aarjaneiro/planetmath huggingface dataset:
proof of dominated convergence theorem

proof of dominated convergence theorem

Define the

functions

h

n

+

and

h

n

-

as follows:

h

n

+

⁢

(

x

)

=

sup

⁡

{

f

m

⁢

(

x

)

:

m

≥

n

}

h

n

-

⁢

(

x

)

=

inf

⁡

{

f

m

⁢

(

x

)

:

m

≥

n

}

These suprema and infima exist because, for every

x

,

|

f

n

⁢

(

x

)

|

≤

g

⁢

(

x

)

. These functions enjoy the following properties:

For every

n

,

|

h

n

±

|

≤

g

The

sequence

h

n

+

is

decreasing

and the sequence

h

n

-

is increasing.

For every

x

,

lim

n

→

∞

⁡

h

n

±

⁢

(

x

)

=

f

⁢

(

x

)

Each

h

n

±

is

measurable

.

The first property follows from immediately from the definition of

supremum

. The second property follows from the fact that the
supremum or

infimum

is being taken over a larger set to define

h

n

±

⁢

(

x

)

than to define

h

m

±

⁢

(

x

)

when

n

>

m

. The third
property is a simple

consequence

of the fact that, for any sequence
of

real numbers

, if the sequence

converges

, then the sequence has an

upper limit

and a

lower limit

which equal each other and equal the
limit. As for the fourth statement, it means that, for every real
number

y

and every

integer

n

, the sets

{

x

∣

h

n

-

⁢

(

x

)

≥

y

}

⁢

and

⁢

{

x

∣

h

n

+

⁢

(

x

)

≤

y

}

are measurable. However, by the definition of

h

n

±

, these sets
can be expressed as

⋃

m

≤

n

{

x

∣

f

n

⁢

(

x

)

≤

y

}

⁢

and

⁢

⋃

m

≥

n

{

x

∣

f

n

⁢

(

x

)

≤

y

}

respectively. Since each

f

n

is assumed to be measurable, each set
in either union is measurable. Since the union of a

countable

number of

measurable sets

is itself measurable, these unions are
measurable, and hence the functions

h

n

±

are measurable.

Because of properties 1 and 4 above and the

assumption

that

g

is

integrable

, it follows that each

h

n

±

is integrable. This

conclusion

and property 2 mean that the

monotone convergence theorem

is applicable so one can conclude that

f

is integrable and that

lim

n

→

∞

⁡

∫

h

n

±

⁢

(

x

)

⁢

𝑑

μ

⁢

(

x

)

=

∫

lim

n

→

∞

⁡

h

n

±

⁢

(

x

)

⁢

d

⁢

μ

⁢

(

x

)

By property 3, the right hand side equals

∫

f

⁢

(

x

)

⁢

𝑑

μ

⁢

(

x

)

.

By construction,

h

n

-

≤

f

n

≤

h

n

+

and hence

∫

h

n

-

≤

∫

f

n

≤

∫

h

n

+

Because the outer two terms in the above

inequality

tend towards the
same limit as

n

→

∞

, the middle term is squeezed into
converging to the same limit. Hence

lim

n

→

∞

⁡

∫

f

n

⁢

(

x

)

⁢

𝑑

μ

⁢

(

x

)

=

∫

f

⁢

(

x

)

⁢

𝑑

μ

⁢

(

x

)

Title

proof of dominated convergence theorem

Canonical name

ProofOfDominatedConvergenceTheorem1

Date of creation

2013-03-22 14:33:58

Last modified on

2013-03-22 14:33:58

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

5

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 28A20

Related topic

ProofOfDominatedConvergenceTheorem

Generated on Fri Feb 9 20:08:09 2018 by

LaTeXML
proof of Egorov’s theorem

proof of Egorov’s theorem

Let

E

i

,

j

=

{

x

∈

E

:

|

f

j

⁢

(

x

)

-

f

⁢

(

x

)

|

<

1

/

i

}

.

