Measure

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Definition

[math]\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}} }[/math][math]\newcommand{\udot}{\cup\mkern-12.5mu\cdot\mkern6.25mu\!}[/math][math]\require{AMScd}\newcommand{\d}[1][]{\mathrm{d}^{#1} }[/math]A (positive)[Note 1] measure[1] on a measurable space [ilmath](X,\mathcal{A})[/ilmath] (where recall [ilmath]X[/ilmath] is a set and [ilmath]\mathcal{A} [/ilmath] is a [ilmath]\sigma[/ilmath]-algebra on that set) is a mapping:

  • [math]\mu:\mathcal{A}\rightarrow[0,\infty][/math]

That satisfies:

  1. [ilmath]\mu(\emptyset)=0[/ilmath]
  2. For any finite sequence of pairwise disjoint sets [ilmath](A_i)_{i=1}^n\subseteq\mathcal{A}[/ilmath] we have [ilmath]\mu\left(\udot_{i=1}^nA_i\right)=\sum^n_{i=1}\mu(A_i)[/ilmath]
  3. For any countably infinite sequence of pairwise disjoint sets [ilmath](A_n)_{n=1}^\infty\subseteq\mathcal{A}[/ilmath] we have [ilmath]\mu\left(\udot_{n=1}^\infty A_n\right)=\sum_{n=1}^\infty\mu(A_n)[/ilmath]

Terminology

TODO: Find references

Of sets

Term Definition Comment
Finite A set [ilmath]A\in\mathcal{A} [/ilmath] is finite if [ilmath]\mu(A)<\infty[/ilmath] - we say "[ilmath]A[/ilmath] has finite measure"
[ilmath]\sigma[/ilmath]-finite A set [ilmath]A\in\mathcal{A} [/ilmath] is [ilmath]\sigma[/ilmath]-finite if [math]\exists(A_n)_{n=1}^\infty:[A\subseteq\cup^\infty_{n=1}A_n\wedge(\forall A_n,\ \mu(A_n)<\infty)][/math]

Of measures

Term Definition Comment
Complete measure [ilmath]\forall A\in\mathcal{A} [/ilmath] we have [math][\mu(A)=0\wedge B\subset A]\implies B\in \mathcal{A}[/math]
Finite measure [ilmath]\mu[/ilmath] is a finite measure if every set [ilmath]\in\mathcal{A} [/ilmath] is finite.
[ilmath]\sigma[/ilmath]-finite measure [ilmath]\mu[/ilmath] is [ilmath]\sigma[/ilmath]-finite if every set [ilmath]\in\mathcal{A} [/ilmath] is [ilmath]\sigma[/ilmath]-finite

Contrast with pre-measure

Note: the family [math]A_n[/math] must be pairwise disjoint

Property Measure Pre-measure
[math]\mu:\mathcal{A}\rightarrow[0,\infty][/math] [math]\mu_0:R\rightarrow[0,\infty][/math]
[math]\mu(\emptyset)=0[/math] [math]\mu_0(\emptyset)=0[/math]
Finitely additive [math]\mu\left(\bigudot^n_{i=1}A_i\right)=\sum^n_{i=1}\mu(A_i)[/math] [math]\mu_0\left(\bigudot^n_{i=1}A_i\right)=\sum^n_{i=1}\mu_0(A_i)[/math]
Countably additive [math]\mu\left(\bigudot^\infty_{n=1}A_n\right)=\sum^\infty_{n=1}\mu(A_n)[/math] If [math]\bigudot^\infty_{n=1}A_n\in R[/math] then [math]\mu_0\left(\bigudot^\infty_{n=1}A_n\right)=\sum^\infty_{n=1}\mu_0(A_n)[/math]

Properties

Here [ilmath](X,\mathcal{A},\mu)[/ilmath] is a measure space, and [ilmath]A,B\in\mathcal{A} [/ilmath]

  • 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 2] 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

  • Subadditive: [ilmath]\mu_0(A\cup B)\le\mu_0(A)+\mu_0(B)[/ilmath]


TODO: Again - be bothered

Related theorems

See also

Notes

  1. ↑ What else is there? Measures, Integrals and Martingales mentions this
  2. ↑ Sometimes stated as monotone (it is monotone in Measures, Integrals and Martingales in fact!)

