what is the associated Borel set of a Borel measurable function on the extended real line? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T04:56:24Z http://mathoverflow.net/feeds/question/23826 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/23826/what-is-the-associated-borel-set-of-a-borel-measurable-function-on-the-extended-r what is the associated Borel set of a Borel measurable function on the extended real line? zzzhhh 2010-05-07T07:21:12Z 2010-05-08T11:18:32Z <p>This question comes from Theorem 19.B in page 81 of Halmos' "Measure Theory", as the image below shows.</p> <p><img src="http://i43.tinypic.com/bhgaj7.png" alt="alt text"></p> <p>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:</p> <p>1)measurable space and measurable set (page 73): <img src="http://i42.tinypic.com/wltbpy.png" alt="alt text"></p> <p>That is, we must make sure two conditions are met: a)<strong>S</strong> is a sigma-ring, b)$\bigcup{\bf S}=X$</p> <p>2)measurable function (page 76-77): <img src="http://i42.tinypic.com/2v1nu3k.png" alt="alt text"></p> <p>This definition shows that a measurable function must be defined on a measurable space, that is, a whole space <em>X</em> together with a sigma-ring <strong>S</strong>, otherwise we cannot check if $N(f)\cap f^{-1}(M)$ is measurable or not.</p> <p>3)Borel measurable function (page 77-78): <img src="http://i40.tinypic.com/xng4ko.png" alt="alt text"></p> <p>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 <code>$\{+\infty\}$</code> and <code>$\{-\infty\}$</code>, that is, <code>${\bf B'=B}\cup \{\{+\infty\}\}\cup\{\{-\infty\}\}$</code>, ${\bf B'}$ is not a sigma-ring since, e.g. <code>$[a,+\infty]=[a,+\infty)\cup\{+\infty\}$</code> is not in ${\bf B'}$. How to solve this problem? Thanks!</p> http://mathoverflow.net/questions/23826/what-is-the-associated-borel-set-of-a-borel-measurable-function-on-the-extended-r/23830#23830 Answer by Kevin Ventullo for what is the associated Borel set of a Borel measurable function on the extended real line? Kevin Ventullo 2010-05-07T08:13:03Z 2010-05-07T08:13:03Z <p>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.</p> http://mathoverflow.net/questions/23826/what-is-the-associated-borel-set-of-a-borel-measurable-function-on-the-extended-r/23928#23928 Answer by zzzhhh for what is the associated Borel set of a Borel measurable function on the extended real line? zzzhhh 2010-05-08T11:18:32Z 2010-05-08T11:18:32Z <p>Two definitions of the extended Borel set:</p> <p>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 $\mathbb R^*$.</p> <p>Bottom-up: Let <strong>B</strong> be the class of all real Borel sets (Borel algebra); then the class of extended Borel sets is the union of <strong>B</strong> with classes of the form <code>${\bf E}_1=\{E\cup\{+\infty\}|E\in\bf B\}, E_2=\{E\cup\{-\infty\}|E\in\bf B\}$</code> and <code>${\bf E}_3=\{E\cup\{+\infty,-\infty\}|E\in\bf B\}$</code>.</p> <p>These two definitions are equivalent.</p> <p>Proof: Let <code>$\bf U^*$</code> denotes the class of all open sets of the extended real number system <code>$\mathbb R^*$</code>, <code>$\bf S(U^*)$</code> the $\sigma$-ring generated by <code>$\bf U^*$</code> and <code>$\bf S^*=B\cup E_1\cup E_2\cup E_3$</code> as in the Bottom-up definition; we propose to prove <code>$\bf S(U^*)=S^*$</code>. The basis of <code>$\mathbb R^*$</code> in order topology consists of nonempty bounded open intervals <code>$(a,b), [-\infty,a)=(-\infty,a)\cup\{-\infty\}$</code> and <code>$(a,+\infty]=(a,+\infty)\cup\{+\infty\}$</code>[1], so any open sets in <code>$\mathbb R^*$</code>, as an arbitrary union of these basis elements, has a form of a union of an open set in <code>$\mathbb R$</code> with possibly <code>$\{+\infty\}$</code> or <code>$\{-\infty\}$</code> or <code>$\{+\infty,-\infty\}$</code>. Since any real open set is a real Borel set, <code>$\bf S^*$</code> contains <code>$\bf U^*$</code>. It is clear <code>$\bf S^*$</code> is a $\sigma$-ring (actually a $\sigma$-algebra), so we get <code>$\bf S^*\supseteq S(U^*)$</code>.</p> <p>Since every real open set is also open in <code>$\mathbb R^*$</code>, we have <code>$\bf U\subseteq U^*$</code> where <strong>U</strong> is the class of all real open sets, and in turn <code>$\bf B=S(U)\subseteq S(U^*)$</code>. Note in addition that <code>$\{+\infty\}=\bigcap \limits_{n = 1}^\infty (n,+\infty]$</code>, so <code>$\{+\infty\}$</code> is a member of the $\sigma$-ring <code>$\bf S(U^*)$</code>. As a result, each element of <code>$\bf S^*$</code>, being a union of real Borel set with possibly inifinities, is still an element of the ring <code>$\bf S(U^*)$</code>, that is, <code>$\bf S^*\subseteq S(U^*)$</code>, which establishes the converse inclusion and completes the whole proof.</p> <p>If there is any error in the proof, please kindly point out. Thanks!</p> <p>[1]Munkres' "Topology", page 84.</p> <p>[2]Apostol's "Mathematical Analysis", page 51.</p>