# Measure theory

## Table of Contents

## Notation

- and are used to denote the
**indicator**or**characteristic function**

## Definition

### Motivation

The motivation behind defining such a thing is related to the Banach-Tarski paradox, which says that it is possible to decompose the 3-dimensional solid unit ball into finitely many pieces and, using only rotations and translations, reassemble the pieces into two solid balls each with the same volume as the original. The pieces in the decomposition, constructed using the axiom of choice, are **non-measurable** sets.

Informally, the axiom of choice, says that given a collecions of bins, each containing at least one object, it's possible to make a selection of exactly one object from each bin.

### Measure space

If is a set with the sigma-algebra and the measure , then we have a **measure space** .

### Product measure

Given two *measurable spaces* and measures on them, one can obtain a **product measurable space** and a **product measure** on that space.

A **product measure** is defined to be a measure on the measurable space , where we've let be the algebra on the Cartesian product . This sigma-algebra is called the *tensor-product sigma-algebra* on the product space.

A **product measure** is defined to be a measure on the measurable space satisfying the property

### and

Let be a sequence of extended real numbers.

The **limit inferior** is defined

The **limit supremum** is defined

### Premeasure

Given a space , and a collection of sets is an **algebra of sets** on if

- If , then
- If and are in , then

Thus, a **algebra of sets** allow only *finite* unions, unlike σ-algebras where we allow *countable* unions.

Given a space and an algebra , a **premeasure** is a function such that

For every

*finite or countable*collection of*disjoint*sets with , if then

*Observe that the last property says that IF this "possibly large" union is in the algebra, THEN that sum exists.*

A **premeasure space** is a triple where is a space, is an algebra, and a premeasure .

### Complete measure

A **complete measure** (or, more precisely, a **complete measure space** ) is a measure space in which every subset of every null set is measurable (having measure zero).

More formally, is **complete** if and only if

If is a premeasure space, then there is a complete measure space such that

- we have

If is σ-finite, then is the only measure on that is equal to on .

## Sobolev space

### Notation

- is an open subset of
- denotes a infinitively differentiable function with compact support
is a multi-index of order , i.e.

### Definition

Vector space of functions equipped with a norm that is a *combination of norms of the function itself and its derivatoves to a given order*.

Intuitively, a **Sobolev space** is a space of functions with sufficiently many derivatives for some application domain, e.g. PDEs, and equipped with a norm that measures *both* size and regularity of a function.

The **Sobolev space** spaces combine the concepts of weak differentiability and Lebesgue norms (i.e. spaces).

For a proper definition for different cases of dimension of the space , have a look at Wikipedia.

### Motivation

Integration by parst yields that for every where , and for all infinitively differentiable functions with compact support :

Observe that LHS only makes sense if we assume to be *locally integrable*. If there exists a locally integrable function , such that

we call the **weak -th partial derivative of **. If this exists, then it is *uniquely defined almost everywhere*, and thus it is uniquely determined as an element of a *Lebesgue space* (i.e. function space).

On the other hand, if , then the classical and the weak derivative coincide!

Thus, if , we may denote it by .

#### Example

is not continuous at zero, and not differentiable at −1, 0, or 1. Yet the function

satisfies the definition of being the weak derivative of , which then qualifies as being in the Sobolev space (for any allowed ).

## Lebesgue measure

### Notation

- denotes the collection of all measurable sets

### Stuff

Given a subset , with the length of a closed interval given by , the **Lebesgue outer measure** is defined as

Lebesgue outer-measure has the following properties:

- Idea: Cover by .
(Monotinicy) if , then

Idea: a cover of is a cover of .

(Countable subadditivity) For every set and every sequence of sets if then

Idea: construct a cover of each , such that :

- Every point in is in
*one*of the

- Every point in is in

**Q:** Is it possible for every to find a cover such that ?
**A:** No. Consider . Given , consider .
This is a cover of so .
If is a cover by open intervals of , then there is at least one such that is a nonempty open interval, so it has a strictly positive lenght, and

If , then

**Idea:** , so .
For reverse, cover by intervals giving a sum within .
Then cover and by intervals of length .
Put the 2 new sets at at the start of the sequence, to get a cover of , and sum of the lengths is *at most* . Hence,

If is an *open* interval, then .

