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21-355: Real Analysis 1

Carnegie Mellon University

Professor Ian Tice - Fall 2013

Project LATEXd by Ivan Wang

Last Updated: November 15, 2013

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CONTENTS 21-355 Notes

Contents

1 The Number Systems 3

1.1 The Natural Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.1 Positivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.1.2 Order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.1.3 Multiplication. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2 The Integers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2.1 Properties of Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2.2 Algebraic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.3 The Rationals and Ordered Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.3.1 Fields and Orders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.4 Problems with Q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.4.1 Bounds (Infimum and Supremum) . . . . . . . . . . . . . . . . . . . . . . . . 8

1.5 The Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.5.1 Defining the Real Numbers: Dedekind Cuts . . . . . . . . . . . . . . . . . . . 9

1.5.2 Defining the Real Numbers: The Least Upper Bound Property . . . . . . . . 10

1.5.3 Defining the Real Numbers: Addition . . . . . . . . . . . . . . . . . . . . . . 10

1.5.4 Defining the Real Numbers: Multiplication . . . . . . . . . . . . . . . . . . . 11

1.5.5 Defining the Real Numbers: Distributivity. . . . . . . . . . . . . . . . . . . . 11

1.5.6 Defining the Real Numbers: Archimedean . . . . . . . . . . . . . . . . . . . . 12

1.6 Properties ofR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.6.1 Absolute Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2 Sequences 13

2.1 Convergence and Bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.1.1 Squeeze Lemma. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.2 Monotonicity and limsup, liminf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.3 Subsequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.3.1 Limsup Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.4 Special Sequences. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 Series 17

3.1 Convergence Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.1.1 Cauchy Criterion Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.1.2 Logarithm. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.2 The numbere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

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CONTENTS 21-355 Notes

3.3 More Convergence Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.4 Algebra of Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.5 Absolute Convergence and Rearrangements . . . . . . . . . . . . . . . . . . . . . . . 22

4 Topology ofR 24

4.1 Open and Closed Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4.1.1 Open Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4.1.2 Closed Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4.1.3 Limit Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4.1.4 Closure, Interior, and Boundary Sets . . . . . . . . . . . . . . . . . . . . . . . 26

4.2 Compact Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.2.1 Heine-Borel Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.3 Connected Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

5 Continuity 305.1 Limits of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

5.1.1 Divergence Criteria. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

5.2 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

5.3 Compactness and Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

5.4 Continuity and Connectedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

5.5 Discontinuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

5.6 Monotone Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

6 Differentiation 35

6.1 The Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

6.2 Mean Value Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

6.3 Darbouxs Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

6.4 LHopitals Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

6.5 Higher Derivatives and Taylors Theorem . . . . . . . . . . . . . . . . . . . . . . . . 38

7 Riemann-Stieltjes Integration 39

7.1 The R-S Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397.2 Integrability Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

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21-355 Notes

1 The Number Systems

1.1 The Natural Numbers

Theorem (existence ofN): There exists a set N satisfying the following properties, known as thePeano Axioms:

PA1 0 N

PA2 There exists a function S : N N called the successor function. In particular,S(n) N.

PA3 n N. S(n)= 0

PA4 S(n) =S(m) = n= m (S is injective, one-to-one)

PA5 [Axiom of Induction] Let P(n) be a property associated to each n N.IfP(0) is true, and P(n) = P(S(n)), then P(n) is true n N.

Definition: PA1 = 0 N. PA2 = S(0) N.

Define 1 =S(0), 2 =S(1), 3 =S(2), etc.PA2guarantees that {0, 1, 2, } N.

PA3prevents wraparound: no successor can map to a negative number.

PA4prevents stagnation: the cycle does not terminate.

Theorem: N ={0, 1, 2, }

Proof: We know that {0, 1, 2, } N, so it suffices to prove that N {0, 1, 2, }.

LetP(n) denote the proposition that n {0, 1, 2, }. Clearly P(0) is true.

Suppose P(n) is true; then n {0, 1, 2, } = S(n) {0, 1, 2, }by construction.

Hence,P(S(n)) is true. By induction, PA5 guarantees that P(n) is true n N.It follows that N {0, 1, 2, }.

Definition: For any m N, we define 0 +m= m.Then ifn+m is defined for n N, we set S(n) +m= S(n+m).

Proposition (Properties of Addition):

1. n N. n+ 0 =n (0 is the additive identity)

2. m, n N. n+S(m) =S(n+m)

3. m, n N. m+n= n+m (commutativity)

4. k,m,n N. k+ (m+n) = (k+m) +n (associativity)5. k,m,n N. n+k = n+m = k= m (cancelation)

Proof:

1. LetP(n) be n+ 0 =n.P(0) is true because 0 + 0 = 0 by definition.Note P(n) = S(n) + 0 = S(n+ 0) = S(n), so P(S(n)) is true. By induction,(1) is true.

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1.1 The Natural Numbers 21-355 Notes

2. Fixm N. LetP(n) denote n +S(m) =S(n+m).P(0) is true because 0 +S(m) =S(m) =S(0 +m).P(n) = S(n)+ S(m) =S(n + S(m)) =S(S(n + m)) =S(S(n) + m), soP(S(n))is true. By induction, since m N was arbitrary, (2) is true.

3. Letm be fixed and P(n) denote n+m= m+n.P(0) is true since 0 + m= m by definition, and m + 0 =m by 1, so 0 + m= m =m+ 0.Suppose P(n); then S(n) + m= S(n + m) =S(m + n) =m + S(n), soP(S(n)) istrue. By induction and arbitrary choice ofm, (3) is true.

4. Fix k, m N and let P(n) denote k+ (m+n) = (k+m) +n.P(0) is true as k+ (m+ 0) =k +m= (k+m) + 0.SupposeP(n); thenk+(m+S(n)) =k+S(m+n) =S(k+(m+n)) =S(k+m)+n=(k+m) +S(n) by (2). By induction and arbitrary choice, (4) is true.

5. Fix m, n N and let P(k) denote proposition 5.P(0) is true because n + 0 =n = n+m = m= 0 = k= m.Suppose P(k); also, supposem + S(k) =n + S(k). ThenS(m + k) =m + S(k) =n+S(k) = S(n+k) = m+k = n+k = m = n (by 4). By the axiom of

induction, (5) is true.

1.1.1 Positivity

Definition: We say that n N is positive ifn = 0.

Proposition (Properties of Positivity):

1. n, m N, ifm is positive, then m +n is positive.

2. n, m N, ifm +n= 0, then m = n = 0.

3. n N, ifn is positive, then there exists a unique m N such that n= S(m).

1.1.2 Order

Definition: For allm, n N, m n or n m iffn = m+p for some p N.m < nor n > m iffm n m=n. The relation provides what is called an order on N.

Proposition (Properties of Order):

Letj, k,m,n N. Then:

1. n n (reflexitivity)

2. m n k m = k n (transitivity)

3. m n m n = m= n (anti-symmetry)

4. jk m n = j+m k +n (order preservation)

5. m < n S(m) n

6. m < n n= m+p for some positive p N.

7. n m S(n)> m

8. n= 0 0< n

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1.2 The Integers 21-355 Notes

Theorem (Trichotomy of Order): Let m, n N. Then exactly one of the following is true:

m < n m= n m > n

Proof: Show that no two can be true simultaneously (by definition of< and>), and then at leastone must be true (by induction on n).

1.1.3 Multiplication

Definition: Fix m N. Define 0 m = 0. Now, if n m is defined for some n N, we defineS(n) m= n m+m.

Proposition (Properties of Multiplication):

Fix k ,m, n N. Then:

1. m n= n m (commutativity)

2. m, n are positive = mn is positive

3. m n= 0 m= 0 n= 0 (no zero divisors)

4. k (m n) = (k m) n(associativity)

5. k m= k n k is positive = m= n (cancelation)

6. k (m+n) = (m+n) k= k m+k n (distributivity)

7. m < n k l k, l are positive = m k < n l

1.2 The Integers

Consider the following relation on the set N N:

(m, n) (m, n) m+n = m+n

Lemma: is an equivalence relation.

Proof:

Reflexivity: m+n= m+n = (m, n) (m, n)

Symmetry: (m, n) (m, n) = m+ n = m +n = m +n = m+ n =(m, n) (m, n)

Transitivity: Suppose (m, n) (m, n) (m, n) (m, n). Then:

m+n = m+n m+n = m+n

=m+n = m+n=(m, n) (m, n)

Definition: Write the equivalence class of (m, n) as [(m, n)] ={(p, q)| (p, q) (m, n)}.Define the integersZ ={[(m, n)]}.

Lemma: Suppose (m, n) (m, n), (p, q) (p, q). Then:

1. (m+p, n+q) (m+p, n+q)

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1.2 The Integers 21-355 Notes

2. (mp+nq,mq+np) (mp+nq, mq+np)

Proof: Consider equalities (a) :m+n = m+n and (b) :p+q=p+q(by definition of).

Using linear combinations of (a) and (b), we derive the two rules of the lemma:

1. (a) + (b)

2. (a)(p+q) + (b)(m+n)

Definition: Let [(m, n)], [(p, q)] Z. Then:

1. [(m, n)] + [(p, q)] = [(m+p, n+q)] (addition of integers)

2. [(m, n)] [(p, q)] = [(mp+nq,mq+np)] (multiplication of integers)

By the lemma, these are well-defined operations.

