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# 7.7: Topologies. Borel Sets. Borel Measures

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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.$$\quad \square$$