must a measure-preserving and order-preserving bijection have measurable inverse? Let $E$ and $F$ be measurable subsets of $\mathbb{R}$ endowed with the Lebesgue measure $\lambda$.  We say that $m:E\to F$ 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 $x\leq y\in E$ implies $m(x)\leq m(y)$ in $F$.
Conjecture 1. If $m:E\to F$ 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 $m:E\to F$ is a bijection which is both measure-preserving and order-preserving, then $m$ and $m^{-1}$ are both strictly order-preserving, i.e. $x<y\in E$ 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 $F\cap[a,b]$ with countably many measurable sets removed.  Note that $C$ cannot have measure zero since it has nonmeasurable subsets.  It follows that $A\cup m^{-1}(B)$ 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 $(\inf D,\sup D)\cap C=(\inf D,\sup D)\cap m(A)=D\setminus\{\inf D,\sup D\}$, which is impossible since $(\inf D,\sup D)\cap C$ is measurable whereas $D\setminus\{\inf D,\sup D\}$ 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 $g\in G$ and $h\in H$ 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 $x<y\in m(A)$ then there is $z\in B$ 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 $A\cup m^{-1}(B)$ and $m(A)\cup B$ both have positive measure, neither $m(A)$ nor $B$ contain any subsets of positive measure, and for every $x<y\in m(A)$ there is $z\in B$ with $x<z<y$, and an analogous fact holds for $A$ and $m^{-1}(B)$, i.e. for any $x<y\in A$ there is $z\in m^{-1}(B)$ with $x<z<y$.
Is there any way to use those order properties to achieve a contradiction?
 A: 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.
