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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).

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1 Answers1

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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.

Del
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  • How do you get the second equality? –  Jan 01 '17 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 01 '17 at 14:27
  • Thanks for your comment. That is exactly where I am confused. Would you elaborate the identity in your comment? –  Jan 01 '17 at 14:35
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    @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 01 '17 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? –  Jan 01 '17 at 14:44
  • I have posted a follow-up question. –  Jan 01 '17 at 14:51
  • I see your point now. Thanks! –  Jan 01 '17 at 15:15
  • @Jack You're welcome! – Del Jan 01 '17 at 15:19