## Full text

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WEN-CHINGLIEN

Department of Mathematics National Cheng Kung University

2009

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## Ch12: Integration on R

n

### 12.1: Jordan Regions

Notations:

(1) R = [a1,b1] × . . . × [an,bn], an n-dimensional rectangle.

(2) A grid G = {R1, . . . ,Rp} on R is a collection of n- dimensional rectangles obtained by subdividing the sides of R.

(3) The volume of R: |R| = (b1− a1) . . . (bn− an).

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Definition (1)

Let E be a given set. The outer sum of E with respect to a grid G on a rectangle R is

V (E ; G) := X

Rj∩E6=∅

|Rj|.

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Remark (12.1)

Let R be an n-dimensional rectangle.

(i) Let E be a subset of R, and let G, H be grids on R. If G is finer than H, then

V (E ; G) ≤ V (E ; H).

(ii) If A and B are subsets of R and A ⊆ B, then V (A; G) ≤ V (B; G).

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Remark (12.1)

Let R be an n-dimensional rectangle.

(i) Let E be a subset of R, and let G, H be grids on R. If G is finer than H, then

V (E ; G) ≤ V (E ; H).

(ii) If A and B are subsets of R and A ⊆ B, then V (A; G) ≤ V (B; G).

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Remark (12.1)

Let R be an n-dimensional rectangle.

(i) Let E be a subset of R, and let G, H be grids on R. If G is finer than H, then

V (E ; G) ≤ V (E ; H).

(ii) If A and B are subsets of R and A ⊆ B, then V (A; G) ≤ V (B; G).

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Proof.

(i) Since G is finer than H, each Q ∈ H is a finite union of Rj’s in G.If Q ∩ E 6= ∅, then some of the Rj’s in Q intersect E and others might not(see Figure 12.3, when the darker lines represent the grid H, the lighter lines represent G\H, and the Rj’s that intersect E are shaded).

Let I1= {R ∈ G : G ∩ E 6= ∅} and I2 = {R ∈ G\I1:R ⊆ Q for some Q ∈ H with Q ∩ E 6= ∅}. Then

V (E ; H) = X

R∈I1

|R| + X

R∈I2

|R| ≥ X

R∈I1

|R| = V (E; G).

(ii) If A ⊆ B, then A ⊆ B (see Exercise 3, p.254). Thus, every rectangle that appears in the sum V (A; G) also appears in the sum V (B; G). Since all |Rj|’s are nonnegative, it follows that V (A; G) ≤ V (B; G).

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Proof.

(i) Since G is finer than H, each Q ∈ H is a finite union of Rj’s in G. If Q ∩ E 6= ∅,then some of the Rj’s in Q intersect E and others might not (see Figure 12.3, when the darker lines represent the grid H, the lighter lines represent G\H, and the Rj’s that intersect E are shaded).

Let I1= {R ∈ G : G ∩ E 6= ∅} and I2 = {R ∈ G\I1:R ⊆ Q for some Q ∈ H with Q ∩ E 6= ∅}. Then

V (E ; H) = X

R∈I1

|R| + X

R∈I2

|R| ≥ X

R∈I1

|R| = V (E; G).

(ii) If A ⊆ B, then A ⊆ B (see Exercise 3, p.254). Thus, every rectangle that appears in the sum V (A; G) also appears in the sum V (B; G). Since all |Rj|’s are nonnegative, it follows that V (A; G) ≤ V (B; G).

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Proof.

(i) Since G is finer than H, each Q ∈ H is a finite union of Rj’s in G. If Q ∩ E 6= ∅, then some of the Rj’s in Q intersect E and others might not(see Figure 12.3, when the darker lines represent the grid H, the lighter lines represent G\H, and the Rj’s that intersect E are shaded).

Let I1= {R ∈ G : G ∩ E 6= ∅} and I2 = {R ∈ G\I1:R ⊆ Q for some Q ∈ H with Q ∩ E 6= ∅}.Then

V (E ; H) = X

R∈I1

|R| + X

R∈I2

|R| ≥ X

R∈I1

|R| = V (E; G).

(ii) If A ⊆ B, then A ⊆ B (see Exercise 3, p.254). Thus, every rectangle that appears in the sum V (A; G) also appears in the sum V (B; G). Since all |Rj|’s are nonnegative, it follows that V (A; G) ≤ V (B; G).

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Proof.

(i) Since G is finer than H, each Q ∈ H is a finite union of Rj’s in G. If Q ∩ E 6= ∅, then some of the Rj’s in Q intersect E and others might not (see Figure 12.3, when the darker lines represent the grid H, the lighter lines represent G\H, and the Rj’s that intersect E are shaded).

Let I1= {R ∈ G : G ∩ E 6= ∅} and I2 = {R ∈ G\I1:R ⊆ Q for some Q ∈ H with Q ∩ E 6= ∅}. Then

V (E ; H)= X

R∈I1

|R| + X

R∈I2

|R| ≥ X

R∈I1

|R| = V (E; G).

