aarjaneiro/planetmath huggingface dataset:
Kolmogorov zero-one law

Kolmogorov zero-one law

Kolomogorov zero-one lawFernando Sanz Gamiz

Theorem

(Kolmogorov).

Let

Ω

be a set,

F

a

sigma-algebra

of subsets of

Ω

and

P

a

probability measure

. Given the

independent

random
variables

{

X

n

,

n

∈

N

}

, defined on

(

Ω

,

F

,

P

)

, it
happens that

P

⁢

(

A

)

=

0

⁢

or

⁢

P

⁢

(

A

)

=

1

,

A

∈

ℱ

∞

,

i.e.,the

probability

of any

tail event

is 0 or 1.

Proof.

Define

ℱ

n

=

σ

⁢

(

X

1

,

X

2

,

…

,

X

n

)

. As any event in

σ

⁢

(

X

n

+

1

,

X

n

+

2

,

…

)

is independent of any event in

σ

⁢

(

X

1

,

X

2

,

…

,

X

n

)

1

1

this assertion should be proved
actually, because independence of random variables is defined for
every finite number of them and we are dealing with events involving
an

infinite

number. By two successive applications of the

Monotone
Class Theorem

, one can readily prove this is in fact correct

, any
event in the tail

σ

-

algebra

ℱ

∞

is independent of
any event in

⋃

n

=

1

∞

ℱ

n

; hence, any event in

ℱ

∞

is independent of any event in

σ

⁢

(

⋃

n

=

1

∞

ℱ

n

)

2

2

again by application
of the Monotone Class Theorem

. But

ℱ

∞

⊂

σ

⁢

(

⋃

n

=

1

∞

ℱ

n

)

3

3

because

ℱ

∞

⊂

σ

⁢

(

X

1

,

X

2

,

…

)

=

σ

⁢

(

⋃

n

=

1

∞

ℱ

n

)

,
this last equality being easily proved

, so any tail event is
independent of itself, i.e.,

P

⁢

(

A

)

=

P

⁢

(

A

∩

A

)

=

P

⁢

(

A

)

⁢

P

⁢

(

A

)

which
implies

P

⁢

(

A

)

=

0

or

P

⁢

(

A

)

=

1

.
∎

Title

Kolmogorov zero-one law

Canonical name

KolmogorovZerooneLaw

Date of creation

2013-03-22 17:07:21

Last modified on

2013-03-22 17:07:21

Owner

fernsanz (8869)

Last modified by

fernsanz (8869)

Numerical id

10

Author

fernsanz (8869)

Entry type

Definition

Classification

msc 28A05

Related topic

TailEvent

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

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limit superior of sets

limit superior of sets

Let

A

1

,

A

2

,

…

be a sequence of sets.
The

limit superior of sets

is defined by

lim sup

⁡

A

n

=

⋂

n

=

1

∞

⋃

k

=

n

∞

A

k

.

It is easy to see that

x

∈

lim sup

⁡

A

n

if and only if

x

∈

A

n

for infinitely many values of

n

.
Because of this, in probability theory the notation

[

A

n

⁢

i

.

o

.

]

is often used to refer to

lim sup

⁡

A

n

, where i.o. stands for

infinitely often

.

The limit inferior of sets is defined by

lim inf

⁡

A

n

=

⋃

n

=

1

∞

⋂

k

=

n

∞

A

k

,

and it can be shown that

x

∈

lim inf

⁡

A

n

if and only if

x

belongs to

A

n

for all but finitely many values of

n

.

Title

limit superior of sets

Canonical

name

LimitSuperiorOfSets

Date of creation

2013-03-22 13:13:22

Last modified on

2013-03-22 13:13:22

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

8

Author

Koro (127)

Entry type

Definition

Classification

msc 28A05

Classification

msc 60A99

Defines

limit inferior of sets

Defines

infinitely often

Defines

i.o.

