1
$\begingroup$

I need some help on following question. Any help is greatly appreciated.

Let $p$ be a sigma finite measure on measurable space $( [0,1], M )$ where $M$ is the sigma algebra of Lebesgue measurable subsets of $[0,1]$. For $x$ in $[0,1]$, define $F(x) = p ([0,x])$.

Show that $p$ is absolutely continuous with respect to Lebesgue measure $m$ on $[0,1]$ iff $F$ is an absolutely continuous function on $[0,1]$ satisfying $F(0) = 0$.

$\endgroup$
  • $\begingroup$ Do you have any thoughts and can share those? Regards $\endgroup$ – Amzoti Aug 8 '13 at 17:19
  • 1
    $\begingroup$ I actually worked on it but not sure about this. Is it correct that F(b) - F(a) = p([a,b]) for 0<a<b<1 regarding measure p? I am not sure what p does..Is it an aritrary measure or it is just same as lebesgue measure on [0,x]? $\endgroup$ – Kushan Aug 8 '13 at 17:25
  • $\begingroup$ Thanks for sharing your attempt in the comment. You might like to edit your question with the details of what you've tried so far. It makes it a lot easier for someone to know what part of the problem you're having trouble with, and also lets us know you're not just here to get easy answers to a homework problem (it's unfortunate but it happens :-/ ). $\endgroup$ – Dan Rust Aug 8 '13 at 17:33
  • $\begingroup$ Thanks for the comments. I am working on prelim questions...I did work it out, but only part I was not sure about was what I have mentioned earlier since I used it.I guess I would have mentioned it with the question..Sorry for that. $\endgroup$ – Kushan Aug 8 '13 at 17:38
1
$\begingroup$

It seems, from your comments, that the part you're asking about is the following statement, which I've made slightly more precise than you wrote.

Claim: For every $0<a<b<1$, we have $F(b)-F(a) = p\big( (a,b] \big)$.

Proof: Note that $[0,b] = [0,a] \cup (a,b]$ and that this is a disjoint union. Since $p$ is a measure, we have $p\big( [0,b] \big) = p\big( [0,a] \big) + p\big( (a,b] \big)$. So by the definition of $F$, we have $F(b) = F(a) + p\big( (a,b] \big)$ as claimed.

(Note that it is entirely possible for $p\big( (a,b] \big) \ne p\big( [a,b] \big)$ - for example, $p$ could equal $1$ on every set containing $a$ and $0$ on all other sets. Of course that doesn't happen for absolutely continuous measures, since the Lebesgue measure of $\{a\}$ equals $0$; but this question requires thinking about all measures.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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