**Idea:** *lower* bound from .
Only bounded nonempty intervals are interesting.
Take the closure to get a compact set. Given a countable cover by open intervals, reduce to a finite subcover.
Then arrange a finite collection of intervals in something like increasing order, possibly dropping unnecessary sets.
Call these new intervals and let be the number of such intervals, and such that

i.e. left-most interval cover the starting-point, and right-most interval cover the end-point. Then

Taking the infimum,

The **Lebesgue measure** is then defined on the Lebesgue sigma-algebra, which is the *collection of all the sets which satisfy the condition that, for every*

For any set in the Lebesgue sigma-algrebra, its **Lebesgue measure** is given by its Lebesgue outer measure .

IMPORTANT!!! This is not necessarily related to the Lebesgue integral! It CAN be be, but the integral is more general than JUST over some Lebesgue measure.

### Intuition

- First part of definition states that the subset is reduced to its outer measure by coverage by sets of closed intervals
- Each set of intervals covers in the sense that when the intervals are combined together by union, they contain
- Total length of any covering interval set can easily overestimate the measure of , because is a subset of the union of the intervals, and so the intervals include points which are not in

**Lebesgue outer measure** emerges as the *greatest lower bound* (infimum) of the lengths from among all possible such sets. Intuitively, it is the total length of those interval sets which fit most tightly and do not overlap.

**In my own words:** Lebesgue outer measure is smallest sum of the lengths of subintervals s.t. the union of these subintervals completely "covers" (i.e. are equivalent to) .

If you take an a real interval , then the Lebesge outer measure is simply .

### Properties

#### Notation

For and , we let

#### Stuff

The collection of Lebesgue measurable sets is a sigma-algebra.

Easy to see is in this collection:

Closed under complements is clear: let be

*Lebesgue measurable*, thenhence this is also true for , and so is

*Lebesgue measurable*.- Closed under countable unions:
**Finite case:**. Consider both*Lebesgue measurable*and some set . Since is L. measurable:Since is L. measurable:

which allows us to rewrite the above equation for :

Observe that

By subadditivity:

Hence,

Then this follows for

*all*finite cases by*induction*.**Countable disjoint case:**Let , and . Further, let .Hence is L. measurable. Thus,

Since the are disjoint and :

Let and note that . Thus, by indiction

Thus,

Taking :

Thus, is L. measurable if the are

*disjoint*and L. measurable!**Countable (not-necessarily-disjoint) case:**If are*not*disjoint, let and let , which gives a sequence of*disjoint*sets, hence the above proof applies.

Every open interval is Lebesgue measurable, and the Borel sigma-algebra is a subset of the sigma-algebra of Lebesgue measurable sets.

Want to prove measurability of intervals of the form .

Idea:

- split any set into the left and right part
- split any cover in the same way
- extend covers by to make them
*open*

is a measure space, and for al intervals , the measure is the *length*.

#### Cantor set

Define

For , with being identity, and

Let and . Then the **Cantor set** is defined

The Cantor set has a Lebesgue measure zero.

We make the following observations:

- Scaled and shifted closed sets are closed
- is a
*finite*union of closed intervals and so is in the Borel sigma-algebra - σ-algebras are closed under countable intersections, hence Cantor set is
*in*the Borel σ-algebra - Finally, Borel σ-algebra is a
*subset*of Lebesgue measurable sets, hence the**Cantor set is Lebesuge measurable**!

Since Lebesgue measure satisfy for any Lebesgue measurable set with finite measure and any with . Since Lebesgue measure is *subadditive*, we have for any

Since , by induction, it follows that

Taking the infimum of over , we have that the Cantor set has measure zero:

##### Cardinality of the Cantor set

Let .

The **terniary expansion** is a sequence with such that

The Cantor set is *uncountable*.

