aarjaneiro/planetmath huggingface dataset:
product σ-algebra

product

σ

-algebra

Given

measurable spaces

(

E

,

ℱ

)

and

(

F

,

𝒢

)

, the

product

σ

-algebra

ℱ

×

𝒢

is defined to be the

σ

-algebra on the

Cartesian product

E

×

F

generated by

sets of the form

A

×

B

for

A

∈

ℱ

and

B

∈

𝒢

.

ℱ

×

𝒢

=

σ

(

A

×

B

:

A

∈

ℱ

,

B

∈

𝒢

)

.

More generally, the product

σ

-algebra can be defined for an arbitrary number of measurable spaces

(

E

i

,

ℱ

i

)

, where

i

runs over an

index set

I

. The product

∏

i

ℱ

i

is the

σ

-algebra on the

generalized cartesian product

∏

i

E

i

generated by sets of the form

∏

i

A

i

where

A

i

∈

ℱ

i

for all

i

, and

A

i

=

E

i

for all but finitely many

i

.
If

π

j

:

∏

i

E

i

→

E

j

are the projection maps, then this is the smallest

σ

-algebra with respect to which each

π

j

is measurable (

http://planetmath.org/MeasurableFunctions

).

Title

product

σ

-algebra

Canonical name

Productsigmaalgebra

Date of creation

2013-03-22 18:47:21

Last modified on

2013-03-22 18:47:21

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

6

Author

gel (22282)

Entry type

Definition

Classification

msc 28A60

Synonym

product sigma-algebra

Related topic

ProductMeasure

Related topic

InfiniteProductMeasure

Generated on Fri Feb 9 20:09:33 2018 by

LaTeXML
σ-algebra

σ

-algebra

Let

X

be a set. A

σ

-algebra is a

collection

M

of subsets of

X

such that

•

X

∈

M

•

If

A

∈

M

then

X

-

A

∈

M

.

•

If

A

1

,

A

2

,

A

3

,

…

is a

countable

subcollection of

M

, that is,

A

j

∈

M

for

j

=

1

,

2

,

3

,

…

(the subcollection can be finite) then the union of all of them is also in

M

:

⋃

j

=

1

∞

A

i

∈

M

.

Title

σ

-algebra

Canonical name

sigmaalgebra

Date of creation

2013-03-22 14:00:04

Last modified on

2013-03-22 14:00:04

Owner

drini (3)

Last modified by

drini (3)

Numerical id

7

Author

drini (3)

Entry type

Definition

Classification

msc 28A60

Generated on Fri Feb 9 20:09:35 2018 by

LaTeXML
σ-algebra

σ

-algebra

Introduction

When defining a measure for a set

E

we usually cannot hope to make every subset of

E

measurable.
Instead we must usually restrict our attention
to a specific

collection

of subsets of

E

,
requiring that this collection be

closed under

operations

that we would expect to preserve measurability.
A

σ

-algebra is such a collection.

Definition

Given a set

E

, a

σ

-algebra

in

E

is a collection

ℱ

of subsets of

E

such that:

•

∅

∈

ℱ

.

•

Any union of countably many

elements of

ℱ

is an element of

ℱ

.

•

The

complement

of any element of

ℱ

in

E

is an element of

ℱ

.

Notes

It follows from the definition that any

σ

-algebra

ℱ

in

E

also

satisfies

the properties:

•

E

∈

ℱ

.

•

Any

intersection

of countably many elements of

ℱ

is an element of

ℱ

.

Note that a

σ

-algebra is a

field of sets

that is closed under

countable

unions and countable intersections
(rather than just finite unions and finite intersections).

Given any collection

C

of subsets of

E

,
the

σ

-algebra

σ

⁢

(

C

)

generated by

C

is defined to be the smallest

σ

-algebra in

E

such that

C

⊆

σ

⁢

(

C

)

.
This is well-defined,
as the intersection of any non-empty collection of

σ

-algebras in

E

is also a

σ

-algebra in

E

.

Examples

For any set

E

,
the

power set

𝒫

⁢

(

E

)

is a

σ

-algebra in

E

,
as is the set

{

∅

,

E

}

.

A more interesting example is the
Borel

σ

-algebra (

http://planetmath.org/BorelSigmaAlgebra

) in

ℝ

,
which is the

σ

-algebra generated by the

open subsets

of

ℝ

,
or, equivalently,
the

σ

-algebra generated by the

compact subsets

of

ℝ

.

