Borel vs measure for all Borel measures - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T14:52:26Z http://mathoverflow.net/feeds/question/22839 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22839/borel-vs-measure-for-all-borel-measures Borel vs measure for all Borel measures Matthew Daws 2010-04-28T12:07:32Z 2010-04-28T14:15:37Z <p>Let X be locally compact and Hausdorff, and let $f:X\rightarrow\mathbb R$ be a function. Suppose that for all finite regular (positive) Borel measures $\mu$, we know that $f$ is $\mu$-measurable. Does it follow that $f$ is Borel? If not, what's a good counter-example?</p> <p>Definitions: The Borel sigma-algebra is generated by the open sets. So $f$ is Borel if $f^{-1}(U)$ is Borel for each open $U\subseteq\mathbb R$.</p> <p>I think the definition of $\mu$-measurable is that for each open $U\subseteq\mathbb R$, we have that $f^{-1}(U)$ is in the completed sigma-algebra for $\mu$. That is, we can find Borel sets $A$ and $B$ with $A \subseteq f^{-1}(U) \subseteq B$ with $\mu(B\setminus A)=0$.</p> <p>Remark: The "obvious" measures are the point mass measures, but then the completed sigma-algebra is $2^X$, so all functions are measurable! So the question is, in some sense, whether $X$ supports "enough" finite regular Borel measures.</p> <p>Vague motivation: This old paper of Barry Johnson: Separate continuity and measurability. Proc. Amer. Math. Soc. 20 1969 420--422 see <a href="http://www.jstor.org/stable/2035668" rel="nofollow">http://www.jstor.org/stable/2035668</a> But Barry's paper clearly gives me enough for what I want, so really this question is out of curiosity, not an attempt to understand the paper better!</p> http://mathoverflow.net/questions/22839/borel-vs-measure-for-all-borel-measures/22840#22840 Answer by Simon Thomas for Borel vs measure for all Borel measures Simon Thomas 2010-04-28T12:16:31Z 2010-04-28T12:43:41Z <p>Let $f: \mathbb{R} \to \mathbb{R}$ be the characteristic function of a subset $A \subseteq \mathbb{R}$ which is analytic but not Borel. Then $f$ is universally measurable but not Borel.</p> http://mathoverflow.net/questions/22839/borel-vs-measure-for-all-borel-measures/22852#22852 Answer by Gerald Edgar for Borel vs measure for all Borel measures Gerald Edgar 2010-04-28T13:48:45Z 2010-04-28T14:15:37Z <p>The collection of Borel sets (or analytic sets, etc.) in $\mathbb{R}$ has power $c = 2^{\aleph_0}$. But the collection of universally measurable sets has power $2^c$. This would follow from the existence of a <strong>universal null set</strong> of power $c$.</p>