(ii) If A ⊆ B, then A ⊆ B (see Exercise 3, p.254). Thus, every rectangle that appears in the sum V (A; G) also appears in the sum V (B; G). Since all |Rj|’s are nonnegative, it follows that V (A; G) ≤ V (B; G).

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Proof.

(i) Since G is finer than H, each Q ∈ H is a finite union of Rj’s in G. If Q ∩ E 6= ∅, then some of the Rj’s in Q intersect E and others might not (see Figure 12.3, when the darker lines represent the grid H, the lighter lines represent G\H, and the Rj’s that intersect E are shaded).

Let I1= {R ∈ G : G ∩ E 6= ∅} and I2 = {R ∈ G\I1:R ⊆ Q for some Q ∈ H with Q ∩ E 6= ∅}.Then

V (E ; H) = X

R∈I1

|R| + X

R∈I2

|R|≥ X

R∈I1

|R| = V (E; G).

(ii) If A ⊆ B, then A ⊆ B (see Exercise 3, p.254). Thus, every rectangle that appears in the sum V (A; G) also appears in the sum V (B; G). Since all |Rj|’s are nonnegative, it follows that V (A; G) ≤ V (B; G).

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Proof.

(i) Since G is finer than H, each Q ∈ H is a finite union of Rj’s in G. If Q ∩ E 6= ∅, then some of the Rj’s in Q intersect E and others might not (see Figure 12.3, when the darker lines represent the grid H, the lighter lines represent G\H, and the Rj’s that intersect E are shaded).

Let I1= {R ∈ G : G ∩ E 6= ∅} and I2 = {R ∈ G\I1:R ⊆ Q for some Q ∈ H with Q ∩ E 6= ∅}. Then

V (E ; H)= X

R∈I1

|R| + X

R∈I2

|R| ≥ X

R∈I1

|R|=V (E ; G).

(ii) If A ⊆ B, then A ⊆ B (see Exercise 3, p.254). Thus, every rectangle that appears in the sum V (A; G) also appears in the sum V (B; G). Since all |Rj|’s are nonnegative, it follows that V (A; G) ≤ V (B; G).

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Proof.

(i) Since G is finer than H, each Q ∈ H is a finite union of Rj’s in G. If Q ∩ E 6= ∅, then some of the Rj’s in Q intersect E and others might not (see Figure 12.3, when the darker lines represent the grid H, the lighter lines represent G\H, and the Rj’s that intersect E are shaded).

Let I1= {R ∈ G : G ∩ E 6= ∅} and I2 = {R ∈ G\I1:R ⊆ Q for some Q ∈ H with Q ∩ E 6= ∅}. Then

V (E ; H) = X

R∈I1

|R| + X

R∈I2

|R|≥ X

R∈I1

|R| = V (E; G).

(ii) If A ⊆ B, then A ⊆ B (see Exercise 3, p.254). Thus, every rectangle that appears in the sum V (A; G) also appears in the sum V (B; G). Since all |Rj|’s are nonnegative, it follows that V (A; G) ≤ V (B; G).

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Proof.

(i) Since G is finer than H, each Q ∈ H is a finite union of Rj’s in G. If Q ∩ E 6= ∅, then some of the Rj’s in Q intersect E and others might not (see Figure 12.3, when the darker lines represent the grid H, the lighter lines represent G\H, and the Rj’s that intersect E are shaded).

Let I1= {R ∈ G : G ∩ E 6= ∅} and I2 = {R ∈ G\I1:R ⊆ Q for some Q ∈ H with Q ∩ E 6= ∅}. Then

V (E ; H) = X

R∈I1

|R| + X

R∈I2

|R| ≥ X

R∈I1

|R|=V (E ; G).

(ii) If A ⊆ B,then A ⊆ B (see Exercise 3, p.254). Thus, every rectangle that appears in the sum V (A; G) also appears in the sum V (B; G). Since all |Rj|’s are nonnegative, it follows that V (A; G) ≤ V (B; G).

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Proof.

(i) Since G is finer than H, each Q ∈ H is a finite union of Rj’s in G. If Q ∩ E 6= ∅, then some of the Rj’s in Q intersect E and others might not (see Figure 12.3, when the darker lines represent the grid H, the lighter lines represent G\H, and the Rj’s that intersect E are shaded).

Let I1= {R ∈ G : G ∩ E 6= ∅} and I2 = {R ∈ G\I1:R ⊆ Q for some Q ∈ H with Q ∩ E 6= ∅}. Then

V (E ; H) = X

R∈I1

|R| + X

R∈I2

|R| ≥ X

R∈I1

|R| = V (E; G).

(ii) If A ⊆ B, then A ⊆ B (see Exercise 3, p.254).Thus, every rectangle that appears in the sum V (A; G) also appears in the sum V (B; G). Since all |Rj|’s are nonnegative, it follows that V (A; G) ≤ V (B; G).

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Proof.

(i) Since G is finer than H, each Q ∈ H is a finite union of Rj’s in G. If Q ∩ E 6= ∅, then some of the Rj’s in Q intersect E and others might not (see Figure 12.3, when the darker lines represent the grid H, the lighter lines represent G\H, and the Rj’s that intersect E are shaded).

