7.7.E: Problems on Topologies, Borel Sets, and Regular Measures
Show that \(\mathcal{G}\) is a topology in \(S\) (in \((a)-(c),\) describe \(\mathcal{B}\) also), given
(a) \(\mathcal{G}=2^{S}\);
(b) \(\mathcal{G}=\{\emptyset, S\}\);
(c) \(\mathcal{G}=\{\emptyset \text { and all sets in } S, \text { containing a fixed point } p\};\) or
(d) \(S=E^{*} ; \mathcal{G}\) consists of all possible unions of sets of the form \((a, b), (a, \infty],\) and \([-\infty, b),\) with \(a, b \in E^{1}.\)
\((S, \rho)\) is called a pseudometric space (and \(\rho\) is a pseudometric) iff the metric laws (i)-(iii) of Chapter 3, s 11 hold, but (i) is weakened to
\[\rho(x, x)=0\]
(so that \(\rho(x, y)\) may be \(0\) even if \(x \neq y\)).
(a) Define "globes," "interiors," and "open sets" (i.e., \(\mathcal{G})\) as in Chapter 3, §12; then show that \(\mathcal{G}\) is a topology for \(S.\)
(b) Let \(S=E^{2}\) and
\[\rho(\overline{x}, \overline{y})=\left|x_{1}-y_{1}\right|,\]
where \(\overline{x}=\left(x_{1}, x_{2}\right)\) and \(\overline{y}=\left(y_{1}, y_{2}\right).\) Show that \(\rho\) is a pseudometric but not a metric (the Hausdorff properly fails!).
Define "neighborhood," "interior, "cluster point," "closure," and "function limit" for topological spaces. Specify some notions (e.g., "diameter," "uniform continuity") that do not carry over (they involve distances).
In a topological space \((S, \mathcal{G}),\) detine
\[\mathcal{G}^{0}=\mathcal{G}, \mathcal{G}^{1}=\mathcal{G}_{\delta}, \mathcal{G}^{2}=\mathcal{G}_{\delta \sigma}, \ldots\]
and
\[\mathcal{F}^{0}=\mathcal{F}, \mathcal{F}^{1}=\mathcal{F}_{\sigma}, \mathcal{F}^{2}=\mathcal{F}_{\sigma \delta}, \mathcal{F}^{3}=\mathcal{F}_{\sigma \delta \sigma}, \text { etc.}\]
(Give an inductive definition.) Then prove by induction that
(a) \(\mathcal{G}^{n} \subseteq \mathcal{B}, \mathcal{F}^{n} \subseteq \mathcal{B}\);
(b) \(\mathcal{G}^{n-1} \subseteq \mathcal{G}^{n}, \mathcal{F}^{n-1} \subseteq \mathcal{F}^{n}\);
(c) \((\forall X \subseteq S) X \in \mathcal{F}^{n}\) iff \(-X \in \mathcal{G}^{n}\);
(d) \(\left(\forall X, Y \in \mathcal{F}^{n}\right) X \cap Y \in \mathcal{F}^{n}, X \cup Y \in \mathcal{F}^{n};\) same for \(\mathcal{G}^{n}\);
(e) \(\left(\forall X \in \mathcal{G}^{n}\right)\left(\forall Y \in \mathcal{F}^{n}\right) X-Y \in \mathcal{G}^{n}\) and \(Y-X \in \mathcal{F}^{n}\).
[Hint: \(X-Y=X \cap-Y\).]
For metric and pseudometric spaces (see Problem 2 ) prove that
\[\mathcal{F}^{n} \subseteq \mathcal{G}^{n+1} \text { and } \mathcal{G}^{n} \subseteq \mathcal{F}^{n+1}\]
(cf. Problem 4).
[Hint for \(\mathcal{F} \subseteq \mathcal{G}_{\delta}:\) Let \(F \in \mathcal{F}.\) Set
\[G_{n}=\bigcup_{p \in F} G_{p}\left(\frac{1}{n}\right);\]
so
\[(\forall n) \quad F \subseteq G_{n} \in \mathcal{G}.\]
Hence
\[F \subseteq \bigcap_{n} G_{n} \in \mathcal{G}_{\delta}.\]
Also,
\[\bigcap_{n} G_{n}=\overline{F}=F\]
by Theorem 3 in Chapter 3, §16. Hence deduce that
\[(\forall F \in \mathcal{F}) \quad F \in \mathcal{G}_{\delta},\]
so \(\mathcal{F} \subseteq \mathcal{G}_{\delta};\) hence \(\mathcal{G} \subseteq \mathcal{F}_{\sigma}\) by Problem 4(c). Now use induction.]
If \(m\) is as in Definition 5, then prove the following.
(i) \(m\) is regular.
(ii) \((\forall A \in \mathcal{M}) m A=\sup \{m X | A \supseteq X \in \mathcal{M} \cap \mathcal{F}\}\).
(iii) The latter implies strong regularity if \(m<\infty\) and \(S \in \mathcal{M}\).
