recursal/Fanatic-Fandom huggingface dataset:
9-6

**Question:** Prove Theorem 9.4, that is, prove that if

f
∈
L
(

R

n

)

{\displaystyle f\in L(\mathbb {R} ^{n})}

 and

K

{\displaystyle K}

 is bounded and uniformly continuous on

R

n

{\displaystyle \R^n}

, then

f
∗
K

{\displaystyle f\*K}

 is bounded and uniformly continuous on

R

n

{\displaystyle \R^n}

.

**Answer**: Since

K

{\displaystyle K}

 is bounded on

R

n

{\displaystyle \R^n}

, there is a constant

M
>
0

{\displaystyle M > 0}

 such that

K
(

x

)
≤
M

{\displaystyle K(\mathbf {x} )\leq M}

 for all

x

∈

R

n

{\displaystyle \mathbf x \in \mathbb R^n}

. Then for all

x

∈

R

n

{\displaystyle \mathbf x \in \mathbb R^n}

,

(
f
∗
K
)
(

x

)

=

∫

R

n

f
(

x

−

t

)
K
(

t

)

d

t

≤

∫

R

n

f
(

x

−

t

)

d

t

M

≤
‖
f

‖

1

M

<
∞
.

This shows that

f
∗
K

{\displaystyle f\*K}

 is bounded on

R

n

{\displaystyle \R^n}

. Let

ε
>
0

{\displaystyle \varepsilon > 0}

. Since

K

{\displaystyle K}

 is uniformly continuous on

R

n

{\displaystyle \R^n}

, there exists

δ
>
0

{\displaystyle \delta > 0}

 such that for any

x

,

y

∈

R

n

{\displaystyle \mathbf {x} ,\mathbf {y} \in \mathbb {R} ^{n}}

, if

‖

x

−

y

‖
<
δ

{\displaystyle \|\mathbf {x} -\mathbf {y} \|<\delta }

, then

K
(

x

)
−
K
(

y

)

<

ε

‖
f

‖

1

1

.

Hence, for all

‖

x

−

y

‖
<
δ

{\displaystyle \|\mathbf {x} -\mathbf {y} \|<\delta }

, we have

(
f
∗
K
)
(

x

)
−
(
f
∗
K
)
(

y

)

=

∫

R

n

f
(

t

)
K
(

x

−

t

)

d

t

−

∫

R

n

f
(

t

)
K
(

y

−

t

)

d

t

=

∫

R

n

f
(

t

)
[
K
(

x

−

t

)
−
K
(

y

−

t

)
]

d

t

≤

ε

‖
f

‖

1

1

∫

R

n

f
(

t

)

d

t

=

ε

‖
f

‖

1

1

‖
f

‖

1

<
ε
.

This proves that

f
∗
K

{\displaystyle f\*K}

 is uniformly continuous.
Ex2-8

Assume

f
∈
V
(
−
∞
,
∞
)

{\displaystyle f\in V(-\infty,\infty)}

. Let

v
(
x
# )
V
(
−
∞
,
x
)

{\displaystyle v(x)=V(-\infty,x)}

 for

x
∈
(
−
∞
,
∞
)

{\displaystyle x \in (- \infty , \infty) }

. It is clear that

f
(
x
# )
v
(
x
)
−
[
v
(
x
)
−
f
(
x
)
]

{\displaystyle f(x)=v(x)-[v(x)-f(x)]}

Then we need to show that

v
(
x
)

{\displaystyle v(x)}

 and

v
(
x
)
−
f
(
x
)

{\displaystyle v(x)-f(x)}

 is increasing function. For any

x
1

<

x
2

{\displaystyle x\_1 < x\_2}

, we have

v
(

x
2

)
−
v
(

x
1

# )
V
[
f
;

x
1

,

x
2

]
≥
0

{\displaystyle v(x\_2)-v(x\_1)=V[f;x\_1,x\_2]\geq 0}

Since

f
(
x
)

{\displaystyle f(x)}

 is of bounded variation, this implies

v
(
x
)

{\displaystyle v(x)}

is increasing bounded function. Moreover,

f
(

x
2

)
−
f
(

x
1

)
≤

f
(

x
2

)
−
f
(

x
1

)

≤
V
[
f
;

x
1

,

x
2

# ]
v
(

x
2

)
−
v
(

x
1

)

{\displaystyle f(x\_2)-f(x\_1)\leq |f(x\_2)-f(x\_1)|\leq V[f;x\_1,x\_2]=v(x\_2)-v(x\_1)}

So we find that

v
(

x
1

)
−
f
(

x
1

)
≤
v
(

x
2

)
−
f
(

x
2

)

{\displaystyle v(x\_1)-f(x\_1)\leq v(x\_2)-f(x\_2)}

Since

f
(
x
)

{\displaystyle f(x)}

 and

v
(
x
)

{\displaystyle v(x)}

 both are of bounded variation,

v
(
x
)
−
f
(
x
)

{\displaystyle v(x)-f(x)}

 is incresing bounded function. Hence

f
(
x
)

{\displaystyle f(x)}

 is the difference of two increasing bounded functions.
Presentation du livre

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Ex4-16

Question: Prove that

f
k

→

m

f

{\displaystyle f\_k\xrightarrow{m}f}

 on

E

{\displaystyle E}

 if and only if given

ϵ
>
0

{\displaystyle \epsilon > 0}

, there exists

K

{\displaystyle K}

 such that

{

f
−

f

k

>
ϵ
}

<
ϵ

 if

k
>
K

{\displaystyle k>K}

. Give an analogous Cauchy criterion.

Answer:

(
⟹
)

{\displaystyle (\Longrightarrow )}

Assume

f
k

→

m

f

{\displaystyle f\_k\xrightarrow{m}f}

 on

E

{\displaystyle E}

. Let

ϵ
>
0

{\displaystyle \epsilon > 0}

 be given. By definition of convergence in measure,

lim

x
→
∞

{
x
∈
E
:

f
(
x
)
−

f

k

(
x
)

>
ϵ
}

=
0

.

By definition of limit, there exists

K

{\displaystyle K}

 such that

{

f
−

f

k

>
ϵ
}

<
ϵ

 if

k
>
K

{\displaystyle k>K}

.

(
⟸
)

{\displaystyle (\Longleftarrow )}

Assume that given any

ϵ
>
0

{\displaystyle \epsilon > 0}

, there exists

K

{\displaystyle K}

 such that

{

f
−

f

k

>
ϵ
}

<
ϵ

 if

k
>
K

{\displaystyle k>K}

.

Let

δ
>
0

{\displaystyle \delta > 0}

. Given any

ϵ
>
0

{\displaystyle \epsilon > 0}

, we wish to show that there exists

K

{\displaystyle K}

 such that

{

f
−

f

k

>
δ
}

<
ϵ

 if

k
<
K

{\displaystyle k<K}

. This would imply

lim

k
→
∞

{

f
−

f

k

>
δ
}

=
0

, and thus

f
k

→

m

f

{\displaystyle f\_k\xrightarrow{m}f}

.

**case 1** :

ϵ
≥
δ

{\displaystyle \epsilon \geq \delta }

.

By assumption, there exists

K

{\displaystyle K}

 such that

{

f
−

f

k

>
δ
}

<
δ
≤
ϵ

 if

k
>
K

{\displaystyle k>K}

.

**case 2** :

ϵ
<
δ

{\displaystyle \epsilon <\delta }

.

Since

{

f
−

f

k

>
δ
}
⊆
{

f
−

f

k

>
ϵ
}

, thus

{

f
−

f

k

>
δ
}

≤

{

f
−

f

k

>
ϵ
}

<
ϵ

 if

k
>
K

{\displaystyle k>K}

.

**Give an analogous Cauchy criterion**:

f
k

→

m

f

{\displaystyle f\_k\xrightarrow{m}f}

 on

E

{\displaystyle E}

 if and only if given

ϵ
>
0

{\displaystyle \epsilon > 0}

, there exists

K

{\displaystyle K}

 such that

{

f

k

−

f

l

>
ϵ
}

<
ϵ

 if

k
,
l
>
K

{\displaystyle k,l>K}

.

