recursal/Fanatic-Fandom huggingface dataset:
Les élfes

Les **Elfes**, surnommés le "**Beau Peuple**", sont des êtres dépassant en tous points les autres espèces, notamment à cause de leurs immenses savoirs. Ils sont originaires d'une terre appelée Alalëa, ayant émigrés en Alagaësia en 5217 A.C.

## culture

Comme le dit Arya, les elfes sont très courtois, car une querelle pourrait rester éternellement à cause de leur longévité. Pour cela, à chaque fois qu'un elfe en croise un autre, ils échangent des formules de politesse. C'est toujours celui qui a un rang moins élevé qui parle en premier. Ils accolent aussi des titres honorifiques au nom de leur interlocuteur, qui change selon les personnes: par exemple, *Svit-kona* pour une personne de grande sagesse, ou encore *Elda* qui est très élogieux.

## compétences

Les elfes possédaient des pouvoirs surhumains, telles que la résistance, la longévité, la vitesse et l'agilité. Il a été mentionné par Brom que même l'elfe le plus faible peut vaincre facilement l'humain le plus vaillant. Les elfes préfèrent essentiellement utiliser des arcs, des flèches, des lances et des épées (un grand nombre d'épées ont été conçues et créés par l'elfe Rhunön), et sont rarement "dérangés" d'utiliser un tout autre type d'armes.

Ils étaient des combattants à l'épée exceptionnels et pendant le Siège de Feinster, deux elfes ont eu comme adversaire plus d'une centaine d'humains épéistes. Les elfes ont des compétences particulières dans le travail du métal, mais ils ne les partagent jamais avec une autre espèce. La langue que les elfes utilisent couramment est l'Ancien langage, ainsi ils ne peuvent pas mentir et pour cela utilisent une technique qui consiste à ne dire qu'une partie de la vérité.

Tout les elfes avaient des capacités étonnantes avec le chant. Ils chantaient en Ancien langage, ils pouvaient l'utiliser pour tisser un sort, modeler une plante, ou tout simplement pour le plaisir, mais leurs chansons ont toujours tenues un sens ou de la puissance.

## quelque elfe notable

* Gëda
* Narí
* Nëya
* Ládin
* Adarë
* Dusan
* Earnë
* Fiolr
* Iduna
* Nuala
* Vrael
* Vanir
* Yaela
* Fäolin
* Anurin
* Alanna
* Celdin
* Edurna
* Laetri
* Lifaen
* Linnëa
* Liotha
* Niduen
* Naudra
* Oromis
* Rhunön
* Wyrden
* Vándil
* Othíara
* Tunivor
* Glaerun
* Formora
* Damítha
* Däthedr
* Erisdar
* Evandar
* Invidia
* Umhodan
* Vestarí
* Enduriel
* Eragon I
* Dellanir
* Acallamh
* Analísia
* Berundal
* Glenwing
* Haldthin
* Kialandí
* Maerzadí
* Tarmunora
* Ceranthor
* Islanzadí
* Gilderien
* Lupusänghren
* Ildrid le Beau
* Gnaevaldrskald
Genres

Genres are a classification of art or creative work. Some genres have formal criteria that must be met, while others are more open to interpretation.

Genres can be classified by:

* The content of the work.
* Its general artistic style, which may also reflect on other aspects such as gender roles and cultural backgrounds. Music genres are often divided into subgenres that represent a style based around one instrument or a type of song.
* The geographic origin of that genre, such as regionality and ethnicity.
* Its popularity (or lack thereof) in society at large.

## Types of genres

Throughout every type of popular medium of entertainment, genres can generally be categorized into three groups.

**Music Genres**

Music genres are a classification of music style. You can find these in:

* Music, in general
* Anything with a score or soundtrack, such as film, television, and video games

**Narrative Genres**

Narrative genres are the types of stories told, or narrative modes. You can find these in any medium that has a plot, such as:

* Film
* Television
* Literature or any print media
* Video games (for the most part)

**Gameplay Genres**

Gameplay genres are types of games, and are classified by the player's goal or motivation. You can find these in:

* Any type of game, but these ones usually refer to video games (the only popular medium to have all three)

## Evolution of genres

An import part of this wiki is examining how each genre evolves throughout the years. This is why each genre is divided into smaller subgenres, styles, or movements - each subgenre will, for the most part, have a decade(s) it was most popular, or was most associated with. For example, Disco, a subgenre of R&B music, will be categorized as having peak popularity in the 1970s.

Numerous factors will be taken into account to determine a subgenre's peak popularity.
Jazz Fusion

Jazz fusion is a style of Jazz music that arose in the late 1960s, predicated on the usage of electric instrumentation and the incorporation of elements of various other styles of music, such as the Rock and Funk music popular at the time, Rhythm & Blues, or Traditional Folk Music.

Jazz fusion's sound is rooted firmly in the jazz tradition through its instrumentation and basis in improvisation with a focus on distinct soloists; however there are many characteristics by which it may be sonically distinguished. One such is its near-universal inclusion of electric instrumentation, especially electric guitars and keyboards, which were uncommon in jazz (although not unheard of) up until the genre's birth. Another is fusion's relative harmonic simplicity compared to other types of jazz, which again reflects its "fusing" with generally harmonically simpler genres. A third is the replacement of many traditional compositional elements of jazz with those of rock, funk, or other styles. For example, in fusion, rhythm sections tend to support soloists through "vamping", or repeating smaller composed phrases, while other jazz styles tend towards "comping", which is largely improvised, and much more reactive towards a soloist. Fusion's overarching song forms, however, are generally more open and improvisational.
Modal Jazz and Post-Bop, jazz styles focused on formal openness and an eschewing of many traditional elements of jazz composition and harmony, paved the way for fusion. Working from a background in these styles, trumpeter Miles Davis's experiments on albums such as *In a Silent Way* and *Bitches Brew* were among the earliest to work with the sonic vocabulary described above. Not only did these albums sell very well (especially for jazz music at the time), but virtually all of the sidemen who played on them went on to make highly commercially successful and influential jazz fusion as bandleaders. Jazz fusion would be one of the last styles of jazz to achieve such success.

Through the 1970s, numerous other approaches to fusion were developed from this basis, from the avant-garde "free fusion" of Ornette Coleman & Prime Time to the synthesizer-focused approach of Herbie Hancock. Fusion evolved in parallel with Jazz-Rock (a term originally synonymous with fusion which grew to refer to a distinct, non-jazz style), with many artists making music in both or a combination of the two styles. It was a basis for Jazz-Funk's development, particularly through Hancock's work throughout the 1970s. As the genre entered the 1980s, growing influence on it from Pop and rhythm & blues gave birth to Smooth Jazz. Fusion has also been influential in the development of Chamber Jazz.
Ex4-1

Poblem: Prove the following:

(i) Corollary 4.2: If

f

{\displaystyle f}

 is measurable, then

{
f
>
−
∞
}

{\displaystyle \{f>-\infty \}}

,

{
f
<
∞
}

{\displaystyle \{f<+\infty \}}

,

{
# f
∞
}

{\displaystyle \{f=+\infty \}}

,

{
a
≤
f
≤
b
}

{\displaystyle \{a \le f \le b\}}

,

{
# f
a
}

{\displaystyle \{ f = a\}}

 etc. are all measurable, Moreover,

f

{\displaystyle f}

 is measurable if and only if

{
a
<
f
<
∞
}

{\displaystyle \{a<f<+\infty \}}

 is measurable for every finite

a

{\displaystyle a}

.

(ii) Theorem 4.8: If

f

{\displaystyle f}

 is measurable and

λ

{\displaystyle \lambda}

 is any real number, then

f
λ
,

and

λ
f

{\displaystyle f+\lambda ,{\text{and}}\lambda f}

 are measurable.

