Upper Derivative and Increasing Function on a Compact Interval 
Definition. For a real valued function $f$ and an interior point $x$ of its domain, the upper derivative of $f$ at $x$ denoted by $\overline{D}f(x)$ is defined as follows: $$\overline{D}f(x)=\lim_{h\rightarrow0}\left[ \sup \left \{\frac{f(x+t)-f(x)}{t}: 0<|t|\leq h \right \} \right]$$

I am working through Royden and Fitzpatrick's proof of the following lemma:

Lemma. Let $f$ be an increasing function on the closed, bounded interval $[a,b]$. Then for each $\alpha>0$, $$m^*\{x\in (a,b) : \overline{D}f(x) \geq 
\alpha \} \leq \frac{1}{\alpha}[f(b)-f(a)].$$

Here is the relevant part of the proof giving me trouble.

Let $\alpha>0$. Define $E_{\alpha}:=\{x\in (a,b): \overline{D}f(x)\geq\alpha \}$. Choose $\alpha' \in (0,\alpha)$. Let $\mathscr{F}$ be the collection of closed, bounded intervals $[c,d]$ contained in $(a,b)$ for which $f(d)-f(c)\geq \alpha ' (d-c)$. Since $\overline{D}f\geq \alpha$ on $E_{\alpha}$, $\mathscr{F}$ is a Vitali covering for $E_{\alpha}$.

I realize that the questions am I about to pose have been asked here (indeed, I copied-&-pasted the relevant parts from that post), but астон вілла олоф мэллбэрг's answer is much too brief for my liking and the long string of comments, which appear to contain part of the answer to Kurome's question, are way to jumbled to get anything out of them. I have the same question as the OP in the link: why is $\scr{F} \neq \emptyset$ and why is $\scr{F}$ a Vitali covering of $E_\alpha$. Specifically, I don't understand the implication
$$t<\delta \implies\frac{f(x+t)-f(x)}{t}\geq\alpha'$$
holds (I tried unpacking the definition of the upper derivative, but I couldn't see it); nor do I understand how астон вілла олоф мэллбэрг is able to choose $d \in (a,b)$ such that $d-x < \delta$. And why does does $[x,d]$ having a length less than $\delta$ imply $\scr{F}$ is a Vitali covering of $E_\alpha$? That $\delta$ wasn't arbitrary. For ease of references, here is астон вілла олоф мэллбэрг's answer:

[астон вілла олоф мэллбэрг's answer]: Take any $x \in E_\alpha$. Now, since $\overline{D}f(x)\geq\alpha$, it follows that for some small $\delta$, $t<\delta \implies\frac{f(x+t)-f(x)}{t}\geq\alpha'$.


(The definition for the upper derivative above is slightly wrong, I will edit it)


Putting $t=d-x$, this means that $t<\delta \implies f(d)-f(x) \geq \alpha'(d-x)$. The interval $[d,x]$ is in $\mathscr{F}$ for every $d$ close enough to $x$,for arbitrary $x$ in $E_\alpha$. This makes $\mathscr{F}$ a Vitali covering for $E_\alpha$.

 A: Based on definition of upper derivative the following implication is wrong:
$$\exists \delta>0,0<|t|<\delta\implies \frac{f(x+t) - f(x)} {t} \geq\alpha'\tag{1}$$ The correct implication is that $$\exists \delta>0,0<h<\delta\implies\sup\left\{\frac{f(x+t)-f(x)}{t},0<|t|<h\right\}>\alpha'\tag{2}$$ This is obtained by using $\epsilon=\alpha-\alpha'$ in the limit definition for $\overline{D} f(x) $.
By choosing arbitrarily small values of $h$ this gives you infinitely many values of $t$ of arbitrarily small modulus such that ratio $(f(x+t) - f(x)) /t>\alpha' $ (why? If $\sup S>s$ then there are some members in $S$ definitely greater than $s$). Thus the collection $\mathscr{F} $ contains intervals of the form $[x, x+t] $ or $[x+t, x] $ depending on whether $t>0$ or $t<0$ and $\mathscr{F} $ is non-empty. Since there is an infinite supply of such $t$ with arbitrarily small modulus, it follows that $\mathscr{F} $ is a Vitali covering for $E_{\alpha} $.

I have to admit that the definition of upper derivative is somewhat difficult to decode and the proof in your book is very terse ("since $\overline{D} f(x) \geq \alpha$ on $E_{\alpha}$, $\mathscr{F} $ is a Vitali covering for $E_{\alpha} $"). Also the answer you have linked contains a wrong implication $(1)$. It also appears that there are many people who are facing problems with this particular proof.

On a closer inspection the definition of upper derivative as given in your post matches that of a limit superior $$\overline{D} f(x) =\limsup_{h\to 0}\frac{f(x+h)-f(x)}{h}=\limsup_{y\to x} \frac {f(y) - f(x)} {y-x} $$ and then the conclusion is obvious from the properties of $\limsup$.
Remember the two fundamental / defining properties of $\limsup$. If $M=\limsup_{x\to a} f(x) $ then


*

*for any number $A<M$ there are infinitely many values of $x$ arbitrarily close to $a$ such that $f(x) > A$. In order to make this even more confusing a number $A<M$ is written as $A=M-\epsilon$ where $\epsilon>0$. 

*for any number $B>M$ we have $f(x) <B$ for all values of $x$ sufficiently close to $a$. A number $B>M$ is usually written as $B=M+\epsilon $ where $\epsilon>0$.


Some books try to replace "arbitrarily" and "sufficiently" using another number $\delta>0$.
For the current question one needs only the first property of $\limsup$. 