Proof is analogous to the previous part, using Theorem 4.23 (Cauchy criterion for convergence in measure).
7-5

Note that g is absolutely continuous. So g is of bounded variation on

[
a
,
b
]

{\displaystyle [a,b]}

. Then we know

# h
f
−
g

{\displaystyle h=f-g}

 is also of bounded variation on

[
a
,
b
]

{\displaystyle [a,b]}

. Then

∫

a

b

ϕ
d
f

=

lim

P
→
0

∑
ϕ
(
ϵ
)
(
f
(

x

i

)
−
f
(

x

i
−
1

)
)

=

lim

P
→
0

∑
ϕ
(
ϵ
)
(
g
(

x

i

)
h
(

x

i

)
−
g
(

x

i
−
1

)
−
h
(

x

i
−
1

)
)

=

lim

P
→
0

∑
ϕ
(
ϵ
)
(
g
(

x

i

)
−
g
(

x

i
−
1

)
)

lim

P
→
0

∑
ϕ
(
ϵ
)
(
h
(

x

i

)
−
h
(

x

i
−
1

)
)

=

∫

a

b

ϕ
d
g

∫

a

b

ϕ
d
h
.

{\displaystyle {\begin{alignedat}{2}\int \_{a}^{b}\phi df&=\lim \_{P\to 0}\sum \phi (\epsilon )(f(x\_{i})-f(x\_{i-1}))\\&=\lim \_{P\to 0}\sum \phi (\epsilon )(g(x\_{i})+h(x\_{i})-g(x\_{i-1})-h(x\_{i-1}))\\&=\lim \_{P\to 0}\sum \phi (\epsilon )(g(x\_{i})-g(x\_{i-1}))+\lim \_{P\to 0}\sum \phi (\epsilon )(h(x\_{i})-h(x\_{i-1}))\\&=\int \_{a}^{b}\phi dg+\int \_{a}^{b}\phi dh.\end{alignedat}}}

Then by Theorem 7.32 we get

∫

a

b

ϕ
d
# g

∫

a

b

ϕ

g
′

d
# x

∫

a

b

ϕ

f
′

d
x

{\displaystyle \int \_{a}^{b}\phi dg=\int \_{a}^{b}\phi g'dx=\int \_{a}^{b}\phi f'dx}

since

f
′

=

g
′

h
′

=

g
′

{\displaystyle f'=g'+h'=g'}

 almost everywhere on

[
a
,
b
]

{\displaystyle [a,b]}

.

Combine these two results, we have

∫

a

b

ϕ
d
# f

∫

a

b

ϕ

f
′

d
x

∫

a

b

ϕ
d
h

{\displaystyle \int \_{a}^{b}\phi df=\int \_{a}^{b}\phi f'dx+\int \_{a}^{b}\phi dh}

This complete the proof.
5-1

Question: If

f

{\displaystyle f}

 is a simple measurable function (not necessarily nonnegative) taking values

a
j

{\displaystyle a\_j}

 on

E
j

{\displaystyle E\_j}

,

# j
1
,
2
,
3...
N

{\displaystyle j=1,2,3...N}

. Show that

∫

E

# j
1

N

a

j

E

j

{\displaystyle \int \_{E}f=\sum \_{j=1}^{N}a\_{j}|E\_{j}|}

.(Use Theorem 5.24. )

Proof:

Theorem 5.24: If

∫
E

f

{\displaystyle \int\_E f}

 exists and

# E

⋃

k

E

k

{\displaystyle E=\bigcup \_{k}^{}E\_{k}}

 is the countable union of disjoint measurable sets

E
k

{\displaystyle E\_k}

, then

∫

E

# f

∑

k

∫

E

k

f

{\displaystyle \int \_{E}f=\sum \_{k}^{}\int \_{E\_{k}}f}

 .

Since

# k
1

N

a

k

χ

E

k

{\displaystyle f=\sum \_{k=1}^{N}a\_{k}\chi \_{E\_{k}}}

 is a simple function,

f

{\displaystyle f}

 is measurable and its Lebesgue integral exists. From theorem 5.24, we could write the Lebesgue integral in this way:

∫

E

f

=

∑

k

∫

E

k

f

=

∑

k

∫

E

k

∑

# i
1

N

a

i

χ

E

i

=

∑

k

∑

# i
1

N

a

i

∫

E

k

χ

E

i

{\displaystyle {\begin{aligned}\int \_{E}f&=\sum \_{k}^{}\int \_{E\_{k}}f\\&=\sum \_{k}^{}\int \_{E\_{k}}\sum \_{i=1}^{N}a\_{i}\chi \_{E\_{i}}\\&=\sum \_{k}^{}\sum \_{i=1}^{N}a\_{i}\int \_{E\_{k}}\chi \_{E\_{i}}\end{aligned}}}

When

k
≠
i

{\displaystyle k \ne i }

,

∫

E

k

χ

E

i

=

∫

E

k

# 0
0

{\displaystyle \int \_{E\_{k}}\chi \_{E\_{i}}=\int \_{E\_{k}}0=0}

; When

# k
i

{\displaystyle k=i}

,

∫

E

k

χ

E

i

=

∫

E

k

# 1

E

k

{\displaystyle \int \_{E\_{k}}\chi \_{E\_{i}}=\int \_{E\_{k}}1=|E\_{k}|}

.

Hence,

∫

E

# f

∑

k

∑

# i
1

N

a

i

∫

E

k

χ

E

i

=

∑

# j
1

N

a

j

E

j

{\displaystyle \int \_{E}f=\sum \_{k}^{}\sum \_{i=1}^{N}a\_{i}\int \_{E\_{k}}\chi \_{E\_{i}}=\sum \_{j=1}^{N}a\_{j}|E\_{j}|}

.
Ex3-18

**Question.** Prove that the outer measure is *translation invariant*; that is, if

E

h

=
{

x

h

∈

R

n

∣

x

∈
E
}

{\displaystyle E\_{\mathbf{h}} = \{\mathbf{x} + \mathbf{h} \in \R^n \mid \mathbf{x} \in E\}}

 is the translate of

E

{\displaystyle E}

 by

h

∈

R

n

{\displaystyle \mathbf{h} \in \R^n}

, show that

E

h

e

=

E

e

. If

E

{\displaystyle E}

 is measurbale, show that

E

h

{\displaystyle E\_\mathbf{h}}

 is also measurable.

**Answer.** Given a countable cover

# S
{

I
i

}

i
∈

Z

{\displaystyle S = \{I\_i\}\_{i \in \Z^+}}

 of

E

{\displaystyle E}

 by closed intervals, observe that

S

h

=

{

(

I
i

)

h

}

i
∈

Z

{\displaystyle S\_{\mathbf{h}} = \left\{(I\_i)\_\mathbf{h}\right\}\_{i \in \Z^+}}

 is a countable cover of

E

h

{\displaystyle E\_\mathbf{h}}

 by closed intervals, where

(

I
i

)

h

{\displaystyle (I\_i)\_{\mathbf{h}}}

 is

I
i

{\displaystyle I\_i}

 translated by

h

{\displaystyle \mathbf{h}}

 for each

i
∈

Z

{\displaystyle i \in \Z^+}

. Clearly

σ
(
S
# )
σ
(

S

h

)

{\displaystyle \sigma(S) = \sigma(S\_\mathbf{h})}

 since

v
(

I
i

# )
v
(
(

I
i

)

h

)

{\displaystyle v(I\_i) = v((I\_i)\_\mathbf{h})}

. Thus,

S

=
{
σ
(
S
)
∣
S

is a countable cover of
E
 by closed intervals
}

⊆
{
σ
(

S

h

)
∣

S

h

 is a countable cover of

E

h

 by closed intervals
# }

S

h

,

{\displaystyle \begin{align}
\mathcal{S} = \{\sigma(S) \mid S \ & \text{is a countable cover of } E \text{ by closed intervals}\} \\
&\sube \{\sigma(S\_\mathbf{h}) \mid S\_{\mathbf{h}} \text{ is a countable cover of } E\_\mathbf{h} \text{ by closed intervals}\} = \mathcal{S}\_\mathbf{h},
\end{align}}

so

inf
(

S

)
≥
inf
(

S

h

)
.

{\displaystyle \inf(\mathcal{S}) \ge \inf(\mathcal{S}\_\mathbf{h}).}

Repeating the argument for

E

h

{\displaystyle E\_\mathbf{h}}

 and

(

E

h

)

−

h

=
E

{\displaystyle (E\_\mathbf{h})\_{-\mathbf{h}} = E}

 gives us the opposite inequality, therefore,

E

e

=
inf
(

S

# )
inf
(

S

h

# )

E

h

e

,

which proves the first part.

