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 $\cali{M}=\cali{B}_{\cl{\R}}$.$\square$