Proof: (i)

First note that

{
f
>
−
∞

# k
1

∞

{
f
>
−
k
}

{\displaystyle \{f>-\infty \}=\cup \_{k=1}^{\infty }\{f>-k\}}

 and

{
f
<
∞

# k
1

∞

{
f
<
k
}

{\displaystyle \{f<+\infty \}=\cup \_{k=1}^{\infty }\{f<k\}}

, both of these sets are measurable each of them is the union of the countable measurable set. Now, note that the set

{
# f
∞
# }
{
f
<
∞

}

c

{\displaystyle \{f=+\infty \}=\{f<+\infty \}^{c}}

 is measurable as the complement of a measurable set. After that, we also have

{
a
≤
f
≤
b
# }
(

∩

# k
1

∞

{
f
>
a
−

1
k

}
)
∩
{
f
>
b

}

c

{\displaystyle \{a\leq f\leq b\}=(\cap \_{k=1}^{\infty }\{f>a-{\frac {1}{k}}\})\cap \{f>b\}^{c}}

this shows that the set is measurable since it is intersection of countable union of measurable set. Lastly, we have

{
# }
{
a
≤
f
≤
a
}

{\displaystyle \{f=a\}=\{a\leq f\leq a\}}

 is measurable.

For the second statement, the forward implication is implied by the above, since

{
a
<
f
<
∞
# }
{
f
>
a
}
∩
{
f
<
∞
}

{\displaystyle \{a<f<+\infty \}=\{f>a\}\cap \{f<+\infty \}}

is measurable . Now, for another direction, assume that all the sets

{
a
<
f
<
∞
}

{\displaystyle \{a<f<+\infty \}}

 are measurable: then

∪

# k
1

∞

{
−
k
<
f
<
∞
# }
{
f
<
∞
}

{\displaystyle \cup \_{k=1}^{\infty }\{-k<f<+\infty \}=\{f<+\infty \}}

is measurable, so

{
# f
∞
# }
{
f
<
∞

}

c

{\displaystyle \{f=+\infty \}=\{f<+\infty \}^{c}}

is measurable, so

{
f
>
a
# }
{
a
<
f
<
∞
}
∪
{
# f
∞
}

{\displaystyle \{f>a\}=\{a<f<+\infty \}\cup \{f=+\infty \}}

is measurable. Thus,

f

{\displaystyle f}

 is measurable.

(ii) If

f

{\displaystyle f}

 is measurable, then

{
f
>
a
−
λ
# }
{
f
λ
>
a
}

{\displaystyle \{f>a-\lambda \}=\{f+\lambda >a\}}

 is a measurable set for all

a

{\displaystyle a}

. If

λ
>
0

{\displaystyle \lambda > 0}

, then

{
f
>
a

/

λ
# }
{
λ
f
>
a
}

{\displaystyle \{f>a/\lambda \}=\{\lambda f>a\}}

 is measurable for all

a

{\displaystyle a}

. If

λ
<
0

{\displaystyle \lambda < 0 }

, then

{
f
<
a

/

λ
# }
{
λ
f
>
a
}

{\displaystyle \{f<a/\lambda \}=\{\lambda f>a\}}

 is measurable for all

a

{\textstyle a}

. If

# λ
0

{\displaystyle \lambda=0}

, then

λ
f
≡
0

{\displaystyle \lambda f\equiv 0}

 identically, which is measurable. Thus

f
λ

{\displaystyle f+\lambda }

 and

λ
f

{\displaystyle \lambda f}

 are measurable function for all

λ
∈

R

{\displaystyle \lambda \in \R }

.
Gravity Falls Theme Song "Made Me Realize"

### About the original

The "**Gravity Falls Main Title Theme**," composed by Brad Breeck, is the opening theme song played over the title sequence to *Gravity Falls*. The official theme song is available on iTunes for $0.99. Beginning with "Northwest Mansion Mystery," a shortened version of the theme is used in episodes that are longer than usual. The "Weirdmageddon" episodes use a corrupted version of the song. The main theme is often mistaken for the song "**You Made Me Realize**," due to its strong similarities to the real "You Made Me Realize" song Breeck wrote for MTV's *Awkward*. "Made Me Realize" is commonly mistaken for the extended version of the *Gravity Falls* theme song. It is also stated on Breeck's website that the two songs are indeed different, and not related to each other.

## About Walt's Cover

Walt started with the original 30 seconds of the song and extended it into a 4 minute song.

Walt's most viewed youtube video is the release of this song.

Walt's cover of Gravity Falls was used in KreatureKid's "The Making of Grunkle Stan" video.
Modal Jazz

A style of Jazz that utilizes musical modes as harmonic framework over traditionally used chord progressions. It took its start in the late 1950s when artists such as George Russell began experimentation with a modal approach to their music. Since then, some of the most well known jazz artists began using modal progressions in their music, including John Coltrane, Miles Davis, and Herbie Hancock. Modal jazz is often cited as a precursor to Avant-Garde Jazz.
Walt Ribeiro Wiki

**Welcome to the Walt Ribeiro Wiki!**
We're a collaborative community website about Walt Ribeiro that anyone, including you, can build and expand. Wikis like this one depend on readers getting involved and adding content. Click the "ADD NEW PAGE" or "EDIT" button at the top of any page to get started!

## Important articles

PlaceholderA Crucial ItemMade Me Realize - Gravity Falls

*Need help building out this community?*

* Rules of this wiki
* Getting Started
* How to Contribute
* Managing your new community
* Guides
* All Help articles

You can also be part of the larger Fandom family of communities. Visit Fandom's Community Central!

*Community Founders*: Write a good and paragraph-length description for your welcome section about your topic. Let your readers know what your topic is about and add some general information about it. Then you should visit the admin dashboard for more tips.
Reality TV GameShows Wiki

```
***About us***

```

**Welcome to Reality TV GameShows**, a group based on gameshows such as total drama, big brother and survivor and other reality tv shows, We host daily and we reward you for joining sessions with a good experience and if you win you can receive a prize that could be a rank up or robux! We are creating multiple games along with the ones we already have, so join for a enjoyable experience! We also create longterms and you can view them on the youtube! -cj\_eh and the staff team

```
***Longterm Seasons***

```

Total Drama Island - Total Drama Island

```
***Longterm Winners***

```

S1| TDWT - Dogeatpizza

S2| TDA - Thomasreedboi

S2.5| TDRR - Snipver and Shawdiaz94004

S3| In progress
Wiktoria Szpak

## Wygląd

### Dalsza rodzina

Marianna Mistrzowska (Babka Maryśka) i Lucjan Orzełek - dziadkowie (rodzice matki)Włodek Lis i Franciszka Lis - dziadkowie (rodzice ojca)

Dziewczyn posiada złote włosy, na których znajduje się różowa pasemka. Posiada jasnoniebieskie oczy, posiada również żółte źrenice w kształcie

gwiazdy. Na jednym oku widnieja opaska. Ponieważ nie widzi na jedno oko. Nosi na szyji wisiorek, (który dostała od Kingi Zdrowyjskiej)
Ex3-10

**Chapter 3 Question 10**

We suppose

E

1

<
∞
,

E

2

<
∞

, otherwise both sides of the equality are unbounded. Now let

# A

E

1

−

E

2

,
# B

E

1

∩

E

2

,
# C

E

2

−

E

1

{\displaystyle A=E\_{1}-E\_{2},B=E\_{1}\cap E\_{2},C=E\_{2}-E\_{1}}

, so

(

E

1

∪

E

2

)

=
A
∪
C
∪
B

{\displaystyle {\begin{aligned}\left(E\_{1}\cup E\_{2}\right)=A\cup C\cup B\end{aligned}}}

then since

A
,
B
,
C

{\displaystyle A,B,C}

 are measurable and disjoint, Theorem 3.23 gives

E

1

∪

E

2

=

A
∪
C
∪
B

=

A

C

B

Since

# A

E

1

−
B
,
# C

E

2

−
B

{\displaystyle A=E\_{1}-B,C=E\_{2}-B}

 and from Corollary 3.25,

B
⊂

E

1

{\displaystyle B\subset E\_{1}}

 and

B
⊂

E

2

{\displaystyle B\subset E\_{2}}

, we get

A

C

B

=

E

1

−
B

E

2

−
B

B

=

E

1

−

B

E

2

−

B

B

=

E

1

E

2

−

B

Thus

E

1

∪

E

2

=

E

1

E

2

−

E

1

∩

E

2

This completes the proof.
Bractwo Piór

Bractwo Piór - Grupa psów działająca przeciwko Psiemu Zabójcy, oraz przeciw hyclom. Założycielką oraz Alfą jest Pierzasta.

