ (x , y ) ∈ [2⁻ⁿ,2⁻ⁿ⁺¹) × [2⁻ⁿ,2⁻ⁿ⁺¹), n ∈N,

−2²ⁿ⁺¹ (x , y ) ∈ [2⁻ⁿ⁻¹,2⁻ⁿ) × [2⁻ⁿ,2⁻ⁿ⁺¹), n ∈N,

0 otherwise.

There exists a function f :R²→ R such that f (x, ·) and f (·, y ) are both integrable on [0, 1], but iterated integrals are not equal.

Proof.

Set

f (x , y ) =







2²ⁿ (x , y ) ∈ [2⁻ⁿ,2⁻ⁿ⁺¹) × [2⁻ⁿ,2⁻ⁿ⁺¹), n ∈N,

−2²ⁿ⁺¹ (x , y ) ∈ [2⁻ⁿ⁻¹,2⁻ⁿ) × [2⁻ⁿ,2⁻ⁿ⁺¹), n ∈N,

0 otherwise.

Notice that for each fixed y0∈ [0, 1), f (x, y₀)takes on only two nonzero values and is integrable on [0, 1) in x .For example, if y0 ∈ [2⁻ⁿ,2⁻ⁿ⁺¹), then f (x , y0) =2²ⁿ for x ∈ [2⁻ⁿ,2⁻ⁿ⁺¹), and f (x , y0) = −2²ⁿ⁺¹for x ∈ [2⁻ⁿ⁻¹, 2⁻ⁿ);hence, f (x , y0)is bounded on [0, 1), and

(26) Z 1

0

f (x , y0)dx =

Z 2⁻ⁿ⁺¹ 2⁻ⁿ

2²ⁿdx − Z 2⁻ⁿ

2⁻ⁿ⁻¹

2²ⁿ⁺¹dx

=2ⁿ− 2ⁿ =0.

the same is true for f (x0,y ) when x0∈ [0,¹₂), but when x0 ∈ [¹₂,1), f (x0,y ) takes on only nonzero value, namely, f (x0,y ) = 4 when y ∈ [¹₂,1), and equals zero otherwise. It follows that

Notice that for each fixed y0∈ [0, 1), f (x, y₀)takes on only two nonzero values and is integrable on [0, 1) in x . For example, if y0 ∈ [2⁻ⁿ,2⁻ⁿ⁺¹),then f (x , y0) =2²ⁿ for x ∈ [2⁻ⁿ,2⁻ⁿ⁺¹), and f (x , y0) = −2²ⁿ⁺¹for x ∈ [2⁻ⁿ⁻¹, 2⁻ⁿ); hence, f (x , y0)is bounded on [0, 1), and

(26) Z 1

0

f (x , y0)dx =

Z 2⁻ⁿ⁺¹ 2⁻ⁿ

2²ⁿdx − Z 2⁻ⁿ

2⁻ⁿ⁻¹

2²ⁿ⁺¹dx

=2ⁿ− 2ⁿ =0.

the same is true for f (x0,y ) when x0∈ [0,¹₂), but when x0 ∈ [¹₂,1), f (x0,y ) takes on only nonzero value, namely, f (x0,y ) = 4 when y ∈ [¹₂,1), and equals zero otherwise. It follows that

Notice that for each fixed y0∈ [0, 1), f (x, y₀)takes on only two nonzero values and is integrable on [0, 1) in x . For example, if y0 ∈ [2⁻ⁿ,2⁻ⁿ⁺¹), then f (x , y0) =2²ⁿ for x ∈ [2⁻ⁿ,2⁻ⁿ⁺¹), and f (x , y0) = −2²ⁿ⁺¹for x ∈ [2⁻ⁿ⁻¹, 2⁻ⁿ);hence, f (x , y0)is bounded on [0, 1), and

(26) Z 1

0

f (x , y0)dx =

Z 2⁻ⁿ⁺¹ 2⁻ⁿ

2²ⁿdx − Z 2⁻ⁿ

2⁻ⁿ⁻¹

2²ⁿ⁺¹dx

=2ⁿ− 2ⁿ =0.

the same is true for f (x0,y ) when x0∈ [0,¹₂), but when x0 ∈ [¹₂,1), f (x0,y ) takes on only nonzero value, namely, f (x0,y ) = 4 when y ∈ [¹₂,1), and equals zero otherwise. It follows that

Notice that for each fixed y0∈ [0, 1), f (x, y₀)takes on only two nonzero values and is integrable on [0, 1) in x . For example, if y0 ∈ [2⁻ⁿ,2⁻ⁿ⁺¹), then f (x , y0) =2²ⁿ for x ∈ [2⁻ⁿ,2⁻ⁿ⁺¹), and f (x , y0) = −2²ⁿ⁺¹for x ∈ [2⁻ⁿ⁻¹, 2⁻ⁿ); hence, f (x , y0)is bounded on [0, 1), and

