Measure

Stub grade: A

This page is a stub

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:

Requires further expansion

$$\newcommand{\bigudot}{ \mathchoice{\mathop{\bigcup\mkern-15mu\cdot\mkern8mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}} }$$ $$\newcommand{\udot}{\cup\mkern-12.5mu\cdot\mkern6.25mu\!}$$ $$\require{AMScd}\newcommand{\d}[1][]{\mathrm{d}^{#1} }$$

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

Definition

A (positive) measure, $\mu$ is a set function from a $\sigma$-ring, $\mathcal{R}$, to the positive extended real values, $\bar{\mathbb{R} }_{\ge 0}$:

• $\mu:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge0}$

Such that:

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

Entirely in words a (positive) measure, $\mu$ is:

• An extended real valued countably additive set function from a $\sigma$-ring, $\mathcal{R}$; $\mu:\mathcal{R}\rightarrow\bar{\mathbb{R} }$.

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

Terminology

For a set

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

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

Of a measure

We may say a measure, $\mu$ is:

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

Of a measure on a $\sigma$-algebra

If $\mu:\mathcal{A}\rightarrow\bar{\mathbb{R} }_{\ge0}$ for a $\sigma$-algebra $\mathcal{A}$ then we can define:

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

Immediate properties

Grade: C

This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.

Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
The message provided is:

Trivial

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

[Expand]

Claim: $\mu(\emptyset)=0$

---

PUT PROOF HERE

Properties

---

TODO: Countable subadditivity and so forth

---

In common with a pre-measure

[Expand]

• Finitely additive: if $A\cap B=\emptyset$ then $\mu_0(A\udot B)=\mu_0(A)+\mu_0(B)$

---

Follows immediately from definition (property 2)

[Expand]

• Monotonic: if $A\subseteq B$ then $\mu_0(A)\le\mu_0(B)$

---

---

TODO: Be bothered to write out

---

[Expand]

• If $A\subseteq B$ and $\mu_0(A)<\infty$ then $\mu_0(B-A)=\mu_0(B)-\mu(A)$

---

---

TODO: Be bothered, note the significance of the finite-ness of $A$ - see Extended real value

---

[Expand]

• Strongly additive: $\mu_0(A\cup B)=\mu_0(A)+\mu_0(B)-\mu_0(A\cap B)$

---

---

TODO: Be bothered

---

[Expand]

• Subadditive: $\mu_0(A\cup B)\le\mu_0(A)+\mu_0(B)$

---

---

TODO: Again - be bothered

---

Related theorems

• A function is a measure iff it measures the empty set as 0, disjoint sets add, and it is continuous from below (with equiv. conditions)

Examples

• Dirac measure
• Counting measure
• Discrete probability measure
• Lebesgue measure

Trivial measures

Here $\mathcal{R}$ is a $\sigma$-ring

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 $+\infty$ it would not be a measure, as take any non-empty $A,B\in\mathcal{R}$ with $A\cap B=\emptyset$, for a measure we would have:
     • $\mu(A\cup B)=\mu(A)+\mu(B)$, which will yield $v=2v\implies v=0$ contradicting that $\mu$ maps non-empty sets to finite non-zero values
2. $$\mu:\mathcal{R}\rightarrow\{0\}$$ by $$\mu:A\mapsto 0$$ is the trivial measure.

(Unknown grade)

This page requires references, it is on a to-do list for being expanded with them.

Please note that this does not mean the content is unreliable, it just means that the author of the page doesn't have a book to hand, or remember the book to find it, which would have been a suitable reference.
The message provided is:

That this is the trivial measure

See also

• Pre-measure
• Outer-measure
• Constructing a measure from a pre-measure
• Measurable space
• Measure space

References

Note: Inline with the Measure theory terminology doctrine the references do not define a measure exactly as such, only an object that fits the place we have named measure. This sounds like a huge discrepancy but as is detailed on that page, it isn't.