aarjaneiro/planetmath huggingface dataset:
L^∞⁢(X,μ)

L

∞

⁢

(

X

,

μ

)

Let

X

be a

nonempty set

and

𝒜

be a

σ

-algebra on

X

. Also, let

μ

be a non-negative

measure

defined on

𝒜

.
Two complex valued

functions

f

and

g

are said to be equal almost everywhere on

X

(denoted as

f

=

g

a.e. if

μ

⁢

{

x

∈

X

:

f

⁢

(

x

)

≠

g

⁢

(

x

)

}

=

0

.

The

relation

of being equal almost everywhere on

X

defines an

equivalence relation

.
It is a common practice in the integration theory to denote the

equivalence class

containing

f

by

f

itself.
It is easy to see that if

f

1

,

f

2

are

equivalent

and

g

1

,

g

2

are equivalent, then

f

1

+

g

1

,

f

2

+

g

2

are equivalent, and

f

1

⁢

g

1

,

f

2

⁢

g

2

are equivalent.
This naturally defines

addition

and multiplication among the equivalent classes of such functions.
For a measureable

f

:

X

→

ℂ

, we define

∥

f

∥

ess

=

inf

⁡

{

M

>

0

:

μ

⁢

{

x

:

|

f

⁢

(

x

)

|

>

M

}

=

0

}

,

called the

essential supremum

of

|

f

|

on

X

.
Now we define,

L

∞

⁢

(

X

,

μ

)

=

{

f

:

X

→

ℂ

:

∥

f

∥

ess

<

∞

}

.

Here the elements of

L

∞

⁢

(

X

,

μ

)

are equivalence classes.

Properties of

L

∞

⁢

(

X

,

μ

)

1.

The space

L

∞

⁢

(

X

,

μ

)

is a

normed linear space

with the
norm

∥

⋅

∥

ess

. Also, the metric defined by
the norm is

complete

, making

L

∞

⁢

(

X

,

μ

)

, a

Banach space

.

2.

L

∞

⁢

(

X

,

μ

)

is the dual of

L

1

⁢

(

X

,

μ

)

if

X

is

σ

-finite.

3.

L

∞

⁢

(

X

,

μ

)

is

closed under

pointwise

multiplication, and
with this multiplication it becomes an algebra.
Further,

L

∞

⁢

(

X

,

μ

)

is also a

C

*

-algebra (

http://planetmath.org/CAlgebra

) with the involution defined by

f

*

⁢

(

x

)

=

f

⁢

(

x

)

¯

. Since this

C

*

-algebra is also a dual of some Banach space, it is called

von Neumann algebra

.

Title

L

∞

⁢

(

X

,

μ

)

Canonical name

LinftyXmu

Date of creation

2013-03-22 13:59:46

Last modified on

2013-03-22 13:59:46

Owner

ack (3732)

Last modified by

ack (3732)

Numerical id

11

Author

ack (3732)

Entry type

Definition

Classification

msc 28A25

Generated on Fri Feb 9 20:08:49 2018 by

LaTeXML
L^p-norm is dual to L^q

L

p

-norm is dual to

L

q

If

(

X

,

𝔐

,

μ

)

is any

measure space

and

1

≤

p

,

q

≤

∞

are Hölder conjugates (

http://planetmath.org/ConjugateIndex

) then, for

f

∈

L

p

, the following

linear function

can be defined

Φ

f

:

L

q

→

ℂ

,

g

↦

Φ

f

⁢

(

g

)

≡

∫

f

⁢

g

⁢

𝑑

μ

.

The Hölder inequality (

http://planetmath.org/HolderInequality

) shows that this gives a well defined and

bounded linear map

. Its

operator norm

is given by

∥

Φ

f

∥

=

{

∥

f

⁢

g

∥

1

:

g

∈

L

q

,

∥

g

∥

q

=

1

}

.

The following theorem shows that the operator norm of

Φ

f

is equal to the

L

p

-norm of

f

.

Theorem

.

