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 $t_n<x$ 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 $a<x<b$ 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 $f'(x)=0.$

  • $\begingroup$ 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 $i\to\infty$. $\endgroup$ – flyingwith Dec 3 at 1:39

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