Note that for all m, n N:

[(m, 0)] = [(n, 0)] m+ 0 =n+ 0 m= n

[(m, 0)] + [(n, 0)] = [(m+n, 0)]

[(m, 0)] [(n, 0)] = [(mn, 0)]

As such, the set {[(n, 0)]| n N} Z behaves exactly like a copy ofN.

Definition: For n N we set n Z to be n := [(n, 0)].

For x = [(m, n)] Z we define x= [(n, m)].

1.2.1 Properties of Integers

(We can see that every integer x Z can be represented as x := m nwhere x = [(m, n)].)

Theorem: Every x Z satisfies exactly one of the following:

1. x= n for some n N\{0}

2. x= 0

3. x= nfor some n N\{0}

Proof: Writex = [(p, q)] for some p, q N. By trichotomy of order on N we know that p < qorp= qor p > q. Each of these correlates to one of the three properties.

Corollary: Z ={0, 1, 2, . . .} { 1, 2, 3, . . .}

1.2.2 Algebraic Properties

Proposition: Let x, y, z Z. Then the following hold:

1. x+y = y +x

2. x+ (y+z) = (x+y) +z

3. x+ 0 = 0 +x= x

4. x+ (x) = (x) +x= 0

5. xy= yx

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1.4 Problems withQ 21-355 Notes

1. x, y Eexactly one of the following is true: x < yor x = y or y < x (trichotomy)

2. x,y,z E, x < y y < z = x < z (transitivity)

Definition: Let F be a field. Then we define x y= x+ (y) and xy =xy1 (for y = 0).

Theorem: Q is an ordered field with order 0. Thenthe well-ordering principle implies that !n S(q). n= min S(q).

Since n S(q), we know that q = mn for some m Z. Then q2 = (mn)

2 = m2

n2 =m2 = 2n2 = m2 is even. We claim that m is also even (proof is exercise to reader).

Then l Z. m = 2l. Then 4l2

= (2l)2

= m2

= 2n2

= n2

= 2l2

= n2

is even= n is even = n= 2p for some p N+.

Hence q= mn = 2l2p =

lp = p S(q). But clearly p < n, which contradicts the fact

that n is the minimal element. By contradiction, the theorem must be true.

1.4.1 Bounds (Infimum and Supremum)

Informally, Q has holes:

Definition: Let Ebe an ordered set with order

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1.5 The Real Numbers 21-355 Notes

Theorem: Q does not satisfy the least upper bound property.

Proof : Consider the set A= {x Q |x >0, x2 2}.

Note that 0 < 1 = 12 2 = 1 A, so A is non-empty. Also, 2 4 = 22 implies(x A = 0< x2 2 and using the same logic as

before, we can choose n large enough such that pq 1n is an upper bound ofA. But

pq

1n 0 = (n N). x < n.

Lemma: IfQ is Archimedean, then (p < q Q).(r Q). p < r < q.(Proofs in HW 2.)

1.5.1 Defining the Real Numbers: Dedekind Cuts

Definition: We say that C P(Q) is a cut(Dedekind cut) iff the following hold:

(C1) =C, C = Q

(C2) Ifp Cand q Q with q < p, then q C.

(C3) Ifp C, (r Q). p < r r C.

Lemma: Suppose C is a cut. Then:

1. p C, q / C = p < q

2. r / C, r < s = s / C

3. Cis bounded above

Lemma: Let q Q. Then {p Q |p < q}is a cut.

Proof: Call the set C. We prove the 3 properties of a cut:

(C1) q 1 C = C = ; q+ 1 / C = C = Q.

(C2) Ifp Cand r Q such that r < p, then r < p < q = r < q = r C.

(C3) Letp Cwhere p < q. Since Q is Archimedean,(r Q). p < r < q = r C.

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Definition: Given q Q we write Cq ={p Q |p < q}. By the above lemma, Cq is a cut.

Definition: We write R ={C P(Q)| C is a cut} = .

Lemma: The following hold:

1. A, B R, exactly one of the following holds: A B, A= B, B A.

2. A, B, C R,A B B C = A C.

Definition: IfA, B R we say that A< B A B, and A B A B. This definesan order on R by the above lemma.

1.5.2 Defining the Real Numbers: The Least Upper Bound Property

Lemma: Suppose =E R is bounded above. Then B:=AEA R.

Theorem: R satisfies the least upper bound property.

Proof: Let =E R be bounded above and set B=AEA R. We claim B= sup E.

First, we show that Bis an upper bound ofE. Let A E. Then A B = A B

(by definition). This is true for all A E, so B is an upper bound.We claim that for C R. C C sinceotherwiseA C = b C, b / C. Hence C q). p / A}. ThenR, +, 0R = C0 = {p Q | p < 0}satisfy the field axioms.

Proof:

(A1) A + B R by previous lemma.

(A2) A + B= {a+b}= {b+a}= B+ A.

(A3) A + (B+ C) ={a+ (b+c)}= {(a+b) +c}= (A + B) + C.

(A4) ShowA R.0R+ A= A.

(A5) Show thatA R, then A + (A) = 0R using Archimedean property.

Theorem (Ordered Field): Let A, B, C R. IfA< Bthen A + C

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1.5.4 Defining the Real Numbers: Multiplication

Lemma: Let A, B R,A, B> 0R. Then C={q Q |q 0} {a b| a A, b B, a, b >0} R.

Proof:

(C1) 0 C = C = . A, Bare bounded above by, say M1, M2, so M1 M2+ 1 / Cand C = Q.

(C2) Letp Cand q < p. Ifq 0 then q C by definition. Ifq >0 then 0 < q < p,but then 0 0. Then 0< q < a b =qa 0, then p = a b fora A, b B, a, b >0. Choose s A such that a < s, t Bsuch that t > b. Thenp= a b < s t S, sor = s tproves the claim.

Definition of Multiplication: Let A, B R.

1. IfA > 0, B> 0 we set A B= {q Q |q 0} {a b| a A, b B, a, b >0} R.

2. IfA = 0 or B= 0, we set A B= 0R.3. IfA > 0 and B< 0, let A B= (A (B)).

4. IfA < 0 and B> 0, let A B= ((A) B).

5. IfA < 0 and B< 0, let A B= (A) (B).

Theorem: R, satisfies (M1-M5) with 1R = C1, andA> 0 = A1 ={q Q |q 0} {q Q |q >0, p > q. p1 / A} R;A< 0 = A1 =(A)1.

Proof : HW3 (similar to addition).

Theorem: IfA, B> 0, thenA B> 0.

Proof : By definition C0 A B = 0 A B . Equality is impossible since A, B> 0.

1.5.5 Defining the Real Numbers: Distributivity

Theorem: LetA, B, C R. Then A (B+ C) =A B+ A C.

Proof : We prove the case where all are positive. The other cases are in HW.

Letp A(B+ C). Ifp 0 then p A B + A Bis trivial (both products contain theinterval less than 0).

Ifp >0, p = a(b+c) for a A, b B, c C fora >0, b+c >0.

Regardless of sign ofb or c,a b A B , a c A C . Hence p = a(b + c) =a b + a cA B+ A C. SoA(B+ C) A B + A C.

Finally, we show the converse is true; let p A B+ A C = p = r +s forr A B , s A C . Case on positivity ofp,r,s to show p A(B+ C).

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1.6 Properties ofR 21-355 Notes

1.5.6 Defining the Real Numbers: Archimedean

Theorem: For p, q Q, the following are true:

1. Cp+q =Cp+ Cq

2. Cp= Cp3. Cpq =CpCq

4. Ifp = 0 then Cp1 = (Cp)15. p < q Q Cp < Cq R

Proof: HW.

Definition: For q Q we sayCq R. Then Q R.

Theorem: There exists an ordered field satisfying the least upper bound property; R is unique (forany ordered field F satisfying these properties, F = R up to isomorphism; and R is Archimedean.

Proof : The basic assertion is Steps (0)-(4). Step (5) proves 1, Step (6) proves 3.

1.6 Properties ofR

Notation: think ofR as numbers, not cut notation.

Proposition: R satisfies the following:

Theorem: For p, q Q, the following are true:

1. R is Archimedean: x R, x >0.n N. x < n

2. N R is not bounded above

3. inf{ 1n |n N, n 1}= 0

4. x R the set B(x) ={m Z |x < m}has a minimum in Z.

5. x, y R, x < y.q Q. x < q < y

Remarks:

1. (5) is interpreted as the density ofQ R. Any element x R can be approxi-mated to arbitrary accuracy by elements ofQ.

2. (4) allows us to define the integer part of anyx R. We can setx= min B(x) 1 Z. Then x x 0 and n N, n 1. Then !y R. y >0 yn =x.

Proof: The case n= 1 is trivial so assume n 2.

SetE= {z R |z >0 zn < x}. We want to showE= and is bounded above. Sett= x1+x ; then 0< t 0.We claim that yn < xand yn > xare both impossible (proof is exercise), so yn =x.

Definition: Let n 1; for x R, x >0, we write x1n =y where yn =x. We set 0

1n = 0.

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1.6.1 Absolute Value

For x R, we define the function | |: R {r R |r 0}:

|x|=

x ifx >0

0 ifx = 0

x ifx 0.|an|< M (n l).Lemma: If a sequence converges, then it is bounded.

Definition: Given{an}, {bn} Rwe define{an+ bn} Rto be the sequence whose elements arean+bn. We similary define {can}for a fixed c R, {anbn}, and {an/bn}where bn= 0, n l.

Theorem (algebra of convergence): Let{an}, {bn} R, c R, and assume that an a, bn bas n . Then the following hold:

1. an+bna+b as n

2. can ca as n

3. anbn ab as n

4. Ifbn= 0 and b= 0, then an/bna/b as n .

Proof: (1), (2) are in next weeks HW.

(3): Note that |anbn ab| = |anbn abn +abn ab| |anbn abn|+ |abn ab| =|bn||an a| + |a||bn b|. Since bn b we know that M >0.|bn|< M(n l).

Let > 0. Since an a and bn b we may choose N1 such that n N1 =|an a| 0. Since an a as n , we knowNl. n N = |an a|< . We claim Kl. k K = (k) N.

If not, then (k) < N(k l); but k (k) < N(k l) is a contradiction. Thenthe claim is true, and k K = (k) N = |a(k) a| < . Since > 0 wasarbitrary, we deduce {a(k)} a as k .

Remark: Converse fails. Example: an = (1)n;a2n= +1 +1, but a2n+1=1 1.

2.3.1 Limsup Theorem

Theorem: Let{an} R be bounded. The following hold:

1. Every subsequence of{an} is bounded.

2. If{ank} is a subsequence, then lim supk ank lim supn an.

3. If{ank} is a subsequence, then lim infn an lim infk ank .

4. There exists a subsequence{ank}such that limk ank = lim supn an.

5. There exists a subsequence{ank}such that limk ank = lim infn an (= (4)).Proof:

1. Trivial.

2. Sincek (k),{a(n)|n k} {an | n k}for every order-preserving. HenceSk = sup{a(n)} |n k} sup{an | n k}= Tk. But:limsupn a(n) = limsupk{a(n) | n k} limsupk{an | n k} =limsupn an.

3. Similar to (2); exercise to reader.

4. Too lazy to LATEX; exercise to reader.

5. Exercise to reader.

Theorem: Suppose {an} R; the following are equivalent:

1. an a as n

2. {an} is bounded, and every convergent subsequence converges to a.

3. {an} is bounded, and lim supn an = lim infn an.

Proof: (1) = (2) proven already.

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2.4 Special Sequences 21-355 Notes

(2) = (3)Limsup theorem (4,5) = {a(k)}, {a(k)} subsequences such thata(k) lim supn an, a(k)liminfn an as k . By (2) the limits must agree.

(3) = (1)Limsup theorem (1-3) = {a(k)}. lim infn anlim infk a(k) lim supk a(k)limsupn an. As the first and last are equal, by transitivity it follows all subsequences

satisfy lim infk a(k)= lim supk a(k). Asanis a subsequence of itself, it thereforeconverges to some a as n .

Theorem (Bolzano-Weierstrass): If{an} R is bounded then there exists a convergent subse-quence. Proof from (4) or (5) of Limsup Theorem.

2.4 Special Sequences

Definition: Given an R for 0 k n, n N we definen

k=0an = a0+a1+ +an.

Lemma (Binomial Theorem): Let x, y R and n N. Then (x+ y)n =n

k=0 nk

xkynk, wherenk

:= n!k!(nk)! N.Theorem: In the following assuming thatn 1:

1. Letx R, x >0. Then an = 1nx 0 as n .

2. Letx R, x >0. Then an = x1/n 1 as n .

3. Letan = n1/n; then an 1 as n .

4. Leta, x R, x >0. Then na

(1+x)a 0 as n .

5. Letx R, |x|< 1. Then an = xn 0 as n .

3 Series

Definition: Let{an}n=l R; for p < qwe write

qn=pan = (ap+ +aq).

1. We define, for each n l, Sn =n

k=l

ak R to be the nth partial sum of{an}

n=l.

2. If s R. Sn s as n , then

n=lan = s. We say the infinite seriesn=lan converges.

3. If the series does not converge, it diverges.

Examples

1. Let an = xn for n 0, x R. Then Sn =

nk=0xk. Notice that (1 x)Sn =

nk=0xk n

k=0xk+1 =

nk=0x

k n+1

k=1xk = 1 xn+1.

SoSn =n

k=0xk = (1x

n+1

1x ). If|x|< 1 then Sn 11x by special seq (5).

2. Suppose{bn}n=0 Rwherebn b as n . Setan = bn+1 bn forn 0. Then the series

n=0an converges and in fact

n=0= b b0.

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3.1 Convergence Results

We develop tools that will let us deduce the convergence of a series without knowing its value.

Theorem: Suppose

n=lan converges. Then an 0 as n .

Proof: Notice thatan = Sn Sn1 and so limn an = limn(Sn Sn1) =S S= 0.

Corollary:n=0(1)n and n=0n diverge, as neither sequences converge to 0.Corollary: The series

n=0x

n converges |x|< 1.

Proof: |x| 1 = |xn|= |x|n 1(n N). The converse was proved last time.

Next, we provide a characterization of convergence in terms of the size of the tails of the series.

Theorem:

n=lan converges >0.Nl. m k N = |m

n=kan|< .

Proof:

n=lan converges Sk =k

n=lan converges {Sk} is Cauchy.

This is useful in practice because we can guarantee a series converges without knowing its value.

Theorem:1. Ifn k.|an| bn for some k l, and

n=l

bn converges, thenn=l

an converges.

2. Ifn k.0 an bn for some k l, andn=l

an diverges, thenn=l

bn diverges.

Proof: (1) Let > 0 and prove with previous theorem and induction on triangle inequality. (2)follows from contrapositive.

Examples:

1.

n=0

(1)n2n converges because |

(1)n2n |=

12n and

n=0

12n converges (

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On the other hand, ifm 2k,Sm a1 + + a2k =a1 +a2 + (a3 + a4) + + (a2k11 + +a2k)

12a1+a2+ + 2

k1a2k = 12Tk.

Now, if

n=02na2n converges, then Tn T as n and so Sm limn Tm =T,

which means {Sm}is bounded and

n=1an converges.

Similarly, ifn=1an converges, then Tk 2limn Sn = {Tk} is bounded =n=02na2n converges.

Theorem: Letp R. Then

n=11np converges p >1.

Proof:

Ifp 0 the result is trivial since 1np 1 (the sequences converges to 0). Assume thatp >0. Then 1(n+1)p

1np , so we can apply the Cauchy criterion:

n=1

1

np converges

n=0

2n

(2n)p converges.

But n=0 2n

(2n)p = n=0 1(2p1)n , and this series converges 12p1 1.Notice

n=1

1n is divergent, but

n=1

1n1+r converges r > 0. To try to find intermediate series,

we need the logarithm.

3.1.2 Logarithm

Definition: From Supplemental Reading 3, for every 1 0} R such that

1. blogbx =x (x >0)

2. logb(1) = 0, logbb= 1

3. 0< x < y logbx 0, z R)

5. logb is a bijection

6. limn

logbnnr = 0 (r R, r >0)

Then from (6), for large nand p >0 we know:n n(logbn)

p n np =n1+p = 1n1+p 1

n(logbn)p

1n .

So 1n(logbn)pis such an intermediate series.

Theorem: Letb >1.

n=2

1n(logbn)

p converges p >1. (n 2 = logbn >0)

Proof:n=2

1n(logbn)

p converges n=1

2n

2n(logb2n)p converges by Cauchy criterion, but

n=1

1(logb2)

pnp = 1(logb2)

p

n=1

1np converges p > 1.

In particular,n=2

1n logbn

is divergent.

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3.2 The numbere 21-355 Notes

3.2 The number e

Lemma:

n=01n! converges.

Proof: Ifn 2 then:

Sn=

n

k=0

1

k!= 1 + 1 +

1

2 1+ +

1

n(n 1) 2 1

1 + 1 +1

2+

1

2 2+ +

1

2n1

1 +k=0

1

2k = 1 + 2 = 3

Since Sn is increasing and bounded, we know that

n=01n! converges.

Definition: We set e =n=0

1n! . Note that e >1.

Theorem: e= limn(1 + 1n)n.

Proof: Let Sn=n

k=01k! , Tn = (1 +

1n)

n

. Then by the Binomial Theorem:

Tn = (1 +1

n)n =

nk=0

n!

k!(n k)!

1

nk

= 1 + 1 + 1

2!

n(n 1)

n2 + +

1

n!

n(n 1) 1

nn

= 1 + 1 + 1

2!(1

1

n) +

1

3!(1

1

n)(1

2

n) + +

1

n!(1

1

n) (1

n 1

n )

1 + 1 + 1

2!+ +

1

n!=Sn

Hence, lim supn Tn lim supn Sn = limn Sn = e.

OTOH, fix m N. Then for n m:

Tn 1 + 1 + 1

2!(1

1

n) + +

1

m!(1

1

n) (1

m 1

n )

= lim infn Tn lim infn RHS 1 + 1 +

1

2!lim infn (1

1

n) + +

1

m!lim infn (1

1

n (1

m 1

n ) =

Then, lettingm , e = limm Sm lim infn Tn.

Thus, e lim infn Tn lim supn Tne = limn Tn = e.

Theorem: n 1.0 < e Sn < 1nn! . Also,e R\Q is irrational.

Proof: Since Sn is increasing, 0< e Sn is clear. The other side can be seen from algebra.

Now, suppose e Q; then e = pq forp, q N, p , q 1.

Then 0< q!(e Sq)< 1q (q 1). Notice that q!e= q!

pq = (q 1)!p N and

q!(1 + 12! + + 1q!) N.

