Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

7.5.E: Problems on Premeasures and Related Topics

  • Page ID
    24444
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    Exercise \(\PageIndex{1}\)

    Fill in the missing details in the proofs, notes, and examples of this section.

    Exercise \(\PageIndex{2}\)

    Describe \(m^{*}\) on \(2^{S}\) induced by a premeasure \(\mu : \mathcal{C} \rightarrow E^{*}\) such that each of the following hold.
    (a) \(\mathcal{C}=\{S, \emptyset\}, \mu S=1\).
    (b) \(\mathcal{C}=\{S, \emptyset, \text {and all singletons}\}; \mu S=\infty, \mu\{x\}=1\).
    (c) \(\mathcal{C}\) as in (b), with \(S\) uncountable; \(\mu S=1,\) and \(\mu X=0\) otherwise.
    (d) \(\mathcal{C}=\{\text {all proper subsets of } S\}; \mu X=1\) when \(\emptyset \subset X \subset S; \mu \emptyset=0\).

    Exercise \(\PageIndex{3}\)

    Show that the premeasures
    \[v^{\prime} : \mathcal{C}^{\prime} \rightarrow[0, \infty]\]
    induce one and the same (Lebesgue) outer measure \(m^{*}\) in \(E^{n},\) with \(v^{\prime}=v\) (volume, as in §2):
    (a) \(\mathcal{C}^{\prime}=\{\text {open intervals}\}\);
    (b) \(\mathcal{C}^{\prime}=\{\text {half-open intervals}\}\);
    (c) \(\mathcal{C}^{\prime}=\{\text {closed intervals}\}\);
    (d) \(\mathcal{C}^{\prime}=\mathcal{C}_{\sigma}\);
    (e) \(\mathcal{C}^{\prime}=\{\text {open sets}\}\);
    (f) \(\mathcal{C}^{\prime}=\{\text {half-open cubes}\}\);
    [Hints: (a) Let \(m^{\prime}\) be the \(v^{\prime}\)-induced outer measure; let \(\mathcal{C}=\{\text {all intervals}\}.\) As \(\mathcal{C}^{\prime} \subseteq \mathcal{C}, m^{\prime} A \geq m^{*} A.\) (Why?) Also,
    \[(\forall \varepsilon>0)\left(\exists\left\{B_{k}\right\} \subseteq \mathcal{C}\right) \quad A \subseteq \bigcup_{k} B_{k} \text { and } \sum v B_{k} \leq m^{*} A+\varepsilon.\]
    (Why?) By Lemma 1 in §2,
    \[\left(\exists\left\{C_{k}\right\} \subseteq \mathcal{C}^{\prime}\right) \quad B_{k} \subseteq C_{k} \text { and } v B_{k}+\frac{\varepsilon}{2^{k}}>v^{\prime} C_{k}.\]
    Deduce that \(m^{*} A \geq m^{\prime} A, m^{*}=m^{\prime}\). Similarly for (b) and (c). For (d), use Corollary 1 and Note 3 in §1. For (e), use Lemma 2 in §2. For (f), use Problem 2 in §2.]

    Exercise \(\PageIndex{3'}\)

    Do Problem 3(a)-(c), with \(m^{*}\) replaced by the Jordan outer content \(c^{*}\) (Note 6).

    Exercise \(\PageIndex{4}\)

    Do Problem 3, with \(v\) and \(m^{*}\) replaced by the LS premeasure and outer measure. (Use Problem 7 in §4.)

    Exercise \(\PageIndex{5}\)

    Show that a set \(A \subseteq E^{n}\) is bounded iff its outer Jordan content is finite.

    Exercise \(\PageIndex{6}\)

    Find a set \(A \subseteq E^{1}\) such that
    (i) its Lebesgue outer measure is \(0\) \(\left(m^{*} A=0\right),\) while its Jordan outer content \(c^{*} A=\infty\);
    (ii) \(m^{*} A=0, c^{*} A=1\) (see Corollary 6 in §2).

