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$f()$ is continuously differentiable on $[0,1]$. The set of $x$ such that $f(x) = 0$ is of measure zero. Can we conclude that there is a finite number of points such that $f(x) = 0$ ?

I think if $f()$ is just continuous, this would not be true: e.g. $f(x) = x sin(1/x)$ if $x \neq 0$ and $f(0) = 0$. This function is continuous but has infinitely many separated zeros (on a set of measure zero). The set of measure zero property only excludes ‘flatness’ of $f(x)$, but it does not exclude the possibility of infinitely many separated points.

However, with continuously differentiable functions, I think this is indeed true. The continuous differentiability implies some smoothness of the function. And since I know that the set of $x$ such that $f(x) = 0$ is of null measure, it gives some implicit restrictions on $f’(x)$ (i.e. the set such that $f(x) = 0$ and $f’(x) = 0$ is empty or at least of null measure). In which case I should get back the finite number of zeros following more traditional lines (e.g. this question).

But I’m not sure this is correct. And I cannot find any counterexample. And maybe I need something less restrictive than “continuously differentiable” to get to the property (e.g. just Lipschitz continuous?).

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If $F\subseteq [0,1]$ is a closed set, then there exists $C^{\infty}$ function $f:[0,1]\rightarrow \mathbb{R}$ such that $f(x)=0$ for $x\in F$. This even holds if you replace $[0,1]$ by a paracompact $C^{\infty}$ manifold $M$.

So for instance you can have smooth function which vanishes precisely on the standard Cantor set $C\subseteq [0,1]$.

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It's not true. You can build a counter-example similar to the function you considered.

Let $f(x)=x^3\sin(\frac{1}{x})$ and note that $f'(x)=3x^2 \sin(\frac{1}{x})-\frac{x}{2}\cos(\frac{1}{x})$ which extends continuously to $x=0$ by setting $f'(0)=0$. By the mean value theorem, we get that $f$ is differentiable at $0$. However, $f$ has exactly the same zeroes as $x\sin(1/x)$.

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