Let

(

X

,

M

,

μ

)

be a

σ

-

finite measure space

and

p

,

q

be Hölder conjugates. Then, any

measurable function

f

:

X

→

C

has

L

p

-norm

∥

f

∥

p

=

sup

⁡

{

∥

f

⁢

g

∥

1

:

g

∈

L

q

,

∥

g

∥

q

=

1

}

.

(1)

Furthermore, if either

p

<

∞

and

∥

f

∥

p

<

∞

or

p

=

1

then

μ

is not required to be

σ

-finite.

Note that the

σ

-finite condition is required, except in the cases mentioned. For example, if

μ

is the

measure

satisfying

μ

⁢

(

A

)

=

∞

for every nonempty set

A

, then

L

p

⁢

(

μ

)

=

{

0

}

for

p

<

∞

and it is easily checked that equality (

1

) fails whenever

f

=

1

and

p

>

1

.

Title

L

p

-norm is dual to

L

q

Canonical name

LpnormIsDualToLq

Date of creation

2013-03-22 18:38:13

Last modified on

2013-03-22 18:38:13

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

5

Author

gel (22282)

Entry type

Theorem

Classification

msc 28A25

Classification

msc 46E30

Related topic

LpSpace

Related topic

HolderInequality

Related topic

BoundedLinearFunctionalsOnLinftymu

Related topic

BoundedLinearFunctionalsOnLpmu

Generated on Fri Feb 9 20:08:51 2018 by

LaTeXML
proof of Clarkson inequality

proof of Clarkson inequality

Suppose

2

≤

p

<

∞

⁢

and

⁢

f

,

g

∈

L

p

.

∥

f

+

g

2

∥

p

p

+

∥

f

-

g

2

∥

p

p

=

∫

|

f

+

g

2

|

p

⁢

𝑑

μ

+

∫

|

f

-

g

2

|

p

⁢

𝑑

μ

(1)

=

1

2

p

⁢

(

∫

|

f

+

g

|

p

⁢

𝑑

μ

+

∫

|

f

-

g

|

p

⁢

𝑑

μ

)

.

(2)

By the

triangle inequality

, we have the following two

inequalities

|

f

+

g

|

p

≤

|

f

|

p

+

|

g

|

p

and

|

f

-

g

|

p

≤

|

f

|

p

+

|

g

|

p

,

and

summing

the two inequalities we get

|

f

+

g

|

p

+

|

f

-

g

|

p

≤

2

⁢

(

|

f

|

p

+

|

g

|

p

)

.

This means that

expression

(2) above is less than or equal to

1

2

p

-

1

⁢

∫

(

|

f

|

p

+

|

g

|

p

)

⁢

𝑑

μ

.

(3)

Hence it follows that

∥

f

+

g

2

∥

p

p

+

∥

f

-

g

2

∥

p

p

≤

1

2

p

-

1

⁢

(

∫

|

f

|

p

⁢

𝑑

μ

+

∫

|

g

|

p

⁢

𝑑

μ

)

=

1

2

p

-

1

⁢

(

∥

f

∥

p

p

+

∥

g

∥

p

p

)

,

which since

p

≥

2

directly implies the desired inequality.

Title

proof of Clarkson inequality

Canonical

name

ProofOfClarksonInequality

Date of creation

2013-03-22 16:24:46

Last modified on

2013-03-22 16:24:46

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

8

Author

CWoo (3771)

Entry type

Proof

Classification

msc 28A25

Generated on Fri Feb 9 20:08:58 2018 by

LaTeXML
summable function

summable function

A

measurable function

f

:

Ω

→

ℝ

where

(

Ω

,

𝒜

,

μ

)

is a

measure space

is said to be

summable

if the

Lebesgue integral

of the

absolute value

of

f

exists and is finite,

∫

Ω

|

f

|

⁢

𝑑

μ

<

+

∞

An alternative way of expressing this condition is to assert that

f

∈

L

1

⁢

(

Ω

)

.

Note that some authors distinguish between integrable and summable: an integrable function is one for which the above

integral

exists; a

summable function

is one for which the integral exists and is finite.