Generated on Fri Feb 9 20:05:11 2018 by

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measurability of analytic sets

measurability of analytic sets

Analytic

subsets (

http://planetmath.org/AnalyticSet2

) of a

measurable space

(

X

,

ℱ

)

do not, in general, have to be

measurable

. See, for example, a

Lebesgue measurable but non-Borel set

(

http://planetmath.org/ALebesgueMeasurableButNonBorelSet

). However, the following result is true.

Theorem

.

All analytic subsets of a measurable space are

universally measurable

.

Therefore for a universally complete measurable space

(

X

,

ℱ

)

all

ℱ

-

analytic sets

are themselves in

ℱ

and, in particular, this applies to any

complete

σ

-

finite measure space

(

http://planetmath.org/SigmaFinite

)

(

X

,

ℱ

,

μ

)

. For example, analytic subsets of the

real numbers

ℝ

are

Lebesgue measurable

.

The proof of the

theorem

follows as a

consequence

of the

capacitability theorem

.
Suppose that

A

is an

ℱ

-analytic set. Then, for any finite

measure

μ

on

(

X

,

ℱ

)

, let

μ

*

be the

outer measure generated by

μ

. This is an

ℱ

-

capacity

and, by the capacitability theorem,

A

is

(

ℱ

,

μ

*

)

-capacitable, hence is in the

completion

of

ℱ

with respect to

μ

(see capacity

generated by

a measure (

http://planetmath.org/CapacityGeneratedByAMeasure

)). As this is true for all such finite measures,

A

is universally measurable.

Title

measurability of analytic sets

Canonical name

MeasurabilityOfAnalyticSets

Date of creation

2013-03-22 18:47:24

Last modified on

2013-03-22 18:47:24

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

6

Author

gel (22282)

Entry type

Theorem

Classification

msc 28A05

Generated on Fri Feb 9 20:05:13 2018 by

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measurable projection theorem

measurable projection theorem

The projection of a

measurable set

from the

product

X

×

Y

of two

measurable spaces

need not itself be measurable. See a

Lebesgue measurable but non-Borel set

for an example. However, the following result can be shown. The notation

ℱ

×

ℬ

refers to the product

σ

-algebra (

http://planetmath.org/ProductSigmaAlgebra

).

Theorem

.

Let

(

X

,

F

)

be a measurable space and

Y

be a

Polish space

with Borel

σ

-algebra

B

.
Then the projection (

http://planetmath.org/ProjectionMap

) of any

S

∈

F

×

B

onto

X

is

universally measurable

.

In particular, if

ℱ

is universally complete then the projection of

S

will be in

ℱ

, and this applies to all

complete

σ

-finite (

http://planetmath.org/SigmaFinite

)

measure spaces

(

X

,

ℱ

,

μ

)

. For example, the projection of any

Borel set

in

ℝ

n

onto

ℝ

is

Lebesgue measurable

.

The

theorem

is a direct

consequence

of the properties of

analytic sets

(

http://planetmath.org/AnalyticSet2

), following from the result that

projections of analytic sets are analytic

and the fact that analytic sets are universally measurable (

http://planetmath.org/MeasurabilityOfAnalyticSets

).
Note, however, that the theorem itself does not refer at all to the

concept

of analytic sets.

The

measurable projection theorem

has important applications to the theory of continuous-time stochastic processes. For example, the début theorem, which says that the first time at which a

progressively measurable

stochastic process

enters a given measurable set is a

stopping time

, follows easily.
Also, if

(

X

t

)

t

∈

ℝ

+

is a jointly measurable process defined on a measurable space

(

Ω

,

ℱ

)

, then the maximum process

X

t

*

=

sup

s

≤

t

⁡

X

s

will be universally measurable since,

{

ω

∈

Ω

:

X

t

*

>

K

}

=

π

Ω

⁢

(

{

(

s

,

ω

)

:

s

≤

t

,

X

s

>

K

}

)

is universally measurable, where

π

Ω

:

Ω

×

ℝ

+

→

Ω

is the

projection map

.
Furthermore, this also shows that the

topology

of

ucp convergence

is well defined on the space of jointly measurable processes.

