MAT337
Lecture Notes
Yuchen Wang
April 14, 2020
Contents
1 Real Numbers 3
1.1 Discussion: The Irrationality of
√
2 . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 The axiom of completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.4 Consequences of Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.5 Cardinality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.5.1 1-1 Correspondence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.5.2 Countable Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.6 Cantor’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2 Sequences and Series 9
2.1 The Limit of a Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.2 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3 Metric Spaces and the Baire Category Theorem 14
3.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.2 Topology on Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.3 Compactness and Bolzano-Weierstrass Theorem . . . . . . . . . . . . . . . . . . 17
3.4 Completeness of Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.5 Perfect Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.6 Separated and Connected Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.7 Baire’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.8 The Baire Category Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4 Functional Limits and Continuity 19
4.1 Functional Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
4.2 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
4.3 Continuous Functions on Compact Sets . . . . . . . . . . . . . . . . . . . . . . 22
4.3.1 Uniform Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
4.4 Sets of Discontinuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
1
CONTENTS 2
5 the Derivative 24
5.1 Derivatives and the Intermediate Value Property . . . . . . . . . . . . . . . . . 24
5.2 the Mean Value Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
6 Sequences and Series of Functions 27
6.1 Uniform Convergence of a Sequence of Functions . . . . . . . . . . . . . . . . . 27
6.2 Uniform Convergence and Differentiation . . . . . . . . . . . . . . . . . . . . . 29
6.3 Series of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
6.4 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
7 The Riemann Integral 33
7.1 The Definition of the Riemann Integral . . . . . . . . . . . . . . . . . . . . . . . 33
7.1.1 Partitions, Upper Sums, and Lower Sums . . . . . . . . . . . . . . . . . 33
7.1.2 Integrability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
7.1.3 Criteria for Integrability . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
7.2 Integrating Functions with Discontinuities . . . . . . . . . . . . . . . . . . . . . 35
7.3 Properties of the Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
7.3.1 Uniform Convergence and Integration . . . . . . . . . . . . . . . . . . . 36
7.4 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . 37
7.5 Lebesgue’s Criterion for Riemann Integrability . . . . . . . . . . . . . . . . . . 38
7.5.1 Sets of Measure Zero . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
7.5.2 α-continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
7.5.3 Lebesgue’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
1 REAL NUMBERS 3
1 Real Numbers
1.1 Discussion: The Irrationality of
√
2
If we make natural numbers N closed under subtraction, we obtain
Z = {. . . ,−1, 0, 1, . . .}
If we take the closure of Z under division by non-zero numbers, we obtain
Q = {m
n
: m ∈ Z, n ∈ N, (m,n) = 1}
Remark 1.1. (m,n) = 1 means that if d ∈ N divides both m and n, then d = 1.
Theorem 1.1. There is no r ∈ Q s.t. r2 = 2.
Proof. Assume for contradiction that there are m ∈ Z.n ∈ N s.t. mn =
√
2 and (m,n) = 1.
Then m2 = 2n2 so that m2 is an even complete square.
Suppose m = p1 . . . pr where pis are prime numbers. Then 2n
2 = m2 = p21 . . . p
2
r =⇒ p2i = 22.
Then 4|m2 and 2|n2, so n has to be even. Therefore both m and n are even.
Then 2|m and 2|n, which leads to a contradiction that if d ∈ N divides both m and n, then
d = 1.
1.2 Preliminaries
Definition 1.1 (set). A set is any collection of objects.
Definition 1.2 (function). Given two sets A and B, a function from A to B is a rule or
mapping that takes each element x ∈ A and associates with it a single element of B. In this
case, we write (f : A→ B). It is the set of pairs (A,B) ∈ A×B s.t.
1. If (x, y1) ∈ f and (x, y2) ∈ f , then y1 = y2.
2. For all x ∈ A, there is some y ∈ B s.t. f(x) = y.
The set A is said to be the domain of f . The range of f is not necessarily equal to B but
refers to the subset of B given by {y ∈ B : y = f(x) for some x ∈ A}.
Example 1.1 (absolute value function). For every x,
|x| =
{
x x ≥ 0
−x x < 0
Theorem 1.2 (triangle inequality).
|x+ y| ≤ |x|+ |y|
1 REAL NUMBERS 4
Proof.
(x+ y)2 = x2 + y2 + 2xy
≤ |x|2 + |y|2 + 2|x||y|
= (|x|+ |y|)2
=⇒ |x+ y| =
√
(x+ y)2
≤
√
(|x|+ |y|)2
= ||x|+ |y||
= |x|+ |y|

