Math Problems Vault

Problem 3.5.1

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 $\sigma (\mathscr {P})$ .

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Problem 3.5.2

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 $\mu_{1}=\mu_{2}$ .

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Problem 3.5.3

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.

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(a)

$\mathcal{A}$ is neither a $\pi$ -system nor a semiring, and $\sigma (\mathcal{A})=2^{\Omega }$ .

(b)

$\mu$ is a measure.

(c)

Define two new distinct measures $v$ and $\xi$ on the $\sigma$ -field $2^{\Omega }$ by the following: $v\left(\left\{ \omega_{1}\right\} \right)=v\left(\left\{ \omega_{4}\right\} \right)=1, v\left(\left\{ \omega_{2}\right\} \right)=v\left(\left\{ \omega_{3}\right\} \right)=2, \xi \left(\left\{ \omega_{2}\right\} \right)=\xi \left(\left\{ \omega_{3}\right\} \right)=1$ , and $\xi \left(\left\{ \omega_{1}\right\} \right)=\xi \left(\left\{ \omega_{4}\right\} \right)=2$ . Then $v$ and $\xi$ are distinct measure extensions of $\mu$ from $\mathcal{A}$ to $\sigma (\mathcal{A})=2^{\Omega }$ .

(d)

Let $\mu^{*}: 2^{\Omega } \rightarrow \mathbb {R}$ denote the outer measure induced by $\mu$ . Then $\mu^{*}=\mu$ on $\mathcal{A}$ and $\mu^{*}$ is a measure.

(e)

$\mu^{*}, v$ , and $\xi$ are distinct measures.

Problem 3.5.4

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 $\varnothing \in \mathcal{A}$ .

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(a)

$\mathcal{A}$ is a semiring and $\sigma (\mathcal{A})=2^{\Omega }$ .

(b)

Define $\mu : \mathcal{A} \rightarrow \overline{\mathbb {R}}$ by stipulating that $\mu (A)$ equals the number of points in $A$ if $A$ is finite and $\mu (A)=+\infty$ if $A$ is infinite. Then $\mu (A)=+\infty$ for all $A \in \mathcal{A}-\left\{ \varnothing \right\}$ , hence $\mu$ fails to be $\sigma$ -finite on $\mathcal{A}$ .

(c)

Define $v: 2^{\Omega } \rightarrow \overline{\mathbb {R}}$ by stating that $v(A)$ equals the number of points in $A$ if $A$ is finite and $v(A)=+\infty$ if $A$ is infinite. Next, define $\xi$ by writing $\xi (A)=2 v(A)$ for all $A \subseteq \Omega$ . Then $v$ and $\xi$ are distinct $\sigma$ -finite measure extensions of $\mu$ from $\mathcal{A}$ to $\sigma (\mathcal{A})$ . Therefore, when $\mu$ fails to be $\sigma$ -finite on $\mathcal{A}$ , there may be more than one measure extension of $\mu$ from $\mathcal{A}$ to $\sigma (\mathcal{A})$ .

(d)

The outer measure $\mu^{*}: 2^{\Omega } \rightarrow \overline{\mathbb {R}}$ induced by $\mu$ is such that $\mu^{*}(A)=+\infty$ for any nonempty $A \subseteq \Omega$ , and $\mu^{*}$ is a non- $\sigma$ -finite measure. Therefore, while we have two distinct $\sigma$ -finite measure extensions of $\mu$ to $\sigma (\mathcal{A})$ , we also have a non- $\sigma$ -finite extension of $\mu$ to $\sigma (\mathcal{A})$ as well.

(e)

$\mathcal{M}\left(\mu^{*}\right)=2^{\Omega }=\sigma (\mathcal{A})$ .

Problem 3.5.5

(a)(b)(c)

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.

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(a)

If $B \in \sigma (\mathcal{F})$ , then $\nu (B) \leqslant \mu^{*}(B)$ .

(b)

If $F \in \sigma (\mathcal{F})$ and $\mu^{*}(F)<+\infty$ , then $\nu (F)=\mu^{*}(F)$ .

(c)

If $\mu$ is $\sigma$ -finite on $\mathcal{F}$ , then $\nu (E)=\mu^{*}(E)$ for all $E \in \sigma (\mathcal{F})$ . This gives the uniqueness of the measure extension whose existence is guaranteed by Exercise 3.4.6.

Problem 3.5.6

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.

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(a)

$\mathscr {D}$ is a $\pi$ -system.

(b)

U is a $\pi$ -system.

(c)

$U$ is closed under complementation.

(d)

$U$ coincides with the minimal field containing the lattice $\mathscr {L}$ .

(e)

Let $\mathcal{A}$ denote a $\sigma$ -field on $\Omega$ that contains $\mathscr {L}$ . Suppose that $\mu$ and $v$ are measures with domain $\mathcal{A}$ such that $\mu =\nu$ on $\mathscr {L}$ . Furthermore, suppose that $\Omega =\cup A_{n}$ , where $A_{n} \in \mathscr {L}$ and $\mu \left(A_{n}\right)<+\infty$ for each $n \in \mathbb {N}$ . Then $\mu =v$ on $\sigma (\mathscr {L})$ .

Problem 3.5.7

(a)(b)(c)(d)

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 $\mathbb {R}^{k}$ .

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(a)

Show that $\mu =v$ by appealing to the Uniqueness Theorem.

(b)

Show that $\mu =v$ by alternatively appealing to Exercise 3.5.6.

(c)

Repeat parts (a) and (b) by replacing the word open with closed.

(d)

Repeat parts (a) and (b) by replacing the word open with compact.

Remark.

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Part (a) is the important part of the exercise.