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.

Suppose you have two finite Borel measures $\mu$ and $\nu$ on $(0,\infty)$. I would like to show that there exists a finite Borel measure $\omega$ such that

$$\int_0^{\infty} f(z) d\omega(z) = \int_0^{\infty}\int_0^{\infty} f(st) d\mu(s)d\nu(t).$$

I could try to use a change of variable formula, but the two integration domains are not diffeomorphic. So I really don't know how to start. Any help would be appreciated! This is not an homework, I am currently practising for an exam.

Thanks!

share|improve this question
    
By the way, the change of variable formula you are referring to (the one with the Jacobian) is proper of the Lebesgue measure. It does not hold with other measures. Take for example a Dirac delta concentrated at the origin. If you apply a dilation to the space, the measure stays the same regardless of the magnitude of the dilation. –  Giuseppe Negro Dec 10 '12 at 21:39
    
I see. Thank you, I did not realize that! –  stephen Dec 11 '12 at 1:36

2 Answers 2

up vote 1 down vote accepted

When we have no idea about the problem, the question we have to ask ourselves is: "if a measure $\omega$ works, what should it have to satisfy?".

We know that for a Borel measure that it's important to know them on intervals of the form $(0,a]$, $a>0$ (because we can deduce their value on $(a,b]$ for $a<b$, and on finite unions of these intervals). So we are tempted to define $$\omega((0,a]):=\int_{(0,+\infty)^2}\chi_{(0,a]}(st)d\mu(s)d\nu(t)=\int_{(0,+\infty)}\mu(0,At^{-1}])d\nu(t).$$

Note that if the collection $\{(a_i,b_i]\}_{i=1}^N\}$ consists of pairwise disjoint elements, so are for each $t$ the $(a_it^{-1},b_it^{-1}]$, which allows us to define $\omega$ over the ring which consists of finite disjoint unions of elements of the form $(a,b]$, $0<a<b<+\infty$. Then we extend it to Borel sets by Caratheodory's extension theorem.

As the involved measures are finite, $\omega$ is actually uniquely determined.

share|improve this answer
1  
Thanks! The way you put it is really intuitive! –  stephen Dec 10 '12 at 21:11

This operation on measures is called convolution for measures on the locally compact abelian group $(0,+\infty)$ under multiplication.

Using exp and log we may convert this to the usual convolution on the LCA group $(-\infty,+\infty)$ under addition. In that case, the convolution $\omega = \mu \ast \nu$ satisfies $$ \int_{-\infty}^{+\infty} f(z)\; \omega(dz) = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} f(x+y) \;\mu(dx)\;\nu(dy) $$

share|improve this answer

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.