## Terytoria

### Wygląd obozu

Płytkie wgłębienie w ziemi, zasłonione krzewami oraz drzewami. jest kilka nor w ziemi

## Członkowie

**Alfa: Pierzasta**

**Doradca: Horyzont,** szary pies w ciemniejsze łaty

**Sierżant: Lis,** rudy pies z białym brzuchem, klatką i szyją.

**Konsyliarz: Kropka,** suka rasy dalmatyńczyk

**Walczący:**

- Kamyk, brązowy pies z szarą szyją i ogonem

- Muchomor biały pies średniej wielkości w brązowe i czarne łaty, praktykant ***MIm***

- Jad, brązowy muskularny pies

- Rdza brązowa suka z oklapniętymi uszami , praktykant ***Blizna***

- Osa, pręgowana suka

- Brzoza, brązowy pies z białą końcówką pyska i pręgą na brzuchu i klatce.

**Polujący:**

- Samuel, tricolor rasy shih-tzu

- Samira, brązowa suczka rasy shih-tzu praktykantka ***Mżawka***

- Lotka, suka rasy Border Collie

- Jeleń, muskularny brązowy pies z łatami z tyłu

**Praktykanci:**

- Mżawka, brązowa suczka w czarne kropeczki

- Blizna, brązowy pies bez ogona

- Mim, głuchy czarny pies rasy labrador

**Rodzicielki:**

- Sasza, smukła kremowa suczka

- Myszka, suczka rasy Owczarek Szkocki

**Przyszli:**

- Mgła, mała jasna suczka. ( Córka Saszy )

- Smog, muskularny szary pies. ( Syn Saszy )

- Ostrokrzew, czarna zielonooka suka. ( Córka Saszy )

- Kiełek, brązowy pręgowany pies. ( Syn Saszy )

- Szron, szara suka. ( Córka Saszy )

- Mech, rudy pies. ( Syn Myszki )

- Mak, brązowa suka. ( Córka Myszki )

**Seniorzy:**

- Michał, kościsty brązowy pies

- Miki, duży biały pies

gi
--

**Alfa** - Rządzi i utrzymuje porządek w Bractwie.

**Doradca** - Po śmierci Alfy przejmuje jego miejsce. Jego słowo też jest bardzo ważne.

**Sierżant** - Organizuje patrole bitewne i łowieckie. Ustanawia również plan bitwy.

**Konsyliarz** - Posiada zdolności lecznicze.

**Rodzicielka** - Suka oczekująca młodych, bądź opiekująca się szczeniakami.

**Praktykant** - Pies uczący się zostać Walczącym lub Polującym.

**Walczący** - Walczy i broni Bractwa przed napastnikami.

**Polujący** - Poluje.

**Senior** - Jest zbyt zmęczony podeszłym wiekiem, aby dalej sprawować swoje obowiązki.

**Przyszły** - Pies za mały aby uczyć się służyć Bractwu.

## Historia Rang

Alfa

Doradca
 Grad | przed: "Niebezpieczna Tajemnic"
 Mak | "Niebezpieczna Tajemnica"
 Horyzont | "Powrót Zabójcy Psów"

Sierżant

Konsyliarz
 Kropka | "Niebezpieczna Tajemnica"
Ex3-20

**Question.** Show that there exist disjoint

E

1

,

E

2

,
…
,

E

k

,
…

{\displaystyle E\_{1},E\_{2},\dots ,E\_{k},\dots }

 such that

⋃

k

E

k

e

<

∑

k

E

k

e

 with strict inequality. (Let

E

{\displaystyle E}

 be a non-measurable subset of

[
0
,
1
]

{\displaystyle [0,1]}

 whose rational translates are disjoint. Consider the translates of

E

{\displaystyle E}

 by all rational numbers

r
,
0
<
r
<
1

{\displaystyle r,0<r<1}

.)

**Answer.** By hint, let

E

{\displaystyle E}

 be a non-measurable subset of

[
0
,
1
]

{\displaystyle [0,1]}

 whose rational translates are disjoint. Consider the translates of

E

{\displaystyle E}

 by all rational numbers

r
,
0
<
r
<
1

{\displaystyle r,0<r<1}

, and denote

E

r

=
{
x
r
∈

R

∣
x
∈
E
}
.

{\displaystyle E\_{r}=\{x+r\in \mathbb {R} \mid x\in E\}.}

Since

0
≤
x
≤
1

{\displaystyle 0 \leq x \leq 1}

 and

0
<
r
<
1

{\displaystyle 0 < r < 1}

, we have

0
≤
x
r
≤
2

{\displaystyle 0\leq x+r\leq 2}

 so that

⋃

r
∈

Q

∩
(
0
,
1
)

E

r

⊆
[
0
,
2
]

{\displaystyle \displaystyle \bigcup \_{r\in \mathbb {Q} \cap (0,1)}E\_{r}\subseteq [0,2]}

 and by the monotonicty of outer measure,

⋃

r
∈

Q

∩
(
0
,
1
)

E

r

e

≤

[
0
,
2
]

=

By Exercise 18, we have

E

r

e

=

E

e

. Moreover,

E

e

>
0

{\displaystyle |E|\_{e}>0}

 since

E

{\displaystyle E}

 is non-measurable. This implies that

∑

r
∈

Q

∩
(
0
,
1
)

E

r

e

=
∞

{\displaystyle \displaystyle \sum \_{r\in \mathbb {Q} \cap (0,1)}|E\_{r}|\_{e}=\infty }

. Therefore we find that

⋃

r
∈

Q

∩
(
0
,
1
)

E

r

e

<

∑

r
∈

Q

∩
(
0
,
1
)

E

r

e

.
Chapter 3

Ex3-1

✓

{\displaystyle \checkmark}

Ex3-2 (only do part (a))

Ex3-3

✓

{\displaystyle \checkmark}

Ex3-7

✓

{\displaystyle \checkmark}

Ex3-8

✓

{\displaystyle \checkmark}

Ex3-9

✓

{\displaystyle \checkmark}

Ex3-10✓

{\displaystyle \checkmark}

Ex3-11✓

{\displaystyle \checkmark}

Ex3-12

EX3-13✓

{\displaystyle \checkmark}

Ex3-15

Ex3-18

Ex3-20

Ex3-28 ✓

{\displaystyle \checkmark}

Ex3-29

Ex3-34
Altinades motors

altinades es la marca de autos,camionetas y clasicos ,el consesionario se ubica en el barrio chino al lado delBrick painter,el almacen se obtiene como recompensa por robar a los motoristas,Esta empresa está basada en Dodge ya que sus logotipos son bastante similares y el parecido de sus modelos con los reales. Consiste en un rostro de un toro en un fondo rojo bordado en plata.

sus autos mas reconocidos son:

altinades comet

trophy

Altinades monster

Altinades corsarius

Altinades skull

Altinades cargo

altinades departamental
Hana Zielinski

Hana Zielinski's former PiffleNet profile picture, taken early 2020

Hana Zielinski is a humanoid synthbio morph serving a lead ballistics engineer for the Moria-based design collective Pozhar Tactical Armaments. Since late 2018, she has also run a production and distribution outlet for largely Pozhar based armaments, mostly concentrated on NSEarth.
Dan Kuso

* BB
* NV Casual
* NV
* GI
* MSA1
* MSA2

15 (New Vestroia)

16 (Gundalian Invaders)

17 (Mechtanium Surge)

18 (BioTechnic Metal)

19 (Master Quest)

27 (Hyperion Leonidas)

### Voice Actor

Scott McCord (English) (Nelvana Network)Bill Rogers (English) (Capcom Co., Ltd.)

