
# 7.10.E: Problems on Generalized Measures


## Exercise $$\PageIndex{1}$$

Complete the proofs of Theorems 1,4, and 5.

## Exercise $$\PageIndex{1'}$$

Do it also for the lemmas and Corollary 3.

## Exercise $$\PageIndex{2}$$

Verify the following.
(i) In Definition 2, one can equivalently replace "countable $$\left\{X_{i}\right\}$$" by "finite $$\left\{X_{i}\right\}$$."
(ii) If $$\mathcal{M}$$ is a ring, Note 1 holds for finite sequences $$\left\{X_{i}\right\}$$.
(iii) If $$s : \mathcal{M} \rightarrow E$$ is additive on $$\mathcal{M},$$ a semiring, so is $$v_{s}$$.
[Hint: Use Theorem 1 from §4.]

## Exercise $$\PageIndex{3}$$

For any set functions $$s, t$$ on $$\mathcal{M},$$ prove that
(i) $$\quad v_{|s|}=v_{s},$$ and
(ii) $$v_{s t} \leq a v_{t},$$ provided $$s t$$ is defined and
$a=\sup \{|s X| | X \in \mathcal{M}\}.$

## Exercise $$\PageIndex{4}$$

Given $$s, t : \mathcal{M} \rightarrow E,$$ show that
(i) $$v_{s+t} \leq v_{s}+v_{t}$$;
(ii) $$v_{k s}=|k| v_{s}$$ ((\k\) as in Corollary 2); and
(iii) if $$E=E^{n}\left(C^{n}\right)$$ and
$s=\sum_{k=1}^{n} s_{k} \overline{e}_{k},$
then
$v_{s_{k}} \leq v_{s} \leq \sum_{k=1}^{n} v_{s k}.$
[Hints: (i) If
$A \supseteq \bigcup X_{i} \text { (disjoint),}$
with $$A_{i}, X_{i} \in \mathcal{M},$$ verify that
$\begin{array}{c}{\left|(s+t) X_{i}\right| \leq\left|s X_{i}\right|+\left|t X_{i}\right|,} \\ {\sum\left|(s+t) X_{i}\right| \leq v_{s} A+v_{t} A, \text { etc.;}}\end{array}$
(ii) is analogous.
(iii) Use (ii) and (i), with $$\left|\overline{e}_{k}\right|=1$$.]

## Exercise $$\PageIndex{5}$$

If $$g \uparrow, h \uparrow,$$ and $$\alpha=g-h$$ on $$E^{1}$$, can one define the signed LS measure $$s_{\alpha}$$ by simply setting $$s_{\alpha}=m_{g}-m_{h}$$ (assuming $$m_{h}<\infty$$)?
[Hint: the domains of $$m_{g}$$ and $$m_{h}$$ may be different. Give an example. How about taking their intersection?]

## Exercise $$\PageIndex{6}$$

Find an LS measure $$m_{\alpha}$$ such that $$\alpha$$ is continuous and one-to-one, but $$m_{\alpha}$$ is not $$m$$-finite ($$m=$$Lebesgue measure).
[Hint: Take
$\alpha(x)=\left\{\begin{array}{ll}{\frac{x^{3}}{|x|},} & {x \neq 0,} \\ {0,} & {x=0,}\end{array}\right.$
and

$\left.A=\bigcup_{n=1}^{\infty}\left(n, n+\frac{1}{n^{2}}\right] .\right]$

## Exercise $$\PageIndex{7}$$

Construct complex and vector-valued LS measures $$s_{\alpha}: \mathcal{M}_{\alpha}^{*} \rightarrow E^{n}\left(C^{n}\right)$$ in $$E^{1}.$$

## Exercise $$\PageIndex{8}$$

Show that if $$s : \mathcal{M} \rightarrow E^{n}\left(C^{n}\right)$$ is additive and bounded on $$\mathcal{M},$$ a ring, so is $$v_{s}$$.
[Hint: By Problem 4(iii), reduce all to the real case.
Use Problem 2. Given a finite disjoint sequence $$\left\{X_{i}\right\} \subseteq \mathcal{M},$$ let $$U^{+}\left(U^{-}\right)$$ be the union of those $$X_{i}$$ for which $$s X_{i} \geq 0 (s X_{i}<0,$$ respectively). Show that
$\left.\sum s X_{i}=s U^{+}-s U^{-} \leq 2 \sup |s|<\infty.\right]$