Let I1= {R ∈ G : G ∩ E 6= ∅} and I2 = {R ∈ G\I1:R ⊆ Q for some Q ∈ H with Q ∩ E 6= ∅}. Then

V (E ; H) = X

R∈I1

|R| + X

R∈I2

|R| ≥ X

R∈I1

|R| = V (E; G).

(ii) If A ⊆ B,then A ⊆ B (see Exercise 3, p.254). Thus, every rectangle that appears in the sum V (A; G) also appears in the sum V (B; G).Since all |Rj|’s are nonnegative, it follows that V (A; G) ≤ V (B; G).

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Proof.

(i) Since G is finer than H, each Q ∈ H is a finite union of Rj’s in G. If Q ∩ E 6= ∅, then some of the Rj’s in Q intersect E and others might not (see Figure 12.3, when the darker lines represent the grid H, the lighter lines represent G\H, and the Rj’s that intersect E are shaded).

Let I1= {R ∈ G : G ∩ E 6= ∅} and I2 = {R ∈ G\I1:R ⊆ Q for some Q ∈ H with Q ∩ E 6= ∅}. Then

V (E ; H) = X

R∈I1

|R| + X

R∈I2

|R| ≥ X

R∈I1

|R| = V (E; G).

(ii) If A ⊆ B, then A ⊆ B (see Exercise 3, p.254).Thus, every rectangle that appears in the sum V (A; G) also appears in the sum V (B; G). Since all |Rj|’s are nonnegative,it follows that V (A; G) ≤ V (B; G).

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Proof.

(i) Since G is finer than H, each Q ∈ H is a finite union of Rj’s in G. If Q ∩ E 6= ∅, then some of the Rj’s in Q intersect E and others might not (see Figure 12.3, when the darker lines represent the grid H, the lighter lines represent G\H, and the Rj’s that intersect E are shaded).

Let I1= {R ∈ G : G ∩ E 6= ∅} and I2 = {R ∈ G\I1:R ⊆ Q for some Q ∈ H with Q ∩ E 6= ∅}. Then

V (E ; H) = X

R∈I1

|R| + X

R∈I2

|R| ≥ X

R∈I1

|R| = V (E; G).

(ii) If A ⊆ B, then A ⊆ B (see Exercise 3, p.254). Thus, every rectangle that appears in the sum V (A; G) also appears in the sum V (B; G).Since all |Rj|’s are nonnegative, it follows that V (A; G) ≤ V (B; G).

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Proof.

(i) Since G is finer than H, each Q ∈ H is a finite union of Rj’s in G. If Q ∩ E 6= ∅, then some of the Rj’s in Q intersect E and others might not (see Figure 12.3, when the darker lines represent the grid H, the lighter lines represent G\H, and the Rj’s that intersect E are shaded).

Let I1= {R ∈ G : G ∩ E 6= ∅} and I2 = {R ∈ G\I1:R ⊆ Q for some Q ∈ H with Q ∩ E 6= ∅}. Then

V (E ; H) = X

R∈I1

|R| + X

R∈I2

|R| ≥ X

R∈I1

|R| = V (E; G).

(ii) If A ⊆ B, then A ⊆ B (see Exercise 3, p.254). Thus, every rectangle that appears in the sum V (A; G) also appears in the sum V (B; G). Since all |Rj|’s are nonnegative,it follows that V (A; G) ≤ V (B; G).

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Proof.

(i) Since G is finer than H, each Q ∈ H is a finite union of Rj’s in G. If Q ∩ E 6= ∅, then some of the Rj’s in Q intersect E and others might not (see Figure 12.3, when the darker lines represent the grid H, the lighter lines represent G\H, and the Rj’s that intersect E are shaded).

Let I1= {R ∈ G : G ∩ E 6= ∅} and I2 = {R ∈ G\I1:R ⊆ Q for some Q ∈ H with Q ∩ E 6= ∅}. Then

V (E ; H) = X

R∈I1

|R| + X

R∈I2

|R| ≥ X

R∈I1

|R| = V (E; G).

(ii) If A ⊆ B, then A ⊆ B (see Exercise 3, p.254). Thus, every rectangle that appears in the sum V (A; G) also appears in the sum V (B; G). Since all |Rj|’s are nonnegative, it follows that V (A; G) ≤ V (B; G).

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Example (12.2)

If R = [0, 1] × [0, 1], A = {(x , y ) : x , y ∈Q ∩ [0, 1]}, and B = R\A, then V (A; G) + V (B; G) = 2V (R; G) no matter how fine G is.

Proof.

Let G = {R1, . . . ,Rp} be a grid on R.Since each Rj is nondegenerate,it is clear by the Density of Rationals (Theorem 1.24) that Rj∩ A 6= ∅ for all j ∈ [1, p]. Hence V (A; G) = |R| = 1. Similarly, the Density of the Irrationals (Exercise 3, p.23) implies V (B; G) = |R| = 1.

