## Advanced Calculus (II)

W^{EN}-C^{HING}L^{IEN}

Department of Mathematics National Cheng Kung University

2009

## Ch9: Convergence in R

^{n}

### 9.5: Applications

Theorem (9.40 Dini)

Suppose that H is a compact subset of**R**^{n}and fk :H →**R**
is a pointwise monotone sequence of continuous

functions. If fk → f pointwise on H as k → ∞ and f is continuous on H, then fk → f uniformly on H. In particular, if φk is a pointwise monotone sequence of functions continuous on an interval [a,b] that converges pointwise to a continuous function, then

k →∞lim Z b

a

φ_{k}(t)dt =
Z b

a

k →∞lim φ_{k}(t)

dt.

Definition (9.41)

(i) A set E ⊆**R is said to be of measure zero if and only if**
for every ε > 0 there is a countable collection of intervals
{I_{j}}_{j∈N}that covers E such that

∞

X

j=1

|I_{j}| ≤ ε.

(ii) A function f : [a, b] →**R is said to be almost**

everywhere continuous on [a, b] if and only if the set of points x ∈ [a, b] where f is discontinuous is a set of measure zero.

Remark (9.43)

If E1,E2, . . . is a sequence of sets of measure zero, then

E =

∞

[

k =1

Ek

is also a set of measure zero.

Proof.

Let ε > 0.By hypothesis, given k ∈**N we can choose a**
collection of intervals {I_{j}^{(k )}}j∈Nthat covers Ek such that

∞

X

j=1

|I_{j}^{(k )}| < ε
2^{k}.

then the collection {I_{j}^{(k )}}k ,j∈Nis countable, covers E , and

∞

X

k =1

∞

X

j=1

|I_{j}^{(k )}| ≤

∞

X

j=1

ε
2^{k} = ε.

Consequently, E is of measure zero.

Proof.

Let ε > 0. By hypothesis, given k ∈**N**we can choose a
collection of intervals {I_{j}^{(k )}}j∈Nthat covers Ek such that

∞

X

j=1

|I_{j}^{(k )}| < ε
2^{k}.

then the collection {I_{j}^{(k )}}k ,j∈Nis countable, covers E , and

∞

X

k =1

∞

X

j=1

|I_{j}^{(k )}| ≤

∞

X

j=1

ε
2^{k} = ε.

Consequently, E is of measure zero.

Proof.

Let ε > 0. By hypothesis, given k ∈**N we can choose a**
collection of intervals {I_{j}^{(k )}}j∈Nthat covers Ek such that

∞

X

j=1

|I_{j}^{(k )}| < ε
2^{k}.

then the collection {I_{j}^{(k )}}k ,j∈Nis countable, covers E ,and

∞

X

k =1

∞

X

j=1

|I_{j}^{(k )}| ≤

∞

X

j=1

ε
2^{k} = ε.

Consequently, E is of measure zero.

Proof.

Let ε > 0. By hypothesis, given k ∈**N we can choose a**
collection of intervals {I_{j}^{(k )}}j∈Nthat covers Ek such that

∞

X

j=1

|I_{j}^{(k )}| < ε
2^{k}.

then the collection {I_{j}^{(k )}}k ,j∈Nis countable, covers E , and

∞

X

k =1

∞

X

j=1

|I_{j}^{(k )}| ≤

∞

X

j=1

ε
2^{k} = ε.

Consequently, E is of measure zero.

Proof.

**N we can choose a**
collection of intervals {I_{j}^{(k )}}j∈Nthat covers Ek such that

∞

X

j=1

|I_{j}^{(k )}| < ε
2^{k}.

then the collection {I_{j}^{(k )}}k ,j∈Nis countable, covers E ,and

∞

X

k =1

∞

X

j=1

|I_{j}^{(k )}| ≤

∞

X

j=1

ε
2^{k} = ε.

Consequently, E is of measure zero.

Proof.

**N we can choose a**
collection of intervals {I_{j}^{(k )}}j∈Nthat covers Ek such that

∞

X

j=1

|I_{j}^{(k )}| < ε
2^{k}.

then the collection {I_{j}^{(k )}}k ,j∈Nis countable, covers E , and

∞

X

k =1

∞

X

j=1

|I_{j}^{(k )}| ≤

∞

X

j=1

ε
2^{k} = ε.

Consequently, E is of measure zero.

Proof.

**N we can choose a**
collection of intervals {I_{j}^{(k )}}j∈Nthat covers Ek such that

∞

X

j=1

|I_{j}^{(k )}| < ε
2^{k}.

then the collection {I_{j}^{(k )}}k ,j∈Nis countable, covers E , and

∞

X

k =1

∞

X

j=1

|I_{j}^{(k )}| ≤

∞

X

j=1

ε
2^{k} = ε.