Definition 1.3 (maximum and minimum). Assume set X ⊆ R. Then the maximum (mini-
mum) of X is an element a ∈ X s.t. for all x ∈ X,x ≤ a(x ≥ a).
Definition 1.4 (least upper bound / supremum). The least upper bound of X (denoted by
sup(X)) is a real number a ∈ R s.t.
1. For all x ∈ X,x ≤ a (this means that a is an upper bound for X)
2. If b is an upper bound for X, then a ≤ b
Example 1.2.
max([0, 1]) = 1
min([0, 1]) = 0
sup((0, 1)) = 1
sup(R), sup(N)DNE
1.3 The axiom of completeness
Definition 1.5 (initial segment). X ⊆ Q is said to be an initial segment if
1. X 6= ∅
2. For all x, y ∈ Q, if x < y and y ∈ X, then x ∈ X.
3. X 6= Q
Alternative definition: Let (A,≤) be a well-ordered set. Then the set
{a ∈ A : a < k}
for some k ∈ A is called an initial segment of A.
Definition 1.6 (real numbers). R = {sup(X) : X is an initial segment of Q}
1 REAL NUMBERS 5
Lemma 1.1 (supremum). Suppose A ⊆ R and s ∈ R is an upper bound for A. If ∀ > 0, ∃a ∈
A, a+ > s, then s = sup(A)
Proof. (⇐= ) Assume for contradiction that t ∈ R is an upper bound for A and t < s.
Let = s−t2 . Obviously > 0.
But then ∀a ∈ A, a+ ≤ t+ < s, which is a contradiction.
( =⇒ ) Assume for contradiction that 0 > 0 and ∀a ∈ A, a+ ≤ S)
Then ∀a ∈ A, a ≤ S − 0.
So s− 0 is an upper bound for A, which is a contradiction that a+ > s.
Theorem 1.3 (the Axiom of Completeness). If X ⊂ R is bounded above, then X has a least
upper bound.
Proof. For x ∈ X, let Ax be the initial segment of Q corresponding to x.
Since X is bounded above, pick b ∈ R s.t. ∀x ∈ X,x < b. Then b /∈ ∪
x∈X
Ax. Note that ∪
x∈X
Ax
is an initial segment of Q. Then sup( ∪
x∈X
Ax) is sup(X).
1.4 Consequences of Completeness
Definition 1.7 (nested sequence of sets). Assume 〈An : n ∈ N〉 is a sequence of sets.
〈An : n ∈ N〉 is said to be nested if
An+1 ⊆ An
Theorem 1.4 (Nested Interval Property). Assume 〈In : n ∈ N〉 is a nested sequence of closed
intervals of R. Then
∩
n
In 6= ∅
Proof. Let [an, bn] = In where an, bn ∈ R.
Since 〈In|n ∈ N〉 is nested,
an ≤ an+1 ≤ bn+1 ≤ bn (†)
for all n ∈ N
Let A = {an : n ∈ N}.
Note that b1 is an upper bound for A. So A has a supremum in R.
We claim that sup(A) ∈ ∩
n
In.
By (†), for all n ∈ N, sup(A) ≤ bn
Obviously, for all n ∈ N, sup(A) ≥ an
So ∀n ∈ N, an ≤ sup(A) ≤ bn.
Therefore ∀n ∈ N, sup(A) ∈ [an, bn].
Example 1.3.
∩
n∈N
(0,
1
n
) = ∅
∩
n∈N
[0,
1
n
] = {0}
Theorem 1.5 (Archimedian Property). We have
1. For every y ∈ R, there is n ∈ N s.t. y ≤ n.
1 REAL NUMBERS 6
2. For every y > 0, there is n ∈ Ns.t. 1n < y.
Proof. (1) Assume for contradiction that N is bounded in R.
Let α = sup(N). Then there is a natural number n ∈ N s.t. n > α− 1.
But then n+ 1 > (α− 1) + 1 = α, which is a natural number greater than α, contradiction.
(2) Exercise.
Theorem 1.6 (density of Q in R). For every two real numbers a and b with a < b, there
exists a rational number r satisfying a < r < b.
Proof. Let n ∈ N s.t. 1n < b− a, 1 < nb− na.
Let m ∈ Z s.t. na < m < nb.
Then a < mn < b.
Pick r = mn and we are done.
1.5 Cardinality
“The size of a set”
1.5.1 1-1 Correspondence
Definition 1.8 (one-to-one and onto). A function f : A → B is one-to-one (1-1) if a1 6= a2
in A implies that f(a1) 6= f(a2) in B. The function f is onto if, given any b ∈ B, it is possible
to find an element a ∈ A for which f(a) = b.
Proposition 1.1. If f : A→ B and g : B → C is 1-1, then g ◦ f : A→ C is 1-1.
Remark 1.2. If a function f : A→ B is both 1-1 and onto, then there is a 1-1 correspondence
between two sets.
Definition 1.9 (the same cardinality). The set A has the same cardinality as B if there exists
f : A→ B that is 1− 1 and onto. In this case, we write A ∼ B.
Proposition 1.2. If A ∼ B,B ∼ C, then A ∼ C
Proposition 1.3. If Card(A) ≤ Card(B) ≤ Card(C), then Card(A) ≤ Card(C)
1.5.2 Countable Sets
A set A is countable if N ∼ A. An infinite set that is not countable is called an uncountable
set.
Theorem 1.7. The set Q is countable.
Proof. Set A1 = {0} and for each n ≥ 2, let An be the set given by
An = {±p
q
: where p, q ∈ N are in lowest terms withp+ q = n}
e.g. A2 = {11 , −11 }, A3 = {12 , −12 , 21 , −21 }
1 REAL NUMBERS 7
The above correspondence is onto because every rational number appears in the correspon-
dence exactly once. The above correspondence is 1-1 because AN were constructed to be
disjoint so that no rational number appears twice.
Theorem 1.8. The set R is uncountable.
Proof. Assume for contradiction that there does exist a bijection function f : N→ R.
Let x1 = f(1), x2 = f(2) and so on. Then since f is onto, can write
R = {x1, x2, x3, x4, . . .} (1)
and be confident that every real number appears somewhere on the list.
We will now use the Nested Interval Property to produce a real number that is not there.
Let I1 be a closed interval that does not contain x1. given an interval In, construct In+1 to
satisfy In+1 ⊆ In and xn+1 /∈ In+1.
If xn0 is some real number from the list in (1), then we have xn0 /∈ In0 , and it follows that
xn0 /∈ ∩∞n=1In
Since we are assuming that the list in (1) contains every real number, then
∩∞n=1In = ∅
However, the NIP asserts that ∩∞n=1In 6= ∅, which is a contradiction.
Theorem 1.9. If A ⊆ B and B is countable, then A is either countable or finite.
Theorem 1.10. We have
(i) If A1, A2, . . . , Am are countable sets, then the union A1 ∪A2 ∪ . . . ∪Am is countable.
(ii) If An is a countable set for each n ∈ N, then ∪∞n=1An is countable.
Theorem 1.11. The open interval (0, 1) = {x ∈ R : 0 < x < 1} is uncountable.
1.6 Cantor’s Theorem
Notation 1.1. Given a set A, the power set P (A) refers to the collection of all subsets of A.
Theorem 1.12 (Cantor’s Theorem). Given any set A, there does not exist a function f :
A→ P (A) that is onto.
1 REAL NUMBERS 8
Proof. Assume, for contradiction, that f : A→ P (A) is onto. For each element a ∈ A, f(a) is
a particular subset of A. The assumption that f is onto means that every subset of A appears
as f(a) for some a ∈ A. To arrive at a contradiction, we will produce a subset B ⊆ A that is
not equal to f(a) for any a ∈ A.
Construct B using the following rule. For each element a ∈ A, consider the subset f(a). This
subset of A may contain the element a or it may not. This depends on the function f . If f(a)
does not contain a, then we include a in our set B: Let
B = {a ∈ A : a /∈ f(a)}
Since we have assumed that our function f : A → P (A) is onto, it must be that B = f(a′)
for some a′ ∈ A.
Case 1 a′ ∈ B
Then a′ /∈ f(a′) = B, a contradiction.
Case 2 a′ /∈ B
Then a′ ∈ f(a′) = B, a contradiction.
Theorem 1.13 (Schro¨der-Bernstein Theorem). If there are 1-1 functions f : A → B and
h : B → A, then there is a bijection g : A→ B.
Proof. Claim: the statement of the theorem is equivalent to the following:
If B ⊆ A and f : A→ B is 1− 1, then there is a bijection g : A→ B. (*)
proof of claim: theorem =⇒ (*):
Take h : X → Y with h(x) = x, then X ⊆ Y .
(*) =⇒ theorem:
Let f : A → B and h : B → A be 1-1 functions, as in the theorem. We need to show that
there is bijection g : A→ B.
Notice that A ⊆ h(B) and h ◦ f : A→ h(B) is a 1-1 function. So by (*), there is a bijection
g0 : A→ h(B).
But h : B → h(B) is also a bijection. So g = h−1 ◦ g0 : A → B is a bijection (using the fact
that bijections are closed under compositions).
Now it suffices to prove (*).
Assume set X ⊆ Y and f : Y → X. Let W = ⋃∞n=0 fn(Y \X).
Define g : Y → X by:
• If y ∈W , then g(y) = f(y)
• If y ∈ Z := Y \W , then g(y) = y
We need to show that g : Y → X is a well-defined bijection.
Since f is 1-1, for all m < n, fm(Y \X) ∩ fn(Y \X) = ∅
2 SEQUENCES AND SERIES 9
Note that
Y \W = Y \
∞⋃
n=0
fn(Y \X)
= [Y \ (Y \X)] \
∞⋃
n=1
fn(Y \X)
= X \
∞⋃
n=1
fn(Y \X)
Therefore for all y ∈ Y, g(y) ∈ X.
(Show g is 1-1) Now assume y1, y2 ∈ Y and g(y1) = g(y2). We show that y1 = y2.
Case 1 y1, y2 ∈W
Then g(y1) = g(y2) =⇒ f(y1) = f(y2) =⇒ y1 = y2.
Case 2 y1 ∈W but y2 ∈ Y \W
Then g(y1) = g(y2) =⇒ f(y1) = y2
Note that if y1 ∈W , then for some n ≥ 0, y1 ∈ fn(Y \X)
Then y2 ∈ fn+1(Y \X) ⊆W
So y2 ∈W , which leads to a contradiction.
Case 3 y1, y2 are both in Z := Y \W
Then g(y1) = g(y2) =⇒ y1 = y2.
Therefore by case 1,2,3, g is 1-1.
(Show g is onto) Let x ∈ X. We need to find y ∈ Y s.t. g(y) = X.
If x ∈ Z, take y = x.
If x ∈ ⋃∞n=1 fn(Y \X), then fix n ∈ N s.t. x ∈ fn(Y \X).
But fn(Y \X) = f(fn−1(Y \X))
Pick y ∈ fn−1(Y \X) s.t. f(y) = x.
Then y ∈W and g(y) = x. Therefore g is onto.
2 Sequences and Series
2.1 The Limit of a Sequence
Definition 2.1 (sequence). A sequence is a function whose domain is N.
Definition 2.2. Let (X, d) be a metric space. A sequence (Xn) ⊆ X converges to an element
x ∈ X if ∀ > 0, ∃N ∈ N, n ≥ N =⇒ d(xn, x) < .
Key property: If lim
n→∞xn = x, limn→∞xn = y, then x = y.
Proof. WTS d(x, y) = 0
Let > 0. We will show that d(x, y) < .
Since lim
n→∞xn = x, then ∃N1,∀n ≥ N1, d(xn, x) <

2
Since lim
n→∞xn = y, then ∃N2,∀n ≥ N2, d(xn, y) <

2
Take n ≥ max(N1, N2), then d(x, y) ≤ d(xn, x) + d(xn, y) < 2 + 2 = .
2 SEQUENCES AND SERIES 10
Proposition 2.1. Suppose (X, d) is a metric space, (X, τ) is a topological space, and F ⊆ X.
If lim
n→∞xn = x, (xn) ⊆ F and F is closed, then x ∈ F .
Proof. Suppose x /∈ F , i.e., x ∈ X \ F .
Since F is closed, then X \ F is open, so there is > 0 s.t. B(x) ⊆ X \ F .
Let N be such that ∀n ≥ N, d(xn, x) < .
Then xn ∈ B(x), which implies that (xn) ⊆ X \ F , a contradiction.
Proposition 2.2. Suppose (X, d) is a metric space and F ⊆ X. If F is not closed, then there
exists (xn) ⊆ F and x /∈ F s.t. lim
n→∞xn = x.
Proof. If F is not closed, then X \ F is not open, so there is x ∈ X \ F s.t. B(x) 6⊆ X \ F
for all > 0.
Take xn ∈ B1/n(x) \ (X \ F ) = B1/n(x) ∩ F for each n ∈ N, then (xn) ⊆ F and lim
n→∞xn = x.

Definition 2.3 (Cauchy sequence). A sequence (xn) in a metric space (xn) in a metric space
(X, d) is a Cauchy sequence if ∀ > 0, ∃N ∈ N,m, n ≥ N =⇒ d(xm, xn) < .
Proposition 2.3. A convergent sequence is Cauchy.
Proof. Let (xn) be a convergent sequence, so that lim
n→∞xn = x. To check (xn) is Cauchy, let
> 0. We need to find N s.t. ∀m,n ≥ N, d(xn, xm) < .
Apply lim
n→∞xn = x to

2 , we get N s.t. ∀n ≥ N, d(x, xn) < 2 .
Notice that N works for Cauchy:
Take m,n ≥ N , then
d(xn, xm) ≤ d(xn, x) + d(x, xm) <
2
+

2
=
Remark 2.1. When X = R with the usual metric, A Cauchy sequence is convergent (the
converse is true).
In general not true. For example, X = R \ {0}, d(x, y) = |x− y|, (xn) = 1n .

Definition 2.4 (monotone sequence). (xn) ⊆ R is monotone if either xn ≤ xm, n ≤ m, or
xn ≥ xm, n ≤ m.
Theorem 2.1 (Monotone Subsequence Theorem). Every sequence (xn) ⊆ R has a monotone
subsequence. prove this
Fact 2.1. If an ≤ bn for all n, a = lim
n→∞an, b = limn→∞bn, then
a ≤ b
Proof. Suppose for contradiction that a > b. Let = a−b2 .
We know ∃N1 s.t. an ∈ B(a) for n ≥ N1 and ∃N2 s.t. bn ∈ B(b) for n ≥ N2. Take
n > max(N1, N2), then we have
bn <
a+ b
2
< an
which is a contradiction.
2 SEQUENCES AND SERIES 11
Theorem 2.2 (Algebraic limit theorem). Suppose a = lim
n→∞an, b = limn→∞bn, then:
1. a+ b = lim
n→∞(an + bn)
2. ab = lim
n→∞anbn
3. ab = limn→∞
an
bn
, and b 6= 0.
Fact 2.2. Monotone bounded sequence (xn) converges to its supremum or infimum.
Proof. We only prove the supremum case.
Fix > 0, let s = sup{xn : n ∈ N}. We have s− < s and thus s− is not an upper bound
of (xn). Therefore, there is N s.t. xN > s− .
Take n ≥ N , then we have
xn ≥ xN > s−
Therefore, we have |xn − s| < .
Definition 2.5 (limit supremum). We define
lim sup
n→∞
xn = inf{ym : m ∈ N}
where ym = sup{xn : n ≥ m}.
Alternatively,
lim sup
n→∞
xn = lim
m→∞ supn≥m
xn
Definition 2.6 (limit infimum).
lim inf
n→∞ xn = sup{zm : m ∈ N}
where zm = inf{xn : n ≥ m}.
Alternatively,
lim inf
n→∞ xn = limm→∞ infn≥m
xn
2.2 Series
Definition 2.7. We define
Sn =
n∑
k=1
ak, lim
n→∞Sn =
∞∑
k=1
ak
We call
∑∞
k=1 ak a summable series if the limit exists, i.e.,
∃A, ∀ > 0, ∃Ns.t.∀n ≥ N, |Sn −A| <
Property 2.1 (Cauchy criterion for series).
∑∞
k=1 is summable iff
∀ > 0, ∃N s.t. ∀n ≥ m ≥ N, |Sn − Sm| =
∣∣∣∣∣
n∑
k=m+1
ak
∣∣∣∣∣ <
2 SEQUENCES AND SERIES 12
Corollary 2.1. If
∑∞
k=1 ak is summable, then |ak| → 0.
Proof. We have |ak| = |sk − sk−1| < for k > N .
Example 2.1.
∑∞
k=1
1
k2
is summable.
Proof.
Sm = 1 +
1
4
+
1
9
+ . . .+
1
m2
< 1 +
1
2 · 1 +
1
3 · 2 + . . .+
1
m(m− 1)
= 1 + (1− 1
2
) + (
1
2
− 1
3
) + . . .+ (
1
m− 1 −
1
m
)
= 1 + 1− 1
m
< 2