Title

summable function

Canonical name

SummableFunction

Date of creation

2013-03-22 18:12:14

Last modified on

2013-03-22 18:12:14

Owner

ehremo (15714)

Last modified by

ehremo (15714)

Numerical id

8

Author

ehremo (15714)

Entry type

Definition

Classification

msc 28A25

Related topic

LebesgueIntegrable

Generated on Fri Feb 9 20:08:59 2018 by

LaTeXML
𝕃^p vs 𝕃^q

𝕃

p

vs

𝕃

q

Let

(

X

,

𝒜

,

μ

)

be a

measure space

and

1

≤

p

,

q

≤

∞

.
Generally there is no connection between

𝕃

p

⁢

(

μ

)

and

𝕃

q

⁢

(

μ

)

as sets.
However, for some special

measures

,
there is an interesting relationship between them.
A few examples:

1.

If

λ

n

is the

Lebesgue measure

on

ℝ

n

and

p

≠

q

,
then

𝕃

p

⁢

(

λ

n

)

⊈

𝕃

q

⁢

(

λ

n

)

for all

n

∈

ℕ

+

.
Here is an example for

n

=

1

and

1

≤

p

<

q

<

∞

.
Let

f

⁢

(

x

)

:=

{

x

-

1

p

,

x

>

1

0

,

x

≤

1

and

g

⁢

(

x

)

:=

{

x

-

1

q

,

x

∈

(

0

,

1

)

0

,

x

∉

(

0

,

1

)

.

This gives

∥

f

∥

q

q

=

p

q

-

p

,

∥

g

∥

p

p

=

q

q

-

p

and

∥

f

∥

p

=

∥

g

∥

q

=

∞

.
So

f

∈

𝕃

q

∖

𝕃

p

and

g

∈

𝕃

p

∖

𝕃

q

.
For the

∞

-norm,

χ

ℝ

∈

𝕃

∞

∖

𝕃

q

,
where

χ

is the

characteristic function

,
and also

f

∉

𝕃

∞

.

2.

If

p

<

q

then

l

p

⊆

l

q

.
This is

trivial

if

q

=

∞

.
Now let

x

=

(

x

0

,

x

1

,

…

)

∈

l

p

and

q

<

∞

.
Then

∥

x

∥

q

q

=

∑

n

=

0

∞

|

x

n

|

q

=

∑

n

=

0

∞

|

x

n

|

p

⁢

|

x

n

|

q

-

p

≤

∑

n

=

0

∞

|

x

n

|

p

⁢

∥

x

∥

∞

q

-

p

=

∥

x

∥

∞

q

-

p

⁢

∥

x

∥

p

p

<

∞

,

so

x

∈

l

q

.

3.

If

μ

⁢

(

X

)

is finite and

p

<

q

,
then

𝕃

q

⊆

𝕃

p

.
This is easy if

q

=

∞

,
because

|

f

|

≤

∥

f

∥

∞

almost everywhere,
so

∥

f

∥

p

p

=

∫

|

f

|

p

⁢

𝑑

μ

≤

∫

∥

f

∥

∞

p

⁢

𝑑

μ

=

∥

f

∥

∞

p

⁢

μ

⁢

(

X

)

<

∞

.
Now let

q

<

∞

,
thus

∥

f

∥

p

p

=

∫

|

f

|

p

⁢

𝑑

μ

=

∫

|

f

|

>

1

|

f

|

p

⁢

𝑑

μ

+

∫

|

f

|

≤

1

|

f

|

p

⁢

𝑑

μ

≤

∫

|

f

|

>

1

|

f

|

q

⁢

𝑑

μ

+

∫

|

f

|

≤

1

𝑑

μ

≤

∥

f

∥

q

q

+

μ

⁢

(

X

)

<

∞

.

