9
$\begingroup$

Let $f:\mathbb{R}\to \mathbb{R}$ be $C^1$ function and $K = \{x : f'(x) = 0 \} $. Show that $\mu \left(f\left(K\right)\right) = 0$, where $\mu$ is Lebesgue measure.

My attempt was following: $$\mu \left(f\left(K\right)\right)= \int_{f(k)} 1 dy \stackrel{(*)}{=} \int_{K}f'(x) dx = \mu\left(K\right) \cdot 0 = 0$$ but we cannot substitute $y = f(x)$ at $(*)$ like that. I was told that there exists quite elementary proof (not using Sard's theorem) so I'm looking for it.

$\endgroup$
1
  • $\begingroup$ What am I missing? It seems like f(x)=0 would be a counterexample since the set of points at which f'(x)=0 is everything (measure 1) $\endgroup$
    – Alex Li
    Jan 27, 2022 at 17:41

1 Answer 1

3
$\begingroup$

The trick is something like this: Suppose you have some point $x$ where the derivative is 0. Then, because of the continuity of the derivative, you can find a small interval around $x$ of length at most $\delta$, and on which $f'$ is at most $\frac{1}{n}$. Now, the measure of the image of that interval has to be no more than $\frac{\delta}{n}$. This follows from simple properties of the derivative of a real function.

Now supposing that you covered all the points of $[0,1]$ which have derivative of $f$ equal to $0$ with such intervals, and that they are non-overlapping. The measure of all the intervals is no more than $1$. The measure of all the intervals in the image space is no more than $\frac{1}{n}$, and this contains the image of all the points where $f'$ is zero. As $n$ is arbitrary you are done.

You have to make sure that you can cover the set by such intervals, but this is not too tricky. If you proved the theorem for domain $[0, 1]$, you of course can also prove it for $\mathbb{R}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .