what is the associated Borel set of a Borel measurable function on the extended real line?

Question

This question comes from Theorem 19.B in page 81 of Halmos' "Measure Theory", as the image below shows.

In this theorem, we are given a function $\phi$ which is a Borel measurable function on the extended real line $\mathbb R^*$. But I can't figure out what the associated Borel set is with regard to which $\phi$ becomes a Borel measurable function. Related definitions of this book are as follows:

1)measurable space and measurable set (page 73):

That is, we must make sure two conditions are met: a)S is a sigma-ring, b)$\bigcup{\bf S}=X$

2)measurable function (page 76-77):

This definition shows that a measurable function must be defined on a measurable space, that is, a whole space X together with a sigma-ring S, otherwise we cannot check if $N(f)\cap f^{-1}(M)$ is measurable or not.

3)Borel measurable function (page 77-78):

In 3), Borel measurable function is defined only for the real line $\mathbb R$, but in Theorem 19.B, $\phi$ is a Borel measurable function on the extended real line $\mathbb R^\ast$. What is the sigma-ring of the measurable space involved in the above definitions? If it is just the real Borel set $\bf B$, we do not have $\bigcup\bf B=\mathbb R^*$, which violates the definition of measurable space in 1). If it is $\bf B$ along with $\{+\infty\}$ and $\{-\infty\}$, that is, ${\bf B'=B}\cup \{\{+\infty\}\}\cup\{\{-\infty\}\}$, ${\bf B'}$ is not a sigma-ring since, e.g. $[a,+\infty]=[a,+\infty)\cup\{+\infty\}$ is not in ${\bf B'}$. How to solve this problem? Thanks!

Answer 1

The Borel sets of $\mathbb{R}^{\ast}$ are just the real Borel sets together with any real Borel set union one or both infinities. In other words, a subset of $\mathbb{R}^{\ast}$ is Borel if and only if its intersection with $\mathbb{R}$ is Borel.

Answer 2

Two definitions of the extended Borel set:

Top-down: The class of extended Borel sets (extended Borel algebra) is the $\sigma$-ring generated by the class of all open sets of the extended real number system $\mathbb R^*$.

Bottom-up: Let B be the class of all real Borel sets (Borel algebra); then the class of extended Borel sets is the union of B with classes of the form ${\bf E}_1=\{E\cup\{+\infty\}|E\in\bf B\}, E_2=\{E\cup\{-\infty\}|E\in\bf B\}$ and ${\bf E}_3=\{E\cup\{+\infty,-\infty\}|E\in\bf B\}$.

These two definitions are equivalent.

Proof: Let $\bf U^*$ denotes the class of all open sets of the extended real number system $\mathbb R^*$, $\bf S(U^*)$ the $\sigma$-ring generated by $\bf U^*$ and $\bf S^*=B\cup E_1\cup E_2\cup E_3$ as in the Bottom-up definition; we propose to prove $\bf S(U^*)=S^*$. The basis of $\mathbb R^*$ in order topology consists of nonempty bounded open intervals $(a,b), [-\infty,a)=(-\infty,a)\cup\{-\infty\}$ and $(a,+\infty]=(a,+\infty)\cup\{+\infty\}$[1], so any open sets in $\mathbb R^*$, as an arbitrary union of these basis elements, has a form of a union of an open set in $\mathbb R$ with possibly $\{+\infty\}$ or $\{-\infty\}$ or $\{+\infty,-\infty\}$. Since any real open set is a real Borel set, $\bf S^*$ contains $\bf U^*$. It is clear $\bf S^*$ is a $\sigma$-ring (actually a $\sigma$-algebra), so we get $\bf S^*\supseteq S(U^*)$.

Since every real open set is also open in $\mathbb R^*$, we have $\bf U\subseteq U^*$ where U is the class of all real open sets, and in turn $\bf B=S(U)\subseteq S(U^*)$. Note in addition that $\{+\infty\}=\bigcap \limits_{n = 1}^\infty (n,+\infty]$, so $\{+\infty\}$ is a member of the $\sigma$-ring $\bf S(U^*)$. As a result, each element of $\bf S^*$, being a union of real Borel set with possibly inifinities, is still an element of the ring $\bf S(U^*)$, that is, $\bf S^*\subseteq S(U^*)$, which establishes the converse inclusion and completes the whole proof.

If there is any error in the proof, please kindly point out. Thanks!

[1]Munkres' "Topology", page 84.

[2]Apostol's "Mathematical Analysis", page 51.