Title

σ

-algebra

Canonical name

sigmaalgebra

Date of creation

2013-03-22 12:00:28

Last modified on

2013-03-22 12:00:28

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

16

Author

yark (2760)

Entry type

Definition

Classification

msc 28A60

Synonym

sigma-algebra

Synonym

sigma algebra

Synonym

σ

algebra

Synonym

Borel structure

Synonym

σ

-field

Synonym

sigma-field

Synonym

sigma field

Synonym

σ

field

Related topic

Algebra2

Related topic

BorelSigmaAlgebra

Related topic

MathcalFMeasurableFunction

Related topic

RingOfSets

Defines

generated by

Generated on Fri Feb 9 20:09:36 2018 by

LaTeXML
centre of mass of half-disc

centre of mass of half-disc

Let

E

be the upper half-disc of the disc

x

2

+

y

2

≦

R

in

ℝ

2

with a surface-density 1. By the symmetry, its

centre of mass

lies on its medium radius, and therefore we only have to calculate the ordinate

Y

of the centre of mass. For doing that, one can use the

double integral

Y

=

1

ν

⁢

(

E

)

⁢

∬

E

y

⁢

𝑑

x

⁢

𝑑

y

,

where

ν

⁢

(

E

)

=

π

⁢

R

2

2

is the area of the half-disc. The region of integration is defined by

E

=

{

(

x

,

y

)

∈

ℝ

2

⋮

-

R

≦

x

≦

R

,

0

≦

y

≦

R

2

-

x

2

}

.

Accordingly we may write

Y

=

2

π

⁢

R

2

⁢

∫

-

R

R

𝑑

x

⁢

∫

0

R

2

-

x

2

y

⁢

𝑑

y

=

2

π

⁢

R

2

⁢

∫

-

R

R

R

2

-

x

2

2

⁢

𝑑

x

=

2

π

⁢

R

2

⁢

/

x

=

-

R

R

⁡

(

R

2

⁢

x

2

-

x

3

6

)

=

4

⁢

R

3

⁢

π

.

Thus the centre of mass is the point

(

0

,

4

⁢

R

3

⁢

π

)

.

Title

centre of mass of half-disc

Canonical name

CentreOfMassOfHalfdisc

Date of creation

2013-03-22 17:20:57

Last modified on

2013-03-22 17:20:57

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Example

Classification

msc 28A75

Classification

msc 26B15

Synonym

center of mass of half-disc

Synonym

centroid of half-disc

Related topic

SubstitutionNotation

Related topic

CentreOfMassOfPolygon

Related topic

CenterOfGravityOfCircularSector

Related topic

AreaOfSphericalZone

Generated on Fri Feb 9 20:09:42 2018 by

LaTeXML
computation of moment of spherical shell

computation of moment of spherical shell

In using the formula for area integration over a sphere derived in the

http://planetmath.org/node/6668

last example, we need to keep in mind that to every point in

x

⁢

y

plane, there correspond two points on the sphere, which are obtained by taking the two signs of the square root. The importance of this fact in obtaining a correct answer is illustrated by our next example, the calculation of the

moment of inertia

of a

spherical shell

.

The moment of a spherical shell is given by the

integral

I

=

∫

S

x

2

⁢

d

2

⁢

A

.

While we could compute this by first converting to

spherical coordinates

and then using the result of

http://planetmath.org/node/6664

example 1, we can avoid the trouble of changing

coordinates

by treating the sphere as a graph. Using the result of the previous example, our integral becomes

∫

S

x

2

⁢

d

2

⁢

A

=

2

⁢

∫

x

2

+

y

2

<

r

2

r

⁢

x

2

r

2

-

x

2

-

y

2

⁢

𝑑

x

⁢

𝑑

y

,

where the factor of 2 takes into account the observation of the preceding paragraph that two points of the sphere correspond to each point of the

x

⁢

y

plane. Computing this integral, we find

2

⁢

∫

-

r

+

r

∫

-

r

2

-

y

2

+

r

2

-

y

2

r

⁢

x

2

r

2

-

x

2

-

y

2

⁢

𝑑

x

⁢

𝑑

y

=

2

⁢

r

⁢

∫

-

r

+

r

(

-

1

2

⁢

x

⁢

r

2

-

x

2

-

y

2

+

1

2

⁢

(

r

2

-

y

2

)