Henceq!(e Sq) Z; but this yields an integer between 0 and 1, a contradiction. So eis irrational.

Remark: In fact, e is transcendental.

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3.3 More Convergence Results 21-355 Notes

3.3 More Convergence Results

Theorem (Root Test): Suppose {an}n=l R and {|an|

1/n} is bounded. Let 0 = lim supn |an|1/n.Then the following holds:

1. If 1, then n=lan diverges.

3. if= 1, both convergence and divergence are possible.

Theorem (Ratio Test): Let{an}n=l R. Then

n=lan:

1. converges if{|an+1an |}n=l is bounded and lim supn

|an+1||an|

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Definition: The seriesn=0

cn, where cn=n

k=0

akbnk, is called the Cauchy product

of the seriesn=0

an,n=0

bn.

Remark: If

an,

bnconverge,

cndoes not necessarily converge if neither series has convergentabsolute values.

3.5 Absolute Convergence and Rearrangements

Proposition: If

n=l |an|converges, then

n=lan converges. Proof is trivial.

Definition: Suppose

n=lan converges. If

n=l |an|converges, the series convergesabsolutely. If|an|diverges, the series is conditionally convergent.

Example:

n=1(1)n

n is conditionally convergent, while

n=1(1)nn2 is absolutely convergent.

Lets try to manipulate the series without being careful.

=

n=1

(1)

n+1

n = 1 12+ 13 14+

= limk

(Sk =k

n=0

(1)n+1

n ) = lim

k(S2k =

2kn=0

(1)n+1

n )

but: S2k = (1 1

2) + (

1

3

1

4+ + (

1

2k 1

1

2k)> 0

Hence, >0. But the next step is questionable:

2=

n=1(2)(1)n+1

n?

=

k=02

2k+ 1

k=12

2k

?=

k=0

2

2k+ 1

k=1

1

k =

k=0

1

2k+ 1

k=1

1

2k =

= 2= >0 a contradiction!

Problem: rearrangement is a delicate issue.

Definition: Let :{m Z| m l} {m Z| m l} be a bijection. The series

n=la(n) iscalled a rearrangement of

n=lan.

Theorem: If

n=lan is absolutely convergent, then every rearrangement converges to

n=lan.

Proof: Let >0.

Since

n=lan converges absolutely, Nl. k m N =k

n=m |an|< 2 .

Letk :

n=m |an| 2 < .

Now choose M N such that {l, l + 1, . . . , N } {(l), (l+ 1), . . . , (M)}. Thenm M = |

mn=lan

mn=la(n)|

n=N|an|< .

Hence limm(m

n=lan

n=la(n)) = 0 and from this we deducelimm

mn=la(n)= limm

mn=lan =

n=lan.

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When a series is only conditionally convergent, the situation is vastly worse.

Theorem: Suppose

n=0an is conditionally convergent. Let c R.There exists a rearrangement (bijection) : N N such that

n=0a(n)= c.

Lemma: Suppose

n=0an is conditionally convergent and set:

bn

= an ifan > 00 ifan0

cn

= an ifan < 00 ifan 0

Then

n=0bn and

n=0cn both diverge.

Proof: Suppose not; one of the series is convergent. If

bn converges, then cn = bn an =cn =

bn

an; but |an|= bn+cnand so

|an|=

bn+

cnis convergent, a contradiction.

A similar argument holds if

cn converges.

Rearrangement Theorem Proof:

Let{a+n }n=0 denote the subsequence of{bn |bn > 0 or bn = 0 an = 0}. Let{a

n }

n=0

denote the subsequence of{cn | cn>0}(from last lemma). Note:

1. a+n 0, an 0 since an0 = bn 0, cn 0.

2.

a+n and

an both diverge because they differ by 0 from

bn,

cnrespectively.

Setm0= n0= 1. Since

a+n diverges we may use the well-ordering principle: m1=min{k N |

kn=0a

+n > c}. Similarly, n1= min{k N |

m1n=0a

+n

kn=0a

n < c}.

Next, ifmp and np are known, we set:

mp+1= min

k N |

p1l=0

mlj=1+ml

a+j

p1l=0

nlj=1+nl

aj +k

j=1+mp

a+j > c

np+1= mink N |

p1l=0

mlj=1+ml

a+j

p1l=0

nlj=1+nl

aj +

mp+1j=1+mp

a+j

kj=1+np

aj < c

Consider the series (a+1 + + a+m1) (a

1 + + a

n+1) + (a

+1+m1

+ + a+m+2) (a1+n1

+ +an2) + . This is clearly a rearrangement of

n=0an.

Write Ap =mp+1

l=1+mpa+l , A

p =

np+1l=1+np

al , and let Sj denote the jth partial sum of

the rearrangement.By construction, lim supj Sj = lim supp(

p+1l=0 A

+l

pl=0A

l ) and

liminfj Sj = liminfp(p

l=0A+l +

pl=0A

l ).

Also,c < p+1l=0 A

+l

pl=0A

l < c+a

+mp+1 and c a

np+1 <

p+1l=0 A

+l

p+1l=0 A

l < c.

Thus, by the squeeze lemma, limp(

p+1l=0 A+l

pl=0Al ) = limp(

pl=0A+l pl=0A

l ) =c, and so limj Sj =c =

n=0a(n) = c.

Remark: One can also rearrange such that

a(n) = .

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21-355 Notes

4 Topology ofR

Our goal in Section 4 is to develop some tools for understanding the topology ofR, which is asort of generalized qualitative geometry.

4.1 Open and Closed Sets

4.1.1 Open Sets

Definition:

1. Fora, b R with a b, we define:

(a, b) ={x R |a < x < b} [a, b) ={x R |a x < b}

(a, b] ={x R |a < x b} [a, b] ={x R |a x b}

2. Forx R and >0, we set B(x, ) = (x , x+) and B[x, ] = [x , x+].We call the set B(x, ) a neighborhood ofxor a ball of radius centered at x.

3. A set E R is open ifx E. >0. B(x, ) E.In other words, every point in Ehas a neighborhood contained in E.

Examples:

1. is vacuously open.

2. R is open because x R. B(x, 1) R.

3. Ifa 0. Thena x < x 0. a /[a, b) and hence B (a, )[a, b).5. [a, b] is not open, nor is (a, b] by previous argument.

6. E= {a} is not open.

7. E= { 1n |n N, n 1} is not open: >0. B(1, )E.

Lemma: IfE R is open A (some index set), then

AE is open.

Proof : Let x

AE. Thenx E0 for some 0 A. Since E0 is open, > 0. B(x, ) E0

AE.

Lemma: IfEi R is open for i [n], n N, then

ni=1Ei is open.

Remark: Infinite intersections of open sets need not be open. Let En = (1

n ,

1

n), n 1.Then

n=1En = {0} which is closed.

4.1.2 Closed Sets

Definition: We say E R is closed iffEc = R\E is open.

Lemma: E is open Ec is closed (by definition).

Examples:

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4.1 Open and Closed Sets 21-355 Notes

1. is closed because c = R is open.

2. R is closed because Rc = is open.

3. [a, b] is closed because [a, b]c = (, a) (b, ) is the union of open sets, and thusopen.

4. [a, b) and (a, b] are not closed because [a, b)c = (, a) [b, ) and B(b, ) [a, b)c ( >0).

5. {a}is closed since {a}c = (, a) (a, ), both open sets.

6. SupposeE R is finite. Write E= {ai| i [n]} where a1< a2< .. . < an. ThenEc = (, a1) (a1, a2) (an1, an) (an, ), all of which are open.

7. E= { 1n |n N, n 1}is not closed. Ec = (, 0]

n=1(

1n+1 ,

1n) (1, ) is not

open because B(0, ) E= { 1n | 1 < n} = = B(0, ) /E

c.

8. E= {0} { 1n |n 1}is closed, asEc = (, 0)

n=1(

1n+1 ,

1n) (1, ) is open.

Lemma:

1. IfE R is closed A, then

AE is closed.

2. IfEi R is closed i [n] then ni=1Ei is closed.Proof : The complement is the union ofEc (open by claim), which is open by previous lemma.

Remark: Example (7) shows that infinite unions of closed sets need not be closed.

4.1.3 Limit Points

Definition: Let E R.

1. A pointx R is a limit point ofE iff. >0.(B(x, ) E)\{x} = .

2. A pointx E is called isolatedif it is not a limit point.

Example: E={ 1n |n 1}. 0 is a limit point, but 1n Eis isolated, since B( 1n , 1n(n+1)) E={ 1n}.

Theorem: LetE R. E is closed every limit point ofEis contained in E.

Proof:

=:AssumeE is closed and x R is a limit point ofE. Ifx Ec then, since Ec is open, > 0. B(x, ) Ec = B(x, ) E= . But this contradicts the fact that x is alimit point ofE; thus x E.

=:Suppose Eis not closed; then Ec is not open and so >0.x Ec. B(x, ) E= .

Since x Ec, (B(x, ) E)\{x}= B (x, ) E= and hence x is a limit point ofE.Then x E Ec, a contradiction; and Eis closed.

Definition: Let{xn}n=l Sfor some set S. We say {xn} is eventually constant if

Nl. xn = xN (n N).

Proposition: LetE R. Thenx is a limit point ofE {xn}n=1 Esuch that the sequence

is not eventually constant and xn x as n .

