Extensions of Measures

Let $(\Omega , \mathcal{A}, \mu )$ denote a measure space. Pick $E \in \mathcal{A}$ and define $\mathcal{A}_{E}=\left\{ F \in \mathcal{A}: F \subseteq E\right\}$. Then $\mathcal{A}_{E}$ is a $\sigma$-field on $E, \mathcal{A}_{E}=\left\{ A \cap E: A \in \mathcal{A}\right\}$, and the restriction $\mu_{E}$ of $\mu$ [from $\mathcal{A}$ ] to $\mathcal{A}_{E}$ is a measure. That is, $\left(E, \mathcal{A}_{E}, \mu_{E}\right)$ is a measure space and $\mu_{E}=\mu$ on $\mathcal{A}_{E}$.

Prove Claim 1 and 2.

Claim 1 Assume the notation of the definition. If $\mu$ is $\sigma$-finite on $\mathscr {G}$, then $v$ is $\sigma$-finite on $\mathscr {G}$ as well.

Claim 2 Suppose that $\varnothing \in \mathscr {G}$, and let $\mathscr {G} \subseteq \mathscr {H} \subseteq 2^{\Omega }$. Let $v: \mathscr {H} \rightarrow \overline{\mathbb {R}}$ denote a measure. Then the restriction of $v$ to $\mathscr {G}$ is a measure.

Let $(\Omega , \mathcal{A}, \mu )$ denote an arbitrary measure space. Define $v$ on $2^{\Omega }$ by writing $v(B)=\inf \left\{ \mu (A): B \subseteq A, A \in \mathcal{A}\right\}$ for each $B \subseteq \Omega$. Then $v$ is an outer measure.

Let $v: 2^{\Omega } \rightarrow \overline{\mathbb {R}}$ be an outer measure, and suppose in addition that $v$ is finitely additive: $v(A \cup B)=v(A)+v(B)$, where $A, B \subseteq \Omega$ are disjoint. Then $v$ is a measure. That is, $\left(\Omega , 2^{\Omega }, v\right)$ is a measure space.

Suppose that $\varphi$ and $\xi$ are outer measures [relative to some common background set $\Omega$ ], and suppose that we define a new function $v: 2^{\Omega } \rightarrow \overline{\mathbb {R}}$ for all $A \subseteq \Omega$ by writing $v(A)=\max \left\{ \varphi (A), \xi (A)\right\}$. Then $v$ is an outer measure [relative to $\Omega$ ].

Let $v$ denote an outer measure, and let $A \subseteq \Omega$. Define a new set function $\nu_{A}$ on $2^{\Omega }$ by writing $\nu_{A}(B)=\nu (B \cap A)$ for each $B \subseteq \Omega$. Then $\nu_{A}$ is an outer measure relative to $\Omega$.

Let $\left\{ v_{n}\right\}_{n=1}^{\infty }$ denote a sequence of outer measures [relative to some common $\Omega$ ], and let $\left\{ a_{n}\right\}_{n=1}^{\infty }$ denote a sequence of nonnegative numbers. For each $A \subseteq \Omega$, let $v(A)=\sum_{n=1}^{\infty } a_{n} \cdot v_{n}(A)$. Then $v$ is an outer measure relative to $\Omega$.

Show directly that if $A, B \in \mathcal{M}(v)$, then $A \cup B, A \backslash B \in \mathcal{M}(v)$.

Suppose that $\Omega$ may be written as $\bigcup_{n=1}^{\infty } A_{n}$, where $\left\{ A_{n}\right\}_{n=1}^{\infty }$ is a nondecreasing sequence of subsets of $\Omega$. If $A \subseteq \Omega$ is such that $A \cap A_{k} \in \mathcal{M}(v)$ for all $k$ exceeding some constant $k_{A}$, then $A \in \mathcal{M}(v)$.

Let $v$ denote an outer measure such that $v(\Omega )<+\infty$, and further suppose that if $A \subseteq \Omega$ with $v(A)<+\infty$, then there exists $B \in \mathcal{M}(v)$ such that $A \subseteq B$ and $v(A)=v(B)$. Then $E \in \mathcal{M}(v)$ iff $v(\Omega )=v(E)+v\left(E^{c}\right)$.