    Exercise \(\PageIndex{7}\)

    Let
    \[\mu_{1}, \mu_{2} : \mathcal{C} \rightarrow[0, \infty]\]
    be two premeasures in \(S\) and let \(m_{1}^{*}\) and \(m_{2}^{*}\) be the outer measures induced by them.
    Prove that if \(m_{1}^{*}=m_{2}^{*}\) on \(\mathcal{C},\) then \(m_{1}^{*}=m_{2}^{*}\) on all of \(2^{S}\).

    Exercise \(\PageIndex{8}\)

    With the notation of Definition 3 and Note 6, prove the following.
    (i) If \(A \subseteq B \subseteq S\) and \(m^{*} B=0,\) then \(m^{*} A=0;\) similarly for \(c^{*}\).
    [Hint: Use monotonicity.]
    (ii) The set family
    \[\left\{X \subseteq S | c^{*} A=0\right\}\]
    is a hereditary set ring, i.e., a ring \(\mathcal{R}\) such that
    \[(\forall B \in \mathcal{R})(\forall A \subseteq B) \quad A \in \mathcal{R}.\]
    (iii) The set family
    \[\left\{X \subseteq S | m^{*} X=0\right\}\]
    is a hereditary \(\sigma\)-ring.
    (iv) So also is
    \[\mathcal{H}=\{\text {those } X \subseteq S \text { that have basic coverings}\};\]
    thus \(\mathcal{H}\) is the hereditary \(\sigma\)-ring generated by \(\mathcal{C}\) (see Problem 14 in §3).

    Exercise \(\PageIndex{9}\)

    Continuing Problem 8(iv), prove that if \(\mu\) is \(\sigma\)-finite (Definition 4), so is \(m^{*}\) when restricted to \(\mathcal{H}.\)
    Show, moreover, that if \(\mathcal{C}\) is a semiring, then each \(X \in \mathcal{H}\) has a basic covering \(\left\{Y_{n}\right\},\) with \(m^{*} Y_{n}<\infty\) and with all \(Y_{n}\) disjoint.
    [Hint: Show that
    \[X \subseteq \bigcup_{n=1}^{\infty} \bigcup_{k=1}^{\infty} B_{n k}\]
    for some sets \(B_{n k} \in \mathcal{C},\) with \(\mu B_{n k}<\infty.\) Then use Note 4 in §5 and Corollary 1 of §1.]

    Exercise \(\PageIndex{10}\)

    Show that if
    \[s : \mathcal{C} \rightarrow E^{*}\]
    is \(\sigma\)-finite and additive on \(\mathcal{C},\) a semiring, then the \(\sigma\)-ring \(\mathcal{R}\) generated by \(\mathcal{C}\) equals the \(\sigma\)-ring \(\mathcal{R}^{\prime}\) generated by
    \[\mathcal{C}^{\prime}=\{X \in \mathcal{C}| | s X |<\infty\}\]
    (cf. Problem 6 in §4).
    [Hint: By \(\sigma\)-finiteness,
    \[(\forall X \in \mathcal{C})\left(\exists\left\{A_{n}\right\} \subseteq \mathcal{C}| | s A_{n} |<\infty\right) \quad X \subseteq \bigcup_{n} A_{n};\]
    so
    \[X=\bigcup_{n}\left(X \cap A_{n}\right), \quad X \cap A_{n} \in \mathcal{C}^{\prime}.\]
    (Use Lemma 3 in §4.)
    Thus \((\forall X \in \mathcal{C}) X\) is a countable union of \(\mathcal{C}^{\prime}\)-sets; so \(\mathcal{C} \subseteq \mathcal{R}^{\prime}.\) Deduce \(\mathcal{R} \subseteq \mathcal{R}^{\prime}\). Proceed.]

    Exercise \(\PageIndex{11}\)

    With all as in Theorem 3, prove that if \(A\) has basic coverings, then
    \[\left(\exists B \in \mathcal{A}_{\delta}\right) \quad A \subseteq B \text { and } m^{*} A=m^{*} B.\]
    [Hint: By formula (4),
    \[(\forall n \in N)\left(\exists X_{n} \in \mathcal{A} | A \subseteq X_{n}\right) \quad m^{*} A \leq m X_{n} \leq m^{*} A+\frac{1}{n}.\]
    (Explain!) Set
    \[B=\bigcap_{n=1}^{\infty} X_{n} \in \mathcal{A}_{\delta}.\]
    Proceed. For \(\mathcal{A}_{\delta},\) see Definition 2(b) in §3.]

