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I would like to find a "simple" proof of the following statement:

Let $f\in C(\mathbb{R})$. Let $A$ be a closed and countable subset of $\mathbb{R}$. Suppose $f'(x)=0$ for all $x\in\mathbb{R}\setminus A$. Show $f(x)$ must be constant.

Here, a simple proof to me can assume knowledge of Folland's Real Analysis Chapters 1-7. There are other questions on here and here asking for the same thing. This is a qualifying exam question and genuinely, I am stuck. Some thought suggests that $\mathbb{R}\setminus A$ might be dense and so I could try to argue that $f'(x)=0$ for $x\in A$ from there. I couldn't get this to work.

The Cantor-Lebesgue function shows that if $A$ being countable is replaced by $m(A)=0$, then the result is false. So, any proof must rely on $A$ being countable.

Thanks for any ideas.

  • See also https://math.stackexchange.com/questions/3311385/how-far-can-we-push-the-fundamental-theorem-of-calculus-for-riemann-integral?noredirect=1&lq=1 (not saying this gives a "simple proof") – Calvin Khor Sep 25 '21 at 05:50
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    What exactly is missing in Folland's Real Analysis' Chapters 1-7 that linked proofs use? – Conifold Sep 25 '21 at 06:00
  • @Conifold One of the linked proofs used Cousin's lemma which isn't something in Folland. I was hoping there's a proof that uses material in Folland which makes it quicker/simpler. Unless of course, the proof in that second link is the best one can hope for. –  Sep 25 '21 at 06:20
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    On Folland I find the following sentence: It is a highly nontrivial theorem that if F is continuous on [a, b], $F'(x)$ exists for every $x \in [a,b]\ A$ where A is countable, and $F'\in L^1[a,b]$, then F is absolutely continuous and hence can be recovered by integration – Richard Chen Sep 25 '21 at 08:16
  • I guess since Folland stated it highly nontrivial, there wouldn't be an easy proof(The proof in Rudin where $A=\varnothing$ uses some decomposition of functions or spaces – Richard Chen Sep 25 '21 at 08:22

1 Answers1

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Isn't it immediate from a cardinality argument?

Suppose $f$ is not constant. So there exists $a<b$ such that $\{a,b\}\subseteq f(\mathbb{R})$.

By IVT, $f(\mathbb{R})\supseteq[a,b]$.

Since $A$ is closed, the complement $\mathbb{R}-A$ is countably many open intervals. On each of these open intervals we have $f'=0$ so $f$ is constant by MVT. So $f(\mathbb{R}-A)$ is countable.

That leaves uncountably many values in $[a,b]$ that need to come from the countable set $A$, contradiction.

user10354138
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    Now I understand why those other proofs are so convoluted. They do not assume that $A$ is closed, only that it is countable. – Conifold Sep 25 '21 at 06:56
  • This is a very nice proof. I'm impressed that it did not rely on any measure theory or even knowledge of analysis at the level of Folland. Thank you! –  Sep 25 '21 at 16:45