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RN where Ri and Rj are nonoverlapping rectangles for all i 6= j, we define the area of S to be A(S

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1. Definition of areas

Definition 1.1. The area of a rectangle R of length a and of width b is defined to be A(R) = ab.

A simple plane region is a union of finitely many rectangles. A simple plane region can be decom- posed into a finite union of non-overlapping rectangles. Two rectangles are called non-overlapping if their intersection is contained in their boundaries.

Definition 1.2. If S is a simple region and S = R1∪ · · · ∪ RN where Ri and Rj are nonoverlapping rectangles for all i 6= j, we define the area of S to be

A(S) = A(R1) + · · · + A(RN).

Let Ω be a bounded plane region. We define the inner area of Ω to be the real number A(Ω) = sup{A(S) : S is a simple region, S ⊆ Ω}

and the outer area of Ω to be the real number

A+(Ω) = inf{A(S0) : S0 is a simple region, S0 ⊇ Ω}.

One can see that

A(Ω) ≤ A+(Ω).

Definition 1.3. Let Ω be a bounded plane region. We say that Ω has an area if A+(Ω) = A(Ω).

In this case, we call the number A+(Ω) = A(Ω) the area of Ω denoted by A(Ω).

Example 1.1. Let Ω = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ x2}. Show that Ω has an area and find its area.

Let n be a natural number and for each 1 ≤ i ≤ n − 1, we consider the rectangle Ri =



(x, y) : i

n ≤ x ≤ i + 1

n , 0 ≤ y ≤ i2 n2

 .

Then S = R1∪ · · · ∪ Rn−1 is a simple region contained in Ω. The area of Ri is given by i2/n3 and hence the area of S is

A(S) = A(R1) + · · · + A(Rn−1) =

n−1

X

i=1

i2

n3 = (n − 1)(2n − 1)

6n2 .

By definition, A(S) ≤ A(Ω) and hence

(n − 1)(2n − 1)

6n2 ≤ A(Ω).

Similarly, we consider the rectangles R0i=



(x, y) : i − 1

n ≤ x ≤ i

n, 0 ≤ y ≤ i2 n2



for each 1 ≤ i ≤ n. Then S0= R01∪ · · · ∪ R0n is a simple region containing Ω. Then, A(S0) = A(R01) + · · · + A(R0n) =

n

X

i=1

i3

n3 = (n + 1)(2n + 1) 6n2 . By definition A+(Ω) ≤ A(S0) and therefore

A+(Ω) ≤ (n + 1)(2n + 1)

6n2 .

Since A(Ω) ≤ A+(Ω), we obtain that for each n ≥ 1, (n − 1)(2n − 1)

6n2 ≤ A(Ω) ≤ A+(Ω) ≤ (n + 1)(2n + 1) 6n2 . By taking n → ∞,

1

3 ≤ A(Ω) ≤ A+(Ω) ≤ 1 3.

1

(2)

2

We find that

A+(Ω) = A(Ω) = 1 3. Hence we find that Ω has an area and its area equals to 1/3.

1.1. Appendix: Least upper bound and the greatest lower bound. Let K be a collection of real numbers.

Definition 1.4. We say that K is bounded above if there exists a real number U such that x ≤ U for all x ∈ K.

In this case, U is called an upper bound for K. The smallest upper bound for K denoted by sup K is called the least upper bound for K. (sup K is an upper bound for K and if U is an upper bound for K, then sup K ≤ U.)

Definition 1.5. We say that K is bounded below if there exists a real number L such that x ≥ L for all x ∈ K.

In this case, L is called a lower bound for K. The largest lower bound for K denoted by inf K is called the greatest lower bound for K. (inf K is a lower bound for K and if L is a lower bound for K, then inf K ≥ L.)

Example 1.2. Let K = (0, 1) the set of all real numbers x such that 0 < x < 1. 2 is an upper bound for K since x < 2 for all x ∈ K. 2 is not the least upper bound for K since 1 < 2 and 1 is an upper bound for K. In fact,

sup K = 1, inf K = 0.

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