# Show that a differentiable everywhere function $f\colon \mathbb{R}\to\mathbb{R}$ has a derivative which is Borel measurable

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

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Show first that a pointwise limit of Borel measurable functions is Borel measurable. The hint exhibits $f'$ as a pointwise limit of the Borel measurable functions $f_n$. –  t.b. Oct 15 '11 at 20:10
I haven't done it yet, but I see how $f'$ can be written as $f' = \alpha\,(f_1 - f_2) = \alpha\,(f_1 + (-1)\,f_2)$. There are Propositions from Cohn's book I can use to do these things. So that is what you suggest first, to show that $f'$ is Borel measurable if $f_1$ and $f_2$ are Borel measurable? Thanks –  nate Oct 15 '11 at 20:23
I'm not sure if I understand what you ask in this comment. You seem to try to argue why $f_n$ is Borel measurable, but that should be clear by elementary manipulations: It should be clear that the function $g_{n}(x) = f(x+(1/n))$ is Borel measurable and so is $f_n(x) = n(g_{n}(x) - f(x))$, no? Then come back to my first comment and write $$f'(x) = \lim_{n\to\infty} f_{n}(x) = \lim_{n\to\infty} \frac{f(x+(1/n)) - f(x)}{1/n}$$ where the rightmost limit exists (and is a real number) by the hypothesis that $f$ is differentiable everywhere. Thus, $f'$ is a pointwise limit of Borel functions. –  t.b. Oct 15 '11 at 20:28
Yes, I understand the elementary manipulations you spoke of. I also see how the algebraic limit theorem is used. I guess I just would like clarification on saying that the pre-image of the mapping is in the Borel $\sigma$-algebra, $f'^{-1}(x) \in \mathcal{B}(\overline{\mathbb{R}})$, because in this case the derivative is the mapping: $(\overline{\mathbb{R}},\mathcal{B}(\overline{\mathbb{R}})) \stackrel{f'}{\longrightarrow} (\overline{\mathbb{R}},\mathcal{B}(\overline{\mathbb{R}}))$. –  nate Oct 15 '11 at 20:56
I don't know why you insist on extended real-valued functions. You only need to consider functions $(\mathbb{R},\mathcal{B}(\mathbb{R})) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$ here. Also, when you write $f'^{-1}(x) \in \mathcal{B}(\overline{\mathbb{R}})$ that's not really what you want, you want to show $\{x \in \mathbb{R}\,:\,f'(x) \gt a\} \in \mathcal{B}(\mathbb{R})$ for all $a \in \mathbb{R}$ (for example) to prove Borel measurability of $f'$. –  t.b. Oct 15 '11 at 21:02