# Rudin RCA Theorem 7.1

In Rudin's Real and Complex Analysis, Theorem 7.1 is provided without a proof. Unfortunately, I have a difficulty in proving it by myself. It says:

Suppose $$\mu$$ is a complex Borel measure in $$R^1$$ and $$\begin{equation*} f(x) = \mu((-\infty,x)) \quad\quad (x\in R^1). \end{equation*}$$ If $$x\in R^1$$ and $$A$$ is a complex number, each of the following two statements implies the other:
(a) $$f$$ is differentiable at $$x$$ and $$f'(x) = A$$.
(b) To every $$\epsilon>0$$ corresponds a $$\delta>0$$ such that $$\begin{equation*} \left|\frac{\mu(I)}{m(I)} - A \right| < \epsilon \end{equation*}$$ for every open segment $$I$$ that contains $$x$$ and whose length is less than $$\delta$$. Here $$m$$ denotes Lebesgue measure on $$R^1$$.

Here is my trial to prove that (a) imples (b).

Assume (a) and let $$\epsilon>0$$. Then, there exists $$\delta>0$$ satisfying $$\begin{equation*} \frac{|f(x') - f(x) - A(x'-x)|}{|x'-x|} < \frac{\epsilon}{2} \end{equation*}$$ for all $$0 < |x'-x| < \delta$$. Let $$x-\delta < \alpha < x < \beta < x+\delta$$ such that $$\beta-\alpha < \delta$$. Then, $$\begin{equation*} \frac{|f(\alpha) - f(x) - A(\alpha - x)|}{|\alpha - x|} = \left| \frac{\mu([\alpha,x))}{x - \alpha} - A \right| < \frac{\epsilon}{2} \end{equation*}$$ and $$\begin{equation*} \frac{|f(\beta) - f(x) - A(\beta - x)|}{|\beta - x|} = \left| \frac{\mu([x,\beta))}{\beta-x} - A \right| < \frac{\epsilon}{2}. \end{equation*}$$ Thus, $$\begin{equation*} \left|\frac{\mu([\alpha,\beta))}{\beta-\alpha} - A\right| \le \left| \frac{\mu([\alpha,x))}{x - \alpha} - A \right| + \left| \frac{\mu([x,\beta))}{\beta-x} - A \right| < \epsilon. \end{equation*}$$

Here is the point I stuck. I could not make $$\mu([\alpha,\beta))$$ to $$\mu((\alpha,\beta))$$. I guess that either $$\mu(\{\alpha\}) = 0$$ or $$\mu([\alpha_i,\beta)) \to \mu((\alpha,\beta))$$ as $$\alpha_i\to\alpha$$ is necessary. Any help will be appreciated.

You need to use sequences to remedy your problem. In the first place, you may as well assume $$A=0.$$ Then, there is a $$\delta>0$$ such that $$|y-x|<\delta\Rightarrow |f(y)-f(x)|<\epsilon |y-x|.$$ Choose $$(a,b)$$ containing $$x$$ such that $$|b-a|<\delta$$ and a sequence $$(t_n)$$ such that $$t_n for each $$n$$ and such that $$(t_n)$$ decreases to $$a$$. Then, $$\mu([t_n,b))=|f(b)-f(t_n)|\le |f(b)-f(x)|+|f(x)-f(t_n)|\le \epsilon|b-t_n|\le \epsilon |b-a|$$. And since $$\bigcup_n[t_n,b)=(a,b)$$, we have $$\mu((a,b))=\mu\left (\bigcup_n[t_n,b)\right)=\lim \mu([t_n,b))\le \epsilon\lim |b-a|=\epsilon|b-a|.$$ This finishes the proof.
For the reverse implication, first note that the hypothesis implies that $$\mu(\{x\})=0$$ so $$f$$ is continuous at $$x$$. Now choose $$a such that $$|b-a|<\delta.$$ Then, $$|f(b)-f(a)|=\mu([a,b))$$ and by hypothesis, if $$n$$ is large enough, we have $$|\mu((a-1/n,b))|\le \epsilon|(b-a)+1/n|$$ so since $$\mu([a,b))=\lim|\mu((a-1/n,b))|,$$ it follows that $$|f(b)-f(a)|\le \epsilon|b-a|$$. To finish, let $$a\to x$$ and use continuity of $$f$$ at $$x$$ to conclude that $$f'(x)=0.$$
• Thanks for your answer. I just realized that Theorem 1.19 can be extended to the complex measure as well. So, $\mu([\alpha_i,\beta)) \to \mu((\alpha,\beta))$ if $\beta > \alpha_1 > \alpha_2 > \cdots > \alpha$ and $\alpha_i \to \alpha$ as $i\to\infty$. – flyingwith Dec 3 at 1:39