Difference between revisions of "Measurable map"
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{{End Example Body}} | {{End Example Body}} | ||
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| + | ==See also== | ||
| + | * [[Random Variable]] | ||
| + | * [[Probability space]] | ||
| + | * [[Measurable space]] | ||
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| + | ==References== | ||
| + | <references/> | ||
{{Definition|Measure Theory}} | {{Definition|Measure Theory}} | ||
Revision as of 22:41, 2 May 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 if
[math]T^{-1}(A')\in\mathcal{A},\ \forall A'\in\mathcal{A}'[/math]
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.
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]
