Difference between revisions of "Measurable map"
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==Definition== | ==Definition== | ||
| − | Let {{M|(X,\mathcal{A})}} and {{M|(X',\mathcal{A}')}} be [[Measurable space|measurable spaces]] then a map <math>T:X\rightarrow X'</math> is called '''<math>\mathcal{A}/\mathcal{A}'</math>-measurable''' if: | + | Let {{M|(X,\mathcal{A})}} and {{M|(X',\mathcal{A}')}} be [[Measurable space|measurable spaces]] then a map <math>T:X\rightarrow X'</math> is called '''<math>\mathcal{A}/\mathcal{A}'</math>-measurable''', or '''<math>\mathcal{A}-\mathcal{A}'</math>-measurable'''<ref name="PTACC">Probability Theory - A Comprehensive Course - Second Edition - Achim Klenke</ref> if: |
* <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> | ||
Note that the function is on {{M|X}} not on {{M|\mathcal{A} }}, I'll write it again to be explicit: | Note that the function is on {{M|X}} not on {{M|\mathcal{A} }}, I'll write it again to be explicit: | ||
Revision as of 14:23, 16 June 2015
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, or [math]\mathcal{A}-\mathcal{A}'[/math]-measurable[1] if:
- [math]T^{-1}(A')\in\mathcal{A},\ \forall A'\in\mathcal{A}'[/math]
Note that the function is on [ilmath]X[/ilmath] not on [ilmath]\mathcal{A} [/ilmath], I'll write it again to be explicit:
- [ilmath]T:X\rightarrow X'[/ilmath]
Notation
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]
See also
References
- ↑ Probability Theory - A Comprehensive Course - Second Edition - Achim Klenke