Yu Kobayashi (Japanese)

### Guardian Bakugan

Olyphia Dragonoid (Drago)Apollonir (Temporary) (Deceased)

Dragonoid Colossus (Deceased)

Reptak (Temporary)

### Home Palace

Wardington City, Earth (BB-NV)Bayview City, Earth (GI-MSA1)

Bakugan City, (MSA2 Formerly)

### Faction

Battle BrawlersBakugan Battle Brawlers Resistance

Order of the Castle Knights (Formerly)

### Other Appearances

Mega Man Grey 2: Reploids AdventMega Man Grey 3: New Vestroia

Mega Man Grey 4: Gundalian Invaders

Mega Man Grey 5: Mechtanium Surge Arc 1

Mega Man Grey 6: Mechtanium Surge Arc 2

Mega Man Grey 7: BioTechnic Metal

Mega Man Grey 8: Master Quest

### Side

Rebel

*For the main character of the Battle Planet series, see Dan Kouzo.*

### Biography

Drago tries to make him stop fighting with him. After the vision was over, Drago and Fear Ripper appeared in the field and Drago was able to defeat him, giving Dan the victory. Dan was surprised hearing Drago talking to Fear Ripper and he decides that he was imagining that he was talking. Masquerade who was sending people's Bakugan to the Doom Dimension, Dan decides to battle him. When he went to the meeting point, he found Shuji and Akira waiting for him for another rematch. Dan unwillingly accepted and defeated Shuji's Ventus Bakugan with ease. After the battle, Grey ran into Masquerade and battled him. After losing two of his Bakugan to Reaper, Dan sends out Drago and he was losing the battle until Dan tried to activate an Ability Card, which Drago was able to exceed and cause the battle to end in a draw. However, as Masquerade left, he revealed that he was holding back, making Dan realize that Masquerade was much stronger than he had anticipated, and that he could have lost Drago.

Dan goes with Grey's team, when they gonna defeat Masquerade.

Although constantly arguing with each other, he's helping Grey to give his power until the end of the series.

His Teamates with Wynton and Shun.
Ex2-18

f
(
z

# k
0

∞

a
k

z

k

{\displaystyle f(z)=\sum\_{k=0}^\infty a\_kz^{k}}

 be a power series. Show that if

∑

# k
0

∞

a
k

<
∞

{\displaystyle \sum\_{k=0}^\infty |a\_k|<\infty}

, then

f
(
z
)

{\displaystyle f(z)}

 is of bounded variation on every radius of the circle

z

=
1

{\displaystyle |z|=1}

. (If, e.g., the radius is

0
≤
x
≤
1

{\displaystyle 0 \leq x \leq 1}

 and the

a
k

{\displaystyle a\_k}

 are real, then

f
(
x

# k
0

∞

a
k

x
k

−

∑

# k
0

∞

a
k
−

x
k

{\displaystyle f(x)=\sum\_{k=0}^\infty a\_k^+x^k-\sum\_{k=0}^\infty a\_k^-x^k}

).

# Γ
{

z
0

,
⋯
,

z
m

}

{\displaystyle \Gamma=\{z\_0,\cdots,z\_m\}}

 be a partition of the ray joining

# z
0

{\displaystyle z=0}

 and

# z

e

i
θ

{\displaystyle z=e^{i \theta}}

 on the unit circle for some

θ
∈
[
0
,
2
π
)

{\displaystyle \theta \in [0, 2\pi)}

. Define

z
k

=

x
k

e

i
θ

{\displaystyle z\_k = x\_k e^{i \theta}}

 with real numbers

# 0

x
0

<

x
1

<
⋯
<

x
n

=
1

{\displaystyle 0=x\_0<x\_1<\cdots<x\_n=1}

. It follows that

S
Γ

=

∑

# j
1

n

f
(

z
j

)
−
f
(

z

j
−
1

)

=

∑

# j
1

n

∑

# k
0

∞

a
k

z
j
k

−

a

k

z

j
−
1

k

≤

∑

# k
0

∞

a
k

∑

# j
1

n

z
j
k

−

z

j
−
1

k

≤

∑

# k
0

∞

a
k

∑

# j
1

n

(

x
j
k

−

x

j
−
1

k

)

e

i
θ

=

∑

# k
0

∞

a
k

<
∞

{\displaystyle \begin{align}
S\_\Gamma&=\sum^n\_{j=1}\biggr |f(z\_j)-f(z\_{j-1}) \biggr| \\
&=\sum^n\_{j=1}\sum^\infty\_{k=0}\biggr | a\_kz\_j^k-a\_{k}z\_{j-1}^k \biggr|\\
&= \sum^\infty\_{k=0} |a\_k|\\
&<\infty
\end{align}}

since

∑

# j
1

n

(

x
j
k

−

x

j
−
1

k

)

=

x

n

k

−

x
0
k

=
1

{\displaystyle \sum^n\_{j=1}\left(x\_j^k-x\_{j-1}^k\right)=x^k\_{n}-x^k\_0=1}

 . As

S
Γ

<
∞

{\displaystyle S\_\Gamma<\infty}

, this implies

V
<
∞

{\displaystyle V<\infty}

 and hence

f
(
z
)

{\displaystyle f(z)}

 is of bounded variation on every radius of the circle

z

=
1

{\displaystyle |z|=1}

.

◼

{\displaystyle \blacksquare }
5-10

If

p
>
0

{\displaystyle p > 0}

 and

∫

E

f
−

f
k

p

→
0
,

{\displaystyle \int\_{E}|f-f\_k|^p\to 0,}

and

∫

E

f
k

p

≤
M

{\displaystyle \int\_{E}|f\_k|^p\leq M }

for all k, show that

∫

E

f

p

≤
M
.

{\displaystyle \int\_{E}|f|^p\leq M.}

Solution, Using the previous problem, we know that

f
k

→

m

f
;

{\displaystyle f\_k\to^{m} f;}

 in particular, there exist a subsequent

f

k
j

→
f

{\displaystyle f\_{k\_j}\to f}

 a.e. in

E

{\displaystyle E}

. Then by Fatou's Lemma,

∫

E

f

p

=

∫

E

lim
inf

f

k
j

p

≤
lim
inf

∫

E

f

k
j

p

≤
M
.
Ex2-31

**Chapter 2, Question 31**

Take

ϕ
(
x
# )
1

{\displaystyle \phi(x)=1}

 on

[
a
,
b
]

{\displaystyle [a, b]}

, since

ϕ

{\displaystyle \phi}

 is constant,

∫
a
b

f
d
ϕ

{\displaystyle \int\_a^b fd\phi}

 exists and

∫
a
b

f
d
# 1
0

{\displaystyle \int\_a^bfd1=0}

By theorem 2.21,

∫
a
b

1
d
f

{\displaystyle \int\_a^b1df}

 exists and

∫
a
b

1
d
# f
f
(
b
)
ϕ
(
b
)
−
f
(
a
)
ϕ
(
a
)
−

∫
a
b

f
d
# 1
f
(
b
)
−
f
(
a
)

{\displaystyle \int\_a^b1df=f(b)\phi(b)-f(a)\phi(a)-\int\_a^bfd1=f(b)-f(a)}

Suppose

f
′

{\displaystyle f'}

 exists and is Riemann-integrable on

[
a
,
b
]

{\displaystyle [a, b]}

, then let

# I

∫
a
b

f
′

d
x

{\displaystyle I=\int\_a^b f'dx}

 and

# J
f
(
b
)
−
f
(
a
# )

∫
a
b

d
f

{\displaystyle J=f(b)-f(a)=\int\_a^b df}

, we want to prove

# I
J

{\displaystyle I = J}

. By definition of Riemann-integrable, given

ε
>
0

{\displaystyle \varepsilon > 0}

 we can choose

δ
>
0

{\displaystyle \delta > 0}

 so that for each partition with

Γ

<
δ

{\displaystyle |\Gamma|<\delta}

and any choice of

ξ
i

∈
[

x

i
−
1

,

x
i

]

{\displaystyle \xi\_i \in [x\_{i-1}, x\_i]}

, we have

R
Γ

−
I

<
ε

 where

R
Γ

=

∑

# i
1

k

f
′

(

ξ
i

)
(

x
i

−

x

i
−
1

)
.

