TITLE: Is the derivative of differentiable function $f:\mathbb{R}\to\mathbb{R}$ measurable on $\mathbb{R}$?
QUESTION [9 upvotes]: Suppose we have a bounded differentiable function $f:\mathbb{R}\to[a,b]\subset\mathbb{R}$. Hence $f$ is continuous and measurable (in terms of standart Lebesgue measure) on $\mathbb{R}$.
I want $f$ to have bounded measurable derivative $f'$. Derivative $f'$ doesn't have to be bounded by default, so I add additional constrain on $f$, saying that $f$ is bounded differentiable function with bounded derivative. Now do I have to explicitly postulate that $f'$ is measurable, or do I have it automatically from already mentioned conditions?

REPLY [14 votes]: Assuming merely that $f$ is differentiable, $f'$ is measurable because $\{x\mid f'(x)>L\}$ is Borel for every $L$:
Since we know $f$ to be differentiable everywhere, the $x$ for which $f'(x)>L$ are those where $\dfrac{f(x+1/n)-f(x)}{1/n}> L$ for all sufficiently large $n$. And once we choose an $n$, the set
$$ A_n = \left\{ x \;\Big|\; \frac{f(x+1/n)-f(x)}{1/n} > L \right\} $$
is open, so
$$\{x\mid f'(x)>a \} = \bigcup_{k=1}^\infty \bigcap_{n=k}^\infty A_n $$
is Borel.