Since

f

n

→

f

almost everywhere, there is a set

S

with

μ

⁢

(

S

)

=

0

such that, given

i

∈

ℕ

and

x

∈

E

-

S

, there is

m

∈

ℕ

such that

j

>

m

implies

|

f

j

⁢

(

x

)

-

f

⁢

(

x

)

|

<

1

/

i

. This can be expressed by

E

-

S

⊂

⋃

m

∈

ℕ

⋂

j

>

m

E

i

,

j

,

or, in other words,

⋂

m

∈

ℕ

⋃

j

>

m

(

E

-

E

i

,

j

)

⊂

S

.

Since

{

⋃

j

>

m

(

E

-

E

i

,

j

)

}

m

∈

ℕ

is a decreasing nested

sequence

of sets, each of which has finite measure, and such that its

intersection

has measure

0

, by continuity from above (

http://planetmath.org/PropertiesForMeasure

) we know that

μ

⁢

(

⋃

j

>

m

(

E

-

E

i

,

j

)

)

→

m

→

∞

0

.

Therefore, for each

i

∈

ℕ

, we can choose

m

i

such that

μ

⁢

(

⋃

j

>

m

i

(

E

-

E

i

,

j

)

)

<

δ

2

i

.

Let

E

δ

=

⋃

i

∈

ℕ

⋃

j

>

m

i

(

E

-

E

i

,

j

)

.

Then

μ

⁢

(

E

δ

)

≤

∑

i

=

1

∞

μ

⁢

(

⋃

j

>

m

i

(

E

-

E

i

,

j

)

)

<

∑

i

=

1

∞

δ

2

i

=

δ

.

We claim that

f

n

→

f

uniformly on

E

-

E

δ

. In fact, given

ε

>

0

, choose

n

such that

1

/

n

<

ε

. If

x

∈

E

-

E

δ

, we have

x

∈

⋂

i

∈

ℕ

⋂

j

>

m

i

E

i

,

j

,

which in particular implies that, if

j

>

m

n

,

x

∈

E

n

,

j

; that is,

|

f

j

⁢

(

x

)

-

f

⁢

(

x

)

|

<

1

/

n

<

ε

.
Hence, for each

ε

>

0

there is

N

(which is given by

m

n

above) such that

j

>

N

implies

|

f

j

⁢

(

x

)

-

f

⁢

(

x

)

|

<

ε

for each

x

∈

E

-

E

δ

, as required. This the proof.

Title

proof of Egorov’s

theorem

Canonical name

ProofOfEgorovsTheorem

Date of creation

2013-03-22 13:47:59

Last modified on

2013-03-22 13:47:59

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

7

Author

Koro (127)

Entry type

Proof

Classification

msc 28A20

Generated on Fri Feb 9 20:08:11 2018 by

LaTeXML
proof of existence of the essential supremum

proof of existence of the essential supremum

Suppose that

(

Ω

,

ℱ

,

μ

)

is a

σ

-

finite measure space

and

𝒮

is a

collection

of

measurable functions

f

:

Ω

→

ℝ

¯

. We show that the

essential supremum

of

𝒮

exists and furthermore, if it is nonempty then there is a

sequence

f

n

∈

𝒮

such that

ess

⁢

sup

⁡

𝒮

=

sup

n

⁡

f

n

.

As any

σ

-finite

measure

is

equivalent

to a probability measure (

http://planetmath.org/AnySigmaFiniteMeasureIsEquivalentToAProbabilityMeasure

), we may suppose without loss of generality that

μ

is a probability measure. Also, without loss of generality, suppose that

𝒮

is nonempty, and let

𝒮

′

consist of the collection of

maximums

of

finite sequences

of

functions

in

𝒮

. Then choose any

continuous

and

strictly increasing

θ

:

ℝ

¯

→

ℝ

. For example, we can take

θ

⁢

(

x

)

=

{

x

/

(

1

+

|

x

|

)

,

if

⁢

|

x

|

<

∞

,

1

,

if

⁢

x

=

∞

,

-

1

,

if

⁢

x

=

-

∞

.

