I want to show that if $F$ is continuous on $[a,b]$ then
$$\limsup_{h \rightarrow0, h>0} \frac{F(x+h)-F(x)}{h}$$
is measurable.
By the definition of $\limsup$ we can write \begin{align} &\limsup_{h \rightarrow0, h>0} \frac{F(x+h)-F(x)}{h}\\ &=\lim_{\delta\rightarrow 0}\left(\sup_{0<h<\delta} \ \frac{F(x+h)-F(x)}{h}\right)\\ &=\lim_{n\rightarrow \infty}\left(\sup_{0<h<\frac{1}{n}} \ \frac{F(x+h)-F(x)}{h}\right). \end{align} If we show $\sup_{0<h<\frac{1}{n}} \ \frac{F(x+h)-F(x)}{h}$ is measurable then we can use the fact that limit of a countable sequence of measurable functions is also measurable.
I want to show that $$\sup_{0<h<\delta}\frac{F(x+h)-F(x)}{h}=\sup_{0<h<\delta,h \in \mathbb{Q}}\frac{F(x+h)-F(x)}{h}$$ It suffices to show that $$\sup_{0<h<\delta,h \in \mathbb{Q}}\frac{F(x+h)-F(x)}{h}\geq \sup_{0<h<\delta}\frac{F(x+h)-F(x)}{h}$$. Given $\epsilon>0$, there exists $0<h_0<\delta$ such that $$\frac{F(x+h_0)-F(x)}{h_0}>\sup_{0<h<\delta}\frac{F(x+h)-F(x)}{h}-\epsilon$$ Let's fix $x$. Since $\frac{F(x+h)-F(x)}{h}$ is a continuous function with variable $h$, there exists a rational $0<q<\delta$ close enough to $h_0$ such that $$\left|\frac{F(x+q)-F(x)}{q}-\frac{F(x+h_0)-F(x)}{h_0}\right|<\epsilon$$ Therefore, $$\frac{F(x+q)-F(x)}{q}>\sup_{0<h<\delta}\frac{F(x+h)-F(x)}{h}-2\epsilon$$ Since $\epsilon$ is arbitrary, $$\sup_{0<h<\delta,h \in \mathbb{Q}}\frac{F(x+h)-F(x)}{h}\geq \sup_{0<h<\delta}\frac{F(x+h)-F(x)}{h}$$
A much easier way of proving that $$\sup_{0<h<\delta,h\in \mathbb Q} \frac {F(x+h)-F(x)} h =\sup_{0<h<\delta} \frac {F(x+h)-F(x)} h$$ is to note that if $$\frac {F(x+h)-F(x)} {h} \leq M$$ for $$0<h<\delta,h\in \mathbb Q$$ then $$\frac {F(x+h)-F(x)} {h} \leq M$$ for $$0<h<\delta .$$ Hence the two suprema are the same . [Take $M$ to be the RHS and note that each side of the equation is less than or equal to the other]