Difference between revisions of "Measurable map"
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* <math>T^{-1}(A')\in\mathcal{A},\ \forall A'\in\mathcal{A}'</math> | * <math>T^{-1}(A')\in\mathcal{A},\ \forall A'\in\mathcal{A}'</math> | ||
| + | '''See also:''' | ||
| + | * (Theorem) [[Conditions for a map to be a measurable map]] | ||
| + | * (Theorem) [[A map from two sigma-algebras, A and B, is measurable if and only if for some generator of B (call it G) we have the inverse image of S is in A for every S in G|A map, {{M|f:(A,\mathcal{A})\rightarrow(F,\mathcal{F})}}, is {{M|\mathcal{A}/\mathcal{F} }} measurable ''iff'' for some generator {{M|\mathcal{F}_0}} of {{M|\mathcal{F} }} we have {{M|\forall S\in\mathcal{F}_0[f^{-1}(S)\in\mathcal{A}]}}]] | ||
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==Notation== | ==Notation== | ||
| + | I like the ''{{M|\mathcal{A}/\mathcal{B} }}-measurable'' notation because it reads nicely for composition. As [[Composition of measurable maps is measurable|a composition of measurable maps is measurable]]<ref name="PAS"/> we see something like, if: | ||
| + | * {{M|f:(A,\mathcal{A})\rightarrow(B,\mathcal{B})}} is measurable (same as saying: {{M|f:A\rightarrow B}} is {{M|\mathcal{A}/\mathcal{B} }}-measurable) and | ||
| + | * {{M|g:(B,\mathcal{B})\rightarrow(C,\mathcal{C})}} is measurable | ||
| + | then: | ||
| + | * {{M|g\circ f:(A,\mathcal{A})\rightarrow(C,\mathcal{C})}} is measurable. | ||
| + | |||
| + | In effect: | ||
| + | * {{M|\mathcal{A}/\mathcal{B} }}-measurable followed by {{M|\mathcal{B}/\mathcal{C} }} measurable {{M|1==}} {{M|\mathcal{A}/\mathcal{C} }}-measurable | ||
| + | |||
| + | ==(OLD) Notation== | ||
{{Todo|Confirm this - it could just be me getting ahead of myself}} | {{Todo|Confirm this - it could just be me getting ahead of myself}} | ||
A given a [[Measure space|measure space]] (a measurable space equipped with a measure) {{M|(X,\mathcal{A},\mu)}} with a measurable map on the following mean the same thing: | A given a [[Measure space|measure space]] (a measurable space equipped with a measure) {{M|(X,\mathcal{A},\mu)}} with a measurable map on the following mean the same thing: | ||
Revision as of 00:14, 3 August 2015
Note: Sometimes called a measurable fuction[1]
Definition
Let [ilmath](X,\mathcal{A})[/ilmath] and [ilmath](X',\mathcal{A}')[/ilmath] be measurable spaces then a map:
- [math]T:X\rightarrow X'[/math]
is called [math]\mathcal{A}/\mathcal{A}'[/math]-measurable[2], or [math]\mathcal{A}-\mathcal{A}'[/math]-measurable[3], or Measurable relative to [ilmath]\mathcal{A} [/ilmath] and [ilmath]\mathcal{A}'[/ilmath][1] if:
- [math]T^{-1}(A')\in\mathcal{A},\ \forall A'\in\mathcal{A}'[/math]
See also:
- (Theorem) Conditions for a map to be a measurable map
- (Theorem) A map, [ilmath]f:(A,\mathcal{A})\rightarrow(F,\mathcal{F})[/ilmath], is [ilmath]\mathcal{A}/\mathcal{F} [/ilmath] measurable iff for some generator [ilmath]\mathcal{F}_0[/ilmath] of [ilmath]\mathcal{F} [/ilmath] we have [ilmath]\forall S\in\mathcal{F}_0[f^{-1}(S)\in\mathcal{A}][/ilmath]
Notation
I like the [ilmath]\mathcal{A}/\mathcal{B} [/ilmath]-measurable notation because it reads nicely for composition. As a composition of measurable maps is measurable[1] we see something like, if:
- [ilmath]f:(A,\mathcal{A})\rightarrow(B,\mathcal{B})[/ilmath] is measurable (same as saying: [ilmath]f:A\rightarrow B[/ilmath] is [ilmath]\mathcal{A}/\mathcal{B} [/ilmath]-measurable) and
- [ilmath]g:(B,\mathcal{B})\rightarrow(C,\mathcal{C})[/ilmath] is measurable
then:
- [ilmath]g\circ f:(A,\mathcal{A})\rightarrow(C,\mathcal{C})[/ilmath] is measurable.
In effect:
- [ilmath]\mathcal{A}/\mathcal{B} [/ilmath]-measurable followed by [ilmath]\mathcal{B}/\mathcal{C} [/ilmath] measurable [ilmath]=[/ilmath] [ilmath]\mathcal{A}/\mathcal{C} [/ilmath]-measurable
(OLD) Notation
TODO: Confirm this - it could just be me getting ahead of myself
A given a measure space (a measurable space equipped with a measure) [ilmath](X,\mathcal{A},\mu)[/ilmath] with a measurable map on the following mean the same thing:
- [math]T:(X,\mathcal{A},\mu)\rightarrow(X',\mathcal{A}',\bar{\mu})[/math] (if [ilmath](X',\mathcal{A}')[/ilmath] is also equipped with a measure)
- [math]T:(X,\mathcal{A},\mu)\rightarrow(X',\mathcal{A}')[/math]
- [math]T:(X,\mathcal{A})\rightarrow(X',\mathcal{A}')[/math]
We would write [math]T:(X,\mathcal{A},\mu)\rightarrow(X',\mathcal{A}')[/math] simply to remind ourselves of the measure we are using, it is not important to the concept of the measurable map.
As usual, the function is on the first thing in the bracket. (see function for more details)
Motivation
From the topic of random variables - which a special case of measurable maps (where the domain can be equipped with a probability measure, a measure where [ilmath]X[/ilmath] has measure 1).
Consider: [math]X:(\Omega,\mathcal{A},\mathbb{P})\rightarrow(V,\mathcal{U})[/math], we know that given a [ilmath]U\in\mathcal{U} [/ilmath] that [ilmath]T^{-1}\in\mathcal{A} [/ilmath] which means we can measure it using [ilmath]\mathbb{P} [/ilmath], which is something we'd want to do.
Example using sum of two die RV
Take [math]\Omega=\{(a,b)|a,b\in\mathbb{N}\, a,b\in[1,6]\}[/math] and [math]\mathcal{A}=\sigma(\Omega)=\mathcal{P}(\Omega)[/math], define [math]\mathbb{P}:\mathcal{P}(\Omega)\rightarrow[0,1]\subset\mathbb{R}[/math] by [math]\mathbb{P}(A)\mapsto \frac{1}{36}|A|[/math]
Take the random variable [math]X:(\Omega,\mathcal{P}(\Omega),\mathbb{P})\rightarrow(\{2,\cdots,12\},\mathcal{P}(\{2,\cdots,12\}))[/math] which assigns each [ilmath](a,b)[/ilmath] to [ilmath]a+b[/ilmath] - the sum of the scores.
It is clear for example that only [math]\{(1,2),(2,1)\}[/math] thus the probability of getting 3 as the sum is 2 out of 36 or [ilmath]\frac{1}{18} [/ilmath]
