7.4.E: Problems on Set Functions
- Page ID
- 24443
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Prove Theorem 2 in detail for semirings.
[Hint: We know that
\[X_{n}-X_{n-1}=\bigcup_{i=1}^{m_{n}} Y_{ni} \text { (disjoint)}\]
for some \(Y_{ni} \in \mathcal{C},\) so
\[\overline{s}\left(X_{n}-X_{n-1}\right)=\sum_{i=1}^{m_{n}} s Y_{ni},\]
with \(\overline{s}\) as in Theorem 1.]
Let \(s\) be additive on \(\mathcal{M},\) a ring. Prove that \(s\) is also \(\sigma\)-additive provided \(s\) is either
(i) left continuous, or
(ii) finite on \(\mathcal{M}\) and right-continuous at \(\emptyset;\) i.e.,
\[\lim _{n \rightarrow \infty} s X_{n}=0\]
when \(X_{n} \searrow \emptyset\) \(\left(X_{n} \in \mathcal{M}\right)\).
[Hint: Let
\[A=\bigcup_{n} A_{n} \text { (disjoint)}, \quad A, A_{n} \in \mathcal{M}.\]
Set
\[X_{n}=\bigcup_{k=1}^{n} A_{k}, Y_{n}=A-X_{n}.\]
Verify that \(X_{n}, Y_{n} \in \mathcal{M}, X_{n} \nearrow A, Y_{n} \searrow \emptyset\).
In case (i),
\[s A=\lim s X_{n}=\sum_{k=1}^{\infty} s A_{k}.\]
(Why?)
For (ii), use the \(Y_{n}\).]
Let
\[\mathcal{M}=\left\{\text {all intervals in the rational field } R \subset E^{1}\right\}.\]
Let
\[s X=b-a\]
if \(a, b\) are the endpoints of \(X \in \mathcal{M}\) \((a, b \in R, a \leq b).\) Prove that
(i) \(\mathcal{M}\) is a semiring;
(ii) \(s\) is continuous;
(iii) \(s\) is additive but not \(\sigma\)-additive; thus Problem 2 fails for semirings.
[Hint: \(R\) is countable. Thus each \(X \in \mathcal{M}\) is a countable union of singletons \(\{x\}=[x, x];\) hence \(s X=0\) if \(s\) were \(\sigma\)-additive.]
Let \(N=\) {naturals}. Let
\[\mathcal{M}=\{\text {all finite subsets of } N \text { and their complements in } N\}.\]
If \(X \in \mathcal{M},\) let \(s X=0\) if \(X\) is finite, and \(s X=\infty\) otherwise. Show that
(i) \(\mathcal{M}\) is a set field;
(ii) \(s\) is right continuous and additive, but not \(\sigma\)-additive.
Thus Problem 2 (ii) fails if \(s\) is not finite.
Let
\[\mathcal{C}=\left\{\text {finite and infinite intervals in } E^{1}\right\}.\]
If \(a, b\) are the endpoints of an interval \(X\) \(\left(a, b \in E^{*}, a<b\right),\) set
\[s X=\left\{\begin{array}{ll}{b-a,} & {a<b,} \\ {0,} & {a=b.}\end{array}\right.\]
Show that \(s\) is \(\sigma\)-additive on \(\mathcal{C},\) a semiring.
Let
\[X_{n}=(n, \infty);\]
so \(s X_{n}=\infty-n=\infty\) and \(X_{n} \searrow \emptyset.\) (Verify!) Yet
\[\lim s X_{n}=\infty \neq s \emptyset.\]
Does this contradict Theorem 2?
Fill in the missing proof details in Theorem 1.
Let \(s\) be additive on \(\mathcal{M}.\) Prove the following.
(i) If \(\mathcal{M}\) is a ring or semiring, so is
\[\mathcal{N}=\{X \in \mathcal{M}| | s X |<\infty\}\]
if \(\mathcal{N} \neq \emptyset\).
