1 |f| and f 2 are integrable when f is integrable

1 **|f|** **and** f 2 **are** **integrable** **when** f **is** **integrable**

Lemma 1.1. Let f : [a, b] → R be a bounded function **and** let P = {x 0 , x 1 , . . . , x n } be a partition of

[a, b]. Then for each i ∈ {1, 2, . . . , n}, M i (f) − m i (f) = sup{|f(x) − f(y)| : x, y ∈ [x i−1 , x i ]}.

Proof. Let x, y ∈ [x i−1 , x i ]. Without loss of generality assume that f(x) ≥ f(y) **and** observe that

M i (f) ≥ f(x) **and** m i (f) ≤ f(y). These inequalities imply that M i (f) − m i (f) ≥ f(x) − f(y). It now

follows that

M i (f) − m i (f) ≥ sup{|f(x) − f(y)| : x, y ∈ [x i−1 , x i ]}. (1)

Let ɛ > 0 be given. There ex**is**t x, y ∈ [x i−1 , x i ] such that f(x) > M i (f)− ɛ 2 **and** f(y) < m i(f)+ ɛ 2 .

So f(x) − f(y) > M i (f) − m i (f) − ɛ, **and** therefore, |f(x) − f(y)| > M i (f) − m i (f) − ɛ. It now follows

that sup{|f(x) − f(y)| : x, y ∈ [x i−1 , x i ]} > M i (f) − m i (f) − ɛ. Since th**is** holds for any ɛ > 0, we

have

sup{|f(x) − f(y)| : x, y ∈ [x i−1 , x i ]} ≥ M i (f) − m i (f). (2)

The inequalities (1) **and** (2) imply the desired equality.

Theorem 1.2. Suppose that f : [a, b] → R **is** an **integrable** function. Then **|f|** **is** also **integrable** on

[a, b].

Proof. Let ɛ > 0 be given. Since f **is** **integrable**, there ex**is**ts a partition P = {x 0 , x 1 , . . . , x n } of

[a, b] such that U(f, P ) − L(f, P ) < ɛ. For any i ∈ {1, 2, . . . , n} **and** all x, y ∈ [x i−1 , x i ], we have

| |f(x)| − |f(y)| | ≤ |f(x) − f(y)|. So M i (f) − m i (f) **is** an upper bound for the set {| |f(x)| − |f(y)| | :

x, y ∈ [x i−1 , x i ]}, which implies that M i (**|f|**) − m i (**|f|**) ≤ M i (f) − m i (f). Now,

U(**|f|**, P ) − L(**|f|**, P ) =

n∑

n∑

(M i (**|f|**) − m i (**|f|**))∆x i ≤ (M i (f) − m i (f))∆x i

i=1

i=1

= U(f, P ) − L(f, P ) < ɛ.

Th**is** shows that **|f|** **is** **integrable** on [a, b].

Theorem 1.3. Suppose that f : [a, b] → R **is** an **integrable** function. Then f 2 **is** also **integrable** on

[a, b].

Proof. Since f **is** bounded on [a, b], there ex**is**ts a B > 0 such that |f(x)+f(y)| < B for all x, y ∈ [a, b.]

Now let ɛ > 0 be given. Since f **is** **integrable**, there ex**is**ts a partition P = {x 0 , x 1 , . . . , x n } of

[a, b] such that U(f, P ) − L(f, P ) < ɛ B . For any i ∈ {1, 2, . . . , n} **and** all x, y ∈ [x i−1, x i ], we have

| (f(x)) 2 − (f(y)) 2 | = |f(x) + f(y)| |f(x) − f(y)| < B|f(x) − f(y)|. So B(M i (f) − m i (f)) **is** an upper

bound for the set {| (f(x)) 2 − (f(y)) 2 | : x, y ∈ [x i−1 , x i ]}, which implies that M i (f 2 ) − m i (f 2 ) ≤

B(M i (f) − m i (f)). Now,

U(f 2 , P ) − L(f 2 , P ) =

n∑

n∑

(M i (f 2 ) − m i (f 2 ))∆x i ≤ B(M i (f) − m i (f))∆x i

i=1

i=1

= B(U(f, P ) − L(f, P )) < B ɛ B = ɛ.

Th**is** shows that f 2 **is** **integrable** on [a, b].

2 Integration for continuous function

Theorem 2.1. Let f : [a, b] → R be continuous on [a, b] **and** let P n = {x 0 = a, x 1 = a + (b−a)

n , x 2 =

a + 2 (b−a)

n

, . . . , x n = b}. Then

∫ b

a

f = lim

n→∞ U(f, P n) = lim

n→∞ L(f, P n).

Proof. It suffices to show that lim (U(f, P n)−L(f, P n )) = 0 since exerc**is**e 29.5 in [1] will then imply

n→∞

the result. Let ɛ > 0 be given. Since f **is** uniformly continuous on [a, b], there ex**is**ts a δ > 0 such that

ɛ

b−a

**when** |x − y| < δ, |f(x) − f(y)| < . Also, continuity of f implies that for each i ∈ {1, 2, . . . , n},

there ex**is**t points s i , t i ∈ [x i−1 , x i ] such that M i = f(t i ) **and** m i = f(s i ). Now if ∆x i = (b−a)

n

< δ,

Since lim

n→∞

U(f, P n ) − L(f, P n ) =

(b−a)

n

n∑

(M i − m i )∆x i =

i=1

n∑

(f(t i ) − f(s i ))∆x i <

i=1

ɛ

b − a

n∑

∆x i

i=1

= ɛ (b − a) = ɛ.

b − a

= 0, there ex**is**ts a N ∈ R such that **when** n > N, we have (b−a)

n

n > N, we get U(f, P n ) − L(f, P n ) < ɛ, which implies that lim

n→∞ (U(f, P n) − L(f, P n )) = 0.

< δ. So **when**

Corollary 2.2. Suppose that f : [a, b] → R **is** continuous on [a, b]. Let P n = {x 0 = a, x 1 = a +

(b−a)

n

, x 2 = a + 2 (b−a)

n

, . . . , x n = b} **and** for each i ∈ {1, 2, . . . , n}, let x ∗ i ∈ [x i−1, x i ] be sample points.

∫ b

n∑

Then f = lim f(x ∗

n→∞

i )∆x i .

a

i=1

Proof. For each i ∈ {1, 2, . . . , n}, m i ≤ f(x ∗ i ) ≤ M i. So

L(f, P n ) =

n∑

m i ∆x i ≤

i=1

n∑

f(x ∗ i )∆x i ≤

i=1

n∑

M i ∆x i = U(f, P n ).

i=1

Since lim U(f, P n ) = lim L(f, P n ), the Squeeze Theorem implies that

∫ b

a

f = lim

n→∞

n∑

i=1

f(x ∗ i )∆x i = lim

n→∞ U(f, P n) = lim

n→∞ L(f, P n).

References

[1] S. Lay, Analys**is** with an introduction to proof, Prentice Hall, Inc., Englewood Cliffs, NJ, 1986.