{\displaystyle |R\_\Gamma - I |<\varepsilon\quad\text{ where } \quad R\_\Gamma=\sum\_{i=1}^k f'(\xi\_i)(x\_i-x\_{i-1}).}

Note that in fact, if we let

S
Γ

{\displaystyle S\_\Gamma}

 denote the Riemann-Stieltjes sum, then

S
Γ

=

∑

# i
1

k

1
⋅
(
f
(

x
i

)
−
f
(

x

i
−
1

)
# )
f
(
b
)
−
f
(
a
)
,

{\displaystyle S\_\Gamma=\sum\_{i=1}^k 1\cdot (f(x\_i)-f(x\_{i-1}))=f(b)-f(a),}

therefore, we know that

S
Γ

=
J

{\displaystyle S\_\Gamma=J}

 for every partition

Γ

{\displaystyle \Gamma}

. By mean value theorem, there are some

ξ
i

∈
[

x

i
−
1

,

x
i

]

{\displaystyle \xi\_i \in [x\_{i-1}, x\_i]}

 for

# i
1
,
2
,
…
,
k

{\displaystyle i=1,2,\ldots,k}

 such that

f
(

x
i

)
−
f
(

x

i
−
1

# )

f
′

(

ξ
i

)
(

x
i

−

x

i
−
1

)
.

{\displaystyle f(x\_i)-f(x\_{i-1})=f'(\xi\_i)(x\_i-x\_{i-1}).}

Moreover, by restricting

Γ

<
δ

{\displaystyle |\Gamma|<\delta}

 with these choices of

ξ
i

{\displaystyle \xi\_i}

, we have

R
Γ

=

S
Γ

=
J

{\displaystyle R\_\Gamma=S\_\Gamma=J}

, thus

I
−
J

<
ε

{\displaystyle |I-J|<\varepsilon}

. By choosing

ε

{\displaystyle \varepsilon}

 arbitrarily small, we proved

# I
J

{\displaystyle I = J}

.
Ex2-14

Let

[
a
,
b
# ]
[
−
1
,
1
]

{\displaystyle [a,b]=[-1,1]}

 and take

f
(
x
# )

{

0

 if

−
1
≤
x
<
0
,

1

 if

0
≤
x
≤

ϕ
(
x
# )

{

0

 if

−
1
≤
x
≤
0
,

1

 if

0
<
x
≤

{\displaystyle f(x)={\begin{cases}0&{\text{ if }}-1\leq x<0,\\1&{\text{ if }}0\leq x\leq 1.\end{cases}}\,\,\,\,\,\,\,\,\,\,\,\,\phi (x)={\begin{cases}0&{\text{ if }}-1\leq x\leq 0,\\1&{\text{ if }}0<x\leq 1.\end{cases}}}

and we can calculate

∫

−
1

0

f
d
# ϕ

lim

Γ

→
0

∑
f
(

ξ

i

)
[
ϕ
(

x

i

)
−
ϕ
(

x

i
−
1

)
# ]

lim

Γ

→
0

∑
f
(

ξ

i

)
(
0
−
0
# )
0

{\displaystyle \int \_{-1}^{0}fd\phi =\lim \_{\left\vert \Gamma \right\vert \to 0}\sum f(\xi \_{i})[\phi (x\_{i})-\phi (x\_{i-1})]=\lim \_{\left\vert \Gamma \right\vert \to 0}\sum f(\xi \_{i})(0-0)=0}

∫

0

1

f
d
# ϕ

lim

Γ

→
0

∑
f
(

ξ

i

)
[
ϕ
(

x

i

)
−
ϕ
(

x

i
−
1

)
# ]
f
(

ξ

i

)
⋅
(
1
−
0
# )
1

{\displaystyle \int \_{0}^{1}fd\phi =\lim \_{\left\vert \Gamma \right\vert \to 0}\sum f(\xi \_{i})[\phi (x\_{i})-\phi (x\_{i-1})]=f(\xi \_{i})\cdot (1-0)=1}

The second equation exist at the point 0, also in this situation we can't get the

∫

−
1

1

f
d
ϕ

{\displaystyle \int \_{-1}^{1}fd\phi }

, because these two functions are both discontinuous (jump discontinuity of first kind) at point 0. In this condition, the different choice of partitions give different bound for

R

Γ

{\displaystyle R\_{\Gamma }}

, as it will give value of 0 or 1 conditionally, so we can't find the limit

∫

−
1

1

f
d
ϕ

{\displaystyle \int \_{-1}^{1}fd\phi }

.

For the Theorem 2.17, we can use contrapositive of the theorem, which is:

> If
>
>
>
>
> ∫
>
> a
>
>
> c
>
>
> f
> d
> ϕ
>
>
> {\displaystyle \int \_{a}^{c}fd\phi }
>
>  or
>
>
>
>
> ∫
>
> c
>
>
> b
>
>
> f
> d
> ϕ
>
>
> {\displaystyle \int \_{c}^{b}fd\phi }
>
>  does not exist, then
>
>
>
>
> ∫
>
> a
>
>
> b
>
>
> f
> d
> ϕ
>
>
> {\displaystyle \int \_{a}^{b}fd\phi }
>
>  does not exist for
>
>
>
> a
> <
> c
> <
> b
>
>
> {\displaystyle a<c<b}
>
> .
>
>

Then the statement above yields

∫

a

c

f
d
ϕ

{\displaystyle \int \_{a}^{c}fd\phi }

 and

∫

c

b

f
d
ϕ

{\displaystyle \int \_{c}^{b}fd\phi }

 may both exist for

a
<
c
<
b

{\displaystyle a<c<b}

, but

∫

a

b

f
d
ϕ

{\displaystyle \int \_{a}^{b}fd\phi }

 does not exist.
6-6

f
∈
L

(

R

)

{\displaystyle f\in L\left(\mathbb{R}\right)}

, define the Fourier transform

f
^

{\displaystyle \widehat{f}}

 of

f

{\displaystyle f}

 by

f
^

(
x
# )

1

2
π

∫

−
∞

∞

f
(
t
)

e

−
i
x
t

d
t

(

x
∈

R

)

.

{\displaystyle \widehat{f}(x)=\dfrac{1}{2\pi}\int^\infty\_{-\infty} f(t)e^{-ixt} \ dt \ \ \ \ \ \ \ \left(x\in\mathbb{R}\right).}

(For a complex-valued function

# F

F
0

i

F
1

{\displaystyle F=F\_0+iF\_1}

 whose real and imaginary parts

F
0

{\displaystyle F\_0}

 and

F
1

{\displaystyle F\_1}

 are integrable, we define

∫
# F
∫

F
0

i
∫

F
1

{\displaystyle \int F=\int F\_0+i\int F\_1}

.) Show that if

f

{\displaystyle f}

 and

g

{\displaystyle g}

 belong to

L

(

R

)

{\displaystyle L\left(\mathbb{R}\right)}

, then

f
∗
g

^

(
x
# )
2
π

f
^

(
x
)

g
^

(
x
)
.

{\displaystyle \widehat{f \ast g}(x)=2\pi \widehat{f}(x)\widehat{g}(x).}

### Solution.

f
∗
g

^

(
x
)

=

1

2
π

∫

R

[

∫

R

f

(

u
−
t

)

g
(
t
)

d
t

]

e

−
i
x
u

d
u

=

1

2
π

∫

R

g
(
t
)

e

−
i
x
t

[

∫

R

f
(
u
−
t
)

e

−
i
x
(
u
−
t
)

d
u

]

d
t

{\displaystyle \begin{align}
\widehat{f\ast g}(x)
&=\dfrac{1}{2\pi}\int\_{\mathbb{R}}\left[\int\_{\mathbb{R}} f\left(u-t\right)g(t) \ dt\right]e^{-ixu} \ du\\
&=\dfrac{1}{2\pi}\int\_{\mathbb{R}} g(t)e^{-ixt} \left[\int\_{\mathbb{R}}f(u-t)e^{-ix(u-t)} \ du \right]\ dt\\
\end{align} }

Denote the change of variable

# v
u
−
t

{\displaystyle v=u-t }

, we have

f
∗
g

^

(
x
)

=

1

2
π

∫

R

g
(
t
)

e

−
i
x
t

[

∫

R

f
(
v
)

e

−
i
x
v

d
v

]

d
t

=

1

2
π

∫

R

g
(
t
)

e

−
i
x
t

d
t
⋅

∫

R

f
(
v
)

e

−
i
x
v

d
v

=
2
π
⋅

[

1

2
π

∫

R

g
(
t
)

e

−
i
x
t

d
t

]

⋅

[

1

2
π

∫

R

f
(
v
)

e

−
i
x
v

d
v

]

=
2
π

f
^

(
x
)
∗

g
^

(
x
)
,

{\displaystyle \begin{align}
\widehat{f\ast g}(x)
&=\dfrac{1}{2\pi}\int\_{\mathbb{R}} g(t)e^{-ixt} \left[\int\_{\mathbb{R}}f(v)e^{-ixv} \ dv \right]\ dt\\
&=\dfrac{1}{2\pi}\int\_{\mathbb{R}} g(t)e^{-ixt} \ dt \cdot \int\_{\mathbb{R}}f(v)e^{-ixv} \ dv\\
&=2\pi\cdot \left[\dfrac{1}{2\pi}\int\_{\mathbb{R}} g(t)e^{-ixt} \ dt\right] \cdot \left[\dfrac{1}{2\pi}\int\_{\mathbb{R}}f(v)e^{-ixv} \ dv \right]\\
&=2\pi\widehat{f}(x)\ast \widehat{g}(x),
\end{align} }

where the second line holds using Fubini's theorem, as desired.