Example 2.2.
∑∞
k=1
1
k =∞
Proof. We have
∞∑
k=1
1
k
= (1/2) + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + . . .
= 1 + (1/2) + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + . . .
= 1 + 1/2 + 1/2 + 1/2 + . . .
→∞

Theorem 2.3 (Algebraic limit theorem for series). Suppose
∑∞
k=1 ak = A,
∑∞
k=1 bk = B, c ∈
R, then
1.
∑∞
k=1 cak = cA
2.
∑∞
k=1(ak + bk) = A+B
Proof. (1) We want to show ∀ > 0,∃N s.t. ∀n ≥ N, |∑∞k=1 cak − cA| < .
We know ∀0 > 0,∃N0 s.t. ∀n ≥ N0 , |
∑∞
k=1 ak −A| < 0.
Take 0 =

|c| , then we have∣∣∣∣∣
∞∑
k=1
cak − cA
∣∣∣∣∣ = |c|
∣∣∣∣∣
∞∑
k=1
ak −A
∣∣∣∣∣ < |c| · |c| =

Property 2.2 (Order comparison test). Suppose bk ≥ ak ≥ 0, ∀k.
2 SEQUENCES AND SERIES 13
1. If
∑∞
k=1 bk <∞ , then
∑∞
k=1 ak <∞.
2. If
∑∞
k=1 ak =∞ , then
∑∞
k=1 bk =∞.
Definition 2.8 (geometric series). We call a series a geometric series if it is of the form
∞∑
k=1
ark
Note that the geometric series converges to a1−r whenever r
m → 0 iff |r| < 1.
Definition 2.9 (absolutely convergence).
∑∞
k=1 ak is absolutely convergent if
∑∞
k=1 |ak| <∞.
Definition 2.10 (conditionally convergence).
∑∞
k=1 ak is conditionally convergent if
∑∞
k=1 ak <
∞, but ∑∞k=1 |ak| =∞
Example 2.3 (alternating series).
∑∞
k=1
(−1)k+1
k <∞ but
∑∞
k=1
∣∣∣ (−1)k+1k ∣∣∣ = ∑∞k=1 1k =∞
Property 2.3 (Absolute convergence test). If
∑∞
k=1 |ak| <∞, then
∑∞
k=1 ak <∞.
Proof. We use Cauchy criterion for
∑∞
k=1 ak: we want to show
∀ > 0,∃N s.t. ∀n ≥ m ≥ N,
∣∣∣∣∣
n∑
k=m+1
ak
∣∣∣∣∣ <
Let > 0.
Since
∑∞
k=1 |ak| <∞, then we know that ∃N s.t. ∀n ≥ m ≥ N ,∣∣∣∣∣
n∑
k=1
|ak| −
m∑
k=1
|ak|
∣∣∣∣∣ <
Then ∣∣∣∣∣
n∑
k=m+1
ak
∣∣∣∣∣ =
∣∣∣∣∣
n∑
k=1
ak −
m∑
k=1
ak
∣∣∣∣∣
≤
n∑
k=1
|ak| −
m∑
k=1
|ak|
≤
∣∣∣∣∣
n∑
k=1
|ak| −
m∑
k=1
|ak|
∣∣∣∣∣
<

Property 2.4 (Alternating series test). Suppose a1 ≥ a2 ≥ . . . ≥ 0, lim
k→∞
ak = 0, then∑∞
k=1(−1)k+1ak <∞.
3 METRIC SPACES AND THE BAIRE CATEGORY THEOREM 14
Proof. We want to show {Sn} = {
∑n
k=1(−1)k+1ak} is Cauchy:
∀ > 0,∃N s.t. ∀m,n ≥ N, |Sn − Sm| <
Let > 0.
Suppose n > m, then |Sn − Sm| = |am+1 − am+2 + . . .+ (−1)n−m+1an|.
Since (an) is a non-negative decreasing sequence, then
am+1 − am+2 + . . .+ (−1)n−m−1an = am+1 − (am+2 − am+3)− (am+4 − am+5)− . . .
≤ am+1
Since lim
k→∞
ak = 0,∃N s.t. ∀m+ 1 ≥ N, am+1 < .
Thus 0 ≤ |Sn − Sm| ≤ am+1 < .

