derivative of a continuously differentiable function 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.
 A: 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? 
A: 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 $m(D_j)>0$.