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Example (12.2)

If R = [0, 1] × [0, 1], A = {(x , y ) : x , y ∈Q ∩ [0, 1]}, and B = R\A, then V (A; G) + V (B; G) = 2V (R; G) no matter how fine G is.

Proof.

Let G = {R1, . . . ,Rp} be a grid on R.Since each Rj is nondegenerate, it is clear by the Density of Rationals (Theorem 1.24) that Rj∩ A 6= ∅ for all j ∈ [1, p].Hence V (A; G) = |R| = 1. Similarly, the Density of the Irrationals (Exercise 3, p.23) implies V (B; G) = |R| = 1.

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Example (12.2)

If R = [0, 1] × [0, 1], A = {(x , y ) : x , y ∈Q ∩ [0, 1]}, and B = R\A, then V (A; G) + V (B; G) = 2V (R; G) no matter how fine G is.

Proof.

Let G = {R1, . . . ,Rp} be a grid on R. Since each Rj is nondegenerate,it is clear by the Density of Rationals (Theorem 1.24) that Rj∩ A 6= ∅ for all j ∈ [1, p]. Hence V (A; G) = |R| = 1.Similarly, the Density of the Irrationals (Exercise 3, p.23) implies V (B; G) = |R| = 1.

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Example (12.2)

If R = [0, 1] × [0, 1], A = {(x , y ) : x , y ∈Q ∩ [0, 1]}, and B = R\A, then V (A; G) + V (B; G) = 2V (R; G) no matter how fine G is.

Proof.

Let G = {R1, . . . ,Rp} be a grid on R. Since each Rj is nondegenerate, it is clear by the Density of Rationals (Theorem 1.24) that Rj∩ A 6= ∅ for all j ∈ [1, p].Hence V (A; G) = |R| = 1. Similarly, the Density of the Irrationals (Exercise 3, p.23) implies V (B; G) = |R| = 1.

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Example (12.2)

If R = [0, 1] × [0, 1], A = {(x , y ) : x , y ∈Q ∩ [0, 1]}, and B = R\A, then V (A; G) + V (B; G) = 2V (R; G) no matter how fine G is.

Proof.

Let G = {R1, . . . ,Rp} be a grid on R. Since each Rj is nondegenerate, it is clear by the Density of Rationals (Theorem 1.24) that Rj∩ A 6= ∅ for all j ∈ [1, p]. Hence V (A; G) = |R| = 1.Similarly, the Density of the Irrationals (Exercise 3, p.23) implies V (B; G) = |R| = 1.

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Example (12.2)

If R = [0, 1] × [0, 1], A = {(x , y ) : x , y ∈Q ∩ [0, 1]}, and B = R\A, then V (A; G) + V (B; G) = 2V (R; G) no matter how fine G is.

Proof.

Let G = {R1, . . . ,Rp} be a grid on R. Since each Rj is nondegenerate, it is clear by the Density of Rationals (Theorem 1.24) that Rj∩ A 6= ∅ for all j ∈ [1, p]. Hence V (A; G) = |R| = 1. Similarly, the Density of the Irrationals (Exercise 3, p.23) implies V (B; G) = |R| = 1.

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Definition (12.3)

Let E be a subset ofRn. Then E is said to be a Jordan region if and only if given ε > 0 there is rectangle R ⊇ E , and a grid G = {R1, . . . ,Rp} on R, such that

V (∂E ; G) := X

Rj∩∂E6=∅

|Rj| < ε.

(The last sum IS the outer sum of ∂E since ∂E = ∂E by Theorem 8.36.)

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Definition (12.4)

Let E be a Jordan region inRn and let R be an n- dimensional rectangle that satisfies E ⊆ R. The volume (or Jordan content) of E is defined by

Vol(E ) := inf

G V (E ; G)

:=inf{V (E ; G) : G ranges over all grids on R}

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Remark (12.5)

If R is an n-dimensional rectangle, then R is a Jordan region inRn and Vol(R) = |R|.

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Proof.

Let ε > 0 and suppose that

R = [a1,b1] × . . . × [an,bn].

Since bj − aj− 2δ → bj − aj as δ → 0, we can choose δ >0 so small that if

Q = [a1+ δ,b1− δ] × . . . × [an+ δ,bn− δ], then |R| − |Q| < ε.

Let G0:= {H1, . . . ,Hq} be the grid on R determined by Pj(G) = {aj,aj+ δ,bj − δ, bj}.

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Proof.

Let ε > 0 and suppose that

R = [a1,b1] × . . . × [an,bn].

Since bj − aj− 2δ → bj − aj as δ → 0,we can choose δ >0 so small that if

Q = [a1+ δ,b1− δ] × . . . × [an+ δ,bn− δ], then |R| − |Q| < ε.

Let G0:= {H1, . . . ,Hq} be the grid on R determined by Pj(G) = {aj,aj+ δ,bj − δ, bj}.

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Proof.

Let ε > 0 and suppose that

R = [a1,b1] × . . . × [an,bn].

Since bj − aj− 2δ → bj − aj as δ → 0, we can choose δ >0 so small that if

Q = [a1+ δ,b1− δ] × . . . × [an+ δ,bn− δ], then |R| − |Q| < ε.