Consequently, E is of measure zero.

Theorem (9.51 Closed Graph Theorem)

Let I be a closed interval and f : I →**R. Then f is**

continuous on I if and only if the graph of f is closed and
connected in**R**^{2}.

Proof.

For any interval J ⊆ I, let G(J) represent the graph of
y = f (x ) for x ∈ J.Suppose that f is continuous on I. The
function x 7→ (x , f (x )) is continuous from I into**R**^{2}, and I is
connected in**R. Thus G(I) is connected in R**^{2}by Theorem
9.30. To prove that G(I) is closed, we shall use Theorem
9.8. Let xk ∈ I and (xk,f (xk)) → (x , y ) as k → ∞. Then
xk → x and f (x_{k}) →y , as k → ∞. Hence, x ∈ I and since
f is continuous, f (xk) →f (x ). In particular, the graph of f
is closed.

Theorem (9.51 Closed Graph Theorem)

Let I be a closed interval and f : I →**R. Then f is**

continuous on I if and only if the graph of f is closed and
connected in**R**^{2}.

Proof.

For any interval J ⊆ I,let G(J) represent the graph of
y = f (x ) for x ∈ J. Suppose that f is continuous on I.The
function x 7→ (x , f (x )) is continuous from I into**R**^{2}, and I is
connected in**R. Thus G(I) is connected in R**^{2}by Theorem
9.30. To prove that G(I) is closed, we shall use Theorem
9.8. Let xk ∈ I and (xk,f (xk)) → (x , y ) as k → ∞. Then
xk → x and f (x_{k}) →y , as k → ∞. Hence, x ∈ I and since
f is continuous, f (xk) →f (x ). In particular, the graph of f
is closed.

Theorem (9.51 Closed Graph Theorem)

Let I be a closed interval and f : I →**R. Then f is**

continuous on I if and only if the graph of f is closed and
connected in**R**^{2}.

Proof.

For any interval J ⊆ I, let G(J) represent the graph of
y = f (x ) for x ∈ J.Suppose that f is continuous on I. The
function x 7→ (x , f (x )) is continuous from I into**R**^{2}, and I is
connected in**R.**Thus G(I) is connected in**R**^{2}by Theorem
9.30. To prove that G(I) is closed, we shall use Theorem
9.8. Let xk ∈ I and (xk,f (xk)) → (x , y ) as k → ∞. Then
xk → x and f (x_{k}) →y , as k → ∞. Hence, x ∈ I and since
f is continuous, f (xk) →f (x ). In particular, the graph of f
is closed.

Theorem (9.51 Closed Graph Theorem)

Let I be a closed interval and f : I →**R. Then f is**

continuous on I if and only if the graph of f is closed and
connected in**R**^{2}.

Proof.

For any interval J ⊆ I, let G(J) represent the graph of
y = f (x ) for x ∈ J. Suppose that f is continuous on I.The
function x 7→ (x , f (x )) is continuous from I into**R**^{2}, and I is
connected in**R. Thus G(I) is connected in R**^{2}by Theorem
9.30.To prove that G(I) is closed, we shall use Theorem
9.8. Let xk ∈ I and (xk,f (xk)) → (x , y ) as k → ∞. Then
xk → x and f (x_{k}) →y , as k → ∞. Hence, x ∈ I and since
f is continuous, f (xk) →f (x ). In particular, the graph of f
is closed.

Theorem (9.51 Closed Graph Theorem)

Let I be a closed interval and f : I →**R. Then f is**

continuous on I if and only if the graph of f is closed and
connected in**R**^{2}.

Proof.

For any interval J ⊆ I, let G(J) represent the graph of
y = f (x ) for x ∈ J. Suppose that f is continuous on I. The
function x 7→ (x , f (x )) is continuous from I into**R**^{2}, and I is
connected in**R.**Thus G(I) is connected in**R**^{2}by Theorem
9.30. To prove that G(I) is closed, we shall use Theorem
9.8.Let xk ∈ I and (xk,f (xk)) → (x , y ) as k → ∞. Then
xk → x and f (x_{k}) →y , as k → ∞. Hence, x ∈ I and since
f is continuous, f (xk) →f (x ). In particular, the graph of f
is closed.

Theorem (9.51 Closed Graph Theorem)

Let I be a closed interval and f : I →**R. Then f is**

continuous on I if and only if the graph of f is closed and
connected in**R**^{2}.

Proof.