Property 2.5 (Ratio test). Given
∑∞
k=1 ak s.t. ak 6= 0 for all k.
If lim
n→∞
∣∣∣an+1an ∣∣∣ = r < 1, then ∑∞k=1 |ak| <∞
Proof. Define S := {n ∈ N :
∣∣∣an+1an ∣∣∣ ≥ r′}, then S contains finitely many elements of N. (If S
were to be infinite set, if we take = r′− r, then
∣∣∣an+1an ∣∣∣− r ≥ r′− r for infinitely many terms
which contradicts that r is the point of convergence.
Therefore, S′ = {n ∈ N :
∣∣∣an+1an < r′∣∣∣ contains all but finitely many elements of N. Let
N = 1 + maxS, then ∀n ≥ N,
∣∣∣an+1an < r′∣∣∣ < r′ =⇒ |an+1| < r′|an|.
Since 0 < r′ < 1,
∑∞
n=1(r
′)n converges which implies |aN |
∑∞
n=1(r
′)n converges. We have∑∞
n=1 |an| =
∑N
n=1 |an|+
∑∞
n=N+1 |an| < C + |aN |
∑∞
n=N+1(r
′)n−N converges, by comparison
test. Hence
∑∞
n=1 |an| converges.
understand the
last two lines
of the proof
Definition 2.11 (rearrangement). Let
∑∞
k=1 ak be a series. A series
∑∞
k=1 bk is called a
rearrangement of
∑∞
k=1 ak if ∀n, !∃k s.t. bk = an.
3 Metric Spaces and the Baire Category Theorem
3.1 Basic Definitions
Definition 3.1 (metric and metric space). Given a set X, a function d : X × X → R is a
metric on X if for all x, y ∈ X:
1. d(x, y) ≥ 0 with d(x, y) = 0 if and only if x = y;
2. d(x, y) = d(y, x);
3. for all z ∈ X, d(x, y) ≤ d(x, z) + d(z, y)
A metric space is a set X together with a metric d.
Example 3.1. The set R considered with d : R2 → [0,∞), (x, y) 7→ |x− y| is a metric space.
3 METRIC SPACES AND THE BAIRE CATEGORY THEOREM 15
Example 3.2. In general, Rn considered with the Euclidean distance is a metric space.
d(x,y) =
√√√√ n∑
i=1
(xi − yi)2
Example 3.3. Let x be a set. The discrete metric d on X is defined by
d(x, y) =
{
0 x = y
1 x 6= y
Fact If (X, d) is a metric space, d′(x, y) = max{1, d(x, y)} for all x, y ∈ X, then (X, d′) is
also a metric space.
Example 3.4. Let X = {f : A→ R}
d(f, g) = sup{|f(x)− g(x)| : x ∈ A}
if the supremum exists.
Definition 3.2. Let (X, d1) and (Y, d2) be metric spaces. A function f : X → Y is
continuous at x ∈ X if ∀ > 0,∃δ > 0, d1(x, y) < δ =⇒ d2(f(x), f(y)) < .
3.2 Topology on Metric Spaces
Definition 3.3 (open ball). An open ball (or -neighbourhood) with radius r and center x is
Br(x) = {y ∈ X : d(x, y) < r}
Definition 3.4 (open set). A set U ⊆ X is open iff
∀x ∈ U,∃ > 0 s.t. B(x) ⊆ U
Example 3.5. B(x) is open.
Proof. Fix x ∈ X and > 0. We want to show: ∀y ∈ B(x),∃δ > 0 s.t. Bδ(y) ⊆ B(x).
Take y ∈ B(x), then d(x, y) < . Take δ = − d(x, y) > 0. Take any z ∈ Bδ(y), we have
d(x, z) ≤ d(x, y) + d(y, z) < d(x, y) + − d(x, y) =
Thus z ∈ B(x) so Bδ(y) ⊆ B(x).
Definition 3.5 (topological space). A topological space is a pair (X, τ), where X is a set and
τ a subset of the power set of X which we call open such that
1. ∅, X ∈ τ
2. U1, . . . , Un ∈ τ =⇒
⋂n
i=1 Ui ∈ τ
3. U1, . . . , Un ∈ τ =⇒
⋃n
i=1 Ui ∈ τ
Example 3.6. (X, {∅, X})
3 METRIC SPACES AND THE BAIRE CATEGORY THEOREM 16
Example 3.7. (X,P (X)) is a discrete topological space, where P (X) is the power set of X.
Example 3.8. Given (X, d) a metric space, define τd : a set U ∈ τd ⇐⇒ ∀x ∈ U,∃ >
0, B(x) ⊆ U . Then τd is a topology.
Proof. (1) First, ∅, X ∈ τd since ∀x ∈ ∅, B1(x) ⊆ ∅ and ∀x ∈ X,B1(x) ⊆ X.
Then suppose U1, . . . , Un ∈ τd.
(2) we want to show:
U =
n⋂
i=1
Ui ∈ τd ⇐⇒ ∀x ∈ U,∃ > 0 s.t. B(x) ⊆ U
Since x ∈ U , then ∀i = 1, . . . , n, x ∈ Ui : ∃i > 0 s.t. Bi(x) ⊆ Ui.
Take = min
1≤i≤n
i, thus B(x) ⊆ Ui ∀i. Hence B(x) ⊆ Ui ⊆ U .
(3) We also want to show:
n⋃
i=1
Ui ∈ τd ⇐⇒ ∀x ∈ U,∃ > 0 s.t. B(x) ⊆ U
Let x ∈ U , then there is some Ui s.t. x ∈ Ui. Since Ui ∈ τd, then ∃ > 0 s.t. B(x) ⊆ Ui ⊆ U .
Therefore, τd is a topology.
Definition 3.6. A subset F of a topological space (X, τ) is closed if X \ F is open.
Property 3.1. Given a topological space (X, τ) and a subset F of it, we have:
1. ∅, X are closed
2. If F1, . . . , Fn are closed, then
⋃n
i=1 Fi is closed
3. If F1, . . . , Fn are closed, then
⋂n
i=1 Fi is closed
Definition 3.7 (topological closure and interior). Given a topological space (X, τ), where
τ ⊆ P (X), and a set F ⊆ X, the topological closure of F is the minimal closed superset of
F , i.e.,
F¯ =
⋂
{H : H is closed, H ⊇ F}
The interior of F is the maximal open subset of F , i.e.,
F ◦ =
⋃
{U : U is open, U ⊆ F}
Example 3.9. Given (X, d) a metric space, define τd : a set U ∈ τd ⇐⇒ ∀x ∈ U,∃ >
0, B(x) ⊆ U . Suppose F ⊆ X, then
F¯ = {x ∈ X : ∀ > 0, B(x) ∩ F 6= ∅} = { lim
n→∞xn : (xn) ⊆ F, limn→∞xn exists}
and
F ◦ = {x ∈ X : ∃ > 0, B(x) ⊆ F} =
⋃
{B(x) : > 0, x ∈ F,B(x) ⊆ F}
3 METRIC SPACES AND THE BAIRE CATEGORY THEOREM 17
3.3 Compactness and Bolzano-Weierstrass Theorem
Definition 3.8 (compactness). A subset K of a metric space (X, d) is compact if every
sequence in K has a convergent subsequence that converges to a limit in K.
Example 3.10. (R, |x− y|) is not compact (e.g. (xn) = n)
Example 3.11. ([0, 1], |x− y|) is compact.
Property 3.2. If (X, d) is compact, then it is bounded, i.e. ∃M s.t. x, y ∈ X, d(x, y) ≤M .
Property 3.3. If Y ⊆ X, (X, d) is a metric space, and (Y, d) is compact, then Y is closed in
X.
Property 3.4. IfK1 ⊇ K2 ⊇ . . . are compact and nonempty subsets ofX, thenK =
⋂∞
n=1Kn
is compact and nonempty.
Theorem 3.1 (Bolzano-Weierstrass theorem). A subset Y of R is compact iff closed and
bounded.
Alternative formation: Every bounded subsequence contains a convergent subsequence.
Remark 3.1. The theorem is true for Rn but is false for infinite dimension.
Theorem 3.2 (Heine-Borel Theorem). Let K be a subset of a metric space (X, d). The
following statements are equivalent:
1. K is compact.
2. K is closed and bounded.
3. Every open cover K ⊆ ⋃i∈I Ui for K has a finite subcover K ⊆ ⋃nl=1 Uil .
3.4 Completeness of Metric Spaces
Definition 3.9 (completeness of metric spaces). A metric space (X, d) is complete if every
Cauchy sequence in X converges to an element of X.
Example 3.12. R, d(x, y) = |x− y|
Example 3.13. (X, d), d discrete metric.
Example 3.14. C[0, 1], d(f, g) = sup
x∈[0,1]
|f(x)− g(x)| = ||f − g||∞
Example 3.15. (NN, d), d((xn), (yn)) = 1min{n:xn 6=yn}
where NN = {x : N→ N}.
3 METRIC SPACES AND THE BAIRE CATEGORY THEOREM 18
3.5 Perfect Sets
Definition 3.10 (perfect set). Let (X, d) be a metric space. P ⊆ X is perfect if it is closed,
nonempty, and for every open U ⊆ X,U ∩ P is not empty and has at least two elements.
Example 3.16. S = [0, 1] ∪ {32} ∪ [2, 3] is not perfect.
Property 3.5. Perfect subsets P of a complete metric space are not countable.
Example 3.17 (Cantor set). Let C0 be the closed interval [0, 1], and define C1 to be the set
that results when the open middle third is removed; that is,
C1 = C0 \ (1
3
,
2
3
) = [0,
1
3
] ∪ [2
3
, 1]
Now construct C2 in a similar way by removing the open middle third of each of the two
components of C1:
C2 = ([0,
1
9
] ∪ [2
9
,
1
3
]) ∪ ([2
3
,
7
9
] ∪ [8
9
, 1])
Continue this process inductively. For each n = 0, 1, 2, . . ., we get a set Cn consisting of
2n closed intervals each having length (13)
n. Finally, we define the Cantor set C to be the
intersection
C =
∞⋂
n=0
Cn
Remark 3.2. As follows
• Since we are always removing open middle thirds, then at each stage, endpoints are
never removed. Thus, C at least contains the endpoints of all of the intervals that make
up each of the sets Cn.
• The Cantor set has zero length.
• The Cantor set is uncountable, with cardinality equal to the cardinality of R.
3.6 Separated and Connected Sets
Definition 3.11 (separated sets). Let (X, d) be a metric space, A 6= ∅, B ⊆ X. A and B are
separated if A¯ ∩B = B¯ ∩A = ∅.
Definition 3.12 (connected sets). A set C ⊆ X is connected if for every decomposition
C = A ∪B s.t. A,B 6= ∅, A and B are not separated, i.e. A¯ ∩B 6= ∅ or B¯ ∩A 6= ∅.
Property 3.6. C ⊆ R is connected iff
∀a, b ∈ C, [a, b] ⊆ C
Proof. Let C = A ∪ B, a0 ∈ A, b0 ∈ B, a0 < b0. We define I0 = [a0, b0], c0 = a0+b02 . Define
I1 = [a0, c0], . . .. We have x ∈ A¯ ∩B or B¯ ∩A. Is this com-
plete?
4 FUNCTIONAL LIMITS AND CONTINUITY 19
3.7 Baire’s Theorem
Definition 3.13 (dense). A set A ⊆ X is dense in the metric space (X, d) if A¯ = X.
Definition 3.14 (nowhere-dense). A subset E of a metric space (X, d) is nowhere-dense in
X if E¯◦ is empty.
i.e., A nowhere-dense set of a metric space is a set whose closure has empty interior.
Remark 3.3. It is a set whose elements are not tightly clustered anywhere.
Example 3.18. Z is nowhere-dense in R.
Example 3.19. S = { 1n : n ∈ N} is nowhere-dense in R.
S¯ = S ∪ {0}, which has empty interior.
Theorem 3.3 (Baire’s Theorem). The set of real numbers R cannot be written as the count-
able union of nowhere-dense sets.
Remark 3.4. Baire’s Theorem asserts that the only way to make R from a countable union
of arbitrary sets is for the closure of at least one of these sets to contain an interval.
3.8 The Baire Category Theorem
Theorem 3.4. Let (X, d) be a complete metric space, and let {On} be a countable collection
of dense, open subsets of X. Then,
⋂∞
n=1{On} is not empty. prove this
Theorem 3.5 (Baire Category Theorem). A complete metric space cannot be written as the
countable union of nowhere-dense sets. prove this
Remark 3.5. This result is called the Baire Category Theorem because it creates two cate-
gories of size for subsets in a metric space:
1. A set of “first category” is one that can be written as a countable union of nowhere-dense
sets. These are the small, intuitively “thin” subsets of a metric space.
2. If our metric space is complete, then it is necessarily of “second category”, meaning it
cannot be written as a countable union of nowhere-dense sets.
Theorem 3.6. The set
D = {f ∈ C[0, 1] : f ′(x) exists for some x ∈ [0, 1]}
is a set of first category in C[0, 1].
4 Functional Limits and Continuity
4.1 Functional Limits
Definition 4.1. Let A ⊆ R, a ∈ A \ {a} (a is an accumulation point of A). Let f : A → R,
define lim
x→af(x) = L iff
∀ > 0, ∃δ > 0 s.t. 0 < |x− a| < δ =⇒ |f(x)− L| <
4 FUNCTIONAL LIMITS AND CONTINUITY 20
Property 4.1 (Sequential criterion for functional limits). a ∈ A \ {a}, f : A → R. The
following are equivalent:
1. lim
x→af(x) = L
2. ∀(xn) ⊆ A \ {a}, xn → a =⇒ f(xn)→ L
Proof. We prove (1) =⇒ (2):
Assume lim
x→af(x) = L, take arbitrary (xn) ⊆ A \ {a} s.t. xn → a.
Let > 0, then ∃δ > 0 s.t. 0 < |x− a| < δ =⇒ |f(x)− L| < .
Also, ∃N s.t. n ≥ N =⇒ |xn − a| < δ.
Therefore, if |xn − a| < δ, then |f(xn)− L| < .
Theorem 4.1 (Algebraic Limit Theorem for functional limits). Suppose f, g : A → R, a ∈
A \ {a}.
Suppose lim
x→af(x) = L, limx→ag(x) = M . Then we have
1. lim
x→acf(x) = cL
2. lim
x→a(f(x) + g(x)) = L+M
3. lim
x→a(f(x)g(x)) = LM
4. lim
x→a(
f(x)
g(x) ) =
L
M when M 6= 0.
Property 4.2 (Divergence criterion). Suppose f : A → R, a ∈ A \ {a} lim
x→af(x) does not
exist if there are two sequences (xn), (yn) ⊆ A \ {a} s.t. xn → a, yn → a, lim
n→∞f(xn) =
L, lim
n→∞f(yn) = M exist but L 6= M .
Example 4.1. Let A = R+, f(x) = sin( 1x). Let an =
1
2npi , bn =
1
2npi+pi
2
.
Then we have an, bn → 0. Besides, lim
n→∞f(an) = 0, limn→∞f(bn) = 1. Hence limx→0+
sin( 1x) does
not exist.
Definition 4.2. Suppose f : A→ R, x ∈ A \ {a}. We define lim
x→af(x) =∞ iff
∀M > 0, ∃δ > 0 s.t. 0 < |x− a| < δ =⇒ f(x) > M
4.2 Continuous Functions
Definition 4.3 (continuity). Suppose (X, dX), (Y, dY ) are metric spaces. f : X → Y is
continuous at a ∈ X if
∀ > 0, ∃δ > 0 s.t. x ∈ BXδ (a) =⇒ f(x) ∈ BY (f(a))
Remark 4.1. Note that for X = Y = R, d(x, y) = |x− y|, so that we can write
∀ > 0,∃δ > 0 s.t. |x− a| < δ =⇒ |f(x)− f(a)| <
i.e.
lim
x→af(x) = f(a)
4 FUNCTIONAL LIMITS AND CONTINUITY 21
Definition 4.4 (continuous function). f : X → Y is continuous if it is continuous at every
point a ∈ X.
Property 4.3. The following are equivalent:
1. f is continuous at a
2. lim
x→af(x) = f(a)
3. ∀(xn) ⊆ A, xn → a =⇒ f(xn)→ f(a).
Corollary 4.1. f is discontinuous at a if there is a sequence (xn)→ a s.t. lim
n→∞f(xn) 6= f(a).
Remark 4.2. Note that we may have lim
x→af(x) exists but f is discontinuous at a.
Theorem 4.2 (Algebraic Continuity Theorem). Suppose f, g : A → R are continuous at
a ∈ A, c ∈ R. We have
1. cf(x) is continuous at a
2. f(x)± g(x) is continuous at a
3. f(x)g(x) is continuous at a
4. f(x)g(x) is continuous at a if g(a) 6= 0
Theorem 4.3. Suppose f : A→ B ⊆ R, g : B → R.
(g ◦ f)(x) = g(f(x)) is continuous at a ∈ A whenever f is continuous at a and g is continuous
at f(a).
Theorem 4.4. Suppose (X, dX), (Y, dY ) are metric spaces and f : X → Y is continuous.
If K ⊆ X is compact, then its image f [K] = {f(x) : x ∈ K} is compact.
Theorem 4.5. Suppose (X, dX), (Y, dY ) are metric spaces. If F ⊆ Y is closed in Y , then
f−1(F ) is closed in X.
Theorem 4.6 (Extreme Value Theorem). If f : K → R is continuous, K is compact, then
∃x1, x2 ∈ K s.t. ∀x ∈ K,
f(x1) ≤ f(x) ≤ f(x2)
Proof. Let H = f [K] = {f(x) : x ∈ K} ⊆ R, which is compact. Since compact subsets of R
are bounded, then let y2 = sup(H).
We have y ≤ y2 for all y ∈ H and ∀ > 0, ∃y ∈ H s.t. y2 − < y ≤ y2.
Take = 1n , then we have some zn ∈ H s.t. y2 − 1n < zn ≤ y2.
as Now we find an ∈ k s.t. f(an) = zn, n = 1, 2, . . .
By theorem, we have ank → x2, then f(x2) = lim
k→∞
f(ank) = y2. Which theo-
rem?
4 FUNCTIONAL LIMITS AND CONTINUITY 22
4.3 Continuous Functions on Compact Sets
4.3.1 Uniform Continuity
Definition 4.5 (uniform continuity). We say function f : A→ R is uniformly continuous on
A if
∀ > 0,∃δ > 0, x, y ∈ A ∧ |x− y| < δ =⇒ |f(x)− f(y)| <
Example 4.2. f(x) = x2 is not uniformly continuous.
Proof. WTS ∃ > 0, ∀δ > 0,∃x, y ∈ R s.t. |x− y| < δ and |f(x)− f(y)| ≥ .
Let = 1, δ > 0.
Choose y = x+ 12δ, so that |x− y| < δ.
f(y)− f(x) = y2 − x2
= (x+
1
2
δ)2 − x2
= x2 + δx+
1
4
δ2 − x2
= δx+
1
4
δ2
If x > 1δ , then f(y)− f(x) > 1.
Property 4.4 (). Function f : A→ R fails to be uniformly continuous iff ∃0 > 0,∃(xn), (yn) ⊆
A s.t. lim
n→∞|xn − yn| = 0 ∧ ∀n, |f(xn)− f(yn)| ≥ 0.
Proof. (⇐) Obvious.
(⇒) Assume f is not uniformly continuous.
Then ∃0 > 0 s.t. ∀δ > 0,∃xn, yn ∈ R s.t. |xn − yn| < δ and |f(xn)− f(yn)| ≥ 0.
Then this is true for δ ∈ N as well.
For each n ∈ N, let δ = 1n , and pick xn, yn as above. Then it is obvious that limn→∞|xn−yn| = 0
and ∀n, |f(xn)− f(yn)| ≥ 0.
Property 4.5 (Continuous functions on compact sets are uniformly continuous). Assume
f : K → R is continuous and K is compact, then f is uniformly continuous on K.
Proof. Assume for a contradiction that f : K → R is continuous and K is compact, but f is
not uniformly continuous. Then by Property 4.4, ∃0 > 0, (xn), (yn) ⊆ K s.t. lim
n→∞|xn−yn| = 0
and ∀n, |f(xn)− f(yn)| ≥ 0.
Since K is compact, then (xn) has a subsequence (xnk) s.t. xnk → x ∈ K.
Moreover, (ynk) has a subsequence (ynkm ) s.t. ynkm → y ∈ K.
Let x′m = xnkm , y
′
m = ynkm , then x
′
m → x, y′m → y.
Since lim
m→∞|x
′
m − y′m| = 0, thus x = y.
4 FUNCTIONAL LIMITS AND CONTINUITY 23
Then
|f(x′m)− f(y′m)| ≥ 0
=⇒ lim
m→∞|f(x
′
m)− f(y′m)| ≥ 0
=⇒ |f(x)− f(y)| ≥ 0
=⇒ 0 ≥ 0
which is a contradiction.
Definition 4.6. A function f : A→ R is said to be Lipschitz if ∃M ∈ N s.t. ∀x 6= y ∈ A,∣∣∣∣f(x)− f(y)x− y
∣∣∣∣ < M
Property 4.6. Lipschitz functions are uniformly continuous.
Proof. Let f : A→ R be Lipschitz on A. Then for every > 0, take δ < M .
Then if |x− y| < δ, then
|f(x)− f(y)| < M |x− y|
< M