Let G0:= {H1, . . . ,Hq} be the grid on R determined by Pj(G) = {aj,aj+ δ,bj − δ, bj}.

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Proof.

Let ε > 0 and suppose that

R = [a1,b1] × . . . × [an,bn].

Since bj − aj− 2δ → bj − aj as δ → 0, we can choose δ >0 so small that if

Q = [a1+ δ,b1− δ] × . . . × [an+ δ,bn− δ], then |R| − |Q| < ε.

Let G0:= {H1, . . . ,Hq} be the grid on R determined by Pj(G) = {aj,aj+ δ,bj − δ, bj}.

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Proof.

Then it is clear that an Hj ∈ G intersects ∂R if and only if Hj 6= Q.Hence,

V (∂R; G) := X

Hj∩∂R6=∅

|Hj| = |R| − |Q| < ε.

This proves that R is a Jordan region.

To compute the volume of R by Definition 12.4, let G = {R1, . . . ,Rp} be any grid on R. Since Rj ∩ R 6= ∅ for all Rj ∈ G, it follows from definition that V (R; G) = |R|.

Taking the infimum of this identity over all grids G on R, we find that Vol(R) = |R|.

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Proof.

Then it is clear that an Hj ∈ G intersects ∂R if and only if Hj 6= Q. Hence,

V (∂R; G) := X

Hj∩∂R6=∅

|Hj| = |R| − |Q| < ε.

This proves that R is a Jordan region.

To compute the volume of R by Definition 12.4,let G = {R1, . . . ,Rp} be any grid on R. Since Rj ∩ R 6= ∅ for all Rj ∈ G, it follows from definition that V (R; G) = |R|.

Taking the infimum of this identity over all grids G on R, we find that Vol(R) = |R|.

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Proof.

Then it is clear that an Hj ∈ G intersects ∂R if and only if Hj 6= Q. Hence,

V (∂R; G) := X

Hj∩∂R6=∅

|Hj| = |R| − |Q| < ε.

This proves that R is a Jordan region.

To compute the volume of R by Definition 12.4, let G = {R1, . . . ,Rp} be any grid on R.Since Rj ∩ R 6= ∅ for all Rj ∈ G, it follows from definition that V (R; G) = |R|.

Taking the infimum of this identity over all grids G on R, we find that Vol(R) = |R|.

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Proof.

Then it is clear that an Hj ∈ G intersects ∂R if and only if Hj 6= Q. Hence,

V (∂R; G) := X

Hj∩∂R6=∅

|Hj| = |R| − |Q| < ε.

This proves that R is a Jordan region.

To compute the volume of R by Definition 12.4,let G = {R1, . . . ,Rp} be any grid on R. Since Rj ∩ R 6= ∅ for all Rj ∈ G,it follows from definition that V (R; G) = |R|.

Taking the infimum of this identity over all grids G on R, we find that Vol(R) = |R|.

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Proof.

Then it is clear that an Hj ∈ G intersects ∂R if and only if Hj 6= Q. Hence,

V (∂R; G) := X

Hj∩∂R6=∅

|Hj| = |R| − |Q| < ε.

This proves that R is a Jordan region.

To compute the volume of R by Definition 12.4, let G = {R1, . . . ,Rp} be any grid on R.Since Rj ∩ R 6= ∅ for all Rj ∈ G, it follows from definition that V (R; G) = |R|.

Taking the infimum of this identity over all grids G on R, we find that Vol(R) = |R|.

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Proof.

Then it is clear that an Hj ∈ G intersects ∂R if and only if Hj 6= Q. Hence,

V (∂R; G) := X

Hj∩∂R6=∅

|Hj| = |R| − |Q| < ε.

This proves that R is a Jordan region.

To compute the volume of R by Definition 12.4, let G = {R1, . . . ,Rp} be any grid on R. Since Rj ∩ R 6= ∅ for all Rj ∈ G,it follows from definition that V (R; G) = |R|.

Taking the infimum of this identity over all grids G on R, we find that Vol(R) = |R|.

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Proof.

Then it is clear that an Hj ∈ G intersects ∂R if and only if Hj 6= Q. Hence,

V (∂R; G) := X

Hj∩∂R6=∅

|Hj| = |R| − |Q| < ε.

This proves that R is a Jordan region.

To compute the volume of R by Definition 12.4, let G = {R1, . . . ,Rp} be any grid on R. Since Rj ∩ R 6= ∅ for all Rj ∈ G, it follows from definition that V (R; G) = |R|.

Taking the infimum of this identity over all grids G on R, we find that Vol(R) = |R|.

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Proof.

Then it is clear that an Hj ∈ G intersects ∂R if and only if Hj 6= Q. Hence,

V (∂R; G) := X

Hj∩∂R6=∅

|Hj| = |R| − |Q| < ε.

This proves that R is a Jordan region.

To compute the volume of R by Definition 12.4, let G = {R1, . . . ,Rp} be any grid on R. Since Rj ∩ R 6= ∅ for all Rj ∈ G, it follows from definition that V (R; G) = |R|.