Proof:

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4.1 Open and Closed Sets 21-355 Notes

=:Suppose x is a limit point ofE, i.e. > 0. (B(x, ) E)\{x} = . Set r1 = 1 andchoose x1 Esuch that x1(B(x, r) E)\{x}.Setrn = min(

1n , |x xn1|) and choose xn (B(x1, rn) E)\{x}.

Thenn 1.{xn}n=1 Eand|x xn1|< |x xn|and |x xn| 0. N 1. n N = |x xn| < . Then {xn | n N} B(x, ) E.If {xn | n N} = {x} then {xn} is eventually constant, a contradiction. Hence={xn |n N}\{x} (B(x, ) E)\{x} = (B(x, ) E)\{x} = , and hence xis a limit point.

Corollary: Let E R. The following are equivalent (proof follows from last theorem):

1. Eis closed.

2. Ifx R is a limit point ofE,x E.

3. If{xn}n=lEis such that xn x as n , then x E.

Corollary: Let E R and E= . Suppose Eis closed.

1. IfEis bounded above, then sup E E, i.e. sup E= max E.

2. IfE is bounded below, then infEE, i.e. infE= min E.

4.1.4 Closure, Interior, and Boundary Sets

Definition: Let E R.

1. LetO(E) ={V R |V Eand V is open} P(R)

C(E) ={C R |E Cand C is closed} P(R).

Note that O(E) and R C(E).2. We defineE0 =

VO(E)V, and call this set the interior ofE.

We define E=

CC(E)C, and call this set the closure ofE.

3. We defineE= E\E0 to be the boundary ofE.

Theorem: LetE R. The following hold:

1. E0 E E

2. E0 is open and E,Eare closed.

3. For every x E,x E0 x E.

4. E= {x R | >0. B(x, ) E= and B (x, ) Ec

= }.5. E is open E=E0, E is closed E= E.

Proof:

1. Trivial.

2. E0 is an arbitrary union of open sets and thus open; E is an arbitrary intersectionof closed sets, so its closed. E= E\E0 = E (R\E0) is the intersection of twoclosed sets, so its closed.

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4.2 Compact Sets 21-355 Notes

3. Trivial.

4. Supposex E. Show the two properties of the set are satisfied via contradiction.Next, assume x in the set, and show that x E.

5. Trivial.

Corollary: Let E R. Then Eis closed EE.

Proof: E is closed = E = E = E E E. On the other hand, if E E thenE E=E0 EE, soE= E.

Theorem (Bolzano-Weierstass, Part 2): Let E R be infinite and bounded. ThenEhas alimit point.

Proof: Since E is infinite we may construct a non-eventually-constant sequence {xn}n=0E. We

do so by choosing x0 E arbitrarily, and xn E\{x0, . . . , xn1} for any n N+. Since E isbounded, the sequence is too, so B-W implies there exists a convergent subsequence{xnk}

k=0E.

This subsequence is not eventually constant by construction, so its limit is a limit point.

4.2 Compact Sets

Definition:

1. Let A be some index set and assume A. V R. We writeV={V}A forthe collection of all of these subsets.

2. IfE R andE

AV, then we say V is a cover ofE.

3. IfV R is open A and V is a cover ofE, we say V is an open cover.

4. LetVbe a cover ofE. We sayW={V}A is asubcover ofE ifA A and Wis a cover ofE.

5. LetVbe a cover ofE. IfA is finite, then W={V}A is afinite subcover ofE,ifW is a subcover ofE.

Examples:

1. EveryE R admits a cover: E=

xE{x}.

2. EveryE R admits an open cover: E

xEB(x, ) for >0.

3. If E is finite and V is an open cover, we claim there is a finite open subcover.Indeed, write E = {ai | 1 i n} and choose Vi such that ai Vi . ThenE

ni=1Vi and {Vi}

ni=1 {V}A. Hence every open cover of a finite set

admits a finite open subcover.

4. E = { 1n | n 1}. V = {B(1n ,

1n(n+1)}

n=1 is an open cover of E. Note that

B(

1

n ,

1

n(n+1)) E=

1

n , so there does not exist a finite subcover.5. E= {0}{ 1n |n 1}. Suppose Vis an open cover ofE. Since 0 E,0 A.0

V0. Since V0 is open, >0. B(0, ) V0. Then B(0, ) E= {1n ||; nN}

where N = min{n N | n 1 }. Hence E\B(0, ) = {1n | 1 n N}. There

existVn for n [N] such that 1n Vn . Then E

Nn=0Vn and Ehas a finite

subcover.

6. Let a < b and E= (a, b). Then V={(a+ 1n+1 , b 1n+1)}nN is an open cover of

E. Since these intervals are nested, there cannot be a finite subcover.

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Definition: Let E R. We say that E is compact if every open cover of E admits a finitesubcover.

Examples:

1. is trivially compact.

2. Ris not compact becauseV={B(0, n)}nN is an open cover that clearly does notadmit a finite subcover ofR.

3. Any finite setE R is compact.

4. (a, b) for a < b is not compact.

5. { 1n |n 1} is not compact.

6. {0} { 1n |n 1} is compact.

Notice in each of our examples of compact sets that the set is closed and bounded.

4.2.1 Heine-Borel Theorem

Theorem: LetK R. Then Kis compact Kis closed and bounded.

Proof:

= Suppose Kis compact.

Notice that

n=1B(0, n) = R (since R is Archimedean) and so K R =

n=1B(0, n).Then {B(0, n)}n=1 is an open cover ofK. Since K is compact, a finite subcover :K

mi=1B(0, ni) for some m N.

Setr= maxi[m]ni. Then Km

i=1B(0, ni) B(0, r) = Kis bounded.

Now we showK is closed. Let x KC. For eachy Kwe setry = 12 |x y|> 0. Then

B(y, ry) B(x, ry) = (y K). Also, {B(y, ry)}yK is an open cover.

K compact = a finite subcover: K n

i=1B(yi, ryi). Set r = mini[n]ri > 0and notice that B(yi, ryi) B(y, r) = . Hence

ni=1B(yi, ryi) B(x, r) = =

K B(x, r) = = B(x, r) KC. This means that KC is open and so Kis closed.

= (Heine-Borel) Suppose K is closed and bounded. IfK= were done, so suppose K= .

Notice that K bounded = infK, sup K R, and K closed = infK, sup K K.In particular, sup K= max K, inf = min K. Let Vbe an open cover ofK.

Let E = {x K | Vadmits a finite subcover ofK[infK, x]} K. Notice thatK [infK, infK] = {infK} is a finite set and hence compact; thus V admits a finitesubcover ofK [infK, infK]. Hence infKE, and so E=. Clearly E is boundedabove by sup K. By LUB property, sup E R and sup E sup K.

We want to show sup E= sup K = max E. Notice thatn 1. xn E K suchthat sup E 1n < xn sup E. Then xn sup E as n , and so sup E K (sinceK is closed).

Write V = {V}A. Since sup E K, 0 A such that sup E V0 . But V0 isopen so > 0. B(sup E, ) V0. By definition, x E. sup E < x sup E.HenceVadmits a finite subcoverofK [infK, x], i.e. K [infK, x]

ni=1Vi . Then

K [infK, sup E]n

i=0Vi = sup EE = sup E= max E.

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Assume for sake of contradiction that max E < max K. Let K = K\n

i=0Vi . K

is closed since its the intersection of closed sets. K = since otherwise K ni=0Vi = max E= max K.

Let y = infK = min K (since K is closed) and note that y > max E. Then K[infK, y] =K[infK, min K]

ni=0Vi{y}. But sincey K

K, Vn+1 Vsuchthaty Vn+1. HenceK [infK, y]

n+1i=0 Vi = y E = max E < y max E,

a contradiction. We then deduce that max E= max K = K=K [min K, max K]is covered by a finite subcover ofV; thus, Kis compact.

Corollary:

1. IfK R is compact and E R is closed, then E K is compact.

2. IfK R is compact and E Kis closed, then Eis compact.

3. IfKi R is compact for i [n], thenn

i=1Ki is compact.

4. IfK R is compact A, then

AK is compact.

4.3 Connected Sets

Definition: We say two sets A, B R are separated ifA B= A B = .A setE R is disconnected ifE= A B such that a= , B= andA, B are separated.If a set is E R is not disconnected, we say its connected.

Examples:

1. (0, 1) and [1, 2) are not separated, though they are disjoint, since (0, 1) [1, 2) =[0, 1] [1, 2) ={1} = .

2. (a, b) and (b, c) fora < b < c are separated, since (a, b) (b, c) = = (a, b) (b, c).Then (a, c)\{b} is disconnected, since (a, c)\{b}= (a, b) (b, c).

3. Similarly, a R.(, a) an (a, ) are separated.

Then R\{a}= (, a) (a, ) is disconnected.

Theorem: LetE R. Then E is connected (x, y Eandx < z < y = zE).

Proof:2 = 1:

If (2) is false then x, y E and z (x, y) such that z / E. ThenE = Lz Rz forLz =E (, z) and Rz =E (z, ). Since x Lz, y Rz, and Lz (, z) andRz (z, ), it follows thatLz and Rz are separated. Hence E is disconnected.

1 = 2

Suppose E is disconnected. Write E= A B with A, B= and A B= A B= .Letx A andy B. Without loss of generality, we assume x < y.

Let z = sup(A [x, y]). Clearly z A and so z / B = z = y = x z y . Ifz /A thenz =x = x < z < y andz /A B=E. Otherwise, ifz A, thenz / B.B is closed, so BC is open; and hence we can find w such that z < w < y, w /B, andw /A. Then x < w < y and w /A B=E. In all cases, then, 2 is true.

