Calculus Riemann integrable functions

Calculus Riemann integrable functions November 6, 2018

Definitions Let f be a function defined on an interval I. Then (a) f is said to be continuous on I if

for each  > 0 and for each x ∈ I, there exists δ = δ(x) > 0, such that if y ∈ I and if |y − x| < δ then |f (y) − f (x)| < .

(b) f is said to be uniformly continuous on I if for each  > 0, there exists δ > 0,

such that if x, y ∈ I and if |y − x| < δ then |f (y) − f (x)| < .

(c) f is not uniformly continuous on I if

there exists ₀ > 0, such that for all n ∈ N = {1, 2, . . .}, there exist x_n, y_n ∈ I satisfying |y_n− x_n| < 1

n, but |f (y_n) − f (x_n)| ≥ ₀.

Examples

(a) Let f be defined by f (x) = 1

x for x ∈ I = (0, 1).

For each a ∈ I, since

x→alim f (x) = f (a), f is continuous on I.

However, for any sufficiently small, fixed δ > 0, since |x − (x − δ/2)| = δ/2 < δ and

|f (x) − f (x − δ/2)| = |1

x − 1

x − δ/2| = δ/2

x|x − δ/2| → ∞ as x → 0⁺, f is not uniformly continuous on I.

(b) Let f be defined by f (x) = x² for x ∈ R = (−∞, ∞).

For each a ∈ R, since

x→alim f (x) = f (a), f is continuous on R.

However, for any fixed δ > 0, since |(x + δ/2) − x| = δ/2 < δ and

|f (x + δ/2) − f (x)| = |(x + δ/2)²− x²| = |x δ + δ²/4| → ∞ as x → ∞, f is not uniformly continuous on R.

TheoremIf f is continuous on a closed interval I = [a, b], then f is uniformly continuous on I.

Fact (Bolzano Weierstrass Theorem) Let {x_n| n ∈ N} be a sequence of points in I. Then there exists a convergent subsequence {x_nk | k ∈ N} of {xn| n ∈ N}, i.e

{xn_k | k ∈ N} ⊂ {xⁿ | n ∈ N} and lim

k→∞ xn_k = x for some x ∈ I.

Calculus Riemann integrable functions (Continued) November 6, 2018

Proof Suppose that f is not uniformly continuous on I. Then there exists ₀ > 0 such that for any n = 1, 2, . . . there exist x_n, y_n ∈ I such that

|y_n− x_n| < 1

n but |f (x_n) − f (y_n)| ≥ ₀ > 0.

Using the Bonzano Weierstrass Theorem, we may assume that {x_nk | k ∈ N} and {ynk | k ∈ N}

are subsequences of {x_n | n ∈ N} and {yn | n ∈ N}, respectively, that converges to some x ∈ I.

Thus we have

k→∞lim x_nk = x = lim

k→∞ y_nk, and

0 = |f (x) − f (x)| = lim

k→∞ |f (x_nk) − f (y_nk)| ≥ ₀ > 0 which is a contradiction. Hence f is uniformly continuous on I.

Definitions

(a) Let I = [a, b] ⊂ R. A set P = {x0, x1, . . . , xn} ⊂ I is called a partition of I if a = x0 < x1 < . . . < xn= b.

(b) Let P be a partition of I. Define

∆x_i = x_i − x_i−1 = the width of the i^th interval [x_i−1, x_i] kP k = max

1≤i≤n ∆xi = max {∆xi | 1 ≤ i ≤ n} = the norm of partition P (c) Let f be a function defined on I and let P be a partition of I. Then

R(f, P ) =

n

X

i=1

f (x^∗_i) ∆x_i, where x^∗_i ∈ [x_i−1, x_i] for each 1 ≤ i ≤ n,

is calleda Riemann sum of f with respect to the partition P of I.

(d) Let f be a function defined on I = [a, b]. The definite integral of f from a to b, denoted Z b

a

f (x) dx is defined to be Z b

a

f (x) dx = lim

kP k→0 n

X

i=1

f (x^∗_i) ∆x_i provided that the limit exists,

where x^∗_i ∈ [x_i−1, x_i] and ∆x_i = x_i− x_i−1 for each 1 ≤ i ≤ n.

(e) Let P = {a = x₀ < x₁ < . . . < x_n = b}, Q = {a = t₀ < t₁ < . . . < t_m = b} be partitions of I. Q is said to be finer than P if

P = {a = x0 < x1 < . . . < xn= b} ⊆ {a = t0 < t1 < . . . < tm = b} = Q

=⇒ kQk = max

1≤j≤m ∆t_j ≤ kP k = max

1≤i≤n ∆x_i.

Page 2

Calculus Riemann integrable functions (Continued) November 6, 2018 (f) Let f be a bounded function defined on I and let P be a partition of I. Then

U (f, P ) =

n

X

i=1

sup

x∈[xi−1,xi]

f (x) ∆x_i, where sup

x∈[xi−1,xi]

f (x) = lub {f (x) | x ∈ [x_i−1, x_i]},

is calledan upper (Riemann) sum of f with respect to the partition P of I, and

L(f, P ) =

n

X

i=1

inf

x∈[xi−1,xi]f (x) ∆x_i, where inf

x∈[xi−1,xi]f (x) = glb{f (x) | x ∈ [x_i−1, x_i]}, is calleda lower (Riemann) sum of f with respect to the partition P of I.

