7.10.E: Problems on Generalized Measures
- Page ID
- 24465
Complete the proofs of Theorems 1,4, and 5.
Do it also for the lemmas and Corollary 3.
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.]
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
\[a=\sup \{|s X| | X \in \mathcal{M}\}.\]
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 \(\left|\overline{e}_{k}\right|=1\).]
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?]
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
\[\left.A=\bigcup_{n=1}^{\infty}\left(n, n+\frac{1}{n^{2}}\right] .\right]\]
Construct complex and vector-valued LS measures \(s_{\alpha}: \mathcal{M}_{\alpha}^{*} \rightarrow E^{n}\left(C^{n}\right)\) in \(E^{1}.\)
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
\[\left.\sum s X_{i}=s U^{+}-s U^{-} \leq 2 \sup |s|<\infty.\right]\]
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
\[\left.2 s^{+} A \geq 2 s U^{+} \geq 2 s^{\prime} A-\varepsilon.\right]\]
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 \(v_{s}(G-F)\).]
(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 \(\mathcal{N}\).]
Do Problem 11 for semirings \(\mathcal{M}\).
[Hint: Use Problem 10(ii).]