Suppose that $v$ is an arbitrary outer measure, and let $A, B \subseteq \Omega$ with $A \in \mathcal{M}(v)$. Show that $v(A \cup B)+v(A \cap B)=v(A)+v(B)$.

Let $v$ denote an outer measure, and let $\left\{ A_{n}\right\}_{n=1}^{\infty }$ denote a nondecreasing sequence of $\mathcal{M}(v)$-sets. Show that $v\left(\lim \left(A \cap A_{n}\right)\right)=\lim v(A \cap A_{n})$ for any $A \subseteq \Omega$. State and prove an analogous result for nonincreasing sequences of $\mathcal{M}(v)$-sets.

Let $v$ denote an outer measure such that the following holds: if $A \subseteq \Omega$ with $\nu (A)<+\infty$, then there is $B \in \mathcal{M}(\nu )$ with $A \subseteq B$ and $v(A)=v(B)$. Then, for any nondecreasing sequence $\left\{ A_{n}\right\}_{n=1}^{\infty }$ of subsets of $\Omega$, we have $\nu \left(\lim A_{n}\right)=\lim \nu \left(A_{n}\right)$.

In each of the following parts, (i) describe the outer measure $\mu^{*}$ on $2^{\Omega }$ induced by the given $\mu$, (ii) describe the collection $\mathcal{M}\left(\mu^{*}\right)$ and determine if $\mathcal{M}\left(\mu^{*}\right)$ is a $\sigma$-field, and (iii) check to see whether $\mu^{*}=\mu$ on the given collection $\mathcal{A}$.

Let $k \in \mathbb {N}$ and refer to the measure $\lambda_{k}: \mathscr {B}^{k} \rightarrow \overline{\mathbb {R}}$ that assigns the value $\prod_{i=1}^{k}\left(b_{i}-a_{i}\right)$ to every $(\boldsymbol {a}, \boldsymbol {b}] \in \mathcal{A}_{k}$, as given in this section's example. For this measure $\lambda_{k}$, we have the following;

This problem reviews the Extension Theorem.

Consider the Extension Theorem framework. If we have $\mu (A)<+\infty$ for each $A \in \mathcal{A}$, it might not be the case that the measure extension $\mu_{\sigma (\mathscr {A})}^{*}$ assigns finite measure to every set in $\sigma (\mathscr {A})$. However, if $\mu$ is $\sigma$-finite on $\mathscr {A}$, then the measure extension $\mu_{\sigma (\mathscr {A})}^{*}$ is $\sigma$-finite on $\sigma (\mathscr {A})$, and the measure extension $\mu_{\mathcal{M}\left(\mu^{*}\right)}^{*}$ is $\sigma$-finite on $\mathcal{M}\left(\mu^{*}\right)$.

There is a measure $v: B \rightarrow \overline{\mathbb {R}}$ with $v(\mathbb {R})=1$ and

where the integral is the familiar Riemann integral from calculus.

Consider the Extension Theorem framework again.

Let $\mathcal{F}$ denote a field on $\Omega$, and let $\mu : \mathcal{F} \rightarrow \overline{\mathbb {R}}$ denote a measure. Let $\mu^{*}: 2^{\Omega } \rightarrow \overline{\mathbb {R}}$ be given by

Then $\mu^{*}$ is an outer measure, and the restriction of $\mu^{*}$ to the $\sigma$-field $\mathcal{M}\left(\mu^{*}\right)$ is a measure. With these facts, we have that $\mu^{*}=\mu$ on $\mathcal{F}$ and $\mathcal{F} \subseteq \mathcal{M}\left(\mu^{*}\right)$. Finally, there exists a measure extension of $\mu$ to $\sigma (\mathcal{F})$.