Corollary: R, (, a), (, a], (a, ), [a, ), (a, b), (a, b], [a, b),and[a, b] are all connected.

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5 Continuity

5.1 Limits of Functions

Definition: Let E R, f :E R, and p R be a limit point. Let q R.We say limxpf(x) = q or f(x) q as x p iff > 0. > 0. x E 0 < |x p| < =|f(x) q|< .

Examples:

1. E= [0, 1], f(x) =x. Let p = 12 . limx 12 f(x) = 12 .

Proof: Let > 0; choose = > 0. Thenx [0, 1] and 0 < |x 12 | < S =|f(x) 12 |< .

2. E= [0, 1], f(x) =x (for x = 12), f(x) = 37 (for x= 12).

By the proof of (1), the claim still holds.

3. f(x) = xn on E = (0, 1) for 2 n N. 0 is a limit point of E; we claimlimx0xn = 0.Proof : Let >0; choose =1/n >0. Thenx(0, 1) and 0 < |x 0|< =

0< x < = 0< xn < n = = |f(x) 0|= xn < .

4. limxpx= p whenever p is a limit point ofE.

5. Ifx E. f(x) = 1 then limxpf(x) = 1 whenever p is a limit point ofE.

6. Let E = R and f(x) = cos(x). From HW6, | cos(x) 1| x2ex2

. We claimlimx0cos(x) = 1.Proof : Let > 0. Choose= min(1,

/e) > 0. Then for x R, 0 < |x 0| 0. x E, 0 < |x 0| < =|f(x) q|< 1. But x E, |x|< = x= 1n ,

1 < n, and|f(x) q|= |

11/n q|=

|n q|< 1, which is a contradiction.

Definition: Let f : E R for some E R. IfA Ewe define f(A) = {f(x)|x A} R asthe image ofA under f. IfB R we define f1(B) ={x E|f(x) B} as the pre-image ofBunderf.

Lemma: Suppose f :E R. Then A B E = f(A) f(B), andA B R = f1(A) f1(B) E.

5.1.1 Divergence Criteria

Theorem (Divergence Criteria): Let E R, f :E R, p be a limit point ofE, q R. Thefollowing are equivalent:

1. limxpf(x) =q

2. For every open set V R such thatq V, an open set U R with p U suchthat f(U E\{p}) V. (Topological characterization)

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3. If{xn}n=l E satisfies xn = p (n l) and xn p as n , the sequence

{f(xn)}n=l R converges andf(xn) qas n . (Sequential characterization)

Proof:

(1) = (2) :Assume (1) and let V R be open with qV . Since V is open, >0. B(q, ) V.Since limx

pf(x) = q, > 0. x E 0 < |x p| < = |f(x) q| < . Let

U = B(p, ) (an open set). Then x U E\{p} = x E |x p| < =|f(x) q|< = f(x) B(q, ) V. So f(U E\{p}) V as desired.

(2) = (3):Assume (2) and let {xn}

n=l E satisfyxn=p, xn p. Let >0 and set V =B(q, )

(open). From (2), open U such that f(U E\{p}) V and p U. Since U is open, > 0. B(p, ) U. Since xn p as n , N l. n N = ) < |xn p| < = xn U E\{p} = f(xn) V =B(q, ). Hence n N = |f(xn) q|< ,and f(x) qas n .

(1) = (3):Suppose (1) is false; then > 0. > 0. x E with 0 < |x p| < such that|f(x) q| . For n N, n 1, set = 1n to find xn Esuch that 0< |xn p|0. >0. x E |x p|< = |f(x) f(p)|<

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Iff :E R is continuous at each p Ewe say f is continuous on E.

Remarks:

1. In order to be continuous at p E, f must be defined at p. Contrast this tolimxpf(x), in which case p need only be a limit point ofE.

2. Informally one can think of continuous functions as those approximated well nearp byf(p), i.e. f(x) f(p) when x p.

3. In the definition, the value of may depend on the point p. If a function iscontinuous on Ethen for a given >0 the = (p) may vary greatly as p varies.

4. Ifp E is isolated (not a limit point ofE), then f is vacuously continuous at p:x E, |x p|< for small enough = x= p.

Example:

We saw last time that limxpP(x) = P(p) for all polynomials P : R R. Hence >0. >0. x R, 0< |x p|< = |P(x) P(p)|< . Hence P is continuousat p.

Theorem: LetE R, f :E R, p Ebe a limit point ofE. Then:

f is continuous at p limxp f(x) =f(p)

Corollary (Algebra of Continuity): LetE R, f , g : E R, andp E. Assume thatf , g arecontinuous atp. Then the following hold:

1. If, R then f+ g is continuous at p.

2. f g is continuous at p.

3. Ifg (p)= 0 then fg :E\g1({0}) R is well-defined and continuous at p.

Proof: Ifp is isolated, the claim is vacuously true. Assume p is not isolated, i.e. p is a limit pointofE. Then the last theorem and algebra of limits gives the result.

Corollary: LetE R, f , g : E R. Iff , g are continuous on E, then:

1. If, R then f+ g is continuous on E.

2. f g is continuous on E.

3. Ifg (x)= 0 (x E), then fg is continuous on E.

Theorem: Let E, F R, f : E R, g : F R. Assumef(E) F, f is continuous at p E,andg is continuous at f(p) F. Theng f :E R(where (g f)(x) =g(f(x))) is continuous atp. Moreover, iff is continuous on Eand g is continuous on F, then g f is continuous on E.

Proof: Let >0.

Sinceg is continuous atf(p), >0. y Fand |y f(p)|< = |g(y)g(f(p))|< .Since f is continuous at p, >0. x E , |x p|< = |f(x) f(p)|< .

Since f(E) Fwe know that x E, |x p|< = f(x) F, |f(x) f(p)|< =|g(f(x)) g(f(p))|< . Hence, g f is continuous by definition.

Examples:

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1. exp, cos, sin : R Rare continuous onR (proof in HW). YAlso, log : (0, ) Ris continuous on (0, ).

2. Let R and set f : (0, ) R via f(x) =x. Notice thatf(x) = exp( log x).Since log and exp are continuous, f(x) =x is continuous.

Definition: LetE Rand A E. We say A is relatively open in E iffA = U Efor some opensetU R. Similarly, we say A is relatively closed in E iffA = C Efor some closed C R.

Proposition: Let A E R. The following hold:

1. A is relatively open in E x A. >0. B(x, ) A E.

2. A is relatively closed in E A= BC Efor some relatively open B E.

Theorem (Continuity Criteria): Let E R, f :E R. The following are equivalent:

1. f is continuous on E.

2. Ifp Eis a limit point ofE, then limxpf(x) =f(p).3. Ifp E is a limit point ofE and {xn}

n=l E satisfies xn p as n , then

f(xn) f(p) as n .

4. IfV R is open, then f1(V) Eis relatively open in E.5. IfC R is closed, then f1(C) Eis relatively closed in E.

Proof:

(1) (2) (3) follows from the sequential criterion of limits, previous theorem.(4) (5) follows since f1(VC) = (f1(V))C E.

(1) = (4):Let V R be open and choose pf1(V). SinceV is open, >0. B(f(p), ) V.It suffices to show, via previous proposition, that >0. B(p, ) Ef1(V). Sincef is continuous on E, > 0. x E, |x p| < = |f(x) f(p)| < . Thatis, x B(p, ) E = |f(x) f(p)| < = f(x) B(f(p), ) V. HenceB(p, ) E f1(V).

(4) = (1):Let p E, >0, and V =B (f(p), ). Thenf1(B(f(p), ))E is relatively open inE = (by previous proposition) >0. B(p, ) Ef1(B(f(p), )). Then x Eand|x p|< = f(x) B(f(p), ) = |f(x) f(p)|< . Since, pwere arbitrary,we deduce fis continuous on E.

5.3 Compactness and Continuity

Theorem: SupposeK R is compact andf :K R is continuous onK. Thenf(K) is compact.

Proof:

Note that for E R, f(f1(E)) E and E f1(f(E)). Let {V}A be an opencover of f(K). Since f is continuous and V is open, f

1(V) is relatively open inK = f1(V) =U K for some open U R.

Since{V}A coverf(K), we see that{f1(V)}Ais a cover ofK. Then{U}A isan open cover ofK. Since Kis compact, there exists a finite subcover: K

ni=1Ui .

Then Kn

i=1Ui K=n

i=1f1(Vi) = f(K)

ni=1f(f

1(Vi))n

i=1Vi .As we have extracted a finite open subcover off(K), f(K) is compact.

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Extreme Value Theorem: Let K R be compact and f : K R be continuous. Thenx0, x1K such that f(x0) f(x) f(x1) (x K). That is, f(x0) = minxKf(x) = min f(K)and f(x1) = maxxKf(x) = max f(K).

Proof: From last theorem, we know f(K) is compact, so its closed and bounded. From a pre-vious theorem, closed and bounded sets contain their infimum and supremum (and thus min, max).

Definition: Let E R and f :E R. We sayf is uniformly continuous on E iff:

>0. >0. x, y E |x y|< = |f(x) f(y)|<

Remarks:

1. f is uniformly continuous on E = f is continuous on E.