We observe that if the *first elements of the expansion* for are in , then . But *importantly*, observe that some numbers have more than *one* terniary expansion, i.e.

in the terniary expansion. One can show that a number if and only if has a terniary expansion with no 1 digits. Hence, the Cantor set is *uncountable*!

One can see that if and only if terniary expansion with no 1 digits, since such an would land in the "gaps" created by the construction of the Cantor set.

##### Uncountable Lebesgue measurable set

There exists *uncountable* Lebesgue measurable sets.

#### Menger sponge

- Generalization of Cantor set to

#### Vitali sets

Let if and only if .

- There are uncountable many equivalence classes, with each equivalence class being countable (as a set).
- By axiom of choice, we can pick one element from each equivalence class.
- Can assume each representative picked is in , and this set we denote

Suppose, for the sake of contradiction, that is *measurable*.

Observe if , then there is a and s.t. , i.e.

Then, by countable additivity

where we've used

Hence, we have our contradiction and so this set, the **Vitali set**, is *not* measurable!

There exists a subset of that is *not* measurable wrt. Lebesgue measure.

### Lebesgue Integral

The **Lebesgue integral** of a function over a measure space is written

which means we're taking the integral wrt. the measure .

#### Special case: non-negative real-valued function

Suppose that is a non-negative real-valued function.

Using the "partitioning of range of " philosophy, the integral of should be the sum over of the elementary area contained in the thin horizontal strip between and , which is just

Letting

The **Lebesgue integral** of is then defined by

where the integral on the right is an ordinary improper Riemann integral. For the set of *measurable* functions, this defines the **Lebesgue integral**.

### Radon measure

- Hard to find a good notion of measure on a topological space that is compatible with the topology in some sense
- One way is to define a measure on the Borel set of the topological space

Let be a measure on the sigma-algebra of Borel sets of a Hausdorff topological space .

- is called
**inner regular**or**tight**if, for any Borel set , is the supremum of over all compact subsets of of - is called
**outer regular**if, for any Borel set , is the infimum of over all open sets*containing* - is called
**locally finite**if every point of has a neighborhood for which is*finite*(if is locally finite, then it follows that is finite on compact sets)

The measure is called a **Radon measure** if it is *inner regular* and *locally finite*.

Suppose and are two measures on a measures on a measurable space and is absolutely continuous wrt. .

Then there exists a non-negative, measurable function on such that

The function is called the **density** or **Radon-Nikodym derivative** of wrt. .

If a Radon-Nikodym derivative of wrt. exists, then denotes the *equivalence class* of measurable functions that are Radon-Nikodym derivatives of wrt. .

is often used to denote , i.e. is just *in* the equivalence class of measurable functions such that this is the case.

This comes from the fact that we have

Suppose and are Radon-Nikodym derivatives of wrt. iff .

The *δ measure* cannot have a Radon-Nikodym derivative since integrating gives us zero for all measurable functions.

### Continuity of measure

Suppose and are two sigma-finite measures on a measure space .

Then we say that is **absolutely continuous** wrt. if

We say that and are **equivalent** if each measure is **absolutely continuous** wrt. to the other.

### Density

Suppose and are two sigma-finite measures on a measure space and that is absolutely continuous wrt. . Then there exists a non-negative, measurable function on such that

### Measure-preserving transformation

is a **measure-preserving transformation** is a transformation on the measure-space if

## Measure

A **measure** on a set is a systematic way of defining a number to each *subset* of that set, intuitively interpreted as *size*.

In this sense, a **measure** is a generalization of the concepts of length, area, volume, etc.

Formally, let be a of subsets of .

Suppose is a function. Then is a measure if

Whenever are pairwise

*disjoint*subsets of in , then

### Properties

Let be a measure space, and such that .

Then .

Let

Then , and by finite additivity property of a measure:

since by definition of a measure.

If are subsets of , then

We know for a sequence of *disjoint* sets we have

So we just let

Then,

Thus,

Concluding our proof!

Let be an *increasing* sequence of measurable sets.

Then

Let be sets from some .