Title

measurable projection theorem

Canonical name

MeasurableProjectionTheorem

Date of creation

2013-03-22 18:48:04

Last modified on

2013-03-22 18:48:04

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

8

Author

gel (22282)

Entry type

Theorem

Classification

msc 28A05

Classification

msc 60G07

Related topic

ProjectionsOfAnalyticSetsAreAnalytic

Generated on Fri Feb 9 20:05:14 2018 by

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measure zero

measure zero

If

(

X

,

M

,

μ

)

is a

measure space

, and

A

∈

M

, then

A

is said to be of

measure zero

if

μ

⁢

(

A

)

=

0

.

Title

measure zero

Canonical name

MeasureZero

Date of creation

2013-03-22 14:00:38

Last modified on

2013-03-22 14:00:38

Owner

bwebste (988)

Last modified by

bwebste (988)

Numerical id

6

Author

bwebste (988)

Entry type

Definition

Classification

msc 28A05

Generated on Fri Feb 9 20:05:18 2018 by

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measure zero in ℝ^n

measure zero in

ℝ

n

0.1

Measure Zero in

ℝ

n

In

ℝ

n

with the

Lebesgue measure

m

there is a

characterization

of the sets that have

measure zero

.

Theorem

-

A subset

X

⊆

ℝ

n

has zero Lebesgue measure if and only if for every

ϵ

>

0

there is a

sequence

of

compact rectangles

{

R

i

}

i

∈

ℕ

that cover

X

and such that

∑

i

m

⁢

(

R

i

)

<

ϵ

.

0.2

Measure Zero Avoiding Measure Theory

In some circumstances one may want to avoid the whole of Lebesgue measure and integration and still be interested in having a notion of measure zero, like for example when studying

Riemann integrals

or constructing the Lebesgue measure from an historical of view. Another interesting example where sets of measure zero arise and there is no reason to introduce

measures

or

integrals

is when studying the of sets where a

function of bounded variation

is not

differentiable

(this sets have always measure zero).

Nevertheless, the notion of measure zero is not lost in this situation. Since the Lebesgue measure of compact rectangles can be easily calculated and defined from the start (see

Jordan content

of an

n

-cell (

http://planetmath.org/JordanContentOfAnNCell

) for example), the condition stated in the previous theorem can be taken as the

definition

of measure zero.

Definition -

A set

X

in

ℝ

n

is said to have

measure zero

if for every

ϵ

>

0

there is a sequence of compact rectangles

{

R

i

}

i

∈

ℕ

that cover

X

and such that

∑

i

m

⁢

(

R

i

)

<

ϵ

, where

m

is the Jordan content (

http://planetmath.org/JordanContentOfAnNCell

).

Similarly, the notion of

almost everywhere

remains essentially the same. One just has to work with the previous definition.

Title

measure zero in

ℝ

n

Canonical

name

MeasureZeroInmathbbRn

Date of creation

2013-03-22 17:57:11

Last modified on

2013-03-22 17:57:11

Owner

asteroid (17536)

Last modified by

asteroid (17536)

Numerical id

6

Author

asteroid (17536)

Entry type

Theorem

Classification

msc 28A05

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

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monotone class

monotone class

Monotone ClassFernando Sanz GÃÂ¡miz

Definition

.

A

collection

ℳ

of subsets of

Ω

is a

monotone class

if it is

closed under

increasing and

decreasing sequences

of sets,
i.e.

1.

A

1

⊆

A

2

⊆

A

3

,

…

∈

ℳ

⇒

⋃

A

n

∈

ℳ

2.

A

1

⊇

A

2

⊇

A

3

,

…

∈

ℳ

⇒

⋂

A

n

∈

ℳ

Title

monotone class

Canonical

name

MonotoneClass

Date of creation

2013-03-22 17:07:25

Last modified on

2013-03-22 17:07:25

Owner

fernsanz (8869)

Last modified by

fernsanz (8869)

Numerical id

5

Author

fernsanz (8869)

Entry type

Definition

Classification

msc 28A05

Related topic

SigmaAlgebra

Related topic

MonotoneClassTheorem

Generated on Fri Feb 9 20:05:21 2018 by

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paved space

paved space

A

paving

on a set

X

is any

collection

𝒜

of subsets of

X

, and

(

X

,

𝒜

)

is said to be a

paved space

.
Given any two paved spaces

(

X

,

𝒜

)

and

(

Y

,

ℬ

)

, the product paving

𝒜

×

ℬ

is defined as

𝒜

×

ℬ

=

{

A

×

B

:

A

∈

𝒜

,

B

∈

ℬ

}

.