As

θ

⁢

(

f

)

is a

bounded

and measurable function for all

f

∈

𝒮

′

, we can set

α

=

sup

⁡

{

∫

θ

⁢

(

f

)

⁢

𝑑

μ

:

f

∈

𝒮

′

}

.

Then choose a sequence

g

n

in

𝒮

′

such that

∫

θ

⁢

(

g

n

)

⁢

𝑑

μ

→

α

. By replacing

g

n

by the maximum of

g

1

,

…

,

g

n

if necessary, we may assume that

g

n

+

1

≥

g

n

for each

n

. Set

f

=

sup

n

⁡

g

n

.

Also, every

g

n

is the maximum of a finite sequence of functions

g

n

,

1

,

…

,

g

n

,

m

n

in

𝒮

. Therefore, there exists a sequence

f

n

∈

𝒮

such that

{

f

1

,

f

2

,

…

}

=

{

g

n

,

m

:

n

≥

1

,

1

≤

m

≤

m

n

}

.

Then,

f

=

sup

n

⁡

f

n

.

It only remains to be shown that

f

is indeed the essential supremum of

𝒮

.
First, by

continuity

of

θ

and the

dominated convergence theorem

,

∫

θ

⁢

(

f

)

⁢

𝑑

μ

=

lim

n

→

∞

⁡

∫

θ

⁢

(

g

n

)

⁢

𝑑

μ

=

α

.

Similarly, for any

g

∈

𝒮

,

∫

θ

⁢

(

f

∨

g

)

=

lim

n

→

∞

⁡

∫

θ

⁢

(

g

n

∨

g

)

⁢

𝑑

μ

≤

α

.

It follows that

θ

⁢

(

f

∨

g

)

-

θ

⁢

(

f

)

is a nonnegative function with nonpositive

integral

, and so is equal to zero

μ

-almost everywhere. As

θ

is strictly increasing,

f

∨

g

=

f

and therefore

f

≥

g

μ

-almost everywhere.

Finally, suppose that

g

:

Ω

→

ℝ

¯

satisfies

g

≥

h

(

μ

-a.e.) for all

h

∈

𝒮

. Then,

g

≥

f

n

and,

g

≥

sup

n

⁡

f

n

=

f

μ

-a.e., as required.

Title

proof of

existence

of the essential supremum

Canonical name

ProofOfExistenceOfTheEssentialSupremum

Date of creation

2013-03-22 18:39:25

Last modified on

2013-03-22 18:39:25

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

7

Author

gel (22282)

Entry type

Proof

Classification

msc 28A20

Generated on Fri Feb 9 20:08:13 2018 by

LaTeXML
proof of Fatou-Lebesgue theorem

proof of Fatou-Lebesgue theorem

Since

|

∫

g

⁢

𝑑

μ

|

≤

∫

|

g

|

⁢

𝑑

μ

≤

∫

Φ

⁢

𝑑

μ

<

∞

, we have that

∫

g

⁢

𝑑

μ

>

-

∞

. Similarly,

∫

h

⁢

𝑑

μ

<

∞

.

The

inequality

lim inf

n

→

∞

⁡

∫

f

n

⁢

𝑑

μ

≤

lim sup

n

→

∞

⁡

∫

f

n

⁢

𝑑

μ

is

obvious

by

definition

of

lim inf

and

lim sup

.

Define a sequence of functions

k

n

:

X

→

ℝ

by

k

n

⁢

(

x

)

=

f

n

⁢

(

x

)

+

Φ

⁢

(

x

)

. Then each

k

n

is nonnegative (since

-

f

n

≤

|

f

n

|

≤

Φ

) and

integrable

(since

k

n

≤

|

f

n

|

+

Φ

≤

2

⁢

Φ

), as is

k

:=

lim inf

n

→

∞

⁡

k

n

. Fatou’s lemma yields that

∫

k

⁢

𝑑

μ

≤

lim inf

n

→

∞

⁡

∫

k

n

⁢

𝑑

μ

. Thus:

∫

g

⁢

𝑑

μ

+

∫

Φ

⁢

𝑑

μ

=

∫

(

g

+

Φ

)