Let $\mathcal{F}$ denote a field on $\Omega$. Suppose that $v: \mathcal{F} \rightarrow \overline{\mathbb {R}}$ is such that (i) $v(A) \geqslant 0$ for all $A \in \mathcal{F}$, (ii) $v$ is finitely additive, and (iii) if $\left\{ A_{n}\right\}_{n=1}^{\infty }$ is a nonincreasing sequence of $\mathcal{F}$-sets with $\lim A_{n}=\varnothing$, the $\lim \nu \left(A_{n}\right)=0$. Define $\nu^{*}: 2^{\Omega } \rightarrow \overline{\mathbb {R}}$ for all $A \subseteq \Omega$ by writing

If $\mu_{1}$ and $\mu_{2}$ are finite measures with domain $\sigma (\mathcal{P})$ (where $\mathcal{P}$ denotes a $\pi$-system on $\Omega$ ), if $\Omega$ can be expressed as an amc union of $\mathscr {P}$-sets, and if $\mu_{1}=\mu_{2}$ on $\mathscr {P}$, then $\mu_{1}=\mu_{2}$ on $\sigma (\mathscr {P})$.

Let $\mu_{1}$ and $\mu_{2}$ denote finite measures with domain $\sigma (\mathcal{P})$, where $\mathcal{P}$ is a $\pi$-system on $\Omega$, and further suppose that $\mu_{1}=\mu_{2}$ on $\mathcal{P}$. Then $\mu_{1}=\mu_{2}$.

Let $\Omega =\left\{ \omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}\right\}$, let $\mathcal{A}$ consist of $\varnothing ,\left\{ \omega_{1}, \omega_{2}\right\}$, $\left\{ \omega_{1}, \omega_{3}\right\} ,\left\{ \omega_{2}, \omega_{4}\right\} ,\left\{ \omega_{3}, \omega_{4}\right\}$, and $\Omega$, and let $\mu : \mathcal{A} \rightarrow \mathbb {R}$ be defined as follows: $\mu (\Omega )=6, \mu (\varnothing )=0$, and $\mu \left(\left\{ \omega_{1}, \omega_{2}\right\} \right)=\mu \left(\left\{ \omega_{1}, \omega_{3}\right\} \right)=\mu \left(\left\{ \omega_{2}, \omega_{4}\right\} \right)=\mu \left(\left\{ \omega_{3}, \omega_{4}\right\} \right)=$ 3.

Let $\Omega =\mathbb {Q}$, and let $\mathcal{A}$ consist of the sets $\left\{ \omega \in \mathbb {Q}: a<\omega \leqslant b\right\}$, where $a<b$ and $a, b \in \mathbb {Q}$. Also, let $\varnothing \in \mathcal{A}$.

We assume the setup of Exercise 3.4.6. The aim of this exercise is to show that if $v$ is a measure with domain $\sigma (\mathcal{F})$ such that $v=\mu$ on the field $\mathcal{F}$, then $\nu$ coincides with the measure extension of $\mu$ to $\sigma (\mathcal{F})$ guaranteed by Exercise 3.4.6.

A lattice on $\Omega$ is a collection $\mathscr {L} \subseteq 2^{\Omega }$ such that (i) $\Omega \in \mathscr {L}$, (ii) $\varnothing \in \mathscr {L}$, (iii) $\mathscr {L}$ is closed under (finite) unions, and (iv) $\mathscr {L}$ is a $\pi$-system. We also define the following two collections: $D=\left\{ B \backslash A: A, B \in \mathscr {L}, A \subseteq B\right\}$, and $\mathcal{U}$ will denote the collection of all finite disjoint unions of $\mathfrak {D}$-sets.

Let $\mu$ and $v$ denote two $\sigma$-finite measures with domain $\mathcal{B}^{k}$ such that $\mu =v$ on the subcollection of open subsets of $\mathbb {R}^{k}$.

Let $k \geqslant 2$, and let $\lambda_{k}: \mathscr {B}^{k} \rightarrow \overline{\mathbb {R}}$ denote the unique measure with domain $\mathscr {B}^{k}$ that assigns the value $\prod_{i=1}^{k}\left(b_{i}-a_{i}\right)$ to each $k$-dimensional rsc rectangle ( $\boldsymbol {a}, \boldsymbol {b}$ ]. Let $A=\left\{ \boldsymbol {x} \in \mathbb {R}^{k}: x_{2}=\cdots =x_{k}=0\right\}$. Then $A \in \mathscr {B}^{k}, A$ is uncountable, and $\lambda_{k}(A)=0$, hence $\left|\overline{\mathscr {B}^{k}}\right|=2^{\text{c }}$.

