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Mathematics LibreTexts

7.7: Topologies. Borel Sets. Borel Measures

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    21648
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    I. Our theory of set families leads quite naturally to a generalization of metric spaces. As we know, in any such space \((S, \rho),\) there is a family \(\mathcal{G}\) of open sets, and a family \(\mathcal{F}\) of all closed sets. In Chapter 3, §12, we derived the following two properties.

    (i) \(\mathcal{G}\) is closed under any (even uncountable) unions and under finite intersections (Chapter 3, §12, Theorem 2). Moreover,

    \[\emptyset \in \mathcal{G} \text { and } S \in \mathcal{G},\]

    (ii) \(\mathcal{F}\) has these properties, with "unions" and "intersections" interchanged (Chapter 3, §12, Theorem 3). Moreover, by definition,

    \[A \in \mathcal{F} \text { iff }-A \in \mathcal{G}.\]

    Now, quite often, it is not so important to have distances (i.e., a metric) defined in \(S,\) but rather to single out two set families, \(\mathcal{G}\) and \(\mathcal{F},\) with properties (i) and (ii), in a suitable manner. For examples, see Problems 1 to 4 below. Once \(\mathcal{G}\) and \(\mathcal{F}\) are given, one does not need a metric to define such notions as continuity, limits, etc. (See Problems 2 and 3.) This leads us to the following definition.

    Definition 1

    A topology for a set \(S\) is any set family \(\mathcal{G} \subseteq 2^{S},\) with properties (i).

    The pair \((S, \mathcal{G})\) then is called a topological space. If confusion is unlikely, we simply write \(S\) for \((S, \mathcal{G}).\)

    \(\mathcal{G}\)-sets are called open sets; their complements form the family \(\mathcal{F}\) (called cotopology) of all closed sets in \(S; \mathcal{F}\) satisfies (ii) (the proof is as in Theorem 3 of Chapter 3, §12).

    Any metric space may be treated as a topological one (with \(\mathcal{G}\) defined as in Chapter 3, §12), but the converse is not true. Thus \((S, \mathcal{G})\) is more general.

    Note 1. By Problem 15 in Chapter 4, §2, a map

    \[f :(S, \rho) \rightarrow\left(T, \rho^{\prime}\right)\]

    is continuous iff \(f^{-1}[B]\) is open in \(S\) whenever \(B\) is open in \(T\).

    We adopt this as a definition, for topological spaces \(S, T\).

    Many other notions (neighborhoods, limits, etc.) carry over from metric spaces by simply treating \(G_{p}\) as "an open set containing \(p.\)" (See Problem 3.)

    Note 2. By (i), \(\mathcal{G}\) is surely closed under countable unions. Thus by Note 2 in §3,

    \[\mathcal{G}=\mathcal{G}_{\sigma}.\]

    Also, \(\mathcal{G}=\mathcal{G}_{d}\) and

    \[\mathcal{F}_{\delta}=\mathcal{F}=\mathcal{F}_{s},\]

    but not

    \[\mathcal{G}=\mathcal{G}_{\delta} \text { or } \mathcal{F}=\mathcal{F}_{\sigma}\]

    in general.

    \(\mathcal{G}\) and \(\mathcal{F}\) need not be rings or \(\sigma\)-rings (closure fails for differences). But by Theorem 2 in §3, \(\mathcal{G}\) and \(\mathcal{F}\) can be "embedded" in a smallest \(\sigma\)-ring. We name it in the following definition.

    Definition 2

    The \(\sigma\)-ring \(\mathcal{B}\) generated by a topology \(\mathcal{G}\) in \(S\) is called the Borel field in \(S.\) (It is a \(\sigma\)-field, as \(S \in \mathcal{G} \subseteq \mathcal{B}.)\)

    Equivalently, \(\mathcal{B}\) is the least \(\sigma\)-ring \(\supseteq \mathcal{F}.\) (Why?)

    \(\mathcal{B}\)-sets are called Borel sets in \((S, \mathcal{G})\).