Finally, we prove an interesting property for

p

-norms:
if

X

is a

finite measure space

,
then for any

measurable function

f

on

X

the equality

lim

p

→

∞

⁡

∥

f

∥

p

=

∥

f

∥

∞

holds.
We have already seen that

∥

f

∥

p

≤

∥

f

∥

∞

⁢

μ

⁢

(

X

)

1

p

.
Now for any

ε

∈

(

0

,

∥

f

∥

∞

)

define

A

ε

:=

{

x

∈

X

:

|

f

⁢

(

x

)

|

≥

∥

f

∥

∞

-

ε

}

,

δ

ε

:=

μ

⁢

(

A

ε

)

>

0

and

g

:=

(

∥

f

∥

∞

-

ε

)

⁢

χ

A

ε

.
Since

|

g

|

≤

|

f

|

,
we have

∥

g

∥

p

=

(

∥

f

∥

∞

-

ε

)

⁢

δ

ε

1

p

≤

∥

f

∥

p

≤

∥

f

∥

∞

⁢

μ

⁢

(

X

)

1

p

.
Now we take

lim inf

on the left and

lim sup

on the right side:

∥

f

∥

∞

-

ε

≤

lim inf

p

→

∞

⁡

∥

f

∥

p

≤

lim sup

p

→

∞

⁡

∥

f

∥

p

≤

∥

f

∥

∞

.
Taking

ε

↓

0

gives

lim

p

→

∞

⁡

∥

f

∥

p

=

∥

f

∥

∞

.

Title

𝕃

p

vs

𝕃

q

Canonical name

mathbbLpVsmathbbLq

Date of creation

2013-03-22 15:22:05

Last modified on

2013-03-22 15:22:05

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

11

Author

yark (2760)

Entry type

Topic

Classification

msc 28A25

Generated on Fri Feb 9 20:09:03 2018 by

LaTeXML
Koch curve

Koch curve

A

Koch curve

is a

fractal

generated by

a replacement rule. This rule is, at each step, to replace the middle

1

/

3

of each

line segment

with two sides of a

right triangle

having sides of length equal to the replaced segment. Two applications of this rule on a single line segment gives us:

To generate the Koch curve, the rule is applied indefinitely, with a starting line segment. Note that, if the length of the initial line segment is

l

, the length

L

K

of the Koch curve at the

n

th step will be

L

K

=

(

4

3

)

n

⁢

l

This quantity increases without bound; hence the Koch curve has

infinite

length. However, the curve still bounds a finite area. We can prove this by noting that in each step, we add an amount of area equal to the area of all the

equilateral triangles

we have just created. We can bound the area of each

triangle

of side length

s

by

s

2

(the square containing the triangle.) Hence, at step

n

, the area

A

K

“under” the Koch curve (assuming

l

=

1

) is

A

K

<

(

1

3

)

2

+

3

⁢

(

1

9

)

2

+

9

⁢

(

1

27

)

2

+

⋯

=

∑

i

=

1

n

1

3

i

-

1

but this is a geometric series of ratio less than one, so it

converges

. Hence a Koch curve has infinite length and bounds a finite area.

A

Koch snowflake

is the figure generated by applying the Koch replacement rule to an equilateral triangle indefinitely.

Title

Koch curve

Canonical

name

KochCurve

Date of creation

2013-03-22 12:05:34

Last modified on

2013-03-22 12:05:34

Owner

akrowne (2)

Last modified by

akrowne (2)

Numerical id

8

Author

akrowne (2)

Entry type

Definition

Classification

msc 28A33

Classification

msc 28A80

Synonym

Koch snowflake

Generated on Fri Feb 9 20:09:05 2018 by

LaTeXML
measurable space

measurable space

A

measurable space

is a set

E

together with a

collection

ℬ

of subsets of

E

which is a

sigma algebra

.

The elements of

ℬ

are called

measurable sets

.

A measurable space is the correct object on which to define a

measure

;

ℬ

will be the collection of sets which actually have a measure. We normally want to ensure that

ℬ

contains all the sets we will ever want to use. We usually cannot take

ℬ

to be the collection of all subsets of

E

because the

axiom of choice

often allows one to construct sets that would lead to a

contradiction

if we gave them a measure (even zero). For the

real numbers

, Vitali’s

theorem

states that

ℬ

cannot be the collection of all subsets if we hope to have a measure that returns the length of an

open interval

.

