# Measure

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Properties (Positive) Measure [ilmath]\mu:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge0} [/ilmath]For a [ilmath]\sigma[/ilmath]-ring, [ilmath]\mathcal{R} [/ilmath] [ilmath]\forall\overbrace{(A_n)_{n=1}^\infty }^{\begin{array}{c}\text{pairwise}\\\text{disjoint}\end{array} }\subseteq\mathcal{R}[\mu\left(\bigudot_{n=1}^\infty A_n\right)=\sum^\infty_{n=1}\mu(A_n)][/ilmath]

## Definition

A (positive) measure, [ilmath]\mu[/ilmath] is a set function from a [ilmath]\sigma[/ilmath]-ring, [ilmath]\mathcal{R} [/ilmath], to the positive extended real values[Note 1], [ilmath]\bar{\mathbb{R} }_{\ge 0} [/ilmath][1][2][3]:

• [ilmath]\mu:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge0} [/ilmath]

Such that:

• [ilmath]\forall(A_n)_{n=1}^\infty\subseteq\mathcal{R}\text{ pairwise disjoint }[\mu\left(\bigudot_{n=1}^\infty A_n\right)=\sum_{n=1}^\infty\mu(A_n)][/ilmath] ([ilmath]\mu[/ilmath] is a countably additive set function)
• Recall that "pairwise disjoint" means [ilmath]\forall i,j\in\mathbb{N}[i\ne j\implies A_i\cap A_j=\emptyset][/ilmath]

Entirely in words a (positive) measure, [ilmath]\mu[/ilmath] is:

Remember that every [ilmath]\sigma[/ilmath]-algebra is a [ilmath]\sigma[/ilmath]-ring, so this definition can be applied directly (and should be in the reader's mind) to [ilmath]\sigma[/ilmath]-algebras

## Terminology

### For a set

We may say a set [ilmath]A\in\mathcal{R} [/ilmath] (for a [ilmath]\sigma[/ilmath]-ring [ilmath]\mathcal{R} [/ilmath]) is:

Term Meaning Example
Finite[1] if [ilmath]\mu(A)<\infty [/ilmath]
• [ilmath]A[/ilmath] is finite
• [ilmath]A[/ilmath] is of finite measure
[ilmath]\sigma[/ilmath]-finite[1] if [ilmath]\exists(A_n)_{n=1}^\infty\subseteq\mathcal{R}\forall i\in\mathbb{N}[A\subseteq\bigcup_{n=1}^\infty A_n\wedge \mu(A_i)<\infty][/ilmath]
• In words: if there exists a sequence of sets in [ilmath]\mathcal{R} [/ilmath] such that [ilmath]A[/ilmath] is in their union and each set has finite measure.
• [ilmath]A[/ilmath] is [ilmath]\sigma[/ilmath]-finite
• [ilmath]A[/ilmath] is of [ilmath]\sigma[/ilmath]-finite measure

### Of a measure

We may say a measure, [ilmath]\mu[/ilmath] is:

Term Meaning Example
Finite[1] If every set in the [ilmath]\sigma[/ilmath]-ring the measure is defined on is of finite measure
• Symbolically, if: [ilmath]\forall A\in\mathcal{R}[\mu(A)<\infty][/ilmath]
• [ilmath]\mu[/ilmath] is a finite measure
[ilmath]\sigma[/ilmath]-finite[1] If every set in the [ilmath]\sigma[/ilmath]-ring the measure is defined on is of [ilmath]\sigma[/ilmath]-finite measure
• Symbolically, if: [ilmath]\forall A\in\mathcal{R}\exists(A_n)_{n=1}^\infty\subseteq\mathcal{R}\forall i\in\mathbb{N}[A\subseteq\bigcup_{n=1}^\infty A_n\wedge \mu(A_i)<\infty][/ilmath]
• [ilmath]\mu[/ilmath] is a [ilmath]\sigma[/ilmath]-finite measure
Complete if [ilmath]\forall A\in\mathcal{R}\forall B\in\mathcal{P}(A)[(\mu(A)=0)\implies(B\in\mathcal{R})][/ilmath]
• In words: for every set of measure 0 in [ilmath]\mathcal{R} [/ilmath] every subset of that set is also in [ilmath]\mathcal{R} [/ilmath]
• [ilmath]\mu[/ilmath] is a complete measure

