4.3 Borel $\s$-Algebra

Proof. Let $\cali{F}\suf 2^{X}$ be the smallest subset of $2^{X}$ satisfying (1,2,3) then $\cali{F}\suf\cali{B}_{X}$. OTOH, let

$\emp,X\in \cali{F}_{0}$ since $\cali{F}$ satisfies (1). By definition $\cali{F}_{0}$ is closed under complements. Let $\curl{A_n }_{n\in\N^+}\suf \cali{F}_{0}$ then $\curl{A_n^c }_{n\in\N^+}\suf \cali{F}_{0}$. Observe

Then $\caps{}{n\in\N^+}A_{n}^{c}\in\cali{F}$ by (2) and $\cups{}{n\in\N^+}A_{n}\in \cali{F}$ by (3) and previous observation. Thus $\cups{}{n\in\N^+}A_{n}\in\cali{F}_{0}$ as well. By (1), $\cali{F}_{0}$ is a $\s$-algebra containing all open subsets of $X$ so $\cali{F}\supf \cali{F}_{0}\supf \cali{B}_{X}$.$\square$

Proof. It suffices to show that the $\s$-algebra generated by any of the above families coincide and the resulting $\s$-algebra is the Borel $\s$-algebra on $\R$. For any $a,b\in\R$
(1)$\to$(2):$(a,\infty)=(-\infty,a)^{c}$
(2)$\to$(3):$[a,\infty)=\caps{}{n\in\N^+}(a-1/n,\infty)$
(3)$\to$(4):$(-\infty,a)=[a,\infty)^{c}$
(4)$\to$(5):$[a,b)=(-\infty,b)\cap(-\infty,a)^{c}$
(5)$\to$(6):$[a,b]=\caps{}{n\in\N^+}(a-1/n,b]$
(6)$\to$(7):$(a,b]=\cups{}{n\in\N^+}[a+1/n,b]$
(7)$\to$(8):$(a,b)=\cups{}{n\in\N^+}(a,b-1/n]$
(8)$\to$(1):$(-\infty,a]=\cups{}{n\in\N^+}\caps{}{k\in\N^+}(-n,a+1/k]$
Here we can always take $n>>1$ sufficiently large. For any $U\suf \R$ open and $q\in U\cap \Q$, there exists $r_{q}>0$ such that $(q-r_{q},q+r_{q})\suf U$. Hence

so (8) generates all open sets in $\R$. Conversely (8) consists of open sets so the $\s$-algebra generated by (8) is the Borel $\s$-algebra on $\R$.$\square$

Proof. As in Proposition 4.3.4. Observe (1)$\to$(2):$(a,\infty]=[-\infty,a]^{c}$
(2)$\to$(3):$[a,\infty]=\caps{}{n\in\N^+}(a-1/n,\infty]$
(3)$\to$(4):$[-\infty,a)=[a,\infty]^{c}$
(4)$\to$(1):$[-\infty,a]=\caps{}{n\in\N^+}[-\infty,a+1/n)$
By Proposition 4.3.4, all elements of (1),(2),(3), and (4) belong to $\cali{B}_{\cl{\R}}$. Conversely let $\cali{M}$ be the $\s$-algebra generated by (1),(2),(3), and (4) then

are elements of $\cali{M}$. For any $a\in\R$ we have $(a,\infty)=(a,\infty]\del\curl{\infty}$, so $\cali{M}\supf\cali{B}_{\R}$ by Proposition 4.3.4. OTOH, for any $E\in\cali{B}_{\cl{\R}}$

Since $\curl{-\infty,\infty}\cup \cali{B}_{\R}\suf \cali{M}$ we have $E\in\cali{M}$. Thus $\cali{M}=\cali{B}_{\cl{\R}}$.$\square$

Proof. (1)$\to$(2): Let $U\suf X$ be open then there exists a countable dense subset $S\suf U$. For each $x\in S$, let $r_{x}>0$ such that $B(x,r)\suf U$, then $U=\cups{}{x\in S}B(x,r_{x})$ is the countable union of open balls.
(2)$\to$(3): For any $x\in X$ and $r>0$, $B(x,r)=\cups{}{n\in\N^+}\cl{B(x,r-1/n)}$
(3)$\to$(1):For any $x\in X$ and $r>0$, $\cl{B(x,r)}$ is closed.
$\square$

Proof. Since $\curl{A\in\cali{B}_X:A\cap Y\in\cali{B}_Y}$ is the $\s$-algebra that contains all open sets in $X$, $\cali{B}_{Y}$ contains the induced $\s$-algebra.

OTOH, the induced $\s$-algeba contains all open sets in $Y$, so it contains $\cali{B}_{Y}$.$\square$