⁢

𝑑

μ

=

∫

k

⁢

𝑑

μ

≤

lim inf

n

→

∞

⁡

∫

k

n

⁢

𝑑

μ

=

lim inf

n

→

∞

⁡

∫

(

f

n

+

Φ

)

⁢

𝑑

μ

=

lim inf

n

→

∞

⁡

(

∫

f

n

⁢

𝑑

μ

+

∫

Φ

⁢

𝑑

μ

)

=

lim inf

n

→

∞

⁡

∫

f

n

⁢

𝑑

μ

+

lim inf

n

→

∞

⁡

∫

Φ

⁢

𝑑

μ

=

lim inf

n

→

∞

⁡

∫

f

n

⁢

𝑑

μ

+

∫

Φ

⁢

𝑑

μ

Since

∫

Φ

⁢

𝑑

μ

<

∞

, it follows that

∫

g

⁢

𝑑

μ

≤

lim inf

n

→

∞

⁡

∫

f

n

⁢

𝑑

μ

.

Note that

|

-

f

n

|

=

|

f

n

|

≤

Φ

. Thus,

-

∫

h

⁢

𝑑

μ

=

∫

-

h

⁢

d

⁢

μ

=

∫

-

lim sup

n

→

∞

⁡

f

n

⁢

d

⁢

μ

=

∫

lim inf

n

→

∞

⁡

(

-

f

n

)

⁢

d

⁢

μ

≤

lim inf

n

→

∞

⁡

∫

-

f

n

⁢

d

⁢

μ

by a previous ,

=

lim inf

n

→

∞

⁡

(

-

∫

f

n

⁢

𝑑

μ

)

=

-

lim sup

n

→

∞

⁡

∫

f

n

⁢

𝑑

μ

.

Hence,

lim sup

n

→

∞

⁡

∫

f

n

⁢

𝑑

μ

≤

∫

h

⁢

𝑑

μ

. It follows that

-

∞

<

∫

g

⁢

𝑑

μ

≤

lim inf

n

→

∞

⁡

∫

f

n

⁢

𝑑

μ

≤

lim sup

n

→

∞

⁡

∫

f

n

⁢

𝑑

μ

≤

∫

h

⁢

𝑑

μ

<

∞

.

∎

Title

proof of Fatou-Lebesgue theorem

Canonical

name

ProofOfFatouLebesgueTheorem

Date of creation

2013-03-22 15:58:50

Last modified on

2013-03-22 15:58:50

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

13

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 28A20

Generated on Fri Feb 9 20:08:16 2018 by

LaTeXML
proof of Fatou’s lemma

proof of Fatou’s lemma

Let

f

⁢

(

x

)

=

lim inf

n

→

∞

⁡

f

n

⁢

(

x

)

and let

g

n

⁢

(

x

)

=

inf

k

≥

n

⁡

f

k

⁢

(

x

)

so that we have

f

⁢

(

x

)

=

sup

n

⁡

g

n

⁢

(

x

)

.

As

g

n

is an

increasing sequence

of

measurable

nonnegative functions we can apply the

monotone convergence Theorem

to obtain

∫

X

f

⁢

𝑑

μ

=

lim

n

→

∞

⁡

∫

X

g

n

⁢

𝑑

μ

.

On the other hand, being

g

n

≤

f

n

, we conclude by observing

lim

n

→

∞

⁡

∫

X

g

n

⁢

𝑑

μ

=

lim inf

n

→

∞

⁡

∫

X

g

n

⁢

𝑑

μ

≤

lim inf

n

→

∞

⁡

∫

X

f

n

⁢

𝑑

μ

.

Title

proof of Fatou’s lemma

Canonical name

ProofOfFatousLemma

Date of creation

2013-03-22 13:29:59

Last modified on

2013-03-22 13:29:59

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

4

Author

paolini (1187)

Entry type

Proof

Classification

msc 28A20

Generated on Fri Feb 9 20:08:18 2018 by

LaTeXML
proof of monotone convergence theorem

proof of monotone convergence theorem

It is enough to prove the following

Theorem 1

Let

(

X

,

μ

)

be a

measurable space

and let

f

k

:

X

→

R

∪

{

+

∞

}

be a

monotone increasing

sequence

of

positive

measurable functions

(i.e.