Title

measurable space

Canonical name

MeasurableSpace

Date of creation

2013-03-22 11:57:30

Last modified on

2013-03-22 11:57:30

Owner

djao (24)

Last modified by

djao (24)

Numerical id

11

Author

djao (24)

Entry type

Definition

Classification

msc 28A33

Defines

measurable set

Generated on Fri Feb 9 20:09:07 2018 by

LaTeXML
counter-example of Fubini’s theorem for the Lebesgue integral

counter-example of Fubini’s theorem for the Lebesgue integral

The following observation demonstrates the

necessity

of the
integrability

assumption

in Fubini’s

theorem

. Let

Q

=

{

(

x

,

y

)

∈

ℝ

2

:

x

≥

0

,

y

≥

0

}

denote the upper, right quadrant.
Let

R

⊂

Q

be the region in the quadrant bounded by the lines

y

=

x

,

y

=

x

-

1

, and let let

S

⊂

Q

be a

similar

region, but this
time bounded by the lines

y

=

x

-

1

,

y

=

x

-

2

. Let

f

=

χ

S

-

χ

R

,

where

χ

denotes a

characteristic function

.

Observe that the

Lebesgue measure

of

R

and of

S

is

infinite

.
Hence,

f

is not a Lebesgue-integrable

function

. However for every

x

≥

0

the function

g

⁢

(

x

)

=

∫

0

∞

f

⁢

(

x

,

y

)

⁢

𝑑

y

is

integrable

.
Indeed,

g

⁢

(

x

)

=

{

-

x

for

⁢

0

≤

x

≤

1

,

x

-

2

for

⁢

1

≤

x

≤

2

,

0

for

⁢

x

≥

2

.

Similarly, for

y

≥

0

, the function

h

⁢

(

y

)

=

∫

0

∞

f

⁢

(

x

,

y

)

⁢

𝑑

x

is integrable. Indeed,

h

⁢

(

y

)

=

0

,

y

≥

0

.

Hence, the values of the

iterated integrals

∫

0

∞

g

⁢

(

x

)

⁢

𝑑

x

=

-

1

,

∫

0

∞

h

⁢

(

y

)

⁢

𝑑

y

=

0

,

are finite, but do not agree. This does not contradict Fubini’s
theorem because the value of the planar

Lebesgue integral

∫

Q

f

⁢

(

x

,

y

)

⁢

𝑑

μ

⁢

(

x

,

y

)

,

where

μ

⁢

(

x

,

y

)

is the planar Lebesgue measure, is not defined.

Title

counter-example of Fubini’s theorem for the Lebesgue integral

Canonical name

CounterexampleOfFubinisTheoremForTheLebesgueIntegral

Date of creation

2013-03-22 18:18:15

Last modified on

2013-03-22 18:18:15

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

6

Author

rmilson (146)

Entry type

Example

Classification

msc 28A35

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

LaTeXML
counter-example to Tonelli’s theorem

counter-example to Tonelli’s theorem

The following observation demonstrates the

necessity

of the

σ

-finite

assumption

in Tonelli’s and Fubini’s

theorem

. Let

X

denote the closed unit interval

[

0

,

1

]

equipped with

Lebesgue measure

and

Y

the same set, but this time equipped with

counting measure

ν

. Let

f

⁢

(

x

,

y

)

=

{

1

if

⁢

x

=

y

,

0

otherwise

.

Observe that

∫

Y

(

∫

X

f

⁢

(

x

,

y

)

⁢

𝑑

μ

⁢

(

x

)

)

⁢

𝑑

ν

⁢

(

y

)

=

0

,

while

∫

X

(

∫

Y

f

⁢

(

x

,

y

)

⁢

𝑑

ν

⁢

(

y

)

)

⁢

𝑑

μ

⁢

(

x

)

=

1

.