For any interval J ⊆ I, let G(J) represent the graph of
y = f (x ) for x ∈ J. Suppose that f is continuous on I. The
function x 7→ (x , f (x )) is continuous from I into**R**^{2}, and I is
connected in**R. Thus G(I) is connected in R**^{2}by Theorem
9.30.To prove that G(I) is closed, we shall use Theorem
9.8. Let xk ∈ I and (xk,f (xk)) → (x , y ) as k → ∞.Then
xk → x and f (x_{k}) →y , as k → ∞. Hence, x ∈ I and since
f is continuous, f (xk) →f (x ). In particular, the graph of f
is closed.

Theorem (9.51 Closed Graph Theorem)

Let I be a closed interval and f : I →**R. Then f is**

continuous on I if and only if the graph of f is closed and
connected in**R**^{2}.

Proof.

For any interval J ⊆ I, let G(J) represent the graph of
y = f (x ) for x ∈ J. Suppose that f is continuous on I. The
function x 7→ (x , f (x )) is continuous from I into**R**^{2}, and I is
connected in**R. Thus G(I) is connected in R**^{2}by Theorem
9.30. To prove that G(I) is closed, we shall use Theorem
9.8.Let xk ∈ I and (xk,f (xk)) → (x , y ) as k → ∞. Then
xk → x and f (x_{k}) →y , as k → ∞.Hence, x ∈ I and since
f is continuous, f (xk) →f (x ). In particular, the graph of f
is closed.

Theorem (9.51 Closed Graph Theorem)

Let I be a closed interval and f : I →**R. Then f is**

continuous on I if and only if the graph of f is closed and
connected in**R**^{2}.

Proof.

For any interval J ⊆ I, let G(J) represent the graph of
y = f (x ) for x ∈ J. Suppose that f is continuous on I. The
function x 7→ (x , f (x )) is continuous from I into**R**^{2}, and I is
connected in**R. Thus G(I) is connected in R**^{2}by Theorem
9.30. To prove that G(I) is closed, we shall use Theorem
9.8. Let xk ∈ I and (xk,f (xk)) → (x , y ) as k → ∞.Then
xk → x and f (x_{k}) →y , as k → ∞. Hence, x ∈ I and since
f is continuous, f (xk) →f (x ). In particular, the graph of f
is closed.

Theorem (9.51 Closed Graph Theorem)

Let I be a closed interval and f : I →**R. Then f is**

continuous on I if and only if the graph of f is closed and
connected in**R**^{2}.

Proof.

For any interval J ⊆ I, let G(J) represent the graph of
y = f (x ) for x ∈ J. Suppose that f is continuous on I. The
function x 7→ (x , f (x )) is continuous from I into**R**^{2}, and I is
connected in**R. Thus G(I) is connected in R**^{2}by Theorem
9.30. To prove that G(I) is closed, we shall use Theorem
9.8. Let xk ∈ I and (xk,f (xk)) → (x , y ) as k → ∞. Then
xk → x and f (x_{k}) →y , as k → ∞.Hence, x ∈ I and since
f is continuous, f (xk) →f (x ). In particular, the graph of f
is closed.

Theorem (9.51 Closed Graph Theorem)

Let I be a closed interval and f : I →**R. Then f is**

continuous on I if and only if the graph of f is closed and
connected in**R**^{2}.

Proof.

For any interval J ⊆ I, let G(J) represent the graph of
y = f (x ) for x ∈ J. Suppose that f is continuous on I. The
function x 7→ (x , f (x )) is continuous from I into**R**^{2}, and I is
connected in**R. Thus G(I) is connected in R**^{2}by Theorem
9.30. To prove that G(I) is closed, we shall use Theorem
9.8. Let xk ∈ I and (xk,f (xk)) → (x , y ) as k → ∞. Then
xk → x and f (x_{k}) →y , as k → ∞. Hence, x ∈ I and since
f is continuous, f (xk) →f (x ). In particular, the graph of f
is closed.

Proof.

Conversely, suppose that the graph of f is closed and
connected in**R**^{2}.We first show that f satisfies the

Intermediate Value Theorem on I. Indeed, suppose to the contrary that there exist x1<x2in I with f (x1) 6=f (x2)and a value y0between f (x1)and f (x2)such that f (t) 6= y0for all t ∈ [x1,x2].Suppose for simplicity that f (x1) <f (x2).

Since f (t) 6= y0for any t ∈ [x1,x2], the open sets U = {(x , y ) : x < x1} ∪ {(x, y ) : x < x2,y < y0},

V = {(x , y ) : x > x2} ∪ {(x, y ) : x > x_{1},y > y0}
separate G(I), a contradiction. Therefore, f satisfies the
Intermediate Value Theorem on I.

