Measurable map

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.

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Example using sum of two die RV

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Take $$\Omega=\{(a,b)|a,b\in\mathbb{N}\, a,b\in[1,6]\}$$ and $$\mathcal{A}=\sigma(\Omega)=\mathcal{P}(\Omega)$$ , define $$\mathbb{P}:\mathcal{P}(\Omega)\rightarrow[0,1]\subset\mathbb{R}$$ by $$\mathbb{P}(A)\mapsto \frac{1}{36}|A|$$

Take the random variable $$X:(\Omega,\mathcal{P}(\Omega),\mathbb{P})\rightarrow(\{2,\cdots,12\},\mathcal{P}(\{2,\cdots,12\}))$$ which assigns each $(a,b)$ to $a+b$ - the sum of the scores.

It is clear for example that only $$\{(1,2),(2,1)\}$$ thus the probability of getting 3 as the sum is 2 out of 36 or $\frac{1}{18}$