## trivial map on $\sigma-$algebra $\mod{}0$ is trivial

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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...

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 $0$.

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You will need some countability-type hypotheses on the measure space.  For example, it is a standard measure space.  If not, build a counterexample in product space $[0,1]^{[0,1]}$.   – Gerald Edgar Apr 25 2011 at 13:20
Gerald, thanks, I've corrected the question. – David Berman Apr 25 2011 at 13:22

## 1 Answer

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.

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Great! thank you, Mark!  – David Berman Apr 25 2011 at 16:33
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