I am trying to follow this proof that has F absolutely continuous on $[a,b]$ and $x$ is such that $F'(x)=0$ on a measurable subset $E$ of $[a,b]$ with measure of $E= b-a$. Anyways, here is the part that's confusing me.
Now let $E$ be the set of those $x \in (a, b)$ where $F'(x)$ exists and is zero. By our assumption $m(E) = b − a$. Next, momentarily fix $\epsilon > 0$. Since for each $x \in E$ we have $\lim_{h\to 0} \frac{F(x+h)-F(x)}{h}=0$ then for each $\eta > 0$ we have an open interval $I = (a_\eta, b_\eta) \subset [a, b]$ containing $x$, with $|F(b_\eta) − F(a_\eta)| \leq \epsilon(b_\eta − a_\eta)$ and $b_\eta − a_\eta < \eta$.
How is this possible? I thought given any $\epsilon$, there exists a $\delta$ such that $\frac{F(x+h)-F(x)}{h}<\epsilon$ if $|h|<\delta$. But here it says its true for any $\eta$.. How is this possible? Please someone explain. I would really appreciate it.
Let $\varepsilon > 0$ be arbitrary. Now from the definition of the derivative at $x$ (which is zero), we get a $\delta > 0$ such for every $h < \delta$ we have \begin{equation}\tag{1} \left| \frac{F(x+h) - F(x)}h \right| < 0 \end{equation}
Now to prove, based on (1), that for any $\eta > 0$ we can find an interval $(a_\eta,b_\eta)$ with $b_\eta - a_\eta < \eta$ and $$ \left| \frac{F(b_\eta) - F(a_\eta)}{b_\eta - a_\eta} \right| < 0\text. $$
First consider the case $\eta \le \delta$. By the above, we can set $a_\eta = x$, $b_\eta = x + \eta/2$ and have (1) apply with $h = \eta/2 \le \delta/2 < \delta$ while $b_\eta - a_\eta = \eta/2 < \eta$.
Now consider the case $\eta > \delta$. This time, we set $a_\eta = x$, $b_\eta = x + \delta/2$. Again, (1) applies with $h = \delta/2 < \delta$ and we have $b_\eta - a_\eta = \delta/2 < \delta < \eta$.