lebesgue measure and differentiability
Hi there
I've got the following two exercises to solve:

Consider the measure space $\displaystyle \([0,\infty\),B,\lambda\)$ where B denotes the Borelsigmaalgebra and $\displaystyle \lambda$ the Lebesguemeasure.
(1)
Let $\displaystyle \mu << \lambda $ and $\displaystyle \frac{d\mu}{d\lambda}$ continuous. Show that
$\displaystyle f(x)=\mu([0,x]) $ is differentiable and that
$\displaystyle f'(x)=\frac{d\mu}{d\lambda}(x)$
(2)
If $\displaystyle \frac{d\mu}{d\lambda}$ is $\displaystyle \lambda$ almost everywhere continuous only is then
$\displaystyle f'(x)=\frac{d\mu}{d\lambda}(x)$ $\displaystyle \lambda$almost everywhere as well?
Why? Prove your answer!

(1)
According to the theorem of RadonNikodym for $\displaystyle g(x)=\frac{d\mu}{d\lambda}(x)$ then
$\displaystyle f(x)= \int_{[0,x]} g d\lambda(x) $
Obviously the derivative of this is g but how can one show this in a proper way?
$\displaystyle f'(x)= \int_{[0,x]} g \frac{d\lambda(x)}{d\lambda(x)} =\int_{[0,x]} g 1 $????
I also tried it this way:
$\displaystyle f'(x)=\frac{d\mu}{d\lambda}(x)\frac{d\mu}{d\lambda}(0) $ but what then is x and 0? Definitely not a onepoint set... cause then it was all zero.
(2)
I'd say that this is right because where
$\displaystyle f'(x) \neq \frac{d\mu}{d\lambda} $ is a Lebesguezero set but how to show?
Could please someone help me to solve this exercise? Any hint would be appreciated.
Regards
Huberscher
1 Attachment(s)
Re: lebesgue measure and differentiability
I think you're trying to make an easy problem hard. Here's a solution:
Attachment 26576
Re: lebesgue measure and differentiability
Thank you very much.
Concerning 2)
A kind of "prove" or argument hmmm
assuming g is continuous at $\displaystyle x_0$ and $\displaystyle x_0$ is contained in a nonzeroset of $\displaystyle \lambda$ then this proof works also, ok.
But when x satisfies $\displaystyle xx_0<\delta$ and x is on the left of $\displaystyle x_0$...doesn't that mean that at x the function g is continuous from the right since the $\displaystyle \delta$ is arbitraty ? That's right yeah?
Re: lebesgue measure and differentiability
Hi there
Well I understand the proof but I still don't understand why this is also valid for a.e. continuity. Can someone please explain me that?
Cause then I have
$\displaystyle \int_{[x_0,x_0+h]}g(x)g(x_0)  dx$ is 0 because the set $\displaystyle [x_0,x_0+h]$ is a zero set but then I still have this $\displaystyle \frac{1}{h}$ which goes to infinity when h goes to 0.
Is there anyone who can answer me that please?
Regards