References

  1. ↑ Measures, Integrals and Martingales - Rene L. Schilling

Old page

Not to be confused with Pre-measure


Definition

A [ilmath]\sigma[/ilmath]-ring [ilmath]\mathcal{A} [/ilmath] and a countably additive, extended real valued. non-negative set function [math]\mu:\mathcal{A}\rightarrow[0,\infty][/math] is a measure. That is:

  • [ilmath]\mu(\emptyset)=0[/ilmath]
  • [math]\mu\left(\bigudot^\infty_{n=1}A_n\right)=\sum^\infty_{n=1}\mu(A_n)[/math]
  • [math]\mu(S)\ge 0\ \forall S\in\mathcal{A}[/math]

Contrast with pre-measure

Note: the family [math]A_n[/math] must be pairwise disjoint

Property Measure Pre-measure
[math]\mu:\mathcal{A}\rightarrow[0,\infty][/math] [math]\mu_0:R\rightarrow[0,\infty][/math]
[math]\mu(\emptyset)=0[/math] [math]\mu_0(\emptyset)=0[/math]
Finitely additive [math]\mu\left(\bigudot^n_{i=1}A_i\right)=\sum^n_{i=1}\mu(A_i)[/math] [math]\mu_0\left(\bigudot^n_{i=1}A_i\right)=\sum^n_{i=1}\mu_0(A_i)[/math]
Countably additive [math]\mu\left(\bigudot^\infty_{n=1}A_n\right)=\sum^\infty_{n=1}\mu(A_n)[/math] If [math]\bigudot^\infty_{n=1}A_n\in R[/math] then [math]\mu_0\left(\bigudot^\infty_{n=1}A_n\right)=\sum^\infty_{n=1}\mu_0(A_n)[/math]

Terminology

These terms apply to pre-measures to, rather [ilmath]\mathcal{A} [/ilmath] you would use the ring the pre-measure is defined on.

Complete measure

A measure is complete if for [ilmath]A\in\mathcal{A} [/ilmath] we have [math][\mu(A)=0\wedge B\subset A]\implies B\in \mathcal{A}[/math]

Finite

A set [ilmath]A\in\mathcal{A} [/ilmath] is finite if [ilmath]\mu(A)<\infty[/ilmath] - we say "[ilmath]A[/ilmath] has finite measure"

Finite measure

[ilmath]\mu[/ilmath] is a finite measure if every set [ilmath]\in\mathcal{A} [/ilmath] is finite.

Sigma-finite

A set [ilmath]A\in\mathcal{A} [/ilmath] is [ilmath]\sigma[/ilmath]-finite if [math]\exists(A_n)_{n=1}^\infty:[A\subseteq\cup^\infty_{n=1}A_n\wedge(\forall A_n,\ \mu(A_n)<\infty)][/math]

Sigma-finite measure

[ilmath]\mu[/ilmath] is [ilmath]\sigma[/ilmath]-finite if every set [ilmath]\in\mathcal{A} [/ilmath] is [ilmath]\sigma[/ilmath]-finite

Total

If [ilmath]\mathcal{A} [/ilmath] is a [ilmath]\sigma[/ilmath]-algebra rather than a ring (that is [ilmath]X\in\mathcal{A} [/ilmath] where [ilmath]X[/ilmath] is the space) then we use

Totally finite measure

If [ilmath]X[/ilmath] is finite

Totally sigma-finite measure

If [ilmath]X[/ilmath] is [ilmath]\sigma[/ilmath]-finite

Examples

Trivial measures

Given the Measurable space [ilmath](X,\mathcal{A})[/ilmath] we can define:

[math]\mu:\mathcal{A}\rightarrow\{0,+\infty\}[/math] by [math]\mu(A)=\left\{\begin{array}{lr} 0 & \text{if }A=\emptyset \\ +\infty & \text{otherwise} \end{array}\right.[/math]

Another trivial measure is:

[math]v:\mathcal{A}\rightarrow\{0\}[/math] by [math]v(A)=0[/math] for all [math]A\in\mathcal{A}[/math]

See also

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