If , then

### Examples of measures

Let

- be a space

The **δ-measure (at )** is

## Sigma-algebra

### Definition

Let be some set, and let be its power set. Then the subset is a called a σ-algebra on if it satisfies the following three properties:

- is closed under complement: if
- is closed under countable unions: if

These properties also imply the following:

- is closed under countable intersections: if

### Generated σ-algebras

Given a space and a collection of subsets , the **σ-algebra generated by **, denoted , is defined to be the *intersection* of all σ-algebras on that contain , i.e.

where

Let be a measurable space and a function from some space to .

The **σ-algebra generated by ** is

*Observe that though this is similar to σ-algebra generated by MEASURABLE function, the definition differs in a sense that the preimage does not have to be measurable. In particular, the σ-algebra generated by a measurable function can be defined as above, where is measurable by definition of being a measurable function, hence corresponding exactly to the other definition.*

Let and be measure spaces and a measurable function.

The **σ-algebra generated by ** is

Let be a space.

If is a collection of σ-algebras, then is *also* a σ-algebra.

### σ-finite

A measure or premeasure space is **finite** if .

A measure on a measure space is said to be sigma-finite if can be written as a countable union of measurable sets of finite measure.

#### Example: counting measure on uncountable set is *not* σ-finite

Let be a space.

The **counting measure** is defined to be such that

On any uncountable set, the counting measure is *not* σ-finite, since if a set has *finite* counting measure it has *countably* many elements, and a countable union of finite sets is countable.

### Properties

Let be a of subsets of a set . Then

If , then

- If then

### Borel sigma-algebra

Any set in a topological space that can be formed from the *open sets* through the operations of:

- countable union
- countable intersection
- complement

is called a **Borel set**.

Thus, for some topological space , the collection of all Borel sets on forms a σ-algebra, called the **Borel algebra** or **Borel σ-algebra** .

More compactly, the **Borel σ-algebra** on is

where is the σ-algebra generated by the standard topology on .

**Borel sets** are important in measure theory, since any measure defined on the open sets of a space, or on the closed sets of a space, must also be defined on all Borel sets of that space.

Any measure defined on the Borel sets is called a **Borel measure**.

### Lebesgue sigma-algebra

Basically the same as the Borel sigma-algebra but the **Lebesgue sigma-algebra** forms a complete measure.

#### Note to self

Suppose we have a Lebesgue mesaure on the real line, with measure space .

Suppose that is *non-measurable* subset of the real line, such as the Vitali set. Then the measure of is **not** defined, but

and this larger set ( ) does have measure zero, i.e. it's not *complete* !

#### Motivation

Suppose we have constructed Lebesgue measure on the real line: denote this measure space by . We now wish to construct some two-dimensional Lebesgue measure on the plane as a product measure.

Naïvely, we could take the sigma-algebra on to be , the smallest sigma-algebra containing all measureable "rectangles" for .

While this approach does define a measure space, it has a flaw: since every singleton set has one-dimensional Lebesgue measure zero,

for any subset of .

What follows is the important part!

However, suppose that is non-measureable subset of the real line, such as the Vitali set. Then the measure of is not defined (since we just supposed that is non-measurable), but

and this larger set ( ) does have measure zero, i.e. it's not *complete* !

#### Construction

Given a (possible incomplete) measure space , there is an extension of this measure space that is *complete* .

The smallest such extension (i.e. the smallest sigma-algebra ) is called the **completion** of the measure space.

It can be constructed as follows:

- Let be the set of all measure zero subsets of (intuitively, those elements of that are not already in are the ones preventing completeness from holding true)
- Let be the sigma-algebra generated by and (i.e. the smallest sigma-algreba that contains every element of
**and**of ) - has an extension to (which is unique if is sigma-finite), called the
*outer measure*of , given by the infimum

Then is a **complete measure space**, and is the completion of .

What we're saying here is:

- For the "multi-dimensional" case we need to take into account the zero-elements in the resulting sigma-algebra due the product between the 1D zero-element and some element NOT in our original sigma-algebra
- The above point means that we do NOT necessarily get completeness, despite the sigma-algebras defined on the sets individually prior to taking the Cartesian product being complete
- To "fix" this, we construct a
*outer measure*on the sigma-algebra where we have included all those zero-elements which are "missed" by the naïve approach,

## Measurable functions

Let and be measurable spaces.

