The Analysis of Data, volume 1

Measurability of Functions

E.6. Measurability of Functions*

Definition E.6.1. Let $(\Omega_1,\mathcal{F}_1)$ and $(\Omega_2,\mathcal{F}_2)$ be two measurable spaces. A function $f:\Omega_1\to\Omega_2$ is $\mathcal{F}_1/\mathcal{F}_2$-measurable if \[ A\in\mathcal{F}_2\qquad \text{implies} \qquad f^{-1}(A)\in\mathcal{F}_1. \]

If $\Omega_2=\R^d$ we implicitly assume (unless stated otherwise) that $\mathcal{F}_2=\mathcal{B}(\R^d)$. Similarly, if $\Omega_1=\R^d$ we implicitly assume (unless stated otherwise) that $\mathcal{F}_1=\mathcal{B}(\R^d)$.

Proposition E.6.1. Let $(\Omega_1,\mathcal{F}_1)$ and $(\Omega_2,\sigma(\mathcal{C}))$ be two measurable spaces. A function $f:\Omega_1\to\Omega_2$ is $\mathcal{F}_1/\mathcal{F}_2$-measurable if \[ A\in\mathcal{C} \qquad \text{implies} \qquad f^{-1}(A)\in\mathcal{F}_1. \]
Proof. We have $f^{-1}(\Omega_2\setminus A) = \Omega_2\setminus f^{-1}(A)$ and $f^{-1}(\cup_n A_n)=\cup_n f^{-1}(A_n)$ (see Proposition A.2.1). This, together with the fact that $\mathcal{F}_1$ is a $\sigma$-algebra proves that $\{A: f^{-1}(A)\in\mathcal{F}_1\}$ is a $\sigma$-algebra that contains $\mathcal{C}$. Since $\sigma(\mathcal{C})$ is the smallest $\sigma$-algebra containing $\mathcal{C}$, the proposition follows.
Proposition E.6.2. The composition of two measurable functions is a measurable function.

In the proposition above, there are three measurable spaces $(\Omega_i,\mathcal{F}_i)$, $i=1,2,3$. If $f:\Omega_1\to\Omega_2$ is $\mathcal{F}_1/\mathcal{F}_2$-measurable and $g:\Omega_2\to\Omega_3$ is $\mathcal{F}_2/\mathcal{F}_3$-measurable, the proposition states that $f\circ g:\Omega_1\to\Omega_3$ is $\mathcal{F}_1/\mathcal{F}_3$-measurable.

Proof. Let $A\in\mathcal{F}_3$. The measurability of $g$ implies that $g^{-1}(A)\in\mathcal{F}_2$, and together with the measurability of $f$ this implies $f^{-1}(g^{-1}(A))\in\mathcal{F}_1$.
Proposition E.6.3. If a function $f:\R^m\to\R^n$ is continuous, then it is measurable.
Proof. The $\sigma$-algebra $\mathcal{B}(\R^n)$ is generated by the set of all open sets. By Proposition E.6.1 it is then sufficient to show that for every open set $A$, $f^{-1}(A)\in\mathcal{B}(\R^m)$. Proposition B.3.4 states that $f^{-1}(A)$ is open, implying that $f^{-1}(A)\in\mathcal{B}(\R^m)$.

The set of measurable functions is strictly larger than the set of continuous functions. For example, the indicator function $I_A:\R\to\R$, defined as $I_A(x)=1$ if $x\in A$ and 0 otherwise, is not continuous (assuming $A\neq \Omega$ and $A\neq \emptyset$) but is measurable if $A\in\mathcal{B}(\R)$.

Proposition E.6.4. Let $f_n, n\in\mathbb{N}$ be measurable functions from $(\Omega,\mathcal{F})$ to $\R$. Then
  1. The functions $\sup_n f_n$, $\inf_n f_n$, $\limsup_n f_n$, $\liminf_n f_n$ are measurable.
  2. If $\lim_n f_n$ exists everywhere then $\lim_n f_n$ is measurable.
Proof. Since $f_n$ are measurable, $(-\infty,x]\in\mathcal{B}(\R)$, and $\mathcal{F}$ is closed under countable intersections, we have \[\left\{\omega:\sup_{n\in\mathbb{N}} f_n(\omega)\leq x\right\}=\bigcap_{n\in\mathbb{N}} \{\omega: f_n(\omega)\leq x\}=\bigcap_{n\in\mathbb{N}} f_n^{-1}((-\infty,x]) \in \mathcal{F}. \] Note that this holds even for $x=+\infty$ or $x=-\infty$. The equation above, together with Proposition E.6.1 and Proposition E.5.2, imply that $\sup_n f_n$ is measurable. The proof for measurability of $\inf_n f_n$ is similar. The functions $\liminf$ and $\limsup$ are compositions of $\inf$ and $\sup$ functions (Definition B.2.6) which are measurable, and thus by Proposition E.6.2 $\liminf$ and $\limsup$ are measurable as well. If $\lim_n f_n$ exists, $\liminf_n f_n=\limsup_n f_n$ establishing that $\lim_n f_n$ is measurable.