M
=
So f is uniformly continuous.
Remark 4.3. The converse does not hold.
Property 4.7 (Continuous image of connected sets is connected). If f : E → R is continuous
and E is connected, then f(E) is connected.
4.4 Sets of Discontinuity
Let f : R→ R, Df = {x ∈ R : f is not continuous at x}.
Example 4.3 (Df = ∅). f is continuous
Example 4.4 (Df = R). f(x) =
{
1, x ∈ Q
0, x ∈ R \Q
Example 4.5. Given a countable set A = {a1, . . .}, define f(an) := 1n and f(x) = 0, ∀x /∈ A.
Then we have Df = A.
Fact 4.1. There is no f : R→ R s.t. Df = R \Q.
Definition 4.7 (Fσ-set). A subset F of R is a Fσ-set if F =
⋃∞
n=1 Fn s.t. Fn is closed for all
n.
5 THE DERIVATIVE 24
Definition 4.8 (α-continuity). Let α > 0, f : R→ R, a ∈ R. f is α-continuous at a if
∃δ > 0 s.t. x, y ∈ (a− δ, a+ δ) =⇒ |f(x)− f(y)| < α
Note that f is continuous at a iff f is α-continuous at a for all a > 0.
Property 4.8. For every f : R→ R, the set Df is Fσ-set of R. red parts
Definition 4.9. Let f : R→ R.
f is removable discontinuous if lim
x→af(x) exists but does not equal f(a).
f has a jump at a if lim
x→a−
f(x) 6= lim
x→a+
f(x).
If lim
x→af(x) does not exist for other reasons, we say f is essential discontinuous.
Definition 4.10 (monotonicity). f : R → R is monotone if either x ≤ y =⇒ f(x) ≤ f(y)
or x ≤ y =⇒ f(x) ≥ f(y).
Property 4.9. Discontinuity of a monotone function f is a jump. Moreover, Df is countable.
5 the Derivative
5.1 Derivatives and the Intermediate Value Property
Definition 5.1 (derivative). Let f : R→ R, c ∈ R. Define the derivative of f at c:
f ′(c) = lim
x→c
f(x)− f(c)
x− c
If f ′(c) exists, we say that f is differentiable at c. If f ′(a) exists for all a ∈ R, we say that g
is differentiable on R.
Property 5.1. If f is differentiable at c, then f is continuous at c.
Proof. We have
lim
x→c(f(x)− f(c)) = limx→c
f(x)− f(c)
x− c · (x− c) = f
′(c) · 0 = 0