A function is a **measurable function** if

where denotes the preimage of the for the measurable set .

Let .

We define the **indicator function** of to be the function given by

Let . Then is measurable if and only if .

Let be a measure space or a probability space.

Let be a sequence of measurable functions.

- For each , the function is
*measurable* - The function is
*measurable* - Thus, if converge
*pointwise*, is*measurable*.

Let be a measurable space, and let .

The following statements are **equivalent**:

- is measurable.
- we have .
- we have .
- we have .
- we have .

A function is measurable if

We also observe that by Proposition proposition:equivalent-statements-to-being-a-measurable-function, it's sufficient to prove

so that's what we set out to do.

For and , consider the following *equivalent* statements:

Thus,

so

Recall that for each , the sequence is an *increasing* sequence in . Therefore, similarily, the following are *equivalent*:

Thus,

Hence,

concluding our proof!

Basically says the same as Prop. proposition:limits-of-measurable-functions-are-measurable, but a bit more "concrete".

Let be a of subsets of a set , and let with be a sequence of measurable functions.

Furthermore, let

Then is a measurable function.

### Simple functions

Let be a of subsets of a set .

A function is called a **simple function** if

- it is measurable
- only takes a finite number of values

Let be a of subsets of a set .

Let be a *nonnegative* measurable function.

Then there exists a sequence of simple functions such that

- for all
Converges to :

Define a function as follows. Let

and let

Then the function

obeys the required properties!

### Almost everywhere and almost surely

Let be a measure or probability space.

Let be a sequence of measurable functions

- For each the function is
*measurable* - The function is
*measurable* - Thus, if the converge pointwise, then is
*measurable*

Let be a measure space. Let be a *condition* in oe variable.

**holds almost everywhere (a.e.)** if

Let be a probability space and be a condition in one variable, then **holds almost surely (a.e.)** if

also denoted

Let be a complete measure space.

- If is measurable and if a.e. then is measurable.
- Being equal a.e. is an
*equivalence relation*on measurable functions.

### Convergence theorems for nonnegative functions

### Problems

Clearly if with s.t. , then

hence

Therefore it's sufficient to prove that if , then there exists a non-degenerate open interval s.t. . (first I said *contained in* , but that is a unecessarily strong statement; if *contained* then what we want would hold, but what we want does not imply *containment*).

As we know, for every there exists such that and

Which implies

which implies

Letting , this implies that there exists an open cover s.t.

and

(this fact that this is true can be seen by considering for all and see that this would imply not being a cover of , and if , then since there exists a "smaller" cover).

Thus,

Hence, letting be s.t.

we have

as wanted!

, we have for *almost every* if and only if for *almost every* , for all .

This is equivalent to saying

if and only if

i.e. is a set of measure zero.

Then clearly

by the assumption.

Follows by the same logic:

This concludes our proof.

## Integration

### Notation

We let

where

### Stuff

Let

where are a set of positive values.

Then the **integral** of over wrt. is given by

Let be a sequence of *nonnegative* measurable functions on . Assume that

- for each
- for each .

Then, we write *pointwise*.

Then is measurable, and

Let . By Proposition proposition:limit-of-measurable-functions-is-measurable, is measurable.

Since each satisfies , we know .

- If , then since and for all we have , and .

Let and .

**Step 1:** Approximate by a simple function.

Let be a simple function such that and . Such an exists by definition of Lebesgue integral. Thus, there are such that , and disjoint mesurable sets such that

If any , it doesn't contribute to the integral, so we may ignore it and assume that there are no such sets.

**Step 2**: Find sets of large measure where the convergence is controlled.

Note that for all we have

That is, for each and ,

For and , let

And since it's easier to work with *disjoint* sets,

Observe that,

Then,

We don't have a "rate of convergence" on , but on we know that we are close, and so we can "control" the convergence.

**Step 3**: Approximate from below.

For each if , then let be such that

and otherwise, let be such that

Let , and let .

