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I'm trying to prove the following theorem.

Let $f:\mathbb{R} \to \mathbb{R}$ be a function and define $$\mathcal A:=\{a\in\mathbb{R}: f\text{ is differentiable in } a \text{ and } f'(a)=0\},$$ then $\mathcal A$ is a Borel set and $f(\mathcal A)$ has measure $0$.

I managed to prove that $\mathcal A$ is a Borel set but I'm struggling to prove that $f(\mathcal A)$ has measure zero. To do so I can use the following theorem:

Let $\epsilon>0$, $\mathcal Y$ be a Lebesgue-integrable subset of $\mathbb{R}$ and $f:\mathbb{R} \to \mathbb{R}$ a function s.t. $\forall\,y\in\mathcal Y$, $f$ is differentiable in $y$ and $|f'(y)| < \epsilon$, then $\exists\,\mathcal B\subset\mathbb{R}$ Borel s.t. $f(\mathcal Y)\subset\mathcal B$ and $\lambda(\mathcal B)\leq \epsilon \lambda(\mathcal Y)$.

Now I assume I have to use this theorem for $\mathcal Y=\mathcal A$, so that $\exists\mathcal B\subset\mathbb{R}$ s.t. $f(\mathcal A)\subset\mathcal B \text{ and } \lambda(\mathcal B) \leq \lambda(\mathcal A)$, and thus $\lambda(f(\mathcal A)) \leq \lambda(\mathcal B) \leq \epsilon \lambda(\mathcal A) \ \ \forall \epsilon>0$ so that $f(\mathcal A)$ has measure $0$ if $\lambda(\mathcal A)<\infty$.

However to use the theorem mentioned above I need $\mathcal A$ to be Lebesgue-measurable, yet I don't know how to prove this. Also, I assume if $\mathcal A$ is Lebesgue-measurable $\lambda(A)<\infty$ so that my proof is correct; but even if $\mathcal A$ is Lebesgue-measurable; is that true?

Ice Tea
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Wessel
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  • Boreal measurable sets are always Lebesgue measurable. To get $\lambda(A)<\infty$, look at $A \cap [-n,n]$ instead. – PhoemueX Feb 02 '15 at 20:51
  • Thanks! Do you mean $\lambda(A\cap[-n,n])=0 \ \ \forall n\in\mathbb{N}$ and $A=\cup_{n\in\mathbb{N}}A\cap[-n,n]$ s.t. $\lambda(A)=\lambda(\cup_{n\in\mathbb{N}}(A\cap[-n,n])\leq \sum_{n\in\mathbb{N}} \lambda(A\cap[-n,n])=\sum_{n\in\mathbb{N}}$ $0 = 0$? – Wessel Feb 03 '15 at 13:59
  • More or less. In the end, you do not want to show that $A$ has measure zero, but that $f(A)$ has measure zero, so you will use that $f(A) = \bigcup_n f(A \cap [-n,n])$ is a null-set as a countable union of null-sets. – PhoemueX Feb 03 '15 at 14:34
  • Well, by the theorem I mentioned in the original post (which I'm supposed to use, according to this exercise) it is enough to show $\lambda(A)$ to be finite, so that $\lambda(f(A))$ is zero, as $\lambda(A)\leq\epsilon\lambda(f(A))$ holds $\forall \epsilon>0$. I think I just correctly proved $\lambda(A)$ to be finite (zero, in fact), but I must admit I do not directly see a way to prove $\lambda(A\cap[-n,n]) = 0$, which I used to do so. – Wessel Feb 03 '15 at 14:56
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    My idea was to apply the theorem to $A \cap [-n,n]$ instead of $A$. You then just have to show $\lambda(A \cap [-n,n]) < \infty$, which is trivial. In the end, you can use my union argument from the previous comment to conclude that $f(A)$ is a null-set, because $f(A \cap[-n,n])$ is, for each $n$. – PhoemueX Feb 03 '15 at 15:02
  • Oh I see what you mean now, thank you. Forget the question I just posted in this comment haha, I figured it out already so I'm editing this. Thank you! – Wessel Feb 03 '15 at 15:23
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    Is this not just Sard's Theorem? – Pete Caradonna Sep 15 '15 at 09:15
  • @PeteCaradonna : Here, Sard's theorem requires that $f$ is once continuously differentiable. We only have that $f$ is differentiable at various points, but we have no promise that this derivative is continuous. In fact, differentiable but not continuously differentiable is the case for "most" functions, but not the functions we usually write down. See https://math.stackexchange.com/a/3338788/123905 for an example. – Eric Towers Oct 24 '21 at 22:54

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