A.e. vanishing function has a.e. vanishing derivative?

Question

Let $f:[a,b]\rightarrow\mathbb{R}$ be differentiable almost everywhere and $f(x)=0$ almost everywhere. Does this imply $f'(x)=0$ almost everywhere?

Answer 1

Proof of your statement:

Take any $x$ from the set $\{x|f'(x) \text{ exists}\}$, which has measure 1. Consider the set with positive measure $[x-\frac 1 n,x+\frac 1 n]\cap [a,b]$. Due to $f=0$ almost everywhere, there exists an $x_n$ in this set with $f(x_n)=0$. By the Squeeze Theorem we have $x_n\rightarrow x$.

As existence of $f'(x)$ implies continuity at $x$ we have $f(x)=0$ and there therefore for any $n$:

$$\frac{f(x_n)-f(x)}{a_n-x}=0 $$

As the limit $f'(x)$ equals any sequential limit, this concludes the proof.