Now, suppose that

E

{\displaystyle E}

 is measurable. Then for every

ε
>
0

{\displaystyle \varepsilon > 0}

, there exists an open set

G

{\displaystyle G}

 with

E
⊆
G

{\displaystyle E \sube G}

 such that

G
∖
E

e

<
ε

{\displaystyle |G \setminus E|\_e < \varepsilon}

. Then

G

h

{\displaystyle G\_\mathbf{h}}

 is an open set such that

G

h

∖

E

h

=

(
G
∖
E

)

h

e

=

G
∖
E

<
ε
.

This proves that

E

h

{\displaystyle E\_\mathbf{h}}

 is also measurable.
Bossa nova

Bossa nova was conceived as a new style to perform original Samba, with major American Jazz music influences. The term "bossa" could mean "way", "mode" or "style", giving significance to its name. Originating in the 1950s, the genre developed in brief performances of syncopated samba in social meetings of the Rio de Janeiro middle-class, with the slight infuence of Bebop and Cool Jazz, musical genres which were introduced in Brazil thanks to the strong influence of American post-war culture during the Juscelino Kubitschek government. The style, during its early stages of development, spread to other significant cultural areas of Rio de Janeiro (universities, pubs), giving it more presence into the local and Brazilian culture.

The year of 1959, with the release of *Chega de saudade* by the artist João Gilberto, is considered the official birth date of bossa nova. The first generation of the genre was represented by Antônio Carlos Jobim, some of whose compositions were performed by João Gilberto, and Vinicius de Moraes, one of the main lyricists of the inception of bossa nova. The duo of Jobim and de Moraes later composed another famous representative bossa nova work, "Garota de Ipanema". Carlos Lyra and Nara Leão were other great names of the first generation.

In the mid-1960s, the genre suffered a cultural split, motivated by artists like Edu Lobo and Marcos Valle and by Brazilian culture preservation organizations. Inspired by a nationalistic and left-wing ideology, these groups criticized the American jazz influence in the music, suggesting Brazilian Music-rooted influences instead. The second generation of the genre was conceived with bossa nova performers associating and collaborating with other great names of other Brazilian genres, giving a more "Brazilian" sound to the genre. In 1966 the bossa nova sound, incorporating strong native Brazilian musical influences, had moved away in style, composition and musical elements from the original genre, marking its end and the conception of the MPB movement, which had great initial influences of the late Brazilian-focused bossa nova sound.
5-6

We let

(

h
n

)

{\displaystyle (h\_n)}

 be family such that

lim

n
→
∞

h
n

=
0

{\displaystyle \lim\_{n\to \infty}h\_n=0}

 and for any

x
≥
0

{\displaystyle x \ge 0}

, define

f
n

(
y
# )

f
(
x

h
n

,
y
)
−
f
(
x
,
y
)

h
n

{\displaystyle f\_n(y)=\frac{f(x+h\_n,y)-f(x,y)}{h\_n}}

Since

∫
f
(
x
,
y
)
d
y

{\displaystyle \int f(x,y)dy}

 exists for all

x
≥
0

{\displaystyle x \ge 0}

,

f
(
x
,
y
)

{\displaystyle f(x, y)}

 is measurable function of

y

{\displaystyle y}

Also, addition of measurable functions and scalar multiplication of measurable function is measurable,

f
n

(
y
)

{\displaystyle f\_n(y)}

 are measurable for all

n

{\displaystyle n}

Now, for any

x
≥
0

{\displaystyle x \ge 0}

lim

n
→
∞

f
n

(
y
# )

lim

n
→
∞

f
(
x

h
n

,
y
)
−
f
(
x
,
y
)

h
n

=

lim

h
→
0

f
(
x
h
,
y
)
−
f
(
x
,
y
)

h

=

∂
f

∂
x

(
x
,
y
)

{\displaystyle \lim\_{n\to \infty}f\_n(y)=\lim\_{n\to\infty}\frac{f(x+h\_n,y)-f(x,y)}{h\_n}=
\lim\_{h\to 0}\frac{f(x+h,y)-f(x,y)}{h}=\frac{\partial f}{\partial x}(x,y)}

The family

(

f
n

)

{\displaystyle (f\_n)}

 is pointwise convergence to

∂
f

∂
x

(
x
,
y
)

{\displaystyle \frac{\partial f}{\partial x}(x,y)}

 hence

∂
f

∂
x

(
x
,
y
)

{\displaystyle \frac{\partial f}{\partial x}(x,y)}

 is measurable on

y

{\displaystyle y}

. This proof the first claim.

For the second assertion, for any

ϵ
>
0

{\displaystyle \epsilon > 0}

, there is some

N
>
0

{\displaystyle N > 0}

 such that for all

n
≥
N

{\displaystyle n \geq N}

, we have

f
n

(
y
)

−

∂
f

∂
x

(
x
,
y
)

≤

f
n

(
y
)
−

∂
f

∂
x

(
x
,
y
)

<
ϵ

⟹

{\displaystyle \implies}

f
n

(
y
)

<
ϵ

∂
f

∂
x

(
x
,
y
)

<
ϵ
M

,

M

{\displaystyle M}

comes from the hypothesis that

∂
f

∂
x

(
x
,
y
)

{\displaystyle \frac{\partial f}{\partial x}(x,y)}

 is bounded. Since constants are integrable, by Lebesgue dominated convergence theorem,

lim

n
≥
N
,
n
→
∞

∫

f
n

(
y
)
d
# y
∫

∂
f

∂
x

(
x
,
y
)
d
y

{\displaystyle \lim\_{n\geq N,n\to\infty}\int f\_n(y)dy=\int{\frac{\partial f}{\partial x}}(x,y)dy}

. But

lim

n
≥
N
,
n
→
∞

∫

f
n

(
y
)
d
# y

lim

n
≥
N
,
n
→
∞

1

h
n

∫
f
(
x

h
n

,
y
)
−
f
(
x
,
y
)
d
# y

{\displaystyle \lim\_{n\geq N,n\to\infty}\int f\_n(y)dy=
\lim\_{n\geq N,n\to\infty}\frac{1}{h\_n}\int f(x+h\_n,y)-f(x,y)dy=

}

lim

h
→
0

1
h

∫
f
(
x
h
,
y
)
−
f
(
x
,
y
)
d
# y

lim

h
→
0

∫
f
(
x
h
,
y
)
d
y
−
∫
f
(
x
,
y
)
d
y

h

=

{\displaystyle \lim\_{h\to 0}\frac{1}{h}\int f(x+h,y)-f(x,y)dy=
\lim\_{h\to 0}\frac{\int f(x+h,y)dy-\int f(x,y)dy}{h}=
}

d

d
x

∫
f
(
x
,
y
)
d
y

{\displaystyle \frac{d}{dx}\int f(x,y)dy
}

, The equality follows.
Ex4-3

### Problem 4.3

Theorem 4.3 can be used to define measurability for vector-valued (e.g., complex-valued) functions. Suppose, for example, that

f

{\textstyle f}

 and

g

{\textstyle g}

 are real-valued and finite in

R

n

{\displaystyle \R^n}

, and let

F
(
x
# )

(

f
(
x
)
,
g
(
x
)

)

{\textstyle F(x)=\left(f(x),g(x)\right)}

. Then

F

{\displaystyle F}

 is said to be measurable if

F

−
1

(
G
)

{\displaystyle F^{-1}(G)}

 is measurable for every open

G
⊂

R

2

{\displaystyle G \subset \R^2}

. Prove that

F

{\displaystyle F}

 is measurable if and only if both

f

{\textstyle f}

 and

g

{\textstyle g}

 are measurable in

R

2

{\displaystyle \mathbb R^{2}}

.