    Exercise \(\PageIndex{12}\)

    Let \((S, \mathcal{C}, \mu)\) and \(m^{*}\) be as in Definition 3. Show that if \(\mathcal{C}\) is a \(\sigma\)-field in \(S,\) then
    \[(\forall A \subseteq S)(\exists B \in \mathcal{C}) \quad A \subseteq B \text { and } m^{*} A=\mu B.\]
    [Hint: Use Problem 11 and Note 3.]

    Exercise \(\PageIndex{13}\)

    \(\Rightarrow^{*}\) Show that if
    \[s : \mathcal{C} \rightarrow E\]
    is \(\sigma\)-finite and \(\sigma\)-additive on \(\mathcal{C},\) a semiring, then \(s\) has at most one \(\sigma\)-additive extension to the \(\sigma\)-ring \(\mathcal{R}\) generated by \(\mathcal{C}.\)
    (Note that \(s\) is automatically \(\sigma\)-finite if it is finite, e.g., complex or vector valued.)
    [Outline: Let
    \[s^{\prime}, s^{\prime \prime} : \mathcal{R} \rightarrow E\]
    be two \(\sigma\)-additive extensions of \(s.\) By Problem 10, \(\mathcal{R}\) is also generated by
    \[\mathcal{C}^{\prime}=\{X \in \mathcal{C}| | s X |<\infty\}.\]
    Now set
    \[\mathcal{R}^{*}=\left\{X \in \mathcal{R} | s^{\prime} X=s^{\prime \prime} X\right\}.\]
    Show that \(\mathcal{R}^{*}\) satisfies properties (i)-(iii) of Theorem 3 in §3, with \(\mathcal{C}\) replaced by \(\mathcal{C}^{\prime};\) so \(\mathcal{R}=\mathcal{R}^{*}\).]

    Exercise \(\PageIndex{14}\)

    Let \(m_{n}^{*}(n=1,2, \ldots)\) be outer measures in \(S\) such that
    \[(\forall X \subseteq S)(\forall n) \quad m_{n}^{*} X \leq m_{n+1}^{*} X.\]
    Set
    \[\mu^{*}=\lim _{n \rightarrow \infty} m_{n}^{*}.\]
    Show that \(\mu^{*}\) is an outer measure in \(S\) (see Note 5).

    Exercise \(\PageIndex{15}\)

    An outer measure \(m^{*}\) in a metric space \((S, \rho)\) is said to have the Carathéodory property (CP) iff
    \[m^{*}(X \cup Y) \geq m^{*} X+m^{*} Y\]
    whenever \(\rho(X, Y)>0,\) where
    \[\rho(X, Y)=\inf \{\rho(x, y) | x \in X, y \in Y\}.\]
    For such \(m^{*},\) prove that
    \[m^{*}\left(\bigcup_{k} X_{k}\right)=\sum_{k} m^{*} X_{k}\]
    if \(\left\{X_{k}\right\} \subseteq 2^{S}\) and
    \[\rho\left(X_{i}, X_{k}\right)>0 \quad(i \neq k).\]
    [Hint: For finite unions, use the CP, subadditivity, and induction. Deduce that
    \[(\forall n) \sum_{k=1}^{n} m^{*} X_{k} \leq m^{*} \bigcup_{k=1}^{\infty} X_{k}.\]
    Let \(n \rightarrow \infty.\) Proceed.]

