Let $f$ be continuously differentiable on $\left[a, b\right]$, and let $E$ be a measurable set. Prove that if $λ(f(E)) = 0$ then its derivative vanishes on $E$ almost everywhere.
$\begingroup$
$\endgroup$
2
-
$\begingroup$ What is $\lambda$? $\endgroup$– Ruvi LecamwasamJun 19, 2015 at 9:25
-
$\begingroup$ Sorry.it means Lebesgue measure $\endgroup$– SaraJun 19, 2015 at 9:27
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
5
Let $e\in E$ such that $f'(e)\ne 0$ (if no such $e$ exists then then $f'$ vanishes everywhere on $E$ and we are done). As the derivative of $f$ is continuous, this means we can choose an interval $C=[a,b]\ni e$ contained in $E$ such that $f'$ is non-zero on this set. On $C$, $f'$ is either positive or negative, we can without loss of generality suppose it is positive.
Think about $\lambda(f([a,b]))$. Does that help?
-
-
$\begingroup$ I think there is not a set such C.For example if E be the Cantor set,we can not find C $\endgroup$– SaraJun 20, 2015 at 6:31
-
$\begingroup$ The Cantor set has measure 0, so the problem is trivially correct in that case. I suppose a more annoying case would be a subset of R\Q. I think in this case the question doesn't make sense. Continuity and differentiability can't be defined just at a point, you need a small open neighbourhood of the point. $\endgroup$ Jun 20, 2015 at 7:54
-
$\begingroup$ I was thinking about it a bit more. The main point is that $f$ is continuously differentiable on $[a,b]$, and $E$ is contained in that set. Thus any $e\in E$ is contained in $[a,b]$, so you can choose some $C$ containing $e$. It doesn't matter if $C$ is contained in $E$ or not but if $E$ has some positive measure you can find an $e$ and $C$ such that $C\cap E$ has positive measure. $\endgroup$ Jun 21, 2015 at 2:00
-
$\begingroup$ What makes you think you can choose an interval $C = [a,b]\ni e$ contained in $E$? Besides $a$, $b$ were a poor choice since it was used in the problem formulation. It's not mentioned that $E$ has to be open. Let's for example say that $E = \{{a+b\over2}\}$ and $f(x)=x$. now good luck finding an interval round ${a+b\over2}$ contained in $E$. You can't even find such that $E\cap C$ has positive measure. $\endgroup$– skykingSep 14, 2015 at 14:28
$\begingroup$
$\endgroup$
I'll need a lemma that is proved after the main proof.
Suppose that $f'$ doesn't vanish everywhere, that is the set $D=\{x: f'(x)\ne0\}$ has positive measure. Now since $m(D) > 0$ there exist a $c\in D$ such that for every neighborhood $\Omega_c$ of $c$ we have $m(D\cup\Omega_c)>0$. Since $f'(c) \ne 0$ we have for some open interval $I$ round $c$ such that $f'$ has strictly the same sign as $f'(c)$.
Now we can see by variable substitution that $\int_{f(I)}\chi_{f(E\cap I)}~dy = \int_I \chi_{E\cap I}|f'| dx > 0$. This means that $m(f(E\cap I))>0$, but that contradicts our assumption that $m(f(E))=0$.
Lemma:
If $D$ be subset of $X$ such that $X$ can be covered by an countable set of measurable compact sets ($\mathbb R$ fulfils this requirement of $X$). If $D$ is measurable with $m(D)>0$: Then there is a $c \in D$ such that for every neighborhood $\Omega_c$ of $c$ we have $m(\Omega_c) > 0$.
Proof:
Now since $X=\bigcup C_j$ where $C_j$ are measurable compact sets we have that $m(D) = m(D\cap\bigcup C_j) = \sum m(D\cup C_j)$ which means that $m(D\cup C_j)>0$ for some $j$, let's call $D_j = D\cup C_j$. So it's enough to show that the claim is true for $D_j$.
Now consider the closure $\overline{D_j}$ of $D_j$. Since $\overline D \cap C_j\supseteq D_j$ is closed and also $\overline D\cap C_j \subseteq C_j$ we hav eth at $\overline{D_j} \subseteq C_j$ and therefore compact.
Now assume that the claim was false, that is for each $c\in D_j$ there was an neighborhood $\Omega_c$ such that $m(\Omega_c)=0$. Now $\bigcup_{c\in D_j}\Omega_c\subseteq D_j$ means that since $D_j$ is compact we can pick a finite subset $N\subseteq D_j$ such that $\bigcup_{c\in N}\Omega_c\subseteq D_j$.
But since $m(\Omega_c)=0$ then $0 = m(\bigcup_{c\in N}\Omega_c) \ge m(D_j)$ which contradicts that $m(D_j)>0$.
