most recent 30 from http://mathoverflow.net 2013-05-24T01:31:48Z http://mathoverflow.net/feeds/question/19402 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19402/how-to-show-that-x-y-is-lebesgue-lebesgue-measurable Nicolò 2010-03-26T10:29:12Z 2010-03-27T17:25:20Z

<p>Which is the cleanest way to show that the difference, $d:R^n\times R^n\rightarrow R^n$, $d(x,y)= x-y$, is Lebesgue-Lebesgue measurable? (i.e. foreach A lebesgue measurable in $R^n$, $d^{-1}(A)$ is Lebesgue measurable in $R^n\times R^n$). Thanks in advance.</p>

http://mathoverflow.net/questions/19402/how-to-show-that-x-y-is-lebesgue-lebesgue-measurable/19416#19416 Colin McQuillan 2010-03-26T12:57:42Z 2010-03-26T13:25:41Z

<p>Unitary matrices preserve measure. A diagonal matrix of full rank is a Lesbesgue-Lesbesgue measurable transformation. Linear maps over the reals have a singular value decomposition.</p>

http://mathoverflow.net/questions/19402/how-to-show-that-x-y-is-lebesgue-lebesgue-measurable/19421#19421 Robin Chapman 2010-03-26T13:31:34Z 2010-03-26T13:31:34Z

<p>Nicolo is asking about functions where the inverse image of a Lebesgue measurable set is Lebesgue measurable. This is stronger than the usual definition of measurability where it is required only the inverse image of each Borel set must be Lebesgue measurable. Continuous functions need not be measurable by this stronger criterion. If $B$ has zero Lebesgue measure and $A=f^{-1}(B)$ has nonzero measure then each subset of $B$ is Lebesgue measurable but its inverse image may be non-measurable. A simple example is given by $f:x\mapsto (x,0)$ from $\mathbb{R}$ to $\mathbb{R}^2$. Taking $A$ to be a non-measurable subset of $\mathbb{R}$ and $B=f(A)$ we see this $f$ is not Lebesgue-Lebesgue measurable. More interesting examples occur on the real line when there are continuous homeomorphisms from $\mathbb{R}$ to itself taking Cantor sets of positive measure to Cantor sets of zero measure.</p> <p>To return to Nicolo's example. Each surjective linear map from $\mathbb{R}^m\to\mathbb{R}^n$ is Lebesgue-Lebesgue measurable as it can be decomposed as a composition of linear bijections and the projection map $\mathbb{R}^m\to\mathbb{R}^n$ mapping onto the first $n$ coordinates (both these types of maps can be seen to be Lebesgue-Lebesgue measurable). By definition, the class Lebesgue-Lebesgue measurable maps is closed under composition (unlike the class of Lebesgue-measurable maps!).</p>