Theorem 5.1 (Algebraic Differentiability Theorem). Suppose f, g are differentiable, a, c ∈ R.
We have
1. (cf)′(a) = cf ′(a)
2. (f + g)′(a) = f ′(a) + g′(a)
3. (f · g)′(a) = f ′(a)g(a) + f(a)g′(a)
4.
(
f
g
)′
(a) = f
′(a)g(a)−f(a)g′(a)
[g(a)]2
5 THE DERIVATIVE 25
Theorem 5.2 (Chain Rule). Let f : A→ B, g : B → R, f(A) ⊆ B so that g ◦ f is defined. If
f is differentiable at c and g is differentiable at f(c), then g ◦ f is differentiable at a with
(g ◦ f)′(c) = g′(f(c)) · f ′(c)
Theorem 5.3 (Interior Extremum Theorem). If f is differentiable on (a, b), f attains maxi-
mum at some c ∈ (a, b), then f ′(c) = 0.
Proof. We have
f ′(c) = lim
x→c−
f(x)− f(c)
x− c ≤ 0
and
f ′(c) = lim
x→c+
f(x)− f(c)
x− c ≥ 0
then f ′(c) = 0.
Theorem 5.4 (Darboux’s Theorem). If f is differentiable on [a, b] and f ′(a) < α < f ′(b) or
f ′(a) > α > f ′(b), then ∃c ∈ (a, b) s.t. f ′(c) = α.
5.2 the Mean Value Theorems
Theorem 5.5 (Rolle’s Theorem). Let f : [a, b]→ R be continuous on [a, b] and differentiable
on (a, b). If f(a) = f(b), then ∃c ∈ (a, b) s.t. f ′(c) = 0.
Proof. By EVT, since f is continuous on a compact set, then f attains a maximum and a
minimum. If both extremums occur at the endpoints, then f is necessarily a constant function
and f ′(x) = 0 on (a, b).
If either the maximum or minimum occurs at some point c ∈ (a, b), then it follows from the
Interior Extremum Theorem that f ′(c) = 0.
Theorem 5.6 (Mean Value Theorem). If f : [a, b] → R is continuous on [a, b] and differen-
tiable on (a, b), then ∃c ∈ (a, b) s.t.
f ′(c) =
f(b)− f(a)
b− a
Proof. Consider
d(x) = f(x)−
[(
f(b)− f(a)
b− a
)
(x− a) + f(a)
]
We know d is continuous on [a, b] and differentiable on (a, b). Also, d(a) = d(b) = 0.
By Rolle’s Theorem, ∃c ∈ (a, b) s.t. d′(c) = 0 =⇒ f ′(c) = f(b)−f(a)b−a .
Corollary 5.1. If f : (a, b) → R is differentiable and f ′(x) = 0 for all x ∈ (a, b), then f is
constant on (a, b).
Proof. Assume x, y ∈ (a, b) and x < y. We set c ∈ (x, y), then by Mean Value Theorem,
0 = f ′(c) =
f(y)− f(x)
y − x =⇒ f(y)− f(x) = 0

5 THE DERIVATIVE 26
Corollary 5.2. If f : (a, b) → R is differentiable and f ′(x) = g′(x) for all x ∈ (a, b), then
f(x) = g(x) + c for some c ∈ R.
Proof. Apply the previous corollary to the function h(x) = f(x)− g(x).
Theorem 5.7 (Generalized Mean Value Theorem). If f, g : [a, b]→ R are continuous on [a, b]
and differentiable on (a, b), then ∃c ∈ (a, b) s.t.
[f(b)− f(a)]g′(c) = [g(b)− g(a)]f ′(c)
If g′ is never zero on (a, b), then
f ′(c)
g′(c)
=
f(b)− f(a)
g(b)− g(a)
Proof. Apply the Mean Value Theorem to the function h(x) = [f(b) − f(a)]g(x) − [g(b) −
g(a)]f(x).
Theorem 5.8 (L’Hospital’s Rule: 0/0 case). Suppose f, g are continuous on I with a ∈ I
and are differentiable on I \ {a}. If f(a) = g(a) = 0 and ∀x 6= a, g′(x) 6= 0, then
lim
x→a
f ′(x)
g′(x)
= L =⇒ lim
x→a
f(x)
g(x)
= L
Proof. Since lim
x→a
f ′(x)
g′(x) = L, then for all > 0,∃δ > 0 s.t.
x ∈ (a− δ, a+ δ) =⇒
∣∣∣∣f ′(x)g′(x) − L
∣∣∣∣ <
By the Generalized Mean Value Theorem, for every y ∈ (a, a+ δ),∃x ∈ (a, y) s.t.
f ′(c)
g′(c)
=
f(b)− f(a)
g(b)− g(a) =
f(y)
g(y)
and thus ∣∣∣∣f(y)g(y) − L
∣∣∣∣ = ∣∣∣∣f ′(x)g′(x) − L
∣∣∣∣ <

Theorem 5.9 (L’Hospital’s Rule: ∞/∞ case). Suppose f, g are differentiable on (a, b) and
g′(x) 6= 0 for all x ∈ (a, b). If lim
x→ag(x) =∞ or −∞, then
lim
x→a
f ′(x)
g′(x)
= L =⇒ lim
x→a
f(x)
g(x)
= L
6 SEQUENCES AND SERIES OF FUNCTIONS 27
6 Sequences and Series of Functions
6.1 Uniform Convergence of a Sequence of Functions
Definition 6.1 (pointwise convergence). For each n ∈ N, let fn be a function defined on a
set A ⊆ R. If ∀x ∈ A, fn(x) → f(x) for some function f , then sequence (fn) of functions
converges pointwise on A to f .
We can write fn → f, lim fn = f , or lim
n→∞fn(x) = f(x).
Example 6.1. Consider fn : R→ R
fn(x) =
x2 + nx
n
We can compute
lim
n→∞fn(x) = limn→∞
x2 + nx
n
= lim
n→∞
x2
n
+ x = x
Thus, (fn) converges pointwise to f(x) = x on R.
Example 6.2. Consider fn : [0, 1]→ R
fn(x) = x
n
If 0 ≤ x < 1, xn → 0. If x = 1, xn → 1. It follows that fn → f pointwise on [0, 1] where
f(x) =
{
0, 0 ≤ x < 1
1, x = 1
Note that pointwise convergent sequence of continuous functions may converge to a non-
continuous function.
Definition 6.2 (uniformly convergence). Let (fn) be a sequence of functions defined on a set
A ⊆ R, then (fn) converges uniformly on A to a limit function f defined on A if
∀ > 0,∃N s.t. ∀n ≥ N, ∀x ∈ A, |f(x)− fn(x)| <
Remark 6.1. This is a stronger notion of convergence.
Example 6.3. Consider fn : R→ R
fn(x) =
x2 + nx
n
which converges pointwise on R to f(x) = x. But the convergence is not uniform, since
|fn(x)− f(x)| =
∣∣∣∣x2 + nxn − x
∣∣∣∣ = x2n
In order to force |fn(x) − f(x)| < , we need N < x2 . Although it is possible to do for each
x ∈ R, there is no way to choose a single value of N that will work for all values of x at the
same time.
On the other hand, we can show that fn → f uniformly on the set [−b, b].
6 SEQUENCES AND SERIES OF FUNCTIONS 28
Property 6.1 (Cauchy Criterion for Uniform Convergence). A sequence of functions (fn)
defined on a set A ⊆ R converges uniformly on A iff
∀ > 0,∃N s.t. ∀x ∈ A,∀m,n ≥ N, |fn(x)− fm(x)| <
Theorem 6.1 (Continuous Limit Theorem). Let (fn) be a sequence of functions defined on
A ⊆ R that converges uniformly on A to a function f . If each fn is continuous at c ∈ A, then
f is continuous at c.
Proof. Let > 0 and fix c ∈ A. Choose N s.t.
|fN (x)− f(x)| <
3
,∀x ∈ A
Since fN is continuous, then ∃δ > 0 s.t.
|x− c| < δ =⇒ |fN (x)− fN (c)| <
3
Thus,
|f(x)− f(c)| = |f(x)− fN (x) + fN (x)− fN (c) + fN (c)− f(c)|
≤ |f(x)− fN (x)|+ |fN (x)− fN (c)|+ |fN (c)− f(x)|
<

3
+

3
+

3
=

Hence f is continuous at c ∈ A.
Property 6.2. (Algebraic Limit Theorem for Uniform Convergence) Suppose (fn), (gn) are
uniformly convergent on A, then
1. (cfn + gn) is uniformly convergent on A
2. If ∃M > 0 s.t. |fn| ≤M and |gn| ≤M , then (fngn) is uniformly convergent.
Proof. (1) Obvious.
(2) Let > 0. Since (fn), (gn) are uniformly convergent on A, then ∃N s.t. ∀m,n ≥ N, |fn(x)−
fm(x)| < 2M and gn(x)− gm(x) < 2M . Using Cauchy criterion, we have
|fm(x)gm(x)− fn(x)gn(x)| = |fm(x)gm(x)− fm(x)gn(x) + fm(x)gn(x)− fn(x)gn(x)|
≤ |fm(x)||gm(x)− gn(x)|+ |gn(x)||fm(x)− fn(x)|
≤M(|gm(x)− gn(x)|+ |fm(x)− fn(x)|
< M(

M
)
=
So (fngn) is uniformly convergent.
6 SEQUENCES AND SERIES OF FUNCTIONS 29
6.2 Uniform Convergence and Differentiation
Theorem 6.2 (Differentiable Limit Theorem). Let fn → f pointwisely on [a, b] and assume
each fn is differentiable. If (f
′
n) converges uniformly on [a, b] to a function g, then the function
f is differentiable and f ′ = g.
Theorem 6.3. Let (fn) be a sequence of differentiable functions defined on [a, b] and assume
(f ′n) converges uniformly on [a, b]. If ∃x0 ∈ [a, b] s.t. fn(x0) is convergent, then (fn) converges
uniformly on [a, b].
Theorem 6.4 (stronger form of Differentiable Limit Theorem). Let (fn) be a sequence of
differentiable functions defined on [a, b] and assume (f ′n) converges uniformly on [a, b] to a
function g. If ∃x0 ∈ [a, b] s.t. fn(x0) is convergent, then (fn) converges uniformly on [a, b].
Moreover, the limit function f = lim fn is differentiable and f
′ = g.
6.3 Series of Functions
Definition 6.3 (pointwise convergence). For each n ∈ N, let fn and f be functions defined
on a set A ⊆ R. The infinite series
∞∑
n=1
fn(x) = f1(x) + f2(x) + f3(x) + . . .
converges pointwise on A to f(x) if the sequence sk(x) of partial sums defined by
sk(x) = f1(x) + f2(x) + . . .+ fk(x)
converges pointwise to f(x).
Definition 6.4 (uniform convergence). The series converges uniformly on A to f if the se-
quence sk(x) converges uniformly on A to f(x).
In either case, we write f =
∑∞
n=1 fn or f(x) =
∑∞
n=1 fn(x).
Theorem 6.5 (Term-by-term Continuity Theorem). Let fn be continuous functions defined
on a set A ⊆ R, and assume ∑∞n=1 fn converges uniformly on A to a function f . Then, f is
continuous on A.
Proof. Apply the Continuous Limit Theorem 6.1 to the partial sums sk = f1 + f2 + . . .+ fk.

