Difference between revisions of "Measure"
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Latest revision as of 14:39, 16 August 2016
(Positive) Measure | |
[ilmath]\mu:\mathcal{R}\rightarrow\bar{\mathbb{R} }_{\ge0} [/ilmath] For a [ilmath]\sigma[/ilmath]-ring, [ilmath]\mathcal{R} [/ilmath] | |
Properties | |
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[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] |
Contents
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:
- An extended real valued countably additive set function from a [ilmath]\sigma[/ilmath]-ring, [ilmath]\mathcal{R} [/ilmath]; [ilmath]\mu:\mathcal{R}\rightarrow\bar{\mathbb{R} } [/ilmath].
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 |
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Finite[1] | if [ilmath]\mu(A)<\infty [/ilmath] |
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[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]
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Of a measure
We may say a measure, [ilmath]\mu[/ilmath] is:
Term | Meaning | Example |
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Finite[1] | If every set in the [ilmath]\sigma[/ilmath]-ring the measure is defined on is of finite measure
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[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
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Complete | if [ilmath]\forall A\in\mathcal{R}\forall B\in\mathcal{P}(A)[(\mu(A)=0)\implies(B\in\mathcal{R})][/ilmath]
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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 |
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Totally finite[1] | if the measure of [ilmath]X[/ilmath] is finite
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Totally [ilmath]\sigma[/ilmath]-finite[1] | if [ilmath]X[/ilmath] is of [ilmath]\sigma[/ilmath]-finite measure
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Immediate properties
The message provided is:
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
- Subadditive: [ilmath]\mu_0(A\cup B)\le\mu_0(A)+\mu_0(B)[/ilmath]
TODO: Again - be bothered
Related theorems
Examples
Trivial measures
Here [ilmath]\mathcal{R} [/ilmath] is a [ilmath]\sigma[/ilmath]-ring[Note 4]
- [math]\mu:\mathcal{R}\rightarrow\{0,+\infty\}[/math] by [math]\mu(A)=\left\{\begin{array}{lr}
0 & \text{if }A=\emptyset \\
+\infty & \text{otherwise}
\end{array}\right.[/math]
- 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
- 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:
- [math]\mu:\mathcal{R}\rightarrow\{0\}[/math] by [math]\mu:A\mapsto 0[/math] is the trivial measure.
The message provided is:
See also
Notes
- ↑ Recall [ilmath]\bar{\mathbb{R} }_{\ge0} [/ilmath] is [ilmath]\mathbb{R}_{\ge0}\cup\{+\infty\} [/ilmath]
- ↑ Remember a sigma-algebra is just a sigma-ring containing the entire space.
- ↑ Sometimes stated as monotone (it is monotone in Measures, Integrals and Martingales in fact!)
- ↑ Remember every [ilmath]\sigma[/ilmath]-algebra is a [ilmath]\sigma[/ilmath]-ring, so [ilmath]\mathcal{R} [/ilmath] could just as well be a [ilmath]\sigma[/ilmath]-algebra
- ↑ Unless [ilmath]\mathcal{R} [/ilmath] was a trivial [ilmath]\sigma[/ilmath]-algebra consisting of the empty set and another set.
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.
- ↑ 1.0 1.1 1.2 1.3 1.4 1.5 1.6 Measure Theory - Paul R. Halmos
- ↑ Measures, Integrals and Martingales - René L. Schilling
- ↑ Measure Theory - Volume 1 - V. I. Bogachev
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