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?