The

iterated integrals

do not give the same value, this despite the
fact that the integrand is a non-negative

function

.

Also observe that there does not exist a

simple function

on

X

×

Y

that is dominated by

f

. Hence,

∫

X

×

Y

f

(

x

,

y

)

d

(

μ

(

x

)

×

ν

(

y

)

=

0

.

Therefore, the integrand is

L

1

integrable

relative to the

product measure

. However, as we observed above, the
iterated integrals do not agree. This observation illustrates the need for the

σ

-finite assumption for Fubini’s theorem.

Title

counter-example to Tonelli’s theorem

Canonical name

CounterexampleToTonellisTheorem

Date of creation

2013-03-22 18:16:36

Last modified on

2013-03-22 18:16:36

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

4

Author

rmilson (146)

Entry type

Example

Classification

msc 28A35

Generated on Fri Feb 9 20:09:11 2018 by

LaTeXML
infinite product measure

infinite product measure

Let

(

E

i

,

ℬ

i

,

μ

i

)

be

measure spaces

, where

i

∈

I

an

index set

, possibly

infinite

. We define the

product

of

(

E

i

,

ℬ

i

,

μ

i

)

as follows:

1.

let

E

=

∏

E

i

, the

Cartesian product

of

E

i

,

2.

let

ℬ

=

σ

⁢

(

(

ℬ

i

)

i

∈

I

)

, the smallest

sigma algebra

containing subsets of

E

of the form

∏

B

i

where

B

i

=

E

i

for all but a finite number of

i

∈

I

.

Then

(

E

,

ℬ

)

is a

measurable space

. The next task is to define a

measure

μ

on

(

E

,

ℬ

)

so that

(

E

,

ℬ

,

μ

)

becomes in addition a measure space. Before proceeding to define

μ

, we make the

assumption

that

each

μ

i

is a

totally finite measure

, that is,

μ

i

⁢

(

E

i

)

<

∞

.

In fact, we can now turn each

(

E

i

,

ℬ

i

,

μ

i

)

into a

probability space

by introducing for each

i

∈

I

a new measure:

μ

¯

i

=

μ

i

μ

i

⁢

(

E

i

)

.

With the assumption that each

(

E

i

,

ℬ

i

,

μ

i

)

is a probability space, it can be shown that there is a

unique

measure

μ

defined on

ℬ

such that, for any

B

∈

ℬ

expressible as a product of

B

i

∈

ℬ

i

with

B

i

=

E

i

for all

i

∈

I

except on a finite subset

J

of

I

:

μ

⁢

(

B

)

=

∏

j

∈

J

μ

j

⁢

(

B

j

)

.

Then

(

E

,

ℬ

,

μ

)

becomes a measure space, and in particular, a probability space.

μ

is sometimes written

∏

μ

i

.

Remarks

.

•

If

I

is infinite, one sees that the total finiteness of

μ

i

can not be dropped. For example, if

I

is the set of positive integers, assume

μ

1

⁢

(

E

1

)

<

∞

and

μ

2

⁢

(

E

2

)

=

∞

. Then

μ

⁢

(

B

)

for

B

:=

B

1

×

∏

i

>

1

E

i

=

B

1

×

E

2

×

∏

i

>

2

E

i

⁢

, where

⁢

B

1

∈

ℬ

1

would not be well-defined (on the one hand, it is

μ

1

⁢

(

B

1

)

<

∞

, but on the other it is

μ

1

⁢

(

B

1

)

⁢

μ

2

⁢

(

E

2

)

=

∞

).

•

The above construction agrees with the result when

I

is finite (see finite

product measure

(

http://planetmath.org/ProductMeasure

)).

