Difference between revisions of "Measurable map"

Revision as of 15:37, 18 March 2015 (view source) Alec (Talk | contribs) m ← Older edit		Revision as of 10:18, 22 March 2015 (view source) Alec (Talk | contribs) m Newer edit →
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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>
		+
		+	==Notation==
		+	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:
		+	* <math>T:(X,\mathcal{A},\mu)\rightarrow(X',\mathcal{A}',\bar{\mu})</math> (if {{M|(X',\mathcal{A}')}} 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 variable|random variables]] - which a special case of measurable maps (where the domain can be equipped with a probability measure, a measure where {{M|X}} has measure 1).
		+
		+
		+	Consider: <math>X:(\Omega,\mathcal{A},\mathbb{P})\rightarrow(V,\mathcal{U})</math>, we know that given a {{M|U\in\mathcal{U} }} that {{M|T^{-1}\in\mathcal{A} }} which means we can measure it using {{M|\mathbb{P} }}, which is something we'd want to do.
		+
		+	{{Begin Example}}
		+	Example using sum of two die RV
		+	{{Begin Example Body}}
		+	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 {{M|(a,b)}} to {{M|a+b}} - 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 {{M|\frac{1}{18} }}
		+	{{End Example Body}}
		+	{{End Example}}
		+
	{{Definition|Measure Theory}}		{{Definition|Measure Theory}}

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Revision as of 10:18, 22 March 2015

Definition

Let $(X,\mathcal{A})$ and $(X',\mathcal{A}')$ be measurable spaces

Then a map

$$T:X\rightarrow X'$$ is called $$\mathcal{A}/\mathcal{A}'$$ -measurable if

$$T^{-1}(A')\in\mathcal{A},\ \forall A'\in\mathcal{A}'$$

Notation

A given a measure space (a measurable space equipped with a measure) $(X,\mathcal{A},\mu)$ with a measurable map on the following mean the same thing:

• $$T:(X,\mathcal{A},\mu)\rightarrow(X',\mathcal{A}',\bar{\mu})$$ (if $(X',\mathcal{A}')$ is also equipped with a measure)
• $$T:(X,\mathcal{A},\mu)\rightarrow(X',\mathcal{A}')$$
• $$T:(X,\mathcal{A})\rightarrow(X',\mathcal{A}')$$

We would write

$$T:(X,\mathcal{A},\mu)\rightarrow(X',\mathcal{A}')$$ 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 $X$ has measure 1).

Consider:

$$X:(\Omega,\mathcal{A},\mathbb{P})\rightarrow(V,\mathcal{U})$$ , we know that given a $U\in\mathcal{U}$ that $T^{-1}\in\mathcal{A}$ which means we can measure it using $\mathbb{P}$, which is something we'd want to do.