Theorem 6.6 (Term-by-term Differentiability Theorem). Let fn be differentiable functions
defined on an interval A, and assume
∑∞
n=1 f
′
n(x) converges uniformly to a limit g(x) on A.
If there exists a point x0 ∈ [a, b] where
∑∞
n=1 fn(x0) converges, then the series
∑∞
n=1 fn(x)
converges uniformly to a differentiable function f(x) satisfying f ′(x) = g(x) on A. In other
words,
f(x) =
∞∑
n=1
fn(x) and f
′(x) = g(x)
6 SEQUENCES AND SERIES OF FUNCTIONS 30
Proof. Apply the stronger form of the Differentiable Limit Theorem 6.4 to the partial sums
sk = f1 + f2 + . . .+ fk.
Theorem 6.7 (Cauchy Criterion for Uniform Convergence of Series). A series
∑∞
n=1 fn con-
verges uniformly on A ⊆ R if and only if
∀ > 0,∃N ∈ N s.t. ∀n > m ≥ N, ∀x ∈ A, |sn − sm| =
∣∣∣∣∣
n∑
i=m+1
fi(x)
∣∣∣∣∣ <
Remark 6.2. The benefit of the Cauchy Criterion is that it does not depend on the value of
the limit.
Corollary 6.1 (Weierstrass M-Test). For each n ∈ N, let fn be a function defined on a set
A ⊆ R, and let Mn > 0 be a real number satisfying that
sup
x∈A
|fn(x)| ≤Mn
If
∑∞
n=1Mn converges, then
∑∞
n=1 fn converges uniformly on A.
Proof. Let > 0. Choose N that satisfies the Cauchy Criterion. Let m > n ≥ N .
Then by Cauchy Criterion for Uniform Convergence of Series,
Mm+1 + . . .+Mn <
Then for n > m ≥ N and all x ∈ A,
|fm+1(x) + . . .+ fn(x)| ≤ |fm+1(x)|+ . . .+ |fn(x)|
≤Mm+1 + . . .+Mn
<

Remark 6.3. The reverse is not true.
Example 6.4. If fn(x) = (−1)n 1n , then
∑∞
n=1 fn(x) is uniformly convergent, but the M-test
fails because if Mn =
1
n (the smallest Mn possible), then
∑∞
n=1
1
n is not convergent.
Corollary 6.2. If
∑∞
n=1 fn converges uniformly on A ⊆ R, then the sequence (fn) converges
uniformly on A to 0.
Proof. WTS ∀ > 0, ∃N ∈ N, ∀x ∈ A, |fn(x)| < .
Let > 0. Since
∑∞
n=1 fn converges uniformly, then by Cauchy Criterion,
∃N ∈ N s.t. ∀n > m ≥ N, ∀x ∈ A, |fm+1(x) + . . .+ fn(x)| <
Let n = m+ 1, then
|fn(x)| <
as wanted.
6 SEQUENCES AND SERIES OF FUNCTIONS 31
Corollary 6.3. Suppose ∀n ∈ N,∀x ∈ A, gn(x) ≥ fn(x) ≥ 0. If
∑∞
n=1 gn converge uniformly
on A, then
∑∞
n=1 fn converges uniformly on A.
Proof. Let > 0. Apply Cauchy Criterion for
∑∞
n=1 gn, we get N ∈ N s.t. for n > m ≥ N
and x ∈ A,
|fm+1(x) + . . .+ fn(x)| = fm(x) + . . .+ fn(x)
≤ gm+1(x) + . . .+ gn(x)
= |gm+1(x) + . . .+ gn(x)|
<
So
∑∞
n=1 fn converges uniformly on A.
6.4 Power Series
Theorem 6.8. If a power series
∑∞
n=0 anx
n converges at some point x0 ∈ R, then it converges
absolutely for any x satisfying |x| < |x0|.
Proof. If
∑∞
n=0 anx
n
0 converges, then (anx
n
0 ) is bounded and → 0.
Let M > 0 be s.t. |anxn0 | ≤M for all n ∈ N. If x ∈ R satisfies |x| < |x0|, then
|anxn| = |anxn0 |
∣∣∣∣ xx0
∣∣∣∣n ≤M ∣∣∣∣ xx0
∣∣∣∣n
But |x/x0| < 1, so the geometric series
∞∑
n=0
M
∣∣∣∣ xx0
∣∣∣∣n
is convergent. By the Comparison Test,
∑∞
n=0 anx
n converges absolutely.
Theorem 6.9. If a power series
∑∞
n=0 anx
n converges absolutely at a point x0, then it
converges uniformly on the closed interval [−c, c], where c = |x0|.
Proof. For n ∈ N, let Mn = |an| · |x0|n.
Note that supx∈[−c,c] |anxn| ≤ |an| · |x0|n = Mn.
Since
∑∞
n=0Mn is convergent by assumption, then by Weierstrass M-Test 6.1,
∑∞
n=0 anx
n
converges uniformly on [−c, c].
Remark 6.4. If the power series g(x) =
∑∞
n=0 anx
n converges conditionally at x = R, then
it is possible for it to diverge when x = −R.
Example 6.5.
∞∑
n=1
(−1)nxn
n
6 SEQUENCES AND SERIES OF FUNCTIONS 32
Lemma 6.1 (Abel’s Lemma). Let bn satisfy b1 ≥ b2 ≥ b3 ≥ . . . ≥ 0, and let
∑∞
n=1 an be a
series for which the partial sums are bounded. In other words, assume there exists A > 0
such that
|a1 + a2 + . . .+ an| ≤ A
for all n ∈ N. Then for all n ∈ N,
|a1b1 + a2b2 + a3b3 + . . .+ anbn| ≤ Ab1
Proof. ∣∣∣∣∣
n∑
k=1
akbk
∣∣∣∣∣ =
∣∣∣∣∣snbn+1 +
n∑
k=1
sk(bk − bk+1)
∣∣∣∣∣ by summation-by-parts formula
≤ |snbn+1|+
∣∣∣∣∣
n∑
k=1
sk(bk − bk+1)
∣∣∣∣∣ by Triangle Inequality
≤ Abn+1 +
n∑
k=1
A(bk − bk+1)
= Abn+1 + (Ab1 −Abn+1)
= Ab1

Theorem 6.10 (Abel’s Theorem). Let g(x) =
∑∞
n=0 anx
n be a power series that converges
at the point x = R > 0. Then the series converges uniformly on the interval [0, R]. A similar
result holds if the series converges at x = −R.
Proof. To set the stage for Abel’s Lemma 6.1, we first write
g(x) =
∞∑
n=0
anx
n =
∞∑
n=0
(anR
n)
( x
R
)n
Let > 0. Since we are assuming that
∑∞
n=0 anR
n converges, then by the Cauchy Criterion
for Uniform Convergence of Series 6.7, ∃N ∈ N s.t. if n > m ≥ N , then
|am+1Rm+11 + am+2Rm+2 + . . .+ anRn| <
Now for any fixed m ∈ N, we apply Abel’s Lemma 6.1 to the sequence ∑∞i=1 am+iRm+i.
Since x ∈ [0, R], then we have ( x
R
)m+1 ≥ ( x
R
)m+2 ≥ . . . ≥ 0
Then∣∣(am+1Rm+1)( x
R
)m+1
+ (am+2R
m+2)
( x
R
)m+2
+ . . .+ (anR
n)
( x
R
)n∣∣ ≤ ( x
R
)m+1 ≤
Therefore the series converges uniformly on the interval [0, R].
7 THE RIEMANN INTEGRAL 33
Theorem 6.11. If
∑∞
n=0 anx
n converges for all x ∈ (−R,R), then the differentiated series∑∞
n=1 nanx
n−1 converges at each x ∈ (−R,R) as well. Consequently, the convergence is
uniform on closed intervals in (−R,R). prove this
Theorem 6.12. Assume f(x) =
∑∞
n=0 anx
n converges on an interval A ⊆ R. The function
f is continuous on A and differentiable on any open interval (−R,R) ⊆ A. The derivative is
given by
f ′(x) =
∞∑
n=1
nanx
n−1
Moreover, f is infinitely differentiable on (−R,R), and the successive derivatives can be
obtained via term-by-term differentiation of the appropriate series:
f (k)(x) =
∞∑
n=k
n(n− 1) . . . (n− k + 1)xn−k
Corollary 6.4. If
∑∞
n=0 anx
n,
∑∞
n=0 bnx
n exist and equal for all x ∈ (−R,R), then it must
be the case that an = bn for all n ∈ N.
7 The Riemann Integral
7.1 The Definition of the Riemann Integral
7.1.1 Partitions, Upper Sums, and Lower Sums
Definition 7.1 (partition). A partition P of [a, b] is a finite set of points from [a, b] that
includes both a and b. The notational convention is to always list the points of a partition
P = {x0, x1, x2, . . . , xn} in increasing order; thus
a = x0 < x1 < x2 < . . . < xn = b
Definition 7.2 (lower sum and upper sum). For each subinterval [xk−1, xk] of P , let
mk = inf{f(x) : x ∈ [xk−1, xk]} and Mk = sup{f(x) : x ∈ [xk−1, xk]}
The lower sum of f with respect to P is given by
L(f, P ) =
n∑
k=1
mk(xk − xk−1)
Likewise, we define the upper sum of f with respect to P by
U(f, P )
n∑
k=1
Mk(xk − xk−1)
Fact 7.1. For a particular partition P , it is clear that U(f, P ) ≥ L(f, P ).
Definition 7.3 (refinement). A partition Q is a refinement of a partition P if Q contains all
of the points of P ; that is, if P ⊆ Q.
7 THE RIEMANN INTEGRAL 34
Lemma 7.1. If P ⊆ Q, then L(f, P ) ≤ L(f,Q), and U(f, P ) ≥ U(f,Q).
Proof. Consider what happens when we refine P by adding a single point z to some subinterval
[xk−1, xk] of P . Focusing on the lower sum, we have
mk(xk − xk−1) = mk(xk − z) +mk(z − xk−1)
≤ m′k(xk − z) +mkkk (z − xk−1)
where
m′k = inf{f(x) : x ∈ [z, xk]} and m
′′
k = inf{f(x) : x ∈ [xk−1, z]}
are each necessarily as large or larger than mk.
By induction, we have L(f, P ) ≤ L(f,Q), and an analogous argument holds for the upper
sums.
Lemma 7.2. If P1 and P2 are any two partitions of [a, b], then L(f, P1) ≤ U(f, P2).
Proof. Let Q = P1 ∪ P2. Because P1 ⊆ Q and P2 ⊆ Q, it follows that
L(f, P1) ≤ L(f,Q) ≤ U(f,Q) ≤ U(f, P2)

