prove that this function has Lebesgue measurable image Denote by $\lambda$ the standard Lebesgue measure.
Let $E$ be a Lebesgue-measurable subset of $\mathbb{R}$ with $\lambda(E)<\infty$.  
By an initial segment of $E$ we mean a set $E'\subseteq E$ satisfying $E'<E\setminus E'$ (in the sense that $x<y$ for all $x\in E'$ and $y\in E\setminus E'$).  Note that initial segments of $E$ must always have the form $E\cap(-\infty,y)$ or $E\cap(-\infty,y]$ for some $y\in[-\infty,\infty]$.
It can be shown that for each $t\in[0,\lambda(E)]$, there exists an initial segment $E_t$ of $E$ with $\lambda(E_t)=t$.
Let us define the function $m:E\to[0,\lambda(E)]$ by the rule
$$m(x)=\inf\{t\in[0,\lambda(E)]:x\in E_t\}.$$
Conjecture 1.  The image $m(E)$ is a Lebesgue-measurable set.
Discussion.
(i)  Clearly, $m$ is order-preserving (i.e., nondecreasing) in the sense that $x\leq y$ if and only if $m(x)\leq m(y)$.  
(ii)  It can be shown that $m$ is measure-preserving in the following sense:  If $A\subseteq[0,\lambda(E)]$ is Lebesgue-measurable then $m^{-1}(A)$ is also Lebesgue-measurable with $\lambda(A)=\lambda[m^{-1}(A)]$.
(iii)  If $F\subseteq E$ and $m(F)$ is measurable then $\lambda[m(F)]=\lambda(F)$.
(iv)   There are definite counter-examples showing that $m$ need not be surjective. In fact, $[0,\lambda(E)]\setminus m(E)$ may even be uncountable.
Sorry to keep asking so many similar questions.  I keep running into these technical, seemingly obvious facts which are resistant to a simple proof (that I can find, anyway).
 A: We can prove that, for any $x \in E$, $m(x) = \lambda((-\infty,x] \cap E))$. Proof: 
For any $x \in E$ and any initial segment $E_t$ such that $x \in E_t$, we have that $(-\infty,x] \cap E \subseteq E_t$. Since $(-\infty,x] \cap E$ is an initial segment, it follows that $m(x) = \lambda((-\infty,x] \cap E))$.
Now, define for all $x \in \mathbb{R}$, $M(x) =  \lambda((-\infty,x] \cap E))$. It is clear that $M$ extends $m$ and that $M(E)=m(E)$. So we must prove that $M(E)$ is Lebesgue measurable. 
Note that, given any $x, y \in \mathbb{R}$, suppose without loss of generality that $y\geqslant x$, so we have:
$$ |M(y)-M(x)| = M(y)-M(x) = \lambda((x,y] \cap E))\leqslant y-x=|y-x|$$
So $M$ is a Lipschitz function. So the image by $M$ of any Lebesgue measurable set is Lebesgue measurable. So $M(E)$ is Lebesgue measurable. 
In more detail: 
Since Lebesgue measure is inner regular, there exist a set $N$ of measure zero and a sequence of compact sets $\{K_n\}_{n\ge 1}$, such that 
$$E=N\cup(\cup_{n=1}^\infty K_n)$$ Since Lebesgue measure is outer regular and $M$ is a Lipschitz function, $M(N)$ has measure zero. Since $M$ is continuous, $M(K_n)$ is compact for every $n\ge 1$. Therefore,
$$M(E)=M(N)\cup\big(\cup_{n=1}^\infty M(K_n)\big)$$ 
is Lebesgue measurable. 
