# 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!

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Also, either you should say 'for all $x$' in 1.) or '$x\in I$' in 2.) (Rather the second one..) – Berci Mar 28 '13 at 19:32
Presumably, you need to know that $f(x)$ is finite. The standard measure on $\mathbb R$ doesn't yield $f(x)$ finite for any $x$, and therefore it is unclear what "differentiable" would mean there. – Thomas Andrews Mar 28 '13 at 19:36
A complex measure is always finite, at least in this book. – Michael Mar 28 '13 at 19:36
Ah, that would appear to be true in the Wikipedia definition, too. Never seen complex measures, so that hadn't occured to me that the standard measure wouldn't also be a complex measure @Michael – Thomas Andrews Mar 28 '13 at 19:39

## 1 Answer

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.

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But $f(x)-f(y)=[y,x)$. Why would singeltons have to be null sets for a borel measure? – Michael Mar 28 '13 at 19:40
if singletons have positive mass, then $f$ is discontinuous at this point and a fortiori not differentiable – Glougloubarbaki Mar 28 '13 at 19:42
but your remark is absolutely correct ! – Glougloubarbaki Mar 28 '13 at 19:43
Thank you for the comment. I am still confused because $x$ is supposed to be an interior point of $I$, since $I$ are assumed to be open intervals. – Michael Mar 28 '13 at 19:47
OK, I'll give a little more details – Glougloubarbaki Mar 28 '13 at 20:03