⁢

arcsin

⁡

x

r

2

-

y

2

)

|

-

r

2

-

y

2

+

r

2

-

y

2

⁢

d

⁢

y

=

2

⁢

r

⁢

∫

-

r

+

r

π

2

⁢

(

r

2

-

y

2

)

⁢

𝑑

y

=

4

3

⁢

π

⁢

r

4

Quick links:

•

http://planetmath.org/node/6660

main entry

•

http://planetmath.org/node/6668

previous example

Title

computation of moment of spherical shell

Canonical name

ComputationOfMomentOfSphericalShell

Date of creation

2013-03-22 14:58:11

Last modified on

2013-03-22 14:58:11

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

13

Author

rspuzio (6075)

Entry type

Example

Classification

msc 28A75

Generated on Fri Feb 9 20:09:44 2018 by

LaTeXML
derivation of surface area measure on sphere

derivation of surface area measure on sphere

The sphere of radius

r

can be described parametrically by

spherical coordinates

(what else ;) ) as follows:

x

=

r

⁢

sin

⁡

u

⁢

sin

⁡

v

y

=

r

⁢

sin

⁡

u

⁢

cos

⁡

v

z

=

r

⁢

cos

⁡

u

Then, using

trigonometric identities

to simplify

expressions

we find that

∂

⁡

(

x

,

y

)

∂

⁡

(

u

,

v

)

=

|

r

⁢

cos

⁡

u

⁢

sin

⁡

v

r

⁢

sin

⁡

u

⁢

cos

⁡

v

r

⁢

cos

⁡

u

⁢

cos

⁡

v

-

r

⁢

sin

⁡

u

⁢

sin

⁡

v

|

=

-

r

2

⁢

cos

⁡

u

⁢

sin

⁡

u

∂

⁡

(

y

,

z

)

∂

⁡

(

u

,

v

)

=

|

r

⁢

cos

⁡

u

⁢

cos

⁡

v

-

r

⁢

sin

⁡

u

⁢

sin

⁡

v

-

r

⁢

sin

⁡

u

0

|

=

-

r

2

⁢

sin

2

⁡

u

⁢

sin

⁡

v

∂

⁡

(

z

,

x

)

∂

⁡

(

u

,

v

)

=

|

-

r

⁢

sin

⁡

u

0

r

⁢

cos

⁡

u

⁢

sin

⁡

v

r

⁢

sin

⁡

u

⁢

cos

⁡

v

|

=

r

2

⁢

sin

2

⁡

u

⁢

cos

⁡

v

and hence, using more trigonometric identities, we find that

(

∂

⁡

(

x

,

y

)

∂

⁡

(

u

,

v

)

)

2

+

(

∂

⁡

(

y

,

z

)

∂

⁡

(

u

,

v

)

)

2

+

(

∂

⁡

(

z

,

x

)

∂

⁡

(

u

,

v

)

)

2

=

r

4

⁢

cos

2

⁡

u

⁢

sin

2

⁡

u

+

r

4

⁢

sin

4

⁡

u

⁢

sin

2

⁡

v

+

r

4

⁢

sin

4

⁡

u

⁢

cos

2

⁡

v

=

r

2

⁢

sin

⁡

u

.

This means that, on a sphere

d

2

⁢

A

=

r

2

⁢

sin

⁡

u

⁢

d

⁢

u

⁢

d

⁢

v

.

Note that in the case of a

unit sphere

, (

r

=

1

) this agrees with the formula presented in the second paragraph of subsection 2 of the main entry.

To return to the main entry

http://planetmath.org/node/6660

click here

Title

derivation of surface area measure on sphere

Canonical

name

DerivationOfSurfaceAreaMeasureOnSphere

Date of creation

2013-03-22 14:57:55

Last modified on

2013-03-22 14:57:55

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

6

Author

rspuzio (6075)

Entry type

Derivation

Classification

msc 28A75

Generated on Fri Feb 9 20:10:08 2018 by

LaTeXML
example of integration over sphere with respect to surface area

example of integration over sphere with respect to surface area

As an example of how to use the formula derived in

http://planetmath.org/node/6664

example 1, let us consider the following example:

∫

S

(

sin

⁡

u

⁢

cos

⁡

v

-

sin

⁡

v

)

2

⁢

d

2

⁢

A

=

∫

0

2

⁢

π

∫

0

π

(

sin

⁡

u

⁢

cos

⁡

v

-

sin

⁡

v

)