Taking the infimum of this identity over all grids G on R, we find that Vol(R) = |R|.

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Proof.

Then it is clear that an Hj ∈ G intersects ∂R if and only if Hj 6= Q. Hence,

V (∂R; G) := X

Hj∩∂R6=∅

|Hj| = |R| − |Q| < ε.

This proves that R is a Jordan region.

To compute the volume of R by Definition 12.4, let G = {R1, . . . ,Rp} be any grid on R. Since Rj ∩ R 6= ∅ for all Rj ∈ G, it follows from definition that V (R; G) = |R|.

Taking the infimum of this identity over all grids G on R, we find that Vol(R) = |R|.

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Remark (12.6)

Suppose that E is a bounded subset ofRn.

(i) E is a Jordan region of volume zero if and only if there is an absolute constant C, that does not depend on E , such that for each ε > 0 one can find a grid G that satisfies V (E ; G) < Cε.

(ii) E is a Jordan region if and only if Vol(∂E ) = 0.

(iii) If E is a set of volume zero and A ⊆ E , then A is a Jordan region and Vol(A) = 0.

(44)

Remark (12.6)

Suppose that E is a bounded subset ofRn.

(i) E is a Jordan region of volume zero if and only if there is an absolute constant C, that does not depend on E , such that for each ε > 0 one can find a grid G that satisfies V (E ; G) < Cε.

(ii) E is a Jordan region if and only if Vol(∂E ) = 0.

(iii) If E is a set of volume zero and A ⊆ E , then A is a Jordan region and Vol(A) = 0.

(45)

Remark (12.6)

Suppose that E is a bounded subset ofRn.

(i) E is a Jordan region of volume zero if and only if there is an absolute constant C, that does not depend on E , such that for each ε > 0 one can find a grid G that satisfies V (E ; G) < Cε.

(ii) E is a Jordan region if and only if Vol(∂E ) = 0.

(iii) If E is a set of volume zero and A ⊆ E , then A is a Jordan region and Vol(A) = 0.

(46)

Remark (12.6)

Suppose that E is a bounded subset ofRn.

(i) E is a Jordan region of volume zero if and only if there is an absolute constant C, that does not depend on E , such that for each ε > 0 one can find a grid G that satisfies V (E ; G) < Cε.

(ii) E is a Jordan region if and only if Vol(∂E ) = 0.

(iii) If E is a set of volume zero and A ⊆ E , then A is a Jordan region and Vol(A) = 0.

(47)

Proof.

By Definition 12.3 and 12.4, and Remark 12.1ii, it suffices to prove (i).Let E be a Jordan region of volume zero, and let ε > 0. By the Approximation Property for Infima, there is a grid G such that V (E ; G) < ε.Hence set C = 1.

Conversely, let ε > 0 and suppose that there is a grid G such that V (E ; G) < Cε. Then ∂E = E \Eo ⊂ E implies

0 ≤ α := inf

G V (∂E ; G) ≤ β := V (E ; G) ≤ Cε.

Since ε > 0 was arbitrary, it follows that α = β = 0. Since α =0, we can use the Approximation Property for Infima to choose a grid H such that V (∂E ; H) < ε. Thus E is a Jordan region. Since β = 0, we conclude by Definition 12.4 that Vol(E ) = 0.

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Proof.

By Definition 12.3 and 12.4, and Remark 12.1ii, it suffices to prove (i). Let E be a Jordan region of volume zero, and let ε > 0.By the Approximation Property for Infima, there is a grid G such that V (E ; G) < ε. Hence set C = 1.

Conversely, let ε > 0 and suppose that there is a grid G such that V (E ; G) < Cε. Then ∂E = E \Eo ⊂ E implies

0 ≤ α := inf

G V (∂E ; G) ≤ β := V (E ; G) ≤ Cε.

Since ε > 0 was arbitrary, it follows that α = β = 0. Since α =0, we can use the Approximation Property for Infima to choose a grid H such that V (∂E ; H) < ε. Thus E is a Jordan region. Since β = 0, we conclude by Definition 12.4 that Vol(E ) = 0.

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Proof.

By Definition 12.3 and 12.4, and Remark 12.1ii, it suffices to prove (i). Let E be a Jordan region of volume zero, and let ε > 0. By the Approximation Property for Infima, there is a grid G such that V (E ; G) < ε.Hence set C = 1.

Conversely, let ε > 0 and suppose that there is a grid G such that V (E ; G) < Cε.Then ∂E = E \Eo ⊂ E implies

0 ≤ α := inf

G V (∂E ; G) ≤ β := V (E ; G) ≤ Cε.

Since ε > 0 was arbitrary, it follows that α = β = 0. Since α =0, we can use the Approximation Property for Infima to choose a grid H such that V (∂E ; H) < ε. Thus E is a Jordan region. Since β = 0, we conclude by Definition 12.4 that Vol(E ) = 0.

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Proof.

By Definition 12.3 and 12.4, and Remark 12.1ii, it suffices to prove (i). Let E be a Jordan region of volume zero, and let ε > 0. By the Approximation Property for Infima, there is a grid G such that V (E ; G) < ε. Hence set C = 1.

