Measurability

Let $f: \Omega \rightarrow \overline{\mathbb {R}}$ be $\mathcal{F} / \mathcal{B}^{*}$-measurable. Let $y \in \overline{\mathbb {R}}$, and let $h: \Omega \rightarrow \overline{\mathbb {R}}$ be such that

Then $h$ is $\mathcal{F} / \mathfrak {B}^{*}$-measurable.

There exists a continuous function $f: \mathbb {R} \rightarrow \mathbb {R}$ and a subset $A \subseteq \mathbb {R}$ such that $A \in \mathscr {L}$ but $f^{-1}(A) \notin \mathscr {L}$.

Using the notation and framework of Section 4.7 and the previous exercise, we have $I_{B}=I_{\phi^{-1}(B)} \circ \phi^{-1}$ and $I_{\phi^{-1}(B)}$ is $\mathcal{L} / \mathfrak {B}^{*}$-measurable, while $I_{B}$ is not $\mathcal{L} / \mathfrak {B}^{*}$-measurable. Therefore, an $\mathcal{L} / \mathfrak {B}^{*}$-measurable function of a homeomorphism is not necessarily $\mathcal{L} / \mathfrak {B}^{*}$-measurable.

A Lebesgue measurable function is not necessarily Borel measurable.

Suppose that $f: \Omega \rightarrow \overline{\mathbb {R}}$ is $\mathcal{F} / \mathcal{B}^{*}$-measurable.

If $f: \Omega \rightarrow \overline{\mathbb {R}}$ is such that $f^{-1}(\left\{ x\right\} ) \in \mathcal{F}$ for every $x \in \overline{\mathbb {R}}$, then $f$ is not necessarily $\mathcal{F} / \mathfrak {B}^{*}$-measurable.

If $A \subseteq \mathbb {R}$ is any type of interval and $f: A \rightarrow \overline{\mathbb {R}}$ is monotone, then $f$ is both Borel and Lebesgue measurable.

Let $f: \Omega \rightarrow \overline{\mathbb {R}}^{k}$ be given by the rule $f(\omega )=\left(f_{1}(\omega ), \cdots , f_{k}(\omega )\right)$, where we have that $f_{i}: \Omega \rightarrow \overline{\mathbb {R}}$ is a function, $i=1, \cdots , k$. Then $f^{-1}(B) \in \mathcal{F}$ for every $B \in \mathfrak {B}^{k*}$ iff $f_{i}$ is $\mathcal{F} / \mathfrak {B}^{*}$-measurable for $i=1, \cdots , k$.

$\mathfrak {B}^{k}$ is the minimal $\sigma$-field $\mathcal{F}$ on $\mathbb {R}^{k}$ such that every continuous $f: \mathbb {R}^{k} \rightarrow \overline{\mathbb {R}}$ is $\mathcal{F} / \mathfrak {B}^{*}$-measurable.

Let $f: \Omega \rightarrow \overline{\mathbb {R}}$ be $\sigma (\mathcal{A}) / \mathfrak {B}^{*}$-measurable, where $\mathcal{A} \subseteq 2^{\Omega }$. There exists an amc subcollection $\mathcal{A}_{f}$ of $\mathcal{A}$ such that $f$ is $\sigma (\mathcal{A}_{f}) / \mathfrak {B}^{*}$-measurable.

Let $f: \Omega \rightarrow \overline{\mathbb {R}}$, and suppose that $\Omega =\bigcup_{n=1}^{\infty } A_{n}$, where $A_{1}, A_{2}, \ldots$ are disjoint $\mathcal{F}$-sets [ $\mathcal{F}$ is a $\sigma$-field on $\Omega$ ]. Let $\mathcal{F}_{n}=\left\{ A \in \mathcal{F}: A \subset A_{n}\right\}$ for each $n \in \mathbb {N}$. Then $\mathcal{F}_{n}$ is a $\sigma$-field for each $n \in \mathbb {N}$. Let $f_{n}$ denote the restriction of $f$ from $\Omega$ to $A_{n}, n \in \mathbb {N}$. Then $f$ is $\mathcal{F} / \mathscr {B}^{*}$-measurable iff $f_{n}$ is $\mathcal{F}_{n} / \mathscr {B}^{*}$-measurable for each $n \in \mathbb {N}$.

Show that the function $\varphi : \overline{\mathbb {R}} \rightarrow \overline{\mathbb {R}}$ given in Example 4 is $\mathfrak {B}^{*} / \mathfrak {B}^{*}$-measurable by suitably appealing to (MF6).

Example 4: Define $\varphi : \overline{\mathbb {R}} \rightarrow \overline{\mathbb {R}}$ by writing $\varphi (x) = 1/x$ for $x \in \overline{\mathbb {R}} \setminus \left\{ 0\right\}$ and $\varphi (0) = 0$ (recall the convention $1/(\pm \infty ) = 0$, so $\varphi$ is defined at $\pm \infty$ too).

(MF6): Let $A \in \mathcal{F} - \left\{ \varnothing \right\}$, and let $A_{1}, \ldots , A_{n}$ denote nonempty disjoint $\mathcal{F}$-sets with union $\Omega$. Suppose that $f_{i}: A_{i} \rightarrow \overline{\mathbb {R}}$ is $\mathcal{F} / \mathfrak {B}^{*}$-measurable for $i = 1, \ldots , n$. If $f: A \rightarrow \overline{\mathbb {R}}$ is such that $f(\omega ) = f_{i}(\omega )$ iff $\omega \in A_{i}$, then $f$ is $\mathcal{F} / \mathfrak {B}^{*}$-measurable. (This also holds for a countable collection $A_{1}, A_{2}, \ldots$ of nonempty disjoint $\mathcal{F}$-sets with union $\Omega$.)

The minimal $\sigma$-field $\mathcal{F}$ on $\Omega$ such that $f: \Omega \rightarrow \overline{\mathbb {R}}$ is $\mathcal{F} / \mathfrak {B}^{*}$-measurable is $f^{-1}\left(\mathfrak {B}^{*}\right)$.

The word continuous in (MF4) may be replaced by either of lower semicontinuous and upper semicontinuous.

(MF4): Let $\Omega = \mathbb {R}^{k}$, and let $\mathcal{F}$ denote an arbitrary $\sigma$-field on $\mathbb {R}^{k}$ that contains the open subsets of $\mathbb {R}^{k}$. Then for any nonempty $A \in \mathcal{F}$ and every continuous $f: A \rightarrow \overline{\mathbb {R}}$, $f$ is $\mathcal{F} / \mathfrak {B}^{*}$-measurable. (Nothing in the proof is special to Euclidean space: $\Omega$ may be any metric or topological space, and $\mathcal{F}$ any $\sigma$-field on $\Omega$ containing the open sets.)