(26) Z 1

0

f (x , y0)dx =

Z 2⁻ⁿ⁺¹ 2⁻ⁿ

2²ⁿdx − Z 2⁻ⁿ

2⁻ⁿ⁻¹

2²ⁿ⁺¹dx

=2ⁿ− 2ⁿ =0.

the same is true for f (x0,y ) when x0∈ [0,¹₂), but when x0 ∈ [¹₂,1), f (x0,y ) takes on only nonzero value, namely, f (x0,y ) = 4 when y ∈ [¹₂,1), and equals zero otherwise. It follows that

Notice that for each fixed y0∈ [0, 1), f (x, y₀)takes on only two nonzero values and is integrable on [0, 1) in x . For example, if y0 ∈ [2⁻ⁿ,2⁻ⁿ⁺¹), then f (x , y0) =2²ⁿ for x ∈ [2⁻ⁿ,2⁻ⁿ⁺¹), and f (x , y0) = −2²ⁿ⁺¹for x ∈ [2⁻ⁿ⁻¹, 2⁻ⁿ); hence, f (x , y0)is bounded on [0, 1), and

(26) Z 1

0

f (x , y0)dx =

Z 2⁻ⁿ⁺¹ 2⁻ⁿ

2²ⁿdx − Z 2⁻ⁿ

2⁻ⁿ⁻¹

2²ⁿ⁺¹dx

=2ⁿ− 2ⁿ =0.

the same is true for f (x0,y ) when x0∈ [0,¹₂),but when x0 ∈ [¹₂,1), f (x0,y ) takes on only nonzero value, namely, f (x0,y ) = 4 when y ∈ [¹₂,1), and equals zero otherwise. It follows that

Notice that for each fixed y0∈ [0, 1), f (x, y₀)takes on only two nonzero values and is integrable on [0, 1) in x . For example, if y0 ∈ [2⁻ⁿ,2⁻ⁿ⁺¹), then f (x , y0) =2²ⁿ for x ∈ [2⁻ⁿ,2⁻ⁿ⁺¹), and f (x , y0) = −2²ⁿ⁺¹for x ∈ [2⁻ⁿ⁻¹, 2⁻ⁿ); hence, f (x , y0)is bounded on [0, 1), and

(26) Z 1

0

f (x , y0)dx =

Z 2⁻ⁿ⁺¹ 2⁻ⁿ

2²ⁿdx − Z 2⁻ⁿ

2⁻ⁿ⁻¹

2²ⁿ⁺¹dx

=2ⁿ− 2ⁿ =0.

the same is true for f (x0,y ) when x0∈ [0,¹₂), but when x0 ∈ [¹₂,1), f (x0,y ) takes on only nonzero value, namely, f (x0,y ) = 4 when y ∈ [¹₂,1), and equals zero otherwise. It follows that

Notice that for each fixed y0∈ [0, 1), f (x, y₀)takes on only two nonzero values and is integrable on [0, 1) in x . For example, if y0 ∈ [2⁻ⁿ,2⁻ⁿ⁺¹), then f (x , y0) =2²ⁿ for x ∈ [2⁻ⁿ,2⁻ⁿ⁺¹), and f (x , y0) = −2²ⁿ⁺¹for x ∈ [2⁻ⁿ⁻¹, 2⁻ⁿ); hence, f (x , y0)is bounded on [0, 1), and

(26) Z 1

0

f (x , y0)dx =

Z 2⁻ⁿ⁺¹ 2⁻ⁿ

2²ⁿdx − Z 2⁻ⁿ

2⁻ⁿ⁻¹

2²ⁿ⁺¹dx

=2ⁿ− 2ⁿ =0.

the same is true for f (x0,y ) when x0∈ [0,¹₂),but when x0 ∈ [¹₂,1), f (x0,y ) takes on only nonzero value, namely, f (x0,y ) = 4 when y ∈ [¹₂,1), and equals zero otherwise. It follows that

Notice that for each fixed y0∈ [0, 1), f (x, y₀)takes on only two nonzero values and is integrable on [0, 1) in x . For example, if y0 ∈ [2⁻ⁿ,2⁻ⁿ⁺¹), then f (x , y0) =2²ⁿ for x ∈ [2⁻ⁿ,2⁻ⁿ⁺¹), and f (x , y0) = −2²ⁿ⁺¹for x ∈ [2⁻ⁿ⁻¹, 2⁻ⁿ); hence, f (x , y0)is bounded on [0, 1), and

(26) Z 1

0

f (x , y0)dx =

Z 2⁻ⁿ⁺¹ 2⁻ⁿ

2²ⁿdx − Z 2⁻ⁿ

2⁻ⁿ⁻¹

2²ⁿ⁺¹dx

=2ⁿ− 2ⁿ =0.

the same is true for f (x0,y ) when x0∈ [0,¹₂), but when x0 ∈ [¹₂,1), f (x0,y ) takes on only nonzero value, namely, f (x0,y ) = 4 when y ∈ [¹₂,1), and equals zero otherwise. It follows that

Z 1 0

Z 1 0

f (x , y )dy dx = Z 1

1 2

Z 1

1 2

4dy dx = 1.