A paved space

(

K

,

𝒦

)

is said to be

compact

if every subcollection of

𝒦

satisfying the

finite intersection property

has nonempty

intersection

. Equivalently, if any

𝒦

′

⊆

𝒦

has empty intersection then there is a finite

𝒦

′′

⊆

𝒦

′

with empty intersection. Then,

𝒦

is said to be a

compact paving

, and

K

is

compactly paved

by

𝒦

.
An example of compact pavings is given by the collection of all compact subsets (

http://planetmath.org/Compact

) of a

Hausdorff topological space

.

For any paving

𝒜

, the notation

𝒜

σ

is often used to denote

countable

unions of

elements of

𝒜

,

𝒜

σ

≡

{

⋃

n

=

1

∞

A

n

:

A

n

∈

𝒜

⁢

for all

⁢

n

∈

ℕ

}

.

Similarly,

𝒜

δ

denotes the countable intersections of elements of

𝒜

,

𝒜

δ

≡

{

⋂

n

=

1

∞

A

n

:

A

n

∈

𝒜

⁢

for all

⁢

n

∈

ℕ

}

.

These

operations

can be combined in any order so that, for example,

𝒜

σ

⁢

δ

=

(

𝒜

σ

)

δ

is the collection of countable intersections of countable unions of elements of

𝒜

.

Note: In the definition of a paved space, some authors additionally require a paving

𝒦

to contain the

empty set

.

References

1

K. Bichteler,

Stochastic integration

with jumps

. Encyclopedia of Mathematics and its Applications, 89. Cambridge University Press, 2002.

2

Claude Dellacherie, Paul-André Meyer,

Probabilities

and potential

. North-Holland Mathematics Studies, 29. North-Holland Publishing Co., 1978.

3

Sheng-we He, Jia-gang Wang, Jia-an Yan,

Semimartingale

theory and stochastic calculus.

Kexue Chubanshe (Science Press), CRC Press, 1992.

4

M.M. Rao,

Measure theory and integration

. Second edition. Monographs and Textbooks in Pure and Applied Mathematics, 265. Marcel Dekker Inc., 2004.

Title

paved space

Canonical name

PavedSpace

Date of creation

2013-03-22 18:44:50

Last modified on

2013-03-22 18:44:50

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

6

Author

gel (22282)

Entry type

Definition

Classification

msc 28A05

Synonym

paving

Synonym

paved set

Related topic

F_sigmaSet

Related topic

G_deltaSet

Related topic

AnalyticSet2

Defines

paving

Defines

compact paving

Defines

compactly paved by

Defines

product paving

Generated on Fri Feb 9 20:05:26 2018 by

LaTeXML
products of compact pavings are compact

products of compact pavings are compact

Suppose that

(

K

i

,

𝒦

i

)

is a

paved space

for each

i

in

index set

I

. The

product

(

http://planetmath.org/GeneralizedCartesianProduct

)

∏

i

∈

I

K

i

is the set of all

functions

x

:

I

→

⋃

i

K

i

such that

x

i

∈

K

i

for each

i

, and the product of subsets

S

i

⊆

K

i

is

∏

i

∈

I

S

i

=

{

x

∈

∏

i

∈

I

K

i

:

x

i

∈

S

i

⁢

for each

⁢

i

∈

I

}

.

Then, the product paving is defined by

∏

i

∈

I

𝒦

i

=

{

∏

i

∈

I

S

i

:

S

i

∈

𝒦

i

⁢

for each

⁢

i

∈

I

}

.

Theorem 1

.

Let

(

K

i

,

K

i

)

be

compact

paved spaces for

i

∈

I

. Then,

∏

i

K

i

is a compact paving on

∏

i

K

i

.

