7.7: Topologies. Borel Sets. Borel Measures

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.\(\quad \square\)