Lemma

Suppose that f is increasing on [a,b].

(i)Ifx0 ∈[a,b), thenf(x0+) exists andf(x0)≤f(x0+).

(ii) Ifx0 ∈(a,b], thenf(x0−)exists andf(x0−)≤f(x0).

Lemma:

Suppose that f is increasing on [a,b].

(i) Ifx0 ∈[a,b), thenf(x0+) exists andf(x0)≤f(x0+).

(ii) Ifx0 ∈(a,b], thenf(x0−)exists andf(x0−)≤f(x0).

Lemma:

Suppose that f is increasing on [a,b].

(i)Ifx0 ∈[a,b), thenf(x0+) exists andf(x0)≤f(x0+).

(ii)Ifx0 ∈(a,b], thenf(x0−)exists andf(x0−)≤f(x0).

Lemma:

Suppose that f is increasing on [a,b].

(i) Ifx0 ∈[a,b), thenf(x0+) exists andf(x0)≤f(x0+).

(ii) Ifx0 ∈(a,b], thenf(x0−)exists andf(x0−)≤f(x0).

Lemma:

Suppose that f is increasing on [a,b].

(i) Ifx0 ∈[a,b), thenf(x0+) exists andf(x0)≤f(x0+).

(ii)Ifx0 ∈(a,b], thenf(x0−)exists andf(x0−)≤f(x0).

Lemma:

Suppose that f is increasing on [a,b].

(i) Ifx0 ∈[a,b), thenf(x0+) exists andf(x0)≤f(x0+).

(ii) Ifx0 ∈(a,b], thenf(x0−)exists andf(x0−)≤f(x0).

Proof:

Fixx0 ∈(a,b]. By symmetry it suffices to show thatf(x0−) exists and satisfiesf(x0−)≤f(x0). Set

E ={f(x) :a<x <x0}ands=supE. Since f is increasing,f(x0)is an upper bound of E. Hence, s is a finite real number that satisfiess≤f(x0). Given >0, choose by the Approximation Property anx1 ∈(a,x0) such thats− <f(x1)≤s. Since f is increasing,

s− <f(x1)≤f(x)≤s

for allx1<x <x0. Therefore,f(x0−)exists and satisfies f(x −) = s≤f(x ).2

Proof:

Fixx0 ∈(a,b]. By symmetry it suffices to show thatf(x0−) exists and satisfiesf(x0−)≤f(x0). Set

E ={f(x) :a<x <x0}ands=supE. Since f is increasing,f(x0)is an upper bound of E. Hence, s is a finite real number that satisfiess≤f(x0). Given >0, choose by the Approximation Property anx1 ∈(a,x0) such thats− <f(x1)≤s. Since f is increasing,

s− <f(x1)≤f(x)≤s

for allx1<x <x0. Therefore,f(x0−)exists and satisfies f(x −) = s≤f(x ).2

Proof:

Fixx0 ∈(a,b]. By symmetry it suffices to show thatf(x0−) exists and satisfiesf(x0−)≤f(x0). Set

E ={f(x) :a<x <x0}ands=supE. Since f is increasing,f(x0)is an upper bound of E. Hence, s is a finite real number that satisfiess≤f(x0). Given >0, choose by the Approximation Property anx1 ∈(a,x0) such thats− <f(x1)≤s. Since f is increasing,

s− <f(x1)≤f(x)≤s

for allx1<x <x0. Therefore,f(x0−)exists and satisfies f(x −) = s≤f(x ).2

Proof:

Fixx0 ∈(a,b]. By symmetry it suffices to show thatf(x0−) exists and satisfiesf(x0−)≤f(x0). Set

E ={f(x) :a<x <x0}ands=supE. Since f is increasing,f(x0)is an upper bound of E.Hence, s is a finite real number that satisfiess≤f(x0). Given >0, choose by the Approximation Property anx1 ∈(a,x0) such thats− <f(x1)≤s. Since f is increasing,

s− <f(x1)≤f(x)≤s

for allx1<x <x0. Therefore,f(x0−)exists and satisfies f(x −) = s≤f(x ).2

Proof:

Fixx0 ∈(a,b]. By symmetry it suffices to show thatf(x0−) exists and satisfiesf(x0−)≤f(x0). Set

E ={f(x) :a<x <x0}ands=supE. Since f is increasing,f(x0)is an upper bound of E. Hence,s is a finite real number that satisfiess≤f(x0). Given >0, choose by the Approximation Property anx1 ∈(a,x0) such thats− <f(x1)≤s. Since f is increasing,

s− <f(x1)≤f(x)≤s

for allx1<x <x0. Therefore,f(x0−)exists and satisfies f(x −) = s≤f(x ).2

Proof:

Fixx0 ∈(a,b]. By symmetry it suffices to show thatf(x0−) exists and satisfiesf(x0−)≤f(x0). Set

E ={f(x) :a<x <x0}ands=supE. Since f is increasing,f(x0)is an upper bound of E.Hence, s is a finite real number that satisfiess≤f(x0). Given >0, choose by the Approximation Property anx1 ∈(a,x0) such thats− <f(x1)≤s. Since f is increasing,

s− <f(x1)≤f(x)≤s

for allx1<x <x0. Therefore,f(x0−)exists and satisfies f(x −) = s≤f(x ).2

Proof:

Fixx0 ∈(a,b]. By symmetry it suffices to show thatf(x0−) exists and satisfiesf(x0−)≤f(x0). Set

E ={f(x) :a<x <x0}ands=supE. Since f is increasing,f(x0)is an upper bound of E. Hence,s is a finite real number that satisfiess≤f(x0). Given >0, choose by the Approximation Property anx1 ∈(a,x0) such thats− <f(x1)≤s. Since f is increasing,

s− <f(x1)≤f(x)≤s

for allx1<x <x0. Therefore,f(x0−)exists and satisfies f(x −) = s≤f(x ).2

Proof:

Fixx0 ∈(a,b]. By symmetry it suffices to show thatf(x0−) exists and satisfiesf(x0−)≤f(x0). Set

E ={f(x) :a<x <x0}ands=supE. Since f is increasing,f(x0)is an upper bound of E. Hence, s is a finite real number that satisfiess≤f(x0). Given >0, choose by the Approximation Property anx1 ∈(a,x0) such thats− <f(x1)≤s. Since f is increasing,

s− <f(x1)≤f(x)≤s

for allx1<x <x0. Therefore,f(x0−)exists and satisfies f(x −) = s≤f(x ).2

Proof:

Fixx0 ∈(a,b]. By symmetry it suffices to show thatf(x0−) exists and satisfiesf(x0−)≤f(x0). Set

E ={f(x) :a<x <x0}ands=supE. Since f is increasing,f(x0)is an upper bound of E. Hence, s is a finite real number that satisfiess≤f(x0). Given >0, choose by the Approximation Property anx1 ∈(a,x0) such thats− <f(x1)≤s. Since f is increasing,

s− <f(x1)≤f(x)≤s

for allx1<x <x0. Therefore,f(x0−)exists and satisfies f(x −) = s≤f(x ).2

Proof:

Fixx0 ∈(a,b]. By symmetry it suffices to show thatf(x0−) exists and satisfiesf(x0−)≤f(x0). Set

E ={f(x) :a<x <x0}ands=supE. Since f is increasing,f(x0)is an upper bound of E. Hence, s is a finite real number that satisfiess≤f(x0). Given >0, choose by the Approximation Property anx1 ∈(a,x0) such thats− <f(x1)≤s. Since f is increasing,

s− <f(x1)≤f(x)≤s

for allx1<x <x0. Therefore,f(x0−)exists and satisfies f(x −) = s≤f(x ).2

Proof:

Fixx0 ∈(a,b]. By symmetry it suffices to show thatf(x0−) exists and satisfiesf(x0−)≤f(x0). Set

E ={f(x) :a<x <x0}ands=supE. Since f is increasing,f(x0)is an upper bound of E. Hence, s is a finite real number that satisfiess≤f(x0). Given >0, choose by the Approximation Property anx1 ∈(a,x0) such thats− <f(x1)≤s. Since f is increasing,

s− <f(x1)≤f(x)≤s

for allx1<x <x0. Therefore,f(x0−)exists and satisfies f(x −) = s≤f(x ).2

Proof:

Fixx0 ∈(a,b]. By symmetry it suffices to show thatf(x0−) exists and satisfiesf(x0−)≤f(x0). Set

E ={f(x) :a<x <x0}ands=supE. Since f is increasing,f(x0)is an upper bound of E. Hence, s is a finite real number that satisfiess≤f(x0). Given >0, choose by the Approximation Property anx1 ∈(a,x0) such thats− <f(x1)≤s. Since f is increasing,

s− <f(x1)≤f(x)≤s

for allx1<x <x0. Therefore,f(x0−)exists and satisfies f(x −) = s≤f(x ).2

Proof:

Fixx0 ∈(a,b]. By symmetry it suffices to show thatf(x0−) exists and satisfiesf(x0−)≤f(x0). Set

E ={f(x) :a<x <x0}ands=supE. Since f is increasing,f(x0)is an upper bound of E. Hence, s is a finite real number that satisfiess≤f(x0). Given >0, choose by the Approximation Property anx1 ∈(a,x0) such thats− <f(x1)≤s. Since f is increasing,

s− <f(x1)≤f(x)≤s

for allx1<x <x0. Therefore,f(x0−)exists and satisfies f(x −) = s≤f(x ).2

Theorem

If f is monotone on an open interval I, then f has at most countably many points of discontinuity on I.

Theorem

If f is monotone on an open interval I, then f has at most countably many points of discontinuity on I.

Theorem (Intermediate Value Theorem For Derivatives) Suppose that f is differentiable on [a,b] with f⁰(a)6=f⁰(b).

If y0 is a real number that lies between f⁰(a)and f⁰(b), then there is an x0∈(a,b)such that f⁰(x0) =y0.

Theorem (Intermediate Value Theorem For Derivatives) Suppose that f is differentiable on [a,b] with f⁰(a)6=f⁰(b).

If y0 is a real number that lies between f⁰(a)and f⁰(b), then there is an x0∈(a,b)such that f⁰(x0) =y0.