For each , and we have

Thus, , and ,

If there is a such that , then

Otherwise (if the integral is finite), then

For every and , there is an such that

For every such that

Therefore

Thus,

as wanted.

Let be any nonnegative measurable functions on .

Then

Let and observe are pointwise increasing

### Properties of integrals

Let be a measure space.

If is a nonnegative measurable function, then there is an increasing sequence of simple functions such that

Given as above and for , let

and

Or a bit more explicit (and maybe a bit clearer),

For each , is a cover of . On each we have , hence on entirety of .

Consider . If , then for which in turn implies

Hence .

Finally, if , then and for all take on values

Hence, for all cases.

Furthermore, for any and , there is the nesting property

so on we have .

(This can be seen by observing that what we're really doing here is dividing the values takes on into a grid, and observing that if we're in then we're either in or ).

For , then

so again and is pointwise increasing.

Let be a measure space.

Let

- be nonnegative, measurable functions
s.t.

is defined

- be a sequence of nonnegative measurable functions.

Then

Finite sum

Scalar multiplication

Infinte sums

Let and be increasing sequence of simple functions converging to , , respectively.

Note is aslo increasing to .

By monotone convergence theorem

The argument is similar for products.

Finally, is an increasing sequence of nonnegative measurable functions, since sums of measurable functions is a measurable function.

Thus, by monotone convergence and the result for *finite* sums

### Integrals on sets

Let be a measure or probability space.

If is a sequence of disjoint measurable sets then

Let be a measure or probability space.

If is a simple function and is a measurable set, then is a *simple function*.

Let be a measure or probability space.

Let be a nonnegative measurable function and .

The **integral of on ** is defined to be

Let be a measure or probability space.

Let be a nonnegative measurable function.

If and are

*disjoint*measurable sets, thenIf are

*disjoint*measurable sets, then

Let be a measure or probability space.

If is a nonnegative measurable function, then defined by :

is a measure on .

If , then defined by :

The **(real) Gaussian measure** on is defined as:

where denotes the Lebesgue measure.

A **Gaussian probability measure** can also be defined for an arbitrary Banach space as follows:

Then, we say is a **Gaussian probability measure on** if and only if is a Borel measure, i.e.

such that is a *real* Gaussian probability measure on for *every* linear functional , i.e. .

Here we have used the notation , defined

where denotes the Borel measures on .

### Integrals of general functions

Let be a measure or probability space.

If is a measurable function, then the positive and negative parts are defined by

Note: and are nonnegative.

Let be a measure or probability space.

If is a measurable function, then and are *measurable* functions.

Let be a measure or probability space.

- A
*nonnegative*function is defined to be**integrable**if it is measurable and . - A function is defined to be
**integrable**if it is measurable and is*integrable*.

For an *integrable* function , the **integral of ** is defined to be

On a set , the integral is defined to be

Note that , but in the actual definition of the integral, we use .

Let be a measure or probability space.

If and are real-valued integrable functions and , then

(Scalar multiplication)

(Additive)

Let be a measure or probability space.

Let and be measurable functions s.t.

If is integrable then is integrable.

#### Examples

##### Consider with Lebesgue measure. Is integrable?

And

and

therefore

Thus, is integrable.

### Lebesge dominated convergence theorem

Let be a measure or probability space.

Let be a nonnegative integrable function and let be a sequence of (not necessarily nonnegative!) measurable functions.

Asssume and all are *real-valued*.

If and such that

and the *pointwise* limit

exists.

Then

That is, if there exists a "dominating function" , then we can "move" the limit into the integral.

Since and such that , we find that and are *nonnegative*.

Consider

From Fatou's lemma, we have

Therefore

Consider , then

(this looks very much like Fatou's lemma, but it ain't; does not necessarily have to be nonnegative as in Fatou's lemma)

Consider

Therefore,

Which implies

Since , we then have exists and is equal to .

#### Examples of failure of dominated convergence

##### Where dominated convergence does not work

On with Lebesgue measure, consider

such that instead of as "usual" with .

Both of these are nonnegative sequences that converge to *pointwise*.

