"Constant" points of an increasing function

Asked Mar 01 '18 at 04:19

Active Mar 01 '18 at 04:19

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I just learned that the discontinuity points of an increasing function are countable. Then what about the "constant" points?

Let $f:(a,b)\rightarrow R$ be an increasing function on a bounded interval, i.e., $f(x)\leq f(y)$ if $a<x<y<b$. Further assume that $f$ is differentiable on $(a,b)$. Let $C=\{x\in (a,b)| \ f'(x)=0\}$. Then, is it true that $f(C)$ is countable, or at least $m(f(C))=0$ where $m$ is the Lebesgue measure on the real line?

I suspect it is true since the "inverse" of $f$ is also increasing so that its discontinuity point is countable. How should I show it with rigor? Any help would be really appreciated. Thanks!

asked Mar 01 '18 at 04:19

Mhr

The way you've defined allows constant functions. – Ovi Mar 01 '18 at 04:21
@Ovi Any problem arises when $f$ is constant? – Mhr Mar 01 '18 at 04:22
Well for constant functions the derivative is $0$ everywhere so $C = (a, b)$, which is uncountable. – Ovi Mar 01 '18 at 04:23
@Ovi yes but $f(C)$ is a singleton so it is countable. – Mhr Mar 01 '18 at 04:24
Oh sorry my mistake I read the question wrong. – Ovi Mar 01 '18 at 04:25
See Monotone functions and non-vanishing of derivative. – dxiv Mar 01 '18 at 04:35
OK. I solved it. By Sard's theorem, $m(f(C))=0$ if we further assume that $f$ is $C^1$. – Mhr Mar 01 '18 at 04:41

"Constant" points of an increasing function

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