Title

infinite product measure

Canonical name

InfiniteProductMeasure

Date of creation

2013-03-22 16:23:14

Last modified on

2013-03-22 16:23:14

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

13

Author

CWoo (3771)

Entry type

Definition

Classification

msc 28A35

Classification

msc 60A10

Related topic

ProductSigmaAlgebra

Defines

totally finite measure

Generated on Fri Feb 9 20:09:25 2018 by

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product measure

product measure

Let

(

E

1

,

ℬ

1

⁢

(

E

1

)

)

and

(

E

2

,

ℬ

2

⁢

(

E

2

)

)

be two

measurable spaces

, with

measures

μ

1

and

μ

2

. Let

ℬ

1

×

ℬ

2

be the

sigma algebra

on

E

1

×

E

2

generated by subsets of the form

B

1

×

B

2

, where

B

1

∈

ℬ

1

⁢

(

E

1

)

and

B

2

∈

ℬ

2

⁢

(

E

2

)

.

The

product measure

μ

1

×

μ

2

is defined to be the unique measure on the measurable space

(

E

1

×

E

2

,

ℬ

1

×

ℬ

2

)

satisfying the property

μ

1

×

μ

2

⁢

(

B

1

×

B

2

)

=

μ

1

⁢

(

B

1

)

⁢

μ

2

⁢

(

B

2

)

⁢

for all

⁢

B

1

∈

ℬ

1

⁢

(

E

1

)

,

B

2

∈

ℬ

2

⁢

(

E

2

)

.

Title

product measure

Canonical name

ProductMeasure

Date of creation

2013-03-22 12:00:33

Last modified on

2013-03-22 12:00:33

Owner

djao (24)

Last modified by

djao (24)

Numerical id

7

Author

djao (24)

Entry type

Definition

Classification

msc 28A35

Generated on Fri Feb 9 20:09:26 2018 by

LaTeXML
proof of Fubini’s theorem for the Lebesgue integral

proof of Fubini’s theorem for the Lebesgue integral

Let

μ

x

and

μ

y

be

measures

on

X

and

Y

respectively, let

μ

be the

product measure

μ

x

⊗

μ

y

, and let

f

⁢

(

x

,

y

)

be

μ

-

integrable

on

A

⊂

X

×

Y

. Then

∫

A

f

⁢

(

x

,

y

)

⁢

𝑑

μ

=

∫

X

(

∫

A

x

f

⁢

(

x

,

y

)

⁢

𝑑

μ

y

)

⁢

𝑑

μ

x

=

∫

Y

(

∫

A

y

f

⁢

(

x

,

y

)

⁢

𝑑

μ

x

)

⁢

𝑑

μ

y

where

A

x

=

{

y

∣

(

x

,

y

)

∈

A

}

,

A

y

=

{

x

∣

(

x

,

y

)

∈

A

}

Proof

: Assume for now that

f

⁢

(

x

,

y

)

≥

0

. Consider the set

U

=

X

×

Y

×

ℝ

equipped with the measure

μ

u

=

μ

x

⊗

μ

y

⊗

μ

1

=

μ

⊗

μ

1

=

μ

x

⊗

λ

where

μ

1

is ordinary

Lebesgue measure

and

λ

=

μ

y

⊗

μ

1

. Also consider the set

W

⊂

U

defined by

W

=

{

(

x

,

y

,

z

)

∣

(

x

,

y

)

∈

A

,

0

≤

z

≤

f

⁢

(

x

,

y

)

}

Then

μ

u

⁢

(

W

)

=

∫

A

f

⁢

(

x

,

y

)

⁢

𝑑

μ

And

μ

u

⁢

(

W

)

=

∫

X

λ

⁢

(

W

x

)

⁢

𝑑

μ

x

where

W

x

=

{

(

y

,

z

)

∣

(

x

,

y

,

z

)

∈

W

}

However, we also have that

λ

⁢

(

W

x

)

=

∫

A

x

f

⁢

(

x

,

y

)

⁢

𝑑

μ

y

Combining the last three equations gives us Fubini’s

theorem

. To remove the

restriction

that

f

⁢

(

x

,

y

)

be nonnegative, write

f

as

f

⁢

(

x

,

y

)

=

f

+

⁢

(

x

,

y

)

-

f

-

⁢

(

x

,

y

)

where

f

+

⁢

(

x

,

y

)

=

|

f

⁢

(

x

,

y

)

|

+

f

⁢

(

x

,

y

)

2

,

f

-

⁢

(

x

,

y

)

=

|

f

⁢

(

x

,

y

)

|

-

f

⁢

(

x

,

y

)

2

are both nonnegative.