2

⁢

sin

⁡

u

⁢

d

⁢

u

⁢

d

⁢

v

=

∫

0

2

⁢

π

∫

0

π

(

sin

3

⁡

u

⁢

cos

2

⁡

v

-

2

⁢

sin

2

⁡

u

⁢

sin

⁡

v

⁢

cos

⁡

v

+

sin

⁡

u

⁢

sin

2

⁡

v

)

⁢

𝑑

u

⁢

𝑑

v

=

∫

0

2

⁢

π

(

2

⁢

cos

2

⁡

v

-

2

3

⁢

cos

2

⁡

v

-

π

⁢

sin

⁡

v

⁢

cos

⁡

v

+

2

⁢

sin

2

⁡

v

)

⁢

𝑑

v

=

2

⁢

π

-

2

⁢

π

3

-

0

+

2

⁢

π

=

10

⁢

π

3

To return to the main entry

http://planetmath.org/node/6660

click here.

Title

example of integration over sphere with respect to

surface area

Canonical name

ExampleOfIntegrationOverSphereWithRespectToSurfaceArea

Date of creation

2013-03-22 14:57:58

Last modified on

2013-03-22 14:57:58

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

5

Author

rspuzio (6075)

Entry type

Example

Classification

msc 28A75

Generated on Fri Feb 9 20:10:10 2018 by

LaTeXML
example of integration with respect to surface area of a paraboloid

example of integration with respect to surface area of a paraboloid

In this example we examine the

paraboloid

given by the equation

z

=

x

2

+

3

⁢

y

2

. Putting

g

⁢

(

x

,

y

)

=

x

2

+

3

⁢

y

2

, we have

1

+

(

∂

⁡

g

∂

⁡

x

)

2

+

(

∂

⁡

g

∂

⁡

y

)

2

=

1

+

(

2

⁢

x

)

2

+

(

6

⁢

y

)

2

=

1

+

4

⁢

x

2

+

36

⁢

y

2

and hence

∫

S

f

⁢

(

x

,

y

)

⁢

d

2

⁢

A

=

∫

f

⁢

(

x

,

y

)

⁢

1

+

4

⁢

x

2

+

36

⁢

y

2

⁢

𝑑

x

⁢

𝑑

y

.

•

http://planetmath.org/node/6660

main entry

•

http://planetmath.org/node/6669

previous example

•

http://planetmath.org/node/6673

next example

Title

example of integration with respect to

surface area

of a paraboloid

Canonical name

ExampleOfIntegrationWithRespectToSurfaceAreaOfAParaboloid

Date of creation

2013-03-22 14:58:20

Last modified on

2013-03-22 14:58:20

Owner

yark (2760)

Last modified by

yark (2760)

Numerical id

15

Author

yark (2760)

Entry type

Example

Classification

msc 28A75

Generated on Fri Feb 9 20:10:12 2018 by

LaTeXML
example of integration with respect to surface area on a helicoid

example of integration with respect to surface area on a helicoid

In this example, we shall consider itegration with respect to

surface area

on the

helicoid

.

The helicoid may be parameterized as follows:

x

=

u

⁢

sin

⁡

v

y

=

u

⁢

cos

⁡

v

z

=

c

⁢

v

(The constant

c

may be thought of as the “pitch of the screw”.) Computing

derivatives

and appying

trigonometric identities

, we obtain

∂

⁡

(

x

,

y

)

∂

⁡

(

u

,

v

)

=

|

sin

⁡

v

u

⁢

cos

⁡

v

cos

⁡

v

-

u

⁢

sin

⁡

v

|

=

-

u

∂

⁡

(

y

,

z

)

∂

⁡

(

u

,

v

)

=

|

cos

⁡

v

-

u

⁢

sin

⁡

v

0

c

|

=

c

⁢

cos

⁡

v

∂

⁡

(

z

,

x

)

∂

⁡

(

u

,

v

)

=

|

0

c

sin

⁡

v

u

⁢

cos

⁡

v

|

=

-

c

⁢

sin

⁡

v

.

