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I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
The following exercise is Exercise 18 on p.39 in Exercises 2B in this book.

Exercise 18
Suppose $f:\mathbb{R}\to\mathbb{R}$ is differentiable at every element of $\mathbb{R}$. Prove that $f'$ is a Borel measurable function from $\mathbb{R}$ to $\mathbb{R}$.

I was not able to solve Exercise 18.
I found the following answer:

https://math.stackexchange.com/a/1803668/384082

I cannot understand this answer.

Let $A_n:=\{x:\frac{f(x+\frac{1}{n})-f(x)}{\frac{1}{n}}>a\}.$
hmakholm left over Monica wrote the following equality:

$\{x:f'(x)>a\}=\bigcup_{k=1}^\infty \bigcap_{n=k}^\infty A_n.$

I cannot prove that $\{x:f'(x)>a\}\supset\bigcup_{k=1}^\infty \bigcap_{n=k}^\infty A_n.$
Let $f(x):=\frac{1}{3}x^3$.
Let $a:=0$.
Then, $0\in A_n$ for any $n\in\{1,2,\dots\}$.
So, $0\in\bigcup_{k=1}^\infty \bigcap_{n=k}^\infty A_n$.
But, $\{x:f'(x)>a\}=\{x:x^2>0\}\not\ni 0$.

So, I guess hmakholm left over Monica's answer needs to be modified.
Am I wrong?

tchappy ha
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    You’re right that identity is incorrect. Instead you should use pointwise limits of Borel functions are Borel plus the other facts like quotients, differences, translations of Borel functions are Borel. – Nick Mar 13 '23 at 03:57
  • @Nick Thank you very much for your answer. – tchappy ha Mar 13 '23 at 04:00
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    An easy way to prove the result is to correct the answer from hmakholm left over Monica:

    $$ A_{n,m} = \left{ x ;\Big|; \frac{f(x+1/n)-f(x)}{1/n} > a + \frac{1}{m} \right} $$ is open, so $${x\mid f'(x)>a } = \bigcup_{m=1}^\infty \bigcup_{k=1}^\infty \bigcap_{n=k}^\infty A_{n,m} $$ is Borel.

     – Ramiro Mar 14 '23 at 00:29
  • @Ramiro Thank you very much for your answer. – tchappy ha Mar 14 '23 at 01:58

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