Take the 2-minute tour ×

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

alt text

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): alt text

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): alt text

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): alt text

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!

share|improve this question
how to input mathbb R with a superscript asterisk? I believe \mathbb R^* is correct, but the latex formular alway displays it incorrectly. In addition, how to input brace { and }? I think \{ and \} should work, but the latex formulate eat them, why? –  zzzhhh May 7 '10 at 7:43
The problem is that markdown, being unaware of math formulas, mess with underscores, asterisks and a few other things inside them. Surrounding the formulas by backquotes is the standard cure. –  Harald Hanche-Olsen May 7 '10 at 14:27
There are complications that come because Halmos does everything using sigma-ring, rather than just sigma-algebra. I think almost everyone simply uses sigma-algebra today. –  Gerald Edgar May 7 '10 at 15:03
to Harald Hanche-Olsen: what is backquotes? to Gerald Edgar: I have noticed that although Halmos' book is a classic, its notions and definitions are not widely accepted, leading to difficulties of understanding in discussion on measure theory as if I'm speaking a foreign language. Could you please recommend a modern textbook whose notions and definitions are widely accepted and at the same time as clear and in-depth as Halmos' book, and that, in particular, contains Kevin's definition of the extended Borel set? Thanks! –  zzzhhh May 8 '10 at 1:24
Some call them backticks, apparently. Look in the box “how to write math” in the right margin. –  Harald Hanche-Olsen May 8 '10 at 4:24

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer
Thank you. This is a sigma-ring that can constitute a measurable space together with $\mathbb B^\ast$, and consideration of only $f^{-1}({+\infty})$ and $f^{-1}({-\infty})$ is sufficient for all Borel sets containing infinity thus defined, but could you please tell me where this definition of extended Borel set is located in this book, or appear in other textbook? –  zzzhhh May 7 '10 at 9:47
In general, the Borel sets of any topological space are just the sigma-algebra generated by the opens. See, for example, pg. 22 of Folland's Real Analysis. Another good reference is "The Elements of Integration and Lebesgue Measure" by Bartle. On page 7 he describes precisely the example of the extended real line. –  Kevin Ventullo May 8 '10 at 6:49
Thank you very much Kevin! I found these references and managed to prove the equivalence of the top-down definition in Folland and Bottom-up definition in Bartle. I post the proof below for my future reference. Today is really a fruitful day, thank you all! PS: Should I continue Halmos or turn to Bartle for a better effect of self-study? –  zzzhhh May 8 '10 at 10:35

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.

share|improve this answer

Your Answer