2. The key difference is that for uniform continuity, >0 works for all points in E.

Examples:

1. Let E= (0, 1) and f(x) = 1x . Its trivial thatf is continuous on E, but it is notuniformly continuous.

Proof: Suppose it is; then for = 12 , > 0. x,y (0, 1) |x y | < =|f(x) f(y)|< 12 . ChooseN N such that

1

< N. Thenx = 1n , y= 1n+1 satisfy

|x y|= 1n(n+1) 1n2 < ifn N. Then

12 >|f(x) f(y)|= |n (n + 1)|= 1, a

contradiction.

Definition: A function f :E R is Lipschitz ifx, y E.k >0.|f(x) f(y)| k|x y|.

Claim: Iff is Lipschitz, it is uniformly continuous. Proof: let = k .

Theorem: LetK Rbe compact and f :K R be continuous. Then f is uniformly continuouson K.

5.4 Continuity and Connectedness

Theorem: Let E R be connected and f :E R be continuous on E. IfXE is connected,then f(X) is connected.

Intermediate Value Theorem: Let a c > f(a) for some c R, then x (a, b). f(x) =c.

5.5 Discontinuities

Lemma: If p is a limit point of E R then p is a limit point of E+

p

= E(p, ) or Ep

=E (, p).

Definition: Let E R, f :E R,pbe a limit point ofE,q R.

1. Ifp is a limit point ofEp, we say limxpf(x) =q >0. >0. x Ep,00. >0. x E+p,0< x p < = |f(x) q|< .

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Proposition: Ifp is not a limit point ofE+p then limxpf(x) = limxpf(x). Ifp is not a limitpoint ofEp then limxpf(x) = limxp+f(x).

Proposition: Ifp is both a limit point of either E+p or Ep, then

limxp f(x) =q limxp

f(x) = limxp+

f(x) =q

Definition: SupposeE R,f :E R,p Eis a limit point ofE. Suppose further that p is nota point of continuity off.

1. We sayfhas a simple discontinuity ofp ifp is not a limit point ofE+p and limxpf(x) exists,p is not a limit point ofEp and limxp+f(x) exists, orp is a limit point ofE+p andE

p and limxp+f(x), limxpf(x) both exist.

2. Otherwise, we sayfhas an essential discontinuity ofp.

5.6 Monotone Functions

Definition: Let E R and f :E R. We say:

f is non-decreasing (increasing) ifx, y Eand x < y = f(x) f(y) (f(x)< f(y)), andf is non-increasing (decreasing) ifx, y Eandx < y = f(y) f(x) (f(y)< f(x)).Iff is non-increasing or non-decreasing, f is monotone.

Theorem: Suppose f : (a, b) R is monotone, and let p (a, b). Then limxpf(x) andlimxp+f(x) both exist. Moreover, iffis non-decreasing, then

limxp

f(x) = sup f((a, p)) f(p) inff((p, b)) = limxp+

f(x)

Corollary: Iff : (a, b) R is monotone, then fhas no essential discontinuities.

Example: f(x) =x is non-decreasing and fhas countably many simple discontinuities.

Theorem: Iff : (a, b) R is monotone, thenfhas at most countably many simple discontinuities.

6 Differentiation

6.1 The Derivative

Definition: Assumef : [a, b] R for a < b R. For allx [a, b], the function : (a, b)\{x} R

via (t) = f(t)f(x)

tx is well-defined, and x is a limit point of (a, b)\{x}. If limtx(t) exists wewrite f(x) = limtx(t) and say that f is differentiable at x.

We define f :{x [a, b]| x is differentiable at x} R to be the derivative off. Iff is differen-

tiable x E[a, b], we say fis differentiable on E.Definition (General): Let E R, f : E R, and x E be a limit point of E. Define

: E\{x} R by(t) = f(t)f(x)tx . If limtx(t) exists we say fis differentiable at x, and writef(x) = limtx(t).

Proposition (locality of derivative): Supposef : E R, g : F R, x E F is a limiitpoint ofE F, and that f and g are differentiable at x. Iff = g on E F then f(x) = g(x).This shows that f(x) only depends on the value off near x.

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Proposition (Newtonian approximation): Let f : E R, xEbe a limit point ofE, andL R. Then the following are equivalent:

1. f is differentiable at x and f(x) =L

2. >0. >0. t E |x t|< = |f(t) (f(x) +L(t x))|< |t x|

Proof follows from definition of limtx(t). Newtons approximation says differentiablefunctions are those that can be well-approximated by affine functions +x. Contin-

uous functions are those well-approximated by constants, while differentiable functionsare well-approximated by the next simplest function.

Theorem: Supposef :E R,x Eis a limit point ofE, andf is differentiable at x. Then f iscontinuous atx.

Proof: By definition, ift E\{x} thenf(t)f(x) =(t)(tx). Thenf(t) =f(x)+(t)(tx) andhence limtxf(t) =f(x)+limtx(t)(t x) =f(x) + f(x)0 =f(x). By the limit chracterizationof continuity, we deduce that fis continuous at x.

Remark: The converse fails. Let f(x) = |x| on R. Since ||x| |y| | |x y |, f is Lipschitz

and hence uniformly continuous. However, for x = 0, t > 0 = (t) = |t|0t0 = 1 and

t < 0 = (t) = t0t0 = 1. Then limt0(t) = 1 = limt0+(t) = 1, so f(0) doesnot exist.

Theorem (Algebra of Derivatives): Letf, g: E R be differentiable at x E. Then:

1. f+g : E R is differentiable at x and (f+ g)(x) =f(x) +g(x)

2. f g: E R is differentiable at xand (f g)(x) =f(x)g(x) +f(x)g(x)

3. If g(x) = 0 then fg : E\g1({0}) R is differentiable at x and ( fg )

(x) =g(x)f(x)f(x)g(x)

g(x)2

Examples:

1. f(x) =+x on R = f(x) = limtxf(t)f(x)

tx =(x R).

2. f(x) =xn forn N = f(x) =nxn1. Proof by induction.

3. Every polynomialP(x) =N

n=0anxn is differentiable, andP(x) =

Nn=0nanx

n1.

4. R(x) = P(x)Q(x) is dfiferentiable when P, Q are polynomials at points p R where

Q(p)= 0.

Theorem (Chain Rule): Supposef :E R is differentiable atx E,f(E) F, andg : F Ris differentiable atf(x) F. Thengf :E R is differentiable atxand (gf)(x) =g (f(x))f(x).

6.2 Mean Value Theorems

Definition: Let f : E R. We say thatf has a local maximum at x E if > 0. t E and|x t|< = f(t) f(x). We say fhas a local minimumatx E iffhas a local maximum.

Iffhas either a local max or min at x E, we sayfhas a local extremumat x.

Theorem: Supposef : E R is differentiable at x E and x is a limit point of both E+x andEx. Iffhas a local extremum at x, then f(x) = 0.

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Proof: It suffices to assume that f has a local max at x. Let > 0 such that t E and|x t|< = f(t) f(x). Then t E , 0< x t < = f(t)f(x)tx 0 and 0< t x < =f(t)f(x)

tx 0. So limtxf(t)f(x)

tx =f(x) 0, limtx

f(t)f(x)tx =f

(x) 0, and thus f(x) = 0.

Remark: The result is false ifx is not a limit point of either E+x or Ex. Consider f(x) = x on

E= [0, 1]; fhas a local min at x = 0, local max at x = 1, but f(x) = 1x [0, 1].

Theorem (Monotonicity part 1): Letf :E R be differentiable at x E.

1. Iff is non-decreasing on E, then f(x) 0.

2. Iffis non-increasing on E, then f(x) 0.

Cauchys Mean Value Theorem: Suppose thatf, g: [a, b] R are continuous on [a, b], differ-entiable on (a, b). Then x (a, b).(g(b) g(a))f(x) = (f(b) f(a))g(x).

Proof:

Consider h: [a, b] R via h(x) = (g(b) g(a))f(x) (f(b) f(a))g(x). To prove theresult, it suffices to findx (a, b) such that h(x) = 0 (since h is cont, diff on [a, b] and(a, b)).

Notice thath(a) =g(b)f(a) g(a)f(b) =h(b).

Ifh is constant, thenh(x) = 0 trivially and were done. Assumeh is not constant; thent (a, b). h(t)> h(a) or h(t)< h(a).

If h(t) h(a), then Extreme Value Theorem guarantees that x (a, b). h(x) =max h([a, b]) and Local Extremum Theorem = h(x) = 0.

If h(t) < h(a) then EVT guarantees x (a, b). h(x) = min h([a, b]) and LET =h(x) = 0.

Corollary (Mean Value Theorem): If f : [a, b] R is cont and diff on [a, b], (a, b) thenx (a, b). f(b) f(a) =f

(x)(b a). Proof: Set g(x) =x.

Corollary (Monotonicity part 2): Suppose f : (a, b) R is differentiable on (a, b). Thefollowing hold:

1. x (a, b). f(x)> 0 = f is increasing.

2. x (a, b). f(x) 0 = f is non-decreasing.

3. x (a, b). f(x) = 0 = f is constant.

4. x (a, b). f(x) 0 = f is non-increasing.

5. x (a, b). f(x)< 0 = f is decreasing.

Proof: by MVT, ifa < x1 < x2 < b, then f(x2) f(x1) =f(x)(x2 x1).

6.3 Darbouxs Theorem

Definition: We say a function g : R R is periodic with periodp >0 ifg(x +p) =g(x) (x R).

Theorem (Darboux): Supposef : [a, b] R is differentiable on [a, b] and f(a) < < f (b).Thenx (a, b). f(x) =.

