Section 3.9

A Further Description of M(μ∗)\mathcal{M}\left(\mu^{*}\right)

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Problem 3.9.1

Countable superadditivity: If A1,A2,…⊆ΩA_{1}, A_{2}, \ldots \subseteq \Omega are disjoint, then μ∗(⋃n=1∞An)⩾∑n=1∞μ∗(An)\mu_{*}\left(\bigcup_{n=1}^{\infty } A_{n}\right) \geqslant \sum_{n=1}^{\infty } \mu_{*}\left(A_{n}\right).

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Problem 3.9.2

For any A⊆Ω,μ∗(A)=inf⁡{μM(μ∗)∗(B):A⊆B∈M(μ∗)}=μ∗∗(A)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).

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Problem 3.9.3

For any A⊆Ω,μ∗(A)=sup⁡{μM(μ∗)∗(B):B⊆A,B∈M(μ∗)}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\}.

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Problem 3.9.4

For any A⊆ΩA \subseteq \Omega, there is E∈σ(A)E \in \sigma (\mathcal{A}) such that E⊆AE \subseteq A and μσ(A)∗(E)=μ∗(A)\mu_{\sigma (\mathscr {A})}^{*}(E)=\mu_{*}(A).

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Problem 3.9.5

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

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Problem 3.9.6

Let A⊆ΩA \subseteq \Omega and let {An}\left\{ A_{n}\right\} denote a disjoint sequence of M(μ∗)\mathcal{M}\left(\mu^{*}\right)-sets. Then we have μ∗(⋃n=1∞(A∩An))=∑n=1∞μ∗(An∩A)\mu_{*}\left(\bigcup_{n=1}^{\infty }\left(A \cap A_{n}\right)\right)=\sum_{n=1}^{\infty } \mu_{*}\left(A_{n} \cap A\right).

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Problem 3.9.7

If A,B⊆ΩA, B \subseteq \Omega are disjoint, then μ∗(A∪B)⩽μ∗(A)+μ∗(B)⩽μ∗(A∪B)\mu_{*}(A \cup B) \leqslant \mu_{*}(A)+ \mu^{*}(B) \leqslant \mu^{*}(A \cup B).

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Problem 3.9.8

If A∈M(μ∗)A \in \mathcal{M}\left(\mu^{*}\right) and B⊆ΩB \subseteq \Omega, then μM(μ∗)∗(A)=μ∗(B∩A)+μ∗(Bc∩A)\mu_{\mathcal{M}\left(\mu^{*}\right)}^{*}(A)=\mu_{*}(B \cap A)+\mu^{*}\left(B^{c} \cap A\right).

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Problem 3.9.9

Let Ω=(0,1]×(0,1]\Omega =(0,1] \times (0,1], and let F={B×(0,1]:B∈B,B⊆(0,1]}\mathcal{F}=\left\{ B \times (0,1]: B \in \mathcal{B}, B \subseteq (0,1]\right\}. Define μ:F→R‾\mu : \mathcal{F} \rightarrow \overline{\mathbb {R}} by requiring that μ(B×(0,1])=v(B)\mu (B \times (0,1])=v(B) whenever B∈BB \in \mathcal{B} and B⊆(0,1]B \subseteq (0,1], where vv is the unique measure on {B∈B:B⊆(0,1]}\left\{ B \in \mathcal{B}: B \subseteq (0,1]\right\} that assigns measure b−ab-a to each (a,b]⊆(0,1](a, b] \subseteq (0,1].

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(a)

F\mathcal{F} is a σ\sigma-field.

(b)

μ\mu is well-defined and is a measure on F\mathcal{F}.

(c)

Let C=(0,1]×{12}C=(0,1] \times \left\{ \frac{1}{2}\right\}. Then C∉FC \notin \mathcal{F}, v∗(C)=0v_{*}(C)=0, and v∗(C)=1v^{*}(C)=1.

Problem 3.9.10

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

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Remark.
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Claim 3 (Section 3.9): We have the following: (a) if A∈M(μ∗)A \in \mathcal{M}\left(\mu^{*}\right), then μ∗(A)=μ∗(A)=μ∗∗(A)\mu_{*}(A)=\mu^{*}(A)=\mu^{* *}(A), and (b) if A⊆ΩA \subseteq \Omega is such that μ∗(A)=μ∗(A)=μ∗∗(A)<+∞\mu_{*}(A)=\mu^{*}(A)=\mu^{* *}(A)<+\infty, then A∈M(μ∗)A \in \mathcal{M}\left(\mu^{*}\right).