Let $u:[0,1]\to\mathbb{R}$ be a absolutely continuous function. It is know that $u'(x)$ exist almost everywhere and $u'\in L^1(0,1)$.

Let $A=\{s\in [0,1]:\ u'(s)\ \mbox{exist and}\ u'(s)\neq 0 \}$ and $B\subset A$. Is it true that if $m(u(B))=0$ then $m(B)=0$, where $m$ denote Lebesgue measure?

  • $\begingroup$ I would say yes, in that I think a differentiable function "stretches" (measure-wise) sets by the absolute value of the Jacobian, here the derivative, but I don't have anything more rigorous. Maybe you can use the proof that a.c functions send sets of measure zero to sets of measure zero to prove this (if true). $\endgroup$ – user99680 Jun 18 '14 at 19:15
  • $\begingroup$ @user99680, yes it is true and I alread had tried to use it, with no success. Maybe there is some trick way to use it, however, I could not find it. $\endgroup$ – Tomás Jun 18 '14 at 19:50
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  • $\begingroup$ @Tomás why is it true that $V_i=f(U_i)$ for some open set $U_i\subset[0,1]$? $\endgroup$ – Alex Schiff Jun 20 '14 at 1:54
  • $\begingroup$ Yes, you are right, this does not seem to be true @AlexSchiff. I will delete the update. $\endgroup$ – Tomás Jun 20 '14 at 13:29

Here goes one solution for this problem, it is lemma 3.45 from Leoni's book, A first course in Sobolev spaces.

Theorem: $I\subset\mathbb{R}$ an interval and $u:I\to \mathbb{R}$. Assume that that $u$ has derivative (finite or infinite) on a set $E\subset I$ (not necessarily measurable), with $m(u(E))=0$. Then $u'=0$ a.e. $x\in E$.

Proof: We have to prove that $E^{\star}=\{x\in E:\ |u'(x)|>0\}$ has zero measure. Write $$E^{\star}_k=\left\{x\in E^{\star}:\ |u(x)-u(y)|\ge \frac{|x-y|}{k},\ \forall\ y\in (x-1/k,x+1/k)\cap I\right\}$$

Note that $E^{\star}=\cup_{k=1}^\infty E^{\star}_k$, therefore, is it sufficiently to prove that $m(E^{\star}_k)=0$ for all $k$. To prove it, fix $k$ and let $J$ be any inteval, with length less than $1/k$. Define $F=J\cap E_k^{\star}$. We will prove that $m(F)=0$ for all $J$.

Indeed, note that, as $m(u(E))=0$ and $F\subset E$ then, for any $\epsilon>0$ there exist a sequence of disjoint open intervals $I_n$ such that $$m(u(F))\le \sum _{n=1}^\infty m(I_n)<\epsilon. \tag{1}$$

Let $F_n=u^{-1}(I_n)\cap F$ and note that $F_n$ cover $F$. Hence

\begin{eqnarray} m(F) &\le& \sum_{n=1}^\infty m(F_n) \nonumber \\ &\le & \sum _{n=1}^\infty \sup _{x,y\in F_n} |x-y| \nonumber \\ &\le& k\sum _{n=1}^\infty \sup_{x,y\in F_n} |u(x)-u(y)|, \tag{2} \end{eqnarray}

where in the last inequality we used the fact that $F_n\subset J\cap E_k^{\star}$. To conclude, note that as $x,y$ varies in $F_n$, $u(x),u(y)$ varies in $u(F)$. Then, combining $(1)$ and $(2)$ we get that $$m(u(F))\le k\epsilon,$$

which proves that $m(u(F))=0$.

  • $\begingroup$ This is very interesting, as it is known that if $u:I\to\Bbb R$ is absolutely continuous and increasing, $u$ maps sets of measure zero to measure zero. See, for example, Rudin and Royden. $\endgroup$ – Alex Schiff Jun 20 '14 at 17:31
  • $\begingroup$ Also, do you mean $E^*=\bigcap_{k=1}^\infty E_k^*$? $\endgroup$ – Alex Schiff Jun 20 '14 at 17:33
  • $\begingroup$ Hi @AlexSchiff, every $E^\star$ is contained in $E$, so it is indeed the union. $\endgroup$ – Tomás Jun 20 '14 at 17:49
  • $\begingroup$ I guess you're right, I dunno why I got confused. $\endgroup$ – Alex Schiff Jun 20 '14 at 22:43

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