### Solution.

Suppose that

F

{\displaystyle F}

 is measurable, then

{

f
>
a

}

=

F

−
1

(

(

a
,
∞

)

×

R

)

{\displaystyle \left\{f>a\right\}=F^{-1}\left(\left(a,\infty\right)\times \mathbb{R}\right)}

 and

{

g
>
a

}

=

F

−
1

(

R

×

(

a
,
∞

)

)

{\displaystyle \left\{g>a\right\}=F^{-1}\left(\mathbb{R}\times\left(a,\infty\right)\right)}

 are measurable for all

a
∈

R

{\displaystyle a \in \R}

 since both

(

a
,
∞

)

×

R

⊂

R

2

{\displaystyle \left(a,\infty\right)\times\mathbb{R}\subset \mathbb{R}^2}

 and

R

×

(

a
,
∞

)

⊂

R

2

{\displaystyle \mathbb{R}\times\left(a,\infty\right)\subset \mathbb{R}^2}

 are open subsets of

R

2

{\displaystyle \mathbb R^{2}}

. This shows that both

f

{\textstyle f}

 and

g

{\textstyle g}

 are measurable in

R

2

{\displaystyle \mathbb R^{2}}

. Conversely, if both

f

{\textstyle f}

 and

g

{\textstyle g}

 are measurable in

R

2

{\displaystyle \mathbb R^{2}}

, then for every open set

G
⊂

R

2

{\displaystyle G \subset \R^2}

 that can be written as the union of closed rectangles whose corners have rational coordinates, we have

F

−
1

(
G
# )

F

−
1

(

⋃
k

[

a
k

,

b
k

]

×

[

c
k

,

d
k

]

)

=

⋃
k

F

−
1

(

[

a
k

,

b
k

]

×

[

c
k

,

d
k

]

)

=

⋃
k

[

{

a
k

≤
f
≤

b
k

}

∩

{

c
k

≤
g
≤

d
k

}

]

{\displaystyle F^{-1}(G)=F^{-1}\left(\bigcup\_k \left[a\_k,b\_k\right]\times \left[c\_k,d\_k\right]\right)=\bigcup\_k F^{-1}\left(\left[a\_k,b\_k\right]\times \left[c\_k,d\_k\right]\right)=\bigcup\_k\left[\left\{a\_k\leq f\leq b\_k\right\}\cap \left\{c\_k\leq g\leq d\_k\right\}\right]}

which is a countable union of measurable sets, thus

F

{\displaystyle F}

 is a measurable set.

◼

{\textstyle \blacksquare}
8-1

i)

∫
E

f

{\displaystyle \int\_E f}

 is finite, if and only if each component is finite, i.e.

∫
E

f
1

{\displaystyle \int\_E f\_1}

and

∫
E

f
2

{\displaystyle \int\_E f\_2}

 is finite. This is if and only if

f
1

{\displaystyle f\_1}

,

f
2

{\displaystyle ,f\_2}

 is finite a.e. in

E

{\displaystyle E}

, hence if and only if

∫
E

f

=

∫
E

f
1
2

f
2
2

 is finite
.

{\displaystyle \int\_E |f|=\int\_E \sqrt{f\_1^2+f\_2^2} \text{ is finite}.}

Before that, we will need a trigonometric formula

a
cos
⁡
α
b
sin
⁡
# α

(

a
2

b
2

)
sin
⁡
(
θ
α
)

{\displaystyle a\cos \alpha+b \sin \alpha=\sqrt(a^2+b^2)\sin(\theta+\alpha)

}

 , with

# θ
arctan
⁡

a
b

{\textstyle \theta = \arctan \frac{a}{b}
}

.

To show the inequality, using the hints we obtain

∫
E

f

=
[
(

∫
E

f
1

)
2

(

∫
E

f
2

)
2

]

1
2

=

∫
E

f
1

cos
⁡
α

∫
E

f
2

sin
⁡
α

{\textstyle \text{To show the inequality, using the hints we obtain }

}

, with

# α
arctan
⁡
(

b
a

)

{\textstyle \alpha=\arctan(\frac{b}{a})
}

.

Where

# b

∫
E

f
1

,
# a

∫
E

f
2

{\displaystyle b=\int\_E f\_1, a=\int\_E f\_2
}

.

Now

∫
E

f

=
(
a
cos
⁡
α
b
sin
⁡
α
)

=
R
e
(
a
cos
⁡
α
b
sin
⁡
α
)

=

∫
E

R
e
(

f
1

cos
⁡
α

f
2

sin
⁡
α
)

≤

∫
E

f
1

cos
⁡
α

f
2

sin
⁡
α

≤

∫
E

(

f
1
2

f
2
2

)

=

∫
E

f

.

{\displaystyle \begin{align}
\end{align}

}
McDonalds.exe

**McDonalds.exe** was a screamer program that spread across the Internet in the early 2000s. It was designed by Xu Haiyi and developed by the Ye Yu Software Company in September 2000, in China.

The program icon is the McDonald's logo on a red background. Even though one would think otherwise, this has nothing to do with the McDonald's fast-food franchise.

When the user launches the program, the screen turns black and some green Chinese text appears, which is a horror story. A few seconds later, the text disappears and the screen returns to its original state. Shortly after, a dialog window appears with some Chinese text, an e-mail address (xu\_yyfy@yeah.net), and a phone number. After a short delay from when the user closes the window, a black and white picture of a girl with no eyes will pop up along with a short scream.

The image will continue to be displayed at two-minute intervals until the process is terminated in Task Manager.
EX3-13

Problem: Define the inner measure of

E

{\displaystyle E}

 by

E

i

=
sup

F

, where the supremum is taken over all closed subset

F

{\displaystyle F}

 of

E

{\displaystyle E}

. Show that

(i)

E

i

≤

E

e

, and

(ii) If

E

e

<
∞

{\displaystyle |E|\_{e}<+\infty}

, then

E

{\displaystyle E}

 is measurable if and only if

E

i

=

E

e

.

Proof:

(i) By monotonicity of outer measure,

F

=

F

e

≤

E

e

 for all closed subset

F
⊆
E
.

{\displaystyle F\subseteq E.}

 Thus, we have

sup

F

=

E

i

≤

E

e

.

(ii) Suppose

E

{\displaystyle E}

 is measurable: then for every

ϵ
>
0

{\displaystyle \epsilon > 0}

, there exists a closed subset

F
⊆
E

{\displaystyle F\subseteq E}

 such that

E
−
F

e

<
ϵ

{\displaystyle |E-F|\_{e}<\epsilon}

. Since

E

{\displaystyle E}

 is measurable and

F

≤

E

<
∞
,

E
−
F

e

=

E

−

F

<
ϵ
.

 Thus,

E

e

=

E

≤
sup

F

# ϵ

E

i

ϵ

 for all

ϵ
>
0

{\displaystyle \epsilon > 0}

, so

E

e

≤

E

i

,

 which yields the equality by part (i).

Conversely, assume that

E

e

≐

E

i

.

Then given

ε
>
0

{\displaystyle \varepsilon > 0}

, there exist an open set

G

{\displaystyle G}

 and a closed set

F

{\displaystyle F}

 with which

F
⊆
E
⊆
G

{\displaystyle F \subseteq E \subseteq G}

 and such that

G

<
ε

E

e

and

E

e

<

F

ε
,

or equivalently,

G

−

E

e

<
ε

and

E

e

−

F

<
ε
.

Then since

(
E
∖
F
)
⊆
(
G
∖
F
)

{\displaystyle (E \setminus F) \subseteq (G \setminus F)}

, the monotonicity of outer measure says that

E
∖
F

e

≤

G
∖
F

=

G

−

F

<
2
ε
,

where in the second equality we have employed Corollary 3.25. We therefore employ Lemma 3.22 to conclude that

E

{\displaystyle E}

 is measurable.
8-18

Prove the parallelogram law for

L
2

{\displaystyle L^2}

:

‖
f
g

‖

2

‖
f
−
g

‖

2

=
2
‖
f

‖

2

2
‖
g

‖

2

Is this true for

L
p

{\displaystyle L^p}

 when p≠2? The geometric interpretation is that the sum of the squares of the lengths of the diagonals of a parallelogram equals the sum of the squares of the edge lengths.