From this we have

(

∂

⁡

(

x

,

y

)

∂

⁡

(

u

,

v

)

)

2

+

(

∂

⁡

(

y

,

z

)

∂

⁡

(

u

,

v

)

)

2

+

(

∂

⁡

(

z

,

x

)

∂

⁡

(

u

,

v

)

)

2

=

u

2

+

c

2

⁢

cos

2

⁡

v

+

c

2

⁢

sin

2

⁡

v

=

u

2

+

c

2

so we can compute area integrals over helicoids as follows

∫

S

f

⁢

(

u

,

v

)

⁢

d

2

⁢

A

=

∫

f

⁢

(

u

,

v

)

⁢

c

2

+

u

2

⁢

𝑑

u

⁢

𝑑

v

To return to the main entry

http://planetmath.org/node/6660

click here

Title

example of integration with respect to surface area on a helicoid

Canonical name

ExampleOfIntegrationWithRespectToSurfaceAreaOnAHelicoid

Date of creation

2013-03-22 14:58:01

Last modified on

2013-03-22 14:58:01

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

7

Author

rspuzio (6075)

Entry type

Example

Classification

msc 28A75

Generated on Fri Feb 9 20:10:14 2018 by

LaTeXML
integration with respect to surface area on a helicoid

integration with respect to surface area on a helicoid

To illustrate the result derived in

http://planetmath.org/node/6666

example 3, let us compute the area of a portion of

helicoid

of height

h

and radius

r

. (This calculation will tell us how much material is needed to make an .) The integral we need to compute in this case is

A

=

∫

d

2

⁢

A

=

∫

0

h

/

c

∫

0

r

c

2

+

u

2

⁢

𝑑

u

⁢

𝑑

v

=

∫

0

h

/

c

(

1

2

⁢

r

⁢

c

2

+

r

2

+

c

2

2

⁢

log

⁡

{

r

c

+

1

+

(

r

c

)

2

}

)

⁢

𝑑

v

=

r

⁢

h

2

⁢

1

+

(

r

c

)

2

+

c

⁢

h

2

⁢

log

⁡

{

r

c

+

1

+

(

r

c

)

2

}

As a second illustration, let us compute the second moment of a helicoid about the axis of

rotation

. In mechanics, this would be called the

moment of inertia

of the helicoid and determines how much energy is needed to make the screw rotate. This is determined as follows:

∫

(

x

2

+

y

2

)

⁢

d

2

⁢

A

=

∫

0

h

/

c

∫

0

r

u

2

⁢

c

2

+

u

2

⁢

𝑑

u

⁢

𝑑

v

=

∫

0

h

/

c

(

r

⁢

(

2

⁢

r

2

+

c

2

)

8

⁢

c

2

+

r

2

-

c

4

8

⁢

log

⁡

{

r

c

+

1

+

(

r

c

)

2

}

)

⁢

𝑑

v

=

r

⁢

h

⁢

(

2

⁢

r

2

+

c

2

)

8

⁢

1

+

(

r

c

)

2

-

h

⁢

c

3

8

⁢

log

⁡

{

r

c

+

1

+

(

r

c

)

2

}

http://planetmath.org/node/6660

main entry

http://planetmath.org/node/6666

previous example

http://planetmath.org/node/6668

next example

Title

integration with respect to

surface area

on a helicoid

Canonical name

IntegrationWithRespectToSurfaceAreaOnAHelicoid

Date of creation

2013-03-22 14:58:04

Last modified on

2013-03-22 14:58:04

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

14

Author

rspuzio (6075)

Entry type

Example

Classification

msc 28A75

Generated on Fri Feb 9 20:10:20 2018 by

LaTeXML
invariance of formula for surface integration with respect to area under change of variables

invariance of formula for surface integration with respect to area under change of variables

First, we can use the

chain rule

for

Jacobians

to see how one of the terms in parentheses transforms:

∂

⁡

(

x

,

y

)

∂

⁡

(

u

,

v

)

=

∂

⁡

(

x

,

y

)

∂

⁡

(

u

′

,

v

′

)

⁢

∂

⁡

(

u

′

,

v

′

)

∂

⁡

(

u

,

v

)

A similar story holds for the other two factors. Combining them, we conclude that

(

∂

⁡

(

x

,

y

)

∂

⁡

(

u

,

v

)

)

2

+

(

∂

⁡

(

y

,

z

)

∂

⁡

(

u

,

v

)

)

2

+

(

∂

⁡

(

z

,

x

)

∂

⁡

(

u

,

v

)

)

2

=

(

∂

⁡

(

x

,

y

)

∂

⁡

(

u

′

,

v

′

)

⁢

∂

⁡

(

u

′

,

v

′

)

∂

⁡

(

u

,

v

)

)

2

+

(

∂

⁡

(

y

,

z

)

