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Let $E$ and $F$ be measurable subsets of $mathbb{R}$ endowed with the Lebesgue measure $lambda$. We say that $m:Eto F$ is measure-preserving if for each measurable subset $A$ of $F$, the set $m^{-1}(A)$ is also measurable, and $lambda[m^{-1}(A)]=lambda(A)$. We say that it is order-preserving if $xleq yin E$ implies $m(x)leq m(y)$ in $F$.

Conjecture 1. If $m:Eto F$ is a bijection which is both measure-preserving and order-preserving, then $m^{-1}$ is measurable w.r.t. the Lebesgue measure, i.e. for each Lebesgue-measurable subset $A$ of $E$, the image $m(A)$ is also Lebesgue-measurable.

Fact 1. If $m:Eto F$ is a bijection which is both measure-preserving and order-preserving, then $m$ and $m^{-1}$ are both strictly order-preserving, i.e. $x<yin E$ if and only if $m(x)<m(y)$ in $F$.

Fact 2. If $A$ is a measurable subset of $E$ and $m(A)$ is a measurable subset of $F$ then $lambda(A)=lambda[m(A)]$.

My idea was to attempt an argument from contradiction. Assume $A$ is a measurable subset of $E$ with $m(A)$ nonmeasurable. By removing at most countably many sets of positive measure if necessary, we can assume that $m(A)$ contains no sets of positive measure. We may also assume that $m(A)$ is contained in some finite interval $[a,b]$. Now $B:=([a,b]cap F)setminus m(A)$ also is nonmeasurable, with $m^{-1}(B)$ disjoint from $A$. Let us remove any sets of positive measure from $B$ as well. Now $C:=m(A)cup B$ is still measurable, because it is equal to $Fcap[a,b]$ with countably many measurable sets removed. Note that $C$ cannot have measure zero since it has nonmeasurable subsets. It follows that $Acup m^{-1}(B)$ has positive measure.

Now let's see if we can bring order into it. Consider the case where there exists a nonmeasurable subset $D$ of $m(A)$ such that $m^{-1}(B)$ does not intersect the interval $(inf D,sup D)$. Then $(inf D,sup D)cap C=(inf D,sup D)cap m(A)=Dsetminus{inf D,sup D}$, which is impossible since $(inf D,sup D)cap C$ is measurable whereas $Dsetminus{inf D,sup D}$ is not. A similar argument holds for $B$. It follows that we can assume without loss of generality that there exists a subset $G$ of $m(A)$ and a subset $H$ of $B$ such that for each $gin G$ and $hin H$ there are measure-zero subsets $A_g$ of $m(A)$ and $B_h$ of $B$ such that $A_g<B_h$ if and only if $g<h$, and $B_h<A_g$ otherwise. As only countably many of the $A_g$'s and $B_h$'s have positive diameter, we may assume without loss of generality that they are all singletons. Now if $x<yin m(A)$ then there is $zin B$ with $x<z<y$. An analogous fact holds for $A$ and $m^{-1}(B)$.

So, to summarize, we have a measurable subset $A$ of $E$, of finite diameter, and a nonmeasurable subset $B$ of $F$ which does not intersect $m(A)$, also of finite diameter, such that $m^{-1}(B)$ is measurable and of finite diameter and does not intersect $A$, $m(A)$ is nonmeasurable and of finite diameter, that $Acup m^{-1}(B)$ and $m(A)cup B$ both have positive measure, neither $m(A)$ nor $B$ contain any subsets of positive measure, and for every $x<yin m(A)$ there is $zin B$ with $x<z<y$, and an analogous fact holds for $A$ and $m^{-1}(B)$, i.e. for any $x<yin A$ there is $zin m^{-1}(B)$ with $x<z<y$.

Is there any way to use those order properties to achieve a contradiction?

It is easy to see that every order-preserving map $f:E to F$ is measurable, because the preimage of any interval is an intersection of an interval with $E$, so it is measurable. Since the inverse of an order-preserving bijection $m$ is order-preserving, it is measurable. The assumption of preserving Lebesgue measure is not needed.

Added: The above solution assumes the standard definition of measurability for real-valued functions: $f$ is Lebesgue-measurable if every sub-level set ${ x in E: f(x)<alpha }$ is Lebesgue-measurable, or equivalently if every preimage of a Borel set is Lebesgue measurable. Using the alternative definition that $f$ is measurable if every preimage of a Lebesgue-measurable set is Lebesgue-measurable, one indeed needs at least a weak form of the measure-preserving property. Since the Lebesgue $sigma$-algebra is the completion of the Borel $sigma$-algebra with respect to Lebesgue measure, a set $A$ is Lebesgue measurable iff there exists a Borel set $B$ with $lambda (A Delta B) = 0$, where $A Delta B = (A setminus B) cup (B setminus A)$ is the symmetric difference of $A$ and $B$. Then, using $f=m^{-1}$, we have $f^{-1}(B) = m(B)$ is a Borel set by the argument above, and $f^{-1}(A) Delta f^{-1}(B) = m(A)Delta m(B) = m(A Delta B)$ is a Lebesgue null set because $m$ preserves Lebesgue measure. (We really only need that $m$-images of null sets are null sets here, which follows from the fact that every Lebesgue null set is contained in a Borel null set, whose image is then again a Borel null set.) This shows that $f^{-1}(B)$ is Lebesgue measurable, so that $f=m^{-1}$ is measurable with respect to this definition, too.