Solution:

‖
f
g

‖

2

2

=

∫

E

f
g

2

=

∫

E

(

f
g

)

2

,

‖
f
−
g

‖

2

2

=

∫

E

f
−
g

2

=

∫

E

(

f
−
g

)

2

,

so, we have

‖
f
g

‖

2

2

‖
f
−
g

‖

2

2

=

∫

E

(

f
g

)

2

∫

E

(

f
−
g

)

2

=

∫

E

(

f
g

)

2

(

f
−
g

)

2

=

∫

E

2

f

2

2

g

2

=
2

∫

E

f

2

2

∫

E

g

2

=
2
(

∫

E

f

2

)

1
2

×
2

2
(

∫

E

g

2

)

1
2

×
2

=
2
‖
f

‖

2

2

2
‖
g

‖

2

2

{\displaystyle

\begin{aligned}
&=\int\_{E}^{} \left ( f+g \right ) ^{2}+\left ( f-g \right ) ^{2}\\
&=\int\_{E}^{} 2f^{2}+2g^{2} \\
&=2\int\_{E}^{} f^{2}+ 2\int\_{E}^{} g^{2} \\
&=2(\int\_{E}^{} f^{2})^{\frac{1}{2} \times 2}+ 2(\int\_{E}^{} g^{2})^{\frac{1}{2} \times 2}\\
 \end{aligned}}

The proof is done now for p=2.

Now we claim it's not true for p≠2.

Assume WLOG that

E

{\displaystyle E}

 is finite (if it is not, then replace the set

E

{\displaystyle E}

 with a finite subset). Let us write

# E

E

1

∪

E

2

{\displaystyle E=E\_{1} \cup E\_{2}}

, where

E
1

,

E
2

{\displaystyle E\_1, E\_2}

 are non-overlapping measurable subsets of

E

{\displaystyle E}

 so that

E
2

=

E
1

=

E

/

2

. We define

# f

χ

E

1

,
# g

χ

E

2

{\displaystyle f=\chi\_{E\_{1}}, g=\chi\_{E\_{2}}}

.

Then

f
# g

χ

E

1

χ

E

2

=

χ

E

{\displaystyle f+g =\chi\_{E\_{1}}+\chi\_{E\_{2}}=\chi\_{E}}

 and

f
−
# g

χ

E

1

−

χ

E

2

=

{

1

,
x
∈

E

1

−
1

,
x
∈

E

2

{\displaystyle f-g =\chi\_{E\_{1}}-\chi\_{E\_{2}}= \left\{\begin{matrix}
 1 \ ,x \in E\_{1}\\
 -1 \ ,x \in E\_{2}\\
\end{matrix}\right.}

. Note that

f
g

=

f
−
g

=

χ

E

.

When

# p
∞

{\displaystyle p=\infty}

, we check that

‖
f

‖

∞

=
‖
g

‖

∞

=
‖
f
g

‖

∞

=
‖
f
−
g

‖

∞

=

e
s
s

sup

E

f

=
1

{\displaystyle ess \ \sup\_{E} |f| =1}

.

Obviously, the identity doesn't hold.

Let us assume now that p is finite. According to the definition of

L
p

{\displaystyle L\_p}

 norm, we have

‖
f
g

‖

p

2

=
‖
f
−
g

‖

p

2

=

E

2

/

p

,
‖
f

‖

p

2

=

E

1

2

/

p

=

E
2

2

/

p

,
‖
g

‖

p

2

=

E

2

2

/

p

=

E
2

2

/

p

,

Therefore,

2
‖
f

‖

p

2

2
‖
g

‖

p

2

=
2

(

E

1

2

/

p

E

2

2

/

p

)

=
4

E
2

2

/

p

and

‖
f
g

‖

p

2

‖
f
−
g

‖

p

2

=
2

E

2

/

p

They are equal if and only if

4
×
(

1
2

)

2

/

p

=
2

{\displaystyle 4 \times (\frac{1}{2})^{2 / p}=2}

, which means

# p
2

{\displaystyle p = 2}

.

Hence this formula is not true for

L
p

{\displaystyle L^p}

 when p≠2, only when

# p
2

{\displaystyle p = 2}

 can it be true.

\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
7-10

(a) Show that if f is absolutely continuous on [a, b] and Z is a subset of [a, b] of measure zero, then the image set defined by f(Z) = {w : w = f(z), z ∈ Z} also has measure zero. Deduce that the image under f of any measurable subset of [a, b] is measurable. (Compare Theorem 3.33.) (Hint: use the fact that the image of an interval [ai, bi] is an interval of length at most V (bi) − V (ai).)

Solution:
5-3

Question 5.3:

Let

{

f
k

}

{\displaystyle \{ f\_k \}}

 be a sequence of nonnegative measurable functions defined on

E

{\displaystyle E}

. If

f
k

→
f

{\displaystyle f\_k \to f}

 and

f
k

≤
f

{\displaystyle f\_k \leq f}

 a.e. on

E

{\displaystyle E}

, show that

∫
E

f
k

→

∫
E

f
.

{\displaystyle \int\_E f\_k \to \int\_E f.}

Solutions:

By Theorem 4.12, if

{

f
k

}

{\displaystyle \{ f\_k \}}

 is a sequence of measurable functions and

lim

k
→
∞

f
k

{\displaystyle \lim\_{k\to \infty} f\_k}

 exists a.e., it is measurable. Since we are given that

lim

k
→
∞

f
k

=
f

{\displaystyle \lim\_{k\to \infty} f\_k=f}

, we can say that

f

{\displaystyle f}

 is indeed a nonnegative measurable function.

Case 1: If

∫
E

f

{\displaystyle \int\_E f}

 is finite, then we can let

# ϕ
f

{\displaystyle \phi=f}

 and apply the **Lebesgue's Dominated Convergence Theorem for Nonnegative Functions** (Theorem 5.19). Since

f

{\displaystyle f}

 is a measurable function such that

f
k

≤
f

{\displaystyle f\_k \leq f}

 a.e. for all

k

{\displaystyle k}

 and

∫
E

f

{\displaystyle \int\_E f}

 is finite, we can conclude that

∫
E

f
k

→

∫
E

f
.

{\displaystyle \int\_E f\_k \to \int\_E f.}

Case 2: If

∫
E

# f
∞

{\displaystyle \int\_E f=\infty}

, then the fact that

f
k

→
f

{\displaystyle f\_k \to f}

 implies that

∫
E

f
k

→
∞

{\displaystyle \int\_E f\_k \to \infty}

. Thus, we show that

∫
E

f
k

→

∫
E

f
.

{\displaystyle \int\_E f\_k \to \int\_E f.}

This concludes the proof.
5-5

**Question 5.5**. Use Egorov's theorem to prove the bounded convergence theorem.

**Solution**.

Suppose

{

f

k

}

{\displaystyle \left\{f\_{k}\right\}}

 is a sequence of measurable functions such that

f

k

→
f

{\displaystyle f\_{k} \rightarrow f}

  a.e. in  E , where

E

<
∞

{\displaystyle |E|<+\infty}

 , and so that

f

k

(
x
)

≤
M

{\displaystyle \left|f\_{k}(x)\right| \leq M}

  for all

x
∈
E

{\displaystyle x \in E}

 . Given

ε
>
0

{\displaystyle \varepsilon > 0}

, by Egorov's theorem there exists a closed set

F
⊆
E

{\displaystyle F\subseteq E}

 such that

E
−
F

<
ε

{\displaystyle |E - F|<\varepsilon}

  and on which

f

k

→
f

{\displaystyle f\_{k} \rightarrow f}

  uniformly. That is, there exists

K
>
0

{\displaystyle K>0}

 such that every

k

{\displaystyle k}

 for

k
>
K

{\displaystyle k>K}

,

f

k

−
f

<
ε

{\displaystyle \left|f\_{k}-f\right|<\varepsilon}

  on

F

{\displaystyle F}

.

Then we can have

∫

E

f

k

−
f

≤

∫

E
−
F

f

k

−
f

∫

F

f

k

−
f

≤

∫

E
−
F

2
M

∫

F

ε
≤
2
M

E
−
F

ε

E

<
(
2
M

E

)
ε

}

.