◼

{\displaystyle \blacksquare }
Assassin's Creed Fanstories Wiki

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Jesteśmy społecznością, która z chęcią przyjmie różne ciekawe pomysły na fanowską opowieść osadzoną w uniwersum Assassin's Creed. Przy tworzeniu nowej historii powiadom tylko administratora, czy chcesz, aby dostęp do edycji opowieści posiadała cała społeczność, czy tylko ty. Z góry dziękujemy za wszelki wkład :)

## Nasze opowieści

### Spis opowieści

Lista wszystkich opowieści fanowskich na naszej Wiki znajduje się **tutaj**.
8-10

Prove that

L
∞

(
E
)

{\displaystyle L^\infty(E)}

 is not separable for any

E

{\displaystyle E}

 where

E
>
0

{\displaystyle \vert E\vert >0}

.

Definition

A metric space is separable if it has countable dense subset.

Proof

Define

f

E

j

(
x
# )

χ

E

j

(
x
)

{\displaystyle f\_{E\_{j}}(x)=\chi \_{E\_{j}}(x)}

for

E

j

⊂
E

{\displaystyle E\_{j}\subset E}

 such that

j
∈

R

{\displaystyle j\in \mathbb {R} }

. We can do this, because

E

{\displaystyle E}

is uncountable since it has positive measure, and so clearly it has uncountably many subsets. Choose any two sets

E
1

,

E
2

{\displaystyle E\_1, E\_2}

, we have

‖

f

E

1

−

f

E

2

‖

∞

=
1

{\displaystyle \Vert f\_{E\_{1}}-f\_{E\_{2}}\Vert \_{\infty }=1}

Let

{

g

j

}

{\displaystyle \{g\_{j}\}}

 be countable dense subset of

L

∞

(

R

)

{\displaystyle L^{\infty }(\mathbb {R} )}

, this implies that for any

f

{\displaystyle f}

, and any

ϵ

{\displaystyle \epsilon}

, there exist a

g
j

{\displaystyle g\_j}

 such that

‖

g

j

−
f

‖

∞

<
ϵ

{\displaystyle \Vert g\_{j}-f\Vert \_{\infty }<\epsilon }

Let consider balls of radius

1
3

{\displaystyle \dfrac{1}{3}}

 for each

f

E

j

{\displaystyle f\_{E\_{j}}}

, then we have

g
j

{\displaystyle g\_j}

 for each

f

E

j

{\displaystyle f\_{E\_{j}}}

. Yet be noticed that

g

i

≠

g

j

{\displaystyle g\_{i}\neq g\_{j}}

 since each ball has diameter at most 2/3, and so it can only contain at most one of the elements of

{

g

j

}

{\displaystyle \{g\_{j}\}}

. This leads to the contradiction that

{

g

j

}

{\displaystyle \{g\_{j}\}}

 is countable set since

{

f

E

j

}

{\displaystyle \{f\_{E\_{j}}\}}

 is uncountable. Hence

L
∞

(
E
)

{\displaystyle L^\infty(E)}

 is not separable for any

E

{\displaystyle E}

 where

E
>
0

{\displaystyle \vert E\vert >0}

.
Ex2-6

**Question:** Let

f
(
x
# )

x
2

s
i
n
(
1

/

x
)

{\displaystyle f(x)=x^2sin(1/x)
}

 for

0
<
x
≤
1

{\displaystyle 0<x \leq 1}

 and

f
(
0
# )
0

{\displaystyle f(0) = 0}

. Show that

V
[
f
;
0
,
1
]
<
∞

{\displaystyle V[ f; 0, 1] <+\infty}

. (Examine the graph of f, or use Exercise 5 and Corollary 2.10.)

**Solution:** By the hint given by the Author, we will try to proof it using Ex2.5 and corollary 2.10.

Firstly, since

sin

{\displaystyle \sin}

 function is

≤
1

{\displaystyle \leq 1}

, we know that

f
(
x
# )

x
2

s
i
n
(
1

/

x
)
≤
1
,

{\displaystyle f(x)=x^2sin(1/x) \leq 1,}

 when

x
∈
(
0
,
1
]

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

and

f
(
0
# )
0
<
1

{\displaystyle f(0)=0<1}

. Thus,

f
(
x
)

{\displaystyle f(x)}

 is finite on

[
0
,
1
]

{\displaystyle [0,1]}

.

Next, we need to show

f
(
x
# )

x
2

s
i
n
(
1

/

x
)

{\displaystyle f(x)=x^2sin(1/x)
}

 of bounded variation on every interval

[
0
ϵ
,
1
]

{\displaystyle [0+\epsilon ,1] }

,
ϵ
>
0

{\displaystyle , \epsilon >0}

 with

V
[
f
;
a
ϵ
,
b
]
≤
M
<
∞

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

.

By Corollary 2.10, since

f
′

(
x
# )
2
x
s
i
n
(
1

/

x
)
−
c
o
s
(
1

/

x
)

{\displaystyle f'(x)=2xsin(1/x)-cos(1/x)
}

 is continuous on

[
ϵ
,
1
]
,
ϵ
>
0

{\displaystyle [\epsilon ,1], \epsilon >0}

, then we have

V

=

∫

ϵ

1

f
′

(
x
)

d
x

=

∫

ϵ

1

2
x
s
i
n
(
1

/

x
)
−
c
o
s
(
1

/

x
)

≤

∫

ϵ

1

3
d
x

=
3
(
1
−
ϵ
)
≤
3

 & \leq \int\_{\epsilon}^1 3 dx \\ &= 3(1-\epsilon) \leq 3 \end{align}
}

where we use trigo inequality and change the

x
s
i
n
(
1

/

x
)
,
c
o
s
(
1

/

x
)

{\displaystyle xsin(1/x), cos(1/x)}

 to 1, as

x
∈
[
ϵ
,
1
]

{\displaystyle x \in [\epsilon, 1] }

.

Thus, f is of bounded variation on

[
0
ϵ
,
1
]
,
∀
ϵ
>
0

{\displaystyle [0+\epsilon,1], \forall \epsilon >0}

,

and

V
[
f
;
a
ϵ
,
b
]
≤
3
<
∞

{\displaystyle V[ f; a+\epsilon, b] \leq 3 < +\infty}

. Also

f
(
x
)

{\displaystyle f(x)}

 is continuous at x=0. By Exercise 2.5,

V
[
f
;
0
,
1
]
<
∞
.

{\displaystyle V[f; 0 , 1 ] < +\infty.
}
Ex3-34

Let

E

{\displaystyle E}

 and

Z

{\displaystyle Z}

 be two sets in

R

n

{\displaystyle \R^n}

 such that

Z

{\displaystyle Z}

 has measure 0. If

E
∪
Z

{\displaystyle E\cup Z}

 is measurable, show that

E

{\displaystyle E}

 is measurable and

E
∪
Z

=

E

.

