TITLE: For any arbitrary $\mu$ there is an arbitrary interval $[x,x+h]$ where $\left| f(x+h) - f(x) \right| < \mu h$
QUESTION [2 upvotes]: In Royden's textbook, this is taken for granted in the proof that if $f$ is absolutely continuous and $f'(x) = 0$ almost everywhere, then $f$ is constant. He says the following

We aim to show that $f(a) = f(c)$ for any $c \in [a,b]$. Let $E \in (a,c)$ be the set of measure $c-a$ where $f'(x) = 0$. Let $\mu$ be an arbitrary positive number. To each $x$ in $E$ there is an arbitrarily small interval $[x,x+h]$ where
  $$\left| f(x+h) - f(x) \right| < \mu h$$

Why is this inequality true? And does this only hold where $f$ is a continuous function with $f'(x) = 0$? I would also appreciate some hint at a formal proof.

REPLY [0 votes]: If $h$ is infinitesimal then the ratio $\frac{f(x+h)-f(x)}{h}$ must be infinitely close to the derivative by definition, and since the derivative is assumed to be zero, the ratio must be infinitesimal. Therefore it is smaller than every real $\mu>0$ as required. For a nice introduction to elementary calculus based on infinitesimals see Keisler.
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