Math Problems Vault

Problem 2.1.1

(a)(b)

This problem deals with some other variants of properties (M1)-(M3).

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(a)

Some define a probability measure $P$ on a $\sigma$ -field $\mathcal{A}$ of subsets of $\Omega$ by stipulating that (i) $0 \leqslant P(A) \leqslant 1$ for all $A \in \mathcal{A}$ , (ii) $P(\Omega )=1$ , and (iii) $P$ is countably additive. This is a special case of our definition of a measure.

(b)

If (M1) and (M3) hold for a set function $\mu$ defined on a field $\mathcal{A}$ with $\mu (A)<+\infty$ for some $A \in \mathcal{A}$ , then $\mu$ is a measure on $\mathcal{A}$ .

Problem 2.1.2

Let $\Omega =\left\{ \omega_{1}, \ldots , \omega_{n}\right\}$ , and let $p_{1}, \ldots , p_{n} \in [0,+\infty ]$ . Define $\mu$ on $2^{\Omega }$ as in Example 2. Then ( $\Omega , 2^{\Omega }, \mu$ ) is a measure space, and $\mu$ is $\sigma$ -finite iff $p_{n}<+\infty$ for each $n \in \mathbb {N}$ .

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Problem 2.1.3

Let $\mathcal{A}=\left\{ \varnothing , \Omega \right\} , \mu (\varnothing )=0$ , and $\mu (\Omega )=+\infty$ . Then $(\Omega , \mathcal{A}, \mu )$ is a measure space, but $\mu$ fails to be $\sigma$ -finite.

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Problem 2.1.4

Let $\Omega$ be uncountable. Let $\mathcal{A}=\left\{ A \subseteq \Omega : A \text{ is amc or } A^{c} \text{ is amc}\right\}$ . Write $\mu (A)=0$ if $A$ is amc and $\mu (A)=+\infty$ if $A^{c}$ is amc. Then ( $\Omega , \mathcal{A}, \mu$ ) is a measure space, and $\mu$ is not $\sigma$ -finite.

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Problem 2.1.5

(a)(b)

Let $\Omega$ be arbitrary, and let $\mathcal{A}=\left\{ A \subseteq \Omega : A \text{ is amc or } A^{c} \text{ is amc}\right\}$ . Define $\mu$ over $\mathcal{A}$ by stating that $\mu (A)=0$ if $A$ is amc, and $\mu (A)=1$ if $A^{c}$ is amc.

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(a)

$\mu$ is not well-defined if $\Omega$ is amc, but $\mu$ is well-defined if $\Omega$ is uncountable.

(b)

$\mu$ is $\sigma$ -finite measure on the $\sigma$ -field $A$ when $\Omega$ is uncountable.

Problem 2.1.6

Suppose that $\mathcal{A}$ is a finite $\sigma$ -field on $\Omega$ . Suppose that $\mu$ is defined on $\mathcal{A}$ such that (M1), (M2), and (M4) hold. Then ( $\Omega , \mathcal{A}, \mu$ ) is a measure space.

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Problem 2.1.7

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

Let $\mathcal{A}=\left\{ A \subseteq \Omega : A \text{ is finite or } A^{c} \text{ is finite}\right\}$ . Define $\mu$ on $\mathcal{A}$ by

\mu (A)= \begin{cases} 0 & \text{ if } A \text{ is finite } \\ 1 & \text{ if } A^{c} \text{ is finite } .\end{cases}

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(a)

$\mu$ fails to be well-defined when $\Omega$ is finite.

(b)

If $\Omega$ is infinite, then $\mu$ satisfies (M1), (M2), and (M4).

(c)

Let $\operatorname {card}(\Omega )=\aleph_{0}$ . Then $\mu$ is finitely additive but not countably subadditive.

(d)

When $\Omega$ is uncountable, $\mu$ is a measure. Is $\mu \sigma$ -finite?

Problem 2.1.8

Let $\operatorname {card}(\Omega )=\aleph_{0}$ and $\mathcal{A}=2^{\Omega }$ . Let

\mu (A)= \begin{cases} 0 & \text{ if } A \text{ is finite } \\ +\infty & \text{ if } A \text{ is infinite } .\end{cases}

Then $\mu$ is well-defined, $\mu$ satisfies (M1), (M2), and (M4), and that (M3) fails. Also, $\mu$ is $\sigma$ -finite.

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Problem 2.1.9

(M5) is not true if the hypothesis $\mu (A)<+\infty$ is omitted.