Since

E

<
∞

{\displaystyle |E|<+\infty}

, we conclude that for every

ε
>
0

{\displaystyle \varepsilon > 0}

, there exists

K
>
0

{\displaystyle K>0}

 such that

∫

E

f

k

−
f

<
c
ε

{\displaystyle \left|\int\_{E} f\_{k}-f\right|< c \varepsilon}

 for every

k
>
K

{\displaystyle k>K}

, where *c* is a constant, so

∫

E

f

k

→

∫

E

f

{\displaystyle \int\_{E} f\_{k} \rightarrow \int\_{E} f }

 as

k
→
∞

{\displaystyle k \rightarrow \infty}

.
Sandbox

* **We can type things here just like word**
* This is the second item in this list
* x

2

y

2

=

z

2

{\displaystyle x^{2}+y^{2}=z^{2}}

 (Pythagorean Theorem)
* arctan
⁡
π

2

=
0

{\displaystyle {\frac {\arctan \pi }{2}}=0}
* 2

{\displaystyle {\sqrt {2}}}

 is irrational, but

1
2

{\displaystyle {\frac {1}{2}}}

 is not.
* 1

1

3
π

>
0

{\displaystyle {\dfrac {1}{1+{\dfrac {3}{\pi }}}}>0}
* tan
⁡

π
2

=
1

{\displaystyle \tan {\dfrac {\pi }{2}}=1}
* 1
# 1
2

{\displaystyle 1+1=2}
* Let

μ

{\displaystyle \mu }

 be a measure on a measurable space. A measurable space is just a set

X

{\displaystyle X}

 equipped with a sigma algebra

σ

{\displaystyle \sigma }

.
* ζ
(
s

# n
1

∞

1

n

s

{\displaystyle \zeta (s)=\sum \_{n=1}^{\infty }{\frac {1}{n^{s}}}}
* ∀
ε
>
0
,
∃
N
∈

N

 s.t.

f
(
x
)
−
L

<
ε

 whenever

n
>
N
.

{\displaystyle \forall \varepsilon >0,\exists N\in \mathbb {N} {\text{ s.t. }}\left\vert f(x)-L\right\vert <\varepsilon {\text{ whenever }}n>N.}
* Euler's formula:

e

i
π

# 1
0

{\displaystyle e^{i\pi }+1=0}
# * △

b

2

−
4
a
c
.

{\displaystyle \triangle =b^{2}-4ac.}
* d

d
x

(

x

2

# )
2
x

{\displaystyle {\frac {d}{dx}}(x^{2})=2x}
* The exterior measure:

E

e

=

inf

S

V
(
S
# )

inf

S

∑

# n
1

∞

V
(

I

i

)

{\displaystyle |E|\_{e}=\inf \_{S}V(S)=\inf \_{S}\sum \_{n=1}^{\infty }V(I\_{i})}

, where

I

i

{\displaystyle I\_{i}}

 are intervals in

S

{\displaystyle S}

.

/
Comment: It would be helpful for all answers to be preceded by the questions.
Ex3-8

**Chapter 3, Question 8**

Problem: Show that the smallest

σ

{\displaystyle \sigma}

algebra containing all closed sets of Euclidean space is identical to the Borel

σ

{\displaystyle \sigma}

algebra.

Proof: Let

A

{\displaystyle \mathcal{A}}

 be the smallest

σ

{\displaystyle \sigma}

algebra containing all the closed sets of Euclidean space, and

B

{\displaystyle \mathcal{B}}

 the Borel

σ

{\displaystyle \sigma}

algebra, which by definition is the smallest

σ

{\displaystyle \sigma}

algebra containing all the open sets of Euclidean space. We want to show

A

=

B

{\displaystyle \mathcal{A} = \mathcal{B}}

. We do this by showing

A

⊂

B

{\displaystyle \mathcal{A} \subset \mathcal{B}}

 and

B

⊂

A

{\displaystyle \mathcal{B} \subset \mathcal{A}}

. We start with

B

⊂

A

{\displaystyle \mathcal{B} \subset \mathcal{A}}

. Suppose

U
⊂

R

n

{\displaystyle U \subset \mathbf{R}^n}

 is an arbitrary open set. We want to show that

U
⊂

A

{\displaystyle U \subset \mathcal{A}}

. Since

R

n

∖
U

{\displaystyle \mathbf{R}^n \setminus U}

 is closed, we have that

R

n

∖
U
∈

A

{\displaystyle \mathbf{R}^n \setminus U \in \mathcal{A}}

. Furthermore,

A

{\displaystyle \mathcal{A}}

 is closed under complements, so

U
∈

A

{\displaystyle U \in \mathcal{A}}

. This shows that

A

{\displaystyle \mathcal{A}}

 contains all the open sets of Euclidean space. Thus we have

B

⊂

A

{\displaystyle \mathcal{B} \subset \mathcal{A}}

.

Now we show

A

⊂

B

{\displaystyle \mathcal{A} \subset \mathcal{B}}

. Let

C

{\displaystyle C}

 be an arbitrary closed set in Euclidean space. Then by definition

R

n

∖
C

{\displaystyle \mathbf{R}^n \setminus C}

 is open. This means that

R

n

∖
C
∈

B

{\displaystyle \mathbf{R}^n \setminus C \in \mathcal{B}}

. Again, since

B

{\displaystyle \mathcal{B}}

 is closed under complements,

C
∈

B

{\displaystyle C \in \mathcal{B}}

. This shows that

B

{\displaystyle \mathcal{B}}

 contains all the closed sets of Euclidean space. Since we assumed

A

{\displaystyle \mathcal{A}}

 is the smallest

σ

{\displaystyle \sigma}

algebra with such property, we have

A

⊂

B

{\displaystyle \mathcal{A} \subset \mathcal{B}}

, as desired.
5-7

Question: Give an example of an

f

{\displaystyle f}

 that is not integrable, but whose improper Riemann integral exists and is finite.

Solution:

Define a function

f
:
E
→

R

{\displaystyle f: E \to \R}

 where

# E
[
0
,
∞
]

{\displaystyle E=[0,\infty ]}

 is measurable such that

f
(
x
# )

sin
⁡

x

x

{\displaystyle f(x)={\frac {\sin {x}}{x}}}

 is a measurable function.

Then we have

∫

0

∞

sin
⁡

x

x

d
# x

lim

t
→
∞

∫

0

t

sin
⁡

x

x

d
# x

π
2

{\displaystyle \int \_{0}^{\infty }{\frac {\sin {x}}{x}}\,dx=\lim \_{t\to \infty }\int \_{0}^{t}{\frac {\sin {x}}{x}}\,dx={\frac {\pi }{2}}}

. Hence there exists as an improper Riemann integral such as

∫

0

∞

sin
⁡

x

x

d
# x

π
2

{\displaystyle \int \_{0}^{\infty }{\frac {\sin {x}}{x}}\,dx={\frac {\pi }{2}}}

, which is finite.

From Theorem 5.21, since

∫

0

∞

sin
⁡

x

x

d
x
∉
L
(
E
)

{\displaystyle \int \_{0}^{\infty }\left|{\frac {\sin {x}}{x}}\right|\,dx\notin L(E)}

, which is not finite, so it is not integrable. This implies

∫

0

∞

sin
⁡

x

x

d
x
∉
L
(
E
)

{\displaystyle \int \_{0}^{\infty }{\frac {\sin {x}}{x}}\,dx\notin L(E)}

 by Theorem 5.21.

(Note: Theorem 5.21 says, let

f

{\displaystyle f}

 be measurable on

E

{\displaystyle E}

. Then

f

{\displaystyle f}

 is integrable over

E

{\displaystyle E}

 iff

f

{\displaystyle |f |}

 is. )
6-2

Let

F
(
x
,
y
# )
f
(
x
)

{\textstyle F(x,y)=f(x)}

 and

G
(
x
,
y
# )
g
(
y
)

{\displaystyle G(x,y)=g(y)}

 for all

x
,
y
∈

R

n

{\displaystyle x, y \in \R^n}

. Note that

{
(
x
,
y
)
∈

R

n

×

R

n

F
(
x
,
y
)
>
α
# }
{
x
∈

R

n

f
(
x
)
>
α
}
×

R

n

{\displaystyle \{(x,y)\in\mathbb{R}^n\times\mathbb{R}^n|F(x,y)>\alpha\}=\{x\in\mathbb{R}^n|f(x)>\alpha\}\times\mathbb{R}^n}

is measurable by Lemma 5.2. Same reason can apply to

G
(
x
,
y
)

{\displaystyle G(x,y)}

. So both

F
(
x
,
y
)

and

G
(
x
,
y
)

{\displaystyle F(x,y) \;\text{and}\;G(x,y)}

 are measurable in

R

n

×

R

n

{\displaystyle \mathbb{R}^n \times \mathbb{R}^n}

. Hence

F
(
x
,
y
)
G
(
x
,
y
# )
f
(
x
)
g
(
y
)

{\displaystyle F(x,y)G(x,y)=f(x)g(y)}

 is measurable in

R

n

×

R

n

{\displaystyle \mathbb{R}^n \times \mathbb{R}^n}

.

