# How to show that the Dini derivatives of a continuous function is measurable?

Suppose $F:[a,b]\to\mathbb{R}$ is continuous. Show that $$D^+(F)(x)=\limsup_{h\to 0+}\frac{F(x+h)-F(x)}{h}$$ is measurable.

This question is related to this one. But specifically I would like to follow the hint in Stein-Shakarchi's Real Analysis:

the continuity of $F$ allows one to restrict to countably many $h$ in taking the $\limsup$.

I don't quite understand the hint. I guess one might aim at getting $$D^+(F)(x)=\lim_{m\to\infty}\sup_{n\geq m}\biggr[F\bigr(x+\dfrac1n\bigr)-F(x)\biggr]\cdot n\tag{1}$$ But I don't see how to use the continuity of $F$ to get (1).

Using continuity you can prove that $$D^+(F)(x)=\limsup_{h\to 0^+,\, h\in \mathbb Q} \frac{F(x+h)-F(x)}{h}=\lim_{n \to\infty} \sup_{h\in\mathbb Q\cap (0, \frac1n) }\frac{F(x+h)-F(x)}{h}$$ and measurability follows from taking countable supremum and infimum.

• How do you get the second equality?
– user9464
Jan 1, 2017 at 13:19
• @Jack It is the definition for me. The point is that $\lim_{n\to\infty} \sup_{h\in (0,1/n)} g(h)=\lim_{n\to\infty} \sup_{h\in (0,1/n)\cap\mathbb Q}g(h)$ whenever $g$ is continuous, and the first is the definition of $\limsup_{h\to 0^+}$.
– Del
Jan 1, 2017 at 14:27
• Thanks for your comment. That is exactly where I am confused. Would you elaborate the identity in your comment?
– user9464
Jan 1, 2017 at 14:35
• @Jack For every $n$, from the continuity of $g$ it follows that $\sup_{h\in (0,1/n)} g(h)=\sup_{h\in (0,1/n)\cap\mathbb Q}g(h)$ because $(0,1/n)\cap\mathbb Q$ is dense in $(0,1/n)$. In particular, the equality still holds after passing to the limit (which exists by monotonicity).
– Del
Jan 1, 2017 at 14:38
• I see. If I understand correctly, one essentially needs to show that for each $n$, the map $x\mapsto\sup_{h\in\mathbb{Q}\cap(0,1/n)}\frac{F(x+h)-F(x)}{h}$ is measurable. How did you get to that?
– user9464
Jan 1, 2017 at 14:44