Notice there is no integrable *dominating* function for either of these sequences:

- would require a dominating function to have infinite integral, therefore no dominating integrable function exists.
- on the right, and so a dominating function would have to be above on some interval which would lead to infinite integral.

Thus, Lebesgue dominated convergence does not apply

##### Noncummtative limits: simple case

##### Noncommutative limits: another one

Consider with Lebesgue measure and

Consider and $ b > 1$ and

Note that , so is *not* integrable.

Consider

#### Commutative limits

Consider

We know that is *integrable* and for all and ,

By multiple applications of LDCT

Showing that in this case the limits do in fact commute.

#### Riemann integrable functions are measurable

All Riemann integrable functions are *measurable*.

For any Riemann integrable function, the Riemann integral and the Lebesgue integral are *equal*.

### Almost everywhere and Lp spaces

If is a nonnegative, measurable function, and , then .

For , let

Observe the are *disjoint* and

Suppose that . This implies that on a set of *positive* measure, i.e.

but this implies that

Thus,

which is a *contradiction*, hence .

Let and be integrable.

is the set of all equivalence classes of integrable functions wrt. the equivalence relation given by a.e. equality, i.e.

If is an integrable function, the norm is

If and , the integral and norm are defined to be

If , then , and

is a real vector space with addition and scalar multiplication given pointwise almost everywhere.

Functions taking on on a set of *zero measure* are fine!

These functions are still the almost everywhere equal to some *integrable* function (even those these infinite-valued functions are integrable), hence these are in .

Let be a Cauchy sequence. Since the are integrable, we may assume we choose valued representatives.

For , let be such that for ,

and .

Thus,

and

Thus, is *finite* almost everywhere. Thus, this series is infinite on a set of measure zero, so we may assume the representatives are zero there and the sum is finite at each .

Thus, converges everywhere.

Let

(observe that the last part is just rewriting the ).

By monotone convergence theorem

Observe that pointwise

## Applications to Probability

### Notation

- is a
*probability space* - Random variable is a measurable function
- denotes the Borel sigma-algebra on
- denotes the probability distribution measure for
- be a sequence of random events
- be a sequence of
*finitely*many random events

### Probability and cumulative distributions

An **elementary event** is an element of .

A **random event** is an element of

A **random variable** is a measurable function from to .

The **probability distribution measure of **, denoted , is defined

The **cumulative distribution function of **, denoted , is defined by

The probability distribution measure is a probability measure on the Borel sets .

If is a *disjoint* sequence of sets in , then

so satisfies countable additivity and is a measure.

Finally,

so is a *probability* measure.

- is
*increasing* - and
is

*right*continuous (i.e. continuous*from the right*)

If , then

Consider the limit as . Let

so

Then,

which, since is

*increasing*impliesLet and . Let

The are

*nested*, and similarily are*nested*.Thus, given , there exists such that

Let so

### Radon-Nikodym derivatives and expectations

Let

- be a rv.
- its probability distribution measure
- its cumulative distribution function
- a Borel measureable function

The following are equivalent:

- is a Radon-Nikodym derivative for wrt. (the Lebesgue measure but
*restricted to Borel measurable sets*)

(2) and (3) are immediately equivalent:

iff (2) or (3) holds when considering only sets of the form .

This statement is also equivalent to (1).

Thus (1) is equivalent to (2) or (3) restricted to sets of the form .

However, sets of the form generate , so from the Carathéodory extension theorem this gives .

To prove more rigorously, let

for s.t. and none of these intervals *overlap*. That is all *finite* unions of left-closed, right-open, disjoint intervals.

Also let

Observe that

and that

One can show that is a premeasure space. Therefore, by the Carathéodory extension theorem, there is a measure on s.t.

Furthermore, since , is *unique*! But both the measures and satisfy these properties, thus

which is the definition of being a Radon-Nikodym derivative of wrt. Lebesgue measure restricted to the Borel σ-algebra, as wanted.

A function is a **probability density function** for if is a Radon-Nikodym derivative of the probability distribution measure , wrt. Lebesgue measure restricted to Borel sets, i.e.

