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Let $f\in C^2\{[0,1],\mathbb{R})$ and $m\{x\in [0,1]:f(x)=0\}>0$. Prove that $$m\{x\in [0,1]:f'(x)=0\}>0,$$ where $m$ denotes Lebesgue measure.

I don’t have any clue to solve this exercise..

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One can also do this without using knowledge about Lebesgue points as follows:

The idea is that we can remove a countable number of points from a set so that the remaining set is “dense in itself”, as follows:

Let

$$

A := \{x \in [0,1] \mid f(x) = 0\}

$$

and

$$

N := \{x \in A \mid \exists \varepsilon_x > 0 \text{ such that } B_{\varepsilon_x} (x) \cap A \text{ is countable}\}.

$$

The **open** covering $(B_{\varepsilon_x}(x))_{x \in N}$ of $N$ has a countable subcover $(B_{\varepsilon_{x_n}}) (x_n)_n$, because $\Bbb{R}$ is second countable.

But this implies that $N \subset \bigcup_n [B_{\varepsilon_{x_n}}(x_n) \cap A]$ is countable as a countable union of countable sets.

Thus, $A \setminus N$ still has positive measure and for each $x \in A \setminus N$, the definition of $N$ shows that $B_\varepsilon (x) \cap (A \setminus \{x\})$ is nonempty (even uncountable), so that there is a sequence $(x_n)_n$ in $A \setminus \{x\}$ with $x_n \to x$.

I leave it to you to conclude $f'(x) = 0$ for all $x \in A \setminus N$. Hence, even

$$

m(\{x \mid f'(x) = 0\}) \geq m(A \setminus N) = m(A) > 0.

$$

Here is another answer, inspired by PhoemueX’s answer: Like PhoemuX, define

$$A := \{x \in [0,1] \mid f(x) = 0\}.$$

Now note (as a direct consequence of the definition of the derivative) that if $x\in A$ and $f'(x)\ne0$, then $x$ is an *isolated* point in $A$. No set of real numbers can have more than a countable set of isolated points, and a countable set has measure zero, of course. So $f'(x)=0$ for all $x\in A$ except for a $x$ in a set of measure zero.

This proof only requires the measurability of $f$ and the existence of $f'(x)$ for almost every $x\in[0,1]$ – the stated assumptions in the problem are much stronger than necessary.

**Hint:** Consider points of density of the set $\{x\in [0,1]:f(x)=0\}$.

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