How to show that the Dini Derivatives of a measurable function is measurable? Let $f:(0,1)\to\mathbb R$ be measurable. Then, the (right upper) Dini derivative
$$ D^+ f(x) = \limsup_{h\to 0^+} \frac{f(x+h) - f(x)}{h} $$
is also measurable (a well known result of Banach). 
Can someone give me a source in English (or German) or a proof sketch? If it makes thing much easier, we can assume $f$ is monotone (in that case please don't argue that $f$ is differentiable a.e. That feels like cheating :)
 A: A proof for Dini derivatives of measurable functions can be found in [1]. (Thanks to PhoemueX for the tip, I cannot upvote because I am too new here.) But the proof for monotone functions is much simpler, so it makes sense to present it here. I'll show that the upper right-hand Dini derivative $D^+ f$ of a monotone function $f$ is measurable. The other three Dini derivatives can be treated similarly. Define 
$$f_h(x) := \sup_{0<\delta<h}\frac{f(x+\delta)-f(x)}{\delta}$$
Then $D^+f(x) = \lim_{n\to\infty} f_{1/n}(x)$. So it suffices to show that $f_h$ is measurable for any $h$. By basic measure theory, it suffices to show that the set 
$$A := \{x:f_h(x)>c\}$$ 
is measurable for every real number $c$. So let $x_0\in A$. This means that there is a $\delta\in(0,h)$ such that 
$$\frac{f(x_0+\delta)-f(x_0)}{\delta} > c$$ 
Now suppose $f$ is increasing. Choose a $\delta'\in(\delta,h)$ such that
$$\frac{f(x_0+\delta)-f(x_0)}{\delta'} > c \qquad (1)$$ 
Choose $x\in[x_0+\delta-\delta',x_0]$. Then
$$
c < \frac{f(x_0+\delta)-f(x_0)}{\delta'}
\leq \frac{f(x+\delta')-f(x)}{\delta'} \qquad (2)
$$
where the second inequality holds because $f$ is increasing. So $x\in A$, and therefore $[x_0+\delta-\delta',x_0]\subseteq A$. 
Now suppose $f$ is decreasing. Then choose $\delta'\in(0,\delta)$ such that (1) holds, and choose $x\in[x_0,x_0+\delta-\delta']$. Then (2) holds again, this time because $f$ is decreasing. So $[x_0,x_0+\delta-\delta']\subseteq A$.
Either way it follows that $A$ is a union of positive-length intervals. So by a standard argument $A$ is a countable (disjoint) union of intervals, and therefore measurable. This completes our proof.
[1] Kannan, R. and Krueger, C. K., Advanced analysis on the real line (Universitext, Springer-Verlag, 1996).