#### Expectation via distributions

**Expectation** of a random variable is

If is a nonnegative function that is measurable, then

If is the characterstic function, then, if ,

so

Multiplying by constants and summing over different characteristic functions, we get the result to be true for *any* simple function.

Given a *nonnegative* function , let be an increasing sequence of *simple* functions converging pointwise to .

Note is the increasing limit of . By two applications of Monotone Convergence

*This techinque, of going from characterstic function → simple functions → general functions, is used heavily, not just in probability theory.*

### Independent events & Borel-Cantelli theorem

A collection of random events are **independent events** if for every finite collection of distinct indeices ,

A random event **occurs** at if .

The **probability that the event occurs** is .

If are independent then are also independent.

Prove that are independent.

Consider , we want to prove

RHS can be written

which is equal to LHS above, and implies that the complement is indeed independent.

The condition that infinitively many of the events occurs at is

This is equivalent to

where we have converted the and .

Furthermore, is itself a *random event*.

If then probability of infinitely many of the events occuring is 0, i.e.

If the are independent and , then probability of infinitely many of the events occuring is 1, i.e.

Suppose .

Suppose are now

*independent*and that . Fix . Then

### Chebyshev's inequality

Let be a probability space.

If is a random variable with mean and variance , then

Let

Then everywhere, so

Hence,

### Independent random variables

Let

- be a probability space.

A collection of σ-algebras , where for all , is *independent* if for every collection of events s.t for all , then is a set of independent events.

A collection of random variables is **independent** if the collection of σ-algebras they generate is independent.

A *sequence* of random variables is **independent and identically distributed (i.i.d)** if the yare independent variables and for we have

where is the cumulative distribution function for .

Let and be independent.

- We have
- If or then
If and , then

Furthermore, if and , then

Consider

- first nonnegative functions
- subcase

Since is nonnegative

Thus, so .

Now consider the subcase where and .

Let and be the σ-algebras generated by and .

Observe that and are measure spaces. Let be an increasing sequence of simple functions that are measurable wrt. and similarily simple increasing to and measurable.

As simple functions, these can be written as

Then,

Since increases to , by MCT

Dividing into positive & negative parts & summing gives .

### Strong Law of Large numbers

#### Notation

are i.i.d. random variables, and we will assume

#### Stuff

Let be a probability space and be a sequence of i.i.d. random variables with

Then the sequence of random variables converges *almost surely* to , i.e.

This is equivalent to occuring with probability 0, and this is the approach we will take.

First consider .

For and , let

and

Since are i.i.d. we have

and since variance rescales quadratically,

Using Chebyshev's inequality

Observe then that with , we have

And so by Borel-Cantelli, since this is a sequence of *independent* random variables, we have

In particular, for any , there are *almost surely* only finitely many with

**Step**: showing that we can do this for any .

Consider . Observe that by *countable subadditivity*,

Now let , which occurs *almost surely* from the above. For any , let

Since , there are only *finitely* many s.t.

as found earlier (the parenthesis are indeed different here, compared to before). Therefore

is arbitrary, so this is true for all . Hence,

This proves that there is a *subsequential* limit almost surely.

**Step**: *subsequential* limit to "sequential" limit.
Given , let be such that . Since are *nonnegative*

and therefore

and since ,

Since the first and the last expressions converge to ,t by the squeeze theorem we have

**Step**: Relaxing nonnegativity assumption on .

Suppose is not necessarily *nonnegative*. Since, by assumption, has finite expectation, is *integrable*. Therefore we know that the positive and negative parts of , denoted , are also integrable. Therefore we can compute the expectations

Similarily, we have that the variance of is *finite*, which allows us to the apply the result we found for being nonnegative to both and :

Let be the set where the mean of the positive / negative part converges. Since

(since otherwise the limit would not converge almost surely). We then have

Thus, almost surely, , and on this we have convergence, so

Concluding our proof.

## Ergodic Theory

Let be a measure-preserving transformation on a measure space with , i.e. it's a *probability space*.

Then is **ergodic** if for every we have