    As \(\mathcal{B}\) is closed under countable unions and intersections, we have not only

    \[\mathcal{B} \supseteq \mathcal{G} \text { and } \mathcal{B} \supseteq \mathcal{F},\]

    but also

    \[\mathcal{B} \supseteq \mathcal{G}_{\delta}, \mathcal{B} \supseteq \mathcal{F}_{\sigma}, \mathcal{B} \supseteq \mathcal{G}_{\delta \sigma}\left[\text { i.e. },\left(\mathcal{G}_{\delta}\right)_{\sigma}\right], \mathcal{B} \supseteq \mathcal{F}_{\sigma \delta}, \text { etc.}\]

    Note that

    \[\mathcal{G}_{\delta \delta}=\mathcal{G}_{\delta}, \mathcal{F}_{\sigma \sigma}=\mathcal{F}_{\sigma}, \text { etc. (Why?)}\]

    II. Special notions apply to measures in metric and topological spaces.

    Definition 3

    A measure \(m : \mathcal{M} \rightarrow E^{*}\) in \((S, \mathcal{G})\) is called topological iff \(\mathcal{G} \subseteq \mathcal{M},\) i.e., all open sets are measurable; \(m\) is a Borel measure iff \(\mathcal{M}=\mathcal{B}\).

    Note 3. If \(\mathcal{G} \subseteq \mathcal{M}\) (a \(\sigma\)-ring), then also \(\mathcal{B} \subseteq \mathcal{M}\) since \(\mathcal{B}\) is, by definition, the least \(\sigma\)-ring \(\supseteq \mathcal{G}.\)

    Thus \(m\) is topological iff \(\mathcal{B} \subseteq \mathcal{M}\) (hence surely \(\mathcal{F} \subseteq \mathcal{M}, \mathcal{G}_{\delta} \subseteq \mathcal{M}, \mathcal{F}_{\sigma} \subseteq \mathcal{M}\), etc.).

    It also follows that any topological measure can be restricted to \(\mathcal{B}\) to obtain a Borel measure, called its Borel restriction.

    Definition 4

    A measure \(m : \mathcal{M} \rightarrow E^{*}\) in \((S, \mathcal{G})\) is called regular iff it is regular with respect to \(\mathcal{M} \cap \mathcal{G},\) the measurable open sets; i.e.,

    \[(\forall A \in \mathcal{M}) \quad m A=\inf \{m X | A \subseteq X \in \mathcal{M} \cap \mathcal{G}\}.\]

    If \(m\) is topological \((\mathcal{G} \subseteq \mathcal{M}),\) this simplifies to

    \[m A=\inf \{m X | A \subseteq X \in \mathcal{G}\},\]

    i.e., \(m\) is \(\mathcal{G}\)-regular (Definition 5 in §5).

    Definition 5

    A measure \(m\) is strongly regular iff for any \(A \in \mathcal{M}\) and \(\varepsilon>0,\) there is an open set \(G \in \mathcal{M}\) and a closed set \(F \in \mathcal{M}\) such that

    \[F \subseteq A \subseteq G, \text { with } m(A-F)<\varepsilon \text { and } m(G-A)<\varepsilon;\]

    thus \(A\) can be "approximated" by open supersets and closed subsets, both measurable. As is easily seen, this implies regularity.

    A kind of converse is given by the following theorem.

    Theorem \(\PageIndex{1}\)

    If a measure \(m : \mathcal{M} \rightarrow E^{*}\) in \((S, \mathcal{G})\) is regular and \(\sigma\)-finite (see Definition 4 in §5), with \(S \in \mathcal{M},\) then \(m\) is also strongly regular.

    Proof

    Fix \(\varepsilon>0\) and let \(m A<\infty\).

    By regularity,

    \[m A=\inf \{m X | A \subseteq X \in \mathcal{M} \cap \mathcal{G}\};\]

    so there is a set \(X \in \mathcal{M} \cap \mathcal{G}\) (measurable and open), with

    \[A \subseteq X \text { and } m X<m A+\varepsilon.\]

    Then

    \[m(X-A)=m X-m A<\varepsilon,\]

    and \(X\) is the open set \(G\) required in (2).