0

≤

f

1

≤

f

2

≤

…

). Then

f

⁢

(

x

)

=

lim

k

→

∞

⁡

f

k

⁢

(

x

)

is

measurable

and

lim

n

→

∞

⁡

∫

X

f

k

⁢

𝑑

μ

=

∫

X

f

⁢

(

x

)

⁢

𝑑

μ

.

First of all by the monotonicity of the sequence we have

f

⁢

(

x

)

=

sup

k

⁡

f

k

⁢

(

x

)

hence we know that

f

is measurable. Moreover being

f

k

≤

f

for all

k

, by the monotonicity of the

integral

, we immediately get

sup

k

⁡

∫

X

f

k

⁢

𝑑

μ

≤

∫

X

f

⁢

(

x

)

⁢

𝑑

μ

.

So take any simple measurable function

s

such that

0

≤

s

≤

f

. Given also

α

<

1

define

E

k

=

{

x

∈

X

:

f

k

⁢

(

x

)

≥

α

⁢

s

⁢

(

x

)

}

.

The sequence

E

k

is an

increasing sequence

of

measurable sets

. Moreover the union of all

E

k

is the whole space

X

since

lim

k

→

∞

⁡

f

k

⁢

(

x

)

=

f

⁢

(

x

)

≥

s

⁢

(

x

)

>

α

⁢

s

⁢

(

x

)

. Moreover it holds

∫

X

f

k

⁢

𝑑

μ

≥

∫

E

k

f

k

⁢

𝑑

μ

≥

α

⁢

∫

E

k

s

⁢

𝑑

μ

.

Since

s

is a simple measurable function it is easy to check that

E

↦

∫

E

s

⁢

𝑑

μ

is a

measure

and hence

sup

k

⁡

∫

X

f

k

⁢

𝑑

μ

≥

α

⁢

∫

X

s

⁢

𝑑

μ

.

But this last

inequality

holds for every

α

<

1

and for all simple measurable functions

s

with

s

≤

f

. Hence by the definition of

Lebesgue integral

sup

k

⁡

∫

X

f

k

⁢

𝑑

μ

≥

∫

X

f

⁢

𝑑

μ

which

completes

the proof.

Title

proof of monotone convergence theorem

Canonical name

ProofOfMonotoneConvergenceTheorem

Date of creation

2013-03-22 13:29:56

Last modified on

2013-03-22 13:29:56

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

7

Author

paolini (1187)

Entry type

Proof

Classification

msc 28A20

Classification

msc 26A42

Generated on Fri Feb 9 20:08:23 2018 by

LaTeXML
regarding the sets A_n from the traveling hump sequence

regarding the sets

A

n

from the traveling hump sequence

In this entry,

⌊

⋅

⌋

denotes the

floor function

.

Following is a proof that, for every positive integer

n

,

[

n

-

2

⌊

log

2

⁡

n

⌋

2

⌊

log

2

⁡

n

⌋

,

n

-

2

⌊

log

2

⁡

n

⌋

+

1

2

⌊

log

2

⁡

n

⌋

]

⊆

[

0

,

1

]

.

Proof.

Note that this is

equivalent

(

http://planetmath.org/Equivalent

) to showing that, for every positive integer

n

,

n

-

2

⌊

log

2

⁡

n

⌋

≥

0

and

n

-

2

⌊

log

2

⁡

n

⌋

+

1

≤

2

⌊

log

2

⁡

n

⌋

. This in turn is equivalent to showing that, for every positive integer

n

,

2

⌊

log

2

⁡

n

⌋

≤

n

and

n

+

1

≤

2

⌊

log

2

⁡

n

⌋

+

1

.

The first

inequality

is easy to prove: For every positive integer

n

,

2

⌊

log

2

⁡

n

⌋

≤

2

log

2

⁡

n

=

n

.

