# lebesgue measure and differentiability

• Jan 14th 2013, 12:22 PM
huberscher
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 Borel-sigma-algebra and $\displaystyle \lambda$ the Lebesgue-measure.
(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?
-

(1)
According to the theorem of Radon-Nikodym 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 one-point 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 Lebesgue-zero set but how to show?

Could please someone help me to solve this exercise? Any hint would be appreciated.

Regards
Huberscher
• Jan 15th 2013, 07:12 PM
johng
Re: lebesgue measure and differentiability
I think you're trying to make an easy problem hard. Here's a solution:
Attachment 26576
• Jan 16th 2013, 04:03 AM
huberscher
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 non-zero-set of $\displaystyle \lambda$ then this proof works also, ok.

But when x satisfies $\displaystyle |x-x_0|<\delta$ 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?
• Feb 1st 2013, 02:24 PM
huberscher
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.