On the other hand, by (26) we have Z 1

0

Z 1 0

f (x , y )dx dy = 0.

Thus the iterated integrals of f are not equal.(Of course, by Fubini’s Theorem, f itself cannot be Riemann

integrable on [0, 1) × [0, 1). In fact, f is not even bounded.)

Z 1 0

Z 1 0

f (x , y )dy dx = Z 1

1 2

Z 1

1 2

4dy dx = 1.

On the other hand, by (26) we have Z 1

0

Z 1 0

f (x , y )dx dy = 0.

Thus the iterated integrals of f are not equal. (Of course, by Fubini’s Theorem, f itself cannot be Riemann

integrable on [0, 1) × [0, 1). In fact, f is not even bounded.)

Z 1 0

Z 1 0

f (x , y )dy dx = Z 1

1 2

Z 1

1 2

4dy dx = 1.

On the other hand, by (26) we have Z 1

0

Z 1 0

f (x , y )dx dy = 0.

Thus the iterated integrals of f are not equal.(Of course, by Fubini’s Theorem, f itself cannot be Riemann

integrable on [0, 1) × [0, 1). In fact, f is not even bounded.)

Z 1 0

Z 1 0

f (x , y )dy dx = Z 1

1 2

Z 1

1 2

4dy dx = 1.

On the other hand, by (26) we have Z 1

0

Z 1 0

f (x , y )dx dy = 0.

Thus the iterated integrals of f are not equal. (Of course, by Fubini’s Theorem, f itself cannot be Riemann

integrable on [0, 1) × [0, 1). In fact, f is not even bounded.)

Let R = [a1,b1] × . . . × [an,bn]be an n-dimensional

rectangle and let f : R →R be integrable on R. If, for each x := (x1, . . . ,xn−1) ∈Rn := [a1,b1] × . . . × [an−1,bn−1], the function f (x.·) is integrable on [an,bn], then

Z bn

an

f (x, t)dt

is integrable on Rn, and

(27)

Z

R

f (x, t)d (x, t) = Z

Rn

Z bn

an

f (x, t)dt d x.

By repeating the argument of Lemma 12.30,we have

(L) Z

R

f (x, t)d (x, t) ≤ (L) Z

Rn

Z bn

an

f (x, t)dt d x

≤ (U) Z

Rn

Z bn

an

f (x, t)dt d x

≤ (U) Z

R

f (x, t)d (x, t)

for any bounded f . Since f is integrable on R, it follows that (27) holds.

By repeating the argument of Lemma 12.30, we have

(L) Z

R

f (x, t)d (x, t)≤ (L) Z

Rn

Z bn

an

f (x, t)dt d x

≤ (U) Z

Rn

Z bn

an

f (x, t)dt d x

≤ (U) Z

R

f (x, t)d (x, t)

for any bounded f . Since f is integrable on R, it follows that (27) holds.

By repeating the argument of Lemma 12.30, we have

(L) Z

R

f (x, t)d (x, t) ≤ (L) Z

Rn

Z bn

an

f (x, t)dt d x

≤ (U) Z

Rn

Z bn

an

f (x, t)dt d x

≤ (U) Z

R

f (x, t)d (x, t)

for any bounded f .Since f is integrable on R, it follows that (27) holds.

By repeating the argument of Lemma 12.30, we have

(L) Z

R

f (x, t)d (x, t) ≤ (L) Z

Rn

Z bn

an

f (x, t)dt d x

≤ (U) Z

Rn

Z bn

an

f (x, t)dt d x

≤ (U) Z

R

f (x, t)d (x, t)

for any bounded f . Since f is integrable on R, it follows that (27) holds.

By repeating the argument of Lemma 12.30, we have

(L) Z

R

f (x, t)d (x, t) ≤ (L) Z

Rn

Z bn

an

f (x, t)dt d x

≤ (U) Z

Rn

Z bn

an

f (x, t)dt d x

≤ (U) Z

R

f (x, t)d (x, t)

for any bounded f .Since f is integrable on R, it follows that (27) holds.

By repeating the argument of Lemma 12.30, we have

(L) Z

R

f (x, t)d (x, t) ≤ (L) Z

Rn

Z bn

an

f (x, t)dt d x

≤ (U) Z

Rn

Z bn

an

f (x, t)dt d x

≤ (U) Z

R

f (x, t)d (x, t)

for any bounded f . Since f is integrable on R, it follows that (27) holds.

Let E be a projectable region inRⁿ generated by j, H, φ, and ψ. Then E is a Jordan region inRⁿ. Moreover, if f : E →R is continuous on E , then

(28) Z

E

f (x)d x

= Z

H

 Z ψ(x₁,...,bx_j,...,xn) φ(x1,...,bx_j,...,xn)

f (x1, . . . ,xn)dxj



d (x1, . . . ,xbj, . . . ,xn).

Thank you.