7.1.2 Integrability
Definition 7.4 (upper integral and lower integral). Let P be the collection of all possible
partitions of the interval [a, b]. The upper integral of f is defined to be
U(f) = inf{U(f, P ) : P ∈ P}
Similarly, we define the lower integral of f by
L(f) = sup{L(f, P ) : P ∈ P}
Lemma 7.3. For any bounded function f on [a, b], it is always the case that
U(f) ≥ L(f)
Definition 7.5 (Riemann Integrability). A bounded function f defined on the interval [a, b]
is Riemann-integrable if U(f) = L(f). In this case, we define
´ b
a f or
´ b
a f(x) dx to be this
common value; namely, ˆ b
a
f = U(f) = L(f)
7 THE RIEMANN INTEGRAL 35
7.1.3 Criteria for Integrability
Theorem 7.1 (Integrability Criterion). A bounded function f is integrable on [a, b] if and
only if, for every > 0, ∃ a partition P of [a, b] such that
U(f, P)− L(f, P) <
prove this
Theorem 7.2. If f is continuous on [a, b], then it is integrable. prove this
Definition 7.6 (tagged partition). A tagged partition (P, {ck}) is one where in addition to
a partition P we choose a sampling point ck in each of the subintervals[xk−1, xk].
P = [x0, x1, . . . , xn]
ck ∈ [xk−1, xk], 0 < k ≤ n
Definition 7.7 (Riemann sum).
R(f, P, {ck}) =
n∑
k=1
f(ck) · (xk − xk−1)
Definition 7.8 (Riemann’s Original Definition of the Integral). A bounded function f is
integrable on [a, b] with
´ b
a f = A if for all > 0, ∃δ > 0 such that for any tagged partition
(P, {ck}) satisfying δxk < δ for all k¡ it follows that
|R(f, P, {ck})−A| <
Remark 7.1. This definition is equivalent to our definition.
7.2 Integrating Functions with Discontinuities
Fact 7.2. Suppose two functions f, g : [a, b] → R are both bounded. and f is integrable.
Suppose there are finitely many points y1, y2, . . . , yl ∈ [a, b] s.t. f(x) = g(x) for x 6= yk for
k = 1, 2, . . . , l.
Then g is integrable and ˆ b
a
g =
ˆ b
a
f
prove this
Theorem 7.3. If f : [a, b]→ R is bounded, and f is integrable on [c, b] for all c ∈ (a, b), then
f is integrable on [a, b]. An analogous result holds at the other endpoint. prove this
Example 7.1 (Dirichlet’s function).
g(x) =
{
1 for x rational
0 for x irrational
If P is some partition of [0, 1], then the density of the rationals in R implies that every
subinterval of P will contain a point where g(x) = 1 as well as a point where g(y) = 0.. It
follows that U(g, P ) = 1 and L(g, P ) = 0. Because this is the case for every partition P , we
see that U(f) = 1, L(f) = 0. The two are not equal, so we conclude that Dirichlet’s function
is not integrable.
7 THE RIEMANN INTEGRAL 36
7.3 Properties of the Integral
Theorem 7.4. Assume f : [a, b]→ R is bounded, and let c ∈ (a, b). Then, f is integrable on
[a, b] if and only if f is integrable on [a, c] and [c, b]. In this case, we have
ˆ b
a
f =
ˆ c
a
f +
ˆ b
c
f
prove this
Theorem 7.5. Assume f and g are integrable functions on the interval [a, b].
1. The function f + g is integrable on [a, b] with
´ b
a (f + g) =
´ b
a f +
´ b
a g.
2. For k ∈ R, the function is kf is integrable with ´ ba kf = k
´ b
a f .
3. If m ≤ f(x) ≤M on [a, b], then m(b− a) ≤ ´ ba f ≤M(b− a).
4. If f(x) ≤ g(x) on [a, b], then ´ ba f ≤
´ b
a g.
5. If f(x) ≤ g(x) on [a, b], then ´ ba f ≤
´ b
a g.
6. The function |f | is integrable an | ´ ba f | ≤
´ b
a |f |.
Definition 7.9. If f is integrable on the interval [a, b], define
ˆ a
b
f = −
ˆ b
a
f
Also for c ∈ [a, b], define ˆ c
c
f = 0
Fact 7.3. If f : [a, b]→ R is integrable, then
|f |(x) = |f(x)|
is also integrable, and ∣∣∣∣ˆ b
a
f
∣∣∣∣ ≤ ˆ b
a
|f |
prove this
7.3.1 Uniform Convergence and Integration
Theorem 7.6 (Integrable Limit Theorem). Assume that fn → f uniformly on [a, b] and that
each fn is integrable. Then, f is integrable and
lim
n→∞
ˆ b
a
fn =
ˆ b
a
f
prove this
7 THE RIEMANN INTEGRAL 37
7.4 The Fundamental Theorem of Calculus
Theorem 7.7 (Fundamental Theorem of Calculus). We have
1. If f : [a, b]→ R is integrable, and F : [a, b]→ R satisfies F ′(x) = f(x) for all x ∈ [a, b],
then ˆ b
a
f = F (b)− F (a)
2. Let g : [a, b]→ R be integrable, and for x ∈ [a, b], define
G(x) =
ˆ x
a
g
Then G is continuous on [a, b]. If g is continuous at some point c ∈ [a, b], then G is
differentiable at c and G′(c) = g(c).
prove this
Example 7.2. f(x) = |x| on [−1, 1].
Define F (x) =
´ x
−1 f(x) dx, then F is continuous and differentiable on [−1, 1].
On [−1, 0], F (x) = −12x2 + 12 .
On [0, 1], F (x) = 12x
2 + 12 .
So combining the two, we keep the relationship F ′(x) = f(x)(= |x|).
Example 7.3. f : [a, b]→ R continuous. Let F (x) = ´ xa f : [a, b]→ R.
We know from FTC that F ′(x) = f(x) for all x ∈ [a, b].
Assume F (x) =
´ x
a f = 0 for all x ∈ [a, b]. Then f(x) = 0 for all x ∈ [a, b].
Example 7.4. T or F:
1. h′ = g does not imply continuity of g. (True)
2. If g is continuous on [a, b], then there is a differentiable h s.t. h′ = g. (True)
3. If H(x) =
´ x
a h is differentiable at c ∈ (a, b), then h is continuous at c. (False)
Counterexample for (3): h : [0, 1]→ R.
h(x) =
{
0, x 6= 1/2
1, x = 1/2
=⇒ H(x) = 0
Example 7.5. fn → 0 pointwise on [0, 1], but lim
n→∞
´
fn does not exist.
fn =
{
xn, 0 ≤ x < 1
0, x = 1
For every x ∈ [0, 1], ´ 10 fn = n→∞, But fn(x)→ 0.
Fact 7.4. If f : [a, b]→ R is integrable and there is M ∈ R s,t, |f(x)| ≤ M for all x ∈ [a, b],
then f2(x) = (f(x))2 is integrable.
Fact 7.5. If f, g are integrable, then so is fg.
7 THE RIEMANN INTEGRAL 38
7.5 Lebesgue’s Criterion for Riemann Integrability
7.5.1 Sets of Measure Zero
Definition 7.10 (measure zero). A set A ⊆ R has measure zero if, for all > 0, there exists
a countable collection of open intervals On with the property that A is contained in the union
of all the intervals On and the sum of the lengths of all of the intervals is less than or equal
to . More precisely, if |On| refers to the length of the interval On, then we have
A ⊆
∞⋃
n=1
On and
∞∑
n=1
|On| ≤
Example 7.6. Consider a finite set A = {a1, a2, . . . , aN}. To show that A has measure zero,
let > 0¿ For each 1 ≤ n ≤ N , construct the interval
Gn =
(
an −
2N
, an +

2N
)
Clearly, A is contained in the union of these intervals, and
N∑
n=1
|Gn| =
N∑
n=1

N
=
Theorem 7.8. Countable sets have measure zero.
Proof. If a countable set A = {a1, a2, . . . , an, . . .}, then define
On = (an −
2n+1
, an +

2n+1
Then |On| = 2n , and A ⊆
⋃
n = 1∞On. Then∑
n = 1∞|On| =
∞∑
n=1

2n
= ·
∞∑
n=1
1
2n
= · 1 =

Theorem 7.9. The canter sets has measure zero.
Fact 7.6. If two sets A and B both have measure zero, then set A ∪B has measure zero.
Fact 7.7. If a sequence of sets (An) all have measure zero, then
⋃
n = 1∞An has measure
zero.
7.5.2 α-continuity
Definition 7.11 (α-continuity). Let f be defined on [a, b], and let α > 0. The function f is
α-continuous at x ∈ [a, b] if there exists δ > 0 such that for all y, z ∈ (x− δ, x+ δ), it follows
that |f(y)− f(z)| < α.
Let f be a bounded function on [a, b]. For each α > 0, define Dα to be the set of points in
[a, b] where the function f fails to be α-continuous; that is,
Dα = {x ∈ [a, b] : f is not α-continuous at x}
7 THE RIEMANN INTEGRAL 39
Fact 7.8. Let Df = {x ∈ [a, b] : f is not continuous at x}. Then Df =
⋃
α>0D
α
f .
Fact 7.9. If α < α′¡ then Dα′ ⊆ Dα.
Fact 7.10. For a fixed α > 0, the set Dα is closed and therefore compact.
7.5.3 Lebesgue’s Theorem
Theorem 7.10 (Lebesgue’s Theorem). Let f be a bounded function defined on the interval
[a, b]. Then. f is Riemann-integrable if and only if the set of points where f is not continuous
has measure zero.
Corollary 7.1. If functions f, g are bounded on [a, b], f is continuous on [a, b] and
D = {x ∈ [a, b] : g(x) 6= f(x)}
has measure zero. Then g is integrable and
ˆ b
a
f =
ˆ b
a
g
prove this

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