Measurable Maps

[math] \newcommand{\R}{\mathbb{R}} \newcommand{\A}{\mathcal{A}} \newcommand{\B}{\mathcal{B}} \newcommand{\N}{\mathbb{N}} \newcommand{\C}{\mathbb{C}} \newcommand{\Rbar}{\overline{\mathbb{R}}} \newcommand{\Bbar}{\overline{\mathcal{B}}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\E}{\mathbb{E}} \newcommand{\p}{\mathbb{P}} \newcommand{\one}{\mathds{1}} \newcommand{\0}{\mathcal{O}} \newcommand{\mat}{\textnormal{Mat}} \newcommand{\sign}{\textnormal{sign}} \newcommand{\CP}{\mathcal{P}} \newcommand{\CT}{\mathcal{T}} \newcommand{\CY}{\mathcal{Y}} \newcommand{\F}{\mathcal{F}} \newcommand{\mathds}{\mathbb}[/math]

Measure theory and the notion of integration require special structures on functions, which need to satisfy different properties, such as being measurable or bounded. The notion of a measurable map is important for the study of the integration with respect to a certain measure and the fact that we can only consider integration with respect to a measure if the integrating function satisfies measurability. It is now important to use the [math]\sigma[/math]-Algebras of the underlying spaces similar to the topological notion of continuity where the topologies of the underlying spaces are used. Let us therefore define a measurable map.

Definition (Measurable Map)

Let [math](E,\A)[/math] and [math](F,\B)[/math] be two measurable spaces and let [math]f:E\longrightarrow F[/math] be a map. We say that [math]f[/math] is measurable, if for all [math]B\in\B[/math] we get [math]f^{-1}(B)\in\A[/math].

Proposition

Let [math](X,\A),(Y,\B)[/math] and [math](Z,\mathcal{G})[/math] be measurable space and consider the composition

[[math]] (X,\A)\xrightarrow{f}(X,\B)\xrightarrow{g}(Z,\mathcal{G}). [[/math]]

If [math]f[/math] and [math]g[/math] are both measurable, then [math]g\circ f[/math] is also measurable.

Show Proof

Exercise.

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Proposition

Let [math](E,\A)[/math] and [math](F,\B)[/math] be two measurable spaces and let [math]f:E\longrightarrow F[/math] be a map. Moreover, assume that there exists [math]\mathcal{C}\subset\mathcal{P}(F)[/math] such that [math]\sigma(\mathcal{C})=\B[/math]. Then [math]f[/math] is measurable if and only if for all [math]C\in\mathcal{C}[/math] we have [math]f^{-1}(C)\in\A[/math].

Show Proof

Let us first define the [math]\sigma[/math]-Algebra [math]\mathcal{G}[/math] by

[[math]] \mathcal{G}:=\{B\in\B\mid f^{-1}(B)\in\A\}\supset \mathcal{C}. [[/math]]

Now, since [math]\mathcal{G}[/math] is a [math]\sigma[/math]-Algebra, we get that [math]\sigma(\mathcal{C})\subset\mathcal{G}[/math] and thus [math]\mathcal{G}=\B[/math], which proves the claim.

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Example

Let [math](F,\B)=(\R,\B(\R))[/math]. To show that [math]f[/math] is measurable, it is enough to show either that [math]f^{-1}((a,b))\in\A[/math] or [math]f^{-1}((-\infty,a))\in\A[/math], for [math]a,b\in\R[/math] with [math]a \lt b[/math].

Example

[continuous maps are measurable] Assume that [math]E[/math] and [math]F[/math] are two metric spaces (or topological spaces), endowed with their Borel [math]\sigma[/math]-Algebra respectively. Then [math]f[/math] is measurable if for every open set [math]O[/math] of [math]F[/math] we have [math]f^{-1}(O)\in\B(E)[/math]. In particular we can say that continuous maps are measurable maps.

Example

Let [math]A\subset E[/math] be a subset of [math]E[/math]. Then the map [math]\one_A[/math] is measurable if and only if [math]A\in\A[/math].