(g) Let f be a function defined on I = [a, b]. Then f is said to be Riemann integrable on I if Z b

a

f (x) dx = lim

kP k→0 R(f, P )= lim

1≤i≤nmax ∆xi→0 n

X

i=1

f (x^∗_i) ∆x_i exists.

Remark Let f be a continuous function defined on I = [a, b]. Let S be a subset of R defined by

S = {

n

X

i=1

f (x^∗_i) ∆x_i | a = x₀ < x₁ < . . . < x_n= b, x^∗_i ∈ [x_i−1, x_i] and ∆x_i = x_i−x_i−1 for 1 ≤ i ≤ n}.

It is easy to see that

(a) P = {x₀ = a < b = x₁} is a partition of I =⇒ f (x)(b − a) ∈ S ∀ x ∈ I =⇒ S 6= ∅.

(b) by the Extreme Value Theorem, ∃ y, z ∈ [a, b] such that f (y) = max

x∈[a,b] f (x) = max {f (x) | x ∈ [a, b]}, f (z) = min

x∈[a,b] f (x) = min {f (x) | x ∈ [a, b]} which implies that f (y)(b − a), f (z)(b − a) ∈ S

S is bounded from above and below by f (y)(b − a) and f (z)(b − a), respectively.

(c) similarly, for any partition P = {a = x₀ < x₁ < . . . < x_n = b} of I,

U (f, P ) =

n

X

i=1

f (yi) ∆xi ∈ S, where yi ∈ [xi−1, xi] andf (yi) = max

x∈[xi−1,xi]f (x) for 1 ≤ i ≤ n, L(f, P ) =

n

X

i=1

f (z_i) ∆x_i ∈ S, where z_i ∈ [x_i−1, x_i] andf (z_i) = min

x∈[xi−1,xi]

f (x) for 1 ≤ i ≤ n.

Hence we have

f (z)(b − a) ≤ L(f, P ) ≤ R(f, P ) ≤ U (f, P ) ≤ f (y)(b − a) for any partition P of I = [a, b].

(d) Let P = {a = x₀ < x₁ < . . . < x_n = b}, Q = {a = t₀ < t₁ < . . . < t_m = b} be partitions of I. Suppose that Q is finer than P. For each 1 ≤ j ≤ m, sinceP ⊆ Q, there exists 1 ≤ i ≤ n such that

[tj−1, tj] ⊆ [xi−1, xi] =⇒ inf

x∈[xi−1,xi]f (x) ≤ inf

x∈[tj−1,tj]f (x) ≤ sup

x∈[tj−1,tj]

f (x) ≤ sup

x∈[xi−1,xi]

f (x)

Page 3

Calculus Riemann integrable functions (Continued) November 6, 2018 Hence we have

f (z)(b − a) ≤ L(f, P ) ≤ L(f, Q) ≤ U (f, Q) ≤ U (f, P )≤ f (y)(b − a).

Let {P_n| n ∈ N} be a sequence of partitions of I such that Pn⊆ Pn+1 for each n = 1, 2, . . . and lim

n→∞ kPnk = 0.

By the completeness of R, there exist L, U ∈ R such that

n→∞lim L(f, P_n) = L ≤ U = lim

n→∞ U (f, P_n).

(e) If P, Q are partitions of I, then P U Q is also a partition of I and since P ⊆ P ∪ Q and Q ⊆ P ∪ Q,

we have

L(f, P ) ≤ L(f, P ∪ Q) ≤ U (f, P ∪ Q) ≤ U (f, Q).

Hence

sup {L(f, P ) | P is a partition of I} ≤ inf {U (f, Q) | Q is a partition of I}.

(f) Let f be a function defined on I = [a, b]. Then f is said to be Riemann integrable on I if and only if

sup {L(f, P ) | P is a partition of I} = L = U = inf {U (f, P ) | P is a partition of I}.

Hence f is said to be Riemann integrable on I if and only if for each  > 0 there exists a partition P of I such that

0 ≤ U (f, P ) − L(f, P ) = |U (f, P ) − L(f, P )| < .

Theorem If f is continuous on a closed interval I = [a, b], then f is Riemann integrable on I.

Proof Since f is continuous on a closed interval I = [a, b], f is uniformly continuous on I = [a, b].

Given  > 0, there exists δ > 0 such that

if x, y ∈ I and if |y − x| < δ then |f (y) − f (x)| <  b − a. Let P = {a = x0 < x1 < . . . < xn = b} be a partition of I = [a, b] such that

kP k = max

1≤i≤n ∆x_i < δ.

Since

0 ≤ U (f, P ) − L(f, P ) =

n

X

i=1

max

x∈[xi−1,xi]f (x) − min

x∈[xi−1,xi]f (x)
∆x_i < 

b − a · (b − a) = , f is Riemann integrable on I.

Page 4