Title

proof of Fubini’s theorem for the

Lebesgue integral

Canonical name

ProofOfFubinisTheoremForTheLebesgueIntegral

Date of creation

2013-03-22 15:21:52

Last modified on

2013-03-22 15:21:52

Owner

azdbacks4234 (14155)

Last modified by

azdbacks4234 (14155)

Numerical id

4

Author

azdbacks4234 (14155)

Entry type

Proof

Classification

msc 28A35

Generated on Fri Feb 9 20:09:29 2018 by

LaTeXML
Tonelli’s theorem

Tonelli’s theorem

Here denote

L

+

⁢

(

X

)

as the space of

measurable functions

X

→

[

0

,

∞

]

. Furthermore all

integrals

are

Lebesgue integrals

.

Theorem

(Tonelli).

Suppose

(

X

,

M

,

μ

)

and

(

Y

,

N

,

ν

)

are

σ

-finite (

http://planetmath.org/SigmaFinite

)

measure spaces

. If

f

∈

L

+

⁢

(

X

×

Y

)

, then the

functions

x

↦

∫

Y

f

⁢

(

x

,

y

)

⁢

𝑑

ν

⁢

(

y

)

and

y

↦

∫

X

f

⁢

(

x

,

y

)

⁢

𝑑

μ

⁢

(

x

)

are in

L

+

⁢

(

X

)

and

L

+

⁢

(

Y

)

respectively, and furthermore if we denote by

μ

×

ν

the

product measure

, then

∫

X

×

Y

f

⁢

d

⁢

(

μ

×

ν

)

=

∫

X

[

∫

Y

f

⁢

(

x

,

y

)

⁢

𝑑

ν

⁢

(

y

)

]

⁢

𝑑

μ

⁢

(

x

)

=

∫

Y

[

∫

X

f

⁢

(

x

,

y

)

⁢

𝑑

μ

⁢

(

x

)

]

⁢

𝑑

ν

⁢

(

y

)

.

Basically this says that you can switch the of integrals, or integrate over the product space as long as everything is positive and the spaces are

σ

-finite. Do note that we allow the functions to take on the value of

infinity

with the standard conventions used in Lebesgue integration. That is,

0

⋅

∞

=

0

, so that if a function is

infinite

on a set of

measure

0, then this does not contribute anything to the value of the integral.
See the entry on extended real numbers for further discussion.

If we take the

counting measure

on

ℕ

, then one can the

Tonelli theorem

for sums.

Theorem

(Tonelli for sums).

Suppose that

f

i

⁢

j

≥

0

for all

i

,

j

∈

N

, then

∑

i

,

j

∈

ℕ

f

i

⁢

j

=

∑

i

=

1

∞

∑

j

=

1

∞

f

i

⁢

j

=

∑

j

=

1

∞

∑

i

=

1

∞

f

i

⁢

j

.

In the above

theorem

we have used

ℕ

as our set for simplicity and familiarity of notation.
If you would have an

uncountable

number of non-zero

elements

f

i

⁢

j

then
all the sums would be infinite and the result would be

trivial

.
So the theorem for arbitrary
sets just reduces to the above case.

References

1

Gerald B. Folland.

. John Wiley & Sons, Inc., New York, New York, 1999

Title

Tonelli’s theorem

Canonical name

TonellisTheorem

Date of creation

2013-03-22 14:15:49

Last modified on

2013-03-22 14:15:49

Owner

jirka (4157)

Last modified by

jirka (4157)

Numerical id

7

Author

jirka (4157)

Entry type

Theorem

Classification

msc 28A35

Related topic

FubinisTheorem

Related topic

FubinisTheoremForTheLebesgueIntegral

Generated on Fri Feb 9 20:09:31 2018 by

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