    If, however, \(m A=\infty,\) use \(\sigma\)-finiteness to obtain

    \[A \subseteq \bigcup_{k=1}^{\infty} X_{k}\]

    for some sets \(X_{k} \in \mathcal{M}, m X_{k}<\infty;\) so

    \[A=\bigcup_{k}\left(A \cap X_{k}\right).\]

    Put

    \[A_{k}=A \cap X_{k} \in \mathcal{M}.\]

    (Why?) Then

    \[A=\bigcup_{k} A_{k},\]

    and

    \[m A_{k} \leq m X_{k}<\infty.\]

    Now, by what was proved above, for each \(A_{k}\) there is an open measurable \(G_{k} \supseteq A_{k},\) with

    \[m\left(G_{k}-A_{k}\right)<\frac{\varepsilon}{2^{k}},\]

    Set

    \[G=\bigcup_{k=1}^{\infty} G_{k}.\]

    Then \(G \in \mathcal{M} \cap \mathcal{G}\) and \(G \supseteq A.\) Moreover,

    \[G-A=\bigcup_{k} G_{k}-\bigcup_{k} A_{k} \subseteq \bigcup_{k}\left(G_{k}-A_{k}\right).\]

    (Verify!) Thus by \(\sigma\)-subadditivity,

    \[m(G-A) \leq \sum_{k} m\left(G_{k}-A_{k}\right)<\sum_{k=1}^{\infty} \frac{\varepsilon}{2^{k}}=\varepsilon,\]

    as required.

    To find also the closed set \(F,\) consider

    \[-A=S-A \in \mathcal{M}.\]

    As shown above, there is an open measurable set \(G^{\prime} \supseteq-A,\) with

    \[\varepsilon>m\left(G^{\prime}-(-A)\right)=m\left(G^{\prime} \cap A\right)=m\left(A-\left(-G^{\prime}\right)\right).\]

    Then

    \[F=-G^{\prime} \subseteq A\]

    is the desired closed set, with \(m(A-F)<\varepsilon. \quad \square\)

    Theorem \(\PageIndex{2}\)

    If \(m : \mathcal{M} \rightarrow E^{*}\) is a strongly regular measure in \((S, \mathcal{G}),\) then for any \(A \in \mathcal{M},\) there are measurable sets \(H \in \mathcal{F}_{\sigma}\) and \(K \in \mathcal{G}_{\delta}\) such that

    \[H \subseteq A \subseteq K \text { and } m(A-H)=0=m(K-A);\]

    hence

    \[m A=m H=m K.\]

    Proof

    Let \(A \in \mathcal{M}.\) By strong regularity, given \(\varepsilon_{n}=1 / n,\) one finds measurable sets

    \[G_{n} \in \mathcal{G} \text { and } F_{n} \in \mathcal{F}, \quad n=1,2, \ldots,\]

    such that

    \[F_{n} \subseteq A \subseteq G_{n}\]

    and

    \[m\left(A-F_{n}\right)<\frac{1}{n} \text { and } m\left(G_{n}-A\right)<\frac{1}{n}, \quad n=1,2, \ldots.\]

    Let

    \[H=\bigcup_{n=1}^{\infty} F_{n} \text { and } K=\bigcap_{n=1}^{\infty} G_{n}.\]

    Then \(H, K \in \mathcal{M}, H \in \mathcal{F}_{\sigma}, K \in \mathcal{G}_{\delta},\) and

    \[H \subseteq A \subseteq K.\]

    Also, \(F_{n} \subseteq H\) and \(G_{n} \supseteq K\).

    Hence

    \[A-H \subseteq A-F_{n} \text { and } K-A \subseteq G_{n}-A;\]

    so by (4),

    \[m(A-H)<\frac{1}{n} \rightarrow 0 \text { and } m(K-A)<\frac{1}{n} \rightarrow 0.\]

    Finally,

    \[m A=m(A-H)+m H=m H,\]

    and similarly \(m A=m K\).

    Thus all is proved.\(\quad \square\)