Now for the second inequality. Let

n

be a positive integer. Let

k

be the unique positive integer such that

2

k

-

1

≤

n

≤

2

k

-

1

. Then

n

+

1

≤

2

k

=

2

k

-

1

+

1

=

2

⌊

k

-

1

⌋

+

1

=

2

⌊

log

2

⁡

2

k

-

1

⌋

+

1

≤

2

⌊

log

2

⁡

n

⌋

+

1

.
∎

Title

regarding the sets

A

n

from the

traveling hump sequence

Canonical name

RegardingTheSetsAnFromTheTravelingHumpSequence

Date of creation

2013-03-22 16:14:28

Last modified on

2013-03-22 16:14:28

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

6

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 28A20

Generated on Fri Feb 9 20:08:29 2018 by

LaTeXML
traveling hump sequence

traveling hump sequence

In this entry,

⌊

⋅

⌋

denotes the

floor function

and

m

denotes

Lebesgue measure

.

For every positive integer

n

, let

A

n

=

[

n

-

2

⌊

log

2

⁡

n

⌋

2

⌊

log

2

⁡

n

⌋

,

n

-

2

⌊

log

2

⁡

n

⌋

+

1

2

⌊

log

2

⁡

n

⌋

]

. Then every

A

n

is a subset of

[

0

,

1

]

(click here (

http://planetmath.org/RegardingTheSetsA_nFromTheTravelingHumpSequence

) to see a proof) and is Lebesgue measurable (clear from the fact that each of them is closed (

http://planetmath.org/Closed

)).

For every positive integer

n

, define

f

n

:

[

0

,

1

]

→

ℝ

by

f

n

=

χ

A

n

, where

χ

S

denotes the

characteristic function

of the set

S

. The

sequence

{

f

n

}

is called the

traveling hump sequence

. This colorful name arises from the sequence of the graphs of these

functions

: A “hump” seems to travel from

[

0

,

1

2

k

]

to

[

2

k

-

1

2

k

,

1

]

, then shrinks by half and starts from the very left again.

The traveling hump sequence is an important sequence for at least two reasons. It provides a

counterexample

for the following two statements:

•

Convergence in measure

implies convergence almost everywhere with respect to

m

.

•

L

1

⁢

(

m

)

convergence (

http://planetmath.org/L1muConvergence

) implies convergence almost everywhere with respect to

m

.

Note that

{

f

n

}

is a sequence of

measurable functions

that does not

converge

pointwise

(

http://planetmath.org/PointwiseConvergence

). For every

x

∈

[

0

,

1

]

, there exist infinitely many positive integers

a

such that

f

a

⁢

(

x

)

=

0

, and there exist infinitely many positive integers

b

such that

f

b

⁢

(

x

)

=

1

.

On the other hand,

{

f

n

}

converges in measure to

0

and converges in

L

1

⁢

(

m

)

(

http://planetmath.org/ConvergesInL1mu

) to

0

.

Title

traveling hump sequence

Canonical name

TravelingHumpSequence

Date of creation

2013-03-22 16:14:08

Last modified on

2013-03-22 16:14:08

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

14

Author

Wkbj79 (1863)

Entry type

Definition

Classification

msc 28A20

Related topic

ModesOfConvergenceOfSequencesOfMeasurableFunctions

Generated on Fri Feb 9 20:08:32 2018 by

LaTeXML
Vitali convergence theorem

Vitali convergence theorem

Let

f

1

,

f

2

,

…

be

𝐋

p

-integrable functions on some

measure space

, for

1

≤

p

<

∞

.

The

sequence

{

f

n

}

converges

in

𝐋

p

to a

measurable function

f

if and and only if

i

the sequence

{

f

n

}

converges to

f

in

measure

;

ii

the

functions

{

|

f

n

|

p

}

are

uniformly integrable

; and

iii

for every

ϵ

>

0

, there exists a set

E

of finite measure, such that

∫

E

c

|

f

n

|

p

<

ϵ

for all

n

.