Conversely, let ε > 0 and suppose that there is a grid G such that V (E ; G) < Cε. Then ∂E = E \Eo ⊂ E implies

0 ≤ α := inf

G V (∂E ; G) ≤ β := V (E ; G) ≤ Cε.

Since ε > 0 was arbitrary, it follows that α = β = 0. Since α =0, we can use the Approximation Property for Infima to choose a grid H such that V (∂E ; H) < ε. Thus E is a Jordan region. Since β = 0, we conclude by Definition 12.4 that Vol(E ) = 0.

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Proof.

By Definition 12.3 and 12.4, and Remark 12.1ii, it suffices to prove (i). Let E be a Jordan region of volume zero, and let ε > 0. By the Approximation Property for Infima, there is a grid G such that V (E ; G) < ε. Hence set C = 1.

Conversely, let ε > 0 and suppose that there is a grid G such that V (E ; G) < Cε.Then ∂E = E \Eo ⊂ E implies

0 ≤ α := inf

G V (∂E ; G) ≤ β := V (E ; G) ≤ Cε.

Since ε > 0 was arbitrary, it follows that α = β = 0.Since α =0, we can use the Approximation Property for Infima to choose a grid H such that V (∂E ; H) < ε. Thus E is a Jordan region. Since β = 0, we conclude by Definition 12.4 that Vol(E ) = 0.

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Proof.

By Definition 12.3 and 12.4, and Remark 12.1ii, it suffices to prove (i). Let E be a Jordan region of volume zero, and let ε > 0. By the Approximation Property for Infima, there is a grid G such that V (E ; G) < ε. Hence set C = 1.

Conversely, let ε > 0 and suppose that there is a grid G such that V (E ; G) < Cε. Then ∂E = E \Eo ⊂ E implies

0 ≤ α := inf

G V (∂E ; G) ≤ β := V (E ; G) ≤ Cε.

Since ε > 0 was arbitrary, it follows that α = β = 0. Since α =0, we can use the Approximation Property for Infima to choose a grid H such that V (∂E ; H) < ε.Thus E is a Jordan region. Since β = 0, we conclude by Definition 12.4 that Vol(E ) = 0.

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Proof.

By Definition 12.3 and 12.4, and Remark 12.1ii, it suffices to prove (i). Let E be a Jordan region of volume zero, and let ε > 0. By the Approximation Property for Infima, there is a grid G such that V (E ; G) < ε. Hence set C = 1.

Conversely, let ε > 0 and suppose that there is a grid G such that V (E ; G) < Cε. Then ∂E = E \Eo ⊂ E implies

0 ≤ α := inf

G V (∂E ; G) ≤ β := V (E ; G) ≤ Cε.

Since ε > 0 was arbitrary, it follows that α = β = 0.Since α =0, we can use the Approximation Property for Infima to choose a grid H such that V (∂E ; H) < ε. Thus E is a Jordan region.Since β = 0, we conclude by Definition 12.4 that Vol(E ) = 0.

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Proof.

By Definition 12.3 and 12.4, and Remark 12.1ii, it suffices to prove (i). Let E be a Jordan region of volume zero, and let ε > 0. By the Approximation Property for Infima, there is a grid G such that V (E ; G) < ε. Hence set C = 1.

Conversely, let ε > 0 and suppose that there is a grid G such that V (E ; G) < Cε. Then ∂E = E \Eo ⊂ E implies

0 ≤ α := inf

G V (∂E ; G) ≤ β := V (E ; G) ≤ Cε.

Since ε > 0 was arbitrary, it follows that α = β = 0. Since α =0, we can use the Approximation Property for Infima to choose a grid H such that V (∂E ; H) < ε.Thus E is a Jordan region. Since β = 0, we conclude by Definition 12.4 that Vol(E ) = 0.

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Proof.

By Definition 12.3 and 12.4, and Remark 12.1ii, it suffices to prove (i). Let E be a Jordan region of volume zero, and let ε > 0. By the Approximation Property for Infima, there is a grid G such that V (E ; G) < ε. Hence set C = 1.

Conversely, let ε > 0 and suppose that there is a grid G such that V (E ; G) < Cε. Then ∂E = E \Eo ⊂ E implies

0 ≤ α := inf

G V (∂E ; G) ≤ β := V (E ; G) ≤ Cε.

Since ε > 0 was arbitrary, it follows that α = β = 0. Since α =0, we can use the Approximation Property for Infima to choose a grid H such that V (∂E ; H) < ε. Thus E is a Jordan region.Since β = 0, we conclude by Definition 12.4 that Vol(E ) = 0.

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Proof.

By Definition 12.3 and 12.4, and Remark 12.1ii, it suffices to prove (i). Let E be a Jordan region of volume zero, and let ε > 0. By the Approximation Property for Infima, there is a grid G such that V (E ; G) < ε. Hence set C = 1.

Conversely, let ε > 0 and suppose that there is a grid G such that V (E ; G) < Cε. Then ∂E = E \Eo ⊂ E implies

0 ≤ α := inf

G V (∂E ; G) ≤ β := V (E ; G) ≤ Cε.

