$f()$ 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?).