Remarks

This

theorem

can be used as a replacement for the more
well-known

dominated convergence theorem

, when a
dominating cannot be found for the functions

f

n

to be integrated.
(If this theorem is known, the dominated convergence theorem
can be derived as a special case.)

In a

finite measure space

, condition (iii) is

trivial

.
In fact, condition (iii) is the tool used to reduce considerations
in the general case to the case of a finite measure space.

In probability , the definition of “uniform integrability”
is slightly different from its definition in general

measure theory

;
either definition may be used in the statement of this theorem.

References

1

Gerald B. Folland.

Real Analysis: Modern Techniques and Their Applications

, second ed. Wiley-Interscience, 1999.

2

Jeffrey S. Rosenthal.

A First Look at

Rigorous

Probability Theory

.
World Scientific, 2003.

Title

Vitali convergence theorem

Canonical name

VitaliConvergenceTheorem

Date of creation

2013-03-22 16:17:10

Last modified on

2013-03-22 16:17:10

Owner

stevecheng (10074)

Last modified by

stevecheng (10074)

Numerical id

9

Author

stevecheng (10074)

Entry type

Theorem

Classification

msc 28A20

Synonym

uniform-integrability convergence theorem

Related topic

ModesOfConvergenceOfSequencesOfMeasurableFunctions

Related topic

UniformlyIntegrable

Related topic

DominatedConvergenceTheorem

Generated on Fri Feb 9 20:08:38 2018 by

LaTeXML
bounded linear functionals on L^∞⁢(μ)

bounded linear functionals on

L

∞

⁢

(

μ

)

For any

measure space

(

X

,

𝔐

,

μ

)

and

g

∈

L

1

⁢

(

μ

)

, the following linear map can be defined

Φ

g

:

L

∞

⁢

(

μ

)

→

ℝ

,

f

↦

Φ

g

⁢

(

f

)

≡

∫

f

⁢

g

⁢

𝑑

μ

.

It is easily shown that

Φ

g

is

bounded

(

http://planetmath.org/OperatorNorm

), so is a member of the

dual space

of

Ł

∞

⁢

(

μ

)

. However, unless the measure space consists of a

finite set

of atoms, not every element of the dual of

L

∞

⁢

(

μ

)

can be written like this. Instead, it is necessary to restrict to linear maps satisfying a bounded convergence property.

Theorem

.

Let

(

X

,

M

,

μ

)

be a

σ

-finite (

http://planetmath.org/SigmaFinite

) measure space and

V

be the space of

bounded linear maps

Φ

:

L

∞

⁢

(

μ

)

→

R

satisfying bounded convergence. That is, if

|

f

n

|

≤

1

are in

L

∞

⁢

(

μ

)

and

f

n

⁢

(

x

)

→

0

for almost every

x

∈

X

, then

Φ

⁢

(

f

n

)

→

0

.

Then

g

↦

Φ

g

gives an

isometric isomorphism

from

L

1

⁢

(

μ

)

to

V

.

Proof.

First, the

operator norm

∥

Φ

g

∥

is equal to the

L

1

-norm of

g

(see

L

p

-norm is dual to

L

q

(

http://planetmath.org/LpNormIsDualToLq

)), so the map

g

↦

Φ

g

gives an

isometric embedding

from

L

1

into the dual of

L

∞

. Furthermore, dominated convergence implies that

Φ

g

satisfies

bounded convergence so

Φ

g

∈

V

. We just need to show that

g

↦

Φ

g

maps onto

V

.

So, suppose that

Φ

∈

V

. It needs to be shows that

Φ

=

Φ

g

for some

g

∈

L

1

.
Defining an additive set function (

http://planetmath.org/

Additive

)

ν

:

𝔐

→

ℝ

by

ν

⁢

(

A

)

=

Φ

⁢

(

1

A

)

for every set

A

∈

𝔐

, the bounded convergence property for

Φ

implies that

ν

is

countably additive

and is therefore a finite

signed measure

. So, the

Radon-Nikodym theorem

gives a

g

∈

L

1

such that

ν

⁢

(

A

)