Corollary: Iff : [a, b] R is differentiable on [a, b], then f has no simple discontinuities.

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6.4 LHopitals Rule 21-355 Notes

6.4 LHopitals Rule

Theorem: Suppose f , g: [a, b] R are continuous on [a, b], differentiable on (a, b), and g(x)= 0(x (a, b)). Assume that limxa

f(x)g(x) =L. Iff(a) =g(a) = 0, then limxa

f(x)g(x) =L.

Proof:

We claim first that g(x)= 0 for x (a, b]. Otherwise, g(x) = 0 for some x (a, b] =

0 = g(x)

g(a)

xa = g(z) for some z (a, x), a contradiction. So fg : (a, b] R is well-

defined.

Let{xn}n=l (a, b] satisfyxn a as n . We claim that limn

f(xn)g(xn)

=L. Oncethis is established, the sequential characterization of limits yields the desired result.

To prove the claim, we apply Cauchys Mean Value Theorem on [ a, xn]: yn (a, xn)such that f(yn)g(xn) = f(yn)(g(xn) g(a)) = g(xn)(f(xn) f(a)) = g(xn)f(xn).Then n l.

f(xn)g(xn)

= f(xn)g(xn)

. Since a < yn < xn, the squeeze lemma implies yn a.

Hence limnf(xn)g(xn)

= limnf(xn)g(xn)

= limxaf(x)g(x) =L.

Remarks:

1. The theorem is also true if we take limits att.2. If f, g : (a, b] R and limxaf(x) = limxag(x) = 0, then the theorem still

works.

6.5 Higher Derivatives and Taylors Theorem

Definition: Suppose f : E R is differentiable at x E, and x is a limit point of {y E | f(x) exists}. We say f is twice differentiable at x if f : {y E | f(y) exists} R isdifferentiable atx; and f(x) =f(2)(x) = (f)(x). Similarly, for n N with n >2, we say f is n-times differentiable atx ifx is a limit point of{y E|f(n1)(y) exists}and f(n1) is differentiableat x, in which case f(n)(x) = (f(n1))(x).

Iff(n) exists n N,n 1 we say fis infinitely differentiable at x.

Theorem (Taylor): Supposef : [a, b] R. Assumef(n1) is continuous on [a, b] andf(n) existson (a, b). Let x, y [a, b] with x=y. Thenz (min{x, y}, max{x, y}) such that

f(y) =n1k=0

f(k)(x)

k! (y x)k +

f(n)(z)

n! (y x)n

(called the Taylor polynomial or Taylor approximation).

Proof:

Supposex < y(y < xis handled without loss of generality). LetP(t) =

n1k=0

f(k)(x)k! (t

x)k

, and set M= f(y)

P(y)

(yx)n . It suffices to prove thatM= f(n)(z)

n! for somez (x, y).Defineg(t) =f(t) P(t) M(t x)n, and notice thatg(n)(t) =f(n)(t) n!M. As such,it suffices to show that g (n)(z) = 0 for some z (x, y).

By construction, g(k)(x) = 0 (k = 0, . . . , n 1), and g(y) = 0 (by choice of M).

By Mean Value Theorem, xi (x, y). g(x1) =

g(y)g(x)yx = 0. Similarly, x2

(x, x1). g(x2) =

g(x1)g(x)x1x = 0. Iterating, we eventually findxn1 (x, y). g

(n1)(xn1) =

0. Then 0 =g(n)(z) = g(n1)(xn1)g(n1)(x)

xn1x = 0 for some z (x, xn1).

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21-355 Notes

7 Riemann-Stieltjes Integration

7.1 The R-S Integral

Definition: Let a, b R with a b. A partition of [a, b] is a finite ordered set P ={x0, . . . , xn}such that a= x0 x1 xn =b. Write [a, b] ={P |Pis a partition of [a, b]}. For brevitywell write = [a, b].

Universal Assumptions: Throughout7 we will always assume that:

1. f : [a, b] R is bounded: x [a, b]. m f(x) M, wherem = inff([a, b]), M=sup f([a, b])

2. : [a, b] R (the integrator or weight function) is non-decreasing (in particular, is also bounded)

Definition: For each P [a, b] we associate to fthe following quantities (P ={x0, . . . , xn}):

1. mi= inf{f(x)| x [xi1, xi]} for i [n]2. Mi= sup{f(x)| x [xi1, xi]}fori [n]

3. i = (xi) (xi1) 0 fori [n]We write U(P,f ,) =

ni=1Mii, L(P,f ,) =

ni=1mii.

Uis the upper Riemann-Stieltjes sum, and L is the lower R-S sum.

Remark: Clearly m((b) (a)) =n

i=1m((xi) (xi1)) =n

i=1mi n

i=1Mii Mn

i=1i =M((b) (a)). HenceP [a, b]. m((b) (a))L(P,f ,)U(P,f ,)M((b) (a)).

Definition of Integral: We define

ba f d = sup{L(P,f ,) | P [a, b]}, and

ba f d = inf{U(P,f ,) | P [a, b]}.

Both are well-defined by the remark.Ifba f d=

ba f d then we say fis R-S integrable with respect to , and writeb

af d=ba f d=

ba f d.

We write R([a, b]; ) ={f : [a, b] R |f is bounded,fis R-S integrable with respect to }.

When(x) =x,baf dx is the Riemann integral and we write R([a, b]).

Heuristics: The function assigns different weights to different points in [a, b]. The

intuition is thatbaf d is a weighted Riemann integral. If is continuous then we

have a geometric interpretation of

baf d: consider the curve in R

2 parameterized by

(x(t), y(t)) = ((t), f(t)); ba

f d= area under this curve.

Lemma: Let f(x) =C (x [a, b]). Then f R([a, b]; ) andba f d= C((b) (a)).

Proof: For any P [a, b] we have mi=Mi = C. Hence U(P,f ,) =L(P,f ,) =n

i=1Ci =

C((b) (a)). Soba f d= sup{(L(P,f ,)}= C((b) (a)) = inf{U(P,f ,}=

ba f d.

Definition: IfP, P [a, b] and every point in P is in P, we say P is a refinement ofP.IfP1, P2 [a, b] we define the common refinement P1#P2[a, b] byP1#P2= P1 P2, orderedappropriately.

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7.2 Integrability Criteria 21-355 Notes

Proposition: If P, P [a, b] and P is a refinement of P, then L(P,f ,) L(P, f ) U(P, f , ) U(P,f ,).

Theorem:ba f d

ba f d.

Proof: Let P1, P2 [a, b]. Then L(P1, f , ) L(P1#P2, f , ) U(P1#P2, f , ) U(P 2, f , )

by last proposition. Hence ba f d= sup{L(P1, f , )| P1 [a, b]} U(P2, f , ). Then ba f dinf{U(P2, f , )| P2 [a, b]}=

ba f d.

7.2 Integrability Criteria

Theorem (Riemann)*: f R([a, b]; ) >0.P [a, b]. U(P,f ,) L(P,f ,)< .

Proof:

(1) = (2):

Let >0. By definition,P1, P2 [a, b].ba f d

2 < L(P1, f , ), and U(P2, f , )

ba f d+ 2 . Let P = P1#P2. Then U(P1, f , ) U(P2, f , ) < ba f d+ 2 . HenceU(P,f ,) L(P,f ,)< .

(2) = (1):

For any partition P [a, b] we knowba f d U(P1, f , ) and L(P,f ,)

ba f d.

We also knowba f d

ba f d. Then (2) implies that for > 0 we have P

[a, b]. U(P1, f , ) L(P1, f , ) < . But then 0 ba f d

ba f d < ( > 0)

and soba f d=

ba f d = f R([a, b]; ).

Lemma: Let P [a, b]. The following are true:

1. IfP is a refinement ofP, then U(P, f , ) L(P, f , ) U(P,f ,) L(P,f ,).2. If si, ti [xi1, xi] (i [n]), then 0

ni=1 |f(si) f(ti)|i U(P,f ,)

L(P,f ,).

3. Iff R([a, b]; ), thenti[xi1, xi] = |n

i=1f(ti)ibaf d| U(P,f ,)

L(P,f ,).

Theorem: Iffis continuous on [a, b], then f R([a, b]; ).

Proof:

Note that EVT impliesf is bounded. Let >0 and choose k >0 sok((b) (a))< .Since fis cont on compact [a, b], f is uniformly continuous. Then >0. x, y [a, b]and |x y|< = |f(x) f(y)|< k.

Choose a partition P [a, b] such that xi xi1< (i [n]). Then by EVT,si, ti [xi1, xi] such that mi = f(si), Mi = f(ti). Then U(P,f ,) L(P,f ,) =n

i=1(Mi mi)i=n

i=1(f(ti) f(si))in

i=1ki= kn

k=1i=k((b) (a)) < . Since > 0 was arbitrary, we deduce Riemanns Theorem thatf R([a, b]; ).

Theorem: Suppose that f is monotone and is continuous on [a, b]; then f R([a, b]; ).

Remarks:

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7.2 Integrability Criteria 21-355 Notes

1. If = x, then all monotone functionsfare inR([a, b]), i.e. all monotone functionsare Riemann integrable.

2. Monotone functions dont have to be continuous, sofwith simple discontinuitiesin R([a, b]; ) when is continuous.

3. Monotone functions can have countably infinite sets of discontinuity; R-S integralscan handle infinite discontinuities.

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21-355 Notes