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Problem 2.1.10

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

Let $\mu$ denote a measure on a $\sigma$ -field $\mathcal{A}$ , and let $A, A_{1}, A_{2}, \ldots \in \mathcal{A}$ .

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(a)

$\mu (A)=\sum_{k=1}^{\infty } \mu \left(A \cap A_{k}\right)$ when the $A_{k}$ 's are disjoint with $\bigcup_{k=1}^{\infty } A_{k}=\Omega$ .

(b)

$\mu \left(A_{1} \Delta A_{2}\right)=0$ iff $\mu \left(A_{1}\right)=\mu \left(A_{2}\right)=\mu \left(A_{1} \cap A_{2}\right)$ .

(c)

$\mu \left(A_{2}\right)=0$ forces both $\mu \left(A_{1} \cup A_{2}\right)=\mu \left(A_{1}\right)$ and $\mu \left(A_{1} \Delta A_{2}\right)=0$ .

(d)

$\mu \left(A_{2}\right)=0$ forces $\mu \left(A_{1} \backslash A_{2}\right)=\mu \left(A_{1}\right)$ .

Problem 2.1.11

(a)(b)

Let $(\Omega , \mathcal{A}, \mu )$ be a measure space such that there is $B \in \mathcal{A}$ with $0<\mu (B)<+\infty$ . Fix such a $B$ , and define $\mu_{B}: \mathcal{A} \rightarrow \overline{\mathbb {R}}$ by the formula $\mu_{B}(A)=\mu (A \cap B) / \mu (B)$ .

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(a)

$\left(\Omega , \mathcal{A}, \mu_{B}\right)$ is a measure space.

(b)

Suppose in addition that $\Omega$ is the disjoint union of an amc collection of sets $B_{n} \in \mathcal{A}$ such that each $B_{n}$ has finite measure, and suppose that $\mu$ is finite. Then for all $A \in \mathcal{A}$ we have $\mu (A)=\sum_{n} \mu_{B_{n}}(A) \cdot \mu \left(B_{n}\right)$ . Also, for each $i \in \mathbb {N}$ we have

\mu _{B_{i}}(A)=\frac{\mu _{A}\left(B_{i}\right) \cdot \mu (A)}{\sum _{n} \mu _{A}\left(B_{n}\right) \cdot \mu (A)}

This formula is known as Bayes' Rule.

Problem 2.1.12

(a)(b)(c)

Let $S=\left\{ s_{1}, \ldots , s_{m}\right\}$ , and let $\left\{ p_{u}: u \in S\right\}$ denote a collection of nonnegative numbers with $\sum_{u \in S} p_{u}=1$ . Let $\Omega$ denote the set of sequences of $S$ . For each $\omega \in \Omega$ , write $\omega =\left(z_{1}(\omega ), z_{2}(\omega ), \ldots \right)$ . Given $n \in \mathbb {N}$ and $H \subseteq S^{n}$ , let

C_{n}(H)=\left\{ \omega \in \Omega :\left(z_{1}(\omega ), \ldots , z_{n}(\omega )\right) \in H\right\} .

Such a set is called a cylinder of rank $n$ . Let $\mathcal{F}=\left\{ C_{n}(H): n \in \mathbb {N}, H \subseteq S^{n}\right\}$ , so that $\mathcal{F}$ consists of all cylinders of all ranks. Define $\mu : \mathcal{F} \rightarrow \overline{\mathbb {R}}$ as follows: if for some $n \in \mathbb {N}$ and $H \subseteq S^{n}$ we have $A=\left\{ \omega \in \Omega :\left(z_{1}(\omega ), \ldots , z_{n}(\omega )\right) \in H\right\}$ , write $\mu (A)=\sum \left\{ p_{i_{1}} \cdots p_{i_{n}}:\left(i_{1}, \ldots , i_{n}\right) \in H\right\}$ .

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(a)

$\mu$ is well-defined.

(b)

$\mu$ satisfies (M1) and (M2), and is finitely additive.

(c)

Accept without proof, or recall, a special case of Tychonoff's Theorem: let $A_{1}, A_{2}, \ldots$ denote a sequence of cylinders with $A_{1} \supseteq A_{2} \supseteq \cdots$ . If $\bigcap_{n=1}^{\infty } A_{n}=\varnothing$ , then $A_{N}=\varnothing$ for some $N \in \mathbb {N}$ . Any function $\nu$ on $\mathcal{F}$ that satisfies (M1), (M2), and (M4) also satisfies (M3). Therefore, $\mu$ is a measure, often referred to as product measure.