Let $(\Omega , \mathcal{F}, v)$ denote a measure space. If $A, B \in \mathcal{F}$ with $A \subseteq E \subseteq B$ and $\nu (B \backslash A)=0$, then $E \in \overline{\mathcal{F}}$ and $\bar{\nu }(E)=\nu (A)=\nu (B)$.

Let ( $\Omega , \mathcal{F}, v$ ) denote a measure space. Furthermore, let $\mathcal{F}_{1}$ denote a sub- $\sigma$-field of $\mathcal{F}^{1}$. Then there exists a minimal $\sigma$-field $\mathcal{F}_{2}$ such that $\mathcal{F}_{1} \subseteq \mathcal{F}_{2} \subseteq \mathcal{F}$ and $\mathcal{N}(v) \subseteq \mathcal{F}_{2}$. Also, $A \in \mathcal{F}_{2}$ iff there exists $B \in \mathcal{F}_{1}$ with $A \Delta B \in \mathcal{N}(v)$.

Let $\Omega$ be uncountable, let $\mathcal{A}$ denote the $\sigma$-field $\left\{ A \subseteq \Omega : A \text{ is amc or } A^{c} \text{ is amc}\right\}$, and define $\mu : \mathcal{A} \rightarrow \overline{\mathbb {R}}$ by stipulating that $\mu (A)$ denotes the number of points in $A$ if $A$ is finite and $\mu (A)=+\infty$ if $A$ is infinite.

The assumption $v(B)=+\infty$ is not superfluous in Claim 4, and the assumption $v(A)<+\infty$ is not superfluous in Claim 6.

Let $v: \mathscr {B}^{k} \rightarrow \overline{\mathbb {R}}$ be nonnegative and finitely additive with $v\left(\mathbb {R}^{k}\right)<+\infty$. Suppose that v(A)=\sup \left\{ v(K): K \subseteq A, K compact for each $A \in \mathscr {B}^{k}$. Then $v$ is a finite measure.

Let $v: \mathcal{B}^{k} \rightarrow \overline{\mathbb {R}}$ be a measure such that $v(B)<+\infty$ for every bounded $B \in \mathcal{B}^{k}$. Let $A \subseteq \mathbb {R}^{k}$, where $A$ is not necessarily in $\mathcal{B}^{k}$. There exist sets $F$ and $G$ such that $F \subseteq A \subseteq G$, $F$ is a countable union of closed sets, and $G$ is a countable intersection of open sets, and $v(G-F)=0$.

Suppose that $v: \mathcal{B}^{k} \rightarrow \mathbb {R}^{k}$ is a finite measure such that if $A \in \mathcal{B}^{k}$ and $v(A)>0$, then there exists $B \in \mathcal{B}^{k}$ with $B \subseteq A$ and $0<v(B)<v(A)$. Then there exists an uncountable compact set $K$ such that $K \subseteq A$ and $v(K)=0$.

Let $\mathcal{A}$ denote a field on $\Omega$, and let $\mu$ and $v$ denote measures with domain $\sigma (\mathcal{A})$ that are $\sigma$-finite on $\mathcal{A}$. Let $E \in \sigma (\mathcal{A})$ be such that $\mu (E), v(E)<+\infty$, and let $\varepsilon >0$ be given. Then there exists $E_{0} \in \mathcal{A}$ such that $\mu \left(E \triangle E_{0}\right), v\left(E \triangle E_{0}\right)<\varepsilon$.

Countable superadditivity: If $A_{1}, A_{2}, \ldots \subseteq \Omega$ are disjoint, then $\mu_{*}\left(\bigcup_{n=1}^{\infty } A_{n}\right) \geqslant \sum_{n=1}^{\infty } \mu_{*}\left(A_{n}\right)$.