#### Of a measure on a [ilmath]\sigma[/ilmath]-algebra

If [ilmath]\mu:\mathcal{A}\rightarrow\bar{\mathbb{R} }_{\ge0} [/ilmath] for a [ilmath]\sigma[/ilmath]-algebra [ilmath]\mathcal{A} [/ilmath][Note 2] then we can define:

Term Meaning Example
Totally finite[1] if the measure of [ilmath]X[/ilmath] is finite
• Symbolically, if [ilmath]\mu(X)<\infty[/ilmath]
• [ilmath]\mu[/ilmath] is totally finite
Totally [ilmath]\sigma[/ilmath]-finite[1] if [ilmath]X[/ilmath] is of [ilmath]\sigma[/ilmath]-finite measure
• Symbolically, if: [ilmath]\exists(A_n)_{n=1}^\infty\subseteq\mathcal{R}\forall i\in\mathbb{N}[X=\bigcup_{n=1}^\infty A_n\wedge \mu(A_i)<\infty][/ilmath]
• [ilmath]\mu[/ilmath] is totally [ilmath]\sigma[/ilmath]-finite

## Immediate properties

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Trivial

This proof has been marked as an page requiring an easy proof

Claim: [ilmath]\mu(\emptyset)=0[/ilmath]

PUT PROOF HERE

## Properties

TODO: Countable subadditivity and so forth

### In common with a pre-measure

• Finitely additive: if [ilmath]A\cap B=\emptyset[/ilmath] then [ilmath]\mu_0(A\udot B)=\mu_0(A)+\mu_0(B)[/ilmath]

Follows immediately from definition (property 2)

• Monotonic: [Note 3] if [ilmath]A\subseteq B[/ilmath] then [ilmath]\mu_0(A)\le\mu_0(B)[/ilmath]

TODO: Be bothered to write out

• If [ilmath]A\subseteq B[/ilmath] and [ilmath]\mu_0(A)<\infty[/ilmath] then [ilmath]\mu_0(B-A)=\mu_0(B)-\mu(A)[/ilmath]

TODO: Be bothered, note the significance of the finite-ness of [ilmath]A[/ilmath] - see Extended real value

• Strongly additive: [ilmath]\mu_0(A\cup B)=\mu_0(A)+\mu_0(B)-\mu_0(A\cap B)[/ilmath]

TODO: Be bothered

TODO: Again - be bothered

## Examples

### Trivial measures

Here [ilmath]\mathcal{R} [/ilmath] is a [ilmath]\sigma[/ilmath]-ring[Note 4]

1. $\mu:\mathcal{R}\rightarrow\{0,+\infty\}$ by $\mu(A)=\left\{\begin{array}{lr} 0 & \text{if }A=\emptyset \\ +\infty & \text{otherwise} \end{array}\right.$
• Note that if we'd chosen a finite and non-zero value instead of [ilmath]+\infty[/ilmath] it would not be a measure[Note 5], as take any non-empty [ilmath]A,B\in\mathcal{R} [/ilmath] with [ilmath]A\cap B=\emptyset[/ilmath], for a measure we would have:
• [ilmath]\mu(A\cup B)=\mu(A)+\mu(B)[/ilmath], which will yield [ilmath]v=2v\implies v=0[/ilmath] contradicting that [ilmath]\mu[/ilmath] maps non-empty sets to finite non-zero values
2. $\mu:\mathcal{R}\rightarrow\{0\}$ by $\mu:A\mapsto 0$ is the trivial measure.
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That this is the trivial measure