I need help solving this interesting result:
Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be differentiable. Prove that $f'$ Borel measurable.
I tried to start with, if $f$ is differentiable then $f'(x)$ exists for all $x∈ \mathbb{R}$.
Thank you in advance.
On
You could also say that: f=differentiable => f=continuous. Since f=differentiable => $\exists$f '(x) = $\lim_{h\to 0}$$\frac{f(x+h)-f(x)}{h}$ => f '(x) = lim$_n$$\frac{f(x+1/n)-f(x)}{1/n}$.
Then from Heine's deffinition (?)* $\forall$(x$_n$)$_n$$\in$$\mathbb{R}$ and $\epsilon$>0: x$_n$ $\rightarrow$ x$_0$ $\Rightarrow$ $\sigma$(f(x$_n$),f(x)) < $\epsilon$.
Suppose f$_n$(x) = $\frac{f(x+1/n)-f(x)}{1/n}$, f=continuous (so f(x+1/n)=continuous $\forall$n). Therefore so far we have: $\forall$x>0 and $\forall$$\epsilon$>0 $\exists$n$_0$$\in$$\mathbb{N}$ so that as n$\geq$n$_0$ then $\sigma$(f(x$_n$),f(x)) < $\epsilon$ $\Rightarrow$ (f$_n$)$_n$ $\rightarrow$ f p.w.
Then f=Baire-1 => f=Borel-meas. (this because f=lim$_n$f$_n$ $\Rightarrow$ f=limsup$_n$f$_n$, for (f$_n$)$_n$ = seqeuence of meassurable functions)
*I dont know how it is called in English. In Measure Theory or Real Analysis we call it in Greek "Arxi/Archi tis Metaforas" (= principle of transport).
PS: I'm an undergraduate student (mathematics), therefore there may be many mistakes (in english and in mathematical structure) I hope i helped.
$f'$ is the pointwise limit of the sequence:
$$f_n(x) = \frac{f(x+1/n) - f(x)}{1/n}$$