Dtyjlui

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 $E'<Esetminus E'$ (in the sense that $x<y$ for all $xin E'$ and $yin Esetminus E'$). Note that initial segments of $E$ must always have the form $Ecap(-infty,y)$ or $Ecap(-infty,y]$ for some $yin[-infty,infty]$.

It can be shown that for each $tin[0,lambda(E)]$, there exists an initial segment $E_t$ of $E$ with $lambda(E_t)=t$.

Let us define the function $m:Eto[0,lambda(E)]$ by the rule
$$m(x)=inf{tin[0,lambda(E)]:xin E_t}.$$

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 $xleq y$ if and only if $m(x)leq m(y)$.

(ii) It can be shown that $m$ is measure-preserving in the following sense: If $Asubseteq[0,lambda(E)]$ is Lebesgue-measurable then $m^{-1}(A)$ is also Lebesgue-measurable with $lambda(A)=lambda[m^{-1}(A)]$.

(iii) If $Fsubseteq E$ 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).

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 $ygeqslant x$, 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 ${K_n}_{nge 1}$, such that
$$E=Ncup(cup_{n=1}^infty K_n)$$ 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 $nge 1$. Therefore,
$$M(E)=M(N)cupbig(cup_{n=1}^infty M(K_n)big)$$
is Lebesgue measurable.