Note that this result is a version of

Tychonoff

’s

theorem

applying to paved spaces and, together with the fact that all compact pavings are

closed subsets

of a

compact space

, is easily seen to be

equivalent

to Tychonoff’s theorem.

Theorem

1

is simple to prove directly. Suppose that

{

A

j

:

j

∈

J

}

is a subset of

∏

i

𝒦

i

satisfying the

finite intersection property

. Writing

A

j

=

∏

i

∈

I

S

i

⁢

j

for

S

i

⁢

j

∈

𝒦

i

gives

⋂

j

∈

J

′

A

j

=

∏

i

∈

I

(

⋂

j

∈

J

′

S

i

⁢

j

)

(1)

for any

J

′

⊆

J

. By the finite intersection property, this is nonempty whenever

J

′

is finite, so

⋂

j

∈

J

′

S

i

⁢

j

≠

∅

. Consequently,

{

S

i

⁢

j

:

j

∈

J

}

⊆

𝒦

i

satisfies

the finite intersection property and, by

compactness

of

𝒦

i

, the

intersection

⋂

j

∈

J

S

i

⁢

j

is nonempty. So equation (

1

) shows that

⋂

j

∈

J

A

j

is nonempty.

Title

products of compact pavings are compact

Canonical name

ProductsOfCompactPavingsAreCompact

Date of creation

2013-03-22 18:45:09

Last modified on

2013-03-22 18:45:09

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

5

Author

gel (22282)

Entry type

Theorem

Classification

msc 28A05

Related topic

SumsOfCompactPavingsAreCompact

Generated on Fri Feb 9 20:05:29 2018 by

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proof of compact pavings are closed subsets of a compact space

proof of compact pavings are closed subsets of a compact space

Let

(

K

,

𝒦

)

be a

compact

paved space (

http://planetmath.org/paved space

). We use the

ultrafilter

lemma (

http://planetmath.org/EveryFilterIsContainedInAnUltrafilter

) to show that there is a compact paving

𝒦

′

containing

𝒦

that is

closed under

arbitrary

intersections

and finite unions.

We first show that the paving

𝒦

1

consisting of all finite unions of elements of

𝒦

is compact.
Let

ℱ

⊆

𝒦

1

satisfy

the

finite intersection property

. It then follows that the

collection

of finite intersections of

ℱ

is a filter (

http://planetmath.org/Filter

). The ultrafilter lemma says that

ℱ

is

contained

in an ultrafilter

𝒰

.

By definition, the ultrafilter satisfies the finite intersection property. So, the

compactness

of

𝒦

implies that

ℱ

′

≡

𝒰

∩

𝒦

has nonempty intersection.
Also, every

element

S

of

ℱ

is a union of finitely many elements of

𝒦

, one of which must be in

𝒰

(see

alternative characterization of ultrafilter

(

http://planetmath.org/AlternativeCharacterizationOfUltrafilter

)). In particular,

S

contains the intersection of

ℱ

′

and,

⋂

ℱ

⊇

⋂

ℱ

′

≠

∅

.

Consequently,

𝒦

1

is compact.

Finally, we let

𝒦

′

be the set of arbitrary intersections of

𝒦

1

. This is closed under all arbitrary intersections and finite unions. Furthermore, if

ℱ

⊆

𝒦

′

satisfies the finite intersection property then so does

ℱ

′

≡

{

A

∈

𝒦

1

:

B

⊆

A

⁢

for some

⁢

B

∈

ℱ

}

.

The compactness of

𝒦

1

gives

⋂

ℱ

=

⋂

ℱ

′

≠

∅

as required.