Solution:

1. In order to get

E

{\displaystyle E}

 from

E
∪
Z

{\displaystyle E\cup Z}

, we can remove the elements of

Z

{\displaystyle Z}

 from

E
∪
Z

{\displaystyle E\cup Z}

 first. Since this operation will also remove the elements of

E

{\displaystyle E}

 that are in

Z

{\displaystyle Z}

, we need to add the elements of

E
∩
Z

{\displaystyle E\cap Z}

 back. Hence, we get

E

=
(
(
E
∪
Z
)
−
Z
)
∪
(
E
∩
Z
)
.

{\displaystyle {\begin{aligned}E&=((E\cup Z)-Z)\cup (E\cap Z).\end{aligned}}}

By using some set operations,

E

=
(
(
E
∪
Z
)
−
Z
)
∪
(
E
∩
Z
)

=
(
(
E
∪
Z
)
∪
(
E
∩
Z
)
)
∩
(

Z

C

∪
(
E
∩
Z
)
)

=
(
E
∪
Z
)
∩
(

Z

C

∪
E
)

=
(
E
∪
Z
)
∩
(
Z
∩

E

C

)

C

.

{\displaystyle {\begin{aligned}E&=((E\cup Z)-Z)\cup (E\cap Z)\\&=((E\cup Z)\cup (E\cap Z))\cap (Z^{C}\cup (E\cap Z))\\&=(E\cup Z)\cap (Z^{C}\cup E)\\&=(E\cup Z)\cap (Z\cap E^{C})^{C}.\end{aligned}}}

Now, since

Z
∩

E

C

{\displaystyle Z\cap E^{C}}

 is a subset of

Z

{\displaystyle Z}

, it has measure 0, hence measurable. This implies that the complement of

Z
∩

E

C

{\displaystyle Z\cap E^{C}}

 is measurable. Thus,

E

{\displaystyle E}

 being the intersection of two measurable sets is measurable.
2. To prove

E
∪
Z

=

E

, since

E

{\displaystyle E}

 and

Z

{\displaystyle Z}

 are measurable,

E
∪
Z

=

E

Z

−

E
∩
Z

=

E

(since

E

 and

E
∩
Z

 have mesaure zero.)
5-26

Prove the following variant of Lebesgue's dominated convergence theorem: if

{

f
k

}

{\displaystyle \{ f\_k \}}

 satisfies

f
k

⟶
m

f

{\displaystyle f\_k {\overset{m}{\longrightarrow}} f}

 on

E

{\displaystyle E}

 and

f
k

≤
ϕ
∈
L
(
E
)

{\displaystyle |f\_k| \le \phi \in L(E)}

, then

f
∈
L
(
E
)

{\displaystyle f \in L(E)}

 and

∫
E

f
k

→

∫
E

f

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

. (Show that every subsequence of

{

f
k

}

{\displaystyle \{ f\_k \}}

 has a subsequence

{

f

k
j

}

{\displaystyle \{ f\_{k\_j} \}}

 such that

∫
E

f

k
j

→

∫
E

f

{\displaystyle \displaystyle\int\_E f\_{k\_j} \to \int\_E f}

.)

We prove by contradiction, that is, suppose

∫
E

f
k

↛

∫
E

f

{\displaystyle \displaystyle \int\_E f\_k \not\to \int\_E f}

. Then there exists

ε
>
0

{\displaystyle \varepsilon > 0}

 and a subsequece

{

f

k
j

}

{\displaystyle \{ f\_{k\_j} \}}

 of

{

f
k

}

{\displaystyle \{ f\_k \}}

 such that

∫
E

f

k
j

−

∫
E

f

≥
ε

{\displaystyle \left| \int\_E f\_{k\_j} - \int\_E f \right| \ge \varepsilon}

for all

j
∈

Z

{\displaystyle j \in \Z^+}

.

Since

f
k

⟶
m

f

{\displaystyle f\_k {\overset{m}{\longrightarrow}} f}

 on

E

{\displaystyle E}

, then so is

f

k
j

⟶
m

f

{\displaystyle f\_{k\_j} {\overset{m}{\longrightarrow}} f}

 on

E

{\displaystyle E}

. By Theorem 4.22, there is a subsequence

{

f

k

j
l

}

{\displaystyle \{f\_{k\_{j\_l}}\}}

 of

{

f

k
j

}

{\displaystyle \{ f\_{k\_j} \}}

 such that

f

k

j
l

→
f

{\displaystyle f\_{k\_{j\_l}} \to f}

 a.e. in

E

{\displaystyle E}

. Thus by applying Theorem 5.36 to

{

f

k

j
l

}

{\displaystyle \{f\_{k\_{j\_l}}\}}

, we have

∫
E

f

k

j
l

→

∫
E

f

{\displaystyle \displaystyle\int\_E f\_{k\_{j\_l}} \to \int\_E f}

, a contradiction.
Ex4-15

Question: Let

{

f
k

}

{\displaystyle \{ f\_k \}}

 be a sequence of measurable functions defined on a measurable

E

{\displaystyle E}

 with

E

<
∞

{\displaystyle |E|<+\infty}

. If

f
k

(

x

)

≤

M

x

<
∞

{\displaystyle |f\_k(\mathbf{x})|\leq M\_\mathbf{x}<+\infty}

 for all

k

{\displaystyle k}

 for each

x

∈
E

{\displaystyle \mathbf{x} \in E}

, show that given

ϵ
>
0

{\displaystyle \epsilon > 0}

, there is a closed

F
⊂
E

{\displaystyle F\subset E}

 and a finite

M

{\displaystyle M}

 such that

E
−
F

<
ϵ

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

 and

f
k

(

x

)

≤
M

{\displaystyle |f\_k(\mathbf{x})|\leq M}

 for all

k

{\displaystyle k}

 and all

x

∈
F

{\displaystyle \mathbf{x}\in F}

.

Answer:

Let

ϵ
>
0

{\displaystyle \epsilon > 0}

. For each

n
∈

N

{\displaystyle n \in \N}

, define

E
n

:=
{
x
∈
E
:

f
k

(
x
)

≤
n

for all

k

# k
1

∞

{
x
∈
E
:

f
k

(
x
)

≤
n
}

.

Note that each

E

n

{\displaystyle E\_{n}}

 is measurable since each

f
k

{\displaystyle f\_k}

 is measurable.

Since each

M
x

<
∞

{\displaystyle M\_x<\infty}

, we have that

E
n

↗
E

{\displaystyle E\_n\nearrow E}

. By the Monotone Convergence Theorem for measure,

lim

n
→
∞

E
n

=

E

<
∞

. Thus, there exists

N

{\displaystyle N}

 such that

E

−

E
N

=

E
∖

E
N

<
ϵ

/

2

.

Let

F

{\displaystyle F}

 be a closed set contained in

E
N

{\displaystyle E\_N}

 such that

E
N

∖
F

<
ϵ

/

2

{\displaystyle |E\_N\backslash F|<\epsilon/2}

.

Then

E
∖
F

=

E
∖

E
N

E
N

∖
F

<
ϵ

 and

f
K

(
x
)

≤
N

{\displaystyle |f\_K(x)|\leq N}

for all

k

{\displaystyle k}

 and all

x
∈
F

{\displaystyle x\in F}

.
Ex4-7

**Chapter 4, Question 7**

Question: Let

f

{\displaystyle f}

be usc and less than

∞

{\displaystyle +\infty }

on a compact set

E

{\displaystyle E}

. Show that

f

{\displaystyle f}

 is bounded above on

E

{\displaystyle E}

. Show also that

f

{\displaystyle f}

assumes its maximum on

E

{\displaystyle E}

, that is, that there exists

x
0

∈
E

{\displaystyle x\_0\in E }

 such that

f
(

x
0

)
≥
f
(
x
)

{\displaystyle f(x\_0)\geq f(x)}

 for all

x
∈
E

{\displaystyle x \in E}

.

Proof: Assume the contrary that

f

{\displaystyle f}

 is unbounded above. Then there is a sequence

{

x
n

}

{\displaystyle \{ x\_n \}}

 in

E

{\displaystyle E}

 such that

{
f
(

x
n

)
}

{\displaystyle \{ f(x\_n) \}}

is strictly increasing and

lim

n
→
∞

f
(

x
n

# )
∞
.

{\displaystyle \lim\_{n\to\infty} f(x\_n)=+\infty. }

Since a compact set is sequentially compact, we can choose a convergent subsequence

{

x

n
k

}

{\displaystyle \{x\_{n\_k}\}}

 which converges to

x
∈
E
.

