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 ${+\infty}$ \{+\infty\}$ and ${-\infty}$, \{-\infty\}$, that is, ${\bf B'=B}\cup${{$+\infty$}}$\cup${{$-\infty$}}B'=B}\cup \{\{+\infty\}\}\cup\{\{-\infty\}\}$, ${\bf B'}$ is not a sigma-ring since, e.g. $[a,+\infty]=[a,+\infty)\cup{+\infty}$ [a,+\infty]=[a,+\infty)\cup\{+\infty\}$ is not in ${\bf B'}$. How to solve this problem? Thanks!