Title

proof of compact pavings are

closed subsets

of a

compact space

Canonical name

ProofOfCompactPavingsAreClosedSubsetsOfACompactSpace

Date of creation

2013-03-22 18:45:07

Last modified on

2013-03-22 18:45:07

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

4

Author

gel (22282)

Entry type

Proof

Classification

msc 28A05

Generated on Fri Feb 9 20:05:36 2018 by

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proof of countable unions and intersections of analytic sets are analytic

proof of countable unions and intersections of analytic sets are analytic

Let

(

X

,

ℱ

)

be a

paved space

and

(

A

n

)

n

∈

ℕ

be a

sequence

of

ℱ

-

analytic sets

(

http://planetmath.org/AnalyticSet2

). We show that their union and

intersection

is also

analytic

.

From the

definition

of

ℱ

-analytic sets, there exist compact paved spaces

(

K

n

,

𝒦

n

)

and

S

n

∈

(

ℱ

×

𝒦

n

)

σ

⁢

δ

such that

A

n

=

{

x

∈

X

:

(

x

,

y

)

∈

S

n

⁢

for some

⁢

y

∈

K

n

}

.

We start by showing that

⋂

n

A

n

is analytic. Let

K

=

∏

n

K

n

and

𝒦

=

∏

n

𝒦

n

be the product paving, and

π

n

:

K

→

K

n

be the

projection map

.
Then

x

∈

⋂

n

A

n

if and only if for each

n

there is a

y

n

∈

K

n

with

(

x

,

y

n

)

∈

S

n

. Equivalently, setting

y

=

(

y

1

,

y

2

,

⋯

)

, then

(

x

,

y

)

∈

⋂

n

π

n

-

1

⁢

(

S

n

)

. However, this is in

(

ℱ

×

𝒦

n

)

σ

⁢

δ

and we can write,

⋂

n

A

n

=

π

X

⁢

(

⋂

n

π

n

-

1

⁢

(

S

n

)

)

,

where

π

X

:

X

×

K

→

X

is the projection map.
As

products of compact pavings are compact

,

(

K

,

𝒦

)

is compact and it follows from the definition that

⋂

n

A

n

is

ℱ

-analytic.

We now show that

⋃

n

A

n

is analytic. Let

K

=

∑

n

K

n

and

𝒦

=

∑

n

𝒦

n

be the

direct sum

paving, which is compact (

http://planetmath.org/SumsOfCompactPavingsAreCompact

). Also, write

S

n

=

⋂

m

=

1

∞

T

m

,

n

for

T

m

,

n

∈

(

ℱ

×

𝒦

n

)

σ

.
We identify

K

n

with a subset of

K

, so that

K

is the union of the

disjoint sets

K

n

.
Then

x

∈

⋃

n

A

n

if and only if

(

x

,

y

)

∈

S

n

for some

n

and some

y

∈

K

,

⋃

n

A

n

=

π

X

⁢

(

⋃

n

S

n

)

.

However, the fact that

K

n

1

,

K

n

2

are

disjoint

for

n

1

≠

n

2

says that

T

m

,

n

1

,

T

m

,

n

2

are disjoint and, therefore,

⋃

n

S

n

=

⋃

n

⋂

m

T

m

,

n

=

⋂

m

⋃

n

T

m

,

n

∈

(

ℱ

×

𝒦

)

σ

⁢

δ

.

So

⋃

n

A

n

is

ℱ

-analytic.

Title

proof of countable unions and intersections of analytic sets are analytic

Canonical

name

ProofOfCountableUnionsAndIntersectionsOfAnalyticSetsAreAnalytic

Date of creation

2013-03-22 18:46:19

Last modified on

2013-03-22 18:46:19

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

4

Author

gel (22282)

Entry type

Proof

Classification

msc 28A05

Generated on Fri Feb 9 20:05:38 2018 by

LaTeXML
proof of equivalent definitions of analytic sets for measurable spaces

proof of equivalent definitions of analytic sets for measurable spaces

Let

(

X

,

ℱ

)

be a

measurable space

and

A

be a subset of

X

. For any

uncountable

Polish space

Y

with Borel

σ

-algebra (

http://planetmath.org/BorelSigmaAlgebra

)

ℬ

, we show that the following are

equivalent

.

1.

A

is

ℱ

-

analytic

(

http://planetmath.org/AnalyticSet2

).

2.

