I am trying to show that if $F:[a,b]\rightarrow\mathbb{R}$ is continuous and of bounded variation then $g(x)=\limsup_{h\rightarrow 0, h>0} \frac{F(x+h)-F(x)}{h}$ is a Lebesgue measurable function.
Here's what I tried. WTS that for all $\alpha\in\mathbb{R}$ we have the set $$\left\{x\in[a,b]|\limsup_{h\rightarrow 0, h>0} \frac{F(x+h)-F(x)}{h}<\alpha\right\}$$ is Lebesgue measurable. By definition of $\limsup$, the above says $$\left\{x\in[a,b]\;|\;\lim_{\epsilon\rightarrow 0}(\sup\{ \frac{F(x+h)-F(x)}{h}|h\in (0,\epsilon]\}) <\alpha\right\}$$ is Lebesgue measurable (1). I first show that for all $\epsilon$, the set $\{x\in[a,b]\;|\;\sup\{ \frac{F(x+h)-F(x)}{h}|h\in (0,\epsilon]\}<\alpha\}$ is measurable. Let $\{r_1,r_2\cdots\}=(0,\epsilon] \cap \mathbb{Q}$. Then $$\{x\in[a,b]\;|\;\sup\{ \frac{F(x+h)-F(x)}{h}|h\in (0,\epsilon]\}<\alpha\}=\cap_{n\in\mathbb{N}}\{x\in[a,b]\;|\;\ \frac{F(x+r_n)-F(x)}{r_n}<\alpha\}.$$ Each set in LHS is measurable since $F$ is continuous, therefore measurable. Hence LHS is countable intersection of measurable sets, therefore measurable. To show (1), I just write use the limit as $n\rightarrow\infty$ of $\frac{1}{n}$ instead of $\epsilon\rightarrow 0$.