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I'm having some struggles with an aspect about something apparently trivial about Sard's theorem, but couldn't find anything online.

Let $f$ be a polynomial.

According to Sard's theorem, the image $f(Z)$ of the set of critical values $$Z = \{a \in X : f'(a) = 0\}$$ has measure zero.

What if I want to show that the set $Z$ itself has measure zero in the domain of $f$?

I feel like it's so simple but i just can't get behind it.

Captain Lama
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Nicola Zaugg
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  • The set of zeroes of a non-zero polynomial always has measure $0$. Apply that to $f'$. – Captain Lama Mar 13 '20 at 23:06
  • What measure is there on $X$? Are we in the context where $X\subseteq \Bbb R^n$ is Lebesgue measurable and $\mu(A)$ is just the Lebesgue measure of $A$ as a subset of $\Bbb R^n$? –  Mar 13 '20 at 23:12
  • @Gae.S. Good point -- since the OP refers to Sard's theorem, we should probably assume that $X$ is a manifold. In this case, "measure zero" is unambiguous without defining a measure on $X$, and may be different than "has Lebesgue measure zero with respect to some embedding $X \to \mathbb{R}^n$". I think the result should hold as long as $f : X \to \mathbb{R}$ is nonconstant and is induced by the composition of a proper embedding $X \to \mathbb{R}^n$ and a polynomial $\mathbb{R}^n \to \mathbb{R}$, although I could easily be wrong; I don't see immediately how to prove this. – diracdeltafunk Mar 13 '20 at 23:34

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In this case Sard's theorem is not needed. If $p$ is a nonzero polynomial, its set of zeros is finite by induction on its degree. If $f$ is not constant, then the set $Z$ of critical points is therefore also finite (it is the set of zeros of the polynomial $f'$), so it has measure zero!

Edit: @Captain Lama has suggested that you may be dealing with a polynomial in more than one variable. In this case, we cannot deduce that the set of zeros of $f'$ is finite, but we can still show that it has measure zero. You can do this with Fubini's theorem and induction on the number of variables – see here for an algebraic proof. You do need to make sure that $f$ is not constant, of course!

diracdeltafunk
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