Proof.

Conversely, suppose that the graph of f is closed and
connected in**R**^{2}. We first show that f satisfies the

Intermediate Value Theorem on I.Indeed, suppose to the contrary that there exist x1<x2in I with f (x1) 6=f (x2)and a value y0between f (x1)and f (x2)such that f (t) 6= y0for all t ∈ [x1,x2]. Suppose for simplicity that f (x1) <f (x2).

Since f (t) 6= y0for any t ∈ [x1,x2], the open sets U = {(x , y ) : x < x1} ∪ {(x, y ) : x < x2,y < y0},

V = {(x , y ) : x > x2} ∪ {(x, y ) : x > x_{1},y > y0}
separate G(I), a contradiction. Therefore, f satisfies the
Intermediate Value Theorem on I.

Proof.

Conversely, suppose that the graph of f is closed and
connected in**R**^{2}. We first show that f satisfies the

Intermediate Value Theorem on I. Indeed, suppose to the contrary that there exist x1<x2in I with f (x1) 6=f (x2)and a value y0between f (x1)and f (x2)such that f (t) 6= y0for all t ∈ [x1,x2].Suppose for simplicity that f (x1) <f (x2).

Since f (t) 6= y0for any t ∈ [x1,x2],the open sets U = {(x , y ) : x < x1} ∪ {(x, y ) : x < x2,y < y0},

V = {(x , y ) : x > x2} ∪ {(x, y ) : x > x_{1},y > y0}
separate G(I), a contradiction. Therefore, f satisfies the
Intermediate Value Theorem on I.

Proof.

Conversely, suppose that the graph of f is closed and
connected in**R**^{2}. We first show that f satisfies the

Intermediate Value Theorem on I. Indeed, suppose to the contrary that there exist x1<x2in I with f (x1) 6=f (x2)and a value y0between f (x1)and f (x2)such that f (t) 6= y0for all t ∈ [x1,x2]. Suppose for simplicity that f (x1) <f (x2).

Since f (t) 6= y0for any t ∈ [x1,x2], the open sets U = {(x , y ) : x < x1} ∪ {(x, y ) : x < x2,y < y0},

V = {(x , y ) : x > x2} ∪ {(x, y ) : x > x_{1},y > y0}
separate G(I), a contradiction.Therefore, f satisfies the
Intermediate Value Theorem on I.

Proof.

**R**^{2}. We first show that f satisfies the

Intermediate Value Theorem on I. Indeed, suppose to the contrary that there exist x1<x2in I with f (x1) 6=f (x2)and a value y0between f (x1)and f (x2)such that f (t) 6= y0for all t ∈ [x1,x2]. Suppose for simplicity that f (x1) <f (x2).

Since f (t) 6= y0for any t ∈ [x1,x2],the open sets U = {(x , y ) : x < x1} ∪ {(x, y ) : x < x2,y < y0},

_{1},y > y0}
separate G(I), a contradiction. Therefore, f satisfies the
Intermediate Value Theorem on I.

Proof.

**R**^{2}. We first show that f satisfies the

Intermediate Value Theorem on I. Indeed, suppose to the contrary that there exist x1<x2in I with f (x1) 6=f (x2)and a value y0between f (x1)and f (x2)such that f (t) 6= y0for all t ∈ [x1,x2]. Suppose for simplicity that f (x1) <f (x2).

V = {(x , y ) : x > x2} ∪ {(x, y ) : x > x_{1},y > y0}
separate G(I), a contradiction.Therefore, f satisfies the
Intermediate Value Theorem on I.

Proof.

**R**^{2}. We first show that f satisfies the

_{1},y > y0}
separate G(I), a contradiction. Therefore, f satisfies the
Intermediate Value Theorem on I.

Proof.

If f is not continuous on I,then there exist numbers x0 ∈ I, ε0>0, and xk ∈ I such that xk → x0 and

|f (x_{k}) −f (x0)| > ε_{0}. By symmetry, we may suppose that
f (xk) >f (x0) + ε_{0} for infinitely many k ’s,say

f (xk_{j}) > f (x0) + ε0>f (x0), j ∈ **N.**

By the Intermediate Value Theorem, choose cj between
xkj and x0such that f (cj) = f (x0) + ε_{0}. By construction,
(cj,f (cj)) → (x0,f (x0) + ε0)and cj → x0as j → ∞. Hence,
the graph of f on I is not closed.

Proof.

If f is not continuous on I, then there exist numbers x0 ∈ I, ε0>0, and xk ∈ I such that xk → x0 and

|f (x_{k}) −f (x0)| > ε_{0}. By symmetry, we may suppose that
f (xk) >f (x0) + ε_{0} for infinitely many k ’s, say

f (xk_{j}) > f (x0) + ε0>f (x0), j ∈ **N.**

By the Intermediate Value Theorem, choose cj between
xkj and x0such that f (cj) = f (x0) + ε_{0}. By construction,
(cj,f (cj)) → (x0,f (x0) + ε0)and cj → x0as j → ∞. Hence,
the graph of f on I is not closed.

Proof.

If f is not continuous on I, then there exist numbers x0 ∈ I, ε0>0, and xk ∈ I such that xk → x0 and

|f (x_{k}) −f (x0)| > ε_{0}. By symmetry, we may suppose that
f (xk) >f (x0) + ε_{0} for infinitely many k ’s,say

f (xk_{j}) > f (x0) + ε0>f (x0), j ∈ **N.**

By the Intermediate Value Theorem,choose cj between
xkj and x0such that f (cj) = f (x0) + ε_{0}. By construction,
(cj,f (cj)) → (x0,f (x0) + ε0)and cj → x0as j → ∞. Hence,
the graph of f on I is not closed.

Proof.

If f is not continuous on I, then there exist numbers x0 ∈ I, ε0>0, and xk ∈ I such that xk → x0 and

|f (x_{k}) −f (x0)| > ε_{0}. By symmetry, we may suppose that
f (xk) >f (x0) + ε_{0} for infinitely many k ’s, say

f (xk_{j}) > f (x0) + ε0>f (x0), j ∈ **N.**

By the Intermediate Value Theorem, choose cj between
xkj and x0such that f (cj) = f (x0) + ε_{0}.By construction,
(cj,f (cj)) → (x0,f (x0) + ε0)and cj → x0as j → ∞. Hence,
the graph of f on I is not closed.

Proof.

If f is not continuous on I, then there exist numbers x0 ∈ I, ε0>0, and xk ∈ I such that xk → x0 and

|f (x_{k}) −f (x0)| > ε_{0}. By symmetry, we may suppose that
f (xk) >f (x0) + ε_{0} for infinitely many k ’s, say

f (xk_{j}) > f (x0) + ε0>f (x0), j ∈ **N.**

By the Intermediate Value Theorem,choose cj between
xkj and x0such that f (cj) = f (x0) + ε_{0}. By construction,
(cj,f (cj)) → (x0,f (x0) + ε0)and cj → x0as j → ∞.Hence,
the graph of f on I is not closed.

Proof.

If f is not continuous on I, then there exist numbers x0 ∈ I, ε0>0, and xk ∈ I such that xk → x0 and

_{k}) −f (x0)| > ε_{0}. By symmetry, we may suppose that
f (xk) >f (x0) + ε_{0} for infinitely many k ’s, say

f (xk_{j}) > f (x0) + ε0>f (x0), j ∈ **N.**

By the Intermediate Value Theorem, choose cj between
xkj and x0such that f (cj) = f (x0) + ε_{0}.By construction,
(cj,f (cj)) → (x0,f (x0) + ε0)and cj → x0as j → ∞. Hence,
the graph of f on I is not closed.

Proof.

If f is not continuous on I, then there exist numbers x0 ∈ I, ε0>0, and xk ∈ I such that xk → x0 and

_{k}) −f (x0)| > ε_{0}. By symmetry, we may suppose that
f (xk) >f (x0) + ε_{0} for infinitely many k ’s, say

f (xk_{j}) > f (x0) + ε0>f (x0), j ∈ **N.**

By the Intermediate Value Theorem, choose cj between
xkj and x0such that f (cj) = f (x0) + ε_{0}. By construction,
(cj,f (cj)) → (x0,f (x0) + ε0)and cj → x0as j → ∞.Hence,
the graph of f on I is not closed.

Proof.

If f is not continuous on I, then there exist numbers x0 ∈ I, ε0>0, and xk ∈ I such that xk → x0 and

_{k}) −f (x0)| > ε_{0}. By symmetry, we may suppose that
f (xk) >f (x0) + ε_{0} for infinitely many k ’s, say

f (xk_{j}) > f (x0) + ε0>f (x0), j ∈ **N.**

By the Intermediate Value Theorem, choose cj between
xkj and x0such that f (cj) = f (x0) + ε_{0}. By construction,
(cj,f (cj)) → (x0,f (x0) + ε0)and cj → x0as j → ∞. Hence,
the graph of f on I is not closed.