So my homework problem is to show that if a function $f: \mathbb{R} \to \mathbb{R}$ is differentiable everywhere, then its derivative f' is Borel measurable.
What I have for something to be Borel measurable is this:
The mapping $(X,\mathcal{M}) \stackrel{f}{\longrightarrow} (Y,\mathcal{N}), $ is measurable w.r.t. $\mathcal{M}$ and $\mathcal{N}$ if $f^{-1}(E) \in \mathcal{M}$ whenever $E \in \mathcal{N}$, for these two measurable spaces. Assuming $\overline{\mathbb{R}}$ instead of $\mathbb{R}$, $(\overline{\mathbb{R}},\mathcal{B}(\overline{\mathbb{R}})) \stackrel{f}{\longrightarrow} (\overline{\mathbb{R}},\mathcal{N}),$ is more specific to the problem, and now $f$ is Borel measurable function if $f^{-1}(E) \in \mathbb{B}(\overline{\mathbb{R}})$ whenever $E \in \mathcal{N}$.
This is great except my question, from Cohn's book Measure Theory, does not explicitly state the $\sigma$-algebras $\mathcal{M}$ and $\mathcal{N}$. Furthermore, I was hinted about using the derivative definition $f_n(x) = \frac{f(x+1/n)-f(x)}{1/n},$ for each fixed $n$. To me this seems like maybe constructing an infinite sequence of real-valued functions, $f_n(x)$, and then somehow showing $f_n(x)^{-1}(x) \in \mathbb{B}(\overline{\mathbb{R}})$ by equating $f$ with $f_n$?????
Alternatively, if I just assume $f$ is an `automorphism' (???) on the measurable space $(\overline{\mathbb{R}},\mathcal{B}(\overline{\mathbb{R}}))$, then for each fixed $n$, the definition of the derivative is a difference in measurable functions, with a multiplicative factor ($1/n$) on the front. But all this seems to show is that the derivative is measurable at each fixed $n$, whether or not it is part of a infinite sequence. Or am I to show that this limit in this sequence exists???
Any guidance would be greatly appreciated!
nate