10

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 :)

user251257
  • 9,229

1 Answers1

7

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).

Carsten Führmann
  • 171
  • how can we show "A is a union of positive-length intervals. So by a standard argument  is a countable (disjoint) union of intervals" ? – Focus May 16 '18 at 15:32
  • @izimath First note that, if $A$ is a union of positive-length intervals from some set $I$, then $A$ is a union of disjoint positive-length intervals: the maximal intervals resulting from unions of intervals from $I$. Now we can put a different rational number into each of the disjoint intervals. Since the rationals are countable, so are the intervals. – Carsten Führmann May 18 '18 at 21:05
  • Why is the choice of $\delta'$ possible? – Aniruddha Deshmukh Dec 12 '20 at 12:40
  • 1
    @AniruddhaDeshmukh For increasing $f$, the numerator is non-negative. So we can make $\delta$ a little greater while staying above $c$. For decreasing $f$, the numerator is non-positive. So we can make $\delta$ a little smaller while staying above $c$. This can be formalized by stating that the fraction is equal to $c + \epsilon$ for some $\epsilon > 0$, and finding a positive resp. negative $\epsilon'$ such that $\delta' = \delta + \epsilon'$ and $\delta'$ satisfies the requirement. – Carsten Führmann Dec 12 '20 at 21:07