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

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

2010-05-07T07:21:12Z

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 by Kevin Ventullo for what is the associated Borel set of a Borel measurable function on the extended real line?

2010-05-07T08:13:03Z

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

2010-05-08T11:18:32Z

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.