∂

⁡

(

u

′

,

v

′

)

⁢

∂

⁡

(

u

′

,

v

′

)

∂

⁡

(

u

,

v

)

)

2

+

(

∂

⁡

(

z

,

x

)

∂

⁡

(

u

′

,

v

′

)

⁢

∂

⁡

(

u

′

,

v

′

)

∂

⁡

(

u

,

v

)

)

2

=

∂

⁡

(

u

′

,

v

′

)

∂

⁡

(

u

,

v

)

⁢

(

∂

⁡

(

x

,

y

)

∂

⁡

(

u

′

,

v

′

)

)

2

+

(

∂

⁡

(

y

,

z

)

∂

⁡

(

u

′

,

v

′

)

)

2

+

(

∂

⁡

(

z

,

x

)

∂

⁡

(

u

′

,

v

′

)

)

2

Since the factor in parentheses in front of the square root is the Jacobi determinant, we can apply the rule change of variables in multidimensional

integrals

to conclude that

∫

f

⁢

(

u

,

v

)

⁢

(

∂

⁡

(

x

,

y

)

∂

⁡

(

u

,

v

)

)

2

+

(

∂

⁡

(

y

,

z

)

∂

⁡

(

u

,

v

)

)

2

+

(

∂

⁡

(

z

,

x

)

∂

⁡

(

u

,

v

)

)

2

⁢

𝑑

u

⁢

𝑑

v

=

∫

f

⁢

(

u

′

,

v

′

)

⁢

(

∂

⁡

(

x

,

y

)

∂

⁡

(

u

′

,

v

′

)

)

2

+

(

∂

⁡

(

y

,

z

)

∂

⁡

(

u

′

,

v

′

)

)

2

+

(

∂

⁡

(

z

,

x

)

∂

⁡

(

u

′

,

v

′

)

)

2

⁢

𝑑

u

′

⁢

𝑑

v

′

,

which shows that our formula gives the same answer for

∫

S

f

⁢

(

u

,

v

)

⁢

d

2

⁢

A

, no matter how we choose to parameterize

S

.

Title

invariance of formula for surface integration with respect to area under change of variables

Canonical name

InvarianceOfFormulaForSurfaceIntegrationWithRespectToAreaUnderChangeOfVariables

Date of creation

2013-03-22 15:07:32

Last modified on

2013-03-22 15:07:32

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

5

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 28A75

Generated on Fri Feb 9 20:10:23 2018 by

LaTeXML
Lebesgue density theorem

Lebesgue density theorem

Let

μ

be the

Lebesgue measure

on

ℝ

n

, and for a

measurable set

A

⊂

ℝ

n

define the density of

A

in

ϵ

-

neighborhood

of

x

∈

ℝ

n

by

d

ϵ

⁢

(

x

)

=

μ

⁢

(

A

∩

B

ϵ

⁢

(

x

)

)

μ

⁢

(

B

ϵ

⁢

(

x

)

)

where

B

ϵ

⁢

(

x

)

denotes the ball of radius

ϵ

centered at

x

.

The

Lebesgue density theorem

asserts that for almost every point of

A

the density

d

⁢

(

x

)

=

lim

ϵ

→

0

⁡

d

ϵ

⁢

(

x

)

exists and is equal to

1

.

In other words, for every measurable set

A

the density of

A

is

0

or

1

almost everywhere. However, it is a curious fact that
if

μ

⁢

(

A

)

>

0

and

μ

⁢

(

ℝ

n

∖

A

)

>

0

, then there are
always points of

ℝ

n

where the density is neither

0

nor

1

[

1

, Lemma 4]

.

References

1

Hallard T. Croft.

Three lattice-point problems of Steinhaus.

Quart. J. Math. Oxford (2)

, 33:71–83, 1982.

http://www.emis.de/cgi-bin/zmen/ZMATH/en/quick.html?type=html&an=0499.10035

Zbl
0499.10035.