## Exercise $$\PageIndex{9}$$

For any $$s : \mathcal{M} \rightarrow E^{*}$$ and $$A \in \mathcal{M},$$ set
$s^{+} A=\sup \{s X | A \supseteq X \in \mathcal{M}\}$
and
$s^{-} A=\sup \{-s X | A \supseteq X \in \mathcal{M}\}.$
Prove that if $$s$$ is additive and bounded on $$\mathcal{M},$$ a ring, so are $$s^{+}$$ and $$s^{-};$$ furthermore,
\begin{aligned} s^{+} &=\frac{1}{2}\left(v_{s}+s\right) \geq 0, \\ s^{-} &=\frac{1}{2}\left(v_{s}-s\right) \geq 0, \\ s &=s^{+}-s^{-}, \text{ and } \\ v_{s} &=s^{+}+s^{-}. \end{aligned}
[Hints: Use Problem 8. Set
$s^{\prime}=\frac{1}{2}\left(v_{s}+s\right).$
Then $$(\forall X \in \mathcal{M} | X \subseteq A)$$
\begin{aligned} 2 s X=s A+s X-s(A-X) & \leq s A+(|s X|+|s(A-X)|) \\ & \leq s A+v_{s} A=2 s^{\prime} A. \end{aligned}
Deduce that $$s^{+} A \leq s^{\prime} A$$.
To prove also that $$s^{\prime} A \leq s^{+} A,$$ let $$\varepsilon>0.$$ By Problems 2 and 8, fix $$\left\{X_{i}\right\} \subseteq \mathcal{M}$$, with
$A=\bigcup_{i=1}^{n} X_{i} \text { (disjoint)}$
and
$v_{s} A-\varepsilon<\sum_{i=1}^{n}\left|s X_{i}\right|.$
Show that
$2 s^{\prime} A-\varepsilon=v_{s} A+s A-\varepsilon \leq s U^{+}-s U^{-}+s \bigcup_{i=1}^{n} X_{i}=2 s U^{+}$
and
$\left.2 s^{+} A \geq 2 s U^{+} \geq 2 s^{\prime} A-\varepsilon.\right]$

## Exercise $$\PageIndex{10}$$

Let
$\mathcal{K}=\{\text {compact sets in a topological space }(S, \mathcal{G})\}$
(adopt Theorem 2 in Chapter 4, §7, as a definition). Given
$s : \mathcal{M} \rightarrow E, \quad \mathcal{M} \subseteq 2^{S},$
we call $$s$$ compact regular (CR) iff
\begin{aligned}(\forall \varepsilon>0) &(\forall A \in \mathcal{M})(\exists F \in \mathcal{K})(\exists G \in \mathcal{G}) \\ F, G & \in \mathcal{M}, F \subseteq A \subseteq G, \text { and } v_{s} G-\varepsilon \leq v_{s} A \leq v_{s} F+\varepsilon. \end{aligned}
Prove the following.
(i) If $$s, t : \mathcal{M} \rightarrow E$$ are $$\mathrm{CR},$$ so are $$s \pm t$$ and $$k s$$ ($$k$$ as in Corollary 2).
(ii) If $$s$$ is additive and CR on $$\mathcal{M},$$ a semiring, so is its extension to the ring $$\mathcal{M}_{s}$$ (Theorem 1 in §4 and Theorem 4 of §3).
(iii) If $$E=E^{n}\left(C^{n}\right)$$ and $$v_{s}<\infty$$ on $$\mathcal{M},$$ a ring, then $$s$$ is CR iff its components $$s_{k}$$ are, or in the case $$E=E^{1},$$ iff $$s^{+}$$ and $$s^{-}$$ are (see Problem 9).
[Hint for (iii): Use (i) and Problem 4(iii). Consider $$v_{s}(G-F)$$.]

## Exercise $$\PageIndex{11}$$

(Aleksandrov.) Show that if $$s : \mathcal{M} \rightarrow E$$ is CR (see Problem 10) and additive on $$\mathcal{M},$$ a ring in a topological space $$S,$$ and if $$v_{s}<\infty$$ on $$\mathcal{M}$$, then $$v_{s}$$ and $$s$$ are $$\sigma$$-additive, and $$v_{s}$$ has a unique $$\sigma$$-additive extension $$\overline{v}_{s}$$ to the $$\sigma$$-ring $$\mathcal{N}$$ generated by $$\mathcal{M}.$$
The latter holds for $$s,$$ too, if $$S \in \mathcal{M}$$ and $$E=E^{n}\left(C^{n}\right)$$.
[Proof outline: The $$\sigma$$-additivity of $$v_{s}$$ results as in Theorem 1 of §2 (first check Lemma 1 in §1 for $$v_{s}$$).
For the $$\sigma$$-additivity of $$s,$$ let
$A=\bigcup_{i=1}^{\infty} A_{i} \text { (disjoint)}, \quad A, A_{i} \in \mathcal{M};$
then
$\left|s A-\sum_{i=1}^{r-1} s A_{i}\right| \leq \sum_{i=r}^{\infty} v_{s} A_{i} \rightarrow 0$
as $$r \rightarrow \infty,$$ for
$\sum_{i=1}^{\infty} v_{s} A_{i}=v_{s} \bigcup_{i=1}^{\infty} A_{i}<\infty.$
(Explain!) Now, Theorem 2 of §6 extends $$v_{s}$$ to a measure on a $$\sigma$$-field
$\mathcal{M}^{*} \supseteq \mathcal{N} \supseteq \mathcal{M}$
(use the minimality of $$\mathcal{N}$$). Its restriction to $$\mathcal{N}$$ is the desired $$\overline{v}_{s}$$ (unique by Problem 15 in §6).
A similar proof holds for $$s,$$ too, if $$s : \mathcal{M} \rightarrow[0, \infty).$$ The case $$s : \mathcal{M} \rightarrow E^{n}\left(C^{n}\right)$$ results via Theorem 5 and Problem 10(iii) provided $$S \in \mathcal{M};$$ for then by Corollary 1, $$v_{s} S<\infty$$ ensures the finiteness of $$v_{s}, s^{+},$$ and $$s^{-}$$ even on $$\mathcal{N}$$.]

## Exercise $$\PageIndex{12}$$

Do Problem 11 for semirings $$\mathcal{M}$$.
[Hint: Use Problem 10(ii).]

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