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Definition. For a real valued function $f$ and an interior point $x$ of its domain, the uppper derivative of $f$ at $x$ denoted by $overlineDf(x)$ is defined as follows: $$overlineDf(x)=lim_hrightarrow0left[ sup left fracf(x+t)-f(x)t: 0< right]$$

A Lemma in Royden and Fitzpatrick's Real Analysis book says:

Lemma. Let $f$ be an increasing function on the closed, bounded interval $[a,b]$. Then for each $alpha>0$, $$m^*xin (a,b) : overlineDf(x) geq
alpha leq frac1alpha[f(b)-f(a)].$$

The book proceeds to prove this by:

Let $alpha>0$. Define $E_alpha:=xin (a,b): overlineDf(x)geqalpha $. Choose $alpha' in (0,alpha)$. Let $mathscrF$ be the collection of closed, bounded intervals $[c,d]$ contained in $(a,b)$ for which $f(d)-f(c)geq alpha ' (d-c)$. Since $overlineDfgeq alpha$ on $E_alpha$, $mathscrF$ is a Vitali covering for $E_alpha$.

I have asked a question about this before, but I still not have understood how exactly can we say that $mathscrF$ is a Vitali covering for $E_alpha$. I have been told that we can use the limit definition of the upper derivative of $f$ to show that there exists a $delta>0$ such that $t<delta implies fracf(x+t)-f(x)tgeq alpha '$. My attempt to do this is as follows (which results in an exact opposite inequality): first, let $xin E_alpha$, and by limit definition, there exists a $delta>0$ such that for $hin(0,infty)$, $$h<delta implies left |sup_t left fracf(x+t)-f(x)t right -overlineDf(x) right |< fracalpha '2.$$ Now, since $overlineDf(x)geq alpha > alpha ' > frac12alpha '$, then $$h<delta implies -alpha < sup_t left fracf(x+t)-f(x)t right < alpha '.$$ If we take $0<|t|leq h<delta$, then $$fracf(x+t)-f(x)tleq sup_t left fracf(x+t)-f(x)t right <alpha'. $$ This is the complete opposite of what I want to show. Am I completely wrong in ways I don't even see? Anyone care to help me?

Maybe this viewpoint will help although it is a rather elaborate tutorial intended not to solve your immediate problem, but to get across an important idea. It is critical in being able to follow Vitali arguments that this becomes completely transparent to you and you see the connection between Vitali coverings and limits as immediate (not something to puzzle over). (Your computations that you suggest do not clarify this at all.)

Let $mathcal I$ be the collection of all
closed intervals $[u,v]$.

F. A collection of intervals $mathcal F subset mathcal I$ is a full cover of a set $E$ if for every point $xin E$ there is a
$delta>0$ so that all intervals $[u,v]$ that contain $x$ and are
smaller than $delta$ belong to $mathcal F$.

V. Dual to this is the Vitali cover: A collection of intervals $mathcal Vsubset mathcal I$ is a Vitali cover of a set $E$ if for
every point $xin E$ and any $delta>0$ there is at least one interval
$[u,v]$ belonging to $mathcal V$ that contains $x$ and is smaller
than $delta$.

The connection between limsup and liminf limits and F and V is, I think, obvious.
But, if not, just think for a while about why the collection
$$
mathcal C =left[u,v]: fracF(v)-F(u)v-u>alpha right$$
must be a full cover of the set where $underline D F(x)>alpha$.
But think too about why this same $mathcal C$ is a Vitali cover
of the larger set where $overline D F(x)>alpha$.

In vaguer words: at a point $x$ where $underline D F(x)>alpha$
all sufficiently small intervals $[u,v]$ that contain $x$ belong
to $mathcal C$ [that is why it is full]. While at a point $x$ where $overline D F(x)>alpha$
there are arbitrarily small intervals $[u,v]$ that contain $x$ and belong
to $mathcal C$ [that is why it is Vitali].

The connection between F and V is this duality: If $mathcal C$ fails to be full at a point
$x$ then it is because the
complement $
mathcal C' = mathcal I setminus mathcal C $
is a Vitali cover at $x$. If $mathcal C$ fails to be Vitali at a point
$x$ then it is because the
complement $
mathcal C' = mathcal I setminus mathcal C $
is full at $x$.

[Final comments: the question asks about the set where
$overline D F(x)geq alpha$. You can't use a Vitali argument immediately so you consider instead the set where
$overline D F(x)geq alpha >alpha'$. This is just a detail, but it can be annoying nonetheless making the argument even more obscure at first.

I am using the definition for upper/lower derivates as limsup and liminf of the ratio $$fracF(v)-F(u)v-u$$ for $u,vto x$ with $uleq x leq v$ and $0<v-u$. The OP version is a bit different. So some minor details would change.
]

You are supposed to take $epsilon=alpha-alpha'$ in the limit definition for $overlineD f(x) $. This will give you a $delta>0$ such that $$0<h<deltaimpliessupleftfracf(x+t)-f(x)t,0<>alpha'$$