Title

Lebesgue density theorem

Canonical name

LebesgueDensityTheorem

Date of creation

2013-03-22 13:21:02

Last modified on

2013-03-22 13:21:02

Owner

bbukh (348)

Last modified by

bbukh (348)

Numerical id

7

Author

bbukh (348)

Entry type

Theorem

Classification

msc 28A75

Generated on Fri Feb 9 20:10:25 2018 by

LaTeXML
area formula

area formula

Let

ℋ

m

denote the

Hausdorff measure

. Let

m

≤

n

and consider a

Lipschitz function

f

:

ℝ

m

→

ℝ

n

.
If

A

⊂

ℝ

m

is a Lebesgue measurable set, the equality

∫

A

J

f

⁢

(

x

)

⁢

𝑑

x

=

∫

ℝ

n

ℋ

0

⁢

(

f

-

1

⁢

(

{

y

}

)

∩

A

)

⁢

𝑑

ℋ

m

⁢

y

holds, where

J

f

⁢

(

x

)

=

det

⁡

(

D

⁢

f

⁢

(

x

)

⋅

D

⁢

f

⁢

(

x

)

*

)

is the Jacobian determinant of

f

in the point

x

and represent the

m

-volume of the image of the unit cube under the linear map

D

⁢

f

⁢

(

x

)

.

If

u

∈

L

1

⁢

(

ℝ

m

)

then one has

∫

ℝ

m

u

⁢

(

x

)

⁢

J

f

⁢

(

x

)

⁢

𝑑

x

=

∫

ℝ

n

∑

x

∈

f

-

1

⁢

(

{

y

}

)

u

⁢

(

x

)

⁢

d

⁢

ℋ

m

⁢

y

.

Notice that this

formula

is a

generalization

of the change of variables in integrals on

ℝ

n

.

Title

area formula

Canonical name

AreaFormula

Date of creation

2013-03-22 14:27:36

Last modified on

2013-03-22 14:27:36

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

12

Author

paolini (1187)

Entry type

Theorem

Classification

msc 28A78

Related topic

ChangeOfVariablesInIntegralOnMathbbRn

Generated on Fri Feb 9 20:10:30 2018 by

LaTeXML
Barnsley fern

Barnsley fern

The

Barnsley fern

F

is the only non-empty

compact subset

of

ℝ

2

satisfying
the

relation

F

=

⋃

i

=

1

4

T

i

⁢

(

F

)

where

T

i

:

ℝ

2

→

ℝ

2

are the following linear mappings:

T

1

⁢

(

x

,

y

)

=

(

0.85

⁢

x

+

0.04

⁢

y

,

-

0.04

⁢

x

+

0.85

⁢

y

+

1.6

)

,

T

2

⁢

(

x

,

y

)

=

(

0.2

⁢

x

-

0.26

⁢

y

,

0.23

⁢

x

+

0.22

⁢

y

+

1.6

)

,

T

3

⁢

(

x

,

y

)

=

(

-

0.15

⁢

x

+

0.28

⁢

y

,

0.26

⁢

x

+

0.24

⁢

y

+

0.44

)

,

T

4

⁢

(

x

,

y

)

=

(

0

,

0.16

⁢

y

)

.

Title

Barnsley fern

Canonical name

BarnsleyFern

Date of creation

2013-03-22 16:05:33

Last modified on

2013-03-22 16:05:33

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

7

Author

paolini (1187)

Entry type

Example

Classification

msc 28A80

Generated on Fri Feb 9 20:10:34 2018 by

LaTeXML
closure and interior of Cantor set

closure and interior of Cantor set

The

Cantor set

is closed and its interior is empty.

To prove the first assertion, note that each of the sets

C

0

,

C

1

,

C

2

,

…

, being the union of a finite number of

closed intervals

is closed. Since the Cantor set is the

intersection

of all these sets
and intersections of

closed sets

are closed, it follows that the
Cantor set is closed.

To prove the second assertion, it suffices to show that given any

open
interval

I

, no matter how small, at least one point of that interval
will not belong to the Cantor set. To accomplish this, the ternary

characterization

of the Cantor set is useful. Because

rational
numbers

whose denominators are powers of 3 are dense, there exists a
rational number

n

/

3

m

contained

in

I

. Expressed in base 3, this
rational number has a finite expansion. If this expansion contains
the digit “1”, then our number does not belong to Cantor set, and we
are done. If not, since

I

is open, there must exist a number

k

>

m

such that

n

/

3

m

+

1

/

3

k

∈

I

. Now, the last digit of the ternary
expansion of this number is “1” by construction, so we also find a
number not belonging to the interval in this case.

Title

closure and interior of Cantor set

Canonical name

ClosureAndInteriorOfCantorSet

Date of creation

2013-03-22 15:43:21

Last modified on

2013-03-22 15:43:21

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

4

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 28A80

Generated on Fri Feb 9 20:10:40 2018 by

LaTeXML