Measurable map

Note: Sometimes called a measurable fuction

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, or $$\mathcal{A}-\mathcal{A}'$$ -measurable, or Measurable relative to $\mathcal{A}$ and $\mathcal{A}'$ if:

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

See also:

• (Theorem) Conditions for a map to be a measurable map
• (Theorem) A map, $f:(A,\mathcal{A})\rightarrow(F,\mathcal{F})$, is $\mathcal{A}/\mathcal{F}$ measurable iff for some generator $\mathcal{F}_0$ of $\mathcal{F}$ we have $\forall S\in\mathcal{F}_0[f^{-1}(S)\in\mathcal{A}]$

Notation

I like the $\mathcal{A}/\mathcal{B}$-measurable notation because it reads nicely for composition. As a composition of measurable maps is measurable we see something like, if:

• $f:(A,\mathcal{A})\rightarrow(B,\mathcal{B})$ is measurable (same as saying: $f:A\rightarrow B$ is $\mathcal{A}/\mathcal{B}$-measurable) and
• $g:(B,\mathcal{B})\rightarrow(C,\mathcal{C})$ is measurable

then:

• $g\circ f:(A,\mathcal{A})\rightarrow(C,\mathcal{C})$ is measurable.

In effect:

• $\mathcal{A}/\mathcal{B}$-measurable followed by $\mathcal{B}/\mathcal{C}$ measurable $=$ $\mathcal{A}/\mathcal{C}$-measurable

Terminology

• $\mathcal{A}$-measurable is short hand for $\mathcal{A}/\mathcal{B}(\overline{\mathbb{R} })$-measurable - where $\mathcal{B}$ denotes the Borel $\sigma$-algebra and $\overline{\mathbb{R} }=\mathbb{R}\cup\{-\infty,+\infty\}$
• A Borel function is a mapping from the Borel $\sigma$-algebra of a topological space, $(X,\mathcal{J})$, then the $\mathcal{B}((X,\mathcal{J}))$-measurable functions are called Borel functions

(OLD) Notation

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TODO: Confirm this - it could just be me getting ahead of myself

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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.

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 $X$ 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 $\frac{1}{18}$

See also

• Random variable
• Probability space
• Measurable space
• Conditions for a map to be a measurable map