Since ε > 0 was arbitrary, it follows that α = β = 0. Since α =0, we can use the Approximation Property for Infima to choose a grid H such that V (∂E ; H) < ε. Thus E is a Jordan region. Since β = 0, we conclude by Definition 12.4 that Vol(E ) = 0.

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Definition (12.7)

Let E := {E`}`∈N be a collection of subsets ofRn.

(i) E is said to be nonoverlapping if and only if Ej∩ Ek is of volume zero for j 6= k .

(ii) E is said to be pairwise disjoint if and only if Ej ∩ Ek

= ∅for j 6= k .

Notice that since ∅ is of volume zero, every collection of pairwise disjoint sets is nonoverlapping.

(58)

Definition (12.7)

Let E := {E`}`∈N be a collection of subsets ofRn.

(i) E is said to be nonoverlapping if and only if Ej∩ Ek is of volume zero for j 6= k .

(ii) E is said to be pairwise disjoint if and only if Ej ∩ Ek

= ∅for j 6= k .

Notice that since ∅ is of volume zero, every collection of pairwise disjoint sets is nonoverlapping.

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Theorem (12.8)

Let E be a subset ofRn. Then E is a Jordan region of volume zero if and only if for every ε > 0 there is a finite collection of cubes Qk of the same size, i.e., all with sides of length s, such that

E ⊂

q

[

k =1

Qk and

q

X

k =1

|Qk| < ε.

(60)

Corollary (12.9)

If E1 and E2are Jordan regions, then E1∪ E2 is a Jordan region and

Vol(E1∪ E2) ≤Vol(E1) +Vol(E2).

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Corollary (12.10)

Suppose that V is a bounded, open set inRnand that φ :V →Rn is 1-1 and C1on V with ∆φ6= 0.

(i) If E is of volume zero and E ⊂ V , then φ(E ) is of volume zero.

(ii) If {Ek}k ∈Nis a nonoverlapping collection of sets inRn with Ek ⊂ V for all k ∈ N, then {φ(Ek)}k ∈N is a

nonoverlapping collection of sets inRn.

(iii) If E is a Jordan region and E ⊂ V , then φ(E ) is a Jordan region.

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Corollary (12.10)

Suppose that V is a bounded, open set inRnand that φ :V →Rn is 1-1 and C1on V with ∆φ6= 0.

(i) If E is of volume zero and E ⊂ V , then φ(E ) is of volume zero.

(ii) If {Ek}k ∈Nis a nonoverlapping collection of sets inRn with Ek ⊂ V for all k ∈ N, then {φ(Ek)}k ∈N is a

nonoverlapping collection of sets inRn.

(iii) If E is a Jordan region and E ⊂ V , then φ(E ) is a Jordan region.

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Corollary (12.10)

Suppose that V is a bounded, open set inRnand that φ :V →Rn is 1-1 and C1on V with ∆φ6= 0.

(i) If E is of volume zero and E ⊂ V , then φ(E ) is of volume zero.

(ii) If {Ek}k ∈Nis a nonoverlapping collection of sets inRn with Ek ⊂ V for all k ∈ N, then {φ(Ek)}k ∈N is a

nonoverlapping collection of sets inRn.

(iii) If E is a Jordan region and E ⊂ V , then φ(E ) is a Jordan region.

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Corollary (12.10)

Suppose that V is a bounded, open set inRnand that φ :V →Rn is 1-1 and C1on V with ∆φ6= 0.

(i) If E is of volume zero and E ⊂ V , then φ(E ) is of volume zero.

(ii) If {Ek}k ∈Nis a nonoverlapping collection of sets inRn with Ek ⊂ V for all k ∈ N, then {φ(Ek)}k ∈N is a

nonoverlapping collection of sets inRn.

(iii) If E is a Jordan region and E ⊂ V , then φ(E ) is a Jordan region.

(65)

Definition (2)

The inner sum of E with E with respect to G is υ(E ; G) := X

Rj⊂Eo

|Rj|.

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Remark (12.11)

Let R be an n-dimensional rectangle, let E be a subset of R, and let G, H be grids on R. If G is finer than H, then

0 ≤ υ(E ; H) ≤ υ(E ; G) ≤ V (E ; G) ≤ V (E ; H).

This leads us to the following fundamental principle.

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Remark (12.12)

Let R be an n-dimensional rectangle and E be a subset of R. If G and H are grids on R, then

0 ≤ υ(E ; G) ≤ V (E ; H).

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Definition (12.13)

Let E be a bounded subset ofRn and let R be an n- dimensional rectangle that satisfies E ⊆ R. The inner volume of E is defined by

Vol(E ) := sup{υ(E ; G) : G ranges over all grids on R}, and the outer volume of E is defined by

Vol(E ) := inf{V (E ; G) : G ranges over all grids on R}.

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Theorem (12.14)

Let E be a bounded subset ofRn. Then E is a Jordan region if and only if Vol(E ) = Vol(E ).

(70)

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