=

∫

A

g

⁢

𝑑

μ

for every

A

∈

𝔐

.
Then, the

equality

Φ

⁢

(

f

⁢

h

)

=

∫

f

⁢

g

⁢

𝑑

μ

is satisfied for

f

=

1

A

with any

A

∈

𝔐

and the

functional monotone class theorem

extends this to any bounded and measurable

f

:

X

→

ℂ

, giving

Φ

g

=

Φ

.
∎

Title

bounded

linear functionals

on

L

∞

⁢

(

μ

)

Canonical name

BoundedLinearFunctionalsOnLinftymu

Date of creation

2013-03-22 18:38:08

Last modified on

2013-03-22 18:38:08

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

5

Author

gel (22282)

Entry type

Theorem

Classification

msc 28A25

Related topic

BoundedLinearFunctionalsOnLpmu

Related topic

RadonNikodymTheorem

Related topic

LpNormIsDualToLq

Generated on Fri Feb 9 20:08:41 2018 by

LaTeXML
Clarkson inequality

Clarkson inequality

The

Clarkson inequality

says that for all

f

,

g

∈

L

p

, for

2

≤

p

<

∞

we have:

∥

f

+

g

2

∥

p

p

+

∥

f

-

g

2

∥

p

p

≤

1

2

⁢

(

∥

f

∥

p

p

+

∥

g

∥

p

p

)

.

The

inequality

can be used to prove that

L

p

space is

uniformly convex

for

2

≤

p

<

∞

.

Remark

.
If

1

<

p

<

2

, then the Clarkson inequality becomes:

∥

f

+

g

2

∥

p

q

+

∥

f

-

g

2

∥

p

q

≤

(

1

2

⁢

∥

f

∥

p

p

+

1

2

⁢

∥

g

∥

p

p

)

1

p

-

1

.

for functions

f

,

g

∈

L

p

, where

q

=

p

p

-

1

.

Title

Clarkson inequality

Canonical name

ClarksonInequality

Date of creation

2013-03-22 16:04:59

Last modified on

2013-03-22 16:04:59

Owner

georgiosl (7242)

Last modified by

georgiosl (7242)

Numerical id

13

Author

georgiosl (7242)

Entry type

Theorem

Classification

msc 28A25

Generated on Fri Feb 9 20:08:45 2018 by

LaTeXML
Hardy-Littlewood maximal operator

Hardy-Littlewood maximal operator

The

Hardy-Littlewood maximal operator

in

ℝ

n

is an

operator

defined on

L

loc

1

⁢

(

ℝ

n

)

(the space of

locally integrable functions

in

ℝ

n

with the

Lebesgue measure

) which maps each locally integrable function

f

to another

function

M

⁢

f

, defined for each

x

∈

ℝ

n

by

M

⁢

f

⁢

(

x

)

=

sup

Q

⁡

1

m

⁢

(

Q

)

⁢

∫

Q

|

f

⁢

(

y

)

|

⁢

𝑑

y

,

where the

supremum

is taken over all cubes

Q

containing

x

.
This function is

lower semicontinuous

(and hence

measurable

), and it is called the

Hardy-Littlewood maximal function

of

f

.

The operator

M

is sublinear, which means that

M

⁢

(

a

⁢

f

+

b

⁢

g

)

≤

|

a

|

⁢

M

⁢

f

+

|

b

|

⁢

M

⁢

g

for each pair of locally integrable functions

f

,

g

and scalars

a

,

b

.

Title

Hardy-Littlewood maximal operator

Canonical name

HardyLittlewoodMaximalOperator

Date of creation

2013-03-22 13:27:30

Last modified on

2013-03-22 13:27:30

Owner

azdbacks4234 (14155)

Last modified by

azdbacks4234 (14155)

Numerical id

8

Author

azdbacks4234 (14155)

Entry type

Definition

Classification

msc 28A25

Classification

msc 28A15

Related topic

HardyLittlewoodMaximalTheorem

Defines

Hardy-Littlewood maximal function

Generated on Fri Feb 9 20:08:47 2018 by

LaTeXML