Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
2  
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
add comment

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.

share|improve this answer
1  
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
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.