(ii) If \(\mathcal{M}\) is generated by a set family \(\mathcal{C},\) with \(|s|<\infty\) on \(\mathcal{C},\) then \(|s|<\infty\) on \(\mathcal{M}.\)
[Hint: Use Problem 16 in §3.]
\(\Rightarrow\) (Lebesgue-Stieltjes set functions.) Let \(\alpha\) and \(s_{\alpha}\) be as in Example (d). Prove the following.
(i) \(s_{\alpha} \geq 0\) on \(\mathcal{C}\) iff \(\alpha \uparrow\) on \(E^{1}\) (see Theorem 2 in Chapter 4, §5).
(ii) \(s_{\alpha}\{p\}=s_{\alpha}[p, p]=0\) iff \(\alpha\) is continuous at \(p\).
(iii) \(s_{\alpha}\) is additive.
[Hint: If
\[A=\bigcup_{i=1}^{n} A_{i} \text { (disjoint),}\]
the intervals \(A_{i-1}, A_{i}\) must be adjacent. For two such intervals, consider all cases like
\[(a, b] \cup(b, c),[a, b) \cup[b, c], \text { etc.}\]
Then use induction on \(n\).]
(iv) If \(\alpha\) is right continuous at \(a\) and \(b,\) then
\[s_{\alpha}(a, b]=\alpha(b)-\alpha(b).\]
If \(\alpha\) is continuous at \(a\) and \(b,\) then
\[s_{\alpha}[a, b]=s_{\alpha}(a, b]=s_{\alpha}[a, b)=s_{\alpha}(a, b).\]
(v) If \(\alpha \uparrow\) on \(E^{1}\), then \(s_{\alpha}\) satisfies Lemma 1 and Corollary 2 in §1 (same proof), as well as Lemma 1, Theorem 1, Corollaries 1-4, and Note 3 in §2 (everything except Corollaries 5 and 6).
[Hint: Use (i) and (iii). For Lemma 1 in §2, take first a half-open \(B=(a, b];\) use the definition of a right-side limit along with Theorems 1 and 2 in Chapter 4, §5, to prove
\[(\forall \varepsilon>0)(\exists c>b) \quad 0 \leq \alpha(c-)-\alpha(b+)<\varepsilon;\]
then set \(C=(a, c).\) Similarly for \(B=[a, b),\) etc. and for the closed interval \(A \subseteq B\).]
(vi) If \(\alpha(x)=x\) then \(s_{\alpha}=v,\) the volume (or length) function in \(E^{1}\).
Construct LS set functions (Example (d)), with \(\alpha \uparrow\) (see Problem 7(v)), so that
(i) \(s_{\alpha}[0,1] \neq s_{\alpha}[1,2]\);
(ii) \(s_{\alpha} E^{1}=1\) (after extending \(s_{\alpha}\) to \(\mathcal{C}_{\sigma}-sets in \(E^{1}\));
(ii') \(s_{\alpha} E^{1}=c\) for a fixed \(c \in(0, \infty)\);
(iii) \(s_{\alpha}\{0\}=1\) and \(s_{\alpha}[0,1]>s_{\alpha}(0,1]\).
Describe \(s_{\alpha}\) if \(\alpha(x)=[x]\) (the integral part of \(x\)).
[Hint: See Figure 16 in Chapter 4, §1.]
For an arbitrary \(\alpha : E^{1} \rightarrow E^{1},\) define \(\sigma_{\alpha} : \mathcal{C} \rightarrow E^{1}\) by
\[\sigma_{\alpha}[a, b]=\sigma_{\alpha}(a, b]=\sigma_{\sigma}[a, b)=\sigma_{\alpha}(a, b)=\alpha(b)-\alpha(a)\]
(the original Stieltjes method). Prove that \(\sigma_{\alpha}\) is additive but not \(\sigma\)-additive unless \(\alpha\) is continuous (for Theorem 2 fails).