For any $A \subseteq \Omega , \mu^{*}(A)=\inf \left\{ \mu_{\mathcal{M}\left(\mu^{*}\right)}^{*}(B): A \subseteq B \in \mathcal{M}\left(\mu^{*}\right)\right\} =\mu^{* *}(A)$.

For any $A \subseteq \Omega , \mu_{*}(A)=\sup \left\{ \mu_{\mathcal{M}\left(\mu^{*}\right)}^{*}(B): B \subseteq A, B \in \mathcal{M}\left(\mu^{*}\right)\right\}$.

For any $A \subseteq \Omega$, there is $E \in \sigma (\mathcal{A})$ such that $E \subseteq A$ and $\mu_{\sigma (\mathscr {A})}^{*}(E)=\mu_{*}(A)$.

The infimum that defines $\mu^{*}(A)$ and the supremum that defines $\mu_{*}(A)$ are achieved for each $A \subseteq \Omega$. That is, there exist $C, B \in \sigma (\mathcal{A})$ with $C \subseteq A \subseteq B$ and $\mu_{\sigma (\mathscr {A})}^{*}(C)=\mu_{*}(A)$ and $\mu_{\sigma (\mathscr {A})}^{*}(B)=\mu^{*}(A)$.

Let $A \subseteq \Omega$ and let $\left\{ A_{n}\right\}$ denote a disjoint sequence of $\mathcal{M}\left(\mu^{*}\right)$-sets. Then we have $\mu_{*}\left(\bigcup_{n=1}^{\infty }\left(A \cap A_{n}\right)\right)=\sum_{n=1}^{\infty } \mu_{*}\left(A_{n} \cap A\right)$.

If $A, B \subseteq \Omega$ are disjoint, then $\mu_{*}(A \cup B) \leqslant \mu_{*}(A)+ \mu^{*}(B) \leqslant \mu^{*}(A \cup B)$.

If $A \in \mathcal{M}\left(\mu^{*}\right)$ and $B \subseteq \Omega$, then $\mu_{\mathcal{M}\left(\mu^{*}\right)}^{*}(A)=\mu_{*}(B \cap A)+\mu^{*}\left(B^{c} \cap A\right)$.

Let $\Omega =(0,1] \times (0,1]$, and let $\mathcal{F}=\left\{ B \times (0,1]: B \in \mathcal{B}, B \subseteq (0,1]\right\}$. Define $\mu : \mathcal{F} \rightarrow \overline{\mathbb {R}}$ by requiring that $\mu (B \times (0,1])=v(B)$ whenever $B \in \mathcal{B}$ and $B \subseteq (0,1]$, where $v$ is the unique measure on $\left\{ B \in \mathcal{B}: B \subseteq (0,1]\right\}$ that assigns measure $b-a$ to each $(a, b] \subseteq (0,1]$.

Is the hypothesis of finiteness essential in part (b) of Claim 3?

Let $\mu : \mathcal{B}^{k} \rightarrow \overline{\mathbb {R}}$ denote a measure that assigns finite measure to all bounded $\mathcal{B}^{k}$-sets. There does not exist an $F: \mathbb {R}^{k} \rightarrow \mathbb {R}$ such that $F \notin \mathfrak {S}_{k}$ and $\mu =\Delta_{F}$ on $\mathcal{A}_{k}$.

Let $\mu : \mathcal{B}^{k} \rightarrow \overline{\mathbb {R}}$ denote a measure such that $\mu \left(\mathbb {R}^{k}\right)=1$. There is a unique $F \in \mathfrak {S}_{k}$ that simultaneously has the following properties:

1. $\lim_{\boldsymbol {x} \rightarrow -\infty } F(\boldsymbol {x})=0$,

2. $\lim_{\boldsymbol {x} \rightarrow +\infty } F(\boldsymbol {x})=1$, and

3. $\mu$ is the unique measure extension of $\Delta_{F}$ from $\mathcal{A}_{k}$ to $\mathcal{B}^{k}$.

There exists a set $B \subseteq \mathbb {R}$ that fails to be in $\mathcal{M}\left(\Delta_{F}^{*}\right)$ for every $F \in \mathfrak {S}$ that is continuous. Follow this outline.