    Exercise \(\PageIndex{16}\)

    Let \((S, \mathcal{C}, \mu)\) and \(m^{*}\) be as in Definition 3, with \(\rho\) a metric for \(S.\) Let \(\mu_{n}\) be the restriction of \(\mu\) to the family \(\mathcal{C}_{n}\) of all \(X \in \mathcal{C}\) of diameter
    \[d X \leq \frac{1}{n}.\]
    Let \(m_{n}^{*}\) be the \(\mu_{n}\)-induced outer measure in \(S.\)
    Prove that
    (i) \(\left\{m_{n}^{*}\right\} \uparrow\) as in Problem 14;
    (ii) the outer measure
    \[\mu^{*}=\lim _{n \rightarrow \infty} m_{n}^{*}\]
    has the CP (see Problem 15), and
    \[\mu^{*} \geq m^{*} \text { on } 2^{S}.\]
    [Outline: Let \(\rho(X, Y)>\varepsilon>0(X, Y \subseteq S)\).
    If for some \(n, X \cup Y\) has no basic covering from \(\mathcal{C}_{n},\) then
    \[\mu^{*}(X \cup Y) \geq m_{n}^{*}(X \cup Y)=\infty \geq \mu^{*} X+\mu^{*} Y,\]
    and the CP follows. (Explain!)
    Thus assume
    \[\left(\forall n>\frac{1}{\varepsilon}\right)(\forall k)\left(\exists B_{n k} \in \mathcal{C}_{n}\right) \quad X \cup Y \subseteq \bigcup_{k=1}^{\infty} B_{n k}.\]
    One can choose the \(B_{n k}\) so that
    \[\sum_{k=1}^{\infty} \mu B_{n k} \leq m_{n}^{*}(X \cup Y)+\varepsilon.\]
    (Why?) As
    \[d B_{n k} \leq \frac{1}{n}<\varepsilon,\]
    some \(B_{n k}\) cover \(X\) only, others \(Y\) only. (Why?) Deduce that
    \[\left(\forall n>\frac{1}{\varepsilon}\right) \quad m_{n}^{*} X+m_{n}^{*} Y \leq \sum_{k=1}^{\infty} \mu_{n} B_{n k} \leq m_{n}^{*}(X \cup Y)+\varepsilon.\]
    Let \(\varepsilon \rightarrow 0\) and then \(n \rightarrow \infty\).
    Also, \(m^{*} \leq m_{n}^{*} \leq \mu^{*}.\) (Why?)]

    Exercise \(\PageIndex{17}\)

    Continuing Problem 16, suppose that
    \((\forall \varepsilon>0)(\forall n, k)(\forall B \in \mathcal{C})\left(\exists B_{n k} \in \mathcal{C}_{n}\right)\)
    \[B \subseteq \bigcup_{k=1}^{\infty} B_{n k} \text { and } \mu B+\varepsilon \geq \sum_{k=1}^{\infty} \mu B_{n k}.\]
    Show that
    \[m^{*}=\lim _{n \rightarrow \infty} \mu_{n}^{*}=\mu^{*},\]
    so \(m^{*}\) itself has the CP.
    [Hints: It suffices to prove that \(m^{*} A \geq \mu^{*} A\) if \(m^{*} A<\infty.\) (Why?)
    Now, given \(\varepsilon>0, A\) has a covering
    \[\left\{B_{i}\right\} \subseteq c\]
    such that
    \[m^{*} A+\varepsilon \geq \sum \mu B_{i}.\]
    (Why?) By assumption,
    \[(\forall n) \quad B_{i} \subseteq \bigcup_{k=1}^{\infty} B_{n k}^{i} \in \mathcal{C}_{n} \text { and } \mu B_{i}+\frac{\varepsilon}{2^{i}} \geq \sum_{k=1}^{\infty} \mu B_{n k}^{i}.\]
    Deduce that
    \[m^{*} A+\varepsilon>\sum \mu B_{i} \geq \sum_{i=1}^{\infty}\left(\sum_{k=1}^{\infty} \mu B_{n k}^{i}-\frac{\varepsilon}{2^{i}}\right)=\sum_{i, k} \mu B_{n k}^{i}-\varepsilon \geq m_{n}^{*} A-\varepsilon.\]
    Let \(\varepsilon \rightarrow 0;\) then \(n \rightarrow \infty\).]

    Exercise \(\PageIndex{18}\)

    Using Problem 17, show that the Lebesgue and Lebesgue-Stieltjes outer measures have the CP.

    • Was this article helpful?