A

is the projection (

http://planetmath.org/GeneralizedCartesianProduct

) of a set

S

∈

ℱ

⊗

ℬ

onto

X

.

Here,

ℱ

⊗

ℬ

denotes the

product

σ

-algebra (

http://planetmath.org/ProductSigmaAlgebra

) of

ℱ

and

ℬ

.

(

1

) implies (

2

)

:
Let

𝒢

denote the paving consisting of the

closed subsets

of

Y

. If

A

is

ℱ

-analytic then there exists a set

S

∈

(

ℱ

×

𝒢

)

σ

⁢

δ

such that

A

=

π

X

⁢

(

S

)

, where

π

X

:

X

×

Y

→

X

is the

projection map

(see

proof of equivalent definitions of analytic sets for paved spaces

).
In particular,

𝒢

⊆

ℬ

implies that

S

is

contained

in the

σ

-algebra

ℱ

⊗

ℬ

.

(

2

) implies (

1

)

:
This is an immediate

consequence

of the result that

projections of analytic sets are analytic

.

Title

proof of equivalent definitions of analytic sets for measurable spaces

Canonical name

ProofOfEquivalentDefinitionsOfAnalyticSetsForMeasurableSpaces

Date of creation

2013-03-22 18:48:41

Last modified on

2013-03-22 18:48:41

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

4

Author

gel (22282)

Entry type

Proof

Classification

msc 28A05

Generated on Fri Feb 9 20:05:40 2018 by

LaTeXML
proof of Vitali’s Theorem

proof of Vitali’s Theorem

Consider the

equivalence relation

in

[

0

,

1

)

given by

x

∼

y

⇔

x

-

y

∈

ℚ

and let

ℱ

be the family of all

equivalence classes

of

∼

.
Let

V

be a of

ℱ

i.e. put in

V

an

element

for each equivalence class of

∼

(notice that we are using the

axiom
of choice

).

Given

q

∈

ℚ

∩

[

0

,

1

)

define

V

q

=

(

(

V

+

q

)

∩

[

0

,

1

)

)

∪

(

(

V

+

q

-

1

)

∩

[

0

,

1

)

)

that is

V

q

is obtained translating

V

by a quantity

q

to the right and then cutting the piece which goes beyond the point

1

and putting it on the left, starting from

0

.

Now notice that given

x

∈

[

0

,

1

)

there exists

y

∈

V

such that

x

∼

y

(because

V

is a

section

of

∼

) and hence there exists

q

∈

ℚ

∩

[

0

,

1

)

such that

x

∈

V

q

. So

⋃

q

∈

ℚ

∩

[

0

,

1

)

V

q

=

[

0

,

1

)

.

Moreover all the

V

q

are

disjoint

. In fact if

x

∈

V

q

∩

V

p

then

x

-

q

(modulus

[

0

,

1

)

) and

x

-

p

are both in

V

which is not possible since they differ by a

rational

quantity

q

-

p

(or

q

-

p

+

1

).

Now if

V

is

Lebesgue measurable

, clearly also

V

q

are

measurable

and

μ

⁢

(

V

q

)

=

μ

⁢

(

V

)

. Moreover by the

countable additivity

of

μ

we have

μ

⁢

(

[

0

,

1

)

)

=

∑

q

∈

ℚ

∩

[

0

,

1

)

μ

⁢

(

V

q

)

=

∑

q

μ

⁢

(

V

)

.

So if

μ

⁢

(

V

)

=

0

we had

μ

⁢

(

[

0

,

1

)

)

=

0

and if

μ

⁢

(

V

)

>

0

we had

μ

⁢

(

[

0

,

1

)

)

=

+

∞

.

So the only possibility is that

V

is not Lebesgue measurable.

Title

proof of Vitali’s

Theorem

Canonical name

ProofOfVitalisTheorem

Date of creation

2013-03-22 13:45:50

Last modified on

2013-03-22 13:45:50

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

7

Author

paolini (1187)

Entry type

Proof

Classification

msc 28A05

Related topic

ProofOfPsuedoparadoxInMeasureTheory

Generated on Fri Feb 9 20:05:50 2018 by

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