Let \(\mu : \mathcal{B} \rightarrow E^{*}\) be a Borel measure in a metric space \((S, \rho).\) Set
\[(\forall A \subseteq S) \quad n^{*} A=\inf \{\mu X | A \subseteq X \in \mathcal{G}\}.\]
Prove that
(i) \(n^{*}\) is an outer measure in \(S\);
(ii) \(n^{*}=\mu\) on \(\mathcal{G}\);
(iii) the \(n^{*}\)-induced measure, \(n : \mathcal{N}^{*} \rightarrow E^{*},\) is topological (so \(\mathcal{B} \subseteq \mathcal{N}^{*}\));
(iv) \(n \geq \mu\) on \(\mathcal{B}\);
(v) \((\forall A \subseteq S)\left(\exists H \in \mathcal{G}_{\delta}\right) A \subseteq H\) and \(\mu H=n^{*} A\).
[Hints: (iii) Using Problem 15 in §5 and Problem 12 in §6, let
\[\rho(X, Y)>\varepsilon>0, \quad U=\bigcup_{x \in X} G_{x}\left(\frac{1}{2} \varepsilon\right), \quad V=\bigcup_{y \in Y} G_{y}\left(\frac{1}{2} \varepsilon\right).\]
Verify that \(U, V \in \mathcal{G}, U \supseteq X, V \supseteq Y, U \cap V=\emptyset\).
By the definition of \(n^{*}\),
\[(\exists G \in \mathcal{G}) \quad G \supseteq X \cup Y \text { and } n^{*} G \leq n^{*}(X \cup Y)+\varepsilon;\]
also, \(X \subseteq G \cap U\) and \(Y \subseteq G \cap V.\) Thus by (ii),
\[n^{*} X \leq \mu(G \cap U) \text { and } n^{*} Y \leq \mu(G \cap V).\]
Hence
\[n^{*} X+n^{*} Y \leq \mu(G \cap U)+\mu(G \cap V)=\mu((G \cap U) \cup(G \cap V)) \leq \mu G=n^{*} G \leq n^{*}(X \cup Y)+\varepsilon.\]
Let \(\varepsilon \rightarrow 0\) to get the \(\mathrm{CP} : n^{*} X+n^{*} Y \leq n^{*}(X \cup Y)\).
(iv) We have \((\forall A \in \mathcal{B})\)
\[n A=n^{*} A=\inf \{\mu X | A \subseteq X \in \mathcal{G}\} \geq \inf \{\mu X | A \subseteq X \in \mathcal{B}\}=\mu A.\]
(Why?)
(v) Use the hint to Problem 11 in §5.]
From Problem 7 with \(m=\mu,\) prove that if
\[A \subseteq G \in \mathcal{G},\]
with \(m G<\infty\) and \(A \in \mathcal{B},\) then \(m A=n A\).
[Hint: \(A, G,\) and \((G-A) \in \mathcal{B}.\) By Problem 7(iii), \(\mathcal{B} \subseteq N^{*}\) and \(n\) is additive on \(\mathcal{B}\); so by Problem 7(ii)(iv),
\[n A=n G-n(G-A) \leq m G-m(G-A)=m A \leq n A.\]
Thus \(m A=n A.\) Explain all!]
Let \(m, n,\) and \(n^{*}\) be as in Problems 7 and 8. Suppose
\[S=\bigcup_{n=1}^{\infty} G_{n},\]
with \(G_{n} \in \mathcal{G}\) and \(m G_{n}<\infty\) (this is called \(\sigma^{0}\)-finiteness).
Prove that
(i) \(m=n\) on \(\mathcal{B},\) and
(ii) \(m\) and \(n\) are strongly regular.
[Hints: Fix \(A \in \mathcal{B}.\) Show that
\[A=\bigcup A_{n} \text { (disjoint)}\]
for some Borel sets \(A_{n} \subseteq G_{n}\) (use Corollary 1 in §1). By Problem 8, \(m A_{n}=n A_{n}\) since
\[A_{n} \subseteq G_{n} \in \mathcal{G}\]
and \(m G_{n}<\infty.\) Now use \(\sigma\)-additivity to find \(m A=n A\).
(ii) Use \(\mathcal{G}\)-regularity, part (i), and Theorem 1.]
Continuing Problems 8 and 9, show that \(n\) is the Lebesgue extension of \(m\) (see Theorem 2 in §6 and Note 3 in §6).
Thus every \(\sigma^{0}\)-finite Borel measure \(m\) in \((S, \rho)\) and its Lebesgue extension are strongly regular.
[Hint: \(m\) induces an outer measure \(m^{*},\) with \(m^{*}=m\) on \(\mathcal{B}.\) It suffices to show that \(m^{*}=n^{*}\) on \(2^{S}.\) (Why?)
So let \(A \subseteq S.\) By Problem 7(v),
\[(\exists H \in \mathcal{B}) A \subseteq H \text { and } n^{*} A=m H=m^{*} H.\]
Also,
\[(\exists K \in \mathcal{B}) A \subseteq K \text { and } m^{*} A=m K\]
(Problem 12 in §5). Deduce that
\[n^{*} A \leq n(H \cap K)=m(H \cap K) \leq m H=n^{*} A\]
and
\(n^{*} A=m(H \cap K)=m^{*} A\).]