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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.

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    $\begingroup$ Hint: $f'$ is a limit of measurable functions. $\endgroup$
    – J.R.
    Commented Nov 12, 2016 at 17:29

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$f'$ is the pointwise limit of the sequence:

$$f_n(x) = \frac{f(x+1/n) - f(x)}{1/n}$$

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  • $\begingroup$ Thank you, but how do you prove that $f_n(x)$ is measurable? And with that you have proved that $f'$ is measurable but not Borel measurable... $\endgroup$ Commented Nov 16, 2016 at 9:50
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    $\begingroup$ @JennaTaylor the composition of Borel functions is a Borel function, this justifies that $f_n$ is Borel. Now the pointwise limit of Borel functions is also Borel. $\endgroup$
    – user384138
    Commented Nov 16, 2016 at 12:38
  • $\begingroup$ Yes, the composition of Borel measurable functions is Borel measurable. But how do you prove that each $f_n$ is Borel measurable? That is what I got left... $\endgroup$ Commented Nov 16, 2016 at 14:10
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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.

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  • $\begingroup$ What you call "Archi tis Metaforas" is called "Sequential Criteria" in English texts. $\endgroup$
    – AKP2002
    Commented Feb 17 at 16:39
  • $\begingroup$ Thank you, I didn't know that! $\endgroup$
    – Giwrgos K
    Commented Feb 21 at 14:04

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