trivial map on $\sigma-$algebra $\mod{}0$ is trivial - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T11:08:57Z http://mathoverflow.net/feeds/question/62916 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62916/trivial-map-on-sigma-algebra-mod0-is-trivial trivial map on $\sigma-$algebra $\mod{}0$ is trivial David Berman 2011-04-25T12:17:11Z 2011-04-25T14:15:49Z <p>Hi everyone! I am currently studying the basic theory of measurable actions and need the following result, which I am not able to prove myself. It is stated without a proof, so probably it should not be hard, but I am lost... </p> <blockquote> <p>Question: Suppose $T$ is an invertible measure-preserving map of standard probability measure space $(X,\mu)$. Suppose that $TA=A$ for all measurable subsets $A\subset{}X$, where the equality is up to sets of measure $0$. Prove that the set of those $x$ where $Tx\neq{}x$ has measure $0$.</p> </blockquote> http://mathoverflow.net/questions/62916/trivial-map-on-sigma-algebra-mod0-is-trivial/62927#62927 Answer by Mark Schwarzmann for trivial map on $\sigma-$algebra $\mod{}0$ is trivial Mark Schwarzmann 2011-04-25T14:15:49Z 2011-04-25T14:15:49Z <p>If $X$ is a standard probability space then we may assume it to be the disjoint union of an interval with Lebesgue measure and a countable set of atoms. If $p$ is an atom, then by assumption, $T(p) = p$, since $p$ has positive measure. So none of the atoms can be "bad" points. So we may assume that there are no atoms, so that $X=[0,1]$ and the measure is Lebesgue measure. Now consider all subintervals $I$ of $[0,1]$ with rational endpoints and gather all points in $TI$ which are outside $I$. We get a countable union of measure zero sets, hence a measure zero set. Denote it by $Z$. Now let $x \notin Z$. This implies that $Tx$ belongs to arbitrarily small intervals around $x$ and is thus equal to $x$. The desired result follows.</p>