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

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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 exist 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 this 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 exists 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.

This 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 exists 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 exists 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.

This 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 exercise 29.5 in [1] will then imply

n→∞

the result. Let ɛ > 0 be given. Since f is uniformly continuous on [a, b], there exists 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 exist 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 exists 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, Analysis with an introduction to proof, Prentice Hall, Inc., Englewood Cliffs, NJ, 1986.

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