Let

E
1

and

E
2

{\displaystyle E\_1 \; \text{and}\; E\_2}

 be measurable subsets of

R

n

{\displaystyle \R^n}

. Then

χ

E
1

{\displaystyle \chi\_{E\_1}}

 and

χ

E
2

{\displaystyle \chi\_{E\_2}}

 are measurable in

R

n

{\displaystyle \R^n}

. Then by what we proof above,

χ

E
1

(
x
)

{\displaystyle \chi\_{E\_1}(x)}

χ

E
2

(
y
)

{\displaystyle \chi\_{E\_2}(y)}

 is measurable in

R

n

×

R

n

{\displaystyle \mathbb{R}^n \times \mathbb{R}^n}

. Since

χ

E
1

(
x
)

{\displaystyle \chi\_{E\_1}(x)}

χ

E
2

(
y
# )

χ

E
1

×

E
2

(
x
,
y
)

{\displaystyle \chi\_{E\_2}(y)=\chi\_{E\_1\times E\_2}(x,y)}

, so

E
1

×

E
2

{\displaystyle E\_1\times E\_2}

 is measurable in

R

n

×

R

n

{\displaystyle \mathbb{R}^n \times \mathbb{R}^n}

.

Lastly, we have

E
1

×

E
2

=

∬

R

n

×

R

n

χ

E
1

×

E
2

(
x
,
y
)
d
x
d
y

=

∬

R

n

×

R

n

χ

E
1

χ

E
2

(
x
,
y
)
d
x
d
y

=

∫

R

n

χ

E
1

(
x
)
d
x

∫

R

n

χ

E
2

(
y
)
d
y

=

E
1

E
2

.

{\displaystyle \begin{align}
\\&=\iint\limits\_{\mathbb R^n\times\mathbb R^n} \chi\_{E\_1}\chi\_{E\_2}(x,y)dxdy
\\&=\int\_{\mathbb R^n}\chi\_{E\_1}(x)dx\int\_{\mathbb R^n}\chi\_{E\_2}(y)dy
\end{align}
}
7-6

Show that if

α
>
0

{\displaystyle \alpha>0}

, then

f
(
x
# )

x
α

{\displaystyle f(x) = x^\alpha}

is absolutely continuous on every bounded subinterval of

[
0
,
∞
)

{\displaystyle [0,\infty)}

.

Proof:

Let

g
(
x
# )

x

α
−
1

{\displaystyle g(x)=x^{\alpha-1}}

, then

f
(
x
# )
α

∫
a
x

g

a
α

{\displaystyle f(x)=\alpha\int\_a^x g\,\,\,+a^\alpha}

.

We want to show

∫
a
x

g

{\displaystyle \int\_a^x g}

 is absolutely continuous.

∫
a
x

g

{\displaystyle \int\_a^x g}

 is absolutely continuous if

g
∈
L
(
[
a
,
b
]
)

{\displaystyle g\in L([a,b])}

, where

[
a
,
b
]
∈
[
0
,
∞
)

{\displaystyle [a,b]\in [0,\infty)}

. We see that

∫
a
b

g

{\displaystyle \int\_a^b g}

 is bounded for

α
≥
1

{\displaystyle \alpha\ge1}

. For

α
<
1

{\displaystyle \alpha < 1}

 and

# a
0

{\displaystyle a = 0}

, The improper Riemann integral

∫
0
b

g
(
t
)
d
t

{\displaystyle \int\_0^b g(t) dt}

 exists, so the Lebesgue integral

∫
0
b

g

{\displaystyle \int\_0^b g}

 exists and is equal to Riemann integral. These show us that

g
∈
L
(
[
a
,
b
]
)

{\displaystyle g\in L([a,b])}

. Since

f
(
x
)

{\displaystyle f(x)}

 is combination of multiple of absolutely continuous function and constant, therefore it is absolutely continuous.
Ex2-5

**Question:**

Suppose

f

{\displaystyle f}

 is finite on

[
a
,
b
]

{\displaystyle [a, b]}

and of bounded variation on every interval

[
a
ϵ
,
b
]

{\displaystyle [a+\epsilon ,b] }

,

ϵ
>
0

{\displaystyle \epsilon > 0}

, with

V
[
f
;
a
ϵ
,
b
]
≤
M
<
∞

{\displaystyle V[ f; a+\epsilon, b] \leq M < +\infty}

. Show that

V
[
f
;
a
,
b
]
<
∞

{\displaystyle V[f; a , b ] < +\infty
}

. Is

V
[
f
;
a
,
b
]
≤
M

{\displaystyle V[f; a , b ] \leq M
}

? If not, what additional assumption will make it so?

**Solution part 1:** (Show that

V
[
f
;
a
,
b
]
<
∞

{\displaystyle V[f; a , b ] < +\infty
}

. Is

V
[
f
;
a
,
b
]
≤
M

{\displaystyle V[f; a , b ] \leq M
}

)

Note

f

{\displaystyle f}

 is finite of

[
a
,
b
]

{\displaystyle [a, b]}

, this implies that

f
(
a
)

{\displaystyle f(a)}

 is finite.

By the definition of variation, consider any partition of

[
a
,
b
]

{\displaystyle [a, b]}

# Γ
{

x
0

,

x
1

,
.
.
.
,

x
n

}

{\displaystyle \Gamma = \{x\_0,x\_1,...,x\_n\} }

.

ϵ
>
0

{\displaystyle \epsilon > 0}

 can be chosen arbitrary small, implies we can always choose

# ϵ

x
1

−

x
0

{\displaystyle \epsilon = x\_1 - x\_0 }

. which equivalence to have

a
1

=

x
1

{\displaystyle a\_1=x\_1}

.

define a new partition on

[
a
ϵ
,
b
]

{\displaystyle [a+\epsilon ,b] }

,

Γ
′

=
Γ
−
{

x
0

# }
{

x
1

,

x
2

,
.
.
.
,

x
n

}

{\displaystyle \Gamma '= \Gamma - \{x\_0 \} = \{x\_1,x\_2,...,x\_n\} }

.

V
[
f
;
a
,
b
]

=

sup
Γ

∑

# j
1

n

f
(

x
j

)
−
f
(

x

j
−
1

)

=

sup
Γ

f
(

x
1

)
−
f
(

x
0

)

sup
Γ

∑

# j
2

n

f
(

x
j

)
−
f
(

x

j
−
1

)

=

sup
Γ

f
(

x
1

)
−
f
(

x
0

)

sup

Γ
′

∑

# j
2

n

f
(

x
j

)
−
f
(

x

j
−
1

)

=

sup
Γ

f
(

x
1

)
−
f
(

x
0

)

V
[
f
;
a
ϵ
,
b
]

{\displaystyle \begin{align}
V[f;a,b]&=\sup\_\Gamma \sum ^n\_{j=1} |f(x\_j)-f(x\_{j-1})|\\
&= \sup\_\Gamma |f(x\_1)-f(x\_0)| + V[f; a+\epsilon , b ]
\end{align}}

Recall that

f

{\displaystyle f}

 is finite function on

[
a
,
b
]

{\displaystyle [a, b]}

, with bounded variation on

[
a
−
ϵ
,
b
]

{\displaystyle [a-\epsilon ,b] }

. This shows the sum is finite

We can now conclude that f has bounded variation on

[
a
,
b
]

{\displaystyle [a, b]}

,

V
[
f
;
a
,
b
]
<
∞

{\displaystyle V[f; a , b ] < +\infty
}

.

**Solution part 2:** (Is

V
[
f
;
a
,
b
]
≤
M

{\displaystyle V[f; a , b ] \leq M
}

? If not, what additional assumption will make it so?)

Is

V
[
f
;
a
,
b
]
≤
M

{\displaystyle V[f; a , b ] \leq M
}

? Generally no. From above we have equation

V
[
f
;
a
,
b
# ]

sup
Γ

f
(

x
1

)
−
f
(

x
0

)

V
[
f
;
a
ϵ
,
b
]

{\displaystyle V[f;a , b ] = \sup\_\Gamma |f(x\_1)-f(x\_0)| + V[f; a+\epsilon , b]}

.

