Suppose $F$ is continuous on $[a,b]$.Show that $D^+F(x)=\limsup_{h\to 0\,h\gt0} \frac{F(x+h)-F(x)}{h} $is measurable. 
Suppose $F$ is continuous on $[a,b]$.Show that $$D^+F(x)=\limsup_{h\to
 0\,h\gt0} \frac{F(x+h)-F(x)}{h} $$is measurable.
  [Hint:the continuity of F allows one to restrict to countably many h in taking the limsup.]  

By the hint,if I can show$$D^+F(x)=\limsup_{n\to
 \infty} \frac{F(x+\frac{1}{n})-F(x)}{\frac{1}{n}} $$
Then let $F_n(x)=$$\frac{F(x+\frac{1}{n})-F(x)}{\frac{1}{n}}$ and we have$$D^+F(x)=\lim_{n\to \infty}\lim_{m\to \infty}\max\{F_n(x),F_{n+1}(x),...,F_{n+m}(x)\}$$
So $D^+F(x)$ is measurable.However,I am quite confused about the hint,why the continuity of F allows one to restrict to countably many h in taking the limsup.  

EDIT:
         How to prove$$D^+(F)(x) = \lim_{\delta \to 0} \sup_{0<h<\delta} \frac{F(x+h) - F(x)}{h} = \lim_{n\to\infty} \sup_{0<h<\frac{1}{n}} \frac{F(x+h) - F(x)}{h}$$
  and I have learnt A continuous mapping is determined by its values on a dense set,however,why this implies we can restrict $h$ in $Q$.

 A: Quick answer: Restrict $h$ in $\mathbb{Q}^+ := \mathbb{Q}\cap(0,\infty)$. If I have time, I will write down the full proof at later time.
Full answer:
Let $f:[a,b]\rightarrow\mathbb{R}$ be a continuous function. Define
$F:[a,b)\rightarrow\bar{\mathbb{R}}$ by 
$$
F(x)=\limsup_{h\rightarrow0+}\frac{f(x+h)-f(x)}{h},
$$
where $\bar{\mathbb{R}}=[-\infty,\infty]$. We go to show that $F$
is Borel mesurable.
Notation: Let $\mathbb{Q}^{+}=\mathbb{Q}\cap(0,\infty)$. Extend the
domain of $f$ to $\mathbb{R}$ by setting $f(x)=f(b)$ whenever $x\in(b,\infty)$
and $f(x)=f(a)$ whenever $x\in(-\infty,a)$. We still denote that
function after extension by $f$. Define $g:\mathbb{R}\rightarrow\bar{\mathbb{R}}$
by 
$$
g(x)=\limsup_{h\rightarrow0+}\frac{f(x+h)-f(x)}{h}.
$$
We assert that for each $x\in\mathbb{R}$, 
$$
g(x)=\lim_{n\rightarrow\infty}\sup_{h\in(0,\frac{1}{n})\cap\mathbb{Q}}\frac{f(x+h)-f(x)}{h}.
$$
Let $x\in\mathbb{R}$ be fixed. To make the demonstration precise,
let us recall the definition of $\limsup$:
$$
g(x)=\lim_{\delta\rightarrow0+}\sup_{h\in(0,\delta)}\frac{f(x+h)-f(x)}{h}.
$$
For each $\delta>0$, let
$$
\alpha(\delta)=\sup_{h\in(0,\delta)}\frac{f(x+h)-f(x)}{h}\in\bar{\mathbb{R}}.
$$
Clearly $\alpha(\delta_{1})\leq\alpha(\delta_{2})$ whenever $\delta_{1}<\delta_{2}$,
so $\lim_{\delta\rightarrow0+}\alpha(\delta)$ always exists (in $\bar{\mathbb{R}})$.
In particular, $\lim_{\delta\rightarrow0+}\alpha(\delta)=\lim_{n\rightarrow\infty}\alpha(\frac{1}{n})$.
It follows that $g(x)$ can be written as 
$$
g(x)=\lim_{n\rightarrow\infty}\sup_{h\in(0,\frac{1}{n})}\frac{f(x+h)-f(x)}{h}.
$$
We go to show that: For each $n\in\mathbb{N}$, 
$$
\sup_{h\in(0,\frac{1}{n})\cap\mathbb{Q}}\frac{f(x+h)-f(x)}{h}=\sup_{h\in(0,\frac{1}{n})}\frac{f(x+h)-f(x)}{h}.
$$
Obviously, $LHS\leq RHS$. Let $\varepsilon>0$ be arbitrary. On the
other hand, let $h\in(0,\frac{1}{n}$) be fixed. Denote $l=\frac{f(x+h)-f(x)}{h}$.
Observe that the function $t\mapsto\frac{f(x+t)-f(x)}{t}$ is continuous
at $t=h$, so there exists $\delta>0$ such that 
$$
|\frac{f(x+t)-f(x)}{t}-l|<\varepsilon
$$
whenever $t\in(h-\delta,h+\delta)$. By density of $\mathbb{Q}$,
we may choose a $t\in(0,\frac{1}{n})\cap\mathbb{Q}\cap(h-\delta,h+\delta)$.
Then 
\begin{eqnarray*}
l & \leq & \frac{f(x+t)-f(x)}{t}+\varepsilon\\
 & \leq & LHS+\varepsilon.
\end{eqnarray*}
Hence $RHS\leq LHS+\varepsilon$. It follows that $RHS\leq LHS$ becasuse
$\varepsilon$ is arbitrary. Now, it is clear that $g(x)=\lim_{n\rightarrow\infty}\sup_{h\in(0,\frac{1}{n})\cap\mathbb{Q}}\frac{f(x+h)-f(x)}{h}$.
Finally, for each $h\in\mathbb{Q^{+}},$let $f_{h}:\mathbb{R}\rightarrow\mathbb{R}$
be a function defined by 
$$
f_{h}(x)=\frac{f(x+h)-f(x)}{h}.
$$
Clearly $f_{h}$ is continuous and hence Borel. Note that for each
$n\in\mathbb{N}$, the family $\{f_{h}\mid h\in\mathbb{Q}\cap(0,\frac{1}{n})\}$
is at most countable, so 
$$
\sup_{h\in\mathbb{Q}\cap(0,\frac{1}{n})}f_{h}
$$
 is also a Borel function. Finally, 
$$
g=\lim_{n\rightarrow\infty}\sup_{h\in\mathbb{Q}\cap(0,\frac{1}{n})}f_{h},
$$
so it is also a Borel function. Note that $F$ is just a restriction of $g$.