{\displaystyle x\in E.}

By definition, one has

f
(
x
)
≥

lim sup

u
∈
E
;
u
→
x

f
(
u
)
≥

lim

k
→
∞

f
(

x

n
k

# )
∞
,

{\displaystyle f(x)\geq \limsup\_{u\in E; u\to x}f(u)\geq \lim\_{k\to\infty}f(x\_{n\_k})=+\infty,}

the last inequality is valid because

x

n
k

→
x

{\displaystyle x\_{n\_k}\to x}

. This means

f
(
x
# )
∞

{\displaystyle f(x)=+\infty}

, contradicts the assumption that

f
<
∞

{\displaystyle f<+\infty}

 on

E

{\displaystyle E}

.

Now let

# M

sup

x
∈
E

f
(
x
)

{\displaystyle M=\sup\_{x\in E}f(x) }

, we prove the supremum is attainable. For each

n
∈

N

{\displaystyle n \in \N}

 we choose

x
n

∈
E

{\displaystyle x\_n\in E}

 such that

f
(

x
n

)
>
M
−
1

/

n

{\displaystyle f(x\_n)>M-1/n}

. There is a convergent subsequence

{

x

n
k

}

{\displaystyle \{x\_{n\_k}\}}

 of

E

{\displaystyle E}

 which converges to

x
∈
E

{\displaystyle x \in E}

. Therefore, we have

M
≥
f
(
x
)
≥

lim sup

u
→
x
;
u
∈
E

f
(
u
)
≥

lim

k
→
∞

f
(

x

n
k

)
≥
M
.

{\displaystyle M\geq f(x)\geq \limsup\_{u\to x; u\in E}f(u)\geq \lim\_{k\to\infty} f(x\_{n\_k})\geq M.}

Therefore, we have

f
(
x
# )
M

{\displaystyle f(x)=M}

.
Ex2-3

**Solution to Chapter 2, Exercise 3**

Suppose

Γ
′

{\displaystyle \Gamma '}

is a partition of

[

a
′

,

b
′

]

{\displaystyle [a',b']}

. Then

# Γ

Γ
′

∪
{
a
,
b
}

{\displaystyle \Gamma =\Gamma '\cup \{a,b\}}

is a partition of

[
a
,
b
]

{\displaystyle [a,b]}

.

P

Γ

[
a
,
b
# ]
[
f
(

x

0

)
−
f
(
a
)

]

∑

# i
1

m

[
f
(

x

i

)
−
f
(

x

i
−
1

)

]

[
f
(
b
)
−
f
(

x

m

)

]

{\displaystyle P\_{\Gamma }[a,b]=[f(x\_{0})-f(a)]^{+}+\sum \_{i=1}^{m}[f(x\_{i})-f(x\_{i-1})]^{+}+[f(b)-f(x\_{m})]^{+}}

Since

[
f
(

x

0

)
−
f
(
a
)

]

≥
0

{\displaystyle [f(x\_{0})-f(a)]^{+}\geq 0}

and

[
f
(
b
)
−
f
(

x

m

)

]

≥
0

{\displaystyle [f(b)-f(x\_{m})]^{+}\geq 0}

, we have

P

Γ

[
a
,
b
]
≥

∑

# i
1

m

[
f
(

x

i

)
−
f
(

x

i
−
1

)

]

=

P

Γ
′

[

a
′

,

b
′

]
.

{\displaystyle P\_{\Gamma }[a,b]\geq \sum \_{i=1}^{m}[f(x\_{i})-f(x\_{i-1})]^{+}=P\_{\Gamma '}[a',b'].}

Then we have

P

Γ
′

[

a
′

,

b
′

]
≤

P

Γ

[
a
,
b
]
≤

V

P

[
a
,
b
]
.

{\displaystyle P\_{\Gamma '}[a',b']\leq P\_{\Gamma }[a,b]\leq V\_{P}[a,b].}

Therefore,

V

P

[

a
′

,

b
′

# ]

sup

Γ
′

P

Γ
′

[

a
′

,

b
′

]
≤

V

P

[
a
,
b
]
.

{\displaystyle V\_{P}[a',b']=\sup \_{\Gamma '}P\_{\Gamma '}[a',b']\leq V\_{P}[a,b].}

Similarly,

N

Γ

[
a
,
b
# ]
[
f
(

x

0

)
−
f
(
a
)

]

−

∑

# i
1

m

[
f
(

x

i

)
−
f
(

x

i
−
1

)

]

−

[
f
(
b
)
−
f
(

x

m

)

]

−

{\displaystyle N\_{\Gamma }[a,b]=[f(x\_{0})-f(a)]^{-}+\sum \_{i=1}^{m}[f(x\_{i})-f(x\_{i-1})]^{-}+[f(b)-f(x\_{m})]^{-}}

Since

[
f
(

x

0

)
−
f
(
a
)

]

−

≥
0

{\displaystyle [f(x\_{0})-f(a)]^{-}\geq 0}

and

[
f
(
b
)
−
f
(

x

m

)

]

−

≥
0

{\displaystyle [f(b)-f(x\_{m})]^{-}\geq 0}

, we have

N

Γ

[
a
,
b
]
≥

∑

# i
1

m

[
f
(

x

i

)
−
f
(

x

i
−
1

)

]

=

N

Γ
′

[

a
′

,

b
′

]
.

{\displaystyle N\_{\Gamma }[a,b]\geq \sum \_{i=1}^{m}[f(x\_{i})-f(x\_{i-1})]^{+}=N\_{\Gamma '}[a',b'].}

Then we have

N

Γ
′

[

a
′

,

b
′

]
≤

N

Γ

[
a
,
b
]
≤

V

N

[
a
,
b
]
.

{\displaystyle N\_{\Gamma '}[a',b']\leq N\_{\Gamma }[a,b]\leq V\_{N}[a,b].}

Therefore,

V

N

[

a
′

,

b
′

# ]

sup

Γ
′

N

Γ
′

[

a
′

,

b
′

]
≤

V

N

[
a
,
b
]
.

{\displaystyle V\_{N}[a',b']=\sup \_{\Gamma '}N\_{\Gamma '}[a',b']\leq V\_{N}[a,b].}
5-9

If

p
>
0

{\displaystyle p > 0}

, and

∫

E

f
−

f
k

p

→
0

{\displaystyle \int\_{E}|f-f\_k|^p\to 0}

as

k
→
∞

{\displaystyle k\to \infty}

, show that

f
k

→

m

f

{\displaystyle f\_k\to^{m}f}

 on

E

{\displaystyle E}

 (and thus there is a subsequence

f

k
j

→
f

{\displaystyle f\_{k\_j}\to f}

 a.e. in

E

{\displaystyle E}

).

Solution: We prove by contrapositive. Suppose there is

ϵ
>
0

{\displaystyle \epsilon > 0}

 such that

X
k

≥
ϵ

{\displaystyle |X\_k|\geq\epsilon}

 for all

k

{\displaystyle k}

, where

X
k

=
{

f
−

f
k

>
ϵ
}
.

{\displaystyle X\_k=\{|f-f\_k|>\epsilon\}.}

 Then

∫

E

f
−

f
k

p

≥

∫

X
k

f
−

f
k

p

≥

∫

X
k

ϵ
p

≥

ϵ

p
1

>
0

for all

k

{\displaystyle k}

. This contradicts the assumption that

∫

E

f
−

f
k

p

→

{\displaystyle \int\_{E}|f-f\_k|^p\to 0.}
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5-21

Question 21: If

∫
A

# f
0

{\displaystyle \int\_A f=0}

 for every measurable subset, prove that

# f
0

{\displaystyle f = 0}

 almost everyway.

Answer:

Note that

∫

R

f

=

∫

x
>
0

f
−

∫

x
≤
0

# f
0

{\displaystyle \int\_\mathcal{R}|f|=\int\_{x>0}f-\int\_{x\leq 0}f =0}

in the case that if

f
≠
0

{\displaystyle f \ne 0}

 in a nonzero measure set, this couldn't happen, this proves the theorem.