It is only possible when

sup
Γ

f
(

x
1

)
−
f
(

x
0

)

=
0

{\displaystyle \sup\_\Gamma |f(x\_1)-f(x\_0)|=0 }

, equivalence saying as

x
1

{\displaystyle x\_1}

 get closer to

x
0

{\displaystyle x\_0}

, then

f
(

x
1

)

{\displaystyle f(x\_1)}

 get closer to

f
(

x
0

)

{\displaystyle f(x\_0)}

.

This happens when

f

{\displaystyle f}

 is continuous around

x
0

=
a

{\displaystyle x\_0=a}

 from positive direction. And this is the additional condition required.
Bloodlust 2 Factions

* Ubuanto - Has its own fast travel point. Hideout entrance in Downtown. Travel from Chinatown through sewer (South-west entrance) to "Downtown South" then run north-west to "Downtown North". Enter the "Computer Lab" then into the "Data Center" (needs ID Pass received via the quest "Faster is better"), box is close to the door.
* Vi Tao - Has its own fast travel point. Hideout entrance is in Asylum Sewers. Travel from Warehouse to Asylum sewers go up the stairs between two Vi Tao banners, you can run past the two guards and enter through the manhole to the right of them. Beware of the Wraith which can spawn in the area just after the manhole.
* Zenchi - Has its own fast travel point. Hideout in Downtown Hotel (travel from Chinatown through sewer to Downtown and enter hotel then unlock door in hotel lobby past Zenchi guards to main dining/dance hall room)
* Wesp - Exit the asylum and go through the yard to the catacombs. Go down the steps and go left and you'll be in a huge room where you can see different levels. On the left are a pair of winding stairs that lead up to a door. Go up there and there will be a huge room with the leader of WESP on the right, and the box is on the left. Wesp guards will not attack if the player is wearing the corresponding Tattoo others a very tough fight. There is a Quest for the leader which will also get them off your back.

Modified Tattoos prevent **Non-restricted** Zenchi, Vi Tao, Wesp guards will not attack if the player is wearing the corresponding Tattoo.

Reputation gains will allow you to train higher levels of some skills.

Max Reputation for returned Items is 10 rep

Can interrogate Vi Tao Guards (only in Sewer Dungeon)
Free Ships

There are several free ships in X4, their locations can be read about at the Egosoft Forum https://forum.egosoft.com/viewtopic.php?t=405524 these are my notes with regard to finding and taking them.

### Callisto Sentinal

The first Mission in the Hatikvah Trade Revolution Mission chain is to deliver 250 Medical Supplies, they provide the goods and a Callisto Sentinel Courier to deliver them in. You get the ship for free. That Mission chain is offered when you are close to Hatikvah's Choice I, look in Mission Offers. It might seem like a useless ship but it is worth about 500,000 Credits.

If you are starting the game with a Custom start in a Spacesuit it might be possible to ride the Highway until you are close enough to land at the HAT Trading Station.

### Odysseus Destroyer (Faulty Logic 7)

There are two routes to Faulty Logic VII. The second route will avoid most of the mines in the Sector.

Holy Vision ... Faulty Logic I ... Faulty Logic VII

The Void ... Frontier Edge ... Atiyas Misfortune I ... Atiyas Misfortune III ... Faulty Logic VII

From the bottom left gate of Faulty Logic 1 just head straight up the map or head straight left from the gate at the top right of the sector, it should be close to a group of 3 asteroids but on one occasion it was alone so maybe they are random. Usually there is no station closeby but I had no problem avoiding the one in the picture.

Go there with a ship capable of fighting a few Xenon ships which sometimes hang around the gates. Take the ship using the "F" menu and "Claim", you will need a Marine for that. Send a few Engineers across, the ship might be damaged during the Claim and they will repair it. It has 2 or 3 repair drones but they tend to stay still and do nothing. The ship has a lot of mines etc which can be sold.

Faulty Logic 1 is full of mines and the ship is big so avoiding them is hard if you take the short route (right) into HoP (Paranid) space, I read somewhere that if you fly high enough or low enough you can avoid them. I fly through the gate at the bottom left then up through a few more Xenon sectors until I reach The Void.
Ex2-1

**Solution to Chapter 2, Exercise 1**

*This proof seems correct!*

Let

f
(
x
# )
x
sin
⁡

1
x

{\displaystyle f(x)=x\sin\frac1x}

 for

x
∈
(
0
,
1
]

{\displaystyle x\in(0,1]}

 and

f
(
0
# )
0

{\displaystyle f(0) = 0}

.

Boundedness: We find that

f
(
x
)

≤

x

≤
1

 on

[
0
,
1
]

{\displaystyle [0,1]}

, thus

f

{\displaystyle f}

 is bounded.

Continuity: For each

x
0

∈
(
0
,
1
]

{\displaystyle x\_0\in(0,1]}

, we see

1

/

x
0

{\displaystyle 1/x\_0}

 is finite, and so the function is continuous.

Now we consider the case

x
0

=
0

{\displaystyle x\_0 = 0}

. For

x
>
0

{\displaystyle x>0}

 we have

−
x
≤
x
sin
⁡

1
x

≤
x
,

{\displaystyle -x\leq x\sin\frac1x\leq x,}

hence when

x
→
0

{\displaystyle x \to 0}

,

f
(
x
)
→
# 0
f
(
0
)

{\displaystyle f(x)\to 0=f(0) }

 by Squeeze theorem. This proves

f

{\displaystyle f}

 is continuous on

[
0
,
1
]
.

{\displaystyle [0,1].}

Not of bounded variation: For each odd natural number

N

{\displaystyle N}

 we choose a partition

# Γ
{

x

N
1

,

x
N

,
…
,

x
1

,

x
0

}

{\displaystyle \Gamma=\{x\_{N+1},x\_N,\ldots,x\_1,x\_0\} }

 (consists of

N
2

{\displaystyle N+2}

 elements) of

[
0
,
1
]

{\displaystyle [0,1]}

 as:

x

N
1

=
0
,

x
0

=
1

{\displaystyle x\_{N+1}=0, x\_0=1 }

, and

x
n

=

2

n
π

 for
# n
1
,
2
,
…
,
N
.

{\displaystyle x\_n=\frac2{n\pi} \quad\text{ for }n=1,2,\ldots,N.}

**Our purpose will be realize the sum

S
Γ

=

∑

# i
1

N
1

f
(

x
i

)
−
f
(

x

i
−
1

)

{\displaystyle S\_\Gamma=\sum\_{i=1}^{N+1} |f(x\_i)-f(x\_{i-1})| }

 as a partial sum of harmonic series, thus proving it is unbounded.**

We note that for

# i
1
,
2
,
…
,
N
,
N
1
,

{\displaystyle i=1,2,\ldots,N,N+1,}

f
(

x
i

)

=

{

0

 if i is even,

2

i
π

 if i is odd.

{\displaystyle |f(x\_i)|=\begin{cases}0&\text{ if i is even,}\\\frac2{i\pi}&\text{ if i is odd.}\end{cases}}

This implies

S
Γ

=

∑

# i
1

N
1

f
(

x
i

)
−
f
(

x

i
−
1

)

≥

∑

# i
2

N
1

f
(

x
i

)
−
f
(

x

i
−
1

)

=

2
π

2

3
π

2

3
π

2

5
π

2

5
π

⋯

2

N
π

2

N
π

>

4
π

∑

# n
1

(
N
−
1
)

/

2

1

2
n
1

.

{\displaystyle \begin{align}S\_\Gamma&=\sum\_{i=1}^{N+1}|f(x\_i)-f(x\_{i-1})|\\
&\geq \sum\_{i=2}^{N+1}|f(x\_i)-f(x\_{i-1})|\\
&=\frac2\pi +\frac2{3\pi}+\frac2{3\pi} +\frac2{5\pi}+\frac2{5\pi}+\cdots+\frac2{N\pi}+\frac2{N\pi}\\
&>\frac4\pi \sum\_{n=1}^{(N-1)/2} \frac1{2n+1}.\end{align}}

As

N
→
∞

{\displaystyle N\to \infty}

, we find that

S
Γ

{\displaystyle S\_\Gamma}

 becomes unbounded, hence

f
(
x
)

{\displaystyle f(x)}

 is not of bounded variation.