1. ## Borel measure

Hi,

I need to prove that their exists a Borel measure v on $\displaystyle \mathbb{R}^+$ which is absolutely continuous with respect to Lebesgue measure and which satisfies $\displaystyle v([a,b])=b^2-a^2$ with $\displaystyle a< b$

I know that the usual Lebesgue measure would satisfy $\displaystyle v([a,b])=b-a$, but I'm terrible at coming up with examples. If I could find one that satisfies the requirement, I'm sure I could prove it is absolutely continuous. Any hints?

Everything I was trying put me in $\displaystyle \mathbb{R}^2$, and I need to be in $\displaystyle \mathbb{R}$. Thanks.

*Note I had a typo and edited that the Lebesque measure satisfies $\displaystyle v([a,b])=b-a$

2. Originally Posted by Benmath
Hi,

I need to prove that their exists a Borel measure v on $\displaystyle \mathbb{R}^+$ which is absolutely continuous with respect to Lebesgue measure and which satisfies $\displaystyle v([a,b])=b^2-a^2$ with $\displaystyle a< b$

I know that the usual Lebesgue measure would satisfy $\displaystyle v([a,b])=b-a$, but I'm terrible at coming up with examples. If I could find one that satisfies the requirement, I'm sure I could prove it is absolutely continuous. Any hints? [/tex]
It looks as though the measure v should be defined by $\displaystyle v(E) = \int_E2x\,dx$ for each Borel set E, where dx means integral with respect to Lebesgue measure. Is that enough of a hint?

3. Thank you Opalg! That is incredibly obvious! I am usually terrible with the very obvious things. Again, thanks!