Sections measure zero imply set is measure zero? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T03:44:59Z http://mathoverflow.net/feeds/question/89375 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89375/sections-measure-zero-imply-set-is-measure-zero Sections measure zero imply set is measure zero? Julie 2012-02-24T09:16:50Z 2012-02-24T13:24:18Z <p>I have a subset $B\subset\mathbb{R}^n\times\mathbb{R}^m$ that I want to show has measure zero. I know that the sections $B^x = {y : (x,y)\in B}$ all have measure zero. I do not know if $B$ is measurable. Is this enough to conclude that $B$ is a measurable set with measure zero?</p> <p>EDIT : I would like to say that when randomly selecting $a\in\mathbb{R}^n\times\mathbb{R}^m$, I am almost always not in $B$. Since I know that $B^x = {y : (x,y)\in B}$ all have measure zero, it seems like a reasonable conclusion. I don't know enough probability or measure theory to put this in a rigorous way, so any suggestions would be great.</p> http://mathoverflow.net/questions/89375/sections-measure-zero-imply-set-is-measure-zero/89378#89378 Answer by Michael Greinecker for Sections measure zero imply set is measure zero? Michael Greinecker 2012-02-24T09:52:10Z 2012-02-24T10:13:13Z <p>Sierpinski gave an <a href="http://matwbn.icm.edu.pl/ksiazki/fm/fm1/fm1113.pdf" rel="nofollow">example</a> of a nonmeasurable subset of $\mathbb{R}^2$ such that all sections are singletons and hence null sets. So the answer is in general no. </p> http://mathoverflow.net/questions/89375/sections-measure-zero-imply-set-is-measure-zero/89394#89394 Answer by Stefan Geschke for Sections measure zero imply set is measure zero? Stefan Geschke 2012-02-24T13:24:18Z 2012-02-24T13:24:18Z <p>Since the Sierpinski article is in French an uses slightly old-fashioned notation, let me sketch a proof of the result.</p> <p>Theorem. There is a function $f:\mathbb R\to\mathbb R$ whose graph is not a measurable subset of $\mathbb R^2$.</p> <p>Proof. We first show that a set $A\subseteq\mathbb R$ of size $&lt;2^{\aleph_0}$ cannot have a complement of measure zero (in fact, if such a set is measurable, then it is of measure zero, but we don't need this). To see this, we enlarge $A$ so that it becomes a subgroup of $(\mathbb R,+)$ that is still of size $&lt;2^{\aleph_0}$.<br> Now choose $x\in\mathbb R\setminus A$. The coset $A+x$ is disjoint from $A$. Hence $\mathbb R$ is the union of $\mathbb R\setminus A$ and $\mathbb R\setminus(A+x)$.<br> It follows that not both of these sets can be of measure zero.<br> However, if $A$ is measurable, they have the same measure. So the complement of $A$ is not of measure zero.</p> <p>Now let $(B_\alpha)_{\alpha&lt;2^{\aleph_0}}$ be an enumeration of all Borel subsets of $\mathbb R^2$ of measure zero.<br> For each $\alpha&lt;2^{\aleph_0}$ choose a pair $(x_\alpha,y_\alpha)\in\mathbb R^2$ such that $x_\alpha\not\in\{x_\beta:\beta&lt;\alpha\}$ and $(x_\alpha,y_\alpha)\not\in B_\alpha$. This is possible since for each $\alpha$ the set $$((\mathbb R\setminus \{x_\beta:\beta&lt;\alpha\})\times\mathbb R)\setminus B_\alpha$$ is not of measure zero by the remark at the beginning of the proof.</p> <p>Now $\{(x_\alpha,y_\alpha):\alpha&lt;2^{\aleph_0}\}$ is a partial function from $\mathbb R$ to $\mathbb R$ that is not contained in any measure zero Borel subset of the plane. It follows that this partial function is not of measure zero.<br> Extend the function to any total function $f:\mathbb R\to\mathbb R$.<br> The extended function still is not of measure zero. But since the sections are singletons, the graph of this function cannot be measurable by Fubini's theorem. \qed </p>