The notion of measurability of a map [math]f:E\longrightarrow F[/math], between two measurable spaces [math](E,\A)[/math] and [math](F,\B)[/math], means that [math]f^{-1}(\B)\in \A[/math]. The smallest [math]\sigma[/math]-Algebra on [math]E[/math] which makes [math]f[/math] measurable is given by [math]f^{-1}(\B)[/math] and we denote it by [math]\sigma(f)[/math]. Moreover, we want to emphasize that we can write [math]\{f\in B\}[/math] for [math]f^{-1}(B)=\{x\in E\mid f(x)\in B\}[/math]. Hence we can write [math]\{f\geq b\}[/math] instead of [math]f^{-1}([b,\infty))[/math] or [math]\{ f=b\}[/math] instead of [math]f^{-1}(\{b\})[/math]. If [math]f[/math] is constant and if for some [math]C\in F[/math], we have [math]f(x)=C[/math] for all [math]x\in E[/math], then [math]f[/math] is always measurable, since [math]f^{-1}(\B)=\{\varnothing, E\}[/math].

Lemma

Let [math](E,\A),(F_1,\B_1)[/math] and [math](F_2,\B_2)[/math] be measurable spaces and let [math]f_1:E\longrightarrow F_1[/math] and [math]f_2:E\longrightarrow F_2[/math] be two measurable maps. Then the map

[[math]] \begin{align*} f:(E,\A)&\longrightarrow (F_1\times F_2,\B_1\otimes \B_2)\\ x&\longmapsto (f_1(x),f_2(x)) \end{align*} [[/math]]

is measurable.

Show Proof

Let us define [math]\mathcal{C}:=\{B_1\times B_2\mid B_1\in\mathcal{B}_1,B_2\in\mathcal{B}_2\}[/math]. Then we get, by definition of the product [math]\sigma[/math]-Algebra, that [math]\sigma(\mathcal{C})=\mathcal{B}_1\otimes\mathcal{B}_2[/math]. Now, for [math]B_1\times B_2\in \mathcal{C}[/math] we get that [math]f^{-1}(B_1\times B_2)=\underbrace{\underbrace{f^{-1}(B_1)}_{\in\mathcal{A}}\cap \underbrace{f^{-1}(B_2)}_{\in\mathcal{A}}}_{\in\mathcal{A}}\in\A[/math]. Therefore, it follows that [math]f[/math] is measurable.

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Consider [math]f_1=\pi_1\circ f[/math] and [math]f_2=\pi_2\circ f[/math] with

[[math]] \begin{align*} \pi_i:F_1\times F_2&\longrightarrow F_i\\ (y_1,y_2)&\longmapsto y_i \end{align*} [[/math]]

for [math]i\in\{1,2\}[/math] with [math]\pi_1[/math] and [math]\pi_2[/math] are measurable. Then [math]f_1[/math] and [math]f_2[/math] are measurable.

Corollary

Let [math](E,\A)[/math] be a measurable space and let [math]f,g:E\longrightarrow \R[/math] be two measurable maps, where [math]\R[/math] is endowed with its Borel [math]\sigma[/math]-Algebra [math]\B(\R)[/math]. Then

• [math]f+g[/math]
• [math]f\cdot g[/math]
• [math]f^+=\max\{f,0\}[/math]
• [math]f^-=\max\{-f,0\}[/math]
• [math]\vert f\vert[/math]

are measurable, where [math]f = f^+-f^-[/math] and [math]\vert f\vert =f^++f^-[/math].

Show Proof

We will only show [math](i)[/math] and leave the other points as an exercise for the reader. The map [math]f+g[/math] is a composition of the map [math]h:x\longmapsto (f(x),g(x))[/math] and [math]r:(a,b)\longmapsto a+b[/math]. The map [math]h[/math] is clearly measurable, since [math]f[/math] and [math]g[/math] are measurable and the map [math]r[/math] is clearly continuous and thus measurable. As we have seen, the composition of two measurable maps is again measurable, which shows that [math]f+g[/math] is measurable. The proof of the other points is similar.

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Let us consider the field of complex numbers [math]\C[/math] and make the identification [math]\C\simeq \R^2[/math]. Then we can naturally make sense of the measurability of the map [math]f:E\longrightarrow \C[/math], where [math]\C[/math] is endowed with its Borel [math]\sigma[/math]-Algebra [math]\B(\C)[/math], by saying that [math]f[/math] is measurable if and only if [math]Re(f)[/math] and [math]Im(f)[/math] are measurable.

General references

Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].