Rudin Real and complex analysis question[Differentiation] At the beginning of the chapter on differentiation, the following theorem is stated without proof. Apparently it is so trivial that it does not require justification. I however don't find it so trivial and would appreciate if someone could assist me in proving it. The theorem is the following:
Let $\mu$ be a complex borel measure on $\Bbb R$ and define $f(x):=\mu((-\infty,x))$ . Following statements are equivalent:
1.) $f$ is differentiable at $x$ and $f'(x)=A$
2.) For every $\epsilon>0$ there exists $\delta>0$ so that for all open segments $I$ containing $x$ with length $<\delta$ the inequality $|\frac{\mu(I)}{m(I)}-A|<\epsilon$ ,where $m$ denotes lebesgue measure.
Thanks in advance!
 A: Suppose $f$ is differentiable at $x$ and let $\epsilon>0$. By definition, there is $\delta>0$ such that for all $y$ with $|x-y| \leq \delta$, $|\frac{f(x)-f(y)}{x-y}-A| \leq \epsilon.$
Then note that if $y \leq x$, $x-y$ is $m(I)$ where $I=[x,y]$, and $f(x)-f(y)$ is $\mu(I)$.
EDIT
Let $\epsilon >0$. Let $y_1<x<y_2$, and $I=(y_1,y_2)$.
Claim : suppose $f$ is differentiable at $x$. Then there is a $\delta >0$ such that if $|y_1-y_2|=m(I) \leq \delta$, then 
$|\frac{f(y_1)-f(y_2)}{y_1-y_2} - f'(x)| \leq \epsilon$.
Proof of the claim : 
Write $f(y_i)=f(x)+f'(x)(y_i-x)+o(y_i-x)$, where $i=1,2$. Therefore, 
$f(y_2)-f(y_1) = f'(x)(y_2-y_1) + o(y_1-y_2)$ by substracting one equality from the other. That's exactly what the claim says.
Proof of "differentiable implies the $\epsilon-\delta$ property" : just note that $y_2 - y_1 =m(I)$ and $f(y_2)-f(y_1)=\mu(I)$.
Conversely, if the property is true, then you have : for all $\epsilon>0$ there is a $\delta>0$ such that for all $y_1<x <y_2$, if $|y_1 - y_2 | \leq \delta$, then 
$|\frac{f(y_2)-f(y_1)}{y_2-y_1}-A| \leq \epsilon$, which almost implies that $f$ is differentiable at $x$ and $f'(x)=A$. It will be true if $f$ is continuous at $x$ (by letting $y_1 \rightarrow x$, for example). And $f$ is continuous at $x$ if and only if $\mu({x})=0$, which must be the case here. 
