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.
 A: 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 $f'(x)=0.$
